針對補償混合卜松隨機過程或碎形布朗運動的橋

國

立

交

通

大

學

應用數學系

碩

士

論

文

針對補償混合卜松隨機過程或碎形布朗運動的橋

A Bridge with Respect to the Compensated Compound Poisson Process

or the Fractional Brownian Motion

研 究 生：吳姿慧

指導教授：吳慶堂 副教授

針對補償混合卜松隨機過程或碎形布朗運動的橋

A Bridge with Respect to the Compensated Compound Poisson Process

or the Fractional Brownian Motion

研 究 生：吳姿慧 Student：Tzu-Hui Wu

指導教授：吳慶堂 Advisor：Ching-Tang Wu

國 立 交 通 大 學

應 用 數 學 系

碩 士 論 文

Submitted to Department of Applied Mathematics College of Science

National Chiao Tung University in partial Fulfillment of the Requirements

for the Degree of Master

in

Applied Mathematics June 2010

Hsinchu, Taiwan, Republic of China

i

針對補償混合卜松隨機過程或碎形布朗運動的橋

學生：吳姿慧

指導教授

吳慶堂

國立交通大學應用數學系碩士班

摘 要

由 Föllmer, Wu, Yor, (1999) 中我們知道特定的隨機微分方程式的解會

是一個布朗運動。在本論文中，我們討論有哪些隨機微分方程它們的解會是

一個補償混合卜松過程。藉此，我們可以製造出新的補償混合卜松過程。同

時，我們也討論一些隨機微分方程的解，觀察它們是不是碎形布朗運動。

ii

A Bridge with Respect to the

Compensated Compound Poisson Process

or the Fractional Brownian Motion

Student：Tzu-Hui Wu

Advisors：Dr. Ching-Tang Wu

Department of Applied Mathematics

National Chiao Tung University

ABSTRACT

From Föllmer, Wu, Yor(1999) we know when the Brownian motion with

nonzero linear drift is again a Brownian motion. In this thesis, instead of

Brownian motion we discuss the case of compensated Poisson processes with

nonzero. So we can construct new compensated compound Poisson processes. We

also discuss whether the solutions of some particular form of stochastic

differential equations are fractional Brownian motions.

iii

誌 謝

完成此篇論文，我最感謝的是我的指導教授─吳慶堂教授，最初衝著

對機率的一份熱忱跟隨老師，老師除了教導機率的相關知識，也教導我財務

數學方面的知識，讓我對機率與財務數學更增添幾分興趣，老師固定每週與

我們meeting，使我學習不間斷，感謝老師不辭辛勞的教導我，讓我順利的完

成口試與論文。

感謝韓傳祥教授和陳冠宇教授費心的審閱我的論文，並且提供我許多

意見，讓我的論文更加完善以及順利通過口試。

感謝育慈學姊，不管在課業上或是研究上，學姊總是耐心的與我討論，

經由討論的過程，也讓我有更多新的想法，有更多方向可以著手去做研究。

感謝佳穎同學平日的照顧和幫助，讓我的研究生活增添更多的歡笑，感謝淑

娟學妹，在我口試時，幫我打理一些事情。

感謝我的媽媽，沒有她的支持與陪伴，我無法順利的完成學業以及研

究，感謝我的哥哥提供我求學上許多寶貴的意見，感謝所有幫助過我的人，

以及關心我的人。

iv

目 錄

中文提要

i

英文提要

ii

誌謝

iii

目錄

iv

Chapter 1.

Introduction

1

Chapter 2.

A Bridge with Respect to the Compensated Poisson Process

5

2.1 Poisson

Process

5

2.2

A Bridge Starts from Zero to a Fixed Point

6

2.3

A Bridge between two Independent Compensated Poisson Processes

11

Chapter 3.

Construction of a New Compensated Compound Poisson Process

15

3.1

Construction of a New Compensated Poisson Process

15

3.2

Compensated Compound Poisson Process

22

3.3

Construction of a New Compensated Compound Poisson Process

23

Chapter 4.

A Bridge with Respect to the Fractional Brownian Motion

35

4.1 Fractional

Brownian

Motion

35

4.2

_{Wiener Integrals for the Fractional Brownian Motion for}

2 1 

H

37

4.3

A Bridge with Respect to the Fractional Brownian Motion

39

CHAPTER 1

Introduction

We consider some generalization of Brownian bridge. We want to change the Brownian motion which is in a Brownian bridge to a compensated Poisson process or a compensated compound Poisson process. We also try to change the Brownian motion which is in a Brownian bridge to a fractional Brownian motion. Then we discuss about what are the properties of these processes. We consider the process

Xt= Bt− tB1, for 0 ≤ t ≤ 1 (1.1)

where B is a Brownian motion is a Brownian bridge from 0 to 0 on [0, 1] (see Shreve [14] Definition 4.7.4). Denote by (FB

t ) the filtration generated by B. The Brownian bridge

(Xt)0≤t≤1 is not adapted to the filtration (FtB)0≤t≤1. In the following we consider the

Brownian Bridge which is adapted to the filtration (FB

t )0≤t≤1. Consider the stochastic

differential equation

dXt = dBt+

−Xt

1 − tdt (1.2) with the initial value X0 = 0. The solution (Xt)0≤t<1 of (1.2) is given by

Xt= (1 − t)

Z t

0

1

1 − sdBs, for 0 ≤ t < 1.

Then the process (Xt)0≤t<1 is a Brownian bridge from 0 to 0 on [0, 1) and it has the same

law of the Brownian bridge which is in (1.1) (see Shreve [14] Section 4.7). The process (Xt)0≤t<1is adapted to the filtration (FtB)0≤t<1. Now we consider two independent

Brow-nian motions (Bt)t≥0 and ( ˜Bt)t≥0. The solution (Xt)0≤t≤1 of the stochastic differential

equation dXt= dBt+ ˜ B1− Xt 1 − t dt 1

with the initial value X0 = 0 is given by Xt= (1 − t) Z t 0 1 1 − sdBs+ ˜B1t, for 0 ≤ t < 1. (1.3) The process (Xt)0≤t<1 in (1.3) is a standard Brownian motion with respect to the

filtra-tion (FX

t )0≤t<1 which is the filtration generated by (Xt)0≤t<1 and the process (Xt)0≤t<1

converges to the final value ˜B1 (cf., for example, Jeulin-Yor [7]). The following solution

(Xt)0≤t<1 of the stochastic differential equation

dXt= dBt+

˜ Bt− Xt

1 − t dt with the initial value X0 = 0 is given by

Xt= (1 − t) Z t 0 1 1 − sdBs+ (1 − t) Z t 0 ˜ Bs (1 − s)2ds, for 0 ≤ t < 1. (1.4)

The process (Xt)0≤t<1 in (1.4) converges to ˜B1 P-a.s. as t → 1 and (Xt)0≤t<1 is no longer

a Brownian motion (see F¨ollmer, H. [5]).

In Shreve [14] we see the introduction about compensated Poisson process and com-pensated compound Poisson process. We know their basic properties from Shreve [14]. In Chapter 2 we change the Brownian motion which is in a Brownian bridge to a compen-sated Poisson process. We will see some bridges with respect to the compencompen-sated Poisson process start from zero to fixed points and see a bridge between two independent com-pensated Poisson processes. In Chapter 3 we will construct a new comcom-pensated Poisson process and a new compensated compound Poisson process.

In the last chapter we introduce the fractional Brownian motion and its basic prop-erties. The fractional Brownian motion was first introduced by Kolmogorov [10]. Man-delbrot and Van Ness [11] established the integral representation for fractional Brownian motion on the whole real line. By the approach of [12], we have the integral representa-tion for fracrepresenta-tional Brownian morepresenta-tion on a finite interval. These integral representarepresenta-tions are all integrals of deterministic integrands with respect to the Brownian motion. Then we

1. INTRODUCTION 3

know that the fractional Brownian motion is adapted to the filtration which is generated by the Brownian motion. Gani, Heyde, Jagers, Kurtz [6] tell us a fractional Brownian motion is not a semimartingale, so we can’t use Itˆo stochastic calculus which is defined for semimartingales to define the stochastic integral with respect to the fractional Brownian motion. In Section 4.2 we have the definition for the Wiener integral of a deterministic integrand with respect to the fractional Brownian motion for the Hurst index H > 1 2 (see Gani, Heyde, Jagers, Kurtz [6]). From the definition we know that the Brownian motion which is in the integral representation for the fractional Brownian motion can be represented by an integral with respect to the fractional Brownian motion. Hence, we know that the Brownian motion and the fractional Brownian motion generate the same filtration. In Section 4.3 we will see a bridge with respect to the fractional Brownian motion starts from zero to a fixed point and a bridge between the fractional Brownian motion and a random variable.

CHAPTER 2

A Bridge with Respect to the Compensated Poisson Process

Let (Ω, F , P) be a probability space. In this chapter we consider the properties of the stochastic process if the Brownian bridge is driven by a compensated Poisson process instead of the Brownian motion. First we would introduce some basic properties of Poisson process.

