Convergence of the $\theta$-Euler-Maruyama method for a class of Stochastic Volterra Integro-Differential Equations

Abstract.

This paper addresses the convergence analysis of the $\theta$-Euler-Maruyama method for a class of stochastic Volterra integro-differential equations (SVIDEs). At first, we discuss the existence, uniqueness, boundedness and Hölder continuity of the theoretical solution. Subsequently, the strong convergence order of the $\theta$-Euler-Maruyama approach for SVIDEs is established. Finally, we provided numerical examples to illustrate the theoretical results.

1. Introduction

Stochastic differential equations (SDEs) have attracted significant attention and are currently emerging as a modeling tool in various scientific fields, including but not limited to telecommunications (see [15]), economics, finance (see [5]), biology, chemistry, and quantum field theory.

The Volterra integral equations (VIEs) were proposed by Vito Volterra, which Traian Lalescu later studied in his 1908 thesis ”Sur les équations de Volterra”, written under the direction of Émile Picard. Volterra integral equations find application in viscoelastic materials, fluid mechanics, and demography (see, e.g., [10],[3], [17],[8],[7]). Stochastic Volterra integral equations (SVIEs) are an extension of ordinary Volterra integral equations to include random noise, making them suitable for modeling systems with stochastic components. SVIEs find applications in various fields, including mathematical finance, biology, physics and engineering. For example, in finance, SVIEs are used to model the evolution of financial asset prices over time, taking into account the stochastic nature of market movements. In biology, they can be used to describe population dynamics subject to random environmental factors. Therefore, in recent years, SVIEs have attracted the attention of many researchers. For instance (see [10],[17],[8],[7]). The exact methods for solving stochastic differential equations (SDEs) involve addressing current challenges in the field (cf. [6], [16], [3], [11], [4], [17], [13], [7]). While there are many analytical methods available, the complexity of these equations makes it difficult to obtain exact solutions. Among the numerical methods are the Milstein method, Runge-Kutta method see [1], Euler-Maruyama method, stochastic theta method, and others (see [10], [9],[12],[2], [17]). Zong et al. [18] conducted a study on two classes of theta-Milstein schemes for stochastic differential equations in terms of convergence and stability.

Recently, Deng et al. [4] examined the semi-implicit Euler method for non-linear time changed stochastic diferential equations. Wang et al. [14] investigated the stochastic theta methods (STMs) for stochastic differential equations (SDEs) with non-global Lipschitz drift and diffusion coefficients. Zhang et al. [17] examined the Euler-Maruyama (EM) method’s numerical analysis of the following generalized SVIDEs:

$dY(t)=$ $f(Y(t),{\displaystyle {\int }_0^t}{K}_1(t,s)Y(s)𝑑s,{\displaystyle {\int }_0^t}{\sigma }_1(t,s)Y(s)𝑑B(s))dt$

$+g(Y(t),{\displaystyle {\int }_0^t}{K}_2(t,s)Y(s)𝑑s,{\displaystyle {\int }_0^t}{\sigma }_2(t,s)Y(s)𝑑B(s))dB(t).$

Lan et al. [7] presented the $\theta$-EM method corresponding to the following SVIDEs:

$$dX(t)=f(X(t),{\int }_0^{t}G(t-s)X(s)𝑑s)dt+g(X(t),{\int }_0^{t}H(t-s)X(s)𝑑s)dB(t).$$

Inspired and motivated by the above works [4, 14, 17, 7], in this paper, we study the strong convergence of the $\theta$-Euler-Maruyama method for a class of stochastic Volterra integro-differential equations (SVIDEs) as follows:

(1.1) $$dY(t)=f(Y(t),{\int }_0^t{\sigma }_1(t,s)Y(s)𝑑B(s))dt+g(Y(t),{\int }_0^t{\sigma }_2(t,s)Y(s)𝑑s)dB(t),t\in [0,T],$$

with initial condition $Y(0)=Y_0\in ℝ$, where $f:ℝ\times ℝ\to ℝ$ and $g:ℝ\times ℝ\to ℝ$ are given functions. The kernels ${\sigma }_i:D\to ℝ$ are continuous on $D:=\{(t,s):0\le s\le t\le T\}$ with the norm ${\parallel {\sigma }_i\parallel }_{\mathrm{\infty }}={\max }_{(t,s)\in D}|{\sigma }_i(t,s)|$ for $i=1,2$. $Y(t)$ is a stochastic processes defined on probability space $(\mathrm{\Omega },ℱ,\mathbb{P})$, $B(t)$ is a standard Brownian motion (1-dimensional Brownian motion) defined on the same probability space. And .

The structure of this paper is as follows: We introduce some fundamental notations and preliminaries in Section 2. We then present the definition of the solution of equation (1.1) and investigate the existence, uniqueness, boundedness and Hölder continuity of the analytic solution in Section 3. The $\theta$-EM method for equation (1.1) is presented, and its order of convergence is taken into account in Section 4. Finally, we provide numerical examples in Section 5 to illustrate the theoretical results.

2. Preliminaries

Let $(\mathrm{\Omega },ℱ,{({ℱ}_t)}_{t\ge 0},\mathbb{P})$ be a complete probability space with a filtration ${({ℱ}_t)}_{t\ge 0}$ satisfying the usual conditions , and let $𝔼$ denote the expectation corresponding to $\mathbb{P}$. A 1-dimensional Brownian motion defined on the probability space is denoted by $B(t)$. Let $L^2([0,T],ℝ)$ by the family of Borel measurable functions $\mathrm{\Phi }:[0,T]\to ℝ$ such that for every We denote ${ℒ}^2([0,T],ℝ)$ the family of $ℝ$-valued ${ℱ}_t$-adapted processes ${\{\mathrm{\Phi }(t)\}}_{t\in [0,T]}$ such that ${ℳ}^2([0,T],ℝ)$ be the family of processes ${ℱ}_t-$adapted ${\{\mathrm{\Phi }(t)\}}_{t\in [0,T]}\in {ℒ}^2([0,T],ℝ)$ such that . For $a,b\in ℝ,a\wedge b:=\min \{a,b\},\text{and}a\vee b:=\max \{a,b\}$. If $𝔻$ is a subset of $\mathrm{\Omega }$, the indicator function is denoted by $1_{𝔻}$. We can write the integral of equation (1.1) as follows:

$Y(t)$ $=Y(0)+{\displaystyle {\int }_0^t}f(Y(u),{\displaystyle {\int }_0^u}{\sigma }_1(u,s)Y(s)𝑑B(s))𝑑u$

(2.2) $+{\displaystyle {\int }_0^t}g(Y(u),{\displaystyle {\int }_0^u}{\sigma }_2(u,s)Y(s)𝑑s)𝑑B(u).$

We will introduce the definition of the solution, we assume that $Y(.)\in {ℒ}^2([0,T],ℝ)$ where

$F(t):=$ $f(Y(t),{\displaystyle {\int }_0^t}{\sigma }_1(t,s)Y(s)𝑑B(s)),$

$G(t):=$ $g(Y(t),{\displaystyle {\int }_0^t}{\sigma }_2(t,s)Y(s)𝑑s).$

Definition 2.1.

A solution of (2) is a continuous stochastic process ${\{Y(t)\}}_{t\in [0,T]}$ with values in $ℝ$ satisfies the following conditions:

1. $Y(.)\in {ℒ}^2([0,T],ℝ),F(.)\in {ℒ}^1([0,T],ℝ)$ and $G(.)\in {ℒ}^2([0,T],ℝ)$,

2. (2) holds for all $t\in [0,T]$ with probability 1. The solution $\{Y(t)\}$ is said to be unique if there is exists other solution $\{\overline{Y}(t)\}$ such that $\{Y(t)\}=\{\overline{Y}(t)\}$ for all $t\in [0,T]$, i.e.,

   $$\mathbb{P}\left\{Y(t)=\overline{Y}(t)\text{for all}t\in [0,T]\right\}=1.$$

Definition 2.2.

Let . A stochastic process $Y(t,\omega ):[0,T]\times \mathrm{\Omega }\to ℝ$ is referred to as Hölder continuous with exponent $\delta >0$ if a constant M exists such that

$$𝔼{|Y(t)-Y(r)|}^2\le M{|t-r|}^{2\delta },\forall t,r\in [0,T].$$

In this article, we propose the following hypotheses.

1. ($H_1$) (Lipschiz condition). Assume that there exist a positive constant $K$ such that

   $${\left|f(x,y)-f(x^{\prime },y^{\prime })\right|}^2\vee {\left|g(x,y)-g(x^{\prime },y^{\prime })\right|}^2\le K\left({\left|x-x^{\prime }\right|}^2+{\left|y-y^{\prime }\right|}^2\right),$$

   for $x,y,x^{\prime },y^{\prime }\in ℝ$.

2. ($H_2$) (Linear growth condition). For $x,y\in ℝ$

   $${\left|f(x,y)\right|}^2\vee {\left|g(x,y)\right|}^2\le K^{\prime }\left(1+{\left|x\right|}^2+{\left|y\right|}^2\right),$$

   where $K^{\prime }=2\left(K\vee {|f(0,0)|}^2\vee {|g(0,0)|}^2\right)$.

3. ($H_3$) (Mean value theorem). Assuming that the coefficients ${\sigma }_i\in C^1(D),$ for $i=1,2.$ of (1.1) satisfy

   $${\left|{\sigma }_i^{\prime }(t,s)\right|}^2\le K^{\prime \prime },$$

   with $K^{\prime \prime }>0,$ for all $(t,s)\in D,$ and

   $${\left|{\sigma }_i(u,s)-{\sigma }_i(u_h,s)\right|}^2={\left|{\sigma }_i({\xi }_i,s)(u,u_h)\right|}^2\le K^{\prime \prime }{h}^2,$$

   where ${\xi }_i\in (u,u_h)$.

