# Yamada-Watanabe Theorem for Stochastic Evolution Equation Driven by Poisson Random Measure

Authors: Huiyan Zhao
Journal: ISRN Probability and Statistics
 (2014)

---

## Abstract

The purpose of this paper is to give a detailed proof of Yamada-Watanabe theorem for stochastic evolution equation driven by pure Poisson random measure.


---

## Body

## 1. Introduction

The main purpose of this paper is to establish the Yamada-Watanabe theory of uniqueness and existence of solutions of stochastic evolution equation driven by pure Poisson random measure in the variational approach. The classical paper [1] has initiated a comprehensive study of relations between different types of uniqueness and existence (e.g., strong solutions, weak solutions, pathwise uniqueness, uniqueness, and joint uniqueness in law) arising in the study of SDEs (see, e.g., [2–4]) and the study is still alive today. New papers are published (see, e.g., [2, 3, 5–7]). In this paper we are concerned with the similar question for stochastic evolution equation driven by Poisson random measure by using the method of Yamada and Watanabe.Yamada and Watanabe's initial work [1] proved that weak existence and pathwise uniqueness imply strong existence and weak uniqueness. For n-dimensional case, see [8, 9]. For infinite dimensional stochastic differential equation, Ondreját [6] proved similar result for stochastic evolution equation in Banach space driven by cylindrical Wiener process, where the solutions are understood in the mild sense. Lately, Röckner et al. [7] proved similar result for stochastic evolution equation in Banach space driven by cylindrical Wiener process under the variational framework. On the other hand, Kurtz [2, 3] obtained a pleasant version of Yamada-Watanabe and Engelbert theorem in an abstract form, which covered most of the work mentioned above. However, we will consider the following concrete stochastic evolution equation by using a different method.In this paper, we will consider the following stochastic evolution equation driven by pure Poisson random measure under the variational framework:(1)X(t)=X(0)+∫0t‍b(s,X)ds+∫0t+‍∫Z‍f(s,z,X)N~(ds,dz).
This type of equations can be applied to many SPDEs, for example, stochastic Burgers equation, stochastic porous media equation, and stochastic Navier-Stokes equation (see, e.g., [9–13]). We will introduce the above equation precisely in Section 2. Our aim is to obtain this jump-case Yamada-Watanabe theorem; that is, weak existence and strong uniqueness (which will be stated in Section 2) imply strong existence and weak uniqueness and vice versa. We note that there are some differences between the jump-case case and the Brownian motion case. It is well known that a Brownian motion can be treated as a canonical map on C([0,∞);ℝn) or C([0,∞);H) (for some Hilbert space H), while for jump-case we have to work on the configuration space ℕ (see Section 2) for Poisson random measure.The structure of the present paper is as follows. In Section2 we introduce the framework and definitions of strong solution and weak solution. In Section 3 we give and prove our main results.


