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Ruin Probability in a Threshold Insurance Risk Model

Isaac K. M. Kwan and Hailiang Yang1

Abstract. This paper considers the ruin probability under

a threshold insurance risk model. We assume that the claim

size of an insurance business depends on the claim time. The

integro-differential equations satisfied by the ruin probability

are derived. A Lundberg type upper bound for the ruin proba-

bility is obtained. In some special cases, closed form solutions

for the ruin probability are obtained. Some numerical exam-

ples are included.

Keywords: Threshold insurance risk model, Ruin probabil-

ity, Lundberg inequality, exponential claim distribution, ODE

with advanced argument.

1 Introduction

The Cramer-Lundberg model was introduced in the early

20th century. The main results on ruin probability include the

Lundberg inequality and Cramer-Lundberg approximation. It

is very difficult or even impossible to obtain the closed form

solution for the ruin probability except for some special cases

such as mixed-exponential or phase-type claim size distribu-

tions. (See, Rolski et al., 1999)

As the assumption of the Cramer-Lundberg model is con-

sidered too restrictive to be applicable, a lot of work hasbeen done on developing more general and realistic models

based on it. Popular models are the Sparre-Andersen Model,

the Markov modulated risk model and the diffusion-perturbed

risk model.

In the Cramer-Lundberg model, the inter-arrival times of

claims independently and identically follow an exponential

distribution. This assumption has been criticized for its lack

of applicability in real situations. In the real world, the depen-

dent structure is a very common phenomenon. Recently, de-

pendent structure in claim sizes sequence has been considered

in the insurance risk model by many authors. Discrete time au-

toregressive model and its extension-linear model are the sim-plest cases (Bowers, et al., 1986; Gerber, 1982). The common

shock component can describe wide scale loss due to natural

disasters. Cossette and Marceau (2000) studied the discrete-

time common shock risk models with different classes of busi-

ness. For more references and detailed discussions on the de-

pendent insurance risk models, we refer the readers to the pa-

pers by Dhaene and Denuit (1999), Dhaene and Goovaerts

(1996) and Dhaene et al. (2002a, 2002b) and the references

therein.

1 Department of Statistics and Actuarial Science, The University of HongKong, Pokfulam Road, Hong Kong, e-mails: h0008303@hkusua.hku.hk(Kwan), hlyang@hkusua.hku.hk (Yang)

In a recent interesting paper by Albrecher and Boxma

(2004), the model with dependence between claim size and

inter-arrival time was first introduced. In their model, the

change of the inter-arrival time distribution depends on the

previous claim size through a threshold structure. The ultimate

ruin probability can be written as the inversion of Laplace

transform of a known function form. However, its closed form

cannot always be found. Albrecher and Boxma (2005) studied

the Gerber-Shiu function in the model with dependence be-

tween claim size and inter-arrival time.

In this paper, a new dependent model which reverses the

dependent structure of the model by Albrecher and Boxma(2004) is introduced. From the accounting point of view, if

losses are inspected and aggregately reported at an arbitrary

time, then the longer the time that comes between the in-

spections, the larger the aggregate losses should be. The new

model can be applied to this situation. The model can also be

related to the Markov-modulated risk model, but the regime-

switching process is caused by an endogenous factor rather

than some exogenous factors. The model can be reformulated

as a random walk process with independent increments sub-

ject to a regime-switching process with 2 states. The Markov

modulated risk model, first introduced by Asmussen (1989,

2000), assumes the parameters of the classical model follow

a Markov process. In that model, premium rate and claim dis-

tribution vary according to some exogenous changes of eco-

nomic states. Yang and Yin (2004) obtained a coupled sys-

tem of integro-differential equations for the ruin probability

for this model. Ng and Yang (2005, 2006) considered the joint

density of surplus prior to and after ruin under this model.

Their work has contributed to solving the ruin problem of

an insurance company subject to exogenous economic cycle

switching.

2 The Insurance Risk Model

In our model, the distribution of claim size depends on thelength of inter-arrival time. Let Xk be the size of the k-th

claim,Tk be the inter-arrival time between the (k-1)-th claim

and the k-th claim. IfTk is less than a threshold level a, then

Xk follows a distribution F1. Otherwise, it follows another

distributionF2. We assume inter-arrival time independently

and identically follows an exponential distribution.

LetUt(u) be the surplus process at time t given the initialsurplus u, the dynamic ofUt(u)is given by:

Ut(u) =u + ct St,where St =

N(t)k=1 Xk and c is a constant and denotes the

premium rate.

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We assume the following net profit condition is satisfied:

ct= (1 + )E[St], >0.