The existence and uniqueness of the solution to (1.1) under hypothesis ($H_2$) are demonstrated by the following theorem. The proof of this theorem is similar to the proof in [17], which pertains to the case where $\mathrm{\Theta }=0.$

3. Theoretical analysis of the class of SVIDE

In this section, we present the theoretical results. The existence and uniqueness of the solution to (2) have been established. Additionally, we verified the Hölder continuity condition for the analytical solutions.

3.1. The existence and uniqueness of the analytical solution

We now discuss the theory of existence and uniqueness of the solution of the equation (2). We first present the following lemma.

Lemma 3.1.

Assume that ($H_2$) is satisfied. If $Y(t)\in {ℒ}^2([0,T],ℝ)$ is a solution of SVIDEs (1.1), and such that

(3.3) $$𝔼{\left[Y(t)\right]}^2\le M_0,\forall t\in [0,T],$$

where $M_0$ depends on ${\sigma }_1,{\sigma }_2,T,K^{\prime }$ and $Y_0$.

Proof.

For every integer $n\ge 1,{\tau }_n$ be a stopping time such that

$${\tau }_n=T\wedge inf\{t\in [0,T]:|Y(t)|\ge n\}.$$

Evidently, ${\tau }_n\to T$ a.s. by letting $n\to \mathrm{\infty }.$ Define $Y_n(t):=Y(t\wedge {\tau }_n)$ for $t\in [0,T]$. It can be verified that $Y_n(t)$ satisfies.

$Y_n(t)$ $=Y(0)+{\displaystyle {\int }_0^t}f(Y_n(u),{\displaystyle {\int }_0^u}{\sigma }_1(u,s)Y_n(s){\mathrm{𝟏}}_{[0,{\tau }_n]}(s)𝑑B(s)){\mathrm{𝟏}}_{[0,{\tau }_n]}(u)𝑑u$

$+{\displaystyle {\int }_0^t}g(Y_n(u),{\displaystyle {\int }_0^u}{\sigma }_2(u,s)Y_n(s){\mathrm{𝟏}}_{[0,{\tau }_n]}(s)𝑑s){\mathrm{𝟏}}_{[0,{\tau }_n]}(u)𝑑B(u),t\in [0,T].$

By Cauchy’s inequality, the Itô isometry, and the inequality ${(x+y+z)}^2\le 3(x^2+y^2+z^2)$, we obtain that

${|Y_n(t)|}^2$ $\le 3{|Y_0|}^2+3t{\displaystyle {\int }_0^t}{\left|f(Y_n(u),{\displaystyle {\int }_0^u}{\sigma }_1(u,s)Y_n(s){\mathrm{𝟏}}_{[0,{\tau }_n]}(s)𝑑B(s)){\mathrm{𝟏}}_{[0,{\tau }_n]}(u)\right|}^2𝑑u$

$+3{\displaystyle {\int }_0^t}|g(Y_n(u),{\displaystyle {\int }_0^u}{\sigma }_2(u,s)Y_n(s){\mathrm{𝟏}}_{[0,{\tau }_n]}(s)ds,){\mathrm{𝟏}}_{[0,{\tau }_n]}(u){|}^{2}du.$

Using Cauchy’s inequality and the Itô isometry, we show that

(3.4) $$𝔼{|Y_n(t)|}^2\le 3𝔼|Y_0|+3TF+3G,$$

where

$$F:=𝔼\left[{\int }_0^t{\left|f(Y_n(u),{\int }_0^u{\sigma }_1(u,s)Y_n(s){\mathrm{𝟏}}_{[0,{\tau }_n]}(s)𝑑B(s)){\mathrm{𝟏}}_{[0,{\tau }_n]}(u)\right|}^2𝑑u\right],$$

and

$$G:=𝔼\left[{\int }_0^t\right|g(Y_n(u),{\int }_0^u{\sigma }_2(u,s)Y_n(s){\mathrm{𝟏}}_{[0,{\tau }_n]}(s)ds,){\mathrm{𝟏}}_{[0,{\tau }_n]}(u){|}^{2}du].$$

First, we calculate $F$, using Cauchy’s inequality and assumption ($H_2$), we get

(3.5)

Second, we calculate $G$, by ($H_2$) and Itô isometry, we get

(3.6)

By substituting (3.5) and (3.6) into (3.4), we have

$𝔼{|Y_n(t)|}^2$ $\le 3𝔼{|Y_0|}^2+3T\left(K^{\prime }T+K^{\prime }(1+{\parallel {\sigma }_1\parallel }_{\mathrm{\infty }}^{2}T){\displaystyle {\int }_0^t}𝔼{|Y_n(s)|}^2𝑑s\right)$

$+3\left(K^{\prime }T+K^{\prime }(1+{\parallel {\sigma }_2\parallel }_{\mathrm{\infty }}^2{T}^2){\displaystyle {\int }_0^t}𝔼{|Y_n(s)|}^2𝑑s\right)$

$\le 3𝔼{|Y_0|}^2+3K^{\prime }T(T+1)+3K^{\prime }T(1+{\parallel {\sigma }_1\parallel }_{\mathrm{\infty }}^{2}T){\displaystyle {\int }_0^t}𝔼{|Y_n(s)|}^2𝑑s$

$+3K^{\prime }(1+{\parallel {\sigma }_2\parallel }_{\mathrm{\infty }}^2{T}^2){\displaystyle {\int }_0^t}𝔼{|Y_n(s)|}^2𝑑s$

$=3𝔼{|Y_0|}^2+3K^{\prime }T(T+1)+3K^{\prime }(C_{1}T+C_2){\displaystyle {\int }_0^t}𝔼{|Y_n(s)|}^2𝑑s,$

where $C_1:=1+{\parallel {\sigma }_1\parallel }_{\mathrm{\infty }}^{2}T$ and $C_2:=1+{\parallel {\sigma }_2\parallel }_{\mathrm{\infty }}^2{T}^2$.

By using Gronwall’s inequality, we have

$$𝔼{|Y_n(t)|}^2\le M_0,$$

where

$$M_0:=3\left(𝔼{|Y_0|}^2+C_3\right)\exp \left(3TC\right),$$

and

$$C:=K^{\prime }(C_{1}T+C_2),C_3=K^{\prime }T(T+1).$$

Since $𝔼{|Y(t\wedge {\tau }_n)|}^2\le M_0,$ for $t\ge 0$, letting $t⟶\mathrm{\infty }$, we conclude that ${\tau }_{\mathrm{\infty }}=\mathrm{\infty }$, that is,

$$𝔼{\left[Y(t)\right]}^2\le M_0.$$

Theorem 3.1.

Suppose ($H_1$) holds. Then, there exists a unique solution $Y(t)\in {ℳ}^2([0,T],ℝ)$ to (1.1) and

$$𝔼{\left[Y(t)\right]}^2\le M_1\text{for}t\in [0,T],$$

where

$$M_1:=3\left((1+CT)𝔼{|Y_0|}^2+C_3\right)\exp (3CT).$$

Proof.

We will divide the proof into two fundamental steps.

Step I. Uniqueness:

Let $Y_1(t)$ and $Y_2(t)$ be two solutions of (1.1). From Lemma 3.1 we have, $Y_1(t),Y_2(t)\in {ℳ}^2([0,T];ℝ)$. By ($H_1$), Cauchy’s inequality and Itô isometry, we show that

$𝔼{|Y_1(t)-Y_2(t)|}^2=2𝔼|{\displaystyle {\int }_0^t}f(Y_1(u),{\displaystyle {\int }_0^u}{\sigma }_1(u,s)Y_1(s)𝑑B(s))𝑑u$

$-{\displaystyle {\int }_0^t}f(Y_2(u),{\displaystyle {\int }_0^u}{\sigma }_1(u,s)Y_2(s)dB(s))du{|}^2$

$+2𝔼|{\displaystyle {\int }_0^t}g(Y_1(u),{\displaystyle {\int }_0^u}{\sigma }_2(u,s)Y_1(s)𝑑s)𝑑B(u)-$

$-{\displaystyle {\int }_0^t}g(Y_2(u),{\displaystyle {\int }_0^u}{\sigma }_2(u,s)Y_2(s)ds)dB(u){|}^2$

$\le 2T{\displaystyle {\int }_0^t}𝔼|f\left(Y_1(u){\displaystyle {\int }_0^u}{\sigma }_1(u,s)Y_1(s)𝑑B(s)\right)$

$-f(Y_2(u),{\displaystyle {\int }_0^u}{\sigma }_1(u,s)Y_2(s)dB(s)){|}^{2}du$

$+2{\displaystyle {\int }_0^t}𝔼{\left|g(Y_1(u),{\displaystyle {\int }_0^u}{\sigma }_2(u,s)Y_1(s)𝑑s)-g(Y_2(u),{\displaystyle {\int }_0^u}{\sigma }_2(u,s)Y_2(s)𝑑s)\right|}^2𝑑u$

$\le 2TK{\displaystyle {\int }_0^t}\left(𝔼{|Y_1(u)-Y_2(u)|}^2+{\displaystyle {\int }_0^u}𝔼{\left|{\sigma }_1(u,s)\left(Y_1(s)-Y_2(s)\right)\right|}^2𝑑s\right)𝑑u$

$+2K{\displaystyle {\int }_0^t}\left(𝔼{|Y_1(u)-Y_2(u)|}^2+u{\displaystyle {\int }_0^u}𝔼{\left|{\sigma }_2(u,s)\left(Y_1(s)-Y_2(s)\right)\right|}^2𝑑s\right)𝑑u$

$\le 2TK{\displaystyle {\int }_0^t}\left(𝔼{|Y_1(u)-Y_2(u)|}^2+{\parallel {\sigma }_1\parallel }_{\mathrm{\infty }}^2{\displaystyle {\int }_0^u}𝔼{|Y_1(s)-Y_2(s)|}^2𝑑s\right)𝑑u$

$+2K{\displaystyle {\int }_0^t}\left(𝔼{|Y_1(u)-Y_2(u)|}^2+T{\parallel {\sigma }_2\parallel }_{\mathrm{\infty }}^2{\displaystyle {\int }_0^u}𝔼{|Y_1(s)-Y_2(s)|}^2𝑑s\right)𝑑u$

$\le 2K\left(T{C}_1+C_2\right){\displaystyle {\int }_0^t}𝔼{|Y_1(s)-Y_2(s)|}^2𝑑s$

$\le C{\displaystyle {\int }_0^t}𝔼{|Y_1(s)-Y_2(s)|}^2𝑑s.$

Finally, from Gronwall’s inequality we conclude that

$$𝔼{|Y_1(t)-Y_2(t)|}^2=0,$$

which proves that $Y_1(t)=Y_2(t)$ for every $t\in [0,T]$.