## 2. Framework and Definitions

LetH be a separable Hilbert space with inner product 〈·,·〉H and norm ∥·∥H. Let V and E be separable Banach space with norms ∥·∥V and ∥·∥E, such that
(2)V⊂H⊂E
continuously and densely.LetD(ℝ+,H) be the space of all càdlàg functions: ℝ+→H. Set ∥x∥V:=∞ if x∈H∖V; we extend ∥·∥V to a function on H. Note that this extension is ℬ(H)-measurable and lower semicontinuous. Hence, the following path space is well defined:
(3)𝔻:={x∈D(ℝ+,H):∫0T‍∥x(t)∥Vdt<∞foranyT∈[0,∞)},
equipped with the metric
(4)d(x1,x2):=∑k=1∞‍2-k{[∫0k‍∥x1(t)-x2(t)∥Vdt+dk(x1,x2)]∧1},
where dk(x1,x2) is the Skorokhod distance on D([0,k],H); see [14]. It is easy to see that (𝔻,d) is a Polish space. Let ℬt(𝔻) denote the σ-algebra generated by all the maps πs:𝔻→H, s∈[0,t], where πs(ω):=ωs, ω∈𝔻.Letℤ¯+={0,1,2,…,+∞} and let (𝒳,ℬ(𝒳)) be a measurable space. We denote by 𝒫ℤ¯+(𝒳) the space of all ℤ¯+-valued measures on (𝒳,ℬ(𝒳)) and consider the measurable space (𝒫ℤ¯+(𝒳),𝒜), where 𝒜 is the σ-algebra generated by the mappings 𝒫ℤ¯+(𝒳)∋n(A)∈ℤ¯+, A∈ℬ(𝒳).Letλ be a σ-finite measure on (𝒳,ℬ(𝒳)). We recall that a Poisson random measure N on 𝒳 with intensity λ is a measurable mapping from a probability space (Ω,ℱ,P) to (𝒫ℤ¯+(𝒳),𝒜) such that the following two conditions hold:(i)
for everyA∈ℬ(𝒳), the random variable N(A) has the Poisson distribution with the mean λ(A);(ii)
for any disjointA1,A2,…,An∈ℬ(𝒳), the random variables N(A1),N(A2),…,N(An) are independent.LetQ be the distribution of N; then (𝒫ℤ¯+(𝒳),𝒜,Q) is a probability space, and the canonical map
(5)N-:𝒫ℤ¯+(𝒳)∋n⟶n∈𝒫ℤ¯+(𝒳)
is a Poisson random measure with intensity λ on this probability space. Let ℕ:=𝒫ℤ¯+(𝒳). Note that ℕ is a Polish space (cf. [15]).In this paper, let𝒳:=[0,∞)×Z, where Z is an arbitrary locally compact Hausdorff topological space with countable base, and dλ=dtν(dz), where ν is a σ-finite measure on Z. We denote by ℬt(ℕ) the σ-algebra generated by the mappings: ℕ∋n→n((0,s]×A)∈ℤ¯+, for all 0⩽s⩽t, for all A∈ℬ(Z).Letb:ℝ+×𝔻→E and f:ℝ+×Z×𝔻→H be ℬ(ℝ+)⊗ℬ(𝔻)/ℬ(E) and ℬ(ℝ+)⊗ℬ(Z)⊗ℬ(𝔻)/ℬ(H)-measurable, respectively, such that for each t∈ℝ+ and z∈Z(6)b(t,·)isℬt(𝔻)/ℬ(E)-measurable,f(t,z,·)isℬt(𝔻)/ℬ(H)-measurable,f(·,z,·)ispredictable.LetN~(dt,dz):=N(dt,dz)-dtν(dz); we consider the following stochastic evolution equation:
(7)dXt=b(t,X)dt+∫Z‍f(t,z,X)N~(dt,dz),
with X0=x∈H. For more results about this type of equation see [9, 11, 13] and the references therein.Definition 1.
Assume that(Ω,ℱ,P,(ℱt)) is a filtered probability space and N is a Poisson random measure defined as above. Then N is called a (ℱt)-Poisson random measure with intensity dtν(dz) if,(i)
