Strong Convergence of the Truncated Euler–Maruyama Method for Nonlinear Stochastic Differential Equations with Jumps

Abstract

The truncated Euler–Maruyama (EM) method for stochastic differential equations with Poisson jumps (SDEwPJs) has been proposed by Deng et al. in 2019. Although the finite-time $L^r$-convergence theory has been established, the strong convergence theory remains absent. In this paper, the strong convergence refers to the use of an $L^2$ measure and places the supremum over time inside the expectation operation. Our version can be used to justify the method within Monte Carlo simulations that compute the expected payoff of financial products. Noting that the conditions imposed are too strict, this paper presents an existence and uniqueness theorem for SDEwPJs under general conditions and proves the convergence of the truncated EM method for these equations. Finally, two examples are considered to illustrate the application of the truncated EM method in option price calculation.

1. Introduction

Stochastic differential equations (SDEs) have become a potent tool applied to mathematics, physics, chemistry, biology, medical science, finance, stochastic control, and so on (see, e.g., [1,2,3]). In real life, however, most SDEs have no explicit solution as their coefficients do not satisfy the global Lipschitz condition. Several authors have found some methods to solve this issue when studying the numerical methods of stochastic differential equations, including the EM method, the Milstein method, the tamed EM method, the tamed Milstein method, the stopped EM, the backward EM, and the backward–forward EM. For the background of these methods, we refer the reader to the papers [4,5,6,7,8,9,10,11,12,13] and the book [14]. The convergence of numerical solutions must be considered, especially the strong convergence, most of which, however, require that the coefficients of SDEs satisfy the global Lipschitz condition. Therefore, it is of great significance to study the strong convergence theory of nonlinear SDEs whose coefficients do not satisfy the global Lipschitz condition.

Mao [15] proposed the truncated EM method to establish the strong convergence theory of nonlinear SDE, a new explicit method. A d-dimensional SDE

$$dx\left(t\right)=f\left(x\right(t\left)\right)dt+g\left(x\right(t\left)\right)dB\left(t\right)$$

(1)

was studied, but f and g only satisfy the local Lipschitz condition. A Khasminskii-type condition

$$2x^{\top }f\left(x\right)+{|g\left(x\right)|}^2\le {K(1+|x|}^2)$$

(2)

was added to guarantee a global solution (see [16,17]). Afterward, the truncated EM method for SDEs was reported. One can refer to [18,19,20,21,22,23], to name a few examples.

Although Deng et al. [23] have studied the finite-time convergence for SDEwPJs under some additional conditions using the truncated EM method, their technical requirements remain overly restrictive (see Remark 1). Furthermore, the authors did not establish stronger convergence properties beyond fundamental results. To address these limitations, this work develops innovative mathematical methodologies to derive both convergence and strong convergence criteria under substantially weakened assumptions. Notably, while our convergence analysis focuses specifically on the $L^2$ framework, this level of convergence proves fully adequate for practical implementations in quantitative finance, as demonstrated in existing applications by Higham and Mao [24].

Hu et al. [20] compared the truncated EM method with some modified EM methods for SDEs without jumps. To achieve the same accuracy, the runtime of the truncated EM method is much shorter than that of the implicit EM method. Although the runtime of the truncated EM method and of the tamed EM method are almost equivalent, the step size of the tamed EM method needs to be smaller to achieve the same accuracy. Thus, the truncated EM method has obvious advantages in applications. In this paper we intend to consider two examples to illustrate the application of the truncated EM method in option price calculation. It turns out that the truncated EM method is not only convenient for computations, but also has an acceptable convergence rate. Furthermore, these two examples illustrate that our main results could cover many nonlinear SDEwPJs.

The rest of the paper is organized as follows. Section 2 outlines some assumptions and details of the truncated EM method. Section 3 examines the moment bounds of the truncated EM solutions. Section 4 shows the convergence of the truncated EM solutions. Section 5 represents some strong convergence results. Finally, Section 6 provides some simulated examples to illustrate the theoretical results.

