uwaterloo.ca · Ruin Analysis of a Threshold Strategy in a Discrete-time Sparre Andersen Model...

27
Ruin Analysis of a Threshold Strategy in a Discrete-time Sparre Andersen Model Steve DREKIC and Ana Maria MERA Department of Statistics and Actuarial Science University of Waterloo 200 University Avenue West Waterloo, Ontario, Canada N2L 3G1 Abstract In this paper, we extend the methodology of Alfa and Drekic (2007) to analyze a discrete-time, delayed Sparre Andersen insurance risk model featuring a single thresh- old level and randomized dividend payments. Using matrix analytic techniques, we construct a set of computational procedures enabling one to calculate probability dis- tributions associated with fundamental ruin-related quantities of interest, namely the time of ruin, the surplus immediately prior to ruin, and the deficit at ruin. Special cases of the general model, including the ordinary and stationary Sparre Andersen variants, are examined in several numerical examples. KEYWORDS: Sparre Andersen model, discrete time, matrix analytic methods, randomized dividends, threshold level. 1

Transcript of uwaterloo.ca · Ruin Analysis of a Threshold Strategy in a Discrete-time Sparre Andersen Model...

Page 1: uwaterloo.ca · Ruin Analysis of a Threshold Strategy in a Discrete-time Sparre Andersen Model Steve DREKIC and Ana Maria MERA Department of Statistics and Actuarial Science University

Ruin Analysis of a Threshold Strategy in a

Discrete-time Sparre Andersen Model

Steve DREKIC and Ana Maria MERA

Department of Statistics and Actuarial Science

University of Waterloo

200 University Avenue West

Waterloo Ontario Canada N2L 3G1

Abstract

In this paper we extend the methodology of Alfa and Drekic (2007) to analyze a

discrete-time delayed Sparre Andersen insurance risk model featuring a single threshshy

old level and randomized dividend payments Using matrix analytic techniques we

construct a set of computational procedures enabling one to calculate probability disshy

tributions associated with fundamental ruin-related quantities of interest namely the

time of ruin the surplus immediately prior to ruin and the deficit at ruin Special cases

of the general model including the ordinary and stationary Sparre Andersen variants

are examined in several numerical examples

KEYWORDS Sparre Andersen model discrete time matrix analytic methods randomized

dividends threshold level

1

1 Introduction and notation

Alfa and Drekic (2007) recently considered a general delayed Sparre Andersen insurance

risk model in discrete time and analyzed it as a doubly-infinite Markov chain to establish

a computational procedure for calculating the (trivariate) joint distribution of the time of

ruin the surplus immediately prior to ruin and the deficit at ruin In this paper we extend

their essential approach to a more general model which now features an embedded threshold

dividend strategy and randomized dividend payments This paper generalizes the model

considered by Tan and Yang (2006) in which a discrete-time process based on the compound

binomial model (ie geometrically distributed interclaim times) is analyzed wherein a random

dividend of 0 or 1 is paid whenever the surplus process exceeds a specified (fixed) threshold

level In addition this paper serves as a contribution to the current state of knowledge

regarding discrete-time risk models with a randomized decision rule for paying dividends a

topic which has garnered some recent attention in the ruin theoretic literature (eg see Bao

(2007) Landriault (2008) and Kim et al (2008))

In this paper we assume that the number of claims process Nt t = 0 1 is a modishy

fied discrete-time renewal process with independent positive interclaim times W1W2

where W1 is the duration from time 0 until the first claim occurs and Wi i = 2 3 is

the time between the (i minus 1)-th and i-th claims We also remark that W2W3 forms

an independent and identically distributed (iid) sequence of positive random variables with

common probability mass funtion (pmf) aj = PrWi = j j = 1 2 na and correspondshy

ing survival function Aj = PrWi gt j = 1 minus L

k

j

=1 ak Throughout the paper we assume

that na lt infin (ie the interclaim time distribution of Wi i = 2 3 has finite support)

In the ordinary Sparre Andersen model a claim is assumed to have occurred at time 0minus

so that W1 has the same distribution as the ordinary interclaim times W2W3 On the

other hand if W1 is not a ldquofullrdquo interclaim time then asymptotically in time the limiting Lnadistribution of this forward recurrence time is defined via the pmf aj = Ajminus1 k=1 Akminus1

j = 1 2 na (eg see Karlin and Taylor (1975 pp 192-193)) This yields another

important risk model namely the stationary Sparre Andersen model in which W1 has pmf aj

rather than aj As a means of accommodating other possible specifications we assume that

W1 has a more general pmf denoted by rj = PrW1 = j j = 1 2 nr where nr lt infin

Let Rj = PrW1 gt j = 1 minus L

k

j

=1 rk denote its survival function Through appropriate

choice of rj it is clear that both the ordinary and stationary Sparre Andersen variants are

special cases of this more general model referred to as the delayed Sparre Andersen risk

model

In what follows let Z denote the set of all integers Zminus the set of negative integers Z+

2

the set of positive integers and N = 0 cup Z+ For t isin Z+ we define Ut as the insurerrsquos

amount of surplus at time t In actual fact Ut represents the amount of surplus at the end

of time interval (t minus 1 t] at which point any premiumsclaims corresponding to this time

interval have been receivedpaid out Specifically with respect to the time interval (t minus 1 t]

we adopt the convention that premiums are received at (t minus 1)+ and any claims are paid

out at tminus We incorporate the notion of a threshold level Z isin Z+ on the insurerrsquos surplus

affecting the amount of premium being received at a given point in time More precisely we

assume that premiums corresponding to the time interval (t t + 1] t isin N are collected at t+

according to (random) rate pt such that

c if Ut lt Z pt =

Xt if Ut ge Z

where di = PrXt = i i = c1 c1 + 1 c2 We refer to c isin Z+ as the pure (constant) L

premium and assume that c1 c2 isin 0 1 c with c1 le c2 and c2 di = 1 Clearly c1i=c1

and c2 are the respective lower and upper support values of the distribution of the random

premium amount at time t Xt Correspondingly we interpret c minus pt as the amount of

(randomized) dividends paid to insureds at time t+ Note that by assumption the probability

distribution of Xt is identical for all values of t

Beginning at time 0 with an initial reserve of u isin N the insurerrsquos amount of surplus at

time t is expressible as tminus1 Nt L L

U t = u + pi minus Yi t isin N i=0 i=1

where individual claim amounts Y1 Y2 are assumed to form an iid sequence of positive

random variables with common pmf αj j = 1 2 mα and corresponding survival function LjΛj = 1 minus k=1 αk Unlike the interclaim time distributions defined above we remark that

the claim amount distribution can be either of finite or infinite support (ie mα le infin) Let

the time of ruin T be defined as T = mint isin Z+|Ut lt 0 with T = infin if Ut ge 0 forall t isin Z+

If ruin does occur let |UT | denote the deficit at ruin and UTminus = UTminus1 + pTminus1 the surplus

immediately prior to ruin Clearly T = infin if mα le minc Z + c1 Conversely if mα gt

minc Z + c1 it is not difficult to verify that |UT | isin 1 2 mα minus minc Z + c1 and

UTminus isin minc Z + c1 minc Z + c1 + 1 mα minus 1 Throughout the remainder of the

paper we assume that mα gt minc Z + c1

Following this introduction Section 2 sets up the underlying Markov chain structure in

our model and derives some necessary preliminaries In Section 3 we further exploit this

structure to construct a set of computational procedures enabling one to calculate probashy

bility distributions associated with the random variables T |UT | and UTminus with specific

3

consideration to the joint pmfrsquos

Z+ωni(u) = Pr T = n UTminus = i | U0 = u n isin i = minc Z+c1 minc Z+c1+1 mαminus1

φnj(u) = Pr T = n |UT | = j | U0 = u n isin Z+ j = 1 2 mαminusminc Z +c1

and

ψnij(u) = Pr T = n UTminus = i |UT | = j | U0 = u n isin Z+

i = minc Z +c1 minc Z +c1 + 1 mαminus1

j = 1 2 3 mαminusi

Finally we implement our various procedures and present some numerical examples in Section

4

2 Model formulation and preliminaries

Adopting the same methodology employed by Alfa and Drekic (2007) let τj =

Pr W gt j|W gt j minus 1 = AjAjminus1 j = 1 2 na where W denotes an arbitrary Wi

i = 2 3 Clearly τ1 = A1 and τna = 0 If we next define the 1 times na row vector

e1 = (1 0 0) the na times na probability transition matrix

0 τ1 0 middot middot middot 0

0 0 τ2

0

S =

0 0 middot middot middot 0 τnaminus1

0 0 middot middot middot 0 0

and the na times 1 column vector

1 minus τ1 1 minus τ2 s = 1 minus τnaminus1

1

it is known that aj = e1Sjminus1 s j = 1 2 na (eg see Alfa (2004)) In other words W

can be expressed as a discrete phase-type random variable (of dimension na) with the above

parameterization implying that s contains the absorption probabilities to claim occurrence

associated with each of the na possible interclaim time values Alfa (2004) refers to this

4

characterization as the ldquoelapsed timerdquo phase-type representation of the pmf aj We remark

that the above pmf formula also holds true for j gt na as it is readily verified that Sjminus1 = Ona

an na times na matrix of zeros for such values of j

In what follows we proceed to set up our delayed Sparre Andersen model as a bivariate

Markov chain conditional on W1 = k isin 1 2 nr Specifically assuming that W1 = k we

construct the bivariate stochastic process (Ut Lt) t = k k + 1 such that Ut represents

the amount of surplus at time t and Lt denotes a specific counter at time t which measures the

ldquoelapsed timerdquo (in the sense described above) length of the next claim occurrence We refer to

the Ut component as the level of the process and the Lt component as the phase of the process

Two important observations are particularly noteworthy regarding this construction First of

all Lk = 1 since we are assuming that the very first claim occurs at time kminus More generally

Lt = 1 if a claim happens to occur at time tminus Secondly the delayed process reverts to the

ordinary process (having interclaim time distribution defined by the pmf aj) upon occurrence

of the first claim As a result the bivariate stochastic process (Ut Lt) t = k k + 1

possesses the following Markovian relationship

(Ut + pt Lt + 1) if there is no claim at time (t + 1)minus (Ut+1 Lt+1) =

(Ut + pt minus Y 1) if there is a claim of amount Y at time (t + 1)minus

Note that the state space for this Markov chain is given by Δ = Z times 1 2 na Furthershy

more the probability transition matrix associated with this Markov chain is given by

middot middot middot minus1 10 middot middot middot Zminus2 Zminus1 Z Z+1 Z+2 middot middot middot

middot middot middot

middot middot middot minus1 middot middot middot Bc Bcminus1 Bcminus2 middot middot middot BcminusZ+1 BcminusZ BcminusZminus1 BcminusZminus2 BcminusZminus3 middot middot middot

0

1

middot middot middot Bc+1 Bc Bcminus1 middot middot middot BcminusZ+2 BcminusZ+1 BcminusZ BcminusZminus1 BcminusZminus2 middot middot middot middot middot middot Bc+2 Bc+1 Bc middot middot middot BcminusZ+3 BcminusZ+2 BcminusZ+1 BcminusZ BcminusZminus1 middot middot middot

middot middot middot

middot middot middot

P = Zminus2 middot middot middot BZ+cminus1 BZ+cminus2 BZ+cminus3 middot middot middot Bc Bcminus1 Bcminus2 Bcminus3 Bcminus4 middot middot middot (21)

Zminus1 middot middot middot BZ+c BZ+cminus1 BZ+cminus2 middot middot middot Bc+1 Bc Bcminus1 Bcminus2 Bcminus3 middot middot middot

Z middot middot middot AZ+1 AZ AZminus1 middot middot middot A2 A1 A0 Aminus1 Aminus2 middot middot middot Z+1 middot middot middot AZ+2 AZ+1 AZ middot middot middot A3 A2 A1 A0 Aminus1 middot middot middot

Z+2 middot middot middot AZ+3 AZ+2 AZ+1 middot middot middot A4 A3 A2 A1 A0 middot middot middot

middot middot middot

middot middot middot

where Ona

if i isin Zminus

Bi = S if i = 0 (22) se1αi if i isin Z+

5

and c2 L

Ai = djBi+j (23) j=c1

We give a brief indication of the reasoning leading to the form of P in (21) For example

consider the transition ldquoUt = 0 rarr Ut+1 = 1rdquo which has block element Bcminus1 containing the

transition probabilities governing the ldquoLt rarr Lt+1 rdquo process At a surplus level of 0 at time

t a premium of pt = c is certain to be received Therefore to arrive at surplus level 1 at

the next time unit a ldquoso-calledrdquo claim of size c minus 1 must occur to neutralize the premium

just received Thus the size of the ldquoclaimrdquo required for this transition to occur determines

the value of i in Bi so that Bcminus1 is obtained and its form subsequently determined by (22)

according to the value of the index c minus 1 Note that only when c gt 1 does an actual claim

occur in this particular scenario We mention that for transitions which begin at or above the

threshold level Z a similar line of reasoning would also apply except now one must condition

on the amount of the random premium being earned as indicated by the form of (23)

Note that this formulation gives rise to a doubly-infinite Markov chain with finite blocks

of size na In the special case of the threshold-free model (obtained by setting c1 = c2 = c

so that dc = 1) we remark that (21) is in agreement with the matrix P in Alfa and Drekic

(2007 p 296) Also since αi = 0 forall i gt mα we note that Bi = Ona if i gt mα

For the computation of ruin-related quantities of interest it is useful to partition the

state space Δ into two mutually exclusive state spaces namely Δ1 = N times 1 2 na

and Δ2 = Zminus times 1 2 na In addition we introduce (one-step) transition matrices

C Δ1 rarr Δ1 and D Δ1 rarr Δ2 which correspond to mapping ldquonon-ruinedrdquo states in Δ1 to

either ldquonon-ruinedrdquo states in Δ1 or ldquoruinedrdquo states in Δ2 respectively In other words the

matrices C and D are essentially submatrices of P and their forms are readily given by

210 middot middot middot Z minus 1 Z Z + 1 middot middot middot

Bc Bcminus1 Bcminus2 middot middot middot BcminusZ+1 BcminusZ BcminusZminus1 middot middot middot Bc+1 Bc Bcminus1 middot middot middot BcminusZ+2 BcminusZ+1 BcminusZ middot middot middot

Bc+2 Bc+1 Bc middot middot middot BcminusZ+3 BcminusZ+2 BcminusZ+1 middot middot middot

middot middot middot

middot middot middot

0

1

2 Z minus 2 BZ+cminus2 BZ+cminus3 BZ+cminus4 middot middot middot Bcminus1 Bcminus2 Bcminus3 middot middot middot (24) C =

Z minus 1 BZ+cminus1 BZ+cminus2 BZ+cminus3 middot middot middot Bc Bcminus1 Bcminus2 middot middot middot

Z AZ AZminus1 AZminus2 middot middot middot A1 A0 Aminus1 middot middot middot

Z + 1 AZ+1 AZ AZminus1 middot middot middot A2 A1 A0 middot middot middot

Z + 2 AZ+2 AZ+1 AZ middot middot middot A3 A2 A1 middot middot middot

middot middot middot

6

and

minus1 minus2 minus3 middot middot middot

0 Bc+1 Bc+2 Bc+3 middot middot middot 1 Bc+2 Bc+3 Bc+4 middot middot middot

2 Bc+3 Bc+4 Bc+5 middot middot middot middot middot middot

Z minus 2 BZ+cminus1 BZ+c BZ+c+1 middot middot middot D = (25) Z minus 1 BZ+c BZ+c+1 BZ+c+2 middot middot middot

Z AZ+1 AZ+2 AZ+3 middot middot middot

Z + 1 AZ+2 AZ+3 AZ+4 middot middot middot Z + 2 AZ+3 AZ+4 AZ+5 middot middot middot

Let us now specify some further definitions required in upcoming derivations First of all

we introduce the function

0 if u ge Z zt =

maxi isin 1 2 t|u + c(i minus 1) lt Z if u lt Z

or equivalently

0 if u ge Z zt =

minlfloorZminuscuminus1 rfloor + 1 t if u lt Z

where the floor function denoted lfloorxrfloor yields the largest integer less than or equal to x We

define dm0 = δm0 where δij in general denotes the Kronecker delta function of i and j

Also define

dm if m = c1 c1 + 1 c2dm1 =

0 elsewhere

and for n = 2 3

Lc2 dj1dmminusjnminus1 if m = nc1 nc1 + 1 nc2dmn = j=c1

0 elsewhere

In other words dmn represents the pmf of the n-fold convolution of dm with itself

Recalling that W1 = k note that zk + ( k+) determines the earliest time point in the set =

0+ 1+ (k minus1)+ at which a random premium is received Note also that zk = k implies

that no such point exists in this set (ie the threshold level Z has not yet been reached when

the premium at time (k minus1)+ is to be received) Therefore if we assume for the moment that

zk k we may condition on the total premiums earned at times zk + (zk + 1)+ (k minus 1)+=

7

to obtain the following expression for b(k) the initial (ie starting at time k) probability row

vector of the above Markov chain corresponding only to the states in Δ1 namely

(kminuszk)c2

b(k) L

= dmkminuszk(αu+czk+me1 αu+czk+mminus1e1 α2e1 α1e1 0 0 ) (26)

m=(kminuszk)c1

where 0 denotes the 1 times na row vector of zeros By further defining the function vn(t) =

b(k)u + ct + (n minus t)c2 we note that the i-th level of is given by the 1 times na row vector L(kminuszk)c2 b(k)αu+czk+mminusie1 for each i isin Ωk = 0 1 vk(zk)minus1 Hence contains

m=(kminuszk)c1 dmkminuszk

zeros from level vk(zk) onwards Moreover (26) holds true even in the case when zk = k

simplifying to give b(k) = (αu+cke1 αu+ckminus1e1 α2e1 α1e1 0 0 )

At this juncture we introduce two additional row vectors namely

(k) (k) (k) (k) b(k)gn = (gn0 gn1 gn2 ) = Cn n isin N (27)

and

h(k) (k) (k) (k) b(k)Cnminus1D =

(k) Z

+ n = (h nminus2 hnminus3 ) = gnminus1D n isin (28) nminus1 h

(k)We remark that gn contains the probabilities of being in the various ldquonon-ruinedrdquo states

at time k + n (ie after claims have been paid out) without having visited a ldquoruinedrdquo state

during the previous n minus 1 transitions given that Uk isin Ωk sub Δ1 In a similar fashion h(nk)

contains the probabilities of being in the various ldquoruinedrdquo states for the first time at time

k + n given that Uk isin Ωk sub Δ1 Note that the probability of being in ruined state ldquo minus jrdquo is (k) (k) (k) (k)

characterized by the row vector h = (h h h ) with the third subscript nminusj nminusj1 nminusj2 nminusjna

representing the value of the elapsed time counter Lk+n Upon further reflection however (k)

=it must be the case that h = 0 for i 1 as ruin can only occur at claim instants nminusji

which in this case implies Lk+n = 1 with probability 1 Thus it immediately follows that (k) (k) (k)

the structure of the 1 times na row vector hnminusj is simply given by hnminusj = (φnj(u) 0 0) (k)

where φ (u) = Pr T = k + n |UT | = j | Consequently as in the analysis with nj Uk isin Ωk

no threshold on the surplus level (see Alfa and Drekic (2007 p 299)) we obtain

(k) (k) prime φ (u) = h (29) nj nminusje1

where e1 prime denotes the transpose of e1

A similar line of probabilistic logic can also be applied to obtain a representation for the (k)

trivariate function ψ (u) = Pr T = k + n UTminus = i |UT | = j | Uk isin Ωk In order for ruin nij

to occur at time k + n with a surplus immediately prior to ruin equal to i we observe that

(i) none of the previous n minus 1 transitions must have included a visit to any state in Δ2

and

8

(ii) the surplus level at time k + n minus 1 must necessarily be equal to i minus pk+nminus1

Note that pk+nminus1 denotes the premium corresponding to the time interval (k + n minus 1 k + n]

which is received at (k + n minus 1)+ Hence the quantity corresponding to points (i) and (ii) (k)

above is the 1 times na row vector g Then at the next time unit a claim must nminus1iminuspk+nminus1

necessarily occur but not before a premium of pk+nminus1 is first collected thereby raising the

surplus level to i Since s contains the absorption probabilities (to claim occurrence) from

the na possible phase states and the claim causing ruin must be of size i + j in order to

ensure that the deficit at ruin is equal to j it follows that

(k) (k)ψ (u) = g s αi+j (210) nij nminus1iminuspk+nminus1

(k)The following corollary provides the means of calculating the quantity g in nminus1iminuspk+nminus1

(210)

Corollary 1 If minc Z + c1 = c then

(k)

g if i = c c+1 Z +c1 minus1 nminus1iminusc

(k) (k) Lminc2iminusZ (k)g = g + j=c1

if i = Z +c1 Z +c1 +1 minmα Z +cminus1nminus1iminuspk+nminus1 nminus1iminusc djgnminus1iminusj

Lc2 (k)

j=c1 djgnminus1iminusj if i = Z +c Z +c+1 mαminus1

(211)

Conversely if minc Z + c1 = Z + c1 then

Lminc2iminusZ (k)

djg if i = Z +c1 Z +c1 +1 cminus1nminus1iminusj j=c1

(k) (k) Lminc2iminusZ (k)gnminus1iminuspk+nminus1

= gnminus1iminusc + j=c1 djgnminus1iminusj if i = c c+1 minmα Z +cminus1

Lc2 (k)djg if i = Z +c Z +c+1 mαminus1j=c1 nminus1iminusj

(212)

Proof The justification of (211) is as follows If c is smaller than Z + c1 the only way of

reaching a surplus level of i isin c c+1 Z +c1 minus1 (prior to ruin) is by receiving a premium

of c since i minus j for any j = c1 c1 +1 c2 would result in a surplus level below Z where the

random premium rate is not applicable Next for fixed i isin Z +c1 Z +c1 +1 Z +c minus1

there are two distinct ways of reaching this surplus level First of all note that the value of

i minus j is guaranteed to produce a surplus amount greater than or equal to Z provided that

j isin c1 c1 + 1 minc2 i minus Z Secondly note that reducing each of value of i by c also

yields a non-negative surplus value less than Z Hence both pure and random premiums

are plausible under this scenario For i isin Z + c Z + c + 1 note that decreasing this

amount by any j = c1 c1 + 1 c2 results in a surplus value greater than or equal to Z as

desired However reducing it by the amount c also has the same impact thus rendering the

pure premium option inadmissible in this case Lastly we incorporate into the range of i the

9

fact that the surplus immediately prior to ruin cannot exceed mα minus1 as indicated in Section

1 The justification of (212) when c is larger than Z + c1 mirrors the analysis above and is

left to the reader D

Recall that in the analysis up to this point we have assumed that W1 = k isin

1 2 3 nr If we now condition on the value of W1 the Law of Total Probability

yields the following two expressions for φnj(u) and ψnij(u) namely

nr L

= rkφ (u) n = nr + 1 nr + 2 φnj(u) n

(kminus) kj

k=1

and nr L

(k)ψnij(u) = rkψ (u) n = nr + 1 nr + 2 nminuskij

k=1

In addition however it is quite possible that T = n for n = 1 2 nr and this can happen

as a result of either

(i) the first claim which causes ruin occurring at time nminus from a (conditional) surplus Lnminus1level of u + czn + m (where m) or i=zn

Xi =

(ii) the first claim which does not cause ruin occurring at some time kminus k isin 1 2 nminus

1 and ruin subsequently occurring n minus k time units later

If we now combine these outcomes with those for n = nr + 1 nr + 2 (and recall that

rn = 0 forall n gt nr) the general expression for φnj(u) becomes

minnminus1nr (nminuszn)c2 L

(k) L

φnj(u)= rkφ (u) + rn dmnminuszn αu+czn+m+j n isin Z+ (213) nminuskj

k=1 m=(nminuszn)c1

Concerning the function ψnij(u) when n le nr and point (i) above is considered we note Lnminus1that i must necessarily equal u + czn + m (again conditional on i=zn

Xi = m) for ruin to

occur in this manner Hence in a similar fashion we acquire

minnminus1nr (nminuszn)c2 L L

ψnij(u) = rkψn

(kminus) kij(u) + rn dmnminuszn

δiu+czn+mαu+czn+m+j

k=1 m=(nminuszn)c1

minnminus1nr L

(k) Z

+ = rkψnminuskij(u) + rndiminusuminuscznnminuszn αi+j n isin (214)

k=1

Finally for fixed i isin minc Z + c1 minc Z + c1 + 1 mα minus 1 summing (214) from

j = 1 to mα minus i yields the following expression for the bivariate pmf ωni(u) namely

minnminus1nr mαminusi mαminusi L L

(k) L

ωni(u) = rk ψnminuskij(u) + rndiminusuminuscznnminusznαi+j

k=1 j=1 j=1

10

minnminus1nr L

(k) Z

+ = Λi rkg s + rndiminusuminuscznnminuszn n isin (215) nminuskminus1iminuspnminus1

k=1

where (210) was used to establish the last equality In the special case when c1 = c2 = c it

is a straightforward exercise to show that equations (213) through (215) simplify to produce

equations (215) through (217) of Alfa and Drekic (2007 p 300)

3 Computational considerations

In order to compute the joint pmfrsquos defined by (213) (214) and (215) it is clear from (k)

equations (29) through (212) that we need to be able to compute the quantities gn and

h(k) n defined by (27) and (28) respectively In what follows we capitalize on the underlying

Markov chain structure in our model to develop a set of computational procedures which

ultimately enable one to calculate these key row vectors

First of all recall that g0(k)

= b(k) contains zeros from a certain level onwards namely (k)

level vk(zk) = u + czk + c2(k minus zk) In general one can easily deduce that gn will also

contain zeros from a certain level onwards to be denoted by lk(n) We begin this section

with a determination of lk(n) which varies in form depending on the value of vk(zk) minus 1

representing the maximum surplus level at time k after the smallest possible claim of size 1

has been applied (at time kminus)

On one hand it may be the case that the value of vk(zk)minus1 lies at or above the threshold

level Z In this situation we remark that zk can take on any value in the set 0 1 k (k)

Since g0 contains zeros from level lk(0) = vk(zk) onwards we observe that at time k + 1

it may be possible to achieve a higher surplus value than vk+1(zk) minus 1 representing the

maximum surplus level assuming there was no drop below Z due to the claim at time kminus

Specifically if the surplus had fallen to level Z minus 1 at time kminus then gained c at time k+ a

bounded level of Z + c minus1 would have been reached at time k +1 and this level is potentially

larger than vk+1(zk) minus 1 As a result we have that lk(1) = maxvk+1(zk) Z + c If we

extend this argument inductively we readily obtain lk(n) = maxvk+n(zk) Z +c +c2(n minus1)

n isin Z+ However since vk+n(zk) = u + czk + c2(k +n minus zk) = vk(zk)+ c2 + c2(n minus 1) we may

alternatively re-express lk(n) as

lk(n) = maxvk+n(zk) Z + c + c2(n minus 1) = maxvk(zk) + c2 Z + c + c2(n minus 1)

Thus

Z+lk(n) = maxlk(0) + c2 Z + c + c2(n minus 1) n isin (31)

11

or equivalently

Z+lk(n) = lk(1) + c2(n minus 1) n isin

Clearly lk(0) gt Z and lk(n) ge Z + c for n isin Z+

The other remaining case to consider involves the situation when the maximum surplus

level at time k is strictly below the threshold level Z Under this scenario we further

distinguish between two cases involving the value of c2 First of all suppose that c2 = 0 so

that vk(zk)minus1 = u +czk minus1 lt Z or equivalently u +czk le Z If u +czk happens to equal Z

then it is possible for zk isin 0 1 k With c2 = 0 however no further premiums can be

earned until after the first claim takes place (which is assumed to occur at time kminus) thereby

implying that Z = u + ci for all i = zk zk + 1 k Hence we know that u + ck = Z

in this case Moreover if u + czk happens to be strictly less than Z then we clearly have

that zk = k Even in the case when c2 gt 0 one can readily verify that zk = k so that

vk(zk) minus 1 = u + ck minus 1 lt Z Thus we proceed with the analysis assuming that zk = k

c2 ge 0 and the inequality u + ck le Z holds true

lowast k

Define the ceiling function denoted lceilxrceil to yield the smallest integer greater than or

equal to x Because a random premium cannot be received until level Z is at least achieved

it is essential to find the point at which the threshold level Z is reached (ie the point at Zminus(u+ck)+1 which u + ck minus1+ ci ge Z) Isolating this inequality for i one obtains i ge

c Hence

assuming that W1 = k the elapsed time after k at which a random premium would begin to

lceilZminus(u+ck)+1 be received is t lowast = rceil Note that t lowast isin Z+ k c k

With zk = k g0(k)

contains zeros from level u + ck onwards so that lk(0) = u + ck

Immediately following this a premium of c is guaranteed to be received at time k+ as the

threshold level Z has not yet been reached Consequently g1(k)

contains zeros from level (k)

b(k)u + c(k + 1) onwards which is c levels further than that in g0 = Continuing this (k)

process inductively we can readily establish that gn contains zeros from level u + c(k + n)

onwards for n = 1 2 t lowast k so that lk(n) = u + c(k + n) for n = 0 1 t lowast k

In the case of n = t lowast + 1 t lowast + 2 gt

(k) is first looked at in isolation At time k +k k +1

t lowast k + 1 the threshold level has been crossed giving rise to a maximum surplus value of either (k)

c(k + t lowast k contains zeros from level lk(t lowast ku + ) + c2 minus 1 or Z + c minus 1 Therefore g + 1) = t lowast k+1

+1(k + t lowast kmaxvk+t lowast k

) Z + c onwards Carrying out the same inductive process as before it is

readily seen that g(nk)

contains zeros from level lk(n) = maxvk+n(k+t lowast k) Z +c+c2(nminust lowast kminus1)

onwards for n = t lowast k + 1 t lowast k + 2 Combining both cases we obtain

u + c(k + n) if n = 0 1 t lowast lk(n) = k (32)

maxlk(t lowast) + c2 Z + c + c2(n minus t lowast minus 1) if n = t lowast + 1 t lowast + 2 k k k k

Note that by construction lk(t lowast k) gt Z and lk(n) le Z for n = 0 1 t lowast k minus1 Interestingly

12

if we extend the definition tk lowast by introducing the quantity ζk

lowast = maxtklowast 0 we remark that

(31) and (32) can actually be combined into a single formula for lk(n) namely

vk+n(zk + n) if n = 0 1 ζklowast

lk(n) = (33) maxlk(ζ

lowast) + c2 Z + c + c2(n minus ζlowast minus 1) if n = ζlowast + 1 ζlowast + 2 k k k k

Clearly we have that lk(n) ge Z + c for n = ζk lowast +1 ζk

lowast +2 Moreover if c1 = c2 = c (33)

reduces to give lk(n) = u + c(k + n) forall n isin N which is in agreement with the results in Alfa

and Drekic (2007 Section 3)

Remark In our above analysis we purposely ignored the possibility of any future claims

occurring after time k in our determination of lk(n) However we recognize that this is not

realistic and future claims would indeed occur after time k thereby further reducing the value

of the maximum surplus level lk(n) minus 1 at time k + n To account for this we imagine all

future interclaim times (ie after time k) to be of maximum length na and all corresponding

claim sizes to be of minimum amount 1 Consequently in the case when c2 gt 0 we should

subtract from formula (33) the amount of lfloornnarfloor where lfloormiddotrfloor again denotes the floor function

Obviously when c2 = 0 (ie no premiums are collected whenever the surplus level is at or

above Z) lk(n) = Z + c for n = ζk lowast + 1 ζk

lowast + 2 (k)

We now turn our attention to the development of a general algorithm for computing gni

(k) (k)for i = 0 1 lk(n) minus 1 With the aid of (24) we apply the key identity gn = gnminus1C

(which can be inferred from (27)) to first obtain

minZlk(nminus1)minus1 lk(nminus1)minus1 c2 L L L

(k) (k) (k)gni = g Bjminusi+c + gnminus1j dmBjminusi+m n isin Z+ (34) nminus1j

j=0 j=Z m=c1

However (22) states that Bi = Ona for i isin Zminus and so (34) becomes (for the same range of

i)

minZlk(nminus1)minus1 c2 lk(nminus1)minus1 L L L

(k) (k) (k)gni = g Bjminusi+c + dm gnminus1jBjminusi+m n isin Z+ (35) nminus1j

j=max0iminusc m=c1 j=maxZiminusm

Taking the form of (33) into consideration (35) yields the following two formulae

lk(nminus1)minus1 L

(k) (k)gni = gnminus1jBjminusi+c n = 1 2 ζ k

lowast (36) j=max0iminusc

and

Zminus1 c2 lk(nminus1)minus1 L L L

(k) (k) (k)g = g + dm g n = ζ lowast + 1 ζ lowast + 2 ni nminus1jBjminusi+c nminus1jBjminusi+m k k

j=max0iminusc m=c1 j=maxZiminusm

(37)

13

For n = 1 2 ζk lowast where ζk

lowast isin Z+ we note that lk(n) ge c + 1 Therefore with the aid of

(22) (36) simplifies to give

Llk(nminus1)minus1 (k) (k) j=0 gnminus1jBjminusi+c if i = 0 1 c minus 1

gni = Llk(nminus1)minus1 (k)

g if i = c c + 1 lk(n) minus 1j=iminusc nminus1jBjminusi+c

Llk(nminus1)minus1 (k)

nminus1js 0 0 if i = 0 1 c minus 1j=0 αjminusi+cg= (38)

(k) Llk(nminus1)minus1 (k) g S+ αjminusi+cgnminus1js 0 0 if i = c c+1 lk(n)minus1 nminus1iminusc j=iminusc+1

In the case of n = ζk lowast + 1 ζk

lowast + 2 we first reiterate that lk(n) ge Z + c We next break

(37) into several cases based on the value of i and the value of minc Z + c1 as follows

Case 1 Suppose minc Z + c1 = c

For i isin 0 1 c minus 1

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+mni nminus1j nminus1j

j=0 m=c1 j=Z

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k)= αjminusi+cg s + dm αjminusi+mg s 0 0 (39) nminus1j nminus1j

j=0 m=c1 j=Z

For i isin c c + 1 Z + c1 minus 1

Zminus1 c2 lk(nminus1)minus1 L L L

(k) (k) (k)g = g + dm gni nminus1j Bjminusi+c nminus1j Bjminusi+m

j=iminusc m=c1 j=Z

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k) (k)= g S + αjminusi+cg dm αjminusi+mgnminus1j s 0 0 (310) nminus1iminusc nminus1j s +

j=iminusc+1 m=c1 j=Z

For i isin Z + c1 Z + c1 + 1 Z + c minus 1

Zminus1 minc2iminusZ lk(nminus1)minus1 c2 lk(nminus1)minus1 L L L L L

(k) (k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+m + dm g Bjminusi+mni nminus1j nminus1j nminus1j

j=iminusc m=c1 j=iminusm m=iminusZ+1 j=Z

minc2iminusZ L

(k) (k)= g S + dmg S nminus1iminusc nminus1iminusm

m=c1

Zminus1 minc2iminusZ lk(nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k)+ αjminusi+cg dm αjminusi+mg dm αjminusi+mgnminus1j s 0 0 nminus1j s + nminus1js +

j=iminusc+1 m=c1 j=iminusm+1 m=iminusZ+1 j=Z

(311)

For i isin Z + c Z + c + 1 lk(n) minus 1

c2 lk(nminus1)minus1 L L

(k) (k)g = dm g Bjminusi+mni nminus1j

m=c1 j=iminusm

c2 c2 lk (nminus1)minus1 L L L

(k) (k)= dmg S + dm αjminusi+mgnminus1j s 0 0 (312) nminus1iminusm

m=c1 m=c1 j=iminusm+1

14

Case 2 Suppose minc Z + c1 = Z + c1

For i isin 0 1 Z + c1 minus 1

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+mni nminus1j nminus1j

j=0 m=c1 j=Z

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k)= αjminusi+cg s + dm αjminusi+mg s 0 0 (313) nminus1j nminus1j

j=0 m=c1 j=Z

For i isin Z + c1 Z + c1 + 1 c minus 1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+m + dm g Bjminusi+mni nminus1j nminus1j nminus1j

j=0 m=c1 j=iminusm m=iminusZ+1 j=Z

minc2iminusZ L

(k)= dmg S nminus1iminusm

m=c1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k)+ αjminusi+cg s + dm αjminusi+mg s + dm αjminusi+mg s 0 0 nminus1j nminus1j nminus1j

j=0 m=c1 j=iminusm+1 m=iminusZ+1 j=Z

(314)

For i isin c c + 1 Z + c minus 1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k) (k)g = g + g + gni nminus1j Bjminusi+c dm nminus1j Bjminusi+m dm nminus1j Bjminusi+m

j=iminusc m=c1 j=iminusm m=iminusZ+1 j=Z

minc2iminusZ L

(k) (k)= g S + dmg S nminus1iminusc nminus1iminusm

m=c1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k)+

nminus1j s 0 0 αjminusi+cgnminus1j s + dm αjminusi+mgnminus1j s + dm αjminusi+mg j=iminusc+1 m=c1 j=iminusm+1 m=iminusZ+1 j=Z

(315)

For i isin Z + c Z + c + 1 lk(n) minus 1

c2 lk (nminus1)minus1 L L

(k) (k)g = dm g Bjminusi+mni nminus1j

m=c1 j=iminusm

c2 c2 lk (nminus1)minus1 L L L

(k) (k)= dmg S + dm αjminusi+mgnminus1j s 0 0 (316) nminus1iminusm

m=c1 m=c1 j=iminusm+1

It is not difficult to verify that equations (38) through (316) simplify to produce equations

(32) and (33) of Alfa and Drekic (2007 pp 301-302) when c1 = c2 = c With equations (38) (k)

through (316) providing the recursive means of calculating gn we next turn our attention

15

to a determination of h(k) defined by (28) For n = 1 2 ζlowast we use (22) and (25) to n k

obtain

lk(nminus1)minus1 lk(nminus1)minus1

h(k) nminusj =

L g

(k) nminus1ℓBc+j+ℓ =

L αc+j+ℓg

(k) nminus1ℓs 0 0 (317)

ℓ=0 ℓ=0

For n = ζk lowast + 1 ζk

lowast + 2 we similarly acquire

Zminus1 lk(nminus1)minus1 L L

(k) (k) (k)hnminusj = gnminus1ℓBc+j+ℓ + gnminus1ℓAj+ℓ

ℓ=0 ℓ=Z

Zminus1 c2 lk(nminus1)minus1 L L L

(k) (k)= αc+j+ℓg s + dm αm+j+ℓg s 0 0 (318) nminus1ℓ nminus1ℓ

ℓ=0 m=c1 ℓ=Z

Note that substituting (317) and (318) into (29) immediately yields

Llk(nminus1)minus1 (k) (k) ℓ=0 αc+j+ℓgnminus1ℓs if n = 1 2 ζk

lowast φ (u) = nj LZminus1 (k) Lc2

Llk(nminus1)minus1 (k) ζlowast s + dm αm+j+ℓg n = + 1 ζlowast + 2 ℓ=0 αc+j+ℓgnminus1ℓ m=c1 ℓ=Z nminus1ℓs if k k

(319)

Clearly in the special case when c1 = c2 = c (319) yields equation (35) of Alfa and Drekic

(2007 p 302) Finally we remark that (319) can be substituted into (213) to obtain a

computational formula for φnj(u)

4 Numerical examples and findings

In this section we illustrate the application of our proposed algorithm with several numerical

examples All calculations in this section were performed using Mathematica (Version 7) on

a Dell Precision T7400 workstation with a 3 Ghz CPU and 4 GB of RAM

In our first illustration (labelled Example 1) we consider the ordinary Sparre Andersen

model in which interclaim times have pmf given by

(211)(911)jminus1 if j = 1 2 na minus 1 aj = (41)

(911)naminus1 if j = na

Note that the pmf (41) is that of a truncated geometric distribution with all the probability

mass on na na +1 assigned to the support value na Clearly as na becomes larger the

closer (41) approximates this particular geometric distribution having mean 55 In what Lna

Lminus1 nafollows we fix na = 25 so that aj = 0992 and jaj = 546 Let the individual j=1 j=1

claim amount distribution be given by the pmf

minus4 minus4j minus 1 j

αj = 1 + minus 1 + j isin Z+ (42) 30 30

16

We remark that the pmf (42) represents a discretized version of the Pareto distribution with

mean 10 Also note that mα = infin thereby implying that the random variables |UT | and

UTminus both have infinite support

Table 1 displays values (rounded to 6 decimal places) of the trivariate cumulative distrishy

bution function

n x y L L L

Ψnxy(u) = Pr T le n UTminus le x |UT | le y | U0 = u = ψℓij(u) ℓ=1 i=mincZ+c1 j=1

corresponding to the above discrete-time risk process with u = 10 Z = 50 c = 5 and

varying premium amount distributions according to the following set of scenarios

(1) d2 = 1

(2) d1 = 25 d2 = 15 d3 = 25

G

5)

(3) di = i

(25)i(35)5minusi i = 0 1 5

(4) di = 15 i = 0 1 4

(5) d1 = 34 d5 = 14

(6) d0 = 35 d5 = 25

We remark that while these six premium amount distributions all have mean equal to 2

they possess different shapes and we investigate the impact of their variability on the joint

distribution of T |UT | and UTminus In particular the variances corresponding to the premium

amount distributions of scenarios (1) through (6) are 0 08 12 2 3 and 6 respectively

Based on the results in Table 1 it would appear that greater variability in the premium

amount distribution generally leads to higher ruin probabilities Perhaps somewhat surprising

though the differences are subtle and not particularly striking as one compares the various

ruin probabilities from scenarios (1) through (6) Obviously we observe more of a difference

between scenarios as the values of n x and y increase

As a follow-up illustration (labelled Example 2) we consider a model in which ordinary

interclaim times have a discrete uniform distribution with pmf aj = 110 j = 1 2 10

Clearly the mean of this distribution is 55 which for comparative purposes is identical to

the ordinary interclaim time mean considered in Example 1 In addition we assume that

the pmf of the individual claim amount distribution is once again given by (42) Table 2

displays values (rounded to 6 decimal places) of Ψnxy(10) for several variants of the Sparre

Andersen model (again with Z = 50 and c = 5) having the above ordinary interclaim time

and claim amount distributions namely

17

(1) ordinary model under scenario (2) from Example 1

(2) ordinary model under scenario (5) from Example 1

(3) stationary model under scenario (2) from Example 1

(4) stationary model under scenario (5) from Example 1

(5) delayed model (with W1 = 1 with probability 1) under scenario (2) from Example 1

(6) delayed model (with W1 = 1 with probability 1) under scenario (5) from Example 1

Concerning the results in Table 2 we first compare the values of Ψnxy(10) in entries (2)

and (5) of Table 1 to their corresponding counterparts in entries (1) and (2) of Table 2 In all

cases we observe that ruin is more likely to occur when interclaim times have the truncated

geometric distribution introduced in our first example as opposed to the discrete uniform

distribution of our second example Since mean interclaim times are equal in both examshy

ples this behaviour must be attributable to the greater variability inherent in the truncated

geometric distribution We also note that under the stationary model assumption the pmf

of W1 is easily verified to be rj = (11 minus j)55 j = 1 2 10 which has mean 4 lt 55

As a result the values of Ψnxy(10) for stationary models (3) and (4) are always larger than

the corresponding values for ordinary models (1) and (2) Finally in the absence of any

knowledge regarding which distribution is in effect for the first interclaim time W1 models

(5) and (6) can be regarded as ldquoworst-case scenariordquo models as they serve to provide upper

bounds on the values of Ψnxy(10)

In conclusion we state that the values in Tables 1 and 2 were generated by first impleshy

menting the recursive procedures (38) through (316) and then summing the appropriate

bivariatetrivariate probabilities computed via (213) (214) or (215) In situations when

x = y = infin however we computed the resulting finite-time ruin probabilities using an altershy

native (albeit standard) method (eg see Cossette et al (2006 Section 3)) Specifically if

we define the survival probability σ(n u) = Pr T gt n | U0 = u = 1minusΨninfininfin(u) as well as

the auxiliary function

0 if w ge Z ηwt =

minlfloorZminuswc minus1 rfloor + 1 t if w lt Z

then by conditioning on the time and amount of the first claim (and recalling that the delayed

process reverts to the ordinary process following this claim) we obtain the expression

minnnr (kminuszk)c2 u+czk+ℓ L L L

σ(u n) = Rn + rk dℓkminuszk αjγ(u + czk + ℓ minus j n minus k)

k=1 ℓ=(kminuszk)c1 j=1

18

where the function γ(w m) for w m isin N is computed recursively via

minmna (kminusηwk)c2 u+cηwk+ℓ L L L

γ(w m) = Am + ak dℓkminusηwk αjγ(u + cηwk + ℓ minus j m minus k)

k=1 j=1 ℓ=(kminusηwk)c1

with starting values γ(w 0) = 1 forall w isin N

Acknowledgements

The authors would like to thank Dr David Landriault for his valuable suggestions which

helped to improve an earlier version of this work This research was supported by the Natural

Sciences and Engineering Research Council of Canada

References

Alfa AS 2004 Markov chain representations of discrete distributions applied to queueing

models Computers amp Operations Research 31 2365-2385

Alfa AS Drekic S 2007 Algorithmic analysis of the Sparre Andersen model in discrete

time ASTIN Bulletin 37 293-317

Bao Z 2007 A note on the compound binomial model with randomized dividend strategy

Applied Mathematics amp Computation 194 276-286

Cossette H Landriault D Marceau E 2006 Ruin probabilities in the discrete time

renewal risk model Insurance Mathematics amp Economics 38 309-323

Karlin S Taylor HM 1975 A First Course in Stochastic Processes 2nd edition Acashy

demic Press New York

Kim B Kim H-S Kim J 2008 A risk model with paying dividends and random

environment Insurance Mathematics amp Economics 42 717-726

Landriault D 2008 Randomized dividends in the compound binomial model with a general

premium rate Scandinavian Actuarial Journal 1-15

Tan J Yang X 2006 The compound binomial model with randomized decisions on

paying dividends Insurance Mathematics amp Economics 39 1-18

19

Table 1 ndash Values of Ψnxy(10) corresponding to Example 1

(a) x = 10

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0005357 0006188 0007163 0008278 0009481

(2) 0005359 0006195 0007175 0008296 0009507

(3) 0005361 0006198 0007181 0008305 0009519

(4) 0005363 0006205 0007193 0008323 0009544

(5) 0005366 0006212 0007208 0008346 0009576

(6) 0005374 0006236 0007249 0008410 0009664

y = 25 (1) 0007631 0008815 0010202 0011789 0013503

(2) 0007634 0008824 0010220 0011816 0013539

(3) 0007636 0008828 0010228 0011829 0013557

(4) 0007639 0008838 0010245 0011854 0013593

(5) 0007643 0008849 0010266 0011887 0013638

(6) 0007656 0008882 0010325 0011978 0013764

y = 50 (1) 0008477 0009792 0011333 0013095 0014998

(2) 0008481 0009802 0011352 0013125 0015039

(3) 0008483 0009807 0011361 0013139 0015059

(4) 0008486 0009817 0011380 0013168 0015098

(5) 0008491 0009830 0011403 0013204 0015149

(6) 0008504 0009866 0011470 0013305 0015288

y = infin (1) 0008777 0010138 0011733 0013558 0015528

(2) 0008781 0010149 0011753 0013588 0015570

(3) 0008783 0010154 0011763 0013603 0015591

(4) 0008787 0010164 0011782 0013633 0015632

(5) 0008791 0010177 0011807 0013671 0015684

(6) 0008805 0010215 0011875 0013775 0015828

20

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0058559 0061875 0065704 0070048 0074714

(2) 0058569 0061903 0065756 0070128 0074825

(3) 0058574 0061917 0065782 0070167 0074880

(4) 0058584 0061945 0065833 0070245 0074987

(5) 0058596 0061979 0065896 0070344 0075124

(6) 0058632 0062079 0066078 0070621 0075506

y = 25 (1) 0089002 0093986 0099739 0106262 0113270

(2) 0089017 0094028 0099817 0106382 0113437

(3) 0089025 0094049 0099856 0106442 0113519

(4) 0089040 0094091 0099932 0106560 0113681

(5) 0089057 0094142 0100028 0106708 0113887

(6) 0089113 0094294 0100302 0107127 0114463

y = 50 (1) 0102786 0108506 0115106 0122590 0130629

(2) 0102804 0108555 0115196 0122728 0130820

(3) 0102812 0108579 0115241 0122797 0130914

(4) 0102830 0108627 0115329 0122932 0131101

(5) 0102850 0108686 0115439 0123103 0131337

(6) 0102914 0108860 0115754 0123584 0132000

y = infin (1) 0108895 0114933 0121901 0129799 0138283

(2) 0108914 0114985 0121996 0129945 0138485

(3) 0108923 0115010 0122043 0130018 0138585

(4) 0108941 0115061 0122136 0130161 0138782

(5) 0108962 0115123 0122252 0130341 0139032

(6) 0109030 0115307 0122585 0130849 0139732

21

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0079856 0086792 009464 0103440 0112831

(2) 0079885 0086869 0094778 0103650 0113118

(3) 0079953 0086978 0094936 0103864 0113393

(4) 0080081 0087188 0095243 0104282 0113929

(5) 0079962 0087075 0095149 0104213 0113890

(6) 0080805 0088366 0096951 0106588 0116875

y = 25 (1) 0124591 0135680 0148207 0162245 0177217

(2) 0124639 0135806 0148433 0162586 0177683

(3) 0124755 0135993 0148703 0162950 0178148

(4) 0124978 0136354 0149226 0163658 0179054

(5) 0124763 0136141 0149035 0163500 0178935

(6) 0126238 0138381 0152148 0167588 0184060

y = 50 (1) 0146757 0160079 0175111 0191945 0209894

(2) 0146816 0160232 0175385 0192360 0210460

(3) 0146965 0160469 0175725 0192817 0211043

(4) 0147249 0160925 0176384 0193706 0212178

(5) 0146968 0160639 0176117 0193470 0211980

(6) 0148860 0163500 0180077 0198658 0218474

y = infin (1) 0158013 0172606 0189059 0207476 0227106

(2) 0158078 0172776 0189362 0207934 0227732

(3) 0158250 0173047 0189751 0208455 0228395

(4) 0158579 0173571 0190503 0209467 0229685

(5) 0158247 0173226 0190172 0209161 0229410

(6) 0160443 0176532 0194735 0215127 0236869

22

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0088667 0104403 0123329 0145277 0169184

(2) 0088711 0104518 0123538 0145598 0169627

(3) 0088733 0104574 0123641 0145756 0169846

(4) 0088777 0104687 0123844 0146066 0170274

(5) 0088829 0104828 0124105 0146470 0170832

(6) 0088988 0105229 0124822 0147560 0172331

y = 25 (1) 0140528 0167839 0201007 0239726 0282102

(2) 0140598 0168021 0201338 0240234 0282804

(3) 0140632 0168111 0201502 0240486 0283151

(4) 0140701 0168289 0201822 0240976 0283827

(5) 0140785 0168515 0202241 0241622 0284718

(6) 0141036 0169148 0203370 0243335 0287071

y = 50 (1) 0167826 0203005 0246173 0296954 0352855

(2) 0167908 0203218 0246561 0297548 0353672

(3) 0167949 0203324 0246753 0297841 0354075

(4) 0168030 0203532 0247128 0298412 0354859

(5) 0168129 0203800 0247622 0299174 0355907

(6) 0168423 0204539 0248936 0301159 0358620

y = infin (1) 0183325 0224926 0276881 0338957 0408257

(2) 0183412 0225150 0277283 0339564 0409079

(3) 0183455 0225260 0277481 0339864 0409485

(4) 0183540 0225478 0277869 0340446 0410270

(5) 0183645 0225760 0278386 0341234 0411339

(6) 0183954 0226529 0279737 0343244 0414037

23

Table 2 ndash Values of Ψnxy(10) corresponding to Example 2

(a) x = 10

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0001901 0002276 0002724 0003240 0003799

(2) 0001904 0002286 0002742 0003269 0003840

(3) 0002651 0003013 0003442 0003936 0004471

(4) 0002655 0003022 0003460 0003964 0004510

(5) 0005792 0006098 0006458 0006871 0007317

(6) 0005795 0006106 0006473 0006894 0007350

y = 25 (1) 0002708 0003243 0003880 0004615 0005411

(2) 0002713 0003257 0003906 0004656 0005469

(3) 0003778 0004293 0004904 0005608 0006369

(4) 0003782 0004306 0004929 0005647 0006424

(5) 0008255 0008690 0009203 0009792 0010427

(6) 0008259 0008702 0009224 0009824 0010473

y = 50 (1) 0003008 0003603 0004310 0005127 0006011

(2) 0003014 0003618 0004340 0005173 0006076

(3) 0004197 0004769 0005448 0006230 0007075

(4) 0004202 0004784 0005476 0006273 0007137

(5) 0009173 0009656 0010226 0010879 0011585

(6) 0009177 0009669 0010249 0010916 0011637

y = infin (1) 0003115 0003730 0004463 0005308 0006224

(2) 0003121 0003746 0004493 0005356 0006291

(3) 0004346 0004938 0005641 0006450 0007325

(4) 0004351 0004954 0005670 0006496 0007389

(5) 0009499 0009999 0010589 0011265 0011996

(6) 0009503 0010012 0010613 0011303 0012050

24

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0031176 0032815 0034733 0036924 0039282

(2) 0031193 0032862 0034821 0037061 0039475

(3) 0050168 0051747 0053587 0055684 0057939

(4) 0050184 0051792 0053671 0055815 0058124

(5) 0136884 0138219 0139763 0141517 0143399

(6) 0136896 0138256 0139832 0141626 0143553

y = 25 (1) 0047622 0050092 0052981 0056280 0059829

(2) 0004765 0050163 0053113 0056487 0060121

(3) 0076570 0078950 0081720 0084878 0088272

(4) 0076593 0079017 0081847 0085076 0088551

(5) 0205774 0207786 0210111 0212752 0215585

(6) 0205793 0207842 0210216 0212917 0215818

y = 50 (1) 0055170 0058009 0061329 0065119 0069196

(2) 0055200 0058091 0061481 0065357 0069533

(3) 0088650 0091385 0094569 0098197 0102096

(4) 0088677 0091463 0094715 0098425 0102417

(5) 0235815 0238128 0240800 0243834 0247089

(6) 0235837 0238192 0240920 0244023 0247356

y = infin (1) 0058565 0061565 0065073 0069077 0073384

(2) 0058597 0061652 0065234 0069329 0073740

(3) 0094061 0096952 0100316 0104149 0108267

(4) 0094090 0097035 0100470 0104390 0108607

(5) 0248472 0250917 0253740 0256945 0260383

(6) 0248495 0250985 0253867 0257145 0260667

25

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0048598 0052524 0056984 0061998 0067342

(2) 0048654 0052673 0057253 0062411 0067916

(3) 0070192 0073979 0078257 0083057 0088167

(4) 0070243 0074119 0078513 0083451 0088714

(5) 0151774 0154981 0158572 0162586 0166852

(6) 0151815 0155096 0158784 0162913 0167308

y = 25 (1) 0076895 0083230 0090413 0098481 0107073

(2) 0076987 0083474 0090854 0099156 0108010

(3) 0110093 0116205 0123095 0130818 0139033

(4) 0110177 0116435 0123515 0131462 0139928

(5) 0230674 0235848 0241632 0248091 0254950

(6) 0230739 0236038 0241980 0248626 0255695

y = 50 (1) 0091496 0099156 0107829 0117563 0127924

(2) 0091610 0099455 0108370 0118387 0129068

(3) 0130130 0137520 0145839 0155156 0165064

(4) 0130233 0137801 0146352 0155942 0166156

(5) 0266599 0272857 0279840 0287632 0295904

(6) 0266680 0273088 0280265 0288286 0296814

y = infin (1) 0099312 0107746 0117285 0127984 0139369

(2) 0099439 0108078 0117883 0128899 0140639

(3) 0140458 0148595 0157745 0167986 0178872

(4) 0140572 0148908 0158314 0168859 0180084

(5) 0282886 0289776 0297457 0306022 0315111

(6) 0282976 0290034 0297929 0306748 0316120

26

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0059096 0072505 0088907 0108122 0129137

(2) 0059223 0072829 0089500 0109052 0130456

(3) 0079757 0092662 0108388 0126778 0146871

(4) 0079875 0092970 0108956 0127669 0148134

(5) 0159788 0170682 0183871 0199249 0216024

(6) 0159881 0170936 0184343 0199990 0217075

y = 25 (1) 0095883 0119741 0149240 0184073 0222390

(2) 0096087 0120262 0150197 0185573 0224520

(3) 0127403 0150359 0178640 0211977 0248613

(4) 0127592 0150855 0179557 0213414 0250653

(5) 0245161 0264533 0288250 0316124 0346709

(6) 0245311 0264942 0289010 0317319 0348407

y = 50 (1) 0116598 0147920 0187100 0233784 0285503

(2) 0116839 0148537 0188231 0235554 0288015

(3) 0153023 0183153 0220713 0265390 0314840

(4) 0153249 0183742 0221798 0267087 0317248

(5) 0285742 0311159 0342651 0380005 0421287

(6) 0285921 0311644 0343551 0381418 0423292

y = infin (1) 0129470 0167217 0215383 0273841 0339734

(2) 0129725 0167865 0216560 0275661 0342282

(3) 0167978 0204275 0250442 0306382 0369383

(4) 0168217 0204895 0251572 0308128 0371826

(5) 0305868 0336470 0375169 0421935 0474525

(6) 0306058 0336980 0376106 0423388 0476560

27

Page 2: uwaterloo.ca · Ruin Analysis of a Threshold Strategy in a Discrete-time Sparre Andersen Model Steve DREKIC and Ana Maria MERA Department of Statistics and Actuarial Science University

1 Introduction and notation

Alfa and Drekic (2007) recently considered a general delayed Sparre Andersen insurance

risk model in discrete time and analyzed it as a doubly-infinite Markov chain to establish

a computational procedure for calculating the (trivariate) joint distribution of the time of

ruin the surplus immediately prior to ruin and the deficit at ruin In this paper we extend

their essential approach to a more general model which now features an embedded threshold

dividend strategy and randomized dividend payments This paper generalizes the model

considered by Tan and Yang (2006) in which a discrete-time process based on the compound

binomial model (ie geometrically distributed interclaim times) is analyzed wherein a random

dividend of 0 or 1 is paid whenever the surplus process exceeds a specified (fixed) threshold

level In addition this paper serves as a contribution to the current state of knowledge

regarding discrete-time risk models with a randomized decision rule for paying dividends a

topic which has garnered some recent attention in the ruin theoretic literature (eg see Bao

(2007) Landriault (2008) and Kim et al (2008))

In this paper we assume that the number of claims process Nt t = 0 1 is a modishy

fied discrete-time renewal process with independent positive interclaim times W1W2

where W1 is the duration from time 0 until the first claim occurs and Wi i = 2 3 is

the time between the (i minus 1)-th and i-th claims We also remark that W2W3 forms

an independent and identically distributed (iid) sequence of positive random variables with

common probability mass funtion (pmf) aj = PrWi = j j = 1 2 na and correspondshy

ing survival function Aj = PrWi gt j = 1 minus L

k

j

=1 ak Throughout the paper we assume

that na lt infin (ie the interclaim time distribution of Wi i = 2 3 has finite support)

In the ordinary Sparre Andersen model a claim is assumed to have occurred at time 0minus

so that W1 has the same distribution as the ordinary interclaim times W2W3 On the

other hand if W1 is not a ldquofullrdquo interclaim time then asymptotically in time the limiting Lnadistribution of this forward recurrence time is defined via the pmf aj = Ajminus1 k=1 Akminus1

j = 1 2 na (eg see Karlin and Taylor (1975 pp 192-193)) This yields another

important risk model namely the stationary Sparre Andersen model in which W1 has pmf aj

rather than aj As a means of accommodating other possible specifications we assume that

W1 has a more general pmf denoted by rj = PrW1 = j j = 1 2 nr where nr lt infin

Let Rj = PrW1 gt j = 1 minus L

k

j

=1 rk denote its survival function Through appropriate

choice of rj it is clear that both the ordinary and stationary Sparre Andersen variants are

special cases of this more general model referred to as the delayed Sparre Andersen risk

model

In what follows let Z denote the set of all integers Zminus the set of negative integers Z+

2

the set of positive integers and N = 0 cup Z+ For t isin Z+ we define Ut as the insurerrsquos

amount of surplus at time t In actual fact Ut represents the amount of surplus at the end

of time interval (t minus 1 t] at which point any premiumsclaims corresponding to this time

interval have been receivedpaid out Specifically with respect to the time interval (t minus 1 t]

we adopt the convention that premiums are received at (t minus 1)+ and any claims are paid

out at tminus We incorporate the notion of a threshold level Z isin Z+ on the insurerrsquos surplus

affecting the amount of premium being received at a given point in time More precisely we

assume that premiums corresponding to the time interval (t t + 1] t isin N are collected at t+

according to (random) rate pt such that

c if Ut lt Z pt =

Xt if Ut ge Z

where di = PrXt = i i = c1 c1 + 1 c2 We refer to c isin Z+ as the pure (constant) L

premium and assume that c1 c2 isin 0 1 c with c1 le c2 and c2 di = 1 Clearly c1i=c1

and c2 are the respective lower and upper support values of the distribution of the random

premium amount at time t Xt Correspondingly we interpret c minus pt as the amount of

(randomized) dividends paid to insureds at time t+ Note that by assumption the probability

distribution of Xt is identical for all values of t

Beginning at time 0 with an initial reserve of u isin N the insurerrsquos amount of surplus at

time t is expressible as tminus1 Nt L L

U t = u + pi minus Yi t isin N i=0 i=1

where individual claim amounts Y1 Y2 are assumed to form an iid sequence of positive

random variables with common pmf αj j = 1 2 mα and corresponding survival function LjΛj = 1 minus k=1 αk Unlike the interclaim time distributions defined above we remark that

the claim amount distribution can be either of finite or infinite support (ie mα le infin) Let

the time of ruin T be defined as T = mint isin Z+|Ut lt 0 with T = infin if Ut ge 0 forall t isin Z+

If ruin does occur let |UT | denote the deficit at ruin and UTminus = UTminus1 + pTminus1 the surplus

immediately prior to ruin Clearly T = infin if mα le minc Z + c1 Conversely if mα gt

minc Z + c1 it is not difficult to verify that |UT | isin 1 2 mα minus minc Z + c1 and

UTminus isin minc Z + c1 minc Z + c1 + 1 mα minus 1 Throughout the remainder of the

paper we assume that mα gt minc Z + c1

Following this introduction Section 2 sets up the underlying Markov chain structure in

our model and derives some necessary preliminaries In Section 3 we further exploit this

structure to construct a set of computational procedures enabling one to calculate probashy

bility distributions associated with the random variables T |UT | and UTminus with specific

3

consideration to the joint pmfrsquos

Z+ωni(u) = Pr T = n UTminus = i | U0 = u n isin i = minc Z+c1 minc Z+c1+1 mαminus1

φnj(u) = Pr T = n |UT | = j | U0 = u n isin Z+ j = 1 2 mαminusminc Z +c1

and

ψnij(u) = Pr T = n UTminus = i |UT | = j | U0 = u n isin Z+

i = minc Z +c1 minc Z +c1 + 1 mαminus1

j = 1 2 3 mαminusi

Finally we implement our various procedures and present some numerical examples in Section

4

2 Model formulation and preliminaries

Adopting the same methodology employed by Alfa and Drekic (2007) let τj =

Pr W gt j|W gt j minus 1 = AjAjminus1 j = 1 2 na where W denotes an arbitrary Wi

i = 2 3 Clearly τ1 = A1 and τna = 0 If we next define the 1 times na row vector

e1 = (1 0 0) the na times na probability transition matrix

0 τ1 0 middot middot middot 0

0 0 τ2

0

S =

0 0 middot middot middot 0 τnaminus1

0 0 middot middot middot 0 0

and the na times 1 column vector

1 minus τ1 1 minus τ2 s = 1 minus τnaminus1

1

it is known that aj = e1Sjminus1 s j = 1 2 na (eg see Alfa (2004)) In other words W

can be expressed as a discrete phase-type random variable (of dimension na) with the above

parameterization implying that s contains the absorption probabilities to claim occurrence

associated with each of the na possible interclaim time values Alfa (2004) refers to this

4

characterization as the ldquoelapsed timerdquo phase-type representation of the pmf aj We remark

that the above pmf formula also holds true for j gt na as it is readily verified that Sjminus1 = Ona

an na times na matrix of zeros for such values of j

In what follows we proceed to set up our delayed Sparre Andersen model as a bivariate

Markov chain conditional on W1 = k isin 1 2 nr Specifically assuming that W1 = k we

construct the bivariate stochastic process (Ut Lt) t = k k + 1 such that Ut represents

the amount of surplus at time t and Lt denotes a specific counter at time t which measures the

ldquoelapsed timerdquo (in the sense described above) length of the next claim occurrence We refer to

the Ut component as the level of the process and the Lt component as the phase of the process

Two important observations are particularly noteworthy regarding this construction First of

all Lk = 1 since we are assuming that the very first claim occurs at time kminus More generally

Lt = 1 if a claim happens to occur at time tminus Secondly the delayed process reverts to the

ordinary process (having interclaim time distribution defined by the pmf aj) upon occurrence

of the first claim As a result the bivariate stochastic process (Ut Lt) t = k k + 1

possesses the following Markovian relationship

(Ut + pt Lt + 1) if there is no claim at time (t + 1)minus (Ut+1 Lt+1) =

(Ut + pt minus Y 1) if there is a claim of amount Y at time (t + 1)minus

Note that the state space for this Markov chain is given by Δ = Z times 1 2 na Furthershy

more the probability transition matrix associated with this Markov chain is given by

middot middot middot minus1 10 middot middot middot Zminus2 Zminus1 Z Z+1 Z+2 middot middot middot

middot middot middot

middot middot middot minus1 middot middot middot Bc Bcminus1 Bcminus2 middot middot middot BcminusZ+1 BcminusZ BcminusZminus1 BcminusZminus2 BcminusZminus3 middot middot middot

0

1

middot middot middot Bc+1 Bc Bcminus1 middot middot middot BcminusZ+2 BcminusZ+1 BcminusZ BcminusZminus1 BcminusZminus2 middot middot middot middot middot middot Bc+2 Bc+1 Bc middot middot middot BcminusZ+3 BcminusZ+2 BcminusZ+1 BcminusZ BcminusZminus1 middot middot middot

middot middot middot

middot middot middot

P = Zminus2 middot middot middot BZ+cminus1 BZ+cminus2 BZ+cminus3 middot middot middot Bc Bcminus1 Bcminus2 Bcminus3 Bcminus4 middot middot middot (21)

Zminus1 middot middot middot BZ+c BZ+cminus1 BZ+cminus2 middot middot middot Bc+1 Bc Bcminus1 Bcminus2 Bcminus3 middot middot middot

Z middot middot middot AZ+1 AZ AZminus1 middot middot middot A2 A1 A0 Aminus1 Aminus2 middot middot middot Z+1 middot middot middot AZ+2 AZ+1 AZ middot middot middot A3 A2 A1 A0 Aminus1 middot middot middot

Z+2 middot middot middot AZ+3 AZ+2 AZ+1 middot middot middot A4 A3 A2 A1 A0 middot middot middot

middot middot middot

middot middot middot

where Ona

if i isin Zminus

Bi = S if i = 0 (22) se1αi if i isin Z+

5

and c2 L

Ai = djBi+j (23) j=c1

We give a brief indication of the reasoning leading to the form of P in (21) For example

consider the transition ldquoUt = 0 rarr Ut+1 = 1rdquo which has block element Bcminus1 containing the

transition probabilities governing the ldquoLt rarr Lt+1 rdquo process At a surplus level of 0 at time

t a premium of pt = c is certain to be received Therefore to arrive at surplus level 1 at

the next time unit a ldquoso-calledrdquo claim of size c minus 1 must occur to neutralize the premium

just received Thus the size of the ldquoclaimrdquo required for this transition to occur determines

the value of i in Bi so that Bcminus1 is obtained and its form subsequently determined by (22)

according to the value of the index c minus 1 Note that only when c gt 1 does an actual claim

occur in this particular scenario We mention that for transitions which begin at or above the

threshold level Z a similar line of reasoning would also apply except now one must condition

on the amount of the random premium being earned as indicated by the form of (23)

Note that this formulation gives rise to a doubly-infinite Markov chain with finite blocks

of size na In the special case of the threshold-free model (obtained by setting c1 = c2 = c

so that dc = 1) we remark that (21) is in agreement with the matrix P in Alfa and Drekic

(2007 p 296) Also since αi = 0 forall i gt mα we note that Bi = Ona if i gt mα

For the computation of ruin-related quantities of interest it is useful to partition the

state space Δ into two mutually exclusive state spaces namely Δ1 = N times 1 2 na

and Δ2 = Zminus times 1 2 na In addition we introduce (one-step) transition matrices

C Δ1 rarr Δ1 and D Δ1 rarr Δ2 which correspond to mapping ldquonon-ruinedrdquo states in Δ1 to

either ldquonon-ruinedrdquo states in Δ1 or ldquoruinedrdquo states in Δ2 respectively In other words the

matrices C and D are essentially submatrices of P and their forms are readily given by

210 middot middot middot Z minus 1 Z Z + 1 middot middot middot

Bc Bcminus1 Bcminus2 middot middot middot BcminusZ+1 BcminusZ BcminusZminus1 middot middot middot Bc+1 Bc Bcminus1 middot middot middot BcminusZ+2 BcminusZ+1 BcminusZ middot middot middot

Bc+2 Bc+1 Bc middot middot middot BcminusZ+3 BcminusZ+2 BcminusZ+1 middot middot middot

middot middot middot

middot middot middot

0

1

2 Z minus 2 BZ+cminus2 BZ+cminus3 BZ+cminus4 middot middot middot Bcminus1 Bcminus2 Bcminus3 middot middot middot (24) C =

Z minus 1 BZ+cminus1 BZ+cminus2 BZ+cminus3 middot middot middot Bc Bcminus1 Bcminus2 middot middot middot

Z AZ AZminus1 AZminus2 middot middot middot A1 A0 Aminus1 middot middot middot

Z + 1 AZ+1 AZ AZminus1 middot middot middot A2 A1 A0 middot middot middot

Z + 2 AZ+2 AZ+1 AZ middot middot middot A3 A2 A1 middot middot middot

middot middot middot

6

and

minus1 minus2 minus3 middot middot middot

0 Bc+1 Bc+2 Bc+3 middot middot middot 1 Bc+2 Bc+3 Bc+4 middot middot middot

2 Bc+3 Bc+4 Bc+5 middot middot middot middot middot middot

Z minus 2 BZ+cminus1 BZ+c BZ+c+1 middot middot middot D = (25) Z minus 1 BZ+c BZ+c+1 BZ+c+2 middot middot middot

Z AZ+1 AZ+2 AZ+3 middot middot middot

Z + 1 AZ+2 AZ+3 AZ+4 middot middot middot Z + 2 AZ+3 AZ+4 AZ+5 middot middot middot

Let us now specify some further definitions required in upcoming derivations First of all

we introduce the function

0 if u ge Z zt =

maxi isin 1 2 t|u + c(i minus 1) lt Z if u lt Z

or equivalently

0 if u ge Z zt =

minlfloorZminuscuminus1 rfloor + 1 t if u lt Z

where the floor function denoted lfloorxrfloor yields the largest integer less than or equal to x We

define dm0 = δm0 where δij in general denotes the Kronecker delta function of i and j

Also define

dm if m = c1 c1 + 1 c2dm1 =

0 elsewhere

and for n = 2 3

Lc2 dj1dmminusjnminus1 if m = nc1 nc1 + 1 nc2dmn = j=c1

0 elsewhere

In other words dmn represents the pmf of the n-fold convolution of dm with itself

Recalling that W1 = k note that zk + ( k+) determines the earliest time point in the set =

0+ 1+ (k minus1)+ at which a random premium is received Note also that zk = k implies

that no such point exists in this set (ie the threshold level Z has not yet been reached when

the premium at time (k minus1)+ is to be received) Therefore if we assume for the moment that

zk k we may condition on the total premiums earned at times zk + (zk + 1)+ (k minus 1)+=

7

to obtain the following expression for b(k) the initial (ie starting at time k) probability row

vector of the above Markov chain corresponding only to the states in Δ1 namely

(kminuszk)c2

b(k) L

= dmkminuszk(αu+czk+me1 αu+czk+mminus1e1 α2e1 α1e1 0 0 ) (26)

m=(kminuszk)c1

where 0 denotes the 1 times na row vector of zeros By further defining the function vn(t) =

b(k)u + ct + (n minus t)c2 we note that the i-th level of is given by the 1 times na row vector L(kminuszk)c2 b(k)αu+czk+mminusie1 for each i isin Ωk = 0 1 vk(zk)minus1 Hence contains

m=(kminuszk)c1 dmkminuszk

zeros from level vk(zk) onwards Moreover (26) holds true even in the case when zk = k

simplifying to give b(k) = (αu+cke1 αu+ckminus1e1 α2e1 α1e1 0 0 )

At this juncture we introduce two additional row vectors namely

(k) (k) (k) (k) b(k)gn = (gn0 gn1 gn2 ) = Cn n isin N (27)

and

h(k) (k) (k) (k) b(k)Cnminus1D =

(k) Z

+ n = (h nminus2 hnminus3 ) = gnminus1D n isin (28) nminus1 h

(k)We remark that gn contains the probabilities of being in the various ldquonon-ruinedrdquo states

at time k + n (ie after claims have been paid out) without having visited a ldquoruinedrdquo state

during the previous n minus 1 transitions given that Uk isin Ωk sub Δ1 In a similar fashion h(nk)

contains the probabilities of being in the various ldquoruinedrdquo states for the first time at time

k + n given that Uk isin Ωk sub Δ1 Note that the probability of being in ruined state ldquo minus jrdquo is (k) (k) (k) (k)

characterized by the row vector h = (h h h ) with the third subscript nminusj nminusj1 nminusj2 nminusjna

representing the value of the elapsed time counter Lk+n Upon further reflection however (k)

=it must be the case that h = 0 for i 1 as ruin can only occur at claim instants nminusji

which in this case implies Lk+n = 1 with probability 1 Thus it immediately follows that (k) (k) (k)

the structure of the 1 times na row vector hnminusj is simply given by hnminusj = (φnj(u) 0 0) (k)

where φ (u) = Pr T = k + n |UT | = j | Consequently as in the analysis with nj Uk isin Ωk

no threshold on the surplus level (see Alfa and Drekic (2007 p 299)) we obtain

(k) (k) prime φ (u) = h (29) nj nminusje1

where e1 prime denotes the transpose of e1

A similar line of probabilistic logic can also be applied to obtain a representation for the (k)

trivariate function ψ (u) = Pr T = k + n UTminus = i |UT | = j | Uk isin Ωk In order for ruin nij

to occur at time k + n with a surplus immediately prior to ruin equal to i we observe that

(i) none of the previous n minus 1 transitions must have included a visit to any state in Δ2

and

8

(ii) the surplus level at time k + n minus 1 must necessarily be equal to i minus pk+nminus1

Note that pk+nminus1 denotes the premium corresponding to the time interval (k + n minus 1 k + n]

which is received at (k + n minus 1)+ Hence the quantity corresponding to points (i) and (ii) (k)

above is the 1 times na row vector g Then at the next time unit a claim must nminus1iminuspk+nminus1

necessarily occur but not before a premium of pk+nminus1 is first collected thereby raising the

surplus level to i Since s contains the absorption probabilities (to claim occurrence) from

the na possible phase states and the claim causing ruin must be of size i + j in order to

ensure that the deficit at ruin is equal to j it follows that

(k) (k)ψ (u) = g s αi+j (210) nij nminus1iminuspk+nminus1

(k)The following corollary provides the means of calculating the quantity g in nminus1iminuspk+nminus1

(210)

Corollary 1 If minc Z + c1 = c then

(k)

g if i = c c+1 Z +c1 minus1 nminus1iminusc

(k) (k) Lminc2iminusZ (k)g = g + j=c1

if i = Z +c1 Z +c1 +1 minmα Z +cminus1nminus1iminuspk+nminus1 nminus1iminusc djgnminus1iminusj

Lc2 (k)

j=c1 djgnminus1iminusj if i = Z +c Z +c+1 mαminus1

(211)

Conversely if minc Z + c1 = Z + c1 then

Lminc2iminusZ (k)

djg if i = Z +c1 Z +c1 +1 cminus1nminus1iminusj j=c1

(k) (k) Lminc2iminusZ (k)gnminus1iminuspk+nminus1

= gnminus1iminusc + j=c1 djgnminus1iminusj if i = c c+1 minmα Z +cminus1

Lc2 (k)djg if i = Z +c Z +c+1 mαminus1j=c1 nminus1iminusj

(212)

Proof The justification of (211) is as follows If c is smaller than Z + c1 the only way of

reaching a surplus level of i isin c c+1 Z +c1 minus1 (prior to ruin) is by receiving a premium

of c since i minus j for any j = c1 c1 +1 c2 would result in a surplus level below Z where the

random premium rate is not applicable Next for fixed i isin Z +c1 Z +c1 +1 Z +c minus1

there are two distinct ways of reaching this surplus level First of all note that the value of

i minus j is guaranteed to produce a surplus amount greater than or equal to Z provided that

j isin c1 c1 + 1 minc2 i minus Z Secondly note that reducing each of value of i by c also

yields a non-negative surplus value less than Z Hence both pure and random premiums

are plausible under this scenario For i isin Z + c Z + c + 1 note that decreasing this

amount by any j = c1 c1 + 1 c2 results in a surplus value greater than or equal to Z as

desired However reducing it by the amount c also has the same impact thus rendering the

pure premium option inadmissible in this case Lastly we incorporate into the range of i the

9

fact that the surplus immediately prior to ruin cannot exceed mα minus1 as indicated in Section

1 The justification of (212) when c is larger than Z + c1 mirrors the analysis above and is

left to the reader D

Recall that in the analysis up to this point we have assumed that W1 = k isin

1 2 3 nr If we now condition on the value of W1 the Law of Total Probability

yields the following two expressions for φnj(u) and ψnij(u) namely

nr L

= rkφ (u) n = nr + 1 nr + 2 φnj(u) n

(kminus) kj

k=1

and nr L

(k)ψnij(u) = rkψ (u) n = nr + 1 nr + 2 nminuskij

k=1

In addition however it is quite possible that T = n for n = 1 2 nr and this can happen

as a result of either

(i) the first claim which causes ruin occurring at time nminus from a (conditional) surplus Lnminus1level of u + czn + m (where m) or i=zn

Xi =

(ii) the first claim which does not cause ruin occurring at some time kminus k isin 1 2 nminus

1 and ruin subsequently occurring n minus k time units later

If we now combine these outcomes with those for n = nr + 1 nr + 2 (and recall that

rn = 0 forall n gt nr) the general expression for φnj(u) becomes

minnminus1nr (nminuszn)c2 L

(k) L

φnj(u)= rkφ (u) + rn dmnminuszn αu+czn+m+j n isin Z+ (213) nminuskj

k=1 m=(nminuszn)c1

Concerning the function ψnij(u) when n le nr and point (i) above is considered we note Lnminus1that i must necessarily equal u + czn + m (again conditional on i=zn

Xi = m) for ruin to

occur in this manner Hence in a similar fashion we acquire

minnminus1nr (nminuszn)c2 L L

ψnij(u) = rkψn

(kminus) kij(u) + rn dmnminuszn

δiu+czn+mαu+czn+m+j

k=1 m=(nminuszn)c1

minnminus1nr L

(k) Z

+ = rkψnminuskij(u) + rndiminusuminuscznnminuszn αi+j n isin (214)

k=1

Finally for fixed i isin minc Z + c1 minc Z + c1 + 1 mα minus 1 summing (214) from

j = 1 to mα minus i yields the following expression for the bivariate pmf ωni(u) namely

minnminus1nr mαminusi mαminusi L L

(k) L

ωni(u) = rk ψnminuskij(u) + rndiminusuminuscznnminusznαi+j

k=1 j=1 j=1

10

minnminus1nr L

(k) Z

+ = Λi rkg s + rndiminusuminuscznnminuszn n isin (215) nminuskminus1iminuspnminus1

k=1

where (210) was used to establish the last equality In the special case when c1 = c2 = c it

is a straightforward exercise to show that equations (213) through (215) simplify to produce

equations (215) through (217) of Alfa and Drekic (2007 p 300)

3 Computational considerations

In order to compute the joint pmfrsquos defined by (213) (214) and (215) it is clear from (k)

equations (29) through (212) that we need to be able to compute the quantities gn and

h(k) n defined by (27) and (28) respectively In what follows we capitalize on the underlying

Markov chain structure in our model to develop a set of computational procedures which

ultimately enable one to calculate these key row vectors

First of all recall that g0(k)

= b(k) contains zeros from a certain level onwards namely (k)

level vk(zk) = u + czk + c2(k minus zk) In general one can easily deduce that gn will also

contain zeros from a certain level onwards to be denoted by lk(n) We begin this section

with a determination of lk(n) which varies in form depending on the value of vk(zk) minus 1

representing the maximum surplus level at time k after the smallest possible claim of size 1

has been applied (at time kminus)

On one hand it may be the case that the value of vk(zk)minus1 lies at or above the threshold

level Z In this situation we remark that zk can take on any value in the set 0 1 k (k)

Since g0 contains zeros from level lk(0) = vk(zk) onwards we observe that at time k + 1

it may be possible to achieve a higher surplus value than vk+1(zk) minus 1 representing the

maximum surplus level assuming there was no drop below Z due to the claim at time kminus

Specifically if the surplus had fallen to level Z minus 1 at time kminus then gained c at time k+ a

bounded level of Z + c minus1 would have been reached at time k +1 and this level is potentially

larger than vk+1(zk) minus 1 As a result we have that lk(1) = maxvk+1(zk) Z + c If we

extend this argument inductively we readily obtain lk(n) = maxvk+n(zk) Z +c +c2(n minus1)

n isin Z+ However since vk+n(zk) = u + czk + c2(k +n minus zk) = vk(zk)+ c2 + c2(n minus 1) we may

alternatively re-express lk(n) as

lk(n) = maxvk+n(zk) Z + c + c2(n minus 1) = maxvk(zk) + c2 Z + c + c2(n minus 1)

Thus

Z+lk(n) = maxlk(0) + c2 Z + c + c2(n minus 1) n isin (31)

11

or equivalently

Z+lk(n) = lk(1) + c2(n minus 1) n isin

Clearly lk(0) gt Z and lk(n) ge Z + c for n isin Z+

The other remaining case to consider involves the situation when the maximum surplus

level at time k is strictly below the threshold level Z Under this scenario we further

distinguish between two cases involving the value of c2 First of all suppose that c2 = 0 so

that vk(zk)minus1 = u +czk minus1 lt Z or equivalently u +czk le Z If u +czk happens to equal Z

then it is possible for zk isin 0 1 k With c2 = 0 however no further premiums can be

earned until after the first claim takes place (which is assumed to occur at time kminus) thereby

implying that Z = u + ci for all i = zk zk + 1 k Hence we know that u + ck = Z

in this case Moreover if u + czk happens to be strictly less than Z then we clearly have

that zk = k Even in the case when c2 gt 0 one can readily verify that zk = k so that

vk(zk) minus 1 = u + ck minus 1 lt Z Thus we proceed with the analysis assuming that zk = k

c2 ge 0 and the inequality u + ck le Z holds true

lowast k

Define the ceiling function denoted lceilxrceil to yield the smallest integer greater than or

equal to x Because a random premium cannot be received until level Z is at least achieved

it is essential to find the point at which the threshold level Z is reached (ie the point at Zminus(u+ck)+1 which u + ck minus1+ ci ge Z) Isolating this inequality for i one obtains i ge

c Hence

assuming that W1 = k the elapsed time after k at which a random premium would begin to

lceilZminus(u+ck)+1 be received is t lowast = rceil Note that t lowast isin Z+ k c k

With zk = k g0(k)

contains zeros from level u + ck onwards so that lk(0) = u + ck

Immediately following this a premium of c is guaranteed to be received at time k+ as the

threshold level Z has not yet been reached Consequently g1(k)

contains zeros from level (k)

b(k)u + c(k + 1) onwards which is c levels further than that in g0 = Continuing this (k)

process inductively we can readily establish that gn contains zeros from level u + c(k + n)

onwards for n = 1 2 t lowast k so that lk(n) = u + c(k + n) for n = 0 1 t lowast k

In the case of n = t lowast + 1 t lowast + 2 gt

(k) is first looked at in isolation At time k +k k +1

t lowast k + 1 the threshold level has been crossed giving rise to a maximum surplus value of either (k)

c(k + t lowast k contains zeros from level lk(t lowast ku + ) + c2 minus 1 or Z + c minus 1 Therefore g + 1) = t lowast k+1

+1(k + t lowast kmaxvk+t lowast k

) Z + c onwards Carrying out the same inductive process as before it is

readily seen that g(nk)

contains zeros from level lk(n) = maxvk+n(k+t lowast k) Z +c+c2(nminust lowast kminus1)

onwards for n = t lowast k + 1 t lowast k + 2 Combining both cases we obtain

u + c(k + n) if n = 0 1 t lowast lk(n) = k (32)

maxlk(t lowast) + c2 Z + c + c2(n minus t lowast minus 1) if n = t lowast + 1 t lowast + 2 k k k k

Note that by construction lk(t lowast k) gt Z and lk(n) le Z for n = 0 1 t lowast k minus1 Interestingly

12

if we extend the definition tk lowast by introducing the quantity ζk

lowast = maxtklowast 0 we remark that

(31) and (32) can actually be combined into a single formula for lk(n) namely

vk+n(zk + n) if n = 0 1 ζklowast

lk(n) = (33) maxlk(ζ

lowast) + c2 Z + c + c2(n minus ζlowast minus 1) if n = ζlowast + 1 ζlowast + 2 k k k k

Clearly we have that lk(n) ge Z + c for n = ζk lowast +1 ζk

lowast +2 Moreover if c1 = c2 = c (33)

reduces to give lk(n) = u + c(k + n) forall n isin N which is in agreement with the results in Alfa

and Drekic (2007 Section 3)

Remark In our above analysis we purposely ignored the possibility of any future claims

occurring after time k in our determination of lk(n) However we recognize that this is not

realistic and future claims would indeed occur after time k thereby further reducing the value

of the maximum surplus level lk(n) minus 1 at time k + n To account for this we imagine all

future interclaim times (ie after time k) to be of maximum length na and all corresponding

claim sizes to be of minimum amount 1 Consequently in the case when c2 gt 0 we should

subtract from formula (33) the amount of lfloornnarfloor where lfloormiddotrfloor again denotes the floor function

Obviously when c2 = 0 (ie no premiums are collected whenever the surplus level is at or

above Z) lk(n) = Z + c for n = ζk lowast + 1 ζk

lowast + 2 (k)

We now turn our attention to the development of a general algorithm for computing gni

(k) (k)for i = 0 1 lk(n) minus 1 With the aid of (24) we apply the key identity gn = gnminus1C

(which can be inferred from (27)) to first obtain

minZlk(nminus1)minus1 lk(nminus1)minus1 c2 L L L

(k) (k) (k)gni = g Bjminusi+c + gnminus1j dmBjminusi+m n isin Z+ (34) nminus1j

j=0 j=Z m=c1

However (22) states that Bi = Ona for i isin Zminus and so (34) becomes (for the same range of

i)

minZlk(nminus1)minus1 c2 lk(nminus1)minus1 L L L

(k) (k) (k)gni = g Bjminusi+c + dm gnminus1jBjminusi+m n isin Z+ (35) nminus1j

j=max0iminusc m=c1 j=maxZiminusm

Taking the form of (33) into consideration (35) yields the following two formulae

lk(nminus1)minus1 L

(k) (k)gni = gnminus1jBjminusi+c n = 1 2 ζ k

lowast (36) j=max0iminusc

and

Zminus1 c2 lk(nminus1)minus1 L L L

(k) (k) (k)g = g + dm g n = ζ lowast + 1 ζ lowast + 2 ni nminus1jBjminusi+c nminus1jBjminusi+m k k

j=max0iminusc m=c1 j=maxZiminusm

(37)

13

For n = 1 2 ζk lowast where ζk

lowast isin Z+ we note that lk(n) ge c + 1 Therefore with the aid of

(22) (36) simplifies to give

Llk(nminus1)minus1 (k) (k) j=0 gnminus1jBjminusi+c if i = 0 1 c minus 1

gni = Llk(nminus1)minus1 (k)

g if i = c c + 1 lk(n) minus 1j=iminusc nminus1jBjminusi+c

Llk(nminus1)minus1 (k)

nminus1js 0 0 if i = 0 1 c minus 1j=0 αjminusi+cg= (38)

(k) Llk(nminus1)minus1 (k) g S+ αjminusi+cgnminus1js 0 0 if i = c c+1 lk(n)minus1 nminus1iminusc j=iminusc+1

In the case of n = ζk lowast + 1 ζk

lowast + 2 we first reiterate that lk(n) ge Z + c We next break

(37) into several cases based on the value of i and the value of minc Z + c1 as follows

Case 1 Suppose minc Z + c1 = c

For i isin 0 1 c minus 1

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+mni nminus1j nminus1j

j=0 m=c1 j=Z

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k)= αjminusi+cg s + dm αjminusi+mg s 0 0 (39) nminus1j nminus1j

j=0 m=c1 j=Z

For i isin c c + 1 Z + c1 minus 1

Zminus1 c2 lk(nminus1)minus1 L L L

(k) (k) (k)g = g + dm gni nminus1j Bjminusi+c nminus1j Bjminusi+m

j=iminusc m=c1 j=Z

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k) (k)= g S + αjminusi+cg dm αjminusi+mgnminus1j s 0 0 (310) nminus1iminusc nminus1j s +

j=iminusc+1 m=c1 j=Z

For i isin Z + c1 Z + c1 + 1 Z + c minus 1

Zminus1 minc2iminusZ lk(nminus1)minus1 c2 lk(nminus1)minus1 L L L L L

(k) (k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+m + dm g Bjminusi+mni nminus1j nminus1j nminus1j

j=iminusc m=c1 j=iminusm m=iminusZ+1 j=Z

minc2iminusZ L

(k) (k)= g S + dmg S nminus1iminusc nminus1iminusm

m=c1

Zminus1 minc2iminusZ lk(nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k)+ αjminusi+cg dm αjminusi+mg dm αjminusi+mgnminus1j s 0 0 nminus1j s + nminus1js +

j=iminusc+1 m=c1 j=iminusm+1 m=iminusZ+1 j=Z

(311)