2.1. Poisson Process

Definition 1. A random variable τ is said to have exponential distribution if τ is a random variable with the probability density function

f (t) =      λe−λt, if t ≥ 0, 0, if t < 0,

where λ is a positive constant. We also say that τ is an exponential random variable.

Let (τn)n∈N be a sequence of independent exponential random variables, all with the

same parameter λ. Let

Sn= n X k=1 τk, i.e., S1 = τ1, S2 = τ1+ τ2, · · · .

Definition 2. The Poisson Process (Nt) is defined by

Nt= inf {n − 1 : Sn> t} = max {n : Sn≤ t}.

Moreover, we say that (Nt) is a Poisson process with intensity λ.

The Poisson process (Nt) is right-continuous in t and it has stationary independent

increments, i.e., for 0 ≤ t0 < t1 < · · · < tm, the random variables

Nt1, Nt2 − Nt1, · · · , Ntm− Ntm−1

are stationary and independent. The mean and variance of Nt are given by

E [Nt] = λt and Var (Nt) = λt

respectively. The Poisson process is no more a martingale. We consider a martingale which has similar properties of Poisson process.

Definition 3. Let (Nt) be a Poisson process with intensity λ. The stochastic process

defined by

Mt= Nt− λt, t ≥ 0,

is called the compensated Poisson process.

Denote by (F_tN) and (F_tM) the filtrations generated by (Nt) and (Mt), respectively.

From the definition of the compensated Poisson process, we know that FM

t = FtN, for

all t ≥ 0. The compensated Poisson process (Mt) with intensity λ is a martingale with

respect to the filtration (F_tN). In next two sections we will discuss about some models with respect to the compensated Poisson process.

2.2. A Bridge Starts from Zero to a Fixed Point Consider the process

Xt= Mt− tM1, for 0 ≤ t ≤ 1

which is a bridge with respect to the compensated Poisson process from 0 to 0 on [0, 1]. Because the term M1 is in the difinition of Xt, for 0 ≤ t ≤ 1, the bridge Xtis not adapted

2.2. A BRIDGE STARTS FROM ZERO TO A FIXED POINT 7

to the filtration (F_tN). We shall later obtain a different process which is also from 0 to 0 but is adapted to the filtration (FN

t ).

We consider the stochastic differential equation dXt = dMt+

−Xt

1 − tdt (2.1) with the initial value X0 = 0. The equation can be solved by applying the Itˆo’s formula

to the function f (t, x) = x exp Z t 0 1 1 − sds  = x 1 − t. We have ft(t, x) = x (1 − t)2, fx(t, x) = 1 1 − t, fxx(t, x) = ftx(t, x) = fxt(t, x) = 0. The Itˆo’s formula implies

f (t, Xt) = Xt 1 − t = Z t 0 Xs (1 − s)2 ds + Z t 0 1 1 − sdXs. (2.2) From (2.1), we obtain Z t 0 Xt (1 − s)2 ds + Z t 0 1 1 − sdXs= Z t 0 1 1 − sdMs. (2.3) By (2.2) and (2.3), we have that the explicit formula of solution Xt, for 0 ≤ t < 1 is given

by Xt= (1 − t) Z t 0 1 1 − sdMs, for 0 ≤ t < 1. (2.4) Due to (2.4) we see that (Xt) is adapted to the filtration (FtN), for 0 ≤ t < 1. From

Shreve [14] we have the following theorem.

Theorem 4 (Theorem 11.4.5, Shreve [14]). Consider the jump process (Xt) given by

Xt = X0+ Z t 0 ΓsdBs+ Z t 0 Θsds + Jt,

where Γ, Θ are adapted processes, B is an adapted Brownian motion, and J is an adapted, right-continuous pure jump process with J0 = 0 having finitely many jumps on finite

interval. Assume the process (Xt) is a martingale, the integrand Φ is left-continuous and

adapted, and satisfies E Z t 0 Φ2_sΓ2_sds  < ∞, for all t ≥ 0.

Then the stochastic integral Z t

0

ΦsdXs is a martingale.

Since the compensated Poisson process (Mt) is a martingale, the process

Z t

0

1

1 − sdMs is also a martingale. Then Xt has zero mean for all t. Next, we compute the value of

the variance of Xt and use the mean and variance of Xt to see where the process (Xt)

approaches when t → 1−.

Theorem 5. Consider the process (Xt)0≤t<1 which is given by (2.4). For 0 ≤ t < 1,

we have that the variance of Xt is given by

Var(Xt) = −λt2+ λt. (2.5) Proof. For 0 ≤ t < 1, EXt2 = (1 − t)2E " Z t 0 1 1 − sdMs 2# .

We will apply the Itˆo’s formula to Z t

0

1

1 − sdMs 2

, so that we can get the value of E " Z t 0 1 1 − sdMs 2# . We set for 0 ≤ t < 1, Yt = Z t 0 1 1 − sdMs = Z t 0 1 1 − sdNs− λ Z t 0 1 1 − sds. Note that the continuous part of Yt, Ytc, is given by dY

c s =

−λ

1 − sds. Take f (x) = x

2 _so

that f0(x) = 2x, f00(x) = 2. The Itˆo’s formula implies f (Yt) = f (Y0) + Z t 0 f0(Ys) dYsc+ 1 2 Z t 0 f00(Ys) dYscdY c s + X 0<s≤t [f (Ys) − f (Ys−)] .

2.2. A BRIDGE STARTS FROM ZERO TO A FIXED POINT 9 Then we have Y_t2 = Y₀2+ Z t 0 2 YsdYsc+ 1 2 Z t 0 2 dY_scdY_sc+ X 0<s≤t Y2 s − Y 2 s−  = Z t 0 2 Ys  _−λ 1 − s  ds + X 0<s≤t Y2 s − Y 2 s− . (2.6)

Next, we take the expectation of both sides of (2.6) and use Fubini’s theorem

EYt2 = −2λ Z t 0 1 1 − sE [Ys] ds + E " X 0<s≤t Y_s2− Y2 s−
# . Since E [Ys] = 0, we obtain EYt2 = E " X 0<s≤t Y_s2− Y_s2−
# . (2.7) The sum X 0<s≤t Y_s2− Y2

s−
can be transformed to an integral with respect to the Poisson

process (Nt) by using the fact

Ys− Ys− = 1 1 − s∆Ns, and (Ys− Ys−)2 = 1 (1 − s)2∆Ns.

Then we have that

X 0<s≤t Y2 s − Y 2 s−  = X 0<s≤t (Ys− Ys−)2+ 2 (Y_s− Y_s−) Y_s− = X 0<s≤t  1 (1 − s)2 ∆Ns+ 2 1 1 − s∆NsYs−  = Z t 0  1 (1 − s)2 + 2 1 − sYs−  dNs. (2.8)

We change the form of the last integral so that we can get the mean of it easierly. Z t 0  1 (1 − s)2 + 2 1 − sYs−  dNs = Z t 0  1 (1 − s)2 + 2 1 − sYs−  dMs + Z t 0  1 (1 − s)2 + 2 1 − sYs−  λ ds. (2.9)

Since Y_s− is left continuous in s, for 0 ≤ s < 1 and (M_t) is a martingale, the process

Z t 0  1 (1 − s)2 + 2 1 − sYs− 

dMs is also a martingale. So we see that

E Z t 0  1 (1 − s)2 + 2 1 − sYs−  dMs  = 0. (2.10)

Due to (2.8), (2.9), (2.10) and by Fubini’s theorem, we have

E " X 0<s≤t Y_s2− Y2 s−
# = Z t 0  1 (1 − s)2 + 2 1 − sE [Ys−]  λ ds. (2.11)

Now we compute the value of the right side of (2.11). E [Ys−] = 0 since E [Y_s] = 0, for

0 ≤ s < 1. From (2.7) and (2.11), we obtain

EXt2  = (1 − t)2_EY_t2 = (1 − t)2 Z t 0 1 (1 − s)2 λ ds = −λt2+ λt.

Since E [Xt] = 0, we know that Var(Xt) = −λt2+ λt. 

We have known the mean and variance of Xt, then we can use these to see where

the process (Xt)0≤t<1 approaches as time approaches 1. Since E [Xt] = 0 and Var(Xt) =

−λt2_{+ λt which converges to 0 as t → 1}−_{, we obtain that X}

t → 0 P-a.s. as t → 1−.

The process (Xt)0≤t<1 is a bridge from 0 to 0 on [0, 1] and it is adapted to the filtration

FN t

2.3. A BRIDGE BETWEEN TWO INDEPENDENT COMPENSATED POISSON PROCESSES 11

In the following we see a process with respect to the compensated Poisson process starting from 0 to b, for some constant b. Consider the stochastic differential equation

dXt= dMt+

b − Xt

1 − t dt, (2.12) where (Mt) is a compensated Poisson process with intensity λ, b is a constant and X0 = 0.