Step II. Existence:

Let $Y_0(t)=Y_0$ and define the Picard approximation.

(3.7) $Y_n(t)$ $=Y_0+{\displaystyle {\int }_0^t}f(Y_{n-1}(u),{\displaystyle {\int }_0^u}{\sigma }_1(u,s)Y_{n-1}(s)𝑑B(s))𝑑u$

$+{\displaystyle {\int }_0^t}g(Y_{n-1}(u),{\displaystyle {\int }_0^u}{\sigma }_2(u,s)Y_{n-1}(s)𝑑s)𝑑B(u),$

for $t\in [0,T]$ and $n=1,2,\mathrm{\dots }.$ It is evident that $Y_0(.)\in {ℳ}^2([0,T];ℝ)$, and through induction, we also get $Y_n(.)\in {ℳ}^2([0,T],ℝ)$. Using the proof of Lemma 3.1, we have

$𝔼{|Y_n(t)|}^2\le$ $3𝔼{|Y_0|}^2+3𝔼{\left|{\displaystyle {\int }_0^t}f(Y_{n-1}(u),{\displaystyle {\int }_0^u}{\sigma }_1(u,s)Y_{n-1}(s)𝑑B(s))𝑑u\right|}^2$

$+3𝔼{\left|{\displaystyle {\int }_0^t}g(Y_{n-1}(u),{\displaystyle {\int }_0^u}{\sigma }_2(u,s)Y_{n-1}(s)𝑑s)𝑑B(u)\right|}^2$

$\le$ $3𝔼{|Y_0|}^2+3T\left(K^{\prime }T+K^{\prime }(1+{\parallel {\sigma }_1\parallel }_{\mathrm{\infty }}^{2}T){\displaystyle {\int }_0^t}𝔼{|Y_{n-1}(s)|}^2𝑑s\right)$

$+3\left(K^{\prime }T+K^{\prime }(1+{\parallel {\sigma }_2\parallel }_{\mathrm{\infty }}^2{T}^2){\displaystyle {\int }_0^t}𝔼{|Y_{n-1}(s)|}^2𝑑s\right)$

$=$ $3𝔼{|Y_0|}^2+3K^{\prime }T(T+1)+3K^{\prime }(C_{1}T+C_2){\displaystyle {\int }_0^t}𝔼{|Y_n(s)|}^2𝑑s$

$=$ $3𝔼{|Y_n(t)|}^2\le 3\left(𝔼{|Y_0|}^2+C_3\right)+3C{\displaystyle {\int }_0^t}𝔼{|Y_{n-1}(s)|}^2𝑑s,$

where $C$ and $C_3$ are those from Lemma 3.1. Thus, for each $n\ge 1$, we have

$\underset{1\le k\le n}{\max }𝔼{|Y_k(t)|}^2\le$ $3\left(𝔼{|Y_0|}^2+C_3\right)+3C{\displaystyle {\int }_0^t}\underset{1\le k\le n}{\max }𝔼{|Y_{k-1}(s)|}^{2}ds$

$\le$ $3\left(𝔼{|Y_0|}^2+C_3\right)+3C{\displaystyle {\int }_0^t}\left[𝔼{|Y_0|}^2+\underset{1\le k\le n}{\max }𝔼{|Y_k(s)|}^2\right]𝑑s$

$\le$ $3\left((1+CT)𝔼{|Y_0|}^2+C_3\right)+3C{\displaystyle {\int }_0^t}\underset{1\le k\le n}{\max }𝔼{|Y_k(s)|}^{2}ds.$

By the Gronwall’s inequality, we obtain

$$\underset{1\le k\le n}{\max }𝔼{|Y_k(t)|}^2\le M_1,$$

where $M_1:=3\left((1+CT)𝔼{|Y_0|}^2+C_3\right)\exp (3CT).$ Since $k$ is arbitrary, we conclude

(3.8) $$𝔼{|Y_n(t)|}^2\le M_1,\text{for}t\in [0,T],n\ge 1.$$

Similarly to the proof of Lemma 3.1, one has

$𝔼{|Y_1(t)-Y_0(t)|}^2$ $=𝔼{|Y_1(t)-Y_0|}^2$

$\le 2𝔼{\left|{\displaystyle {\int }_0^t}f(Y_0(u),{\displaystyle {\int }_0^u}{\sigma }_1(u,s)Y_0(s)𝑑B(s))𝑑u\right|}^2$

$+2𝔼{\left|{\displaystyle {\int }_0^t}g(Y_0(u),{\displaystyle {\int }_0^u}{\sigma }_2(u,s)Y_0(s)𝑑s)𝑑B(u)\right|}^2$

$\le 2K^{\prime }\left(T(1+T)+\left(T\underset{C_1}{\underbrace{\left(1+T{\parallel {\sigma }_1\parallel }_{\mathrm{\infty }}^2\right)}}+\underset{C_2}{\underbrace{\left(1+T^2{\parallel {\sigma }_2\parallel }_{\mathrm{\infty }}^2\right)}}\right)𝔼{|Y_0|}^2\right):=C_0,$

and

$𝔼{|Y_2(t)-Y_1(t)|}^2$

$\le 2𝔼{\left|{\displaystyle {\int }_0^t}\left(f(Y_1(u),{\displaystyle {\int }_0^u}{\sigma }_1(u,s)Y_1(s)𝑑B(s))-f(Y_0(u),{\displaystyle {\int }_0^u}{\sigma }_1(u,s)Y_0(s)𝑑B(s))\right)𝑑u\right|}^2$

$+2𝔼{\left|{\displaystyle {\int }_0^t}\left(g(Y_1(u),{\displaystyle {\int }_0^u}{\sigma }_2(u,s)Y_1(s)𝑑s)-g(Y_0(u),{\displaystyle {\int }_0^u}{\sigma }_2(u,s)Y_0(s)𝑑s)\right)𝑑B(u)\right|}^2$

$\le 2T{\displaystyle {\int }_0^t}K\left(𝔼{\left|Y_1(u)-Y_0(u)\right|}^2+{\parallel {\sigma }_1\parallel }_{\mathrm{\infty }}^2{\displaystyle {\int }_0^u}𝔼{\left|Y_1(s)-Y_0(s)\right|}^2𝑑s\right)$

$+2K\left(𝔼{\left|Y_1(u)-Y_0(u)\right|}^2+T{\parallel {\sigma }_2\parallel }_{\mathrm{\infty }}^2{\displaystyle {\int }_0^u}𝔼{\left|Y_1(s)-Y_0(s)\right|}^2𝑑s\right)du$

$\le K\left(T{C}_1+C_2\right){\displaystyle {\int }_0^t}𝔼{|Y_1(s)-Y_0(s)|}^2𝑑s$

$\le 2C{\displaystyle {\int }_0^t}{C}_0𝑑s=2C{C}_{0}T.$

We claim that for $n\ge 0$,

(3.9) $$𝔼{|Y_n(t)-Y_{n-1}(t)|}^2\le \frac{C_0{(2Ct)}^{n-1}}{\left(n-1\right)!}.$$

By induction, we need to show that (3.9) still holds for $n+1$. Note that

(3.10) ${|Y_{n+1}(t)-Y_n(t)|}^2$ $\le 2𝔼|{\displaystyle {\int }_0^t}[f(Y_n(u),{\displaystyle {\int }_0^u}{\sigma }_1(u,s)Y_n(s)dB(s))$

$-f(Y_{n-1}(u),{\displaystyle {\int }_0^u}{\sigma }_1(u,s)Y_{n-1}(s)dB(s))\left]du\right|{}^2$

$+2𝔼|{\displaystyle {\int }_0^t}[g(Y_n(u),{\displaystyle {\int }_0^u}{\sigma }_2(u,s)Y_n(s)ds)$

$-g(Y_{n-1}(u),{\displaystyle {\int }_0^u}{\sigma }_2(u,s)Y_{n-1}(s)ds)]dB(u)|{}^2.$

Using (3.9), and similar to the proof of Lemma 3.1, we show that

$𝔼{|Y_{n+1}(t)-Y_n(t)|}^2\le$ $2C{\displaystyle {\int }_0^t}𝔼{|Y_n(s)-Y_{n-1}(s)|}^2𝑑s$

$\le$ $2C{\displaystyle {\int }_0^t}\frac{C_0{(2Cs)}^{n-1}}{\left(n-1\right)!}𝑑s$

$=$ ${2C\frac{C_0{(2C)}^{n-1}{s}^n}{n!}|}_0^t$

$=$ $\frac{C_0{(2Ct)}^n}{n!}.$

By Chebyshev inequality, we get

$$\mathbb{P}\left\{{|Y_n(t)-Y_{n-1}(t)|}^2>\frac{1}{2^n}\right\}\le \frac{C_0{(2CT)}^{n-1}}{\left(n-1\right)!},$$

Since . Thus, applying the Borel-Cantelli lemma, we can show that

$$\forall \omega \in \mathrm{\Omega },\exists n_0=n_0(\omega ),{|Y_n(t)-Y_{n-1}(t)|}^2\le \frac{1}{2^n},\text{for}n\ge n_0.$$

With probability $1$, it follows that,

$$Y_0(t)+\sum _{k=1}^{n-1}|Y_k(t)-Y_{k-1}(t)|=Y_n(t)$$

are convergent uniformly for $t\in [0,T]$. $Y(t)$ denote the limit. $Y(t)$ is obviously continuous and ${ℱ}_t$-adapted. On the other hand, it can be seen from (3.9) that, ${\{Y_n(t)\}}_{n\ge 1}$ is a Cauchy sequence in ${ℒ}^2([0,T],ℝ)$ for any t.