2. The Truncated EM Method

Throughout this paper, ${\mathbb{R}}_{+}$ denotes the family of non-negative real numbers, ${\mathbb{R}}^n$ denotes the n-dimensional Euclidean space, and ${\mathbb{R}}^{n\times m}$ denotes the space of $n\times m$ matrices with real entries. Let $(\Omega ,\mathcal{F},{\left\{{\mathcal{F}}_t\right\}}_{t\ge 0},\mathbb{P})$ be a complete probability space with a filtration ${\left\{{\mathcal{F}}_t\right\}}_{t\ge 0}$ satisfying the usual conditions, that is, it is right, continuous, and increasing while ${\mathcal{F}}_0$ contains all $\mathbb{P}$-null sets. Let $B\left(t\right)$ be an m-dimensional Brownian motion and $\tilde{N}\left(t\right)$ is a compensated Poisson process, which means $\tilde{N}\left(t\right)=N\left(t\right)-\lambda t$ in which $N\left(t\right)$ is a scalar Poisson process with intensity $\lambda >0$. We assume that $B\left(t\right),N\left(t\right),$ and $t\ge 0$ are independent and defined on $(\Omega ,\mathcal{F},{\left\{{\mathcal{F}}_t\right\}}_{t\ge 0},\mathbb{P})$. If G is a vector or matrix, its transpose is denoted by $G^{\top }$. If $x\in {\mathbb{R}}^n$, then $|x|$ is the Euclidean norm. If A is a set, its indicator function is denoted by ${\mathbf{1}}_A$, namely ${\mathbf{1}}_A\left(x\right)=1$ if $x\in A$ and 0 otherwise. We set $inf\varnothing =\infty$ (∅ denotes the empty set). For two real numbers, a and b, we use $a\vee b=\max \{a,b\}$ and $a\wedge b=\min \{a,b\}$. Moreover, let $\left[k\right]$ denote the largest integer which does not exceed k, and let $\mathbb{E}\left[Y\right]$ represent the expectation of the random variable Y.

We consider an n-dimensional nonlinear SDE with Poisson jumps

$$dx\left(t\right)=f\left(x\left(t\right)\right)dt+g\left(x\left(t\right)\right)dB\left(t\right)+u\left(x\left(t^{-}\right)\right)d\tilde{N}\left(t\right)$$

(3)

on $t\ge 0$ with the initial value $x\left(0\right)=x_0\in {\mathbb{R}}^n$, where $x\left(t^{-}\right)$ denotes ${lim}_{s\uparrow t}x\left(s\right)$, f: ${\mathbb{R}}^n\to {\mathbb{R}}^n$, g: ${\mathbb{R}}^n\to {\mathbb{R}}^{n\times m}$, u: ${\mathbb{R}}^n\to {\mathbb{R}}^n$. It should be noted that (3) could also be written as

$$dx\left(t\right)=[f\left(x\left(t\right)\right)-\lambda u\left(x\left(t\right)\right)]dt+g\left(x\left(t\right)\right)dB\left(t\right)+u\left(x\left(t^{-}\right)\right)dN\left(t\right).$$

(4)

In this paper, we propose two standing hypotheses.

Assumption 1 

(Local Lipschitz condition). For any $R>0$, there is a $K_R>0$, such that

$$|f\left(x\right)-f\left(y\right)|\vee |g\left(x\right)-g\left(y\right)|\vee |u\left(x\right)-u\left(y\right)|\le K_R|x-y|$$

(5)

for all $x,y\in {\mathbb{R}}^n$ with $|x|\vee |y|\le R$.

Assumption 2 

(Khasminskii-type condition). There is a $K>0$, such that

$$x^{\top }f\left(x\right)+{\displaystyle \frac{1}{2}}{|g\left(x\right)|}^2+{\displaystyle \frac{\lambda }{2}}{|u\left(x\right)|}^2\le {K(1+|x|}^2)$$

(6)

for all $x\in {\mathbb{R}}^n$.

Remark 1. 

A major difference from Assumptions 3.1 and 3.3 in Deng et al. [23] is that our assumptions are more simplified. Evidently, our Assumptions 1 and 2 are derivable from (3.3) and (3.8) in Deng et al. [23], but not vice versa. Therefore, our assumptions are weaker than those in Deng et al. [23].

The following theorem shows that there exists a unique solution to (3).

Theorem 1. 