Note that, by Walds Identity, we have

ct= (1 + )t(P(T < a)1+ P(T a)2),therefore

c= (1 + )((1 ea)1+ ea2).LetT = inft0{t : Ut(u) < 0} be the time of ruin, then(u) =P{T = |U(0) =u} is the ultimate survival proba-bility, and(u) = 1 (u)is the ultimate ruin probability.

3 Integro-differential Equation and the IntegralEquation

The following theorem gives the integro-differential equation

satisfied by the survival probability.

Theorem3.1. The ultimate survival probability(x)satisfiesthe following integro-differential equation:

cd(u)

du (u)

=ea u+ca0

[f1(x) f2(x)](u + ca x)dx

u0

f1(x)(u x)dx. (1)

Proof.By conditioning on the claim time, we have

(u) = a0

et u+ct

0f1(x)(u + ct x)dxdt

+

a

et u+ct0

f2(x)(u + ct x)dxdt.

Lets = u + ct

c(u) =

u+cau

e(suc )

s0

f1(x)(s x)dxds

+

u+ca

e(suc )

s0

f2(x)(s x)dxds.

(2)

Differentiating both sides with respect to u:

cd(u)

du =ea u+ca0

f1(x)(u + ca x)dx

u0

f1(x)(u x)dx

+

c

u+cau

e(suc )

s0

f1(x)(s x)dxds

ea u+ca0

f2(x)(u + ca x)dx

+

c

u+ca

e(suc )

s

0

f2(x)(s x)dxds.

Substituting (2) into the equation above, we have:

cd(u)

du =ea u+ca0

[f1(x) f2(x)](u + ca x)dx

u

0

f1(x)(u x)dx + (u).

From this we have (1), and the proof of Theorem 3.1 is com-

pleted.

From theorem 3.1, we can obtain the following result.

Theorem3.2. The ultimate ruin probability(u)satisfies thefollowing integral equation:

c(u) =

u

F1(x)dx +

u0

(u x)F1(x)dx

+ ea

u+ca

[F2(x) F1(x)]dx (3)

+ u+ca0

(u + ca x)[F2(x) F1(x)]dx.Proof.Integrating both sides of the equation (1) from0 to u,we have

c[(u) (0)] u0

(y)dy

=ea u0

y+ca0

[f1(x) f2(x)](y+ ca x)dxdy

u

0 y

0

f1(x)(y x)dxdy

=ea u+ca

0

u+caca

[f1(y x) f2(y x)](x)dydx

u+caca

xca

[f1(x) f2(x)](y x)dydx

u0

ux

f1(x)(y x)dydx

=ea u+ca0

(x)

[F1(u + ca x) F1(ca x)]

[F2(u + ca x) F2(ca x)]

dx

u0

uy0

f1(x)(y)dxdy.

Therefore,

c[(u) (0)]

=

u0

(x)F1(u x)dx

+ ea u+ca0

(u + ca x)[F1(x) F2(x)]dx

ea ca

0

(ca x)[F1(x) F2(x)]dx. (4)

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Letu ,c[1 (0)] =1+ ea(2 1)

ea ca0

(ca x)[F1(x) F2(x)]dx

(0) = 1 1

c ea

c (2 1)ea

c

ca0

(ca x)[F1(x) F2(x)]dx.

Substituting this equation into (4) the result is obtained by

substituting(u) = 1 (u).In general, an explicit solution for ultimate ruin probability

cannot be obtained. However, it can be estimated in several

ways. In the following, we will provide an upper bound and

introduce a simulation method. Since the adjustment coeffi-

cient plays an important role here, lets turn to the adjustment

coefficient first.

4 Adjustment Coefficient and Lundberg Inequality

By similar argument as in the case of the classical model, we

have the following result.

Theorem 4.1. Assume that both M1(r) =0

erxf1(x)dx

and M2(r) =0

erxf2(x)dx exist, and that there existsr1, r

2 R{} such that Mi(r) 0 such that E[eR(XcT)] = 1.Ris called the adjustment coefficient.

Similar to the classical model, Lundbergs inequality can be

obtained with the adjustment coefficient as follows:Theorem4.2. (Lundbergs Inequality)

(u) eRu (5)whereu > 0 is the initial surplus, and R is the adjustmentcoefficient defined in Theorem 4.1.

Proof.LetSbe a random walk with an identical and indepen-

dent incrementY =X cT. Then the classical technique ofchanging measure can be applied to our model too. By a simi-

lar argument (see Asmussen, 2000), and defining a new prob-

ability measure by PL(A) = E[eRSn ; A], A Fn, we have

(u) = EL[eRS(u) ], where (u)is the ruin time. The result

follows immediately by noting that(u) = EL[eRS(u) ] =eRuEL[eR(u)] eRu (here (u) is the first overshootamount).