We can solve the stochastic differential equation by the same method as before. Then we obtain the solution Xt, for 0 ≤ t < 1 which is given by

Xt= (1 − t)

Z t 0

1

1 − sdMs+ bt, for 0 ≤ t < 1. (2.13) Then we obtain E [Xt] = bt. By Theorem 5, we have

Var (Xt) = Var  (1 − t) Z t 0 1 1 − sdMs  = −λt2 + λt.

Since we have known that the first term of (2.13) converges to 0 P-a.s as t → 1−, we have that Xt→ b P-a.s. as t → 1−. Hence, we see that the process (Xt) is a bridge from 0 to b

and is adapted to the filtration F_tN
. We have discussed about a bridge with respect to the compensated Poisson process from 0 to a fixed point. In next section, we will discuss about a bridge between two independent compensated Poisson process.

2.3. A Bridge between two Independent Compensated Poisson Processes We consider the stochastic differential equation

dXs = dMs+

˜ Ms− Xs

1 − s ds,

where M is a compensated Poisson process with intensity λ, ˜M is another compensated Poisson process which has the same intensity λ is independent of M and X starts from 0.

We can also use the same method as before to solve the stochastic differential equation and get the solution Xt, for 0 ≤ t < 1, is given by

Xt= (1 − t) Z t 0 1 1 − sdMs+ (1 − t) Z t 0 ˜ Ms (1 − s)2 ds, for 0 ≤ t < 1. (2.14) The process (Xt)0≤t<1 has another form by applying Itˆo product rule to the second term

of (2.14) Xt = (1 − t) Z t 0 1 1 − sdMs+ (1 − t)  ˜ Mt 1 1 − t − Z t 0 1 1 − sd ˜Ms  = (1 − t) Z t 0 1 1 − sdMs+ ˜Mt− (1 − t) Z t 0 1 1 − sd ˜Ms. (2.15) We have known that the process which has the form as (2.4) converges to 0 P-a.s. as t → 1−_{, so the first term and the third term in (2.15) both converge to 0 P-a.s. as t → 1}−. Then Xt → ˜M1 P-a.s. as t → 1−. The process (Xt)0≤t<1 is a bridge from 0 to ˜M1 on

[0, 1].

Remark 6. Suppose that the process (Xt)0≤t<1 is given by

Xt= Mt+ Z t 0 ˜ Ms− Xs 1 − s ds. (2.16) The filtration (F_tX)0≤t<1 which is generated by (Xt)0≤t<1 contains the information about

when (Xt)0≤t<1 jumps. Since the second term of (2.16) is continuous in t, (Xt)0≤t<1

jumps at the same time as (Mt)0≤t<1 jumps. So the filtration (FtX)0≤t<1 contains the

information about when (Mt)0≤t<1 jumps, i.e., (Mt)0≤t<1 is adapted to (FtX)0≤t<1. From

(2.16), we know that ( ˜Mt) is also adapted to (FtX)0≤t<1. Hence, we obtain the

Doob-Meyer decomposition of (Xt)0≤t<1 in its own filtration FtX

0≤t<1. We may regard X as

2.3. A BRIDGE BETWEEN TWO INDEPENDENT COMPENSATED POISSON PROCESSES 13

Consider two independent compensated Poisson processes (Mt)t≥0 and ( ˜Mt)t≥0. The

stochastic process X is given by

Xt= Mt+

Z t

0

Zsds, (2.17)

where M is a Poisson process and the drift Z depends linearly on X and ˜M . We want to characterize those cases where X is again a compensated Poisson process. Since the second term of (2.17) is continuous in t, we have that (Xt) jumps at the same time as

(Mt) jumps. So (Mt) is adapted to the filtration FtX
. Suppose (Xt) is a compensated

Poisson process. Since X and M all start from 0, and they jump at the same time, X is just equal to M . So the process

Z t

0

Zsds is equal to zero. Recall that the process

(Xt)0≤t<1 which is given by (2.16) is a semimartingale in its natural filtration FtX

0≤t<1,

but it is not a compensated Poisson process since the second term of (2.16) is not equal to zero.

The integral with respect to time in (2.17) makes X and M jump simultaneously. In next chapter, we will transform the second term of (2.17) to an integral with respect to the compensated Poisson process which is independent of M . Then the process M and the integral together decide when the process X jumps.

CHAPTER 3

Construction of a New Compensated Compound Poisson

Process

We have discussed about whether X is a compensated Poisson process in the model

Xt= Mt+

Z t 0

Zsds.

We will transform the above integral to an integral with respect to the compensated Poisson process which is independent of M and characterize those cases where X is again a compensated Poisson process. Since the jump size of Poisson process is equal to 1, we will extend the discussion to the compensated compound Poisson process which has random jump sizes.

3.1. Construction of a New Compensated Poisson Process We consider the process X which is given by

Xt = Mt+

Z t

0

f (s) d ˜Ms, (3.1)

where M is a compensated Poisson process with intensity λ, ˜M is another compensated Poisson process with intensity ˜λ is independent of M and f is a deterministic differentiable function. We want to know the form of the moment generating function of Xt, for t ≥ 0,

so that we can see in which cases the process X given by (3.1) is again a compensated Poisson process. The moment generating function of Xt, for t ≥ 0 is given by

ϕXt(u) = E [exp {uMt}] · E

 exp  u Z t 0 f (s) d ˜Ms  . 15

In Shreve [14] we have known that the moment generating function for the compensated Poisson process Mt is given by

E [exp {uMt}] = exp {λt (eu− u − 1)} .

So we only need to focus on the expectation E  exp  u Z t 0 f (s) d ˜Ms  . Lemma 1. Consider the process

Z t

0

f (s) d ˜Ms,

where ˜M is a compensated Poisson process with intensity ˜λ and f is a nonrandom differ-entiable function. Then it’s moment generating function is given by

E  exp  u Z t 0 f (s) d ˜Ms  = exp  ˜ λ Z t 0 euf (s)− 1
ds − u Z t 0 f (s) ds  .

Proof. We will apply the Itˆo’s formula to exp  u Z t 0 f (s) d ˜Ms− ˜λ Z t 0 euf (s)− 1
ds − u Z t 0 f (s) ds  , so that we can know it is a martingale. We set for t ≥ 0,

Yt= u Z t 0 f (s) d ˜Ms− ˜λ Z t 0 euf (s)− 1
ds − u Z t 0 f (s) ds  and Zt= exp {Yt} .

Note that the continuous part of Ys, Ysc, is given by

dY_sc= ˜λ −euf (s)+ 1
ds.

Take f (x) = ex _{so that f}0_{(x) = e}x_{, f}00_{(x) = e}x_{. The Itˆo’s formula implies}

Zt= Z0+ Z t 0 ZsdYsc+ 1 2 Z t 0 ZsdYscdY c s + X 0<s≤t [Zs− Zs−] . (3.2)

3.1. CONSTRUCTION OF A NEW COMPENSATED POISSON PROCESS 17

Since Y_scis the continuous part of Ys, we can change the integrand Zswhich is in the second

term of (3.2) to Zs−. When the Poisson process ˜N jumps at time s, Z_s = Z_s− × euf (s).

When ( ˜Nt) does not jump at time s, Zs= Zs−. So we have

Zs− Zs− = Z_s− euf (s)− 1
∆ ˜N_s. Then Zt = 1 + ˜λ Z t 0 Zs− −euf (s)+ 1
ds + X 0<s≤t h Zs− euf (s)− 1
∆ ˜N_s i = 1 + Z t 0 Zs− euf (s)− 1
d ˜M_s− ˜N_s  + Z t 0 Zs− euf (s)− 1
d ˜N_s = 1 + Z t 0 Zs− euf (s)− 1
d ˜M_s.

Since M is a martingale and Zs− euf (s)− 1
is left continuous in s, the above integral is

also a martingale. So the process (Zt) is a martingale and we have E [Zt] = 1, i.e.,

E  exp  u Z t 0 f (s) d ˜Ms− ˜λ Z t 0 euf (s)− 1
ds − u Z t 0 f (s) ds  = 1. Hence, we obtain E  exp  u Z t 0 f (s) d ˜Ms  = exp  ˜ λ Z t 0 euf (s)− 1
ds − u Z t 0 f (s) ds  .  Moreover, we have that the moment generating function of Xt is given by

ϕXt(u) = exp {λt (e u_{− u − 1)} · exp}  ˜ λ Z t 0 euf (s)− 1
ds − u Z t 0 f (s) ds  = exp  λt (eu− u − 1) + ˜λ Z t 0 euf (s)− 1
ds − ˜λu Z t 0 f (s) ds  . (3.3) Next, we use (3.3) to see in which cases the process X is a compensated Poisson process. If f ≡ 0, then it is obvious that Xt= Mt. In the following proposition we set f 6= 0.

Proposition 2. Let the stochastic process (Xt) satisfy (3.1). Then (Xt) is a

compen-sated Poisson process if and only if f ≡ 1. Moreover, (Xt) has the intensity λ + ˜λ.