For i isin Z + c Z + c + 1 lk(n) minus 1

c2 lk(nminus1)minus1 L L

(k) (k)g = dm g Bjminusi+mni nminus1j

m=c1 j=iminusm

c2 c2 lk (nminus1)minus1 L L L

(k) (k)= dmg S + dm αjminusi+mgnminus1j s 0 0 (312) nminus1iminusm

m=c1 m=c1 j=iminusm+1

14

Case 2 Suppose minc Z + c1 = Z + c1

For i isin 0 1 Z + c1 minus 1

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+mni nminus1j nminus1j

j=0 m=c1 j=Z

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k)= αjminusi+cg s + dm αjminusi+mg s 0 0 (313) nminus1j nminus1j

j=0 m=c1 j=Z

For i isin Z + c1 Z + c1 + 1 c minus 1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+m + dm g Bjminusi+mni nminus1j nminus1j nminus1j

j=0 m=c1 j=iminusm m=iminusZ+1 j=Z

minc2iminusZ L

(k)= dmg S nminus1iminusm

m=c1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k)+ αjminusi+cg s + dm αjminusi+mg s + dm αjminusi+mg s 0 0 nminus1j nminus1j nminus1j

j=0 m=c1 j=iminusm+1 m=iminusZ+1 j=Z

(314)

For i isin c c + 1 Z + c minus 1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k) (k)g = g + g + gni nminus1j Bjminusi+c dm nminus1j Bjminusi+m dm nminus1j Bjminusi+m

j=iminusc m=c1 j=iminusm m=iminusZ+1 j=Z

minc2iminusZ L

(k) (k)= g S + dmg S nminus1iminusc nminus1iminusm

m=c1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k)+

nminus1j s 0 0 αjminusi+cgnminus1j s + dm αjminusi+mgnminus1j s + dm αjminusi+mg j=iminusc+1 m=c1 j=iminusm+1 m=iminusZ+1 j=Z

(315)

For i isin Z + c Z + c + 1 lk(n) minus 1

c2 lk (nminus1)minus1 L L

(k) (k)g = dm g Bjminusi+mni nminus1j

m=c1 j=iminusm

c2 c2 lk (nminus1)minus1 L L L

(k) (k)= dmg S + dm αjminusi+mgnminus1j s 0 0 (316) nminus1iminusm

m=c1 m=c1 j=iminusm+1

It is not difficult to verify that equations (38) through (316) simplify to produce equations

(32) and (33) of Alfa and Drekic (2007 pp 301-302) when c1 = c2 = c With equations (38) (k)

through (316) providing the recursive means of calculating gn we next turn our attention

15

to a determination of h(k) defined by (28) For n = 1 2 ζlowast we use (22) and (25) to n k

obtain

lk(nminus1)minus1 lk(nminus1)minus1

h(k) nminusj =

L g

(k) nminus1ℓBc+j+ℓ =

L αc+j+ℓg

(k) nminus1ℓs 0 0 (317)

ℓ=0 ℓ=0

For n = ζk lowast + 1 ζk

lowast + 2 we similarly acquire

Zminus1 lk(nminus1)minus1 L L

(k) (k) (k)hnminusj = gnminus1ℓBc+j+ℓ + gnminus1ℓAj+ℓ

ℓ=0 ℓ=Z

Zminus1 c2 lk(nminus1)minus1 L L L

(k) (k)= αc+j+ℓg s + dm αm+j+ℓg s 0 0 (318) nminus1ℓ nminus1ℓ

ℓ=0 m=c1 ℓ=Z

Note that substituting (317) and (318) into (29) immediately yields

Llk(nminus1)minus1 (k) (k) ℓ=0 αc+j+ℓgnminus1ℓs if n = 1 2 ζk

lowast φ (u) = nj LZminus1 (k) Lc2

Llk(nminus1)minus1 (k) ζlowast s + dm αm+j+ℓg n = + 1 ζlowast + 2 ℓ=0 αc+j+ℓgnminus1ℓ m=c1 ℓ=Z nminus1ℓs if k k

(319)

Clearly in the special case when c1 = c2 = c (319) yields equation (35) of Alfa and Drekic

(2007 p 302) Finally we remark that (319) can be substituted into (213) to obtain a

computational formula for φnj(u)

4 Numerical examples and findings

In this section we illustrate the application of our proposed algorithm with several numerical

examples All calculations in this section were performed using Mathematica (Version 7) on

a Dell Precision T7400 workstation with a 3 Ghz CPU and 4 GB of RAM

In our first illustration (labelled Example 1) we consider the ordinary Sparre Andersen

model in which interclaim times have pmf given by

(211)(911)jminus1 if j = 1 2 na minus 1 aj = (41)

(911)naminus1 if j = na

Note that the pmf (41) is that of a truncated geometric distribution with all the probability

mass on na na +1 assigned to the support value na Clearly as na becomes larger the

closer (41) approximates this particular geometric distribution having mean 55 In what Lna

Lminus1 nafollows we fix na = 25 so that aj = 0992 and jaj = 546 Let the individual j=1 j=1

claim amount distribution be given by the pmf

minus4 minus4j minus 1 j

αj = 1 + minus 1 + j isin Z+ (42) 30 30

16

We remark that the pmf (42) represents a discretized version of the Pareto distribution with

mean 10 Also note that mα = infin thereby implying that the random variables |UT | and

UTminus both have infinite support

Table 1 displays values (rounded to 6 decimal places) of the trivariate cumulative distrishy

bution function

n x y L L L

Ψnxy(u) = Pr T le n UTminus le x |UT | le y | U0 = u = ψℓij(u) ℓ=1 i=mincZ+c1 j=1

corresponding to the above discrete-time risk process with u = 10 Z = 50 c = 5 and

varying premium amount distributions according to the following set of scenarios

(1) d2 = 1

(2) d1 = 25 d2 = 15 d3 = 25

G

5)

(3) di = i

(25)i(35)5minusi i = 0 1 5

(4) di = 15 i = 0 1 4

(5) d1 = 34 d5 = 14

(6) d0 = 35 d5 = 25

We remark that while these six premium amount distributions all have mean equal to 2

they possess different shapes and we investigate the impact of their variability on the joint

distribution of T |UT | and UTminus In particular the variances corresponding to the premium

amount distributions of scenarios (1) through (6) are 0 08 12 2 3 and 6 respectively

Based on the results in Table 1 it would appear that greater variability in the premium

amount distribution generally leads to higher ruin probabilities Perhaps somewhat surprising

though the differences are subtle and not particularly striking as one compares the various

ruin probabilities from scenarios (1) through (6) Obviously we observe more of a difference

between scenarios as the values of n x and y increase

As a follow-up illustration (labelled Example 2) we consider a model in which ordinary

interclaim times have a discrete uniform distribution with pmf aj = 110 j = 1 2 10

Clearly the mean of this distribution is 55 which for comparative purposes is identical to

the ordinary interclaim time mean considered in Example 1 In addition we assume that

the pmf of the individual claim amount distribution is once again given by (42) Table 2

displays values (rounded to 6 decimal places) of Ψnxy(10) for several variants of the Sparre

Andersen model (again with Z = 50 and c = 5) having the above ordinary interclaim time

and claim amount distributions namely

17

(1) ordinary model under scenario (2) from Example 1

(2) ordinary model under scenario (5) from Example 1

(3) stationary model under scenario (2) from Example 1

(4) stationary model under scenario (5) from Example 1

(5) delayed model (with W1 = 1 with probability 1) under scenario (2) from Example 1

(6) delayed model (with W1 = 1 with probability 1) under scenario (5) from Example 1

Concerning the results in Table 2 we first compare the values of Ψnxy(10) in entries (2)

and (5) of Table 1 to their corresponding counterparts in entries (1) and (2) of Table 2 In all

cases we observe that ruin is more likely to occur when interclaim times have the truncated

geometric distribution introduced in our first example as opposed to the discrete uniform

distribution of our second example Since mean interclaim times are equal in both examshy

ples this behaviour must be attributable to the greater variability inherent in the truncated

geometric distribution We also note that under the stationary model assumption the pmf

of W1 is easily verified to be rj = (11 minus j)55 j = 1 2 10 which has mean 4 lt 55

As a result the values of Ψnxy(10) for stationary models (3) and (4) are always larger than

the corresponding values for ordinary models (1) and (2) Finally in the absence of any

knowledge regarding which distribution is in effect for the first interclaim time W1 models

(5) and (6) can be regarded as ldquoworst-case scenariordquo models as they serve to provide upper

bounds on the values of Ψnxy(10)

In conclusion we state that the values in Tables 1 and 2 were generated by first impleshy

menting the recursive procedures (38) through (316) and then summing the appropriate

bivariatetrivariate probabilities computed via (213) (214) or (215) In situations when

x = y = infin however we computed the resulting finite-time ruin probabilities using an altershy

native (albeit standard) method (eg see Cossette et al (2006 Section 3)) Specifically if

we define the survival probability σ(n u) = Pr T gt n | U0 = u = 1minusΨninfininfin(u) as well as

the auxiliary function

0 if w ge Z ηwt =

minlfloorZminuswc minus1 rfloor + 1 t if w lt Z

then by conditioning on the time and amount of the first claim (and recalling that the delayed

process reverts to the ordinary process following this claim) we obtain the expression

minnnr (kminuszk)c2 u+czk+ℓ L L L

σ(u n) = Rn + rk dℓkminuszk αjγ(u + czk + ℓ minus j n minus k)

k=1 ℓ=(kminuszk)c1 j=1

18

where the function γ(w m) for w m isin N is computed recursively via

minmna (kminusηwk)c2 u+cηwk+ℓ L L L

γ(w m) = Am + ak dℓkminusηwk αjγ(u + cηwk + ℓ minus j m minus k)

k=1 j=1 ℓ=(kminusηwk)c1

with starting values γ(w 0) = 1 forall w isin N

Acknowledgements

The authors would like to thank Dr David Landriault for his valuable suggestions which

helped to improve an earlier version of this work This research was supported by the Natural

Sciences and Engineering Research Council of Canada

References

Alfa AS 2004 Markov chain representations of discrete distributions applied to queueing

models Computers amp Operations Research 31 2365-2385

Alfa AS Drekic S 2007 Algorithmic analysis of the Sparre Andersen model in discrete

time ASTIN Bulletin 37 293-317

Bao Z 2007 A note on the compound binomial model with randomized dividend strategy

Applied Mathematics amp Computation 194 276-286

Cossette H Landriault D Marceau E 2006 Ruin probabilities in the discrete time

renewal risk model Insurance Mathematics amp Economics 38 309-323

Karlin S Taylor HM 1975 A First Course in Stochastic Processes 2nd edition Acashy

demic Press New York

Kim B Kim H-S Kim J 2008 A risk model with paying dividends and random

environment Insurance Mathematics amp Economics 42 717-726

Landriault D 2008 Randomized dividends in the compound binomial model with a general

premium rate Scandinavian Actuarial Journal 1-15

Tan J Yang X 2006 The compound binomial model with randomized decisions on

paying dividends Insurance Mathematics amp Economics 39 1-18

19

Table 1 ndash Values of Ψnxy(10) corresponding to Example 1

(a) x = 10

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0005357 0006188 0007163 0008278 0009481

(2) 0005359 0006195 0007175 0008296 0009507

(3) 0005361 0006198 0007181 0008305 0009519

(4) 0005363 0006205 0007193 0008323 0009544

(5) 0005366 0006212 0007208 0008346 0009576

(6) 0005374 0006236 0007249 0008410 0009664

y = 25 (1) 0007631 0008815 0010202 0011789 0013503

(2) 0007634 0008824 0010220 0011816 0013539

(3) 0007636 0008828 0010228 0011829 0013557

(4) 0007639 0008838 0010245 0011854 0013593

(5) 0007643 0008849 0010266 0011887 0013638

(6) 0007656 0008882 0010325 0011978 0013764

y = 50 (1) 0008477 0009792 0011333 0013095 0014998

(2) 0008481 0009802 0011352 0013125 0015039

(3) 0008483 0009807 0011361 0013139 0015059

(4) 0008486 0009817 0011380 0013168 0015098

(5) 0008491 0009830 0011403 0013204 0015149

(6) 0008504 0009866 0011470 0013305 0015288

y = infin (1) 0008777 0010138 0011733 0013558 0015528

(2) 0008781 0010149 0011753 0013588 0015570

(3) 0008783 0010154 0011763 0013603 0015591

(4) 0008787 0010164 0011782 0013633 0015632

(5) 0008791 0010177 0011807 0013671 0015684

(6) 0008805 0010215 0011875 0013775 0015828

20

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0058559 0061875 0065704 0070048 0074714

(2) 0058569 0061903 0065756 0070128 0074825

(3) 0058574 0061917 0065782 0070167 0074880

(4) 0058584 0061945 0065833 0070245 0074987

(5) 0058596 0061979 0065896 0070344 0075124

(6) 0058632 0062079 0066078 0070621 0075506

y = 25 (1) 0089002 0093986 0099739 0106262 0113270

(2) 0089017 0094028 0099817 0106382 0113437

(3) 0089025 0094049 0099856 0106442 0113519

(4) 0089040 0094091 0099932 0106560 0113681

(5) 0089057 0094142 0100028 0106708 0113887

(6) 0089113 0094294 0100302 0107127 0114463

y = 50 (1) 0102786 0108506 0115106 0122590 0130629

(2) 0102804 0108555 0115196 0122728 0130820

(3) 0102812 0108579 0115241 0122797 0130914

(4) 0102830 0108627 0115329 0122932 0131101

(5) 0102850 0108686 0115439 0123103 0131337

(6) 0102914 0108860 0115754 0123584 0132000

y = infin (1) 0108895 0114933 0121901 0129799 0138283

(2) 0108914 0114985 0121996 0129945 0138485

(3) 0108923 0115010 0122043 0130018 0138585

(4) 0108941 0115061 0122136 0130161 0138782

(5) 0108962 0115123 0122252 0130341 0139032

(6) 0109030 0115307 0122585 0130849 0139732

21

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0079856 0086792 009464 0103440 0112831

(2) 0079885 0086869 0094778 0103650 0113118

(3) 0079953 0086978 0094936 0103864 0113393

(4) 0080081 0087188 0095243 0104282 0113929

(5) 0079962 0087075 0095149 0104213 0113890

(6) 0080805 0088366 0096951 0106588 0116875

y = 25 (1) 0124591 0135680 0148207 0162245 0177217

(2) 0124639 0135806 0148433 0162586 0177683

(3) 0124755 0135993 0148703 0162950 0178148

(4) 0124978 0136354 0149226 0163658 0179054

(5) 0124763 0136141 0149035 0163500 0178935

(6) 0126238 0138381 0152148 0167588 0184060

y = 50 (1) 0146757 0160079 0175111 0191945 0209894

(2) 0146816 0160232 0175385 0192360 0210460

(3) 0146965 0160469 0175725 0192817 0211043

(4) 0147249 0160925 0176384 0193706 0212178

(5) 0146968 0160639 0176117 0193470 0211980

(6) 0148860 0163500 0180077 0198658 0218474

y = infin (1) 0158013 0172606 0189059 0207476 0227106

(2) 0158078 0172776 0189362 0207934 0227732

(3) 0158250 0173047 0189751 0208455 0228395

(4) 0158579 0173571 0190503 0209467 0229685

(5) 0158247 0173226 0190172 0209161 0229410

(6) 0160443 0176532 0194735 0215127 0236869

22

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0088667 0104403 0123329 0145277 0169184

(2) 0088711 0104518 0123538 0145598 0169627

(3) 0088733 0104574 0123641 0145756 0169846

(4) 0088777 0104687 0123844 0146066 0170274

(5) 0088829 0104828 0124105 0146470 0170832

(6) 0088988 0105229 0124822 0147560 0172331

y = 25 (1) 0140528 0167839 0201007 0239726 0282102

(2) 0140598 0168021 0201338 0240234 0282804

(3) 0140632 0168111 0201502 0240486 0283151

(4) 0140701 0168289 0201822 0240976 0283827

(5) 0140785 0168515 0202241 0241622 0284718

(6) 0141036 0169148 0203370 0243335 0287071

y = 50 (1) 0167826 0203005 0246173 0296954 0352855

(2) 0167908 0203218 0246561 0297548 0353672

(3) 0167949 0203324 0246753 0297841 0354075

(4) 0168030 0203532 0247128 0298412 0354859

(5) 0168129 0203800 0247622 0299174 0355907

(6) 0168423 0204539 0248936 0301159 0358620

y = infin (1) 0183325 0224926 0276881 0338957 0408257

(2) 0183412 0225150 0277283 0339564 0409079

(3) 0183455 0225260 0277481 0339864 0409485

(4) 0183540 0225478 0277869 0340446 0410270

(5) 0183645 0225760 0278386 0341234 0411339

(6) 0183954 0226529 0279737 0343244 0414037

23

Table 2 ndash Values of Ψnxy(10) corresponding to Example 2

(a) x = 10

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0001901 0002276 0002724 0003240 0003799

(2) 0001904 0002286 0002742 0003269 0003840

(3) 0002651 0003013 0003442 0003936 0004471

(4) 0002655 0003022 0003460 0003964 0004510

(5) 0005792 0006098 0006458 0006871 0007317

(6) 0005795 0006106 0006473 0006894 0007350

y = 25 (1) 0002708 0003243 0003880 0004615 0005411

(2) 0002713 0003257 0003906 0004656 0005469

(3) 0003778 0004293 0004904 0005608 0006369

(4) 0003782 0004306 0004929 0005647 0006424

(5) 0008255 0008690 0009203 0009792 0010427

(6) 0008259 0008702 0009224 0009824 0010473

y = 50 (1) 0003008 0003603 0004310 0005127 0006011

(2) 0003014 0003618 0004340 0005173 0006076

(3) 0004197 0004769 0005448 0006230 0007075

(4) 0004202 0004784 0005476 0006273 0007137

(5) 0009173 0009656 0010226 0010879 0011585

(6) 0009177 0009669 0010249 0010916 0011637

y = infin (1) 0003115 0003730 0004463 0005308 0006224

(2) 0003121 0003746 0004493 0005356 0006291

(3) 0004346 0004938 0005641 0006450 0007325

(4) 0004351 0004954 0005670 0006496 0007389

(5) 0009499 0009999 0010589 0011265 0011996

(6) 0009503 0010012 0010613 0011303 0012050

24

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0031176 0032815 0034733 0036924 0039282

(2) 0031193 0032862 0034821 0037061 0039475

(3) 0050168 0051747 0053587 0055684 0057939

(4) 0050184 0051792 0053671 0055815 0058124

(5) 0136884 0138219 0139763 0141517 0143399

(6) 0136896 0138256 0139832 0141626 0143553

y = 25 (1) 0047622 0050092 0052981 0056280 0059829

(2) 0004765 0050163 0053113 0056487 0060121

(3) 0076570 0078950 0081720 0084878 0088272

(4) 0076593 0079017 0081847 0085076 0088551

(5) 0205774 0207786 0210111 0212752 0215585

(6) 0205793 0207842 0210216 0212917 0215818

y = 50 (1) 0055170 0058009 0061329 0065119 0069196

(2) 0055200 0058091 0061481 0065357 0069533

(3) 0088650 0091385 0094569 0098197 0102096

(4) 0088677 0091463 0094715 0098425 0102417

(5) 0235815 0238128 0240800 0243834 0247089

(6) 0235837 0238192 0240920 0244023 0247356

y = infin (1) 0058565 0061565 0065073 0069077 0073384

(2) 0058597 0061652 0065234 0069329 0073740

(3) 0094061 0096952 0100316 0104149 0108267

(4) 0094090 0097035 0100470 0104390 0108607

(5) 0248472 0250917 0253740 0256945 0260383

(6) 0248495 0250985 0253867 0257145 0260667

25

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0048598 0052524 0056984 0061998 0067342

(2) 0048654 0052673 0057253 0062411 0067916

(3) 0070192 0073979 0078257 0083057 0088167

(4) 0070243 0074119 0078513 0083451 0088714

(5) 0151774 0154981 0158572 0162586 0166852

(6) 0151815 0155096 0158784 0162913 0167308

y = 25 (1) 0076895 0083230 0090413 0098481 0107073

(2) 0076987 0083474 0090854 0099156 0108010

(3) 0110093 0116205 0123095 0130818 0139033

(4) 0110177 0116435 0123515 0131462 0139928

(5) 0230674 0235848 0241632 0248091 0254950

(6) 0230739 0236038 0241980 0248626 0255695

y = 50 (1) 0091496 0099156 0107829 0117563 0127924

(2) 0091610 0099455 0108370 0118387 0129068

(3) 0130130 0137520 0145839 0155156 0165064

(4) 0130233 0137801 0146352 0155942 0166156

(5) 0266599 0272857 0279840 0287632 0295904

(6) 0266680 0273088 0280265 0288286 0296814

y = infin (1) 0099312 0107746 0117285 0127984 0139369

(2) 0099439 0108078 0117883 0128899 0140639

(3) 0140458 0148595 0157745 0167986 0178872

(4) 0140572 0148908 0158314 0168859 0180084

(5) 0282886 0289776 0297457 0306022 0315111

(6) 0282976 0290034 0297929 0306748 0316120

26

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0059096 0072505 0088907 0108122 0129137

(2) 0059223 0072829 0089500 0109052 0130456

(3) 0079757 0092662 0108388 0126778 0146871

(4) 0079875 0092970 0108956 0127669 0148134

(5) 0159788 0170682 0183871 0199249 0216024

(6) 0159881 0170936 0184343 0199990 0217075

y = 25 (1) 0095883 0119741 0149240 0184073 0222390

(2) 0096087 0120262 0150197 0185573 0224520

(3) 0127403 0150359 0178640 0211977 0248613

(4) 0127592 0150855 0179557 0213414 0250653

(5) 0245161 0264533 0288250 0316124 0346709

(6) 0245311 0264942 0289010 0317319 0348407

y = 50 (1) 0116598 0147920 0187100 0233784 0285503

(2) 0116839 0148537 0188231 0235554 0288015

(3) 0153023 0183153 0220713 0265390 0314840

(4) 0153249 0183742 0221798 0267087 0317248

(5) 0285742 0311159 0342651 0380005 0421287

(6) 0285921 0311644 0343551 0381418 0423292

y = infin (1) 0129470 0167217 0215383 0273841 0339734

(2) 0129725 0167865 0216560 0275661 0342282

(3) 0167978 0204275 0250442 0306382 0369383

(4) 0168217 0204895 0251572 0308128 0371826

(5) 0305868 0336470 0375169 0421935 0474525

(6) 0306058 0336980 0376106 0423388 0476560

27

Page 3: uwaterloo.ca · Ruin Analysis of a Threshold Strategy in a Discrete-time Sparre Andersen Model Steve DREKIC and Ana Maria MERA Department of Statistics and Actuarial Science University

the set of positive integers and N = 0 cup Z+ For t isin Z+ we define Ut as the insurerrsquos

amount of surplus at time t In actual fact Ut represents the amount of surplus at the end

of time interval (t minus 1 t] at which point any premiumsclaims corresponding to this time

interval have been receivedpaid out Specifically with respect to the time interval (t minus 1 t]

we adopt the convention that premiums are received at (t minus 1)+ and any claims are paid

out at tminus We incorporate the notion of a threshold level Z isin Z+ on the insurerrsquos surplus

affecting the amount of premium being received at a given point in time More precisely we

assume that premiums corresponding to the time interval (t t + 1] t isin N are collected at t+

according to (random) rate pt such that

c if Ut lt Z pt =

Xt if Ut ge Z

where di = PrXt = i i = c1 c1 + 1 c2 We refer to c isin Z+ as the pure (constant) L

premium and assume that c1 c2 isin 0 1 c with c1 le c2 and c2 di = 1 Clearly c1i=c1

and c2 are the respective lower and upper support values of the distribution of the random

premium amount at time t Xt Correspondingly we interpret c minus pt as the amount of

(randomized) dividends paid to insureds at time t+ Note that by assumption the probability

distribution of Xt is identical for all values of t

Beginning at time 0 with an initial reserve of u isin N the insurerrsquos amount of surplus at

time t is expressible as tminus1 Nt L L

U t = u + pi minus Yi t isin N i=0 i=1

where individual claim amounts Y1 Y2 are assumed to form an iid sequence of positive

random variables with common pmf αj j = 1 2 mα and corresponding survival function LjΛj = 1 minus k=1 αk Unlike the interclaim time distributions defined above we remark that

the claim amount distribution can be either of finite or infinite support (ie mα le infin) Let

the time of ruin T be defined as T = mint isin Z+|Ut lt 0 with T = infin if Ut ge 0 forall t isin Z+

If ruin does occur let |UT | denote the deficit at ruin and UTminus = UTminus1 + pTminus1 the surplus

immediately prior to ruin Clearly T = infin if mα le minc Z + c1 Conversely if mα gt

minc Z + c1 it is not difficult to verify that |UT | isin 1 2 mα minus minc Z + c1 and

UTminus isin minc Z + c1 minc Z + c1 + 1 mα minus 1 Throughout the remainder of the

paper we assume that mα gt minc Z + c1

Following this introduction Section 2 sets up the underlying Markov chain structure in

our model and derives some necessary preliminaries In Section 3 we further exploit this

structure to construct a set of computational procedures enabling one to calculate probashy

bility distributions associated with the random variables T |UT | and UTminus with specific

3

consideration to the joint pmfrsquos

Z+ωni(u) = Pr T = n UTminus = i | U0 = u n isin i = minc Z+c1 minc Z+c1+1 mαminus1

φnj(u) = Pr T = n |UT | = j | U0 = u n isin Z+ j = 1 2 mαminusminc Z +c1

and

ψnij(u) = Pr T = n UTminus = i |UT | = j | U0 = u n isin Z+

i = minc Z +c1 minc Z +c1 + 1 mαminus1

j = 1 2 3 mαminusi

Finally we implement our various procedures and present some numerical examples in Section

4

2 Model formulation and preliminaries

Adopting the same methodology employed by Alfa and Drekic (2007) let τj =

Pr W gt j|W gt j minus 1 = AjAjminus1 j = 1 2 na where W denotes an arbitrary Wi

i = 2 3 Clearly τ1 = A1 and τna = 0 If we next define the 1 times na row vector

e1 = (1 0 0) the na times na probability transition matrix

0 τ1 0 middot middot middot 0

0 0 τ2

0

S =

0 0 middot middot middot 0 τnaminus1

0 0 middot middot middot 0 0

and the na times 1 column vector

1 minus τ1 1 minus τ2 s = 1 minus τnaminus1

1

it is known that aj = e1Sjminus1 s j = 1 2 na (eg see Alfa (2004)) In other words W

can be expressed as a discrete phase-type random variable (of dimension na) with the above

parameterization implying that s contains the absorption probabilities to claim occurrence

associated with each of the na possible interclaim time values Alfa (2004) refers to this

4

characterization as the ldquoelapsed timerdquo phase-type representation of the pmf aj We remark

that the above pmf formula also holds true for j gt na as it is readily verified that Sjminus1 = Ona

an na times na matrix of zeros for such values of j

In what follows we proceed to set up our delayed Sparre Andersen model as a bivariate

Markov chain conditional on W1 = k isin 1 2 nr Specifically assuming that W1 = k we

construct the bivariate stochastic process (Ut Lt) t = k k + 1 such that Ut represents

the amount of surplus at time t and Lt denotes a specific counter at time t which measures the

ldquoelapsed timerdquo (in the sense described above) length of the next claim occurrence We refer to

the Ut component as the level of the process and the Lt component as the phase of the process

Two important observations are particularly noteworthy regarding this construction First of

all Lk = 1 since we are assuming that the very first claim occurs at time kminus More generally

Lt = 1 if a claim happens to occur at time tminus Secondly the delayed process reverts to the

ordinary process (having interclaim time distribution defined by the pmf aj) upon occurrence

of the first claim As a result the bivariate stochastic process (Ut Lt) t = k k + 1

possesses the following Markovian relationship

(Ut + pt Lt + 1) if there is no claim at time (t + 1)minus (Ut+1 Lt+1) =

(Ut + pt minus Y 1) if there is a claim of amount Y at time (t + 1)minus

Note that the state space for this Markov chain is given by Δ = Z times 1 2 na Furthershy

more the probability transition matrix associated with this Markov chain is given by

middot middot middot minus1 10 middot middot middot Zminus2 Zminus1 Z Z+1 Z+2 middot middot middot

middot middot middot

middot middot middot minus1 middot middot middot Bc Bcminus1 Bcminus2 middot middot middot BcminusZ+1 BcminusZ BcminusZminus1 BcminusZminus2 BcminusZminus3 middot middot middot

0

1

middot middot middot Bc+1 Bc Bcminus1 middot middot middot BcminusZ+2 BcminusZ+1 BcminusZ BcminusZminus1 BcminusZminus2 middot middot middot middot middot middot Bc+2 Bc+1 Bc middot middot middot BcminusZ+3 BcminusZ+2 BcminusZ+1 BcminusZ BcminusZminus1 middot middot middot

middot middot middot

middot middot middot

P = Zminus2 middot middot middot BZ+cminus1 BZ+cminus2 BZ+cminus3 middot middot middot Bc Bcminus1 Bcminus2 Bcminus3 Bcminus4 middot middot middot (21)

Zminus1 middot middot middot BZ+c BZ+cminus1 BZ+cminus2 middot middot middot Bc+1 Bc Bcminus1 Bcminus2 Bcminus3 middot middot middot

Z middot middot middot AZ+1 AZ AZminus1 middot middot middot A2 A1 A0 Aminus1 Aminus2 middot middot middot Z+1 middot middot middot AZ+2 AZ+1 AZ middot middot middot A3 A2 A1 A0 Aminus1 middot middot middot

Z+2 middot middot middot AZ+3 AZ+2 AZ+1 middot middot middot A4 A3 A2 A1 A0 middot middot middot

middot middot middot

middot middot middot

where Ona

if i isin Zminus

Bi = S if i = 0 (22) se1αi if i isin Z+

5

and c2 L

Ai = djBi+j (23) j=c1

We give a brief indication of the reasoning leading to the form of P in (21) For example

consider the transition ldquoUt = 0 rarr Ut+1 = 1rdquo which has block element Bcminus1 containing the

transition probabilities governing the ldquoLt rarr Lt+1 rdquo process At a surplus level of 0 at time

t a premium of pt = c is certain to be received Therefore to arrive at surplus level 1 at

the next time unit a ldquoso-calledrdquo claim of size c minus 1 must occur to neutralize the premium

just received Thus the size of the ldquoclaimrdquo required for this transition to occur determines

the value of i in Bi so that Bcminus1 is obtained and its form subsequently determined by (22)

according to the value of the index c minus 1 Note that only when c gt 1 does an actual claim

occur in this particular scenario We mention that for transitions which begin at or above the

threshold level Z a similar line of reasoning would also apply except now one must condition

on the amount of the random premium being earned as indicated by the form of (23)

Note that this formulation gives rise to a doubly-infinite Markov chain with finite blocks

of size na In the special case of the threshold-free model (obtained by setting c1 = c2 = c

so that dc = 1) we remark that (21) is in agreement with the matrix P in Alfa and Drekic

(2007 p 296) Also since αi = 0 forall i gt mα we note that Bi = Ona if i gt mα

For the computation of ruin-related quantities of interest it is useful to partition the

state space Δ into two mutually exclusive state spaces namely Δ1 = N times 1 2 na

and Δ2 = Zminus times 1 2 na In addition we introduce (one-step) transition matrices

C Δ1 rarr Δ1 and D Δ1 rarr Δ2 which correspond to mapping ldquonon-ruinedrdquo states in Δ1 to

either ldquonon-ruinedrdquo states in Δ1 or ldquoruinedrdquo states in Δ2 respectively In other words the

matrices C and D are essentially submatrices of P and their forms are readily given by

210 middot middot middot Z minus 1 Z Z + 1 middot middot middot

Bc Bcminus1 Bcminus2 middot middot middot BcminusZ+1 BcminusZ BcminusZminus1 middot middot middot Bc+1 Bc Bcminus1 middot middot middot BcminusZ+2 BcminusZ+1 BcminusZ middot middot middot

Bc+2 Bc+1 Bc middot middot middot BcminusZ+3 BcminusZ+2 BcminusZ+1 middot middot middot

middot middot middot

middot middot middot

0

1

2 Z minus 2 BZ+cminus2 BZ+cminus3 BZ+cminus4 middot middot middot Bcminus1 Bcminus2 Bcminus3 middot middot middot (24) C =

Z minus 1 BZ+cminus1 BZ+cminus2 BZ+cminus3 middot middot middot Bc Bcminus1 Bcminus2 middot middot middot

Z AZ AZminus1 AZminus2 middot middot middot A1 A0 Aminus1 middot middot middot

Z + 1 AZ+1 AZ AZminus1 middot middot middot A2 A1 A0 middot middot middot

Z + 2 AZ+2 AZ+1 AZ middot middot middot A3 A2 A1 middot middot middot

middot middot middot

6

and

minus1 minus2 minus3 middot middot middot

0 Bc+1 Bc+2 Bc+3 middot middot middot 1 Bc+2 Bc+3 Bc+4 middot middot middot

2 Bc+3 Bc+4 Bc+5 middot middot middot middot middot middot

Z minus 2 BZ+cminus1 BZ+c BZ+c+1 middot middot middot D = (25) Z minus 1 BZ+c BZ+c+1 BZ+c+2 middot middot middot

Z AZ+1 AZ+2 AZ+3 middot middot middot

Z + 1 AZ+2 AZ+3 AZ+4 middot middot middot Z + 2 AZ+3 AZ+4 AZ+5 middot middot middot

Let us now specify some further definitions required in upcoming derivations First of all

we introduce the function

0 if u ge Z zt =

maxi isin 1 2 t|u + c(i minus 1) lt Z if u lt Z

or equivalently

0 if u ge Z zt =

minlfloorZminuscuminus1 rfloor + 1 t if u lt Z

where the floor function denoted lfloorxrfloor yields the largest integer less than or equal to x We

define dm0 = δm0 where δij in general denotes the Kronecker delta function of i and j

Also define

dm if m = c1 c1 + 1 c2dm1 =

0 elsewhere

and for n = 2 3

Lc2 dj1dmminusjnminus1 if m = nc1 nc1 + 1 nc2dmn = j=c1

0 elsewhere

In other words dmn represents the pmf of the n-fold convolution of dm with itself

Recalling that W1 = k note that zk + ( k+) determines the earliest time point in the set =

0+ 1+ (k minus1)+ at which a random premium is received Note also that zk = k implies

that no such point exists in this set (ie the threshold level Z has not yet been reached when

the premium at time (k minus1)+ is to be received) Therefore if we assume for the moment that

zk k we may condition on the total premiums earned at times zk + (zk + 1)+ (k minus 1)+=

7

to obtain the following expression for b(k) the initial (ie starting at time k) probability row

vector of the above Markov chain corresponding only to the states in Δ1 namely

(kminuszk)c2

b(k) L

= dmkminuszk(αu+czk+me1 αu+czk+mminus1e1 α2e1 α1e1 0 0 ) (26)

m=(kminuszk)c1

where 0 denotes the 1 times na row vector of zeros By further defining the function vn(t) =

b(k)u + ct + (n minus t)c2 we note that the i-th level of is given by the 1 times na row vector L(kminuszk)c2 b(k)αu+czk+mminusie1 for each i isin Ωk = 0 1 vk(zk)minus1 Hence contains

m=(kminuszk)c1 dmkminuszk

zeros from level vk(zk) onwards Moreover (26) holds true even in the case when zk = k

simplifying to give b(k) = (αu+cke1 αu+ckminus1e1 α2e1 α1e1 0 0 )

At this juncture we introduce two additional row vectors namely

(k) (k) (k) (k) b(k)gn = (gn0 gn1 gn2 ) = Cn n isin N (27)

and

h(k) (k) (k) (k) b(k)Cnminus1D =

(k) Z

+ n = (h nminus2 hnminus3 ) = gnminus1D n isin (28) nminus1 h

(k)We remark that gn contains the probabilities of being in the various ldquonon-ruinedrdquo states

at time k + n (ie after claims have been paid out) without having visited a ldquoruinedrdquo state

during the previous n minus 1 transitions given that Uk isin Ωk sub Δ1 In a similar fashion h(nk)

contains the probabilities of being in the various ldquoruinedrdquo states for the first time at time

k + n given that Uk isin Ωk sub Δ1 Note that the probability of being in ruined state ldquo minus jrdquo is (k) (k) (k) (k)

characterized by the row vector h = (h h h ) with the third subscript nminusj nminusj1 nminusj2 nminusjna

representing the value of the elapsed time counter Lk+n Upon further reflection however (k)

=it must be the case that h = 0 for i 1 as ruin can only occur at claim instants nminusji

which in this case implies Lk+n = 1 with probability 1 Thus it immediately follows that (k) (k) (k)

the structure of the 1 times na row vector hnminusj is simply given by hnminusj = (φnj(u) 0 0) (k)

where φ (u) = Pr T = k + n |UT | = j | Consequently as in the analysis with nj Uk isin Ωk

no threshold on the surplus level (see Alfa and Drekic (2007 p 299)) we obtain

(k) (k) prime φ (u) = h (29) nj nminusje1

where e1 prime denotes the transpose of e1

A similar line of probabilistic logic can also be applied to obtain a representation for the (k)

trivariate function ψ (u) = Pr T = k + n UTminus = i |UT | = j | Uk isin Ωk In order for ruin nij

to occur at time k + n with a surplus immediately prior to ruin equal to i we observe that

(i) none of the previous n minus 1 transitions must have included a visit to any state in Δ2

and

8

(ii) the surplus level at time k + n minus 1 must necessarily be equal to i minus pk+nminus1

Note that pk+nminus1 denotes the premium corresponding to the time interval (k + n minus 1 k + n]

which is received at (k + n minus 1)+ Hence the quantity corresponding to points (i) and (ii) (k)

above is the 1 times na row vector g Then at the next time unit a claim must nminus1iminuspk+nminus1

necessarily occur but not before a premium of pk+nminus1 is first collected thereby raising the

surplus level to i Since s contains the absorption probabilities (to claim occurrence) from

the na possible phase states and the claim causing ruin must be of size i + j in order to

ensure that the deficit at ruin is equal to j it follows that

(k) (k)ψ (u) = g s αi+j (210) nij nminus1iminuspk+nminus1

(k)The following corollary provides the means of calculating the quantity g in nminus1iminuspk+nminus1

(210)

Corollary 1 If minc Z + c1 = c then

(k)

g if i = c c+1 Z +c1 minus1 nminus1iminusc

(k) (k) Lminc2iminusZ (k)g = g + j=c1

if i = Z +c1 Z +c1 +1 minmα Z +cminus1nminus1iminuspk+nminus1 nminus1iminusc djgnminus1iminusj

Lc2 (k)

j=c1 djgnminus1iminusj if i = Z +c Z +c+1 mαminus1

(211)

Conversely if minc Z + c1 = Z + c1 then

Lminc2iminusZ (k)

djg if i = Z +c1 Z +c1 +1 cminus1nminus1iminusj j=c1

(k) (k) Lminc2iminusZ (k)gnminus1iminuspk+nminus1

= gnminus1iminusc + j=c1 djgnminus1iminusj if i = c c+1 minmα Z +cminus1

Lc2 (k)djg if i = Z +c Z +c+1 mαminus1j=c1 nminus1iminusj

(212)

Proof The justification of (211) is as follows If c is smaller than Z + c1 the only way of

reaching a surplus level of i isin c c+1 Z +c1 minus1 (prior to ruin) is by receiving a premium

of c since i minus j for any j = c1 c1 +1 c2 would result in a surplus level below Z where the

random premium rate is not applicable Next for fixed i isin Z +c1 Z +c1 +1 Z +c minus1

there are two distinct ways of reaching this surplus level First of all note that the value of

i minus j is guaranteed to produce a surplus amount greater than or equal to Z provided that

j isin c1 c1 + 1 minc2 i minus Z Secondly note that reducing each of value of i by c also

yields a non-negative surplus value less than Z Hence both pure and random premiums

are plausible under this scenario For i isin Z + c Z + c + 1 note that decreasing this

amount by any j = c1 c1 + 1 c2 results in a surplus value greater than or equal to Z as

desired However reducing it by the amount c also has the same impact thus rendering the

pure premium option inadmissible in this case Lastly we incorporate into the range of i the

9

fact that the surplus immediately prior to ruin cannot exceed mα minus1 as indicated in Section

1 The justification of (212) when c is larger than Z + c1 mirrors the analysis above and is

left to the reader D

Recall that in the analysis up to this point we have assumed that W1 = k isin

1 2 3 nr If we now condition on the value of W1 the Law of Total Probability

yields the following two expressions for φnj(u) and ψnij(u) namely

nr L

= rkφ (u) n = nr + 1 nr + 2 φnj(u) n

(kminus) kj

k=1

and nr L

(k)ψnij(u) = rkψ (u) n = nr + 1 nr + 2 nminuskij

k=1

In addition however it is quite possible that T = n for n = 1 2 nr and this can happen

as a result of either

(i) the first claim which causes ruin occurring at time nminus from a (conditional) surplus Lnminus1level of u + czn + m (where m) or i=zn

Xi =

(ii) the first claim which does not cause ruin occurring at some time kminus k isin 1 2 nminus

1 and ruin subsequently occurring n minus k time units later

If we now combine these outcomes with those for n = nr + 1 nr + 2 (and recall that

rn = 0 forall n gt nr) the general expression for φnj(u) becomes

minnminus1nr (nminuszn)c2 L

(k) L

φnj(u)= rkφ (u) + rn dmnminuszn αu+czn+m+j n isin Z+ (213) nminuskj

k=1 m=(nminuszn)c1

Concerning the function ψnij(u) when n le nr and point (i) above is considered we note Lnminus1that i must necessarily equal u + czn + m (again conditional on i=zn

Xi = m) for ruin to

occur in this manner Hence in a similar fashion we acquire

minnminus1nr (nminuszn)c2 L L

ψnij(u) = rkψn

(kminus) kij(u) + rn dmnminuszn

δiu+czn+mαu+czn+m+j

k=1 m=(nminuszn)c1

minnminus1nr L

(k) Z

+ = rkψnminuskij(u) + rndiminusuminuscznnminuszn αi+j n isin (214)

k=1

Finally for fixed i isin minc Z + c1 minc Z + c1 + 1 mα minus 1 summing (214) from

j = 1 to mα minus i yields the following expression for the bivariate pmf ωni(u) namely

minnminus1nr mαminusi mαminusi L L

(k) L

ωni(u) = rk ψnminuskij(u) + rndiminusuminuscznnminusznαi+j

k=1 j=1 j=1

10

minnminus1nr L

(k) Z

+ = Λi rkg s + rndiminusuminuscznnminuszn n isin (215) nminuskminus1iminuspnminus1

k=1

where (210) was used to establish the last equality In the special case when c1 = c2 = c it

is a straightforward exercise to show that equations (213) through (215) simplify to produce

equations (215) through (217) of Alfa and Drekic (2007 p 300)

3 Computational considerations

In order to compute the joint pmfrsquos defined by (213) (214) and (215) it is clear from (k)

equations (29) through (212) that we need to be able to compute the quantities gn and

h(k) n defined by (27) and (28) respectively In what follows we capitalize on the underlying

Markov chain structure in our model to develop a set of computational procedures which

ultimately enable one to calculate these key row vectors

First of all recall that g0(k)

= b(k) contains zeros from a certain level onwards namely (k)

level vk(zk) = u + czk + c2(k minus zk) In general one can easily deduce that gn will also

contain zeros from a certain level onwards to be denoted by lk(n) We begin this section

with a determination of lk(n) which varies in form depending on the value of vk(zk) minus 1

representing the maximum surplus level at time k after the smallest possible claim of size 1

has been applied (at time kminus)

On one hand it may be the case that the value of vk(zk)minus1 lies at or above the threshold

level Z In this situation we remark that zk can take on any value in the set 0 1 k (k)

Since g0 contains zeros from level lk(0) = vk(zk) onwards we observe that at time k + 1

it may be possible to achieve a higher surplus value than vk+1(zk) minus 1 representing the

maximum surplus level assuming there was no drop below Z due to the claim at time kminus

Specifically if the surplus had fallen to level Z minus 1 at time kminus then gained c at time k+ a

bounded level of Z + c minus1 would have been reached at time k +1 and this level is potentially

larger than vk+1(zk) minus 1 As a result we have that lk(1) = maxvk+1(zk) Z + c If we

extend this argument inductively we readily obtain lk(n) = maxvk+n(zk) Z +c +c2(n minus1)

n isin Z+ However since vk+n(zk) = u + czk + c2(k +n minus zk) = vk(zk)+ c2 + c2(n minus 1) we may

alternatively re-express lk(n) as

lk(n) = maxvk+n(zk) Z + c + c2(n minus 1) = maxvk(zk) + c2 Z + c + c2(n minus 1)

Thus

Z+lk(n) = maxlk(0) + c2 Z + c + c2(n minus 1) n isin (31)

11

or equivalently

Z+lk(n) = lk(1) + c2(n minus 1) n isin

Clearly lk(0) gt Z and lk(n) ge Z + c for n isin Z+

The other remaining case to consider involves the situation when the maximum surplus

level at time k is strictly below the threshold level Z Under this scenario we further

distinguish between two cases involving the value of c2 First of all suppose that c2 = 0 so

that vk(zk)minus1 = u +czk minus1 lt Z or equivalently u +czk le Z If u +czk happens to equal Z

then it is possible for zk isin 0 1 k With c2 = 0 however no further premiums can be

earned until after the first claim takes place (which is assumed to occur at time kminus) thereby

implying that Z = u + ci for all i = zk zk + 1 k Hence we know that u + ck = Z

in this case Moreover if u + czk happens to be strictly less than Z then we clearly have

that zk = k Even in the case when c2 gt 0 one can readily verify that zk = k so that

vk(zk) minus 1 = u + ck minus 1 lt Z Thus we proceed with the analysis assuming that zk = k

c2 ge 0 and the inequality u + ck le Z holds true

lowast k

Define the ceiling function denoted lceilxrceil to yield the smallest integer greater than or

equal to x Because a random premium cannot be received until level Z is at least achieved

it is essential to find the point at which the threshold level Z is reached (ie the point at Zminus(u+ck)+1 which u + ck minus1+ ci ge Z) Isolating this inequality for i one obtains i ge

c Hence

assuming that W1 = k the elapsed time after k at which a random premium would begin to

lceilZminus(u+ck)+1 be received is t lowast = rceil Note that t lowast isin Z+ k c k

With zk = k g0(k)

contains zeros from level u + ck onwards so that lk(0) = u + ck

Immediately following this a premium of c is guaranteed to be received at time k+ as the

threshold level Z has not yet been reached Consequently g1(k)

contains zeros from level (k)

b(k)u + c(k + 1) onwards which is c levels further than that in g0 = Continuing this (k)

process inductively we can readily establish that gn contains zeros from level u + c(k + n)

onwards for n = 1 2 t lowast k so that lk(n) = u + c(k + n) for n = 0 1 t lowast k

In the case of n = t lowast + 1 t lowast + 2 gt

(k) is first looked at in isolation At time k +k k +1

t lowast k + 1 the threshold level has been crossed giving rise to a maximum surplus value of either (k)

c(k + t lowast k contains zeros from level lk(t lowast ku + ) + c2 minus 1 or Z + c minus 1 Therefore g + 1) = t lowast k+1

+1(k + t lowast kmaxvk+t lowast k

) Z + c onwards Carrying out the same inductive process as before it is

readily seen that g(nk)

contains zeros from level lk(n) = maxvk+n(k+t lowast k) Z +c+c2(nminust lowast kminus1)

onwards for n = t lowast k + 1 t lowast k + 2 Combining both cases we obtain

u + c(k + n) if n = 0 1 t lowast lk(n) = k (32)

maxlk(t lowast) + c2 Z + c + c2(n minus t lowast minus 1) if n = t lowast + 1 t lowast + 2 k k k k

Note that by construction lk(t lowast k) gt Z and lk(n) le Z for n = 0 1 t lowast k minus1 Interestingly

12

if we extend the definition tk lowast by introducing the quantity ζk

lowast = maxtklowast 0 we remark that

(31) and (32) can actually be combined into a single formula for lk(n) namely

vk+n(zk + n) if n = 0 1 ζklowast

lk(n) = (33) maxlk(ζ

lowast) + c2 Z + c + c2(n minus ζlowast minus 1) if n = ζlowast + 1 ζlowast + 2 k k k k

Clearly we have that lk(n) ge Z + c for n = ζk lowast +1 ζk

lowast +2 Moreover if c1 = c2 = c (33)

reduces to give lk(n) = u + c(k + n) forall n isin N which is in agreement with the results in Alfa

and Drekic (2007 Section 3)

Remark In our above analysis we purposely ignored the possibility of any future claims

occurring after time k in our determination of lk(n) However we recognize that this is not

realistic and future claims would indeed occur after time k thereby further reducing the value

of the maximum surplus level lk(n) minus 1 at time k + n To account for this we imagine all

future interclaim times (ie after time k) to be of maximum length na and all corresponding

claim sizes to be of minimum amount 1 Consequently in the case when c2 gt 0 we should

subtract from formula (33) the amount of lfloornnarfloor where lfloormiddotrfloor again denotes the floor function

Obviously when c2 = 0 (ie no premiums are collected whenever the surplus level is at or

above Z) lk(n) = Z + c for n = ζk lowast + 1 ζk

lowast + 2 (k)

We now turn our attention to the development of a general algorithm for computing gni

(k) (k)for i = 0 1 lk(n) minus 1 With the aid of (24) we apply the key identity gn = gnminus1C

(which can be inferred from (27)) to first obtain

minZlk(nminus1)minus1 lk(nminus1)minus1 c2 L L L

(k) (k) (k)gni = g Bjminusi+c + gnminus1j dmBjminusi+m n isin Z+ (34) nminus1j

j=0 j=Z m=c1

However (22) states that Bi = Ona for i isin Zminus and so (34) becomes (for the same range of

i)

minZlk(nminus1)minus1 c2 lk(nminus1)minus1 L L L

(k) (k) (k)gni = g Bjminusi+c + dm gnminus1jBjminusi+m n isin Z+ (35) nminus1j

j=max0iminusc m=c1 j=maxZiminusm

Taking the form of (33) into consideration (35) yields the following two formulae

lk(nminus1)minus1 L

(k) (k)gni = gnminus1jBjminusi+c n = 1 2 ζ k

lowast (36) j=max0iminusc

and

Zminus1 c2 lk(nminus1)minus1 L L L

(k) (k) (k)g = g + dm g n = ζ lowast + 1 ζ lowast + 2 ni nminus1jBjminusi+c nminus1jBjminusi+m k k

j=max0iminusc m=c1 j=maxZiminusm

(37)

13

For n = 1 2 ζk lowast where ζk

lowast isin Z+ we note that lk(n) ge c + 1 Therefore with the aid of

(22) (36) simplifies to give

Llk(nminus1)minus1 (k) (k) j=0 gnminus1jBjminusi+c if i = 0 1 c minus 1

gni = Llk(nminus1)minus1 (k)

g if i = c c + 1 lk(n) minus 1j=iminusc nminus1jBjminusi+c

Llk(nminus1)minus1 (k)

nminus1js 0 0 if i = 0 1 c minus 1j=0 αjminusi+cg= (38)

(k) Llk(nminus1)minus1 (k) g S+ αjminusi+cgnminus1js 0 0 if i = c c+1 lk(n)minus1 nminus1iminusc j=iminusc+1

In the case of n = ζk lowast + 1 ζk

lowast + 2 we first reiterate that lk(n) ge Z + c We next break

(37) into several cases based on the value of i and the value of minc Z + c1 as follows

Case 1 Suppose minc Z + c1 = c

For i isin 0 1 c minus 1

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+mni nminus1j nminus1j

j=0 m=c1 j=Z

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k)= αjminusi+cg s + dm αjminusi+mg s 0 0 (39) nminus1j nminus1j

j=0 m=c1 j=Z

For i isin c c + 1 Z + c1 minus 1

Zminus1 c2 lk(nminus1)minus1 L L L

(k) (k) (k)g = g + dm gni nminus1j Bjminusi+c nminus1j Bjminusi+m

j=iminusc m=c1 j=Z

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k) (k)= g S + αjminusi+cg dm αjminusi+mgnminus1j s 0 0 (310) nminus1iminusc nminus1j s +

j=iminusc+1 m=c1 j=Z

For i isin Z + c1 Z + c1 + 1 Z + c minus 1

Zminus1 minc2iminusZ lk(nminus1)minus1 c2 lk(nminus1)minus1 L L L L L

(k) (k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+m + dm g Bjminusi+mni nminus1j nminus1j nminus1j

j=iminusc m=c1 j=iminusm m=iminusZ+1 j=Z

minc2iminusZ L

(k) (k)= g S + dmg S nminus1iminusc nminus1iminusm

m=c1

Zminus1 minc2iminusZ lk(nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k)+ αjminusi+cg dm αjminusi+mg dm αjminusi+mgnminus1j s 0 0 nminus1j s + nminus1js +

j=iminusc+1 m=c1 j=iminusm+1 m=iminusZ+1 j=Z

(311)

For i isin Z + c Z + c + 1 lk(n) minus 1

c2 lk(nminus1)minus1 L L

(k) (k)g = dm g Bjminusi+mni nminus1j

m=c1 j=iminusm

c2 c2 lk (nminus1)minus1 L L L

(k) (k)= dmg S + dm αjminusi+mgnminus1j s 0 0 (312) nminus1iminusm

m=c1 m=c1 j=iminusm+1

14

Case 2 Suppose minc Z + c1 = Z + c1

For i isin 0 1 Z + c1 minus 1

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+mni nminus1j nminus1j

j=0 m=c1 j=Z

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k)= αjminusi+cg s + dm αjminusi+mg s 0 0 (313) nminus1j nminus1j

j=0 m=c1 j=Z

For i isin Z + c1 Z + c1 + 1 c minus 1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+m + dm g Bjminusi+mni nminus1j nminus1j nminus1j

j=0 m=c1 j=iminusm m=iminusZ+1 j=Z

minc2iminusZ L

(k)= dmg S nminus1iminusm

m=c1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k)+ αjminusi+cg s + dm αjminusi+mg s + dm αjminusi+mg s 0 0 nminus1j nminus1j nminus1j

j=0 m=c1 j=iminusm+1 m=iminusZ+1 j=Z

(314)

For i isin c c + 1 Z + c minus 1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k) (k)g = g + g + gni nminus1j Bjminusi+c dm nminus1j Bjminusi+m dm nminus1j Bjminusi+m

j=iminusc m=c1 j=iminusm m=iminusZ+1 j=Z

minc2iminusZ L

(k) (k)= g S + dmg S nminus1iminusc nminus1iminusm

m=c1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k)+

nminus1j s 0 0 αjminusi+cgnminus1j s + dm αjminusi+mgnminus1j s + dm αjminusi+mg j=iminusc+1 m=c1 j=iminusm+1 m=iminusZ+1 j=Z

(315)

For i isin Z + c Z + c + 1 lk(n) minus 1

c2 lk (nminus1)minus1 L L

(k) (k)g = dm g Bjminusi+mni nminus1j

m=c1 j=iminusm

c2 c2 lk (nminus1)minus1 L L L

(k) (k)= dmg S + dm αjminusi+mgnminus1j s 0 0 (316) nminus1iminusm

m=c1 m=c1 j=iminusm+1

It is not difficult to verify that equations (38) through (316) simplify to produce equations

(32) and (33) of Alfa and Drekic (2007 pp 301-302) when c1 = c2 = c With equations (38) (k)

through (316) providing the recursive means of calculating gn we next turn our attention

15

to a determination of h(k) defined by (28) For n = 1 2 ζlowast we use (22) and (25) to n k

obtain

lk(nminus1)minus1 lk(nminus1)minus1

h(k) nminusj =

L g

(k) nminus1ℓBc+j+ℓ =

L αc+j+ℓg

(k) nminus1ℓs 0 0 (317)

ℓ=0 ℓ=0

For n = ζk lowast + 1 ζk

lowast + 2 we similarly acquire

Zminus1 lk(nminus1)minus1 L L

(k) (k) (k)hnminusj = gnminus1ℓBc+j+ℓ + gnminus1ℓAj+ℓ

ℓ=0 ℓ=Z

Zminus1 c2 lk(nminus1)minus1 L L L

(k) (k)= αc+j+ℓg s + dm αm+j+ℓg s 0 0 (318) nminus1ℓ nminus1ℓ

ℓ=0 m=c1 ℓ=Z

Note that substituting (317) and (318) into (29) immediately yields

Llk(nminus1)minus1 (k) (k) ℓ=0 αc+j+ℓgnminus1ℓs if n = 1 2 ζk

lowast φ (u) = nj LZminus1 (k) Lc2

Llk(nminus1)minus1 (k) ζlowast s + dm αm+j+ℓg n = + 1 ζlowast + 2 ℓ=0 αc+j+ℓgnminus1ℓ m=c1 ℓ=Z nminus1ℓs if k k

(319)

Clearly in the special case when c1 = c2 = c (319) yields equation (35) of Alfa and Drekic

(2007 p 302) Finally we remark that (319) can be substituted into (213) to obtain a

computational formula for φnj(u)

4 Numerical examples and findings

In this section we illustrate the application of our proposed algorithm with several numerical

examples All calculations in this section were performed using Mathematica (Version 7) on

a Dell Precision T7400 workstation with a 3 Ghz CPU and 4 GB of RAM

In our first illustration (labelled Example 1) we consider the ordinary Sparre Andersen

model in which interclaim times have pmf given by

(211)(911)jminus1 if j = 1 2 na minus 1 aj = (41)

(911)naminus1 if j = na

Note that the pmf (41) is that of a truncated geometric distribution with all the probability

mass on na na +1 assigned to the support value na Clearly as na becomes larger the

closer (41) approximates this particular geometric distribution having mean 55 In what Lna

Lminus1 nafollows we fix na = 25 so that aj = 0992 and jaj = 546 Let the individual j=1 j=1

claim amount distribution be given by the pmf

minus4 minus4j minus 1 j

αj = 1 + minus 1 + j isin Z+ (42) 30 30

16

We remark that the pmf (42) represents a discretized version of the Pareto distribution with

mean 10 Also note that mα = infin thereby implying that the random variables |UT | and

UTminus both have infinite support

Table 1 displays values (rounded to 6 decimal places) of the trivariate cumulative distrishy

bution function

n x y L L L

Ψnxy(u) = Pr T le n UTminus le x |UT | le y | U0 = u = ψℓij(u) ℓ=1 i=mincZ+c1 j=1

corresponding to the above discrete-time risk process with u = 10 Z = 50 c = 5 and

varying premium amount distributions according to the following set of scenarios

(1) d2 = 1

(2) d1 = 25 d2 = 15 d3 = 25

G

5)

(3) di = i

(25)i(35)5minusi i = 0 1 5

(4) di = 15 i = 0 1 4

(5) d1 = 34 d5 = 14

(6) d0 = 35 d5 = 25

We remark that while these six premium amount distributions all have mean equal to 2

they possess different shapes and we investigate the impact of their variability on the joint

distribution of T |UT | and UTminus In particular the variances corresponding to the premium

amount distributions of scenarios (1) through (6) are 0 08 12 2 3 and 6 respectively

Based on the results in Table 1 it would appear that greater variability in the premium

amount distribution generally leads to higher ruin probabilities Perhaps somewhat surprising

though the differences are subtle and not particularly striking as one compares the various

ruin probabilities from scenarios (1) through (6) Obviously we observe more of a difference

between scenarios as the values of n x and y increase

As a follow-up illustration (labelled Example 2) we consider a model in which ordinary

interclaim times have a discrete uniform distribution with pmf aj = 110 j = 1 2 10

Clearly the mean of this distribution is 55 which for comparative purposes is identical to

the ordinary interclaim time mean considered in Example 1 In addition we assume that

the pmf of the individual claim amount distribution is once again given by (42) Table 2

displays values (rounded to 6 decimal places) of Ψnxy(10) for several variants of the Sparre

Andersen model (again with Z = 50 and c = 5) having the above ordinary interclaim time

and claim amount distributions namely

17

(1) ordinary model under scenario (2) from Example 1

(2) ordinary model under scenario (5) from Example 1

(3) stationary model under scenario (2) from Example 1

(4) stationary model under scenario (5) from Example 1

(5) delayed model (with W1 = 1 with probability 1) under scenario (2) from Example 1

(6) delayed model (with W1 = 1 with probability 1) under scenario (5) from Example 1

Concerning the results in Table 2 we first compare the values of Ψnxy(10) in entries (2)

and (5) of Table 1 to their corresponding counterparts in entries (1) and (2) of Table 2 In all

cases we observe that ruin is more likely to occur when interclaim times have the truncated

geometric distribution introduced in our first example as opposed to the discrete uniform

distribution of our second example Since mean interclaim times are equal in both examshy

ples this behaviour must be attributable to the greater variability inherent in the truncated

geometric distribution We also note that under the stationary model assumption the pmf

of W1 is easily verified to be rj = (11 minus j)55 j = 1 2 10 which has mean 4 lt 55

As a result the values of Ψnxy(10) for stationary models (3) and (4) are always larger than

the corresponding values for ordinary models (1) and (2) Finally in the absence of any

knowledge regarding which distribution is in effect for the first interclaim time W1 models

(5) and (6) can be regarded as ldquoworst-case scenariordquo models as they serve to provide upper

bounds on the values of Ψnxy(10)

In conclusion we state that the values in Tables 1 and 2 were generated by first impleshy

menting the recursive procedures (38) through (316) and then summing the appropriate

bivariatetrivariate probabilities computed via (213) (214) or (215) In situations when

x = y = infin however we computed the resulting finite-time ruin probabilities using an altershy

native (albeit standard) method (eg see Cossette et al (2006 Section 3)) Specifically if

we define the survival probability σ(n u) = Pr T gt n | U0 = u = 1minusΨninfininfin(u) as well as

the auxiliary function

0 if w ge Z ηwt =

minlfloorZminuswc minus1 rfloor + 1 t if w lt Z

then by conditioning on the time and amount of the first claim (and recalling that the delayed

process reverts to the ordinary process following this claim) we obtain the expression

minnnr (kminuszk)c2 u+czk+ℓ L L L

σ(u n) = Rn + rk dℓkminuszk αjγ(u + czk + ℓ minus j n minus k)

k=1 ℓ=(kminuszk)c1 j=1

18

where the function γ(w m) for w m isin N is computed recursively via

minmna (kminusηwk)c2 u+cηwk+ℓ L L L

γ(w m) = Am + ak dℓkminusηwk αjγ(u + cηwk + ℓ minus j m minus k)

k=1 j=1 ℓ=(kminusηwk)c1

with starting values γ(w 0) = 1 forall w isin N

Acknowledgements

The authors would like to thank Dr David Landriault for his valuable suggestions which

helped to improve an earlier version of this work This research was supported by the Natural

Sciences and Engineering Research Council of Canada

References

Alfa AS 2004 Markov chain representations of discrete distributions applied to queueing

models Computers amp Operations Research 31 2365-2385

Alfa AS Drekic S 2007 Algorithmic analysis of the Sparre Andersen model in discrete

time ASTIN Bulletin 37 293-317

Bao Z 2007 A note on the compound binomial model with randomized dividend strategy

Applied Mathematics amp Computation 194 276-286

Cossette H Landriault D Marceau E 2006 Ruin probabilities in the discrete time

renewal risk model Insurance Mathematics amp Economics 38 309-323

Karlin S Taylor HM 1975 A First Course in Stochastic Processes 2nd edition Acashy

demic Press New York

Kim B Kim H-S Kim J 2008 A risk model with paying dividends and random

environment Insurance Mathematics amp Economics 42 717-726

Landriault D 2008 Randomized dividends in the compound binomial model with a general

premium rate Scandinavian Actuarial Journal 1-15

Tan J Yang X 2006 The compound binomial model with randomized decisions on

paying dividends Insurance Mathematics amp Economics 39 1-18

19

Table 1 ndash Values of Ψnxy(10) corresponding to Example 1

(a) x = 10

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0005357 0006188 0007163 0008278 0009481

(2) 0005359 0006195 0007175 0008296 0009507

(3) 0005361 0006198 0007181 0008305 0009519

(4) 0005363 0006205 0007193 0008323 0009544

(5) 0005366 0006212 0007208 0008346 0009576

(6) 0005374 0006236 0007249 0008410 0009664

y = 25 (1) 0007631 0008815 0010202 0011789 0013503

(2) 0007634 0008824 0010220 0011816 0013539

(3) 0007636 0008828 0010228 0011829 0013557

(4) 0007639 0008838 0010245 0011854 0013593

(5) 0007643 0008849 0010266 0011887 0013638

(6) 0007656 0008882 0010325 0011978 0013764

y = 50 (1) 0008477 0009792 0011333 0013095 0014998

(2) 0008481 0009802 0011352 0013125 0015039

(3) 0008483 0009807 0011361 0013139 0015059

(4) 0008486 0009817 0011380 0013168 0015098

(5) 0008491 0009830 0011403 0013204 0015149

(6) 0008504 0009866 0011470 0013305 0015288

y = infin (1) 0008777 0010138 0011733 0013558 0015528

(2) 0008781 0010149 0011753 0013588 0015570

(3) 0008783 0010154 0011763 0013603 0015591

(4) 0008787 0010164 0011782 0013633 0015632

(5) 0008791 0010177 0011807 0013671 0015684

(6) 0008805 0010215 0011875 0013775 0015828

20

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0058559 0061875 0065704 0070048 0074714

(2) 0058569 0061903 0065756 0070128 0074825

(3) 0058574 0061917 0065782 0070167 0074880

(4) 0058584 0061945 0065833 0070245 0074987

(5) 0058596 0061979 0065896 0070344 0075124

(6) 0058632 0062079 0066078 0070621 0075506

y = 25 (1) 0089002 0093986 0099739 0106262 0113270

(2) 0089017 0094028 0099817 0106382 0113437

(3) 0089025 0094049 0099856 0106442 0113519

(4) 0089040 0094091 0099932 0106560 0113681

(5) 0089057 0094142 0100028 0106708 0113887

(6) 0089113 0094294 0100302 0107127 0114463

y = 50 (1) 0102786 0108506 0115106 0122590 0130629

(2) 0102804 0108555 0115196 0122728 0130820

(3) 0102812 0108579 0115241 0122797 0130914

(4) 0102830 0108627 0115329 0122932 0131101

(5) 0102850 0108686 0115439 0123103 0131337

(6) 0102914 0108860 0115754 0123584 0132000

y = infin (1) 0108895 0114933 0121901 0129799 0138283

(2) 0108914 0114985 0121996 0129945 0138485

(3) 0108923 0115010 0122043 0130018 0138585

(4) 0108941 0115061 0122136 0130161 0138782

(5) 0108962 0115123 0122252 0130341 0139032

(6) 0109030 0115307 0122585 0130849 0139732

21

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0079856 0086792 009464 0103440 0112831

(2) 0079885 0086869 0094778 0103650 0113118

(3) 0079953 0086978 0094936 0103864 0113393

(4) 0080081 0087188 0095243 0104282 0113929

(5) 0079962 0087075 0095149 0104213 0113890

(6) 0080805 0088366 0096951 0106588 0116875

y = 25 (1) 0124591 0135680 0148207 0162245 0177217

(2) 0124639 0135806 0148433 0162586 0177683

(3) 0124755 0135993 0148703 0162950 0178148

(4) 0124978 0136354 0149226 0163658 0179054

(5) 0124763 0136141 0149035 0163500 0178935

(6) 0126238 0138381 0152148 0167588 0184060

y = 50 (1) 0146757 0160079 0175111 0191945 0209894

(2) 0146816 0160232 0175385 0192360 0210460

(3) 0146965 0160469 0175725 0192817 0211043

(4) 0147249 0160925 0176384 0193706 0212178

(5) 0146968 0160639 0176117 0193470 0211980

(6) 0148860 0163500 0180077 0198658 0218474

y = infin (1) 0158013 0172606 0189059 0207476 0227106

(2) 0158078 0172776 0189362 0207934 0227732

(3) 0158250 0173047 0189751 0208455 0228395

(4) 0158579 0173571 0190503 0209467 0229685

(5) 0158247 0173226 0190172 0209161 0229410

(6) 0160443 0176532 0194735 0215127 0236869

22

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0088667 0104403 0123329 0145277 0169184

(2) 0088711 0104518 0123538 0145598 0169627

(3) 0088733 0104574 0123641 0145756 0169846

(4) 0088777 0104687 0123844 0146066 0170274

(5) 0088829 0104828 0124105 0146470 0170832

(6) 0088988 0105229 0124822 0147560 0172331

y = 25 (1) 0140528 0167839 0201007 0239726 0282102

(2) 0140598 0168021 0201338 0240234 0282804

(3) 0140632 0168111 0201502 0240486 0283151

(4) 0140701 0168289 0201822 0240976 0283827

(5) 0140785 0168515 0202241 0241622 0284718

(6) 0141036 0169148 0203370 0243335 0287071

y = 50 (1) 0167826 0203005 0246173 0296954 0352855

(2) 0167908 0203218 0246561 0297548 0353672

(3) 0167949 0203324 0246753 0297841 0354075

(4) 0168030 0203532 0247128 0298412 0354859

(5) 0168129 0203800 0247622 0299174 0355907

(6) 0168423 0204539 0248936 0301159 0358620

y = infin (1) 0183325 0224926 0276881 0338957 0408257

(2) 0183412 0225150 0277283 0339564 0409079

(3) 0183455 0225260 0277481 0339864 0409485

(4) 0183540 0225478 0277869 0340446 0410270

(5) 0183645 0225760 0278386 0341234 0411339

(6) 0183954 0226529 0279737 0343244 0414037

23

Table 2 ndash Values of Ψnxy(10) corresponding to Example 2

(a) x = 10

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0001901 0002276 0002724 0003240 0003799

(2) 0001904 0002286 0002742 0003269 0003840

(3) 0002651 0003013 0003442 0003936 0004471

(4) 0002655 0003022 0003460 0003964 0004510

(5) 0005792 0006098 0006458 0006871 0007317

(6) 0005795 0006106 0006473 0006894 0007350

y = 25 (1) 0002708 0003243 0003880 0004615 0005411

(2) 0002713 0003257 0003906 0004656 0005469

(3) 0003778 0004293 0004904 0005608 0006369

(4) 0003782 0004306 0004929 0005647 0006424

(5) 0008255 0008690 0009203 0009792 0010427

(6) 0008259 0008702 0009224 0009824 0010473

y = 50 (1) 0003008 0003603 0004310 0005127 0006011

(2) 0003014 0003618 0004340 0005173 0006076

(3) 0004197 0004769 0005448 0006230 0007075

(4) 0004202 0004784 0005476 0006273 0007137

(5) 0009173 0009656 0010226 0010879 0011585

(6) 0009177 0009669 0010249 0010916 0011637

y = infin (1) 0003115 0003730 0004463 0005308 0006224

(2) 0003121 0003746 0004493 0005356 0006291

(3) 0004346 0004938 0005641 0006450 0007325

(4) 0004351 0004954 0005670 0006496 0007389

(5) 0009499 0009999 0010589 0011265 0011996

(6) 0009503 0010012 0010613 0011303 0012050

24

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0031176 0032815 0034733 0036924 0039282

(2) 0031193 0032862 0034821 0037061 0039475

(3) 0050168 0051747 0053587 0055684 0057939

(4) 0050184 0051792 0053671 0055815 0058124

(5) 0136884 0138219 0139763 0141517 0143399

(6) 0136896 0138256 0139832 0141626 0143553

y = 25 (1) 0047622 0050092 0052981 0056280 0059829

(2) 0004765 0050163 0053113 0056487 0060121

(3) 0076570 0078950 0081720 0084878 0088272

(4) 0076593 0079017 0081847 0085076 0088551

(5) 0205774 0207786 0210111 0212752 0215585

(6) 0205793 0207842 0210216 0212917 0215818

y = 50 (1) 0055170 0058009 0061329 0065119 0069196

(2) 0055200 0058091 0061481 0065357 0069533

(3) 0088650 0091385 0094569 0098197 0102096

(4) 0088677 0091463 0094715 0098425 0102417

(5) 0235815 0238128 0240800 0243834 0247089

(6) 0235837 0238192 0240920 0244023 0247356

y = infin (1) 0058565 0061565 0065073 0069077 0073384

(2) 0058597 0061652 0065234 0069329 0073740

(3) 0094061 0096952 0100316 0104149 0108267

(4) 0094090 0097035 0100470 0104390 0108607

(5) 0248472 0250917 0253740 0256945 0260383

(6) 0248495 0250985 0253867 0257145 0260667

25

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0048598 0052524 0056984 0061998 0067342

(2) 0048654 0052673 0057253 0062411 0067916

(3) 0070192 0073979 0078257 0083057 0088167

(4) 0070243 0074119 0078513 0083451 0088714

(5) 0151774 0154981 0158572 0162586 0166852

(6) 0151815 0155096 0158784 0162913 0167308

y = 25 (1) 0076895 0083230 0090413 0098481 0107073

(2) 0076987 0083474 0090854 0099156 0108010

(3) 0110093 0116205 0123095 0130818 0139033

(4) 0110177 0116435 0123515 0131462 0139928

(5) 0230674 0235848 0241632 0248091 0254950

(6) 0230739 0236038 0241980 0248626 0255695

y = 50 (1) 0091496 0099156 0107829 0117563 0127924

(2) 0091610 0099455 0108370 0118387 0129068

(3) 0130130 0137520 0145839 0155156 0165064

(4) 0130233 0137801 0146352 0155942 0166156

(5) 0266599 0272857 0279840 0287632 0295904

(6) 0266680 0273088 0280265 0288286 0296814

y = infin (1) 0099312 0107746 0117285 0127984 0139369

(2) 0099439 0108078 0117883 0128899 0140639

(3) 0140458 0148595 0157745 0167986 0178872

(4) 0140572 0148908 0158314 0168859 0180084

(5) 0282886 0289776 0297457 0306022 0315111

(6) 0282976 0290034 0297929 0306748 0316120

26

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0059096 0072505 0088907 0108122 0129137

(2) 0059223 0072829 0089500 0109052 0130456

(3) 0079757 0092662 0108388 0126778 0146871

(4) 0079875 0092970 0108956 0127669 0148134

(5) 0159788 0170682 0183871 0199249 0216024

(6) 0159881 0170936 0184343 0199990 0217075

y = 25 (1) 0095883 0119741 0149240 0184073 0222390

(2) 0096087 0120262 0150197 0185573 0224520

(3) 0127403 0150359 0178640 0211977 0248613

(4) 0127592 0150855 0179557 0213414 0250653

(5) 0245161 0264533 0288250 0316124 0346709

(6) 0245311 0264942 0289010 0317319 0348407

y = 50 (1) 0116598 0147920 0187100 0233784 0285503

(2) 0116839 0148537 0188231 0235554 0288015

(3) 0153023 0183153 0220713 0265390 0314840

(4) 0153249 0183742 0221798 0267087 0317248

(5) 0285742 0311159 0342651 0380005 0421287

(6) 0285921 0311644 0343551 0381418 0423292

y = infin (1) 0129470 0167217 0215383 0273841 0339734

(2) 0129725 0167865 0216560 0275661 0342282

(3) 0167978 0204275 0250442 0306382 0369383

(4) 0168217 0204895 0251572 0308128 0371826

(5) 0305868 0336470 0375169 0421935 0474525

(6) 0306058 0336980 0376106 0423388 0476560

27

Page 4: uwaterloo.ca · Ruin Analysis of a Threshold Strategy in a Discrete-time Sparre Andersen Model Steve DREKIC and Ana Maria MERA Department of Statistics and Actuarial Science University

consideration to the joint pmfrsquos

Z+ωni(u) = Pr T = n UTminus = i | U0 = u n isin i = minc Z+c1 minc Z+c1+1 mαminus1

φnj(u) = Pr T = n |UT | = j | U0 = u n isin Z+ j = 1 2 mαminusminc Z +c1

and

ψnij(u) = Pr T = n UTminus = i |UT | = j | U0 = u n isin Z+

i = minc Z +c1 minc Z +c1 + 1 mαminus1

j = 1 2 3 mαminusi

Finally we implement our various procedures and present some numerical examples in Section

4

2 Model formulation and preliminaries

Adopting the same methodology employed by Alfa and Drekic (2007) let τj =

Pr W gt j|W gt j minus 1 = AjAjminus1 j = 1 2 na where W denotes an arbitrary Wi

i = 2 3 Clearly τ1 = A1 and τna = 0 If we next define the 1 times na row vector

e1 = (1 0 0) the na times na probability transition matrix

0 τ1 0 middot middot middot 0

0 0 τ2

0

S =

0 0 middot middot middot 0 τnaminus1

0 0 middot middot middot 0 0

and the na times 1 column vector

1 minus τ1 1 minus τ2 s = 1 minus τnaminus1

1

it is known that aj = e1Sjminus1 s j = 1 2 na (eg see Alfa (2004)) In other words W

can be expressed as a discrete phase-type random variable (of dimension na) with the above

parameterization implying that s contains the absorption probabilities to claim occurrence

associated with each of the na possible interclaim time values Alfa (2004) refers to this

4

characterization as the ldquoelapsed timerdquo phase-type representation of the pmf aj We remark

that the above pmf formula also holds true for j gt na as it is readily verified that Sjminus1 = Ona

an na times na matrix of zeros for such values of j

In what follows we proceed to set up our delayed Sparre Andersen model as a bivariate

Markov chain conditional on W1 = k isin 1 2 nr Specifically assuming that W1 = k we

construct the bivariate stochastic process (Ut Lt) t = k k + 1 such that Ut represents

the amount of surplus at time t and Lt denotes a specific counter at time t which measures the

ldquoelapsed timerdquo (in the sense described above) length of the next claim occurrence We refer to

the Ut component as the level of the process and the Lt component as the phase of the process

Two important observations are particularly noteworthy regarding this construction First of

all Lk = 1 since we are assuming that the very first claim occurs at time kminus More generally

Lt = 1 if a claim happens to occur at time tminus Secondly the delayed process reverts to the

ordinary process (having interclaim time distribution defined by the pmf aj) upon occurrence

of the first claim As a result the bivariate stochastic process (Ut Lt) t = k k + 1

possesses the following Markovian relationship

(Ut + pt Lt + 1) if there is no claim at time (t + 1)minus (Ut+1 Lt+1) =

(Ut + pt minus Y 1) if there is a claim of amount Y at time (t + 1)minus

Note that the state space for this Markov chain is given by Δ = Z times 1 2 na Furthershy

more the probability transition matrix associated with this Markov chain is given by

middot middot middot minus1 10 middot middot middot Zminus2 Zminus1 Z Z+1 Z+2 middot middot middot

middot middot middot

middot middot middot minus1 middot middot middot Bc Bcminus1 Bcminus2 middot middot middot BcminusZ+1 BcminusZ BcminusZminus1 BcminusZminus2 BcminusZminus3 middot middot middot

0

1

middot middot middot Bc+1 Bc Bcminus1 middot middot middot BcminusZ+2 BcminusZ+1 BcminusZ BcminusZminus1 BcminusZminus2 middot middot middot middot middot middot Bc+2 Bc+1 Bc middot middot middot BcminusZ+3 BcminusZ+2 BcminusZ+1 BcminusZ BcminusZminus1 middot middot middot

middot middot middot

middot middot middot

P = Zminus2 middot middot middot BZ+cminus1 BZ+cminus2 BZ+cminus3 middot middot middot Bc Bcminus1 Bcminus2 Bcminus3 Bcminus4 middot middot middot (21)

Zminus1 middot middot middot BZ+c BZ+cminus1 BZ+cminus2 middot middot middot Bc+1 Bc Bcminus1 Bcminus2 Bcminus3 middot middot middot

Z middot middot middot AZ+1 AZ AZminus1 middot middot middot A2 A1 A0 Aminus1 Aminus2 middot middot middot Z+1 middot middot middot AZ+2 AZ+1 AZ middot middot middot A3 A2 A1 A0 Aminus1 middot middot middot

Z+2 middot middot middot AZ+3 AZ+2 AZ+1 middot middot middot A4 A3 A2 A1 A0 middot middot middot

middot middot middot

middot middot middot

where Ona

if i isin Zminus

Bi = S if i = 0 (22) se1αi if i isin Z+

5

and c2 L

Ai = djBi+j (23) j=c1

We give a brief indication of the reasoning leading to the form of P in (21) For example

consider the transition ldquoUt = 0 rarr Ut+1 = 1rdquo which has block element Bcminus1 containing the

transition probabilities governing the ldquoLt rarr Lt+1 rdquo process At a surplus level of 0 at time

t a premium of pt = c is certain to be received Therefore to arrive at surplus level 1 at

the next time unit a ldquoso-calledrdquo claim of size c minus 1 must occur to neutralize the premium

just received Thus the size of the ldquoclaimrdquo required for this transition to occur determines

the value of i in Bi so that Bcminus1 is obtained and its form subsequently determined by (22)

according to the value of the index c minus 1 Note that only when c gt 1 does an actual claim

occur in this particular scenario We mention that for transitions which begin at or above the

threshold level Z a similar line of reasoning would also apply except now one must condition

on the amount of the random premium being earned as indicated by the form of (23)

Note that this formulation gives rise to a doubly-infinite Markov chain with finite blocks

of size na In the special case of the threshold-free model (obtained by setting c1 = c2 = c

so that dc = 1) we remark that (21) is in agreement with the matrix P in Alfa and Drekic

(2007 p 296) Also since αi = 0 forall i gt mα we note that Bi = Ona if i gt mα

For the computation of ruin-related quantities of interest it is useful to partition the

state space Δ into two mutually exclusive state spaces namely Δ1 = N times 1 2 na

and Δ2 = Zminus times 1 2 na In addition we introduce (one-step) transition matrices

C Δ1 rarr Δ1 and D Δ1 rarr Δ2 which correspond to mapping ldquonon-ruinedrdquo states in Δ1 to

either ldquonon-ruinedrdquo states in Δ1 or ldquoruinedrdquo states in Δ2 respectively In other words the

matrices C and D are essentially submatrices of P and their forms are readily given by

210 middot middot middot Z minus 1 Z Z + 1 middot middot middot

Bc Bcminus1 Bcminus2 middot middot middot BcminusZ+1 BcminusZ BcminusZminus1 middot middot middot Bc+1 Bc Bcminus1 middot middot middot BcminusZ+2 BcminusZ+1 BcminusZ middot middot middot

Bc+2 Bc+1 Bc middot middot middot BcminusZ+3 BcminusZ+2 BcminusZ+1 middot middot middot

middot middot middot

middot middot middot

0

1

2 Z minus 2 BZ+cminus2 BZ+cminus3 BZ+cminus4 middot middot middot Bcminus1 Bcminus2 Bcminus3 middot middot middot (24) C =

Z minus 1 BZ+cminus1 BZ+cminus2 BZ+cminus3 middot middot middot Bc Bcminus1 Bcminus2 middot middot middot

Z AZ AZminus1 AZminus2 middot middot middot A1 A0 Aminus1 middot middot middot

Z + 1 AZ+1 AZ AZminus1 middot middot middot A2 A1 A0 middot middot middot

Z + 2 AZ+2 AZ+1 AZ middot middot middot A3 A2 A1 middot middot middot

middot middot middot

6

and

minus1 minus2 minus3 middot middot middot

0 Bc+1 Bc+2 Bc+3 middot middot middot 1 Bc+2 Bc+3 Bc+4 middot middot middot

2 Bc+3 Bc+4 Bc+5 middot middot middot middot middot middot

Z minus 2 BZ+cminus1 BZ+c BZ+c+1 middot middot middot D = (25) Z minus 1 BZ+c BZ+c+1 BZ+c+2 middot middot middot

Z AZ+1 AZ+2 AZ+3 middot middot middot

Z + 1 AZ+2 AZ+3 AZ+4 middot middot middot Z + 2 AZ+3 AZ+4 AZ+5 middot middot middot

Let us now specify some further definitions required in upcoming derivations First of all

we introduce the function

0 if u ge Z zt =

maxi isin 1 2 t|u + c(i minus 1) lt Z if u lt Z

or equivalently

0 if u ge Z zt =

minlfloorZminuscuminus1 rfloor + 1 t if u lt Z

where the floor function denoted lfloorxrfloor yields the largest integer less than or equal to x We

define dm0 = δm0 where δij in general denotes the Kronecker delta function of i and j

Also define

dm if m = c1 c1 + 1 c2dm1 =

0 elsewhere

and for n = 2 3

Lc2 dj1dmminusjnminus1 if m = nc1 nc1 + 1 nc2dmn = j=c1

0 elsewhere

In other words dmn represents the pmf of the n-fold convolution of dm with itself

Recalling that W1 = k note that zk + ( k+) determines the earliest time point in the set =

0+ 1+ (k minus1)+ at which a random premium is received Note also that zk = k implies

that no such point exists in this set (ie the threshold level Z has not yet been reached when

the premium at time (k minus1)+ is to be received) Therefore if we assume for the moment that

zk k we may condition on the total premiums earned at times zk + (zk + 1)+ (k minus 1)+=

7

to obtain the following expression for b(k) the initial (ie starting at time k) probability row

vector of the above Markov chain corresponding only to the states in Δ1 namely

(kminuszk)c2

b(k) L

= dmkminuszk(αu+czk+me1 αu+czk+mminus1e1 α2e1 α1e1 0 0 ) (26)

m=(kminuszk)c1

where 0 denotes the 1 times na row vector of zeros By further defining the function vn(t) =

b(k)u + ct + (n minus t)c2 we note that the i-th level of is given by the 1 times na row vector L(kminuszk)c2 b(k)αu+czk+mminusie1 for each i isin Ωk = 0 1 vk(zk)minus1 Hence contains

m=(kminuszk)c1 dmkminuszk

zeros from level vk(zk) onwards Moreover (26) holds true even in the case when zk = k

simplifying to give b(k) = (αu+cke1 αu+ckminus1e1 α2e1 α1e1 0 0 )

At this juncture we introduce two additional row vectors namely

(k) (k) (k) (k) b(k)gn = (gn0 gn1 gn2 ) = Cn n isin N (27)

and

h(k) (k) (k) (k) b(k)Cnminus1D =

(k) Z

+ n = (h nminus2 hnminus3 ) = gnminus1D n isin (28) nminus1 h

(k)We remark that gn contains the probabilities of being in the various ldquonon-ruinedrdquo states

at time k + n (ie after claims have been paid out) without having visited a ldquoruinedrdquo state

during the previous n minus 1 transitions given that Uk isin Ωk sub Δ1 In a similar fashion h(nk)

contains the probabilities of being in the various ldquoruinedrdquo states for the first time at time

k + n given that Uk isin Ωk sub Δ1 Note that the probability of being in ruined state ldquo minus jrdquo is (k) (k) (k) (k)

characterized by the row vector h = (h h h ) with the third subscript nminusj nminusj1 nminusj2 nminusjna

representing the value of the elapsed time counter Lk+n Upon further reflection however (k)

=it must be the case that h = 0 for i 1 as ruin can only occur at claim instants nminusji

which in this case implies Lk+n = 1 with probability 1 Thus it immediately follows that (k) (k) (k)

the structure of the 1 times na row vector hnminusj is simply given by hnminusj = (φnj(u) 0 0) (k)

where φ (u) = Pr T = k + n |UT | = j | Consequently as in the analysis with nj Uk isin Ωk

no threshold on the surplus level (see Alfa and Drekic (2007 p 299)) we obtain

(k) (k) prime φ (u) = h (29) nj nminusje1

where e1 prime denotes the transpose of e1

A similar line of probabilistic logic can also be applied to obtain a representation for the (k)

trivariate function ψ (u) = Pr T = k + n UTminus = i |UT | = j | Uk isin Ωk In order for ruin nij

to occur at time k + n with a surplus immediately prior to ruin equal to i we observe that

(i) none of the previous n minus 1 transitions must have included a visit to any state in Δ2

and

8

(ii) the surplus level at time k + n minus 1 must necessarily be equal to i minus pk+nminus1

Note that pk+nminus1 denotes the premium corresponding to the time interval (k + n minus 1 k + n]

which is received at (k + n minus 1)+ Hence the quantity corresponding to points (i) and (ii) (k)

above is the 1 times na row vector g Then at the next time unit a claim must nminus1iminuspk+nminus1

necessarily occur but not before a premium of pk+nminus1 is first collected thereby raising the

surplus level to i Since s contains the absorption probabilities (to claim occurrence) from

the na possible phase states and the claim causing ruin must be of size i + j in order to

ensure that the deficit at ruin is equal to j it follows that

(k) (k)ψ (u) = g s αi+j (210) nij nminus1iminuspk+nminus1

(k)The following corollary provides the means of calculating the quantity g in nminus1iminuspk+nminus1

(210)

Corollary 1 If minc Z + c1 = c then

(k)

g if i = c c+1 Z +c1 minus1 nminus1iminusc

(k) (k) Lminc2iminusZ (k)g = g + j=c1

if i = Z +c1 Z +c1 +1 minmα Z +cminus1nminus1iminuspk+nminus1 nminus1iminusc djgnminus1iminusj

Lc2 (k)

j=c1 djgnminus1iminusj if i = Z +c Z +c+1 mαminus1

(211)

Conversely if minc Z + c1 = Z + c1 then

Lminc2iminusZ (k)

djg if i = Z +c1 Z +c1 +1 cminus1nminus1iminusj j=c1

(k) (k) Lminc2iminusZ (k)gnminus1iminuspk+nminus1

= gnminus1iminusc + j=c1 djgnminus1iminusj if i = c c+1 minmα Z +cminus1

Lc2 (k)djg if i = Z +c Z +c+1 mαminus1j=c1 nminus1iminusj

(212)

Proof The justification of (211) is as follows If c is smaller than Z + c1 the only way of

reaching a surplus level of i isin c c+1 Z +c1 minus1 (prior to ruin) is by receiving a premium

of c since i minus j for any j = c1 c1 +1 c2 would result in a surplus level below Z where the

random premium rate is not applicable Next for fixed i isin Z +c1 Z +c1 +1 Z +c minus1

there are two distinct ways of reaching this surplus level First of all note that the value of

i minus j is guaranteed to produce a surplus amount greater than or equal to Z provided that

j isin c1 c1 + 1 minc2 i minus Z Secondly note that reducing each of value of i by c also

yields a non-negative surplus value less than Z Hence both pure and random premiums

are plausible under this scenario For i isin Z + c Z + c + 1 note that decreasing this

amount by any j = c1 c1 + 1 c2 results in a surplus value greater than or equal to Z as

desired However reducing it by the amount c also has the same impact thus rendering the

pure premium option inadmissible in this case Lastly we incorporate into the range of i the

9

fact that the surplus immediately prior to ruin cannot exceed mα minus1 as indicated in Section

1 The justification of (212) when c is larger than Z + c1 mirrors the analysis above and is

left to the reader D

Recall that in the analysis up to this point we have assumed that W1 = k isin

1 2 3 nr If we now condition on the value of W1 the Law of Total Probability

yields the following two expressions for φnj(u) and ψnij(u) namely

nr L

= rkφ (u) n = nr + 1 nr + 2 φnj(u) n

(kminus) kj

k=1

and nr L

(k)ψnij(u) = rkψ (u) n = nr + 1 nr + 2 nminuskij

k=1

In addition however it is quite possible that T = n for n = 1 2 nr and this can happen

as a result of either

(i) the first claim which causes ruin occurring at time nminus from a (conditional) surplus Lnminus1level of u + czn + m (where m) or i=zn