Under Assumptions 1 and 2, Equation (3) has a unique global solution $x\left(t\right)$. Moreover, for $\forall T>0$,

$$\underset{0\le t\le T}{sup}{\mathbb{E}|x\left(t\right)|}^2\le (|x_0{|}^2+2KT)e^{2KT}.$$

(7)

Proof. 

Under Assumptions 1 and 2, the existence and uniqueness of the solution follows immediately from Theorem 1 in [25]. To show (7), we may rewrite (4) as

$$x\left(t\right)=x_0+{\int }_0^t[f\left(x\left(s\right)\right)-\lambda u\left(x\left(s\right)\right)]ds+{\int }_0^{t}g\left(x\left(s\right)\right)dB\left(s\right)+{\int }_0^{t}u\left(x\left(s^{-}\right)\right)dN\left(s\right).$$

(8)

By the Itô–Doeblin formula for one jump process (see Chapter 11 of [26]), we derive from (8) that, for $0\le t\le T$,

$$\begin{array}{cc}\hfill {|x\left(t\right)|}^2& = |x_0{|}^2+{\int }_0^t{[2x\left(s\right)}^{\top }f\left(x\left(s\right)\right)-2\lambda x{\left(s\right)}^{\top }u\left(x\left(s\right)\right)+{|g\left(x\left(s\right)\right)|}^2]ds\hfill \\ \hfill & +{\int }_0^{t}2x{\left(s\right)}^{\top }g\left(x\left(s\right)\right)dB\left(s\right)+{\int }_0^t[{(x\left(s^{-}\right)+u\left(x\left(s^{-}\right)\right))}^2-{\left(x\left(s^{-}\right)\right)}^2]dN\left(s\right).\hfill \end{array}$$

By Assumption 2, we derive that

$$\begin{array}{cc}\hfill {\mathbb{E}|x\left(t\right)|}^2& = |x_0{|}^2+\mathbb{E}{\int }_0^t{[2x\left(s\right)}^{\top }f\left(x\left(s\right)\right)-2\lambda x{\left(s\right)}^{\top }u\left(x\left(s\right)\right)+{|g\left(x\left(s\right)\right)|}^2]ds\hfill \\ \hfill & + \lambda \mathbb{E}{\int }_0^t{[|u\left(x\left(s\right)\right)|}^2+2x{\left(s\right)}^{\top }u\left(x\left(s\right)\right)]ds\hfill \\ \hfill & = |x_0{|}^2+\mathbb{E}{\int }_0^t{[2x\left(s\right)}^{\top }f\left(x\left(s\right)\right)+{|g\left(x\left(s\right)\right)|}^2+{\lambda |u\left(x\left(s\right)\right)|}^2]ds\hfill \\ \hfill & \le |x_0{|}^2+2K{\int }_0^t{\mathbb{E}(1+|x\left(s\right)}^2|)ds\le (|x_0{|}^2+2KT)+2K{\int }_0^t\mathbb{E}|x{\left(s\right)}^2|ds.\hfill \end{array}$$

Applying the Gronwall inequality yields the required result. □

In what follows, we adopt the similar notation and functions as introduced in [15]. To define the truncated EM scheme, we first choose a strictly increasing continuous function $\varphi :{\mathbb{R}}_{+}\to {\mathbb{R}}_{+}$, such that $\varphi \left(r\right)\to \infty$ as $r\to \infty$ and

$$\underset{|x|\le r}{sup}\left(|f(x)|\vee |g(x)|\vee |u(x)|\right)\le \varphi \left(r\right)$$

(9)

for all $r\ge 0$. We note that ${\varphi }^{-1}$ (denoting the inverse function of $\varphi$) is a strictly increasing continuous function from $\left[\varphi \right(0),\infty )$ to ${\mathbb{R}}_{+}$, so we can choose a number ${\Delta }^{*}\in (0,1]$ and a strictly decreasing function $l:(0,{\Delta }^{*}]\to (0,\infty )$, such that

$$l\left({\Delta }^{*}\right)\ge \varphi \left(2\right),\underset{\Delta \to 0}{lim}l(\Delta )=\infty \mathrm{and}{\Delta }^{1/4}l(\Delta )\le 1,\forall \Delta \in (0,1].$$