for anyt∈[0,∞), B∈ℬ(Z) with ν(B)<∞, then N((0,t]×B) is ℱt-measurable;(ii)
for anyt,s∈[0,∞), B∈ℬ(Z) with ν(B)<∞, then N((t,t+s]×B) is independent of ℱt.Remark 2.
The canonical Poisson random measure on(ℕ,ℬ(ℕ),Q,(ℬt(ℕ))) is (ℬt(ℕ))-Poisson random measure.Definition 3.
For somep∈[1,2], a pair (X,N), where X=(X(t))t∈[0,∞) is an (ℱt)-adapted process with paths in 𝔻 and N is a (ℱt)-Poisson random measure with intensity dtν(dz) on a stochastic basis (Ω,ℱ,P,(ℱt)), is called a weak solution of (7) if(i)
for anyT∈[0,∞)(8)∫0T‍∥b(t,X)∥Eds+∫0T‍∫Z‍∥f(s,z,X)∥Hpν(dz)ds<∞P-a.e.;(ii)
as a stochastic equation onE one has
(9)X(t)=X(0)+∫0t‍b(s,X)ds+∫0t+‍∫Z‍f(s,z,X)N~(ds,dz)t∈[0,∞)P-a.e.Definition 4.
One says that weak uniqueness holds for (7) if whenever (X,N) and (X′,N′) are two weak solutions with stochastic bases (Ω,ℱ,P,(ℱt)) and (Ω′,ℱ′,P′,(ℱt′)) such that
(10)P∘X(0)-1=P′∘X′(0)-1,
then
(11)P∘X-1=P′∘(X′)-1.Definition 5.
One says that pathwise uniqueness holds for (7) if whenever (X,N) and (X′,N) are two weak solutions on the same stochatic bases (Ω,ℱ,P,(ℱt)) such that X(0)=X′(0)P-a.s.; then P-a.s.:
(12)X(t)=X′(t)t∈[0,∞).In order to define strong solutions one needs to introduce the following classΓ of maps. Let Γ denote the set of all maps F:H×ℕ→𝔻 such that for every probability measure μ on (H,ℬ(H)) there exists a ℬ(H)⊗ℬ(ℕ)¯μ⊗Q/ℬ(𝔻)-measurable map Fμ:H×ℕ→𝔻 such that for μ-a.e. x∈H(13)F(x,n)=Fμ(x,n)forQ-a.e.n∈ℕ.
Here ℬ(H)⊗ℬ(ℕ)¯μ⊗Q denotes the completion of ℬ(H)⊗ℬ(ℕ) with respect to μ⊗Q, where Q denotes the distribution of the Poisson random measure N on space ℕ. Of course, Fμ is uniquely determined by μ⊗Q-a.e.Definition 6.
A weak solution(X,N) to (7) on (Ω,ℱ,P,(ℱt)) is said to be a strong solution if there exists F∈Γ with respect to P∘X(0)-1 such that, for x∈H, n→F(x,n) is ℬt(ℕ)¯Q/ℬt(𝔻)-measurable for every t∈[0,∞) and
(14)X=FP∘X(0)-1(X(0),N)P-a.e.,
where ℬt(ℕ)¯Q denote the completion with respect to Q in ℬ(ℕ).Definition 7.
Equation (7) is said to have a unique strong solution, if there exists F∈Γ satisfying the condition in the above definition such that the following conditions hold.  (1)
For every Poisson random measureN with intensity dtν(dz) on a stochastic basis (Ω,ℱ,P,(ℱt)) and any ℱ0/ℬ(H)-measurable ξ:Ω→H the càdlàg process
(15)X:=FP∘ξ-1(ξ,N)
is (ℱt)-adapted and satisfies (i) and (ii) of Definition 3; that is, (X,N) is a weak solution to (7), and X(0)=ξ  P-a.e.(2)
For any weak solution(X,N) to (7) one has
(16)X:=FP∘X(0)-1(X(0),N)P-a.e.