Xi =

(ii) the first claim which does not cause ruin occurring at some time kminus k isin 1 2 nminus

1 and ruin subsequently occurring n minus k time units later

If we now combine these outcomes with those for n = nr + 1 nr + 2 (and recall that

rn = 0 forall n gt nr) the general expression for φnj(u) becomes

minnminus1nr (nminuszn)c2 L

(k) L

φnj(u)= rkφ (u) + rn dmnminuszn αu+czn+m+j n isin Z+ (213) nminuskj

k=1 m=(nminuszn)c1

Concerning the function ψnij(u) when n le nr and point (i) above is considered we note Lnminus1that i must necessarily equal u + czn + m (again conditional on i=zn

Xi = m) for ruin to

occur in this manner Hence in a similar fashion we acquire

minnminus1nr (nminuszn)c2 L L

ψnij(u) = rkψn

(kminus) kij(u) + rn dmnminuszn

δiu+czn+mαu+czn+m+j

k=1 m=(nminuszn)c1

minnminus1nr L

(k) Z

+ = rkψnminuskij(u) + rndiminusuminuscznnminuszn αi+j n isin (214)

k=1

Finally for fixed i isin minc Z + c1 minc Z + c1 + 1 mα minus 1 summing (214) from

j = 1 to mα minus i yields the following expression for the bivariate pmf ωni(u) namely

minnminus1nr mαminusi mαminusi L L

(k) L

ωni(u) = rk ψnminuskij(u) + rndiminusuminuscznnminusznαi+j

k=1 j=1 j=1

10

minnminus1nr L

(k) Z

+ = Λi rkg s + rndiminusuminuscznnminuszn n isin (215) nminuskminus1iminuspnminus1

k=1

where (210) was used to establish the last equality In the special case when c1 = c2 = c it

is a straightforward exercise to show that equations (213) through (215) simplify to produce

equations (215) through (217) of Alfa and Drekic (2007 p 300)

3 Computational considerations

In order to compute the joint pmfrsquos defined by (213) (214) and (215) it is clear from (k)

equations (29) through (212) that we need to be able to compute the quantities gn and

h(k) n defined by (27) and (28) respectively In what follows we capitalize on the underlying

Markov chain structure in our model to develop a set of computational procedures which

ultimately enable one to calculate these key row vectors

First of all recall that g0(k)

= b(k) contains zeros from a certain level onwards namely (k)

level vk(zk) = u + czk + c2(k minus zk) In general one can easily deduce that gn will also

contain zeros from a certain level onwards to be denoted by lk(n) We begin this section

with a determination of lk(n) which varies in form depending on the value of vk(zk) minus 1

representing the maximum surplus level at time k after the smallest possible claim of size 1

has been applied (at time kminus)

On one hand it may be the case that the value of vk(zk)minus1 lies at or above the threshold

level Z In this situation we remark that zk can take on any value in the set 0 1 k (k)

Since g0 contains zeros from level lk(0) = vk(zk) onwards we observe that at time k + 1

it may be possible to achieve a higher surplus value than vk+1(zk) minus 1 representing the

maximum surplus level assuming there was no drop below Z due to the claim at time kminus

Specifically if the surplus had fallen to level Z minus 1 at time kminus then gained c at time k+ a

bounded level of Z + c minus1 would have been reached at time k +1 and this level is potentially

larger than vk+1(zk) minus 1 As a result we have that lk(1) = maxvk+1(zk) Z + c If we

extend this argument inductively we readily obtain lk(n) = maxvk+n(zk) Z +c +c2(n minus1)

n isin Z+ However since vk+n(zk) = u + czk + c2(k +n minus zk) = vk(zk)+ c2 + c2(n minus 1) we may

alternatively re-express lk(n) as

lk(n) = maxvk+n(zk) Z + c + c2(n minus 1) = maxvk(zk) + c2 Z + c + c2(n minus 1)

Thus

Z+lk(n) = maxlk(0) + c2 Z + c + c2(n minus 1) n isin (31)

11

or equivalently

Z+lk(n) = lk(1) + c2(n minus 1) n isin

Clearly lk(0) gt Z and lk(n) ge Z + c for n isin Z+

The other remaining case to consider involves the situation when the maximum surplus

level at time k is strictly below the threshold level Z Under this scenario we further

distinguish between two cases involving the value of c2 First of all suppose that c2 = 0 so

that vk(zk)minus1 = u +czk minus1 lt Z or equivalently u +czk le Z If u +czk happens to equal Z

then it is possible for zk isin 0 1 k With c2 = 0 however no further premiums can be

earned until after the first claim takes place (which is assumed to occur at time kminus) thereby

implying that Z = u + ci for all i = zk zk + 1 k Hence we know that u + ck = Z

in this case Moreover if u + czk happens to be strictly less than Z then we clearly have

that zk = k Even in the case when c2 gt 0 one can readily verify that zk = k so that

vk(zk) minus 1 = u + ck minus 1 lt Z Thus we proceed with the analysis assuming that zk = k

c2 ge 0 and the inequality u + ck le Z holds true

lowast k

Define the ceiling function denoted lceilxrceil to yield the smallest integer greater than or

equal to x Because a random premium cannot be received until level Z is at least achieved

it is essential to find the point at which the threshold level Z is reached (ie the point at Zminus(u+ck)+1 which u + ck minus1+ ci ge Z) Isolating this inequality for i one obtains i ge

c Hence

assuming that W1 = k the elapsed time after k at which a random premium would begin to

lceilZminus(u+ck)+1 be received is t lowast = rceil Note that t lowast isin Z+ k c k

With zk = k g0(k)

contains zeros from level u + ck onwards so that lk(0) = u + ck

Immediately following this a premium of c is guaranteed to be received at time k+ as the

threshold level Z has not yet been reached Consequently g1(k)

contains zeros from level (k)

b(k)u + c(k + 1) onwards which is c levels further than that in g0 = Continuing this (k)

process inductively we can readily establish that gn contains zeros from level u + c(k + n)

onwards for n = 1 2 t lowast k so that lk(n) = u + c(k + n) for n = 0 1 t lowast k

In the case of n = t lowast + 1 t lowast + 2 gt

(k) is first looked at in isolation At time k +k k +1

t lowast k + 1 the threshold level has been crossed giving rise to a maximum surplus value of either (k)

c(k + t lowast k contains zeros from level lk(t lowast ku + ) + c2 minus 1 or Z + c minus 1 Therefore g + 1) = t lowast k+1

+1(k + t lowast kmaxvk+t lowast k

) Z + c onwards Carrying out the same inductive process as before it is

readily seen that g(nk)

contains zeros from level lk(n) = maxvk+n(k+t lowast k) Z +c+c2(nminust lowast kminus1)

onwards for n = t lowast k + 1 t lowast k + 2 Combining both cases we obtain

u + c(k + n) if n = 0 1 t lowast lk(n) = k (32)

maxlk(t lowast) + c2 Z + c + c2(n minus t lowast minus 1) if n = t lowast + 1 t lowast + 2 k k k k

Note that by construction lk(t lowast k) gt Z and lk(n) le Z for n = 0 1 t lowast k minus1 Interestingly

12

if we extend the definition tk lowast by introducing the quantity ζk

lowast = maxtklowast 0 we remark that

(31) and (32) can actually be combined into a single formula for lk(n) namely

vk+n(zk + n) if n = 0 1 ζklowast

lk(n) = (33) maxlk(ζ

lowast) + c2 Z + c + c2(n minus ζlowast minus 1) if n = ζlowast + 1 ζlowast + 2 k k k k

Clearly we have that lk(n) ge Z + c for n = ζk lowast +1 ζk

lowast +2 Moreover if c1 = c2 = c (33)

reduces to give lk(n) = u + c(k + n) forall n isin N which is in agreement with the results in Alfa

and Drekic (2007 Section 3)

Remark In our above analysis we purposely ignored the possibility of any future claims

occurring after time k in our determination of lk(n) However we recognize that this is not

realistic and future claims would indeed occur after time k thereby further reducing the value

of the maximum surplus level lk(n) minus 1 at time k + n To account for this we imagine all

future interclaim times (ie after time k) to be of maximum length na and all corresponding

claim sizes to be of minimum amount 1 Consequently in the case when c2 gt 0 we should

subtract from formula (33) the amount of lfloornnarfloor where lfloormiddotrfloor again denotes the floor function

Obviously when c2 = 0 (ie no premiums are collected whenever the surplus level is at or

above Z) lk(n) = Z + c for n = ζk lowast + 1 ζk

lowast + 2 (k)

We now turn our attention to the development of a general algorithm for computing gni

(k) (k)for i = 0 1 lk(n) minus 1 With the aid of (24) we apply the key identity gn = gnminus1C

(which can be inferred from (27)) to first obtain

minZlk(nminus1)minus1 lk(nminus1)minus1 c2 L L L

(k) (k) (k)gni = g Bjminusi+c + gnminus1j dmBjminusi+m n isin Z+ (34) nminus1j

j=0 j=Z m=c1

However (22) states that Bi = Ona for i isin Zminus and so (34) becomes (for the same range of

i)

minZlk(nminus1)minus1 c2 lk(nminus1)minus1 L L L

(k) (k) (k)gni = g Bjminusi+c + dm gnminus1jBjminusi+m n isin Z+ (35) nminus1j

j=max0iminusc m=c1 j=maxZiminusm

Taking the form of (33) into consideration (35) yields the following two formulae

lk(nminus1)minus1 L

(k) (k)gni = gnminus1jBjminusi+c n = 1 2 ζ k

lowast (36) j=max0iminusc

and

Zminus1 c2 lk(nminus1)minus1 L L L

(k) (k) (k)g = g + dm g n = ζ lowast + 1 ζ lowast + 2 ni nminus1jBjminusi+c nminus1jBjminusi+m k k

j=max0iminusc m=c1 j=maxZiminusm

(37)

13

For n = 1 2 ζk lowast where ζk

lowast isin Z+ we note that lk(n) ge c + 1 Therefore with the aid of

(22) (36) simplifies to give

Llk(nminus1)minus1 (k) (k) j=0 gnminus1jBjminusi+c if i = 0 1 c minus 1

gni = Llk(nminus1)minus1 (k)

g if i = c c + 1 lk(n) minus 1j=iminusc nminus1jBjminusi+c

Llk(nminus1)minus1 (k)

nminus1js 0 0 if i = 0 1 c minus 1j=0 αjminusi+cg= (38)

(k) Llk(nminus1)minus1 (k) g S+ αjminusi+cgnminus1js 0 0 if i = c c+1 lk(n)minus1 nminus1iminusc j=iminusc+1

In the case of n = ζk lowast + 1 ζk

lowast + 2 we first reiterate that lk(n) ge Z + c We next break

(37) into several cases based on the value of i and the value of minc Z + c1 as follows

Case 1 Suppose minc Z + c1 = c

For i isin 0 1 c minus 1

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+mni nminus1j nminus1j

j=0 m=c1 j=Z

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k)= αjminusi+cg s + dm αjminusi+mg s 0 0 (39) nminus1j nminus1j

j=0 m=c1 j=Z

For i isin c c + 1 Z + c1 minus 1

Zminus1 c2 lk(nminus1)minus1 L L L

(k) (k) (k)g = g + dm gni nminus1j Bjminusi+c nminus1j Bjminusi+m

j=iminusc m=c1 j=Z

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k) (k)= g S + αjminusi+cg dm αjminusi+mgnminus1j s 0 0 (310) nminus1iminusc nminus1j s +

j=iminusc+1 m=c1 j=Z

For i isin Z + c1 Z + c1 + 1 Z + c minus 1

Zminus1 minc2iminusZ lk(nminus1)minus1 c2 lk(nminus1)minus1 L L L L L

(k) (k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+m + dm g Bjminusi+mni nminus1j nminus1j nminus1j

j=iminusc m=c1 j=iminusm m=iminusZ+1 j=Z

minc2iminusZ L

(k) (k)= g S + dmg S nminus1iminusc nminus1iminusm

m=c1

Zminus1 minc2iminusZ lk(nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k)+ αjminusi+cg dm αjminusi+mg dm αjminusi+mgnminus1j s 0 0 nminus1j s + nminus1js +

j=iminusc+1 m=c1 j=iminusm+1 m=iminusZ+1 j=Z

(311)

For i isin Z + c Z + c + 1 lk(n) minus 1

c2 lk(nminus1)minus1 L L

(k) (k)g = dm g Bjminusi+mni nminus1j

m=c1 j=iminusm

c2 c2 lk (nminus1)minus1 L L L

(k) (k)= dmg S + dm αjminusi+mgnminus1j s 0 0 (312) nminus1iminusm

m=c1 m=c1 j=iminusm+1

14

Case 2 Suppose minc Z + c1 = Z + c1

For i isin 0 1 Z + c1 minus 1

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+mni nminus1j nminus1j

j=0 m=c1 j=Z

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k)= αjminusi+cg s + dm αjminusi+mg s 0 0 (313) nminus1j nminus1j

j=0 m=c1 j=Z

For i isin Z + c1 Z + c1 + 1 c minus 1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+m + dm g Bjminusi+mni nminus1j nminus1j nminus1j

j=0 m=c1 j=iminusm m=iminusZ+1 j=Z

minc2iminusZ L

(k)= dmg S nminus1iminusm

m=c1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k)+ αjminusi+cg s + dm αjminusi+mg s + dm αjminusi+mg s 0 0 nminus1j nminus1j nminus1j

j=0 m=c1 j=iminusm+1 m=iminusZ+1 j=Z

(314)

For i isin c c + 1 Z + c minus 1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k) (k)g = g + g + gni nminus1j Bjminusi+c dm nminus1j Bjminusi+m dm nminus1j Bjminusi+m

j=iminusc m=c1 j=iminusm m=iminusZ+1 j=Z

minc2iminusZ L

(k) (k)= g S + dmg S nminus1iminusc nminus1iminusm

m=c1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k)+

nminus1j s 0 0 αjminusi+cgnminus1j s + dm αjminusi+mgnminus1j s + dm αjminusi+mg j=iminusc+1 m=c1 j=iminusm+1 m=iminusZ+1 j=Z

(315)

For i isin Z + c Z + c + 1 lk(n) minus 1

c2 lk (nminus1)minus1 L L

(k) (k)g = dm g Bjminusi+mni nminus1j

m=c1 j=iminusm

c2 c2 lk (nminus1)minus1 L L L

(k) (k)= dmg S + dm αjminusi+mgnminus1j s 0 0 (316) nminus1iminusm

m=c1 m=c1 j=iminusm+1

It is not difficult to verify that equations (38) through (316) simplify to produce equations

(32) and (33) of Alfa and Drekic (2007 pp 301-302) when c1 = c2 = c With equations (38) (k)

through (316) providing the recursive means of calculating gn we next turn our attention

15

to a determination of h(k) defined by (28) For n = 1 2 ζlowast we use (22) and (25) to n k

obtain

lk(nminus1)minus1 lk(nminus1)minus1

h(k) nminusj =

L g

(k) nminus1ℓBc+j+ℓ =

L αc+j+ℓg

(k) nminus1ℓs 0 0 (317)

ℓ=0 ℓ=0

For n = ζk lowast + 1 ζk

lowast + 2 we similarly acquire

Zminus1 lk(nminus1)minus1 L L

(k) (k) (k)hnminusj = gnminus1ℓBc+j+ℓ + gnminus1ℓAj+ℓ

ℓ=0 ℓ=Z

Zminus1 c2 lk(nminus1)minus1 L L L

(k) (k)= αc+j+ℓg s + dm αm+j+ℓg s 0 0 (318) nminus1ℓ nminus1ℓ

ℓ=0 m=c1 ℓ=Z

Note that substituting (317) and (318) into (29) immediately yields

Llk(nminus1)minus1 (k) (k) ℓ=0 αc+j+ℓgnminus1ℓs if n = 1 2 ζk

lowast φ (u) = nj LZminus1 (k) Lc2

Llk(nminus1)minus1 (k) ζlowast s + dm αm+j+ℓg n = + 1 ζlowast + 2 ℓ=0 αc+j+ℓgnminus1ℓ m=c1 ℓ=Z nminus1ℓs if k k

(319)

Clearly in the special case when c1 = c2 = c (319) yields equation (35) of Alfa and Drekic

(2007 p 302) Finally we remark that (319) can be substituted into (213) to obtain a

computational formula for φnj(u)

4 Numerical examples and findings

In this section we illustrate the application of our proposed algorithm with several numerical

examples All calculations in this section were performed using Mathematica (Version 7) on

a Dell Precision T7400 workstation with a 3 Ghz CPU and 4 GB of RAM

In our first illustration (labelled Example 1) we consider the ordinary Sparre Andersen

model in which interclaim times have pmf given by

(211)(911)jminus1 if j = 1 2 na minus 1 aj = (41)

(911)naminus1 if j = na

Note that the pmf (41) is that of a truncated geometric distribution with all the probability

mass on na na +1 assigned to the support value na Clearly as na becomes larger the

closer (41) approximates this particular geometric distribution having mean 55 In what Lna

Lminus1 nafollows we fix na = 25 so that aj = 0992 and jaj = 546 Let the individual j=1 j=1

claim amount distribution be given by the pmf

minus4 minus4j minus 1 j

αj = 1 + minus 1 + j isin Z+ (42) 30 30

16

We remark that the pmf (42) represents a discretized version of the Pareto distribution with

mean 10 Also note that mα = infin thereby implying that the random variables |UT | and

UTminus both have infinite support

Table 1 displays values (rounded to 6 decimal places) of the trivariate cumulative distrishy

bution function

n x y L L L

Ψnxy(u) = Pr T le n UTminus le x |UT | le y | U0 = u = ψℓij(u) ℓ=1 i=mincZ+c1 j=1

corresponding to the above discrete-time risk process with u = 10 Z = 50 c = 5 and

varying premium amount distributions according to the following set of scenarios

(1) d2 = 1

(2) d1 = 25 d2 = 15 d3 = 25

G

5)

(3) di = i

(25)i(35)5minusi i = 0 1 5

(4) di = 15 i = 0 1 4

(5) d1 = 34 d5 = 14

(6) d0 = 35 d5 = 25

We remark that while these six premium amount distributions all have mean equal to 2

they possess different shapes and we investigate the impact of their variability on the joint

distribution of T |UT | and UTminus In particular the variances corresponding to the premium

amount distributions of scenarios (1) through (6) are 0 08 12 2 3 and 6 respectively

Based on the results in Table 1 it would appear that greater variability in the premium

amount distribution generally leads to higher ruin probabilities Perhaps somewhat surprising

though the differences are subtle and not particularly striking as one compares the various

ruin probabilities from scenarios (1) through (6) Obviously we observe more of a difference

between scenarios as the values of n x and y increase

As a follow-up illustration (labelled Example 2) we consider a model in which ordinary

interclaim times have a discrete uniform distribution with pmf aj = 110 j = 1 2 10

Clearly the mean of this distribution is 55 which for comparative purposes is identical to

the ordinary interclaim time mean considered in Example 1 In addition we assume that

the pmf of the individual claim amount distribution is once again given by (42) Table 2

displays values (rounded to 6 decimal places) of Ψnxy(10) for several variants of the Sparre

Andersen model (again with Z = 50 and c = 5) having the above ordinary interclaim time

and claim amount distributions namely

17

(1) ordinary model under scenario (2) from Example 1

(2) ordinary model under scenario (5) from Example 1

(3) stationary model under scenario (2) from Example 1

(4) stationary model under scenario (5) from Example 1

(5) delayed model (with W1 = 1 with probability 1) under scenario (2) from Example 1

(6) delayed model (with W1 = 1 with probability 1) under scenario (5) from Example 1

Concerning the results in Table 2 we first compare the values of Ψnxy(10) in entries (2)

and (5) of Table 1 to their corresponding counterparts in entries (1) and (2) of Table 2 In all

cases we observe that ruin is more likely to occur when interclaim times have the truncated

geometric distribution introduced in our first example as opposed to the discrete uniform

distribution of our second example Since mean interclaim times are equal in both examshy

ples this behaviour must be attributable to the greater variability inherent in the truncated

geometric distribution We also note that under the stationary model assumption the pmf

of W1 is easily verified to be rj = (11 minus j)55 j = 1 2 10 which has mean 4 lt 55

As a result the values of Ψnxy(10) for stationary models (3) and (4) are always larger than

the corresponding values for ordinary models (1) and (2) Finally in the absence of any

knowledge regarding which distribution is in effect for the first interclaim time W1 models

(5) and (6) can be regarded as ldquoworst-case scenariordquo models as they serve to provide upper

bounds on the values of Ψnxy(10)

In conclusion we state that the values in Tables 1 and 2 were generated by first impleshy

menting the recursive procedures (38) through (316) and then summing the appropriate

bivariatetrivariate probabilities computed via (213) (214) or (215) In situations when

x = y = infin however we computed the resulting finite-time ruin probabilities using an altershy

native (albeit standard) method (eg see Cossette et al (2006 Section 3)) Specifically if

we define the survival probability σ(n u) = Pr T gt n | U0 = u = 1minusΨninfininfin(u) as well as

the auxiliary function

0 if w ge Z ηwt =

minlfloorZminuswc minus1 rfloor + 1 t if w lt Z

then by conditioning on the time and amount of the first claim (and recalling that the delayed

process reverts to the ordinary process following this claim) we obtain the expression

minnnr (kminuszk)c2 u+czk+ℓ L L L

σ(u n) = Rn + rk dℓkminuszk αjγ(u + czk + ℓ minus j n minus k)

k=1 ℓ=(kminuszk)c1 j=1

18

where the function γ(w m) for w m isin N is computed recursively via

minmna (kminusηwk)c2 u+cηwk+ℓ L L L

γ(w m) = Am + ak dℓkminusηwk αjγ(u + cηwk + ℓ minus j m minus k)

k=1 j=1 ℓ=(kminusηwk)c1

with starting values γ(w 0) = 1 forall w isin N

Acknowledgements

The authors would like to thank Dr David Landriault for his valuable suggestions which

helped to improve an earlier version of this work This research was supported by the Natural

Sciences and Engineering Research Council of Canada

References

Alfa AS 2004 Markov chain representations of discrete distributions applied to queueing

models Computers amp Operations Research 31 2365-2385

Alfa AS Drekic S 2007 Algorithmic analysis of the Sparre Andersen model in discrete

time ASTIN Bulletin 37 293-317

Bao Z 2007 A note on the compound binomial model with randomized dividend strategy

Applied Mathematics amp Computation 194 276-286

Cossette H Landriault D Marceau E 2006 Ruin probabilities in the discrete time

renewal risk model Insurance Mathematics amp Economics 38 309-323

Karlin S Taylor HM 1975 A First Course in Stochastic Processes 2nd edition Acashy

demic Press New York

Kim B Kim H-S Kim J 2008 A risk model with paying dividends and random

environment Insurance Mathematics amp Economics 42 717-726

Landriault D 2008 Randomized dividends in the compound binomial model with a general

premium rate Scandinavian Actuarial Journal 1-15

Tan J Yang X 2006 The compound binomial model with randomized decisions on

paying dividends Insurance Mathematics amp Economics 39 1-18

19

Table 1 ndash Values of Ψnxy(10) corresponding to Example 1

(a) x = 10

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0005357 0006188 0007163 0008278 0009481

(2) 0005359 0006195 0007175 0008296 0009507

(3) 0005361 0006198 0007181 0008305 0009519

(4) 0005363 0006205 0007193 0008323 0009544

(5) 0005366 0006212 0007208 0008346 0009576

(6) 0005374 0006236 0007249 0008410 0009664

y = 25 (1) 0007631 0008815 0010202 0011789 0013503

(2) 0007634 0008824 0010220 0011816 0013539

(3) 0007636 0008828 0010228 0011829 0013557

(4) 0007639 0008838 0010245 0011854 0013593

(5) 0007643 0008849 0010266 0011887 0013638

(6) 0007656 0008882 0010325 0011978 0013764

y = 50 (1) 0008477 0009792 0011333 0013095 0014998

(2) 0008481 0009802 0011352 0013125 0015039

(3) 0008483 0009807 0011361 0013139 0015059

(4) 0008486 0009817 0011380 0013168 0015098

(5) 0008491 0009830 0011403 0013204 0015149

(6) 0008504 0009866 0011470 0013305 0015288

y = infin (1) 0008777 0010138 0011733 0013558 0015528

(2) 0008781 0010149 0011753 0013588 0015570

(3) 0008783 0010154 0011763 0013603 0015591

(4) 0008787 0010164 0011782 0013633 0015632

(5) 0008791 0010177 0011807 0013671 0015684

(6) 0008805 0010215 0011875 0013775 0015828

20

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0058559 0061875 0065704 0070048 0074714

(2) 0058569 0061903 0065756 0070128 0074825

(3) 0058574 0061917 0065782 0070167 0074880

(4) 0058584 0061945 0065833 0070245 0074987

(5) 0058596 0061979 0065896 0070344 0075124

(6) 0058632 0062079 0066078 0070621 0075506

y = 25 (1) 0089002 0093986 0099739 0106262 0113270

(2) 0089017 0094028 0099817 0106382 0113437

(3) 0089025 0094049 0099856 0106442 0113519

(4) 0089040 0094091 0099932 0106560 0113681

(5) 0089057 0094142 0100028 0106708 0113887

(6) 0089113 0094294 0100302 0107127 0114463

y = 50 (1) 0102786 0108506 0115106 0122590 0130629

(2) 0102804 0108555 0115196 0122728 0130820

(3) 0102812 0108579 0115241 0122797 0130914

(4) 0102830 0108627 0115329 0122932 0131101

(5) 0102850 0108686 0115439 0123103 0131337

(6) 0102914 0108860 0115754 0123584 0132000

y = infin (1) 0108895 0114933 0121901 0129799 0138283

(2) 0108914 0114985 0121996 0129945 0138485

(3) 0108923 0115010 0122043 0130018 0138585

(4) 0108941 0115061 0122136 0130161 0138782

(5) 0108962 0115123 0122252 0130341 0139032

(6) 0109030 0115307 0122585 0130849 0139732

21

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0079856 0086792 009464 0103440 0112831

(2) 0079885 0086869 0094778 0103650 0113118

(3) 0079953 0086978 0094936 0103864 0113393

(4) 0080081 0087188 0095243 0104282 0113929

(5) 0079962 0087075 0095149 0104213 0113890

(6) 0080805 0088366 0096951 0106588 0116875

y = 25 (1) 0124591 0135680 0148207 0162245 0177217

(2) 0124639 0135806 0148433 0162586 0177683

(3) 0124755 0135993 0148703 0162950 0178148

(4) 0124978 0136354 0149226 0163658 0179054

(5) 0124763 0136141 0149035 0163500 0178935

(6) 0126238 0138381 0152148 0167588 0184060

y = 50 (1) 0146757 0160079 0175111 0191945 0209894

(2) 0146816 0160232 0175385 0192360 0210460

(3) 0146965 0160469 0175725 0192817 0211043

(4) 0147249 0160925 0176384 0193706 0212178

(5) 0146968 0160639 0176117 0193470 0211980

(6) 0148860 0163500 0180077 0198658 0218474

y = infin (1) 0158013 0172606 0189059 0207476 0227106

(2) 0158078 0172776 0189362 0207934 0227732

(3) 0158250 0173047 0189751 0208455 0228395

(4) 0158579 0173571 0190503 0209467 0229685

(5) 0158247 0173226 0190172 0209161 0229410

(6) 0160443 0176532 0194735 0215127 0236869

22

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0088667 0104403 0123329 0145277 0169184

(2) 0088711 0104518 0123538 0145598 0169627

(3) 0088733 0104574 0123641 0145756 0169846

(4) 0088777 0104687 0123844 0146066 0170274

(5) 0088829 0104828 0124105 0146470 0170832

(6) 0088988 0105229 0124822 0147560 0172331

y = 25 (1) 0140528 0167839 0201007 0239726 0282102

(2) 0140598 0168021 0201338 0240234 0282804

(3) 0140632 0168111 0201502 0240486 0283151

(4) 0140701 0168289 0201822 0240976 0283827

(5) 0140785 0168515 0202241 0241622 0284718

(6) 0141036 0169148 0203370 0243335 0287071

y = 50 (1) 0167826 0203005 0246173 0296954 0352855

(2) 0167908 0203218 0246561 0297548 0353672

(3) 0167949 0203324 0246753 0297841 0354075

(4) 0168030 0203532 0247128 0298412 0354859

(5) 0168129 0203800 0247622 0299174 0355907

(6) 0168423 0204539 0248936 0301159 0358620

y = infin (1) 0183325 0224926 0276881 0338957 0408257

(2) 0183412 0225150 0277283 0339564 0409079

(3) 0183455 0225260 0277481 0339864 0409485

(4) 0183540 0225478 0277869 0340446 0410270

(5) 0183645 0225760 0278386 0341234 0411339

(6) 0183954 0226529 0279737 0343244 0414037

23

Table 2 ndash Values of Ψnxy(10) corresponding to Example 2

(a) x = 10

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0001901 0002276 0002724 0003240 0003799

(2) 0001904 0002286 0002742 0003269 0003840

(3) 0002651 0003013 0003442 0003936 0004471

(4) 0002655 0003022 0003460 0003964 0004510

(5) 0005792 0006098 0006458 0006871 0007317

(6) 0005795 0006106 0006473 0006894 0007350

y = 25 (1) 0002708 0003243 0003880 0004615 0005411

(2) 0002713 0003257 0003906 0004656 0005469

(3) 0003778 0004293 0004904 0005608 0006369

(4) 0003782 0004306 0004929 0005647 0006424

(5) 0008255 0008690 0009203 0009792 0010427

(6) 0008259 0008702 0009224 0009824 0010473

y = 50 (1) 0003008 0003603 0004310 0005127 0006011

(2) 0003014 0003618 0004340 0005173 0006076

(3) 0004197 0004769 0005448 0006230 0007075

(4) 0004202 0004784 0005476 0006273 0007137

(5) 0009173 0009656 0010226 0010879 0011585

(6) 0009177 0009669 0010249 0010916 0011637

y = infin (1) 0003115 0003730 0004463 0005308 0006224

(2) 0003121 0003746 0004493 0005356 0006291

(3) 0004346 0004938 0005641 0006450 0007325

(4) 0004351 0004954 0005670 0006496 0007389

(5) 0009499 0009999 0010589 0011265 0011996

(6) 0009503 0010012 0010613 0011303 0012050

24

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0031176 0032815 0034733 0036924 0039282

(2) 0031193 0032862 0034821 0037061 0039475

(3) 0050168 0051747 0053587 0055684 0057939

(4) 0050184 0051792 0053671 0055815 0058124

(5) 0136884 0138219 0139763 0141517 0143399

(6) 0136896 0138256 0139832 0141626 0143553

y = 25 (1) 0047622 0050092 0052981 0056280 0059829

(2) 0004765 0050163 0053113 0056487 0060121

(3) 0076570 0078950 0081720 0084878 0088272

(4) 0076593 0079017 0081847 0085076 0088551

(5) 0205774 0207786 0210111 0212752 0215585

(6) 0205793 0207842 0210216 0212917 0215818

y = 50 (1) 0055170 0058009 0061329 0065119 0069196

(2) 0055200 0058091 0061481 0065357 0069533

(3) 0088650 0091385 0094569 0098197 0102096

(4) 0088677 0091463 0094715 0098425 0102417

(5) 0235815 0238128 0240800 0243834 0247089

(6) 0235837 0238192 0240920 0244023 0247356

y = infin (1) 0058565 0061565 0065073 0069077 0073384

(2) 0058597 0061652 0065234 0069329 0073740

(3) 0094061 0096952 0100316 0104149 0108267

(4) 0094090 0097035 0100470 0104390 0108607

(5) 0248472 0250917 0253740 0256945 0260383

(6) 0248495 0250985 0253867 0257145 0260667

25

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0048598 0052524 0056984 0061998 0067342

(2) 0048654 0052673 0057253 0062411 0067916

(3) 0070192 0073979 0078257 0083057 0088167

(4) 0070243 0074119 0078513 0083451 0088714

(5) 0151774 0154981 0158572 0162586 0166852

(6) 0151815 0155096 0158784 0162913 0167308

y = 25 (1) 0076895 0083230 0090413 0098481 0107073

(2) 0076987 0083474 0090854 0099156 0108010

(3) 0110093 0116205 0123095 0130818 0139033

(4) 0110177 0116435 0123515 0131462 0139928

(5) 0230674 0235848 0241632 0248091 0254950

(6) 0230739 0236038 0241980 0248626 0255695

y = 50 (1) 0091496 0099156 0107829 0117563 0127924

(2) 0091610 0099455 0108370 0118387 0129068

(3) 0130130 0137520 0145839 0155156 0165064

(4) 0130233 0137801 0146352 0155942 0166156

(5) 0266599 0272857 0279840 0287632 0295904

(6) 0266680 0273088 0280265 0288286 0296814

y = infin (1) 0099312 0107746 0117285 0127984 0139369

(2) 0099439 0108078 0117883 0128899 0140639

(3) 0140458 0148595 0157745 0167986 0178872

(4) 0140572 0148908 0158314 0168859 0180084

(5) 0282886 0289776 0297457 0306022 0315111

(6) 0282976 0290034 0297929 0306748 0316120

26

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0059096 0072505 0088907 0108122 0129137

(2) 0059223 0072829 0089500 0109052 0130456

(3) 0079757 0092662 0108388 0126778 0146871

(4) 0079875 0092970 0108956 0127669 0148134

(5) 0159788 0170682 0183871 0199249 0216024

(6) 0159881 0170936 0184343 0199990 0217075

y = 25 (1) 0095883 0119741 0149240 0184073 0222390

(2) 0096087 0120262 0150197 0185573 0224520

(3) 0127403 0150359 0178640 0211977 0248613

(4) 0127592 0150855 0179557 0213414 0250653

(5) 0245161 0264533 0288250 0316124 0346709

(6) 0245311 0264942 0289010 0317319 0348407

y = 50 (1) 0116598 0147920 0187100 0233784 0285503

(2) 0116839 0148537 0188231 0235554 0288015

(3) 0153023 0183153 0220713 0265390 0314840

(4) 0153249 0183742 0221798 0267087 0317248

(5) 0285742 0311159 0342651 0380005 0421287

(6) 0285921 0311644 0343551 0381418 0423292

y = infin (1) 0129470 0167217 0215383 0273841 0339734

(2) 0129725 0167865 0216560 0275661 0342282

(3) 0167978 0204275 0250442 0306382 0369383

(4) 0168217 0204895 0251572 0308128 0371826

(5) 0305868 0336470 0375169 0421935 0474525

(6) 0306058 0336980 0376106 0423388 0476560

27

Page 5: uwaterloo.ca · Ruin Analysis of a Threshold Strategy in a Discrete-time Sparre Andersen Model Steve DREKIC and Ana Maria MERA Department of Statistics and Actuarial Science University

characterization as the ldquoelapsed timerdquo phase-type representation of the pmf aj We remark

that the above pmf formula also holds true for j gt na as it is readily verified that Sjminus1 = Ona

an na times na matrix of zeros for such values of j

In what follows we proceed to set up our delayed Sparre Andersen model as a bivariate

Markov chain conditional on W1 = k isin 1 2 nr Specifically assuming that W1 = k we

construct the bivariate stochastic process (Ut Lt) t = k k + 1 such that Ut represents

the amount of surplus at time t and Lt denotes a specific counter at time t which measures the

ldquoelapsed timerdquo (in the sense described above) length of the next claim occurrence We refer to

the Ut component as the level of the process and the Lt component as the phase of the process

Two important observations are particularly noteworthy regarding this construction First of

all Lk = 1 since we are assuming that the very first claim occurs at time kminus More generally

Lt = 1 if a claim happens to occur at time tminus Secondly the delayed process reverts to the

ordinary process (having interclaim time distribution defined by the pmf aj) upon occurrence

of the first claim As a result the bivariate stochastic process (Ut Lt) t = k k + 1

possesses the following Markovian relationship

(Ut + pt Lt + 1) if there is no claim at time (t + 1)minus (Ut+1 Lt+1) =

(Ut + pt minus Y 1) if there is a claim of amount Y at time (t + 1)minus

Note that the state space for this Markov chain is given by Δ = Z times 1 2 na Furthershy

more the probability transition matrix associated with this Markov chain is given by

middot middot middot minus1 10 middot middot middot Zminus2 Zminus1 Z Z+1 Z+2 middot middot middot

middot middot middot

middot middot middot minus1 middot middot middot Bc Bcminus1 Bcminus2 middot middot middot BcminusZ+1 BcminusZ BcminusZminus1 BcminusZminus2 BcminusZminus3 middot middot middot

0

1

middot middot middot Bc+1 Bc Bcminus1 middot middot middot BcminusZ+2 BcminusZ+1 BcminusZ BcminusZminus1 BcminusZminus2 middot middot middot middot middot middot Bc+2 Bc+1 Bc middot middot middot BcminusZ+3 BcminusZ+2 BcminusZ+1 BcminusZ BcminusZminus1 middot middot middot

middot middot middot

middot middot middot

P = Zminus2 middot middot middot BZ+cminus1 BZ+cminus2 BZ+cminus3 middot middot middot Bc Bcminus1 Bcminus2 Bcminus3 Bcminus4 middot middot middot (21)

Zminus1 middot middot middot BZ+c BZ+cminus1 BZ+cminus2 middot middot middot Bc+1 Bc Bcminus1 Bcminus2 Bcminus3 middot middot middot

Z middot middot middot AZ+1 AZ AZminus1 middot middot middot A2 A1 A0 Aminus1 Aminus2 middot middot middot Z+1 middot middot middot AZ+2 AZ+1 AZ middot middot middot A3 A2 A1 A0 Aminus1 middot middot middot

Z+2 middot middot middot AZ+3 AZ+2 AZ+1 middot middot middot A4 A3 A2 A1 A0 middot middot middot

middot middot middot

middot middot middot

where Ona

if i isin Zminus

Bi = S if i = 0 (22) se1αi if i isin Z+

5

and c2 L

Ai = djBi+j (23) j=c1

We give a brief indication of the reasoning leading to the form of P in (21) For example

consider the transition ldquoUt = 0 rarr Ut+1 = 1rdquo which has block element Bcminus1 containing the

transition probabilities governing the ldquoLt rarr Lt+1 rdquo process At a surplus level of 0 at time

t a premium of pt = c is certain to be received Therefore to arrive at surplus level 1 at

the next time unit a ldquoso-calledrdquo claim of size c minus 1 must occur to neutralize the premium

just received Thus the size of the ldquoclaimrdquo required for this transition to occur determines

the value of i in Bi so that Bcminus1 is obtained and its form subsequently determined by (22)

according to the value of the index c minus 1 Note that only when c gt 1 does an actual claim

occur in this particular scenario We mention that for transitions which begin at or above the

threshold level Z a similar line of reasoning would also apply except now one must condition

on the amount of the random premium being earned as indicated by the form of (23)

Note that this formulation gives rise to a doubly-infinite Markov chain with finite blocks

of size na In the special case of the threshold-free model (obtained by setting c1 = c2 = c

so that dc = 1) we remark that (21) is in agreement with the matrix P in Alfa and Drekic

(2007 p 296) Also since αi = 0 forall i gt mα we note that Bi = Ona if i gt mα

For the computation of ruin-related quantities of interest it is useful to partition the

state space Δ into two mutually exclusive state spaces namely Δ1 = N times 1 2 na

and Δ2 = Zminus times 1 2 na In addition we introduce (one-step) transition matrices

C Δ1 rarr Δ1 and D Δ1 rarr Δ2 which correspond to mapping ldquonon-ruinedrdquo states in Δ1 to

either ldquonon-ruinedrdquo states in Δ1 or ldquoruinedrdquo states in Δ2 respectively In other words the

matrices C and D are essentially submatrices of P and their forms are readily given by

210 middot middot middot Z minus 1 Z Z + 1 middot middot middot

Bc Bcminus1 Bcminus2 middot middot middot BcminusZ+1 BcminusZ BcminusZminus1 middot middot middot Bc+1 Bc Bcminus1 middot middot middot BcminusZ+2 BcminusZ+1 BcminusZ middot middot middot

Bc+2 Bc+1 Bc middot middot middot BcminusZ+3 BcminusZ+2 BcminusZ+1 middot middot middot

middot middot middot

middot middot middot

0

1

2 Z minus 2 BZ+cminus2 BZ+cminus3 BZ+cminus4 middot middot middot Bcminus1 Bcminus2 Bcminus3 middot middot middot (24) C =

Z minus 1 BZ+cminus1 BZ+cminus2 BZ+cminus3 middot middot middot Bc Bcminus1 Bcminus2 middot middot middot

Z AZ AZminus1 AZminus2 middot middot middot A1 A0 Aminus1 middot middot middot

Z + 1 AZ+1 AZ AZminus1 middot middot middot A2 A1 A0 middot middot middot

Z + 2 AZ+2 AZ+1 AZ middot middot middot A3 A2 A1 middot middot middot

middot middot middot

6

and

minus1 minus2 minus3 middot middot middot

0 Bc+1 Bc+2 Bc+3 middot middot middot 1 Bc+2 Bc+3 Bc+4 middot middot middot

2 Bc+3 Bc+4 Bc+5 middot middot middot middot middot middot

Z minus 2 BZ+cminus1 BZ+c BZ+c+1 middot middot middot D = (25) Z minus 1 BZ+c BZ+c+1 BZ+c+2 middot middot middot

Z AZ+1 AZ+2 AZ+3 middot middot middot

Z + 1 AZ+2 AZ+3 AZ+4 middot middot middot Z + 2 AZ+3 AZ+4 AZ+5 middot middot middot

Let us now specify some further definitions required in upcoming derivations First of all

we introduce the function

0 if u ge Z zt =

maxi isin 1 2 t|u + c(i minus 1) lt Z if u lt Z

or equivalently

0 if u ge Z zt =

minlfloorZminuscuminus1 rfloor + 1 t if u lt Z

where the floor function denoted lfloorxrfloor yields the largest integer less than or equal to x We

define dm0 = δm0 where δij in general denotes the Kronecker delta function of i and j

Also define

dm if m = c1 c1 + 1 c2dm1 =

0 elsewhere

and for n = 2 3

Lc2 dj1dmminusjnminus1 if m = nc1 nc1 + 1 nc2dmn = j=c1

0 elsewhere

In other words dmn represents the pmf of the n-fold convolution of dm with itself

Recalling that W1 = k note that zk + ( k+) determines the earliest time point in the set =

0+ 1+ (k minus1)+ at which a random premium is received Note also that zk = k implies

that no such point exists in this set (ie the threshold level Z has not yet been reached when

the premium at time (k minus1)+ is to be received) Therefore if we assume for the moment that

zk k we may condition on the total premiums earned at times zk + (zk + 1)+ (k minus 1)+=

7

to obtain the following expression for b(k) the initial (ie starting at time k) probability row

vector of the above Markov chain corresponding only to the states in Δ1 namely

(kminuszk)c2

b(k) L

= dmkminuszk(αu+czk+me1 αu+czk+mminus1e1 α2e1 α1e1 0 0 ) (26)

m=(kminuszk)c1

where 0 denotes the 1 times na row vector of zeros By further defining the function vn(t) =

b(k)u + ct + (n minus t)c2 we note that the i-th level of is given by the 1 times na row vector L(kminuszk)c2 b(k)αu+czk+mminusie1 for each i isin Ωk = 0 1 vk(zk)minus1 Hence contains

m=(kminuszk)c1 dmkminuszk

zeros from level vk(zk) onwards Moreover (26) holds true even in the case when zk = k

simplifying to give b(k) = (αu+cke1 αu+ckminus1e1 α2e1 α1e1 0 0 )

At this juncture we introduce two additional row vectors namely

(k) (k) (k) (k) b(k)gn = (gn0 gn1 gn2 ) = Cn n isin N (27)

and

h(k) (k) (k) (k) b(k)Cnminus1D =

(k) Z

+ n = (h nminus2 hnminus3 ) = gnminus1D n isin (28) nminus1 h

(k)We remark that gn contains the probabilities of being in the various ldquonon-ruinedrdquo states

at time k + n (ie after claims have been paid out) without having visited a ldquoruinedrdquo state

during the previous n minus 1 transitions given that Uk isin Ωk sub Δ1 In a similar fashion h(nk)

contains the probabilities of being in the various ldquoruinedrdquo states for the first time at time

k + n given that Uk isin Ωk sub Δ1 Note that the probability of being in ruined state ldquo minus jrdquo is (k) (k) (k) (k)

characterized by the row vector h = (h h h ) with the third subscript nminusj nminusj1 nminusj2 nminusjna

representing the value of the elapsed time counter Lk+n Upon further reflection however (k)

=it must be the case that h = 0 for i 1 as ruin can only occur at claim instants nminusji

which in this case implies Lk+n = 1 with probability 1 Thus it immediately follows that (k) (k) (k)

the structure of the 1 times na row vector hnminusj is simply given by hnminusj = (φnj(u) 0 0) (k)

where φ (u) = Pr T = k + n |UT | = j | Consequently as in the analysis with nj Uk isin Ωk

no threshold on the surplus level (see Alfa and Drekic (2007 p 299)) we obtain

(k) (k) prime φ (u) = h (29) nj nminusje1

where e1 prime denotes the transpose of e1

A similar line of probabilistic logic can also be applied to obtain a representation for the (k)

trivariate function ψ (u) = Pr T = k + n UTminus = i |UT | = j | Uk isin Ωk In order for ruin nij

to occur at time k + n with a surplus immediately prior to ruin equal to i we observe that

(i) none of the previous n minus 1 transitions must have included a visit to any state in Δ2

and

8

(ii) the surplus level at time k + n minus 1 must necessarily be equal to i minus pk+nminus1

Note that pk+nminus1 denotes the premium corresponding to the time interval (k + n minus 1 k + n]

which is received at (k + n minus 1)+ Hence the quantity corresponding to points (i) and (ii) (k)

above is the 1 times na row vector g Then at the next time unit a claim must nminus1iminuspk+nminus1

necessarily occur but not before a premium of pk+nminus1 is first collected thereby raising the

surplus level to i Since s contains the absorption probabilities (to claim occurrence) from

the na possible phase states and the claim causing ruin must be of size i + j in order to

ensure that the deficit at ruin is equal to j it follows that

(k) (k)ψ (u) = g s αi+j (210) nij nminus1iminuspk+nminus1

(k)The following corollary provides the means of calculating the quantity g in nminus1iminuspk+nminus1

(210)

Corollary 1 If minc Z + c1 = c then

(k)

g if i = c c+1 Z +c1 minus1 nminus1iminusc

(k) (k) Lminc2iminusZ (k)g = g + j=c1

if i = Z +c1 Z +c1 +1 minmα Z +cminus1nminus1iminuspk+nminus1 nminus1iminusc djgnminus1iminusj

Lc2 (k)

j=c1 djgnminus1iminusj if i = Z +c Z +c+1 mαminus1

(211)

Conversely if minc Z + c1 = Z + c1 then

Lminc2iminusZ (k)

djg if i = Z +c1 Z +c1 +1 cminus1nminus1iminusj j=c1

(k) (k) Lminc2iminusZ (k)gnminus1iminuspk+nminus1

= gnminus1iminusc + j=c1 djgnminus1iminusj if i = c c+1 minmα Z +cminus1

Lc2 (k)djg if i = Z +c Z +c+1 mαminus1j=c1 nminus1iminusj

(212)

Proof The justification of (211) is as follows If c is smaller than Z + c1 the only way of

reaching a surplus level of i isin c c+1 Z +c1 minus1 (prior to ruin) is by receiving a premium

of c since i minus j for any j = c1 c1 +1 c2 would result in a surplus level below Z where the

random premium rate is not applicable Next for fixed i isin Z +c1 Z +c1 +1 Z +c minus1

there are two distinct ways of reaching this surplus level First of all note that the value of

i minus j is guaranteed to produce a surplus amount greater than or equal to Z provided that

j isin c1 c1 + 1 minc2 i minus Z Secondly note that reducing each of value of i by c also

yields a non-negative surplus value less than Z Hence both pure and random premiums

are plausible under this scenario For i isin Z + c Z + c + 1 note that decreasing this

amount by any j = c1 c1 + 1 c2 results in a surplus value greater than or equal to Z as

desired However reducing it by the amount c also has the same impact thus rendering the

pure premium option inadmissible in this case Lastly we incorporate into the range of i the

9

fact that the surplus immediately prior to ruin cannot exceed mα minus1 as indicated in Section

1 The justification of (212) when c is larger than Z + c1 mirrors the analysis above and is

left to the reader D

Recall that in the analysis up to this point we have assumed that W1 = k isin

1 2 3 nr If we now condition on the value of W1 the Law of Total Probability

yields the following two expressions for φnj(u) and ψnij(u) namely

nr L

= rkφ (u) n = nr + 1 nr + 2 φnj(u) n

(kminus) kj

k=1

and nr L

(k)ψnij(u) = rkψ (u) n = nr + 1 nr + 2 nminuskij

k=1

In addition however it is quite possible that T = n for n = 1 2 nr and this can happen

as a result of either

(i) the first claim which causes ruin occurring at time nminus from a (conditional) surplus Lnminus1level of u + czn + m (where m) or i=zn

Xi =

(ii) the first claim which does not cause ruin occurring at some time kminus k isin 1 2 nminus

1 and ruin subsequently occurring n minus k time units later

If we now combine these outcomes with those for n = nr + 1 nr + 2 (and recall that

rn = 0 forall n gt nr) the general expression for φnj(u) becomes

minnminus1nr (nminuszn)c2 L

(k) L

φnj(u)= rkφ (u) + rn dmnminuszn αu+czn+m+j n isin Z+ (213) nminuskj

k=1 m=(nminuszn)c1

Concerning the function ψnij(u) when n le nr and point (i) above is considered we note Lnminus1that i must necessarily equal u + czn + m (again conditional on i=zn

Xi = m) for ruin to

occur in this manner Hence in a similar fashion we acquire

minnminus1nr (nminuszn)c2 L L

ψnij(u) = rkψn

(kminus) kij(u) + rn dmnminuszn

δiu+czn+mαu+czn+m+j

k=1 m=(nminuszn)c1

minnminus1nr L

(k) Z

+ = rkψnminuskij(u) + rndiminusuminuscznnminuszn αi+j n isin (214)

k=1

Finally for fixed i isin minc Z + c1 minc Z + c1 + 1 mα minus 1 summing (214) from

j = 1 to mα minus i yields the following expression for the bivariate pmf ωni(u) namely

minnminus1nr mαminusi mαminusi L L

(k) L

ωni(u) = rk ψnminuskij(u) + rndiminusuminuscznnminusznαi+j

k=1 j=1 j=1

10

minnminus1nr L

(k) Z

+ = Λi rkg s + rndiminusuminuscznnminuszn n isin (215) nminuskminus1iminuspnminus1

k=1

where (210) was used to establish the last equality In the special case when c1 = c2 = c it

is a straightforward exercise to show that equations (213) through (215) simplify to produce

equations (215) through (217) of Alfa and Drekic (2007 p 300)

3 Computational considerations

In order to compute the joint pmfrsquos defined by (213) (214) and (215) it is clear from (k)

equations (29) through (212) that we need to be able to compute the quantities gn and

h(k) n defined by (27) and (28) respectively In what follows we capitalize on the underlying

Markov chain structure in our model to develop a set of computational procedures which

ultimately enable one to calculate these key row vectors

First of all recall that g0(k)

= b(k) contains zeros from a certain level onwards namely (k)

level vk(zk) = u + czk + c2(k minus zk) In general one can easily deduce that gn will also

contain zeros from a certain level onwards to be denoted by lk(n) We begin this section

with a determination of lk(n) which varies in form depending on the value of vk(zk) minus 1

representing the maximum surplus level at time k after the smallest possible claim of size 1

has been applied (at time kminus)

On one hand it may be the case that the value of vk(zk)minus1 lies at or above the threshold

level Z In this situation we remark that zk can take on any value in the set 0 1 k (k)

Since g0 contains zeros from level lk(0) = vk(zk) onwards we observe that at time k + 1

it may be possible to achieve a higher surplus value than vk+1(zk) minus 1 representing the

maximum surplus level assuming there was no drop below Z due to the claim at time kminus

Specifically if the surplus had fallen to level Z minus 1 at time kminus then gained c at time k+ a

bounded level of Z + c minus1 would have been reached at time k +1 and this level is potentially

larger than vk+1(zk) minus 1 As a result we have that lk(1) = maxvk+1(zk) Z + c If we

extend this argument inductively we readily obtain lk(n) = maxvk+n(zk) Z +c +c2(n minus1)

n isin Z+ However since vk+n(zk) = u + czk + c2(k +n minus zk) = vk(zk)+ c2 + c2(n minus 1) we may

alternatively re-express lk(n) as

lk(n) = maxvk+n(zk) Z + c + c2(n minus 1) = maxvk(zk) + c2 Z + c + c2(n minus 1)

Thus

Z+lk(n) = maxlk(0) + c2 Z + c + c2(n minus 1) n isin (31)

11

or equivalently

Z+lk(n) = lk(1) + c2(n minus 1) n isin

Clearly lk(0) gt Z and lk(n) ge Z + c for n isin Z+

The other remaining case to consider involves the situation when the maximum surplus

level at time k is strictly below the threshold level Z Under this scenario we further

distinguish between two cases involving the value of c2 First of all suppose that c2 = 0 so

that vk(zk)minus1 = u +czk minus1 lt Z or equivalently u +czk le Z If u +czk happens to equal Z

then it is possible for zk isin 0 1 k With c2 = 0 however no further premiums can be

earned until after the first claim takes place (which is assumed to occur at time kminus) thereby

implying that Z = u + ci for all i = zk zk + 1 k Hence we know that u + ck = Z

in this case Moreover if u + czk happens to be strictly less than Z then we clearly have

that zk = k Even in the case when c2 gt 0 one can readily verify that zk = k so that

vk(zk) minus 1 = u + ck minus 1 lt Z Thus we proceed with the analysis assuming that zk = k

c2 ge 0 and the inequality u + ck le Z holds true

lowast k

Define the ceiling function denoted lceilxrceil to yield the smallest integer greater than or

equal to x Because a random premium cannot be received until level Z is at least achieved

it is essential to find the point at which the threshold level Z is reached (ie the point at Zminus(u+ck)+1 which u + ck minus1+ ci ge Z) Isolating this inequality for i one obtains i ge

c Hence

assuming that W1 = k the elapsed time after k at which a random premium would begin to

lceilZminus(u+ck)+1 be received is t lowast = rceil Note that t lowast isin Z+ k c k

With zk = k g0(k)

contains zeros from level u + ck onwards so that lk(0) = u + ck

Immediately following this a premium of c is guaranteed to be received at time k+ as the

threshold level Z has not yet been reached Consequently g1(k)

contains zeros from level (k)

b(k)u + c(k + 1) onwards which is c levels further than that in g0 = Continuing this (k)

process inductively we can readily establish that gn contains zeros from level u + c(k + n)

onwards for n = 1 2 t lowast k so that lk(n) = u + c(k + n) for n = 0 1 t lowast k

In the case of n = t lowast + 1 t lowast + 2 gt

(k) is first looked at in isolation At time k +k k +1

t lowast k + 1 the threshold level has been crossed giving rise to a maximum surplus value of either (k)

c(k + t lowast k contains zeros from level lk(t lowast ku + ) + c2 minus 1 or Z + c minus 1 Therefore g + 1) = t lowast k+1

+1(k + t lowast kmaxvk+t lowast k

) Z + c onwards Carrying out the same inductive process as before it is

readily seen that g(nk)

contains zeros from level lk(n) = maxvk+n(k+t lowast k) Z +c+c2(nminust lowast kminus1)

onwards for n = t lowast k + 1 t lowast k + 2 Combining both cases we obtain

u + c(k + n) if n = 0 1 t lowast lk(n) = k (32)

maxlk(t lowast) + c2 Z + c + c2(n minus t lowast minus 1) if n = t lowast + 1 t lowast + 2 k k k k

Note that by construction lk(t lowast k) gt Z and lk(n) le Z for n = 0 1 t lowast k minus1 Interestingly

12

if we extend the definition tk lowast by introducing the quantity ζk

lowast = maxtklowast 0 we remark that

(31) and (32) can actually be combined into a single formula for lk(n) namely

vk+n(zk + n) if n = 0 1 ζklowast

lk(n) = (33) maxlk(ζ

lowast) + c2 Z + c + c2(n minus ζlowast minus 1) if n = ζlowast + 1 ζlowast + 2 k k k k

Clearly we have that lk(n) ge Z + c for n = ζk lowast +1 ζk

lowast +2 Moreover if c1 = c2 = c (33)

reduces to give lk(n) = u + c(k + n) forall n isin N which is in agreement with the results in Alfa

and Drekic (2007 Section 3)

Remark In our above analysis we purposely ignored the possibility of any future claims

occurring after time k in our determination of lk(n) However we recognize that this is not

realistic and future claims would indeed occur after time k thereby further reducing the value

of the maximum surplus level lk(n) minus 1 at time k + n To account for this we imagine all

future interclaim times (ie after time k) to be of maximum length na and all corresponding

claim sizes to be of minimum amount 1 Consequently in the case when c2 gt 0 we should

subtract from formula (33) the amount of lfloornnarfloor where lfloormiddotrfloor again denotes the floor function

Obviously when c2 = 0 (ie no premiums are collected whenever the surplus level is at or

above Z) lk(n) = Z + c for n = ζk lowast + 1 ζk

lowast + 2 (k)

We now turn our attention to the development of a general algorithm for computing gni

(k) (k)for i = 0 1 lk(n) minus 1 With the aid of (24) we apply the key identity gn = gnminus1C

(which can be inferred from (27)) to first obtain

minZlk(nminus1)minus1 lk(nminus1)minus1 c2 L L L

(k) (k) (k)gni = g Bjminusi+c + gnminus1j dmBjminusi+m n isin Z+ (34) nminus1j

j=0 j=Z m=c1

However (22) states that Bi = Ona for i isin Zminus and so (34) becomes (for the same range of

i)

minZlk(nminus1)minus1 c2 lk(nminus1)minus1 L L L

(k) (k) (k)gni = g Bjminusi+c + dm gnminus1jBjminusi+m n isin Z+ (35) nminus1j

j=max0iminusc m=c1 j=maxZiminusm

Taking the form of (33) into consideration (35) yields the following two formulae

lk(nminus1)minus1 L

(k) (k)gni = gnminus1jBjminusi+c n = 1 2 ζ k

lowast (36) j=max0iminusc

and

Zminus1 c2 lk(nminus1)minus1 L L L

(k) (k) (k)g = g + dm g n = ζ lowast + 1 ζ lowast + 2 ni nminus1jBjminusi+c nminus1jBjminusi+m k k

j=max0iminusc m=c1 j=maxZiminusm

(37)

13

For n = 1 2 ζk lowast where ζk

lowast isin Z+ we note that lk(n) ge c + 1 Therefore with the aid of

(22) (36) simplifies to give

Llk(nminus1)minus1 (k) (k) j=0 gnminus1jBjminusi+c if i = 0 1 c minus 1

gni = Llk(nminus1)minus1 (k)

g if i = c c + 1 lk(n) minus 1j=iminusc nminus1jBjminusi+c

Llk(nminus1)minus1 (k)

nminus1js 0 0 if i = 0 1 c minus 1j=0 αjminusi+cg= (38)

(k) Llk(nminus1)minus1 (k) g S+ αjminusi+cgnminus1js 0 0 if i = c c+1 lk(n)minus1 nminus1iminusc j=iminusc+1

In the case of n = ζk lowast + 1 ζk

lowast + 2 we first reiterate that lk(n) ge Z + c We next break

(37) into several cases based on the value of i and the value of minc Z + c1 as follows

Case 1 Suppose minc Z + c1 = c

For i isin 0 1 c minus 1

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+mni nminus1j nminus1j

j=0 m=c1 j=Z

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k)= αjminusi+cg s + dm αjminusi+mg s 0 0 (39) nminus1j nminus1j

j=0 m=c1 j=Z

For i isin c c + 1 Z + c1 minus 1

Zminus1 c2 lk(nminus1)minus1 L L L

(k) (k) (k)g = g + dm gni nminus1j Bjminusi+c nminus1j Bjminusi+m

j=iminusc m=c1 j=Z

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k) (k)= g S + αjminusi+cg dm αjminusi+mgnminus1j s 0 0 (310) nminus1iminusc nminus1j s +

j=iminusc+1 m=c1 j=Z

For i isin Z + c1 Z + c1 + 1 Z + c minus 1

Zminus1 minc2iminusZ lk(nminus1)minus1 c2 lk(nminus1)minus1 L L L L L

(k) (k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+m + dm g Bjminusi+mni nminus1j nminus1j nminus1j

j=iminusc m=c1 j=iminusm m=iminusZ+1 j=Z

minc2iminusZ L

(k) (k)= g S + dmg S nminus1iminusc nminus1iminusm

m=c1

Zminus1 minc2iminusZ lk(nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k)+ αjminusi+cg dm αjminusi+mg dm αjminusi+mgnminus1j s 0 0 nminus1j s + nminus1js +

j=iminusc+1 m=c1 j=iminusm+1 m=iminusZ+1 j=Z

(311)

For i isin Z + c Z + c + 1 lk(n) minus 1

c2 lk(nminus1)minus1 L L

(k) (k)g = dm g Bjminusi+mni nminus1j

m=c1 j=iminusm

c2 c2 lk (nminus1)minus1 L L L

(k) (k)= dmg S + dm αjminusi+mgnminus1j s 0 0 (312) nminus1iminusm

m=c1 m=c1 j=iminusm+1

14

Case 2 Suppose minc Z + c1 = Z + c1

For i isin 0 1 Z + c1 minus 1

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+mni nminus1j nminus1j

j=0 m=c1 j=Z

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k)= αjminusi+cg s + dm αjminusi+mg s 0 0 (313) nminus1j nminus1j

j=0 m=c1 j=Z

For i isin Z + c1 Z + c1 + 1 c minus 1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+m + dm g Bjminusi+mni nminus1j nminus1j nminus1j

j=0 m=c1 j=iminusm m=iminusZ+1 j=Z

minc2iminusZ L

(k)= dmg S nminus1iminusm

m=c1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k)+ αjminusi+cg s + dm αjminusi+mg s + dm αjminusi+mg s 0 0 nminus1j nminus1j nminus1j

j=0 m=c1 j=iminusm+1 m=iminusZ+1 j=Z

(314)

For i isin c c + 1 Z + c minus 1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k) (k)g = g + g + gni nminus1j Bjminusi+c dm nminus1j Bjminusi+m dm nminus1j Bjminusi+m

j=iminusc m=c1 j=iminusm m=iminusZ+1 j=Z

minc2iminusZ L

(k) (k)= g S + dmg S nminus1iminusc nminus1iminusm

m=c1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k)+

nminus1j s 0 0 αjminusi+cgnminus1j s + dm αjminusi+mgnminus1j s + dm αjminusi+mg j=iminusc+1 m=c1 j=iminusm+1 m=iminusZ+1 j=Z

(315)

For i isin Z + c Z + c + 1 lk(n) minus 1

c2 lk (nminus1)minus1 L L

(k) (k)g = dm g Bjminusi+mni nminus1j

m=c1 j=iminusm

c2 c2 lk (nminus1)minus1 L L L

(k) (k)= dmg S + dm αjminusi+mgnminus1j s 0 0 (316) nminus1iminusm

m=c1 m=c1 j=iminusm+1

It is not difficult to verify that equations (38) through (316) simplify to produce equations

(32) and (33) of Alfa and Drekic (2007 pp 301-302) when c1 = c2 = c With equations (38) (k)

through (316) providing the recursive means of calculating gn we next turn our attention

15

to a determination of h(k) defined by (28) For n = 1 2 ζlowast we use (22) and (25) to n k

obtain

lk(nminus1)minus1 lk(nminus1)minus1

h(k) nminusj =

L g

(k) nminus1ℓBc+j+ℓ =

L αc+j+ℓg

(k) nminus1ℓs 0 0 (317)

ℓ=0 ℓ=0

For n = ζk lowast + 1 ζk

lowast + 2 we similarly acquire

Zminus1 lk(nminus1)minus1 L L

(k) (k) (k)hnminusj = gnminus1ℓBc+j+ℓ + gnminus1ℓAj+ℓ

ℓ=0 ℓ=Z

Zminus1 c2 lk(nminus1)minus1 L L L

(k) (k)= αc+j+ℓg s + dm αm+j+ℓg s 0 0 (318) nminus1ℓ nminus1ℓ

ℓ=0 m=c1 ℓ=Z

Note that substituting (317) and (318) into (29) immediately yields

Llk(nminus1)minus1 (k) (k) ℓ=0 αc+j+ℓgnminus1ℓs if n = 1 2 ζk

lowast φ (u) = nj LZminus1 (k) Lc2

Llk(nminus1)minus1 (k) ζlowast s + dm αm+j+ℓg n = + 1 ζlowast + 2 ℓ=0 αc+j+ℓgnminus1ℓ m=c1 ℓ=Z nminus1ℓs if k k

(319)

Clearly in the special case when c1 = c2 = c (319) yields equation (35) of Alfa and Drekic

(2007 p 302) Finally we remark that (319) can be substituted into (213) to obtain a

computational formula for φnj(u)

4 Numerical examples and findings

In this section we illustrate the application of our proposed algorithm with several numerical

examples All calculations in this section were performed using Mathematica (Version 7) on

a Dell Precision T7400 workstation with a 3 Ghz CPU and 4 GB of RAM

In our first illustration (labelled Example 1) we consider the ordinary Sparre Andersen

model in which interclaim times have pmf given by

(211)(911)jminus1 if j = 1 2 na minus 1 aj = (41)

(911)naminus1 if j = na

Note that the pmf (41) is that of a truncated geometric distribution with all the probability

mass on na na +1 assigned to the support value na Clearly as na becomes larger the

closer (41) approximates this particular geometric distribution having mean 55 In what Lna

Lminus1 nafollows we fix na = 25 so that aj = 0992 and jaj = 546 Let the individual j=1 j=1

claim amount distribution be given by the pmf

minus4 minus4j minus 1 j

αj = 1 + minus 1 + j isin Z+ (42) 30 30

16

We remark that the pmf (42) represents a discretized version of the Pareto distribution with

mean 10 Also note that mα = infin thereby implying that the random variables |UT | and

UTminus both have infinite support

Table 1 displays values (rounded to 6 decimal places) of the trivariate cumulative distrishy

bution function

n x y L L L

Ψnxy(u) = Pr T le n UTminus le x |UT | le y | U0 = u = ψℓij(u) ℓ=1 i=mincZ+c1 j=1

corresponding to the above discrete-time risk process with u = 10 Z = 50 c = 5 and

varying premium amount distributions according to the following set of scenarios

(1) d2 = 1

(2) d1 = 25 d2 = 15 d3 = 25

G

5)

(3) di = i

(25)i(35)5minusi i = 0 1 5

(4) di = 15 i = 0 1 4

(5) d1 = 34 d5 = 14

(6) d0 = 35 d5 = 25

We remark that while these six premium amount distributions all have mean equal to 2

they possess different shapes and we investigate the impact of their variability on the joint

distribution of T |UT | and UTminus In particular the variances corresponding to the premium

amount distributions of scenarios (1) through (6) are 0 08 12 2 3 and 6 respectively

Based on the results in Table 1 it would appear that greater variability in the premium

amount distribution generally leads to higher ruin probabilities Perhaps somewhat surprising

though the differences are subtle and not particularly striking as one compares the various

ruin probabilities from scenarios (1) through (6) Obviously we observe more of a difference

between scenarios as the values of n x and y increase

As a follow-up illustration (labelled Example 2) we consider a model in which ordinary

interclaim times have a discrete uniform distribution with pmf aj = 110 j = 1 2 10

Clearly the mean of this distribution is 55 which for comparative purposes is identical to

the ordinary interclaim time mean considered in Example 1 In addition we assume that

the pmf of the individual claim amount distribution is once again given by (42) Table 2

displays values (rounded to 6 decimal places) of Ψnxy(10) for several variants of the Sparre

Andersen model (again with Z = 50 and c = 5) having the above ordinary interclaim time

and claim amount distributions namely

17

(1) ordinary model under scenario (2) from Example 1

(2) ordinary model under scenario (5) from Example 1

(3) stationary model under scenario (2) from Example 1

(4) stationary model under scenario (5) from Example 1

(5) delayed model (with W1 = 1 with probability 1) under scenario (2) from Example 1

(6) delayed model (with W1 = 1 with probability 1) under scenario (5) from Example 1

Concerning the results in Table 2 we first compare the values of Ψnxy(10) in entries (2)

and (5) of Table 1 to their corresponding counterparts in entries (1) and (2) of Table 2 In all

cases we observe that ruin is more likely to occur when interclaim times have the truncated

geometric distribution introduced in our first example as opposed to the discrete uniform

distribution of our second example Since mean interclaim times are equal in both examshy

ples this behaviour must be attributable to the greater variability inherent in the truncated

geometric distribution We also note that under the stationary model assumption the pmf

of W1 is easily verified to be rj = (11 minus j)55 j = 1 2 10 which has mean 4 lt 55

As a result the values of Ψnxy(10) for stationary models (3) and (4) are always larger than

the corresponding values for ordinary models (1) and (2) Finally in the absence of any

knowledge regarding which distribution is in effect for the first interclaim time W1 models

(5) and (6) can be regarded as ldquoworst-case scenariordquo models as they serve to provide upper

bounds on the values of Ψnxy(10)

In conclusion we state that the values in Tables 1 and 2 were generated by first impleshy

menting the recursive procedures (38) through (316) and then summing the appropriate

bivariatetrivariate probabilities computed via (213) (214) or (215) In situations when

x = y = infin however we computed the resulting finite-time ruin probabilities using an altershy

native (albeit standard) method (eg see Cossette et al (2006 Section 3)) Specifically if

we define the survival probability σ(n u) = Pr T gt n | U0 = u = 1minusΨninfininfin(u) as well as

the auxiliary function

0 if w ge Z ηwt =

minlfloorZminuswc minus1 rfloor + 1 t if w lt Z

then by conditioning on the time and amount of the first claim (and recalling that the delayed

process reverts to the ordinary process following this claim) we obtain the expression

minnnr (kminuszk)c2 u+czk+ℓ L L L

σ(u n) = Rn + rk dℓkminuszk αjγ(u + czk + ℓ minus j n minus k)

k=1 ℓ=(kminuszk)c1 j=1

18

where the function γ(w m) for w m isin N is computed recursively via

minmna (kminusηwk)c2 u+cηwk+ℓ L L L

γ(w m) = Am + ak dℓkminusηwk αjγ(u + cηwk + ℓ minus j m minus k)

k=1 j=1 ℓ=(kminusηwk)c1

with starting values γ(w 0) = 1 forall w isin N

Acknowledgements

The authors would like to thank Dr David Landriault for his valuable suggestions which

helped to improve an earlier version of this work This research was supported by the Natural

Sciences and Engineering Research Council of Canada

References

Alfa AS 2004 Markov chain representations of discrete distributions applied to queueing

models Computers amp Operations Research 31 2365-2385

Alfa AS Drekic S 2007 Algorithmic analysis of the Sparre Andersen model in discrete

time ASTIN Bulletin 37 293-317

Bao Z 2007 A note on the compound binomial model with randomized dividend strategy

Applied Mathematics amp Computation 194 276-286

Cossette H Landriault D Marceau E 2006 Ruin probabilities in the discrete time

renewal risk model Insurance Mathematics amp Economics 38 309-323

Karlin S Taylor HM 1975 A First Course in Stochastic Processes 2nd edition Acashy

demic Press New York

Kim B Kim H-S Kim J 2008 A risk model with paying dividends and random

environment Insurance Mathematics amp Economics 42 717-726

Landriault D 2008 Randomized dividends in the compound binomial model with a general

premium rate Scandinavian Actuarial Journal 1-15

Tan J Yang X 2006 The compound binomial model with randomized decisions on

paying dividends Insurance Mathematics amp Economics 39 1-18

19

Table 1 ndash Values of Ψnxy(10) corresponding to Example 1

(a) x = 10

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0005357 0006188 0007163 0008278 0009481

(2) 0005359 0006195 0007175 0008296 0009507

(3) 0005361 0006198 0007181 0008305 0009519

(4) 0005363 0006205 0007193 0008323 0009544

(5) 0005366 0006212 0007208 0008346 0009576

(6) 0005374 0006236 0007249 0008410 0009664

y = 25 (1) 0007631 0008815 0010202 0011789 0013503

(2) 0007634 0008824 0010220 0011816 0013539

(3) 0007636 0008828 0010228 0011829 0013557

(4) 0007639 0008838 0010245 0011854 0013593

(5) 0007643 0008849 0010266 0011887 0013638

(6) 0007656 0008882 0010325 0011978 0013764

y = 50 (1) 0008477 0009792 0011333 0013095 0014998

(2) 0008481 0009802 0011352 0013125 0015039

(3) 0008483 0009807 0011361 0013139 0015059

(4) 0008486 0009817 0011380 0013168 0015098

(5) 0008491 0009830 0011403 0013204 0015149

(6) 0008504 0009866 0011470 0013305 0015288

y = infin (1) 0008777 0010138 0011733 0013558 0015528

(2) 0008781 0010149 0011753 0013588 0015570

(3) 0008783 0010154 0011763 0013603 0015591

(4) 0008787 0010164 0011782 0013633 0015632

(5) 0008791 0010177 0011807 0013671 0015684

(6) 0008805 0010215 0011875 0013775 0015828

20

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0058559 0061875 0065704 0070048 0074714

(2) 0058569 0061903 0065756 0070128 0074825

(3) 0058574 0061917 0065782 0070167 0074880

(4) 0058584 0061945 0065833 0070245 0074987

(5) 0058596 0061979 0065896 0070344 0075124

(6) 0058632 0062079 0066078 0070621 0075506

y = 25 (1) 0089002 0093986 0099739 0106262 0113270

(2) 0089017 0094028 0099817 0106382 0113437

(3) 0089025 0094049 0099856 0106442 0113519

(4) 0089040 0094091 0099932 0106560 0113681

(5) 0089057 0094142 0100028 0106708 0113887

(6) 0089113 0094294 0100302 0107127 0114463

y = 50 (1) 0102786 0108506 0115106 0122590 0130629

(2) 0102804 0108555 0115196 0122728 0130820

(3) 0102812 0108579 0115241 0122797 0130914

(4) 0102830 0108627 0115329 0122932 0131101

(5) 0102850 0108686 0115439 0123103 0131337

(6) 0102914 0108860 0115754 0123584 0132000

y = infin (1) 0108895 0114933 0121901 0129799 0138283

(2) 0108914 0114985 0121996 0129945 0138485

(3) 0108923 0115010 0122043 0130018 0138585

(4) 0108941 0115061 0122136 0130161 0138782

(5) 0108962 0115123 0122252 0130341 0139032

(6) 0109030 0115307 0122585 0130849 0139732

21

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0079856 0086792 009464 0103440 0112831

(2) 0079885 0086869 0094778 0103650 0113118

(3) 0079953 0086978 0094936 0103864 0113393

(4) 0080081 0087188 0095243 0104282 0113929

(5) 0079962 0087075 0095149 0104213 0113890

(6) 0080805 0088366 0096951 0106588 0116875

y = 25 (1) 0124591 0135680 0148207 0162245 0177217

(2) 0124639 0135806 0148433 0162586 0177683

(3) 0124755 0135993 0148703 0162950 0178148

(4) 0124978 0136354 0149226 0163658 0179054

(5) 0124763 0136141 0149035 0163500 0178935

(6) 0126238 0138381 0152148 0167588 0184060

y = 50 (1) 0146757 0160079 0175111 0191945 0209894

(2) 0146816 0160232 0175385 0192360 0210460

(3) 0146965 0160469 0175725 0192817 0211043

(4) 0147249 0160925 0176384 0193706 0212178

(5) 0146968 0160639 0176117 0193470 0211980

(6) 0148860 0163500 0180077 0198658 0218474

y = infin (1) 0158013 0172606 0189059 0207476 0227106

(2) 0158078 0172776 0189362 0207934 0227732

(3) 0158250 0173047 0189751 0208455 0228395

(4) 0158579 0173571 0190503 0209467 0229685

(5) 0158247 0173226 0190172 0209161 0229410

(6) 0160443 0176532 0194735 0215127 0236869

22

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0088667 0104403 0123329 0145277 0169184

(2) 0088711 0104518 0123538 0145598 0169627

(3) 0088733 0104574 0123641 0145756 0169846

(4) 0088777 0104687 0123844 0146066 0170274

(5) 0088829 0104828 0124105 0146470 0170832

(6) 0088988 0105229 0124822 0147560 0172331

y = 25 (1) 0140528 0167839 0201007 0239726 0282102

(2) 0140598 0168021 0201338 0240234 0282804

(3) 0140632 0168111 0201502 0240486 0283151

(4) 0140701 0168289 0201822 0240976 0283827

(5) 0140785 0168515 0202241 0241622 0284718

(6) 0141036 0169148 0203370 0243335 0287071

y = 50 (1) 0167826 0203005 0246173 0296954 0352855

(2) 0167908 0203218 0246561 0297548 0353672

(3) 0167949 0203324 0246753 0297841 0354075

(4) 0168030 0203532 0247128 0298412 0354859

(5) 0168129 0203800 0247622 0299174 0355907

(6) 0168423 0204539 0248936 0301159 0358620

y = infin (1) 0183325 0224926 0276881 0338957 0408257

(2) 0183412 0225150 0277283 0339564 0409079

(3) 0183455 0225260 0277481 0339864 0409485

(4) 0183540 0225478 0277869 0340446 0410270

(5) 0183645 0225760 0278386 0341234 0411339

(6) 0183954 0226529 0279737 0343244 0414037

23

Table 2 ndash Values of Ψnxy(10) corresponding to Example 2

(a) x = 10

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0001901 0002276 0002724 0003240 0003799

(2) 0001904 0002286 0002742 0003269 0003840

(3) 0002651 0003013 0003442 0003936 0004471

(4) 0002655 0003022 0003460 0003964 0004510

(5) 0005792 0006098 0006458 0006871 0007317

(6) 0005795 0006106 0006473 0006894 0007350

y = 25 (1) 0002708 0003243 0003880 0004615 0005411

(2) 0002713 0003257 0003906 0004656 0005469

(3) 0003778 0004293 0004904 0005608 0006369

(4) 0003782 0004306 0004929 0005647 0006424

(5) 0008255 0008690 0009203 0009792 0010427

(6) 0008259 0008702 0009224 0009824 0010473

y = 50 (1) 0003008 0003603 0004310 0005127 0006011

(2) 0003014 0003618 0004340 0005173 0006076

(3) 0004197 0004769 0005448 0006230 0007075

(4) 0004202 0004784 0005476 0006273 0007137

(5) 0009173 0009656 0010226 0010879 0011585

(6) 0009177 0009669 0010249 0010916 0011637

y = infin (1) 0003115 0003730 0004463 0005308 0006224

(2) 0003121 0003746 0004493 0005356 0006291

(3) 0004346 0004938 0005641 0006450 0007325

(4) 0004351 0004954 0005670 0006496 0007389

(5) 0009499 0009999 0010589 0011265 0011996

(6) 0009503 0010012 0010613 0011303 0012050

24

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0031176 0032815 0034733 0036924 0039282

(2) 0031193 0032862 0034821 0037061 0039475

(3) 0050168 0051747 0053587 0055684 0057939

(4) 0050184 0051792 0053671 0055815 0058124

(5) 0136884 0138219 0139763 0141517 0143399

(6) 0136896 0138256 0139832 0141626 0143553

y = 25 (1) 0047622 0050092 0052981 0056280 0059829

(2) 0004765 0050163 0053113 0056487 0060121

(3) 0076570 0078950 0081720 0084878 0088272

(4) 0076593 0079017 0081847 0085076 0088551

(5) 0205774 0207786 0210111 0212752 0215585

(6) 0205793 0207842 0210216 0212917 0215818

y = 50 (1) 0055170 0058009 0061329 0065119 0069196

(2) 0055200 0058091 0061481 0065357 0069533

(3) 0088650 0091385 0094569 0098197 0102096

(4) 0088677 0091463 0094715 0098425 0102417

(5) 0235815 0238128 0240800 0243834 0247089

(6) 0235837 0238192 0240920 0244023 0247356

y = infin (1) 0058565 0061565 0065073 0069077 0073384

(2) 0058597 0061652 0065234 0069329 0073740

(3) 0094061 0096952 0100316 0104149 0108267

(4) 0094090 0097035 0100470 0104390 0108607

(5) 0248472 0250917 0253740 0256945 0260383

(6) 0248495 0250985 0253867 0257145 0260667

25

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0048598 0052524 0056984 0061998 0067342

(2) 0048654 0052673 0057253 0062411 0067916

(3) 0070192 0073979 0078257 0083057 0088167

(4) 0070243 0074119 0078513 0083451 0088714

(5) 0151774 0154981 0158572 0162586 0166852

(6) 0151815 0155096 0158784 0162913 0167308

y = 25 (1) 0076895 0083230 0090413 0098481 0107073

(2) 0076987 0083474 0090854 0099156 0108010

(3) 0110093 0116205 0123095 0130818 0139033

(4) 0110177 0116435 0123515 0131462 0139928

(5) 0230674 0235848 0241632 0248091 0254950

(6) 0230739 0236038 0241980 0248626 0255695

y = 50 (1) 0091496 0099156 0107829 0117563 0127924

(2) 0091610 0099455 0108370 0118387 0129068

(3) 0130130 0137520 0145839 0155156 0165064

(4) 0130233 0137801 0146352 0155942 0166156

(5) 0266599 0272857 0279840 0287632 0295904

(6) 0266680 0273088 0280265 0288286 0296814

y = infin (1) 0099312 0107746 0117285 0127984 0139369

(2) 0099439 0108078 0117883 0128899 0140639

(3) 0140458 0148595 0157745 0167986 0178872

(4) 0140572 0148908 0158314 0168859 0180084

(5) 0282886 0289776 0297457 0306022 0315111

(6) 0282976 0290034 0297929 0306748 0316120

26

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0059096 0072505 0088907 0108122 0129137

(2) 0059223 0072829 0089500 0109052 0130456

(3) 0079757 0092662 0108388 0126778 0146871

(4) 0079875 0092970 0108956 0127669 0148134

(5) 0159788 0170682 0183871 0199249 0216024

(6) 0159881 0170936 0184343 0199990 0217075

y = 25 (1) 0095883 0119741 0149240 0184073 0222390

(2) 0096087 0120262 0150197 0185573 0224520

(3) 0127403 0150359 0178640 0211977 0248613

(4) 0127592 0150855 0179557 0213414 0250653

(5) 0245161 0264533 0288250 0316124 0346709

(6) 0245311 0264942 0289010 0317319 0348407

y = 50 (1) 0116598 0147920 0187100 0233784 0285503

(2) 0116839 0148537 0188231 0235554 0288015

(3) 0153023 0183153 0220713 0265390 0314840

(4) 0153249 0183742 0221798 0267087 0317248

(5) 0285742 0311159 0342651 0380005 0421287

(6) 0285921 0311644 0343551 0381418 0423292

y = infin (1) 0129470 0167217 0215383 0273841 0339734

(2) 0129725 0167865 0216560 0275661 0342282

(3) 0167978 0204275 0250442 0306382 0369383

(4) 0168217 0204895 0251572 0308128 0371826

(5) 0305868 0336470 0375169 0421935 0474525

(6) 0306058 0336980 0376106 0423388 0476560

27

Page 6: uwaterloo.ca · Ruin Analysis of a Threshold Strategy in a Discrete-time Sparre Andersen Model Steve DREKIC and Ana Maria MERA Department of Statistics and Actuarial Science University

and c2 L

Ai = djBi+j (23) j=c1

We give a brief indication of the reasoning leading to the form of P in (21) For example

consider the transition ldquoUt = 0 rarr Ut+1 = 1rdquo which has block element Bcminus1 containing the

transition probabilities governing the ldquoLt rarr Lt+1 rdquo process At a surplus level of 0 at time

t a premium of pt = c is certain to be received Therefore to arrive at surplus level 1 at

the next time unit a ldquoso-calledrdquo claim of size c minus 1 must occur to neutralize the premium

just received Thus the size of the ldquoclaimrdquo required for this transition to occur determines

the value of i in Bi so that Bcminus1 is obtained and its form subsequently determined by (22)

according to the value of the index c minus 1 Note that only when c gt 1 does an actual claim

occur in this particular scenario We mention that for transitions which begin at or above the

threshold level Z a similar line of reasoning would also apply except now one must condition

on the amount of the random premium being earned as indicated by the form of (23)

Note that this formulation gives rise to a doubly-infinite Markov chain with finite blocks

of size na In the special case of the threshold-free model (obtained by setting c1 = c2 = c

so that dc = 1) we remark that (21) is in agreement with the matrix P in Alfa and Drekic

(2007 p 296) Also since αi = 0 forall i gt mα we note that Bi = Ona if i gt mα

For the computation of ruin-related quantities of interest it is useful to partition the

state space Δ into two mutually exclusive state spaces namely Δ1 = N times 1 2 na

and Δ2 = Zminus times 1 2 na In addition we introduce (one-step) transition matrices

C Δ1 rarr Δ1 and D Δ1 rarr Δ2 which correspond to mapping ldquonon-ruinedrdquo states in Δ1 to

either ldquonon-ruinedrdquo states in Δ1 or ldquoruinedrdquo states in Δ2 respectively In other words the

matrices C and D are essentially submatrices of P and their forms are readily given by

210 middot middot middot Z minus 1 Z Z + 1 middot middot middot

Bc Bcminus1 Bcminus2 middot middot middot BcminusZ+1 BcminusZ BcminusZminus1 middot middot middot Bc+1 Bc Bcminus1 middot middot middot BcminusZ+2 BcminusZ+1 BcminusZ middot middot middot

Bc+2 Bc+1 Bc middot middot middot BcminusZ+3 BcminusZ+2 BcminusZ+1 middot middot middot

middot middot middot

middot middot middot

0

1

2 Z minus 2 BZ+cminus2 BZ+cminus3 BZ+cminus4 middot middot middot Bcminus1 Bcminus2 Bcminus3 middot middot middot (24) C =

Z minus 1 BZ+cminus1 BZ+cminus2 BZ+cminus3 middot middot middot Bc Bcminus1 Bcminus2 middot middot middot

Z AZ AZminus1 AZminus2 middot middot middot A1 A0 Aminus1 middot middot middot

Z + 1 AZ+1 AZ AZminus1 middot middot middot A2 A1 A0 middot middot middot

Z + 2 AZ+2 AZ+1 AZ middot middot middot A3 A2 A1 middot middot middot

middot middot middot

6

and

minus1 minus2 minus3 middot middot middot

0 Bc+1 Bc+2 Bc+3 middot middot middot 1 Bc+2 Bc+3 Bc+4 middot middot middot

2 Bc+3 Bc+4 Bc+5 middot middot middot middot middot middot

Z minus 2 BZ+cminus1 BZ+c BZ+c+1 middot middot middot D = (25) Z minus 1 BZ+c BZ+c+1 BZ+c+2 middot middot middot

Z AZ+1 AZ+2 AZ+3 middot middot middot

Z + 1 AZ+2 AZ+3 AZ+4 middot middot middot Z + 2 AZ+3 AZ+4 AZ+5 middot middot middot

Let us now specify some further definitions required in upcoming derivations First of all

we introduce the function

0 if u ge Z zt =

maxi isin 1 2 t|u + c(i minus 1) lt Z if u lt Z

or equivalently

0 if u ge Z zt =

minlfloorZminuscuminus1 rfloor + 1 t if u lt Z

where the floor function denoted lfloorxrfloor yields the largest integer less than or equal to x We

define dm0 = δm0 where δij in general denotes the Kronecker delta function of i and j

Also define

dm if m = c1 c1 + 1 c2dm1 =

0 elsewhere

and for n = 2 3

Lc2 dj1dmminusjnminus1 if m = nc1 nc1 + 1 nc2dmn = j=c1

0 elsewhere

In other words dmn represents the pmf of the n-fold convolution of dm with itself

Recalling that W1 = k note that zk + ( k+) determines the earliest time point in the set =

0+ 1+ (k minus1)+ at which a random premium is received Note also that zk = k implies

that no such point exists in this set (ie the threshold level Z has not yet been reached when

the premium at time (k minus1)+ is to be received) Therefore if we assume for the moment that

zk k we may condition on the total premiums earned at times zk + (zk + 1)+ (k minus 1)+=

7

to obtain the following expression for b(k) the initial (ie starting at time k) probability row

vector of the above Markov chain corresponding only to the states in Δ1 namely

(kminuszk)c2

b(k) L

= dmkminuszk(αu+czk+me1 αu+czk+mminus1e1 α2e1 α1e1 0 0 ) (26)

m=(kminuszk)c1

where 0 denotes the 1 times na row vector of zeros By further defining the function vn(t) =

b(k)u + ct + (n minus t)c2 we note that the i-th level of is given by the 1 times na row vector L(kminuszk)c2 b(k)αu+czk+mminusie1 for each i isin Ωk = 0 1 vk(zk)minus1 Hence contains

m=(kminuszk)c1 dmkminuszk

zeros from level vk(zk) onwards Moreover (26) holds true even in the case when zk = k

simplifying to give b(k) = (αu+cke1 αu+ckminus1e1 α2e1 α1e1 0 0 )

At this juncture we introduce two additional row vectors namely

(k) (k) (k) (k) b(k)gn = (gn0 gn1 gn2 ) = Cn n isin N (27)

and

h(k) (k) (k) (k) b(k)Cnminus1D =

(k) Z

+ n = (h nminus2 hnminus3 ) = gnminus1D n isin (28) nminus1 h

(k)We remark that gn contains the probabilities of being in the various ldquonon-ruinedrdquo states

at time k + n (ie after claims have been paid out) without having visited a ldquoruinedrdquo state

during the previous n minus 1 transitions given that Uk isin Ωk sub Δ1 In a similar fashion h(nk)

contains the probabilities of being in the various ldquoruinedrdquo states for the first time at time

k + n given that Uk isin Ωk sub Δ1 Note that the probability of being in ruined state ldquo minus jrdquo is (k) (k) (k) (k)

characterized by the row vector h = (h h h ) with the third subscript nminusj nminusj1 nminusj2 nminusjna

representing the value of the elapsed time counter Lk+n Upon further reflection however (k)

=it must be the case that h = 0 for i 1 as ruin can only occur at claim instants nminusji

which in this case implies Lk+n = 1 with probability 1 Thus it immediately follows that (k) (k) (k)

the structure of the 1 times na row vector hnminusj is simply given by hnminusj = (φnj(u) 0 0) (k)

where φ (u) = Pr T = k + n |UT | = j | Consequently as in the analysis with nj Uk isin Ωk

no threshold on the surplus level (see Alfa and Drekic (2007 p 299)) we obtain

(k) (k) prime φ (u) = h (29) nj nminusje1

where e1 prime denotes the transpose of e1

A similar line of probabilistic logic can also be applied to obtain a representation for the (k)

trivariate function ψ (u) = Pr T = k + n UTminus = i |UT | = j | Uk isin Ωk In order for ruin nij

to occur at time k + n with a surplus immediately prior to ruin equal to i we observe that