${\left(𝔼{|Y_n(t)-Y_m(t)|}^2\right)}^{\frac{1}{2}}=$ ${\parallel Y_n(t)-Y_m(t)\parallel }_{L^2}$

$\le$ ${\displaystyle \sum _{k=m}^{n-1}}{\parallel Y_n(t)-Y_{n-1}(t)\parallel }_{L^2}$

$\le$ ${\displaystyle \sum _{k=m}^{n-1}}{\left(\frac{C_0{(2CT)}^{n-1}}{\left(n-1\right)!}\right)}^{\frac{1}{2}}.$

Letting $n,m\to \mathrm{\infty },$ therefore

$${\left(𝔼{|Y_n(t)-Y_m(t)|}^2\right)}^{\frac{1}{2}}⟶0,$$

${\{Y_n(t)\}}_{n\ge 1}$ is a Cauchy sequence in ${ℒ}^2([0,T],ℝ)$. Therefore, we also have $Y_n(t)⟶Y(t)$ in ${ℒ}^2([0,T],ℝ).$

$$𝔼{|Y(t)|}^2\le M_1,\text{for}t\in [0,T],$$

where $M_1$ depends on ${\sigma }_1,{\sigma }_2,T,K^{\prime }$ and $Y_0$, resulting in $Y(\cdot )\in {ℳ}^2([0,T];ℝ)$. It has to be demonstrated that $Y(t)$ satisfies (2). Note that

$𝔼{\left|{\displaystyle {\int }_0^t}f(Y_n(u),{\displaystyle {\int }_0^u}{\sigma }_1(u,s)Y_n(s)𝑑B(s))𝑑u-{\displaystyle {\int }_0^t}f(Y(u),{\displaystyle {\int }_0^u}{\sigma }_1(u,s)Y_n(s)𝑑B(s))𝑑u\right|}^2$

$+𝔼{\left|{\displaystyle {\int }_0^t}g(Y_n(u),{\displaystyle {\int }_0^u}{\sigma }_2(u,s)Y_n(s)𝑑s)𝑑B(u)-{\displaystyle {\int }_0^t}g(Y(u),{\displaystyle {\int }_0^u}{\sigma }_2(u,s)Y(s)𝑑s)𝑑B(u)\right|}^2$

$\le C{\displaystyle {\int }_0^t}𝔼{|Y_n(s)-Y(s)|}^2𝑑s.$

Letting $n\to \mathrm{\infty }$ in (3.7), we get

(3.11) $Y(t)$ $=Y_0+{\displaystyle {\int }_0^t}f(Y(u),{\displaystyle {\int }_0^u}{\sigma }_1(u,s)Y(s)𝑑B(s))𝑑u$

$+{\displaystyle {\int }_0^t}g(Y(u),{\displaystyle {\int }_0^u}{\sigma }_2(u,s)Y(s)𝑑s)𝑑B(u).$

The proof is complete.

∎

3.2. Hölder continuity of the analytic solutions

We now show Hölder Continuity property by the analytical solution of SVIDEs (2).

Theorem 3.2.

Assume that ($H_2$) holds. Then, the solution Y(t) is Hölder continuous with exponent $\delta =\frac{1}{2}.$

Proof.

For

$Y(t)-Y(r)$ $={\displaystyle {\int }_r^t}f(Y(u),{\displaystyle {\int }_0^u}{\sigma }_1(u,s)Y(s)𝑑B(s))𝑑u$

$+{\displaystyle {\int }_r^t}g(Y(u),{\displaystyle {\int }_0^u}{\sigma }_2(u,s)Y(s)𝑑s)𝑑B(u).$

Using expectation, we obtain

(3.12) $𝔼{\left|Y(t)-Y(r)\right|}^2$ $\le 2𝔼{\left|{\displaystyle {\int }_r^t}f(Y(u),{\displaystyle {\int }_0^u}{\sigma }_1(u,s)Y(s)𝑑B(s))𝑑u\right|}^2$

$+2𝔼{\left|{\displaystyle {\int }_r^t}(Y(u),{\displaystyle {\int }_0^u}{\sigma }_2(u,s)Y(s)𝑑s)𝑑B(u)\right|}^2.$

Using Cauchy’s inequality, Itô isometry and ($H_2$), we have

$𝔼$ ${\left|{\displaystyle {\int }_r^t}f(Y(u),{\displaystyle {\int }_0^u}{\sigma }_1(u,s)Y(s)𝑑B(s))𝑑u\right|}^2\le$

$\le (t-r){\displaystyle {\int }_r^t}{K}^{\prime }\left[1+𝔼{|Y(u)|}^2+𝔼{\left|{\displaystyle {\int }_0^u}{\sigma }_1(u,s)Y(s)𝑑B(s)\right|}^2\right]$

(3.13) $\le (t-r)K^{\prime }\left[(t-r)+{\displaystyle {\int }_r^t}𝔼{|Y(s)|}^2𝑑s+T{\parallel {\sigma }_1\parallel }_{\mathrm{\infty }}^2{\displaystyle {\int }_r^t}𝔼{|Y(s)|}^2𝑑s\right]$

$\le (t-r)K^{\prime }\left[(t-r)+\underset{C_1}{\underbrace{\left(1+T{\parallel {\sigma }_1\parallel }_{\mathrm{\infty }}^2\right)}}{\displaystyle {\int }_0^t}𝔼{|Y(s)|}^2𝑑s\right]$

$\le K^{\prime }T\left(1+C_1{M}_1\right)\left(t-r\right).$

Also, we get

(3.14) $𝔼$ ${\left|{\displaystyle {\int }_r^t}g(Y(u),{\displaystyle {\int }_0^u}{\sigma }_2(u,s)Y(s)𝑑s)𝑑B(u)\right|}^2\le$

$\le {\displaystyle {\int }_r^t}{K}^{\prime }\left[1+𝔼{|Y(u)|}^2+𝔼{\left|{\displaystyle {\int }_0^u}{\sigma }_2(u,s)Y(s)𝑑s\right|}^2\right]𝑑u$

$\le K^{\prime }{\displaystyle {\int }_r^t}\left[1+𝔼{|Y(u)|}^2+T{\parallel {\sigma }_2\parallel }_{\mathrm{\infty }}^2{\displaystyle {\int }_0^t}𝔼{|Y(s)|}^2𝑑s\right]𝑑u$

$\le K^{\prime }{\displaystyle {\int }_r^t}\left[1+\underset{C_2}{\underbrace{\left(1+T^2{\parallel {\sigma }_2\parallel }_{\mathrm{\infty }}^2\right)}}{M}_1\right]𝑑u$

$\le K^{\prime }\left(1+C_2{M}_1\right)\left(t-r\right).$

By substituting (Proof) and (3.14) into (3.12), we have

$$𝔼{|Y(t)-Y(r)|}^2\le M|t-r|,$$

where $M:=2K^{\prime }\left[T(1+C_1{M}_1)+(1+C_2{M}_1)\right].$

Consequently, $Y(t)$, $t\in [0,T]$ is Hölder continuous with exponent $1/2$.

4. Numerical analysis of the class of SVIDE

Let $I_h:=\left\{t_n=nh,n=0,1,\mathrm{\dots },N\right\},I=[0,T]$. For $n=0,1,\mathrm{\dots },N-1$, we have defined the numerical results of SVIDE

4.1. $\theta$-Euler Maruyama method

We apply the $\theta$-EM method to SVIDEs (LABEL:SDEf) (see [1]–[3] and references therein),

(4.1) $X_{n+1}$ $=X_n+h[\mathrm{\Theta }f(X_{n+1},{\displaystyle \sum _{i=0}^{n-1}}{\sigma }_1(t_{n+1},t_{i+1})X_{i+1}\mathrm{\Delta }{B}_{i+1})$

$+(1-\mathrm{\Theta })f(X_n,{\displaystyle \sum _{i=0}^{n-1}}{\sigma }_1(t_n,t_i)X_i\mathrm{\Delta }{B}_i)]$

$+g(X_n,{\displaystyle \sum _{i=0}^{n-1}}{\displaystyle {\int }_{t_i}^{t_{i+1}}}{\sigma }_2(t_n,s)X_i𝑑s)\mathrm{\Delta }{B}_n,$

where $\mathrm{\Theta }\in [0,1]$ with initial data $X_0=Y_0$, where $n\in \mathbb{N}$, $t_n=nh,\mathrm{\Delta }{B}_n=B(t_{n+1})-B(t_n)$. By induction, we rewrite (4.1) in the following form:

(4.2) $X_{n+1}$ $=X_0+{\displaystyle \sum _{j=0}^n}h(\mathrm{\Theta }f(X_{j+1},{\displaystyle \sum _{i=0}^{j-1}}{\sigma }_1(t_{j+1},t_{i+1})X_{i+1}\mathrm{\Delta }{B}_{i+1})$

$+(1-\mathrm{\Theta })f(X_j,{\displaystyle \sum _{i=0}^{j-1}}{\sigma }_1(t_j,t_i)X_i\mathrm{\Delta }{B}_i))$

$+{\displaystyle \sum _{j=0}^n}g(X_j,{\displaystyle \sum _{i=0}^{j-1}}{\displaystyle {\int }_{t_i}^{t_{i+1}}}{\sigma }_2(t_j,s)X_i𝑑s)\mathrm{\Delta }{B}_j.$

Remark 4.1.