In addition, from the above proof, we can see, similar to the

classical case, that we can use (u) = EL[eRS(u) ] to de-

velop a simulation algorithm to simulate the ruin probability.

5 Exponential Claim Size Distribution

Suppose the distributions F1 and F2 are exponential, i.e.

F1(x) = 1 e1x and F2(x) = 1e2x. Then we show, inthis section, that the ruin probability satisfies a second order

linear differential equation with advanced argument.

Theorem 5.1. (u) satisfies the following third order differ-ential equation with advanced argument:

cd3(u)

du3 + (c1+ c2 ) d

2(u)

du2 + 2(c1 ) d(u)

du

ea(1

2)

d(u + ca)

du

= 0. (6)

Proof.From equation (1), we have

c(1)(u) (u)

=ea u+ca0

[1e1x 2e2x](u + ca x)dx

u0

1e1x(u x)dx.

Differentiating both sides with respect to u once and substi-

tuting (1) into the equation, we have

c d2(u)du2

d(u)du

=ea(1 2)(u + ca)

1(u) 1

cd(u)

du (u)

+ ea(2 1) u+ca0

2e2(u+cax)(x)dx.

Rearranging the terms,

cd2(u)

du2 + (c1 ) d(u)

du ea(1 2)(u + ca)

= ea(2

1

) u+ca0

2

e2(u+cax)(x)dx.

Let

h(u) =ea(2 1) u+ca0

2e2(u+cax)(x)dx

Then,

dh(u)

du =ea(2 1)2

(u + ca)

u+ca

0

2e2(u+cax)(x)dx

,

dh(u)

du =ea(2 1)2(u + ca) 2h(u)

and

dh(u)

du + 2h(u) =e

a(2 1)2(u + ca).

Substituting

h(u) =cd2(u)

du2 + (c1 ) d(u)

du

ea(1

2)(u + ca)

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into the above equations and rearranging the term, we have

cd3(u)

du3 + (c1+ c2 ) d

2(u)

du2 + 2(c1 ) d(u)

du

ea(1 2) d(u + ca)du

= 0.

The result follows immediately by substituting(u) = 1 (u)into the above equation.

Corollary5.1. The ultimate ruin probability (u)satisfies thefollowing second order differential equation with advanced ar-

gument:

cd2(u)

du2 + (c1+ c2 ) d(u)

du + 2(c1 )(u)

ea(1 2)(u + ca) = 0. (7)Proof.Since () = 0, () = 0 and () = 0, theresult follows immediately by integrating (6) fromu to .

Theorem5.2. The solution to the differential equation (7) is:

(u) =ni=1

pi(u)eRiu

where Ri C are the roots of the following characteristicequation:

cR2 + (c1+ c2 )R+ 2(c1 ) ea(1 2)eRca = 0 (8)

andpi(u)is al 1-th order polynomial inu only if the mul-tiplicity of the rootRiis l. In addition,n is finite if the region

of the possible roots is confined within any vertical strip in thecomplex plane.

Proof.By Taylors Theorem,

(u + ca) =

k=0

(ca)k(k)(u)

k!

Substituting this into (7):

cd2(u)

du2 + (c1+ c2 ) d(u)

du + 2(c1 )(u)

ea(1

2)

k=0

(ca)k(k)(u)

k! = 0.

Then, it becomes an ordinary differential equation. The corre-

sponding characteristic equation is:

cR2 + (c1+ c2 )R+ 2(c1 )

ea(1 2)k=0

(ca)kRk

k! = 0

or, equivalently,

cR2 + (c1+ c2 )R+ 2(c1 )

ea(1

2)e

Rca = 0.

Since

h(R) =cR2 + (c1+ c2 )R+ 2(c1 ) ea(1 2)eRca

is an entire function, there can be only a finite number of zeros

of h(R) in any compact set. This implies there are only a finite

number of solutions in any vertical strip in the complex plane.

Moreover, the linearity of the equation (7) implies any finite

sum of the functionsukeRiu, k = 0, 1, 2, . . . , l 1whereRiis a root of multiplicity l of (8), is a solution of (7) (see Hale

and Verduyn Lunel, 1993).

The main problem to find the solution of a second order

differential equation with advanced argument is that the non-

linearity of the characteristic equation makes it very difficult,

if not impossible, to find all its roots explicitly. But, for the

current case, the characteristic equation (8) can be solved as

shown in the following subsection.