Proof. “ =⇒ ” : The moment generating function for compensated Poisson process must be as the form expnˆλt (eu− u − 1)o, for some constant ˆλ. Suppose that (Xt) is

a compensated Poisson process with intensity ˆλ. We let the moment generating function of Xt equal to  expnˆλt (eu− u − 1)o, i.e. exp  λt (eu− u − 1) + ˜λ Z t 0 euf (s)− 1
ds − ˜λu Z t 0 f (s) ds  = expnˆλt (eu− u − 1)o. Then we get the equation

λt (eu− u − 1) + ˜λ Z t 0 euf (s)− 1
ds − ˜λu Z t 0 f (s) ds = ˆλt (eu− u − 1) . (3.4) We differentiate with respect to t on both sides of (3.4)

λ (eu − u − 1) + ˜λ euf (t)− 1
− ˜λuf (t) = ˆλ (eu− u − 1) . We differentiate with respect to t again

˜

λueuf (t)f0(t) − ˜λuf0(t) = 0 . Then we get

˜

λuf0(t) euf (t)− 1
= 0 .

This implies that f0(t) = 0 or euf (t) − 1 = 0. So we have that f (t) is a constant. Set f ≡ C, where C is a positive constant. From (3.1), we have

Xt= Mt+ C ˜Mt.

The moment generating function of Xt is given by

ϕX(u) = exp

n

λt (eu − u − 1) + ˜λt euC − uC − C
o .

3.1. CONSTRUCTION OF A NEW COMPENSATED POISSON PROCESS 19

Suppose the following equation holds

expnλt (eu− u − 1) + ˜λt euC − uC − C
o

= expnˆλt (eu− u − 1)o. Then we have

λt (eu− u − 1) + ˜λt euC − uC − C
= ˆλt (eu− u − 1) . We differentiate with respect to u twice, then we get

λeu+ ˜λC2euC = ˆλeu. We multiply e−u on both sides of the above formula

λ + ˜λC2eu(C−1) = ˆλ . Then eu(C−1) = ˆ λ − λ ˜ λC2 .

Hence, we obtain C = 1 and then ˆλ = λ + ˜λ .

“ ⇐= ” : Since f ≡ 1, we have Xt = Mt+ ˜Mt. We will show that the law of X agrees with

the law of a compensated Poisson process which has intensity λ + ˜λ. Denote by F_tM, ˜M the filtration generated by M and ˜M . Let

Zt= exp {uMt− λt (eu− u − 1)} .

By the proof of Lemma 1, we have that the process (Zt) is a martingale with respect to

 F_tM, ˜M. Let ˜ Zt= exp n u ˜Mt− ˜λt (eu− u − 1) o . ( ˜Zt) is also a martingale with respect to



F_tM, ˜M. The two processes (Zt) and ( ˜Zt) are

independent and they are all martingales with respect to the filtration F_tM, ˜M. From (3.3), we know that the moment generating function of Xt is

ϕXt(u) = exp

n

For fixed u ∈ R, the process Vt(u) is defined by

V_t(u) = expnuXt−



λ + ˜λt (eu− u − 1)o = ZtZ˜t.

We will show that V_t(u) is a martingale with respect to the filtration F_tM, ˜M. For 0 < s < t, E  V_t(u)     FM, ˜M s  = E  ZtZ˜t     FM, ˜M s  = E  (Zt− Zs) ˜Zt− ˜Zs  + ZsZ˜t+ ZtZ˜s− ZsZ˜s     FM, ˜M s  = E  (Zt− Zs) ˜Zt− ˜Zs     FM, ˜M s  + E  ZsZ˜t     FM, ˜M s  + E  ZtZ˜s     F_sM, ˜M  − E  ZsZ˜s     F_sM, ˜M  .

Since Zt− Zs, ˜Zt− ˜Zs are independent of FsM, ˜M and Zs, ˜Zs are adapted to FsM, ˜M, we

have E  V_t(u)     F_sM, ˜M  = E h (Zt− Zs) ˜Zt− ˜Zs i + ZsE  ˜ Zt     F_sM, ˜M  + ˜ZsE  Zt     F_sM, ˜M  − ZsZ˜s = E [(Zt− Zs)] · Eh ˜Zt− ˜Zs i + ZsZ˜s+ ZsZ˜s− ZsZ˜s = ZsZ˜s= Vs(u).

So the process V_t(u) is a martingale with respect to F_tM, ˜M. For fixed u2 ∈ R and

0 < t1 < t2, V(u2) t1 = E  V(u2) t2     F_tM, ˜₁ M  .

3.1. CONSTRUCTION OF A NEW COMPENSATED POISSON PROCESS 21

Now fixed u1 ∈ R. Since

V(u1) t1 V(u2) t1 is adapted to F_tM, ˜₁ M, we have V(u1) t1 = E " V(u1) t1 V (u2) t2 V(u2) t1     F_tM, ˜₁ M # = E  exp {u1Xt1 + u2(Xt2 − Xt1)}     F_tM, ˜₁ M  · expn−λ + ˜λt1(eu1 − u1− 1) o · expn−λ + ˜λ(t2− t1) (eu2 − u2 − 1) o . Now we use the martingale property of V(u1)

t and we take expectation of both sides of the

above formula 1 = V(u1) 0 = EhV(u1) t1 i = E [exp {u1Xt1 + u2(Xt2 − Xt1)}] · expn−λ + ˜λ  t1(eu1 − u1− 1) o · expn−λ + ˜λ  (t2− t1) (eu2 − u2− 1) o . So we obtain E [exp {u1Xt1 + u2(Xt2 − Xt1)}] = exp n λ + ˜λ  t1(eu1 − u1 − 1) o · expnλ + ˜λ  (t2 − t1) (eu2 − u2− 1) o . Since the above joint moment generating function factors into the product of moment generating functions, Xt1 and Xt2 − Xt1 must be independent. We also know that the

moment generating function of Xt2 − Xt1 is

ϕX_t2−X_t1(u) = exp

n

λ + ˜λ(t2− t1) (eu− u − 1)

o .

Next, we computer the joint moment generating function of the random variables Xt1,

Poisson process. For 0 < t1 < t2 < · · · < tn, the joint moment generating function of the

random variables Xt1, Xt2, · · · , Xtn is given by

ϕX_t1,X_t2,··· ,Xtn(u1, u2, · · · , un)

= Eexp unXtn+ un−1Xtn−1+ · · · + u1Xt1



= Eexp un Xtn − Xtn−1
+ (un−1+ un) Xtn−1− Xtn−2
+ · · · + (u1+ u2+ · · · un) Xt1



= Eexp un Xtn − Xtn−1
 · E exp (un−1+ un) Xtn−1− Xtn−2
 ·

· · · E [exp {(u1+ u2+ · · · un) Xt1}] .

We have known the form of the moment generating function of increments of X, then we obtain ϕX_t1,X_t2,··· ,Xtn (u1, u2, · · · , un) = expnλ + ˜λ(tn− tn−1) (eun− un− 1) o · expnλ + ˜λ(tn−1− tn−2) e(un−1+un)+ (un−1+ un) − 1
o · · · · expnλ + ˜λt1 e(u1+u2+···+un)− (u1+ u2+ · · · + un) − 1
o .

This is the moment generating function for a compensated Poisson process with intensity λ + ˜λ. This completes the proof. _

3.2. Compensated Compound Poisson Process

Let (Nt) be a Poisson process with intensity λ and let Y1, Y2, . . . be a sequence of

independent, identically distributed random variables with mean β, where β = E [Yi].

The random variables Y1, Y2, . . . are independent of the Poisson process (Nt).

Definition 3. The stochastic process (Qt) defined by

Qt= Nt

X

i=1

3.3. CONSTRUCTION OF A NEW COMPENSATED COMPOUND POISSON PROCESS 23

is called the compound Poisson process.

The compound Poisson process (Qt) jumps at the same time as the Poisson process

(Nt) jumps. The jump sizes of the compound Poisson process are random. The

com-pensated compound Poisson process (Qt− βλt) is a martingale. In this chapter, we only

regard the compound Poisson process which has finitely many possible jump sizes on finite interval. The following theorem says that a compound Poisson process can be regarded as a sum of independent Poisson processes each has fixed jump-size.

Theorem 4 (Shreve [14] Theorem 11.3.3.). Let y1, y2, . . . , yM be a finite set of nonzero

numbers and let p(y1), p(y2),. . . , p(yM) be positive numbers that sum to 1. Let Y1, Y2, . . .

be a sequence of independent, identically distributed random variables with P (Yi = ym) =

p(ym), m = 1, . . . , M . Let (Nt) be a Poisson process with intensity λ and define the

compound Poisson process

Qt= Nt

X

i=1

Yi.

For m = 1, . . . , M , let N_t(m) denote the number of jumps in Q of size ym in [0, t]. Then

Nt= M X m=1 N_t(m) and Qt= M X m=1 ymN (m) t ,

where the process N(1)_{, . . . , N}(M ) _{are independent Poisson processes and each N}(m) _has

intensity λp(ym).

