# Numerical Solutions of Stochastic Differential Delay Equations with Poisson Random Measure under the Generalized Khasminskii-Type Conditions

Authors: Minghui Song; Hui Yu
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

The Euler method is introduced for stochastic differential delay equations (SDDEs) with Poisson random measure under the generalized Khasminskii-type conditions which cover more classes of such equations than before. The main aims of this paper are to prove the existence of global solutions to such equations and then to investigate the convergence of the Euler method in probability under the generalized Khasminskii-type conditions. Numerical example is given to indicate our results.


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## Body

## 1. Introduction

To take into consideration stochastic effects such as corporate defaults, operational failures, market crashes or central bank announcements in financial market, the research on stochastic differential equations (SDEs) with Poisson random measure (see [1, 2]) is important, since Merton initiated the model of such equations in 1976 (see [3]). Due to the rate of change of financial dynamics system depending on its past history, SDDE with Poisson random measure (see [4, 5]), the case we propose and consider in this work, is meaningful.Since there is no explicit solution for an SDDE with Poisson random measure, one needs, in general, numerical methods which can be classified into strong approximations and weak approximations (see [6–8]).We here give an overview of the results on the strong approximations of differential equation driven by Wiener process and Poisson random measure. Platen [9] presented a convergence theorem with order γ∈{0.5,1,1.5,…} and originally introduced the jump-adapted methods which are based on all the jump times. Moreover, Bruti-Liberati and Platen (see [10]) get the jump-adapted order 1.5 scheme, and they also construct the derivative free or implicit jump-adapted schemes with desired order of strong convergence. In [11], for a class of pure jump systems, the order of Taylor schemes is given under weaker conditions than the current literature. In [7, 10], Bruti-Liberati and Platen present the drift-implicit schemes which have order γ∈{0.5,1}. Recently, [8] develops adaptive time stepping algorithms based on a jump augmented Monte Carlo Euler-Maruyama method, which achieve a prescribed precision. Mao [4] presents the convergence of numerical solutions for variable delay differential equations with Poisson random measure. In [12], the improved Runge-Kutta methods are presented to improve the accuracy behaviour of problems with small noise for SDEs driven by Poisson random measure. Clearly, the results above require that the SDEs with poisson random measure satisfy the global Lipschitz conditions and the linear growth conditions. In [5], the Euler scheme is proved to converge to the analytic solution for SDDEs with Wiener process and Poisson random measure under weaker conditions than the global Lipschitz condition and the linear growth condition.However, there are many SDDEs with Poisson random measure, especially highly nonlinear equations, which do not satisfy the above-mentioned conditions and classical Khasminskii-type conditions (see [13–15]). And in Section 5, we give such highly nonlinear equation. Our work is motivated by [16] in which the generalized Khasminskii-type conditions are applied to SDDEs with Wiener process. The main contribution in our paper is to present Euler method for SDDEs with Poisson random measure under the generalized Khasminskii-type conditions which cover more classes of these equations than all the mentioned classical conditions above.Our work is organized as follows. In Section2, the properties of SDDEs with Poisson random measure are given under the generalized Khasminskii-type conditions. In Section 3, Euler method is analyzed under such conditions. In Section 4, we present the convergence in probability of the Euler method. In Section 5, an example is given.


## 2. The Generalized Khasminskii-Type Conditions for SDDEs with Poisson Random Measure

### 2.1. Problem's Setting

Throughout this paper, unless otherwise specified, we use the following notations. Let|·| be the Euclidean norm in Rd,d∈N. Let u1∨u2=max{u1,u2} and u1∧u2=min{u1,u2}. If A is a vector or matrix, its transpose is denoted by AT. If A is a matrix, its trace norm is denoted by |A|=trace(ATA). Let τ>0 and R+=[0,∞). Let C([-τ,0];Rd) denote the family of continuous functions from [-τ,0] to Rd with the norm |φ|=sup-τ≤θ≤0|φ(θ)|. Denote by C(Rd;R+) the family of continuous functions from Rd to R+. Let C2(Rd;R+) denote the family of continuously two times differentiable R+-valued functions from Rd to R+. [z] denotes the largest integer which is less than or equal to z in R. I𝒜 denotes the indicator function of a set 𝒜.The followingd-dimensional SDDE with Poisson random measure is considered in our paper:
(2.1)dx(t)=a(x(t-),x((t-τ)-))dt+b(x(t-),x((t-τ)-))dW(t)+∫ε‍c(x(t-),x((t-τ)-),v)p~ϕ(dv×dt),
for t>0, where p~ϕ(dv×dt):=pϕ(dv×dt)-ϕ(dv)dt. Here x(t-) denotes lims↑tx(s). The initial data of (2.1) is given by
(2.2){x(t):-τ≤t≤0}=ξ(t)∈C([-τ,0];Rd),
where x(-τ-)=x(-τ).The drift coefficienta:Rd×Rd→Rd, the diffusion coefficient b:Rd×Rd→Rd×m0, and the jump coefficient c:Rd×Rd×ε→Rd are assumed to be Borel measurable functions and the coefficients are sufficiently smooth.The randomness in (2.1) is generated by the following (see [8]). An m0-dimensional Wiener process W={W(t)=(W1(t),…,Wm0(t))T} with independent scalar components is defined on a filtered probability space (ΩW,ℱW,(ℱtW)t≥0,PW). A Poisson random measure pϕ(ω,dv×dt) is on ΩJ×ε×[0,∞), where ε⊆Rr∖{0} with r∈N, and its deterministic compensated measure ϕ(dv)dt=λf(v)dvdt.f(v) is a probability density, and we require finite intensity λ=ϕ(ε)<∞. The Poisson random measure is defined on a filtered probability space (ΩJ,ℱJ,(ℱtJ)t≥0,PJ). The process x(t) is thus defined on a product space (Ω,ℱ,(ℱt)t≥0,P), where Ω=ΩW×ΩJ,ℱ=ℱW×ℱJ,(ℱt)t≥0=(ℱtW)t≥0×(ℱtJ)t≥0, P=PW×PJ, and ℱ0 contains all P-null sets. The Wiener process and the Poisson random measure are mutually independent.To state the generalized Khasminskii-type conditions, we define the operatorLV:Rd×Rd→R by
(2.3)LV(x,y)=Vx(x)a(x,y)+12trace(bT(x,y)Vxx(x)b(x,y))+∫ε‍(V(x+c(x,y,v))-V(x)-Vx(x)c(x,y,v))ϕ(dv),
where
(2.4)V∈C2(Rd;R+),Vx=(∂V(x)∂x1,…,∂V(x)∂xd),Vxx=(∂2V(x)∂xi∂xj)d×d.Now the generalized Khasminskii-type conditions are given by the following assumptions.Assumption 2.1.
For each integerk≥1, there exists a positive constant Ck, dependent on k, such that
(2.5)|a(x,y)-a(x¯,y¯)|2∨|b(x,y)-b(x¯,y¯)|2≤Ck(|x-y|2+|x¯-y¯|2),
for x,y∈Rd with |x|∨|y|≤k. And there exists a positive constant C such that
(2.6)∫ε‍|c(x,y,v)-c(x¯,y¯,v)|2ϕ(dv)≤C(|x-y|2+|x¯-y¯|2),
for x,y∈Rd.Assumption 2.2.
There are two functionsV∈C2(Rd;R+) and U∈C(Rd;R+) as well as two positive constants μ1 and μ2 such that
(2.7)lim⁡|x|→∞V(x)=∞,(2.8)LV(x,y)≤μ1(1+V(x)+V(y)+U(y))-μ2U(x),
for all (x,y)∈Rd×Rd.Assumption 2.3.
There exists a positive constantC¯ such that
(2.9)∫ε‍|c(0,0,v)|2ϕ(dv)≤C¯.Assumption 2.4.
There exists a positive constantL such that the initial data (2.2) satisfies
(2.10)|ξ(t)-ξ(s)|≤L|t-s|1/2,for-τ≤t,s≤0.


