Stochastic Calculus for Jump Processes

Chapter 15Stochastic Calculus for Jump Processes

In this chapter we present the construction of processes with jumps and inde-pendent increments, including the Poisson and compound Poisson processes.We also with stochastic integrals and stochastic calculus with jumps, andwith the Girsanov theorem for jump processes, which will be used for pric-ing and the determination of risk-neutral measures in the next chapter, inrelation with market incompleteness.

15.1 The Poisson Process

The most elementary and useful jump process is the standard Poisson process(Nt)t∈R+ which is a counting process, i.e. (Nt)t∈R+ has jumps of size +1 only,and its paths are constant in between two jumps.

t

Nt

0

1

2

3

4

5

6

T1 T2 T3 T4 T5 T6

Fig. 15.1: Sample path of a Poisson process (Nt)t∈R+ .

In other words, the value Nt at time t is given by†

† The notation Nt is not to be confused with the same notation used for numéraireprocesses in Chapter 12.

501

N. Privault

Nt =∞∑k=1

1[Tk,∞)(t), t ∈ R+, (15.1)

where

1[Tk,∞)(t) =

1 if t > Tk,

0 if 0 6 t < Tk,

k > 1, and (Tk)k>1 is the increasing family of jump times of (Nt)t∈R+ suchthat

limk→∞

Tk = +∞.

In addition, (Nt)t∈R+ is assumed to satisfy the following conditions:

1. Independence of increments: for all 0 6 t0 < t1 < · · · < tn and n > 1 theincrements

Nt1 −Nt0 , . . . , Ntn −Ntn−1 ,

are mutually independent random variables. The next figure representsa sample path of a Poisson process.

2. Stationarity of increments: Nt+h − Ns+h has the same distribution asNt −Ns for all h > 0 and 0 6 s 6 t.

The meaning of the above stationarity condition is that for all fixed k ∈ Nwe have

P(Nt+h −Ns+h = k) = P(Nt −Ns = k),

for all h > 0, i.e., the value of the probability

P(Nt+h −Ns+h = k)

does not depend on h > 0, for all fixed 0 6 s 6 t and k ∈ N.

Based on the above assumption, given T > 0 a time value, a natural questionarises:

what is the probability distribution of the random variable NT ?

We already know that Nt takes values in N and therefore it has a discretedistribution for all t ∈ R+.

It is a remarkable fact that the distribution of the increments of (Nt)t∈R+ ,can be completely determined from the above conditions, as shown in thefollowing theorem.

As seen in the next result, cf. [BN96], Nt−Ns has the Poisson distributionwith parameter λ(t− s).

502

This version: December 22, 2017http://www.ntu.edu.sg/home/nprivault/indext.html

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http://en.wikipedia.org/wiki/Poisson_distribution

http://www.ntu.edu.sg/home/nprivault/indext.html

http://en.wikipedia.org/wiki/Simeon_Denis_Poisson

Stochastic Calculus for Jump Processes

Theorem 15.1. Assume that the counting process (Nt)t∈R+ satisfies theabove independence and stationarity Conditions 1 and 2. Then for all fixed0 6 s 6 t we have

P(Nt −Ns = k) = e−λ(t−s) (λ(t− s))kk! , k ∈ N, (15.2)

for some constant λ > 0.

The parameter λ > 0 is called the intensity of the Poisson process (Nt)t∈R+

and it is given byλ := lim

h→0

1hP(Nh = 1). (15.3)

The proof of the above Theorem 15.1 is technical and not included here,cf. e.g. [BN96] for details, and we could in fact take this distribution property(15.2) as one of the hypotheses that define the Poisson process.

Precisely, we could restate the definition of the standard Poisson process(Nt)t∈R+ with intensity λ > 0 as being a process defined by (15.1), which isassumed to have independent increments distributed according to the Poissondistribution, in the sense that for all 0 6 t0 6 t1 < · · · < tn,

(Nt1 −Nt0 , . . . , Ntn −Ntn−1)

is a vector of independent Poisson random variables with respective param-eters

(λ(t1 − t0), . . . , λ(tn − tn−1)).

In particular, Nt has the Poisson distribution with parameter λt, i.e.,

P(Nt = k) = (λt)kk! e−λt, t > 0.

The expected value IE[Nt] of Nt can be computed as

IE[Nt] =∞∑k=0

kP(Nt = k)

= e−λt∞∑k=0

(λt)k(k − 1)!

= λt e−λt∞∑k=0

(λt)kk!

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N. Privault

= λt, (15.4)

cf. Exercise A.1. Similarly, we have

IE[N2t

]=∞∑k=0

k2P(Nt = k)

= e−λt∞∑k=0

k(λt)k

(k − 1)!

= e−λt∞∑k=2

(λt)k(k − 2)! + e−λt

∞∑k=1

(λt)k(k − 1)!

= (λt)2 e−λt∞∑k=0

(λt)kk! + λt e−λt

∞∑k=0

(λt)kk!

= (λt)2 + λt

andVar[Nt] = IE

[N2t

]−(IE[Nt]

)2 = λt.

Short Time Behaviour

From (15.3) above we deduce the short time asymptotics∗P(Nh = 0) = e−hλ = 1− hλ+ o(h), h→ 0,

P(Nh = 1) = hλ e−hλ ' hλ, h→ 0.

By stationarity of the Poisson process we also find more generally that

P(Nt+h −Nt = 0) = e−hλ = 1− hλ+ o(h), h→ 0,

P(Nt+h −Nt = 1) = hλ e−hλ ' hλ, h→ 0,

P(Nt+h −Nt = 2) ' h2λ2

2 = o(h), h→ 0, t > 0,

for all t > 0. This means that within a “short” interval [t, t+h] of length h, theincrement Nt+h−Nt behaves like a Bernoulli random variable with parameterλh. This fact can be used for the random simulation of Poisson process paths.

More generally, for k > 1 we have∗ The notation f(h) = o(h) means limh→0 f(h)/hk = 0, and f(h) ' hk meanslimh→0 f(h)/hk = 1.

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P(Nt+h −Nt = k) ' hk λk

k! , h→ 0, t > 0.

The intensity of the Poisson process can in fact be made time-dependent (e.g.by a time change), in which case we have

P(Nt −Ns = k) = exp(−

w t

sλ(u)du

) (r tsλ(u)du

)kk! , k = 0, 1, 2, . . . .

In particular,

P(Nt+dt −Nt = k) =

e−λ(t)dt = 1− λ(t)dt+ o(h), k = 0,

λ(t) e−λ(t)dtdt ' λ(t)dt, k = 1,

o(dt), k > 2.

The intensity process (λ(t))t∈R+ can also be made random, as in the case ofCox processes.