The theorem tells us the fact that a compound Poisson process can be represented by some independent Poisson processes each has fixed jump-size. We will use this theorem to construct a new compensated compound Poisson process.

3.3. Construction of a New Compensated Compound Poisson Process We consider two independent compound Poisson process which have some conditions as follows. Let y1, y2, . . . , yM be a finite set of nonzero numbers and let p(y1), p(y2),

. . . , p(yM) be positive numbers whose summation is identical to 1. Let Y1, Y2, . . . be

a sequence of independent, identically distributed random variables with P (Yi = ym) =

p(ym), m = 1, . . . , M and E [Yi] = β. Let (Nt) be a Poisson process with intensity λ and

define the compound Poisson process Qt=

Nt

X

i=1

Yi. (3.5)

For m = 1, . . . , M , let N_t(m) denote the number of jumps in Q of size ym in [0, t]. Then

we have Qt= M X m=1 ymNt(m),

where the process N(1), . . . , N(M ) are independent Poisson process, and each N(m) has intensity λp(ym).

Let ˜y1, ˜y2, . . . , ˜y_Mˆ be another finite set of nonzero numbers and let ˜p(˜y1), ˜p(˜y2),

. . . , ˜p(˜y_Mˆ) be positive numbers that sum to 1. Let ˜Y1, ˜Y2, . . . be another sequence of

independent, identically distributed random variables with P ˜Yi = ˜ym



= ˜p(˜ym), m =

1, . . . , ˜M , E[ ˜Yi] = ˜β and ˜Y1, ˜Y2, . . . are independent of the sequence Y1, Y2, . . .. Let ( ˜Nt) be

a Poisson process with intensity ˜λ and it is independent of (Nt). Define another compound

Poisson process ˜ Qt= ˜ Nt X i=1 ˜ Yi. (3.6)

For m = 1, . . . , ˜M , let ˜N_t(m) denote the number of jumps in ˜Q of size ˜ym in [0, t]. Then

we have ˜ Qt= ˜ M X m=1 ˜ ymN˜ (m) t ,

where the process ˜N(1)_{, . . . , ˜N}( ˜M ) _{are independent Poisson process, and each ˜N}(m) _has

intensity ˜λ˜p(˜ym).

Consider the stochastic process X which is given by Xt = (Qt− βλt) + Z t 0 f (s) d ˜Qs− ˜β ˜λs  , (3.7)

3.3. CONSTRUCTION OF A NEW COMPENSATED COMPOUND POISSON PROCESS 25

where f is a nonrandom differentiable function and f 6= 0. We use the similar method in Section 3.1 to see that in which cases X is again a compound Poisson process. First, we want to know the form of the moment generating function of Xt, for t ≥ 0. The moment

generating function of Xt, for t ≥ 0 is given by

ϕXt(u) = E [exp {u (Qt− βλt)}] · E  exp  u Z t 0 f (s) d ˜Qs− ˜β ˜λs  . (3.8)

The following theorem tells us the form of the moment generating function for a compound Poisson process, so that we can get the form of the moment generating function of Xt.

Theorem 5 (Shreve [14] Section 11.3.2). The moment generating function for the compound Poisson process (Qt) defined as (3.5) is given by

ϕQt(u) = exp {λt (ϕY1(u) − 1)} .

By the above theorem, we know that the moment generating function of (Qt− βλt)

is given by ϕ(Qt−βλt)(u) = exp ( λt M X m=1 p(ym) (euym − 1) − uβλt ) . (3.9)

We remain to obtain the form of the moment generating function of Z t

0

f (s) d ˜Qs− ˜β ˜λs

 , so that we can get the form of the moment generating function of Xt.

Theorem 6. Consider the process

Z t

0

f (s) d ˜Qs− ˜β ˜λs

 ,

where ( ˜Qt) is given by (3.6) with intensity ˜λ, ˜β = E[ ˜Yi], and f is a nonrandom

differen-tiable function. Then its moment generating function is given by

E  exp  u Z t 0 f (s) d ˜Qs− ˜β ˜λs  = exp    ˜ λ ˜ M X m=1  ˜ p(˜ym) Z t 0 euf (s)˜ym− 1
ds     · exp  −u ˜β ˜λ Z t 0 f (s) ds  . Proof. E  exp  u Z t 0 f (s) d ˜Qs− ˜β ˜λs  = E  exp  u Z t 0 f (s) d ˜Qs  · exp  −u ˜β ˜λ Z t 0 f (s) ds  . (3.10)

We focus on the first term of (3.10). Using the fact taht ˜N(1), . . . , ˜N( ˜M ) are independent Poisson processes, we know that

E  exp  u Z t 0 f (s) d ˜Qs  = E  exp    u Z t 0 f (s) d   ˜ M X m=1 ˜ ymN˜s(m)        = ˜ M Y m=1 E  exp  u Z t 0 f (s)˜ymd ˜Ns(m)  . (3.11)

From Theorem 1, we have that for 1 ≤ m ≤ ˜M ,

E  exp  u Z t 0 f (s)˜ymd ˜Ns(m)  = exp  ˜ λ˜p(˜ym) Z t 0 euf (s)˜ym − 1
ds  . (3.12)

Due to (3.10), (3.11) and (3.12), we have

E  exp  u Z t 0 f (s) d ˜Qs− ˜β ˜λt  = exp    ˜ λ ˜ M X m=1  ˜ p(˜ym) Z t 0 euf (s)˜ym− 1
ds     · exp  −u ˜β ˜λ Z t 0 f (s) ds  .

3.3. CONSTRUCTION OF A NEW COMPENSATED COMPOUND POISSON PROCESS 27

From (3.8), (3.9) and Theorem 6, we obtain that the moment generating function of Xt is given by ϕXt(u) = exp    λt M X m=1 p(ym) (euym− 1) − uβλt + ˜λ ˜ M X m=1  ˜ p(˜ym) Z t 0 euf (s)˜ym− 1
ds  −u ˜β ˜λ Z t 0 f (s) ds  . (3.13) Next, we want to see in which cases the process X is a compensated compound Poisson process.

Proposition 7. Let the stochastic process (Xt) satisfy (3.7). Then (Xt) is a

com-pensated compound Poisson process if and only if f ≡ C. Moreover, (Xt) has intensity



λ + ˜λ.

Proof. “ =⇒ ” : Suppose that (Xt) is a compensated compound Poisson process

with intensity ˆλ. We let the moment generating function of Xt be equal to

exp    ˆ λt ˆ M X m=1 ˆ p(ˆym) euˆym − 1
− u ˆβ ˆλt    , (3.14)

for some ˆM , ˆβ, ˆym and ˆp(ˆym), for 1 ≤ m ≤ ˆM . Let ˆYi denote the size of the ith

jump for X, for i ≥ 1. Then ˆY1, ˆY2, . . . are independent and from (3.14) we know that

the distribution of finitely many jump sizes of X is given by P ˆYi = ˆym



= ˆp(ˆym), for

1 ≤ m ≤ ˆM . The mean of ˆYi is equal to ˆβ. Suppose that the following equation holds

exp    λt M X m=1 p(ym) (euym − 1) − uβλt + ˜λ ˜ M X m=1  ˜ p(˜ym) Z t 0 euf (s)˜ym− 1
ds  − u ˜β ˜λ Z t 0 f (s) ds    = exp    ˆ λt ˆ M X m=1 ˆ p(ˆym) euˆym − 1
− u ˆβ ˆλt    .

Then we have the equation λt M X m=1 p(ym) (euym− 1) − uβλt + ˜λ ˜ M X m=1  ˜ p(˜ym) Z t 0 euf (s)˜ym− 1
ds  − u ˜β ˜λ Z t 0 f (s) ds = ˆλt ˆ M X m=1 ˆ p(ˆym) euˆym− 1
− u ˆβ ˆλt.

We differentiate with respect to t on both sides of the above equation

λ M X m=1 p(ym) (euym− 1) − uβλ + ˜λ ˜ M X m=1 ˜ p(˜ym) euf (t)˜ym− 1
− u ˜β ˜λf (t) = ˆλ ˆ M X m=1 ˆ p(ˆym) euˆym− 1
− u ˆβ ˆλ .

If we differentiate with respect to t again, then we obtain

u˜λf0(t) ˜ M X m=1 ˜ p(˜ym)˜ym euf (t)˜ym
− u ˜β ˜λf 0 (t) = 0 . Then u˜λf0(t)   ˜ M X m=1 ˜ p(˜ym)˜ym euf (t)˜ym
− ˜β  = 0 . This implies that f0(t) = 0 or

˜ M X m=1 ˜ p(˜ym)˜ym euf (t)˜ym
− ˜β = 0 . (3.15)

We differentiate with respect to t on both sides of (3.15), then we have

uf0(t) ˜ M X m=1 ˜ p(˜ym)(˜ym)2 euf (t)˜ym
= 0 .

So we know that f must be a constant. Set f ≡ C, where C is a constant. From (3.7), we have

Xt= (Qt− βλt) + C ˜Qt− ˜β ˜λt

 .