### 2.2. The Existence of Global Solutions

In this section, we analyze the existence and the property of the global solution to (2.1) under Assumptions 2.1, 2.2, and 2.4.In order to demonstrate the existence of the global solution to (2.1), we redefine the following concepts mainly according to [17, 18].Definition 2.5.
Let{x(t)}t≥-τ be an Rd-valued stochastic process. The process is said to be càdlàg if it is right continuous and for almost all ω∈Ω the left limit lims↑tx(s) exists and is finite for all t>-τ.Definition 2.6.
Letσ∞ be a stopping time such that 0≤σ∞≤T a.s. An Rd-valued, ℱt-adapted, and càdlàg process {x(t):-τ≤t<σ∞} is called a local solution of (2.1) if x(t)=ξ(t) on t∈[-τ,0] and, moreover, there is a nondecreasing sequence {σk}k≥1 of stopping times such that 0≤σk↑σ∞ a.s. and
(2.11)x(t∧σk)=x(0)+∫0t∧σk‍a(x(s-),x((s-τ)-))ds+∫0t∧σk‍b(x(s-),x((s-τ)-))dW(s)+∫0t∧σk‍∫ε‍c(x(s-),x((s-τ)-),v)p~ϕ(dv×ds)
holds for any t∈[0,T] and k≥1 with probability 1. If, furthermore,
(2.12)lim⁡t→σ∞sup⁡|x(t)|=∞wheneverσ∞<T,
then it is called a maximal local solution of (2.1) and σ∞ is called the explosion time. A local solution {x(t):-τ≤t<σ∞} to (2.1) is called a global solution if σ∞=∞.Lemma 2.7.
Under Assumptions2.1 and 2.4, for any given initial data (2.2), there is a unique maximal local solution to (2.1).Proof.
From Assumption2.4, for the initial data (2.2), we have
(2.13)max⁡-τ≤t≤0|ξ(t)|≤max⁡-τ≤t≤0|ξ(t)-ξ(0)|+|ξ(0)|≤Lτ+|ξ(0)|.
For each integer k≥[Lτ+|ξ(0)|]+1, we define
(2.14)z[k]=|z|∧k|z|z,0[k]=0,
for z∈Rd. And then we define the truncation functions
(2.15)ak(x,y)=a(x[k],y[k]),bk(x,y)=b(x[k],y[k]),ck(x,y,v)=c(x,y,v),
for x,y∈Rd and each k≥[Lτ+|ξ(0)|]+1. Moreover, we define the following equation:
(2.16)dxk(t)=ak(xk(t-),xk((t-τ)-))dt+bk(xk(t-),xk((t-τ)-))dW(t)+∫ε‍ck(xk(t-),xk((t-τ)-),v)p~ϕ(dv×dt),
on t∈[0,T] with initial data xk(t)=ξ(t) on t∈[-τ,0]. Obviously, the equation satisfies the global Lipschitz conditions and the linear growth conditions. Therefore according to [4], there is a unique global solution xk(t) to (2.16) and its solution is a càdlàg process (see [17]). We define the stopping time
(2.17)σk=T∧inf⁡{t∈[0,T]:|xk(t)|≥k},
for k≥[Lτ+|ξ(0)|]+1, and
(2.18)σ1=⋯=σ[Lτ+|ξ(0)|]=σ[Lτ+|ξ(0)|]+1,
where we set infϕ=∞ (as usual ϕ denotes the empty set) throughout our paper. We can easily get
(2.19)xk(t)=xk+1(t),-τ≤t≤σk,
which means {σk}k≥1 is a nondecreasing sequence and then let limk→∞σk=σ∞ a.s. Now, we define {x(t):-τ≤t<σ∞} with x(t)=ξ(t) on t∈[-τ,0] and
(2.20)x(t)=xk(t),t∈[σk-1,σk),k≥1,
where σ0=0. And from (2.16) and (2.19), we can also obtain
(2.21)x(t∧σk)=xk(t∧σk)=x0+∫0t∧σk‍a(x(s-),x((s-τ)-))ds+∫0t∧σk‍b(x(s-),x((s-τ)-))dW(s)+∫0t∧σk‍∫ε‍c(x(s-),x((s-τ)-),v)p~ϕ(dv×ds),
for any t∈[0,T] and k≥1 with probability 1. Moreover, if σ∞<T, then
(2.22)lim⁡t→σ∞sup⁡|x(t)|≥lim⁡k→∞sup⁡|x(σk)|=lim⁡k→∞sup⁡|xk(σk)|=∞.
Hence {x(t):-τ≤t<σ∞} is a maximal local solution to (2.1).
To show the uniqueness of the solution to (2.1), let {x¯(t):-τ≤t<σ¯∞} be another maximal local solution. As the same proof as Theorem 2.8 in [17], we infer that
(2.23)P(x(t,ω)=x¯(t,ω),(t,ω)∈[-τ,σk∧σ¯k)×Ω)=1,k≥1.
Hence by k→∞, we get
(2.24)P(x(t,ω)=x¯(t,ω),(t,ω)∈[-τ,σ∞∧σ¯∞)×Ω)=1.
Thereforex(t) is a unique local solution and then it is a unique maximal local solution to (2.1).
So we complete the whole proof.Now, the existence of the global solution to (2.1) is shown in the following theorem.Theorem 2.8.
Under Assumptions2.1, 2.2, and 2.4, for any given initial data (2.2), there is a unique global solution x(t) to (2.1) on t∈[-τ,∞).Proof.
According to Lemma2.7, there exists a unique maximal local solution to (2.1) on [-τ,σ∞). Hence in order to show that this local solution is a global one, we only need to demonstrate σ∞=∞ a.s. Using Itô's formula (see [1]) to V(x(t)), we have
(2.25)dV(x(t))=({12}Vx(x(t-))a(x(t-),x((t-τ)-))+12trace(bT(x(t-),x((t-τ)-))Vxx(x(t-))b(x(t-),x((t-τ)-))))dt+∫ε‍(V(x(t-)+c(x(t-),x((t-τ)-),v))-V(x(t-))-Vx(x(t-))c(x(t-),x((t-τ)-),v))ϕ(dv)dt+Vx(x(t-))b(x(t-),x((t-τ)-))dW(t)+∫ε‍(V(x(t-)+c(x(t-),x((t-τ)-),v))-V(x(t-)))p~ϕ(dv×dt)=LV(x(t-),x((t-τ)-))dt+Vx(x(t-))b(x(t-),x((t-τ)-))dW(t)+∫ε‍(V(x(t-)+c(x(t-),x((t-τ)-),v))-V(x(t-)))p~ϕ(dv×dt),
for t∈[0,σ∞).
Our proof is divided into the following steps.
Step
1. For any integer k≥[Lτ+|ξ(0)|]+1 and 0≤t≤τ, by taking integration and expectations and using Assumption 2.2 to (2.25), we get
(2.26)EV(x(t∧σk))-EV(x(0))≤E∫0t∧σk‍(μ1(1+V(x(s-))+V(x((s-τ)-))+U(x((s-τ)-)))-μ2U(x(s-)))ds,
which means
(2.27)EV(x(t∧σk))≤C1+μ1E∫0t∧σk‍V(x(s-))ds-μ2E∫0t∧σk‍U(x(s-))ds,
where
(2.28)C1=V(x(0))+μ1τ+μ1∫-τ0‍V(ξ(s))ds+μ1∫-τ0‍U(ξ(s))ds<∞.
From (2.27), we obtain
(2.29)EV(x(t∧σk))≤C1+μ1E∫0t∧σk‍V(x(s-))ds≤C1+μ1∫0t‍EV(x(s∧σk-))ds,
by the Gronwall inequality (see [18]), which leads to
(2.30)EV(x(t∧σk))≤C1eμ1τ,
for 0≤t≤τ and k≥[Lτ+|ξ(0)|]+1. Let
(2.31)ςk=inf⁡|x|≥k,0≤t<∞V(x),fork≥[Lτ+|ξ(0)|]+1.
Therefore, from (2.30), we have
(2.32)ςkP(σk≤τ)≤E(V(x(σk))I{σk≤τ})≤EV(x(τ∧σk))≤C1eμ1τ,
by taking k→∞, which gives
(2.33)P(σ∞≤τ)=0.
Hence we get
(2.34)P(σ∞>τ)=1.
It thus follows from (2.30) and (2.34) that
(2.35)EV(x(t))≤C1eμ1τ,0≤t≤τ,
by taking k→∞.
Moreover, from (2.27), we get
(2.36)E∫0τ∧σk‍U(x(s-))ds≤μ2-1(C1+μ1E∫0τ∧σk‍V(x(s-))ds)≤μ2-1(C1+μ1∫0τ‍EV(x(s∧σk-))ds),
by taking k→∞, which gives
(2.37)E∫0τ‍U(x(s))ds≤μ2-1(C1+τμ1C1eμ1τ)<∞,
where (2.34) and (2.35) are used.
Step
2. For any integer k≥[Lτ+|ξ(0)|]+1 and 0≤t≤2τ, the similar analysis as above gives
(2.38)EV(x(t∧σk))≤C2+μ1E∫0t∧σk‍V(x(s-))ds-μ2E∫0t∧σk‍U(x(s-))ds,
where
(2.39)C2=V(x(0))+2μ1τ+μ1∫-τ0‍V(ξ(s))ds+μ1∫-τ0‍U(ξ(s))ds+μ1E∫0τ‍V(x(s))ds+μ1E∫0τ‍U(x(s))ds<∞,
from (2.35) and (2.37).
Thus,(2.40)EV(x(t∧σk))≤C2+μ1E∫0t‍V(x(s∧σk-))ds,
which gives
(2.41)EV(x(t∧σk))≤C2e2μ1τ,
for 0≤t≤2τ and k≥[Lτ+|ξ(0)|]+1. Hence we get
(2.42)ςkP(σk≤2τ)≤EV(x(2τ∧σk))≤C2e2μ1τ,
by taking k→∞, which implies
(2.43)P(σ∞≤2τ)=0,
that is,
(2.44)P(σ∞>2τ)=1.
Moreover, by taking k→∞ to (2.41), we then get
(2.45)EV(x(t))≤C2e2μ1τ,0≤t≤2τ.
Therefore, from (2.38), (2.44), and (2.45), we have
(2.46)E∫02τ‍U(x(s))ds≤μ2-1(C2+2τμ1C1e2μ1τ)<∞.
Step
3. So for any i∈N, we repeat the similar analysis as above and then obtain
(2.47)P(σ∞>iτ)=1,EV(x(t))≤Cieiμ1τ,0≤t≤iτ,E∫0τ‍U(x(s))ds≤μ2-1(Ci+iτμ1Cieiμ1τ)<∞,
where
(2.48)Ci=V(x(0))+μ1E∫-τ(i-1)τ‍(1+V(x(s))+U(x(s)))ds<∞.
So we can getP(σ∞=∞)=1 and the required result follows.In the following lemma, we show that the solution of (2.1) remains in a compact set with a large probability.Lemma 2.9.
Under Assumptions2.1, 2.2, and 2.4, for any pair of ϵ∈(0,1) and T>0, there exists a sufficiently large integer k*, dependent on ϵ and T, such that
(2.49)P(σk≤T)≤ϵ,∀k≥k*,
where σk is defined in Lemma 2.7.Proof.
According to Theorem2.8, we can get
(2.50)EV(x(T∧σk))≤Cieiμ1τ,
for i large enough to iτ≥T and k≥[Lτ+|ξ(0)|]+1. Therefore, we have
(2.51)ςkP(σk≤T)≤E(V(x(σk))I{σk≤T})≤EV(x(T∧σk))≤Cieiμ1τ,
where
(2.52)ςk=inf⁡|x|≥kV(x)fork≥[Lτ+|ξ(0)|]+1.
Under (2.7) in Assumption 2.2, there exists a sufficiently large integer k* such that
(2.53)P(σk≤T)≤Cieiμ1τςk≤ϵ,∀k≥k*.
So we complete the proof.