Poisson Process Jump Times

In order to determine the distribution of the first jump time T1 we note thatwe have the equivalence

T1 > t ⇐⇒ Nt = 0,

which implies

P(T1 > t) = P(Nt = 0) = e−λt, t ∈ R+,

i.e., T1 has an exponential distribution with parameter λ > 0.

In order to prove the next proposition we note that more generally, wehave the equivalence

Tn > t ⇐⇒ Nt 6 n− 1,

for all n > 1. This allows us to compute the distribution of Tn with its density.It coincides with the gamma distribution with integer parameter n > 1, alsoknown as the Erlang distribution in queueing theory.

Proposition 15.2. For all n > 1 the probability distribution of Tn has thegamma probability density function

t 7−→ λn e−λt tn−1

(n− 1)!" 505



N. Privault

on R+, i.e., for all t > 0 the probability P(Tn > t) is given by

P(Tn > t) = λnw∞t

e−λs sn−1

(n− 1)!ds.

Proof. We have

P(T1 > t) = P(Nt = 0) = e−λt, t ∈ R+,

and by induction, assuming that

P(Tn−1 > t) = λw∞t

e−λs (λs)n−2

(n− 2)!ds, n > 2,

we obtain

P(Tn > t) = P(Tn > t > Tn−1) + P(Tn−1 > t)= P(Nt = n− 1) + P(Tn−1 > t)

= e−λt (λt)n−1

(n− 1)! + λw∞t

e−λs (λs)n−2

(n− 2)!ds

= λw∞t

e−λs (λs)n−1

(n− 1)!ds, t ∈ R+,

where we applied an integration by parts to derive the last line.

Random samples of Poisson process jump times can be generated using thefollowing R code.

lambda = 2.0n = 10for (k in 1:n)tauk <- rexp(n)/lambda; Ti <- cumsum(tauk)taukTi

In particular, for all n ∈ Z and t ∈ R+, we have

P(Nt = n) = pn(t) = e−λt (λt)n

n! ,

i.e., pn−1 : R+ → R+, n > 1, is the density function of Tn.

In addition to Proposition 15.2 we could show the following proposition whichrelies on the strong Markov property, see e.g. Theorem 6.5.4 of [Nor98].

Proposition 15.3. The random interjump times

τk := Tk+1 − Tk

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spent in state k ∈ N, with T0 = 0, form a sequence of independent iden-tically distributed random variables having the exponential distribution withparameter λ > 0, i.e.,

P(τ0 > t0, . . . , τn > tn) = e−λ(t0+t1+···+tn), t0, . . . , tn ∈ R+.

As the expectation of the exponentially distributed random variable τk withparameter λ > 0 is given by

IE[τk] = λw∞

0x e−λxdx = 1

λ,

we can check that the higher the intensity λ (i.e., the higher the probabilityof having a jump within a small interval), the smaller is the time spent ineach state k ∈ N on average.

In addition, conditionally to NT = n, the n jump times on [0, T ] ofthe Poisson process (Nt)t∈R+ are independent uniformly distributed randomvariables on [0, T ]n, cf. e.g. § 12.1 of [Pri13]. This fact can be useful for therandom simulation of the Poisson process.

Compensated Poisson Martingale

From (15.4) above we deduce that

IE[Nt − λt] = 0, (15.5)

i.e., the compensated Poisson process (Nt−λt)t∈R+ has centered increments.Since in addition (Nt − λt)t∈R+ also has independent increments, we get thefollowing proposition, cf. e.g. Example 2 page 177. We let

Ft := σ(Ns : s ∈ [0, t]), t ∈ R+,

denote the filtration generated by the Poisson process (Nt)t∈R+ .

Proposition 15.4. The compensated Poisson process

(Nt − λt)t∈R+

is a martingale with respect (Ft)t∈R+ .

Extensions of the Poisson process include Poisson processes with time-dependent intensity, and with random time-dependent intensity (Cox pro-cesses). Poisson processes belong to the family of renewal processes which arecounting processes of the form

Nt =∑n>1

1[Tn,∞)(t), t ∈ R+,

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N. Privault

for which τk := Tk+1 − Tk, k ∈ N, is a sequence of independent identicallydistributed random variables.

15.2 Compound Poisson Processes

The Poisson process itself appears to be too limited to develop realistic pricemodels as its jumps are of constant size. Therefore there is some interest inconsidering jump processes that can have random jump sizes.

Let (Zk)k>1 denote an i.i.d. sequence of square-integrable random vari-ables distributed as the common random variable Z with probability distri-bution ν(dy) on R, independent of the Poisson process (Nt)t∈R+ . We have

P(Zk ∈ [a, b]) = ν([a, b]) =w b

aν(dy), −∞ < a 6 b <∞, k > 1.

Definition 15.5. The process

Yt :=Nt∑k=1

Zk, t ∈ R+, (15.6)

is called a compound Poisson process.∗

In particular, we note that the jump size

∆Yt := Yt − Yt− , t ∈ R+,

of (Yt)t∈R+ at time t is given by the relation

∆Yt = ZNt∆Nt, t ∈ R+, (15.7)

where∆Nt := Nt −Nt− ∈ 0, 1, t ∈ R+,

denotes the jump size of the standard Poisson process (Nt)t∈R+ , and Nt− isthe left limit

Nt− := limst

Ns, t > 0,

For a typical example of a compound Poisson process we can assume thatjump sizes are Gaussian distributed with mean δ and variance η2, in whichcase ν(dy) is given by

ν(dy) = 1√2πη2

e−(y−δ)2/(2η2)dy.

∗ We use the conventionn∑k=1

Zk = 0 if n = 0, so that Y0 = 0.

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The next Figure 15.2 represents a sample path of a compound Poisson pro-cess, with here Z1 = 0.9, Z2 = −0.7, Z3 = 1.4, Z4 = 0.6, Z5 = −2.5, Z6 = 1.5,Z7 = −0.5.

t

Yt

-1

0

1

2

3

T1 T2 T3 T4 T5 T6 T7

Fig. 15.2: Sample path of a compound Poisson process (Yt)t∈R+ .

Given that NT = n, the n jump sizes of (Yt)t∈R+ on [0, T ] are independentrandom variables which are distributed on R according to ν(dx). Based onthis fact, the next proposition allows us to compute the moment generatingfunction (MGF) of the increment YT − Yt.

Proposition 15.6. For any t ∈ [0, T ] we have

IE [exp (α(YT − Yt))] = exp(λ(T − t)

w∞−∞

( eαy − 1)ν(dy)), α ∈ R.