5.1 Explicit Solution

In this section, we try to obtain the analytical formula of(u)from the second order differential equation with advanced ar-

gument (7).

First of all, it is important to see that, since as u ,(u) 0, we do not need to care about the root Ri of thecharacteristic equation (8) with non-negative real part.

Next, we will show that the negative real roots of (8) always

exist.

Lemma5.1. The characteristic equation (8) always has 2 neg-

ative real roots.

Proof.Let

g(x) = cx2 + (c1+ c2 )x + 2(c1 )h(x) = ea(1 2)ecax

So, solving (8) is equivalent to finding the interception points

ofg(x)andh(x). We tackle the problem by looking at differ-ent cases.

Case 1: 1 > 2. For this case, we have 1 < 2. There-

fore, by the net-profit condition, c > 1 c1 > 0.This impliesg(x) = (cx+c1 )(x+2)has 2 negativereal roots1 and 2. Ash(x) > 0, x R,h(x) > g(x)forx

[1, 2]. Moreover, by the net-profit condition,

g(0) h(0) = 2(c1 ) ea(1 2)= 12(c ((1 ea)1+ ea2))> 0.

Note thatg (x) > h(x) as x , g(x) h(x) is strictlydecreasing on x (, 1], and g(x) h(x) is strictly in-creasing onx (2, 0]. We have shown thatg(x) h(x)< 0for x [1, 2]. Therefore it is clear that the characteristicequation has 2 negative real roots.

Case 2: 1 < 2. Now, h(x) < 0,x R. Ifg(x) at itsminimum is less than h(x), then there must be 2 negativereal roots of g(x) = h(x). This can be seen from the fol-lowing reasoning: by the net-profit condition,g(0)> h(0); as

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x ,g(x) > h(x); andg(x) h(x)is strictly decreas-ing whenx < and g(x) h(x) is strictly increasing whenx > , whereis the only point at whichg (x) = h(x), i.e.g(x) h(x)attains its minimum at. The minimum point ofg(x)is c1c22c and its minimum value is:

g

c1 c22c

=( + c2 c1)2

4c .

The value ofh(x)at c1c22c is:

h

c1 c2

2c

=(1 2)e

ac (+c1+c2).

Let r(a) =(+c2c1)2

4c (12)e ac (+c1+c2), then

r(0) =( + c2 c1)2

4c (1 2)

= ( + c1

c2)

2

4c 0,and,

r(a) = (1 2)( + c1+ c2)

c e

ac (+c1+c2) 22A

(B2 4AC)k2 + 4A2Dk2e 2ABk

(B24AC)k2+4A2

2A = 0, for2 > 1(9)

withA = c, B = c1+ c2 ,C = 2(c1 ), D =ea(1 2) and k = ca. In addition, if the conditionabove is satisfied, the real double roots are the one and only

one multiple root.

Proof.Let f(x) = Ax2 + Bx + CDekx and g(x) =Ax2 +Bx +CwhereA, B,C,k are defined before. Sinceg(x)canbe factorized into g(x) = (cx+c1)(x+2), all of its rootsare real. This implies B2 4AC 0. Ifris a multiple root, itmust satisfy the following system of simultaneous equations:

f(r) = 0f(r) = 0

Ar2 + Br + C Dekr = 02Ar+ B Dkekr = 0

Therefore,

Akr2 + (Bk

2A)r+ (Ck

B) = 0. (10)

The determinant of the quadratic equation (10) is:

(Bk 2A)24(Ak)(Ck B) = (B24AC)k2 + 4A2 >0.

Therefore,r must be real. The solution of (10) is:

r=2A

Bk

(B2 4AC)k2 + 4A22Ak .From the proof of Lemma 5.1, if1 > 2, then the roots of

the characteristic equation are outside the setB B2 4AC

2A ,

B+ B2 4AC2A

.

Therefore,

2A Bk +

(B2 4AC)k2 + 4A22Ak

is the only possible double root. Substituting this into 2Ar+B Dkekr = 0, we have

2A

2A Bk +

(B2 4AC)k2 + 4A22Ak

+ B

Dkek

2ABk+(B24AC)k2+4A2

2Ak

= 0,

or

2A +

(B2 4AC)k2 + 4A2

Dk2ek

2ABk+(B24AC)k2+4A2

2Ak = 0.

This is the double root condition for 1 > 2. The proof for

2 > 1 is similar.

It is not possible to have a root of multiplicity more than

2 because the quadratic equation (10) does not have double

roots. Besides real roots, the characteristic equation may have

some complex roots. Now, we want to restrict the possible

range of complex roots.