(i) none of the previous n minus 1 transitions must have included a visit to any state in Δ2

and

8

(ii) the surplus level at time k + n minus 1 must necessarily be equal to i minus pk+nminus1

Note that pk+nminus1 denotes the premium corresponding to the time interval (k + n minus 1 k + n]

which is received at (k + n minus 1)+ Hence the quantity corresponding to points (i) and (ii) (k)

above is the 1 times na row vector g Then at the next time unit a claim must nminus1iminuspk+nminus1

necessarily occur but not before a premium of pk+nminus1 is first collected thereby raising the

surplus level to i Since s contains the absorption probabilities (to claim occurrence) from

the na possible phase states and the claim causing ruin must be of size i + j in order to

ensure that the deficit at ruin is equal to j it follows that

(k) (k)ψ (u) = g s αi+j (210) nij nminus1iminuspk+nminus1

(k)The following corollary provides the means of calculating the quantity g in nminus1iminuspk+nminus1

(210)

Corollary 1 If minc Z + c1 = c then

(k)

g if i = c c+1 Z +c1 minus1 nminus1iminusc

(k) (k) Lminc2iminusZ (k)g = g + j=c1

if i = Z +c1 Z +c1 +1 minmα Z +cminus1nminus1iminuspk+nminus1 nminus1iminusc djgnminus1iminusj

Lc2 (k)

j=c1 djgnminus1iminusj if i = Z +c Z +c+1 mαminus1

(211)

Conversely if minc Z + c1 = Z + c1 then

Lminc2iminusZ (k)

djg if i = Z +c1 Z +c1 +1 cminus1nminus1iminusj j=c1

(k) (k) Lminc2iminusZ (k)gnminus1iminuspk+nminus1

= gnminus1iminusc + j=c1 djgnminus1iminusj if i = c c+1 minmα Z +cminus1

Lc2 (k)djg if i = Z +c Z +c+1 mαminus1j=c1 nminus1iminusj

(212)

Proof The justification of (211) is as follows If c is smaller than Z + c1 the only way of

reaching a surplus level of i isin c c+1 Z +c1 minus1 (prior to ruin) is by receiving a premium

of c since i minus j for any j = c1 c1 +1 c2 would result in a surplus level below Z where the

random premium rate is not applicable Next for fixed i isin Z +c1 Z +c1 +1 Z +c minus1

there are two distinct ways of reaching this surplus level First of all note that the value of

i minus j is guaranteed to produce a surplus amount greater than or equal to Z provided that

j isin c1 c1 + 1 minc2 i minus Z Secondly note that reducing each of value of i by c also

yields a non-negative surplus value less than Z Hence both pure and random premiums

are plausible under this scenario For i isin Z + c Z + c + 1 note that decreasing this

amount by any j = c1 c1 + 1 c2 results in a surplus value greater than or equal to Z as

desired However reducing it by the amount c also has the same impact thus rendering the

pure premium option inadmissible in this case Lastly we incorporate into the range of i the

9

fact that the surplus immediately prior to ruin cannot exceed mα minus1 as indicated in Section

1 The justification of (212) when c is larger than Z + c1 mirrors the analysis above and is

left to the reader D

Recall that in the analysis up to this point we have assumed that W1 = k isin

1 2 3 nr If we now condition on the value of W1 the Law of Total Probability

yields the following two expressions for φnj(u) and ψnij(u) namely

nr L

= rkφ (u) n = nr + 1 nr + 2 φnj(u) n

(kminus) kj

k=1

and nr L

(k)ψnij(u) = rkψ (u) n = nr + 1 nr + 2 nminuskij

k=1

In addition however it is quite possible that T = n for n = 1 2 nr and this can happen

as a result of either

(i) the first claim which causes ruin occurring at time nminus from a (conditional) surplus Lnminus1level of u + czn + m (where m) or i=zn

Xi =

(ii) the first claim which does not cause ruin occurring at some time kminus k isin 1 2 nminus

1 and ruin subsequently occurring n minus k time units later

If we now combine these outcomes with those for n = nr + 1 nr + 2 (and recall that

rn = 0 forall n gt nr) the general expression for φnj(u) becomes

minnminus1nr (nminuszn)c2 L

(k) L

φnj(u)= rkφ (u) + rn dmnminuszn αu+czn+m+j n isin Z+ (213) nminuskj

k=1 m=(nminuszn)c1

Concerning the function ψnij(u) when n le nr and point (i) above is considered we note Lnminus1that i must necessarily equal u + czn + m (again conditional on i=zn

Xi = m) for ruin to

occur in this manner Hence in a similar fashion we acquire

minnminus1nr (nminuszn)c2 L L

ψnij(u) = rkψn

(kminus) kij(u) + rn dmnminuszn

δiu+czn+mαu+czn+m+j

k=1 m=(nminuszn)c1

minnminus1nr L

(k) Z

+ = rkψnminuskij(u) + rndiminusuminuscznnminuszn αi+j n isin (214)

k=1

Finally for fixed i isin minc Z + c1 minc Z + c1 + 1 mα minus 1 summing (214) from

j = 1 to mα minus i yields the following expression for the bivariate pmf ωni(u) namely

minnminus1nr mαminusi mαminusi L L

(k) L

ωni(u) = rk ψnminuskij(u) + rndiminusuminuscznnminusznαi+j

k=1 j=1 j=1

10

minnminus1nr L

(k) Z

+ = Λi rkg s + rndiminusuminuscznnminuszn n isin (215) nminuskminus1iminuspnminus1

k=1

where (210) was used to establish the last equality In the special case when c1 = c2 = c it

is a straightforward exercise to show that equations (213) through (215) simplify to produce

equations (215) through (217) of Alfa and Drekic (2007 p 300)

3 Computational considerations

In order to compute the joint pmfrsquos defined by (213) (214) and (215) it is clear from (k)

equations (29) through (212) that we need to be able to compute the quantities gn and

h(k) n defined by (27) and (28) respectively In what follows we capitalize on the underlying

Markov chain structure in our model to develop a set of computational procedures which

ultimately enable one to calculate these key row vectors

First of all recall that g0(k)

= b(k) contains zeros from a certain level onwards namely (k)

level vk(zk) = u + czk + c2(k minus zk) In general one can easily deduce that gn will also

contain zeros from a certain level onwards to be denoted by lk(n) We begin this section

with a determination of lk(n) which varies in form depending on the value of vk(zk) minus 1

representing the maximum surplus level at time k after the smallest possible claim of size 1

has been applied (at time kminus)

On one hand it may be the case that the value of vk(zk)minus1 lies at or above the threshold

level Z In this situation we remark that zk can take on any value in the set 0 1 k (k)

Since g0 contains zeros from level lk(0) = vk(zk) onwards we observe that at time k + 1

it may be possible to achieve a higher surplus value than vk+1(zk) minus 1 representing the

maximum surplus level assuming there was no drop below Z due to the claim at time kminus

Specifically if the surplus had fallen to level Z minus 1 at time kminus then gained c at time k+ a

bounded level of Z + c minus1 would have been reached at time k +1 and this level is potentially

larger than vk+1(zk) minus 1 As a result we have that lk(1) = maxvk+1(zk) Z + c If we

extend this argument inductively we readily obtain lk(n) = maxvk+n(zk) Z +c +c2(n minus1)

n isin Z+ However since vk+n(zk) = u + czk + c2(k +n minus zk) = vk(zk)+ c2 + c2(n minus 1) we may

alternatively re-express lk(n) as

lk(n) = maxvk+n(zk) Z + c + c2(n minus 1) = maxvk(zk) + c2 Z + c + c2(n minus 1)

Thus

Z+lk(n) = maxlk(0) + c2 Z + c + c2(n minus 1) n isin (31)

11

or equivalently

Z+lk(n) = lk(1) + c2(n minus 1) n isin

Clearly lk(0) gt Z and lk(n) ge Z + c for n isin Z+

The other remaining case to consider involves the situation when the maximum surplus

level at time k is strictly below the threshold level Z Under this scenario we further

distinguish between two cases involving the value of c2 First of all suppose that c2 = 0 so

that vk(zk)minus1 = u +czk minus1 lt Z or equivalently u +czk le Z If u +czk happens to equal Z

then it is possible for zk isin 0 1 k With c2 = 0 however no further premiums can be

earned until after the first claim takes place (which is assumed to occur at time kminus) thereby

implying that Z = u + ci for all i = zk zk + 1 k Hence we know that u + ck = Z

in this case Moreover if u + czk happens to be strictly less than Z then we clearly have

that zk = k Even in the case when c2 gt 0 one can readily verify that zk = k so that

vk(zk) minus 1 = u + ck minus 1 lt Z Thus we proceed with the analysis assuming that zk = k

c2 ge 0 and the inequality u + ck le Z holds true

lowast k

Define the ceiling function denoted lceilxrceil to yield the smallest integer greater than or

equal to x Because a random premium cannot be received until level Z is at least achieved

it is essential to find the point at which the threshold level Z is reached (ie the point at Zminus(u+ck)+1 which u + ck minus1+ ci ge Z) Isolating this inequality for i one obtains i ge

c Hence

assuming that W1 = k the elapsed time after k at which a random premium would begin to

lceilZminus(u+ck)+1 be received is t lowast = rceil Note that t lowast isin Z+ k c k

With zk = k g0(k)

contains zeros from level u + ck onwards so that lk(0) = u + ck

Immediately following this a premium of c is guaranteed to be received at time k+ as the

threshold level Z has not yet been reached Consequently g1(k)

contains zeros from level (k)

b(k)u + c(k + 1) onwards which is c levels further than that in g0 = Continuing this (k)

process inductively we can readily establish that gn contains zeros from level u + c(k + n)

onwards for n = 1 2 t lowast k so that lk(n) = u + c(k + n) for n = 0 1 t lowast k

In the case of n = t lowast + 1 t lowast + 2 gt

(k) is first looked at in isolation At time k +k k +1

t lowast k + 1 the threshold level has been crossed giving rise to a maximum surplus value of either (k)

c(k + t lowast k contains zeros from level lk(t lowast ku + ) + c2 minus 1 or Z + c minus 1 Therefore g + 1) = t lowast k+1

+1(k + t lowast kmaxvk+t lowast k

) Z + c onwards Carrying out the same inductive process as before it is

readily seen that g(nk)

contains zeros from level lk(n) = maxvk+n(k+t lowast k) Z +c+c2(nminust lowast kminus1)

onwards for n = t lowast k + 1 t lowast k + 2 Combining both cases we obtain

u + c(k + n) if n = 0 1 t lowast lk(n) = k (32)

maxlk(t lowast) + c2 Z + c + c2(n minus t lowast minus 1) if n = t lowast + 1 t lowast + 2 k k k k

Note that by construction lk(t lowast k) gt Z and lk(n) le Z for n = 0 1 t lowast k minus1 Interestingly

12

if we extend the definition tk lowast by introducing the quantity ζk

lowast = maxtklowast 0 we remark that

(31) and (32) can actually be combined into a single formula for lk(n) namely

vk+n(zk + n) if n = 0 1 ζklowast

lk(n) = (33) maxlk(ζ

lowast) + c2 Z + c + c2(n minus ζlowast minus 1) if n = ζlowast + 1 ζlowast + 2 k k k k

Clearly we have that lk(n) ge Z + c for n = ζk lowast +1 ζk

lowast +2 Moreover if c1 = c2 = c (33)

reduces to give lk(n) = u + c(k + n) forall n isin N which is in agreement with the results in Alfa

and Drekic (2007 Section 3)

Remark In our above analysis we purposely ignored the possibility of any future claims

occurring after time k in our determination of lk(n) However we recognize that this is not

realistic and future claims would indeed occur after time k thereby further reducing the value

of the maximum surplus level lk(n) minus 1 at time k + n To account for this we imagine all

future interclaim times (ie after time k) to be of maximum length na and all corresponding

claim sizes to be of minimum amount 1 Consequently in the case when c2 gt 0 we should

subtract from formula (33) the amount of lfloornnarfloor where lfloormiddotrfloor again denotes the floor function

Obviously when c2 = 0 (ie no premiums are collected whenever the surplus level is at or

above Z) lk(n) = Z + c for n = ζk lowast + 1 ζk

lowast + 2 (k)

We now turn our attention to the development of a general algorithm for computing gni

(k) (k)for i = 0 1 lk(n) minus 1 With the aid of (24) we apply the key identity gn = gnminus1C

(which can be inferred from (27)) to first obtain

minZlk(nminus1)minus1 lk(nminus1)minus1 c2 L L L

(k) (k) (k)gni = g Bjminusi+c + gnminus1j dmBjminusi+m n isin Z+ (34) nminus1j

j=0 j=Z m=c1

However (22) states that Bi = Ona for i isin Zminus and so (34) becomes (for the same range of

i)

minZlk(nminus1)minus1 c2 lk(nminus1)minus1 L L L

(k) (k) (k)gni = g Bjminusi+c + dm gnminus1jBjminusi+m n isin Z+ (35) nminus1j

j=max0iminusc m=c1 j=maxZiminusm

Taking the form of (33) into consideration (35) yields the following two formulae

lk(nminus1)minus1 L

(k) (k)gni = gnminus1jBjminusi+c n = 1 2 ζ k

lowast (36) j=max0iminusc

and

Zminus1 c2 lk(nminus1)minus1 L L L

(k) (k) (k)g = g + dm g n = ζ lowast + 1 ζ lowast + 2 ni nminus1jBjminusi+c nminus1jBjminusi+m k k

j=max0iminusc m=c1 j=maxZiminusm

(37)

13

For n = 1 2 ζk lowast where ζk

lowast isin Z+ we note that lk(n) ge c + 1 Therefore with the aid of

(22) (36) simplifies to give

Llk(nminus1)minus1 (k) (k) j=0 gnminus1jBjminusi+c if i = 0 1 c minus 1

gni = Llk(nminus1)minus1 (k)

g if i = c c + 1 lk(n) minus 1j=iminusc nminus1jBjminusi+c

Llk(nminus1)minus1 (k)

nminus1js 0 0 if i = 0 1 c minus 1j=0 αjminusi+cg= (38)

(k) Llk(nminus1)minus1 (k) g S+ αjminusi+cgnminus1js 0 0 if i = c c+1 lk(n)minus1 nminus1iminusc j=iminusc+1

In the case of n = ζk lowast + 1 ζk

lowast + 2 we first reiterate that lk(n) ge Z + c We next break

(37) into several cases based on the value of i and the value of minc Z + c1 as follows

Case 1 Suppose minc Z + c1 = c

For i isin 0 1 c minus 1

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+mni nminus1j nminus1j

j=0 m=c1 j=Z

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k)= αjminusi+cg s + dm αjminusi+mg s 0 0 (39) nminus1j nminus1j

j=0 m=c1 j=Z

For i isin c c + 1 Z + c1 minus 1

Zminus1 c2 lk(nminus1)minus1 L L L

(k) (k) (k)g = g + dm gni nminus1j Bjminusi+c nminus1j Bjminusi+m

j=iminusc m=c1 j=Z

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k) (k)= g S + αjminusi+cg dm αjminusi+mgnminus1j s 0 0 (310) nminus1iminusc nminus1j s +

j=iminusc+1 m=c1 j=Z

For i isin Z + c1 Z + c1 + 1 Z + c minus 1

Zminus1 minc2iminusZ lk(nminus1)minus1 c2 lk(nminus1)minus1 L L L L L

(k) (k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+m + dm g Bjminusi+mni nminus1j nminus1j nminus1j

j=iminusc m=c1 j=iminusm m=iminusZ+1 j=Z

minc2iminusZ L

(k) (k)= g S + dmg S nminus1iminusc nminus1iminusm

m=c1

Zminus1 minc2iminusZ lk(nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k)+ αjminusi+cg dm αjminusi+mg dm αjminusi+mgnminus1j s 0 0 nminus1j s + nminus1js +

j=iminusc+1 m=c1 j=iminusm+1 m=iminusZ+1 j=Z

(311)

For i isin Z + c Z + c + 1 lk(n) minus 1

c2 lk(nminus1)minus1 L L

(k) (k)g = dm g Bjminusi+mni nminus1j

m=c1 j=iminusm

c2 c2 lk (nminus1)minus1 L L L

(k) (k)= dmg S + dm αjminusi+mgnminus1j s 0 0 (312) nminus1iminusm

m=c1 m=c1 j=iminusm+1

14

Case 2 Suppose minc Z + c1 = Z + c1

For i isin 0 1 Z + c1 minus 1

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+mni nminus1j nminus1j

j=0 m=c1 j=Z

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k)= αjminusi+cg s + dm αjminusi+mg s 0 0 (313) nminus1j nminus1j

j=0 m=c1 j=Z

For i isin Z + c1 Z + c1 + 1 c minus 1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+m + dm g Bjminusi+mni nminus1j nminus1j nminus1j

j=0 m=c1 j=iminusm m=iminusZ+1 j=Z

minc2iminusZ L

(k)= dmg S nminus1iminusm

m=c1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k)+ αjminusi+cg s + dm αjminusi+mg s + dm αjminusi+mg s 0 0 nminus1j nminus1j nminus1j

j=0 m=c1 j=iminusm+1 m=iminusZ+1 j=Z

(314)

For i isin c c + 1 Z + c minus 1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k) (k)g = g + g + gni nminus1j Bjminusi+c dm nminus1j Bjminusi+m dm nminus1j Bjminusi+m

j=iminusc m=c1 j=iminusm m=iminusZ+1 j=Z

minc2iminusZ L

(k) (k)= g S + dmg S nminus1iminusc nminus1iminusm

m=c1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k)+

nminus1j s 0 0 αjminusi+cgnminus1j s + dm αjminusi+mgnminus1j s + dm αjminusi+mg j=iminusc+1 m=c1 j=iminusm+1 m=iminusZ+1 j=Z

(315)

For i isin Z + c Z + c + 1 lk(n) minus 1

c2 lk (nminus1)minus1 L L

(k) (k)g = dm g Bjminusi+mni nminus1j

m=c1 j=iminusm

c2 c2 lk (nminus1)minus1 L L L

(k) (k)= dmg S + dm αjminusi+mgnminus1j s 0 0 (316) nminus1iminusm

m=c1 m=c1 j=iminusm+1

It is not difficult to verify that equations (38) through (316) simplify to produce equations

(32) and (33) of Alfa and Drekic (2007 pp 301-302) when c1 = c2 = c With equations (38) (k)

through (316) providing the recursive means of calculating gn we next turn our attention

15

to a determination of h(k) defined by (28) For n = 1 2 ζlowast we use (22) and (25) to n k

obtain

lk(nminus1)minus1 lk(nminus1)minus1

h(k) nminusj =

L g

(k) nminus1ℓBc+j+ℓ =

L αc+j+ℓg

(k) nminus1ℓs 0 0 (317)

ℓ=0 ℓ=0

For n = ζk lowast + 1 ζk

lowast + 2 we similarly acquire

Zminus1 lk(nminus1)minus1 L L

(k) (k) (k)hnminusj = gnminus1ℓBc+j+ℓ + gnminus1ℓAj+ℓ

ℓ=0 ℓ=Z

Zminus1 c2 lk(nminus1)minus1 L L L

(k) (k)= αc+j+ℓg s + dm αm+j+ℓg s 0 0 (318) nminus1ℓ nminus1ℓ

ℓ=0 m=c1 ℓ=Z

Note that substituting (317) and (318) into (29) immediately yields

Llk(nminus1)minus1 (k) (k) ℓ=0 αc+j+ℓgnminus1ℓs if n = 1 2 ζk

lowast φ (u) = nj LZminus1 (k) Lc2

Llk(nminus1)minus1 (k) ζlowast s + dm αm+j+ℓg n = + 1 ζlowast + 2 ℓ=0 αc+j+ℓgnminus1ℓ m=c1 ℓ=Z nminus1ℓs if k k

(319)

Clearly in the special case when c1 = c2 = c (319) yields equation (35) of Alfa and Drekic

(2007 p 302) Finally we remark that (319) can be substituted into (213) to obtain a

computational formula for φnj(u)

4 Numerical examples and findings

In this section we illustrate the application of our proposed algorithm with several numerical

examples All calculations in this section were performed using Mathematica (Version 7) on

a Dell Precision T7400 workstation with a 3 Ghz CPU and 4 GB of RAM

In our first illustration (labelled Example 1) we consider the ordinary Sparre Andersen

model in which interclaim times have pmf given by

(211)(911)jminus1 if j = 1 2 na minus 1 aj = (41)

(911)naminus1 if j = na

Note that the pmf (41) is that of a truncated geometric distribution with all the probability

mass on na na +1 assigned to the support value na Clearly as na becomes larger the

closer (41) approximates this particular geometric distribution having mean 55 In what Lna

Lminus1 nafollows we fix na = 25 so that aj = 0992 and jaj = 546 Let the individual j=1 j=1

claim amount distribution be given by the pmf

minus4 minus4j minus 1 j

αj = 1 + minus 1 + j isin Z+ (42) 30 30

16

We remark that the pmf (42) represents a discretized version of the Pareto distribution with

mean 10 Also note that mα = infin thereby implying that the random variables |UT | and

UTminus both have infinite support

Table 1 displays values (rounded to 6 decimal places) of the trivariate cumulative distrishy

bution function

n x y L L L

Ψnxy(u) = Pr T le n UTminus le x |UT | le y | U0 = u = ψℓij(u) ℓ=1 i=mincZ+c1 j=1

corresponding to the above discrete-time risk process with u = 10 Z = 50 c = 5 and

varying premium amount distributions according to the following set of scenarios

(1) d2 = 1

(2) d1 = 25 d2 = 15 d3 = 25

G

5)

(3) di = i

(25)i(35)5minusi i = 0 1 5

(4) di = 15 i = 0 1 4

(5) d1 = 34 d5 = 14

(6) d0 = 35 d5 = 25

We remark that while these six premium amount distributions all have mean equal to 2

they possess different shapes and we investigate the impact of their variability on the joint

distribution of T |UT | and UTminus In particular the variances corresponding to the premium

amount distributions of scenarios (1) through (6) are 0 08 12 2 3 and 6 respectively

Based on the results in Table 1 it would appear that greater variability in the premium

amount distribution generally leads to higher ruin probabilities Perhaps somewhat surprising

though the differences are subtle and not particularly striking as one compares the various

ruin probabilities from scenarios (1) through (6) Obviously we observe more of a difference

between scenarios as the values of n x and y increase

As a follow-up illustration (labelled Example 2) we consider a model in which ordinary

interclaim times have a discrete uniform distribution with pmf aj = 110 j = 1 2 10

Clearly the mean of this distribution is 55 which for comparative purposes is identical to

the ordinary interclaim time mean considered in Example 1 In addition we assume that

the pmf of the individual claim amount distribution is once again given by (42) Table 2

displays values (rounded to 6 decimal places) of Ψnxy(10) for several variants of the Sparre

Andersen model (again with Z = 50 and c = 5) having the above ordinary interclaim time

and claim amount distributions namely

17

(1) ordinary model under scenario (2) from Example 1

(2) ordinary model under scenario (5) from Example 1

(3) stationary model under scenario (2) from Example 1

(4) stationary model under scenario (5) from Example 1

(5) delayed model (with W1 = 1 with probability 1) under scenario (2) from Example 1

(6) delayed model (with W1 = 1 with probability 1) under scenario (5) from Example 1

Concerning the results in Table 2 we first compare the values of Ψnxy(10) in entries (2)

and (5) of Table 1 to their corresponding counterparts in entries (1) and (2) of Table 2 In all

cases we observe that ruin is more likely to occur when interclaim times have the truncated

geometric distribution introduced in our first example as opposed to the discrete uniform

distribution of our second example Since mean interclaim times are equal in both examshy

ples this behaviour must be attributable to the greater variability inherent in the truncated

geometric distribution We also note that under the stationary model assumption the pmf

of W1 is easily verified to be rj = (11 minus j)55 j = 1 2 10 which has mean 4 lt 55

As a result the values of Ψnxy(10) for stationary models (3) and (4) are always larger than

the corresponding values for ordinary models (1) and (2) Finally in the absence of any

knowledge regarding which distribution is in effect for the first interclaim time W1 models

(5) and (6) can be regarded as ldquoworst-case scenariordquo models as they serve to provide upper

bounds on the values of Ψnxy(10)

In conclusion we state that the values in Tables 1 and 2 were generated by first impleshy

menting the recursive procedures (38) through (316) and then summing the appropriate

bivariatetrivariate probabilities computed via (213) (214) or (215) In situations when

x = y = infin however we computed the resulting finite-time ruin probabilities using an altershy

native (albeit standard) method (eg see Cossette et al (2006 Section 3)) Specifically if

we define the survival probability σ(n u) = Pr T gt n | U0 = u = 1minusΨninfininfin(u) as well as

the auxiliary function

0 if w ge Z ηwt =

minlfloorZminuswc minus1 rfloor + 1 t if w lt Z

then by conditioning on the time and amount of the first claim (and recalling that the delayed

process reverts to the ordinary process following this claim) we obtain the expression

minnnr (kminuszk)c2 u+czk+ℓ L L L

σ(u n) = Rn + rk dℓkminuszk αjγ(u + czk + ℓ minus j n minus k)

k=1 ℓ=(kminuszk)c1 j=1

18

where the function γ(w m) for w m isin N is computed recursively via

minmna (kminusηwk)c2 u+cηwk+ℓ L L L

γ(w m) = Am + ak dℓkminusηwk αjγ(u + cηwk + ℓ minus j m minus k)

k=1 j=1 ℓ=(kminusηwk)c1

with starting values γ(w 0) = 1 forall w isin N

Acknowledgements

The authors would like to thank Dr David Landriault for his valuable suggestions which

helped to improve an earlier version of this work This research was supported by the Natural

Sciences and Engineering Research Council of Canada

References

Alfa AS 2004 Markov chain representations of discrete distributions applied to queueing

models Computers amp Operations Research 31 2365-2385

Alfa AS Drekic S 2007 Algorithmic analysis of the Sparre Andersen model in discrete

time ASTIN Bulletin 37 293-317

Bao Z 2007 A note on the compound binomial model with randomized dividend strategy

Applied Mathematics amp Computation 194 276-286

Cossette H Landriault D Marceau E 2006 Ruin probabilities in the discrete time

renewal risk model Insurance Mathematics amp Economics 38 309-323

Karlin S Taylor HM 1975 A First Course in Stochastic Processes 2nd edition Acashy

demic Press New York

Kim B Kim H-S Kim J 2008 A risk model with paying dividends and random

environment Insurance Mathematics amp Economics 42 717-726

Landriault D 2008 Randomized dividends in the compound binomial model with a general

premium rate Scandinavian Actuarial Journal 1-15

Tan J Yang X 2006 The compound binomial model with randomized decisions on

paying dividends Insurance Mathematics amp Economics 39 1-18

19

Table 1 ndash Values of Ψnxy(10) corresponding to Example 1

(a) x = 10

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0005357 0006188 0007163 0008278 0009481

(2) 0005359 0006195 0007175 0008296 0009507

(3) 0005361 0006198 0007181 0008305 0009519

(4) 0005363 0006205 0007193 0008323 0009544

(5) 0005366 0006212 0007208 0008346 0009576

(6) 0005374 0006236 0007249 0008410 0009664

y = 25 (1) 0007631 0008815 0010202 0011789 0013503

(2) 0007634 0008824 0010220 0011816 0013539

(3) 0007636 0008828 0010228 0011829 0013557

(4) 0007639 0008838 0010245 0011854 0013593

(5) 0007643 0008849 0010266 0011887 0013638

(6) 0007656 0008882 0010325 0011978 0013764

y = 50 (1) 0008477 0009792 0011333 0013095 0014998

(2) 0008481 0009802 0011352 0013125 0015039

(3) 0008483 0009807 0011361 0013139 0015059

(4) 0008486 0009817 0011380 0013168 0015098

(5) 0008491 0009830 0011403 0013204 0015149

(6) 0008504 0009866 0011470 0013305 0015288

y = infin (1) 0008777 0010138 0011733 0013558 0015528

(2) 0008781 0010149 0011753 0013588 0015570

(3) 0008783 0010154 0011763 0013603 0015591

(4) 0008787 0010164 0011782 0013633 0015632

(5) 0008791 0010177 0011807 0013671 0015684

(6) 0008805 0010215 0011875 0013775 0015828

20

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0058559 0061875 0065704 0070048 0074714

(2) 0058569 0061903 0065756 0070128 0074825

(3) 0058574 0061917 0065782 0070167 0074880

(4) 0058584 0061945 0065833 0070245 0074987

(5) 0058596 0061979 0065896 0070344 0075124

(6) 0058632 0062079 0066078 0070621 0075506

y = 25 (1) 0089002 0093986 0099739 0106262 0113270

(2) 0089017 0094028 0099817 0106382 0113437

(3) 0089025 0094049 0099856 0106442 0113519

(4) 0089040 0094091 0099932 0106560 0113681

(5) 0089057 0094142 0100028 0106708 0113887

(6) 0089113 0094294 0100302 0107127 0114463

y = 50 (1) 0102786 0108506 0115106 0122590 0130629

(2) 0102804 0108555 0115196 0122728 0130820

(3) 0102812 0108579 0115241 0122797 0130914

(4) 0102830 0108627 0115329 0122932 0131101

(5) 0102850 0108686 0115439 0123103 0131337

(6) 0102914 0108860 0115754 0123584 0132000

y = infin (1) 0108895 0114933 0121901 0129799 0138283

(2) 0108914 0114985 0121996 0129945 0138485

(3) 0108923 0115010 0122043 0130018 0138585

(4) 0108941 0115061 0122136 0130161 0138782

(5) 0108962 0115123 0122252 0130341 0139032

(6) 0109030 0115307 0122585 0130849 0139732

21

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0079856 0086792 009464 0103440 0112831

(2) 0079885 0086869 0094778 0103650 0113118

(3) 0079953 0086978 0094936 0103864 0113393

(4) 0080081 0087188 0095243 0104282 0113929

(5) 0079962 0087075 0095149 0104213 0113890

(6) 0080805 0088366 0096951 0106588 0116875

y = 25 (1) 0124591 0135680 0148207 0162245 0177217

(2) 0124639 0135806 0148433 0162586 0177683

(3) 0124755 0135993 0148703 0162950 0178148

(4) 0124978 0136354 0149226 0163658 0179054

(5) 0124763 0136141 0149035 0163500 0178935

(6) 0126238 0138381 0152148 0167588 0184060

y = 50 (1) 0146757 0160079 0175111 0191945 0209894

(2) 0146816 0160232 0175385 0192360 0210460

(3) 0146965 0160469 0175725 0192817 0211043

(4) 0147249 0160925 0176384 0193706 0212178

(5) 0146968 0160639 0176117 0193470 0211980

(6) 0148860 0163500 0180077 0198658 0218474

y = infin (1) 0158013 0172606 0189059 0207476 0227106

(2) 0158078 0172776 0189362 0207934 0227732

(3) 0158250 0173047 0189751 0208455 0228395

(4) 0158579 0173571 0190503 0209467 0229685

(5) 0158247 0173226 0190172 0209161 0229410

(6) 0160443 0176532 0194735 0215127 0236869

22

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0088667 0104403 0123329 0145277 0169184

(2) 0088711 0104518 0123538 0145598 0169627

(3) 0088733 0104574 0123641 0145756 0169846

(4) 0088777 0104687 0123844 0146066 0170274

(5) 0088829 0104828 0124105 0146470 0170832

(6) 0088988 0105229 0124822 0147560 0172331

y = 25 (1) 0140528 0167839 0201007 0239726 0282102

(2) 0140598 0168021 0201338 0240234 0282804

(3) 0140632 0168111 0201502 0240486 0283151

(4) 0140701 0168289 0201822 0240976 0283827

(5) 0140785 0168515 0202241 0241622 0284718

(6) 0141036 0169148 0203370 0243335 0287071

y = 50 (1) 0167826 0203005 0246173 0296954 0352855

(2) 0167908 0203218 0246561 0297548 0353672

(3) 0167949 0203324 0246753 0297841 0354075

(4) 0168030 0203532 0247128 0298412 0354859

(5) 0168129 0203800 0247622 0299174 0355907

(6) 0168423 0204539 0248936 0301159 0358620

y = infin (1) 0183325 0224926 0276881 0338957 0408257

(2) 0183412 0225150 0277283 0339564 0409079

(3) 0183455 0225260 0277481 0339864 0409485

(4) 0183540 0225478 0277869 0340446 0410270

(5) 0183645 0225760 0278386 0341234 0411339

(6) 0183954 0226529 0279737 0343244 0414037

23

Table 2 ndash Values of Ψnxy(10) corresponding to Example 2

(a) x = 10

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0001901 0002276 0002724 0003240 0003799

(2) 0001904 0002286 0002742 0003269 0003840

(3) 0002651 0003013 0003442 0003936 0004471

(4) 0002655 0003022 0003460 0003964 0004510

(5) 0005792 0006098 0006458 0006871 0007317

(6) 0005795 0006106 0006473 0006894 0007350

y = 25 (1) 0002708 0003243 0003880 0004615 0005411

(2) 0002713 0003257 0003906 0004656 0005469

(3) 0003778 0004293 0004904 0005608 0006369

(4) 0003782 0004306 0004929 0005647 0006424

(5) 0008255 0008690 0009203 0009792 0010427

(6) 0008259 0008702 0009224 0009824 0010473

y = 50 (1) 0003008 0003603 0004310 0005127 0006011

(2) 0003014 0003618 0004340 0005173 0006076

(3) 0004197 0004769 0005448 0006230 0007075

(4) 0004202 0004784 0005476 0006273 0007137

(5) 0009173 0009656 0010226 0010879 0011585

(6) 0009177 0009669 0010249 0010916 0011637

y = infin (1) 0003115 0003730 0004463 0005308 0006224

(2) 0003121 0003746 0004493 0005356 0006291

(3) 0004346 0004938 0005641 0006450 0007325

(4) 0004351 0004954 0005670 0006496 0007389

(5) 0009499 0009999 0010589 0011265 0011996

(6) 0009503 0010012 0010613 0011303 0012050

24

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0031176 0032815 0034733 0036924 0039282

(2) 0031193 0032862 0034821 0037061 0039475

(3) 0050168 0051747 0053587 0055684 0057939

(4) 0050184 0051792 0053671 0055815 0058124

(5) 0136884 0138219 0139763 0141517 0143399

(6) 0136896 0138256 0139832 0141626 0143553

y = 25 (1) 0047622 0050092 0052981 0056280 0059829

(2) 0004765 0050163 0053113 0056487 0060121

(3) 0076570 0078950 0081720 0084878 0088272

(4) 0076593 0079017 0081847 0085076 0088551

(5) 0205774 0207786 0210111 0212752 0215585

(6) 0205793 0207842 0210216 0212917 0215818

y = 50 (1) 0055170 0058009 0061329 0065119 0069196

(2) 0055200 0058091 0061481 0065357 0069533

(3) 0088650 0091385 0094569 0098197 0102096

(4) 0088677 0091463 0094715 0098425 0102417

(5) 0235815 0238128 0240800 0243834 0247089

(6) 0235837 0238192 0240920 0244023 0247356

y = infin (1) 0058565 0061565 0065073 0069077 0073384

(2) 0058597 0061652 0065234 0069329 0073740

(3) 0094061 0096952 0100316 0104149 0108267

(4) 0094090 0097035 0100470 0104390 0108607

(5) 0248472 0250917 0253740 0256945 0260383

(6) 0248495 0250985 0253867 0257145 0260667

25

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0048598 0052524 0056984 0061998 0067342

(2) 0048654 0052673 0057253 0062411 0067916

(3) 0070192 0073979 0078257 0083057 0088167

(4) 0070243 0074119 0078513 0083451 0088714

(5) 0151774 0154981 0158572 0162586 0166852

(6) 0151815 0155096 0158784 0162913 0167308

y = 25 (1) 0076895 0083230 0090413 0098481 0107073

(2) 0076987 0083474 0090854 0099156 0108010

(3) 0110093 0116205 0123095 0130818 0139033

(4) 0110177 0116435 0123515 0131462 0139928

(5) 0230674 0235848 0241632 0248091 0254950

(6) 0230739 0236038 0241980 0248626 0255695

y = 50 (1) 0091496 0099156 0107829 0117563 0127924

(2) 0091610 0099455 0108370 0118387 0129068

(3) 0130130 0137520 0145839 0155156 0165064

(4) 0130233 0137801 0146352 0155942 0166156

(5) 0266599 0272857 0279840 0287632 0295904

(6) 0266680 0273088 0280265 0288286 0296814

y = infin (1) 0099312 0107746 0117285 0127984 0139369

(2) 0099439 0108078 0117883 0128899 0140639

(3) 0140458 0148595 0157745 0167986 0178872

(4) 0140572 0148908 0158314 0168859 0180084

(5) 0282886 0289776 0297457 0306022 0315111

(6) 0282976 0290034 0297929 0306748 0316120

26

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0059096 0072505 0088907 0108122 0129137

(2) 0059223 0072829 0089500 0109052 0130456

(3) 0079757 0092662 0108388 0126778 0146871

(4) 0079875 0092970 0108956 0127669 0148134

(5) 0159788 0170682 0183871 0199249 0216024

(6) 0159881 0170936 0184343 0199990 0217075

y = 25 (1) 0095883 0119741 0149240 0184073 0222390

(2) 0096087 0120262 0150197 0185573 0224520

(3) 0127403 0150359 0178640 0211977 0248613

(4) 0127592 0150855 0179557 0213414 0250653

(5) 0245161 0264533 0288250 0316124 0346709

(6) 0245311 0264942 0289010 0317319 0348407

y = 50 (1) 0116598 0147920 0187100 0233784 0285503

(2) 0116839 0148537 0188231 0235554 0288015

(3) 0153023 0183153 0220713 0265390 0314840

(4) 0153249 0183742 0221798 0267087 0317248

(5) 0285742 0311159 0342651 0380005 0421287

(6) 0285921 0311644 0343551 0381418 0423292

y = infin (1) 0129470 0167217 0215383 0273841 0339734

(2) 0129725 0167865 0216560 0275661 0342282

(3) 0167978 0204275 0250442 0306382 0369383

(4) 0168217 0204895 0251572 0308128 0371826

(5) 0305868 0336470 0375169 0421935 0474525

(6) 0306058 0336980 0376106 0423388 0476560

27

Page 7: uwaterloo.ca · Ruin Analysis of a Threshold Strategy in a Discrete-time Sparre Andersen Model Steve DREKIC and Ana Maria MERA Department of Statistics and Actuarial Science University

and

minus1 minus2 minus3 middot middot middot

0 Bc+1 Bc+2 Bc+3 middot middot middot 1 Bc+2 Bc+3 Bc+4 middot middot middot

2 Bc+3 Bc+4 Bc+5 middot middot middot middot middot middot

Z minus 2 BZ+cminus1 BZ+c BZ+c+1 middot middot middot D = (25) Z minus 1 BZ+c BZ+c+1 BZ+c+2 middot middot middot

Z AZ+1 AZ+2 AZ+3 middot middot middot

Z + 1 AZ+2 AZ+3 AZ+4 middot middot middot Z + 2 AZ+3 AZ+4 AZ+5 middot middot middot

Let us now specify some further definitions required in upcoming derivations First of all

we introduce the function

0 if u ge Z zt =

maxi isin 1 2 t|u + c(i minus 1) lt Z if u lt Z

or equivalently

0 if u ge Z zt =

minlfloorZminuscuminus1 rfloor + 1 t if u lt Z

where the floor function denoted lfloorxrfloor yields the largest integer less than or equal to x We

define dm0 = δm0 where δij in general denotes the Kronecker delta function of i and j

Also define

dm if m = c1 c1 + 1 c2dm1 =

0 elsewhere

and for n = 2 3

Lc2 dj1dmminusjnminus1 if m = nc1 nc1 + 1 nc2dmn = j=c1

0 elsewhere

In other words dmn represents the pmf of the n-fold convolution of dm with itself

Recalling that W1 = k note that zk + ( k+) determines the earliest time point in the set =

0+ 1+ (k minus1)+ at which a random premium is received Note also that zk = k implies

that no such point exists in this set (ie the threshold level Z has not yet been reached when

the premium at time (k minus1)+ is to be received) Therefore if we assume for the moment that

zk k we may condition on the total premiums earned at times zk + (zk + 1)+ (k minus 1)+=

7

to obtain the following expression for b(k) the initial (ie starting at time k) probability row

vector of the above Markov chain corresponding only to the states in Δ1 namely

(kminuszk)c2

b(k) L

= dmkminuszk(αu+czk+me1 αu+czk+mminus1e1 α2e1 α1e1 0 0 ) (26)

m=(kminuszk)c1

where 0 denotes the 1 times na row vector of zeros By further defining the function vn(t) =

b(k)u + ct + (n minus t)c2 we note that the i-th level of is given by the 1 times na row vector L(kminuszk)c2 b(k)αu+czk+mminusie1 for each i isin Ωk = 0 1 vk(zk)minus1 Hence contains

m=(kminuszk)c1 dmkminuszk

zeros from level vk(zk) onwards Moreover (26) holds true even in the case when zk = k

simplifying to give b(k) = (αu+cke1 αu+ckminus1e1 α2e1 α1e1 0 0 )

At this juncture we introduce two additional row vectors namely

(k) (k) (k) (k) b(k)gn = (gn0 gn1 gn2 ) = Cn n isin N (27)

and

h(k) (k) (k) (k) b(k)Cnminus1D =

(k) Z

+ n = (h nminus2 hnminus3 ) = gnminus1D n isin (28) nminus1 h

(k)We remark that gn contains the probabilities of being in the various ldquonon-ruinedrdquo states

at time k + n (ie after claims have been paid out) without having visited a ldquoruinedrdquo state

during the previous n minus 1 transitions given that Uk isin Ωk sub Δ1 In a similar fashion h(nk)

contains the probabilities of being in the various ldquoruinedrdquo states for the first time at time

k + n given that Uk isin Ωk sub Δ1 Note that the probability of being in ruined state ldquo minus jrdquo is (k) (k) (k) (k)

characterized by the row vector h = (h h h ) with the third subscript nminusj nminusj1 nminusj2 nminusjna

representing the value of the elapsed time counter Lk+n Upon further reflection however (k)

=it must be the case that h = 0 for i 1 as ruin can only occur at claim instants nminusji

which in this case implies Lk+n = 1 with probability 1 Thus it immediately follows that (k) (k) (k)

the structure of the 1 times na row vector hnminusj is simply given by hnminusj = (φnj(u) 0 0) (k)

where φ (u) = Pr T = k + n |UT | = j | Consequently as in the analysis with nj Uk isin Ωk

no threshold on the surplus level (see Alfa and Drekic (2007 p 299)) we obtain

(k) (k) prime φ (u) = h (29) nj nminusje1

where e1 prime denotes the transpose of e1

A similar line of probabilistic logic can also be applied to obtain a representation for the (k)

trivariate function ψ (u) = Pr T = k + n UTminus = i |UT | = j | Uk isin Ωk In order for ruin nij

to occur at time k + n with a surplus immediately prior to ruin equal to i we observe that

(i) none of the previous n minus 1 transitions must have included a visit to any state in Δ2

and

8

(ii) the surplus level at time k + n minus 1 must necessarily be equal to i minus pk+nminus1

Note that pk+nminus1 denotes the premium corresponding to the time interval (k + n minus 1 k + n]

which is received at (k + n minus 1)+ Hence the quantity corresponding to points (i) and (ii) (k)

above is the 1 times na row vector g Then at the next time unit a claim must nminus1iminuspk+nminus1

necessarily occur but not before a premium of pk+nminus1 is first collected thereby raising the

surplus level to i Since s contains the absorption probabilities (to claim occurrence) from

the na possible phase states and the claim causing ruin must be of size i + j in order to

ensure that the deficit at ruin is equal to j it follows that

(k) (k)ψ (u) = g s αi+j (210) nij nminus1iminuspk+nminus1

(k)The following corollary provides the means of calculating the quantity g in nminus1iminuspk+nminus1

(210)

Corollary 1 If minc Z + c1 = c then

(k)

g if i = c c+1 Z +c1 minus1 nminus1iminusc

(k) (k) Lminc2iminusZ (k)g = g + j=c1

if i = Z +c1 Z +c1 +1 minmα Z +cminus1nminus1iminuspk+nminus1 nminus1iminusc djgnminus1iminusj

Lc2 (k)

j=c1 djgnminus1iminusj if i = Z +c Z +c+1 mαminus1

(211)

Conversely if minc Z + c1 = Z + c1 then

Lminc2iminusZ (k)

djg if i = Z +c1 Z +c1 +1 cminus1nminus1iminusj j=c1

(k) (k) Lminc2iminusZ (k)gnminus1iminuspk+nminus1

= gnminus1iminusc + j=c1 djgnminus1iminusj if i = c c+1 minmα Z +cminus1

Lc2 (k)djg if i = Z +c Z +c+1 mαminus1j=c1 nminus1iminusj

(212)

Proof The justification of (211) is as follows If c is smaller than Z + c1 the only way of

reaching a surplus level of i isin c c+1 Z +c1 minus1 (prior to ruin) is by receiving a premium

of c since i minus j for any j = c1 c1 +1 c2 would result in a surplus level below Z where the

random premium rate is not applicable Next for fixed i isin Z +c1 Z +c1 +1 Z +c minus1

there are two distinct ways of reaching this surplus level First of all note that the value of

i minus j is guaranteed to produce a surplus amount greater than or equal to Z provided that

j isin c1 c1 + 1 minc2 i minus Z Secondly note that reducing each of value of i by c also

yields a non-negative surplus value less than Z Hence both pure and random premiums

are plausible under this scenario For i isin Z + c Z + c + 1 note that decreasing this

amount by any j = c1 c1 + 1 c2 results in a surplus value greater than or equal to Z as

desired However reducing it by the amount c also has the same impact thus rendering the

pure premium option inadmissible in this case Lastly we incorporate into the range of i the

9

fact that the surplus immediately prior to ruin cannot exceed mα minus1 as indicated in Section

1 The justification of (212) when c is larger than Z + c1 mirrors the analysis above and is

left to the reader D

Recall that in the analysis up to this point we have assumed that W1 = k isin

1 2 3 nr If we now condition on the value of W1 the Law of Total Probability

yields the following two expressions for φnj(u) and ψnij(u) namely

nr L

= rkφ (u) n = nr + 1 nr + 2 φnj(u) n

(kminus) kj

k=1

and nr L

(k)ψnij(u) = rkψ (u) n = nr + 1 nr + 2 nminuskij

k=1

In addition however it is quite possible that T = n for n = 1 2 nr and this can happen

as a result of either

(i) the first claim which causes ruin occurring at time nminus from a (conditional) surplus Lnminus1level of u + czn + m (where m) or i=zn

Xi =

(ii) the first claim which does not cause ruin occurring at some time kminus k isin 1 2 nminus

1 and ruin subsequently occurring n minus k time units later

If we now combine these outcomes with those for n = nr + 1 nr + 2 (and recall that

rn = 0 forall n gt nr) the general expression for φnj(u) becomes

minnminus1nr (nminuszn)c2 L

(k) L

φnj(u)= rkφ (u) + rn dmnminuszn αu+czn+m+j n isin Z+ (213) nminuskj

k=1 m=(nminuszn)c1

Concerning the function ψnij(u) when n le nr and point (i) above is considered we note Lnminus1that i must necessarily equal u + czn + m (again conditional on i=zn

Xi = m) for ruin to

occur in this manner Hence in a similar fashion we acquire

minnminus1nr (nminuszn)c2 L L

ψnij(u) = rkψn

(kminus) kij(u) + rn dmnminuszn

δiu+czn+mαu+czn+m+j

k=1 m=(nminuszn)c1

minnminus1nr L

(k) Z

+ = rkψnminuskij(u) + rndiminusuminuscznnminuszn αi+j n isin (214)

k=1

Finally for fixed i isin minc Z + c1 minc Z + c1 + 1 mα minus 1 summing (214) from

j = 1 to mα minus i yields the following expression for the bivariate pmf ωni(u) namely

minnminus1nr mαminusi mαminusi L L

(k) L

ωni(u) = rk ψnminuskij(u) + rndiminusuminuscznnminusznαi+j

k=1 j=1 j=1

10

minnminus1nr L

(k) Z

+ = Λi rkg s + rndiminusuminuscznnminuszn n isin (215) nminuskminus1iminuspnminus1

k=1

where (210) was used to establish the last equality In the special case when c1 = c2 = c it

is a straightforward exercise to show that equations (213) through (215) simplify to produce

equations (215) through (217) of Alfa and Drekic (2007 p 300)

3 Computational considerations

In order to compute the joint pmfrsquos defined by (213) (214) and (215) it is clear from (k)

equations (29) through (212) that we need to be able to compute the quantities gn and

h(k) n defined by (27) and (28) respectively In what follows we capitalize on the underlying

Markov chain structure in our model to develop a set of computational procedures which

ultimately enable one to calculate these key row vectors

First of all recall that g0(k)

= b(k) contains zeros from a certain level onwards namely (k)

level vk(zk) = u + czk + c2(k minus zk) In general one can easily deduce that gn will also

contain zeros from a certain level onwards to be denoted by lk(n) We begin this section

with a determination of lk(n) which varies in form depending on the value of vk(zk) minus 1

representing the maximum surplus level at time k after the smallest possible claim of size 1

has been applied (at time kminus)

On one hand it may be the case that the value of vk(zk)minus1 lies at or above the threshold

level Z In this situation we remark that zk can take on any value in the set 0 1 k (k)

Since g0 contains zeros from level lk(0) = vk(zk) onwards we observe that at time k + 1

it may be possible to achieve a higher surplus value than vk+1(zk) minus 1 representing the

maximum surplus level assuming there was no drop below Z due to the claim at time kminus

Specifically if the surplus had fallen to level Z minus 1 at time kminus then gained c at time k+ a

bounded level of Z + c minus1 would have been reached at time k +1 and this level is potentially

larger than vk+1(zk) minus 1 As a result we have that lk(1) = maxvk+1(zk) Z + c If we

extend this argument inductively we readily obtain lk(n) = maxvk+n(zk) Z +c +c2(n minus1)

n isin Z+ However since vk+n(zk) = u + czk + c2(k +n minus zk) = vk(zk)+ c2 + c2(n minus 1) we may

alternatively re-express lk(n) as

lk(n) = maxvk+n(zk) Z + c + c2(n minus 1) = maxvk(zk) + c2 Z + c + c2(n minus 1)

Thus

Z+lk(n) = maxlk(0) + c2 Z + c + c2(n minus 1) n isin (31)

11

or equivalently

Z+lk(n) = lk(1) + c2(n minus 1) n isin

Clearly lk(0) gt Z and lk(n) ge Z + c for n isin Z+

The other remaining case to consider involves the situation when the maximum surplus

level at time k is strictly below the threshold level Z Under this scenario we further

distinguish between two cases involving the value of c2 First of all suppose that c2 = 0 so

that vk(zk)minus1 = u +czk minus1 lt Z or equivalently u +czk le Z If u +czk happens to equal Z

then it is possible for zk isin 0 1 k With c2 = 0 however no further premiums can be

earned until after the first claim takes place (which is assumed to occur at time kminus) thereby

implying that Z = u + ci for all i = zk zk + 1 k Hence we know that u + ck = Z

in this case Moreover if u + czk happens to be strictly less than Z then we clearly have

that zk = k Even in the case when c2 gt 0 one can readily verify that zk = k so that

vk(zk) minus 1 = u + ck minus 1 lt Z Thus we proceed with the analysis assuming that zk = k

c2 ge 0 and the inequality u + ck le Z holds true

lowast k

Define the ceiling function denoted lceilxrceil to yield the smallest integer greater than or

equal to x Because a random premium cannot be received until level Z is at least achieved

it is essential to find the point at which the threshold level Z is reached (ie the point at Zminus(u+ck)+1 which u + ck minus1+ ci ge Z) Isolating this inequality for i one obtains i ge

c Hence

assuming that W1 = k the elapsed time after k at which a random premium would begin to

lceilZminus(u+ck)+1 be received is t lowast = rceil Note that t lowast isin Z+ k c k

With zk = k g0(k)

contains zeros from level u + ck onwards so that lk(0) = u + ck

Immediately following this a premium of c is guaranteed to be received at time k+ as the

threshold level Z has not yet been reached Consequently g1(k)

contains zeros from level (k)

b(k)u + c(k + 1) onwards which is c levels further than that in g0 = Continuing this (k)

process inductively we can readily establish that gn contains zeros from level u + c(k + n)

onwards for n = 1 2 t lowast k so that lk(n) = u + c(k + n) for n = 0 1 t lowast k

In the case of n = t lowast + 1 t lowast + 2 gt

(k) is first looked at in isolation At time k +k k +1

t lowast k + 1 the threshold level has been crossed giving rise to a maximum surplus value of either (k)

c(k + t lowast k contains zeros from level lk(t lowast ku + ) + c2 minus 1 or Z + c minus 1 Therefore g + 1) = t lowast k+1

+1(k + t lowast kmaxvk+t lowast k

) Z + c onwards Carrying out the same inductive process as before it is

readily seen that g(nk)

contains zeros from level lk(n) = maxvk+n(k+t lowast k) Z +c+c2(nminust lowast kminus1)

onwards for n = t lowast k + 1 t lowast k + 2 Combining both cases we obtain

u + c(k + n) if n = 0 1 t lowast lk(n) = k (32)

maxlk(t lowast) + c2 Z + c + c2(n minus t lowast minus 1) if n = t lowast + 1 t lowast + 2 k k k k

Note that by construction lk(t lowast k) gt Z and lk(n) le Z for n = 0 1 t lowast k minus1 Interestingly

12

if we extend the definition tk lowast by introducing the quantity ζk

lowast = maxtklowast 0 we remark that

(31) and (32) can actually be combined into a single formula for lk(n) namely

vk+n(zk + n) if n = 0 1 ζklowast

lk(n) = (33) maxlk(ζ

lowast) + c2 Z + c + c2(n minus ζlowast minus 1) if n = ζlowast + 1 ζlowast + 2 k k k k

Clearly we have that lk(n) ge Z + c for n = ζk lowast +1 ζk

lowast +2 Moreover if c1 = c2 = c (33)

reduces to give lk(n) = u + c(k + n) forall n isin N which is in agreement with the results in Alfa

and Drekic (2007 Section 3)

Remark In our above analysis we purposely ignored the possibility of any future claims

occurring after time k in our determination of lk(n) However we recognize that this is not

realistic and future claims would indeed occur after time k thereby further reducing the value

of the maximum surplus level lk(n) minus 1 at time k + n To account for this we imagine all

future interclaim times (ie after time k) to be of maximum length na and all corresponding

claim sizes to be of minimum amount 1 Consequently in the case when c2 gt 0 we should

subtract from formula (33) the amount of lfloornnarfloor where lfloormiddotrfloor again denotes the floor function

Obviously when c2 = 0 (ie no premiums are collected whenever the surplus level is at or

above Z) lk(n) = Z + c for n = ζk lowast + 1 ζk

lowast + 2 (k)

We now turn our attention to the development of a general algorithm for computing gni

(k) (k)for i = 0 1 lk(n) minus 1 With the aid of (24) we apply the key identity gn = gnminus1C

(which can be inferred from (27)) to first obtain

minZlk(nminus1)minus1 lk(nminus1)minus1 c2 L L L

(k) (k) (k)gni = g Bjminusi+c + gnminus1j dmBjminusi+m n isin Z+ (34) nminus1j

j=0 j=Z m=c1

However (22) states that Bi = Ona for i isin Zminus and so (34) becomes (for the same range of

i)

minZlk(nminus1)minus1 c2 lk(nminus1)minus1 L L L

(k) (k) (k)gni = g Bjminusi+c + dm gnminus1jBjminusi+m n isin Z+ (35) nminus1j

j=max0iminusc m=c1 j=maxZiminusm

Taking the form of (33) into consideration (35) yields the following two formulae

lk(nminus1)minus1 L

(k) (k)gni = gnminus1jBjminusi+c n = 1 2 ζ k

lowast (36) j=max0iminusc

and

Zminus1 c2 lk(nminus1)minus1 L L L

(k) (k) (k)g = g + dm g n = ζ lowast + 1 ζ lowast + 2 ni nminus1jBjminusi+c nminus1jBjminusi+m k k

j=max0iminusc m=c1 j=maxZiminusm

(37)

13

For n = 1 2 ζk lowast where ζk

lowast isin Z+ we note that lk(n) ge c + 1 Therefore with the aid of

(22) (36) simplifies to give

Llk(nminus1)minus1 (k) (k) j=0 gnminus1jBjminusi+c if i = 0 1 c minus 1

gni = Llk(nminus1)minus1 (k)

g if i = c c + 1 lk(n) minus 1j=iminusc nminus1jBjminusi+c

Llk(nminus1)minus1 (k)

nminus1js 0 0 if i = 0 1 c minus 1j=0 αjminusi+cg= (38)

(k) Llk(nminus1)minus1 (k) g S+ αjminusi+cgnminus1js 0 0 if i = c c+1 lk(n)minus1 nminus1iminusc j=iminusc+1

In the case of n = ζk lowast + 1 ζk

lowast + 2 we first reiterate that lk(n) ge Z + c We next break

(37) into several cases based on the value of i and the value of minc Z + c1 as follows

Case 1 Suppose minc Z + c1 = c

For i isin 0 1 c minus 1

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+mni nminus1j nminus1j

j=0 m=c1 j=Z

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k)= αjminusi+cg s + dm αjminusi+mg s 0 0 (39) nminus1j nminus1j

j=0 m=c1 j=Z

For i isin c c + 1 Z + c1 minus 1

Zminus1 c2 lk(nminus1)minus1 L L L

(k) (k) (k)g = g + dm gni nminus1j Bjminusi+c nminus1j Bjminusi+m

j=iminusc m=c1 j=Z

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k) (k)= g S + αjminusi+cg dm αjminusi+mgnminus1j s 0 0 (310) nminus1iminusc nminus1j s +

j=iminusc+1 m=c1 j=Z

For i isin Z + c1 Z + c1 + 1 Z + c minus 1

Zminus1 minc2iminusZ lk(nminus1)minus1 c2 lk(nminus1)minus1 L L L L L

(k) (k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+m + dm g Bjminusi+mni nminus1j nminus1j nminus1j

j=iminusc m=c1 j=iminusm m=iminusZ+1 j=Z

minc2iminusZ L

(k) (k)= g S + dmg S nminus1iminusc nminus1iminusm

m=c1

Zminus1 minc2iminusZ lk(nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k)+ αjminusi+cg dm αjminusi+mg dm αjminusi+mgnminus1j s 0 0 nminus1j s + nminus1js +

j=iminusc+1 m=c1 j=iminusm+1 m=iminusZ+1 j=Z

(311)

For i isin Z + c Z + c + 1 lk(n) minus 1

c2 lk(nminus1)minus1 L L

(k) (k)g = dm g Bjminusi+mni nminus1j

m=c1 j=iminusm

c2 c2 lk (nminus1)minus1 L L L

(k) (k)= dmg S + dm αjminusi+mgnminus1j s 0 0 (312) nminus1iminusm

m=c1 m=c1 j=iminusm+1

14

Case 2 Suppose minc Z + c1 = Z + c1

For i isin 0 1 Z + c1 minus 1

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+mni nminus1j nminus1j

j=0 m=c1 j=Z

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k)= αjminusi+cg s + dm αjminusi+mg s 0 0 (313) nminus1j nminus1j

j=0 m=c1 j=Z

For i isin Z + c1 Z + c1 + 1 c minus 1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+m + dm g Bjminusi+mni nminus1j nminus1j nminus1j

j=0 m=c1 j=iminusm m=iminusZ+1 j=Z

minc2iminusZ L

(k)= dmg S nminus1iminusm

m=c1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k)+ αjminusi+cg s + dm αjminusi+mg s + dm αjminusi+mg s 0 0 nminus1j nminus1j nminus1j

j=0 m=c1 j=iminusm+1 m=iminusZ+1 j=Z

(314)

For i isin c c + 1 Z + c minus 1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k) (k)g = g + g + gni nminus1j Bjminusi+c dm nminus1j Bjminusi+m dm nminus1j Bjminusi+m

j=iminusc m=c1 j=iminusm m=iminusZ+1 j=Z

minc2iminusZ L

(k) (k)= g S + dmg S nminus1iminusc nminus1iminusm

m=c1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k)+

nminus1j s 0 0 αjminusi+cgnminus1j s + dm αjminusi+mgnminus1j s + dm αjminusi+mg j=iminusc+1 m=c1 j=iminusm+1 m=iminusZ+1 j=Z

(315)

For i isin Z + c Z + c + 1 lk(n) minus 1

c2 lk (nminus1)minus1 L L

(k) (k)g = dm g Bjminusi+mni nminus1j

m=c1 j=iminusm

c2 c2 lk (nminus1)minus1 L L L

(k) (k)= dmg S + dm αjminusi+mgnminus1j s 0 0 (316) nminus1iminusm

m=c1 m=c1 j=iminusm+1

It is not difficult to verify that equations (38) through (316) simplify to produce equations

(32) and (33) of Alfa and Drekic (2007 pp 301-302) when c1 = c2 = c With equations (38) (k)

through (316) providing the recursive means of calculating gn we next turn our attention

15

to a determination of h(k) defined by (28) For n = 1 2 ζlowast we use (22) and (25) to n k

obtain

lk(nminus1)minus1 lk(nminus1)minus1

h(k) nminusj =

L g

(k) nminus1ℓBc+j+ℓ =

L αc+j+ℓg

(k) nminus1ℓs 0 0 (317)

ℓ=0 ℓ=0

For n = ζk lowast + 1 ζk

lowast + 2 we similarly acquire

Zminus1 lk(nminus1)minus1 L L

(k) (k) (k)hnminusj = gnminus1ℓBc+j+ℓ + gnminus1ℓAj+ℓ

ℓ=0 ℓ=Z

Zminus1 c2 lk(nminus1)minus1 L L L

(k) (k)= αc+j+ℓg s + dm αm+j+ℓg s 0 0 (318) nminus1ℓ nminus1ℓ

ℓ=0 m=c1 ℓ=Z

Note that substituting (317) and (318) into (29) immediately yields

Llk(nminus1)minus1 (k) (k) ℓ=0 αc+j+ℓgnminus1ℓs if n = 1 2 ζk

lowast φ (u) = nj LZminus1 (k) Lc2

Llk(nminus1)minus1 (k) ζlowast s + dm αm+j+ℓg n = + 1 ζlowast + 2 ℓ=0 αc+j+ℓgnminus1ℓ m=c1 ℓ=Z nminus1ℓs if k k

(319)

Clearly in the special case when c1 = c2 = c (319) yields equation (35) of Alfa and Drekic

(2007 p 302) Finally we remark that (319) can be substituted into (213) to obtain a

computational formula for φnj(u)

4 Numerical examples and findings

In this section we illustrate the application of our proposed algorithm with several numerical

examples All calculations in this section were performed using Mathematica (Version 7) on

a Dell Precision T7400 workstation with a 3 Ghz CPU and 4 GB of RAM

In our first illustration (labelled Example 1) we consider the ordinary Sparre Andersen

model in which interclaim times have pmf given by

(211)(911)jminus1 if j = 1 2 na minus 1 aj = (41)

(911)naminus1 if j = na

Note that the pmf (41) is that of a truncated geometric distribution with all the probability

mass on na na +1 assigned to the support value na Clearly as na becomes larger the

closer (41) approximates this particular geometric distribution having mean 55 In what Lna

Lminus1 nafollows we fix na = 25 so that aj = 0992 and jaj = 546 Let the individual j=1 j=1

claim amount distribution be given by the pmf

minus4 minus4j minus 1 j

αj = 1 + minus 1 + j isin Z+ (42) 30 30

16

We remark that the pmf (42) represents a discretized version of the Pareto distribution with

mean 10 Also note that mα = infin thereby implying that the random variables |UT | and

UTminus both have infinite support

Table 1 displays values (rounded to 6 decimal places) of the trivariate cumulative distrishy

bution function

n x y L L L

Ψnxy(u) = Pr T le n UTminus le x |UT | le y | U0 = u = ψℓij(u) ℓ=1 i=mincZ+c1 j=1

corresponding to the above discrete-time risk process with u = 10 Z = 50 c = 5 and

varying premium amount distributions according to the following set of scenarios

(1) d2 = 1

(2) d1 = 25 d2 = 15 d3 = 25

G

5)

(3) di = i

(25)i(35)5minusi i = 0 1 5

(4) di = 15 i = 0 1 4

(5) d1 = 34 d5 = 14

(6) d0 = 35 d5 = 25

We remark that while these six premium amount distributions all have mean equal to 2

they possess different shapes and we investigate the impact of their variability on the joint

distribution of T |UT | and UTminus In particular the variances corresponding to the premium

amount distributions of scenarios (1) through (6) are 0 08 12 2 3 and 6 respectively

Based on the results in Table 1 it would appear that greater variability in the premium

amount distribution generally leads to higher ruin probabilities Perhaps somewhat surprising

though the differences are subtle and not particularly striking as one compares the various

ruin probabilities from scenarios (1) through (6) Obviously we observe more of a difference

between scenarios as the values of n x and y increase

As a follow-up illustration (labelled Example 2) we consider a model in which ordinary

interclaim times have a discrete uniform distribution with pmf aj = 110 j = 1 2 10

Clearly the mean of this distribution is 55 which for comparative purposes is identical to

the ordinary interclaim time mean considered in Example 1 In addition we assume that

the pmf of the individual claim amount distribution is once again given by (42) Table 2

displays values (rounded to 6 decimal places) of Ψnxy(10) for several variants of the Sparre

Andersen model (again with Z = 50 and c = 5) having the above ordinary interclaim time

and claim amount distributions namely

17

(1) ordinary model under scenario (2) from Example 1

(2) ordinary model under scenario (5) from Example 1

(3) stationary model under scenario (2) from Example 1

(4) stationary model under scenario (5) from Example 1

(5) delayed model (with W1 = 1 with probability 1) under scenario (2) from Example 1

(6) delayed model (with W1 = 1 with probability 1) under scenario (5) from Example 1

Concerning the results in Table 2 we first compare the values of Ψnxy(10) in entries (2)

and (5) of Table 1 to their corresponding counterparts in entries (1) and (2) of Table 2 In all

cases we observe that ruin is more likely to occur when interclaim times have the truncated

geometric distribution introduced in our first example as opposed to the discrete uniform

distribution of our second example Since mean interclaim times are equal in both examshy

ples this behaviour must be attributable to the greater variability inherent in the truncated

geometric distribution We also note that under the stationary model assumption the pmf

of W1 is easily verified to be rj = (11 minus j)55 j = 1 2 10 which has mean 4 lt 55

As a result the values of Ψnxy(10) for stationary models (3) and (4) are always larger than

the corresponding values for ordinary models (1) and (2) Finally in the absence of any

knowledge regarding which distribution is in effect for the first interclaim time W1 models

(5) and (6) can be regarded as ldquoworst-case scenariordquo models as they serve to provide upper

bounds on the values of Ψnxy(10)

In conclusion we state that the values in Tables 1 and 2 were generated by first impleshy

menting the recursive procedures (38) through (316) and then summing the appropriate

bivariatetrivariate probabilities computed via (213) (214) or (215) In situations when

x = y = infin however we computed the resulting finite-time ruin probabilities using an altershy

native (albeit standard) method (eg see Cossette et al (2006 Section 3)) Specifically if

we define the survival probability σ(n u) = Pr T gt n | U0 = u = 1minusΨninfininfin(u) as well as

the auxiliary function

0 if w ge Z ηwt =

minlfloorZminuswc minus1 rfloor + 1 t if w lt Z

then by conditioning on the time and amount of the first claim (and recalling that the delayed

process reverts to the ordinary process following this claim) we obtain the expression

minnnr (kminuszk)c2 u+czk+ℓ L L L

σ(u n) = Rn + rk dℓkminuszk αjγ(u + czk + ℓ minus j n minus k)

k=1 ℓ=(kminuszk)c1 j=1

18

where the function γ(w m) for w m isin N is computed recursively via

minmna (kminusηwk)c2 u+cηwk+ℓ L L L

γ(w m) = Am + ak dℓkminusηwk αjγ(u + cηwk + ℓ minus j m minus k)

k=1 j=1 ℓ=(kminusηwk)c1

with starting values γ(w 0) = 1 forall w isin N

Acknowledgements

The authors would like to thank Dr David Landriault for his valuable suggestions which

helped to improve an earlier version of this work This research was supported by the Natural

Sciences and Engineering Research Council of Canada

References

Alfa AS 2004 Markov chain representations of discrete distributions applied to queueing

models Computers amp Operations Research 31 2365-2385

Alfa AS Drekic S 2007 Algorithmic analysis of the Sparre Andersen model in discrete

time ASTIN Bulletin 37 293-317

Bao Z 2007 A note on the compound binomial model with randomized dividend strategy

Applied Mathematics amp Computation 194 276-286

Cossette H Landriault D Marceau E 2006 Ruin probabilities in the discrete time

renewal risk model Insurance Mathematics amp Economics 38 309-323

Karlin S Taylor HM 1975 A First Course in Stochastic Processes 2nd edition Acashy

demic Press New York

Kim B Kim H-S Kim J 2008 A risk model with paying dividends and random

environment Insurance Mathematics amp Economics 42 717-726

Landriault D 2008 Randomized dividends in the compound binomial model with a general

premium rate Scandinavian Actuarial Journal 1-15

Tan J Yang X 2006 The compound binomial model with randomized decisions on

paying dividends Insurance Mathematics amp Economics 39 1-18

19

Table 1 ndash Values of Ψnxy(10) corresponding to Example 1

(a) x = 10

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0005357 0006188 0007163 0008278 0009481

(2) 0005359 0006195 0007175 0008296 0009507

(3) 0005361 0006198 0007181 0008305 0009519

(4) 0005363 0006205 0007193 0008323 0009544

(5) 0005366 0006212 0007208 0008346 0009576

(6) 0005374 0006236 0007249 0008410 0009664

y = 25 (1) 0007631 0008815 0010202 0011789 0013503

(2) 0007634 0008824 0010220 0011816 0013539

(3) 0007636 0008828 0010228 0011829 0013557

(4) 0007639 0008838 0010245 0011854 0013593

(5) 0007643 0008849 0010266 0011887 0013638

(6) 0007656 0008882 0010325 0011978 0013764

y = 50 (1) 0008477 0009792 0011333 0013095 0014998

(2) 0008481 0009802 0011352 0013125 0015039

(3) 0008483 0009807 0011361 0013139 0015059

(4) 0008486 0009817 0011380 0013168 0015098

(5) 0008491 0009830 0011403 0013204 0015149

(6) 0008504 0009866 0011470 0013305 0015288

y = infin (1) 0008777 0010138 0011733 0013558 0015528

(2) 0008781 0010149 0011753 0013588 0015570

(3) 0008783 0010154 0011763 0013603 0015591

(4) 0008787 0010164 0011782 0013633 0015632

(5) 0008791 0010177 0011807 0013671 0015684

(6) 0008805 0010215 0011875 0013775 0015828

20

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0058559 0061875 0065704 0070048 0074714

(2) 0058569 0061903 0065756 0070128 0074825

(3) 0058574 0061917 0065782 0070167 0074880

(4) 0058584 0061945 0065833 0070245 0074987

(5) 0058596 0061979 0065896 0070344 0075124

(6) 0058632 0062079 0066078 0070621 0075506

y = 25 (1) 0089002 0093986 0099739 0106262 0113270

(2) 0089017 0094028 0099817 0106382 0113437

(3) 0089025 0094049 0099856 0106442 0113519

(4) 0089040 0094091 0099932 0106560 0113681

(5) 0089057 0094142 0100028 0106708 0113887

(6) 0089113 0094294 0100302 0107127 0114463

y = 50 (1) 0102786 0108506 0115106 0122590 0130629

(2) 0102804 0108555 0115196 0122728 0130820

(3) 0102812 0108579 0115241 0122797 0130914

(4) 0102830 0108627 0115329 0122932 0131101

(5) 0102850 0108686 0115439 0123103 0131337

(6) 0102914 0108860 0115754 0123584 0132000

y = infin (1) 0108895 0114933 0121901 0129799 0138283

(2) 0108914 0114985 0121996 0129945 0138485

(3) 0108923 0115010 0122043 0130018 0138585

(4) 0108941 0115061 0122136 0130161 0138782

(5) 0108962 0115123 0122252 0130341 0139032

(6) 0109030 0115307 0122585 0130849 0139732

21

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0079856 0086792 009464 0103440 0112831

(2) 0079885 0086869 0094778 0103650 0113118

(3) 0079953 0086978 0094936 0103864 0113393

(4) 0080081 0087188 0095243 0104282 0113929

(5) 0079962 0087075 0095149 0104213 0113890

(6) 0080805 0088366 0096951 0106588 0116875

y = 25 (1) 0124591 0135680 0148207 0162245 0177217

(2) 0124639 0135806 0148433 0162586 0177683

(3) 0124755 0135993 0148703 0162950 0178148

(4) 0124978 0136354 0149226 0163658 0179054

(5) 0124763 0136141 0149035 0163500 0178935

(6) 0126238 0138381 0152148 0167588 0184060

y = 50 (1) 0146757 0160079 0175111 0191945 0209894

(2) 0146816 0160232 0175385 0192360 0210460

(3) 0146965 0160469 0175725 0192817 0211043

(4) 0147249 0160925 0176384 0193706 0212178

(5) 0146968 0160639 0176117 0193470 0211980

(6) 0148860 0163500 0180077 0198658 0218474

y = infin (1) 0158013 0172606 0189059 0207476 0227106

(2) 0158078 0172776 0189362 0207934 0227732

(3) 0158250 0173047 0189751 0208455 0228395

(4) 0158579 0173571 0190503 0209467 0229685

(5) 0158247 0173226 0190172 0209161 0229410

(6) 0160443 0176532 0194735 0215127 0236869

22

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0088667 0104403 0123329 0145277 0169184

(2) 0088711 0104518 0123538 0145598 0169627

(3) 0088733 0104574 0123641 0145756 0169846

(4) 0088777 0104687 0123844 0146066 0170274

(5) 0088829 0104828 0124105 0146470 0170832

(6) 0088988 0105229 0124822 0147560 0172331

y = 25 (1) 0140528 0167839 0201007 0239726 0282102

(2) 0140598 0168021 0201338 0240234 0282804

(3) 0140632 0168111 0201502 0240486 0283151

(4) 0140701 0168289 0201822 0240976 0283827

(5) 0140785 0168515 0202241 0241622 0284718

(6) 0141036 0169148 0203370 0243335 0287071

y = 50 (1) 0167826 0203005 0246173 0296954 0352855

(2) 0167908 0203218 0246561 0297548 0353672

(3) 0167949 0203324 0246753 0297841 0354075

(4) 0168030 0203532 0247128 0298412 0354859

(5) 0168129 0203800 0247622 0299174 0355907

(6) 0168423 0204539 0248936 0301159 0358620

y = infin (1) 0183325 0224926 0276881 0338957 0408257

(2) 0183412 0225150 0277283 0339564 0409079

(3) 0183455 0225260 0277481 0339864 0409485

(4) 0183540 0225478 0277869 0340446 0410270

(5) 0183645 0225760 0278386 0341234 0411339

(6) 0183954 0226529 0279737 0343244 0414037

23

Table 2 ndash Values of Ψnxy(10) corresponding to Example 2

(a) x = 10

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0001901 0002276 0002724 0003240 0003799

(2) 0001904 0002286 0002742 0003269 0003840

(3) 0002651 0003013 0003442 0003936 0004471

(4) 0002655 0003022 0003460 0003964 0004510

(5) 0005792 0006098 0006458 0006871 0007317

(6) 0005795 0006106 0006473 0006894 0007350

y = 25 (1) 0002708 0003243 0003880 0004615 0005411

(2) 0002713 0003257 0003906 0004656 0005469

(3) 0003778 0004293 0004904 0005608 0006369

(4) 0003782 0004306 0004929 0005647 0006424

(5) 0008255 0008690 0009203 0009792 0010427

(6) 0008259 0008702 0009224 0009824 0010473

y = 50 (1) 0003008 0003603 0004310 0005127 0006011

(2) 0003014 0003618 0004340 0005173 0006076

(3) 0004197 0004769 0005448 0006230 0007075

(4) 0004202 0004784 0005476 0006273 0007137

(5) 0009173 0009656 0010226 0010879 0011585

(6) 0009177 0009669 0010249 0010916 0011637

y = infin (1) 0003115 0003730 0004463 0005308 0006224

(2) 0003121 0003746 0004493 0005356 0006291

(3) 0004346 0004938 0005641 0006450 0007325

(4) 0004351 0004954 0005670 0006496 0007389

(5) 0009499 0009999 0010589 0011265 0011996

(6) 0009503 0010012 0010613 0011303 0012050

24

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0031176 0032815 0034733 0036924 0039282

(2) 0031193 0032862 0034821 0037061 0039475

(3) 0050168 0051747 0053587 0055684 0057939

(4) 0050184 0051792 0053671 0055815 0058124

(5) 0136884 0138219 0139763 0141517 0143399

(6) 0136896 0138256 0139832 0141626 0143553

y = 25 (1) 0047622 0050092 0052981 0056280 0059829

(2) 0004765 0050163 0053113 0056487 0060121

(3) 0076570 0078950 0081720 0084878 0088272

(4) 0076593 0079017 0081847 0085076 0088551

(5) 0205774 0207786 0210111 0212752 0215585

(6) 0205793 0207842 0210216 0212917 0215818

y = 50 (1) 0055170 0058009 0061329 0065119 0069196

(2) 0055200 0058091 0061481 0065357 0069533

(3) 0088650 0091385 0094569 0098197 0102096

(4) 0088677 0091463 0094715 0098425 0102417

(5) 0235815 0238128 0240800 0243834 0247089

(6) 0235837 0238192 0240920 0244023 0247356

y = infin (1) 0058565 0061565 0065073 0069077 0073384

(2) 0058597 0061652 0065234 0069329 0073740

(3) 0094061 0096952 0100316 0104149 0108267

(4) 0094090 0097035 0100470 0104390 0108607

(5) 0248472 0250917 0253740 0256945 0260383

(6) 0248495 0250985 0253867 0257145 0260667

25

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0048598 0052524 0056984 0061998 0067342

(2) 0048654 0052673 0057253 0062411 0067916

(3) 0070192 0073979 0078257 0083057 0088167

(4) 0070243 0074119 0078513 0083451 0088714

(5) 0151774 0154981 0158572 0162586 0166852

(6) 0151815 0155096 0158784 0162913 0167308

y = 25 (1) 0076895 0083230 0090413 0098481 0107073

(2) 0076987 0083474 0090854 0099156 0108010

(3) 0110093 0116205 0123095 0130818 0139033

(4) 0110177 0116435 0123515 0131462 0139928

(5) 0230674 0235848 0241632 0248091 0254950

(6) 0230739 0236038 0241980 0248626 0255695

y = 50 (1) 0091496 0099156 0107829 0117563 0127924

(2) 0091610 0099455 0108370 0118387 0129068

(3) 0130130 0137520 0145839 0155156 0165064

(4) 0130233 0137801 0146352 0155942 0166156

(5) 0266599 0272857 0279840 0287632 0295904

(6) 0266680 0273088 0280265 0288286 0296814

y = infin (1) 0099312 0107746 0117285 0127984 0139369

(2) 0099439 0108078 0117883 0128899 0140639

(3) 0140458 0148595 0157745 0167986 0178872

(4) 0140572 0148908 0158314 0168859 0180084

(5) 0282886 0289776 0297457 0306022 0315111

(6) 0282976 0290034 0297929 0306748 0316120

26

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0059096 0072505 0088907 0108122 0129137

(2) 0059223 0072829 0089500 0109052 0130456

(3) 0079757 0092662 0108388 0126778 0146871

(4) 0079875 0092970 0108956 0127669 0148134

(5) 0159788 0170682 0183871 0199249 0216024

(6) 0159881 0170936 0184343 0199990 0217075

y = 25 (1) 0095883 0119741 0149240 0184073 0222390

(2) 0096087 0120262 0150197 0185573 0224520

(3) 0127403 0150359 0178640 0211977 0248613

(4) 0127592 0150855 0179557 0213414 0250653

(5) 0245161 0264533 0288250 0316124 0346709

(6) 0245311 0264942 0289010 0317319 0348407

y = 50 (1) 0116598 0147920 0187100 0233784 0285503

(2) 0116839 0148537 0188231 0235554 0288015

(3) 0153023 0183153 0220713 0265390 0314840

(4) 0153249 0183742 0221798 0267087 0317248

(5) 0285742 0311159 0342651 0380005 0421287

(6) 0285921 0311644 0343551 0381418 0423292

y = infin (1) 0129470 0167217 0215383 0273841 0339734

(2) 0129725 0167865 0216560 0275661 0342282

(3) 0167978 0204275 0250442 0306382 0369383

(4) 0168217 0204895 0251572 0308128 0371826

(5) 0305868 0336470 0375169 0421935 0474525

(6) 0306058 0336980 0376106 0423388 0476560

27

Page 8: uwaterloo.ca · Ruin Analysis of a Threshold Strategy in a Discrete-time Sparre Andersen Model Steve DREKIC and Ana Maria MERA Department of Statistics and Actuarial Science University

to obtain the following expression for b(k) the initial (ie starting at time k) probability row

vector of the above Markov chain corresponding only to the states in Δ1 namely

(kminuszk)c2

b(k) L

= dmkminuszk(αu+czk+me1 αu+czk+mminus1e1 α2e1 α1e1 0 0 ) (26)

m=(kminuszk)c1

where 0 denotes the 1 times na row vector of zeros By further defining the function vn(t) =

b(k)u + ct + (n minus t)c2 we note that the i-th level of is given by the 1 times na row vector L(kminuszk)c2 b(k)αu+czk+mminusie1 for each i isin Ωk = 0 1 vk(zk)minus1 Hence contains

m=(kminuszk)c1 dmkminuszk

zeros from level vk(zk) onwards Moreover (26) holds true even in the case when zk = k

simplifying to give b(k) = (αu+cke1 αu+ckminus1e1 α2e1 α1e1 0 0 )

At this juncture we introduce two additional row vectors namely

(k) (k) (k) (k) b(k)gn = (gn0 gn1 gn2 ) = Cn n isin N (27)

and

h(k) (k) (k) (k) b(k)Cnminus1D =

(k) Z

+ n = (h nminus2 hnminus3 ) = gnminus1D n isin (28) nminus1 h

(k)We remark that gn contains the probabilities of being in the various ldquonon-ruinedrdquo states

at time k + n (ie after claims have been paid out) without having visited a ldquoruinedrdquo state

during the previous n minus 1 transitions given that Uk isin Ωk sub Δ1 In a similar fashion h(nk)

contains the probabilities of being in the various ldquoruinedrdquo states for the first time at time

k + n given that Uk isin Ωk sub Δ1 Note that the probability of being in ruined state ldquo minus jrdquo is (k) (k) (k) (k)

characterized by the row vector h = (h h h ) with the third subscript nminusj nminusj1 nminusj2 nminusjna

representing the value of the elapsed time counter Lk+n Upon further reflection however (k)

=it must be the case that h = 0 for i 1 as ruin can only occur at claim instants nminusji

which in this case implies Lk+n = 1 with probability 1 Thus it immediately follows that (k) (k) (k)

the structure of the 1 times na row vector hnminusj is simply given by hnminusj = (φnj(u) 0 0) (k)

where φ (u) = Pr T = k + n |UT | = j | Consequently as in the analysis with nj Uk isin Ωk

no threshold on the surplus level (see Alfa and Drekic (2007 p 299)) we obtain

(k) (k) prime φ (u) = h (29) nj nminusje1

where e1 prime denotes the transpose of e1

A similar line of probabilistic logic can also be applied to obtain a representation for the (k)

trivariate function ψ (u) = Pr T = k + n UTminus = i |UT | = j | Uk isin Ωk In order for ruin nij

to occur at time k + n with a surplus immediately prior to ruin equal to i we observe that

(i) none of the previous n minus 1 transitions must have included a visit to any state in Δ2

and

8

(ii) the surplus level at time k + n minus 1 must necessarily be equal to i minus pk+nminus1

Note that pk+nminus1 denotes the premium corresponding to the time interval (k + n minus 1 k + n]

which is received at (k + n minus 1)+ Hence the quantity corresponding to points (i) and (ii) (k)

above is the 1 times na row vector g Then at the next time unit a claim must nminus1iminuspk+nminus1

necessarily occur but not before a premium of pk+nminus1 is first collected thereby raising the

surplus level to i Since s contains the absorption probabilities (to claim occurrence) from

the na possible phase states and the claim causing ruin must be of size i + j in order to

ensure that the deficit at ruin is equal to j it follows that

(k) (k)ψ (u) = g s αi+j (210) nij nminus1iminuspk+nminus1

(k)The following corollary provides the means of calculating the quantity g in nminus1iminuspk+nminus1

(210)

Corollary 1 If minc Z + c1 = c then

(k)

g if i = c c+1 Z +c1 minus1 nminus1iminusc

(k) (k) Lminc2iminusZ (k)g = g + j=c1

if i = Z +c1 Z +c1 +1 minmα Z +cminus1nminus1iminuspk+nminus1 nminus1iminusc djgnminus1iminusj

Lc2 (k)

j=c1 djgnminus1iminusj if i = Z +c Z +c+1 mαminus1

(211)

Conversely if minc Z + c1 = Z + c1 then

Lminc2iminusZ (k)

djg if i = Z +c1 Z +c1 +1 cminus1nminus1iminusj j=c1

(k) (k) Lminc2iminusZ (k)gnminus1iminuspk+nminus1

= gnminus1iminusc + j=c1 djgnminus1iminusj if i = c c+1 minmα Z +cminus1

Lc2 (k)djg if i = Z +c Z +c+1 mαminus1j=c1 nminus1iminusj

(212)

Proof The justification of (211) is as follows If c is smaller than Z + c1 the only way of

reaching a surplus level of i isin c c+1 Z +c1 minus1 (prior to ruin) is by receiving a premium

of c since i minus j for any j = c1 c1 +1 c2 would result in a surplus level below Z where the

random premium rate is not applicable Next for fixed i isin Z +c1 Z +c1 +1 Z +c minus1

there are two distinct ways of reaching this surplus level First of all note that the value of

i minus j is guaranteed to produce a surplus amount greater than or equal to Z provided that

j isin c1 c1 + 1 minc2 i minus Z Secondly note that reducing each of value of i by c also

yields a non-negative surplus value less than Z Hence both pure and random premiums

are plausible under this scenario For i isin Z + c Z + c + 1 note that decreasing this

amount by any j = c1 c1 + 1 c2 results in a surplus value greater than or equal to Z as

desired However reducing it by the amount c also has the same impact thus rendering the

pure premium option inadmissible in this case Lastly we incorporate into the range of i the

9

fact that the surplus immediately prior to ruin cannot exceed mα minus1 as indicated in Section

1 The justification of (212) when c is larger than Z + c1 mirrors the analysis above and is

left to the reader D

Recall that in the analysis up to this point we have assumed that W1 = k isin

1 2 3 nr If we now condition on the value of W1 the Law of Total Probability

yields the following two expressions for φnj(u) and ψnij(u) namely

nr L

= rkφ (u) n = nr + 1 nr + 2 φnj(u) n

(kminus) kj

k=1

and nr L

(k)ψnij(u) = rkψ (u) n = nr + 1 nr + 2 nminuskij

k=1

In addition however it is quite possible that T = n for n = 1 2 nr and this can happen

as a result of either

(i) the first claim which causes ruin occurring at time nminus from a (conditional) surplus Lnminus1level of u + czn + m (where m) or i=zn

Xi =

(ii) the first claim which does not cause ruin occurring at some time kminus k isin 1 2 nminus

1 and ruin subsequently occurring n minus k time units later

If we now combine these outcomes with those for n = nr + 1 nr + 2 (and recall that

rn = 0 forall n gt nr) the general expression for φnj(u) becomes

minnminus1nr (nminuszn)c2 L

(k) L

φnj(u)= rkφ (u) + rn dmnminuszn αu+czn+m+j n isin Z+ (213) nminuskj

k=1 m=(nminuszn)c1

Concerning the function ψnij(u) when n le nr and point (i) above is considered we note Lnminus1that i must necessarily equal u + czn + m (again conditional on i=zn

Xi = m) for ruin to

occur in this manner Hence in a similar fashion we acquire

minnminus1nr (nminuszn)c2 L L

ψnij(u) = rkψn

(kminus) kij(u) + rn dmnminuszn

δiu+czn+mαu+czn+m+j

k=1 m=(nminuszn)c1

minnminus1nr L

(k) Z

+ = rkψnminuskij(u) + rndiminusuminuscznnminuszn αi+j n isin (214)

k=1

Finally for fixed i isin minc Z + c1 minc Z + c1 + 1 mα minus 1 summing (214) from

j = 1 to mα minus i yields the following expression for the bivariate pmf ωni(u) namely

minnminus1nr mαminusi mαminusi L L

(k) L

ωni(u) = rk ψnminuskij(u) + rndiminusuminuscznnminusznαi+j

k=1 j=1 j=1

10

minnminus1nr L

(k) Z

+ = Λi rkg s + rndiminusuminuscznnminuszn n isin (215) nminuskminus1iminuspnminus1

k=1

where (210) was used to establish the last equality In the special case when c1 = c2 = c it

is a straightforward exercise to show that equations (213) through (215) simplify to produce

equations (215) through (217) of Alfa and Drekic (2007 p 300)

3 Computational considerations

In order to compute the joint pmfrsquos defined by (213) (214) and (215) it is clear from (k)

equations (29) through (212) that we need to be able to compute the quantities gn and

h(k) n defined by (27) and (28) respectively In what follows we capitalize on the underlying

Markov chain structure in our model to develop a set of computational procedures which

ultimately enable one to calculate these key row vectors

First of all recall that g0(k)

= b(k) contains zeros from a certain level onwards namely (k)

level vk(zk) = u + czk + c2(k minus zk) In general one can easily deduce that gn will also

contain zeros from a certain level onwards to be denoted by lk(n) We begin this section

with a determination of lk(n) which varies in form depending on the value of vk(zk) minus 1

representing the maximum surplus level at time k after the smallest possible claim of size 1

has been applied (at time kminus)

On one hand it may be the case that the value of vk(zk)minus1 lies at or above the threshold

level Z In this situation we remark that zk can take on any value in the set 0 1 k (k)

Since g0 contains zeros from level lk(0) = vk(zk) onwards we observe that at time k + 1

it may be possible to achieve a higher surplus value than vk+1(zk) minus 1 representing the

maximum surplus level assuming there was no drop below Z due to the claim at time kminus