The scheme (4.1) is called the $\mathrm{\Theta }$-EM and the choice $\mathrm{\Theta }=0$ gives Euler Maruyama method in [4]

(4.3) $$X_{n+1}=X_n+hf(X_n,\sum _{i=0}^{n-1}{\sigma }_1(t_n,t_i)X_i\mathrm{\Delta }{B}_i)+g(X_n,\sum _{i=0}^{n-1}{\int }_{t_i}^{t_{i+1}}{\sigma }_2(t_n,s)X_i𝑑s)\mathrm{\Delta }{B}_n.$$

$\theta =\frac{1}{2}$ gives the trapezoidal solver, and $\theta =1$ gives the implicit, or backward Euler method in [2].

Now, we examine the stability properties with respect to SVIDE (4.1) in the following result.

Theorem 4.1.

Assume that ($H_2$) holds. Let $\{X_n\}$ be the numerical solution of the $\theta$-EM method (4.1). Then there exists a positive constant $K_0$, which depends on ${\sigma }_1,{\sigma }_2,T,Y_0$ and $K^{\prime }$, but not on $h$, such that

$$𝔼{|X_n|}^2\le K_0.$$

Proof.

For all $0\le t_{n+1}\le T$, we have

$𝔼{|X_{n+1}|}^2$ $\le 3𝔼|X_0{|}^2+3𝔼|{\displaystyle \sum _{j=0}^n}h[\theta f(X_{j+1},{\displaystyle \sum _{i=0}^{j-1}}{\sigma }_1(t_{j+1},t_{i+1})X_{i+1}\mathrm{\Delta }{B}_{i+1})$

$+(1-\theta )f(X_j,{\displaystyle \sum _{i=0}^{j-1}}{\sigma }_1(t_j,t_i)X_i\mathrm{\Delta }{B}_i)\left]\right|{}^2$

$+3𝔼{\left|{\displaystyle \sum _{j=0}^n}g(X_j,{\displaystyle \sum _{i=0}^{j-1}}{\displaystyle {\int }_{t_i}^{t_{i+1}}}{\sigma }_2(t_j,s)X_i𝑑s)\mathrm{\Delta }{B}_j\right|}^2$

(4.4) $\le 3𝔼{|X_0|}^2+3I_1+3I_2,$

where

$I_1:=$ $𝔼|{\displaystyle \sum _{j=0}^n}h[\theta f(X_{j+1},{\displaystyle \sum _{i=1}^{j-1}}{\sigma }_1(t_{j+1},t_{i+1})X_{i+1}\mathrm{\Delta }{B}_{i+1})$

$+(1-\theta )f(X_j,{\displaystyle \sum _{i=0}^{j-1}}{\sigma }_1(t_j,t_i)X_i\mathrm{\Delta }{B}_i)\left]\right|{}^2,$

and

$$I_2:=𝔼{\left|\sum _{j=1}^{n}g(X_j,\sum _{i=1}^{j-1}{\int }_{t_i}^{t_{i+1}}{\sigma }_2(t_j,s)X_i𝑑s)\mathrm{\Delta }{B}_j\right|}^2.$$

By Cauchy’s inequality, Itô isometry and ($H_2$), we get

(4.5) $I_1=$ $𝔼|{\displaystyle \sum _{j=0}^n}h[\theta f(X_{j+1},{\displaystyle \sum _{i=0}^{j-1}}{\sigma }_1(t_{j+1},t_{i+1})X_{i+1}\mathrm{\Delta }{B}_{i+1})$

$+(1-\theta )f(X_j,{\displaystyle \sum _{i=0}^{j-1}}{\sigma }_1(t_j,t_i)X_i\mathrm{\Delta }{B}_i)\left]\right|{}^2$

$\le 2(n+1){\displaystyle \sum _{j=0}^n}{h}^2\left[{\theta }^2𝔼\right|f(X_{j+1},{\displaystyle \sum _{i=0}^{j-1}}{\sigma }_1(t_{j+1},t_{i+1})X_{i+1}\mathrm{\Delta }{B}_{i+1}){|}^2$

$+{(1-\theta )}^2𝔼\left|f(X_j,{\displaystyle \sum _{i=0}^{j-1}}{\sigma }_1(t_j,t_i)X_i\mathrm{\Delta }{B}_i){|}^2\right]$

$\le \underset{2t_{n+1}h\le 2Th}{\underbrace{2(n+1)h^2}}{K}^{\prime }{\displaystyle \sum _{j=0}^n}[2+𝔼|X_{j+1}{|}^2+𝔼|{\displaystyle \sum _{i=0}^{j-1}}{\sigma }_1(t_{j+1},t_{i+1})X_{i+1}\mathrm{\Delta }{B}_{i+1}{|}^2$

$+𝔼|X_j{|}^2+𝔼|{\displaystyle \sum _{i=0}^{j-1}}{\sigma }_1(t_j,t_i)X_i\mathrm{\Delta }{B}_i{|}^2]$

$\le 2Th{K}^{\prime }[2(n+1)+{\displaystyle \sum _{i=1}^n}𝔼|X_i{|}^2+T{\parallel {\sigma }_1\parallel }_{\mathrm{\infty }}^2{\displaystyle \sum _{i=1}^n}𝔼|X_i{|}^2+{\displaystyle \sum _{i=1}^n}𝔼|X_i{|}^2+T{\parallel {\sigma }_1\parallel }_{\mathrm{\infty }}^2{\displaystyle \sum _{i=1}^n}𝔼|X_i{|}^2$

$+𝔼|X_0{|}^2+T{\parallel {\sigma }_1\parallel }_{\mathrm{\infty }}^2𝔼|X_0{|}^2+𝔼|X_{n+1}{|}^2+T{\parallel {\sigma }_1\parallel }_{\mathrm{\infty }}^2𝔼|X_{n+1}{|}^2]$

$\le 2Th{K}^{\prime }\left[2(n+1)+2C_1{\displaystyle \sum _{i=1}^n}𝔼{|X_i|}^2+C_1\left(𝔼{|X_0|}^2+𝔼{|X_{n+1}|}^2\right)\right]$

$\le 2T^2{K}^{\prime }(2+C_1𝔼{|X_0|}^2)+4Th{K}^{\prime }{C}_1{\displaystyle \sum _{i=1}^n}𝔼{|X_i|}^2+2T^2{K}^{\prime }{C}_1𝔼{|X_{n+1}|}^2.$

Similar to [4].

Using Cauchy’s inequality, Minkowski inequality and ($H_2$), we show that

$I_2=$ ${\displaystyle \sum _{j=0}^n}𝔼{\left|g(X_j,{\displaystyle \sum _{i=0}^{j-1}}{\displaystyle {\int }_{t_i}^{t_{i+1}}}{\sigma }_2(t_j,s)X_i𝑑s)\mathrm{\Delta }{B}_j\right|}^2$

$\times g(X_{i_2},{\displaystyle \sum _{i=0}^{i_2-1}}{\displaystyle {\int }_{t_i}^{t_{i+1}}}{\sigma }_2(t_{i_2},s)X_{i}ds)\mathrm{\Delta }{B}_{i_2}]$

$\le h{K}^{\prime }{\displaystyle \sum _{j=0}^n}\left[1+𝔼{|X_j|}^2+𝔼{\left|{\displaystyle \sum _{i=0}^{j-1}}{\displaystyle {\int }_{t_i}^{t_{i+1}}}{\sigma }_2(t_j,s)X_i𝑑s\right|}^2\right]$

Since $X_{i_1}$ is ${ℱ}_{t_{i_1}}$-adapted, $\mathrm{\Delta }{B}_{i_1}$ is ${ℱ}_{t_{i_1+1}}$-adapted, $X_{i_2}$ is ${ℱ}_{t_{i_2}}$-adapted, and , so $X_{i_1}\mathrm{\Delta }{B}_{i_1}{X}_{i_2}$ is ${ℱ}_{t_{i_2}}$-adapted. Namely, $g_{X_{i_2}}(t_{i_2})\mathrm{\Delta }{B}_{i_1}.g_{X_{i_2}}(t_{i_2})$ is ${ℱ}_{t_{i_2}}$-adapted, therefore,
$g_{X_{i_2}}(t_{i_2})\mathrm{\Delta }{B}_{i_1}.g_{X_{i_2}}(t_{i_2})$ is independent of $\mathrm{\Delta }{B}_{i_2}$, and by $\mathrm{\Delta }{B}_{i_2}\rightsquigarrow N(0,t_{i_2+1}-t_{i_2})$, we obtain

we have

$I_2\le$ $h{\displaystyle \sum _{j=0}^n}𝔼{\left|g(X_j,{\displaystyle \sum _{i=0}^{j-1}}{\displaystyle {\int }_{t_i}^{t_{i+1}}}{\sigma }_2(t_j,s)X_i𝑑s)\right|}^2$