Lemma5.3. IfRiis a complex root of the characteristic equa-

tion, thenRiis an element of the following set:

Case 1:1 > 2

If roots ofg(p) + h(p) = 0 exist,r= p + qi :p (1, 1) (2, 2),

q=

B+ B2 4C2

.

Otherwise,

r= p + qi : p (1, 2), q=

B+ B2 4C

2

.

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Case 2:2 > 1

r= p + qi :p (1, 1) (2, min(2, 0)),

q= B+ B2 4C2

where,

B = (p + 2)2 + (p + (1

c))2,

C= (p + 2)2(p + (1

c))2 (

c)2e2a(1 2)2e2cap,

1and 2 are the 2 negative real roots of the equation g(x) h(x) = 0,1 and 2 are the 2 negative real roots of the equa-tiong(x) + h(x) = 0, where

g(x) = (x + 2)(x + (1 c

))

h(x) = (

c)ea(1 2)ecax.

Proof.Suppose R = p+ q i C : p R, q R\{0} isa complex root of (8). Since the characteristic equation is a

real analytic function, a complex root must exist in the form

of conjugate pairs. LetR be a complex root and R be its con-

jugate. SubstitutingR and R into (8), we have the following

system of simultaneous equations:

cR2 + (c1+ c2

)R+ 2(c1

)

=ea(1 2)eRca

cR2

+ (c1+ c2 )R+ 2(c1 )=ea(1 2)eRca.

(11)

Factorizing the left hand sides, we have (cR+ c1 )(R+ 2) =ea(1 2)eRca(cR+ c1 )(R+ 2) =ea(1 2)eRca.

Multiplying the first equation in (11) by the second one, we

obtain

|R|2 + 2(R+ R) + 22 c2|R|2 + c(c1 )(R+ R) + (c1 )2

=2e2a(1 2)2eca(R+R).SubstitutingR = p + qi, we have

p2 + q2 + 22p + 22

c2(p2 + q2) + 2c(c1 )p + (c1 )2

=2e2a(1 2)2e2cap

Therefore,

q4 + Bq2 + C= 0 (12)

where

B = (p + 2)2 + (p + (1

c))2,

C= (p + 2)2(p + (1

c))2 (

c)2e2a(1 2)2e2cap.

So (12) is a quadratic equation in q2

. From the basic knowl-edge about quadratic equations, since B >0, it is impossiblefor both roots to be greater than zero. So, we must have C h(p) > 0,p (, 1) (2, 0). There-fore, (14) is rejected. So, (13) is the only possible solution,

i.e. p (1, 2). As g(p) is strictly decreasing from posi-tive to negative up to the minimum and then strictly increas-

ing to positive again when p moves from 1 to 2 and h(p)is always positive, there are at most 2 roots of the equation

g(p) + h(p) = 0 or g(p) > h(p),p 0. Let 1 and2 be these 2 roots, if they exist, such that 1 2. Thenthe possible range of a complex root is R = p+ qi, wherep (1, 1)(2, 2) if roots of g(p) + h(p) = 0 exist,p (1, 2), otherwise, and

q=

B+ B2 4C2

withB and Cdefined as before.

The proof of case 2:1 < 2 is similar.

The final step to prove that (8) only have 2 distinct negative

real roots, no root of the characteristic equation is complex, is

by using the Rouches Theorem.

Theorem 5.3. With the conditions imposed on the ultimate

ruin probability: C ([0, ), [0, 1]) and (u) < 0, u0, the solution of the second order differential equation withadvanced argument (7) does not have any oscillatory terms,

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i.e. no root of the characteristic equation is complex. More-

over, the 2 distinct negative real roots are simple roots, i.e. the

double root condition (9) does not exist.

Proof.As g(x) is a quadratic function and h(x) is an expo-nential function, we can always find a closed contour in the

left-half plane such that

|g(x)

| >

|h(x)

|. (For the definition

of g(x) and h(x), please refer to the proof of Lemma 5.1)As g(x) has only 2 roots in the left-half plane, by Rouch esTheorem,g(x) h(x) also has only 2 roots in the left-halfplane. And, Lemma 5.1 shows that there are always 2 distinct

negative real roots. This completes the proof (Rudin, 1987).

Therefore, the solution of (7) is:

(u) = AeR1u + BeR2u, (15)

where Aand B are some real constants, R1 and R2 are the 2negative real roots of the characteristic equation (8).

In order to determine the constants A and B, substituting(15) into (1) and (3), then puttingu = 0, we obtain a system

of simultaneous equations as in (16).By solving these 2 equations with 2 unknowns, we obtain

the constants Aand B.