(15.8)

Proof. Since Nt has a Poisson distribution with parameter t > 0 and isindependent of (Zk)k>1, for all α ∈ R we have, by conditioning on the valueof NT −Nt = n,

IE [exp (α(YT − Yt))] = IE[

exp(α

NT∑k=Nt+1

Zk

)]

= IE[

exp(α

NT−Nt∑k=1

Zk

)]

=∞∑n=0

IE[

exp(α

n∑k=1

Zk

)∣∣∣NT −Nt = n

]P(NT −Nt = n)

= e−λ(T−t)∞∑n=0

λn

n! (T − t)n IE[

exp(α

n∑k=1

Zk

)]

= e−λ(T−t)∞∑n=0

λn

n! (T − t)nn∏k=1

IE [exp (αZk)]

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N. Privault

= e−λ(T−t)∞∑n=0

λn

n! (T − t)n (IE [exp (αZ)])n

= exp (λ(T − t) (IE [exp (αZ)]− 1))

= exp(λ(T − t)

w∞−∞

( eαy − 1)ν(dy)),

since the probability distribution ν(dy) of Z satisfies

IE [exp (αZ)] =w∞−∞

eαyν(dy) andw∞−∞

ν(dy) = 1.

From the moment generating function (15.8) we can compute the expectationof Yt for fixed t as the product of the mean number of jump times IE[Nt] = λtand the mean jump size IE[Z], i.e.,

IE[Yt] = ∂

∂αIE[ eαYt ]|α=0 = λt

w∞−∞

yν(dy) = IE[Nt] IE[Z] = λt IE[Z].(15.9)

Note that the above identity requires to exchange the differentiation and ex-pectation operators, which is possible when the moment generating function(15.8) takes finite values for all α in a certain neighborhood (−ε, ε) of 0.

Relation (15.9) states that the mean value of Yt is the mean jump size IE[Z]times the mean number of jumps IE[Nt]. It can also be directly using seriessummations as

IE[Yt] = IE[Nt∑k=1

Zk

]

=∞∑n=1

IE[

n∑k=1

Zk

∣∣∣Nt = n

]P(Nt = n)

= e−λt∞∑n=1

λntn

n! IE[

n∑k=1

Zk

∣∣∣Nt = n

]

= e−λt∞∑n=1

λntn

n! IE[

n∑k=1

Zk

]

= λt e−λt IE[Z]∞∑n=1

(λt)n−1

(n− 1)!= λt IE[Z].

Regarding the variance, we have510


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IE[Y 2t ] = ∂2

∂α2 IE[ eαYt ]|α=0

= λtw∞−∞

y2ν(dy) + (λ(T − t))2(w∞−∞

yν(dy))2

= λt IE[Z2] + (λt IE[Z])2,

which yields

Var [Yt] = λtw∞−∞

y2ν(dy) = λt IE[|Z|2].

More generally, one can show that for all 0 6 t0 6 t1 · · · 6 tn and α1, . . . , αn ∈R we have

IE[

n∏k=1

eiαk(Ytk−Ytk−1 )

]= exp

(λ

n∑k=1

(tk − tk−1)w∞−∞

( eiαky − 1)ν(dy))

=n∏k=1

exp(λ(tk − tk−1)

w∞−∞

( eiαky − 1)ν(dy))

=n∏k=1

IE[

eiαk(Ytk−Ytk−1 )].

By a multivariate version of Theorem 17.11, the above identity shows thefollowing proposition.

Proposition 15.7. The compound Poisson process

Yt =Nt∑k=1

Zk, t ∈ R+,

has independent increments, i.e. for any finite sequence of times t0 < t1 <· · · < tn, the increments

Yt1 − Yt0 , Yt2 − Yt1 , . . . , Ytn − Ytn−1

are mutually independent random variables.

Since the compensated compound Poisson process also has independent andcentered increments by (15.5) we have the following counterpart of Proposi-tion 15.4, cf. also Example 2 page 177.

Proposition 15.8. The compensated compound Poisson process

Mt := Yt − λt IE[Z], t ∈ R+,

is a martingale.

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N. Privault

By construction, compound Poisson processes only have a finite number ofjumps on any interval. They belong to the family of Lévy processes whichmay have an infinite number of jumps on any finite time interval, cf. [CT04].

15.3 Stochastic Integrals with Jumps

Based on the relationdYt = ZNtdNt,

we define the stochastic integral of a stochastic process (φt)t∈R+ with respectto (Yt)t∈R+ by

w T

0φtdYt =

w T

0φtZNtdNt :=

NT∑k=1

φTkZk. (15.10)

Note that the expression (15.10) ofw T

0φtdYt has a natural financial interpre-

tation as the value at time T of a portfolio containing a (possibly fractional)quantity φt of a risky asset at time t, whose price evolves according to ran-dom returns Zk at random times Tk.

In particular the compound Poisson process (Yt)t∈R+ in (15.5) admits thestochastic integral representation

Yt = Y0 +Nt∑k=1

Zk = Y0 +w t

0ZNsdNs.

The next result is also called the smoothing formula, cf. Theorem 9.2.1 in[Bré99].

Proposition 15.9. Let (φt)t∈R+ be a proces adapted to the filtration gener-ated by (Yt)t∈R+ and such that

IE[w T

0|φt|dt

]<∞, T > 0.

The expected value of the compound Poisson compensated stochastic integralcan be expressed as

IE[w T

0φt−dYt

]= IE

[w T

0φt−ZtdNt

]= λ IE[Z] IE

[w T

0φtdt

],

(15.11)

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where φt− denotes the left limit

φt− := limst

φs, t > 0.

Proof. By Proposition 15.8 the compensated compound Poisson process (Yt−λt IE[Z])t∈R+ is a martingale, and as a consequence the indefinite stochasticintegral

t 7−→w t

0φs−d(Ys − λs IE[Z]) =

w t

0φs−(ZNsdNs − λ IE[Z]ds)

is also a martingale, by an argument similar to that in the proof of Propo-sition 6.1 because the adaptedness of (φt)t∈R+ to the filtration generated by(Yt)t∈R+ , makes (φt−)t>0 predictable, i.e. adapted with respect to the filtra-tion

Ft− := σ(Ys : s ∈ [0, t)

), t > 0.

It remains to use the fact that the expectation of a martingale remains con-stant over time.

Note that while the identity in expectations (15.11) holds for the left limitφt− , it need not hold for φt itself. Taking for example φt = Yt := Nt, wecheck that

λ IE[w T

0Nt−dt

]= λ IE

[w T

0Ntdt

]= λ

w T

0IE [Nt] dt

= λ2w T

0tdt

= (λT )2

2 ,

and w T

0Nt−dNt =

NT∑k=1

(k − 1) = 12NT (NT − 1),

hence

IE[w T

0Nt−dNt

]= 1

2(IE[N2T

]− IE[NT ]

)= (λT )2

2 = λ IE[w T

0Nt−dt

].

On the other hand, we have

w T

0NtdNt =

NT∑k=1

k = 12NT (NT + 1),

hence

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N. Privault

IE[w T

0NtdNt

]= 1

2(IE[N2T

]+ IE[NT ]

)= 1

2((λT )2 + 2λT

)= (λT )2

2 + λT

6= λ IE[w T

0Ntdt

].