3.3. CONSTRUCTION OF A NEW COMPENSATED COMPOUND POISSON PROCESS 29

The moment generating function of Xt is given by

ϕXt(u) = exp    λt M X m=1 p(ym) (euym − 1) − uβλt + ˜λt ˜ M X m=1 ˜ p(˜ym) euC ˜ym− 1
− u ˜β ˜λCt    = exp     λ + ˜λt   M X m=1  λp(ym) λ + ˜λ  (euym_{− 1) +} ˜ M X m=1 ˜_λ˜p(˜y m) λ + ˜λ ! euC ˜ym_{− 1}
  −uλ + ˜λt λβ λ + ˜λ + ˜ λC ˜β λ + ˜λ !) .

This implies that X is a compensated Poisson process with intensity λ + ˜λ and the distribution for finitely many jump sizes of X is given by

P ˆYi = ym  = λp(ym) λ + ˜λ , for 1 ≤ m ≤ M and P ˆYi = C ˜yn  = ˜ λ˜p(˜yn) λ + ˜λ , for 1 ≤ n ≤ ˜M . We also know that the mean of ˆYi, for i ≥ 1 is given by

E[Yˆi] = λβ λ + ˜λ + ˜ λC ˜β λ + ˜λ. “ ⇐= ” : If f ≡ C, then we have Xt = (Qt− βλt) + C ˜Qt− ˜β ˜λt 

. We will show that the law of X agrees with the law of a compensated compound Poisson process which has intensity λ + ˜λ. Set Zt= exp ( u (Qt− βλt) − λt M X m=1 p(ym) (euym − 1) − uβλt !) . We will show that (Zt) is a martingale. Since β =PM_m=1ymp(ym), we obtain

Qt− βλt = M X m=1 ym  N_t(m)− λp(ym)t  . Then we have Zt= exp ( u M X m=1 ym  N_t(m)− λp(ym)t  − λt M X m=1 p(ym) (euym− 1) − uβλt !) .

Set Yt = u M X m=1 ym  N_t(m)− λp(ym)t  − λt M X m=1 p(ym) (euym− 1) − uβλt ! . Note that dY_sc = −λ M X m=1 p(ym) (euym− 1) ! ds.

Take f (x) = ex _{so that f}0_{(x) = e}x_{, f}00_{(x) = e}x_{. The Itˆo’s formula implies}

Zt= Z0+ Z t 0 ZsdYsc+ 1 2 Z t 0 ZsdYscdY c s + X 0<s≤t [Zs− Zs−] . (3.16) Since Yc

s is the continuous part of Ys, we can change the integrand Zs which is in the

second term of (3.16) to Zs−. When the compound Poisson process Q jumps at time s,

the jump size of Q at time s must be equal to one of y1, y2,· · · , yM. If the jump size of Q

at time s is equal to ym, for some m, then we have Zs = Zs−× euym. If Q does not jump

at time s, then Zs = Zs−. So we have

Zs− Zs− = M X m=1 Z_s−(euym− 1) ∆N_s(m). Then Zt = 1 + Z t 0 Zs− −λ M X m=1 p(ym) (euym − 1) ! ds + X 0<s≤t M X m=1 Zs−(euym− 1) ∆N(m) s .

Set M_t(m) = N_t(m) − λ p(ym) t, for 1 ≤ m ≤ M . Then M(m) is a compensated Poisson

process with intensity λp(ym) and M(m) is a martingale. We have

Zt = 1 + M X m=1 Z t 0 Zs−(euym − 1) d M_s(m)− N_s(m)
+ M X m=1 X 0<s≤t Zs−(euym − 1) ∆N_s(m) = 1 + M X m=1 Z t 0 Zs−(euym − 1) d M_s(m)− N_s(m)
+ M X m=1 Z t 0 Zs−(euym− 1) dN_s(m) = 1 + M X m=1 Z t 0 Zs−(euym − 1) dM_s(m).

3.3. CONSTRUCTION OF A NEW COMPENSATED COMPOUND POISSON PROCESS 31

Since M(m)is a martingale and Zs−(euym− 1) is left continuous in s,

Z t

0

Zs−(euym− 1) dM_s(m)

is also a martingale. The sum of finitely many martingales is a martingale, so the pro-cess (Zt) is a martingale. Denote by



F_tN(1),··· , N(M ), ˜N(1),··· , ˜N( ˜M ) the filtration generated by N(1), · · · , N(M ), ˜N(1), · · · , ˜N( ˜M ). The process (Zt) is a martingale with respect to

 F_tN(1),··· , N(M ), ˜N(1),··· , ˜N( ˜M ). Set ˜ Zt= exp    uC ˜Qt− ˜β ˜λt  −  λt˜ ˜ M X m=1 ˜ p(˜ym) euC ˜ym − 1
− uC ˜β ˜λt      .

By the similar method, we have that ˜Zt is also a martingale with respect to



F_tN(1),··· , N(M ), ˜N(1),··· , ˜N( ˜M ). The two processes (Zt) and ( ˜Zt) are independent and they

are all martingales with respect to the filtrationF_tN(1),··· , N(M ), ˜N(1),··· , ˜N( ˜M ). From (3.13), we know that the moment generating function of Xt is given by

ϕXt(u) = exp    λt M X m=1 p(ym) (euym − 1) − uβλt + ˜λt ˜ M X m=1 ˜ p(˜ym) euC ˜ym− 1
− uC ˜β ˜λt    = exp     λ + ˜λt   M X m=1  λp(ym) λ + ˜λ  (euym_{− 1) +} ˜ M X m=1 ˜_λ˜p(˜ym) λ + ˜λ ! euC ˜ym_{− 1}
  −uλ + ˜λt λβ λ + ˜λ + C ˜λ ˜β λ + ˜λ !) . For the sake of simplicity, we let

ηt(u) = exp     λ + ˜λt   M X m=1  λp(ym) λ + ˜λ  (euym_{− 1) +} ˜ M X m=1 ˜_λ˜p(˜y m) λ + ˜λ ! euC ˜ym_{− 1}
  −uλ + ˜λt λβ λ + ˜λ + C ˜λ ˜β λ + ˜λ !) .

For fixed u ∈ R, the process Vt(u) is defined by

V_t(u) = exp {uXt} ηt(u)

Since Zt− Zs, ˜Zt− ˜Zs are independent of FN

(1)_{,··· , N}(M )_{, ˜N}(1)_{,··· , ˜N}( ˜M )

s and Zs, ˜Zsare adapted

to FN(1)_{,··· , N}(M )_{, ˜N}(1)_{,··· , ˜N}( ˜M )

s , we can use the same method as the proof in Proposition 2

to show that V_t(u) = ZtZ˜t is a martingale with respect to

 F_tN(1),··· , N(M ), ˜N(1),··· , ˜N( ˜M ). For fixed u2 ∈ R and 0 < t1 < t2, V (u2) t1 = E  V(u2) t2     F_tM, ˜₁ M 

. Now fixed u1 ∈ R. Since

V(u1) t1 V(u2) t1 is adapted to F_tN₁ (1),··· , N(M ), ˜N(1),··· , ˜N( ˜M ), we have V(u1) t1 = E " V(u1) t1 V (u2) t2 V(u2) t1     F_tN₁ (1),··· , N(M ), ˜N(1),··· , ˜N( ˜M ) # = E  exp {u1Xt1 + u2(Xt2 − Xt1)}     F_tN₁ (1),··· , N(M ), ˜N(1),··· , ˜N( ˜M )  · η−1_t1 (u1) · η−1t2−t1(u2).

Now we use the martingale property of V(u1)

t . We take expectation of both sides of the

above formula 1 = E h V(u1) t1 i = E [exp {u1Xt1 + u2(Xt2 − Xt1)}] · η −1 t1 (u1) · η −1 t2−t1(u2). So we obtain E [exp {u1Xt1 + u2(Xt2 − Xt1)}] = ηt1(u1) · ηt2−t1(u2).

Since the above joint moment generating function factors into the product of moment generating functions, Xt1 and Xt2 − Xt1 must be independent. We also know that the

moment generating function of Xt2 − Xt1 is

ϕX_t2−X_t1(u) = exp     λ + ˜λ(t2− t1)   M X m=1  λp(ym) λ + ˜λ  (eu2ym_{− 1) +} ˜ M X m=1 ˜_λ˜p(˜y m) λ + ˜λ ! eu2C ˜ym _{− 1}
  −u2  λ + ˜λ(t2− t1) λβ λ + ˜λ + C ˜λ ˜β λ + ˜λ !) .