(4.6) $\le$ $h{K}^{\prime }{\displaystyle \sum _{j=0}^n}\left[1+𝔼{|X_j|}^2+𝔼{\left|{\displaystyle \sum _{i=0}^{j-1}}{\displaystyle {\int }_{t_i}^{t_{i+1}}}{\sigma }_2(t_j,s)X_i𝑑s\right|}^2\right]$

$\le$ $h{K}^{\prime }[n+1+{\displaystyle \sum _{i=0}^n}𝔼|X_i{|}^2+T^2\parallel {\sigma }_2{\parallel }_{\mathrm{\infty }}^2{\displaystyle \sum _{i=0}^n}𝔼|X_i{|}^2]$

$\le$ $T{K}^{\prime }+C_{2}T{K}^{\prime }𝔼{|X_0|}^2+h{C}_2{K}^{\prime }{\displaystyle \sum _{i=1}^n}𝔼{|X_i|}^2.$

Substituting (4.5) and (Proof) into (Proof),

$𝔼{|X_{n+1}|}^2\le$

$\le 3𝔼{|X_0|}^2+3\left(2T^2{K}^{\prime }(2+C_1𝔼{|X_0|}^2)+4Th{K}^{\prime }{C}_1{\displaystyle \sum _{i=1}^n}𝔼{|X_i|}^2+2T^2{K}^{\prime }{C}_1𝔼{|X_{n+1}|}^2\right)$

$+3\left(T{K}^{\prime }+C_{2}T{K}^{\prime }𝔼{|X_0|}^2+h{C}_2{K}^{\prime }{\displaystyle \sum _{i=1}^n}𝔼{|X_i|}^2\right)$

$\le 3T{K}^{\prime }((1+4T)+(1+2T{C}_1+C_2)𝔼|X_0{|}^2+3h{K}^{\prime }(2T{C}_1+C_2){\displaystyle \sum _{i=1}^n}𝔼|X_i{|}^2$

$+2T{C}_1𝔼|X_{n+1}{|}^2).$

So only if it’s if $C^{\prime }:=1-6T^2{K}^{\prime }{C}_1\ne 0$

$𝔼{|X_{n+1}|}^2\le$

$\le \frac{3T{K}^{\prime }}{C^{\prime }}\left(\left(1+4T\right)+(1+2T{C}_1+C_2)𝔼{|X_0|}^2+3h{K}^{\prime }\left(2T{C}_1+C_2\right){\displaystyle \sum _{i=1}^n}𝔼{|X_i|}^2\right).$

Through The discrete Gronwall’s inequality, we get

$$𝔼{|X_{n+1}|}^2\le K_0,$$

where

$$K_0:=\frac{3T{K}^{\prime }}{C^{\prime }}\left(\left(1+4T\right)+(1+2T{C}_1+C_2)𝔼{|X_0|}^2\right)\exp \left(\frac{3T{K}^{\prime }}{C^{\prime }}(2T{C}_1+C_2)\right)$$

Similar to [4, 3, 1, 2], the convergence order of the $\theta$-EM method can be enhanced by including more terms in the numerical approximation.

4.2. Strong convergence of the $\theta$-Euler Maruyama method

In order to obtain the convergence result for the $\theta$-Euler Maruyama method (LABEL:theta_num-sul), we now introduce time continuous interpolations of the discrete numerical approximations.

Define $u_n:=t_n$ and $X_h(s):=X_n$, for $s\in [t_n,t_{n+1})$ with $0\le n\le N-1$.

Let $t\in [t_n,t_{n+1})$ with $0\le n\le N-1$ and X(t) be the continuous form of $X_n$ with $X(t_n)=X_n$, we obtain

(4.7) $X(t)$ $=X(t_n)+{\displaystyle {\int }_{t_n}^t}(\theta f(X_h(u_{n+1}),{\displaystyle {\int }_0^{u_h}}{\sigma }_1(u_{n+1},s)X_h(s)dB(s))$

$+(1-\theta )f(X_h(u_n),{\displaystyle {\int }_0^{u_h}}{\sigma }_1(u_n,s)X_h(s)dB(s)))du$

$+{\displaystyle {\int }_{t_n}^t}g(X_h(u_n),{\displaystyle {\int }_0^{u_h}}{\sigma }_2(u_n,s)X_h(s)𝑑s)𝑑B(u)$

$=X(t_0)+{\displaystyle {\int }_0^t}(\theta f(X_h(u_{n+1}),{\displaystyle {\int }_0^{u_h}}{\sigma }_1(u_{n+1},s)X_h(s)dB(s))$

$+(1-\theta )f(X_h(u_n),{\displaystyle {\int }_0^{u_h}}{\sigma }_1(u_n,s)X_h(s)dB(s)))du$

$+{\displaystyle {\int }_0^t}g(X_h(u_n),{\displaystyle {\int }_0^{u_h}}{\sigma }_2(u_n,s)X_h(s)𝑑s)𝑑B(u).$

The following theorem illustrates the convergence order of (4.1), and its proof proceeds similarly to [3] for the situation when $\theta =0.$

Lemma 4.1.

Assume that ($H_2$) holds. Let $\{X_n\}$ be the numerical solution of the $\theta$-Euler Maruyama method (LABEL:theta_num-sul). Then there exists a positive constant $K_1$, which depends on ${\sigma }_1$, ${\sigma }_2,K^{\prime }$ and T, but not on h, such that

$$𝔼{\left|X(t)-X(t_n)\right|}^2\le K_{1}h.$$

Proof.

It is easy to see that

$𝔼{\left|X(t)-X(t_n)\right|}^2$ $\le 2𝔼|{\displaystyle {\int }_{t_n}^t}(\theta f(X_h(u_{n+1}),{\displaystyle {\int }_0^{u_n}}{\sigma }_1(u_{n+1},s)X_h(s)dB(s))$

$+(1-\theta )f(X_h(u_n),{\displaystyle {\int }_0^{u_n}}{\sigma }_1(u_n,s)X_h(s)dB(s))\left)du\right|{}^2$

$+2𝔼{\left|{\displaystyle {\int }_{t_n}^t}g(X_h(u_n),{\displaystyle {\int }_0^{u_n}}{\sigma }_2(u_n,s)X_h(s)𝑑s)𝑑B(u)\right|}^2$

$\le 2J_1+2J_2,$

$J_1:=$ $𝔼|{\displaystyle {\int }_{t_n}^t}(\theta f(X_h(u_{n+1}),{\displaystyle {\int }_0^{u_n}}{\sigma }_1(u_{n+1},s)X_h(s)dB(s))$

$+(1-\theta )f(X_h(u_n),{\displaystyle {\int }_0^{u_n}}{\sigma }_1(u_n,s)X_h(s)dB(s))\left)du\right|{}^2,$

and

$$J_2:=𝔼{\left|{\int }_{t_n}^{t}g(X_h(u_n),{\int }_0^{u_n}{\sigma }_2(u_n,s)X_h(s)𝑑s)𝑑B(u)\right|}^2.$$

By ${(x+y)}^2\le 2x^2+2y^2$, Cauchy’s inequality, Itô isometry and ($H_2$), we get

(4.8) $J_1$ $\le 2h{\displaystyle {\int }_{t_n}^t}\left(\theta 𝔼\right|f(X_h(u_{n+1}),{\displaystyle {\int }_0^{u_n}}{\sigma }_1(u_{n+1},s)X_h(s)dB(s)){|}^2$

$+(1-\theta )𝔼\left|f(X_h(u_n),{\displaystyle {\int }_0^{u_n}}{\sigma }_1(u_n,s)X_h(s)dB(s)){|}^2\right)du$

$\le 2h[{\theta }^2{\displaystyle {\int }_{t_n}^t}\left(𝔼\right|f(X_h(u_{n+1}),{\displaystyle {\int }_0^{u_n}}{\sigma }_1(u_{n+1},s)X_h(s)dB(s)){|}^2)du$

$+{(1-\theta )}^2{\displaystyle {\int }_{t_n}^t}\left(𝔼\right|f(X_h(u_n),{\displaystyle {\int }_0^{u_n}}{\sigma }_1(u_n,s)X_h(s)dB(s)){|}^2)du]$

$\le 2h{\displaystyle {\int }_{t_n}^t}{K}^{\prime }[2+𝔼|X_{n+1}{|}^2+𝔼|{\displaystyle {\int }_0^{t_n}}{\sigma }_1(u_{n+1},s)X_h(s)dB(s){|}^2$

$+𝔼|X_n{|}^2+𝔼|{\displaystyle {\int }_0^{t_n}}{\sigma }_1(u_n,s)X_h(s)dB(s){|}^2]du$

$\le 2h{\displaystyle {\int }_{t_n}^t}{K}^{\prime }\left[2+𝔼{|X_{n+1}|}^2+{\parallel {\sigma }_1\parallel }_{\mathrm{\infty }}^2{\displaystyle {\int }_0^{t_n}}𝔼{\left|X_h(s)\right|}^2𝑑s+𝔼{|X_n|}^2+{\parallel {\sigma }_1\parallel }_{\mathrm{\infty }}^2{\displaystyle {\int }_0^{t_n}}𝔼{\left|X_h(s)\right|}^2𝑑s\right]𝑑u$

$\le 2h^2{K}^{\prime }\left[2+𝔼{|X_{n+1}|}^2+{\parallel {\sigma }_1\parallel }_{\mathrm{\infty }}^2{\displaystyle {\int }_0^{t_n}}𝔼{\left|X_h(s)\right|}^2𝑑s+𝔼{|X_n|}^2+{\parallel {\sigma }_1\parallel }_{\mathrm{\infty }}^2{\displaystyle {\int }_0^{t_n}}𝔼{\left|X_h(s)\right|}^2𝑑s\right]$

$\le 2h^2{K}^{\prime }\left(2+2\left(1+{\parallel {\sigma }_1\parallel }_{\mathrm{\infty }}^{2}T\right)K_0\right):=4h^2{K}^{\prime }\left(1+C_1{K}_0\right).$

By ($H_2$), Cauchy’s inequality and Itô isometry, we obtain

(4.9)

From (4.2) and (4.3) we get

$$𝔼{\left|X(t)-X(t_n)\right|}^2\le K_{1}h,$$

where

$$K_1:=2K^{\prime }\left((2T+1)+(2C_{1}T+C_2)M_0\right).$$

Theorem 4.2.