Under similar conditions, the compound Poisson compensated stochastic in-tegral can be shown to satisfy the Itô isometry (15.12) in the next proposition.

Proposition 15.10. Let (φt)t∈R+ be a process adapted to the filtration gen-erated by (Yt)t∈R+ , and such that

IE[w T

0|φt|2dt

]<∞, T > 0.

The expected value of the squared compound Poisson compensated stochasticintegral can be computed as

IE[(w T

0φt−(dYt − λ IE[Z]dt)

)2]

= λ IE[|Z|2] IE[w T

0|φt|2dt

],

(15.12)

Note that in (15.12), the generic jump size Z is squared but λ is not.

Proof. From the stochastic Fubini type theorem we have(w T


)2(15.13)

= 2w T

0φt−

w t−

0φs−(dYs − λ IE[Z]ds)(dYt − λ IE[Z]dt) (15.14)

+w T

0|φt− |2|ZNt |2dNt, (15.15)

where integration over the diagonal s = t has been excluded in (15.14) asthe inner integral has an upper limit t− rather than t. Next, taking expecta-tion on both sides of (15.13)-(15.15) we find

IE[(w T


)2]

= IE[w T

0|φt− |2|ZNt |2dNt

]= λ IE[|Z|2] IE

[w T

0|φt|2dt

],

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where we used the vanishing of the expectation of the double stochastic in-tegral:

IE[w T

0φt−

w t−

0φs−(dYs − λ IE[Z]ds)(dYt − λ IE[Z]dt)

]= 0,

and the martingale property of the compensated compound Poisson process

t 7−→

(Nt∑k=1|Zk|2

)− λt IE[Z2], t ∈ R+,

as in the proof of Proposition 15.9. The isometry relation (15.12) can also beproved using simple predictable processes, similarly to the proof of Proposi-tion 4.9.

Next, take (Bt)t∈R+ a standard Brownian motion independent of (Yt)t∈R+

and (Xt)t∈R+ a jump-diffusion process of the form

Xt :=w t

0usdBs +

w t

0vsds+ Yt, t ∈ R+,

where (ut)t∈R+ is a process which is adapted to the filtration (Ft)t∈R+ gen-erated by (Bt)t∈R+ and (Yt)t∈R+ , and such that

IE[w T

0|φt|2|ut|2dt

]<∞ and IE

[w T

0|φtvt|dt

]<∞, T > 0.

We define the stochastic integral of (φt)t∈R+ with respect to (Xt)t∈R+ by

w T

0φtdXt :=

w T

0φtutdBt +

w T

0φtvtdt+

NT∑k=1

φTkZk, T > 0.

For the mixed continuous-jump martingale

Xt :=w t

0usdBs + Yt − λt IE[Z], t ∈ R+,

we then have the isometry:

IE[(w T

0φt−dXt

)2]

= IE[w T

0|φt|2|ut|2dt

]+ λ IE[|Z|2] IE

[w T

0|φt|2dt

].

(15.16)

provided that (φs)s∈R+ is adapted to the filtration (Ft)t∈R+ generated by(Bt)t∈R+ and (Yt)t∈R+ . The isometry formula (15.16) will be used in Sec-" 515



N. Privault

tion 16.6 for mean-variance hedging in jump-diffusion models.

More generally, when (Xt)t∈R+ contains an additional drift term,

Xt = X0 +w t

0usdBs +

w t

0vsds+

w t

0ηsdYs, t ∈ R+,

the stochastic integral of (φt)t∈R+ with respect to (Xt)t∈R+ is given byw T

0φsdXs :=

w T

0φsusdBs +

w T

0φsvsds+

w T

0ηsφsdYs

=w T

0φsusdBs +

w T

0φsvsds+

NT∑k=1

φTkηTkZk, T > 0.

15.4 Itô Formula with Jumps

Let us first consider the case of a standard Poisson process (Nt)t∈R+ withintensity λ. We have the telescoping sum

f(Nt) = f(0) +Nt∑k=1

(f(k)− f(k − 1))

= f(0) +w t

0(f(1 +Ns−)− f(Ns−))dNs

= f(0) +w t

0(f(Ns)− f(Ns − 1))dNs

= f(0) +w t

0(f(Ns)− f(Ns−))dNs.

Here, Ns− denotes the left limit

Ns− = limh0

Ns−h.

In particular we have Ns = Ns− + 1 if dNs = 1, and

k = NTk = 1 +NT−k, k > 1.

By the same argument, in the case of the compound Poisson process (Yt)t∈R+

we find the pathwise Itô formula

f(Yt) = f(0) +Nt∑k=1

(f(YTk)− f(YT−k

))

= f(0) +Nt∑k=1

(f(YT−k

+ Zk)− f(YT−k

))

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= f(0) +w t

0(f(Ys− + ZNs)− f(Ys−))dNs

= f(0) +w t

0(f(Ys)− f(Ys−))dNs,

which can be decomposed using a compensated Poisson stochastic integralas

f(Yt) = f(0) +w t

0(f(Ys)− f(Ys−))(dNs − λds) + λ

w t

0(f(Ys)− f(Ys−))ds.

More generally, for a process of the form

Xt = X0 +w t

0usdBs +

w t

0vsds+

w t

0ηsdYs, t ∈ R+,

we find, by combining the Itô formula for Brownian motion with the aboveargument we get

f(Xt) = f(X0) +w t

0usf

′(Xs)dBs + 12

w t

0f ′′(Xs)|us|2ds+

w t

0vsf′(Xs)ds

+NT∑k=1

(f(XT−

k+ ηTkZk

)− f

(XT−

k

))= f(X0) +

w t

0usf

′(Xs)dBs + 12

w t


w t

0vsf′(Xs)ds

+w t

0(f(Xs− + ηsZNs)− f(Xs−))dNs, t ∈ R+,

i.e.,

f(Xt) = f(X0) +w t

0vsf′(Xs)ds+

w t

0usf

′(Xs)dBs + 12

w t

0f ′′(Xs)|us|2ds

+w t

0(f(Xs)− f(Xs−))dNs, t ∈ R+. (15.17)

The integral Itô formula (15.17) can be rewritten in differential notation as

df(Xt) = vtf′(Xt)dt+utf ′(Xt)dBt+

12f′′(Xt)|ut|2dt+(f(Xt)−f(Xt−))dNt,

(15.18)

t ∈ R+. Given an Itô process of the form

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N. Privault

Xt := X0 +w t

0usdBs +

w t

0ηsdYt, t ∈ R+,

the Itô formula with jumps (15.17) can be rewritten as

f(Xt) = f(X0) +w t

0vsf′(Xs−)ds+

w t

0usf

′(Xs−)dBs

+12

w t


w t

0ηsf′(Xs−)dYs

+w t

0(f(Xs)− f(Xs−)−∆Xsf

′(Xs−)) dNs, t ∈ R+,

where we used the relation dXs = ηs∆Ys, which impliesw t

0ηsf′(Xs−)dYs =

w t

0∆Xsf

′(Xs−)dNs, t > 0.