3.3. CONSTRUCTION OF A NEW COMPENSATED COMPOUND POISSON PROCESS 33

For 0 < t1 < t2 < · · · < tn, the joint moment generating function of the random variables

Xt1, Xt2,· · · , Xtn is given by

ϕX_t1,X_t2,··· ,Xtn(u1, u2, · · · , un)

= Eexp un Xtn − Xtn−1
 · E exp (un−1+ un) Xtn−1− Xtn−2
 ·

· · · E [exp {(u1+ u2+ · · · + un) Xt1}] = exp     λ + ˜λ(tn− tn−1)   M X m=1  λp(ym) λ + ˜λ  (eunym_{− 1) +} ˜ M X m=1 ˜_λ˜p(˜y m) λ + ˜λ ! eunC ˜ym_{− 1}
  −un  λ + ˜λ(tn− tn−1) λβ λ + ˜λ + C ˜λ ˜β λ + ˜λ !) · exp (  λ + ˜λ(tn−1− tn−2) " _M X m=1  λp(y_m) λ + ˜λ  e(un−1+un)ym_{− 1}
+ ˜ M X m=1 ˜_λ˜p(˜y m) λ + ˜λ ! e(un−1+un)C ˜ym_{− 1}
  −(un−1+ un)  λ + ˜λ(tn−1− tn−2) λβ λ + ˜λ + C ˜λ ˜β λ + ˜λ !) · · · exp     λ + ˜λt1   M X m=1  λp(ym) λ + ˜λ  e(u1+···+un)ym − 1
+ ˜ M X m=1 ˜_λ˜p(˜y m) λ + ˜λ ! e(u1+···+un)C ˜ym_{− 1}
  −(u1+ · · · + un)  λ + ˜λt1 λβ λ + ˜λ + C ˜λ ˜β λ + ˜λ !) .

This is the moment generating function for a compensated compound Poisson process with intensity λ + ˜λ. This completes the proof. _

CHAPTER 4

A Bridge with Respect to the Fractional Brownian Motion

In this chapter, we introduce the fractional Brownian motion and some properties of this process. We change the Brownian motion which is in the Brownian bridge to a fractional Brownian motion and check if the new process converges.

4.1. Fractional Brownian Motion

Let (Ω, F , P) be a probability space. The process (Xt) is a Gaussian process if for

all 0 ≤ t1 < t2 < · · · < tn, the random variables Xt1, Xt2,· · · , Xtn are jointly normally

distributed. The jointly normally distribution of the random variables Xt1, Xt2,· · · , Xtn

is determined by the means and covariances of these random variables. So the law of a Gaussian process is entirely determined by the mean function E [Xt] and the covariance

function Cov (Xt, Xs), for t, s ≥ 0.

Definition 1. A fractional Brownian motion (B_t(H))t≥0 with Hurst index H ∈ (0, 1)

is a continuous and centered Gaussian process with the covariance function

E h B_t(H)B_s(H)i= RH(t, s) = 1 2 s 2H_{+ t}2H _{− |t − s|}2H
_{. (4.1)}

The fractional Brownian motion was first introduced by Kolmogorov in [10] and stud-ied by Mandelbrot and Van Ness in [11], where a stochastic integral representation of this process in terms of a standard Brownian motion was established. By the above definition we know that a fractional Brownian motion has the following properties.

(1) Self-similarity: From (4.1) we know that the process {a−HB_at(H), t ≥ 0} and {B_t(H), t ≥ 0} have the same law, for any a > 0.

(2) Stationary increments: From (4.1) we have that the increments of the fractional Brownian motion in [s, t] has a normal distribution with zero mean and variance

E   B(H)_t − B(H) s 2 = |t − s|2H. So B_t+s(H)− Bs(H) has the same law of Bt(H), for s, t ≥ 0.

For H = 1

2, the covariance function is R12(t, s) = min(s, t), then the process B

(1₂) _{is a}

standard Brownian motion. However, for H 6= 1

2 , the increments of the process are not independent. Now we discuss the integral representations for the fractional Brownian motion. The integral representation for fractional Brownian motion on the whole real line which is given by B_t(H) = 1 CH Z R h (t − s)+
H− 1 2 _{− (−s)}+
H− 1 2 _dB s i , (4.2) where B is a Brownian motion, H ∈ (0, 1) and

CH = Z ∞ 0  (1 + s)H−12 − sH− 1 2 2 ds + 1 2H 1₂

is obtained by Mandelbrot and Van Ness in [11]. For s ∈ R, t ≥ 0 the function ft(s) =

((t − s)+₎H−12 − ((−s)+₎H− 1

2 _satisfies

Z

R

f_t2(s) ds < ∞, so the stochastic integral on the right side of (4.2) is well defined. The following integral representation for fractional Brownian motion is over a finite interval. By [12], for H > 1

2, the fractional Brownian motion can be represented as

B_t(H) = Z t

0

K_H(1)(t, s) dBs, for t ≥ 0 (4.3)

where (Bt) is a standard Brownian motion and

K_H(1)(t, s) = C_H(1)s12−H Z t s |u − s|H−3_{2 u}H−1_{2 du,} where C_H(1) = " H(2H − 1) β 2 − 2H, H − 1₂
#1₂ and t > s. For H < 1

2, the integral representation on the finite interval is different from the integral representation for H > 1

4.2. WIENER INTEGRALS FOR THE FRACTIONAL BROWNIAN MOTION FOR H >1

2 37

H < 1

2, the fractional Brownian motion can be represented as B_t(H) =

Z t

0

K_H(2)(t, s) dBs, for t ≥ 0,

where (Bt) is a standard Brownian motion and

K_H(2)(t, s) = C_H(2) "  t s H−1₂ (t − s)H−12 −  H − 1 2  s12−H Z t s (u − s)H−12uH− 3 2 du # where C_H(2) =  2H (1 − 2H)β(1 − 2H, H + 1₂) 1₂

and t > s. In Section 1.8 of [6], we know that the fractional brownian motion is not a semimartingale, for H 6= 1

2. So we can not use Itˆo stochastic calculus which is defined for semimartingales to define the stochastic integral with respect to the fractional brownian motion. In next section the definition of the integral of deterministic processes with respect to a fractional brownian motion will be introduced.

4.2. Wiener Integrals for the Fractional Brownian Motion for H > 1 2 The stochastic integrals of deterministic processes with respect to a Gaussian process are called Wiener integrals. Let (B_t(H))t≥0 be a fractional brownian motion with Hurst

index H > 1

2 on the probability space (Ω, F , P). Fix a time interval [0, T ]. For 0 = t0 < t1 < · · · < tn= T the stochastic integral of a step function of the form

ϕt= n X i=1 aiI(ti−1, ti](t) is naturally defined by Z T 0 ϕtdB (H) t = n X i=1 ai  B(H)_ti − B(H)_ti−1.

The integral can be extended to general functions by using the convergence in L2_(Ω).

as hf, gi_H = H(2H − 1) Z T 0 Z T 0 f (r) g(u) |r − u|2H−2du dr. (4.4) Then H is a Hilbert space. Section 2.1 of [6] tells us the mapping ϕ −→

Z T

0

ϕtdB (H) t ,

where ϕ is a step function on [0, T ] can be extended to a linear isometry between H and the Gaussian subspace of L2_{(Ω) which is spanned by the random variables {B}(H)

t ; t ∈ [0, T ]}.

Section 2.1 of [6] also tells us the definition of the Wiener integral of the deterministic function with respect to the fractional Brownian motion.

Definition 2. For H > 1

2and ψ ∈ H, the Wiener integral of the deterministic function ψ with respect to the fractional Brownian motion B(H) is defined as

Z T 0 ψsdBs(H) = Z T 0 (K_H(1)∗ψ)(s) dBs, (4.5)

where (Bt) is a standard Brownian motion and

(K_H(1)∗ψ)(s) = Z T s ψt ∂K_H(1) ∂t (t, s) dt which is a square-integrable function.

The integral on the right side of (4.5) is well defined for ψ ∈ H and we get the representation of the Wiener integral of the deterministic function with respect to the fractional brownian motion in terms of an integral with respect to the Brownian motion. If ψ = (K_H(1)∗)−1I[0, t], then we have Bt = Z t 0  (K_H(1)∗)−1I[0, t]  (s) dB_s(H). (4.6) From (4.3) and (4.6), we know that B(H) _{and B generate the same filtration. Due to the}

isometry property of the mapping ψ −→ Z T 0 ψtdB (H) t , where ψ ∈ H , we have Z T 0 ψsdBs(H) 2 L2_(Ω) = kψk2_H,

4.3. A BRIDGE WITH RESPECT TO THE FRACTIONAL BROWNIAN MOTION 39 i.e., E " Z T 0 ψtdB (H) t 2# = H(2H − 1) Z T 0 Z T 0 ψrψu|r − u|2H−2du dr. (4.7)

The left side of (4.7) is the variance of the Wiener integral of the deterministic function ψ with respect to the fractional brownian motion B(H)_{. In next section we use the variance}

of the Wiener integral to know where the integral converges.

4.3. A Bridge with Respect to the Fractional Brownian Motion Let (B_t(H)) be a fractional brownian motion with Hurst index H > 1

2. Consider the stochastic differential equation

dXt= dB (H) t +

b − Xt

1 − t dt,

with the initial value X0 = 0 and some constant b. The solution (Xt)0≤t≤1 is given by

Xt = (1 − t) Z t 0 1 1 − sdB (H) s + bt. (4.8)

Since the fractional brownian motion B(H) _{is a centered Gaussian process, the process}

(1 − t) Z t 0 1 1 − sdB (H)

s is also a centered Gaussian process. Then we have

E  (1 − t) Z t 0 1 1 − sdB (H) s  = 0.