Remarks: 1. Similar to the classical compound Poisson

model, when the claim size is exponentially distributed, the

ruin probability has an explicit expression.

2. By substituting (15) into equation (6) and using (16), it

is easy to see that (15) is a solution of the original equation.

3. If the claim sizes have an Erlang or hyperexponential

distribution, it should be still possible to obtain an explicit ex-

pression for the ruin probability, but the mathematics becomes

very complex in this case. This is an interesting topic, we will

investigate this in our future research.

5.2 Numerical Illustration

In this numerical illustration, the following values are as-

signed: = 1, 1 = 1, 2 = 0.1, c = 10, and a = 0.1,1.0,5.0. Therefore, the net profit condition is satisfied. In thefollowing numerical illustration, both values obtained by the

simulation method as described in Asmussen (2000) and the

analytical formula described in Section 5.1 are shown. Their

values are compared and the following error measures are

shown in the table: Error = Simulation Value - Analytical An-

swer; ABS(Error) = Absolute Error; Relative Error = Absolute

Error/Simulation Value; 10000 simulations were used for allcases.

From tables 1-3, it is seen that the absolute error is of or-

der104 and the relative error is less than1% for most cases.When the ruin probability is very small, some simulated nu-

merical values have little bigger relative errors (about 3%).This is because the ruin probability is too small and it may

need huge number of simulations to obtain accurate results in

these cases. In view of this, it can be concluded that the simu-

lation and the analytical formula give consistent results.

Acknowledgments.We would like to thank the referee for his

careful reading of the paper and helpful suggestions and com-

ments. This research was supported by the Research GrantsCouncil of the Hong Kong Special Administrative Region,

China (Project No. HKU 7050/05P)

References

[1] Albrecher, H., Boxma, O.J., 2004. A ruin model with dependence be-tween claim sizes and claim intervals. Insurance: Mathematics and

Economics, 35, 245-254.[2] Albrecher, H., Boxma, O.J., 2005. On the discounted penalty function

in a Markov-dependent risk model. Insurance: Mathematics and Eco-nomics, 37, 650-672.

[3] Asmussen, S., 1989. Risk theory in a Markovian environment.Scandi-navian Actuarial Journal, 1989, 69-100.

[4] Asmussen, S., 2000. Ruin Probabilities. World Scientific, Singapore.[5] Bellen,A., Zennaro, M., 2003. Numerical Methods for Delay Differen-

tial Equations. Oxford University Press, New York.[6] Bowers, N.L., Gerber, H.U., Hickman, J.C., Jones, D.A. and Nesbitt,

C.J. ,1986. Actuarial Mathematics. The Society of Actuaries, Schaum-burg, IL.

[7] Cossette, H., Marceau,E., 2000. The discrete-time risk model with cor-related classes of business. Insurance: Mathematics and Economics,

26, 133-149.[8] Dhaene, J. and Denuit, M. Thesafest dependence structure among risks,

Insurance: Mathematics and Economics, 1999,25, 11-21.[9] Dhaene, J., Denuit, M., Goovaerts, M. J., Kaas, R. and Vyncke, D. The

concept of comonotonicity in actuarial science and finance: theory, In-surance: Mathematics and Economics, 2002a,31, 3-33.

[10] Dhaene, J., Denuit, M., Goovaerts, M. J., Kaas, R. and Vyncke, D. Theconcept of comonotonicity in actuarial science and finance: applica-tions,Insurance: Mathematics and Economics, 2002b,31, 133-161.

[11] Dhaene, J. and Goovaerts, M. J. Dependence of risks and stop-loss or-der,ASTIN Bulletin, 1996,26, 201-212.

[12] Driver, R.D., 1977. Ordinary and Delay Differential Equations.Springer-Verlag, New York.

[13] Gerber, H.U., 1982. Ruin theory in the linear model.Insurance: Math-ematics and Economics, 1, 177-184.

[14] Grandell, J., 1991. Aspects of Risk Theory, Springer-Verlag, New York.

[15] Hale, J.K., Verduyn Lunel, S.M., 1993. Introduction to functional dif-ferential equations. Springer-Verlag, New York.[16] Michaud, F., 1996. Estimating the probability of ruin for variable pre-

miums by simulation.ASTIN Bulletin, 26 (1), 93-105.[17] Ng, A.C.Y. and Yang, H., 2005. Lundberg-type bounds for the joint

distribution of surplus immediately before and at ruin under a Markov-modulated tisk model. ASTIN Bulletin, 35 (2), 351-361.

[18] Ng, A.C.Y. and H. Yang, 2006. On the joint distribution of surplus priorand immediately after ruin under a Markovian regime switching modelStochastic Processes and Their Applications, 116 (2), 244-266.

