Construction of a New Compensated Compound Poisson Process

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

X_t= M_t+ Z t

0

Z_sds.

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

X_t = M_t+ Z t

0

f (s) d ˜M_s, (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 X_t, 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 X_t, for t ≥ 0 is given by

ϕ_Xt(u) = E [exp {uMt}] · E

 exp

 u

Z t 0

f (s) d ˜M_s



.

15

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

E [exp {uMt}] = exp {λt (e^u− u − 1)} . So we only need to focus on the expectation

E

Lemma 1. Consider the process Z t

0

f (s) d ˜M_s,

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

Proof. We will apply the Itˆo’s formula to exp

3.1. CONSTRUCTION OF A NEW COMPENSATED POISSON PROCESS 17

Since Y_s^cis the continuous part of Y_s, we can change the integrand Z_swhich is in the second term of (3.2) to Z_s⁻. When the Poisson process ˜N jumps at time s, Z_s = Z_s⁻ × e^{uf (s)}.

Hence, we obtain

E Moreover, we have that the moment generating function of X_t is given by

ϕ_Xt(u) = exp {λt (e^u− u − 1)} · exp

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 X_t= M_t. In the following proposition we set f 6= 0.

Proposition 2. Let the stochastic process (X_t) satisfy (3.1). Then (X_t) is a compen-sated Poisson process if and only if f ≡ 1. Moreover, (X_t) has the intensity λ + ˜λ.

Proof. “ =⇒ ” : The moment generating function for compensated Poisson process must be as the form 

expnˆλt (e^u− u − 1)o

, for some constant ˆλ. Suppose that (X_t) is a compensated Poisson process with intensity ˆλ. We let the moment generating function of X_t equal to 

expnˆλt (e^u− u − 1)o

, i.e.

exp



λt (e^u− u − 1) + ˜λ Z t

0

e^{uf (s)}− 1
ds − ˜λu Z t

0

f (s) ds



= expnˆλt (e^u− u − 1)o .

Then we get the equation

λt (e^u− u − 1) + ˜λ Z t

0

e^{uf (s)}− 1
ds − ˜λu Z t

0

f (s) ds = ˆλt (e^u− u − 1) . (3.4)

We differentiate with respect to t on both sides of (3.4)

λ (e^u − u − 1) + ˜λ e^{uf (t)}− 1
− ˜λuf (t) = ˆλ (e^u− u − 1) .

We differentiate with respect to t again

λue˜ ^{uf (t)}f⁰(t) − ˜λuf⁰(t) = 0 .

Then we get

λuf˜ ⁰(t) e^{uf (t)}− 1
= 0 .

This implies that f⁰(t) = 0 or e^{uf (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

X_t= M_t+ C ˜M_t.

The moment generating function of X_t is given by

ϕ_X(u) = expn

λt (e^u − u − 1) + ˜λt e^uC − uC − C
o .

3.1. CONSTRUCTION OF A NEW COMPENSATED POISSON PROCESS 19

Suppose the following equation holds expn

λt (e^u− u − 1) + ˜λt e^uC − uC − C
o

= expnˆλt (e^u− u − 1)o . Then we have

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

λe^u+ ˜λC²e^uC = ˆλe^u. We multiply e^−u on both sides of the above formula

λ + ˜λC²e^u(C−1) = ˆλ . Then

e^u(C−1) = λ − λˆ

˜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_t^{M, ˜M} the filtration generated by M and ˜M . Let

Z_t= exp {uM_t− λt (e^u− u − 1)} .

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

F_t^{M, ˜M} . Let

Z˜t= exp n

u ˜Mt− ˜λt (e^u− u − 1)o . ( ˜Z_t) is also a martingale with respect to 

F_t^{M, ˜M}

. The two processes (Z_t) and ( ˜Z_t) are independent and they are all martingales with respect to the filtration 

F_t^{M, ˜M}

. From (3.3), we know that the moment generating function of Xt is

ϕ_Xt(u) = expn

λ + ˜λ

t (e^u− u − 1)o .