We will use the variance of the first term of (4.8) to see where the process (Xt) approaches

as t → 1.

Theorem 3. Suppose that the process (Xt)0≤t<1 satisfies (4.8). Then we have that

Xt→ b P-a.s. as t → 1.

Proof. From the formula (4.7), we get E "  (1 − t) Z t 0 1 1 − sdB (H) s 2# = (1 − t)2H(2H − 1) Z t 0 Z t 0 |r − u|2H−2 1 1 − r 1 1 − udu dr. (4.9)

By Fubini’s theorem, we obtain (1 − t)2H(2H − 1) Z t 0 Z t 0 |r − u|2H−2 1 1 − r 1 1 − udu dr = 2H (2H − 1) (1 − t)2 Z t 0 Z r 0 (r − u)2H−2 1 1 − r 1 1 − udu dr. (4.10)

Since 0 ≤ u ≤ r ≤ t ≤ 1, we know that 1 1 − u ≤ 1 1 − r. Then we have 2H (2H − 1) (1 − t)2 Z t 0 Z r 0 (r − u)2H−2 1 1 − r 1 1 − udu dr ≤ 2H(2H − 1)(1 − t)2 Z t 0 Z r 0 (r − u)2H−2 1 (1 − r)2 du dr = 2H(1 − t)2 Z t 0 r2H−1 1 (1 − r)2 dr. Since r ≤ 1, we get 2H(1 − t)2 Z t 0 r2H−1 1 (1 − r)2 dr ≤ 2H(1 − t) 2 Z t 0 1 (1 − r)2 dr = 2Ht(1 − t). Then we obtain E "  (1 − t) Z t 0 1 1 − sdB (H) s 2# ≤ 2Ht(1 − t)

which converges to 0 as t → 1. So we have that the process (1 − t) Z t 0 1 1 − sdB (H) s

converges to 0 P-a.s. as t → 1. Finally, we know that the process (Xt)0≤t≤1 converges

to b P-a.s. as t → 1.  The process (Xt)0≤t≤1 which satisfies (4.8) is a bridge with respect to the fractional

Brownian motion from 0 to fixed point b. Next, we have a bridge with respect to the fractional Brownian motion from 0 to a random variable. Let Y be a random variable. Consider the stochastic differential equation

dXt= dB (H) t +

Y − Xt

4.3. A BRIDGE WITH RESPECT TO THE FRACTIONAL BROWNIAN MOTION 41

with the initial value X0 = 0. The solution (Xt)0≤t≤1 is given by

Xt= (1 − t) Z t 0 1 1 − sdB (H) s + t Y. (4.11)

By the proof of Theorem 3, we have that the process (Xt)0≤t≤1 satisfying (4.11)

con-verges to Y P-a.s. as t → 1.

In the following we want to know whether the process (Xt)0≤t≤1 is a fractional

Brown-ian motion if we let Y be a standard normally distributed random variable with E [Y ] = 0 and Var(Y ) = 1. From (4.11) we know (Xt)0≤t≤1 is a centered Gaussian process. Now

we see whether E [X2t] is equal to t2H. If E [X2t] 6= t2H, then (Xt)0≤t≤1 is not a fractional

Brownian motion. The variance of Xt is given by

EXt2 = (1 − t)2E " Z t 0 1 1 − sdB (H) s 2# + t2_EY2 + 2t(1 − t) E  Y Z t 0 1 1 − sdB (H) s  . B(H) _{and Y are independent and Y ∼ N (0, 1), then we have}

EXt2 = (1 − t) 2 E " Z t 0 1 1 − sdB (H) s 2# + t2 + 2t(1 − t) E[Y ] · E Z t 0 1 1 − sdB (H) s  = (1 − t)2_E " Z t 0 1 1 − sdB (H) s 2# + t2 From (4.9) and (4.10), we have that

EX_t2 = t2+ 2H (2H − 1) (1 − t)2 Z t 0 Z r 0 (r − u)2H−2 1 1 − r 1 1 − udu dr. (4.12) By Taylor’s formula, we have 1

1 − u = ∞ X k=0 uk. Then we obtain Z r 0 (r − u)2H−2 1 1 − udu = ∞ X k=0 Z r 0 (r − u)2H−2ukdu. (4.13)

The first term of (4.13) is Z r 0 (r − u)2H−2du = 1 2H − 1r 2H−1_.

The second term of (4.13) is Z r 0 (r − u)2H−2u du = Z r 0 1 2H − 1 (r − u) 2H−1 du = 1 (2H − 1)(2H)r 2H_.

For all k ∈ Z, for n = 1, we have that Z r 0 (r − u)2H+kundu = Z r 0 (r − u)2H+ku du = Z r 0 1 2H + k + 1 (r − u) 2H+k+1_du = 1 (2H + k + 1)(2H + k + 2)r 2H+k+2_.

Suppose that for all k ∈ Z, for n = m − 1, the following equation holds Z r 0 (r − u)2H+kum−1du = (m − 1)! (2H + k + 1)(2H + k + 2) · · · (2H + k + m)r 2H+k+m_{. (4.14)} For n = m, we have Z r 0 (r − u)2H+kumdu = Z r 0 1 2H + k + 1(r − u) 2H+k+1_{m u}m−1_du = m 2H + k + 1 Z r 0 (r − u)2H+k+1um−1du. Let ˜k = k + 1. Due to (4.14), we obtain

Z r 0 (r − u)2H+kumdu = m 2H + ˜k Z r 0 (r − u)2H+˜kum−1du = m! (2H + ˜k)(2H + ˜k + 1) · · · (2H + ˜k + m)r 2H+˜k+m = m! (2H + k + 1)(2H + k + 2) · · · (2H + k + m + 1)r 2H+k+m+1

4.3. A BRIDGE WITH RESPECT TO THE FRACTIONAL BROWNIAN MOTION 43

By induction on n, the result Z r

0

(r − u)2H+kundu = n!

(2H + k + 1)(2H + k + 2) · · · (2H + k + n + 1)r

2H+k+n+1

holds for all n ∈ N. Then we get Z r 0 (r − u)2H−2 1 1 − udu = ∞ X k=0 k! (2H − 1)(2H) · · · (2H + k − 1)r 2H+k−1_. Then we obtain Z t 0 Z r 0 (r − u)2H−2 1 1 − r 1 1 − udu dr = Z t 0 1 1 − r ∞ X k=0 k! (2H − 1)(2H) · · · (2H + k − 1)r 2H+k−1 ! dr = ∞ X k=0 k! (2H − 1)(2H) · · · (2H + k − 1) Z t 0 1 1 − rr 2H+k−1_dr. Since 1 1 − r = ∞ X j=0 rj, we get Z t 0 Z r 0 (r − u)2H−2 1 1 − r 1 1 − udu dr = ∞ X k=0 k! (2H − 1)(2H) · · · (2H + k − 1) Z t 0 ∞ X j=0 r2H+k+j−1dr = ∞ X k=0 ∞ X j=0 k! (2H − 1)(2H) · · · (2H + k − 1) Z t 0 r2H+k+j−1dr = ∞ X k=0 ∞ X j=0 k! (2H − 1)(2H) · · · (2H + k − 1)· 1 2H + k + j t 2H+k+j_.

From (4.12), we have that the variance of Xt is given by

EXt2  = t2+ (2H)(2H − 1)(1 − t)2 ∞ X k=0 ∞ X j=0 k! (2H − 1)(2H) · · · (2H + k − 1) · 1 2H + k + j t 2H+k+j_. (4.15)

Suppose the following equation holds t2+ (2H)(2H − 1)(1 − t)2 ∞ X k=0 ∞ X j=0 k! (2H − 1)(2H) · · · (2H + k − 1)· 1 2H + k + j t 2H+k+j = t2H. Then (2H)(2H − 1)(1 − t)2 ∞ X k=0 ∞ X j=0 k! (2H − 1)(2H) · · · (2H + k − 1)· 1 2H + k + j t k+j _{= 1 − t}2−2H_.

Let s = 1 − t, then we have that (2H)(2H − 1)s2 ∞ X k=0 ∞ X j=0 k! (2H − 1)(2H) · · · (2H + k − 1)· 1 2H + k + j (1 − s) k+j = 1 − (1 − s)2−2H. (4.16) Since 1 − (1 − s)2−2H = 1 − ∞ X m=0    2 − 2H m   (−s) m = (2 − 2H)s + ∞ X m=2    2 − 2H m   (−s) m_.

But from (4.16), the exponent of s in every term on the left side of (4.16) is larger than two. Then we know that E [X2

t] 6= t2H, for 0 < t ≤ 1. Thus (Xt)0≤t≤1 is not a fractional

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