[19] Rolski, T., Schmidli, H., Schmidt, V., Teugels, J., 1999. Stochastic Pro-cesses for Insurance and Finance. Wiley & Sons, New York.

[20] Rudin, W., 1987. Real And Complex Analysis. McGraw-Hill, Singa-pore.

[21] Sharma, P., 2002. Prudential Supervision of Insurance Undertaking.Conference of The Insurance Supervisory Services of The MemberStates of The European Union, The London Working Group, London.

[22] SparreAnderson, E., 1957. Onthe collectivetheory of riskin the caseofcontagion between claims. Transactions XVth International Congressof Actuaries, New York, II, 219-229.

[23] Thorin, O., 1974. On the asymptotic behavior of the ruin probability foran infinite period when the epochs of claims form a renewal process.Scandinavian Actuarial Journal, 1974, 65-102.

[24] Thorin, O., 1982. Probability of ruin.Scandinavian Actuarial Journal,1982, 65-102.

[25] Tsoi, A.C., 1975. Explicit solution of a class of delay-differential equa-tions.International Journal of Control, 21 (1), 39-48

[26] Yang, H., Yin, G., 2004. Ruin probability for a model under Markovianswitching regime. In:Probability, finance and insurance : proceedingsof a workshop at the University of Hong Kong, Hong Kong, 15-17 July

2002. World Scientific, Singapore, 206-217.

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cR1+ eaeR1ca 1(1e(R1+1)ca)R1+1 2(1e(R1+2)ca)R1+2 A

+

cR2+ eaeR2ca1(1e(R2+1)ca)

R2+1 2(1e

(R2+2)ca)R2+2

B = + ea

e2ca e1ca

c eaeR1ca1e(R1+2)ca

R1+2 1e(R1+1)ca

R1+1

A

+

c eaeR2ca1e(R2+2)ca

R2+2 1e(R2+1)ca

R2+1

B = 1+ e

ae2ca

2 e1ca

1

(16)

u Simulation Analytical Error ABS(Error) Relative Error

0.00 0.9115596118 0.9116696406 (0.0001100288) 0.0001100288 0.0121%

0.50 0.9048371313 0.9054406162 (0.0006034849) 0.0006034849 0.0667%

1.00 0.8985198057 0.8999696713 (0.0014498656) 0.0014498656 0.1614%

1.50 0.8959430657 0.8949803872 0.0009626785 0.0009626785 0.1074%

2.00 0.8921900284 0.8902997967 0.0018902317 0.0018902317 0.2119%

2.50 0.8848555856 0.8858196541 (0.0009640685) 0.0009640685 0.1090%3.00 0.8813273961 0.8814722047 (0.0001448086) 0.0001448086 0.0164%

3.50 0.8773258376 0.8772150256 0.0001108120 0.0001108120 0.0126%

4.00 0.8722445930 0.8730215425 (0.0007769495) 0.0007769495 0.0891%

4.50 0.8685343156 0.8688750959 (0.0003407803) 0.0003407803 0.0392%

5.00 0.8645557028 0.8647652296 (0.0002095268) 0.0002095268 0.0242%

5.50 0.8623378291 0.8606853686 0.0016524605 0.0016524605 0.1916%

6.00 0.8564633668 0.8566313659 (0.0001679991) 0.0001679991 0.0196%

6.50 0.8534838914 0.8526005937 0.0008832977 0.0008832977 0.1035%

7.00 0.8482639014 0.8485913750 (0.0003274736) 0.0003274736 0.0386%

7.50 0.8451824181 0.8446026277 0.0005797904 0.0005797904 0.0686%

8.00 0.8405123096 0.8406336418 (0.0001213322) 0.0001213322 0.0144%

8.50 0.8363292482 0.8366839407 (0.0003546925) 0.0003546925 0.0424%9.00 0.8328753270 0.8327531934 0.0001221336 0.0001221336 0.0147%

9.50 0.8275063948 0.8288411609 (0.0013347661) 0.0013347661 0.1613%

10.00 0.8253310033 0.8249476610 0.0003833423 0.0003833423 0.0464%

100.00 0.3537946754 0.3534812424 0.0003134330 0.0003134330 0.0886%

1000.00 0.0000736495 0.0000737511 (0.0000001015) 0.0000001015 0.1379%

Table 1:Threshold Model:(u)witha = 0.1

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u Simulation Analytical Error ABS(Error) Relative Error