## 2.1. Problem's Setting

Throughout this paper, unless otherwise specified, we use the following notations. Let|·| be the Euclidean norm in Rd,d∈N. Let u1∨u2=max{u1,u2} and u1∧u2=min{u1,u2}. If A is a vector or matrix, its transpose is denoted by AT. If A is a matrix, its trace norm is denoted by |A|=trace(ATA). Let τ>0 and R+=[0,∞). Let C([-τ,0];Rd) denote the family of continuous functions from [-τ,0] to Rd with the norm |φ|=sup-τ≤θ≤0|φ(θ)|. Denote by C(Rd;R+) the family of continuous functions from Rd to R+. Let C2(Rd;R+) denote the family of continuously two times differentiable R+-valued functions from Rd to R+. [z] denotes the largest integer which is less than or equal to z in R. I𝒜 denotes the indicator function of a set 𝒜.The followingd-dimensional SDDE with Poisson random measure is considered in our paper:
(2.1)dx(t)=a(x(t-),x((t-τ)-))dt+b(x(t-),x((t-τ)-))dW(t)+∫ε‍c(x(t-),x((t-τ)-),v)p~ϕ(dv×dt),
for t>0, where p~ϕ(dv×dt):=pϕ(dv×dt)-ϕ(dv)dt. Here x(t-) denotes lims↑tx(s). The initial data of (2.1) is given by
(2.2){x(t):-τ≤t≤0}=ξ(t)∈C([-τ,0];Rd),
where x(-τ-)=x(-τ).The drift coefficienta:Rd×Rd→Rd, the diffusion coefficient b:Rd×Rd→Rd×m0, and the jump coefficient c:Rd×Rd×ε→Rd are assumed to be Borel measurable functions and the coefficients are sufficiently smooth.The randomness in (2.1) is generated by the following (see [8]). An m0-dimensional Wiener process W={W(t)=(W1(t),…,Wm0(t))T} with independent scalar components is defined on a filtered probability space (ΩW,ℱW,(ℱtW)t≥0,PW). A Poisson random measure pϕ(ω,dv×dt) is on ΩJ×ε×[0,∞), where ε⊆Rr∖{0} with r∈N, and its deterministic compensated measure ϕ(dv)dt=λf(v)dvdt.f(v) is a probability density, and we require finite intensity λ=ϕ(ε)<∞. The Poisson random measure is defined on a filtered probability space (ΩJ,ℱJ,(ℱtJ)t≥0,PJ). The process x(t) is thus defined on a product space (Ω,ℱ,(ℱt)t≥0,P), where Ω=ΩW×ΩJ,ℱ=ℱW×ℱJ,(ℱt)t≥0=(ℱtW)t≥0×(ℱtJ)t≥0, P=PW×PJ, and ℱ0 contains all P-null sets. The Wiener process and the Poisson random measure are mutually independent.To state the generalized Khasminskii-type conditions, we define the operatorLV:Rd×Rd→R by
(2.3)LV(x,y)=Vx(x)a(x,y)+12trace(bT(x,y)Vxx(x)b(x,y))+∫ε‍(V(x+c(x,y,v))-V(x)-Vx(x)c(x,y,v))ϕ(dv),
where
(2.4)V∈C2(Rd;R+),Vx=(∂V(x)∂x1,…,∂V(x)∂xd),Vxx=(∂2V(x)∂xi∂xj)d×d.Now the generalized Khasminskii-type conditions are given by the following assumptions.Assumption 2.1.
For each integerk≥1, there exists a positive constant Ck, dependent on k, such that
(2.5)|a(x,y)-a(x¯,y¯)|2∨|b(x,y)-b(x¯,y¯)|2≤Ck(|x-y|2+|x¯-y¯|2),
for x,y∈Rd with |x|∨|y|≤k. And there exists a positive constant C such that
(2.6)∫ε‍|c(x,y,v)-c(x¯,y¯,v)|2ϕ(dv)≤C(|x-y|2+|x¯-y¯|2),
for x,y∈Rd.Assumption 2.2.
There are two functionsV∈C2(Rd;R+) and U∈C(Rd;R+) as well as two positive constants μ1 and μ2 such that
(2.7)lim⁡|x|→∞V(x)=∞,(2.8)LV(x,y)≤μ1(1+V(x)+V(y)+U(y))-μ2U(x),
for all (x,y)∈Rd×Rd.Assumption 2.3.
There exists a positive constantC¯ such that
(2.9)∫ε‍|c(0,0,v)|2ϕ(dv)≤C¯.Assumption 2.4.
There exists a positive constantL such that the initial data (2.2) satisfies
(2.10)|ξ(t)-ξ(s)|≤L|t-s|1/2,for-τ≤t,s≤0.