(10)

Given stepsize $\Delta \in (0,1]$, define the truncated functions

$$f_{\Delta }\left(x\right)=f\left((|x|\wedge {\varphi }^{-1}\left(l(\Delta )\right)){\displaystyle \frac{x}{|x|}}\right),g_{\Delta }\left(x\right)=g\left((|x|\wedge {\varphi }^{-1}\left(l(\Delta )\right)){\displaystyle \frac{x}{|x|}}\right)$$

and

$$u_{\Delta }\left(x\right)=u\left((|x|\wedge {\varphi }^{-1}\left(l(\Delta )\right)){\displaystyle \frac{x}{|x|}}\right)$$

(11)

for all $x\in {\mathbb{R}}^n$, where we set $x/|x|=0$ when $x=0$. We could easily observe that

$$|f_{\Delta }\left(x\right)|\vee |g_{\Delta }\left(x\right)|\vee |u_{\Delta }\left(x\right)|\le \varphi \left({\varphi }^{-1}\left(l(\Delta )\right)\right)=l(\Delta )$$

(12)

for all $x\in {\mathbb{R}}^n$.

We can now define the truncated EM numerical solutions for (3). For a given stepsize $\Delta >0$, define $t_k=k\Delta$ for $k=0,1,\cdots$. The discrete-time truncated EM numerical solutions $X_{\Delta }\left(t_k\right)\approx x\left(t_k\right)$ satisfying the iterative scheme

$$X_{\Delta }\left(t_{k+1}\right)=X_{\Delta }\left(t_k\right)+f_{\Delta }\left(X_{\Delta }\left(t_k\right)\right)\Delta +g_{\Delta }\left(X_{\Delta }\left(t_k\right)\right)\Delta B_k+u_{\Delta }\left(X_{\Delta }\left(t_k\right)\right)\Delta {\tilde{N}}_k$$

(13)

with initial $X_{\Delta }\left(0\right)=x_0$, where $\Delta B_k=B\left(t_{k+1}\right)-B\left(t_k\right)$, $\Delta {\tilde{N}}_k=\tilde{N}\left(t_{k+1}\right)-\tilde{N}\left(t_k\right)$.

In our analysis, it will be more convenient to use continuous-time truncated EM solutions. We hence introduce the step function ${\overline{x}}_{\Delta }\left(t\right)$, which is defined by

$${\overline{x}}_{\Delta }\left(t\right):=X_{\Delta }\left(t_k\right),t\in [t_k,t_{k+1}).$$

(14)

The continuous-time truncated EM approximation $x_{\Delta }\left(t\right)$ is defined by

$$x_{\Delta }\left(t\right):=x_0+{\int }_0^t{f}_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right)ds+{\int }_0^t{g}_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right)dB\left(s\right)+{\int }_0^t{u}_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right)d\tilde{N}\left(s\right)$$

(15)

for $t\ge 0$.

In addition, for $t\in [t_k,t_{k+1})$, the continuous-time truncated EM approximation can be written in the following equivalent form:

$$\begin{array}{cc}\hfill x_{\Delta }\left(t\right)& = x_{\Delta }\left(t_k\right)+f_{\Delta }\left({\overline{x}}_{\Delta }\left(t_k\right)\right)(t-t_k)+g_{\Delta }\left({\overline{x}}_{\Delta }\left(t_k\right)\right)(B\left(t\right)-B\left(t_k\right))\hfill \\ \hfill & + u_{\Delta }\left({\overline{x}}_{\Delta }\left(t_k\right)\right)(\tilde{N}\left(t\right)-\tilde{N}\left(t_k\right)).\hfill \end{array}$$

(16)

Noting that $x_{\Delta }\left(t_k\right)$, ${\overline{x}}_{\Delta }\left(t_k\right)$ and $X_{\Delta }\left(t_k\right)$ coincide with the discrete solution at the gridpoints, we have $x_{\Delta }\left(t_k\right)={\overline{x}}_{\Delta }\left(t_k\right)=X_{\Delta }\left(t_k\right)$ for all $k\ge 0$.