0.00 0.2625050912 0.2601480944 0.0023569968 0.0023569968 0.8979%

0.50 0.2246567269 0.2237709041 0.0008858228 0.0008858228 0.3943%

1.00 0.1977463165 0.1982445034 (0.0004981869) 0.0004981869 0.2519%

1.50 0.1803669841 0.1797307357 0.0006362484 0.0006362484 0.3528%

2.00 0.1664142034 0.1657785099 0.0006356935 0.0006356935 0.3820%

2.50 0.1563116991 0.1548213168 0.0014903823 0.0014903823 0.9535%

3.00 0.1467117461 0.1458568266 0.0008549195 0.0008549195 0.5827%

3.50 0.1376949726 0.1382425851 (0.0005476125) 0.0005476125 0.3977%

4.00 0.1305287148 0.1315657393 (0.0010370245) 0.0010370245 0.7945%

4.50 0.1254713687 0.1255599647 (0.0000885960) 0.0000885960 0.0706%

5.00 0.1196010346 0.1200524915 (0.0004514569) 0.0004514569 0.3775%

5.50 0.1153665576 0.1149303218 0.0004362358 0.0004362358 0.3781%

6.00 0.1097673734 0.1101186841 (0.0003513107) 0.0003513107 0.3201%

6.50 0.1049012689 0.1055672905 (0.0006660216) 0.0006660216 0.6349%

7.00 0.1008715708 0.1012415696 (0.0003699988) 0.0003699988 0.3668%

7.50 0.0971278418 0.0971170724 0.0000107695 0.0000107695 0.0111%

8.00 0.0926352101 0.0931759009 (0.0005406908) 0.0005406908 0.5837%

8.50 0.0899396220 0.0894044274 0.0005351946 0.0005351946 0.5951%

9.00 0.0859221546 0.0857918359 0.0001303187 0.0001303187 0.1517%

9.50 0.0825878080 0.0823291892 0.0002586188 0.0002586188 0.3131%

10.00 0.0791310366 0.0790088300 0.0001222066 0.0001222066 0.1544%

100.00 0.0000484060 0.0000483619 0.0000000441 0.0000000441 0.0911%

1000.00 3.59479E-37 3.57541E-37 1.93769E-39 1.93769E-39 0.5390%

Table 2:Threshold model with exponential claims:(u)witha = 1.0

u Simulation Analytical Error ABS(Error) Relative Error

0.00 0.0998797604 0.1000459789 (0.0001662185) 0.0001662185 0.1664%

0.50 0.0634174570 0.0638083541 0.(0003908971) 0.0003908971) 0.6164%1.00 0.0407301485 0.0407014333 0.0000287152 0.0000287152 0.0705%

1.50 0.0259903835 0.0259670576 0.0000233259 0.0000233259 0.0897%

2.00 0.0164675026 0.0165712891 (0.0001037865) 0.0001037865 0.6303%

2.50 0.0107655853 0.0105796016 0.0001859837 0.0001859837 1.7276%

3.00 0.0068542675 0.0067584853 0.0000957822 0.0000957822 1.3974%

3.50 0.0043268066 0.0043214179 0.0000053887 0.0000053887 0.1245%

4.00 0.0027779508 0.0027668890 0.0000110618 0.0000110618 0.3982%

4.50 0.0017843073 0.0017751202 0.0000091871 0.0000091871 0.5149%

5.00 0.0011643828 0.0011422101 0.0000221727 0.0000221727 1.9042%

5.50 0.0007338959 0.0007381444 (0.0000042485) 0.0000042485 0.5789%

6.00 0.0004807254 0.0004800209 0.0000007045 0.0000007045 0.1465%

6.50 0.0003150775 0.0003149777 0.0000000998 0.0000000998 0.0317%7.00 0.0002129975 0.0002093073 0.0000036902 0.0000036902 1.7325%

7.50 0.0001370107 0.0001415159 (0.0000045052) 0.0000045052 3.2882%

8.00 0.0000958759 0.0000978973 (0.0000020214) 0.0000020214 2.1983%

8.50 0.0000681606 0.0000697112 (0.0000015506) 0.0000015506 2.2750%

9.00 0.0000506812 0.0000513834 (0.0000007022) 0.0000007022 1.3854%

9.50 0.0000393105 0.0000393590 (0.0000000485) 0.0000000485 0.1233%

10.00 0.0000325404 0.0000313702 0.0000011702 0.0000011702 3.5961%

100.00 2.28643E-09 2.34957E-09 (6.31416E-11) 6.31416E-11 2.7616%

1000.00 1.93411E-48 1.93411E-48 (1.17416E-50) 1.17416E-50 0.6108%

Table 3:Threshold model with exponential claims:(u)witha = 5.0

49