Specifically if the surplus had fallen to level Z minus 1 at time kminus then gained c at time k+ a

bounded level of Z + c minus1 would have been reached at time k +1 and this level is potentially

larger than vk+1(zk) minus 1 As a result we have that lk(1) = maxvk+1(zk) Z + c If we

extend this argument inductively we readily obtain lk(n) = maxvk+n(zk) Z +c +c2(n minus1)

n isin Z+ However since vk+n(zk) = u + czk + c2(k +n minus zk) = vk(zk)+ c2 + c2(n minus 1) we may

alternatively re-express lk(n) as

lk(n) = maxvk+n(zk) Z + c + c2(n minus 1) = maxvk(zk) + c2 Z + c + c2(n minus 1)

Thus

Z+lk(n) = maxlk(0) + c2 Z + c + c2(n minus 1) n isin (31)

11

or equivalently

Z+lk(n) = lk(1) + c2(n minus 1) n isin

Clearly lk(0) gt Z and lk(n) ge Z + c for n isin Z+

The other remaining case to consider involves the situation when the maximum surplus

level at time k is strictly below the threshold level Z Under this scenario we further

distinguish between two cases involving the value of c2 First of all suppose that c2 = 0 so

that vk(zk)minus1 = u +czk minus1 lt Z or equivalently u +czk le Z If u +czk happens to equal Z

then it is possible for zk isin 0 1 k With c2 = 0 however no further premiums can be

earned until after the first claim takes place (which is assumed to occur at time kminus) thereby

implying that Z = u + ci for all i = zk zk + 1 k Hence we know that u + ck = Z

in this case Moreover if u + czk happens to be strictly less than Z then we clearly have

that zk = k Even in the case when c2 gt 0 one can readily verify that zk = k so that

vk(zk) minus 1 = u + ck minus 1 lt Z Thus we proceed with the analysis assuming that zk = k

c2 ge 0 and the inequality u + ck le Z holds true

lowast k

Define the ceiling function denoted lceilxrceil to yield the smallest integer greater than or

equal to x Because a random premium cannot be received until level Z is at least achieved

it is essential to find the point at which the threshold level Z is reached (ie the point at Zminus(u+ck)+1 which u + ck minus1+ ci ge Z) Isolating this inequality for i one obtains i ge

c Hence

assuming that W1 = k the elapsed time after k at which a random premium would begin to

lceilZminus(u+ck)+1 be received is t lowast = rceil Note that t lowast isin Z+ k c k

With zk = k g0(k)

contains zeros from level u + ck onwards so that lk(0) = u + ck

Immediately following this a premium of c is guaranteed to be received at time k+ as the

threshold level Z has not yet been reached Consequently g1(k)

contains zeros from level (k)

b(k)u + c(k + 1) onwards which is c levels further than that in g0 = Continuing this (k)

process inductively we can readily establish that gn contains zeros from level u + c(k + n)

onwards for n = 1 2 t lowast k so that lk(n) = u + c(k + n) for n = 0 1 t lowast k

In the case of n = t lowast + 1 t lowast + 2 gt

(k) is first looked at in isolation At time k +k k +1

t lowast k + 1 the threshold level has been crossed giving rise to a maximum surplus value of either (k)

c(k + t lowast k contains zeros from level lk(t lowast ku + ) + c2 minus 1 or Z + c minus 1 Therefore g + 1) = t lowast k+1

+1(k + t lowast kmaxvk+t lowast k

) Z + c onwards Carrying out the same inductive process as before it is

readily seen that g(nk)

contains zeros from level lk(n) = maxvk+n(k+t lowast k) Z +c+c2(nminust lowast kminus1)

onwards for n = t lowast k + 1 t lowast k + 2 Combining both cases we obtain

u + c(k + n) if n = 0 1 t lowast lk(n) = k (32)

maxlk(t lowast) + c2 Z + c + c2(n minus t lowast minus 1) if n = t lowast + 1 t lowast + 2 k k k k

Note that by construction lk(t lowast k) gt Z and lk(n) le Z for n = 0 1 t lowast k minus1 Interestingly

12

if we extend the definition tk lowast by introducing the quantity ζk

lowast = maxtklowast 0 we remark that

(31) and (32) can actually be combined into a single formula for lk(n) namely

vk+n(zk + n) if n = 0 1 ζklowast

lk(n) = (33) maxlk(ζ

lowast) + c2 Z + c + c2(n minus ζlowast minus 1) if n = ζlowast + 1 ζlowast + 2 k k k k

Clearly we have that lk(n) ge Z + c for n = ζk lowast +1 ζk

lowast +2 Moreover if c1 = c2 = c (33)

reduces to give lk(n) = u + c(k + n) forall n isin N which is in agreement with the results in Alfa

and Drekic (2007 Section 3)

Remark In our above analysis we purposely ignored the possibility of any future claims

occurring after time k in our determination of lk(n) However we recognize that this is not

realistic and future claims would indeed occur after time k thereby further reducing the value

of the maximum surplus level lk(n) minus 1 at time k + n To account for this we imagine all

future interclaim times (ie after time k) to be of maximum length na and all corresponding

claim sizes to be of minimum amount 1 Consequently in the case when c2 gt 0 we should

subtract from formula (33) the amount of lfloornnarfloor where lfloormiddotrfloor again denotes the floor function

Obviously when c2 = 0 (ie no premiums are collected whenever the surplus level is at or

above Z) lk(n) = Z + c for n = ζk lowast + 1 ζk

lowast + 2 (k)

We now turn our attention to the development of a general algorithm for computing gni

(k) (k)for i = 0 1 lk(n) minus 1 With the aid of (24) we apply the key identity gn = gnminus1C

(which can be inferred from (27)) to first obtain

minZlk(nminus1)minus1 lk(nminus1)minus1 c2 L L L

(k) (k) (k)gni = g Bjminusi+c + gnminus1j dmBjminusi+m n isin Z+ (34) nminus1j

j=0 j=Z m=c1

However (22) states that Bi = Ona for i isin Zminus and so (34) becomes (for the same range of

i)

minZlk(nminus1)minus1 c2 lk(nminus1)minus1 L L L

(k) (k) (k)gni = g Bjminusi+c + dm gnminus1jBjminusi+m n isin Z+ (35) nminus1j

j=max0iminusc m=c1 j=maxZiminusm

Taking the form of (33) into consideration (35) yields the following two formulae

lk(nminus1)minus1 L

(k) (k)gni = gnminus1jBjminusi+c n = 1 2 ζ k

lowast (36) j=max0iminusc

and

Zminus1 c2 lk(nminus1)minus1 L L L

(k) (k) (k)g = g + dm g n = ζ lowast + 1 ζ lowast + 2 ni nminus1jBjminusi+c nminus1jBjminusi+m k k

j=max0iminusc m=c1 j=maxZiminusm

(37)

13

For n = 1 2 ζk lowast where ζk

lowast isin Z+ we note that lk(n) ge c + 1 Therefore with the aid of

(22) (36) simplifies to give

Llk(nminus1)minus1 (k) (k) j=0 gnminus1jBjminusi+c if i = 0 1 c minus 1

gni = Llk(nminus1)minus1 (k)

g if i = c c + 1 lk(n) minus 1j=iminusc nminus1jBjminusi+c

Llk(nminus1)minus1 (k)

nminus1js 0 0 if i = 0 1 c minus 1j=0 αjminusi+cg= (38)

(k) Llk(nminus1)minus1 (k) g S+ αjminusi+cgnminus1js 0 0 if i = c c+1 lk(n)minus1 nminus1iminusc j=iminusc+1

In the case of n = ζk lowast + 1 ζk

lowast + 2 we first reiterate that lk(n) ge Z + c We next break

(37) into several cases based on the value of i and the value of minc Z + c1 as follows

Case 1 Suppose minc Z + c1 = c

For i isin 0 1 c minus 1

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+mni nminus1j nminus1j

j=0 m=c1 j=Z

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k)= αjminusi+cg s + dm αjminusi+mg s 0 0 (39) nminus1j nminus1j

j=0 m=c1 j=Z

For i isin c c + 1 Z + c1 minus 1

Zminus1 c2 lk(nminus1)minus1 L L L

(k) (k) (k)g = g + dm gni nminus1j Bjminusi+c nminus1j Bjminusi+m

j=iminusc m=c1 j=Z

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k) (k)= g S + αjminusi+cg dm αjminusi+mgnminus1j s 0 0 (310) nminus1iminusc nminus1j s +

j=iminusc+1 m=c1 j=Z

For i isin Z + c1 Z + c1 + 1 Z + c minus 1

Zminus1 minc2iminusZ lk(nminus1)minus1 c2 lk(nminus1)minus1 L L L L L

(k) (k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+m + dm g Bjminusi+mni nminus1j nminus1j nminus1j

j=iminusc m=c1 j=iminusm m=iminusZ+1 j=Z

minc2iminusZ L

(k) (k)= g S + dmg S nminus1iminusc nminus1iminusm

m=c1

Zminus1 minc2iminusZ lk(nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k)+ αjminusi+cg dm αjminusi+mg dm αjminusi+mgnminus1j s 0 0 nminus1j s + nminus1js +

j=iminusc+1 m=c1 j=iminusm+1 m=iminusZ+1 j=Z

(311)

For i isin Z + c Z + c + 1 lk(n) minus 1

c2 lk(nminus1)minus1 L L

(k) (k)g = dm g Bjminusi+mni nminus1j

m=c1 j=iminusm

c2 c2 lk (nminus1)minus1 L L L

(k) (k)= dmg S + dm αjminusi+mgnminus1j s 0 0 (312) nminus1iminusm

m=c1 m=c1 j=iminusm+1

14

Case 2 Suppose minc Z + c1 = Z + c1

For i isin 0 1 Z + c1 minus 1

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+mni nminus1j nminus1j

j=0 m=c1 j=Z

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k)= αjminusi+cg s + dm αjminusi+mg s 0 0 (313) nminus1j nminus1j

j=0 m=c1 j=Z

For i isin Z + c1 Z + c1 + 1 c minus 1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+m + dm g Bjminusi+mni nminus1j nminus1j nminus1j

j=0 m=c1 j=iminusm m=iminusZ+1 j=Z

minc2iminusZ L

(k)= dmg S nminus1iminusm

m=c1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k)+ αjminusi+cg s + dm αjminusi+mg s + dm αjminusi+mg s 0 0 nminus1j nminus1j nminus1j

j=0 m=c1 j=iminusm+1 m=iminusZ+1 j=Z

(314)

For i isin c c + 1 Z + c minus 1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k) (k)g = g + g + gni nminus1j Bjminusi+c dm nminus1j Bjminusi+m dm nminus1j Bjminusi+m

j=iminusc m=c1 j=iminusm m=iminusZ+1 j=Z

minc2iminusZ L

(k) (k)= g S + dmg S nminus1iminusc nminus1iminusm

m=c1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k)+

nminus1j s 0 0 αjminusi+cgnminus1j s + dm αjminusi+mgnminus1j s + dm αjminusi+mg j=iminusc+1 m=c1 j=iminusm+1 m=iminusZ+1 j=Z

(315)

For i isin Z + c Z + c + 1 lk(n) minus 1

c2 lk (nminus1)minus1 L L

(k) (k)g = dm g Bjminusi+mni nminus1j

m=c1 j=iminusm

c2 c2 lk (nminus1)minus1 L L L

(k) (k)= dmg S + dm αjminusi+mgnminus1j s 0 0 (316) nminus1iminusm

m=c1 m=c1 j=iminusm+1

It is not difficult to verify that equations (38) through (316) simplify to produce equations

(32) and (33) of Alfa and Drekic (2007 pp 301-302) when c1 = c2 = c With equations (38) (k)

through (316) providing the recursive means of calculating gn we next turn our attention

15

to a determination of h(k) defined by (28) For n = 1 2 ζlowast we use (22) and (25) to n k

obtain

lk(nminus1)minus1 lk(nminus1)minus1

h(k) nminusj =

L g

(k) nminus1ℓBc+j+ℓ =

L αc+j+ℓg

(k) nminus1ℓs 0 0 (317)

ℓ=0 ℓ=0

For n = ζk lowast + 1 ζk

lowast + 2 we similarly acquire

Zminus1 lk(nminus1)minus1 L L

(k) (k) (k)hnminusj = gnminus1ℓBc+j+ℓ + gnminus1ℓAj+ℓ

ℓ=0 ℓ=Z

Zminus1 c2 lk(nminus1)minus1 L L L

(k) (k)= αc+j+ℓg s + dm αm+j+ℓg s 0 0 (318) nminus1ℓ nminus1ℓ

ℓ=0 m=c1 ℓ=Z

Note that substituting (317) and (318) into (29) immediately yields

Llk(nminus1)minus1 (k) (k) ℓ=0 αc+j+ℓgnminus1ℓs if n = 1 2 ζk

lowast φ (u) = nj LZminus1 (k) Lc2

Llk(nminus1)minus1 (k) ζlowast s + dm αm+j+ℓg n = + 1 ζlowast + 2 ℓ=0 αc+j+ℓgnminus1ℓ m=c1 ℓ=Z nminus1ℓs if k k

(319)

Clearly in the special case when c1 = c2 = c (319) yields equation (35) of Alfa and Drekic

(2007 p 302) Finally we remark that (319) can be substituted into (213) to obtain a

computational formula for φnj(u)

4 Numerical examples and findings

In this section we illustrate the application of our proposed algorithm with several numerical

examples All calculations in this section were performed using Mathematica (Version 7) on

a Dell Precision T7400 workstation with a 3 Ghz CPU and 4 GB of RAM

In our first illustration (labelled Example 1) we consider the ordinary Sparre Andersen

model in which interclaim times have pmf given by

(211)(911)jminus1 if j = 1 2 na minus 1 aj = (41)

(911)naminus1 if j = na

Note that the pmf (41) is that of a truncated geometric distribution with all the probability

mass on na na +1 assigned to the support value na Clearly as na becomes larger the

closer (41) approximates this particular geometric distribution having mean 55 In what Lna

Lminus1 nafollows we fix na = 25 so that aj = 0992 and jaj = 546 Let the individual j=1 j=1

claim amount distribution be given by the pmf

minus4 minus4j minus 1 j

αj = 1 + minus 1 + j isin Z+ (42) 30 30

16

We remark that the pmf (42) represents a discretized version of the Pareto distribution with

mean 10 Also note that mα = infin thereby implying that the random variables |UT | and

UTminus both have infinite support

Table 1 displays values (rounded to 6 decimal places) of the trivariate cumulative distrishy

bution function

n x y L L L

Ψnxy(u) = Pr T le n UTminus le x |UT | le y | U0 = u = ψℓij(u) ℓ=1 i=mincZ+c1 j=1

corresponding to the above discrete-time risk process with u = 10 Z = 50 c = 5 and

varying premium amount distributions according to the following set of scenarios

(1) d2 = 1

(2) d1 = 25 d2 = 15 d3 = 25

G

5)

(3) di = i

(25)i(35)5minusi i = 0 1 5

(4) di = 15 i = 0 1 4

(5) d1 = 34 d5 = 14

(6) d0 = 35 d5 = 25

We remark that while these six premium amount distributions all have mean equal to 2

they possess different shapes and we investigate the impact of their variability on the joint

distribution of T |UT | and UTminus In particular the variances corresponding to the premium

amount distributions of scenarios (1) through (6) are 0 08 12 2 3 and 6 respectively

Based on the results in Table 1 it would appear that greater variability in the premium

amount distribution generally leads to higher ruin probabilities Perhaps somewhat surprising

though the differences are subtle and not particularly striking as one compares the various

ruin probabilities from scenarios (1) through (6) Obviously we observe more of a difference

between scenarios as the values of n x and y increase

As a follow-up illustration (labelled Example 2) we consider a model in which ordinary

interclaim times have a discrete uniform distribution with pmf aj = 110 j = 1 2 10

Clearly the mean of this distribution is 55 which for comparative purposes is identical to

the ordinary interclaim time mean considered in Example 1 In addition we assume that

the pmf of the individual claim amount distribution is once again given by (42) Table 2

displays values (rounded to 6 decimal places) of Ψnxy(10) for several variants of the Sparre

Andersen model (again with Z = 50 and c = 5) having the above ordinary interclaim time

and claim amount distributions namely

17

(1) ordinary model under scenario (2) from Example 1

(2) ordinary model under scenario (5) from Example 1

(3) stationary model under scenario (2) from Example 1

(4) stationary model under scenario (5) from Example 1

(5) delayed model (with W1 = 1 with probability 1) under scenario (2) from Example 1

(6) delayed model (with W1 = 1 with probability 1) under scenario (5) from Example 1

Concerning the results in Table 2 we first compare the values of Ψnxy(10) in entries (2)

and (5) of Table 1 to their corresponding counterparts in entries (1) and (2) of Table 2 In all

cases we observe that ruin is more likely to occur when interclaim times have the truncated

geometric distribution introduced in our first example as opposed to the discrete uniform

distribution of our second example Since mean interclaim times are equal in both examshy

ples this behaviour must be attributable to the greater variability inherent in the truncated

geometric distribution We also note that under the stationary model assumption the pmf

of W1 is easily verified to be rj = (11 minus j)55 j = 1 2 10 which has mean 4 lt 55

As a result the values of Ψnxy(10) for stationary models (3) and (4) are always larger than

the corresponding values for ordinary models (1) and (2) Finally in the absence of any

knowledge regarding which distribution is in effect for the first interclaim time W1 models

(5) and (6) can be regarded as ldquoworst-case scenariordquo models as they serve to provide upper

bounds on the values of Ψnxy(10)

In conclusion we state that the values in Tables 1 and 2 were generated by first impleshy

menting the recursive procedures (38) through (316) and then summing the appropriate

bivariatetrivariate probabilities computed via (213) (214) or (215) In situations when

x = y = infin however we computed the resulting finite-time ruin probabilities using an altershy

native (albeit standard) method (eg see Cossette et al (2006 Section 3)) Specifically if

we define the survival probability σ(n u) = Pr T gt n | U0 = u = 1minusΨninfininfin(u) as well as

the auxiliary function

0 if w ge Z ηwt =

minlfloorZminuswc minus1 rfloor + 1 t if w lt Z

then by conditioning on the time and amount of the first claim (and recalling that the delayed

process reverts to the ordinary process following this claim) we obtain the expression

minnnr (kminuszk)c2 u+czk+ℓ L L L

σ(u n) = Rn + rk dℓkminuszk αjγ(u + czk + ℓ minus j n minus k)

k=1 ℓ=(kminuszk)c1 j=1

18

where the function γ(w m) for w m isin N is computed recursively via

minmna (kminusηwk)c2 u+cηwk+ℓ L L L

γ(w m) = Am + ak dℓkminusηwk αjγ(u + cηwk + ℓ minus j m minus k)

k=1 j=1 ℓ=(kminusηwk)c1

with starting values γ(w 0) = 1 forall w isin N

Acknowledgements

The authors would like to thank Dr David Landriault for his valuable suggestions which

helped to improve an earlier version of this work This research was supported by the Natural

Sciences and Engineering Research Council of Canada

References

Alfa AS 2004 Markov chain representations of discrete distributions applied to queueing

models Computers amp Operations Research 31 2365-2385

Alfa AS Drekic S 2007 Algorithmic analysis of the Sparre Andersen model in discrete

time ASTIN Bulletin 37 293-317

Bao Z 2007 A note on the compound binomial model with randomized dividend strategy

Applied Mathematics amp Computation 194 276-286

Cossette H Landriault D Marceau E 2006 Ruin probabilities in the discrete time

renewal risk model Insurance Mathematics amp Economics 38 309-323

Karlin S Taylor HM 1975 A First Course in Stochastic Processes 2nd edition Acashy

demic Press New York

Kim B Kim H-S Kim J 2008 A risk model with paying dividends and random

environment Insurance Mathematics amp Economics 42 717-726

Landriault D 2008 Randomized dividends in the compound binomial model with a general

premium rate Scandinavian Actuarial Journal 1-15

Tan J Yang X 2006 The compound binomial model with randomized decisions on

paying dividends Insurance Mathematics amp Economics 39 1-18

19

Table 1 ndash Values of Ψnxy(10) corresponding to Example 1

(a) x = 10

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0005357 0006188 0007163 0008278 0009481

(2) 0005359 0006195 0007175 0008296 0009507

(3) 0005361 0006198 0007181 0008305 0009519

(4) 0005363 0006205 0007193 0008323 0009544

(5) 0005366 0006212 0007208 0008346 0009576

(6) 0005374 0006236 0007249 0008410 0009664

y = 25 (1) 0007631 0008815 0010202 0011789 0013503

(2) 0007634 0008824 0010220 0011816 0013539

(3) 0007636 0008828 0010228 0011829 0013557

(4) 0007639 0008838 0010245 0011854 0013593

(5) 0007643 0008849 0010266 0011887 0013638

(6) 0007656 0008882 0010325 0011978 0013764

y = 50 (1) 0008477 0009792 0011333 0013095 0014998

(2) 0008481 0009802 0011352 0013125 0015039

(3) 0008483 0009807 0011361 0013139 0015059

(4) 0008486 0009817 0011380 0013168 0015098

(5) 0008491 0009830 0011403 0013204 0015149

(6) 0008504 0009866 0011470 0013305 0015288

y = infin (1) 0008777 0010138 0011733 0013558 0015528

(2) 0008781 0010149 0011753 0013588 0015570

(3) 0008783 0010154 0011763 0013603 0015591

(4) 0008787 0010164 0011782 0013633 0015632

(5) 0008791 0010177 0011807 0013671 0015684

(6) 0008805 0010215 0011875 0013775 0015828

20

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0058559 0061875 0065704 0070048 0074714

(2) 0058569 0061903 0065756 0070128 0074825

(3) 0058574 0061917 0065782 0070167 0074880

(4) 0058584 0061945 0065833 0070245 0074987

(5) 0058596 0061979 0065896 0070344 0075124

(6) 0058632 0062079 0066078 0070621 0075506

y = 25 (1) 0089002 0093986 0099739 0106262 0113270

(2) 0089017 0094028 0099817 0106382 0113437

(3) 0089025 0094049 0099856 0106442 0113519

(4) 0089040 0094091 0099932 0106560 0113681

(5) 0089057 0094142 0100028 0106708 0113887

(6) 0089113 0094294 0100302 0107127 0114463

y = 50 (1) 0102786 0108506 0115106 0122590 0130629

(2) 0102804 0108555 0115196 0122728 0130820

(3) 0102812 0108579 0115241 0122797 0130914

(4) 0102830 0108627 0115329 0122932 0131101

(5) 0102850 0108686 0115439 0123103 0131337

(6) 0102914 0108860 0115754 0123584 0132000

y = infin (1) 0108895 0114933 0121901 0129799 0138283

(2) 0108914 0114985 0121996 0129945 0138485

(3) 0108923 0115010 0122043 0130018 0138585

(4) 0108941 0115061 0122136 0130161 0138782

(5) 0108962 0115123 0122252 0130341 0139032

(6) 0109030 0115307 0122585 0130849 0139732

21

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0079856 0086792 009464 0103440 0112831

(2) 0079885 0086869 0094778 0103650 0113118

(3) 0079953 0086978 0094936 0103864 0113393

(4) 0080081 0087188 0095243 0104282 0113929

(5) 0079962 0087075 0095149 0104213 0113890

(6) 0080805 0088366 0096951 0106588 0116875

y = 25 (1) 0124591 0135680 0148207 0162245 0177217

(2) 0124639 0135806 0148433 0162586 0177683

(3) 0124755 0135993 0148703 0162950 0178148

(4) 0124978 0136354 0149226 0163658 0179054

(5) 0124763 0136141 0149035 0163500 0178935

(6) 0126238 0138381 0152148 0167588 0184060

y = 50 (1) 0146757 0160079 0175111 0191945 0209894

(2) 0146816 0160232 0175385 0192360 0210460

(3) 0146965 0160469 0175725 0192817 0211043

(4) 0147249 0160925 0176384 0193706 0212178

(5) 0146968 0160639 0176117 0193470 0211980

(6) 0148860 0163500 0180077 0198658 0218474

y = infin (1) 0158013 0172606 0189059 0207476 0227106

(2) 0158078 0172776 0189362 0207934 0227732

(3) 0158250 0173047 0189751 0208455 0228395

(4) 0158579 0173571 0190503 0209467 0229685

(5) 0158247 0173226 0190172 0209161 0229410

(6) 0160443 0176532 0194735 0215127 0236869

22

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0088667 0104403 0123329 0145277 0169184

(2) 0088711 0104518 0123538 0145598 0169627

(3) 0088733 0104574 0123641 0145756 0169846

(4) 0088777 0104687 0123844 0146066 0170274

(5) 0088829 0104828 0124105 0146470 0170832

(6) 0088988 0105229 0124822 0147560 0172331

y = 25 (1) 0140528 0167839 0201007 0239726 0282102

(2) 0140598 0168021 0201338 0240234 0282804

(3) 0140632 0168111 0201502 0240486 0283151

(4) 0140701 0168289 0201822 0240976 0283827

(5) 0140785 0168515 0202241 0241622 0284718

(6) 0141036 0169148 0203370 0243335 0287071

y = 50 (1) 0167826 0203005 0246173 0296954 0352855

(2) 0167908 0203218 0246561 0297548 0353672

(3) 0167949 0203324 0246753 0297841 0354075

(4) 0168030 0203532 0247128 0298412 0354859

(5) 0168129 0203800 0247622 0299174 0355907

(6) 0168423 0204539 0248936 0301159 0358620

y = infin (1) 0183325 0224926 0276881 0338957 0408257

(2) 0183412 0225150 0277283 0339564 0409079

(3) 0183455 0225260 0277481 0339864 0409485

(4) 0183540 0225478 0277869 0340446 0410270

(5) 0183645 0225760 0278386 0341234 0411339

(6) 0183954 0226529 0279737 0343244 0414037

23

Table 2 ndash Values of Ψnxy(10) corresponding to Example 2

(a) x = 10

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0001901 0002276 0002724 0003240 0003799

(2) 0001904 0002286 0002742 0003269 0003840

(3) 0002651 0003013 0003442 0003936 0004471

(4) 0002655 0003022 0003460 0003964 0004510

(5) 0005792 0006098 0006458 0006871 0007317

(6) 0005795 0006106 0006473 0006894 0007350

y = 25 (1) 0002708 0003243 0003880 0004615 0005411

(2) 0002713 0003257 0003906 0004656 0005469

(3) 0003778 0004293 0004904 0005608 0006369

(4) 0003782 0004306 0004929 0005647 0006424

(5) 0008255 0008690 0009203 0009792 0010427

(6) 0008259 0008702 0009224 0009824 0010473

y = 50 (1) 0003008 0003603 0004310 0005127 0006011

(2) 0003014 0003618 0004340 0005173 0006076

(3) 0004197 0004769 0005448 0006230 0007075

(4) 0004202 0004784 0005476 0006273 0007137

(5) 0009173 0009656 0010226 0010879 0011585

(6) 0009177 0009669 0010249 0010916 0011637

y = infin (1) 0003115 0003730 0004463 0005308 0006224

(2) 0003121 0003746 0004493 0005356 0006291

(3) 0004346 0004938 0005641 0006450 0007325

(4) 0004351 0004954 0005670 0006496 0007389

(5) 0009499 0009999 0010589 0011265 0011996

(6) 0009503 0010012 0010613 0011303 0012050

24

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0031176 0032815 0034733 0036924 0039282

(2) 0031193 0032862 0034821 0037061 0039475

(3) 0050168 0051747 0053587 0055684 0057939

(4) 0050184 0051792 0053671 0055815 0058124

(5) 0136884 0138219 0139763 0141517 0143399

(6) 0136896 0138256 0139832 0141626 0143553

y = 25 (1) 0047622 0050092 0052981 0056280 0059829

(2) 0004765 0050163 0053113 0056487 0060121

(3) 0076570 0078950 0081720 0084878 0088272

(4) 0076593 0079017 0081847 0085076 0088551

(5) 0205774 0207786 0210111 0212752 0215585

(6) 0205793 0207842 0210216 0212917 0215818

y = 50 (1) 0055170 0058009 0061329 0065119 0069196

(2) 0055200 0058091 0061481 0065357 0069533

(3) 0088650 0091385 0094569 0098197 0102096

(4) 0088677 0091463 0094715 0098425 0102417

(5) 0235815 0238128 0240800 0243834 0247089

(6) 0235837 0238192 0240920 0244023 0247356

y = infin (1) 0058565 0061565 0065073 0069077 0073384

(2) 0058597 0061652 0065234 0069329 0073740

(3) 0094061 0096952 0100316 0104149 0108267

(4) 0094090 0097035 0100470 0104390 0108607

(5) 0248472 0250917 0253740 0256945 0260383

(6) 0248495 0250985 0253867 0257145 0260667

25

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0048598 0052524 0056984 0061998 0067342

(2) 0048654 0052673 0057253 0062411 0067916

(3) 0070192 0073979 0078257 0083057 0088167

(4) 0070243 0074119 0078513 0083451 0088714

(5) 0151774 0154981 0158572 0162586 0166852

(6) 0151815 0155096 0158784 0162913 0167308

y = 25 (1) 0076895 0083230 0090413 0098481 0107073

(2) 0076987 0083474 0090854 0099156 0108010

(3) 0110093 0116205 0123095 0130818 0139033

(4) 0110177 0116435 0123515 0131462 0139928

(5) 0230674 0235848 0241632 0248091 0254950

(6) 0230739 0236038 0241980 0248626 0255695

y = 50 (1) 0091496 0099156 0107829 0117563 0127924

(2) 0091610 0099455 0108370 0118387 0129068

(3) 0130130 0137520 0145839 0155156 0165064

(4) 0130233 0137801 0146352 0155942 0166156

(5) 0266599 0272857 0279840 0287632 0295904

(6) 0266680 0273088 0280265 0288286 0296814

y = infin (1) 0099312 0107746 0117285 0127984 0139369

(2) 0099439 0108078 0117883 0128899 0140639

(3) 0140458 0148595 0157745 0167986 0178872

(4) 0140572 0148908 0158314 0168859 0180084

(5) 0282886 0289776 0297457 0306022 0315111

(6) 0282976 0290034 0297929 0306748 0316120

26

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0059096 0072505 0088907 0108122 0129137

(2) 0059223 0072829 0089500 0109052 0130456

(3) 0079757 0092662 0108388 0126778 0146871

(4) 0079875 0092970 0108956 0127669 0148134

(5) 0159788 0170682 0183871 0199249 0216024

(6) 0159881 0170936 0184343 0199990 0217075

y = 25 (1) 0095883 0119741 0149240 0184073 0222390

(2) 0096087 0120262 0150197 0185573 0224520

(3) 0127403 0150359 0178640 0211977 0248613

(4) 0127592 0150855 0179557 0213414 0250653

(5) 0245161 0264533 0288250 0316124 0346709

(6) 0245311 0264942 0289010 0317319 0348407

y = 50 (1) 0116598 0147920 0187100 0233784 0285503

(2) 0116839 0148537 0188231 0235554 0288015

(3) 0153023 0183153 0220713 0265390 0314840

(4) 0153249 0183742 0221798 0267087 0317248

(5) 0285742 0311159 0342651 0380005 0421287

(6) 0285921 0311644 0343551 0381418 0423292

y = infin (1) 0129470 0167217 0215383 0273841 0339734

(2) 0129725 0167865 0216560 0275661 0342282

(3) 0167978 0204275 0250442 0306382 0369383

(4) 0168217 0204895 0251572 0308128 0371826

(5) 0305868 0336470 0375169 0421935 0474525

(6) 0306058 0336980 0376106 0423388 0476560

27

Page 9: uwaterloo.ca · Ruin Analysis of a Threshold Strategy in a Discrete-time Sparre Andersen Model Steve DREKIC and Ana Maria MERA Department of Statistics and Actuarial Science University

(ii) the surplus level at time k + n minus 1 must necessarily be equal to i minus pk+nminus1

Note that pk+nminus1 denotes the premium corresponding to the time interval (k + n minus 1 k + n]

which is received at (k + n minus 1)+ Hence the quantity corresponding to points (i) and (ii) (k)

above is the 1 times na row vector g Then at the next time unit a claim must nminus1iminuspk+nminus1

necessarily occur but not before a premium of pk+nminus1 is first collected thereby raising the

surplus level to i Since s contains the absorption probabilities (to claim occurrence) from

the na possible phase states and the claim causing ruin must be of size i + j in order to

ensure that the deficit at ruin is equal to j it follows that

(k) (k)ψ (u) = g s αi+j (210) nij nminus1iminuspk+nminus1

(k)The following corollary provides the means of calculating the quantity g in nminus1iminuspk+nminus1

(210)

Corollary 1 If minc Z + c1 = c then

(k)

g if i = c c+1 Z +c1 minus1 nminus1iminusc

(k) (k) Lminc2iminusZ (k)g = g + j=c1

if i = Z +c1 Z +c1 +1 minmα Z +cminus1nminus1iminuspk+nminus1 nminus1iminusc djgnminus1iminusj

Lc2 (k)

j=c1 djgnminus1iminusj if i = Z +c Z +c+1 mαminus1

(211)

Conversely if minc Z + c1 = Z + c1 then

Lminc2iminusZ (k)

djg if i = Z +c1 Z +c1 +1 cminus1nminus1iminusj j=c1

(k) (k) Lminc2iminusZ (k)gnminus1iminuspk+nminus1

= gnminus1iminusc + j=c1 djgnminus1iminusj if i = c c+1 minmα Z +cminus1

Lc2 (k)djg if i = Z +c Z +c+1 mαminus1j=c1 nminus1iminusj

(212)

Proof The justification of (211) is as follows If c is smaller than Z + c1 the only way of

reaching a surplus level of i isin c c+1 Z +c1 minus1 (prior to ruin) is by receiving a premium

of c since i minus j for any j = c1 c1 +1 c2 would result in a surplus level below Z where the

random premium rate is not applicable Next for fixed i isin Z +c1 Z +c1 +1 Z +c minus1

there are two distinct ways of reaching this surplus level First of all note that the value of

i minus j is guaranteed to produce a surplus amount greater than or equal to Z provided that

j isin c1 c1 + 1 minc2 i minus Z Secondly note that reducing each of value of i by c also

yields a non-negative surplus value less than Z Hence both pure and random premiums

are plausible under this scenario For i isin Z + c Z + c + 1 note that decreasing this

amount by any j = c1 c1 + 1 c2 results in a surplus value greater than or equal to Z as

desired However reducing it by the amount c also has the same impact thus rendering the

pure premium option inadmissible in this case Lastly we incorporate into the range of i the

9

fact that the surplus immediately prior to ruin cannot exceed mα minus1 as indicated in Section

1 The justification of (212) when c is larger than Z + c1 mirrors the analysis above and is

left to the reader D

Recall that in the analysis up to this point we have assumed that W1 = k isin

1 2 3 nr If we now condition on the value of W1 the Law of Total Probability

yields the following two expressions for φnj(u) and ψnij(u) namely

nr L

= rkφ (u) n = nr + 1 nr + 2 φnj(u) n

(kminus) kj

k=1

and nr L

(k)ψnij(u) = rkψ (u) n = nr + 1 nr + 2 nminuskij

k=1

In addition however it is quite possible that T = n for n = 1 2 nr and this can happen

as a result of either

(i) the first claim which causes ruin occurring at time nminus from a (conditional) surplus Lnminus1level of u + czn + m (where m) or i=zn

Xi =

(ii) the first claim which does not cause ruin occurring at some time kminus k isin 1 2 nminus

1 and ruin subsequently occurring n minus k time units later

If we now combine these outcomes with those for n = nr + 1 nr + 2 (and recall that

rn = 0 forall n gt nr) the general expression for φnj(u) becomes

minnminus1nr (nminuszn)c2 L

(k) L

φnj(u)= rkφ (u) + rn dmnminuszn αu+czn+m+j n isin Z+ (213) nminuskj

k=1 m=(nminuszn)c1

Concerning the function ψnij(u) when n le nr and point (i) above is considered we note Lnminus1that i must necessarily equal u + czn + m (again conditional on i=zn

Xi = m) for ruin to

occur in this manner Hence in a similar fashion we acquire

minnminus1nr (nminuszn)c2 L L

ψnij(u) = rkψn

(kminus) kij(u) + rn dmnminuszn

δiu+czn+mαu+czn+m+j

k=1 m=(nminuszn)c1

minnminus1nr L

(k) Z

+ = rkψnminuskij(u) + rndiminusuminuscznnminuszn αi+j n isin (214)

k=1

Finally for fixed i isin minc Z + c1 minc Z + c1 + 1 mα minus 1 summing (214) from

j = 1 to mα minus i yields the following expression for the bivariate pmf ωni(u) namely

minnminus1nr mαminusi mαminusi L L

(k) L

ωni(u) = rk ψnminuskij(u) + rndiminusuminuscznnminusznαi+j

k=1 j=1 j=1

10

minnminus1nr L

(k) Z

+ = Λi rkg s + rndiminusuminuscznnminuszn n isin (215) nminuskminus1iminuspnminus1

k=1

where (210) was used to establish the last equality In the special case when c1 = c2 = c it

is a straightforward exercise to show that equations (213) through (215) simplify to produce

equations (215) through (217) of Alfa and Drekic (2007 p 300)

3 Computational considerations

In order to compute the joint pmfrsquos defined by (213) (214) and (215) it is clear from (k)

equations (29) through (212) that we need to be able to compute the quantities gn and

h(k) n defined by (27) and (28) respectively In what follows we capitalize on the underlying

Markov chain structure in our model to develop a set of computational procedures which

ultimately enable one to calculate these key row vectors

First of all recall that g0(k)

= b(k) contains zeros from a certain level onwards namely (k)

level vk(zk) = u + czk + c2(k minus zk) In general one can easily deduce that gn will also

contain zeros from a certain level onwards to be denoted by lk(n) We begin this section

with a determination of lk(n) which varies in form depending on the value of vk(zk) minus 1

representing the maximum surplus level at time k after the smallest possible claim of size 1

has been applied (at time kminus)

On one hand it may be the case that the value of vk(zk)minus1 lies at or above the threshold

level Z In this situation we remark that zk can take on any value in the set 0 1 k (k)

Since g0 contains zeros from level lk(0) = vk(zk) onwards we observe that at time k + 1

it may be possible to achieve a higher surplus value than vk+1(zk) minus 1 representing the

maximum surplus level assuming there was no drop below Z due to the claim at time kminus

Specifically if the surplus had fallen to level Z minus 1 at time kminus then gained c at time k+ a

bounded level of Z + c minus1 would have been reached at time k +1 and this level is potentially

larger than vk+1(zk) minus 1 As a result we have that lk(1) = maxvk+1(zk) Z + c If we

extend this argument inductively we readily obtain lk(n) = maxvk+n(zk) Z +c +c2(n minus1)

n isin Z+ However since vk+n(zk) = u + czk + c2(k +n minus zk) = vk(zk)+ c2 + c2(n minus 1) we may

alternatively re-express lk(n) as

lk(n) = maxvk+n(zk) Z + c + c2(n minus 1) = maxvk(zk) + c2 Z + c + c2(n minus 1)

Thus

Z+lk(n) = maxlk(0) + c2 Z + c + c2(n minus 1) n isin (31)

11

or equivalently

Z+lk(n) = lk(1) + c2(n minus 1) n isin

Clearly lk(0) gt Z and lk(n) ge Z + c for n isin Z+

The other remaining case to consider involves the situation when the maximum surplus

level at time k is strictly below the threshold level Z Under this scenario we further

distinguish between two cases involving the value of c2 First of all suppose that c2 = 0 so

that vk(zk)minus1 = u +czk minus1 lt Z or equivalently u +czk le Z If u +czk happens to equal Z

then it is possible for zk isin 0 1 k With c2 = 0 however no further premiums can be

earned until after the first claim takes place (which is assumed to occur at time kminus) thereby

implying that Z = u + ci for all i = zk zk + 1 k Hence we know that u + ck = Z

in this case Moreover if u + czk happens to be strictly less than Z then we clearly have

that zk = k Even in the case when c2 gt 0 one can readily verify that zk = k so that

vk(zk) minus 1 = u + ck minus 1 lt Z Thus we proceed with the analysis assuming that zk = k

c2 ge 0 and the inequality u + ck le Z holds true

lowast k

Define the ceiling function denoted lceilxrceil to yield the smallest integer greater than or

equal to x Because a random premium cannot be received until level Z is at least achieved

it is essential to find the point at which the threshold level Z is reached (ie the point at Zminus(u+ck)+1 which u + ck minus1+ ci ge Z) Isolating this inequality for i one obtains i ge

c Hence

assuming that W1 = k the elapsed time after k at which a random premium would begin to

lceilZminus(u+ck)+1 be received is t lowast = rceil Note that t lowast isin Z+ k c k

With zk = k g0(k)

contains zeros from level u + ck onwards so that lk(0) = u + ck

Immediately following this a premium of c is guaranteed to be received at time k+ as the

threshold level Z has not yet been reached Consequently g1(k)

contains zeros from level (k)

b(k)u + c(k + 1) onwards which is c levels further than that in g0 = Continuing this (k)

process inductively we can readily establish that gn contains zeros from level u + c(k + n)

onwards for n = 1 2 t lowast k so that lk(n) = u + c(k + n) for n = 0 1 t lowast k

In the case of n = t lowast + 1 t lowast + 2 gt

(k) is first looked at in isolation At time k +k k +1

t lowast k + 1 the threshold level has been crossed giving rise to a maximum surplus value of either (k)

c(k + t lowast k contains zeros from level lk(t lowast ku + ) + c2 minus 1 or Z + c minus 1 Therefore g + 1) = t lowast k+1

+1(k + t lowast kmaxvk+t lowast k

) Z + c onwards Carrying out the same inductive process as before it is

readily seen that g(nk)

contains zeros from level lk(n) = maxvk+n(k+t lowast k) Z +c+c2(nminust lowast kminus1)

onwards for n = t lowast k + 1 t lowast k + 2 Combining both cases we obtain

u + c(k + n) if n = 0 1 t lowast lk(n) = k (32)

maxlk(t lowast) + c2 Z + c + c2(n minus t lowast minus 1) if n = t lowast + 1 t lowast + 2 k k k k

Note that by construction lk(t lowast k) gt Z and lk(n) le Z for n = 0 1 t lowast k minus1 Interestingly

12

if we extend the definition tk lowast by introducing the quantity ζk

lowast = maxtklowast 0 we remark that

(31) and (32) can actually be combined into a single formula for lk(n) namely

vk+n(zk + n) if n = 0 1 ζklowast

lk(n) = (33) maxlk(ζ

lowast) + c2 Z + c + c2(n minus ζlowast minus 1) if n = ζlowast + 1 ζlowast + 2 k k k k

Clearly we have that lk(n) ge Z + c for n = ζk lowast +1 ζk

lowast +2 Moreover if c1 = c2 = c (33)

reduces to give lk(n) = u + c(k + n) forall n isin N which is in agreement with the results in Alfa

and Drekic (2007 Section 3)

Remark In our above analysis we purposely ignored the possibility of any future claims

occurring after time k in our determination of lk(n) However we recognize that this is not

realistic and future claims would indeed occur after time k thereby further reducing the value

of the maximum surplus level lk(n) minus 1 at time k + n To account for this we imagine all

future interclaim times (ie after time k) to be of maximum length na and all corresponding

claim sizes to be of minimum amount 1 Consequently in the case when c2 gt 0 we should

subtract from formula (33) the amount of lfloornnarfloor where lfloormiddotrfloor again denotes the floor function

Obviously when c2 = 0 (ie no premiums are collected whenever the surplus level is at or

above Z) lk(n) = Z + c for n = ζk lowast + 1 ζk

lowast + 2 (k)

We now turn our attention to the development of a general algorithm for computing gni

(k) (k)for i = 0 1 lk(n) minus 1 With the aid of (24) we apply the key identity gn = gnminus1C

(which can be inferred from (27)) to first obtain

minZlk(nminus1)minus1 lk(nminus1)minus1 c2 L L L

(k) (k) (k)gni = g Bjminusi+c + gnminus1j dmBjminusi+m n isin Z+ (34) nminus1j

j=0 j=Z m=c1

However (22) states that Bi = Ona for i isin Zminus and so (34) becomes (for the same range of

i)

minZlk(nminus1)minus1 c2 lk(nminus1)minus1 L L L

(k) (k) (k)gni = g Bjminusi+c + dm gnminus1jBjminusi+m n isin Z+ (35) nminus1j

j=max0iminusc m=c1 j=maxZiminusm

Taking the form of (33) into consideration (35) yields the following two formulae

lk(nminus1)minus1 L

(k) (k)gni = gnminus1jBjminusi+c n = 1 2 ζ k

lowast (36) j=max0iminusc

and

Zminus1 c2 lk(nminus1)minus1 L L L

(k) (k) (k)g = g + dm g n = ζ lowast + 1 ζ lowast + 2 ni nminus1jBjminusi+c nminus1jBjminusi+m k k

j=max0iminusc m=c1 j=maxZiminusm

(37)

13

For n = 1 2 ζk lowast where ζk

lowast isin Z+ we note that lk(n) ge c + 1 Therefore with the aid of

(22) (36) simplifies to give

Llk(nminus1)minus1 (k) (k) j=0 gnminus1jBjminusi+c if i = 0 1 c minus 1

gni = Llk(nminus1)minus1 (k)

g if i = c c + 1 lk(n) minus 1j=iminusc nminus1jBjminusi+c

Llk(nminus1)minus1 (k)

nminus1js 0 0 if i = 0 1 c minus 1j=0 αjminusi+cg= (38)

(k) Llk(nminus1)minus1 (k) g S+ αjminusi+cgnminus1js 0 0 if i = c c+1 lk(n)minus1 nminus1iminusc j=iminusc+1

In the case of n = ζk lowast + 1 ζk

lowast + 2 we first reiterate that lk(n) ge Z + c We next break

(37) into several cases based on the value of i and the value of minc Z + c1 as follows

Case 1 Suppose minc Z + c1 = c

For i isin 0 1 c minus 1

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+mni nminus1j nminus1j

j=0 m=c1 j=Z

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k)= αjminusi+cg s + dm αjminusi+mg s 0 0 (39) nminus1j nminus1j

j=0 m=c1 j=Z

For i isin c c + 1 Z + c1 minus 1

Zminus1 c2 lk(nminus1)minus1 L L L

(k) (k) (k)g = g + dm gni nminus1j Bjminusi+c nminus1j Bjminusi+m

j=iminusc m=c1 j=Z

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k) (k)= g S + αjminusi+cg dm αjminusi+mgnminus1j s 0 0 (310) nminus1iminusc nminus1j s +

j=iminusc+1 m=c1 j=Z

For i isin Z + c1 Z + c1 + 1 Z + c minus 1

Zminus1 minc2iminusZ lk(nminus1)minus1 c2 lk(nminus1)minus1 L L L L L

(k) (k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+m + dm g Bjminusi+mni nminus1j nminus1j nminus1j

j=iminusc m=c1 j=iminusm m=iminusZ+1 j=Z

minc2iminusZ L

(k) (k)= g S + dmg S nminus1iminusc nminus1iminusm

m=c1

Zminus1 minc2iminusZ lk(nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k)+ αjminusi+cg dm αjminusi+mg dm αjminusi+mgnminus1j s 0 0 nminus1j s + nminus1js +

j=iminusc+1 m=c1 j=iminusm+1 m=iminusZ+1 j=Z

(311)

For i isin Z + c Z + c + 1 lk(n) minus 1

c2 lk(nminus1)minus1 L L

(k) (k)g = dm g Bjminusi+mni nminus1j

m=c1 j=iminusm

c2 c2 lk (nminus1)minus1 L L L

(k) (k)= dmg S + dm αjminusi+mgnminus1j s 0 0 (312) nminus1iminusm

m=c1 m=c1 j=iminusm+1

14

Case 2 Suppose minc Z + c1 = Z + c1

For i isin 0 1 Z + c1 minus 1

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+mni nminus1j nminus1j

j=0 m=c1 j=Z

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k)= αjminusi+cg s + dm αjminusi+mg s 0 0 (313) nminus1j nminus1j

j=0 m=c1 j=Z

For i isin Z + c1 Z + c1 + 1 c minus 1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+m + dm g Bjminusi+mni nminus1j nminus1j nminus1j

j=0 m=c1 j=iminusm m=iminusZ+1 j=Z

minc2iminusZ L

(k)= dmg S nminus1iminusm

m=c1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k)+ αjminusi+cg s + dm αjminusi+mg s + dm αjminusi+mg s 0 0 nminus1j nminus1j nminus1j

j=0 m=c1 j=iminusm+1 m=iminusZ+1 j=Z

(314)

For i isin c c + 1 Z + c minus 1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k) (k)g = g + g + gni nminus1j Bjminusi+c dm nminus1j Bjminusi+m dm nminus1j Bjminusi+m

j=iminusc m=c1 j=iminusm m=iminusZ+1 j=Z

minc2iminusZ L

(k) (k)= g S + dmg S nminus1iminusc nminus1iminusm

m=c1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k)+

nminus1j s 0 0 αjminusi+cgnminus1j s + dm αjminusi+mgnminus1j s + dm αjminusi+mg j=iminusc+1 m=c1 j=iminusm+1 m=iminusZ+1 j=Z

(315)

For i isin Z + c Z + c + 1 lk(n) minus 1

c2 lk (nminus1)minus1 L L

(k) (k)g = dm g Bjminusi+mni nminus1j

m=c1 j=iminusm

c2 c2 lk (nminus1)minus1 L L L

(k) (k)= dmg S + dm αjminusi+mgnminus1j s 0 0 (316) nminus1iminusm

m=c1 m=c1 j=iminusm+1

It is not difficult to verify that equations (38) through (316) simplify to produce equations

(32) and (33) of Alfa and Drekic (2007 pp 301-302) when c1 = c2 = c With equations (38) (k)

through (316) providing the recursive means of calculating gn we next turn our attention

15

to a determination of h(k) defined by (28) For n = 1 2 ζlowast we use (22) and (25) to n k

obtain

lk(nminus1)minus1 lk(nminus1)minus1

h(k) nminusj =

L g

(k) nminus1ℓBc+j+ℓ =

L αc+j+ℓg

(k) nminus1ℓs 0 0 (317)

ℓ=0 ℓ=0

For n = ζk lowast + 1 ζk

lowast + 2 we similarly acquire

Zminus1 lk(nminus1)minus1 L L

(k) (k) (k)hnminusj = gnminus1ℓBc+j+ℓ + gnminus1ℓAj+ℓ

ℓ=0 ℓ=Z

Zminus1 c2 lk(nminus1)minus1 L L L

(k) (k)= αc+j+ℓg s + dm αm+j+ℓg s 0 0 (318) nminus1ℓ nminus1ℓ

ℓ=0 m=c1 ℓ=Z

Note that substituting (317) and (318) into (29) immediately yields

Llk(nminus1)minus1 (k) (k) ℓ=0 αc+j+ℓgnminus1ℓs if n = 1 2 ζk

lowast φ (u) = nj LZminus1 (k) Lc2

Llk(nminus1)minus1 (k) ζlowast s + dm αm+j+ℓg n = + 1 ζlowast + 2 ℓ=0 αc+j+ℓgnminus1ℓ m=c1 ℓ=Z nminus1ℓs if k k

(319)

Clearly in the special case when c1 = c2 = c (319) yields equation (35) of Alfa and Drekic

(2007 p 302) Finally we remark that (319) can be substituted into (213) to obtain a

computational formula for φnj(u)

4 Numerical examples and findings

In this section we illustrate the application of our proposed algorithm with several numerical

examples All calculations in this section were performed using Mathematica (Version 7) on

a Dell Precision T7400 workstation with a 3 Ghz CPU and 4 GB of RAM

In our first illustration (labelled Example 1) we consider the ordinary Sparre Andersen

model in which interclaim times have pmf given by

(211)(911)jminus1 if j = 1 2 na minus 1 aj = (41)

(911)naminus1 if j = na

Note that the pmf (41) is that of a truncated geometric distribution with all the probability

mass on na na +1 assigned to the support value na Clearly as na becomes larger the

closer (41) approximates this particular geometric distribution having mean 55 In what Lna

Lminus1 nafollows we fix na = 25 so that aj = 0992 and jaj = 546 Let the individual j=1 j=1

claim amount distribution be given by the pmf

minus4 minus4j minus 1 j

αj = 1 + minus 1 + j isin Z+ (42) 30 30

16

We remark that the pmf (42) represents a discretized version of the Pareto distribution with

mean 10 Also note that mα = infin thereby implying that the random variables |UT | and

UTminus both have infinite support

Table 1 displays values (rounded to 6 decimal places) of the trivariate cumulative distrishy

bution function

n x y L L L

Ψnxy(u) = Pr T le n UTminus le x |UT | le y | U0 = u = ψℓij(u) ℓ=1 i=mincZ+c1 j=1

corresponding to the above discrete-time risk process with u = 10 Z = 50 c = 5 and

varying premium amount distributions according to the following set of scenarios

(1) d2 = 1

(2) d1 = 25 d2 = 15 d3 = 25

G

5)

(3) di = i

(25)i(35)5minusi i = 0 1 5

(4) di = 15 i = 0 1 4

(5) d1 = 34 d5 = 14

(6) d0 = 35 d5 = 25

We remark that while these six premium amount distributions all have mean equal to 2

they possess different shapes and we investigate the impact of their variability on the joint

distribution of T |UT | and UTminus In particular the variances corresponding to the premium

amount distributions of scenarios (1) through (6) are 0 08 12 2 3 and 6 respectively

Based on the results in Table 1 it would appear that greater variability in the premium

amount distribution generally leads to higher ruin probabilities Perhaps somewhat surprising

though the differences are subtle and not particularly striking as one compares the various

ruin probabilities from scenarios (1) through (6) Obviously we observe more of a difference

between scenarios as the values of n x and y increase

As a follow-up illustration (labelled Example 2) we consider a model in which ordinary

interclaim times have a discrete uniform distribution with pmf aj = 110 j = 1 2 10

Clearly the mean of this distribution is 55 which for comparative purposes is identical to

the ordinary interclaim time mean considered in Example 1 In addition we assume that

the pmf of the individual claim amount distribution is once again given by (42) Table 2

displays values (rounded to 6 decimal places) of Ψnxy(10) for several variants of the Sparre

Andersen model (again with Z = 50 and c = 5) having the above ordinary interclaim time

and claim amount distributions namely

17

(1) ordinary model under scenario (2) from Example 1

(2) ordinary model under scenario (5) from Example 1

(3) stationary model under scenario (2) from Example 1

(4) stationary model under scenario (5) from Example 1

(5) delayed model (with W1 = 1 with probability 1) under scenario (2) from Example 1

(6) delayed model (with W1 = 1 with probability 1) under scenario (5) from Example 1

Concerning the results in Table 2 we first compare the values of Ψnxy(10) in entries (2)

and (5) of Table 1 to their corresponding counterparts in entries (1) and (2) of Table 2 In all

cases we observe that ruin is more likely to occur when interclaim times have the truncated

geometric distribution introduced in our first example as opposed to the discrete uniform

distribution of our second example Since mean interclaim times are equal in both examshy

ples this behaviour must be attributable to the greater variability inherent in the truncated

geometric distribution We also note that under the stationary model assumption the pmf

of W1 is easily verified to be rj = (11 minus j)55 j = 1 2 10 which has mean 4 lt 55

As a result the values of Ψnxy(10) for stationary models (3) and (4) are always larger than

the corresponding values for ordinary models (1) and (2) Finally in the absence of any

knowledge regarding which distribution is in effect for the first interclaim time W1 models

(5) and (6) can be regarded as ldquoworst-case scenariordquo models as they serve to provide upper

bounds on the values of Ψnxy(10)

In conclusion we state that the values in Tables 1 and 2 were generated by first impleshy

menting the recursive procedures (38) through (316) and then summing the appropriate

bivariatetrivariate probabilities computed via (213) (214) or (215) In situations when

x = y = infin however we computed the resulting finite-time ruin probabilities using an altershy

native (albeit standard) method (eg see Cossette et al (2006 Section 3)) Specifically if

we define the survival probability σ(n u) = Pr T gt n | U0 = u = 1minusΨninfininfin(u) as well as

the auxiliary function

0 if w ge Z ηwt =

minlfloorZminuswc minus1 rfloor + 1 t if w lt Z

then by conditioning on the time and amount of the first claim (and recalling that the delayed

process reverts to the ordinary process following this claim) we obtain the expression

minnnr (kminuszk)c2 u+czk+ℓ L L L

σ(u n) = Rn + rk dℓkminuszk αjγ(u + czk + ℓ minus j n minus k)

k=1 ℓ=(kminuszk)c1 j=1

18

where the function γ(w m) for w m isin N is computed recursively via

minmna (kminusηwk)c2 u+cηwk+ℓ L L L

γ(w m) = Am + ak dℓkminusηwk αjγ(u + cηwk + ℓ minus j m minus k)

k=1 j=1 ℓ=(kminusηwk)c1

with starting values γ(w 0) = 1 forall w isin N

Acknowledgements

The authors would like to thank Dr David Landriault for his valuable suggestions which

helped to improve an earlier version of this work This research was supported by the Natural

Sciences and Engineering Research Council of Canada

References

Alfa AS 2004 Markov chain representations of discrete distributions applied to queueing

models Computers amp Operations Research 31 2365-2385

Alfa AS Drekic S 2007 Algorithmic analysis of the Sparre Andersen model in discrete

time ASTIN Bulletin 37 293-317

Bao Z 2007 A note on the compound binomial model with randomized dividend strategy

Applied Mathematics amp Computation 194 276-286

Cossette H Landriault D Marceau E 2006 Ruin probabilities in the discrete time

renewal risk model Insurance Mathematics amp Economics 38 309-323

Karlin S Taylor HM 1975 A First Course in Stochastic Processes 2nd edition Acashy

demic Press New York

Kim B Kim H-S Kim J 2008 A risk model with paying dividends and random

environment Insurance Mathematics amp Economics 42 717-726

Landriault D 2008 Randomized dividends in the compound binomial model with a general

premium rate Scandinavian Actuarial Journal 1-15

Tan J Yang X 2006 The compound binomial model with randomized decisions on

paying dividends Insurance Mathematics amp Economics 39 1-18

19

Table 1 ndash Values of Ψnxy(10) corresponding to Example 1

(a) x = 10

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0005357 0006188 0007163 0008278 0009481

(2) 0005359 0006195 0007175 0008296 0009507

(3) 0005361 0006198 0007181 0008305 0009519

(4) 0005363 0006205 0007193 0008323 0009544

(5) 0005366 0006212 0007208 0008346 0009576

(6) 0005374 0006236 0007249 0008410 0009664

y = 25 (1) 0007631 0008815 0010202 0011789 0013503

(2) 0007634 0008824 0010220 0011816 0013539

(3) 0007636 0008828 0010228 0011829 0013557

(4) 0007639 0008838 0010245 0011854 0013593

(5) 0007643 0008849 0010266 0011887 0013638

(6) 0007656 0008882 0010325 0011978 0013764

y = 50 (1) 0008477 0009792 0011333 0013095 0014998

(2) 0008481 0009802 0011352 0013125 0015039

(3) 0008483 0009807 0011361 0013139 0015059

(4) 0008486 0009817 0011380 0013168 0015098

(5) 0008491 0009830 0011403 0013204 0015149

(6) 0008504 0009866 0011470 0013305 0015288

y = infin (1) 0008777 0010138 0011733 0013558 0015528

(2) 0008781 0010149 0011753 0013588 0015570

(3) 0008783 0010154 0011763 0013603 0015591

(4) 0008787 0010164 0011782 0013633 0015632

(5) 0008791 0010177 0011807 0013671 0015684

(6) 0008805 0010215 0011875 0013775 0015828

20

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0058559 0061875 0065704 0070048 0074714

(2) 0058569 0061903 0065756 0070128 0074825

(3) 0058574 0061917 0065782 0070167 0074880

(4) 0058584 0061945 0065833 0070245 0074987

(5) 0058596 0061979 0065896 0070344 0075124

(6) 0058632 0062079 0066078 0070621 0075506

y = 25 (1) 0089002 0093986 0099739 0106262 0113270

(2) 0089017 0094028 0099817 0106382 0113437

(3) 0089025 0094049 0099856 0106442 0113519

(4) 0089040 0094091 0099932 0106560 0113681

(5) 0089057 0094142 0100028 0106708 0113887

(6) 0089113 0094294 0100302 0107127 0114463

y = 50 (1) 0102786 0108506 0115106 0122590 0130629

(2) 0102804 0108555 0115196 0122728 0130820

(3) 0102812 0108579 0115241 0122797 0130914

(4) 0102830 0108627 0115329 0122932 0131101

(5) 0102850 0108686 0115439 0123103 0131337

(6) 0102914 0108860 0115754 0123584 0132000

y = infin (1) 0108895 0114933 0121901 0129799 0138283

(2) 0108914 0114985 0121996 0129945 0138485

(3) 0108923 0115010 0122043 0130018 0138585

(4) 0108941 0115061 0122136 0130161 0138782

(5) 0108962 0115123 0122252 0130341 0139032

(6) 0109030 0115307 0122585 0130849 0139732

21

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0079856 0086792 009464 0103440 0112831

(2) 0079885 0086869 0094778 0103650 0113118

(3) 0079953 0086978 0094936 0103864 0113393

(4) 0080081 0087188 0095243 0104282 0113929

(5) 0079962 0087075 0095149 0104213 0113890

(6) 0080805 0088366 0096951 0106588 0116875

y = 25 (1) 0124591 0135680 0148207 0162245 0177217

(2) 0124639 0135806 0148433 0162586 0177683

(3) 0124755 0135993 0148703 0162950 0178148

(4) 0124978 0136354 0149226 0163658 0179054

(5) 0124763 0136141 0149035 0163500 0178935

(6) 0126238 0138381 0152148 0167588 0184060

y = 50 (1) 0146757 0160079 0175111 0191945 0209894

(2) 0146816 0160232 0175385 0192360 0210460

(3) 0146965 0160469 0175725 0192817 0211043

(4) 0147249 0160925 0176384 0193706 0212178

(5) 0146968 0160639 0176117 0193470 0211980

(6) 0148860 0163500 0180077 0198658 0218474

y = infin (1) 0158013 0172606 0189059 0207476 0227106

(2) 0158078 0172776 0189362 0207934 0227732

(3) 0158250 0173047 0189751 0208455 0228395

(4) 0158579 0173571 0190503 0209467 0229685

(5) 0158247 0173226 0190172 0209161 0229410

(6) 0160443 0176532 0194735 0215127 0236869

22

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0088667 0104403 0123329 0145277 0169184

(2) 0088711 0104518 0123538 0145598 0169627

(3) 0088733 0104574 0123641 0145756 0169846

(4) 0088777 0104687 0123844 0146066 0170274

(5) 0088829 0104828 0124105 0146470 0170832

(6) 0088988 0105229 0124822 0147560 0172331

y = 25 (1) 0140528 0167839 0201007 0239726 0282102

(2) 0140598 0168021 0201338 0240234 0282804

(3) 0140632 0168111 0201502 0240486 0283151

(4) 0140701 0168289 0201822 0240976 0283827

(5) 0140785 0168515 0202241 0241622 0284718

(6) 0141036 0169148 0203370 0243335 0287071

y = 50 (1) 0167826 0203005 0246173 0296954 0352855

(2) 0167908 0203218 0246561 0297548 0353672

(3) 0167949 0203324 0246753 0297841 0354075

(4) 0168030 0203532 0247128 0298412 0354859

(5) 0168129 0203800 0247622 0299174 0355907

(6) 0168423 0204539 0248936 0301159 0358620

y = infin (1) 0183325 0224926 0276881 0338957 0408257

(2) 0183412 0225150 0277283 0339564 0409079

(3) 0183455 0225260 0277481 0339864 0409485

(4) 0183540 0225478 0277869 0340446 0410270

(5) 0183645 0225760 0278386 0341234 0411339

(6) 0183954 0226529 0279737 0343244 0414037

23

Table 2 ndash Values of Ψnxy(10) corresponding to Example 2

(a) x = 10

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0001901 0002276 0002724 0003240 0003799

(2) 0001904 0002286 0002742 0003269 0003840

(3) 0002651 0003013 0003442 0003936 0004471

(4) 0002655 0003022 0003460 0003964 0004510

(5) 0005792 0006098 0006458 0006871 0007317

(6) 0005795 0006106 0006473 0006894 0007350

y = 25 (1) 0002708 0003243 0003880 0004615 0005411

(2) 0002713 0003257 0003906 0004656 0005469

(3) 0003778 0004293 0004904 0005608 0006369

(4) 0003782 0004306 0004929 0005647 0006424

(5) 0008255 0008690 0009203 0009792 0010427

(6) 0008259 0008702 0009224 0009824 0010473

y = 50 (1) 0003008 0003603 0004310 0005127 0006011

(2) 0003014 0003618 0004340 0005173 0006076

(3) 0004197 0004769 0005448 0006230 0007075

(4) 0004202 0004784 0005476 0006273 0007137

(5) 0009173 0009656 0010226 0010879 0011585

(6) 0009177 0009669 0010249 0010916 0011637

y = infin (1) 0003115 0003730 0004463 0005308 0006224

(2) 0003121 0003746 0004493 0005356 0006291

(3) 0004346 0004938 0005641 0006450 0007325

(4) 0004351 0004954 0005670 0006496 0007389

(5) 0009499 0009999 0010589 0011265 0011996

(6) 0009503 0010012 0010613 0011303 0012050

24

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0031176 0032815 0034733 0036924 0039282

(2) 0031193 0032862 0034821 0037061 0039475

(3) 0050168 0051747 0053587 0055684 0057939

(4) 0050184 0051792 0053671 0055815 0058124

(5) 0136884 0138219 0139763 0141517 0143399

(6) 0136896 0138256 0139832 0141626 0143553

y = 25 (1) 0047622 0050092 0052981 0056280 0059829

(2) 0004765 0050163 0053113 0056487 0060121

(3) 0076570 0078950 0081720 0084878 0088272

(4) 0076593 0079017 0081847 0085076 0088551

(5) 0205774 0207786 0210111 0212752 0215585

(6) 0205793 0207842 0210216 0212917 0215818

y = 50 (1) 0055170 0058009 0061329 0065119 0069196

(2) 0055200 0058091 0061481 0065357 0069533

(3) 0088650 0091385 0094569 0098197 0102096

(4) 0088677 0091463 0094715 0098425 0102417

(5) 0235815 0238128 0240800 0243834 0247089

(6) 0235837 0238192 0240920 0244023 0247356

y = infin (1) 0058565 0061565 0065073 0069077 0073384

(2) 0058597 0061652 0065234 0069329 0073740

(3) 0094061 0096952 0100316 0104149 0108267

(4) 0094090 0097035 0100470 0104390 0108607

(5) 0248472 0250917 0253740 0256945 0260383

(6) 0248495 0250985 0253867 0257145 0260667

25

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0048598 0052524 0056984 0061998 0067342

(2) 0048654 0052673 0057253 0062411 0067916

(3) 0070192 0073979 0078257 0083057 0088167

(4) 0070243 0074119 0078513 0083451 0088714

(5) 0151774 0154981 0158572 0162586 0166852

(6) 0151815 0155096 0158784 0162913 0167308

y = 25 (1) 0076895 0083230 0090413 0098481 0107073

(2) 0076987 0083474 0090854 0099156 0108010

(3) 0110093 0116205 0123095 0130818 0139033

(4) 0110177 0116435 0123515 0131462 0139928

(5) 0230674 0235848 0241632 0248091 0254950

(6) 0230739 0236038 0241980 0248626 0255695

y = 50 (1) 0091496 0099156 0107829 0117563 0127924

(2) 0091610 0099455 0108370 0118387 0129068

(3) 0130130 0137520 0145839 0155156 0165064

(4) 0130233 0137801 0146352 0155942 0166156

(5) 0266599 0272857 0279840 0287632 0295904

(6) 0266680 0273088 0280265 0288286 0296814

y = infin (1) 0099312 0107746 0117285 0127984 0139369

(2) 0099439 0108078 0117883 0128899 0140639

(3) 0140458 0148595 0157745 0167986 0178872

(4) 0140572 0148908 0158314 0168859 0180084

(5) 0282886 0289776 0297457 0306022 0315111

(6) 0282976 0290034 0297929 0306748 0316120

26

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0059096 0072505 0088907 0108122 0129137

(2) 0059223 0072829 0089500 0109052 0130456

(3) 0079757 0092662 0108388 0126778 0146871

(4) 0079875 0092970 0108956 0127669 0148134

(5) 0159788 0170682 0183871 0199249 0216024

(6) 0159881 0170936 0184343 0199990 0217075

y = 25 (1) 0095883 0119741 0149240 0184073 0222390

(2) 0096087 0120262 0150197 0185573 0224520

(3) 0127403 0150359 0178640 0211977 0248613

(4) 0127592 0150855 0179557 0213414 0250653

(5) 0245161 0264533 0288250 0316124 0346709

(6) 0245311 0264942 0289010 0317319 0348407

y = 50 (1) 0116598 0147920 0187100 0233784 0285503

(2) 0116839 0148537 0188231 0235554 0288015

(3) 0153023 0183153 0220713 0265390 0314840

(4) 0153249 0183742 0221798 0267087 0317248

(5) 0285742 0311159 0342651 0380005 0421287

(6) 0285921 0311644 0343551 0381418 0423292

y = infin (1) 0129470 0167217 0215383 0273841 0339734

(2) 0129725 0167865 0216560 0275661 0342282

(3) 0167978 0204275 0250442 0306382 0369383

(4) 0168217 0204895 0251572 0308128 0371826

(5) 0305868 0336470 0375169 0421935 0474525

(6) 0306058 0336980 0376106 0423388 0476560

27

Page 10: uwaterloo.ca · Ruin Analysis of a Threshold Strategy in a Discrete-time Sparre Andersen Model Steve DREKIC and Ana Maria MERA Department of Statistics and Actuarial Science University

fact that the surplus immediately prior to ruin cannot exceed mα minus1 as indicated in Section

1 The justification of (212) when c is larger than Z + c1 mirrors the analysis above and is

left to the reader D

Recall that in the analysis up to this point we have assumed that W1 = k isin

1 2 3 nr If we now condition on the value of W1 the Law of Total Probability

yields the following two expressions for φnj(u) and ψnij(u) namely

nr L

= rkφ (u) n = nr + 1 nr + 2 φnj(u) n

(kminus) kj

k=1

and nr L

(k)ψnij(u) = rkψ (u) n = nr + 1 nr + 2 nminuskij

k=1

In addition however it is quite possible that T = n for n = 1 2 nr and this can happen

as a result of either

(i) the first claim which causes ruin occurring at time nminus from a (conditional) surplus Lnminus1level of u + czn + m (where m) or i=zn

Xi =

(ii) the first claim which does not cause ruin occurring at some time kminus k isin 1 2 nminus

1 and ruin subsequently occurring n minus k time units later

If we now combine these outcomes with those for n = nr + 1 nr + 2 (and recall that

rn = 0 forall n gt nr) the general expression for φnj(u) becomes

minnminus1nr (nminuszn)c2 L

(k) L

φnj(u)= rkφ (u) + rn dmnminuszn αu+czn+m+j n isin Z+ (213) nminuskj

k=1 m=(nminuszn)c1

Concerning the function ψnij(u) when n le nr and point (i) above is considered we note Lnminus1that i must necessarily equal u + czn + m (again conditional on i=zn

Xi = m) for ruin to

occur in this manner Hence in a similar fashion we acquire

minnminus1nr (nminuszn)c2 L L

ψnij(u) = rkψn

(kminus) kij(u) + rn dmnminuszn

δiu+czn+mαu+czn+m+j

k=1 m=(nminuszn)c1

minnminus1nr L

(k) Z

+ = rkψnminuskij(u) + rndiminusuminuscznnminuszn αi+j n isin (214)

k=1

Finally for fixed i isin minc Z + c1 minc Z + c1 + 1 mα minus 1 summing (214) from

j = 1 to mα minus i yields the following expression for the bivariate pmf ωni(u) namely

minnminus1nr mαminusi mαminusi L L

(k) L

ωni(u) = rk ψnminuskij(u) + rndiminusuminuscznnminusznαi+j

k=1 j=1 j=1

10

minnminus1nr L

(k) Z

+ = Λi rkg s + rndiminusuminuscznnminuszn n isin (215) nminuskminus1iminuspnminus1

k=1

where (210) was used to establish the last equality In the special case when c1 = c2 = c it

is a straightforward exercise to show that equations (213) through (215) simplify to produce

equations (215) through (217) of Alfa and Drekic (2007 p 300)

3 Computational considerations

In order to compute the joint pmfrsquos defined by (213) (214) and (215) it is clear from (k)

equations (29) through (212) that we need to be able to compute the quantities gn and

h(k) n defined by (27) and (28) respectively In what follows we capitalize on the underlying

Markov chain structure in our model to develop a set of computational procedures which

ultimately enable one to calculate these key row vectors

First of all recall that g0(k)

= b(k) contains zeros from a certain level onwards namely (k)

level vk(zk) = u + czk + c2(k minus zk) In general one can easily deduce that gn will also

contain zeros from a certain level onwards to be denoted by lk(n) We begin this section

with a determination of lk(n) which varies in form depending on the value of vk(zk) minus 1

representing the maximum surplus level at time k after the smallest possible claim of size 1

has been applied (at time kminus)

On one hand it may be the case that the value of vk(zk)minus1 lies at or above the threshold

level Z In this situation we remark that zk can take on any value in the set 0 1 k (k)

Since g0 contains zeros from level lk(0) = vk(zk) onwards we observe that at time k + 1

it may be possible to achieve a higher surplus value than vk+1(zk) minus 1 representing the

maximum surplus level assuming there was no drop below Z due to the claim at time kminus

Specifically if the surplus had fallen to level Z minus 1 at time kminus then gained c at time k+ a

bounded level of Z + c minus1 would have been reached at time k +1 and this level is potentially

larger than vk+1(zk) minus 1 As a result we have that lk(1) = maxvk+1(zk) Z + c If we

extend this argument inductively we readily obtain lk(n) = maxvk+n(zk) Z +c +c2(n minus1)

n isin Z+ However since vk+n(zk) = u + czk + c2(k +n minus zk) = vk(zk)+ c2 + c2(n minus 1) we may

alternatively re-express lk(n) as

lk(n) = maxvk+n(zk) Z + c + c2(n minus 1) = maxvk(zk) + c2 Z + c + c2(n minus 1)

Thus

Z+lk(n) = maxlk(0) + c2 Z + c + c2(n minus 1) n isin (31)

11

or equivalently

Z+lk(n) = lk(1) + c2(n minus 1) n isin

Clearly lk(0) gt Z and lk(n) ge Z + c for n isin Z+

The other remaining case to consider involves the situation when the maximum surplus

level at time k is strictly below the threshold level Z Under this scenario we further

distinguish between two cases involving the value of c2 First of all suppose that c2 = 0 so

that vk(zk)minus1 = u +czk minus1 lt Z or equivalently u +czk le Z If u +czk happens to equal Z

then it is possible for zk isin 0 1 k With c2 = 0 however no further premiums can be

earned until after the first claim takes place (which is assumed to occur at time kminus) thereby

implying that Z = u + ci for all i = zk zk + 1 k Hence we know that u + ck = Z

in this case Moreover if u + czk happens to be strictly less than Z then we clearly have

that zk = k Even in the case when c2 gt 0 one can readily verify that zk = k so that

vk(zk) minus 1 = u + ck minus 1 lt Z Thus we proceed with the analysis assuming that zk = k

c2 ge 0 and the inequality u + ck le Z holds true

lowast k

Define the ceiling function denoted lceilxrceil to yield the smallest integer greater than or

equal to x Because a random premium cannot be received until level Z is at least achieved

it is essential to find the point at which the threshold level Z is reached (ie the point at Zminus(u+ck)+1 which u + ck minus1+ ci ge Z) Isolating this inequality for i one obtains i ge

c Hence

assuming that W1 = k the elapsed time after k at which a random premium would begin to

lceilZminus(u+ck)+1 be received is t lowast = rceil Note that t lowast isin Z+ k c k

With zk = k g0(k)

contains zeros from level u + ck onwards so that lk(0) = u + ck

Immediately following this a premium of c is guaranteed to be received at time k+ as the

threshold level Z has not yet been reached Consequently g1(k)

contains zeros from level (k)

b(k)u + c(k + 1) onwards which is c levels further than that in g0 = Continuing this (k)

process inductively we can readily establish that gn contains zeros from level u + c(k + n)

onwards for n = 1 2 t lowast k so that lk(n) = u + c(k + n) for n = 0 1 t lowast k

In the case of n = t lowast + 1 t lowast + 2 gt

(k) is first looked at in isolation At time k +k k +1

t lowast k + 1 the threshold level has been crossed giving rise to a maximum surplus value of either (k)

c(k + t lowast k contains zeros from level lk(t lowast ku + ) + c2 minus 1 or Z + c minus 1 Therefore g + 1) = t lowast k+1

+1(k + t lowast kmaxvk+t lowast k

) Z + c onwards Carrying out the same inductive process as before it is

readily seen that g(nk)

contains zeros from level lk(n) = maxvk+n(k+t lowast k) Z +c+c2(nminust lowast kminus1)

onwards for n = t lowast k + 1 t lowast k + 2 Combining both cases we obtain

u + c(k + n) if n = 0 1 t lowast lk(n) = k (32)

maxlk(t lowast) + c2 Z + c + c2(n minus t lowast minus 1) if n = t lowast + 1 t lowast + 2 k k k k

Note that by construction lk(t lowast k) gt Z and lk(n) le Z for n = 0 1 t lowast k minus1 Interestingly

12

if we extend the definition tk lowast by introducing the quantity ζk

lowast = maxtklowast 0 we remark that

(31) and (32) can actually be combined into a single formula for lk(n) namely

vk+n(zk + n) if n = 0 1 ζklowast

lk(n) = (33) maxlk(ζ

lowast) + c2 Z + c + c2(n minus ζlowast minus 1) if n = ζlowast + 1 ζlowast + 2 k k k k

Clearly we have that lk(n) ge Z + c for n = ζk lowast +1 ζk

lowast +2 Moreover if c1 = c2 = c (33)

reduces to give lk(n) = u + c(k + n) forall n isin N which is in agreement with the results in Alfa

and Drekic (2007 Section 3)

Remark In our above analysis we purposely ignored the possibility of any future claims

occurring after time k in our determination of lk(n) However we recognize that this is not

realistic and future claims would indeed occur after time k thereby further reducing the value

of the maximum surplus level lk(n) minus 1 at time k + n To account for this we imagine all

future interclaim times (ie after time k) to be of maximum length na and all corresponding

claim sizes to be of minimum amount 1 Consequently in the case when c2 gt 0 we should

subtract from formula (33) the amount of lfloornnarfloor where lfloormiddotrfloor again denotes the floor function

Obviously when c2 = 0 (ie no premiums are collected whenever the surplus level is at or

above Z) lk(n) = Z + c for n = ζk lowast + 1 ζk

lowast + 2 (k)

We now turn our attention to the development of a general algorithm for computing gni

(k) (k)for i = 0 1 lk(n) minus 1 With the aid of (24) we apply the key identity gn = gnminus1C

(which can be inferred from (27)) to first obtain

minZlk(nminus1)minus1 lk(nminus1)minus1 c2 L L L

(k) (k) (k)gni = g Bjminusi+c + gnminus1j dmBjminusi+m n isin Z+ (34) nminus1j

j=0 j=Z m=c1

However (22) states that Bi = Ona for i isin Zminus and so (34) becomes (for the same range of

i)

minZlk(nminus1)minus1 c2 lk(nminus1)minus1 L L L

(k) (k) (k)gni = g Bjminusi+c + dm gnminus1jBjminusi+m n isin Z+ (35) nminus1j

j=max0iminusc m=c1 j=maxZiminusm

Taking the form of (33) into consideration (35) yields the following two formulae

lk(nminus1)minus1 L

(k) (k)gni = gnminus1jBjminusi+c n = 1 2 ζ k

lowast (36) j=max0iminusc

and

Zminus1 c2 lk(nminus1)minus1 L L L

(k) (k) (k)g = g + dm g n = ζ lowast + 1 ζ lowast + 2 ni nminus1jBjminusi+c nminus1jBjminusi+m k k

j=max0iminusc m=c1 j=maxZiminusm

(37)

13

For n = 1 2 ζk lowast where ζk

lowast isin Z+ we note that lk(n) ge c + 1 Therefore with the aid of

(22) (36) simplifies to give

Llk(nminus1)minus1 (k) (k) j=0 gnminus1jBjminusi+c if i = 0 1 c minus 1

gni = Llk(nminus1)minus1 (k)

g if i = c c + 1 lk(n) minus 1j=iminusc nminus1jBjminusi+c

Llk(nminus1)minus1 (k)

nminus1js 0 0 if i = 0 1 c minus 1j=0 αjminusi+cg= (38)

(k) Llk(nminus1)minus1 (k) g S+ αjminusi+cgnminus1js 0 0 if i = c c+1 lk(n)minus1 nminus1iminusc j=iminusc+1

In the case of n = ζk lowast + 1 ζk

lowast + 2 we first reiterate that lk(n) ge Z + c We next break

(37) into several cases based on the value of i and the value of minc Z + c1 as follows

Case 1 Suppose minc Z + c1 = c

For i isin 0 1 c minus 1

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+mni nminus1j nminus1j

j=0 m=c1 j=Z

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k)= αjminusi+cg s + dm αjminusi+mg s 0 0 (39) nminus1j nminus1j

j=0 m=c1 j=Z

For i isin c c + 1 Z + c1 minus 1

Zminus1 c2 lk(nminus1)minus1 L L L

(k) (k) (k)g = g + dm gni nminus1j Bjminusi+c nminus1j Bjminusi+m

j=iminusc m=c1 j=Z

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k) (k)= g S + αjminusi+cg dm αjminusi+mgnminus1j s 0 0 (310) nminus1iminusc nminus1j s +

j=iminusc+1 m=c1 j=Z

For i isin Z + c1 Z + c1 + 1 Z + c minus 1

Zminus1 minc2iminusZ lk(nminus1)minus1 c2 lk(nminus1)minus1 L L L L L

(k) (k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+m + dm g Bjminusi+mni nminus1j nminus1j nminus1j

j=iminusc m=c1 j=iminusm m=iminusZ+1 j=Z

minc2iminusZ L

(k) (k)= g S + dmg S nminus1iminusc nminus1iminusm

m=c1

Zminus1 minc2iminusZ lk(nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k)+ αjminusi+cg dm αjminusi+mg dm αjminusi+mgnminus1j s 0 0 nminus1j s + nminus1js +

j=iminusc+1 m=c1 j=iminusm+1 m=iminusZ+1 j=Z

(311)

For i isin Z + c Z + c + 1 lk(n) minus 1

c2 lk(nminus1)minus1 L L

(k) (k)g = dm g Bjminusi+mni nminus1j

m=c1 j=iminusm

c2 c2 lk (nminus1)minus1 L L L

(k) (k)= dmg S + dm αjminusi+mgnminus1j s 0 0 (312) nminus1iminusm

m=c1 m=c1 j=iminusm+1

14

Case 2 Suppose minc Z + c1 = Z + c1

For i isin 0 1 Z + c1 minus 1

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+mni nminus1j nminus1j

j=0 m=c1 j=Z

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k)= αjminusi+cg s + dm αjminusi+mg s 0 0 (313) nminus1j nminus1j

j=0 m=c1 j=Z

For i isin Z + c1 Z + c1 + 1 c minus 1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+m + dm g Bjminusi+mni nminus1j nminus1j nminus1j

j=0 m=c1 j=iminusm m=iminusZ+1 j=Z

minc2iminusZ L

(k)= dmg S nminus1iminusm

m=c1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k)+ αjminusi+cg s + dm αjminusi+mg s + dm αjminusi+mg s 0 0 nminus1j nminus1j nminus1j

j=0 m=c1 j=iminusm+1 m=iminusZ+1 j=Z

(314)

For i isin c c + 1 Z + c minus 1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k) (k)g = g + g + gni nminus1j Bjminusi+c dm nminus1j Bjminusi+m dm nminus1j Bjminusi+m

j=iminusc m=c1 j=iminusm m=iminusZ+1 j=Z

minc2iminusZ L

(k) (k)= g S + dmg S nminus1iminusc nminus1iminusm

m=c1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k)+

nminus1j s 0 0 αjminusi+cgnminus1j s + dm αjminusi+mgnminus1j s + dm αjminusi+mg j=iminusc+1 m=c1 j=iminusm+1 m=iminusZ+1 j=Z

(315)

For i isin Z + c Z + c + 1 lk(n) minus 1

c2 lk (nminus1)minus1 L L

(k) (k)g = dm g Bjminusi+mni nminus1j

m=c1 j=iminusm

c2 c2 lk (nminus1)minus1 L L L

(k) (k)= dmg S + dm αjminusi+mgnminus1j s 0 0 (316) nminus1iminusm

m=c1 m=c1 j=iminusm+1

It is not difficult to verify that equations (38) through (316) simplify to produce equations

(32) and (33) of Alfa and Drekic (2007 pp 301-302) when c1 = c2 = c With equations (38) (k)

through (316) providing the recursive means of calculating gn we next turn our attention

15

to a determination of h(k) defined by (28) For n = 1 2 ζlowast we use (22) and (25) to n k

obtain

lk(nminus1)minus1 lk(nminus1)minus1

h(k) nminusj =

L g

(k) nminus1ℓBc+j+ℓ =

L αc+j+ℓg

(k) nminus1ℓs 0 0 (317)

ℓ=0 ℓ=0

For n = ζk lowast + 1 ζk

lowast + 2 we similarly acquire

Zminus1 lk(nminus1)minus1 L L

(k) (k) (k)hnminusj = gnminus1ℓBc+j+ℓ + gnminus1ℓAj+ℓ

ℓ=0 ℓ=Z

Zminus1 c2 lk(nminus1)minus1 L L L

(k) (k)= αc+j+ℓg s + dm αm+j+ℓg s 0 0 (318) nminus1ℓ nminus1ℓ

ℓ=0 m=c1 ℓ=Z

Note that substituting (317) and (318) into (29) immediately yields

Llk(nminus1)minus1 (k) (k) ℓ=0 αc+j+ℓgnminus1ℓs if n = 1 2 ζk

lowast φ (u) = nj LZminus1 (k) Lc2

Llk(nminus1)minus1 (k) ζlowast s + dm αm+j+ℓg n = + 1 ζlowast + 2 ℓ=0 αc+j+ℓgnminus1ℓ m=c1 ℓ=Z nminus1ℓs if k k

(319)

Clearly in the special case when c1 = c2 = c (319) yields equation (35) of Alfa and Drekic

(2007 p 302) Finally we remark that (319) can be substituted into (213) to obtain a

computational formula for φnj(u)

4 Numerical examples and findings

In this section we illustrate the application of our proposed algorithm with several numerical

examples All calculations in this section were performed using Mathematica (Version 7) on

a Dell Precision T7400 workstation with a 3 Ghz CPU and 4 GB of RAM

In our first illustration (labelled Example 1) we consider the ordinary Sparre Andersen

model in which interclaim times have pmf given by

(211)(911)jminus1 if j = 1 2 na minus 1 aj = (41)

(911)naminus1 if j = na

Note that the pmf (41) is that of a truncated geometric distribution with all the probability

mass on na na +1 assigned to the support value na Clearly as na becomes larger the

closer (41) approximates this particular geometric distribution having mean 55 In what Lna

Lminus1 nafollows we fix na = 25 so that aj = 0992 and jaj = 546 Let the individual j=1 j=1

claim amount distribution be given by the pmf

minus4 minus4j minus 1 j

αj = 1 + minus 1 + j isin Z+ (42) 30 30

16

We remark that the pmf (42) represents a discretized version of the Pareto distribution with

mean 10 Also note that mα = infin thereby implying that the random variables |UT | and

UTminus both have infinite support

Table 1 displays values (rounded to 6 decimal places) of the trivariate cumulative distrishy

bution function

n x y L L L

Ψnxy(u) = Pr T le n UTminus le x |UT | le y | U0 = u = ψℓij(u) ℓ=1 i=mincZ+c1 j=1

corresponding to the above discrete-time risk process with u = 10 Z = 50 c = 5 and

varying premium amount distributions according to the following set of scenarios

(1) d2 = 1

(2) d1 = 25 d2 = 15 d3 = 25

G

5)

(3) di = i

(25)i(35)5minusi i = 0 1 5

(4) di = 15 i = 0 1 4

(5) d1 = 34 d5 = 14

(6) d0 = 35 d5 = 25

We remark that while these six premium amount distributions all have mean equal to 2

they possess different shapes and we investigate the impact of their variability on the joint

distribution of T |UT | and UTminus In particular the variances corresponding to the premium

amount distributions of scenarios (1) through (6) are 0 08 12 2 3 and 6 respectively

Based on the results in Table 1 it would appear that greater variability in the premium

amount distribution generally leads to higher ruin probabilities Perhaps somewhat surprising

though the differences are subtle and not particularly striking as one compares the various

ruin probabilities from scenarios (1) through (6) Obviously we observe more of a difference

between scenarios as the values of n x and y increase

As a follow-up illustration (labelled Example 2) we consider a model in which ordinary

interclaim times have a discrete uniform distribution with pmf aj = 110 j = 1 2 10

Clearly the mean of this distribution is 55 which for comparative purposes is identical to

the ordinary interclaim time mean considered in Example 1 In addition we assume that

the pmf of the individual claim amount distribution is once again given by (42) Table 2

displays values (rounded to 6 decimal places) of Ψnxy(10) for several variants of the Sparre

Andersen model (again with Z = 50 and c = 5) having the above ordinary interclaim time

and claim amount distributions namely

17

(1) ordinary model under scenario (2) from Example 1

(2) ordinary model under scenario (5) from Example 1

(3) stationary model under scenario (2) from Example 1

(4) stationary model under scenario (5) from Example 1

(5) delayed model (with W1 = 1 with probability 1) under scenario (2) from Example 1

(6) delayed model (with W1 = 1 with probability 1) under scenario (5) from Example 1

Concerning the results in Table 2 we first compare the values of Ψnxy(10) in entries (2)

and (5) of Table 1 to their corresponding counterparts in entries (1) and (2) of Table 2 In all

cases we observe that ruin is more likely to occur when interclaim times have the truncated

geometric distribution introduced in our first example as opposed to the discrete uniform

distribution of our second example Since mean interclaim times are equal in both examshy

ples this behaviour must be attributable to the greater variability inherent in the truncated

geometric distribution We also note that under the stationary model assumption the pmf

of W1 is easily verified to be rj = (11 minus j)55 j = 1 2 10 which has mean 4 lt 55

As a result the values of Ψnxy(10) for stationary models (3) and (4) are always larger than

the corresponding values for ordinary models (1) and (2) Finally in the absence of any

knowledge regarding which distribution is in effect for the first interclaim time W1 models

(5) and (6) can be regarded as ldquoworst-case scenariordquo models as they serve to provide upper

bounds on the values of Ψnxy(10)