## 3. The Main Result and Its Proof

Let us now state the main result.Theorem 8.
Letb and f be as above. Then (7) has a unique strong solution if and only if both of the following properties hold. (i)
For every probability measureμ on (H,ℬ(H)) there exists a weak solution (X,N) of (7) such that μ is the distribution of X(0).(ii)
Pathwise uniqueness holds for (7).Proof.
Suppose that (7) has a unique strong solution. Then (ii) obviously holds. To show (i) we only have to take the probability space (ℕ,ℬ(ℕ),Q) and consider (H×ℕ,ℬ(H)⊗ℬ(ℕ)¯μ⊗Q,μ⊗Q) with filtration
(17)⋂ε>0‍σ(ℬ(H)⊗ℬt+ε(ℕ),𝒩),t⩾0,
where 𝒩 denotes all μ⊗Q-zero sets in ℬ(H)⊗ℬ(ℕ)¯μ⊗Q. Let ξ:H×ℕ→H and N: H×ℕ→ℕ be the canonical projections. Then (X:=FP∘ξ-1(ξ,N),N) is the desired weak solution in (i).
Now let us suppose that (i) and (ii) hold. We are going to show that there exists a unique strong solution for (7).
Let(X,N) with stochastic basis (Ω,ℱ,P,(ℱt)) be a weak solution to (7) with initial distribution μ. Define a probability measure Pμ on (H×𝔻×ℕ,ℬ(H)⊗ℬ(𝔻)⊗ℬ(ℕ)), by
(18)Pμ:=P∘(X(0),X,N)-1.
We have the following lemma.Lemma 9.
There exists a familyKμ((x,n),dω), x∈H, n∈ℕ, of probability measures on (𝔻,ℬ(𝔻)) having the following properties.(i)
For everyA∈ℬ(𝔻) the map
(19)H×ℕ∋(x,n)⟼Kμ((x,n),A)
is ℬ(H)⊗ℬ(ℕ)-measurable.(ii)
For everyℬ(H)⊗ℬ(𝔻)⊗ℬ(ℕ)-measurable function f:H×𝔻×ℕ→[0,∞), one has
(20)∫H×𝔻×ℕ‍f(x,ω,n)Pμ(dx,dω,dn)=∫H‍∫ℕ‍∫𝔻‍f(x,ω,n)Kμ((x,n),dω)Q(dn)μ(dx).(iii)
Ift∈[0,∞) and f:𝔻→[0,∞) is ℬt(𝔻)-measurable, then
(21)H×ℕ∋(x,n)⟼∫𝔻‍f(ω)Kμ((x,n),dω)
is ℬ(H)⊗ℬt(ℕ)¯μ⊗Q-measurable, where ℬ(H)⊗ℬt(ℕ)¯μ⊗Q denotes the completion with respect to μ⊗Q in ℬ(H)⊗ℬ(ℕ).Proof.
LetΠ:H×𝔻×ℕ→H×ℕ be the canonical projection. Since X(0) is ℱ0-measurable, hence P-independent of N, it follows that
(22)Pμ∘Π-1=P∘(X(0),N)-1=μ⊗Q.
We recall that 𝔻 is a Polish space and, by the existence result on regular conditional distributions the family Kμ((x,n),dω), x∈H, n∈ℕ, exist and satisfy (i) and (ii).
To prove (iii) it suffices to show that fort∈[0,∞) and for all A0∈ℬ(H), A1∈ℬt(𝔻), A∈ℬt(ℕ), B∈ℬ(Z) and
(23)A′:={N((t,r1]×B)∈C1,N((r1,r2]×B)∈C2,…,N((rk-1,rk]×B)∈Ck}×(⊂ℕ),t⩽r1<⋯<rk,C1,…,Ck∈ℬ(ℝ),(24)∫A0‍∫ℕ‍1A∩A′(n)Kμ((x,n),A1)Q(dn)μ(dx)=∫A0‍∫ℕ‍1A∩A′(n)𝔼μ⊗Q(Kμ(·,A1)∣ℬ(H)⊗ℬt(ℕ))×Q(dn)μ(dx).
Since the system of all A∩A′, A∈ℬt(ℕ), A′ as above generate ℬ(ℕ). According to (ii) above, the left side of (24) is equal to
(25)∫H×𝔻×ℕ‍1A0(x)1A∩A′(n)1A1(ω)Pμ(dx,dω,dn)=∫Ω‍1A0(X(0))1A(N)1A′(N)1A1(X)dP.
Note that N is a Poisson random measure on [0,∞)×Z with stochastic basis (Ω,ℱ,P,(ℱt)), so 1A′(N) is P-independent of ℱt; then the right side of (25) is