The above Poisson stochastic integral can be written asw t


′(Xs−)) dNs

=w t

0(f(Xs− + ηs∆Ys)− f(Xs−)− ηs∆Ysf ′(Xs−)) dNs

=w t

0(f(Xs− + ηsZNs)− f(Xs−)− ηsZNsf ′(Xs−)) dNs

=w t

0

(f(Xs− + ηsZ1+Ns− )− f(Xs−)− ηsZNsf ′(Xs−)

)dNs,

and when ηs is a deterministic function it can be compensated into the mar-tingale

w t


′(Xs−)) dNs

−w t

0IE[f(x+ ηsZ)− f(x)− ηsZf ′(x)

]x=Xs−

ds

=w t

0(f(Xs)− f(Xs−)− ηsZf ′(Xs−)) dNs

−λw t

0

w∞−∞

(f(Xs− + ηsy)− f(Xs−)− ηsyf ′(Xs−)) ν(dy)ds.

This above formulation is at the basis of the extension of Itô’s formula toLévy processes with an infinite number of jumps on any interval, using thebound

|f(x+ y)− f(x)− yf ′(x)| 6 Cy2, y ∈ [−1, 1],

that follows from Taylor’s theorem for f a C2(R) function, cf. e.g. Theo-rem 4.4.7 of [App04] in the setting of Poisson random measures. Such pro-cesses, also called “infinite activity Lévy processes” are also useful in financialmodeling, [CT04] and include the gamma process, stable processes, variancegamma processes, inverse Gaussian processes, etc, as in the following illus-

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trations.

1. Gamma process.

0

t

Fig. 15.3: Sample trajectories of a gamma process.

2. Variance Gamma process.

0

t

Fig. 15.4: Sample trajectories of a variance gamma process.

3. Inverse Gaussian process.

0

t

Fig. 15.5: Sample trajectories of an inverse Gaussian process.

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N. Privault

4. Negative Inverse Gaussian process.

0

t

Fig. 15.6: Sample trajectories of a negative inverse Gaussian process.

5. Stable process.

0

t

Fig. 15.7: Sample trajectories of a stable process.

The above sample path of a stable process can be compared to the USD/CNYexchange rate over the year 2015, according to the date retrieved from thefollowing code.

library(quantmod)getSymbols("CNY=X",from="2015-01-01",to="2015-12-6",src="yahoo")rate=Ad(`CNY=X`)chartSeries(rate,up.col="blue",theme="white")

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6.20

6.25

6.30

6.35

6.40

rate [2015−01−01/2015−12−04]

Last 6.3875

Jan 012015

Mar 022015

May 012015

Jul 012015

Sep 012015

Nov 022015

Fig. 15.8: USD/CNY Exchange rate data.

Itô multiplication table with jumps

For a process (Xt)t∈R+ is given by

Xt =w t

0usdBs +

w t

0vsds+

w t

0ηsdNs, t ∈ R+,

the Itô formula with jumps reads

f(Xt) = f(0) +w t

0usf

′(Xs)dBs + 12

w t

0|us|2f ′′(Xs)dBs

+w t

0vsf′(Xs)ds+

w t

0(f(Xs− + ηs)− f(Xs−))dNs

= f(0) +w t

0usf

′(Xs)dBs + 12

w t

0|us|2f ′′(Xs)dBs

+w t

0vsf′(Xs)ds+

w t

0(f(Xs)− f(Xs−))dNs.

Given two Itô processes (Xt)t∈R+ and (Yt)t∈R+ written in differential notationas

dXt = utdBt + vtdt+ ηtdNt, t ∈ R+,

anddYt = atdBt + btdt+ ctdNt, t ∈ R+,

the Itô formula for jump processes also shows that

d(XtYt) = XtdYt + YtdXt + dXt · dYt

where the product dXt ·dYt is computed according to the following extensionof the Itô multiplication Table 4.1:

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N. Privault

· dt dBt dNt

dt 0 0 0dBt 0 dt 0dNt 0 0 dNt

Table 15.1: Itô multiplication table with jumps.

In other words, we have

dXt · dYt = (vtdt+ utdBt + ηtdNt)(btdt+ atdBt + ctdNt)= btvt(dt)2 + btutdt · dBt + btηtdt · dNt + ctvtdt · dNt

+atvtdtdBt + atut(dBt)2 + atηtdBt · dNt+ctutdNt · dBt + ctut(dBt)2 + ctηtdNt · dNt

= atutdt+ ctηtdNt,

sincedNt · dNt = (dNt)2 = dNt,

as ∆Nt ∈ 0, 1. In particular, we have

(dXt)2 = (vtdt+ utdBt + ηtdNt)2 = u2tdt+ η2

t dNt.

15.5 Stochastic Differential Equations with Jumps

Let us start with the simplest example of a constant market return η writtenas

η := St − St−St−

, (15.19)

assuming the presence of a jump at time t, i.e., dNt = 1. Using the relationdSt = St − St− , (15.19) rewrites as

ηdNt = St − St−St−

= dStSt−

, (15.20)

ordSt = ηSt−dNt, (15.21)

which is a stochastic differential equation with respect to the standard Pois-son process, with constant coefficient η ∈ R. Note that the left limit St− in(15.21) occurs naturally from the definition (15.20) of market returns whendividing by the previous index value St− . The use of the left limit St− is alsonecessary when computing pathwise solutions by solving for St from St− .

In the presence of a jump at time t, the equation (15.19) also reads

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St = (1 + η)St− , dNt = 1,

which can be applied by induction at the successive jump times T1, T2, . . . , TNtuntil time t, to derive the solution

St = S0(1 + η)Nt , t ∈ R+,

of (15.21).

Next, consider the case where η is time-dependent, i.e.,

dSt = ηtSt−dNt. (15.22)

At each jump time Tk, Relation (15.22) reads

dSTk = STk − ST−k

= ηTkST−k,

i.e.,STk = (1 + ηTk)ST−

k,

and repeating this argument for all k = 1, . . . , Nt yields the product solution

St = S0

Nt∏k=1

(1 + ηTk) = S0∏

∆Ns=106s6t

(1 + ηs), t ∈ R+.

The equationdSt = µtStdt+ ηtSt−(dNt − λdt), (15.23)

is then solved as

St = S0 exp(w t

0µsds− λ

w t

0ηsds

) Nt∏k=1

(1 + ηTk), t ∈ R+.

A random simulation of the numerical solution of the above equation (15.23)is given in Figure 15.9 for η = 1.29 and constant µ = µt, t ∈ R+.