Suppose($H_1$) and ${\sigma }_i\in C^1(D)$, for $i=1,2$ satisfy ($H_3$). Let X (t) and Y (t) are The numerical solution of the $\theta$-Euler-Maruyama method and the analytical solution (LABEL:solu-EDSf), respectively. Then there exists a positive constant $K_2$, which depends on ${\sigma }_1,{\sigma }_2,K,K^{\prime },$ and T, but not on h, such as

$$𝔼{|X(t)-Y(t)|}^2\le K_{2}h.$$

Proof.

By ($H_1$), Cauchy’s inequality and the Itô isometry, we have

(4.10) $$𝔼{|X(t)-Y(t)|}^2\le 2L_1+2L_2,$$

where

$L_1:=$ $𝔼|{\displaystyle {\int }_0^t}\theta (f(Y(u),{\displaystyle {\int }_0^u}{\sigma }_1(u,s)Y(s)dB(s))$

$-f(X_h(u_{n+1}),{\displaystyle {\int }_0^{u_h}}{\sigma }_1(u_h,s)X_h(s)dB(s)))$

$+(1-\theta )(f(Y(u),{\displaystyle {\int }_0^u}{\sigma }_1(u,s)Y(s)dB(s))$

$-f(X_h(u_n),{\displaystyle {\int }_0^{u_n}}{\sigma }_1(u_h,s)X_h(s)dB(s))\left)du\right|{}^2,$

and

$L_2:=𝔼|$ ${\displaystyle {\int }_0^t}g(Y(u),{\displaystyle {\int }_0^u}{\sigma }_2(u,s)Y(s)𝑑s)𝑑B(u)$

$-{\displaystyle {\int }_0^t}g(X_h(u_h),{\displaystyle {\int }_0^{u_h}}{\sigma }_2(u_h,s)X_h(s)ds))dB(u)|{}^2.$

By Cauchy inequality and Itô isometry, we obtain

$L_1\le$ $T{\displaystyle {\int }_0^t}𝔼|\theta (f(Y(u),{\displaystyle {\int }_0^u}{\sigma }_1(u,s)Y(s)dB(s))$

$-f(X_h(u_{n+1}),{\displaystyle {\int }_0^{u_h}}{\sigma }_1(u_{n+1},s)X_h(s)dB(s)))$

$+(1-\theta )(f(Y(u),{\displaystyle {\int }_0^u}{\sigma }_1(u,s)Y(s)dB(s))$

$-f(X_h(u_n),{\displaystyle {\int }_0^{u_h}}{\sigma }_1(u_n,s)X_h(s)dB(s))\left)\right|{}^{2}du$

$\le 2T\left({\theta }^2{\displaystyle {\int }_0^t}𝔼\right|f(Y(u),{\displaystyle {\int }_0^u}{\sigma }_1(u,s)Y(s)dB(s))$

$-f(X_h(u_{n+1}),{\displaystyle {\int }_0^{u_h}}{\sigma }_1(u_{n+1},s)X_h(s)dB(s)){|}^{2}du$

$+{(1-\theta )}^2{\displaystyle {\int }_0^t}𝔼|f(Y(u),{\displaystyle {\int }_0^u}{\sigma }_1(u,s)Y(s)𝑑B(s))$

$-f(X_h(u_n),{\displaystyle {\int }_0^{u_h}}{\sigma }_1(u_n,s)X_h(s)dB(s)){|}^{2}du).$

Next, using ($H_1$) and ${(x+y)}^2\le x^2+y^2$, one has

$L_1\le 2KT{\displaystyle {\int }_0^t}\left[𝔼\right|Y(u)-X_h(u_{n+1}){|}^2+𝔼|{\displaystyle {\int }_0^u}{\sigma }_1(u,s)Y(s)dB(s)$

$-{\displaystyle {\int }_0^{u_h}}{\sigma }_1(u_{n+1},s)X_h(s)dB(s){|}^2]du$

$+2KT{\displaystyle {\int }_0^t}\left[𝔼\right|Y(u)-X_h(u_n){|}^2+𝔼|{\displaystyle {\int }_0^u}{\sigma }_1(u,s)Y(s)dB(s)$

$-{\displaystyle {\int }_0^{u_n}}{\sigma }_1(u_n,s)X_h(s)dB(s){|}^2]du$

$\le 2KT{\displaystyle {\int }_0^t}\left[𝔼\right|Y(u)-X(u)+X(u)-X_h(u_{n+1}){|}^2$

$\le 𝔼|{\displaystyle {\int }_{u_h}^u}{\sigma }_1(u_{n+1},s)X_h(s)𝑑B(s)$

$+{\displaystyle {\int }_0^u}({\sigma }_1(u,s)Y(s)dB(s)-{\sigma }_1(u_{n+1},s)X_h(s)dB(s){|}^2)$

$+𝔼{\left|Y(u)-X(u)+X(u)-X_h(u_n)\right|}^2$

$+𝔼\left|{\displaystyle {\int }_{u_h}^u}{\sigma }_1(u_n,s)X_h(s)dB(s)+{\displaystyle {\int }_0^u}\left({\sigma }_1(u,s)Y(s)dB(s)-{\sigma }_1(u_n,s)X_h(s)dB(s){|}^2\right)\right]du$

$\le 4KT{\displaystyle {\int }_0^t}\left[𝔼\right|Y(u)-X(u){|}^2+𝔼|X(u)-X_h(u_{n+1}){|}^2+𝔼|{\displaystyle {\int }_{u_h}^u}{\sigma }_1(u_{n+1},s)X_h(s)dB(s){|}^2$

$+𝔼{\left|{\displaystyle {\int }_0^u}\left({\sigma }_1(u,s)Y(s)-{\sigma }_1(u_{n+1},s)X_h(s)\right)𝑑B(s)\right|}^2$

$+𝔼{\left|Y(u)-X(u)\right|}^2+𝔼{\left|X(u)-X_h(u_n)\right|}^2$

$+𝔼\left|{\displaystyle {\int }_{u_h}^u}{\sigma }_1(u_n,s)X_h(s)dB(s){|}^2+𝔼\right|{\displaystyle {\int }_0^u}\left({\sigma }_1(u,s)Y(s)-{\sigma }_1(u_n,s)X_h(s)\right)dB(s){|}^2]du.$

By Hölder’s inequality, Itô isometry, LABEL:bond-thetam and Lemma 4.1, we have

${\displaystyle {\int }_0^t}𝔼{\left|X(u)-X_h(u_n)\right|}^2𝑑u\le$ $T{K}_{1}h,$

${\displaystyle {\int }_0^t}𝔼{\left|X(u)-X_h(u_{n+1})\right|}^2𝑑u\le$ $T{K}_{1}h,$

and

${\displaystyle {\int }_0^t}𝔼{\left|{\displaystyle {\int }_{u_h}^u}{\sigma }_1(u_{n+1},s)X_h(s)𝑑B(s)\right|}^2𝑑u\le$ ${\displaystyle {\int }_0^t}{\displaystyle {\int }_{u_h}^u}{\sigma }_1(u_{n+1},s)𝔼{|X_h(s)|}^2𝑑s𝑑u$

$\le$ ${\displaystyle {\int }_0^t}\parallel \sigma {}_1{\parallel }_{\mathrm{\infty }}^{2}h{K}_{0}du$

$\le$ $\parallel \sigma {}_1{\parallel }_{\mathrm{\infty }}^{2}h{K}_{0}T,$

${\displaystyle {\int }_0^t}𝔼{\left|{\displaystyle {\int }_{u_h}^u}{\sigma }_1(u_{n+1},s)X_h(s)𝑑B(s)\right|}^2𝑑u\le$ $\parallel \sigma {}_1{\parallel }_{\mathrm{\infty }}^{2}h{K}_{0}T.$

By ($H_3$), we show that

${\displaystyle {\int }_0^t}𝔼{\left|{\displaystyle {\int }_0^u}\left({\sigma }_1(u,s)Y(s)-{\sigma }_1(u_n,s)X_h(s)\right)𝑑B(s)\right|}^2𝑑u$

$\le {\displaystyle {\int }_0^t}{\displaystyle {\int }_0^u}𝔼{\left|{\sigma }_1(u,s)Y(s)-{\sigma }_1(u_n,s)X_h(s)\right|}^2𝑑s𝑑u$

$\le {\displaystyle {\int }_0^t}{\displaystyle {\int }_0^u}𝔼{\left|{\sigma }_1(u,s)Y(s)-{\sigma }_1(u_n,s)X_h(s)+{\sigma }_1(u,s)X_h(s)-{\sigma }_1(u,s)X_h(s)\right|}^2𝑑s𝑑u$