(26)∫Ω‍1A′(N)dP·∫Ω‍1A0(X(0))1A(N)1A1(X)dP=Q(A′)∫H×𝔻×ℕ‍1A0(x)1A(n)1A1(ω)Pμ(dx,dω,dn)=Q(A′)∫A0‍∫A‍Kμ((x,n),A1)Q(dn)μ(dx)=Q(A′)∫A0‍∫A‍𝔼μ⊗Q(Kμ(·,A1)∣ℬ(H)⊗ℬt(ℕ))×Q(dn)μ(dx)=∫A0‍∫ℕ‍1A∩A′(n)𝔼μ⊗Q(Kμ(·,A1)∣ℬ(H)⊗ℬt(ℕ))×Q(dn)μ(dx),
since A′ is Q-independent of ℬt(ℕ).Forx∈H define a measure Px on
(27)(H×𝔻×𝔻×ℕ,ℬ(H)⊗ℬ(𝔻)⊗ℬ(𝔻)⊗ℬ(ℕ))
by
(28)Px(A):=∫H‍∫𝔻‍∫𝔻‍∫ℕ‍1A(z,ω1,ω2,n)Kμ((z,n),dω1)Kμ((z,n),dω2)×Q(dn)δx(dz).Define the stochastic basis(29)Ω~:=H×𝔻×𝔻×ℕ,ℱ~x:=ℬ(H)⊗ℬ(𝔻)⊗ℬ(𝔻)⊗ℬ(ℕ)¯Px,ℱ~tx:=⋂ε>0‍σ(ℬ(H)⊗ℬt+ε(𝔻)⊗ℬt+ε(𝔻)⊗ℬt+ε(ℕ),𝒩x),
where
(30)𝒩x:={A∈ℱ~x:Px(A)=0},
and define the maps
(31)Π0:Ω~⟶H,(x,ω1,ω2,n)⟼x,Πi:Ω~⟶𝔻,(x,ω1,ω2,n)⟼ωi∈𝔻,i=1,2,Π3:Ω~⟶ℕ,(x,ω1,ω2,n)⟼n∈ℕ.Then, it is easy to see that(32)Px∘Π0-1=δx,Px∘Π3-1=Q(=P∘N-1).Lemma 10.
There existsA∈ℬ(H) with μ(A)=0 such that, for all x∉A, one has that Π3 is an (ℱ~tx)-Poisson random measure on (Ω~,ℱ~x,Px).Proof.
By definitionΠ3([0,·]×B) is (ℱ~tx)-adapted for every x∈H. We only need to check that condition (ii) of Definition 1 holds for almost all x∈H. Furthermore, for 0⩽s<t, y∈H and A0∈ℬ(H), Ai∈ℬs(𝔻), i=1,2, A3∈ℬs(ℕ), B∈ℬ(Z),
(33)∫H‍𝔼Px([-Π3([0,s]×B)]1A0×A1×A2×A3exp[i〈y,Π3([0,t]×B)-Π3([0,s]×B)〉]1A0×A1×A2×A3)μ(dx)=∫H‍∫ℕ‍exp[i〈y,n([0,t]×B)-n([0,s]×B)〉]×1A0(x)1A3(n)Kμ((x,n),A1)×Kμ((x,n),A2)Q(dn)μ(dx)=∫ℕ‍exp[i〈y,n([0,t]×B)-n([0,s]×B)〉]Q(dn)×∫H‍Px(A0×A1×A2×A3)μ(dx)=∫H‍𝔼Px(exp[i〈y,Π3([0,t]×B)-Π3([0,s]×B)〉])×Px(A0×A1×A2×A3)μ(dx),
which ends the proof.Fix a probability measureμ on (H,ℬ(H)) and let (X,N) with stochastic basis (Ω,ℱ,P,(ℱt)) be a weak solution to (7) with initial distribution μ. Let f:ℝ+×Z×𝔻→H be a predictable process defined as above. In particular we assume that, for some p∈[1,2], ∫0t‍∫Z‍∥f(s,z,ω)∥Hpμ(dz)ds<∞P∘X-1-a.e. for all t∈(0,∞). We aim to give a pointwise definition for stochastic integral ∫0T‍∫Z‍f(s,z,X)N(ds,dz). First of all, we give a pointwise definition for stochastic integral ∫0T‍∫Z‍f(s,z,ω)n(ds,dz) at the fixed time T.Fork⩾1, define τk(ω):=inf{t⩾0:∫0t‍∫Z‍∥f(s,z,ω)∥Hpμ(dz)ds⩾k}; it is easy to see that {ω:τk(ω)⩽t}∈ℬt(𝔻) for t>0. Also note that τk(ω)↑τ(ω):=inf{t⩾0:∫0t‍∫Z‍∥f(s,z,ω)∥Hpμ(dz)ds=∞} pointwise on 𝔻 where the infimum of the empty set is taken to be infinity. Firstly, let g be the elementary predictable process and define the stochastic integral as usual (cf. [12, page 130]). That is, if g=∑j=0m‍gj1(tj,tj+1]1Aj, where 0=t0<t1<⋯<tm=T, A1,…,Am∈ℬ(H) such that μ(Aj)<∞, j=1,…,m and gj are bounded and ℱj-measurable, then
(34)Ig(ω,n)=∫0T‍∫Z‍g(s,z,ω)n(ds,dz):=∑j=0m‍gj(ω)n((tj,tj+1)×Aj).