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

is a martingale with respect to the filtration 

F_t^{M, ˜M}

So the process  V_t^(u)

is a martingale with respect to 

F_t^{M, ˜M}

3.1. CONSTRUCTION OF A NEW COMPENSATED POISSON PROCESS 21

Now we use the martingale property of V_t^(u1) and we take expectation of both sides of the above formula So we obtain

E [exp {u¹Xt1 + u2(Xt2 − Xt1)}] 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 X_t2 − X_t1 is

ϕ_Xt2−X_t1(u) = expn

λ + ˜λ

(t₂− t₁) (e^u− u − 1)o .

Next, we computer the joint moment generating function of the random variables X_t1, X_t2, · · · , X_tn, for 0 < t₁ < t₂ < · · · < t_n, so that we can know whether X is a compensated

Poisson process. For 0 < t₁ < t₂ < · · · < t_n, the joint moment generating function of the random variables X_t1, X_t2, · · · , X_tn is given by

ϕX_t1,X_t2,··· ,X_tn(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+ · · · u_n) X_t1}] .

We have known the form of the moment generating function of increments of X, then we obtain

ϕ_{Xt1,Xt2,··· ,Xtn} (u₁, u₂, · · · , u_n)

= expn

λ + ˜λ

(t_n− t_n−1) (e^un− u_n− 1)o

· expn

λ + ˜λ

(t_n−1− t_n−2) e^(un−1+un)+ (u_n−1+ u_n) − 1
o

·

· · · expn

λ + ˜λ

t₁ e^{(u1+u2+···+un)}− (u₁+ u₂+ · · · + u_n) − 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 (N_t) 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 [Y_i].

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

Definition 3. The stochastic process (Qt) defined by Q_t=

Nt

X

i=1

Y_i, t ≥ 0

3.3. CONSTRUCTION OF A NEW COMPENSATED COMPOUND POISSON PROCESS 23

is called the compound Poisson process.

The compound Poisson process (Q_t) jumps at the same time as the Poisson process (N_t) 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, . . . , y_M be a finite set of nonzero numbers and let p(y₁), p(y₂),. . . , p(y_M) be positive numbers that sum to 1. Let Y₁, Y₂, . . . be a sequence of independent, identically distributed random variables with P (Yi = y_m) = p(y_m), m = 1, . . . , M . Let (N_t) be a Poisson process with intensity λ and define the compound Poisson process

Q_t=

Nt

X

i=1

Y_i.

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

N_t=

M

X

m=1

N_t^(m) and Q_t=

M

X

m=1

y_mN_t^(m),

where the process N⁽¹⁾, . . . , N^{(M )} are independent Poisson processes and each N^(m) has intensity λp(y_m).

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, . . . , y_M be a finite set of nonzero numbers and let p(y₁), p(y₂),

. . . , p(y_M) be positive numbers whose summation is identical to 1. Let Y₁, Y₂, . . . be a sequence of independent, identically distributed random variables with P (Yi = y_m) = p(y_m), m = 1, . . . , M and E [Y_i] = β. Let (N_t) be a Poisson process with intensity λ and define the compound Poisson process

Q_t=

Nt

X

i=1

Y_i. (3.5)

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

Qt=

M

X

m=1

ymN_t^(m),

where the process N⁽¹⁾, . . . , N^{(M )} are independent Poisson process, and each N^(m) has intensity λp(y_m).

Let ˜y₁, ˜y₂, . . . , ˜yMˆ be another finite set of nonzero numbers and let ˜p(˜y₁), ˜p(˜y₂), . . . , ˜p(˜yMˆ) be positive numbers that sum to 1. Let ˜Y₁, ˜Y₂, . . . be another sequence of independent, identically distributed random variables with P ˜Y_i = ˜y_m

= ˜p(˜y_m), m = 1, . . . , ˜M , E[ ˜Y_i] = ˜β and ˜Y₁, ˜Y₂, . . . are independent of the sequence Y₁, Y₂, . . .. Let ( ˜N_t) be a Poisson process with intensity ˜λ and it is independent of (N_t). Define another compound Poisson process

Q˜_t=

N˜t

X

i=1

Y˜_i. (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

Q˜_t=

M˜

X

m=1

˜

y_mN˜_t^(m),

where the process ˜N⁽¹⁾, . . . , ˜N^{( ˜M )} are independent Poisson process, and each ˜N^(m) has intensity ˜λ˜p(˜y_m).