In conclusion we state that the values in Tables 1 and 2 were generated by first impleshy

menting the recursive procedures (38) through (316) and then summing the appropriate

bivariatetrivariate probabilities computed via (213) (214) or (215) In situations when

x = y = infin however we computed the resulting finite-time ruin probabilities using an altershy

native (albeit standard) method (eg see Cossette et al (2006 Section 3)) Specifically if

we define the survival probability σ(n u) = Pr T gt n | U0 = u = 1minusΨninfininfin(u) as well as

the auxiliary function

0 if w ge Z ηwt =

minlfloorZminuswc minus1 rfloor + 1 t if w lt Z

then by conditioning on the time and amount of the first claim (and recalling that the delayed

process reverts to the ordinary process following this claim) we obtain the expression

minnnr (kminuszk)c2 u+czk+ℓ L L L

σ(u n) = Rn + rk dℓkminuszk αjγ(u + czk + ℓ minus j n minus k)

k=1 ℓ=(kminuszk)c1 j=1

18

where the function γ(w m) for w m isin N is computed recursively via

minmna (kminusηwk)c2 u+cηwk+ℓ L L L

γ(w m) = Am + ak dℓkminusηwk αjγ(u + cηwk + ℓ minus j m minus k)

k=1 j=1 ℓ=(kminusηwk)c1

with starting values γ(w 0) = 1 forall w isin N

Acknowledgements

The authors would like to thank Dr David Landriault for his valuable suggestions which

helped to improve an earlier version of this work This research was supported by the Natural

Sciences and Engineering Research Council of Canada

References

Alfa AS 2004 Markov chain representations of discrete distributions applied to queueing

models Computers amp Operations Research 31 2365-2385

Alfa AS Drekic S 2007 Algorithmic analysis of the Sparre Andersen model in discrete

time ASTIN Bulletin 37 293-317

Bao Z 2007 A note on the compound binomial model with randomized dividend strategy

Applied Mathematics amp Computation 194 276-286

Cossette H Landriault D Marceau E 2006 Ruin probabilities in the discrete time

renewal risk model Insurance Mathematics amp Economics 38 309-323

Karlin S Taylor HM 1975 A First Course in Stochastic Processes 2nd edition Acashy

demic Press New York

Kim B Kim H-S Kim J 2008 A risk model with paying dividends and random

environment Insurance Mathematics amp Economics 42 717-726

Landriault D 2008 Randomized dividends in the compound binomial model with a general

premium rate Scandinavian Actuarial Journal 1-15

Tan J Yang X 2006 The compound binomial model with randomized decisions on

paying dividends Insurance Mathematics amp Economics 39 1-18

19

Table 1 ndash Values of Ψnxy(10) corresponding to Example 1

(a) x = 10

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0005357 0006188 0007163 0008278 0009481

(2) 0005359 0006195 0007175 0008296 0009507

(3) 0005361 0006198 0007181 0008305 0009519

(4) 0005363 0006205 0007193 0008323 0009544

(5) 0005366 0006212 0007208 0008346 0009576

(6) 0005374 0006236 0007249 0008410 0009664

y = 25 (1) 0007631 0008815 0010202 0011789 0013503

(2) 0007634 0008824 0010220 0011816 0013539

(3) 0007636 0008828 0010228 0011829 0013557

(4) 0007639 0008838 0010245 0011854 0013593

(5) 0007643 0008849 0010266 0011887 0013638

(6) 0007656 0008882 0010325 0011978 0013764

y = 50 (1) 0008477 0009792 0011333 0013095 0014998

(2) 0008481 0009802 0011352 0013125 0015039

(3) 0008483 0009807 0011361 0013139 0015059

(4) 0008486 0009817 0011380 0013168 0015098

(5) 0008491 0009830 0011403 0013204 0015149

(6) 0008504 0009866 0011470 0013305 0015288

y = infin (1) 0008777 0010138 0011733 0013558 0015528

(2) 0008781 0010149 0011753 0013588 0015570

(3) 0008783 0010154 0011763 0013603 0015591

(4) 0008787 0010164 0011782 0013633 0015632

(5) 0008791 0010177 0011807 0013671 0015684

(6) 0008805 0010215 0011875 0013775 0015828

20

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0058559 0061875 0065704 0070048 0074714

(2) 0058569 0061903 0065756 0070128 0074825

(3) 0058574 0061917 0065782 0070167 0074880

(4) 0058584 0061945 0065833 0070245 0074987

(5) 0058596 0061979 0065896 0070344 0075124

(6) 0058632 0062079 0066078 0070621 0075506

y = 25 (1) 0089002 0093986 0099739 0106262 0113270

(2) 0089017 0094028 0099817 0106382 0113437

(3) 0089025 0094049 0099856 0106442 0113519

(4) 0089040 0094091 0099932 0106560 0113681

(5) 0089057 0094142 0100028 0106708 0113887

(6) 0089113 0094294 0100302 0107127 0114463

y = 50 (1) 0102786 0108506 0115106 0122590 0130629

(2) 0102804 0108555 0115196 0122728 0130820

(3) 0102812 0108579 0115241 0122797 0130914

(4) 0102830 0108627 0115329 0122932 0131101

(5) 0102850 0108686 0115439 0123103 0131337

(6) 0102914 0108860 0115754 0123584 0132000

y = infin (1) 0108895 0114933 0121901 0129799 0138283

(2) 0108914 0114985 0121996 0129945 0138485

(3) 0108923 0115010 0122043 0130018 0138585

(4) 0108941 0115061 0122136 0130161 0138782

(5) 0108962 0115123 0122252 0130341 0139032

(6) 0109030 0115307 0122585 0130849 0139732

21

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0079856 0086792 009464 0103440 0112831

(2) 0079885 0086869 0094778 0103650 0113118

(3) 0079953 0086978 0094936 0103864 0113393

(4) 0080081 0087188 0095243 0104282 0113929

(5) 0079962 0087075 0095149 0104213 0113890

(6) 0080805 0088366 0096951 0106588 0116875

y = 25 (1) 0124591 0135680 0148207 0162245 0177217

(2) 0124639 0135806 0148433 0162586 0177683

(3) 0124755 0135993 0148703 0162950 0178148

(4) 0124978 0136354 0149226 0163658 0179054

(5) 0124763 0136141 0149035 0163500 0178935

(6) 0126238 0138381 0152148 0167588 0184060

y = 50 (1) 0146757 0160079 0175111 0191945 0209894

(2) 0146816 0160232 0175385 0192360 0210460

(3) 0146965 0160469 0175725 0192817 0211043

(4) 0147249 0160925 0176384 0193706 0212178

(5) 0146968 0160639 0176117 0193470 0211980

(6) 0148860 0163500 0180077 0198658 0218474

y = infin (1) 0158013 0172606 0189059 0207476 0227106

(2) 0158078 0172776 0189362 0207934 0227732

(3) 0158250 0173047 0189751 0208455 0228395

(4) 0158579 0173571 0190503 0209467 0229685

(5) 0158247 0173226 0190172 0209161 0229410

(6) 0160443 0176532 0194735 0215127 0236869

22

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0088667 0104403 0123329 0145277 0169184

(2) 0088711 0104518 0123538 0145598 0169627

(3) 0088733 0104574 0123641 0145756 0169846

(4) 0088777 0104687 0123844 0146066 0170274

(5) 0088829 0104828 0124105 0146470 0170832

(6) 0088988 0105229 0124822 0147560 0172331

y = 25 (1) 0140528 0167839 0201007 0239726 0282102

(2) 0140598 0168021 0201338 0240234 0282804

(3) 0140632 0168111 0201502 0240486 0283151

(4) 0140701 0168289 0201822 0240976 0283827

(5) 0140785 0168515 0202241 0241622 0284718

(6) 0141036 0169148 0203370 0243335 0287071

y = 50 (1) 0167826 0203005 0246173 0296954 0352855

(2) 0167908 0203218 0246561 0297548 0353672

(3) 0167949 0203324 0246753 0297841 0354075

(4) 0168030 0203532 0247128 0298412 0354859

(5) 0168129 0203800 0247622 0299174 0355907

(6) 0168423 0204539 0248936 0301159 0358620

y = infin (1) 0183325 0224926 0276881 0338957 0408257

(2) 0183412 0225150 0277283 0339564 0409079

(3) 0183455 0225260 0277481 0339864 0409485

(4) 0183540 0225478 0277869 0340446 0410270

(5) 0183645 0225760 0278386 0341234 0411339

(6) 0183954 0226529 0279737 0343244 0414037

23

Table 2 ndash Values of Ψnxy(10) corresponding to Example 2

(a) x = 10

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0001901 0002276 0002724 0003240 0003799

(2) 0001904 0002286 0002742 0003269 0003840

(3) 0002651 0003013 0003442 0003936 0004471

(4) 0002655 0003022 0003460 0003964 0004510

(5) 0005792 0006098 0006458 0006871 0007317

(6) 0005795 0006106 0006473 0006894 0007350

y = 25 (1) 0002708 0003243 0003880 0004615 0005411

(2) 0002713 0003257 0003906 0004656 0005469

(3) 0003778 0004293 0004904 0005608 0006369

(4) 0003782 0004306 0004929 0005647 0006424

(5) 0008255 0008690 0009203 0009792 0010427

(6) 0008259 0008702 0009224 0009824 0010473

y = 50 (1) 0003008 0003603 0004310 0005127 0006011

(2) 0003014 0003618 0004340 0005173 0006076

(3) 0004197 0004769 0005448 0006230 0007075

(4) 0004202 0004784 0005476 0006273 0007137

(5) 0009173 0009656 0010226 0010879 0011585

(6) 0009177 0009669 0010249 0010916 0011637

y = infin (1) 0003115 0003730 0004463 0005308 0006224

(2) 0003121 0003746 0004493 0005356 0006291

(3) 0004346 0004938 0005641 0006450 0007325

(4) 0004351 0004954 0005670 0006496 0007389

(5) 0009499 0009999 0010589 0011265 0011996

(6) 0009503 0010012 0010613 0011303 0012050

24

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0031176 0032815 0034733 0036924 0039282

(2) 0031193 0032862 0034821 0037061 0039475

(3) 0050168 0051747 0053587 0055684 0057939

(4) 0050184 0051792 0053671 0055815 0058124

(5) 0136884 0138219 0139763 0141517 0143399

(6) 0136896 0138256 0139832 0141626 0143553

y = 25 (1) 0047622 0050092 0052981 0056280 0059829

(2) 0004765 0050163 0053113 0056487 0060121

(3) 0076570 0078950 0081720 0084878 0088272

(4) 0076593 0079017 0081847 0085076 0088551

(5) 0205774 0207786 0210111 0212752 0215585

(6) 0205793 0207842 0210216 0212917 0215818

y = 50 (1) 0055170 0058009 0061329 0065119 0069196

(2) 0055200 0058091 0061481 0065357 0069533

(3) 0088650 0091385 0094569 0098197 0102096

(4) 0088677 0091463 0094715 0098425 0102417

(5) 0235815 0238128 0240800 0243834 0247089

(6) 0235837 0238192 0240920 0244023 0247356

y = infin (1) 0058565 0061565 0065073 0069077 0073384

(2) 0058597 0061652 0065234 0069329 0073740

(3) 0094061 0096952 0100316 0104149 0108267

(4) 0094090 0097035 0100470 0104390 0108607

(5) 0248472 0250917 0253740 0256945 0260383

(6) 0248495 0250985 0253867 0257145 0260667

25

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0048598 0052524 0056984 0061998 0067342

(2) 0048654 0052673 0057253 0062411 0067916

(3) 0070192 0073979 0078257 0083057 0088167

(4) 0070243 0074119 0078513 0083451 0088714

(5) 0151774 0154981 0158572 0162586 0166852

(6) 0151815 0155096 0158784 0162913 0167308

y = 25 (1) 0076895 0083230 0090413 0098481 0107073

(2) 0076987 0083474 0090854 0099156 0108010

(3) 0110093 0116205 0123095 0130818 0139033

(4) 0110177 0116435 0123515 0131462 0139928

(5) 0230674 0235848 0241632 0248091 0254950

(6) 0230739 0236038 0241980 0248626 0255695

y = 50 (1) 0091496 0099156 0107829 0117563 0127924

(2) 0091610 0099455 0108370 0118387 0129068

(3) 0130130 0137520 0145839 0155156 0165064

(4) 0130233 0137801 0146352 0155942 0166156

(5) 0266599 0272857 0279840 0287632 0295904

(6) 0266680 0273088 0280265 0288286 0296814

y = infin (1) 0099312 0107746 0117285 0127984 0139369

(2) 0099439 0108078 0117883 0128899 0140639

(3) 0140458 0148595 0157745 0167986 0178872

(4) 0140572 0148908 0158314 0168859 0180084

(5) 0282886 0289776 0297457 0306022 0315111

(6) 0282976 0290034 0297929 0306748 0316120

26

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0059096 0072505 0088907 0108122 0129137

(2) 0059223 0072829 0089500 0109052 0130456

(3) 0079757 0092662 0108388 0126778 0146871

(4) 0079875 0092970 0108956 0127669 0148134

(5) 0159788 0170682 0183871 0199249 0216024

(6) 0159881 0170936 0184343 0199990 0217075

y = 25 (1) 0095883 0119741 0149240 0184073 0222390

(2) 0096087 0120262 0150197 0185573 0224520

(3) 0127403 0150359 0178640 0211977 0248613

(4) 0127592 0150855 0179557 0213414 0250653

(5) 0245161 0264533 0288250 0316124 0346709

(6) 0245311 0264942 0289010 0317319 0348407

y = 50 (1) 0116598 0147920 0187100 0233784 0285503

(2) 0116839 0148537 0188231 0235554 0288015

(3) 0153023 0183153 0220713 0265390 0314840

(4) 0153249 0183742 0221798 0267087 0317248

(5) 0285742 0311159 0342651 0380005 0421287

(6) 0285921 0311644 0343551 0381418 0423292

y = infin (1) 0129470 0167217 0215383 0273841 0339734

(2) 0129725 0167865 0216560 0275661 0342282

(3) 0167978 0204275 0250442 0306382 0369383

(4) 0168217 0204895 0251572 0308128 0371826

(5) 0305868 0336470 0375169 0421935 0474525

(6) 0306058 0336980 0376106 0423388 0476560

27

Page 11: uwaterloo.ca · Ruin Analysis of a Threshold Strategy in a Discrete-time Sparre Andersen Model Steve DREKIC and Ana Maria MERA Department of Statistics and Actuarial Science University

minnminus1nr L

(k) Z

+ = Λi rkg s + rndiminusuminuscznnminuszn n isin (215) nminuskminus1iminuspnminus1

k=1

where (210) was used to establish the last equality In the special case when c1 = c2 = c it

is a straightforward exercise to show that equations (213) through (215) simplify to produce

equations (215) through (217) of Alfa and Drekic (2007 p 300)

3 Computational considerations

In order to compute the joint pmfrsquos defined by (213) (214) and (215) it is clear from (k)

equations (29) through (212) that we need to be able to compute the quantities gn and

h(k) n defined by (27) and (28) respectively In what follows we capitalize on the underlying

Markov chain structure in our model to develop a set of computational procedures which

ultimately enable one to calculate these key row vectors

First of all recall that g0(k)

= b(k) contains zeros from a certain level onwards namely (k)

level vk(zk) = u + czk + c2(k minus zk) In general one can easily deduce that gn will also

contain zeros from a certain level onwards to be denoted by lk(n) We begin this section

with a determination of lk(n) which varies in form depending on the value of vk(zk) minus 1

representing the maximum surplus level at time k after the smallest possible claim of size 1

has been applied (at time kminus)

On one hand it may be the case that the value of vk(zk)minus1 lies at or above the threshold

level Z In this situation we remark that zk can take on any value in the set 0 1 k (k)

Since g0 contains zeros from level lk(0) = vk(zk) onwards we observe that at time k + 1

it may be possible to achieve a higher surplus value than vk+1(zk) minus 1 representing the

maximum surplus level assuming there was no drop below Z due to the claim at time kminus

Specifically if the surplus had fallen to level Z minus 1 at time kminus then gained c at time k+ a

bounded level of Z + c minus1 would have been reached at time k +1 and this level is potentially

larger than vk+1(zk) minus 1 As a result we have that lk(1) = maxvk+1(zk) Z + c If we

extend this argument inductively we readily obtain lk(n) = maxvk+n(zk) Z +c +c2(n minus1)

n isin Z+ However since vk+n(zk) = u + czk + c2(k +n minus zk) = vk(zk)+ c2 + c2(n minus 1) we may

alternatively re-express lk(n) as

lk(n) = maxvk+n(zk) Z + c + c2(n minus 1) = maxvk(zk) + c2 Z + c + c2(n minus 1)

Thus

Z+lk(n) = maxlk(0) + c2 Z + c + c2(n minus 1) n isin (31)

11

or equivalently

Z+lk(n) = lk(1) + c2(n minus 1) n isin

Clearly lk(0) gt Z and lk(n) ge Z + c for n isin Z+

The other remaining case to consider involves the situation when the maximum surplus

level at time k is strictly below the threshold level Z Under this scenario we further

distinguish between two cases involving the value of c2 First of all suppose that c2 = 0 so

that vk(zk)minus1 = u +czk minus1 lt Z or equivalently u +czk le Z If u +czk happens to equal Z

then it is possible for zk isin 0 1 k With c2 = 0 however no further premiums can be

earned until after the first claim takes place (which is assumed to occur at time kminus) thereby

implying that Z = u + ci for all i = zk zk + 1 k Hence we know that u + ck = Z

in this case Moreover if u + czk happens to be strictly less than Z then we clearly have

that zk = k Even in the case when c2 gt 0 one can readily verify that zk = k so that

vk(zk) minus 1 = u + ck minus 1 lt Z Thus we proceed with the analysis assuming that zk = k

c2 ge 0 and the inequality u + ck le Z holds true

lowast k

Define the ceiling function denoted lceilxrceil to yield the smallest integer greater than or

equal to x Because a random premium cannot be received until level Z is at least achieved

it is essential to find the point at which the threshold level Z is reached (ie the point at Zminus(u+ck)+1 which u + ck minus1+ ci ge Z) Isolating this inequality for i one obtains i ge

c Hence

assuming that W1 = k the elapsed time after k at which a random premium would begin to

lceilZminus(u+ck)+1 be received is t lowast = rceil Note that t lowast isin Z+ k c k

With zk = k g0(k)

contains zeros from level u + ck onwards so that lk(0) = u + ck

Immediately following this a premium of c is guaranteed to be received at time k+ as the

threshold level Z has not yet been reached Consequently g1(k)

contains zeros from level (k)

b(k)u + c(k + 1) onwards which is c levels further than that in g0 = Continuing this (k)

process inductively we can readily establish that gn contains zeros from level u + c(k + n)

onwards for n = 1 2 t lowast k so that lk(n) = u + c(k + n) for n = 0 1 t lowast k

In the case of n = t lowast + 1 t lowast + 2 gt

(k) is first looked at in isolation At time k +k k +1

t lowast k + 1 the threshold level has been crossed giving rise to a maximum surplus value of either (k)

c(k + t lowast k contains zeros from level lk(t lowast ku + ) + c2 minus 1 or Z + c minus 1 Therefore g + 1) = t lowast k+1

+1(k + t lowast kmaxvk+t lowast k

) Z + c onwards Carrying out the same inductive process as before it is

readily seen that g(nk)

contains zeros from level lk(n) = maxvk+n(k+t lowast k) Z +c+c2(nminust lowast kminus1)

onwards for n = t lowast k + 1 t lowast k + 2 Combining both cases we obtain

u + c(k + n) if n = 0 1 t lowast lk(n) = k (32)

maxlk(t lowast) + c2 Z + c + c2(n minus t lowast minus 1) if n = t lowast + 1 t lowast + 2 k k k k

Note that by construction lk(t lowast k) gt Z and lk(n) le Z for n = 0 1 t lowast k minus1 Interestingly

12

if we extend the definition tk lowast by introducing the quantity ζk

lowast = maxtklowast 0 we remark that

(31) and (32) can actually be combined into a single formula for lk(n) namely

vk+n(zk + n) if n = 0 1 ζklowast

lk(n) = (33) maxlk(ζ

lowast) + c2 Z + c + c2(n minus ζlowast minus 1) if n = ζlowast + 1 ζlowast + 2 k k k k

Clearly we have that lk(n) ge Z + c for n = ζk lowast +1 ζk

lowast +2 Moreover if c1 = c2 = c (33)

reduces to give lk(n) = u + c(k + n) forall n isin N which is in agreement with the results in Alfa

and Drekic (2007 Section 3)

Remark In our above analysis we purposely ignored the possibility of any future claims

occurring after time k in our determination of lk(n) However we recognize that this is not

realistic and future claims would indeed occur after time k thereby further reducing the value

of the maximum surplus level lk(n) minus 1 at time k + n To account for this we imagine all

future interclaim times (ie after time k) to be of maximum length na and all corresponding

claim sizes to be of minimum amount 1 Consequently in the case when c2 gt 0 we should

subtract from formula (33) the amount of lfloornnarfloor where lfloormiddotrfloor again denotes the floor function

Obviously when c2 = 0 (ie no premiums are collected whenever the surplus level is at or

above Z) lk(n) = Z + c for n = ζk lowast + 1 ζk

lowast + 2 (k)

We now turn our attention to the development of a general algorithm for computing gni

(k) (k)for i = 0 1 lk(n) minus 1 With the aid of (24) we apply the key identity gn = gnminus1C

(which can be inferred from (27)) to first obtain

minZlk(nminus1)minus1 lk(nminus1)minus1 c2 L L L

(k) (k) (k)gni = g Bjminusi+c + gnminus1j dmBjminusi+m n isin Z+ (34) nminus1j

j=0 j=Z m=c1

However (22) states that Bi = Ona for i isin Zminus and so (34) becomes (for the same range of

i)

minZlk(nminus1)minus1 c2 lk(nminus1)minus1 L L L

(k) (k) (k)gni = g Bjminusi+c + dm gnminus1jBjminusi+m n isin Z+ (35) nminus1j

j=max0iminusc m=c1 j=maxZiminusm

Taking the form of (33) into consideration (35) yields the following two formulae

lk(nminus1)minus1 L

(k) (k)gni = gnminus1jBjminusi+c n = 1 2 ζ k

lowast (36) j=max0iminusc

and

Zminus1 c2 lk(nminus1)minus1 L L L

(k) (k) (k)g = g + dm g n = ζ lowast + 1 ζ lowast + 2 ni nminus1jBjminusi+c nminus1jBjminusi+m k k

j=max0iminusc m=c1 j=maxZiminusm

(37)

13

For n = 1 2 ζk lowast where ζk

lowast isin Z+ we note that lk(n) ge c + 1 Therefore with the aid of

(22) (36) simplifies to give

Llk(nminus1)minus1 (k) (k) j=0 gnminus1jBjminusi+c if i = 0 1 c minus 1

gni = Llk(nminus1)minus1 (k)

g if i = c c + 1 lk(n) minus 1j=iminusc nminus1jBjminusi+c

Llk(nminus1)minus1 (k)

nminus1js 0 0 if i = 0 1 c minus 1j=0 αjminusi+cg= (38)

(k) Llk(nminus1)minus1 (k) g S+ αjminusi+cgnminus1js 0 0 if i = c c+1 lk(n)minus1 nminus1iminusc j=iminusc+1

In the case of n = ζk lowast + 1 ζk

lowast + 2 we first reiterate that lk(n) ge Z + c We next break

(37) into several cases based on the value of i and the value of minc Z + c1 as follows

Case 1 Suppose minc Z + c1 = c

For i isin 0 1 c minus 1

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+mni nminus1j nminus1j

j=0 m=c1 j=Z

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k)= αjminusi+cg s + dm αjminusi+mg s 0 0 (39) nminus1j nminus1j

j=0 m=c1 j=Z

For i isin c c + 1 Z + c1 minus 1

Zminus1 c2 lk(nminus1)minus1 L L L

(k) (k) (k)g = g + dm gni nminus1j Bjminusi+c nminus1j Bjminusi+m

j=iminusc m=c1 j=Z

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k) (k)= g S + αjminusi+cg dm αjminusi+mgnminus1j s 0 0 (310) nminus1iminusc nminus1j s +

j=iminusc+1 m=c1 j=Z

For i isin Z + c1 Z + c1 + 1 Z + c minus 1

Zminus1 minc2iminusZ lk(nminus1)minus1 c2 lk(nminus1)minus1 L L L L L

(k) (k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+m + dm g Bjminusi+mni nminus1j nminus1j nminus1j

j=iminusc m=c1 j=iminusm m=iminusZ+1 j=Z

minc2iminusZ L

(k) (k)= g S + dmg S nminus1iminusc nminus1iminusm

m=c1

Zminus1 minc2iminusZ lk(nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k)+ αjminusi+cg dm αjminusi+mg dm αjminusi+mgnminus1j s 0 0 nminus1j s + nminus1js +

j=iminusc+1 m=c1 j=iminusm+1 m=iminusZ+1 j=Z

(311)

For i isin Z + c Z + c + 1 lk(n) minus 1

c2 lk(nminus1)minus1 L L

(k) (k)g = dm g Bjminusi+mni nminus1j

m=c1 j=iminusm

c2 c2 lk (nminus1)minus1 L L L

(k) (k)= dmg S + dm αjminusi+mgnminus1j s 0 0 (312) nminus1iminusm

m=c1 m=c1 j=iminusm+1

14

Case 2 Suppose minc Z + c1 = Z + c1

For i isin 0 1 Z + c1 minus 1

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+mni nminus1j nminus1j

j=0 m=c1 j=Z

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k)= αjminusi+cg s + dm αjminusi+mg s 0 0 (313) nminus1j nminus1j

j=0 m=c1 j=Z

For i isin Z + c1 Z + c1 + 1 c minus 1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+m + dm g Bjminusi+mni nminus1j nminus1j nminus1j

j=0 m=c1 j=iminusm m=iminusZ+1 j=Z

minc2iminusZ L

(k)= dmg S nminus1iminusm

m=c1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k)+ αjminusi+cg s + dm αjminusi+mg s + dm αjminusi+mg s 0 0 nminus1j nminus1j nminus1j

j=0 m=c1 j=iminusm+1 m=iminusZ+1 j=Z

(314)

For i isin c c + 1 Z + c minus 1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k) (k)g = g + g + gni nminus1j Bjminusi+c dm nminus1j Bjminusi+m dm nminus1j Bjminusi+m

j=iminusc m=c1 j=iminusm m=iminusZ+1 j=Z

minc2iminusZ L

(k) (k)= g S + dmg S nminus1iminusc nminus1iminusm

m=c1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k)+

nminus1j s 0 0 αjminusi+cgnminus1j s + dm αjminusi+mgnminus1j s + dm αjminusi+mg j=iminusc+1 m=c1 j=iminusm+1 m=iminusZ+1 j=Z

(315)

For i isin Z + c Z + c + 1 lk(n) minus 1

c2 lk (nminus1)minus1 L L

(k) (k)g = dm g Bjminusi+mni nminus1j

m=c1 j=iminusm

c2 c2 lk (nminus1)minus1 L L L

(k) (k)= dmg S + dm αjminusi+mgnminus1j s 0 0 (316) nminus1iminusm

m=c1 m=c1 j=iminusm+1

It is not difficult to verify that equations (38) through (316) simplify to produce equations

(32) and (33) of Alfa and Drekic (2007 pp 301-302) when c1 = c2 = c With equations (38) (k)

through (316) providing the recursive means of calculating gn we next turn our attention

15

to a determination of h(k) defined by (28) For n = 1 2 ζlowast we use (22) and (25) to n k

obtain

lk(nminus1)minus1 lk(nminus1)minus1

h(k) nminusj =

L g

(k) nminus1ℓBc+j+ℓ =

L αc+j+ℓg

(k) nminus1ℓs 0 0 (317)

ℓ=0 ℓ=0

For n = ζk lowast + 1 ζk

lowast + 2 we similarly acquire

Zminus1 lk(nminus1)minus1 L L

(k) (k) (k)hnminusj = gnminus1ℓBc+j+ℓ + gnminus1ℓAj+ℓ

ℓ=0 ℓ=Z

Zminus1 c2 lk(nminus1)minus1 L L L

(k) (k)= αc+j+ℓg s + dm αm+j+ℓg s 0 0 (318) nminus1ℓ nminus1ℓ

ℓ=0 m=c1 ℓ=Z

Note that substituting (317) and (318) into (29) immediately yields

Llk(nminus1)minus1 (k) (k) ℓ=0 αc+j+ℓgnminus1ℓs if n = 1 2 ζk

lowast φ (u) = nj LZminus1 (k) Lc2

Llk(nminus1)minus1 (k) ζlowast s + dm αm+j+ℓg n = + 1 ζlowast + 2 ℓ=0 αc+j+ℓgnminus1ℓ m=c1 ℓ=Z nminus1ℓs if k k

(319)

Clearly in the special case when c1 = c2 = c (319) yields equation (35) of Alfa and Drekic

(2007 p 302) Finally we remark that (319) can be substituted into (213) to obtain a

computational formula for φnj(u)

4 Numerical examples and findings

In this section we illustrate the application of our proposed algorithm with several numerical

examples All calculations in this section were performed using Mathematica (Version 7) on

a Dell Precision T7400 workstation with a 3 Ghz CPU and 4 GB of RAM

In our first illustration (labelled Example 1) we consider the ordinary Sparre Andersen

model in which interclaim times have pmf given by

(211)(911)jminus1 if j = 1 2 na minus 1 aj = (41)

(911)naminus1 if j = na

Note that the pmf (41) is that of a truncated geometric distribution with all the probability

mass on na na +1 assigned to the support value na Clearly as na becomes larger the

closer (41) approximates this particular geometric distribution having mean 55 In what Lna

Lminus1 nafollows we fix na = 25 so that aj = 0992 and jaj = 546 Let the individual j=1 j=1

claim amount distribution be given by the pmf

minus4 minus4j minus 1 j

αj = 1 + minus 1 + j isin Z+ (42) 30 30

16

We remark that the pmf (42) represents a discretized version of the Pareto distribution with

mean 10 Also note that mα = infin thereby implying that the random variables |UT | and

UTminus both have infinite support

Table 1 displays values (rounded to 6 decimal places) of the trivariate cumulative distrishy

bution function

n x y L L L

Ψnxy(u) = Pr T le n UTminus le x |UT | le y | U0 = u = ψℓij(u) ℓ=1 i=mincZ+c1 j=1

corresponding to the above discrete-time risk process with u = 10 Z = 50 c = 5 and

varying premium amount distributions according to the following set of scenarios

(1) d2 = 1

(2) d1 = 25 d2 = 15 d3 = 25

G

5)

(3) di = i

(25)i(35)5minusi i = 0 1 5

(4) di = 15 i = 0 1 4

(5) d1 = 34 d5 = 14

(6) d0 = 35 d5 = 25

We remark that while these six premium amount distributions all have mean equal to 2

they possess different shapes and we investigate the impact of their variability on the joint

distribution of T |UT | and UTminus In particular the variances corresponding to the premium

amount distributions of scenarios (1) through (6) are 0 08 12 2 3 and 6 respectively

Based on the results in Table 1 it would appear that greater variability in the premium

amount distribution generally leads to higher ruin probabilities Perhaps somewhat surprising

though the differences are subtle and not particularly striking as one compares the various

ruin probabilities from scenarios (1) through (6) Obviously we observe more of a difference

between scenarios as the values of n x and y increase

As a follow-up illustration (labelled Example 2) we consider a model in which ordinary

interclaim times have a discrete uniform distribution with pmf aj = 110 j = 1 2 10

Clearly the mean of this distribution is 55 which for comparative purposes is identical to

the ordinary interclaim time mean considered in Example 1 In addition we assume that

the pmf of the individual claim amount distribution is once again given by (42) Table 2

displays values (rounded to 6 decimal places) of Ψnxy(10) for several variants of the Sparre

Andersen model (again with Z = 50 and c = 5) having the above ordinary interclaim time

and claim amount distributions namely

17

(1) ordinary model under scenario (2) from Example 1

(2) ordinary model under scenario (5) from Example 1

(3) stationary model under scenario (2) from Example 1

(4) stationary model under scenario (5) from Example 1

(5) delayed model (with W1 = 1 with probability 1) under scenario (2) from Example 1

(6) delayed model (with W1 = 1 with probability 1) under scenario (5) from Example 1

Concerning the results in Table 2 we first compare the values of Ψnxy(10) in entries (2)

and (5) of Table 1 to their corresponding counterparts in entries (1) and (2) of Table 2 In all

cases we observe that ruin is more likely to occur when interclaim times have the truncated

geometric distribution introduced in our first example as opposed to the discrete uniform

distribution of our second example Since mean interclaim times are equal in both examshy

ples this behaviour must be attributable to the greater variability inherent in the truncated

geometric distribution We also note that under the stationary model assumption the pmf

of W1 is easily verified to be rj = (11 minus j)55 j = 1 2 10 which has mean 4 lt 55

As a result the values of Ψnxy(10) for stationary models (3) and (4) are always larger than

the corresponding values for ordinary models (1) and (2) Finally in the absence of any

knowledge regarding which distribution is in effect for the first interclaim time W1 models

(5) and (6) can be regarded as ldquoworst-case scenariordquo models as they serve to provide upper

bounds on the values of Ψnxy(10)

In conclusion we state that the values in Tables 1 and 2 were generated by first impleshy

menting the recursive procedures (38) through (316) and then summing the appropriate

bivariatetrivariate probabilities computed via (213) (214) or (215) In situations when

x = y = infin however we computed the resulting finite-time ruin probabilities using an altershy

native (albeit standard) method (eg see Cossette et al (2006 Section 3)) Specifically if

we define the survival probability σ(n u) = Pr T gt n | U0 = u = 1minusΨninfininfin(u) as well as

the auxiliary function

0 if w ge Z ηwt =

minlfloorZminuswc minus1 rfloor + 1 t if w lt Z

then by conditioning on the time and amount of the first claim (and recalling that the delayed

process reverts to the ordinary process following this claim) we obtain the expression

minnnr (kminuszk)c2 u+czk+ℓ L L L

σ(u n) = Rn + rk dℓkminuszk αjγ(u + czk + ℓ minus j n minus k)

k=1 ℓ=(kminuszk)c1 j=1

18

where the function γ(w m) for w m isin N is computed recursively via

minmna (kminusηwk)c2 u+cηwk+ℓ L L L

γ(w m) = Am + ak dℓkminusηwk αjγ(u + cηwk + ℓ minus j m minus k)

k=1 j=1 ℓ=(kminusηwk)c1

with starting values γ(w 0) = 1 forall w isin N

Acknowledgements

The authors would like to thank Dr David Landriault for his valuable suggestions which

helped to improve an earlier version of this work This research was supported by the Natural

Sciences and Engineering Research Council of Canada

References

Alfa AS 2004 Markov chain representations of discrete distributions applied to queueing

models Computers amp Operations Research 31 2365-2385

Alfa AS Drekic S 2007 Algorithmic analysis of the Sparre Andersen model in discrete

time ASTIN Bulletin 37 293-317

Bao Z 2007 A note on the compound binomial model with randomized dividend strategy

Applied Mathematics amp Computation 194 276-286

Cossette H Landriault D Marceau E 2006 Ruin probabilities in the discrete time

renewal risk model Insurance Mathematics amp Economics 38 309-323

Karlin S Taylor HM 1975 A First Course in Stochastic Processes 2nd edition Acashy

demic Press New York

Kim B Kim H-S Kim J 2008 A risk model with paying dividends and random

environment Insurance Mathematics amp Economics 42 717-726

Landriault D 2008 Randomized dividends in the compound binomial model with a general

premium rate Scandinavian Actuarial Journal 1-15

Tan J Yang X 2006 The compound binomial model with randomized decisions on

paying dividends Insurance Mathematics amp Economics 39 1-18

19

Table 1 ndash Values of Ψnxy(10) corresponding to Example 1

(a) x = 10

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0005357 0006188 0007163 0008278 0009481

(2) 0005359 0006195 0007175 0008296 0009507

(3) 0005361 0006198 0007181 0008305 0009519

(4) 0005363 0006205 0007193 0008323 0009544

(5) 0005366 0006212 0007208 0008346 0009576

(6) 0005374 0006236 0007249 0008410 0009664

y = 25 (1) 0007631 0008815 0010202 0011789 0013503

(2) 0007634 0008824 0010220 0011816 0013539

(3) 0007636 0008828 0010228 0011829 0013557

(4) 0007639 0008838 0010245 0011854 0013593

(5) 0007643 0008849 0010266 0011887 0013638

(6) 0007656 0008882 0010325 0011978 0013764

y = 50 (1) 0008477 0009792 0011333 0013095 0014998

(2) 0008481 0009802 0011352 0013125 0015039

(3) 0008483 0009807 0011361 0013139 0015059

(4) 0008486 0009817 0011380 0013168 0015098

(5) 0008491 0009830 0011403 0013204 0015149

(6) 0008504 0009866 0011470 0013305 0015288

y = infin (1) 0008777 0010138 0011733 0013558 0015528

(2) 0008781 0010149 0011753 0013588 0015570

(3) 0008783 0010154 0011763 0013603 0015591

(4) 0008787 0010164 0011782 0013633 0015632

(5) 0008791 0010177 0011807 0013671 0015684

(6) 0008805 0010215 0011875 0013775 0015828

20

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0058559 0061875 0065704 0070048 0074714

(2) 0058569 0061903 0065756 0070128 0074825

(3) 0058574 0061917 0065782 0070167 0074880

(4) 0058584 0061945 0065833 0070245 0074987

(5) 0058596 0061979 0065896 0070344 0075124

(6) 0058632 0062079 0066078 0070621 0075506

y = 25 (1) 0089002 0093986 0099739 0106262 0113270

(2) 0089017 0094028 0099817 0106382 0113437

(3) 0089025 0094049 0099856 0106442 0113519

(4) 0089040 0094091 0099932 0106560 0113681

(5) 0089057 0094142 0100028 0106708 0113887

(6) 0089113 0094294 0100302 0107127 0114463

y = 50 (1) 0102786 0108506 0115106 0122590 0130629

(2) 0102804 0108555 0115196 0122728 0130820

(3) 0102812 0108579 0115241 0122797 0130914

(4) 0102830 0108627 0115329 0122932 0131101

(5) 0102850 0108686 0115439 0123103 0131337

(6) 0102914 0108860 0115754 0123584 0132000

y = infin (1) 0108895 0114933 0121901 0129799 0138283

(2) 0108914 0114985 0121996 0129945 0138485

(3) 0108923 0115010 0122043 0130018 0138585

(4) 0108941 0115061 0122136 0130161 0138782

(5) 0108962 0115123 0122252 0130341 0139032

(6) 0109030 0115307 0122585 0130849 0139732

21

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0079856 0086792 009464 0103440 0112831

(2) 0079885 0086869 0094778 0103650 0113118

(3) 0079953 0086978 0094936 0103864 0113393

(4) 0080081 0087188 0095243 0104282 0113929

(5) 0079962 0087075 0095149 0104213 0113890

(6) 0080805 0088366 0096951 0106588 0116875

y = 25 (1) 0124591 0135680 0148207 0162245 0177217

(2) 0124639 0135806 0148433 0162586 0177683

(3) 0124755 0135993 0148703 0162950 0178148

(4) 0124978 0136354 0149226 0163658 0179054

(5) 0124763 0136141 0149035 0163500 0178935

(6) 0126238 0138381 0152148 0167588 0184060

y = 50 (1) 0146757 0160079 0175111 0191945 0209894

(2) 0146816 0160232 0175385 0192360 0210460

(3) 0146965 0160469 0175725 0192817 0211043

(4) 0147249 0160925 0176384 0193706 0212178

(5) 0146968 0160639 0176117 0193470 0211980

(6) 0148860 0163500 0180077 0198658 0218474

y = infin (1) 0158013 0172606 0189059 0207476 0227106

(2) 0158078 0172776 0189362 0207934 0227732

(3) 0158250 0173047 0189751 0208455 0228395

(4) 0158579 0173571 0190503 0209467 0229685

(5) 0158247 0173226 0190172 0209161 0229410

(6) 0160443 0176532 0194735 0215127 0236869

22

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0088667 0104403 0123329 0145277 0169184

(2) 0088711 0104518 0123538 0145598 0169627

(3) 0088733 0104574 0123641 0145756 0169846

(4) 0088777 0104687 0123844 0146066 0170274

(5) 0088829 0104828 0124105 0146470 0170832

(6) 0088988 0105229 0124822 0147560 0172331

y = 25 (1) 0140528 0167839 0201007 0239726 0282102

(2) 0140598 0168021 0201338 0240234 0282804

(3) 0140632 0168111 0201502 0240486 0283151

(4) 0140701 0168289 0201822 0240976 0283827

(5) 0140785 0168515 0202241 0241622 0284718

(6) 0141036 0169148 0203370 0243335 0287071

y = 50 (1) 0167826 0203005 0246173 0296954 0352855

(2) 0167908 0203218 0246561 0297548 0353672

(3) 0167949 0203324 0246753 0297841 0354075

(4) 0168030 0203532 0247128 0298412 0354859

(5) 0168129 0203800 0247622 0299174 0355907

(6) 0168423 0204539 0248936 0301159 0358620

y = infin (1) 0183325 0224926 0276881 0338957 0408257

(2) 0183412 0225150 0277283 0339564 0409079

(3) 0183455 0225260 0277481 0339864 0409485

(4) 0183540 0225478 0277869 0340446 0410270

(5) 0183645 0225760 0278386 0341234 0411339

(6) 0183954 0226529 0279737 0343244 0414037

23

Table 2 ndash Values of Ψnxy(10) corresponding to Example 2

(a) x = 10

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0001901 0002276 0002724 0003240 0003799

(2) 0001904 0002286 0002742 0003269 0003840

(3) 0002651 0003013 0003442 0003936 0004471

(4) 0002655 0003022 0003460 0003964 0004510

(5) 0005792 0006098 0006458 0006871 0007317

(6) 0005795 0006106 0006473 0006894 0007350

y = 25 (1) 0002708 0003243 0003880 0004615 0005411

(2) 0002713 0003257 0003906 0004656 0005469

(3) 0003778 0004293 0004904 0005608 0006369

(4) 0003782 0004306 0004929 0005647 0006424

(5) 0008255 0008690 0009203 0009792 0010427

(6) 0008259 0008702 0009224 0009824 0010473

y = 50 (1) 0003008 0003603 0004310 0005127 0006011

(2) 0003014 0003618 0004340 0005173 0006076

(3) 0004197 0004769 0005448 0006230 0007075

(4) 0004202 0004784 0005476 0006273 0007137

(5) 0009173 0009656 0010226 0010879 0011585

(6) 0009177 0009669 0010249 0010916 0011637

y = infin (1) 0003115 0003730 0004463 0005308 0006224

(2) 0003121 0003746 0004493 0005356 0006291

(3) 0004346 0004938 0005641 0006450 0007325

(4) 0004351 0004954 0005670 0006496 0007389

(5) 0009499 0009999 0010589 0011265 0011996

(6) 0009503 0010012 0010613 0011303 0012050

24

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0031176 0032815 0034733 0036924 0039282

(2) 0031193 0032862 0034821 0037061 0039475

(3) 0050168 0051747 0053587 0055684 0057939

(4) 0050184 0051792 0053671 0055815 0058124

(5) 0136884 0138219 0139763 0141517 0143399

(6) 0136896 0138256 0139832 0141626 0143553

y = 25 (1) 0047622 0050092 0052981 0056280 0059829

(2) 0004765 0050163 0053113 0056487 0060121

(3) 0076570 0078950 0081720 0084878 0088272

(4) 0076593 0079017 0081847 0085076 0088551

(5) 0205774 0207786 0210111 0212752 0215585

(6) 0205793 0207842 0210216 0212917 0215818

y = 50 (1) 0055170 0058009 0061329 0065119 0069196

(2) 0055200 0058091 0061481 0065357 0069533

(3) 0088650 0091385 0094569 0098197 0102096

(4) 0088677 0091463 0094715 0098425 0102417

(5) 0235815 0238128 0240800 0243834 0247089

(6) 0235837 0238192 0240920 0244023 0247356

y = infin (1) 0058565 0061565 0065073 0069077 0073384

(2) 0058597 0061652 0065234 0069329 0073740

(3) 0094061 0096952 0100316 0104149 0108267

(4) 0094090 0097035 0100470 0104390 0108607

(5) 0248472 0250917 0253740 0256945 0260383

(6) 0248495 0250985 0253867 0257145 0260667

25

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0048598 0052524 0056984 0061998 0067342

(2) 0048654 0052673 0057253 0062411 0067916

(3) 0070192 0073979 0078257 0083057 0088167

(4) 0070243 0074119 0078513 0083451 0088714

(5) 0151774 0154981 0158572 0162586 0166852

(6) 0151815 0155096 0158784 0162913 0167308

y = 25 (1) 0076895 0083230 0090413 0098481 0107073

(2) 0076987 0083474 0090854 0099156 0108010

(3) 0110093 0116205 0123095 0130818 0139033

(4) 0110177 0116435 0123515 0131462 0139928

(5) 0230674 0235848 0241632 0248091 0254950

(6) 0230739 0236038 0241980 0248626 0255695

y = 50 (1) 0091496 0099156 0107829 0117563 0127924

(2) 0091610 0099455 0108370 0118387 0129068

(3) 0130130 0137520 0145839 0155156 0165064

(4) 0130233 0137801 0146352 0155942 0166156

(5) 0266599 0272857 0279840 0287632 0295904

(6) 0266680 0273088 0280265 0288286 0296814

y = infin (1) 0099312 0107746 0117285 0127984 0139369

(2) 0099439 0108078 0117883 0128899 0140639

(3) 0140458 0148595 0157745 0167986 0178872

(4) 0140572 0148908 0158314 0168859 0180084

(5) 0282886 0289776 0297457 0306022 0315111

(6) 0282976 0290034 0297929 0306748 0316120

26

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0059096 0072505 0088907 0108122 0129137

(2) 0059223 0072829 0089500 0109052 0130456

(3) 0079757 0092662 0108388 0126778 0146871

(4) 0079875 0092970 0108956 0127669 0148134

(5) 0159788 0170682 0183871 0199249 0216024

(6) 0159881 0170936 0184343 0199990 0217075

y = 25 (1) 0095883 0119741 0149240 0184073 0222390

(2) 0096087 0120262 0150197 0185573 0224520

(3) 0127403 0150359 0178640 0211977 0248613

(4) 0127592 0150855 0179557 0213414 0250653

(5) 0245161 0264533 0288250 0316124 0346709

(6) 0245311 0264942 0289010 0317319 0348407

y = 50 (1) 0116598 0147920 0187100 0233784 0285503

(2) 0116839 0148537 0188231 0235554 0288015

(3) 0153023 0183153 0220713 0265390 0314840

(4) 0153249 0183742 0221798 0267087 0317248

(5) 0285742 0311159 0342651 0380005 0421287

(6) 0285921 0311644 0343551 0381418 0423292

y = infin (1) 0129470 0167217 0215383 0273841 0339734

(2) 0129725 0167865 0216560 0275661 0342282

(3) 0167978 0204275 0250442 0306382 0369383

(4) 0168217 0204895 0251572 0308128 0371826

(5) 0305868 0336470 0375169 0421935 0474525

(6) 0306058 0336980 0376106 0423388 0476560

27

Page 12: uwaterloo.ca · Ruin Analysis of a Threshold Strategy in a Discrete-time Sparre Andersen Model Steve DREKIC and Ana Maria MERA Department of Statistics and Actuarial Science University

or equivalently

Z+lk(n) = lk(1) + c2(n minus 1) n isin

Clearly lk(0) gt Z and lk(n) ge Z + c for n isin Z+

The other remaining case to consider involves the situation when the maximum surplus

level at time k is strictly below the threshold level Z Under this scenario we further

distinguish between two cases involving the value of c2 First of all suppose that c2 = 0 so

that vk(zk)minus1 = u +czk minus1 lt Z or equivalently u +czk le Z If u +czk happens to equal Z

then it is possible for zk isin 0 1 k With c2 = 0 however no further premiums can be

earned until after the first claim takes place (which is assumed to occur at time kminus) thereby

implying that Z = u + ci for all i = zk zk + 1 k Hence we know that u + ck = Z

in this case Moreover if u + czk happens to be strictly less than Z then we clearly have

that zk = k Even in the case when c2 gt 0 one can readily verify that zk = k so that

vk(zk) minus 1 = u + ck minus 1 lt Z Thus we proceed with the analysis assuming that zk = k

c2 ge 0 and the inequality u + ck le Z holds true

lowast k

Define the ceiling function denoted lceilxrceil to yield the smallest integer greater than or

equal to x Because a random premium cannot be received until level Z is at least achieved

it is essential to find the point at which the threshold level Z is reached (ie the point at Zminus(u+ck)+1 which u + ck minus1+ ci ge Z) Isolating this inequality for i one obtains i ge

c Hence

assuming that W1 = k the elapsed time after k at which a random premium would begin to

lceilZminus(u+ck)+1 be received is t lowast = rceil Note that t lowast isin Z+ k c k

With zk = k g0(k)

contains zeros from level u + ck onwards so that lk(0) = u + ck

Immediately following this a premium of c is guaranteed to be received at time k+ as the

threshold level Z has not yet been reached Consequently g1(k)

contains zeros from level (k)

b(k)u + c(k + 1) onwards which is c levels further than that in g0 = Continuing this (k)

process inductively we can readily establish that gn contains zeros from level u + c(k + n)

onwards for n = 1 2 t lowast k so that lk(n) = u + c(k + n) for n = 0 1 t lowast k

In the case of n = t lowast + 1 t lowast + 2 gt

(k) is first looked at in isolation At time k +k k +1

t lowast k + 1 the threshold level has been crossed giving rise to a maximum surplus value of either (k)

c(k + t lowast k contains zeros from level lk(t lowast ku + ) + c2 minus 1 or Z + c minus 1 Therefore g + 1) = t lowast k+1

+1(k + t lowast kmaxvk+t lowast k

) Z + c onwards Carrying out the same inductive process as before it is

readily seen that g(nk)

contains zeros from level lk(n) = maxvk+n(k+t lowast k) Z +c+c2(nminust lowast kminus1)

onwards for n = t lowast k + 1 t lowast k + 2 Combining both cases we obtain

u + c(k + n) if n = 0 1 t lowast lk(n) = k (32)

maxlk(t lowast) + c2 Z + c + c2(n minus t lowast minus 1) if n = t lowast + 1 t lowast + 2 k k k k

Note that by construction lk(t lowast k) gt Z and lk(n) le Z for n = 0 1 t lowast k minus1 Interestingly

12

if we extend the definition tk lowast by introducing the quantity ζk

lowast = maxtklowast 0 we remark that

(31) and (32) can actually be combined into a single formula for lk(n) namely

vk+n(zk + n) if n = 0 1 ζklowast

lk(n) = (33) maxlk(ζ

lowast) + c2 Z + c + c2(n minus ζlowast minus 1) if n = ζlowast + 1 ζlowast + 2 k k k k

Clearly we have that lk(n) ge Z + c for n = ζk lowast +1 ζk

lowast +2 Moreover if c1 = c2 = c (33)

reduces to give lk(n) = u + c(k + n) forall n isin N which is in agreement with the results in Alfa

and Drekic (2007 Section 3)

Remark In our above analysis we purposely ignored the possibility of any future claims

occurring after time k in our determination of lk(n) However we recognize that this is not

realistic and future claims would indeed occur after time k thereby further reducing the value

of the maximum surplus level lk(n) minus 1 at time k + n To account for this we imagine all

future interclaim times (ie after time k) to be of maximum length na and all corresponding

claim sizes to be of minimum amount 1 Consequently in the case when c2 gt 0 we should

subtract from formula (33) the amount of lfloornnarfloor where lfloormiddotrfloor again denotes the floor function

Obviously when c2 = 0 (ie no premiums are collected whenever the surplus level is at or

above Z) lk(n) = Z + c for n = ζk lowast + 1 ζk

lowast + 2 (k)

We now turn our attention to the development of a general algorithm for computing gni

(k) (k)for i = 0 1 lk(n) minus 1 With the aid of (24) we apply the key identity gn = gnminus1C

(which can be inferred from (27)) to first obtain

minZlk(nminus1)minus1 lk(nminus1)minus1 c2 L L L

(k) (k) (k)gni = g Bjminusi+c + gnminus1j dmBjminusi+m n isin Z+ (34) nminus1j

j=0 j=Z m=c1

However (22) states that Bi = Ona for i isin Zminus and so (34) becomes (for the same range of

i)

minZlk(nminus1)minus1 c2 lk(nminus1)minus1 L L L

(k) (k) (k)gni = g Bjminusi+c + dm gnminus1jBjminusi+m n isin Z+ (35) nminus1j

j=max0iminusc m=c1 j=maxZiminusm

Taking the form of (33) into consideration (35) yields the following two formulae

lk(nminus1)minus1 L

(k) (k)gni = gnminus1jBjminusi+c n = 1 2 ζ k

lowast (36) j=max0iminusc

and

Zminus1 c2 lk(nminus1)minus1 L L L

(k) (k) (k)g = g + dm g n = ζ lowast + 1 ζ lowast + 2 ni nminus1jBjminusi+c nminus1jBjminusi+m k k

j=max0iminusc m=c1 j=maxZiminusm

(37)

13

For n = 1 2 ζk lowast where ζk

lowast isin Z+ we note that lk(n) ge c + 1 Therefore with the aid of

(22) (36) simplifies to give

Llk(nminus1)minus1 (k) (k) j=0 gnminus1jBjminusi+c if i = 0 1 c minus 1

gni = Llk(nminus1)minus1 (k)

g if i = c c + 1 lk(n) minus 1j=iminusc nminus1jBjminusi+c

Llk(nminus1)minus1 (k)

nminus1js 0 0 if i = 0 1 c minus 1j=0 αjminusi+cg= (38)

(k) Llk(nminus1)minus1 (k) g S+ αjminusi+cgnminus1js 0 0 if i = c c+1 lk(n)minus1 nminus1iminusc j=iminusc+1

In the case of n = ζk lowast + 1 ζk

lowast + 2 we first reiterate that lk(n) ge Z + c We next break

(37) into several cases based on the value of i and the value of minc Z + c1 as follows

Case 1 Suppose minc Z + c1 = c

For i isin 0 1 c minus 1

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+mni nminus1j nminus1j

j=0 m=c1 j=Z

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k)= αjminusi+cg s + dm αjminusi+mg s 0 0 (39) nminus1j nminus1j

j=0 m=c1 j=Z

For i isin c c + 1 Z + c1 minus 1

Zminus1 c2 lk(nminus1)minus1 L L L

(k) (k) (k)g = g + dm gni nminus1j Bjminusi+c nminus1j Bjminusi+m

j=iminusc m=c1 j=Z

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k) (k)= g S + αjminusi+cg dm αjminusi+mgnminus1j s 0 0 (310) nminus1iminusc nminus1j s +

j=iminusc+1 m=c1 j=Z

For i isin Z + c1 Z + c1 + 1 Z + c minus 1

Zminus1 minc2iminusZ lk(nminus1)minus1 c2 lk(nminus1)minus1 L L L L L

(k) (k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+m + dm g Bjminusi+mni nminus1j nminus1j nminus1j

j=iminusc m=c1 j=iminusm m=iminusZ+1 j=Z

minc2iminusZ L

(k) (k)= g S + dmg S nminus1iminusc nminus1iminusm

m=c1

Zminus1 minc2iminusZ lk(nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k)+ αjminusi+cg dm αjminusi+mg dm αjminusi+mgnminus1j s 0 0 nminus1j s + nminus1js +

j=iminusc+1 m=c1 j=iminusm+1 m=iminusZ+1 j=Z

(311)

For i isin Z + c Z + c + 1 lk(n) minus 1

c2 lk(nminus1)minus1 L L

(k) (k)g = dm g Bjminusi+mni nminus1j

m=c1 j=iminusm

c2 c2 lk (nminus1)minus1 L L L

(k) (k)= dmg S + dm αjminusi+mgnminus1j s 0 0 (312) nminus1iminusm

m=c1 m=c1 j=iminusm+1

14

Case 2 Suppose minc Z + c1 = Z + c1

For i isin 0 1 Z + c1 minus 1

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+mni nminus1j nminus1j

j=0 m=c1 j=Z

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k)= αjminusi+cg s + dm αjminusi+mg s 0 0 (313) nminus1j nminus1j

j=0 m=c1 j=Z

For i isin Z + c1 Z + c1 + 1 c minus 1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+m + dm g Bjminusi+mni nminus1j nminus1j nminus1j

j=0 m=c1 j=iminusm m=iminusZ+1 j=Z

minc2iminusZ L

(k)= dmg S nminus1iminusm

m=c1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k)+ αjminusi+cg s + dm αjminusi+mg s + dm αjminusi+mg s 0 0 nminus1j nminus1j nminus1j

j=0 m=c1 j=iminusm+1 m=iminusZ+1 j=Z

(314)

For i isin c c + 1 Z + c minus 1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k) (k)g = g + g + gni nminus1j Bjminusi+c dm nminus1j Bjminusi+m dm nminus1j Bjminusi+m

j=iminusc m=c1 j=iminusm m=iminusZ+1 j=Z

minc2iminusZ L

(k) (k)= g S + dmg S nminus1iminusc nminus1iminusm

m=c1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k)+

nminus1j s 0 0 αjminusi+cgnminus1j s + dm αjminusi+mgnminus1j s + dm αjminusi+mg j=iminusc+1 m=c1 j=iminusm+1 m=iminusZ+1 j=Z

(315)

For i isin Z + c Z + c + 1 lk(n) minus 1

c2 lk (nminus1)minus1 L L

(k) (k)g = dm g Bjminusi+mni nminus1j

m=c1 j=iminusm

c2 c2 lk (nminus1)minus1 L L L

(k) (k)= dmg S + dm αjminusi+mgnminus1j s 0 0 (316) nminus1iminusm

m=c1 m=c1 j=iminusm+1

It is not difficult to verify that equations (38) through (316) simplify to produce equations

(32) and (33) of Alfa and Drekic (2007 pp 301-302) when c1 = c2 = c With equations (38) (k)

through (316) providing the recursive means of calculating gn we next turn our attention

15

to a determination of h(k) defined by (28) For n = 1 2 ζlowast we use (22) and (25) to n k

obtain

lk(nminus1)minus1 lk(nminus1)minus1

h(k) nminusj =

L g

(k) nminus1ℓBc+j+ℓ =

L αc+j+ℓg

(k) nminus1ℓs 0 0 (317)

ℓ=0 ℓ=0

For n = ζk lowast + 1 ζk

lowast + 2 we similarly acquire

Zminus1 lk(nminus1)minus1 L L

(k) (k) (k)hnminusj = gnminus1ℓBc+j+ℓ + gnminus1ℓAj+ℓ

ℓ=0 ℓ=Z

Zminus1 c2 lk(nminus1)minus1 L L L

(k) (k)= αc+j+ℓg s + dm αm+j+ℓg s 0 0 (318) nminus1ℓ nminus1ℓ

ℓ=0 m=c1 ℓ=Z

Note that substituting (317) and (318) into (29) immediately yields

Llk(nminus1)minus1 (k) (k) ℓ=0 αc+j+ℓgnminus1ℓs if n = 1 2 ζk

lowast φ (u) = nj LZminus1 (k) Lc2

Llk(nminus1)minus1 (k) ζlowast s + dm αm+j+ℓg n = + 1 ζlowast + 2 ℓ=0 αc+j+ℓgnminus1ℓ m=c1 ℓ=Z nminus1ℓs if k k

(319)

Clearly in the special case when c1 = c2 = c (319) yields equation (35) of Alfa and Drekic

(2007 p 302) Finally we remark that (319) can be substituted into (213) to obtain a

computational formula for φnj(u)

4 Numerical examples and findings

In this section we illustrate the application of our proposed algorithm with several numerical

examples All calculations in this section were performed using Mathematica (Version 7) on

a Dell Precision T7400 workstation with a 3 Ghz CPU and 4 GB of RAM

In our first illustration (labelled Example 1) we consider the ordinary Sparre Andersen

model in which interclaim times have pmf given by

(211)(911)jminus1 if j = 1 2 na minus 1 aj = (41)

(911)naminus1 if j = na

Note that the pmf (41) is that of a truncated geometric distribution with all the probability

mass on na na +1 assigned to the support value na Clearly as na becomes larger the

closer (41) approximates this particular geometric distribution having mean 55 In what Lna

Lminus1 nafollows we fix na = 25 so that aj = 0992 and jaj = 546 Let the individual j=1 j=1

claim amount distribution be given by the pmf

minus4 minus4j minus 1 j

αj = 1 + minus 1 + j isin Z+ (42) 30 30

16

We remark that the pmf (42) represents a discretized version of the Pareto distribution with

mean 10 Also note that mα = infin thereby implying that the random variables |UT | and

UTminus both have infinite support

Table 1 displays values (rounded to 6 decimal places) of the trivariate cumulative distrishy

bution function

n x y L L L

Ψnxy(u) = Pr T le n UTminus le x |UT | le y | U0 = u = ψℓij(u) ℓ=1 i=mincZ+c1 j=1

corresponding to the above discrete-time risk process with u = 10 Z = 50 c = 5 and

varying premium amount distributions according to the following set of scenarios

(1) d2 = 1

(2) d1 = 25 d2 = 15 d3 = 25

G

5)

(3) di = i

(25)i(35)5minusi i = 0 1 5

(4) di = 15 i = 0 1 4

(5) d1 = 34 d5 = 14

(6) d0 = 35 d5 = 25

We remark that while these six premium amount distributions all have mean equal to 2

they possess different shapes and we investigate the impact of their variability on the joint

distribution of T |UT | and UTminus In particular the variances corresponding to the premium

amount distributions of scenarios (1) through (6) are 0 08 12 2 3 and 6 respectively

Based on the results in Table 1 it would appear that greater variability in the premium

amount distribution generally leads to higher ruin probabilities Perhaps somewhat surprising

though the differences are subtle and not particularly striking as one compares the various

ruin probabilities from scenarios (1) through (6) Obviously we observe more of a difference

between scenarios as the values of n x and y increase

As a follow-up illustration (labelled Example 2) we consider a model in which ordinary

interclaim times have a discrete uniform distribution with pmf aj = 110 j = 1 2 10

Clearly the mean of this distribution is 55 which for comparative purposes is identical to

the ordinary interclaim time mean considered in Example 1 In addition we assume that

the pmf of the individual claim amount distribution is once again given by (42) Table 2

displays values (rounded to 6 decimal places) of Ψnxy(10) for several variants of the Sparre

Andersen model (again with Z = 50 and c = 5) having the above ordinary interclaim time

and claim amount distributions namely

17

(1) ordinary model under scenario (2) from Example 1

(2) ordinary model under scenario (5) from Example 1

(3) stationary model under scenario (2) from Example 1

(4) stationary model under scenario (5) from Example 1

(5) delayed model (with W1 = 1 with probability 1) under scenario (2) from Example 1

(6) delayed model (with W1 = 1 with probability 1) under scenario (5) from Example 1

Concerning the results in Table 2 we first compare the values of Ψnxy(10) in entries (2)

and (5) of Table 1 to their corresponding counterparts in entries (1) and (2) of Table 2 In all

cases we observe that ruin is more likely to occur when interclaim times have the truncated

geometric distribution introduced in our first example as opposed to the discrete uniform

distribution of our second example Since mean interclaim times are equal in both examshy

ples this behaviour must be attributable to the greater variability inherent in the truncated

geometric distribution We also note that under the stationary model assumption the pmf

of W1 is easily verified to be rj = (11 minus j)55 j = 1 2 10 which has mean 4 lt 55

As a result the values of Ψnxy(10) for stationary models (3) and (4) are always larger than

the corresponding values for ordinary models (1) and (2) Finally in the absence of any

knowledge regarding which distribution is in effect for the first interclaim time W1 models

(5) and (6) can be regarded as ldquoworst-case scenariordquo models as they serve to provide upper

bounds on the values of Ψnxy(10)

In conclusion we state that the values in Tables 1 and 2 were generated by first impleshy

menting the recursive procedures (38) through (316) and then summing the appropriate

bivariatetrivariate probabilities computed via (213) (214) or (215) In situations when

x = y = infin however we computed the resulting finite-time ruin probabilities using an altershy

native (albeit standard) method (eg see Cossette et al (2006 Section 3)) Specifically if

we define the survival probability σ(n u) = Pr T gt n | U0 = u = 1minusΨninfininfin(u) as well as

the auxiliary function

0 if w ge Z ηwt =

minlfloorZminuswc minus1 rfloor + 1 t if w lt Z

then by conditioning on the time and amount of the first claim (and recalling that the delayed

process reverts to the ordinary process following this claim) we obtain the expression

minnnr (kminuszk)c2 u+czk+ℓ L L L

σ(u n) = Rn + rk dℓkminuszk αjγ(u + czk + ℓ minus j n minus k)

k=1 ℓ=(kminuszk)c1 j=1

18

where the function γ(w m) for w m isin N is computed recursively via

minmna (kminusηwk)c2 u+cηwk+ℓ L L L

γ(w m) = Am + ak dℓkminusηwk αjγ(u + cηwk + ℓ minus j m minus k)

k=1 j=1 ℓ=(kminusηwk)c1

with starting values γ(w 0) = 1 forall w isin N

Acknowledgements

The authors would like to thank Dr David Landriault for his valuable suggestions which

helped to improve an earlier version of this work This research was supported by the Natural

Sciences and Engineering Research Council of Canada

References

Alfa AS 2004 Markov chain representations of discrete distributions applied to queueing

models Computers amp Operations Research 31 2365-2385

Alfa AS Drekic S 2007 Algorithmic analysis of the Sparre Andersen model in discrete

time ASTIN Bulletin 37 293-317

Bao Z 2007 A note on the compound binomial model with randomized dividend strategy

Applied Mathematics amp Computation 194 276-286

Cossette H Landriault D Marceau E 2006 Ruin probabilities in the discrete time

renewal risk model Insurance Mathematics amp Economics 38 309-323

Karlin S Taylor HM 1975 A First Course in Stochastic Processes 2nd edition Acashy

demic Press New York

Kim B Kim H-S Kim J 2008 A risk model with paying dividends and random

environment Insurance Mathematics amp Economics 42 717-726

Landriault D 2008 Randomized dividends in the compound binomial model with a general

premium rate Scandinavian Actuarial Journal 1-15

Tan J Yang X 2006 The compound binomial model with randomized decisions on

paying dividends Insurance Mathematics amp Economics 39 1-18

19

Table 1 ndash Values of Ψnxy(10) corresponding to Example 1

(a) x = 10

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0005357 0006188 0007163 0008278 0009481

(2) 0005359 0006195 0007175 0008296 0009507

(3) 0005361 0006198 0007181 0008305 0009519

(4) 0005363 0006205 0007193 0008323 0009544

(5) 0005366 0006212 0007208 0008346 0009576

(6) 0005374 0006236 0007249 0008410 0009664

y = 25 (1) 0007631 0008815 0010202 0011789 0013503

(2) 0007634 0008824 0010220 0011816 0013539

(3) 0007636 0008828 0010228 0011829 0013557

(4) 0007639 0008838 0010245 0011854 0013593

(5) 0007643 0008849 0010266 0011887 0013638

(6) 0007656 0008882 0010325 0011978 0013764

y = 50 (1) 0008477 0009792 0011333 0013095 0014998

(2) 0008481 0009802 0011352 0013125 0015039

(3) 0008483 0009807 0011361 0013139 0015059

(4) 0008486 0009817 0011380 0013168 0015098

(5) 0008491 0009830 0011403 0013204 0015149

(6) 0008504 0009866 0011470 0013305 0015288

y = infin (1) 0008777 0010138 0011733 0013558 0015528

(2) 0008781 0010149 0011753 0013588 0015570

(3) 0008783 0010154 0011763 0013603 0015591

(4) 0008787 0010164 0011782 0013633 0015632

(5) 0008791 0010177 0011807 0013671 0015684

(6) 0008805 0010215 0011875 0013775 0015828

20

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0058559 0061875 0065704 0070048 0074714

(2) 0058569 0061903 0065756 0070128 0074825

(3) 0058574 0061917 0065782 0070167 0074880

(4) 0058584 0061945 0065833 0070245 0074987

(5) 0058596 0061979 0065896 0070344 0075124

(6) 0058632 0062079 0066078 0070621 0075506

y = 25 (1) 0089002 0093986 0099739 0106262 0113270

(2) 0089017 0094028 0099817 0106382 0113437

(3) 0089025 0094049 0099856 0106442 0113519

(4) 0089040 0094091 0099932 0106560 0113681

(5) 0089057 0094142 0100028 0106708 0113887

(6) 0089113 0094294 0100302 0107127 0114463

y = 50 (1) 0102786 0108506 0115106 0122590 0130629

(2) 0102804 0108555 0115196 0122728 0130820

(3) 0102812 0108579 0115241 0122797 0130914

(4) 0102830 0108627 0115329 0122932 0131101

(5) 0102850 0108686 0115439 0123103 0131337

(6) 0102914 0108860 0115754 0123584 0132000

y = infin (1) 0108895 0114933 0121901 0129799 0138283

(2) 0108914 0114985 0121996 0129945 0138485

(3) 0108923 0115010 0122043 0130018 0138585

(4) 0108941 0115061 0122136 0130161 0138782

(5) 0108962 0115123 0122252 0130341 0139032

(6) 0109030 0115307 0122585 0130849 0139732

21

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0079856 0086792 009464 0103440 0112831

(2) 0079885 0086869 0094778 0103650 0113118

(3) 0079953 0086978 0094936 0103864 0113393

(4) 0080081 0087188 0095243 0104282 0113929

(5) 0079962 0087075 0095149 0104213 0113890

(6) 0080805 0088366 0096951 0106588 0116875

y = 25 (1) 0124591 0135680 0148207 0162245 0177217

(2) 0124639 0135806 0148433 0162586 0177683

(3) 0124755 0135993 0148703 0162950 0178148

(4) 0124978 0136354 0149226 0163658 0179054

(5) 0124763 0136141 0149035 0163500 0178935

(6) 0126238 0138381 0152148 0167588 0184060

y = 50 (1) 0146757 0160079 0175111 0191945 0209894

(2) 0146816 0160232 0175385 0192360 0210460

(3) 0146965 0160469 0175725 0192817 0211043

(4) 0147249 0160925 0176384 0193706 0212178

(5) 0146968 0160639 0176117 0193470 0211980

(6) 0148860 0163500 0180077 0198658 0218474

y = infin (1) 0158013 0172606 0189059 0207476 0227106

(2) 0158078 0172776 0189362 0207934 0227732

(3) 0158250 0173047 0189751 0208455 0228395

(4) 0158579 0173571 0190503 0209467 0229685

(5) 0158247 0173226 0190172 0209161 0229410

(6) 0160443 0176532 0194735 0215127 0236869

22

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0088667 0104403 0123329 0145277 0169184

(2) 0088711 0104518 0123538 0145598 0169627

(3) 0088733 0104574 0123641 0145756 0169846

(4) 0088777 0104687 0123844 0146066 0170274

(5) 0088829 0104828 0124105 0146470 0170832

(6) 0088988 0105229 0124822 0147560 0172331

y = 25 (1) 0140528 0167839 0201007 0239726 0282102

(2) 0140598 0168021 0201338 0240234 0282804

(3) 0140632 0168111 0201502 0240486 0283151

(4) 0140701 0168289 0201822 0240976 0283827

(5) 0140785 0168515 0202241 0241622 0284718

(6) 0141036 0169148 0203370 0243335 0287071

y = 50 (1) 0167826 0203005 0246173 0296954 0352855

(2) 0167908 0203218 0246561 0297548 0353672

(3) 0167949 0203324 0246753 0297841 0354075

(4) 0168030 0203532 0247128 0298412 0354859

(5) 0168129 0203800 0247622 0299174 0355907

(6) 0168423 0204539 0248936 0301159 0358620

y = infin (1) 0183325 0224926 0276881 0338957 0408257

(2) 0183412 0225150 0277283 0339564 0409079

(3) 0183455 0225260 0277481 0339864 0409485

(4) 0183540 0225478 0277869 0340446 0410270

(5) 0183645 0225760 0278386 0341234 0411339

(6) 0183954 0226529 0279737 0343244 0414037

23

Table 2 ndash Values of Ψnxy(10) corresponding to Example 2

(a) x = 10

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0001901 0002276 0002724 0003240 0003799

(2) 0001904 0002286 0002742 0003269 0003840

(3) 0002651 0003013 0003442 0003936 0004471

(4) 0002655 0003022 0003460 0003964 0004510

(5) 0005792 0006098 0006458 0006871 0007317

(6) 0005795 0006106 0006473 0006894 0007350

y = 25 (1) 0002708 0003243 0003880 0004615 0005411

(2) 0002713 0003257 0003906 0004656 0005469

(3) 0003778 0004293 0004904 0005608 0006369

(4) 0003782 0004306 0004929 0005647 0006424

(5) 0008255 0008690 0009203 0009792 0010427

(6) 0008259 0008702 0009224 0009824 0010473

y = 50 (1) 0003008 0003603 0004310 0005127 0006011

(2) 0003014 0003618 0004340 0005173 0006076

(3) 0004197 0004769 0005448 0006230 0007075

(4) 0004202 0004784 0005476 0006273 0007137

(5) 0009173 0009656 0010226 0010879 0011585

(6) 0009177 0009669 0010249 0010916 0011637

y = infin (1) 0003115 0003730 0004463 0005308 0006224

(2) 0003121 0003746 0004493 0005356 0006291

(3) 0004346 0004938 0005641 0006450 0007325

(4) 0004351 0004954 0005670 0006496 0007389

(5) 0009499 0009999 0010589 0011265 0011996

(6) 0009503 0010012 0010613 0011303 0012050

24

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0031176 0032815 0034733 0036924 0039282

(2) 0031193 0032862 0034821 0037061 0039475

(3) 0050168 0051747 0053587 0055684 0057939

(4) 0050184 0051792 0053671 0055815 0058124

(5) 0136884 0138219 0139763 0141517 0143399

(6) 0136896 0138256 0139832 0141626 0143553

y = 25 (1) 0047622 0050092 0052981 0056280 0059829

(2) 0004765 0050163 0053113 0056487 0060121

(3) 0076570 0078950 0081720 0084878 0088272

(4) 0076593 0079017 0081847 0085076 0088551

(5) 0205774 0207786 0210111 0212752 0215585

(6) 0205793 0207842 0210216 0212917 0215818

y = 50 (1) 0055170 0058009 0061329 0065119 0069196

(2) 0055200 0058091 0061481 0065357 0069533

(3) 0088650 0091385 0094569 0098197 0102096

(4) 0088677 0091463 0094715 0098425 0102417

(5) 0235815 0238128 0240800 0243834 0247089

(6) 0235837 0238192 0240920 0244023 0247356

y = infin (1) 0058565 0061565 0065073 0069077 0073384

(2) 0058597 0061652 0065234 0069329 0073740

(3) 0094061 0096952 0100316 0104149 0108267

(4) 0094090 0097035 0100470 0104390 0108607

(5) 0248472 0250917 0253740 0256945 0260383

(6) 0248495 0250985 0253867 0257145 0260667

25

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0048598 0052524 0056984 0061998 0067342

(2) 0048654 0052673 0057253 0062411 0067916

(3) 0070192 0073979 0078257 0083057 0088167

(4) 0070243 0074119 0078513 0083451 0088714

(5) 0151774 0154981 0158572 0162586 0166852

(6) 0151815 0155096 0158784 0162913 0167308

y = 25 (1) 0076895 0083230 0090413 0098481 0107073

(2) 0076987 0083474 0090854 0099156 0108010

(3) 0110093 0116205 0123095 0130818 0139033

(4) 0110177 0116435 0123515 0131462 0139928

(5) 0230674 0235848 0241632 0248091 0254950

(6) 0230739 0236038 0241980 0248626 0255695

y = 50 (1) 0091496 0099156 0107829 0117563 0127924

(2) 0091610 0099455 0108370 0118387 0129068

(3) 0130130 0137520 0145839 0155156 0165064

(4) 0130233 0137801 0146352 0155942 0166156

(5) 0266599 0272857 0279840 0287632 0295904

(6) 0266680 0273088 0280265 0288286 0296814

y = infin (1) 0099312 0107746 0117285 0127984 0139369

(2) 0099439 0108078 0117883 0128899 0140639

(3) 0140458 0148595 0157745 0167986 0178872

(4) 0140572 0148908 0158314 0168859 0180084

(5) 0282886 0289776 0297457 0306022 0315111

(6) 0282976 0290034 0297929 0306748 0316120

26

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0059096 0072505 0088907 0108122 0129137

(2) 0059223 0072829 0089500 0109052 0130456

(3) 0079757 0092662 0108388 0126778 0146871

(4) 0079875 0092970 0108956 0127669 0148134

(5) 0159788 0170682 0183871 0199249 0216024

(6) 0159881 0170936 0184343 0199990 0217075

y = 25 (1) 0095883 0119741 0149240 0184073 0222390

(2) 0096087 0120262 0150197 0185573 0224520

(3) 0127403 0150359 0178640 0211977 0248613

(4) 0127592 0150855 0179557 0213414 0250653

(5) 0245161 0264533 0288250 0316124 0346709

(6) 0245311 0264942 0289010 0317319 0348407

y = 50 (1) 0116598 0147920 0187100 0233784 0285503

(2) 0116839 0148537 0188231 0235554 0288015

(3) 0153023 0183153 0220713 0265390 0314840

(4) 0153249 0183742 0221798 0267087 0317248

(5) 0285742 0311159 0342651 0380005 0421287

(6) 0285921 0311644 0343551 0381418 0423292

y = infin (1) 0129470 0167217 0215383 0273841 0339734

(2) 0129725 0167865 0216560 0275661 0342282

(3) 0167978 0204275 0250442 0306382 0369383

(4) 0168217 0204895 0251572 0308128 0371826

(5) 0305868 0336470 0375169 0421935 0474525

(6) 0306058 0336980 0376106 0423388 0476560

27

Page 13: uwaterloo.ca · Ruin Analysis of a Threshold Strategy in a Discrete-time Sparre Andersen Model Steve DREKIC and Ana Maria MERA Department of Statistics and Actuarial Science University

if we extend the definition tk lowast by introducing the quantity ζk

lowast = maxtklowast 0 we remark that

(31) and (32) can actually be combined into a single formula for lk(n) namely

vk+n(zk + n) if n = 0 1 ζklowast

lk(n) = (33) maxlk(ζ

lowast) + c2 Z + c + c2(n minus ζlowast minus 1) if n = ζlowast + 1 ζlowast + 2 k k k k

Clearly we have that lk(n) ge Z + c for n = ζk lowast +1 ζk

lowast +2 Moreover if c1 = c2 = c (33)

reduces to give lk(n) = u + c(k + n) forall n isin N which is in agreement with the results in Alfa

and Drekic (2007 Section 3)

Remark In our above analysis we purposely ignored the possibility of any future claims

occurring after time k in our determination of lk(n) However we recognize that this is not

realistic and future claims would indeed occur after time k thereby further reducing the value

of the maximum surplus level lk(n) minus 1 at time k + n To account for this we imagine all

future interclaim times (ie after time k) to be of maximum length na and all corresponding

claim sizes to be of minimum amount 1 Consequently in the case when c2 gt 0 we should

subtract from formula (33) the amount of lfloornnarfloor where lfloormiddotrfloor again denotes the floor function

Obviously when c2 = 0 (ie no premiums are collected whenever the surplus level is at or

above Z) lk(n) = Z + c for n = ζk lowast + 1 ζk

lowast + 2 (k)

We now turn our attention to the development of a general algorithm for computing gni

(k) (k)for i = 0 1 lk(n) minus 1 With the aid of (24) we apply the key identity gn = gnminus1C

(which can be inferred from (27)) to first obtain

minZlk(nminus1)minus1 lk(nminus1)minus1 c2 L L L

(k) (k) (k)gni = g Bjminusi+c + gnminus1j dmBjminusi+m n isin Z+ (34) nminus1j

j=0 j=Z m=c1

However (22) states that Bi = Ona for i isin Zminus and so (34) becomes (for the same range of

i)

minZlk(nminus1)minus1 c2 lk(nminus1)minus1 L L L

(k) (k) (k)gni = g Bjminusi+c + dm gnminus1jBjminusi+m n isin Z+ (35) nminus1j

j=max0iminusc m=c1 j=maxZiminusm

Taking the form of (33) into consideration (35) yields the following two formulae

lk(nminus1)minus1 L

(k) (k)gni = gnminus1jBjminusi+c n = 1 2 ζ k

lowast (36) j=max0iminusc

and

Zminus1 c2 lk(nminus1)minus1 L L L

(k) (k) (k)g = g + dm g n = ζ lowast + 1 ζ lowast + 2 ni nminus1jBjminusi+c nminus1jBjminusi+m k k

j=max0iminusc m=c1 j=maxZiminusm

(37)

13

For n = 1 2 ζk lowast where ζk

lowast isin Z+ we note that lk(n) ge c + 1 Therefore with the aid of

(22) (36) simplifies to give

Llk(nminus1)minus1 (k) (k) j=0 gnminus1jBjminusi+c if i = 0 1 c minus 1

gni = Llk(nminus1)minus1 (k)

g if i = c c + 1 lk(n) minus 1j=iminusc nminus1jBjminusi+c

Llk(nminus1)minus1 (k)

nminus1js 0 0 if i = 0 1 c minus 1j=0 αjminusi+cg= (38)

(k) Llk(nminus1)minus1 (k) g S+ αjminusi+cgnminus1js 0 0 if i = c c+1 lk(n)minus1 nminus1iminusc j=iminusc+1

In the case of n = ζk lowast + 1 ζk

lowast + 2 we first reiterate that lk(n) ge Z + c We next break

(37) into several cases based on the value of i and the value of minc Z + c1 as follows

Case 1 Suppose minc Z + c1 = c

For i isin 0 1 c minus 1

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+mni nminus1j nminus1j

j=0 m=c1 j=Z

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k)= αjminusi+cg s + dm αjminusi+mg s 0 0 (39) nminus1j nminus1j

j=0 m=c1 j=Z

For i isin c c + 1 Z + c1 minus 1

Zminus1 c2 lk(nminus1)minus1 L L L

(k) (k) (k)g = g + dm gni nminus1j Bjminusi+c nminus1j Bjminusi+m

j=iminusc m=c1 j=Z

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k) (k)= g S + αjminusi+cg dm αjminusi+mgnminus1j s 0 0 (310) nminus1iminusc nminus1j s +

j=iminusc+1 m=c1 j=Z

For i isin Z + c1 Z + c1 + 1 Z + c minus 1

Zminus1 minc2iminusZ lk(nminus1)minus1 c2 lk(nminus1)minus1 L L L L L

(k) (k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+m + dm g Bjminusi+mni nminus1j nminus1j nminus1j

j=iminusc m=c1 j=iminusm m=iminusZ+1 j=Z

minc2iminusZ L

(k) (k)= g S + dmg S nminus1iminusc nminus1iminusm

m=c1

Zminus1 minc2iminusZ lk(nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k)+ αjminusi+cg dm αjminusi+mg dm αjminusi+mgnminus1j s 0 0 nminus1j s + nminus1js +

j=iminusc+1 m=c1 j=iminusm+1 m=iminusZ+1 j=Z

(311)

For i isin Z + c Z + c + 1 lk(n) minus 1

c2 lk(nminus1)minus1 L L

(k) (k)g = dm g Bjminusi+mni nminus1j

m=c1 j=iminusm

c2 c2 lk (nminus1)minus1 L L L

(k) (k)= dmg S + dm αjminusi+mgnminus1j s 0 0 (312) nminus1iminusm

m=c1 m=c1 j=iminusm+1

14

Case 2 Suppose minc Z + c1 = Z + c1

For i isin 0 1 Z + c1 minus 1

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+mni nminus1j nminus1j

j=0 m=c1 j=Z

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k)= αjminusi+cg s + dm αjminusi+mg s 0 0 (313) nminus1j nminus1j

j=0 m=c1 j=Z

For i isin Z + c1 Z + c1 + 1 c minus 1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+m + dm g Bjminusi+mni nminus1j nminus1j nminus1j

j=0 m=c1 j=iminusm m=iminusZ+1 j=Z

minc2iminusZ L

(k)= dmg S nminus1iminusm

m=c1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k)+ αjminusi+cg s + dm αjminusi+mg s + dm αjminusi+mg s 0 0 nminus1j nminus1j nminus1j

j=0 m=c1 j=iminusm+1 m=iminusZ+1 j=Z

(314)

For i isin c c + 1 Z + c minus 1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k) (k)g = g + g + gni nminus1j Bjminusi+c dm nminus1j Bjminusi+m dm nminus1j Bjminusi+m

j=iminusc m=c1 j=iminusm m=iminusZ+1 j=Z

minc2iminusZ L

(k) (k)= g S + dmg S nminus1iminusc nminus1iminusm

m=c1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k)+

nminus1j s 0 0 αjminusi+cgnminus1j s + dm αjminusi+mgnminus1j s + dm αjminusi+mg j=iminusc+1 m=c1 j=iminusm+1 m=iminusZ+1 j=Z

(315)

For i isin Z + c Z + c + 1 lk(n) minus 1

c2 lk (nminus1)minus1 L L

(k) (k)g = dm g Bjminusi+mni nminus1j

m=c1 j=iminusm

c2 c2 lk (nminus1)minus1 L L L

(k) (k)= dmg S + dm αjminusi+mgnminus1j s 0 0 (316) nminus1iminusm

m=c1 m=c1 j=iminusm+1

It is not difficult to verify that equations (38) through (316) simplify to produce equations

(32) and (33) of Alfa and Drekic (2007 pp 301-302) when c1 = c2 = c With equations (38) (k)

through (316) providing the recursive means of calculating gn we next turn our attention

15

to a determination of h(k) defined by (28) For n = 1 2 ζlowast we use (22) and (25) to n k

obtain

lk(nminus1)minus1 lk(nminus1)minus1

h(k) nminusj =

L g

(k) nminus1ℓBc+j+ℓ =

L αc+j+ℓg

(k) nminus1ℓs 0 0 (317)

ℓ=0 ℓ=0

For n = ζk lowast + 1 ζk

lowast + 2 we similarly acquire

Zminus1 lk(nminus1)minus1 L L

(k) (k) (k)hnminusj = gnminus1ℓBc+j+ℓ + gnminus1ℓAj+ℓ

ℓ=0 ℓ=Z

Zminus1 c2 lk(nminus1)minus1 L L L

(k) (k)= αc+j+ℓg s + dm αm+j+ℓg s 0 0 (318) nminus1ℓ nminus1ℓ

ℓ=0 m=c1 ℓ=Z

Note that substituting (317) and (318) into (29) immediately yields

Llk(nminus1)minus1 (k) (k) ℓ=0 αc+j+ℓgnminus1ℓs if n = 1 2 ζk

lowast φ (u) = nj LZminus1 (k) Lc2

Llk(nminus1)minus1 (k) ζlowast s + dm αm+j+ℓg n = + 1 ζlowast + 2 ℓ=0 αc+j+ℓgnminus1ℓ m=c1 ℓ=Z nminus1ℓs if k k

(319)

Clearly in the special case when c1 = c2 = c (319) yields equation (35) of Alfa and Drekic

(2007 p 302) Finally we remark that (319) can be substituted into (213) to obtain a

computational formula for φnj(u)

4 Numerical examples and findings

In this section we illustrate the application of our proposed algorithm with several numerical

examples All calculations in this section were performed using Mathematica (Version 7) on

a Dell Precision T7400 workstation with a 3 Ghz CPU and 4 GB of RAM

In our first illustration (labelled Example 1) we consider the ordinary Sparre Andersen

model in which interclaim times have pmf given by

(211)(911)jminus1 if j = 1 2 na minus 1 aj = (41)

(911)naminus1 if j = na

Note that the pmf (41) is that of a truncated geometric distribution with all the probability

mass on na na +1 assigned to the support value na Clearly as na becomes larger the

closer (41) approximates this particular geometric distribution having mean 55 In what Lna

Lminus1 nafollows we fix na = 25 so that aj = 0992 and jaj = 546 Let the individual j=1 j=1

claim amount distribution be given by the pmf

minus4 minus4j minus 1 j

αj = 1 + minus 1 + j isin Z+ (42) 30 30

16

We remark that the pmf (42) represents a discretized version of the Pareto distribution with

mean 10 Also note that mα = infin thereby implying that the random variables |UT | and

UTminus both have infinite support

Table 1 displays values (rounded to 6 decimal places) of the trivariate cumulative distrishy

bution function

n x y L L L

Ψnxy(u) = Pr T le n UTminus le x |UT | le y | U0 = u = ψℓij(u) ℓ=1 i=mincZ+c1 j=1

corresponding to the above discrete-time risk process with u = 10 Z = 50 c = 5 and

varying premium amount distributions according to the following set of scenarios

(1) d2 = 1

(2) d1 = 25 d2 = 15 d3 = 25

G

5)

(3) di = i

(25)i(35)5minusi i = 0 1 5

(4) di = 15 i = 0 1 4

(5) d1 = 34 d5 = 14

(6) d0 = 35 d5 = 25

We remark that while these six premium amount distributions all have mean equal to 2

they possess different shapes and we investigate the impact of their variability on the joint

distribution of T |UT | and UTminus In particular the variances corresponding to the premium

amount distributions of scenarios (1) through (6) are 0 08 12 2 3 and 6 respectively

Based on the results in Table 1 it would appear that greater variability in the premium

amount distribution generally leads to higher ruin probabilities Perhaps somewhat surprising

though the differences are subtle and not particularly striking as one compares the various

ruin probabilities from scenarios (1) through (6) Obviously we observe more of a difference

between scenarios as the values of n x and y increase

As a follow-up illustration (labelled Example 2) we consider a model in which ordinary

interclaim times have a discrete uniform distribution with pmf aj = 110 j = 1 2 10

Clearly the mean of this distribution is 55 which for comparative purposes is identical to

the ordinary interclaim time mean considered in Example 1 In addition we assume that

the pmf of the individual claim amount distribution is once again given by (42) Table 2

displays values (rounded to 6 decimal places) of Ψnxy(10) for several variants of the Sparre

Andersen model (again with Z = 50 and c = 5) having the above ordinary interclaim time

and claim amount distributions namely

17

(1) ordinary model under scenario (2) from Example 1

(2) ordinary model under scenario (5) from Example 1

(3) stationary model under scenario (2) from Example 1

(4) stationary model under scenario (5) from Example 1

(5) delayed model (with W1 = 1 with probability 1) under scenario (2) from Example 1

(6) delayed model (with W1 = 1 with probability 1) under scenario (5) from Example 1

Concerning the results in Table 2 we first compare the values of Ψnxy(10) in entries (2)

and (5) of Table 1 to their corresponding counterparts in entries (1) and (2) of Table 2 In all

cases we observe that ruin is more likely to occur when interclaim times have the truncated

geometric distribution introduced in our first example as opposed to the discrete uniform

distribution of our second example Since mean interclaim times are equal in both examshy

ples this behaviour must be attributable to the greater variability inherent in the truncated

geometric distribution We also note that under the stationary model assumption the pmf

of W1 is easily verified to be rj = (11 minus j)55 j = 1 2 10 which has mean 4 lt 55