Fix k⩾1 and let gmk be the elementary predictable such that
(35)∫0T‍∫Z‍𝔼(∥gmk(s,z,ω)-1(0,τk(ω)]f(s,z,ω)∥p)μ(dz)ds⟶0
as m→∞. By
(36)𝔼∫0T‍∫Z‍∥g(s,z,ω)∥pn(ds,dz)=∫0T‍∫Z‍𝔼∥g(s,z,ω)∥pμ(dz)ds.
Define
(37)Ik(ω,n):=limmIgmk(ω,n),
where the limit exists in Lp(P∘X-1) and P∘X-1 almost everywhere; then put
(38)I(ω,n):=∑k=1∞‍1(τk-1(ω)<T⩽τk(ω))Ik(ω,n).Now we map back to the stochastic basis(Ω,ℱ,P,(ℱ)t); we find that τk(X) is a sequence of (ℱt)-stopping times with τk(X)↑∞  p-a.e. Also gmk(s,X,z) are (ℱt)-predictable processes that converge in Lp([0,T]×Ω×Z,dt⊗dP⊗μ(dz)) to 1(0,τk(X)]f(s,X,z), so that Igmk(X,N) converge in Lp to the stochastic integral ∫0T‍∫Z‍1(0,τk(X)]f(s,X,z)N(ds,dz). Putting these pieces together shows that I(X,N) is a version of the stochastic integral ∫0T‍∫Z‍f(s,X,z)N(ds,dz).By the same argument as in [7], we have the following two lemmas.Lemma 11.
There existsA∈ℬ(H) with μ(A)=0 such that, for x∉A, I(Π1,Π3) is a Px-version of the stochastic integral ∫0T‍∫Z‍f(s,Π1,z)Π3(ds,dz) on the stochastic basis (Ω~,ℱ~x,Px,(ℱ~tx)).Lemma 12.
There existsA′∈H with μ(A′)=0 such that for all x∉A′, (Π1,Π3) and (Π2,Π3) with stochastic basis (Ω~,ℱ~x,Px,(ℱ~tx)) are weak solutions of (7) such that
(39)Π1(0)=Π2(0)Px-a.e.;
therefore, Π1=Π2Px-a.e.Lemma 13.
There exists aℬ(H)⊗ℬ(ℕ)¯μ⊗Q/ℬ(𝔻)-measurable map
(40)Fμ:H×ℕ⟶𝔻
such that for μ⊗Q-a.e. (x,n)∈H×ℕ(41)Kμ((x,n),·)=δFμ(x,n),
where δFμ(x,n) is the Dirac measure on ℬ(𝔻) with mass in Fμ(x,n). Furthermore, Fμ(x,n) is ℬ(H)⊗ℬt(ℕ)¯μ⊗Q/ℬt(𝔻)-measurable for all t∈[0,∞).Proof.
By Lemma12, for all x∉A′, we have
(42)1=Px({Π1=Π2})=∫𝔻‍∫𝔻‍∫ℕ‍1D(ω1,ω2)Kμ((x,n),dω1)×Kμ((x,n),dω2)Q(dn),
where D:={(ω,ω)∈𝔻×𝔻:ω∈𝔻}. Note that 𝔻 is Polish space, by Lemma 2.2 in [7], there exists A∈ℬ(H)×ℬ(ℕ) such that μ×Q(A)=0 and for all (x,n)∉A there exists Fμ(x,n)∈𝔻 such that
(43)Kμ((x,n),dω)=δFμ(x,n)(dω).
Set Fμ(x,n):=0 if (x,n)∈A. Let C∈ℬ(𝔻); then
(44){Fμ∈C}=({Fμ∈C}∩A)∪({Kμ(·,C)=1}∩Ac)
and the other properties of Fμ follow from Lemma 9.Having defined the mappingFμ let us check the conditions of Definitions 6 and 7.Lemma 14.
One has(45)X=Fμ(X(0),N)P-a.e.Proof.
By Lemmas9 and 13, we have
(46)P({X=Fμ(X(0),N)})=∫H‍∫𝔻‍∫ℕ‍1{ω=Fμ(X(0),N)}(x,ω,n)δFμ(X(0),N)×(dω)Q(dn)μ(dx)=1.LetN′ be another (ℱt′)-Poisson random measure on [0,∞)×Z with intensity dtν(dz) on a stochastic basis (Ω′,ℱ′,P′,(ℱt′)) and let ξ:Ω→H be an ℱ0′/ℬ(H)-measurable map and μ:=P′∘ξ-1. Let Fμ be as above and set
(47)X′:=Fμ(ξ,N′).
By the same argument as in [7], we also have the following lemma.Lemma 15.
(
X
′
,
N
′