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 X_t, for t ≥ 0. The moment generating function of X_t, for t ≥ 0 is given by

ϕXt(u) = E [exp {u (Q^t− 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 X_t.

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

ϕ_Qt(u) = exp {λt (ϕ_Y1(u) − 1)} .

By the above theorem, we know that the moment generating function of (Q_t− βλt) is given by

ϕ_(Qt−βλt)(u) = exp (

λt

M

X

m=1

p(y_m) (e^uym − 1) − uβλt )

. (3.9)

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

0

f (s) d ˜Q_s− ˜β ˜λs , so that we can get the form of the moment generating function of X_t.

Theorem 6. Consider the process

Z t 0

f (s) d ˜Qs− ˜β ˜λs

 ,

where ( ˜Q_t) is given by (3.6) with intensity ˜λ, ˜β = E[ ˜Y_i], and f is a nonrandom differen-tiable function. Then its moment generating function is given by

E Poisson processes, we know that

E

This completes the proof. 

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 X_t is given by Next, we want to see in which cases the process X is a compensated compound Poisson process.

Proposition 7. Let the stochastic process (X_t) satisfy (3.7). Then (X_t) is a com-pensated compound Poisson process if and only if f ≡ C. Moreover, (X_t) has intensity



λ + ˜λ .

Proof. “ =⇒ ” : Suppose that (Xt) is a compensated compound Poisson process with intensity ˆλ. We let the moment generating function of X_t be equal to

exp

Then we have the equation

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

λ

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

u˜λf⁰(t)

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

uf⁰(t)

3.3. CONSTRUCTION OF A NEW COMPENSATED COMPOUND POISSON PROCESS 29

The moment generating function of X_t is given by

ϕXt(u) = exp

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 ˆY_i = y_m the law of X agrees with the law of a compensated compound Poisson process which has intensity λ + ˜λ. Set Then we have

Z_t= exp

Set second term of (3.16) to Z_s⁻. 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,· · · , y_M. If the jump size of Q process with intensity λp(y_m) and M^(m) is a martingale. We have

Z_t = 1 +

3.3. CONSTRUCTION OF A NEW COMPENSATED COMPOUND POISSON PROCESS 31

Since M^(m)is a martingale and Z_s⁻(e^uym− 1) is left continuous in s, Z t

0

Z_s⁻(e^uym− 1) dM_s^(m) is also a martingale. The sum of finitely many martingales is a martingale, so the pro-cess (Z_t) is a martingale. Denote by 

F_t^{N(1),··· , N(M ), ˜N(1),··· , ˜N( ˜M )}

the filtration generated by N⁽¹⁾, · · · , N^{(M )}, ˜N⁽¹⁾, · · · , ˜N^{( ˜M )}. The process (Z_t) is a martingale with respect to

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

F_t^{N(1),··· , N(M ), ˜N(1),··· , ˜N( ˜M )}

. The two processes (Z_t) and ( ˜Z_t) are independent and they are all martingales with respect to the filtration

F_t^{N(1),··· , N(M ), ˜N(1),··· , ˜N( ˜M )}

. From (3.13), we know that the moment generating function of X_t is given by

ϕ_Xt(u) = exp

For the sake of simplicity, we let

η_t(u) = exp

Since Z_t− Z_s, ˜Z_t− ˜Z_s are independent of F_s^{N(1),··· , N(M ), ˜N(1),··· , ˜N( ˜M )} and Z_s, ˜Z_sare adapted

Now we use the martingale property of V_t^(u1). We take expectation of both sides of the above formula

So we obtain

E [exp {u1X_t1 + u₂(X_t2 − X_t1)}] = η_t1(u₁) · η_t2−t1(u₂).

Since the above joint moment generating function factors into the product of moment generating functions, X_t1 and X_t2 − X_t1 must be independent. We also know that the moment generating function of X_t2 − X_t1 is

3.3. CONSTRUCTION OF A NEW COMPENSATED COMPOUND POISSON PROCESS 33

This is the moment generating function for a compensated compound Poisson process

with intensity λ + ˜λ. This completes the proof. 

CHAPTER 4

在文檔中 針對補償混合卜松隨機過程或碎形布朗運動的橋 (頁 21-41)