At the end of this section, we state a lemma. The following lemma shows that the truncated functions $f_{\Delta }$, $g_{\Delta }$ and $u_{\Delta }$ preserve Assumption 2 for all $\Delta \in (0,{\Delta }^{*}]$.

Lemma 1. 

Let Assumption 2 hold. Then, for all $\Delta \in (0,{\Delta }^{*}]$, we have

$$x^{\top }{f}_{\Delta }\left(x\right)+{\displaystyle \frac{1}{2}}|g_{\Delta }{\left(x\right)|}^2+{\displaystyle \frac{\lambda }{2}}|u_{\Delta }{\left(x\right)|}^2\le {2K(1+|x|}^2)$$

(17)

for all $x\in {\mathbb{R}}^n$.

Proof. 

Fix any $\Delta \in (0,{\Delta }^{*}]$. For $|x|\le {\varphi }^{-1}\left(l(\Delta )\right)$, by Assumption 2 and (14), we derive that

$$I:=x^{\top }{f}_{\Delta }\left(x\right)+{\displaystyle \frac{1}{2}}|g_{\Delta }{\left(x\right)|}^2+{\displaystyle \frac{\lambda }{2}}|u_{\Delta }{\left(x\right)|}^2=x^{\top }f\left(x\right)+{\displaystyle \frac{1}{2}}{|g\left(x\right)|}^2+{\displaystyle \frac{\lambda }{2}}{|u\left(x\right)|}^2\le {K(1+|x|}^2),$$

and hence, the required assertion (17) follows. If $|x|>{\varphi }^{-1}\left(l(\Delta )\right)$, by Assumption 2, we have

$$\begin{array}{cc}\hfill I& =x^{\top }f\left({\varphi }^{-1}\left(l(\Delta )\right){\displaystyle \frac{x}{|x|}}\right)+{\displaystyle \frac{1}{2}}{\left|g\left({\varphi }^{-1}\left(l(\Delta )\right){\displaystyle \frac{x}{|x|}}\right)\right|}^2+{\displaystyle \frac{\lambda }{2}}{\left|u\left({\varphi }^{-1}\left(l(\Delta )\right){\displaystyle \frac{x}{|x|}}\right)\right|}^2\hfill \\ \hfill & ={\varphi }^{-1}\left(l(\Delta )\right){\displaystyle \frac{x^{\top }}{|x|}}f\left({\varphi }^{-1}\left(l(\Delta )\right){\displaystyle \frac{x}{|x|}}\right)+{\displaystyle \frac{1}{2}}{\left|g\left({\varphi }^{-1}\left(l(\Delta )\right){\displaystyle \frac{x}{|x|}}\right)\right|}^2\hfill \\ \hfill & +{\displaystyle \frac{\lambda }{2}}{\left|u\left({\varphi }^{-1}\left(l(\Delta )\right){\displaystyle \frac{x}{|x|}}\right)\right|}^2+\left({\displaystyle \frac{|x|}{{\varphi }^{-1}\left(l(\Delta )\right)}}-1\right){\varphi }^{-1}\left(l(\Delta )\right){\displaystyle \frac{x^{\top }}{|x|}}f\left({\varphi }^{-1}\left(l(\Delta )\right){\displaystyle \frac{x}{|x|}}\right)\hfill \\ \hfill & \le K(1+{\left[{\varphi }^{-1}\left(l(\Delta )\right)\right]}^2)+\left({\displaystyle \frac{|x|}{{\varphi }^{-1}\left(l(\Delta )\right)}}-1\right){\varphi }^{-1}\left(l(\Delta )\right){\displaystyle \frac{x^{\top }}{|x|}}f\left({\varphi }^{-1}\left(l(\Delta )\right){\displaystyle \frac{x}{|x|}}\right).\hfill \end{array}$$

Since ${\varphi }^{-1}$ is a strictly increasing continuous function and l is a strictly decreasing function, by (10), we have

$${\varphi }^{-1}\left(l(\Delta )\right)\ge {\varphi }^{-1}\left(l\left({\Delta }^{*}\right)\right)\ge 2,\forall \Delta \in (0,{\Delta }^{*}].$$