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N. Privault

Fig. 15.9: Geometric Poisson process.∗

The above simulation can be compared to the real sales ranking data ofFigure 15.10.

Fig. 15.10: Ranking data.

Next, consider the equation

dSt = µtStdt+ ηtSt−(dYt − λ IE[Z]dt)

driven by the compensated compound Poisson process (Yt − λ IE[Z]t)t∈R+ ,also written as

dSt = µtStdt+ ηtSt−(ZNtdNt − λ IE[Z]dt),

with solution∗ The animation works in Acrobat Reader on the entire pdf file.

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St = S0 exp(w t

0µsds− λ IE[Z]

w t

0ηsds

) Nt∏k=1

(1+ηTkZk) t ∈ R+. (15.24)

A random simulation of the geometric compound Poisson process (15.24) isgiven in Figure 15.11.

Fig. 15.11: Geometric compound Poisson process.∗

In the case of a jump-diffusion stochastic differential equation of the form

dSt = µtStdt+ ηtSt−(dYt − λ IE[Z]dt) + σtStdBt,

we get

St = S0 exp(w t

0µsds− λ IE[Z]

w t

0ηsds+

w t

0σsdBs −

12

w t

0|σs|2ds

)×

Nt∏k=1

(1 + ηTkZk), t ∈ R+.

A random simulation of the geometric Brownian motion with compound Pois-son jumps is given in Figure 15.12.∗ The animation works in Acrobat Reader on the entire pdf file.

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N. Privault

Fig. 15.12: Geometric Brownian motion with compound Poisson jumps.∗

By rewriting St as

St = S0 exp(w t

0µsds+

w t

0ηs(dYs − λ IE[Z]ds) +

w t

0σsdBs −

12

w t

0|σs|2ds

)×

Nt∏k=1

( e−ηTk (1 + ηTkZk)),

t ∈ R+, one can extend this jump model to processes with an infinite numberof jumps on any finite time interval, cf. [CT04]. The next Figure 15.13 showsa number of downward and upward jumps occuring in the SMRT historicalshare price data, with a typical geometric Brownian behavior in betweenjumps.

Fig. 15.13: SMRT Share price.

∗ The animation works in Acrobat Reader on the entire pdf file.

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15.6 Girsanov Theorem for Jump Processes

Recall that in its simplest form, the Girsanov theorem for Brownian motionfollows from the calculation

IE[f(BT − µT )] = 1√2πT

w∞−∞

f(x− µT ) e−x2/(2T )dx

= 1√2πT

w∞−∞

f(x) e−(x+µT )2/(2T )dx

= 1√2πT

w∞−∞

f(x) e−µx−µ2T/2 e−x

2/(2T )dx

= IE[f(BT ) e−µBT−µ2T/2]

= IE[f(BT )], (15.25)

for any bounded measurable function f on R, which shows that BT is aGaussian random variable with mean −µT under the probability measureP−µ defined by

dP−µ = e−µBT−µ2T/2dP,

cf. Section 6.2. Equivalently we have

IE[f(BT )] = IE[f(BT + µT )], (15.26)

hence

under the probability measure

dP−µ := e−µBT−µ2T/2dP,

the random variable BT +µT has the centered Gaussian distributionN (0, T ).

More generally, the Girsanov theorem states that (Bt + µt)t∈[0,T ] is a stan-dard Brownian motion under P−µ.

When Brownian motion is replaced with a standard Poisson process(Nt)t∈R+ , the above space shift

Bt 7−→ Bt + µt

may not be used because Nt + µt cannot be a Poisson process, whatever thechange of probability applied, since by construction, the paths of the stan-dard Poisson process has jumps of unit size and remain constant between

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N. Privault

jump times.

The correct way to proceed in order to extend (15.26) to the Poisson caseis to replace the space shift with a time contraction (or dilation) by a certainfactor 1 + c with c > −1, i.e.,

Nt 7−→ Nt/(1+c) or Nt 7−→ N(1+c)t.

Assume that (Nt)t∈R+ is a standard Poisson process with intensity λ undera probability measure Pλ. By analogy with (15.25) we can write

Pλ(N(1+c)T = k) = e−λ(1+c)T (λ(1 + c)T )kk! = e−λcT (1 + c)kPλ(NT = k),

k ∈ N, and for f any bounded function on N we have

IEλ[f(N(1+c)T )] =∞∑k=0

f(k)Pλ(N(1+c)T = k) (15.27)

= e−λcT∞∑k=0

f(k)(1 + c)kPλ(NT = k)

= e−λcT IE[f(NT )(1 + c)NT ]= e−λcT

wΩ

(1 + c)NT f(NT )dPλ

=wΩf(NT )dPλ

= IEλ[f(NT )],

i.e., for f(x) := 1x6n,

Pλ(N(1+c)T 6 n

)= Pλ(NT 6 n), n ∈ N,

orPλ(NT/(1+c) 6 n) = Pλ

(NT 6 n

), n ∈ N,

where the probability measure Pλ is defined by

dPλ := e−λcT (1 + c)NT dPλ.

Consequently,

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dPλ := e−λcT (1 + c)NT dPλ,

the random variable NT has the centered Poisson distribution P(λ(1+c)T )with intensity λ(1 + c)T , i.e., the law of N(1+c)T under Pλ.

Equivalently to (15.27) we have

IE[f(NT )] = IEλ[f(NT/(1+c))],

i.e., under Pλ the law of NT/(1+c) is that of a standard Poisson randomvariable with parameter λT . As a consequence, (Nt/(1+c))t∈R+ is a standardPoisson process with intensity λ under Pλ, and since (Nt/(1+c)− λt)t∈R+ hasindependent increments, the compensated process

Nt/(1+c) − λt, t ∈ R+,

is a martingale under Pλ by (6.2). In addition, we have

Nt/(1+c) =∑n>1

1[Tn,∞)(t/(1 + c)) =∑n>1

1[(1+c)Tn,∞)(t), t ∈ R+,

which shows that the jump times of (Nt/(1+c))t∈[0,T ], given by

((1 + c)Tn)n>1,

and are distributed under Pλ as the jump times of a Poisson process withintensity λ.

Next, taking λ > 0 and letting

c := −1 + λ

λ,

i.e., λ = (1 + c)λ we can rewrite the above by saying that

Pλ(N(1+c)T = k) = e−λcT (1 + c)kPλ(NT = k)

= e−λT (λT )kk!

= Pλ(NT = k),

k ∈ N, and

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N. Privault


dPλ := e−λcT (1 + c)NT dPλ = e−(λ−λ)T(λ

λ

)NTdPλ,

the law of NT is that of a Poisson random variable with intensity

λT = λ(1 + c)T.