As a result the values of Ψnxy(10) for stationary models (3) and (4) are always larger than

the corresponding values for ordinary models (1) and (2) Finally in the absence of any

knowledge regarding which distribution is in effect for the first interclaim time W1 models

(5) and (6) can be regarded as ldquoworst-case scenariordquo models as they serve to provide upper

bounds on the values of Ψnxy(10)

In conclusion we state that the values in Tables 1 and 2 were generated by first impleshy

menting the recursive procedures (38) through (316) and then summing the appropriate

bivariatetrivariate probabilities computed via (213) (214) or (215) In situations when

x = y = infin however we computed the resulting finite-time ruin probabilities using an altershy

native (albeit standard) method (eg see Cossette et al (2006 Section 3)) Specifically if

we define the survival probability σ(n u) = Pr T gt n | U0 = u = 1minusΨninfininfin(u) as well as

the auxiliary function

0 if w ge Z ηwt =

minlfloorZminuswc minus1 rfloor + 1 t if w lt Z

then by conditioning on the time and amount of the first claim (and recalling that the delayed

process reverts to the ordinary process following this claim) we obtain the expression

minnnr (kminuszk)c2 u+czk+ℓ L L L

σ(u n) = Rn + rk dℓkminuszk αjγ(u + czk + ℓ minus j n minus k)

k=1 ℓ=(kminuszk)c1 j=1

18

where the function γ(w m) for w m isin N is computed recursively via

minmna (kminusηwk)c2 u+cηwk+ℓ L L L

γ(w m) = Am + ak dℓkminusηwk αjγ(u + cηwk + ℓ minus j m minus k)

k=1 j=1 ℓ=(kminusηwk)c1

with starting values γ(w 0) = 1 forall w isin N

Acknowledgements

The authors would like to thank Dr David Landriault for his valuable suggestions which

helped to improve an earlier version of this work This research was supported by the Natural

Sciences and Engineering Research Council of Canada

References

Alfa AS 2004 Markov chain representations of discrete distributions applied to queueing

models Computers amp Operations Research 31 2365-2385

Alfa AS Drekic S 2007 Algorithmic analysis of the Sparre Andersen model in discrete

time ASTIN Bulletin 37 293-317

Bao Z 2007 A note on the compound binomial model with randomized dividend strategy

Applied Mathematics amp Computation 194 276-286

Cossette H Landriault D Marceau E 2006 Ruin probabilities in the discrete time

renewal risk model Insurance Mathematics amp Economics 38 309-323

Karlin S Taylor HM 1975 A First Course in Stochastic Processes 2nd edition Acashy

demic Press New York

Kim B Kim H-S Kim J 2008 A risk model with paying dividends and random

environment Insurance Mathematics amp Economics 42 717-726

Landriault D 2008 Randomized dividends in the compound binomial model with a general

premium rate Scandinavian Actuarial Journal 1-15

Tan J Yang X 2006 The compound binomial model with randomized decisions on

paying dividends Insurance Mathematics amp Economics 39 1-18

19

Table 1 ndash Values of Ψnxy(10) corresponding to Example 1

(a) x = 10

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0005357 0006188 0007163 0008278 0009481

(2) 0005359 0006195 0007175 0008296 0009507

(3) 0005361 0006198 0007181 0008305 0009519

(4) 0005363 0006205 0007193 0008323 0009544

(5) 0005366 0006212 0007208 0008346 0009576

(6) 0005374 0006236 0007249 0008410 0009664

y = 25 (1) 0007631 0008815 0010202 0011789 0013503

(2) 0007634 0008824 0010220 0011816 0013539

(3) 0007636 0008828 0010228 0011829 0013557

(4) 0007639 0008838 0010245 0011854 0013593

(5) 0007643 0008849 0010266 0011887 0013638

(6) 0007656 0008882 0010325 0011978 0013764

y = 50 (1) 0008477 0009792 0011333 0013095 0014998

(2) 0008481 0009802 0011352 0013125 0015039

(3) 0008483 0009807 0011361 0013139 0015059

(4) 0008486 0009817 0011380 0013168 0015098

(5) 0008491 0009830 0011403 0013204 0015149

(6) 0008504 0009866 0011470 0013305 0015288

y = infin (1) 0008777 0010138 0011733 0013558 0015528

(2) 0008781 0010149 0011753 0013588 0015570

(3) 0008783 0010154 0011763 0013603 0015591

(4) 0008787 0010164 0011782 0013633 0015632

(5) 0008791 0010177 0011807 0013671 0015684

(6) 0008805 0010215 0011875 0013775 0015828

20

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0058559 0061875 0065704 0070048 0074714

(2) 0058569 0061903 0065756 0070128 0074825

(3) 0058574 0061917 0065782 0070167 0074880

(4) 0058584 0061945 0065833 0070245 0074987

(5) 0058596 0061979 0065896 0070344 0075124

(6) 0058632 0062079 0066078 0070621 0075506

y = 25 (1) 0089002 0093986 0099739 0106262 0113270

(2) 0089017 0094028 0099817 0106382 0113437

(3) 0089025 0094049 0099856 0106442 0113519

(4) 0089040 0094091 0099932 0106560 0113681

(5) 0089057 0094142 0100028 0106708 0113887

(6) 0089113 0094294 0100302 0107127 0114463

y = 50 (1) 0102786 0108506 0115106 0122590 0130629

(2) 0102804 0108555 0115196 0122728 0130820

(3) 0102812 0108579 0115241 0122797 0130914

(4) 0102830 0108627 0115329 0122932 0131101

(5) 0102850 0108686 0115439 0123103 0131337

(6) 0102914 0108860 0115754 0123584 0132000

y = infin (1) 0108895 0114933 0121901 0129799 0138283

(2) 0108914 0114985 0121996 0129945 0138485

(3) 0108923 0115010 0122043 0130018 0138585

(4) 0108941 0115061 0122136 0130161 0138782

(5) 0108962 0115123 0122252 0130341 0139032

(6) 0109030 0115307 0122585 0130849 0139732

21

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0079856 0086792 009464 0103440 0112831

(2) 0079885 0086869 0094778 0103650 0113118

(3) 0079953 0086978 0094936 0103864 0113393

(4) 0080081 0087188 0095243 0104282 0113929

(5) 0079962 0087075 0095149 0104213 0113890

(6) 0080805 0088366 0096951 0106588 0116875

y = 25 (1) 0124591 0135680 0148207 0162245 0177217

(2) 0124639 0135806 0148433 0162586 0177683

(3) 0124755 0135993 0148703 0162950 0178148

(4) 0124978 0136354 0149226 0163658 0179054

(5) 0124763 0136141 0149035 0163500 0178935

(6) 0126238 0138381 0152148 0167588 0184060

y = 50 (1) 0146757 0160079 0175111 0191945 0209894

(2) 0146816 0160232 0175385 0192360 0210460

(3) 0146965 0160469 0175725 0192817 0211043

(4) 0147249 0160925 0176384 0193706 0212178

(5) 0146968 0160639 0176117 0193470 0211980

(6) 0148860 0163500 0180077 0198658 0218474

y = infin (1) 0158013 0172606 0189059 0207476 0227106

(2) 0158078 0172776 0189362 0207934 0227732

(3) 0158250 0173047 0189751 0208455 0228395

(4) 0158579 0173571 0190503 0209467 0229685

(5) 0158247 0173226 0190172 0209161 0229410

(6) 0160443 0176532 0194735 0215127 0236869

22

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0088667 0104403 0123329 0145277 0169184

(2) 0088711 0104518 0123538 0145598 0169627

(3) 0088733 0104574 0123641 0145756 0169846

(4) 0088777 0104687 0123844 0146066 0170274

(5) 0088829 0104828 0124105 0146470 0170832

(6) 0088988 0105229 0124822 0147560 0172331

y = 25 (1) 0140528 0167839 0201007 0239726 0282102

(2) 0140598 0168021 0201338 0240234 0282804

(3) 0140632 0168111 0201502 0240486 0283151

(4) 0140701 0168289 0201822 0240976 0283827

(5) 0140785 0168515 0202241 0241622 0284718

(6) 0141036 0169148 0203370 0243335 0287071

y = 50 (1) 0167826 0203005 0246173 0296954 0352855

(2) 0167908 0203218 0246561 0297548 0353672

(3) 0167949 0203324 0246753 0297841 0354075

(4) 0168030 0203532 0247128 0298412 0354859

(5) 0168129 0203800 0247622 0299174 0355907

(6) 0168423 0204539 0248936 0301159 0358620

y = infin (1) 0183325 0224926 0276881 0338957 0408257

(2) 0183412 0225150 0277283 0339564 0409079

(3) 0183455 0225260 0277481 0339864 0409485

(4) 0183540 0225478 0277869 0340446 0410270

(5) 0183645 0225760 0278386 0341234 0411339

(6) 0183954 0226529 0279737 0343244 0414037

23

Table 2 ndash Values of Ψnxy(10) corresponding to Example 2

(a) x = 10

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0001901 0002276 0002724 0003240 0003799

(2) 0001904 0002286 0002742 0003269 0003840

(3) 0002651 0003013 0003442 0003936 0004471

(4) 0002655 0003022 0003460 0003964 0004510

(5) 0005792 0006098 0006458 0006871 0007317

(6) 0005795 0006106 0006473 0006894 0007350

y = 25 (1) 0002708 0003243 0003880 0004615 0005411

(2) 0002713 0003257 0003906 0004656 0005469

(3) 0003778 0004293 0004904 0005608 0006369

(4) 0003782 0004306 0004929 0005647 0006424

(5) 0008255 0008690 0009203 0009792 0010427

(6) 0008259 0008702 0009224 0009824 0010473

y = 50 (1) 0003008 0003603 0004310 0005127 0006011

(2) 0003014 0003618 0004340 0005173 0006076

(3) 0004197 0004769 0005448 0006230 0007075

(4) 0004202 0004784 0005476 0006273 0007137

(5) 0009173 0009656 0010226 0010879 0011585

(6) 0009177 0009669 0010249 0010916 0011637

y = infin (1) 0003115 0003730 0004463 0005308 0006224

(2) 0003121 0003746 0004493 0005356 0006291

(3) 0004346 0004938 0005641 0006450 0007325

(4) 0004351 0004954 0005670 0006496 0007389

(5) 0009499 0009999 0010589 0011265 0011996

(6) 0009503 0010012 0010613 0011303 0012050

24

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0031176 0032815 0034733 0036924 0039282

(2) 0031193 0032862 0034821 0037061 0039475

(3) 0050168 0051747 0053587 0055684 0057939

(4) 0050184 0051792 0053671 0055815 0058124

(5) 0136884 0138219 0139763 0141517 0143399

(6) 0136896 0138256 0139832 0141626 0143553

y = 25 (1) 0047622 0050092 0052981 0056280 0059829

(2) 0004765 0050163 0053113 0056487 0060121

(3) 0076570 0078950 0081720 0084878 0088272

(4) 0076593 0079017 0081847 0085076 0088551

(5) 0205774 0207786 0210111 0212752 0215585

(6) 0205793 0207842 0210216 0212917 0215818

y = 50 (1) 0055170 0058009 0061329 0065119 0069196

(2) 0055200 0058091 0061481 0065357 0069533

(3) 0088650 0091385 0094569 0098197 0102096

(4) 0088677 0091463 0094715 0098425 0102417

(5) 0235815 0238128 0240800 0243834 0247089

(6) 0235837 0238192 0240920 0244023 0247356

y = infin (1) 0058565 0061565 0065073 0069077 0073384

(2) 0058597 0061652 0065234 0069329 0073740

(3) 0094061 0096952 0100316 0104149 0108267

(4) 0094090 0097035 0100470 0104390 0108607

(5) 0248472 0250917 0253740 0256945 0260383

(6) 0248495 0250985 0253867 0257145 0260667

25

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0048598 0052524 0056984 0061998 0067342

(2) 0048654 0052673 0057253 0062411 0067916

(3) 0070192 0073979 0078257 0083057 0088167

(4) 0070243 0074119 0078513 0083451 0088714

(5) 0151774 0154981 0158572 0162586 0166852

(6) 0151815 0155096 0158784 0162913 0167308

y = 25 (1) 0076895 0083230 0090413 0098481 0107073

(2) 0076987 0083474 0090854 0099156 0108010

(3) 0110093 0116205 0123095 0130818 0139033

(4) 0110177 0116435 0123515 0131462 0139928

(5) 0230674 0235848 0241632 0248091 0254950

(6) 0230739 0236038 0241980 0248626 0255695

y = 50 (1) 0091496 0099156 0107829 0117563 0127924

(2) 0091610 0099455 0108370 0118387 0129068

(3) 0130130 0137520 0145839 0155156 0165064

(4) 0130233 0137801 0146352 0155942 0166156

(5) 0266599 0272857 0279840 0287632 0295904

(6) 0266680 0273088 0280265 0288286 0296814

y = infin (1) 0099312 0107746 0117285 0127984 0139369

(2) 0099439 0108078 0117883 0128899 0140639

(3) 0140458 0148595 0157745 0167986 0178872

(4) 0140572 0148908 0158314 0168859 0180084

(5) 0282886 0289776 0297457 0306022 0315111

(6) 0282976 0290034 0297929 0306748 0316120

26

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0059096 0072505 0088907 0108122 0129137

(2) 0059223 0072829 0089500 0109052 0130456

(3) 0079757 0092662 0108388 0126778 0146871

(4) 0079875 0092970 0108956 0127669 0148134

(5) 0159788 0170682 0183871 0199249 0216024

(6) 0159881 0170936 0184343 0199990 0217075

y = 25 (1) 0095883 0119741 0149240 0184073 0222390

(2) 0096087 0120262 0150197 0185573 0224520

(3) 0127403 0150359 0178640 0211977 0248613

(4) 0127592 0150855 0179557 0213414 0250653

(5) 0245161 0264533 0288250 0316124 0346709

(6) 0245311 0264942 0289010 0317319 0348407

y = 50 (1) 0116598 0147920 0187100 0233784 0285503

(2) 0116839 0148537 0188231 0235554 0288015

(3) 0153023 0183153 0220713 0265390 0314840

(4) 0153249 0183742 0221798 0267087 0317248

(5) 0285742 0311159 0342651 0380005 0421287

(6) 0285921 0311644 0343551 0381418 0423292

y = infin (1) 0129470 0167217 0215383 0273841 0339734

(2) 0129725 0167865 0216560 0275661 0342282

(3) 0167978 0204275 0250442 0306382 0369383

(4) 0168217 0204895 0251572 0308128 0371826

(5) 0305868 0336470 0375169 0421935 0474525

(6) 0306058 0336980 0376106 0423388 0476560

27

Page 14: uwaterloo.ca · Ruin Analysis of a Threshold Strategy in a Discrete-time Sparre Andersen Model Steve DREKIC and Ana Maria MERA Department of Statistics and Actuarial Science University

For n = 1 2 ζk lowast where ζk

lowast isin Z+ we note that lk(n) ge c + 1 Therefore with the aid of

(22) (36) simplifies to give

Llk(nminus1)minus1 (k) (k) j=0 gnminus1jBjminusi+c if i = 0 1 c minus 1

gni = Llk(nminus1)minus1 (k)

g if i = c c + 1 lk(n) minus 1j=iminusc nminus1jBjminusi+c

Llk(nminus1)minus1 (k)

nminus1js 0 0 if i = 0 1 c minus 1j=0 αjminusi+cg= (38)

(k) Llk(nminus1)minus1 (k) g S+ αjminusi+cgnminus1js 0 0 if i = c c+1 lk(n)minus1 nminus1iminusc j=iminusc+1

In the case of n = ζk lowast + 1 ζk

lowast + 2 we first reiterate that lk(n) ge Z + c We next break

(37) into several cases based on the value of i and the value of minc Z + c1 as follows

Case 1 Suppose minc Z + c1 = c

For i isin 0 1 c minus 1

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+mni nminus1j nminus1j

j=0 m=c1 j=Z

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k)= αjminusi+cg s + dm αjminusi+mg s 0 0 (39) nminus1j nminus1j

j=0 m=c1 j=Z

For i isin c c + 1 Z + c1 minus 1

Zminus1 c2 lk(nminus1)minus1 L L L

(k) (k) (k)g = g + dm gni nminus1j Bjminusi+c nminus1j Bjminusi+m

j=iminusc m=c1 j=Z

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k) (k)= g S + αjminusi+cg dm αjminusi+mgnminus1j s 0 0 (310) nminus1iminusc nminus1j s +

j=iminusc+1 m=c1 j=Z

For i isin Z + c1 Z + c1 + 1 Z + c minus 1

Zminus1 minc2iminusZ lk(nminus1)minus1 c2 lk(nminus1)minus1 L L L L L

(k) (k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+m + dm g Bjminusi+mni nminus1j nminus1j nminus1j

j=iminusc m=c1 j=iminusm m=iminusZ+1 j=Z

minc2iminusZ L

(k) (k)= g S + dmg S nminus1iminusc nminus1iminusm

m=c1

Zminus1 minc2iminusZ lk(nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k)+ αjminusi+cg dm αjminusi+mg dm αjminusi+mgnminus1j s 0 0 nminus1j s + nminus1js +

j=iminusc+1 m=c1 j=iminusm+1 m=iminusZ+1 j=Z

(311)

For i isin Z + c Z + c + 1 lk(n) minus 1

c2 lk(nminus1)minus1 L L

(k) (k)g = dm g Bjminusi+mni nminus1j

m=c1 j=iminusm

c2 c2 lk (nminus1)minus1 L L L

(k) (k)= dmg S + dm αjminusi+mgnminus1j s 0 0 (312) nminus1iminusm

m=c1 m=c1 j=iminusm+1

14

Case 2 Suppose minc Z + c1 = Z + c1

For i isin 0 1 Z + c1 minus 1

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+mni nminus1j nminus1j

j=0 m=c1 j=Z

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k)= αjminusi+cg s + dm αjminusi+mg s 0 0 (313) nminus1j nminus1j

j=0 m=c1 j=Z

For i isin Z + c1 Z + c1 + 1 c minus 1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+m + dm g Bjminusi+mni nminus1j nminus1j nminus1j

j=0 m=c1 j=iminusm m=iminusZ+1 j=Z

minc2iminusZ L

(k)= dmg S nminus1iminusm

m=c1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k)+ αjminusi+cg s + dm αjminusi+mg s + dm αjminusi+mg s 0 0 nminus1j nminus1j nminus1j

j=0 m=c1 j=iminusm+1 m=iminusZ+1 j=Z

(314)

For i isin c c + 1 Z + c minus 1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k) (k)g = g + g + gni nminus1j Bjminusi+c dm nminus1j Bjminusi+m dm nminus1j Bjminusi+m

j=iminusc m=c1 j=iminusm m=iminusZ+1 j=Z

minc2iminusZ L

(k) (k)= g S + dmg S nminus1iminusc nminus1iminusm

m=c1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k)+

nminus1j s 0 0 αjminusi+cgnminus1j s + dm αjminusi+mgnminus1j s + dm αjminusi+mg j=iminusc+1 m=c1 j=iminusm+1 m=iminusZ+1 j=Z

(315)

For i isin Z + c Z + c + 1 lk(n) minus 1

c2 lk (nminus1)minus1 L L

(k) (k)g = dm g Bjminusi+mni nminus1j

m=c1 j=iminusm

c2 c2 lk (nminus1)minus1 L L L

(k) (k)= dmg S + dm αjminusi+mgnminus1j s 0 0 (316) nminus1iminusm

m=c1 m=c1 j=iminusm+1

It is not difficult to verify that equations (38) through (316) simplify to produce equations

(32) and (33) of Alfa and Drekic (2007 pp 301-302) when c1 = c2 = c With equations (38) (k)

through (316) providing the recursive means of calculating gn we next turn our attention

15

to a determination of h(k) defined by (28) For n = 1 2 ζlowast we use (22) and (25) to n k

obtain

lk(nminus1)minus1 lk(nminus1)minus1

h(k) nminusj =

L g

(k) nminus1ℓBc+j+ℓ =

L αc+j+ℓg

(k) nminus1ℓs 0 0 (317)

ℓ=0 ℓ=0

For n = ζk lowast + 1 ζk

lowast + 2 we similarly acquire

Zminus1 lk(nminus1)minus1 L L

(k) (k) (k)hnminusj = gnminus1ℓBc+j+ℓ + gnminus1ℓAj+ℓ

ℓ=0 ℓ=Z

Zminus1 c2 lk(nminus1)minus1 L L L

(k) (k)= αc+j+ℓg s + dm αm+j+ℓg s 0 0 (318) nminus1ℓ nminus1ℓ

ℓ=0 m=c1 ℓ=Z

Note that substituting (317) and (318) into (29) immediately yields

Llk(nminus1)minus1 (k) (k) ℓ=0 αc+j+ℓgnminus1ℓs if n = 1 2 ζk

lowast φ (u) = nj LZminus1 (k) Lc2

Llk(nminus1)minus1 (k) ζlowast s + dm αm+j+ℓg n = + 1 ζlowast + 2 ℓ=0 αc+j+ℓgnminus1ℓ m=c1 ℓ=Z nminus1ℓs if k k

(319)

Clearly in the special case when c1 = c2 = c (319) yields equation (35) of Alfa and Drekic

(2007 p 302) Finally we remark that (319) can be substituted into (213) to obtain a

computational formula for φnj(u)

4 Numerical examples and findings

In this section we illustrate the application of our proposed algorithm with several numerical

examples All calculations in this section were performed using Mathematica (Version 7) on

a Dell Precision T7400 workstation with a 3 Ghz CPU and 4 GB of RAM

In our first illustration (labelled Example 1) we consider the ordinary Sparre Andersen

model in which interclaim times have pmf given by

(211)(911)jminus1 if j = 1 2 na minus 1 aj = (41)

(911)naminus1 if j = na

Note that the pmf (41) is that of a truncated geometric distribution with all the probability

mass on na na +1 assigned to the support value na Clearly as na becomes larger the

closer (41) approximates this particular geometric distribution having mean 55 In what Lna

Lminus1 nafollows we fix na = 25 so that aj = 0992 and jaj = 546 Let the individual j=1 j=1

claim amount distribution be given by the pmf

minus4 minus4j minus 1 j

αj = 1 + minus 1 + j isin Z+ (42) 30 30

16

We remark that the pmf (42) represents a discretized version of the Pareto distribution with

mean 10 Also note that mα = infin thereby implying that the random variables |UT | and

UTminus both have infinite support

Table 1 displays values (rounded to 6 decimal places) of the trivariate cumulative distrishy

bution function

n x y L L L

Ψnxy(u) = Pr T le n UTminus le x |UT | le y | U0 = u = ψℓij(u) ℓ=1 i=mincZ+c1 j=1

corresponding to the above discrete-time risk process with u = 10 Z = 50 c = 5 and

varying premium amount distributions according to the following set of scenarios

(1) d2 = 1

(2) d1 = 25 d2 = 15 d3 = 25

G

5)

(3) di = i

(25)i(35)5minusi i = 0 1 5

(4) di = 15 i = 0 1 4

(5) d1 = 34 d5 = 14

(6) d0 = 35 d5 = 25

We remark that while these six premium amount distributions all have mean equal to 2

they possess different shapes and we investigate the impact of their variability on the joint

distribution of T |UT | and UTminus In particular the variances corresponding to the premium

amount distributions of scenarios (1) through (6) are 0 08 12 2 3 and 6 respectively

Based on the results in Table 1 it would appear that greater variability in the premium

amount distribution generally leads to higher ruin probabilities Perhaps somewhat surprising

though the differences are subtle and not particularly striking as one compares the various

ruin probabilities from scenarios (1) through (6) Obviously we observe more of a difference

between scenarios as the values of n x and y increase

As a follow-up illustration (labelled Example 2) we consider a model in which ordinary

interclaim times have a discrete uniform distribution with pmf aj = 110 j = 1 2 10

Clearly the mean of this distribution is 55 which for comparative purposes is identical to

the ordinary interclaim time mean considered in Example 1 In addition we assume that

the pmf of the individual claim amount distribution is once again given by (42) Table 2

displays values (rounded to 6 decimal places) of Ψnxy(10) for several variants of the Sparre

Andersen model (again with Z = 50 and c = 5) having the above ordinary interclaim time

and claim amount distributions namely

17

(1) ordinary model under scenario (2) from Example 1

(2) ordinary model under scenario (5) from Example 1

(3) stationary model under scenario (2) from Example 1

(4) stationary model under scenario (5) from Example 1

(5) delayed model (with W1 = 1 with probability 1) under scenario (2) from Example 1

(6) delayed model (with W1 = 1 with probability 1) under scenario (5) from Example 1

Concerning the results in Table 2 we first compare the values of Ψnxy(10) in entries (2)

and (5) of Table 1 to their corresponding counterparts in entries (1) and (2) of Table 2 In all

cases we observe that ruin is more likely to occur when interclaim times have the truncated

geometric distribution introduced in our first example as opposed to the discrete uniform

distribution of our second example Since mean interclaim times are equal in both examshy

ples this behaviour must be attributable to the greater variability inherent in the truncated

geometric distribution We also note that under the stationary model assumption the pmf

of W1 is easily verified to be rj = (11 minus j)55 j = 1 2 10 which has mean 4 lt 55

As a result the values of Ψnxy(10) for stationary models (3) and (4) are always larger than

the corresponding values for ordinary models (1) and (2) Finally in the absence of any

knowledge regarding which distribution is in effect for the first interclaim time W1 models

(5) and (6) can be regarded as ldquoworst-case scenariordquo models as they serve to provide upper

bounds on the values of Ψnxy(10)

In conclusion we state that the values in Tables 1 and 2 were generated by first impleshy

menting the recursive procedures (38) through (316) and then summing the appropriate

bivariatetrivariate probabilities computed via (213) (214) or (215) In situations when

x = y = infin however we computed the resulting finite-time ruin probabilities using an altershy

native (albeit standard) method (eg see Cossette et al (2006 Section 3)) Specifically if

we define the survival probability σ(n u) = Pr T gt n | U0 = u = 1minusΨninfininfin(u) as well as

the auxiliary function

0 if w ge Z ηwt =

minlfloorZminuswc minus1 rfloor + 1 t if w lt Z

then by conditioning on the time and amount of the first claim (and recalling that the delayed

process reverts to the ordinary process following this claim) we obtain the expression

minnnr (kminuszk)c2 u+czk+ℓ L L L

σ(u n) = Rn + rk dℓkminuszk αjγ(u + czk + ℓ minus j n minus k)

k=1 ℓ=(kminuszk)c1 j=1

18

where the function γ(w m) for w m isin N is computed recursively via

minmna (kminusηwk)c2 u+cηwk+ℓ L L L

γ(w m) = Am + ak dℓkminusηwk αjγ(u + cηwk + ℓ minus j m minus k)

k=1 j=1 ℓ=(kminusηwk)c1

with starting values γ(w 0) = 1 forall w isin N

Acknowledgements

The authors would like to thank Dr David Landriault for his valuable suggestions which

helped to improve an earlier version of this work This research was supported by the Natural

Sciences and Engineering Research Council of Canada

References

Alfa AS 2004 Markov chain representations of discrete distributions applied to queueing

models Computers amp Operations Research 31 2365-2385

Alfa AS Drekic S 2007 Algorithmic analysis of the Sparre Andersen model in discrete

time ASTIN Bulletin 37 293-317

Bao Z 2007 A note on the compound binomial model with randomized dividend strategy

Applied Mathematics amp Computation 194 276-286

Cossette H Landriault D Marceau E 2006 Ruin probabilities in the discrete time

renewal risk model Insurance Mathematics amp Economics 38 309-323

Karlin S Taylor HM 1975 A First Course in Stochastic Processes 2nd edition Acashy

demic Press New York

Kim B Kim H-S Kim J 2008 A risk model with paying dividends and random

environment Insurance Mathematics amp Economics 42 717-726

Landriault D 2008 Randomized dividends in the compound binomial model with a general

premium rate Scandinavian Actuarial Journal 1-15

Tan J Yang X 2006 The compound binomial model with randomized decisions on

paying dividends Insurance Mathematics amp Economics 39 1-18

19

Table 1 ndash Values of Ψnxy(10) corresponding to Example 1

(a) x = 10

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0005357 0006188 0007163 0008278 0009481

(2) 0005359 0006195 0007175 0008296 0009507

(3) 0005361 0006198 0007181 0008305 0009519

(4) 0005363 0006205 0007193 0008323 0009544

(5) 0005366 0006212 0007208 0008346 0009576

(6) 0005374 0006236 0007249 0008410 0009664

y = 25 (1) 0007631 0008815 0010202 0011789 0013503

(2) 0007634 0008824 0010220 0011816 0013539

(3) 0007636 0008828 0010228 0011829 0013557

(4) 0007639 0008838 0010245 0011854 0013593

(5) 0007643 0008849 0010266 0011887 0013638

(6) 0007656 0008882 0010325 0011978 0013764

y = 50 (1) 0008477 0009792 0011333 0013095 0014998

(2) 0008481 0009802 0011352 0013125 0015039

(3) 0008483 0009807 0011361 0013139 0015059

(4) 0008486 0009817 0011380 0013168 0015098

(5) 0008491 0009830 0011403 0013204 0015149

(6) 0008504 0009866 0011470 0013305 0015288

y = infin (1) 0008777 0010138 0011733 0013558 0015528

(2) 0008781 0010149 0011753 0013588 0015570

(3) 0008783 0010154 0011763 0013603 0015591

(4) 0008787 0010164 0011782 0013633 0015632

(5) 0008791 0010177 0011807 0013671 0015684

(6) 0008805 0010215 0011875 0013775 0015828

20

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0058559 0061875 0065704 0070048 0074714

(2) 0058569 0061903 0065756 0070128 0074825

(3) 0058574 0061917 0065782 0070167 0074880

(4) 0058584 0061945 0065833 0070245 0074987

(5) 0058596 0061979 0065896 0070344 0075124

(6) 0058632 0062079 0066078 0070621 0075506

y = 25 (1) 0089002 0093986 0099739 0106262 0113270

(2) 0089017 0094028 0099817 0106382 0113437

(3) 0089025 0094049 0099856 0106442 0113519

(4) 0089040 0094091 0099932 0106560 0113681

(5) 0089057 0094142 0100028 0106708 0113887

(6) 0089113 0094294 0100302 0107127 0114463

y = 50 (1) 0102786 0108506 0115106 0122590 0130629

(2) 0102804 0108555 0115196 0122728 0130820

(3) 0102812 0108579 0115241 0122797 0130914

(4) 0102830 0108627 0115329 0122932 0131101

(5) 0102850 0108686 0115439 0123103 0131337

(6) 0102914 0108860 0115754 0123584 0132000

y = infin (1) 0108895 0114933 0121901 0129799 0138283

(2) 0108914 0114985 0121996 0129945 0138485

(3) 0108923 0115010 0122043 0130018 0138585

(4) 0108941 0115061 0122136 0130161 0138782

(5) 0108962 0115123 0122252 0130341 0139032

(6) 0109030 0115307 0122585 0130849 0139732

21

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0079856 0086792 009464 0103440 0112831

(2) 0079885 0086869 0094778 0103650 0113118

(3) 0079953 0086978 0094936 0103864 0113393

(4) 0080081 0087188 0095243 0104282 0113929

(5) 0079962 0087075 0095149 0104213 0113890

(6) 0080805 0088366 0096951 0106588 0116875

y = 25 (1) 0124591 0135680 0148207 0162245 0177217

(2) 0124639 0135806 0148433 0162586 0177683

(3) 0124755 0135993 0148703 0162950 0178148

(4) 0124978 0136354 0149226 0163658 0179054

(5) 0124763 0136141 0149035 0163500 0178935

(6) 0126238 0138381 0152148 0167588 0184060

y = 50 (1) 0146757 0160079 0175111 0191945 0209894

(2) 0146816 0160232 0175385 0192360 0210460

(3) 0146965 0160469 0175725 0192817 0211043

(4) 0147249 0160925 0176384 0193706 0212178

(5) 0146968 0160639 0176117 0193470 0211980

(6) 0148860 0163500 0180077 0198658 0218474

y = infin (1) 0158013 0172606 0189059 0207476 0227106

(2) 0158078 0172776 0189362 0207934 0227732

(3) 0158250 0173047 0189751 0208455 0228395

(4) 0158579 0173571 0190503 0209467 0229685

(5) 0158247 0173226 0190172 0209161 0229410

(6) 0160443 0176532 0194735 0215127 0236869

22

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0088667 0104403 0123329 0145277 0169184

(2) 0088711 0104518 0123538 0145598 0169627

(3) 0088733 0104574 0123641 0145756 0169846

(4) 0088777 0104687 0123844 0146066 0170274

(5) 0088829 0104828 0124105 0146470 0170832

(6) 0088988 0105229 0124822 0147560 0172331

y = 25 (1) 0140528 0167839 0201007 0239726 0282102

(2) 0140598 0168021 0201338 0240234 0282804

(3) 0140632 0168111 0201502 0240486 0283151

(4) 0140701 0168289 0201822 0240976 0283827

(5) 0140785 0168515 0202241 0241622 0284718

(6) 0141036 0169148 0203370 0243335 0287071

y = 50 (1) 0167826 0203005 0246173 0296954 0352855

(2) 0167908 0203218 0246561 0297548 0353672

(3) 0167949 0203324 0246753 0297841 0354075

(4) 0168030 0203532 0247128 0298412 0354859

(5) 0168129 0203800 0247622 0299174 0355907

(6) 0168423 0204539 0248936 0301159 0358620

y = infin (1) 0183325 0224926 0276881 0338957 0408257

(2) 0183412 0225150 0277283 0339564 0409079

(3) 0183455 0225260 0277481 0339864 0409485

(4) 0183540 0225478 0277869 0340446 0410270

(5) 0183645 0225760 0278386 0341234 0411339

(6) 0183954 0226529 0279737 0343244 0414037

23

Table 2 ndash Values of Ψnxy(10) corresponding to Example 2

(a) x = 10

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0001901 0002276 0002724 0003240 0003799

(2) 0001904 0002286 0002742 0003269 0003840

(3) 0002651 0003013 0003442 0003936 0004471

(4) 0002655 0003022 0003460 0003964 0004510

(5) 0005792 0006098 0006458 0006871 0007317

(6) 0005795 0006106 0006473 0006894 0007350

y = 25 (1) 0002708 0003243 0003880 0004615 0005411

(2) 0002713 0003257 0003906 0004656 0005469

(3) 0003778 0004293 0004904 0005608 0006369

(4) 0003782 0004306 0004929 0005647 0006424

(5) 0008255 0008690 0009203 0009792 0010427

(6) 0008259 0008702 0009224 0009824 0010473

y = 50 (1) 0003008 0003603 0004310 0005127 0006011

(2) 0003014 0003618 0004340 0005173 0006076

(3) 0004197 0004769 0005448 0006230 0007075

(4) 0004202 0004784 0005476 0006273 0007137

(5) 0009173 0009656 0010226 0010879 0011585

(6) 0009177 0009669 0010249 0010916 0011637

y = infin (1) 0003115 0003730 0004463 0005308 0006224

(2) 0003121 0003746 0004493 0005356 0006291

(3) 0004346 0004938 0005641 0006450 0007325

(4) 0004351 0004954 0005670 0006496 0007389

(5) 0009499 0009999 0010589 0011265 0011996

(6) 0009503 0010012 0010613 0011303 0012050

24

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0031176 0032815 0034733 0036924 0039282

(2) 0031193 0032862 0034821 0037061 0039475

(3) 0050168 0051747 0053587 0055684 0057939

(4) 0050184 0051792 0053671 0055815 0058124

(5) 0136884 0138219 0139763 0141517 0143399

(6) 0136896 0138256 0139832 0141626 0143553

y = 25 (1) 0047622 0050092 0052981 0056280 0059829

(2) 0004765 0050163 0053113 0056487 0060121

(3) 0076570 0078950 0081720 0084878 0088272

(4) 0076593 0079017 0081847 0085076 0088551

(5) 0205774 0207786 0210111 0212752 0215585

(6) 0205793 0207842 0210216 0212917 0215818

y = 50 (1) 0055170 0058009 0061329 0065119 0069196

(2) 0055200 0058091 0061481 0065357 0069533

(3) 0088650 0091385 0094569 0098197 0102096

(4) 0088677 0091463 0094715 0098425 0102417

(5) 0235815 0238128 0240800 0243834 0247089

(6) 0235837 0238192 0240920 0244023 0247356

y = infin (1) 0058565 0061565 0065073 0069077 0073384

(2) 0058597 0061652 0065234 0069329 0073740

(3) 0094061 0096952 0100316 0104149 0108267

(4) 0094090 0097035 0100470 0104390 0108607

(5) 0248472 0250917 0253740 0256945 0260383

(6) 0248495 0250985 0253867 0257145 0260667

25

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0048598 0052524 0056984 0061998 0067342

(2) 0048654 0052673 0057253 0062411 0067916

(3) 0070192 0073979 0078257 0083057 0088167

(4) 0070243 0074119 0078513 0083451 0088714

(5) 0151774 0154981 0158572 0162586 0166852

(6) 0151815 0155096 0158784 0162913 0167308

y = 25 (1) 0076895 0083230 0090413 0098481 0107073

(2) 0076987 0083474 0090854 0099156 0108010

(3) 0110093 0116205 0123095 0130818 0139033

(4) 0110177 0116435 0123515 0131462 0139928

(5) 0230674 0235848 0241632 0248091 0254950

(6) 0230739 0236038 0241980 0248626 0255695

y = 50 (1) 0091496 0099156 0107829 0117563 0127924

(2) 0091610 0099455 0108370 0118387 0129068

(3) 0130130 0137520 0145839 0155156 0165064

(4) 0130233 0137801 0146352 0155942 0166156

(5) 0266599 0272857 0279840 0287632 0295904

(6) 0266680 0273088 0280265 0288286 0296814

y = infin (1) 0099312 0107746 0117285 0127984 0139369

(2) 0099439 0108078 0117883 0128899 0140639

(3) 0140458 0148595 0157745 0167986 0178872

(4) 0140572 0148908 0158314 0168859 0180084

(5) 0282886 0289776 0297457 0306022 0315111

(6) 0282976 0290034 0297929 0306748 0316120

26

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0059096 0072505 0088907 0108122 0129137

(2) 0059223 0072829 0089500 0109052 0130456

(3) 0079757 0092662 0108388 0126778 0146871

(4) 0079875 0092970 0108956 0127669 0148134

(5) 0159788 0170682 0183871 0199249 0216024

(6) 0159881 0170936 0184343 0199990 0217075

y = 25 (1) 0095883 0119741 0149240 0184073 0222390

(2) 0096087 0120262 0150197 0185573 0224520

(3) 0127403 0150359 0178640 0211977 0248613

(4) 0127592 0150855 0179557 0213414 0250653

(5) 0245161 0264533 0288250 0316124 0346709

(6) 0245311 0264942 0289010 0317319 0348407

y = 50 (1) 0116598 0147920 0187100 0233784 0285503

(2) 0116839 0148537 0188231 0235554 0288015

(3) 0153023 0183153 0220713 0265390 0314840

(4) 0153249 0183742 0221798 0267087 0317248

(5) 0285742 0311159 0342651 0380005 0421287

(6) 0285921 0311644 0343551 0381418 0423292

y = infin (1) 0129470 0167217 0215383 0273841 0339734

(2) 0129725 0167865 0216560 0275661 0342282

(3) 0167978 0204275 0250442 0306382 0369383

(4) 0168217 0204895 0251572 0308128 0371826

(5) 0305868 0336470 0375169 0421935 0474525

(6) 0306058 0336980 0376106 0423388 0476560

27

Page 15: uwaterloo.ca · Ruin Analysis of a Threshold Strategy in a Discrete-time Sparre Andersen Model Steve DREKIC and Ana Maria MERA Department of Statistics and Actuarial Science University

Case 2 Suppose minc Z + c1 = Z + c1

For i isin 0 1 Z + c1 minus 1

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+mni nminus1j nminus1j

j=0 m=c1 j=Z

Zminus1 c2 lk (nminus1)minus1 L L L

(k) (k)= αjminusi+cg s + dm αjminusi+mg s 0 0 (313) nminus1j nminus1j

j=0 m=c1 j=Z

For i isin Z + c1 Z + c1 + 1 c minus 1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k) (k)g = g Bjminusi+c + dm g Bjminusi+m + dm g Bjminusi+mni nminus1j nminus1j nminus1j

j=0 m=c1 j=iminusm m=iminusZ+1 j=Z

minc2iminusZ L

(k)= dmg S nminus1iminusm

m=c1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k)+ αjminusi+cg s + dm αjminusi+mg s + dm αjminusi+mg s 0 0 nminus1j nminus1j nminus1j

j=0 m=c1 j=iminusm+1 m=iminusZ+1 j=Z

(314)

For i isin c c + 1 Z + c minus 1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k) (k)g = g + g + gni nminus1j Bjminusi+c dm nminus1j Bjminusi+m dm nminus1j Bjminusi+m

j=iminusc m=c1 j=iminusm m=iminusZ+1 j=Z

minc2iminusZ L

(k) (k)= g S + dmg S nminus1iminusc nminus1iminusm

m=c1

Zminus1 minc2iminusZ lk (nminus1)minus1 c2 lk (nminus1)minus1 L L L L L

(k) (k) (k)+

nminus1j s 0 0 αjminusi+cgnminus1j s + dm αjminusi+mgnminus1j s + dm αjminusi+mg j=iminusc+1 m=c1 j=iminusm+1 m=iminusZ+1 j=Z

(315)

For i isin Z + c Z + c + 1 lk(n) minus 1

c2 lk (nminus1)minus1 L L

(k) (k)g = dm g Bjminusi+mni nminus1j

m=c1 j=iminusm

c2 c2 lk (nminus1)minus1 L L L

(k) (k)= dmg S + dm αjminusi+mgnminus1j s 0 0 (316) nminus1iminusm

m=c1 m=c1 j=iminusm+1

It is not difficult to verify that equations (38) through (316) simplify to produce equations

(32) and (33) of Alfa and Drekic (2007 pp 301-302) when c1 = c2 = c With equations (38) (k)

through (316) providing the recursive means of calculating gn we next turn our attention

15

to a determination of h(k) defined by (28) For n = 1 2 ζlowast we use (22) and (25) to n k

obtain

lk(nminus1)minus1 lk(nminus1)minus1

h(k) nminusj =

L g

(k) nminus1ℓBc+j+ℓ =

L αc+j+ℓg

(k) nminus1ℓs 0 0 (317)

ℓ=0 ℓ=0

For n = ζk lowast + 1 ζk

lowast + 2 we similarly acquire

Zminus1 lk(nminus1)minus1 L L

(k) (k) (k)hnminusj = gnminus1ℓBc+j+ℓ + gnminus1ℓAj+ℓ

ℓ=0 ℓ=Z

Zminus1 c2 lk(nminus1)minus1 L L L

(k) (k)= αc+j+ℓg s + dm αm+j+ℓg s 0 0 (318) nminus1ℓ nminus1ℓ

ℓ=0 m=c1 ℓ=Z

Note that substituting (317) and (318) into (29) immediately yields

Llk(nminus1)minus1 (k) (k) ℓ=0 αc+j+ℓgnminus1ℓs if n = 1 2 ζk

lowast φ (u) = nj LZminus1 (k) Lc2

Llk(nminus1)minus1 (k) ζlowast s + dm αm+j+ℓg n = + 1 ζlowast + 2 ℓ=0 αc+j+ℓgnminus1ℓ m=c1 ℓ=Z nminus1ℓs if k k

(319)

Clearly in the special case when c1 = c2 = c (319) yields equation (35) of Alfa and Drekic

(2007 p 302) Finally we remark that (319) can be substituted into (213) to obtain a

computational formula for φnj(u)

4 Numerical examples and findings

In this section we illustrate the application of our proposed algorithm with several numerical

examples All calculations in this section were performed using Mathematica (Version 7) on

a Dell Precision T7400 workstation with a 3 Ghz CPU and 4 GB of RAM

In our first illustration (labelled Example 1) we consider the ordinary Sparre Andersen

model in which interclaim times have pmf given by

(211)(911)jminus1 if j = 1 2 na minus 1 aj = (41)

(911)naminus1 if j = na

Note that the pmf (41) is that of a truncated geometric distribution with all the probability

mass on na na +1 assigned to the support value na Clearly as na becomes larger the

closer (41) approximates this particular geometric distribution having mean 55 In what Lna

Lminus1 nafollows we fix na = 25 so that aj = 0992 and jaj = 546 Let the individual j=1 j=1

claim amount distribution be given by the pmf

minus4 minus4j minus 1 j

αj = 1 + minus 1 + j isin Z+ (42) 30 30

16

We remark that the pmf (42) represents a discretized version of the Pareto distribution with

mean 10 Also note that mα = infin thereby implying that the random variables |UT | and

UTminus both have infinite support

Table 1 displays values (rounded to 6 decimal places) of the trivariate cumulative distrishy

bution function

n x y L L L

Ψnxy(u) = Pr T le n UTminus le x |UT | le y | U0 = u = ψℓij(u) ℓ=1 i=mincZ+c1 j=1

corresponding to the above discrete-time risk process with u = 10 Z = 50 c = 5 and

varying premium amount distributions according to the following set of scenarios

(1) d2 = 1

(2) d1 = 25 d2 = 15 d3 = 25

G

5)

(3) di = i

(25)i(35)5minusi i = 0 1 5

(4) di = 15 i = 0 1 4

(5) d1 = 34 d5 = 14

(6) d0 = 35 d5 = 25

We remark that while these six premium amount distributions all have mean equal to 2

they possess different shapes and we investigate the impact of their variability on the joint

distribution of T |UT | and UTminus In particular the variances corresponding to the premium

amount distributions of scenarios (1) through (6) are 0 08 12 2 3 and 6 respectively

Based on the results in Table 1 it would appear that greater variability in the premium

amount distribution generally leads to higher ruin probabilities Perhaps somewhat surprising

though the differences are subtle and not particularly striking as one compares the various

ruin probabilities from scenarios (1) through (6) Obviously we observe more of a difference

between scenarios as the values of n x and y increase

As a follow-up illustration (labelled Example 2) we consider a model in which ordinary

interclaim times have a discrete uniform distribution with pmf aj = 110 j = 1 2 10

Clearly the mean of this distribution is 55 which for comparative purposes is identical to

the ordinary interclaim time mean considered in Example 1 In addition we assume that

the pmf of the individual claim amount distribution is once again given by (42) Table 2

displays values (rounded to 6 decimal places) of Ψnxy(10) for several variants of the Sparre

Andersen model (again with Z = 50 and c = 5) having the above ordinary interclaim time

and claim amount distributions namely

17

(1) ordinary model under scenario (2) from Example 1

(2) ordinary model under scenario (5) from Example 1

(3) stationary model under scenario (2) from Example 1

(4) stationary model under scenario (5) from Example 1

(5) delayed model (with W1 = 1 with probability 1) under scenario (2) from Example 1

(6) delayed model (with W1 = 1 with probability 1) under scenario (5) from Example 1

Concerning the results in Table 2 we first compare the values of Ψnxy(10) in entries (2)

and (5) of Table 1 to their corresponding counterparts in entries (1) and (2) of Table 2 In all

cases we observe that ruin is more likely to occur when interclaim times have the truncated

geometric distribution introduced in our first example as opposed to the discrete uniform

distribution of our second example Since mean interclaim times are equal in both examshy

ples this behaviour must be attributable to the greater variability inherent in the truncated

geometric distribution We also note that under the stationary model assumption the pmf

of W1 is easily verified to be rj = (11 minus j)55 j = 1 2 10 which has mean 4 lt 55

As a result the values of Ψnxy(10) for stationary models (3) and (4) are always larger than

the corresponding values for ordinary models (1) and (2) Finally in the absence of any

knowledge regarding which distribution is in effect for the first interclaim time W1 models

(5) and (6) can be regarded as ldquoworst-case scenariordquo models as they serve to provide upper

bounds on the values of Ψnxy(10)

In conclusion we state that the values in Tables 1 and 2 were generated by first impleshy

menting the recursive procedures (38) through (316) and then summing the appropriate

bivariatetrivariate probabilities computed via (213) (214) or (215) In situations when

x = y = infin however we computed the resulting finite-time ruin probabilities using an altershy

native (albeit standard) method (eg see Cossette et al (2006 Section 3)) Specifically if

we define the survival probability σ(n u) = Pr T gt n | U0 = u = 1minusΨninfininfin(u) as well as

the auxiliary function

0 if w ge Z ηwt =

minlfloorZminuswc minus1 rfloor + 1 t if w lt Z

then by conditioning on the time and amount of the first claim (and recalling that the delayed

process reverts to the ordinary process following this claim) we obtain the expression

minnnr (kminuszk)c2 u+czk+ℓ L L L

σ(u n) = Rn + rk dℓkminuszk αjγ(u + czk + ℓ minus j n minus k)

k=1 ℓ=(kminuszk)c1 j=1

18

where the function γ(w m) for w m isin N is computed recursively via

minmna (kminusηwk)c2 u+cηwk+ℓ L L L

γ(w m) = Am + ak dℓkminusηwk αjγ(u + cηwk + ℓ minus j m minus k)

k=1 j=1 ℓ=(kminusηwk)c1

with starting values γ(w 0) = 1 forall w isin N

Acknowledgements

The authors would like to thank Dr David Landriault for his valuable suggestions which

helped to improve an earlier version of this work This research was supported by the Natural

Sciences and Engineering Research Council of Canada

References

Alfa AS 2004 Markov chain representations of discrete distributions applied to queueing

models Computers amp Operations Research 31 2365-2385

Alfa AS Drekic S 2007 Algorithmic analysis of the Sparre Andersen model in discrete

time ASTIN Bulletin 37 293-317

Bao Z 2007 A note on the compound binomial model with randomized dividend strategy

Applied Mathematics amp Computation 194 276-286

Cossette H Landriault D Marceau E 2006 Ruin probabilities in the discrete time

renewal risk model Insurance Mathematics amp Economics 38 309-323

Karlin S Taylor HM 1975 A First Course in Stochastic Processes 2nd edition Acashy

demic Press New York

Kim B Kim H-S Kim J 2008 A risk model with paying dividends and random

environment Insurance Mathematics amp Economics 42 717-726

Landriault D 2008 Randomized dividends in the compound binomial model with a general

premium rate Scandinavian Actuarial Journal 1-15

Tan J Yang X 2006 The compound binomial model with randomized decisions on

paying dividends Insurance Mathematics amp Economics 39 1-18

19

Table 1 ndash Values of Ψnxy(10) corresponding to Example 1

(a) x = 10

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0005357 0006188 0007163 0008278 0009481

(2) 0005359 0006195 0007175 0008296 0009507

(3) 0005361 0006198 0007181 0008305 0009519

(4) 0005363 0006205 0007193 0008323 0009544

(5) 0005366 0006212 0007208 0008346 0009576

(6) 0005374 0006236 0007249 0008410 0009664

y = 25 (1) 0007631 0008815 0010202 0011789 0013503

(2) 0007634 0008824 0010220 0011816 0013539

(3) 0007636 0008828 0010228 0011829 0013557

(4) 0007639 0008838 0010245 0011854 0013593

(5) 0007643 0008849 0010266 0011887 0013638

(6) 0007656 0008882 0010325 0011978 0013764

y = 50 (1) 0008477 0009792 0011333 0013095 0014998

(2) 0008481 0009802 0011352 0013125 0015039

(3) 0008483 0009807 0011361 0013139 0015059

(4) 0008486 0009817 0011380 0013168 0015098

(5) 0008491 0009830 0011403 0013204 0015149

(6) 0008504 0009866 0011470 0013305 0015288

y = infin (1) 0008777 0010138 0011733 0013558 0015528

(2) 0008781 0010149 0011753 0013588 0015570

(3) 0008783 0010154 0011763 0013603 0015591

(4) 0008787 0010164 0011782 0013633 0015632

(5) 0008791 0010177 0011807 0013671 0015684

(6) 0008805 0010215 0011875 0013775 0015828

20

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0058559 0061875 0065704 0070048 0074714

(2) 0058569 0061903 0065756 0070128 0074825

(3) 0058574 0061917 0065782 0070167 0074880

(4) 0058584 0061945 0065833 0070245 0074987

(5) 0058596 0061979 0065896 0070344 0075124

(6) 0058632 0062079 0066078 0070621 0075506

y = 25 (1) 0089002 0093986 0099739 0106262 0113270

(2) 0089017 0094028 0099817 0106382 0113437

(3) 0089025 0094049 0099856 0106442 0113519

(4) 0089040 0094091 0099932 0106560 0113681

(5) 0089057 0094142 0100028 0106708 0113887

(6) 0089113 0094294 0100302 0107127 0114463

y = 50 (1) 0102786 0108506 0115106 0122590 0130629

(2) 0102804 0108555 0115196 0122728 0130820

(3) 0102812 0108579 0115241 0122797 0130914

(4) 0102830 0108627 0115329 0122932 0131101

(5) 0102850 0108686 0115439 0123103 0131337

(6) 0102914 0108860 0115754 0123584 0132000

y = infin (1) 0108895 0114933 0121901 0129799 0138283

(2) 0108914 0114985 0121996 0129945 0138485

(3) 0108923 0115010 0122043 0130018 0138585

(4) 0108941 0115061 0122136 0130161 0138782

(5) 0108962 0115123 0122252 0130341 0139032

(6) 0109030 0115307 0122585 0130849 0139732

21

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0079856 0086792 009464 0103440 0112831

(2) 0079885 0086869 0094778 0103650 0113118

(3) 0079953 0086978 0094936 0103864 0113393

(4) 0080081 0087188 0095243 0104282 0113929

(5) 0079962 0087075 0095149 0104213 0113890

(6) 0080805 0088366 0096951 0106588 0116875

y = 25 (1) 0124591 0135680 0148207 0162245 0177217

(2) 0124639 0135806 0148433 0162586 0177683

(3) 0124755 0135993 0148703 0162950 0178148

(4) 0124978 0136354 0149226 0163658 0179054

(5) 0124763 0136141 0149035 0163500 0178935

(6) 0126238 0138381 0152148 0167588 0184060

y = 50 (1) 0146757 0160079 0175111 0191945 0209894

(2) 0146816 0160232 0175385 0192360 0210460

(3) 0146965 0160469 0175725 0192817 0211043

(4) 0147249 0160925 0176384 0193706 0212178

(5) 0146968 0160639 0176117 0193470 0211980

(6) 0148860 0163500 0180077 0198658 0218474

y = infin (1) 0158013 0172606 0189059 0207476 0227106

(2) 0158078 0172776 0189362 0207934 0227732

(3) 0158250 0173047 0189751 0208455 0228395

(4) 0158579 0173571 0190503 0209467 0229685

(5) 0158247 0173226 0190172 0209161 0229410

(6) 0160443 0176532 0194735 0215127 0236869

22

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0088667 0104403 0123329 0145277 0169184

(2) 0088711 0104518 0123538 0145598 0169627

(3) 0088733 0104574 0123641 0145756 0169846

(4) 0088777 0104687 0123844 0146066 0170274

(5) 0088829 0104828 0124105 0146470 0170832

(6) 0088988 0105229 0124822 0147560 0172331

y = 25 (1) 0140528 0167839 0201007 0239726 0282102

(2) 0140598 0168021 0201338 0240234 0282804

(3) 0140632 0168111 0201502 0240486 0283151

(4) 0140701 0168289 0201822 0240976 0283827

(5) 0140785 0168515 0202241 0241622 0284718

(6) 0141036 0169148 0203370 0243335 0287071

y = 50 (1) 0167826 0203005 0246173 0296954 0352855

(2) 0167908 0203218 0246561 0297548 0353672

(3) 0167949 0203324 0246753 0297841 0354075

(4) 0168030 0203532 0247128 0298412 0354859

(5) 0168129 0203800 0247622 0299174 0355907

(6) 0168423 0204539 0248936 0301159 0358620

y = infin (1) 0183325 0224926 0276881 0338957 0408257

(2) 0183412 0225150 0277283 0339564 0409079

(3) 0183455 0225260 0277481 0339864 0409485

(4) 0183540 0225478 0277869 0340446 0410270

(5) 0183645 0225760 0278386 0341234 0411339

(6) 0183954 0226529 0279737 0343244 0414037

23

Table 2 ndash Values of Ψnxy(10) corresponding to Example 2

(a) x = 10

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0001901 0002276 0002724 0003240 0003799

(2) 0001904 0002286 0002742 0003269 0003840

(3) 0002651 0003013 0003442 0003936 0004471

(4) 0002655 0003022 0003460 0003964 0004510

(5) 0005792 0006098 0006458 0006871 0007317

(6) 0005795 0006106 0006473 0006894 0007350

y = 25 (1) 0002708 0003243 0003880 0004615 0005411

(2) 0002713 0003257 0003906 0004656 0005469

(3) 0003778 0004293 0004904 0005608 0006369

(4) 0003782 0004306 0004929 0005647 0006424

(5) 0008255 0008690 0009203 0009792 0010427

(6) 0008259 0008702 0009224 0009824 0010473

y = 50 (1) 0003008 0003603 0004310 0005127 0006011

(2) 0003014 0003618 0004340 0005173 0006076

(3) 0004197 0004769 0005448 0006230 0007075

(4) 0004202 0004784 0005476 0006273 0007137

(5) 0009173 0009656 0010226 0010879 0011585

(6) 0009177 0009669 0010249 0010916 0011637

y = infin (1) 0003115 0003730 0004463 0005308 0006224

(2) 0003121 0003746 0004493 0005356 0006291

(3) 0004346 0004938 0005641 0006450 0007325

(4) 0004351 0004954 0005670 0006496 0007389

(5) 0009499 0009999 0010589 0011265 0011996

(6) 0009503 0010012 0010613 0011303 0012050

24

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0031176 0032815 0034733 0036924 0039282

(2) 0031193 0032862 0034821 0037061 0039475

(3) 0050168 0051747 0053587 0055684 0057939

(4) 0050184 0051792 0053671 0055815 0058124

(5) 0136884 0138219 0139763 0141517 0143399

(6) 0136896 0138256 0139832 0141626 0143553

y = 25 (1) 0047622 0050092 0052981 0056280 0059829

(2) 0004765 0050163 0053113 0056487 0060121

(3) 0076570 0078950 0081720 0084878 0088272

(4) 0076593 0079017 0081847 0085076 0088551

(5) 0205774 0207786 0210111 0212752 0215585

(6) 0205793 0207842 0210216 0212917 0215818

y = 50 (1) 0055170 0058009 0061329 0065119 0069196

(2) 0055200 0058091 0061481 0065357 0069533

(3) 0088650 0091385 0094569 0098197 0102096

(4) 0088677 0091463 0094715 0098425 0102417

(5) 0235815 0238128 0240800 0243834 0247089

(6) 0235837 0238192 0240920 0244023 0247356

y = infin (1) 0058565 0061565 0065073 0069077 0073384

(2) 0058597 0061652 0065234 0069329 0073740

(3) 0094061 0096952 0100316 0104149 0108267

(4) 0094090 0097035 0100470 0104390 0108607

(5) 0248472 0250917 0253740 0256945 0260383

(6) 0248495 0250985 0253867 0257145 0260667

25

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0048598 0052524 0056984 0061998 0067342

(2) 0048654 0052673 0057253 0062411 0067916

(3) 0070192 0073979 0078257 0083057 0088167

(4) 0070243 0074119 0078513 0083451 0088714

(5) 0151774 0154981 0158572 0162586 0166852

(6) 0151815 0155096 0158784 0162913 0167308

y = 25 (1) 0076895 0083230 0090413 0098481 0107073

(2) 0076987 0083474 0090854 0099156 0108010

(3) 0110093 0116205 0123095 0130818 0139033

(4) 0110177 0116435 0123515 0131462 0139928

(5) 0230674 0235848 0241632 0248091 0254950

(6) 0230739 0236038 0241980 0248626 0255695

y = 50 (1) 0091496 0099156 0107829 0117563 0127924

(2) 0091610 0099455 0108370 0118387 0129068

(3) 0130130 0137520 0145839 0155156 0165064

(4) 0130233 0137801 0146352 0155942 0166156

(5) 0266599 0272857 0279840 0287632 0295904

(6) 0266680 0273088 0280265 0288286 0296814

y = infin (1) 0099312 0107746 0117285 0127984 0139369

(2) 0099439 0108078 0117883 0128899 0140639

(3) 0140458 0148595 0157745 0167986 0178872

(4) 0140572 0148908 0158314 0168859 0180084

(5) 0282886 0289776 0297457 0306022 0315111

(6) 0282976 0290034 0297929 0306748 0316120

26

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0059096 0072505 0088907 0108122 0129137

(2) 0059223 0072829 0089500 0109052 0130456

(3) 0079757 0092662 0108388 0126778 0146871

(4) 0079875 0092970 0108956 0127669 0148134

(5) 0159788 0170682 0183871 0199249 0216024

(6) 0159881 0170936 0184343 0199990 0217075

y = 25 (1) 0095883 0119741 0149240 0184073 0222390

(2) 0096087 0120262 0150197 0185573 0224520

(3) 0127403 0150359 0178640 0211977 0248613

(4) 0127592 0150855 0179557 0213414 0250653

(5) 0245161 0264533 0288250 0316124 0346709

(6) 0245311 0264942 0289010 0317319 0348407

y = 50 (1) 0116598 0147920 0187100 0233784 0285503

(2) 0116839 0148537 0188231 0235554 0288015

(3) 0153023 0183153 0220713 0265390 0314840

(4) 0153249 0183742 0221798 0267087 0317248

(5) 0285742 0311159 0342651 0380005 0421287

(6) 0285921 0311644 0343551 0381418 0423292

y = infin (1) 0129470 0167217 0215383 0273841 0339734

(2) 0129725 0167865 0216560 0275661 0342282

(3) 0167978 0204275 0250442 0306382 0369383

(4) 0168217 0204895 0251572 0308128 0371826

(5) 0305868 0336470 0375169 0421935 0474525

(6) 0306058 0336980 0376106 0423388 0476560

27

Page 16: uwaterloo.ca · Ruin Analysis of a Threshold Strategy in a Discrete-time Sparre Andersen Model Steve DREKIC and Ana Maria MERA Department of Statistics and Actuarial Science University

to a determination of h(k) defined by (28) For n = 1 2 ζlowast we use (22) and (25) to n k

obtain

lk(nminus1)minus1 lk(nminus1)minus1

h(k) nminusj =

L g

(k) nminus1ℓBc+j+ℓ =

L αc+j+ℓg

(k) nminus1ℓs 0 0 (317)

ℓ=0 ℓ=0

For n = ζk lowast + 1 ζk

lowast + 2 we similarly acquire

Zminus1 lk(nminus1)minus1 L L

(k) (k) (k)hnminusj = gnminus1ℓBc+j+ℓ + gnminus1ℓAj+ℓ

ℓ=0 ℓ=Z

Zminus1 c2 lk(nminus1)minus1 L L L

(k) (k)= αc+j+ℓg s + dm αm+j+ℓg s 0 0 (318) nminus1ℓ nminus1ℓ

ℓ=0 m=c1 ℓ=Z

Note that substituting (317) and (318) into (29) immediately yields

Llk(nminus1)minus1 (k) (k) ℓ=0 αc+j+ℓgnminus1ℓs if n = 1 2 ζk

lowast φ (u) = nj LZminus1 (k) Lc2

Llk(nminus1)minus1 (k) ζlowast s + dm αm+j+ℓg n = + 1 ζlowast + 2 ℓ=0 αc+j+ℓgnminus1ℓ m=c1 ℓ=Z nminus1ℓs if k k

(319)

Clearly in the special case when c1 = c2 = c (319) yields equation (35) of Alfa and Drekic

(2007 p 302) Finally we remark that (319) can be substituted into (213) to obtain a

computational formula for φnj(u)

4 Numerical examples and findings

In this section we illustrate the application of our proposed algorithm with several numerical

examples All calculations in this section were performed using Mathematica (Version 7) on

a Dell Precision T7400 workstation with a 3 Ghz CPU and 4 GB of RAM

In our first illustration (labelled Example 1) we consider the ordinary Sparre Andersen

model in which interclaim times have pmf given by

(211)(911)jminus1 if j = 1 2 na minus 1 aj = (41)

(911)naminus1 if j = na

Note that the pmf (41) is that of a truncated geometric distribution with all the probability

mass on na na +1 assigned to the support value na Clearly as na becomes larger the

closer (41) approximates this particular geometric distribution having mean 55 In what Lna

Lminus1 nafollows we fix na = 25 so that aj = 0992 and jaj = 546 Let the individual j=1 j=1

claim amount distribution be given by the pmf

minus4 minus4j minus 1 j

αj = 1 + minus 1 + j isin Z+ (42) 30 30

16

We remark that the pmf (42) represents a discretized version of the Pareto distribution with

mean 10 Also note that mα = infin thereby implying that the random variables |UT | and

UTminus both have infinite support

Table 1 displays values (rounded to 6 decimal places) of the trivariate cumulative distrishy

bution function

n x y L L L

Ψnxy(u) = Pr T le n UTminus le x |UT | le y | U0 = u = ψℓij(u) ℓ=1 i=mincZ+c1 j=1

corresponding to the above discrete-time risk process with u = 10 Z = 50 c = 5 and

varying premium amount distributions according to the following set of scenarios

(1) d2 = 1

(2) d1 = 25 d2 = 15 d3 = 25

G

5)

(3) di = i

(25)i(35)5minusi i = 0 1 5

(4) di = 15 i = 0 1 4

(5) d1 = 34 d5 = 14

(6) d0 = 35 d5 = 25

We remark that while these six premium amount distributions all have mean equal to 2

they possess different shapes and we investigate the impact of their variability on the joint

distribution of T |UT | and UTminus In particular the variances corresponding to the premium

amount distributions of scenarios (1) through (6) are 0 08 12 2 3 and 6 respectively

Based on the results in Table 1 it would appear that greater variability in the premium

amount distribution generally leads to higher ruin probabilities Perhaps somewhat surprising

though the differences are subtle and not particularly striking as one compares the various

ruin probabilities from scenarios (1) through (6) Obviously we observe more of a difference

between scenarios as the values of n x and y increase

As a follow-up illustration (labelled Example 2) we consider a model in which ordinary

interclaim times have a discrete uniform distribution with pmf aj = 110 j = 1 2 10

Clearly the mean of this distribution is 55 which for comparative purposes is identical to

the ordinary interclaim time mean considered in Example 1 In addition we assume that

the pmf of the individual claim amount distribution is once again given by (42) Table 2

displays values (rounded to 6 decimal places) of Ψnxy(10) for several variants of the Sparre

Andersen model (again with Z = 50 and c = 5) having the above ordinary interclaim time

and claim amount distributions namely

17

(1) ordinary model under scenario (2) from Example 1

(2) ordinary model under scenario (5) from Example 1

(3) stationary model under scenario (2) from Example 1

(4) stationary model under scenario (5) from Example 1

(5) delayed model (with W1 = 1 with probability 1) under scenario (2) from Example 1

(6) delayed model (with W1 = 1 with probability 1) under scenario (5) from Example 1

Concerning the results in Table 2 we first compare the values of Ψnxy(10) in entries (2)

and (5) of Table 1 to their corresponding counterparts in entries (1) and (2) of Table 2 In all

cases we observe that ruin is more likely to occur when interclaim times have the truncated

geometric distribution introduced in our first example as opposed to the discrete uniform

distribution of our second example Since mean interclaim times are equal in both examshy

ples this behaviour must be attributable to the greater variability inherent in the truncated

geometric distribution We also note that under the stationary model assumption the pmf

of W1 is easily verified to be rj = (11 minus j)55 j = 1 2 10 which has mean 4 lt 55

As a result the values of Ψnxy(10) for stationary models (3) and (4) are always larger than

the corresponding values for ordinary models (1) and (2) Finally in the absence of any

knowledge regarding which distribution is in effect for the first interclaim time W1 models

(5) and (6) can be regarded as ldquoworst-case scenariordquo models as they serve to provide upper

bounds on the values of Ψnxy(10)

In conclusion we state that the values in Tables 1 and 2 were generated by first impleshy

menting the recursive procedures (38) through (316) and then summing the appropriate

bivariatetrivariate probabilities computed via (213) (214) or (215) In situations when

x = y = infin however we computed the resulting finite-time ruin probabilities using an altershy

native (albeit standard) method (eg see Cossette et al (2006 Section 3)) Specifically if

we define the survival probability σ(n u) = Pr T gt n | U0 = u = 1minusΨninfininfin(u) as well as

the auxiliary function

0 if w ge Z ηwt =

minlfloorZminuswc minus1 rfloor + 1 t if w lt Z

then by conditioning on the time and amount of the first claim (and recalling that the delayed

process reverts to the ordinary process following this claim) we obtain the expression

minnnr (kminuszk)c2 u+czk+ℓ L L L

σ(u n) = Rn + rk dℓkminuszk αjγ(u + czk + ℓ minus j n minus k)

k=1 ℓ=(kminuszk)c1 j=1

18

where the function γ(w m) for w m isin N is computed recursively via

minmna (kminusηwk)c2 u+cηwk+ℓ L L L

γ(w m) = Am + ak dℓkminusηwk αjγ(u + cηwk + ℓ minus j m minus k)

k=1 j=1 ℓ=(kminusηwk)c1

with starting values γ(w 0) = 1 forall w isin N

Acknowledgements

The authors would like to thank Dr David Landriault for his valuable suggestions which

helped to improve an earlier version of this work This research was supported by the Natural

Sciences and Engineering Research Council of Canada

References

Alfa AS 2004 Markov chain representations of discrete distributions applied to queueing

models Computers amp Operations Research 31 2365-2385

Alfa AS Drekic S 2007 Algorithmic analysis of the Sparre Andersen model in discrete

time ASTIN Bulletin 37 293-317

Bao Z 2007 A note on the compound binomial model with randomized dividend strategy

Applied Mathematics amp Computation 194 276-286

Cossette H Landriault D Marceau E 2006 Ruin probabilities in the discrete time

renewal risk model Insurance Mathematics amp Economics 38 309-323

Karlin S Taylor HM 1975 A First Course in Stochastic Processes 2nd edition Acashy

demic Press New York

Kim B Kim H-S Kim J 2008 A risk model with paying dividends and random

environment Insurance Mathematics amp Economics 42 717-726

Landriault D 2008 Randomized dividends in the compound binomial model with a general

premium rate Scandinavian Actuarial Journal 1-15

Tan J Yang X 2006 The compound binomial model with randomized decisions on

paying dividends Insurance Mathematics amp Economics 39 1-18

19

Table 1 ndash Values of Ψnxy(10) corresponding to Example 1

(a) x = 10

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0005357 0006188 0007163 0008278 0009481

(2) 0005359 0006195 0007175 0008296 0009507

(3) 0005361 0006198 0007181 0008305 0009519

(4) 0005363 0006205 0007193 0008323 0009544

(5) 0005366 0006212 0007208 0008346 0009576

(6) 0005374 0006236 0007249 0008410 0009664

y = 25 (1) 0007631 0008815 0010202 0011789 0013503

(2) 0007634 0008824 0010220 0011816 0013539

(3) 0007636 0008828 0010228 0011829 0013557

(4) 0007639 0008838 0010245 0011854 0013593

(5) 0007643 0008849 0010266 0011887 0013638

(6) 0007656 0008882 0010325 0011978 0013764

y = 50 (1) 0008477 0009792 0011333 0013095 0014998

(2) 0008481 0009802 0011352 0013125 0015039

(3) 0008483 0009807 0011361 0013139 0015059

(4) 0008486 0009817 0011380 0013168 0015098

(5) 0008491 0009830 0011403 0013204 0015149

(6) 0008504 0009866 0011470 0013305 0015288

y = infin (1) 0008777 0010138 0011733 0013558 0015528

(2) 0008781 0010149 0011753 0013588 0015570

(3) 0008783 0010154 0011763 0013603 0015591

(4) 0008787 0010164 0011782 0013633 0015632

(5) 0008791 0010177 0011807 0013671 0015684

(6) 0008805 0010215 0011875 0013775 0015828

20

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0058559 0061875 0065704 0070048 0074714

(2) 0058569 0061903 0065756 0070128 0074825

(3) 0058574 0061917 0065782 0070167 0074880

(4) 0058584 0061945 0065833 0070245 0074987

(5) 0058596 0061979 0065896 0070344 0075124

(6) 0058632 0062079 0066078 0070621 0075506

y = 25 (1) 0089002 0093986 0099739 0106262 0113270

(2) 0089017 0094028 0099817 0106382 0113437

(3) 0089025 0094049 0099856 0106442 0113519

(4) 0089040 0094091 0099932 0106560 0113681

(5) 0089057 0094142 0100028 0106708 0113887

(6) 0089113 0094294 0100302 0107127 0114463

y = 50 (1) 0102786 0108506 0115106 0122590 0130629

(2) 0102804 0108555 0115196 0122728 0130820

(3) 0102812 0108579 0115241 0122797 0130914

(4) 0102830 0108627 0115329 0122932 0131101

(5) 0102850 0108686 0115439 0123103 0131337

(6) 0102914 0108860 0115754 0123584 0132000

y = infin (1) 0108895 0114933 0121901 0129799 0138283

(2) 0108914 0114985 0121996 0129945 0138485

(3) 0108923 0115010 0122043 0130018 0138585

(4) 0108941 0115061 0122136 0130161 0138782

(5) 0108962 0115123 0122252 0130341 0139032

(6) 0109030 0115307 0122585 0130849 0139732

21

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0079856 0086792 009464 0103440 0112831

(2) 0079885 0086869 0094778 0103650 0113118

(3) 0079953 0086978 0094936 0103864 0113393

(4) 0080081 0087188 0095243 0104282 0113929

(5) 0079962 0087075 0095149 0104213 0113890

(6) 0080805 0088366 0096951 0106588 0116875

y = 25 (1) 0124591 0135680 0148207 0162245 0177217

(2) 0124639 0135806 0148433 0162586 0177683

(3) 0124755 0135993 0148703 0162950 0178148

(4) 0124978 0136354 0149226 0163658 0179054

(5) 0124763 0136141 0149035 0163500 0178935

(6) 0126238 0138381 0152148 0167588 0184060

y = 50 (1) 0146757 0160079 0175111 0191945 0209894

(2) 0146816 0160232 0175385 0192360 0210460

(3) 0146965 0160469 0175725 0192817 0211043

(4) 0147249 0160925 0176384 0193706 0212178

(5) 0146968 0160639 0176117 0193470 0211980

(6) 0148860 0163500 0180077 0198658 0218474

y = infin (1) 0158013 0172606 0189059 0207476 0227106

(2) 0158078 0172776 0189362 0207934 0227732

(3) 0158250 0173047 0189751 0208455 0228395

(4) 0158579 0173571 0190503 0209467 0229685

(5) 0158247 0173226 0190172 0209161 0229410

(6) 0160443 0176532 0194735 0215127 0236869

22

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0088667 0104403 0123329 0145277 0169184

(2) 0088711 0104518 0123538 0145598 0169627

(3) 0088733 0104574 0123641 0145756 0169846

(4) 0088777 0104687 0123844 0146066 0170274

(5) 0088829 0104828 0124105 0146470 0170832

(6) 0088988 0105229 0124822 0147560 0172331

y = 25 (1) 0140528 0167839 0201007 0239726 0282102

(2) 0140598 0168021 0201338 0240234 0282804

(3) 0140632 0168111 0201502 0240486 0283151

(4) 0140701 0168289 0201822 0240976 0283827

(5) 0140785 0168515 0202241 0241622 0284718

(6) 0141036 0169148 0203370 0243335 0287071

y = 50 (1) 0167826 0203005 0246173 0296954 0352855

(2) 0167908 0203218 0246561 0297548 0353672

(3) 0167949 0203324 0246753 0297841 0354075

(4) 0168030 0203532 0247128 0298412 0354859

(5) 0168129 0203800 0247622 0299174 0355907

(6) 0168423 0204539 0248936 0301159 0358620

y = infin (1) 0183325 0224926 0276881 0338957 0408257

(2) 0183412 0225150 0277283 0339564 0409079

(3) 0183455 0225260 0277481 0339864 0409485

(4) 0183540 0225478 0277869 0340446 0410270

(5) 0183645 0225760 0278386 0341234 0411339

(6) 0183954 0226529 0279737 0343244 0414037

23

Table 2 ndash Values of Ψnxy(10) corresponding to Example 2

(a) x = 10

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0001901 0002276 0002724 0003240 0003799

(2) 0001904 0002286 0002742 0003269 0003840

(3) 0002651 0003013 0003442 0003936 0004471

(4) 0002655 0003022 0003460 0003964 0004510

(5) 0005792 0006098 0006458 0006871 0007317

(6) 0005795 0006106 0006473 0006894 0007350

y = 25 (1) 0002708 0003243 0003880 0004615 0005411

(2) 0002713 0003257 0003906 0004656 0005469

(3) 0003778 0004293 0004904 0005608 0006369

(4) 0003782 0004306 0004929 0005647 0006424

(5) 0008255 0008690 0009203 0009792 0010427

(6) 0008259 0008702 0009224 0009824 0010473

y = 50 (1) 0003008 0003603 0004310 0005127 0006011

(2) 0003014 0003618 0004340 0005173 0006076

(3) 0004197 0004769 0005448 0006230 0007075

(4) 0004202 0004784 0005476 0006273 0007137

(5) 0009173 0009656 0010226 0010879 0011585

(6) 0009177 0009669 0010249 0010916 0011637

y = infin (1) 0003115 0003730 0004463 0005308 0006224

(2) 0003121 0003746 0004493 0005356 0006291

(3) 0004346 0004938 0005641 0006450 0007325

(4) 0004351 0004954 0005670 0006496 0007389

(5) 0009499 0009999 0010589 0011265 0011996

(6) 0009503 0010012 0010613 0011303 0012050

24

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0031176 0032815 0034733 0036924 0039282

(2) 0031193 0032862 0034821 0037061 0039475

(3) 0050168 0051747 0053587 0055684 0057939

(4) 0050184 0051792 0053671 0055815 0058124

(5) 0136884 0138219 0139763 0141517 0143399

(6) 0136896 0138256 0139832 0141626 0143553

y = 25 (1) 0047622 0050092 0052981 0056280 0059829

(2) 0004765 0050163 0053113 0056487 0060121

(3) 0076570 0078950 0081720 0084878 0088272

(4) 0076593 0079017 0081847 0085076 0088551

(5) 0205774 0207786 0210111 0212752 0215585

(6) 0205793 0207842 0210216 0212917 0215818

y = 50 (1) 0055170 0058009 0061329 0065119 0069196

(2) 0055200 0058091 0061481 0065357 0069533

(3) 0088650 0091385 0094569 0098197 0102096

(4) 0088677 0091463 0094715 0098425 0102417

(5) 0235815 0238128 0240800 0243834 0247089

(6) 0235837 0238192 0240920 0244023 0247356

y = infin (1) 0058565 0061565 0065073 0069077 0073384

(2) 0058597 0061652 0065234 0069329 0073740

(3) 0094061 0096952 0100316 0104149 0108267

(4) 0094090 0097035 0100470 0104390 0108607

(5) 0248472 0250917 0253740 0256945 0260383

(6) 0248495 0250985 0253867 0257145 0260667

25

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0048598 0052524 0056984 0061998 0067342

(2) 0048654 0052673 0057253 0062411 0067916

(3) 0070192 0073979 0078257 0083057 0088167

(4) 0070243 0074119 0078513 0083451 0088714

(5) 0151774 0154981 0158572 0162586 0166852

(6) 0151815 0155096 0158784 0162913 0167308

y = 25 (1) 0076895 0083230 0090413 0098481 0107073

(2) 0076987 0083474 0090854 0099156 0108010

(3) 0110093 0116205 0123095 0130818 0139033

(4) 0110177 0116435 0123515 0131462 0139928

(5) 0230674 0235848 0241632 0248091 0254950

(6) 0230739 0236038 0241980 0248626 0255695

y = 50 (1) 0091496 0099156 0107829 0117563 0127924

(2) 0091610 0099455 0108370 0118387 0129068

(3) 0130130 0137520 0145839 0155156 0165064

(4) 0130233 0137801 0146352 0155942 0166156

(5) 0266599 0272857 0279840 0287632 0295904

(6) 0266680 0273088 0280265 0288286 0296814

y = infin (1) 0099312 0107746 0117285 0127984 0139369

(2) 0099439 0108078 0117883 0128899 0140639

(3) 0140458 0148595 0157745 0167986 0178872

(4) 0140572 0148908 0158314 0168859 0180084

(5) 0282886 0289776 0297457 0306022 0315111

(6) 0282976 0290034 0297929 0306748 0316120

26

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0059096 0072505 0088907 0108122 0129137

(2) 0059223 0072829 0089500 0109052 0130456

(3) 0079757 0092662 0108388 0126778 0146871

(4) 0079875 0092970 0108956 0127669 0148134

(5) 0159788 0170682 0183871 0199249 0216024

(6) 0159881 0170936 0184343 0199990 0217075

y = 25 (1) 0095883 0119741 0149240 0184073 0222390

(2) 0096087 0120262 0150197 0185573 0224520

(3) 0127403 0150359 0178640 0211977 0248613

(4) 0127592 0150855 0179557 0213414 0250653

(5) 0245161 0264533 0288250 0316124 0346709

(6) 0245311 0264942 0289010 0317319 0348407

y = 50 (1) 0116598 0147920 0187100 0233784 0285503

(2) 0116839 0148537 0188231 0235554 0288015

(3) 0153023 0183153 0220713 0265390 0314840

(4) 0153249 0183742 0221798 0267087 0317248

(5) 0285742 0311159 0342651 0380005 0421287

(6) 0285921 0311644 0343551 0381418 0423292

y = infin (1) 0129470 0167217 0215383 0273841 0339734

(2) 0129725 0167865 0216560 0275661 0342282

(3) 0167978 0204275 0250442 0306382 0369383

(4) 0168217 0204895 0251572 0308128 0371826

(5) 0305868 0336470 0375169 0421935 0474525

(6) 0306058 0336980 0376106 0423388 0476560

27

Page 17: uwaterloo.ca · Ruin Analysis of a Threshold Strategy in a Discrete-time Sparre Andersen Model Steve DREKIC and Ana Maria MERA Department of Statistics and Actuarial Science University

We remark that the pmf (42) represents a discretized version of the Pareto distribution with

mean 10 Also note that mα = infin thereby implying that the random variables |UT | and

UTminus both have infinite support

Table 1 displays values (rounded to 6 decimal places) of the trivariate cumulative distrishy

bution function

n x y L L L

Ψnxy(u) = Pr T le n UTminus le x |UT | le y | U0 = u = ψℓij(u) ℓ=1 i=mincZ+c1 j=1

corresponding to the above discrete-time risk process with u = 10 Z = 50 c = 5 and

varying premium amount distributions according to the following set of scenarios

(1) d2 = 1

(2) d1 = 25 d2 = 15 d3 = 25

G

5)

(3) di = i

(25)i(35)5minusi i = 0 1 5

(4) di = 15 i = 0 1 4

(5) d1 = 34 d5 = 14

(6) d0 = 35 d5 = 25

We remark that while these six premium amount distributions all have mean equal to 2

they possess different shapes and we investigate the impact of their variability on the joint

distribution of T |UT | and UTminus In particular the variances corresponding to the premium

amount distributions of scenarios (1) through (6) are 0 08 12 2 3 and 6 respectively

Based on the results in Table 1 it would appear that greater variability in the premium

amount distribution generally leads to higher ruin probabilities Perhaps somewhat surprising

though the differences are subtle and not particularly striking as one compares the various

ruin probabilities from scenarios (1) through (6) Obviously we observe more of a difference

between scenarios as the values of n x and y increase

As a follow-up illustration (labelled Example 2) we consider a model in which ordinary

interclaim times have a discrete uniform distribution with pmf aj = 110 j = 1 2 10

Clearly the mean of this distribution is 55 which for comparative purposes is identical to

the ordinary interclaim time mean considered in Example 1 In addition we assume that

the pmf of the individual claim amount distribution is once again given by (42) Table 2

displays values (rounded to 6 decimal places) of Ψnxy(10) for several variants of the Sparre

Andersen model (again with Z = 50 and c = 5) having the above ordinary interclaim time

and claim amount distributions namely

17

(1) ordinary model under scenario (2) from Example 1

(2) ordinary model under scenario (5) from Example 1

(3) stationary model under scenario (2) from Example 1

(4) stationary model under scenario (5) from Example 1

(5) delayed model (with W1 = 1 with probability 1) under scenario (2) from Example 1

(6) delayed model (with W1 = 1 with probability 1) under scenario (5) from Example 1

Concerning the results in Table 2 we first compare the values of Ψnxy(10) in entries (2)

and (5) of Table 1 to their corresponding counterparts in entries (1) and (2) of Table 2 In all

cases we observe that ruin is more likely to occur when interclaim times have the truncated

geometric distribution introduced in our first example as opposed to the discrete uniform

distribution of our second example Since mean interclaim times are equal in both examshy

ples this behaviour must be attributable to the greater variability inherent in the truncated

geometric distribution We also note that under the stationary model assumption the pmf

of W1 is easily verified to be rj = (11 minus j)55 j = 1 2 10 which has mean 4 lt 55

As a result the values of Ψnxy(10) for stationary models (3) and (4) are always larger than

the corresponding values for ordinary models (1) and (2) Finally in the absence of any

knowledge regarding which distribution is in effect for the first interclaim time W1 models

(5) and (6) can be regarded as ldquoworst-case scenariordquo models as they serve to provide upper

bounds on the values of Ψnxy(10)

In conclusion we state that the values in Tables 1 and 2 were generated by first impleshy

menting the recursive procedures (38) through (316) and then summing the appropriate

bivariatetrivariate probabilities computed via (213) (214) or (215) In situations when

x = y = infin however we computed the resulting finite-time ruin probabilities using an altershy

native (albeit standard) method (eg see Cossette et al (2006 Section 3)) Specifically if

we define the survival probability σ(n u) = Pr T gt n | U0 = u = 1minusΨninfininfin(u) as well as

the auxiliary function

0 if w ge Z ηwt =

minlfloorZminuswc minus1 rfloor + 1 t if w lt Z

then by conditioning on the time and amount of the first claim (and recalling that the delayed

process reverts to the ordinary process following this claim) we obtain the expression

minnnr (kminuszk)c2 u+czk+ℓ L L L

σ(u n) = Rn + rk dℓkminuszk αjγ(u + czk + ℓ minus j n minus k)

k=1 ℓ=(kminuszk)c1 j=1

18

where the function γ(w m) for w m isin N is computed recursively via

minmna (kminusηwk)c2 u+cηwk+ℓ L L L

γ(w m) = Am + ak dℓkminusηwk αjγ(u + cηwk + ℓ minus j m minus k)

k=1 j=1 ℓ=(kminusηwk)c1

with starting values γ(w 0) = 1 forall w isin N

Acknowledgements

The authors would like to thank Dr David Landriault for his valuable suggestions which

helped to improve an earlier version of this work This research was supported by the Natural

Sciences and Engineering Research Council of Canada

References

Alfa AS 2004 Markov chain representations of discrete distributions applied to queueing

models Computers amp Operations Research 31 2365-2385

Alfa AS Drekic S 2007 Algorithmic analysis of the Sparre Andersen model in discrete

time ASTIN Bulletin 37 293-317

Bao Z 2007 A note on the compound binomial model with randomized dividend strategy

Applied Mathematics amp Computation 194 276-286

Cossette H Landriault D Marceau E 2006 Ruin probabilities in the discrete time

renewal risk model Insurance Mathematics amp Economics 38 309-323

Karlin S Taylor HM 1975 A First Course in Stochastic Processes 2nd edition Acashy

demic Press New York

Kim B Kim H-S Kim J 2008 A risk model with paying dividends and random

environment Insurance Mathematics amp Economics 42 717-726

Landriault D 2008 Randomized dividends in the compound binomial model with a general

premium rate Scandinavian Actuarial Journal 1-15

Tan J Yang X 2006 The compound binomial model with randomized decisions on

paying dividends Insurance Mathematics amp Economics 39 1-18

19

Table 1 ndash Values of Ψnxy(10) corresponding to Example 1

(a) x = 10

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0005357 0006188 0007163 0008278 0009481

(2) 0005359 0006195 0007175 0008296 0009507

(3) 0005361 0006198 0007181 0008305 0009519

(4) 0005363 0006205 0007193 0008323 0009544

(5) 0005366 0006212 0007208 0008346 0009576

(6) 0005374 0006236 0007249 0008410 0009664

y = 25 (1) 0007631 0008815 0010202 0011789 0013503

(2) 0007634 0008824 0010220 0011816 0013539

(3) 0007636 0008828 0010228 0011829 0013557

(4) 0007639 0008838 0010245 0011854 0013593

(5) 0007643 0008849 0010266 0011887 0013638

(6) 0007656 0008882 0010325 0011978 0013764

y = 50 (1) 0008477 0009792 0011333 0013095 0014998

(2) 0008481 0009802 0011352 0013125 0015039

(3) 0008483 0009807 0011361 0013139 0015059

(4) 0008486 0009817 0011380 0013168 0015098

(5) 0008491 0009830 0011403 0013204 0015149

(6) 0008504 0009866 0011470 0013305 0015288

y = infin (1) 0008777 0010138 0011733 0013558 0015528

(2) 0008781 0010149 0011753 0013588 0015570

(3) 0008783 0010154 0011763 0013603 0015591

(4) 0008787 0010164 0011782 0013633 0015632

(5) 0008791 0010177 0011807 0013671 0015684

(6) 0008805 0010215 0011875 0013775 0015828

20

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0058559 0061875 0065704 0070048 0074714

(2) 0058569 0061903 0065756 0070128 0074825

(3) 0058574 0061917 0065782 0070167 0074880

(4) 0058584 0061945 0065833 0070245 0074987

(5) 0058596 0061979 0065896 0070344 0075124

(6) 0058632 0062079 0066078 0070621 0075506

y = 25 (1) 0089002 0093986 0099739 0106262 0113270

(2) 0089017 0094028 0099817 0106382 0113437

(3) 0089025 0094049 0099856 0106442 0113519

(4) 0089040 0094091 0099932 0106560 0113681

(5) 0089057 0094142 0100028 0106708 0113887

(6) 0089113 0094294 0100302 0107127 0114463

y = 50 (1) 0102786 0108506 0115106 0122590 0130629

(2) 0102804 0108555 0115196 0122728 0130820

(3) 0102812 0108579 0115241 0122797 0130914

(4) 0102830 0108627 0115329 0122932 0131101

(5) 0102850 0108686 0115439 0123103 0131337

(6) 0102914 0108860 0115754 0123584 0132000

y = infin (1) 0108895 0114933 0121901 0129799 0138283

(2) 0108914 0114985 0121996 0129945 0138485

(3) 0108923 0115010 0122043 0130018 0138585

(4) 0108941 0115061 0122136 0130161 0138782

(5) 0108962 0115123 0122252 0130341 0139032

(6) 0109030 0115307 0122585 0130849 0139732

21

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0079856 0086792 009464 0103440 0112831