## 2.2. The Existence of Global Solutions

In this section, we analyze the existence and the property of the global solution to (2.1) under Assumptions 2.1, 2.2, and 2.4.In order to demonstrate the existence of the global solution to (2.1), we redefine the following concepts mainly according to [17, 18].Definition 2.5.
Let{x(t)}t≥-τ be an Rd-valued stochastic process. The process is said to be càdlàg if it is right continuous and for almost all ω∈Ω the left limit lims↑tx(s) exists and is finite for all t>-τ.Definition 2.6.
Letσ∞ be a stopping time such that 0≤σ∞≤T a.s. An Rd-valued, ℱt-adapted, and càdlàg process {x(t):-τ≤t<σ∞} is called a local solution of (2.1) if x(t)=ξ(t) on t∈[-τ,0] and, moreover, there is a nondecreasing sequence {σk}k≥1 of stopping times such that 0≤σk↑σ∞ a.s. and
(2.11)x(t∧σk)=x(0)+∫0t∧σk‍a(x(s-),x((s-τ)-))ds+∫0t∧σk‍b(x(s-),x((s-τ)-))dW(s)+∫0t∧σk‍∫ε‍c(x(s-),x((s-τ)-),v)p~ϕ(dv×ds)
holds for any t∈[0,T] and k≥1 with probability 1. If, furthermore,
(2.12)lim⁡t→σ∞sup⁡|x(t)|=∞wheneverσ∞<T,
then it is called a maximal local solution of (2.1) and σ∞ is called the explosion time. A local solution {x(t):-τ≤t<σ∞} to (2.1) is called a global solution if σ∞=∞.Lemma 2.7.
Under Assumptions2.1 and 2.4, for any given initial data (2.2), there is a unique maximal local solution to (2.1).Proof.
From Assumption2.4, for the initial data (2.2), we have
(2.13)max⁡-τ≤t≤0|ξ(t)|≤max⁡-τ≤t≤0|ξ(t)-ξ(0)|+|ξ(0)|≤Lτ+|ξ(0)|.
For each integer k≥[Lτ+|ξ(0)|]+1, we define
(2.14)z[k]=|z|∧k|z|z,0[k]=0,
for z∈Rd. And then we define the truncation functions
(2.15)ak(x,y)=a(x[k],y[k]),bk(x,y)=b(x[k],y[k]),ck(x,y,v)=c(x,y,v),
for x,y∈Rd and each k≥[Lτ+|ξ(0)|]+1. Moreover, we define the following equation:
(2.16)dxk(t)=ak(xk(t-),xk((t-τ)-))dt+bk(xk(t-),xk((t-τ)-))dW(t)+∫ε‍ck(xk(t-),xk((t-τ)-),v)p~ϕ(dv×dt),
on t∈[0,T] with initial data xk(t)=ξ(t) on t∈[-τ,0]. Obviously, the equation satisfies the global Lipschitz conditions and the linear growth conditions. Therefore according to [4], there is a unique global solution xk(t) to (2.16) and its solution is a càdlàg process (see [17]). We define the stopping time
(2.17)σk=T∧inf⁡{t∈[0,T]:|xk(t)|≥k},
for k≥[Lτ+|ξ(0)|]+1, and
(2.18)σ1=⋯=σ[Lτ+|ξ(0)|]=σ[Lτ+|ξ(0)|]+1,
where we set infϕ=∞ (as usual ϕ denotes the empty set) throughout our paper. We can easily get
(2.19)xk(t)=xk+1(t),-τ≤t≤σk,
which means {σk}k≥1 is a nondecreasing sequence and then let limk→∞σk=σ∞ a.s. Now, we define {x(t):-τ≤t<σ∞} with x(t)=ξ(t) on t∈[-τ,0] and
(2.20)x(t)=xk(t),t∈[σk-1,σk),k≥1,
where σ0=0. And from (2.16) and (2.19), we can also obtain
(2.21)x(t∧σk)=xk(t∧σk)=x0+∫0t∧σk‍a(x(s-),x((s-τ)-))ds+∫0t∧σk‍b(x(s-),x((s-τ)-))dW(s)+∫0t∧σk‍∫ε‍c(x(s-),x((s-τ)-),v)p~ϕ(dv×ds),
for any t∈[0,T] and k≥1 with probability 1. Moreover, if σ∞<T, then
(2.22)lim⁡t→σ∞sup⁡|x(t)|≥lim⁡k→∞sup⁡|x(σk)|=lim⁡k→∞sup⁡|xk(σk)|=∞.
Hence {x(t):-τ≤t<σ∞} is a maximal local solution to (2.1).
To show the uniqueness of the solution to (2.1), let {x¯(t):-τ≤t<σ¯∞} be another maximal local solution. As the same proof as Theorem 2.8 in [17], we infer that
(2.23)P(x(t,ω)=x¯(t,ω),(t,ω)∈[-τ,σk∧σ¯k)×Ω)=1,k≥1.
Hence by k→∞, we get
(2.24)P(x(t,ω)=x¯(t,ω),(t,ω)∈[-τ,σ∞∧σ¯∞)×Ω)=1.
Thereforex(t) is a unique local solution and then it is a unique maximal local solution to (2.1).
So we complete the whole proof.Now, the existence of the global solution to (2.1) is shown in the following theorem.Theorem 2.8.
Under Assumptions2.1, 2.2, and 2.4, for any given initial data (2.2), there is a unique global solution x(t) to (2.1) on t∈[-τ,∞).Proof.
According to Lemma2.7, there exists a unique maximal local solution to (2.1) on [-τ,σ∞). Hence in order to show that this local solution is a global one, we only need to demonstrate σ∞=∞ a.s. Using Itô's formula (see [1]) to V(x(t)), we have
(2.25)dV(x(t))=({12}Vx(x(t-))a(x(t-),x((t-τ)-))+12trace(bT(x(t-),x((t-τ)-))Vxx(x(t-))b(x(t-),x((t-τ)-))))dt+∫ε‍(V(x(t-)+c(x(t-),x((t-τ)-),v))-V(x(t-))-Vx(x(t-))c(x(t-),x((t-τ)-),v))ϕ(dv)dt+Vx(x(t-))b(x(t-),x((t-τ)-))dW(t)+∫ε‍(V(x(t-)+c(x(t-),x((t-τ)-),v))-V(x(t-)))p~ϕ(dv×dt)=LV(x(t-),x((t-τ)-))dt+Vx(x(t-))b(x(t-),x((t-τ)-))dW(t)+∫ε‍(V(x(t-)+c(x(t-),x((t-τ)-),v))-V(x(t-)))p~ϕ(dv×dt),
for t∈[0,σ∞).
Our proof is divided into the following steps.
Step
1. For any integer k≥[Lτ+|ξ(0)|]+1 and 0≤t≤τ, by taking integration and expectations and using Assumption 2.2 to (2.25), we get
(2.26)EV(x(t∧σk))-EV(x(0))≤E∫0t∧σk‍(μ1(1+V(x(s-))+V(x((s-τ)-))+U(x((s-τ)-)))-μ2U(x(s-)))ds,
which means
(2.27)EV(x(t∧σk))≤C1+μ1E∫0t∧σk‍V(x(s-))ds-μ2E∫0t∧σk‍U(x(s-))ds,
where
(2.28)C1=V(x(0))+μ1τ+μ1∫-τ0‍V(ξ(s))ds+μ1∫-τ0‍U(ξ(s))ds<∞.
From (2.27), we obtain
(2.29)EV(x(t∧σk))≤C1+μ1E∫0t∧σk‍V(x(s-))ds≤C1+μ1∫0t‍EV(x(s∧σk-))ds,
by the Gronwall inequality (see [18]), which leads to
(2.30)EV(x(t∧σk))≤C1eμ1τ,
for 0≤t≤τ and k≥[Lτ+|ξ(0)|]+1. Let
(2.31)ςk=inf⁡|x|≥k,0≤t<∞V(x),fork≥[Lτ+|ξ(0)|]+1.
Therefore, from (2.30), we have
(2.32)ςkP(σk≤τ)≤E(V(x(σk))I{σk≤τ})≤EV(x(τ∧σk))≤C1eμ1τ,
by taking k→∞, which gives
(2.33)P(σ∞≤τ)=0.
Hence we get
(2.34)P(σ∞>τ)=1.
It thus follows from (2.30) and (2.34) that
(2.35)EV(x(t))≤C1eμ1τ,0≤t≤τ,
by taking k→∞.
Moreover, from (2.27), we get
(2.36)E∫0τ∧σk‍U(x(s-))ds≤μ2-1(C1+μ1E∫0τ∧σk‍V(x(s-))ds)≤μ2-1(C1+μ1∫0τ‍EV(x(s∧σk-))ds),
by taking k→∞, which gives
(2.37)E∫0τ‍U(x(s))ds≤μ2-1(C1+τμ1C1eμ1τ)<∞,
where (2.34) and (2.35) are used.
Step
2. For any integer k≥[Lτ+|ξ(0)|]+1 and 0≤t≤2τ, the similar analysis as above gives
(2.38)EV(x(t∧σk))≤C2+μ1E∫0t∧σk‍V(x(s-))ds-μ2E∫0t∧σk‍U(x(s-))ds,
where
(2.39)C2=V(x(0))+2μ1τ+μ1∫-τ0‍V(ξ(s))ds+μ1∫-τ0‍U(ξ(s))ds+μ1E∫0τ‍V(x(s))ds+μ1E∫0τ‍U(x(s))ds<∞,
from (2.35) and (2.37).
Thus,(2.40)EV(x(t∧σk))≤C2+μ1E∫0t‍V(x(s∧σk-))ds,
which gives
(2.41)EV(x(t∧σk))≤C2e2μ1τ,
for 0≤t≤2τ and k≥[Lτ+|ξ(0)|]+1. Hence we get
(2.42)ςkP(σk≤2τ)≤EV(x(2τ∧σk))≤C2e2μ1τ,
by taking k→∞, which implies
(2.43)P(σ∞≤2τ)=0,
that is,
(2.44)P(σ∞>2τ)=1.
Moreover, by taking k→∞ to (2.41), we then get
(2.45)EV(x(t))≤C2e2μ1τ,0≤t≤2τ.
Therefore, from (2.38), (2.44), and (2.45), we have
(2.46)E∫02τ‍U(x(s))ds≤μ2-1(C2+2τμ1C1e2μ1τ)<∞.
Step
3. So for any i∈N, we repeat the similar analysis as above and then obtain
(2.47)P(σ∞>iτ)=1,EV(x(t))≤Cieiμ1τ,0≤t≤iτ,E∫0τ‍U(x(s))ds≤μ2-1(Ci+iτμ1Cieiμ1τ)<∞,
where
(2.48)Ci=V(x(0))+μ1E∫-τ(i-1)τ‍(1+V(x(s))+U(x(s)))ds<∞.
So we can getP(σ∞=∞)=1 and the required result follows.In the following lemma, we show that the solution of (2.1) remains in a compact set with a large probability.Lemma 2.9.
Under Assumptions2.1, 2.2, and 2.4, for any pair of ϵ∈(0,1) and T>0, there exists a sufficiently large integer k*, dependent on ϵ and T, such that
(2.49)P(σk≤T)≤ϵ,∀k≥k*,
where σk is defined in Lemma 2.7.Proof.
According to Theorem2.8, we can get
(2.50)EV(x(T∧σk))≤Cieiμ1τ,
for i large enough to iτ≥T and k≥[Lτ+|ξ(0)|]+1. Therefore, we have
(2.51)ςkP(σk≤T)≤E(V(x(σk))I{σk≤T})≤EV(x(T∧σk))≤Cieiμ1τ,
where
(2.52)ςk=inf⁡|x|≥kV(x)fork≥[Lτ+|ξ(0)|]+1.
Under (2.7) in Assumption 2.2, there exists a sufficiently large integer k* such that
(2.53)P(σk≤T)≤Cieiμ1τςk≤ϵ,∀k≥k*.
So we complete the proof.