) is a weak solution to (7) with X′(0)=ξ  P′-a.e.To complete the proof we still have to constructF∈Γ and to check the adaptiveness conditions on it. Below we will apply what we have obtained above to δx replacing μ. So, for each x∈H, we have a function Fδx. Now, define
(48)F(x,n):=Fδx(x,n),x∈H,n∈ℕ.The proof is complete by the following lemma.Lemma 16.
Letμ be a probability on (H,ℬ(H)) and Fμ:H×ℕ→𝔻 as constructed in Lemma 13. Then for μ-a.e. x∈H(49)F(x,·)=Fμ(x,·)Q-a.e.
Furthermore, F(x,·) is ℬt(ℕ)¯Q/ℬt(𝔻)-measurable for every x∈H, t∈[0,∞).Proof.
Let(50)Ω-:=H×𝔻×ℕ,ℱ-:=ℬ(H)⊗ℬ(𝔻)⊗ℬ(ℕ),
and fix x∈H. Define a measure P-x on (Ω-,ℱ-) by
(51)P-x(A):=∫H‍∫𝔻‍∫ℕ‍1A(z,ω,n)×Kμ((z,n),dω)Q(dn)δx(dz),
where Kμ is as in Lemma 9. Consider the stochastic basis (Ω-,ℱ-x,P-x,(ℱ-tx)), where
(52)ℱ-x:=ℬ(H)⊗ℬ(𝔻)⊗ℬ(ℕ)¯P-x,ℱ-tx:=⋂ε>0‍σ(ℬ(H)⊗ℬt+ε(𝔻)⊗ℬt+ε(ℕ),𝒩-x),
where
(53)𝒩-x:={A∈ℱ~x:P-x(A)=0}.
Let
(54)Π0:Ω-⟶H,(x,ω,n)⟼x,Π:Ω-⟶𝔻,(x,ω,n)⟼ω∈𝔻,Π3:Ω-⟶ℕ,(x,ω,n)⟼n∈ℕ.
As above, we can show that (Π,Π3) on (Ω-,ℱ-x,P-x,(ℱ-tx)) is a weak solution to (7) with Π(0)=xQ-x-a.e., and (Fδx(x,Π3),Π3) on (Ω-,ℱ-x,P-x,(ℱ-tx)) is also a weak solution to (7) with
(55)Fδx(x,Π3)(0)=x.
Hence, by our pathwise uniqueness assumption (ii), we have
(56)Fδx(x,Π3)=ΠQ-x-a.e.
Hence, for A∈ℱ- and by Lemma 13 and (56)
(57)∫H‍∫𝔻‍∫ℕ‍1A(x,ω,n)δKμ((x,n),dω)Q(dn)μ(dx)=∫H‍P-x(A)μ(dx).
But for each x∈H(58)P-x(A)=∫Ω-‍1A(Π0,Fδx(x,Π3),Π3)dP-x=∫ℕ‍1A(x,Fδx(x,n),n)Q(dn)=∫ℕ‍∫𝔻‍1A(x,ω,n)δFδx(x,n)(dω)Q(dn).
Since x→∫ℕ‍∫𝔻‍1A(x,ω,n)Kμ((x,n),dω)Q(dn) is ℬ(H)¯μ-measurable, so is the right side of (58). Integrating the above equation with respect to μ and we obtain
(59)∫H‍∫ℕ‍∫𝔻‍1A(x,ω,n)δFμ(x,n)(dω)Q(dn)μ(dx)=∫H‍∫ℕ‍∫𝔻‍1A(x,ω,n)δFδx(x,n)(dω)Q(dn)μ(dx),
which proofs the assertion.
Letx∈H, t∈[0,∞), A∈ℬt(𝔻), and define
(60)F-δx:=1{x}×ℕFδx.
Then
(61)F-δx=Fδxδx⊗Q-a.e.;
hence,
(62){F-δx∈A}∈ℬ(H)⊗ℬt(ℕ)¯δx⊗Q.
But
(63){F-δx∈A}={x}×{Fδx(x,·)∈A}∪{H∖{x}×{0∈A}},
so by (62) it follows that
(64){Fδx(x,·)∈A}∈ℬt(ℕ)¯Q.Remark 17.
Our result can be applied to paper [10, page 19] to make sure that there exists a measurable map 𝒢ϵ:𝕄→D([0,T]:Φ-p1) such that 𝒢ϵ(ϵnϵ-1) is the unique strong solution of the equation they considered.


## 4. Conclusions

What we have seen from above is a Yamada-Watanabe theorem for a type of stochastic partial differential equations with jumps, proved by using the original idea of Yamada and Watanabe’s. Comparing with results of Kurtz [2, 3], their results can also imply the same result, where they used another abstract method.



---

*Source: 982190-2014-02-02.xml*