$\le 2{\displaystyle {\int }_0^t}{\displaystyle {\int }_0^u}𝔼{\left|{\sigma }_1(u,s)\right|}^2𝔼{\left|Y(s)-X_h(s)\right|}^2𝑑s𝑑u$

$+2{\displaystyle {\int }_0^t}{\displaystyle {\int }_0^u}𝔼{\left|{\sigma }_1(u,s)-{\sigma }_1(u_n,s)\right|}^2𝔼{\left|X_h(s)\right|}^2𝑑s𝑑u$

$\le 4{\displaystyle {\int }_0^t}{\displaystyle {\int }_0^u}𝔼{\left|{\sigma }_1(u,s)\right|}^2𝔼{\left|Y(s)-X(s)\right|}^2𝑑s𝑑u$

$\le +4{\displaystyle {\int }_0^t}{\displaystyle {\int }_0^u}𝔼{\left|{\sigma }_1(u,s)\right|}^2𝔼{\left|X(s)-X_h(s)\right|}^2𝑑s𝑑u$

$\le +2{\displaystyle {\int }_0^t}{\displaystyle {\int }_0^u}𝔼{\left|{\sigma }_1(u,s)-{\sigma }_1(u_n,s)\right|}^2𝔼{\left|X_h(s)\right|}^2𝑑s𝑑u$

$\le 2K^{\prime \prime }h{K}_0^{\prime }{T}^2+4\parallel \sigma {}_1{\parallel }_{\mathrm{\infty }}^{2}h{K}_1^{\prime }{T}^2+4T\parallel \sigma {}_1{\parallel }_{\mathrm{\infty }}^2{\displaystyle {\int }_0^t}𝔼|Y(s)-X(s){|}^{2}ds.$

Thus

(4.11) $$L_1\le L_{11}hT+L_{12}{\int }_0^u𝔼{\left|Y(s)-X(s)\right|}^2𝑑s,$$

where

$$L_{11}:=8KT\left[\left(K_0^{\prime }+4T{K}_1^{\prime }\right){\parallel {\sigma }_1\parallel }_{\mathrm{\infty }}^2+K_1+2T{K}^{\prime \prime }{K}_0\right],$$

and

$$L_{12}:=8KT\left[1+4T{\parallel {\sigma }_1\parallel }_{\mathrm{\infty }}^2\right].$$

Using LABEL:bond-thetam, we get

$L_2$ $:=𝔼|{\displaystyle {\int }_0^t}g(Y(u),{\displaystyle {\int }_0^u}{\sigma }_2(u,s)Y(s)𝑑s)𝑑B(u)-$

$-{\displaystyle {\int }_0^t}g(X_h(u_h),{\displaystyle {\int }_0^{u_h}}{\sigma }_2(u_h,s)X_h(s)ds))dB(u)|{}^2.$

$L_2$ $\le {\displaystyle {\int }_0^t}𝔼{\left|g(Y(u),{\displaystyle {\int }_0^u}{\sigma }_2(u,s)Y(s)𝑑s)-g(X_h(u_h),{\displaystyle {\int }_0^{u_h}}{\sigma }_2(u_h,s)X_h(s)𝑑s)\right|}^2𝑑u$

$\le K{\displaystyle {\int }_0^t}\left[𝔼{\left|Y(u)-X_h(u_h)\right|}^2+𝔼{\left|{\displaystyle {\int }_0^u}{\sigma }_2(u,s)Y(s)𝑑s-{\displaystyle {\int }_0^{u_h}}{\sigma }_2(u_h,s)X_h(s)𝑑s\right|}^2\right]𝑑u$

$\le K{\displaystyle {\int }_0^t}\left[𝔼\right|Y(u)-X(u)+X(u)-X_h(u_h){|}^2+𝔼|{\displaystyle {\int }_0^u}{\sigma }_2(u,s)Y(s)ds-$

${\displaystyle {\int }_0^{u_h}}{\sigma }_2(u_h,s)X_h(s)ds{|}^2]du$

$\le 2KT{\displaystyle {\int }_0^t}\left[𝔼\right|X(u)-X_h(u_h){|}^2+𝔼|Y(u)-X(u){|}^2$

$+𝔼|{\displaystyle {\int }_{u_h}^u}{\sigma }_2(u_h,s)X_h(s)ds{|}^2+𝔼|{\displaystyle {\int }_0^u}{\sigma }_2(u,s)Y(s)ds-{\displaystyle {\int }_0^u}{\sigma }_2(u_h,s)X_h(s)ds{|}^2]du$

$\le 2KT{\displaystyle {\int }_0^t}\left[𝔼\right|X(u)-X_h(u_h){|}^2+𝔼|Y(u)-X(u){|}^2$

$+𝔼|{\displaystyle {\int }_{u_h}^u}{\sigma }_2(u_h,s)X_h(s)ds{|}^2+𝔼|{\displaystyle {\int }_0^u}({\sigma }_2(u,s)Y(s)-{\sigma }_2(u_h,s)X_h(s))ds{|}^2]du.$

By Cauchy’s inequality, Itô isometry, LABEL:bond-thetam and Lemma 4.1,

${\displaystyle {\int }_0^t}𝔼{\left|{\displaystyle {\int }_{u_h}^u}{\sigma }_2(u_h,s)X_h(s)𝑑s\right|}^2𝑑u\le$ ${\displaystyle {\int }_0^t}\left[{\displaystyle {\int }_{u_h}^u}{|{\sigma }_2(u_h,s)|}^2𝑑s{\displaystyle {\int }_{u_h}^u}𝔼{|X_h(s)|}^2𝑑s\right]𝑑u$

$\le$ ${\displaystyle {\int }_0^t}\left[{\displaystyle {\int }_{u_h}^u}{\parallel {\sigma }_2\parallel }_{\mathrm{\infty }}^2𝑑s{\displaystyle {\int }_{u_h}^u}{K}_0𝑑s\right]𝑑u$

$\le$ $h^{2}T{\parallel {\sigma }_2\parallel }_{\mathrm{\infty }}^2{K}_0,$

by Cauchy’s inequality, LABEL:bond-thetam, Lemma 4.1 and ${(x+y+z)}^2\le 2x^2+2y^2+2z^2$, we can obtain

${\displaystyle {\int }_0^t}$ $\left[𝔼{\left|{\displaystyle {\int }_0^u}\left({\sigma }_2(u_h,s)X_h(s)-{\sigma }_2(u,s)Y(s)\right)𝑑s\right|}^2\right]du$

$={\displaystyle {\int }_0^t}\left[𝔼{\left|{\displaystyle {\int }_0^u}\left({\sigma }_2(u_h,s)X_h(s)-{\sigma }_2(u,s)[Y(s)-X_h(s)+X_h(s)-X(s)+X(s)]\right)𝑑s\right|}^2\right]𝑑u$

$\le 2{\displaystyle {\int }_0^t}{\displaystyle {\int }_0^u}|({\sigma }_2(u_h,s)-{\sigma }_2(u,s){|}^2𝔼|X_h(s){|}^{2}dsdu$

$+2{\displaystyle {\int }_0^t}{\displaystyle {\int }_0^u}{\left|{\sigma }_2(u,s)\right|}^2𝔼{\left|Y(s)-X(s)\right|}^2𝑑s𝑑u$

$+2{\displaystyle {\int }_0^t}{\displaystyle {\int }_0^u}{\left|{\sigma }_2(u,s)\right|}^2𝔼{\left|X(s)-X_h(s)\right|}^2𝑑s𝑑u$

$\le 2{\displaystyle {\int }_0^t}{\displaystyle {\int }_0^u}{K}^{\prime \prime }{h}^2{K}_0𝑑s𝑑u+2{\displaystyle {\int }_0^t}{\displaystyle {\int }_0^u}{\parallel {\sigma }_2\parallel }_{\mathrm{\infty }}^2𝔼{\left|Y(s)-X(s)\right|}^2𝑑s𝑑u$

$+2{\displaystyle {\int }_0^t}{\displaystyle {\int }_0^u}{\parallel {\sigma }_2\parallel }_{\mathrm{\infty }}^2{K}_{1}h𝑑s𝑑u$

$\le 2T^2{h}^2{K}^{\prime \prime }{K}_0+2T^{2}h{\parallel {\sigma }_2\parallel }_{\mathrm{\infty }}^2{K}_1+2T{\parallel {\sigma }_2\parallel }_{\mathrm{\infty }}^2{\displaystyle {\int }_0^u}𝔼{\left|Y(s)-X(s)\right|}^2𝑑s,$

Thus

(4.12) $$L_2\le L_{21}hT+L_{22}{\int }_0^u𝔼{\left|Y(s)-X(s)\right|}^2𝑑s,$$

where

$$L_{21}:=2KT\left[\left(h{K}_0+2T{K}_1\right){\parallel {\sigma }_2\parallel }_{\mathrm{\infty }}^{2}T+2Th{K}_0{K}^{\prime \prime }\right],$$

and

$$L_{22}:=2KT\left[1+2T{\parallel {\sigma }_2\parallel }_{\mathrm{\infty }}^2\right].$$

By compensating (4.5) and (4.6) in (4.4), we have

$$𝔼{|X(t)-Y(t)|}^2\le 2\left(L_{11}+L_{21}\right)hT+2(L_{12}+L_{22}){\int }_0^u𝔼{\left|Y(s)-X(s)\right|}^2𝑑s,$$

$L_{11},L_{12},L_{21}$ and $L_{22}$ knew it before. By Gronwall’s inequality, we have

$$𝔼{|X(t)-Y(t)|}^2\le K_{2}h,$$

where

$$K_2:=2T\left(L_{11}+L_{21}\right)\exp \left(2(L_{12}+L_{22})T\right).$$

$\mathrm{\square }$