## 3. The Euler Method

In this section, we introduce the Euler method to (2.1) under Assumptions 2.1, 2.2, 2.3, and 2.4.Given a step sizeΔt=τ/m∈(0,1),m∈N, the Euler method applied to (2.1) computes approximation Xn≈x(tn), where tn=nΔt for n=-m,-(m-1),…,-1,0,1,…, by setting
(3.1)Xn=ξ(nΔt)forn=-m,-(m-1),…,-1,0,
and forming
(3.2)Xn+1=Xn+a(Xn,Xn-m)Δt+b(Xn,Xn-m)ΔWn+∫tntn+1‍∫ε‍c(Xn,Xn-m,v)p~ϕ(dv×dt),
for n=0,1,…, where ΔWn=W(tn+1)-W(tn).The continuous-time Euler methodX¯(t) on t∈[-τ,∞) is then defined by
(3.3)X¯(t):=ξ(t)fort∈[-τ,0],(3.4)X¯(t):=X0+∫0t‍a(Z(s),Z(s-τ))ds+∫0t‍b(Z(s),Z(s-τ))dW(s)+∫0t‍∫ε‍c(Z(s),Z(s-τ),v)p~ϕ(dv×ds),
for t≥0, where
(3.5)Z(t)=∑n=-m∞‍XnI[nΔt,(n+1)Δt)(t)fort∈[-τ,∞).Actually, we can see in [11] that pϕ={pϕ(t):=pϕ(ε×[0,t])} is a process that counts the number of jumps until some given time. The Poisson random measure pϕ(dv×dt) generates a sequence of pairs {(ιi,ξi),i∈{1,2,…,pϕ(T)}} for a given finite positive constant T if λ<∞. Here {ιi:Ω→R+,i∈{1,2,…,pϕ(T)}} is a sequence of increasing nonnegative random variables representing the jump times of a standard Poisson process with intensity λ, and {ξi:Ω→ε,i∈{1,2,…,pϕ(T)}} is a sequence of independent identically distributed random variables, where ξi is distributed according to ϕ(dv)/ϕ(ε). Then (3.2) can equivalently be of the following form:
(3.6)Xn+1=Xn+(a(Xn,Xn-m)-∫ε‍c(Xn,Xn-m,v)ϕ(dv))Δt+b(Xn,Xn-m)ΔWn+∑i=pϕ(tn)+1pϕ(tn+1)‍c(Xn,Xn-m,ξi).In order to analyze the Euler method, we will give two lemmas.The first lemma shows the close relation between the continuous-time Euler solution (3.4) and its step function Z(t).Lemma 3.1.
Suppose Assumptions2.1 and 2.3 hold. Then for any T>0, there exists a positive constant K1(k), dependent on integer k and independent of Δt, such that for all Δt∈(0,1) the continuous-time Euler method (3.4) satisfies
(3.7)E|X¯(t)-Z(t)|2≤K1(k)Δt,
for 0≤t≤T∧σk∧ρk and k≥[Lτ+|ξ(0)|]+1, where σk is defined in Lemma 2.7 and ρk=inf{t≥0:|X¯(t)|≥k}.Proof.
For0≤t≤T∧σk∧ρk and k≥[Lτ+|ξ(0)|]+1, there is an integer n such that t∈[tn,tn+1). Thus it follows from (3.2) that
(3.8)X¯(t)-Z(t)=Xn+∫tnt‍a(Z(s),Z(s-τ))ds+∫tnt‍b(Z(s),Z(s-τ))dW(s)+∫tnt‍∫ε‍c(Z(s),Z(s-τ),v)p~ϕ(dv×ds)-Xn.
Therefore, by taking expectations and the Cauchy-Schwarz inequality and using the martingale properties ofdW(t) and p~ϕ(dv×dt), we get
(3.9)E|X¯(t)-Z(t)|2≤3E|∫tnt‍a(Z(s),Z(s-τ))ds|2+3E|∫tnt‍b(Z(s),Z(s-τ))dW(s)|2+3E|∫tnt‍∫ε‍c(Z(s),Z(s-τ),v)p~ϕ(dv×ds)|2≤3ΔtE∫tnt‍|a(Z(s),Z(s-τ))|2ds+3E∫tnt‍|b(Z(s),Z(s-τ))|2ds+3E∫tnt‍∫ε‍|c(Z(s),Z(s-τ),v)|2ϕ(dv)ds,
where the inequality |u1+u2+u3|2≤3|u1|2+3|u2|2+3|u3|2 for u1,u2,u3∈Rd is used. Therefore, by applying Assumption 2.1, we get
(3.10)E∫tnt‍|a(Z(s),Z(s-τ))|2ds≤2E∫tnt‍|a(Z(s),Z(s-τ))-a(0,0)|2ds+2E∫tnt‍|a(0,0)|2ds≤2CkE∫tnt‍(|Z(s)|2+|Z(s-τ)|2)ds+2|a(0,0)|2Δt≤4k2CkΔt+2|a(0,0)|2Δt,E∫tnt‍|b(Z(s),Z(s-τ))|2ds≤4k2CkΔt+2|b(0,0)|2Δt,E∫tnt‍∫ε‍|c(Z(s),Z(s-τ),v)|2ϕ(dv)ds≤4k2CΔt+2Δt∫ε‍|c(0,0,v)|2ϕ(dv).
Hence by substituting (3.10) into (3.9), we get
(3.11)E|X¯(t)-Z(t)|2≤Δt(24k2Ck+12k2C+6|a(0,0)|2+6|b(0,0)|2+6∫ε‍|c(0,0,v)|2ϕ(dv)),
for 0≤t≤T∧σk∧ρk and k≥[Lτ+|ξ(0)|]+1.
So from Assumption2.3, we can get the result (3.7) by choosing
(3.12)K1(k)=24k2Ck+12k2C+6|a(0,0)|2+6|b(0,0)|2+6∫ε‍|c(0,0,v)|2ϕ(dv).In the following lemma, we demonstrate that the solution of continuous-time Euler method (3.4) remains in a compact set with a large probability.Lemma 3.2.
Under Assumptions2.1, 2.2, 2.3, and 2.4, for any pair of ϵ∈(0,1) and T>0, there exist a sufficiently large integer k* and a sufficiently small Δt1* such that
(3.13)P(ρk*≤T)≤ϵ,∀Δt≤Δt1*,
where ρk* is defined in Lemma 3.1.Proof.
Our proof is completed by the following steps.
Step
1. Using Itô's formula (see [1]) to V(X¯(t)), for t≥0, we have