Consequently, since

(Nt − (1 + c)λt)t∈R+ = (Nt − λt)t∈R+

has independent increments, the compensated Poisson process

Nt − (1 + c)λt = Nt − λt

is a martingale under Pλ by (6.2), although when c 6= 0 it is not a martingaleunder Pλ.

When µ 6= r, the discounted price process (St)t∈R+ = ( e−rtSt)t∈R+ writtenas

dSt

St= (µ− r)dt+ σ(dNt − λdt)

is not martingale under Pλ, however we can rewrite the equation as

dSt

St= +σ

(dNt −

(λ− µ− r

σ

)dt

)and letting

λ := λ− µ− rσ

= (1 + c)λ

withc := −µ− r

σλ,

we havedSt

St= σ

(dNt − λdt

)hence the discounted price process (St)t∈R+ is martingale under the proba-bility measure Pλ defined as

dPλ := e−λcT (1 + c)NT dPλ = e(µ−r)/σ(

1− µ− rσλ

)NTdPλ.

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We note that ifµ− r < σλ

then the risk-neutral probability measure Pλ exists and is unique, thereforeby Theorems 5.7 and 5.10 the market is without arbitrage and complete.

In the case of compound Poisson processes, the Girsanov theorem can beextended to variations in jump sizes in addition to time variations, and wehave the following more general result.

Theorem 15.11. Let (Yt)t>0 be a compound Poisson process with inten-sity λ > 0 and jump distribution ν(dx). Consider another jump distributionν(dx), and let

ψ(x) := λ

λ

ν(dx)ν(dx) − 1, x ∈ R.

Then,


dPλ,ν := e−(λ−λ)TNT∏k=1

(1 + ψ(Zk))dPλ,ν ,

the process

Yt :=Nt∑k=1

Zk, t ∈ R+,

is a compound Poisson process with

- modified intensity λ > 0, and

- modified jump distribution ν(dx).

Proof. For any bounded measurable function f on R, we extend (15.27) tothe following change of variable

IEλ,ν [f(YT )] = e−(λ−λ)T IEλ,ν

[f(YT )

NT∏i=1

(1 + ψ(Zi))]

= e−(λ−λ)T∞∑k=0

IEλ,ν

[f

(k∑i=1

Zi

)k∏i=1

(1 + ψ(Zi))∣∣∣ NT = k

]Pλ(NT = k)

= e−λT∞∑k=0

(λT )kk! IEλ,ν

[f

(k∑i=1

Zi

)k∏i=1

(1 + ψ(Zi))]

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N. Privault

= e−λT∞∑k=0

(λT )kk!

w∞−∞· · ·

w∞−∞

f(z1 + · · ·+ zk)k∏i=1

(1 + ψ(zi))ν(dz1) · · · ν(dzk)

= e−λT∞∑k=0

(λT )kk!

w∞−∞· · ·

w∞−∞

f(z1 + · · ·+ zk)(

k∏i=1

dν

dν(zi))ν(dz1) · · · ν(dzk)

= e−λT∞∑k=0

(λT )kk!

w∞−∞· · ·

w∞−∞

f(z1 + · · ·+ zk)ν(dz1) · · · ν(dzk).

This shows that under Pλ,ν , YT has the distribution of a compound Poissonprocess with intensity λ and jump distribution ν. We refer to Proposition 9.6of [CT04] for the independence of increments of (Yt)t∈R+ under Pλ,ν .

For example, in case ν ' N (α, σ2) and ν ' N (β, η2) we have

ν(dx)ν(dx) = η

σexp

(− 1

2σ2 (x− α)2 + 12η2 (x− β)2

), x ∈ R.

Note that the compound Poisson process with intensity λ > 0 and jumpdistribution ν can be built as

Xt :=Nλt/λ∑k=1

h(Zk),

provided that ν is the image measure of ν by the function h : R→ R, i.e.,

P(h(Zk) ∈ A) = P(Zk ∈ h−1(A)) = ν(h−1(A)) = ν(A),

for all measurable subset A of R.

Compensated Compound Poisson Martingale

As a consequence of Theorem 15.11, the compensated process

Yt − λt IEν [Z]

becomes a martingale under the probability measure Pλ,ν defined by

dPλ,ν = e−(λ−λ)TNT∏k=1

(1 + ψ(Zk))dPλ,ν .

Finally, the Girsanov theorem can be extended to the linear combination ofa standard Brownian motion (Bt)t∈R+ and an independent compound Pois-son process (Yt)t∈R+ , as in the following result which is a particular case ofTheorem 33.2 of [Sat99].

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Theorem 15.12. Let (Yt)t>0 be a compound Poisson process with intensityλ > 0 and jump distribution ν(dx). Consider another jump distribution ν(dx)and intensity parameter λ > 0, and let

ψ(x) := λ

λ

dν

dν(x)− 1, x ∈ R,

and let (ut)t∈R+ be a bounded adapted process. Then the process(Bt +

w t

0usds+ Yt − λ IEν [Z]t

)t∈R+

is a martingale under the probability measure

dPu,λ,ν = exp(−(λ− λ)T −

w T

0usdBs −

12

w T

0|us|2ds

) NT∏k=1

(1+ψ(Zk))dPλ,ν .

(15.28)

As a consequence of Theorem 15.12, if

Bt +w t

0vsds+ Yt (15.29)

is not a martingale under Pλ,ν , it will become a martingale under Pu,λ,νprovided that u, λ and ν are chosen in such a way that

vs = us − λ IEν [Z], s ∈ R, (15.30)

in which case (15.29) can be rewritten into the martingale decomposition

dBt + utdt+ dYt − λ IEν [Z]dt,

in which both(Bt +

w t

0usds

)t∈R+

and(Yt − λt IEν [Z]

)t∈R+

are martingales

under Pu,λ,ν

When λ = λ = 0, Theorem 15.12 coincides with the usual Girsanov the-orem for Brownian motion, in which case (15.30) admits only one solutiongiven by u = v and there is uniqueness of Pu,0,0. Note that uniqueness occursalso when u = 0 in the absence of Brownian motion with Poisson jumps offixed size a (i.e., ν(dx) = ν(dx) = δa(dx)) since in this case (15.30) also ad-mits only one solution λ = v and there is uniqueness of P0,λ,δa . These remarkswill be of importance for arbitrage pricing in jump models in Chapter 16.

When µ 6= r, the discounted price process (St)t∈R+ = ( e−rtSt)t∈R+ definedby

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N. Privault

dSt

St= (µ− r)dt+ σdBt + η(dYt − λt IEν [Z])

is not martingale under Pλ,ν , however we can rewrite the equation as

dSt

St= +σ(udt+ dBt) + η

(dYt −

(uσ

η+ λ− µ− r

η

)dt

)and choosing u, ν, and λ such that

λ IEν [Z] = uσ

η+ λ− µ− r

η, (15.31)

we havedSt

St= σ(udt+ dBt) + η

(dYt − λ IEν [Z]dt

)hence the discounted price process (St)t∈R+ is martingale under the proba-bility measure Pu,λ,ν , and the market is without arbitrage by Theorem 5.7and the existence of a risk-neutral probability measure Pu,λ,ν . However, themarket is not complete due to the non uniqueness of solutions (u, ν, λ) to(15.31), and Theorem 5.10 does not apply in this situation.