(2) 0079885 0086869 0094778 0103650 0113118

(3) 0079953 0086978 0094936 0103864 0113393

(4) 0080081 0087188 0095243 0104282 0113929

(5) 0079962 0087075 0095149 0104213 0113890

(6) 0080805 0088366 0096951 0106588 0116875

y = 25 (1) 0124591 0135680 0148207 0162245 0177217

(2) 0124639 0135806 0148433 0162586 0177683

(3) 0124755 0135993 0148703 0162950 0178148

(4) 0124978 0136354 0149226 0163658 0179054

(5) 0124763 0136141 0149035 0163500 0178935

(6) 0126238 0138381 0152148 0167588 0184060

y = 50 (1) 0146757 0160079 0175111 0191945 0209894

(2) 0146816 0160232 0175385 0192360 0210460

(3) 0146965 0160469 0175725 0192817 0211043

(4) 0147249 0160925 0176384 0193706 0212178

(5) 0146968 0160639 0176117 0193470 0211980

(6) 0148860 0163500 0180077 0198658 0218474

y = infin (1) 0158013 0172606 0189059 0207476 0227106

(2) 0158078 0172776 0189362 0207934 0227732

(3) 0158250 0173047 0189751 0208455 0228395

(4) 0158579 0173571 0190503 0209467 0229685

(5) 0158247 0173226 0190172 0209161 0229410

(6) 0160443 0176532 0194735 0215127 0236869

22

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0088667 0104403 0123329 0145277 0169184

(2) 0088711 0104518 0123538 0145598 0169627

(3) 0088733 0104574 0123641 0145756 0169846

(4) 0088777 0104687 0123844 0146066 0170274

(5) 0088829 0104828 0124105 0146470 0170832

(6) 0088988 0105229 0124822 0147560 0172331

y = 25 (1) 0140528 0167839 0201007 0239726 0282102

(2) 0140598 0168021 0201338 0240234 0282804

(3) 0140632 0168111 0201502 0240486 0283151

(4) 0140701 0168289 0201822 0240976 0283827

(5) 0140785 0168515 0202241 0241622 0284718

(6) 0141036 0169148 0203370 0243335 0287071

y = 50 (1) 0167826 0203005 0246173 0296954 0352855

(2) 0167908 0203218 0246561 0297548 0353672

(3) 0167949 0203324 0246753 0297841 0354075

(4) 0168030 0203532 0247128 0298412 0354859

(5) 0168129 0203800 0247622 0299174 0355907

(6) 0168423 0204539 0248936 0301159 0358620

y = infin (1) 0183325 0224926 0276881 0338957 0408257

(2) 0183412 0225150 0277283 0339564 0409079

(3) 0183455 0225260 0277481 0339864 0409485

(4) 0183540 0225478 0277869 0340446 0410270

(5) 0183645 0225760 0278386 0341234 0411339

(6) 0183954 0226529 0279737 0343244 0414037

23

Table 2 ndash Values of Ψnxy(10) corresponding to Example 2

(a) x = 10

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0001901 0002276 0002724 0003240 0003799

(2) 0001904 0002286 0002742 0003269 0003840

(3) 0002651 0003013 0003442 0003936 0004471

(4) 0002655 0003022 0003460 0003964 0004510

(5) 0005792 0006098 0006458 0006871 0007317

(6) 0005795 0006106 0006473 0006894 0007350

y = 25 (1) 0002708 0003243 0003880 0004615 0005411

(2) 0002713 0003257 0003906 0004656 0005469

(3) 0003778 0004293 0004904 0005608 0006369

(4) 0003782 0004306 0004929 0005647 0006424

(5) 0008255 0008690 0009203 0009792 0010427

(6) 0008259 0008702 0009224 0009824 0010473

y = 50 (1) 0003008 0003603 0004310 0005127 0006011

(2) 0003014 0003618 0004340 0005173 0006076

(3) 0004197 0004769 0005448 0006230 0007075

(4) 0004202 0004784 0005476 0006273 0007137

(5) 0009173 0009656 0010226 0010879 0011585

(6) 0009177 0009669 0010249 0010916 0011637

y = infin (1) 0003115 0003730 0004463 0005308 0006224

(2) 0003121 0003746 0004493 0005356 0006291

(3) 0004346 0004938 0005641 0006450 0007325

(4) 0004351 0004954 0005670 0006496 0007389

(5) 0009499 0009999 0010589 0011265 0011996

(6) 0009503 0010012 0010613 0011303 0012050

24

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0031176 0032815 0034733 0036924 0039282

(2) 0031193 0032862 0034821 0037061 0039475

(3) 0050168 0051747 0053587 0055684 0057939

(4) 0050184 0051792 0053671 0055815 0058124

(5) 0136884 0138219 0139763 0141517 0143399

(6) 0136896 0138256 0139832 0141626 0143553

y = 25 (1) 0047622 0050092 0052981 0056280 0059829

(2) 0004765 0050163 0053113 0056487 0060121

(3) 0076570 0078950 0081720 0084878 0088272

(4) 0076593 0079017 0081847 0085076 0088551

(5) 0205774 0207786 0210111 0212752 0215585

(6) 0205793 0207842 0210216 0212917 0215818

y = 50 (1) 0055170 0058009 0061329 0065119 0069196

(2) 0055200 0058091 0061481 0065357 0069533

(3) 0088650 0091385 0094569 0098197 0102096

(4) 0088677 0091463 0094715 0098425 0102417

(5) 0235815 0238128 0240800 0243834 0247089

(6) 0235837 0238192 0240920 0244023 0247356

y = infin (1) 0058565 0061565 0065073 0069077 0073384

(2) 0058597 0061652 0065234 0069329 0073740

(3) 0094061 0096952 0100316 0104149 0108267

(4) 0094090 0097035 0100470 0104390 0108607

(5) 0248472 0250917 0253740 0256945 0260383

(6) 0248495 0250985 0253867 0257145 0260667

25

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0048598 0052524 0056984 0061998 0067342

(2) 0048654 0052673 0057253 0062411 0067916

(3) 0070192 0073979 0078257 0083057 0088167

(4) 0070243 0074119 0078513 0083451 0088714

(5) 0151774 0154981 0158572 0162586 0166852

(6) 0151815 0155096 0158784 0162913 0167308

y = 25 (1) 0076895 0083230 0090413 0098481 0107073

(2) 0076987 0083474 0090854 0099156 0108010

(3) 0110093 0116205 0123095 0130818 0139033

(4) 0110177 0116435 0123515 0131462 0139928

(5) 0230674 0235848 0241632 0248091 0254950

(6) 0230739 0236038 0241980 0248626 0255695

y = 50 (1) 0091496 0099156 0107829 0117563 0127924

(2) 0091610 0099455 0108370 0118387 0129068

(3) 0130130 0137520 0145839 0155156 0165064

(4) 0130233 0137801 0146352 0155942 0166156

(5) 0266599 0272857 0279840 0287632 0295904

(6) 0266680 0273088 0280265 0288286 0296814

y = infin (1) 0099312 0107746 0117285 0127984 0139369

(2) 0099439 0108078 0117883 0128899 0140639

(3) 0140458 0148595 0157745 0167986 0178872

(4) 0140572 0148908 0158314 0168859 0180084

(5) 0282886 0289776 0297457 0306022 0315111

(6) 0282976 0290034 0297929 0306748 0316120

26

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0059096 0072505 0088907 0108122 0129137

(2) 0059223 0072829 0089500 0109052 0130456

(3) 0079757 0092662 0108388 0126778 0146871

(4) 0079875 0092970 0108956 0127669 0148134

(5) 0159788 0170682 0183871 0199249 0216024

(6) 0159881 0170936 0184343 0199990 0217075

y = 25 (1) 0095883 0119741 0149240 0184073 0222390

(2) 0096087 0120262 0150197 0185573 0224520

(3) 0127403 0150359 0178640 0211977 0248613

(4) 0127592 0150855 0179557 0213414 0250653

(5) 0245161 0264533 0288250 0316124 0346709

(6) 0245311 0264942 0289010 0317319 0348407

y = 50 (1) 0116598 0147920 0187100 0233784 0285503

(2) 0116839 0148537 0188231 0235554 0288015

(3) 0153023 0183153 0220713 0265390 0314840

(4) 0153249 0183742 0221798 0267087 0317248

(5) 0285742 0311159 0342651 0380005 0421287

(6) 0285921 0311644 0343551 0381418 0423292

y = infin (1) 0129470 0167217 0215383 0273841 0339734

(2) 0129725 0167865 0216560 0275661 0342282

(3) 0167978 0204275 0250442 0306382 0369383

(4) 0168217 0204895 0251572 0308128 0371826

(5) 0305868 0336470 0375169 0421935 0474525

(6) 0306058 0336980 0376106 0423388 0476560

27

Page 18: uwaterloo.ca · Ruin Analysis of a Threshold Strategy in a Discrete-time Sparre Andersen Model Steve DREKIC and Ana Maria MERA Department of Statistics and Actuarial Science University

(1) ordinary model under scenario (2) from Example 1

(2) ordinary model under scenario (5) from Example 1

(3) stationary model under scenario (2) from Example 1

(4) stationary model under scenario (5) from Example 1

(5) delayed model (with W1 = 1 with probability 1) under scenario (2) from Example 1

(6) delayed model (with W1 = 1 with probability 1) under scenario (5) from Example 1

Concerning the results in Table 2 we first compare the values of Ψnxy(10) in entries (2)

and (5) of Table 1 to their corresponding counterparts in entries (1) and (2) of Table 2 In all

cases we observe that ruin is more likely to occur when interclaim times have the truncated

geometric distribution introduced in our first example as opposed to the discrete uniform

distribution of our second example Since mean interclaim times are equal in both examshy

ples this behaviour must be attributable to the greater variability inherent in the truncated

geometric distribution We also note that under the stationary model assumption the pmf

of W1 is easily verified to be rj = (11 minus j)55 j = 1 2 10 which has mean 4 lt 55

As a result the values of Ψnxy(10) for stationary models (3) and (4) are always larger than

the corresponding values for ordinary models (1) and (2) Finally in the absence of any

knowledge regarding which distribution is in effect for the first interclaim time W1 models

(5) and (6) can be regarded as ldquoworst-case scenariordquo models as they serve to provide upper

bounds on the values of Ψnxy(10)

In conclusion we state that the values in Tables 1 and 2 were generated by first impleshy

menting the recursive procedures (38) through (316) and then summing the appropriate

bivariatetrivariate probabilities computed via (213) (214) or (215) In situations when

x = y = infin however we computed the resulting finite-time ruin probabilities using an altershy

native (albeit standard) method (eg see Cossette et al (2006 Section 3)) Specifically if

we define the survival probability σ(n u) = Pr T gt n | U0 = u = 1minusΨninfininfin(u) as well as

the auxiliary function

0 if w ge Z ηwt =

minlfloorZminuswc minus1 rfloor + 1 t if w lt Z

then by conditioning on the time and amount of the first claim (and recalling that the delayed

process reverts to the ordinary process following this claim) we obtain the expression

minnnr (kminuszk)c2 u+czk+ℓ L L L

σ(u n) = Rn + rk dℓkminuszk αjγ(u + czk + ℓ minus j n minus k)

k=1 ℓ=(kminuszk)c1 j=1

18

where the function γ(w m) for w m isin N is computed recursively via

minmna (kminusηwk)c2 u+cηwk+ℓ L L L

γ(w m) = Am + ak dℓkminusηwk αjγ(u + cηwk + ℓ minus j m minus k)

k=1 j=1 ℓ=(kminusηwk)c1

with starting values γ(w 0) = 1 forall w isin N

Acknowledgements

The authors would like to thank Dr David Landriault for his valuable suggestions which

helped to improve an earlier version of this work This research was supported by the Natural

Sciences and Engineering Research Council of Canada

References

Alfa AS 2004 Markov chain representations of discrete distributions applied to queueing

models Computers amp Operations Research 31 2365-2385

Alfa AS Drekic S 2007 Algorithmic analysis of the Sparre Andersen model in discrete

time ASTIN Bulletin 37 293-317

Bao Z 2007 A note on the compound binomial model with randomized dividend strategy

Applied Mathematics amp Computation 194 276-286

Cossette H Landriault D Marceau E 2006 Ruin probabilities in the discrete time

renewal risk model Insurance Mathematics amp Economics 38 309-323

Karlin S Taylor HM 1975 A First Course in Stochastic Processes 2nd edition Acashy

demic Press New York

Kim B Kim H-S Kim J 2008 A risk model with paying dividends and random

environment Insurance Mathematics amp Economics 42 717-726

Landriault D 2008 Randomized dividends in the compound binomial model with a general

premium rate Scandinavian Actuarial Journal 1-15

Tan J Yang X 2006 The compound binomial model with randomized decisions on

paying dividends Insurance Mathematics amp Economics 39 1-18

19

Table 1 ndash Values of Ψnxy(10) corresponding to Example 1

(a) x = 10

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0005357 0006188 0007163 0008278 0009481

(2) 0005359 0006195 0007175 0008296 0009507

(3) 0005361 0006198 0007181 0008305 0009519

(4) 0005363 0006205 0007193 0008323 0009544

(5) 0005366 0006212 0007208 0008346 0009576

(6) 0005374 0006236 0007249 0008410 0009664

y = 25 (1) 0007631 0008815 0010202 0011789 0013503

(2) 0007634 0008824 0010220 0011816 0013539

(3) 0007636 0008828 0010228 0011829 0013557

(4) 0007639 0008838 0010245 0011854 0013593

(5) 0007643 0008849 0010266 0011887 0013638

(6) 0007656 0008882 0010325 0011978 0013764

y = 50 (1) 0008477 0009792 0011333 0013095 0014998

(2) 0008481 0009802 0011352 0013125 0015039

(3) 0008483 0009807 0011361 0013139 0015059

(4) 0008486 0009817 0011380 0013168 0015098

(5) 0008491 0009830 0011403 0013204 0015149

(6) 0008504 0009866 0011470 0013305 0015288

y = infin (1) 0008777 0010138 0011733 0013558 0015528

(2) 0008781 0010149 0011753 0013588 0015570

(3) 0008783 0010154 0011763 0013603 0015591

(4) 0008787 0010164 0011782 0013633 0015632

(5) 0008791 0010177 0011807 0013671 0015684

(6) 0008805 0010215 0011875 0013775 0015828

20

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0058559 0061875 0065704 0070048 0074714

(2) 0058569 0061903 0065756 0070128 0074825

(3) 0058574 0061917 0065782 0070167 0074880

(4) 0058584 0061945 0065833 0070245 0074987

(5) 0058596 0061979 0065896 0070344 0075124

(6) 0058632 0062079 0066078 0070621 0075506

y = 25 (1) 0089002 0093986 0099739 0106262 0113270

(2) 0089017 0094028 0099817 0106382 0113437

(3) 0089025 0094049 0099856 0106442 0113519

(4) 0089040 0094091 0099932 0106560 0113681

(5) 0089057 0094142 0100028 0106708 0113887

(6) 0089113 0094294 0100302 0107127 0114463

y = 50 (1) 0102786 0108506 0115106 0122590 0130629

(2) 0102804 0108555 0115196 0122728 0130820

(3) 0102812 0108579 0115241 0122797 0130914

(4) 0102830 0108627 0115329 0122932 0131101

(5) 0102850 0108686 0115439 0123103 0131337

(6) 0102914 0108860 0115754 0123584 0132000

y = infin (1) 0108895 0114933 0121901 0129799 0138283

(2) 0108914 0114985 0121996 0129945 0138485

(3) 0108923 0115010 0122043 0130018 0138585

(4) 0108941 0115061 0122136 0130161 0138782

(5) 0108962 0115123 0122252 0130341 0139032

(6) 0109030 0115307 0122585 0130849 0139732

21

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0079856 0086792 009464 0103440 0112831

(2) 0079885 0086869 0094778 0103650 0113118

(3) 0079953 0086978 0094936 0103864 0113393

(4) 0080081 0087188 0095243 0104282 0113929

(5) 0079962 0087075 0095149 0104213 0113890

(6) 0080805 0088366 0096951 0106588 0116875

y = 25 (1) 0124591 0135680 0148207 0162245 0177217

(2) 0124639 0135806 0148433 0162586 0177683

(3) 0124755 0135993 0148703 0162950 0178148

(4) 0124978 0136354 0149226 0163658 0179054

(5) 0124763 0136141 0149035 0163500 0178935

(6) 0126238 0138381 0152148 0167588 0184060

y = 50 (1) 0146757 0160079 0175111 0191945 0209894

(2) 0146816 0160232 0175385 0192360 0210460

(3) 0146965 0160469 0175725 0192817 0211043

(4) 0147249 0160925 0176384 0193706 0212178

(5) 0146968 0160639 0176117 0193470 0211980

(6) 0148860 0163500 0180077 0198658 0218474

y = infin (1) 0158013 0172606 0189059 0207476 0227106

(2) 0158078 0172776 0189362 0207934 0227732

(3) 0158250 0173047 0189751 0208455 0228395

(4) 0158579 0173571 0190503 0209467 0229685

(5) 0158247 0173226 0190172 0209161 0229410

(6) 0160443 0176532 0194735 0215127 0236869

22

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0088667 0104403 0123329 0145277 0169184

(2) 0088711 0104518 0123538 0145598 0169627

(3) 0088733 0104574 0123641 0145756 0169846

(4) 0088777 0104687 0123844 0146066 0170274

(5) 0088829 0104828 0124105 0146470 0170832

(6) 0088988 0105229 0124822 0147560 0172331

y = 25 (1) 0140528 0167839 0201007 0239726 0282102

(2) 0140598 0168021 0201338 0240234 0282804

(3) 0140632 0168111 0201502 0240486 0283151

(4) 0140701 0168289 0201822 0240976 0283827

(5) 0140785 0168515 0202241 0241622 0284718

(6) 0141036 0169148 0203370 0243335 0287071

y = 50 (1) 0167826 0203005 0246173 0296954 0352855

(2) 0167908 0203218 0246561 0297548 0353672

(3) 0167949 0203324 0246753 0297841 0354075

(4) 0168030 0203532 0247128 0298412 0354859

(5) 0168129 0203800 0247622 0299174 0355907

(6) 0168423 0204539 0248936 0301159 0358620

y = infin (1) 0183325 0224926 0276881 0338957 0408257

(2) 0183412 0225150 0277283 0339564 0409079

(3) 0183455 0225260 0277481 0339864 0409485

(4) 0183540 0225478 0277869 0340446 0410270

(5) 0183645 0225760 0278386 0341234 0411339

(6) 0183954 0226529 0279737 0343244 0414037

23

Table 2 ndash Values of Ψnxy(10) corresponding to Example 2

(a) x = 10

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0001901 0002276 0002724 0003240 0003799

(2) 0001904 0002286 0002742 0003269 0003840

(3) 0002651 0003013 0003442 0003936 0004471

(4) 0002655 0003022 0003460 0003964 0004510

(5) 0005792 0006098 0006458 0006871 0007317

(6) 0005795 0006106 0006473 0006894 0007350

y = 25 (1) 0002708 0003243 0003880 0004615 0005411

(2) 0002713 0003257 0003906 0004656 0005469

(3) 0003778 0004293 0004904 0005608 0006369

(4) 0003782 0004306 0004929 0005647 0006424

(5) 0008255 0008690 0009203 0009792 0010427

(6) 0008259 0008702 0009224 0009824 0010473

y = 50 (1) 0003008 0003603 0004310 0005127 0006011

(2) 0003014 0003618 0004340 0005173 0006076

(3) 0004197 0004769 0005448 0006230 0007075

(4) 0004202 0004784 0005476 0006273 0007137

(5) 0009173 0009656 0010226 0010879 0011585

(6) 0009177 0009669 0010249 0010916 0011637

y = infin (1) 0003115 0003730 0004463 0005308 0006224

(2) 0003121 0003746 0004493 0005356 0006291

(3) 0004346 0004938 0005641 0006450 0007325

(4) 0004351 0004954 0005670 0006496 0007389

(5) 0009499 0009999 0010589 0011265 0011996

(6) 0009503 0010012 0010613 0011303 0012050

24

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0031176 0032815 0034733 0036924 0039282

(2) 0031193 0032862 0034821 0037061 0039475

(3) 0050168 0051747 0053587 0055684 0057939

(4) 0050184 0051792 0053671 0055815 0058124

(5) 0136884 0138219 0139763 0141517 0143399

(6) 0136896 0138256 0139832 0141626 0143553

y = 25 (1) 0047622 0050092 0052981 0056280 0059829

(2) 0004765 0050163 0053113 0056487 0060121

(3) 0076570 0078950 0081720 0084878 0088272

(4) 0076593 0079017 0081847 0085076 0088551

(5) 0205774 0207786 0210111 0212752 0215585

(6) 0205793 0207842 0210216 0212917 0215818

y = 50 (1) 0055170 0058009 0061329 0065119 0069196

(2) 0055200 0058091 0061481 0065357 0069533

(3) 0088650 0091385 0094569 0098197 0102096

(4) 0088677 0091463 0094715 0098425 0102417

(5) 0235815 0238128 0240800 0243834 0247089

(6) 0235837 0238192 0240920 0244023 0247356

y = infin (1) 0058565 0061565 0065073 0069077 0073384

(2) 0058597 0061652 0065234 0069329 0073740

(3) 0094061 0096952 0100316 0104149 0108267

(4) 0094090 0097035 0100470 0104390 0108607

(5) 0248472 0250917 0253740 0256945 0260383

(6) 0248495 0250985 0253867 0257145 0260667

25

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0048598 0052524 0056984 0061998 0067342

(2) 0048654 0052673 0057253 0062411 0067916

(3) 0070192 0073979 0078257 0083057 0088167

(4) 0070243 0074119 0078513 0083451 0088714

(5) 0151774 0154981 0158572 0162586 0166852

(6) 0151815 0155096 0158784 0162913 0167308

y = 25 (1) 0076895 0083230 0090413 0098481 0107073

(2) 0076987 0083474 0090854 0099156 0108010

(3) 0110093 0116205 0123095 0130818 0139033

(4) 0110177 0116435 0123515 0131462 0139928

(5) 0230674 0235848 0241632 0248091 0254950

(6) 0230739 0236038 0241980 0248626 0255695

y = 50 (1) 0091496 0099156 0107829 0117563 0127924

(2) 0091610 0099455 0108370 0118387 0129068

(3) 0130130 0137520 0145839 0155156 0165064

(4) 0130233 0137801 0146352 0155942 0166156

(5) 0266599 0272857 0279840 0287632 0295904

(6) 0266680 0273088 0280265 0288286 0296814

y = infin (1) 0099312 0107746 0117285 0127984 0139369

(2) 0099439 0108078 0117883 0128899 0140639

(3) 0140458 0148595 0157745 0167986 0178872

(4) 0140572 0148908 0158314 0168859 0180084

(5) 0282886 0289776 0297457 0306022 0315111

(6) 0282976 0290034 0297929 0306748 0316120

26

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0059096 0072505 0088907 0108122 0129137

(2) 0059223 0072829 0089500 0109052 0130456

(3) 0079757 0092662 0108388 0126778 0146871

(4) 0079875 0092970 0108956 0127669 0148134

(5) 0159788 0170682 0183871 0199249 0216024

(6) 0159881 0170936 0184343 0199990 0217075

y = 25 (1) 0095883 0119741 0149240 0184073 0222390

(2) 0096087 0120262 0150197 0185573 0224520

(3) 0127403 0150359 0178640 0211977 0248613

(4) 0127592 0150855 0179557 0213414 0250653

(5) 0245161 0264533 0288250 0316124 0346709

(6) 0245311 0264942 0289010 0317319 0348407

y = 50 (1) 0116598 0147920 0187100 0233784 0285503

(2) 0116839 0148537 0188231 0235554 0288015

(3) 0153023 0183153 0220713 0265390 0314840

(4) 0153249 0183742 0221798 0267087 0317248

(5) 0285742 0311159 0342651 0380005 0421287

(6) 0285921 0311644 0343551 0381418 0423292

y = infin (1) 0129470 0167217 0215383 0273841 0339734

(2) 0129725 0167865 0216560 0275661 0342282

(3) 0167978 0204275 0250442 0306382 0369383

(4) 0168217 0204895 0251572 0308128 0371826

(5) 0305868 0336470 0375169 0421935 0474525

(6) 0306058 0336980 0376106 0423388 0476560

27

Page 19: uwaterloo.ca · Ruin Analysis of a Threshold Strategy in a Discrete-time Sparre Andersen Model Steve DREKIC and Ana Maria MERA Department of Statistics and Actuarial Science University

where the function γ(w m) for w m isin N is computed recursively via

minmna (kminusηwk)c2 u+cηwk+ℓ L L L

γ(w m) = Am + ak dℓkminusηwk αjγ(u + cηwk + ℓ minus j m minus k)

k=1 j=1 ℓ=(kminusηwk)c1

with starting values γ(w 0) = 1 forall w isin N

Acknowledgements

The authors would like to thank Dr David Landriault for his valuable suggestions which

helped to improve an earlier version of this work This research was supported by the Natural

Sciences and Engineering Research Council of Canada

References

Alfa AS 2004 Markov chain representations of discrete distributions applied to queueing

models Computers amp Operations Research 31 2365-2385

Alfa AS Drekic S 2007 Algorithmic analysis of the Sparre Andersen model in discrete

time ASTIN Bulletin 37 293-317

Bao Z 2007 A note on the compound binomial model with randomized dividend strategy

Applied Mathematics amp Computation 194 276-286

Cossette H Landriault D Marceau E 2006 Ruin probabilities in the discrete time

renewal risk model Insurance Mathematics amp Economics 38 309-323

Karlin S Taylor HM 1975 A First Course in Stochastic Processes 2nd edition Acashy

demic Press New York

Kim B Kim H-S Kim J 2008 A risk model with paying dividends and random

environment Insurance Mathematics amp Economics 42 717-726

Landriault D 2008 Randomized dividends in the compound binomial model with a general

premium rate Scandinavian Actuarial Journal 1-15

Tan J Yang X 2006 The compound binomial model with randomized decisions on

paying dividends Insurance Mathematics amp Economics 39 1-18

19

Table 1 ndash Values of Ψnxy(10) corresponding to Example 1

(a) x = 10

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0005357 0006188 0007163 0008278 0009481

(2) 0005359 0006195 0007175 0008296 0009507

(3) 0005361 0006198 0007181 0008305 0009519

(4) 0005363 0006205 0007193 0008323 0009544

(5) 0005366 0006212 0007208 0008346 0009576

(6) 0005374 0006236 0007249 0008410 0009664

y = 25 (1) 0007631 0008815 0010202 0011789 0013503

(2) 0007634 0008824 0010220 0011816 0013539

(3) 0007636 0008828 0010228 0011829 0013557

(4) 0007639 0008838 0010245 0011854 0013593

(5) 0007643 0008849 0010266 0011887 0013638

(6) 0007656 0008882 0010325 0011978 0013764

y = 50 (1) 0008477 0009792 0011333 0013095 0014998

(2) 0008481 0009802 0011352 0013125 0015039

(3) 0008483 0009807 0011361 0013139 0015059

(4) 0008486 0009817 0011380 0013168 0015098

(5) 0008491 0009830 0011403 0013204 0015149

(6) 0008504 0009866 0011470 0013305 0015288

y = infin (1) 0008777 0010138 0011733 0013558 0015528

(2) 0008781 0010149 0011753 0013588 0015570

(3) 0008783 0010154 0011763 0013603 0015591

(4) 0008787 0010164 0011782 0013633 0015632

(5) 0008791 0010177 0011807 0013671 0015684

(6) 0008805 0010215 0011875 0013775 0015828

20

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0058559 0061875 0065704 0070048 0074714

(2) 0058569 0061903 0065756 0070128 0074825

(3) 0058574 0061917 0065782 0070167 0074880

(4) 0058584 0061945 0065833 0070245 0074987

(5) 0058596 0061979 0065896 0070344 0075124

(6) 0058632 0062079 0066078 0070621 0075506

y = 25 (1) 0089002 0093986 0099739 0106262 0113270

(2) 0089017 0094028 0099817 0106382 0113437

(3) 0089025 0094049 0099856 0106442 0113519

(4) 0089040 0094091 0099932 0106560 0113681

(5) 0089057 0094142 0100028 0106708 0113887

(6) 0089113 0094294 0100302 0107127 0114463

y = 50 (1) 0102786 0108506 0115106 0122590 0130629

(2) 0102804 0108555 0115196 0122728 0130820

(3) 0102812 0108579 0115241 0122797 0130914

(4) 0102830 0108627 0115329 0122932 0131101

(5) 0102850 0108686 0115439 0123103 0131337

(6) 0102914 0108860 0115754 0123584 0132000

y = infin (1) 0108895 0114933 0121901 0129799 0138283

(2) 0108914 0114985 0121996 0129945 0138485

(3) 0108923 0115010 0122043 0130018 0138585

(4) 0108941 0115061 0122136 0130161 0138782

(5) 0108962 0115123 0122252 0130341 0139032

(6) 0109030 0115307 0122585 0130849 0139732

21

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0079856 0086792 009464 0103440 0112831

(2) 0079885 0086869 0094778 0103650 0113118

(3) 0079953 0086978 0094936 0103864 0113393

(4) 0080081 0087188 0095243 0104282 0113929

(5) 0079962 0087075 0095149 0104213 0113890

(6) 0080805 0088366 0096951 0106588 0116875

y = 25 (1) 0124591 0135680 0148207 0162245 0177217

(2) 0124639 0135806 0148433 0162586 0177683

(3) 0124755 0135993 0148703 0162950 0178148

(4) 0124978 0136354 0149226 0163658 0179054

(5) 0124763 0136141 0149035 0163500 0178935

(6) 0126238 0138381 0152148 0167588 0184060

y = 50 (1) 0146757 0160079 0175111 0191945 0209894

(2) 0146816 0160232 0175385 0192360 0210460

(3) 0146965 0160469 0175725 0192817 0211043

(4) 0147249 0160925 0176384 0193706 0212178

(5) 0146968 0160639 0176117 0193470 0211980

(6) 0148860 0163500 0180077 0198658 0218474

y = infin (1) 0158013 0172606 0189059 0207476 0227106

(2) 0158078 0172776 0189362 0207934 0227732

(3) 0158250 0173047 0189751 0208455 0228395

(4) 0158579 0173571 0190503 0209467 0229685

(5) 0158247 0173226 0190172 0209161 0229410

(6) 0160443 0176532 0194735 0215127 0236869

22

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0088667 0104403 0123329 0145277 0169184

(2) 0088711 0104518 0123538 0145598 0169627

(3) 0088733 0104574 0123641 0145756 0169846

(4) 0088777 0104687 0123844 0146066 0170274

(5) 0088829 0104828 0124105 0146470 0170832

(6) 0088988 0105229 0124822 0147560 0172331

y = 25 (1) 0140528 0167839 0201007 0239726 0282102

(2) 0140598 0168021 0201338 0240234 0282804

(3) 0140632 0168111 0201502 0240486 0283151

(4) 0140701 0168289 0201822 0240976 0283827

(5) 0140785 0168515 0202241 0241622 0284718

(6) 0141036 0169148 0203370 0243335 0287071

y = 50 (1) 0167826 0203005 0246173 0296954 0352855

(2) 0167908 0203218 0246561 0297548 0353672

(3) 0167949 0203324 0246753 0297841 0354075

(4) 0168030 0203532 0247128 0298412 0354859

(5) 0168129 0203800 0247622 0299174 0355907

(6) 0168423 0204539 0248936 0301159 0358620

y = infin (1) 0183325 0224926 0276881 0338957 0408257

(2) 0183412 0225150 0277283 0339564 0409079

(3) 0183455 0225260 0277481 0339864 0409485

(4) 0183540 0225478 0277869 0340446 0410270

(5) 0183645 0225760 0278386 0341234 0411339

(6) 0183954 0226529 0279737 0343244 0414037

23

Table 2 ndash Values of Ψnxy(10) corresponding to Example 2

(a) x = 10

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0001901 0002276 0002724 0003240 0003799

(2) 0001904 0002286 0002742 0003269 0003840

(3) 0002651 0003013 0003442 0003936 0004471

(4) 0002655 0003022 0003460 0003964 0004510

(5) 0005792 0006098 0006458 0006871 0007317

(6) 0005795 0006106 0006473 0006894 0007350

y = 25 (1) 0002708 0003243 0003880 0004615 0005411

(2) 0002713 0003257 0003906 0004656 0005469

(3) 0003778 0004293 0004904 0005608 0006369

(4) 0003782 0004306 0004929 0005647 0006424

(5) 0008255 0008690 0009203 0009792 0010427

(6) 0008259 0008702 0009224 0009824 0010473

y = 50 (1) 0003008 0003603 0004310 0005127 0006011

(2) 0003014 0003618 0004340 0005173 0006076

(3) 0004197 0004769 0005448 0006230 0007075

(4) 0004202 0004784 0005476 0006273 0007137

(5) 0009173 0009656 0010226 0010879 0011585

(6) 0009177 0009669 0010249 0010916 0011637

y = infin (1) 0003115 0003730 0004463 0005308 0006224

(2) 0003121 0003746 0004493 0005356 0006291

(3) 0004346 0004938 0005641 0006450 0007325

(4) 0004351 0004954 0005670 0006496 0007389

(5) 0009499 0009999 0010589 0011265 0011996

(6) 0009503 0010012 0010613 0011303 0012050

24

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0031176 0032815 0034733 0036924 0039282

(2) 0031193 0032862 0034821 0037061 0039475

(3) 0050168 0051747 0053587 0055684 0057939

(4) 0050184 0051792 0053671 0055815 0058124

(5) 0136884 0138219 0139763 0141517 0143399

(6) 0136896 0138256 0139832 0141626 0143553

y = 25 (1) 0047622 0050092 0052981 0056280 0059829

(2) 0004765 0050163 0053113 0056487 0060121

(3) 0076570 0078950 0081720 0084878 0088272

(4) 0076593 0079017 0081847 0085076 0088551

(5) 0205774 0207786 0210111 0212752 0215585

(6) 0205793 0207842 0210216 0212917 0215818

y = 50 (1) 0055170 0058009 0061329 0065119 0069196

(2) 0055200 0058091 0061481 0065357 0069533

(3) 0088650 0091385 0094569 0098197 0102096

(4) 0088677 0091463 0094715 0098425 0102417

(5) 0235815 0238128 0240800 0243834 0247089

(6) 0235837 0238192 0240920 0244023 0247356

y = infin (1) 0058565 0061565 0065073 0069077 0073384

(2) 0058597 0061652 0065234 0069329 0073740

(3) 0094061 0096952 0100316 0104149 0108267

(4) 0094090 0097035 0100470 0104390 0108607

(5) 0248472 0250917 0253740 0256945 0260383

(6) 0248495 0250985 0253867 0257145 0260667

25

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0048598 0052524 0056984 0061998 0067342

(2) 0048654 0052673 0057253 0062411 0067916

(3) 0070192 0073979 0078257 0083057 0088167

(4) 0070243 0074119 0078513 0083451 0088714

(5) 0151774 0154981 0158572 0162586 0166852

(6) 0151815 0155096 0158784 0162913 0167308

y = 25 (1) 0076895 0083230 0090413 0098481 0107073

(2) 0076987 0083474 0090854 0099156 0108010

(3) 0110093 0116205 0123095 0130818 0139033

(4) 0110177 0116435 0123515 0131462 0139928

(5) 0230674 0235848 0241632 0248091 0254950

(6) 0230739 0236038 0241980 0248626 0255695

y = 50 (1) 0091496 0099156 0107829 0117563 0127924

(2) 0091610 0099455 0108370 0118387 0129068

(3) 0130130 0137520 0145839 0155156 0165064

(4) 0130233 0137801 0146352 0155942 0166156

(5) 0266599 0272857 0279840 0287632 0295904

(6) 0266680 0273088 0280265 0288286 0296814

y = infin (1) 0099312 0107746 0117285 0127984 0139369

(2) 0099439 0108078 0117883 0128899 0140639

(3) 0140458 0148595 0157745 0167986 0178872

(4) 0140572 0148908 0158314 0168859 0180084

(5) 0282886 0289776 0297457 0306022 0315111

(6) 0282976 0290034 0297929 0306748 0316120

26

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0059096 0072505 0088907 0108122 0129137

(2) 0059223 0072829 0089500 0109052 0130456

(3) 0079757 0092662 0108388 0126778 0146871

(4) 0079875 0092970 0108956 0127669 0148134

(5) 0159788 0170682 0183871 0199249 0216024

(6) 0159881 0170936 0184343 0199990 0217075

y = 25 (1) 0095883 0119741 0149240 0184073 0222390

(2) 0096087 0120262 0150197 0185573 0224520

(3) 0127403 0150359 0178640 0211977 0248613

(4) 0127592 0150855 0179557 0213414 0250653

(5) 0245161 0264533 0288250 0316124 0346709

(6) 0245311 0264942 0289010 0317319 0348407

y = 50 (1) 0116598 0147920 0187100 0233784 0285503

(2) 0116839 0148537 0188231 0235554 0288015

(3) 0153023 0183153 0220713 0265390 0314840

(4) 0153249 0183742 0221798 0267087 0317248

(5) 0285742 0311159 0342651 0380005 0421287

(6) 0285921 0311644 0343551 0381418 0423292

y = infin (1) 0129470 0167217 0215383 0273841 0339734

(2) 0129725 0167865 0216560 0275661 0342282

(3) 0167978 0204275 0250442 0306382 0369383

(4) 0168217 0204895 0251572 0308128 0371826

(5) 0305868 0336470 0375169 0421935 0474525

(6) 0306058 0336980 0376106 0423388 0476560

27

Page 20: uwaterloo.ca · Ruin Analysis of a Threshold Strategy in a Discrete-time Sparre Andersen Model Steve DREKIC and Ana Maria MERA Department of Statistics and Actuarial Science University

Table 1 ndash Values of Ψnxy(10) corresponding to Example 1

(a) x = 10

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0005357 0006188 0007163 0008278 0009481

(2) 0005359 0006195 0007175 0008296 0009507

(3) 0005361 0006198 0007181 0008305 0009519

(4) 0005363 0006205 0007193 0008323 0009544

(5) 0005366 0006212 0007208 0008346 0009576

(6) 0005374 0006236 0007249 0008410 0009664

y = 25 (1) 0007631 0008815 0010202 0011789 0013503

(2) 0007634 0008824 0010220 0011816 0013539

(3) 0007636 0008828 0010228 0011829 0013557

(4) 0007639 0008838 0010245 0011854 0013593

(5) 0007643 0008849 0010266 0011887 0013638

(6) 0007656 0008882 0010325 0011978 0013764

y = 50 (1) 0008477 0009792 0011333 0013095 0014998

(2) 0008481 0009802 0011352 0013125 0015039

(3) 0008483 0009807 0011361 0013139 0015059

(4) 0008486 0009817 0011380 0013168 0015098

(5) 0008491 0009830 0011403 0013204 0015149

(6) 0008504 0009866 0011470 0013305 0015288

y = infin (1) 0008777 0010138 0011733 0013558 0015528

(2) 0008781 0010149 0011753 0013588 0015570

(3) 0008783 0010154 0011763 0013603 0015591

(4) 0008787 0010164 0011782 0013633 0015632

(5) 0008791 0010177 0011807 0013671 0015684

(6) 0008805 0010215 0011875 0013775 0015828

20

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0058559 0061875 0065704 0070048 0074714

(2) 0058569 0061903 0065756 0070128 0074825

(3) 0058574 0061917 0065782 0070167 0074880

(4) 0058584 0061945 0065833 0070245 0074987

(5) 0058596 0061979 0065896 0070344 0075124

(6) 0058632 0062079 0066078 0070621 0075506

y = 25 (1) 0089002 0093986 0099739 0106262 0113270

(2) 0089017 0094028 0099817 0106382 0113437

(3) 0089025 0094049 0099856 0106442 0113519

(4) 0089040 0094091 0099932 0106560 0113681

(5) 0089057 0094142 0100028 0106708 0113887

(6) 0089113 0094294 0100302 0107127 0114463

y = 50 (1) 0102786 0108506 0115106 0122590 0130629

(2) 0102804 0108555 0115196 0122728 0130820

(3) 0102812 0108579 0115241 0122797 0130914

(4) 0102830 0108627 0115329 0122932 0131101

(5) 0102850 0108686 0115439 0123103 0131337

(6) 0102914 0108860 0115754 0123584 0132000

y = infin (1) 0108895 0114933 0121901 0129799 0138283

(2) 0108914 0114985 0121996 0129945 0138485

(3) 0108923 0115010 0122043 0130018 0138585

(4) 0108941 0115061 0122136 0130161 0138782

(5) 0108962 0115123 0122252 0130341 0139032

(6) 0109030 0115307 0122585 0130849 0139732

21

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0079856 0086792 009464 0103440 0112831

(2) 0079885 0086869 0094778 0103650 0113118

(3) 0079953 0086978 0094936 0103864 0113393

(4) 0080081 0087188 0095243 0104282 0113929

(5) 0079962 0087075 0095149 0104213 0113890

(6) 0080805 0088366 0096951 0106588 0116875

y = 25 (1) 0124591 0135680 0148207 0162245 0177217

(2) 0124639 0135806 0148433 0162586 0177683

(3) 0124755 0135993 0148703 0162950 0178148

(4) 0124978 0136354 0149226 0163658 0179054

(5) 0124763 0136141 0149035 0163500 0178935

(6) 0126238 0138381 0152148 0167588 0184060

y = 50 (1) 0146757 0160079 0175111 0191945 0209894

(2) 0146816 0160232 0175385 0192360 0210460

(3) 0146965 0160469 0175725 0192817 0211043

(4) 0147249 0160925 0176384 0193706 0212178

(5) 0146968 0160639 0176117 0193470 0211980

(6) 0148860 0163500 0180077 0198658 0218474

y = infin (1) 0158013 0172606 0189059 0207476 0227106

(2) 0158078 0172776 0189362 0207934 0227732

(3) 0158250 0173047 0189751 0208455 0228395

(4) 0158579 0173571 0190503 0209467 0229685

(5) 0158247 0173226 0190172 0209161 0229410

(6) 0160443 0176532 0194735 0215127 0236869

22

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0088667 0104403 0123329 0145277 0169184

(2) 0088711 0104518 0123538 0145598 0169627

(3) 0088733 0104574 0123641 0145756 0169846

(4) 0088777 0104687 0123844 0146066 0170274

(5) 0088829 0104828 0124105 0146470 0170832

(6) 0088988 0105229 0124822 0147560 0172331

y = 25 (1) 0140528 0167839 0201007 0239726 0282102

(2) 0140598 0168021 0201338 0240234 0282804

(3) 0140632 0168111 0201502 0240486 0283151

(4) 0140701 0168289 0201822 0240976 0283827

(5) 0140785 0168515 0202241 0241622 0284718

(6) 0141036 0169148 0203370 0243335 0287071

y = 50 (1) 0167826 0203005 0246173 0296954 0352855

(2) 0167908 0203218 0246561 0297548 0353672

(3) 0167949 0203324 0246753 0297841 0354075

(4) 0168030 0203532 0247128 0298412 0354859

(5) 0168129 0203800 0247622 0299174 0355907

(6) 0168423 0204539 0248936 0301159 0358620

y = infin (1) 0183325 0224926 0276881 0338957 0408257

(2) 0183412 0225150 0277283 0339564 0409079

(3) 0183455 0225260 0277481 0339864 0409485

(4) 0183540 0225478 0277869 0340446 0410270

(5) 0183645 0225760 0278386 0341234 0411339

(6) 0183954 0226529 0279737 0343244 0414037

23

Table 2 ndash Values of Ψnxy(10) corresponding to Example 2

(a) x = 10

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0001901 0002276 0002724 0003240 0003799

(2) 0001904 0002286 0002742 0003269 0003840

(3) 0002651 0003013 0003442 0003936 0004471

(4) 0002655 0003022 0003460 0003964 0004510

(5) 0005792 0006098 0006458 0006871 0007317

(6) 0005795 0006106 0006473 0006894 0007350

y = 25 (1) 0002708 0003243 0003880 0004615 0005411

(2) 0002713 0003257 0003906 0004656 0005469

(3) 0003778 0004293 0004904 0005608 0006369

(4) 0003782 0004306 0004929 0005647 0006424

(5) 0008255 0008690 0009203 0009792 0010427

(6) 0008259 0008702 0009224 0009824 0010473

y = 50 (1) 0003008 0003603 0004310 0005127 0006011

(2) 0003014 0003618 0004340 0005173 0006076

(3) 0004197 0004769 0005448 0006230 0007075

(4) 0004202 0004784 0005476 0006273 0007137

(5) 0009173 0009656 0010226 0010879 0011585

(6) 0009177 0009669 0010249 0010916 0011637

y = infin (1) 0003115 0003730 0004463 0005308 0006224

(2) 0003121 0003746 0004493 0005356 0006291

(3) 0004346 0004938 0005641 0006450 0007325

(4) 0004351 0004954 0005670 0006496 0007389

(5) 0009499 0009999 0010589 0011265 0011996

(6) 0009503 0010012 0010613 0011303 0012050

24

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0031176 0032815 0034733 0036924 0039282

(2) 0031193 0032862 0034821 0037061 0039475

(3) 0050168 0051747 0053587 0055684 0057939

(4) 0050184 0051792 0053671 0055815 0058124

(5) 0136884 0138219 0139763 0141517 0143399

(6) 0136896 0138256 0139832 0141626 0143553

y = 25 (1) 0047622 0050092 0052981 0056280 0059829

(2) 0004765 0050163 0053113 0056487 0060121

(3) 0076570 0078950 0081720 0084878 0088272

(4) 0076593 0079017 0081847 0085076 0088551

(5) 0205774 0207786 0210111 0212752 0215585

(6) 0205793 0207842 0210216 0212917 0215818

y = 50 (1) 0055170 0058009 0061329 0065119 0069196

(2) 0055200 0058091 0061481 0065357 0069533

(3) 0088650 0091385 0094569 0098197 0102096

(4) 0088677 0091463 0094715 0098425 0102417

(5) 0235815 0238128 0240800 0243834 0247089

(6) 0235837 0238192 0240920 0244023 0247356

y = infin (1) 0058565 0061565 0065073 0069077 0073384

(2) 0058597 0061652 0065234 0069329 0073740

(3) 0094061 0096952 0100316 0104149 0108267

(4) 0094090 0097035 0100470 0104390 0108607

(5) 0248472 0250917 0253740 0256945 0260383

(6) 0248495 0250985 0253867 0257145 0260667

25

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0048598 0052524 0056984 0061998 0067342

(2) 0048654 0052673 0057253 0062411 0067916

(3) 0070192 0073979 0078257 0083057 0088167

(4) 0070243 0074119 0078513 0083451 0088714

(5) 0151774 0154981 0158572 0162586 0166852

(6) 0151815 0155096 0158784 0162913 0167308

y = 25 (1) 0076895 0083230 0090413 0098481 0107073

(2) 0076987 0083474 0090854 0099156 0108010

(3) 0110093 0116205 0123095 0130818 0139033

(4) 0110177 0116435 0123515 0131462 0139928

(5) 0230674 0235848 0241632 0248091 0254950

(6) 0230739 0236038 0241980 0248626 0255695

y = 50 (1) 0091496 0099156 0107829 0117563 0127924

(2) 0091610 0099455 0108370 0118387 0129068

(3) 0130130 0137520 0145839 0155156 0165064

(4) 0130233 0137801 0146352 0155942 0166156

(5) 0266599 0272857 0279840 0287632 0295904

(6) 0266680 0273088 0280265 0288286 0296814

y = infin (1) 0099312 0107746 0117285 0127984 0139369

(2) 0099439 0108078 0117883 0128899 0140639

(3) 0140458 0148595 0157745 0167986 0178872

(4) 0140572 0148908 0158314 0168859 0180084

(5) 0282886 0289776 0297457 0306022 0315111

(6) 0282976 0290034 0297929 0306748 0316120

26

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0059096 0072505 0088907 0108122 0129137

(2) 0059223 0072829 0089500 0109052 0130456

(3) 0079757 0092662 0108388 0126778 0146871

(4) 0079875 0092970 0108956 0127669 0148134

(5) 0159788 0170682 0183871 0199249 0216024

(6) 0159881 0170936 0184343 0199990 0217075

y = 25 (1) 0095883 0119741 0149240 0184073 0222390

(2) 0096087 0120262 0150197 0185573 0224520

(3) 0127403 0150359 0178640 0211977 0248613

(4) 0127592 0150855 0179557 0213414 0250653

(5) 0245161 0264533 0288250 0316124 0346709

(6) 0245311 0264942 0289010 0317319 0348407

y = 50 (1) 0116598 0147920 0187100 0233784 0285503

(2) 0116839 0148537 0188231 0235554 0288015

(3) 0153023 0183153 0220713 0265390 0314840

(4) 0153249 0183742 0221798 0267087 0317248

(5) 0285742 0311159 0342651 0380005 0421287

(6) 0285921 0311644 0343551 0381418 0423292

y = infin (1) 0129470 0167217 0215383 0273841 0339734

(2) 0129725 0167865 0216560 0275661 0342282

(3) 0167978 0204275 0250442 0306382 0369383

(4) 0168217 0204895 0251572 0308128 0371826

(5) 0305868 0336470 0375169 0421935 0474525

(6) 0306058 0336980 0376106 0423388 0476560

27

Page 21: uwaterloo.ca · Ruin Analysis of a Threshold Strategy in a Discrete-time Sparre Andersen Model Steve DREKIC and Ana Maria MERA Department of Statistics and Actuarial Science University

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0058559 0061875 0065704 0070048 0074714

(2) 0058569 0061903 0065756 0070128 0074825

(3) 0058574 0061917 0065782 0070167 0074880

(4) 0058584 0061945 0065833 0070245 0074987

(5) 0058596 0061979 0065896 0070344 0075124

(6) 0058632 0062079 0066078 0070621 0075506

y = 25 (1) 0089002 0093986 0099739 0106262 0113270

(2) 0089017 0094028 0099817 0106382 0113437

(3) 0089025 0094049 0099856 0106442 0113519

(4) 0089040 0094091 0099932 0106560 0113681

(5) 0089057 0094142 0100028 0106708 0113887

(6) 0089113 0094294 0100302 0107127 0114463

y = 50 (1) 0102786 0108506 0115106 0122590 0130629

(2) 0102804 0108555 0115196 0122728 0130820

(3) 0102812 0108579 0115241 0122797 0130914

(4) 0102830 0108627 0115329 0122932 0131101

(5) 0102850 0108686 0115439 0123103 0131337

(6) 0102914 0108860 0115754 0123584 0132000

y = infin (1) 0108895 0114933 0121901 0129799 0138283

(2) 0108914 0114985 0121996 0129945 0138485

(3) 0108923 0115010 0122043 0130018 0138585

(4) 0108941 0115061 0122136 0130161 0138782

(5) 0108962 0115123 0122252 0130341 0139032

(6) 0109030 0115307 0122585 0130849 0139732

21

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0079856 0086792 009464 0103440 0112831

(2) 0079885 0086869 0094778 0103650 0113118

(3) 0079953 0086978 0094936 0103864 0113393

(4) 0080081 0087188 0095243 0104282 0113929

(5) 0079962 0087075 0095149 0104213 0113890

(6) 0080805 0088366 0096951 0106588 0116875

y = 25 (1) 0124591 0135680 0148207 0162245 0177217

(2) 0124639 0135806 0148433 0162586 0177683

(3) 0124755 0135993 0148703 0162950 0178148

(4) 0124978 0136354 0149226 0163658 0179054

(5) 0124763 0136141 0149035 0163500 0178935

(6) 0126238 0138381 0152148 0167588 0184060

y = 50 (1) 0146757 0160079 0175111 0191945 0209894

(2) 0146816 0160232 0175385 0192360 0210460

(3) 0146965 0160469 0175725 0192817 0211043

(4) 0147249 0160925 0176384 0193706 0212178

(5) 0146968 0160639 0176117 0193470 0211980

(6) 0148860 0163500 0180077 0198658 0218474

y = infin (1) 0158013 0172606 0189059 0207476 0227106

(2) 0158078 0172776 0189362 0207934 0227732

(3) 0158250 0173047 0189751 0208455 0228395

(4) 0158579 0173571 0190503 0209467 0229685

(5) 0158247 0173226 0190172 0209161 0229410

(6) 0160443 0176532 0194735 0215127 0236869

22

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0088667 0104403 0123329 0145277 0169184

(2) 0088711 0104518 0123538 0145598 0169627

(3) 0088733 0104574 0123641 0145756 0169846

(4) 0088777 0104687 0123844 0146066 0170274

(5) 0088829 0104828 0124105 0146470 0170832

(6) 0088988 0105229 0124822 0147560 0172331

y = 25 (1) 0140528 0167839 0201007 0239726 0282102

(2) 0140598 0168021 0201338 0240234 0282804

(3) 0140632 0168111 0201502 0240486 0283151

(4) 0140701 0168289 0201822 0240976 0283827

(5) 0140785 0168515 0202241 0241622 0284718

(6) 0141036 0169148 0203370 0243335 0287071

y = 50 (1) 0167826 0203005 0246173 0296954 0352855

(2) 0167908 0203218 0246561 0297548 0353672

(3) 0167949 0203324 0246753 0297841 0354075

(4) 0168030 0203532 0247128 0298412 0354859

(5) 0168129 0203800 0247622 0299174 0355907

(6) 0168423 0204539 0248936 0301159 0358620

y = infin (1) 0183325 0224926 0276881 0338957 0408257

(2) 0183412 0225150 0277283 0339564 0409079

(3) 0183455 0225260 0277481 0339864 0409485

(4) 0183540 0225478 0277869 0340446 0410270

(5) 0183645 0225760 0278386 0341234 0411339

(6) 0183954 0226529 0279737 0343244 0414037

23

Table 2 ndash Values of Ψnxy(10) corresponding to Example 2

(a) x = 10

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0001901 0002276 0002724 0003240 0003799

(2) 0001904 0002286 0002742 0003269 0003840

(3) 0002651 0003013 0003442 0003936 0004471

(4) 0002655 0003022 0003460 0003964 0004510

(5) 0005792 0006098 0006458 0006871 0007317

(6) 0005795 0006106 0006473 0006894 0007350

y = 25 (1) 0002708 0003243 0003880 0004615 0005411

(2) 0002713 0003257 0003906 0004656 0005469

(3) 0003778 0004293 0004904 0005608 0006369

(4) 0003782 0004306 0004929 0005647 0006424

(5) 0008255 0008690 0009203 0009792 0010427

(6) 0008259 0008702 0009224 0009824 0010473

y = 50 (1) 0003008 0003603 0004310 0005127 0006011

(2) 0003014 0003618 0004340 0005173 0006076

(3) 0004197 0004769 0005448 0006230 0007075

(4) 0004202 0004784 0005476 0006273 0007137

(5) 0009173 0009656 0010226 0010879 0011585

(6) 0009177 0009669 0010249 0010916 0011637

y = infin (1) 0003115 0003730 0004463 0005308 0006224

(2) 0003121 0003746 0004493 0005356 0006291

(3) 0004346 0004938 0005641 0006450 0007325

(4) 0004351 0004954 0005670 0006496 0007389

(5) 0009499 0009999 0010589 0011265 0011996

(6) 0009503 0010012 0010613 0011303 0012050

24

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0031176 0032815 0034733 0036924 0039282

(2) 0031193 0032862 0034821 0037061 0039475

(3) 0050168 0051747 0053587 0055684 0057939

(4) 0050184 0051792 0053671 0055815 0058124

(5) 0136884 0138219 0139763 0141517 0143399

(6) 0136896 0138256 0139832 0141626 0143553

y = 25 (1) 0047622 0050092 0052981 0056280 0059829

(2) 0004765 0050163 0053113 0056487 0060121

(3) 0076570 0078950 0081720 0084878 0088272

(4) 0076593 0079017 0081847 0085076 0088551

(5) 0205774 0207786 0210111 0212752 0215585

(6) 0205793 0207842 0210216 0212917 0215818

y = 50 (1) 0055170 0058009 0061329 0065119 0069196

(2) 0055200 0058091 0061481 0065357 0069533

(3) 0088650 0091385 0094569 0098197 0102096

(4) 0088677 0091463 0094715 0098425 0102417

(5) 0235815 0238128 0240800 0243834 0247089

(6) 0235837 0238192 0240920 0244023 0247356

y = infin (1) 0058565 0061565 0065073 0069077 0073384

(2) 0058597 0061652 0065234 0069329 0073740

(3) 0094061 0096952 0100316 0104149 0108267

(4) 0094090 0097035 0100470 0104390 0108607

(5) 0248472 0250917 0253740 0256945 0260383

(6) 0248495 0250985 0253867 0257145 0260667

25

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0048598 0052524 0056984 0061998 0067342

(2) 0048654 0052673 0057253 0062411 0067916

(3) 0070192 0073979 0078257 0083057 0088167

(4) 0070243 0074119 0078513 0083451 0088714

(5) 0151774 0154981 0158572 0162586 0166852

(6) 0151815 0155096 0158784 0162913 0167308

y = 25 (1) 0076895 0083230 0090413 0098481 0107073

(2) 0076987 0083474 0090854 0099156 0108010

(3) 0110093 0116205 0123095 0130818 0139033

(4) 0110177 0116435 0123515 0131462 0139928

(5) 0230674 0235848 0241632 0248091 0254950

(6) 0230739 0236038 0241980 0248626 0255695

y = 50 (1) 0091496 0099156 0107829 0117563 0127924

(2) 0091610 0099455 0108370 0118387 0129068

(3) 0130130 0137520 0145839 0155156 0165064

(4) 0130233 0137801 0146352 0155942 0166156

(5) 0266599 0272857 0279840 0287632 0295904

(6) 0266680 0273088 0280265 0288286 0296814

y = infin (1) 0099312 0107746 0117285 0127984 0139369

(2) 0099439 0108078 0117883 0128899 0140639

(3) 0140458 0148595 0157745 0167986 0178872

(4) 0140572 0148908 0158314 0168859 0180084

(5) 0282886 0289776 0297457 0306022 0315111

(6) 0282976 0290034 0297929 0306748 0316120

26

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0059096 0072505 0088907 0108122 0129137

(2) 0059223 0072829 0089500 0109052 0130456

(3) 0079757 0092662 0108388 0126778 0146871

(4) 0079875 0092970 0108956 0127669 0148134

(5) 0159788 0170682 0183871 0199249 0216024

(6) 0159881 0170936 0184343 0199990 0217075

y = 25 (1) 0095883 0119741 0149240 0184073 0222390

(2) 0096087 0120262 0150197 0185573 0224520

(3) 0127403 0150359 0178640 0211977 0248613

(4) 0127592 0150855 0179557 0213414 0250653

(5) 0245161 0264533 0288250 0316124 0346709

(6) 0245311 0264942 0289010 0317319 0348407

y = 50 (1) 0116598 0147920 0187100 0233784 0285503

(2) 0116839 0148537 0188231 0235554 0288015

(3) 0153023 0183153 0220713 0265390 0314840

(4) 0153249 0183742 0221798 0267087 0317248

(5) 0285742 0311159 0342651 0380005 0421287

(6) 0285921 0311644 0343551 0381418 0423292

y = infin (1) 0129470 0167217 0215383 0273841 0339734

(2) 0129725 0167865 0216560 0275661 0342282

(3) 0167978 0204275 0250442 0306382 0369383

(4) 0168217 0204895 0251572 0308128 0371826

(5) 0305868 0336470 0375169 0421935 0474525

(6) 0306058 0336980 0376106 0423388 0476560

27

Page 22: uwaterloo.ca · Ruin Analysis of a Threshold Strategy in a Discrete-time Sparre Andersen Model Steve DREKIC and Ana Maria MERA Department of Statistics and Actuarial Science University

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0079856 0086792 009464 0103440 0112831

(2) 0079885 0086869 0094778 0103650 0113118

(3) 0079953 0086978 0094936 0103864 0113393

(4) 0080081 0087188 0095243 0104282 0113929

(5) 0079962 0087075 0095149 0104213 0113890

(6) 0080805 0088366 0096951 0106588 0116875

y = 25 (1) 0124591 0135680 0148207 0162245 0177217

(2) 0124639 0135806 0148433 0162586 0177683

(3) 0124755 0135993 0148703 0162950 0178148

(4) 0124978 0136354 0149226 0163658 0179054

(5) 0124763 0136141 0149035 0163500 0178935

(6) 0126238 0138381 0152148 0167588 0184060

y = 50 (1) 0146757 0160079 0175111 0191945 0209894

(2) 0146816 0160232 0175385 0192360 0210460

(3) 0146965 0160469 0175725 0192817 0211043

(4) 0147249 0160925 0176384 0193706 0212178

(5) 0146968 0160639 0176117 0193470 0211980

(6) 0148860 0163500 0180077 0198658 0218474

y = infin (1) 0158013 0172606 0189059 0207476 0227106

(2) 0158078 0172776 0189362 0207934 0227732

(3) 0158250 0173047 0189751 0208455 0228395

(4) 0158579 0173571 0190503 0209467 0229685

(5) 0158247 0173226 0190172 0209161 0229410

(6) 0160443 0176532 0194735 0215127 0236869

22

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0088667 0104403 0123329 0145277 0169184

(2) 0088711 0104518 0123538 0145598 0169627

(3) 0088733 0104574 0123641 0145756 0169846

(4) 0088777 0104687 0123844 0146066 0170274

(5) 0088829 0104828 0124105 0146470 0170832

(6) 0088988 0105229 0124822 0147560 0172331

y = 25 (1) 0140528 0167839 0201007 0239726 0282102

(2) 0140598 0168021 0201338 0240234 0282804

(3) 0140632 0168111 0201502 0240486 0283151

(4) 0140701 0168289 0201822 0240976 0283827

(5) 0140785 0168515 0202241 0241622 0284718

(6) 0141036 0169148 0203370 0243335 0287071

y = 50 (1) 0167826 0203005 0246173 0296954 0352855

(2) 0167908 0203218 0246561 0297548 0353672

(3) 0167949 0203324 0246753 0297841 0354075

(4) 0168030 0203532 0247128 0298412 0354859

(5) 0168129 0203800 0247622 0299174 0355907

(6) 0168423 0204539 0248936 0301159 0358620

y = infin (1) 0183325 0224926 0276881 0338957 0408257

(2) 0183412 0225150 0277283 0339564 0409079

(3) 0183455 0225260 0277481 0339864 0409485

(4) 0183540 0225478 0277869 0340446 0410270

(5) 0183645 0225760 0278386 0341234 0411339

(6) 0183954 0226529 0279737 0343244 0414037

23

Table 2 ndash Values of Ψnxy(10) corresponding to Example 2

(a) x = 10

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0001901 0002276 0002724 0003240 0003799

(2) 0001904 0002286 0002742 0003269 0003840

(3) 0002651 0003013 0003442 0003936 0004471

(4) 0002655 0003022 0003460 0003964 0004510

(5) 0005792 0006098 0006458 0006871 0007317

(6) 0005795 0006106 0006473 0006894 0007350

y = 25 (1) 0002708 0003243 0003880 0004615 0005411

(2) 0002713 0003257 0003906 0004656 0005469

(3) 0003778 0004293 0004904 0005608 0006369

(4) 0003782 0004306 0004929 0005647 0006424

(5) 0008255 0008690 0009203 0009792 0010427

(6) 0008259 0008702 0009224 0009824 0010473

y = 50 (1) 0003008 0003603 0004310 0005127 0006011

(2) 0003014 0003618 0004340 0005173 0006076

(3) 0004197 0004769 0005448 0006230 0007075

(4) 0004202 0004784 0005476 0006273 0007137

(5) 0009173 0009656 0010226 0010879 0011585

(6) 0009177 0009669 0010249 0010916 0011637

y = infin (1) 0003115 0003730 0004463 0005308 0006224

(2) 0003121 0003746 0004493 0005356 0006291

(3) 0004346 0004938 0005641 0006450 0007325

(4) 0004351 0004954 0005670 0006496 0007389

(5) 0009499 0009999 0010589 0011265 0011996

(6) 0009503 0010012 0010613 0011303 0012050

24

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0031176 0032815 0034733 0036924 0039282

(2) 0031193 0032862 0034821 0037061 0039475

(3) 0050168 0051747 0053587 0055684 0057939

(4) 0050184 0051792 0053671 0055815 0058124

(5) 0136884 0138219 0139763 0141517 0143399

(6) 0136896 0138256 0139832 0141626 0143553

y = 25 (1) 0047622 0050092 0052981 0056280 0059829

(2) 0004765 0050163 0053113 0056487 0060121

(3) 0076570 0078950 0081720 0084878 0088272

(4) 0076593 0079017 0081847 0085076 0088551

(5) 0205774 0207786 0210111 0212752 0215585

(6) 0205793 0207842 0210216 0212917 0215818

y = 50 (1) 0055170 0058009 0061329 0065119 0069196

(2) 0055200 0058091 0061481 0065357 0069533

(3) 0088650 0091385 0094569 0098197 0102096

(4) 0088677 0091463 0094715 0098425 0102417

(5) 0235815 0238128 0240800 0243834 0247089

(6) 0235837 0238192 0240920 0244023 0247356

y = infin (1) 0058565 0061565 0065073 0069077 0073384

(2) 0058597 0061652 0065234 0069329 0073740

(3) 0094061 0096952 0100316 0104149 0108267

(4) 0094090 0097035 0100470 0104390 0108607

(5) 0248472 0250917 0253740 0256945 0260383

(6) 0248495 0250985 0253867 0257145 0260667

25

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0048598 0052524 0056984 0061998 0067342

(2) 0048654 0052673 0057253 0062411 0067916

(3) 0070192 0073979 0078257 0083057 0088167

(4) 0070243 0074119 0078513 0083451 0088714

(5) 0151774 0154981 0158572 0162586 0166852

(6) 0151815 0155096 0158784 0162913 0167308

y = 25 (1) 0076895 0083230 0090413 0098481 0107073

(2) 0076987 0083474 0090854 0099156 0108010

(3) 0110093 0116205 0123095 0130818 0139033

(4) 0110177 0116435 0123515 0131462 0139928

(5) 0230674 0235848 0241632 0248091 0254950

(6) 0230739 0236038 0241980 0248626 0255695

y = 50 (1) 0091496 0099156 0107829 0117563 0127924

(2) 0091610 0099455 0108370 0118387 0129068

(3) 0130130 0137520 0145839 0155156 0165064

(4) 0130233 0137801 0146352 0155942 0166156

(5) 0266599 0272857 0279840 0287632 0295904

(6) 0266680 0273088 0280265 0288286 0296814

y = infin (1) 0099312 0107746 0117285 0127984 0139369

(2) 0099439 0108078 0117883 0128899 0140639

(3) 0140458 0148595 0157745 0167986 0178872

(4) 0140572 0148908 0158314 0168859 0180084

(5) 0282886 0289776 0297457 0306022 0315111

(6) 0282976 0290034 0297929 0306748 0316120

26

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0059096 0072505 0088907 0108122 0129137

(2) 0059223 0072829 0089500 0109052 0130456

(3) 0079757 0092662 0108388 0126778 0146871

(4) 0079875 0092970 0108956 0127669 0148134

(5) 0159788 0170682 0183871 0199249 0216024

(6) 0159881 0170936 0184343 0199990 0217075

y = 25 (1) 0095883 0119741 0149240 0184073 0222390

(2) 0096087 0120262 0150197 0185573 0224520

(3) 0127403 0150359 0178640 0211977 0248613

(4) 0127592 0150855 0179557 0213414 0250653

(5) 0245161 0264533 0288250 0316124 0346709

(6) 0245311 0264942 0289010 0317319 0348407

y = 50 (1) 0116598 0147920 0187100 0233784 0285503

(2) 0116839 0148537 0188231 0235554 0288015

(3) 0153023 0183153 0220713 0265390 0314840

(4) 0153249 0183742 0221798 0267087 0317248

(5) 0285742 0311159 0342651 0380005 0421287

(6) 0285921 0311644 0343551 0381418 0423292

y = infin (1) 0129470 0167217 0215383 0273841 0339734

(2) 0129725 0167865 0216560 0275661 0342282

(3) 0167978 0204275 0250442 0306382 0369383

(4) 0168217 0204895 0251572 0308128 0371826

(5) 0305868 0336470 0375169 0421935 0474525

(6) 0306058 0336980 0376106 0423388 0476560

27

Page 23: uwaterloo.ca · Ruin Analysis of a Threshold Strategy in a Discrete-time Sparre Andersen Model Steve DREKIC and Ana Maria MERA Department of Statistics and Actuarial Science University

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0088667 0104403 0123329 0145277 0169184

(2) 0088711 0104518 0123538 0145598 0169627

(3) 0088733 0104574 0123641 0145756 0169846

(4) 0088777 0104687 0123844 0146066 0170274

(5) 0088829 0104828 0124105 0146470 0170832

(6) 0088988 0105229 0124822 0147560 0172331

y = 25 (1) 0140528 0167839 0201007 0239726 0282102

(2) 0140598 0168021 0201338 0240234 0282804

(3) 0140632 0168111 0201502 0240486 0283151

(4) 0140701 0168289 0201822 0240976 0283827

(5) 0140785 0168515 0202241 0241622 0284718

(6) 0141036 0169148 0203370 0243335 0287071

y = 50 (1) 0167826 0203005 0246173 0296954 0352855

(2) 0167908 0203218 0246561 0297548 0353672

(3) 0167949 0203324 0246753 0297841 0354075

(4) 0168030 0203532 0247128 0298412 0354859

(5) 0168129 0203800 0247622 0299174 0355907

(6) 0168423 0204539 0248936 0301159 0358620

y = infin (1) 0183325 0224926 0276881 0338957 0408257

(2) 0183412 0225150 0277283 0339564 0409079

(3) 0183455 0225260 0277481 0339864 0409485

(4) 0183540 0225478 0277869 0340446 0410270

(5) 0183645 0225760 0278386 0341234 0411339

(6) 0183954 0226529 0279737 0343244 0414037

23

Table 2 ndash Values of Ψnxy(10) corresponding to Example 2

(a) x = 10

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0001901 0002276 0002724 0003240 0003799

(2) 0001904 0002286 0002742 0003269 0003840

(3) 0002651 0003013 0003442 0003936 0004471

(4) 0002655 0003022 0003460 0003964 0004510

(5) 0005792 0006098 0006458 0006871 0007317

(6) 0005795 0006106 0006473 0006894 0007350

y = 25 (1) 0002708 0003243 0003880 0004615 0005411

(2) 0002713 0003257 0003906 0004656 0005469

(3) 0003778 0004293 0004904 0005608 0006369

(4) 0003782 0004306 0004929 0005647 0006424

(5) 0008255 0008690 0009203 0009792 0010427

(6) 0008259 0008702 0009224 0009824 0010473

y = 50 (1) 0003008 0003603 0004310 0005127 0006011

(2) 0003014 0003618 0004340 0005173 0006076

(3) 0004197 0004769 0005448 0006230 0007075

(4) 0004202 0004784 0005476 0006273 0007137

(5) 0009173 0009656 0010226 0010879 0011585

(6) 0009177 0009669 0010249 0010916 0011637

y = infin (1) 0003115 0003730 0004463 0005308 0006224

(2) 0003121 0003746 0004493 0005356 0006291

(3) 0004346 0004938 0005641 0006450 0007325

(4) 0004351 0004954 0005670 0006496 0007389

(5) 0009499 0009999 0010589 0011265 0011996

(6) 0009503 0010012 0010613 0011303 0012050

24

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0031176 0032815 0034733 0036924 0039282

(2) 0031193 0032862 0034821 0037061 0039475

(3) 0050168 0051747 0053587 0055684 0057939

(4) 0050184 0051792 0053671 0055815 0058124

(5) 0136884 0138219 0139763 0141517 0143399

(6) 0136896 0138256 0139832 0141626 0143553

y = 25 (1) 0047622 0050092 0052981 0056280 0059829

(2) 0004765 0050163 0053113 0056487 0060121

(3) 0076570 0078950 0081720 0084878 0088272

(4) 0076593 0079017 0081847 0085076 0088551

(5) 0205774 0207786 0210111 0212752 0215585

(6) 0205793 0207842 0210216 0212917 0215818

y = 50 (1) 0055170 0058009 0061329 0065119 0069196

(2) 0055200 0058091 0061481 0065357 0069533

(3) 0088650 0091385 0094569 0098197 0102096

(4) 0088677 0091463 0094715 0098425 0102417

(5) 0235815 0238128 0240800 0243834 0247089

(6) 0235837 0238192 0240920 0244023 0247356

y = infin (1) 0058565 0061565 0065073 0069077 0073384

(2) 0058597 0061652 0065234 0069329 0073740

(3) 0094061 0096952 0100316 0104149 0108267

(4) 0094090 0097035 0100470 0104390 0108607

(5) 0248472 0250917 0253740 0256945 0260383

(6) 0248495 0250985 0253867 0257145 0260667

25

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0048598 0052524 0056984 0061998 0067342

(2) 0048654 0052673 0057253 0062411 0067916

(3) 0070192 0073979 0078257 0083057 0088167

(4) 0070243 0074119 0078513 0083451 0088714

(5) 0151774 0154981 0158572 0162586 0166852

(6) 0151815 0155096 0158784 0162913 0167308

y = 25 (1) 0076895 0083230 0090413 0098481 0107073

(2) 0076987 0083474 0090854 0099156 0108010

(3) 0110093 0116205 0123095 0130818 0139033

(4) 0110177 0116435 0123515 0131462 0139928

(5) 0230674 0235848 0241632 0248091 0254950

(6) 0230739 0236038 0241980 0248626 0255695

y = 50 (1) 0091496 0099156 0107829 0117563 0127924

(2) 0091610 0099455 0108370 0118387 0129068

(3) 0130130 0137520 0145839 0155156 0165064

(4) 0130233 0137801 0146352 0155942 0166156

(5) 0266599 0272857 0279840 0287632 0295904

(6) 0266680 0273088 0280265 0288286 0296814

y = infin (1) 0099312 0107746 0117285 0127984 0139369

(2) 0099439 0108078 0117883 0128899 0140639

(3) 0140458 0148595 0157745 0167986 0178872

(4) 0140572 0148908 0158314 0168859 0180084

(5) 0282886 0289776 0297457 0306022 0315111

(6) 0282976 0290034 0297929 0306748 0316120

26

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0059096 0072505 0088907 0108122 0129137

(2) 0059223 0072829 0089500 0109052 0130456

(3) 0079757 0092662 0108388 0126778 0146871

(4) 0079875 0092970 0108956 0127669 0148134

(5) 0159788 0170682 0183871 0199249 0216024

(6) 0159881 0170936 0184343 0199990 0217075

y = 25 (1) 0095883 0119741 0149240 0184073 0222390

(2) 0096087 0120262 0150197 0185573 0224520

(3) 0127403 0150359 0178640 0211977 0248613

(4) 0127592 0150855 0179557 0213414 0250653

(5) 0245161 0264533 0288250 0316124 0346709

(6) 0245311 0264942 0289010 0317319 0348407

y = 50 (1) 0116598 0147920 0187100 0233784 0285503

(2) 0116839 0148537 0188231 0235554 0288015

(3) 0153023 0183153 0220713 0265390 0314840

(4) 0153249 0183742 0221798 0267087 0317248

(5) 0285742 0311159 0342651 0380005 0421287

(6) 0285921 0311644 0343551 0381418 0423292

y = infin (1) 0129470 0167217 0215383 0273841 0339734

(2) 0129725 0167865 0216560 0275661 0342282

(3) 0167978 0204275 0250442 0306382 0369383

(4) 0168217 0204895 0251572 0308128 0371826

(5) 0305868 0336470 0375169 0421935 0474525

(6) 0306058 0336980 0376106 0423388 0476560

27

Page 24: uwaterloo.ca · Ruin Analysis of a Threshold Strategy in a Discrete-time Sparre Andersen Model Steve DREKIC and Ana Maria MERA Department of Statistics and Actuarial Science University

Table 2 ndash Values of Ψnxy(10) corresponding to Example 2

(a) x = 10

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0001901 0002276 0002724 0003240 0003799

(2) 0001904 0002286 0002742 0003269 0003840

(3) 0002651 0003013 0003442 0003936 0004471

(4) 0002655 0003022 0003460 0003964 0004510

(5) 0005792 0006098 0006458 0006871 0007317

(6) 0005795 0006106 0006473 0006894 0007350

y = 25 (1) 0002708 0003243 0003880 0004615 0005411

(2) 0002713 0003257 0003906 0004656 0005469

(3) 0003778 0004293 0004904 0005608 0006369

(4) 0003782 0004306 0004929 0005647 0006424

(5) 0008255 0008690 0009203 0009792 0010427

(6) 0008259 0008702 0009224 0009824 0010473

y = 50 (1) 0003008 0003603 0004310 0005127 0006011

(2) 0003014 0003618 0004340 0005173 0006076

(3) 0004197 0004769 0005448 0006230 0007075

(4) 0004202 0004784 0005476 0006273 0007137

(5) 0009173 0009656 0010226 0010879 0011585

(6) 0009177 0009669 0010249 0010916 0011637

y = infin (1) 0003115 0003730 0004463 0005308 0006224

(2) 0003121 0003746 0004493 0005356 0006291

(3) 0004346 0004938 0005641 0006450 0007325

(4) 0004351 0004954 0005670 0006496 0007389

(5) 0009499 0009999 0010589 0011265 0011996

(6) 0009503 0010012 0010613 0011303 0012050

24

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0031176 0032815 0034733 0036924 0039282

(2) 0031193 0032862 0034821 0037061 0039475

(3) 0050168 0051747 0053587 0055684 0057939

(4) 0050184 0051792 0053671 0055815 0058124

(5) 0136884 0138219 0139763 0141517 0143399

(6) 0136896 0138256 0139832 0141626 0143553

y = 25 (1) 0047622 0050092 0052981 0056280 0059829

(2) 0004765 0050163 0053113 0056487 0060121

(3) 0076570 0078950 0081720 0084878 0088272

(4) 0076593 0079017 0081847 0085076 0088551

(5) 0205774 0207786 0210111 0212752 0215585

(6) 0205793 0207842 0210216 0212917 0215818

y = 50 (1) 0055170 0058009 0061329 0065119 0069196

(2) 0055200 0058091 0061481 0065357 0069533

(3) 0088650 0091385 0094569 0098197 0102096

(4) 0088677 0091463 0094715 0098425 0102417

(5) 0235815 0238128 0240800 0243834 0247089

(6) 0235837 0238192 0240920 0244023 0247356

y = infin (1) 0058565 0061565 0065073 0069077 0073384

(2) 0058597 0061652 0065234 0069329 0073740

(3) 0094061 0096952 0100316 0104149 0108267

(4) 0094090 0097035 0100470 0104390 0108607

(5) 0248472 0250917 0253740 0256945 0260383

(6) 0248495 0250985 0253867 0257145 0260667

25

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0048598 0052524 0056984 0061998 0067342

(2) 0048654 0052673 0057253 0062411 0067916

(3) 0070192 0073979 0078257 0083057 0088167

(4) 0070243 0074119 0078513 0083451 0088714

(5) 0151774 0154981 0158572 0162586 0166852

(6) 0151815 0155096 0158784 0162913 0167308

y = 25 (1) 0076895 0083230 0090413 0098481 0107073

(2) 0076987 0083474 0090854 0099156 0108010

(3) 0110093 0116205 0123095 0130818 0139033

(4) 0110177 0116435 0123515 0131462 0139928

(5) 0230674 0235848 0241632 0248091 0254950

(6) 0230739 0236038 0241980 0248626 0255695

y = 50 (1) 0091496 0099156 0107829 0117563 0127924

(2) 0091610 0099455 0108370 0118387 0129068

(3) 0130130 0137520 0145839 0155156 0165064

(4) 0130233 0137801 0146352 0155942 0166156

(5) 0266599 0272857 0279840 0287632 0295904

(6) 0266680 0273088 0280265 0288286 0296814

y = infin (1) 0099312 0107746 0117285 0127984 0139369

(2) 0099439 0108078 0117883 0128899 0140639

(3) 0140458 0148595 0157745 0167986 0178872

(4) 0140572 0148908 0158314 0168859 0180084

(5) 0282886 0289776 0297457 0306022 0315111

(6) 0282976 0290034 0297929 0306748 0316120

26

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0059096 0072505 0088907 0108122 0129137

(2) 0059223 0072829 0089500 0109052 0130456

(3) 0079757 0092662 0108388 0126778 0146871

(4) 0079875 0092970 0108956 0127669 0148134

(5) 0159788 0170682 0183871 0199249 0216024

(6) 0159881 0170936 0184343 0199990 0217075

y = 25 (1) 0095883 0119741 0149240 0184073 0222390

(2) 0096087 0120262 0150197 0185573 0224520

(3) 0127403 0150359 0178640 0211977 0248613

(4) 0127592 0150855 0179557 0213414 0250653

(5) 0245161 0264533 0288250 0316124 0346709

(6) 0245311 0264942 0289010 0317319 0348407

y = 50 (1) 0116598 0147920 0187100 0233784 0285503

(2) 0116839 0148537 0188231 0235554 0288015

(3) 0153023 0183153 0220713 0265390 0314840

(4) 0153249 0183742 0221798 0267087 0317248

(5) 0285742 0311159 0342651 0380005 0421287

(6) 0285921 0311644 0343551 0381418 0423292

y = infin (1) 0129470 0167217 0215383 0273841 0339734

(2) 0129725 0167865 0216560 0275661 0342282

(3) 0167978 0204275 0250442 0306382 0369383

(4) 0168217 0204895 0251572 0308128 0371826

(5) 0305868 0336470 0375169 0421935 0474525

(6) 0306058 0336980 0376106 0423388 0476560

27

Page 25: uwaterloo.ca · Ruin Analysis of a Threshold Strategy in a Discrete-time Sparre Andersen Model Steve DREKIC and Ana Maria MERA Department of Statistics and Actuarial Science University

(b) x = 25

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0031176 0032815 0034733 0036924 0039282

(2) 0031193 0032862 0034821 0037061 0039475

(3) 0050168 0051747 0053587 0055684 0057939

(4) 0050184 0051792 0053671 0055815 0058124

(5) 0136884 0138219 0139763 0141517 0143399

(6) 0136896 0138256 0139832 0141626 0143553

y = 25 (1) 0047622 0050092 0052981 0056280 0059829

(2) 0004765 0050163 0053113 0056487 0060121

(3) 0076570 0078950 0081720 0084878 0088272

(4) 0076593 0079017 0081847 0085076 0088551

(5) 0205774 0207786 0210111 0212752 0215585

(6) 0205793 0207842 0210216 0212917 0215818

y = 50 (1) 0055170 0058009 0061329 0065119 0069196

(2) 0055200 0058091 0061481 0065357 0069533

(3) 0088650 0091385 0094569 0098197 0102096

(4) 0088677 0091463 0094715 0098425 0102417

(5) 0235815 0238128 0240800 0243834 0247089

(6) 0235837 0238192 0240920 0244023 0247356

y = infin (1) 0058565 0061565 0065073 0069077 0073384

(2) 0058597 0061652 0065234 0069329 0073740

(3) 0094061 0096952 0100316 0104149 0108267

(4) 0094090 0097035 0100470 0104390 0108607

(5) 0248472 0250917 0253740 0256945 0260383

(6) 0248495 0250985 0253867 0257145 0260667

25

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0048598 0052524 0056984 0061998 0067342

(2) 0048654 0052673 0057253 0062411 0067916

(3) 0070192 0073979 0078257 0083057 0088167

(4) 0070243 0074119 0078513 0083451 0088714

(5) 0151774 0154981 0158572 0162586 0166852

(6) 0151815 0155096 0158784 0162913 0167308

y = 25 (1) 0076895 0083230 0090413 0098481 0107073

(2) 0076987 0083474 0090854 0099156 0108010

(3) 0110093 0116205 0123095 0130818 0139033

(4) 0110177 0116435 0123515 0131462 0139928

(5) 0230674 0235848 0241632 0248091 0254950

(6) 0230739 0236038 0241980 0248626 0255695

y = 50 (1) 0091496 0099156 0107829 0117563 0127924

(2) 0091610 0099455 0108370 0118387 0129068

(3) 0130130 0137520 0145839 0155156 0165064

(4) 0130233 0137801 0146352 0155942 0166156

(5) 0266599 0272857 0279840 0287632 0295904

(6) 0266680 0273088 0280265 0288286 0296814

y = infin (1) 0099312 0107746 0117285 0127984 0139369

(2) 0099439 0108078 0117883 0128899 0140639

(3) 0140458 0148595 0157745 0167986 0178872

(4) 0140572 0148908 0158314 0168859 0180084

(5) 0282886 0289776 0297457 0306022 0315111

(6) 0282976 0290034 0297929 0306748 0316120

26

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0059096 0072505 0088907 0108122 0129137

(2) 0059223 0072829 0089500 0109052 0130456

(3) 0079757 0092662 0108388 0126778 0146871

(4) 0079875 0092970 0108956 0127669 0148134

(5) 0159788 0170682 0183871 0199249 0216024

(6) 0159881 0170936 0184343 0199990 0217075

y = 25 (1) 0095883 0119741 0149240 0184073 0222390

(2) 0096087 0120262 0150197 0185573 0224520

(3) 0127403 0150359 0178640 0211977 0248613

(4) 0127592 0150855 0179557 0213414 0250653

(5) 0245161 0264533 0288250 0316124 0346709

(6) 0245311 0264942 0289010 0317319 0348407

y = 50 (1) 0116598 0147920 0187100 0233784 0285503

(2) 0116839 0148537 0188231 0235554 0288015

(3) 0153023 0183153 0220713 0265390 0314840

(4) 0153249 0183742 0221798 0267087 0317248

(5) 0285742 0311159 0342651 0380005 0421287

(6) 0285921 0311644 0343551 0381418 0423292

y = infin (1) 0129470 0167217 0215383 0273841 0339734

(2) 0129725 0167865 0216560 0275661 0342282

(3) 0167978 0204275 0250442 0306382 0369383

(4) 0168217 0204895 0251572 0308128 0371826

(5) 0305868 0336470 0375169 0421935 0474525

(6) 0306058 0336980 0376106 0423388 0476560

27

Page 26: uwaterloo.ca · Ruin Analysis of a Threshold Strategy in a Discrete-time Sparre Andersen Model Steve DREKIC and Ana Maria MERA Department of Statistics and Actuarial Science University

(c) x = 50

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0048598 0052524 0056984 0061998 0067342

(2) 0048654 0052673 0057253 0062411 0067916

(3) 0070192 0073979 0078257 0083057 0088167

(4) 0070243 0074119 0078513 0083451 0088714

(5) 0151774 0154981 0158572 0162586 0166852

(6) 0151815 0155096 0158784 0162913 0167308

y = 25 (1) 0076895 0083230 0090413 0098481 0107073

(2) 0076987 0083474 0090854 0099156 0108010

(3) 0110093 0116205 0123095 0130818 0139033

(4) 0110177 0116435 0123515 0131462 0139928

(5) 0230674 0235848 0241632 0248091 0254950

(6) 0230739 0236038 0241980 0248626 0255695

y = 50 (1) 0091496 0099156 0107829 0117563 0127924

(2) 0091610 0099455 0108370 0118387 0129068

(3) 0130130 0137520 0145839 0155156 0165064

(4) 0130233 0137801 0146352 0155942 0166156

(5) 0266599 0272857 0279840 0287632 0295904

(6) 0266680 0273088 0280265 0288286 0296814

y = infin (1) 0099312 0107746 0117285 0127984 0139369

(2) 0099439 0108078 0117883 0128899 0140639

(3) 0140458 0148595 0157745 0167986 0178872

(4) 0140572 0148908 0158314 0168859 0180084

(5) 0282886 0289776 0297457 0306022 0315111

(6) 0282976 0290034 0297929 0306748 0316120

26

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0059096 0072505 0088907 0108122 0129137

(2) 0059223 0072829 0089500 0109052 0130456

(3) 0079757 0092662 0108388 0126778 0146871

(4) 0079875 0092970 0108956 0127669 0148134

(5) 0159788 0170682 0183871 0199249 0216024

(6) 0159881 0170936 0184343 0199990 0217075

y = 25 (1) 0095883 0119741 0149240 0184073 0222390

(2) 0096087 0120262 0150197 0185573 0224520

(3) 0127403 0150359 0178640 0211977 0248613

(4) 0127592 0150855 0179557 0213414 0250653

(5) 0245161 0264533 0288250 0316124 0346709

(6) 0245311 0264942 0289010 0317319 0348407

y = 50 (1) 0116598 0147920 0187100 0233784 0285503

(2) 0116839 0148537 0188231 0235554 0288015

(3) 0153023 0183153 0220713 0265390 0314840

(4) 0153249 0183742 0221798 0267087 0317248

(5) 0285742 0311159 0342651 0380005 0421287

(6) 0285921 0311644 0343551 0381418 0423292

y = infin (1) 0129470 0167217 0215383 0273841 0339734

(2) 0129725 0167865 0216560 0275661 0342282

(3) 0167978 0204275 0250442 0306382 0369383

(4) 0168217 0204895 0251572 0308128 0371826

(5) 0305868 0336470 0375169 0421935 0474525

(6) 0306058 0336980 0376106 0423388 0476560

27

Page 27: uwaterloo.ca · Ruin Analysis of a Threshold Strategy in a Discrete-time Sparre Andersen Model Steve DREKIC and Ana Maria MERA Department of Statistics and Actuarial Science University

(d) x = infin

n = 25 n = 50 n = 100 n = 200 n = 400

y = 10 (1) 0059096 0072505 0088907 0108122 0129137

(2) 0059223 0072829 0089500 0109052 0130456

(3) 0079757 0092662 0108388 0126778 0146871

(4) 0079875 0092970 0108956 0127669 0148134

(5) 0159788 0170682 0183871 0199249 0216024

(6) 0159881 0170936 0184343 0199990 0217075

y = 25 (1) 0095883 0119741 0149240 0184073 0222390

(2) 0096087 0120262 0150197 0185573 0224520

(3) 0127403 0150359 0178640 0211977 0248613

(4) 0127592 0150855 0179557 0213414 0250653

(5) 0245161 0264533 0288250 0316124 0346709

(6) 0245311 0264942 0289010 0317319 0348407

y = 50 (1) 0116598 0147920 0187100 0233784 0285503

(2) 0116839 0148537 0188231 0235554 0288015

(3) 0153023 0183153 0220713 0265390 0314840

(4) 0153249 0183742 0221798 0267087 0317248

(5) 0285742 0311159 0342651 0380005 0421287

(6) 0285921 0311644 0343551 0381418 0423292

y = infin (1) 0129470 0167217 0215383 0273841 0339734

(2) 0129725 0167865 0216560 0275661 0342282

(3) 0167978 0204275 0250442 0306382 0369383

(4) 0168217 0204895 0251572 0308128 0371826

(5) 0305868 0336470 0375169 0421935 0474525

(6) 0306058 0336980 0376106 0423388 0476560

27