(3.14)dV(X¯(t))=({12}Vx(X¯(t))a(Z(t),Z(t-τ))+12trace(bT(Z(t),Z(t-τ))Vxx(X¯(t))b(Z(t),Z(t-τ))))dt+∫ε‍(V(X¯(t)+c(Z(t),Z(t-τ),v))-V(X¯(t))-Vx(X¯(t))c(Z(t),Z(t-τ),v))ϕ(dv)dt+Vx(X¯(t))b(Z(t),Z(t-τ))dW(t)+∫ε‍(V(X¯(t)+c(Z(t),Z(t-τ),v))-V(X¯(t)))p~ϕ(dv×dt)=LV(X¯(t),X¯(t-τ))dt+f(X¯(t),X¯(t-τ),Z(t),Z(t-τ))dt+Vx(X¯(t))b(Z(t),Z(t-τ))dW(t)+∫ε‍(V(X¯(t)+c(Z(t),Z(t-τ),v))-V(X¯(t)))p~ϕ(dv×dt),
where f(x,y,Z1,Z2):Rd×Rd×Rd×Rd→R is defined by
(3.15)f(x,y,Z1,Z2)=Vx(x)(a(Z1,Z2)-a(x,y))+12trace(bT(Z1,Z2)Vxx(x)b(Z1,Z2))-12trace(bT(x,y)Vxx(x)b(x,y))+∫ε‍(Vx(x)c(x,y,v)-Vx(x)c(Z1,Z2,v))ϕ(dv)+∫ε‍(V(x+c(Z1,Z2,v))-V(x+c(x,y,v)))ϕ(dv).
Moreover, for (x,y,Z1,Z2)∈Rd×Rd×Rd×Rd with |x|∨|y|∨|Z1|∨|Z2|≤k, we have
(3.16)f(x,y,Z1,Z2)=Vx(x)(a(Z1,Z2)-a(x,y))+12trace((bT(Z1,Z2)-bT(x,y))Vxx(x)b(Z1,Z2))+12trace(bT(x,y)Vxx(x)(b(Z1,Z2)-b(x,y)))+∫ε‍Vx(x)(c(x,y,v)-c(Z1,Z2,v))ϕ(dv)+∫ε‍(V(x+c(Z1,Z2,v))-V(x+c(x,y,v)))ϕ(dv)≤Lk(|x-Z1|+|y-Z2|),
where Assumptions 2.1 and 2.2 are used and Lk is a positive constant dependent on integer k, intensity λ and independent of Δt. Therefore, from (3.16), Assumption 2.4, and (3.7) in Lemma 3.1, we obtain
(3.17)E∫0t∧ρk‍f(X¯(s),X¯(s-τ),Z(s),Z(s-τ))ds≤LkE∫0t∧ρk‍|X¯(s)-Z(s)|ds+LkE∫0t∧ρk‍|X¯(s-τ)-Z(s-τ)|ds≤2LkE∫0t∧ρk‍|X¯(s)-Z(s)|ds+Lk∫-τ0‍|ξ(s)-Z(s)|ds≤2Lk∫0t‍(E|X¯(s∧ρk)-Z(s∧ρk)|2)1/2ds+Lk∑n=-m-1‍∫tntn+1‍|ξ(s)-ξ(tn)|ds≤2LkTK1(k)Δt+LkLτΔt,
for 0≤t≤T and k≥[Lτ+|ξ(0)|]+1. Hence by taking expectations and integration to (3.14), applying the martingale properties of dW(t) and p~ϕ(dv×dt), and then using (3.17) and Assumption 2.2, we obtain
(3.18)EV(X¯(t∧ρk))=V(X0)+E∫0t∧ρk‍LV(X¯(s),X¯(s-τ))ds+E∫0t∧ρk‍f(X¯(s),X¯(s-τ),Z(s),Z(s-τ))ds≤V(X0)+2LkTK1(k)Δt+τLkLΔt+E∫0t∧ρk‍(μ1(1+V(X¯(s))+V(X¯(s-τ))+U(X¯(s-τ)))-μ2U(X¯(s)))ds,
for 0≤t≤T and k≥[Lτ+|ξ(0)|]+1.
Step
2. For 0≤t≤τ and k≥[Lτ+|ξ(0)|]+1, it follows from (3.18) that
(3.19)EV(X¯(t∧ρk))≤V(X0)+2LkTK1(k)Δt+τLkLΔt+α1+μ1E∫0t∧ρk‍V(X¯(s))ds-μ2E∫0t∧ρk‍U(X¯(s))ds,
where α1=μ1∫-τ0‍(1+V(ξ(s))+U(ξ(s)))ds. Thus from (3.19), we get
(3.20)EV(X¯(t∧ρk))≤V(X0)+2LkTK1(k)Δt+τLkLΔt+α1+μ1E∫0t‍V(X¯(s∧ρk))ds,
by the Gronwall inequality (see [18]), which gives
(3.21)EV(X¯(t∧ρk))≤(V(X0)+2LkTK1(k)Δt+τLkLΔt+α1)eμ1t,
for 0≤t≤τ and k≥[Lτ+|ξ(0)|]+1. Moreover, from (3.19) and (3.21), we have
(3.22)E∫0τ∧ρk‍U(X¯(s))ds≤μ2-1(V(X0)+2LkTK1(k)Δt+τLkLΔt+α1+μ1∫0τ‍EV(X¯(s∧ρk))ds)≤(V(X0)+2LkTK1(k)Δt+τLkLΔt+α1)eμ1τμ2-1,
for k≥[Lτ+|ξ(0)|]+1.
Step
3. For 0≤t≤2τ and k≥[Lτ+|ξ(0)|]+1, it follows from (3.18) that
(3.23)EV(X¯(t∧ρk))≤V(X0)+2LkTK1(k)Δt+τLkLΔt+α2+μ1E∫0t∧ρk‍V(X¯(s))ds-μ2E∫0t∧ρk‍U(X¯(s))ds.
As the same way as Step 2, we can obtain
(3.24)EV(X¯(t∧ρk))≤(V(X0)+2LkTK1(k)Δt+τLkLΔt+α2)eμ1t,0≤t≤2τ,(3.25)E∫02τ∧ρk‍U(X¯(s))ds≤(V(X0)+2LkTK1(k)Δt+τLkLΔt+α2)e2μ1τμ2-1,
where
(3.26)α2=μ1E∫02τ∧ρk‍(1+V(X¯(s-τ))+U(X¯(s-τ)))ds=μ1E∫-τ2τ∧ρk-τ‍(1+V(X¯(s))+U(X¯(s)))ds≤μ1E∫-ττ∧ρk‍(1+V(X¯(s))+U(X¯(s)))ds=α1+μ1E∫0τ∧ρk‍(1+V(X¯(s))+U(X¯(s)))ds≤α1+μ1τ+μ1∫0τ‍EV(X¯(s∧ρk))ds+μ1E∫0τ∧ρk‍U(X¯(s))ds≤μ1τ+(V(X0)+2LkTK1(k)Δt+τLkLΔt+α1)(eμ1τ+eμ1τμ1μ2-1),
from (3.21) and (3.22). So (3.24) becomes
(3.27)EV(X¯(t∧ρk))≤(V(X0)+2LkTK1(k)Δt+τLkLΔt+α1)β1,0+β2,0,
for 0≤t≤2τ and k≥[Lτ+|ξ(0)|]+1, where
(3.28)β1,0=(1+eμ1τ+eμ1τμ1μ2-1)e2μ1τ,β2,0=(μ1τ-α1)e2μ1τ.
Step
4. By repeating the same way in Steps 2 and 3, we get
(3.29)EV(X¯(T∧ρk))≤(V(X0)+2LkTK1(k)Δt+τLkLΔt+α1)β1+β2,
for k≥[Lτ+|ξ(0)|]+1, where β1 and β2 are two constants dependent on μ1,μ2,τ,T and independent of k and Δt. Therefore, we have
(3.30)P(ρk≤T)≤V(X0)β1+α1β1+β2+2β1LkTK1(k)Δt+β1τLkLΔtϑk,
where
(3.31)ϑk=inf⁡|X¯|≥kV(X¯),k≥[Lτ+|ξ(0)|]+1.
Now, for any ϵ∈(0,1), we can choose sufficiently large integer k* such that
(3.32)V(X0)β1+α1β1+β2ϑk*≤ϵ2
and sufficiently small Δt1* such that
(3.33)2β1Lk*TK1(k*)Δt1*+β1τLk*LΔt1*ϑk*≤ϵ2.
So from (3.30), we can obtain
(3.34)P(ρk*≤T)≤ϵ,∀Δt≤Δt1*.