Exercises

Exercise 15.1 Let (Nt)t∈R+ be a standard Poisson process with intensityλ > 0, started at N0 = 0.a) Solve the stochastic differential equation

dSt = ηSt−dNt − ηλStdt = ηSt−(dNt − λdt).

b) Using the first Poisson jump time T1, solve the stochastic differential equa-tion

dSt = −ληStdt+ dNt, t ∈ (0, T2).

Exercise 15.2 Consider a standard Poisson process (Nt)t∈R+ with intensityλ > 0.a) Solve the stochastic differential equation dXt = αXtdt + σdNt over the

time intervals [0, T1), [T1, T2), [T2, T3), [T3, T4), where X0 = 1.b) Write a differential equation for f(t) := IE[Xt], and solve it for t ∈ R+.

Exercise 15.3 Consider a standard Poisson process (Nt)t∈R+ with intensityλ > 0.534


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a) Solve the stochastic differential equation dXt = σXt−dNt for (Xt)t∈R+ ,where σ > 0 and X0 = 1.

b) Show that the solution (St)t∈R+ of the stochastic differential equation

dSt = rdt+ σSt−dNt,

is given by St = S0Xt + rXt

w t

0X−1s ds.

c) Compute IE[Xt] and IE[Xt/Xs], 0 6 s 6 t.d) Compute IE[St], t ∈ R+.

Exercise 15.4 Let (Nt)t∈R+ be a standard Poisson process with intensityλ > 0, started at N0 = 0.a) Is the process t 7→ Nt − 2λt a submartingale, a martingale, or a su-

permartingale?b) Let r > 0. Solve the stochastic differential equation

dSt = rStdt+ σSt−(dNt − λdt).

c) Is the process t 7→ St of Question (b) a submartingale, a martingale, or asupermartingale?

d) Compute the price at time 0 of the European call option with strike priceK = S0 e(r−λσ)T , where σ > 0.

Exercise 15.5 Affine stochastic differential equation with jumps. Consider astandard Poisson process (Nt)t∈R+ with intensity λ > 0.a) Solve the stochastic differential equation dXt = adNt + σXt−dNt, where

σ > 0, and a ∈ R.b) Compute IE[Xt] for t ∈ R+.

Exercise 15.6 Consider the compound Poisson process Yt :=Nt∑k=1

Zk, where

(Nt)t∈R+ is a standard Poisson process with intensity λ > 0, and (Zk)k>1 isan i.i.d. sequence of N (0, 1) Gaussian random variables. Solve the stochasticdifferential equation

dSt = rStdt+ ηSt−dYt,

where η, r ∈ R.

Exercise 15.7 Show, by direct computation or using the moment generatingfunction (15.8), that the variance of the compound Poisson process Yt withintensity λ > 0 satisfies

Var [Yt] = λt IE[|Z|2] = λtw∞−∞

x2ν(dx).

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Exercise 15.8 Consider an exponential compound Poisson process of theform

St = S0 eµt+σBt+Yt , t ∈ R+,

where (Yt)t∈R+ is a compound Poisson process of the form (15.6).

a) Derive the stochastic differential equation with jumps satisfied by (St)t∈R+ .b) Let r > 0. Find a family (Pu,λ,ν) of probability measures under which the

discounted asset price e−rtSt is a martingale.

Exercise 15.9 Consider (Nt)t∈R+ a standard Poisson process with inten-sity λ > 0 under a probability measure P. Let (St)t∈R+ be defined by thestochastic differential equation

dSt = µStdt+ YNtSt−dNt, (15.32)

where (Yk)k>1 is an i.i.d. sequence of random variables of the form

Yk = eXk − 1, where Xk ' N (0, σ2), k > 1.

a) Solve the equation (15.32).b) We assume that µ and the risk-free rate r > 0 are chosen such that the

discounted process ( e−rtSt)t∈R+ is a martingale under P. What relationdoes this impose on µ and r?

c) Under the relation of Question (b), compute the price at time t of a Euro-pean call option on ST with strike price κ and maturity T , using a seriesexpansion of Black-Scholes functions.

Exercise 15.10 Consider a standard Poisson process (Nt)t∈R+ with intensityλ > 0 under a probability measure P. Let (St)t∈R+ be the mean revertingprocess defined by the stochastic differential equation

dSt = −αStdt+ σ(dNt − βdt), (15.33)

where S0 > 0 and α, β > 0.

a) Solve the equation (15.33) for St.b) Compute f(t) := IE[St] for all t ∈ R+.c) Under which condition on α, β, σ and λ does the process St become a

submartingale?d) Propose a method for the calculation of expectations of the form IE[φ(ST )]

where φ is a payoff function.

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Exercise 15.11 Let (Nt)t∈[0,T ] be a standard Poisson process started atN0 = 0, with intensity λ > 0 under the probability measure Pλ, and considerthe compound Poisson process (Yt)t∈[0,T ] with i.i.d. jump sizes (Zk)k>1 ofdistribution ν(dx).

a) Under the probability measure Pλ, the process t 7→ Yt−λt(t+ IE[Z]) is a:

submartingale| martingale| supermartingale|

b) Consider the process (St)t∈[0,T ] given by

dSt = µStdt+ σSt−dYt.

Find λ such that the discounted process (St)t∈[0,T ] := (e−rtSt)t∈[0,T ] is amartingale under the probability measure Pλ defined by its density

dPλdPλ

:= e−(λ−λ)T(λ

λ

)NT.

with respect to Pλ.c) Price the forward contract with payoff ST − κ.

Exercise 15.12 Consider (Yt)t∈R+ a compound Poisson process written as

Yt =Nt∑k=1

Zk, t ∈ R+,

where (Nt)t∈R+ a standard Poisson process with intensity λ > 0 and (Zk)k>1is an i.i.d family of random variables with probability distribution ν(dx) onR under a probability measure P. Let (St)t∈R+ be defined by the stochasticdifferential equation

dSt = µStdt+ St−dYt. (15.34)

a) Solve the equation (15.34).b) We assume that µ, ν(dx) and the risk-free rate r > 0 are chosen such

that the discounted process ( e−rtSt)t∈R+ is a martingale under P. Whatrelation does this impose on µ, ν(dx) and r?

c) Under the relation of Question (b), compute the price at time t of aEuropean call option on ST with strike price κ and maturity T , using aseries expansion of integrals.

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Stochastic Calculus for Jump Processes

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