## 4. Convergence in Probability

In this section, we show the convergence in probability of the Euler method to (2.1) over a finite time interval [0,T], which is based on the following lemma.Lemma 4.1.
Under Assumptions2.1, 2.3, and 2.4, for any T>0, there exists a positive constant K2(k), dependent on k and independent of Δt, such that for all Δt∈(0,1) the solution of (2.1) and the continuous-time Euler method (3.4) satisfy
(4.1)E(sup⁡0≤t≤T|x(t∧σk∧ρk)-X¯(t∧σk∧ρk)|2)≤K2(k)Δt,
where σk and ρk are defined in Lemmas 2.7 and 3.1, respectively, and k≥[Lτ+|ξ(0)|]+1.Proof.
From (2.1) and (3.4), for any 0≤t′≤T and k≥[Lτ+|ξ(0)|]+1, we have
(4.2)E(sup⁡0≤t≤t′|x(t∧σk∧ρk)-X¯(t∧σk∧ρk)|2)≤3E(sup⁡0≤t≤t′|∫0t∧σk∧ρk‍(a(x(s-),x((s-τ)-))-a(Z(s),Z(s-τ)))ds|2)+3E(sup⁡0≤t≤t′|∫0t∧σk∧ρk‍(b(x(s-),x((s-τ)-))-b(Z(s),Z(s-τ)))dW(s)|2)+3E(sup⁡0≤t≤t′|∫0t∧σk∧ρk‍∫ε‍(c(x(s-),x((s-τ)-),v)-c(Z(s),Z(s-τ),v))p~ϕ(dv×ds)|2),
where the inequality |u1+u2+u3|2≤3|u1|2+3|u2|2+3|u3|2 for u1,u2,u3∈Rd is used. Therefore, by using the Cauchy-Schwarz inequality, Assumptions 2.1 and 2.4, Fubini's Theorem, and Lemma 3.1, we obtain
(4.3)E(sup⁡0≤t≤t′|∫0t∧σk∧ρk‍(a(x(s-),x((s-τ)-))-a(Z(s),Z(s-τ)))ds|2)≤E(sup⁡0≤t≤t′∫0t∧σk∧ρk‍12ds∫0t∧σk∧ρk‍|a(x(s-),x((s-τ)-))-a(Z(s),Z(s-τ))|2ds)≤TE(∫0t′∧σk∧ρk‍|a(x(s-),x((s-τ)-))-a(Z(s),Z(s-τ))|2ds)≤TCkE(∫0t′∧σk∧ρk‍|x(s-)-Z(s)|2ds)+TCkE(∫0t′∧σk∧ρk‍|x((s-τ)-)-Z(s-τ)|2ds)≤2TCkE(∫0t′∧σk∧ρk‍|x(s-)-Z(s)|2ds)+TCk∫-τ0‍|ξ(s)-Z(s)|2ds≤4TCkE(∫0t′∧σk∧ρk‍|X¯(s)-Z(s)|2ds)+4TCkE(∫0t′∧σk∧ρk‍|x(s-)-X¯(s)|2ds)+TCk∑n=-m-1‍∫tntn+1‍|ξ(s)-ξ(tn)|2ds≤4TCk(∫0t′‍E|X¯(s∧σk∧ρk)-Z(s∧σk∧ρk)|2ds)+4TCk∫0t′‍E(sup⁡0≤u≤s|x(u∧σk∧ρk-)-X¯(u∧σk∧ρk)|2)ds+TCkL2τΔt≤(4T2CkK1(k)+TCkL2τ)Δt+4TCk∫0t′‍E(sup⁡0≤u≤s|x(u∧σk∧ρk-)-X¯(u∧σk∧ρk)|2)ds.
Moreover, by using the martingale properties of dW(t) and p~ϕ(dv×dt), Assumptions 2.1 and 2.4, Fubini's Theorem, and Lemma 3.1, we have
(4.4)E(sup⁡0≤t≤t′|∫0t∧σk∧ρk‍(b(x(s-),x((s-τ)-))-b(Z(s),Z(s-τ)))dW(s)|2)≤4E∫0t′∧σk∧ρk‍|b(x(s-),x((s-τ)-))-b(Z(s),Z(s-τ))|2ds≤4CkE∫0t′∧σk∧ρk‍|x(s-)-Z(s)|2ds+4CkE∫0t′∧σk∧ρk‍|x((s-τ)-)-Z(s-τ)|2ds≤8CkE∫0t′∧σk∧ρk‍|x(s-)-Z(s)|2ds+4Ck∫-τ0‍|ξ(s)-Z(s)|2ds≤16CkE(∫0t′∧σk∧ρk‍|X¯(s)-Z(s)|2ds)+16CkE(∫0t′∧σk∧ρk‍|x(s-)-X¯(s)|2ds)+4Ck∑n=-m-1‍∫tntn+1‍|ξ(s)-ξ(tn)|2ds≤(16TCkK1(k)+4CkL2τ)Δt+16Ck∫0t′‍E(sup⁡0≤u≤s|x(u∧σk∧ρk-)-X¯(u∧σk∧ρk)|2)ds,E(sup⁡0≤t≤t′|∫0t∧σk∧ρk‍∫ε‍(c(x(s-),x((s-τ)-),v)-c(Z(s),Z(s-τ),v))p~ϕ(dv×ds)|2)≤4E|∫0t′∧σk∧ρk‍∫ε‍(c(x(s-),x((s-τ)-),v)-c(Z(s),Z(s-τ),v))p~ϕ(dv×ds)|2=4E∫0t′∧σk∧ρk‍∫ε‍|c(x(s-),x((s-τ)-),v)-c(Z(s),Z(s-τ),v)|2ϕ(dv)ds≤(16TCK1(k)+4CL2τ)Δt+16C∫0t′‍E(sup⁡0≤u≤s|x(u∧σk∧ρk-)-X¯(u∧σk∧ρk)|2)ds.
Therefore by substituting (4.3) and (4.4) into (4.2), we get
(4.5)E(sup⁡0≤t≤t′|x(t∧σk∧ρk)-X¯(t∧σk∧ρk)|2)≤Δt(12T2CkK1(k)+3TCkL2τ+48TCkK1(k)+48TCK1(k)+12CkL2τ+12CL2τ)+(12TCk+48Ck+48C)∫0t′‍E(sup⁡0≤u≤s|x(u∧σk∧ρk-)-X¯(u∧σk∧ρk)|2)ds.
So by using the Gronwall inequality (see [18]), we have the result (4.1) by choosing
(4.6)K2(k)=(12T2CkK1(k)+3TCkL2τ+48TCkK1(k)+48TCK1(k)+12CkL2τ+12CL2τ)exp⁡(12T2Ck+48TCk+48TC).Now, we state our main theorem which shows the convergence in probability of the continuous-time Euler method (3.4).Theorem 4.2.
Under Assumptions2.1, 2.2, 2.3, and 2.4, for sufficiently small ϵ,ς∈(0,1), there is a Δt* such that for all Δt<Δt*(4.7)P(sup⁡0≤t≤T|x(t)-X¯(t)|2≥ς)≤ϵ,
for any T>0.Proof.
For sufficiently smallϵ,ς∈(0,1), we define
(4.8)Ω¯={ω:sup⁡0≤t≤T|x(t)-X¯(t)|2≥ς}.
By Lemmas 2.9 and 3.2, there exists a pair of k⋆ and Δt1⋆ such that
(4.9)P(σk⋆≤T)≤ϵ3,P(ρk⋆≤T)≤ϵ3,∀Δt≤Δt1⋆.
We then have
(4.10)P(Ω¯)≤P(Ω¯∩{σk⋆∧ρk⋆>T})+P(σk⋆∧σk⋆≤T)≤P(Ω¯∩{σk⋆∧σk⋆>T})+P(σk⋆≤T)+P(ρk⋆≤T)≤P(Ω¯∩{σk⋆∧ρk⋆>T})+2ϵ3,
for Δt≤Δt1⋆. Moreover, from Lemma 4.1, we have
(4.11)ςP(Ω¯∩{σk⋆∧ρk⋆>T})≤E(I{σk⋆∧ρk⋆>T}sup⁡0≤t≤T|x(t)-X¯(t)|2)≤E(sup⁡0≤t≤T|x(t∧σk⋆∧ρk⋆)-X¯(t∧σk⋆∧ρk⋆)|2)≤K2(k⋆)Δt,
which gives
(4.12)P(Ω¯∩{σk⋆∧ρk⋆>T})≤ϵ3,
for Δt≤Δt2⋆. Hence, from the inequalities above, we get
(4.13)P(Ω¯)≤ϵ,
for Δt≤Δt⋆, where Δt⋆=min{Δt1⋆,Δt2⋆}.
We remark that the continuous-time Euler solutionX¯(t) cannot be computed, since it requires knowledge of the entire Brownian motion and Poisson random measure paths. So the last theorem demonstrates the convergence in probability of the discrete Euler solution (3.2).Theorem 4.3.
Under Assumptions2.1, 2.2, 2.3, and 2.4, for sufficiently small ϵ,ς∈(0,1), there is a Δt* such that for all Δt<Δt*(4.14)P(|x(t)-Z(t)|2≥ς,0≤t≤T)≤ϵ,
for any T>0.Proof.
For sufficiently smallϵ,ς∈(0,1), we define
(4.15)Ω~={ω:|x(t)-Z(t)|2≥ς,0≤t≤T}.
As the same analysis as the proof in Theorem 4.2, we have
(4.16)P(Ω~)≤P(Ω~∩{σk⋆∧ρk⋆>T})+2ϵ3.
Recalling that(4.17)ςP(Ω~∩{σk⋆∧ρk⋆>T})≤E(|x(T)-Z(T)|2I{σk⋆∧ρk⋆>T})≤E|x(T∧σk⋆∧ρk⋆)-Z(T∧σk⋆∧ρk⋆)|2≤2E(sup⁡0≤t≤T|x(t∧σk⋆∧ρk⋆)-X¯(t∧σk⋆∧ρk⋆)|2)+2E|X¯(T∧σk⋆∧ρk⋆)-Z(T∧σk⋆∧ρk⋆)|2≤2K1(k⋆)Δt+2K2(k⋆)Δt
and using Lemmas 3.1 and 4.1, we obtain
(4.18)P(Ω~∩{σk⋆∧ρk⋆>T})≤ϵ3,
for sufficiently small Δt. So the inequalities above demonstrate
(4.19)P(Ω~)≤ϵ,
for all sufficiently small Δt.
Hence we complete the result (4.14).


## 5. Numerical Example

In this section, a numerical example is analyzed under Assumptions2.1, 2.2, 2.3, and 2.4 which cover many highly nonlinear SDDEs driven by Poisson random measure.Now, we consider the following equation:(5.1)dx(t)=(2x2((t-0.05)-)-4x3(t-))dt+3x2((t-0.05)-)dW(t)+∫ε‍vx2((t-0.05)-)p~ϕ(dv×dt),t>0,
with x(t)=t2,t∈[-0.05,0], where d=m=r=1. The compensated measure of the Poisson random measure pϕ(dv×dt) is given by ϕ(dv)dt=λf(v)dvdt, where λ=5 and
(5.2)f(v)=12πvexp⁡(-(ln⁡v)22),0≤v<∞,
is the density function of a lognormal random variable.Clearly, the equation cannot satisfy the global Lipschitz conditions, the linear growth conditions and the classical Khasminskii-type conditions. But, the local Lipschitz conditions are satisfied. On the other hand, forV(x)=|x|2, we have
(5.3)LV(x,y)=2x(2y2-4x3)+9y4+∫ε‍v2y4λ12πvexp⁡(-(ln⁡v)22)dv≤2x2+60y4-8x4=60(1+x2+y2+y4)-8x4,
where U(x)=x4,μ1=60,μ2=8. In other words, the equation satisfies Assumptions 2.1, 2.2, 2.3, and 2.4.So according to Theorem2.8, (5.1) has a unique global solution x(t) on t∈[-0.05,∞).Given the stepsizeΔt=0.0025, we can have the Euler method to (5.1)
(5.4)Xn=(nΔt)2,forn=-20,-19,…,-1,0,Xn+1=Xn+(2Xn-202-4Xn3)Δt+3Xn-202ΔWn+∫tntn+1‍∫ε‍vXn-202p~ϕ(dv×dt),
for n=0,1,…, where ΔWn=W(tn+1)-W(tn).And in Matlab experiment, we actually obtain the discrete Euler(5.5)Xn+1=Xn+(2Xn-202-4Xn3-5eXn-202)Δt+3Xn-202ΔWn+Xn-202∑i=pϕ(tn)+1pϕ(tn+1)‍ξi,
where ξi is distributed according to f(v). Subsequently, we can get the result in Theorems 4.2 and 4.3.



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*Source: 127397-2012-09-02.xml*