We also see from Assumption 2 that $x^{T}f\left(x\right)\le {K(1+|x|}^2)$ for any $x\in {\mathbb{R}}^n$. Therefore, we have

$$\begin{array}{cc}\hfill I& \le K(1+{\left[{\varphi }^{-1}\left(l(\Delta )\right)\right]}^2)+\left({\displaystyle \frac{|x|}{{\varphi }^{-1}\left(l(\Delta )\right)}}-1\right)K(1+{\left[{\varphi }^{-1}\left(l(\Delta )\right)\right]}^2)\hfill \\ \hfill & ={\displaystyle \frac{|x|}{{\varphi }^{-1}\left(l(\Delta )\right)}}K(1+{\left[{\varphi }^{-1}\left(l(\Delta )\right)\right]}^2)\le K|x|(\frac{1}{2}+{\varphi }^{-1}\left(l(\Delta )\right))\hfill \\ \hfill & \le K|x|(\frac{1}{2}+|x|)\le {2K(1+|x|}^2).\hfill \end{array}$$

The proof is complete. □

3. Moment Bounds of the Truncated EM Solutions

In this section, we will examine the moment bounds of the discrete approximation solutions. We first present a lemma which gives an upper bound for the second moment of $x_{\Delta }\left(t\right)$.

Lemma 2. 

For any given $T>0$, we have

$$\underset{0\le t\le T}{sup}\mathbb{E}|x_{\Delta }{\left(t\right)|}^2\le 4{|x_0|}^2+4T(1+T+\lambda ){\left(l(\Delta )\right)}^2.$$

(18)

Proof. 

For any $0\le t\le T$, according to (15), we may easily observe that

$$\begin{array}{cc}\hfill |x_{\Delta }{\left(t\right)|}^2& \le 4|x_0{|}^2+4{\left|{\int }_0^t{f}_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right)ds\right|}^2+4{\left|{\int }_0^t{g}_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right)dB\left(s\right)\right|}^2\hfill \\ \hfill & +4{\left|{\int }_0^t{u}_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right)d\tilde{N}\left(s\right)\right|}^2.\hfill \end{array}$$

By (12), the Cauchy–Schwarz inequality, and the martingale isometry, we have

$$\begin{array}{cc}\hfill \mathbb{E}|x_{\Delta }{\left(t\right)|}^2& \le 4|x_0{|}^2+4t^2{\left(l(\Delta )\right)}^2+4\mathbb{E}{\int }_0^t{\left(l(\Delta )\right)}^{2}ds+4\lambda \mathbb{E}{\int }_0^t{\left(l(\Delta )\right)}^{2}ds\hfill \\ \hfill & \le 4|x_0{|}^2+4t^2{\left(l(\Delta )\right)}^2+4t{\left(l(\Delta )\right)}^2+4\lambda t{\left(l(\Delta )\right)}^2.\hfill \end{array}$$

Taking the supremum t over $[0,T]$ yields the required result. □

Lemma 3. 

For any $\Delta \in (0,{\Delta }^{*}]$ and $t\ge 0$, we have

$$\mathbb{E}|x_{\Delta }\left(t\right)-{\overline{x}}_{\Delta }{\left(t\right)|}^2\le 3{\Delta }^{{\displaystyle \frac{1}{2}}}\left(\Delta +1+\lambda \right).$$

(19)

Consequently,

$$\underset{\Delta \to 0}{lim}\mathbb{E}{|x_{\Delta }\left(t\right)-{\overline{x}}_{\Delta }\left(t\right)|}^2=0.$$

(20)

Proof. 

Fix any $\Delta \in (0,{\Delta }^{*}]$ and $t\ge 0$. For $t\in [t_k,t_{k+1})$, by the Cauchy–Schwarz inequality, the martingale isometry, and (12), we have

$$\begin{array}{cc}\hfill \mathbb{E}|x_{\Delta }\left(t\right)-{\overline{x}}_{\Delta }{\left(t\right)|}^2& =\mathbb{E}|x_{\Delta }\left(t\right)-x_{\Delta }\left(t_k\right){|}^2\hfill \\ \hfill & =\mathbb{E}{\left|{\int }_{t_k}^t{f}_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right)ds+{\int }_{t_k}^t{g}_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right)dB\left(s\right)+{\int }_{t_k}^t{u}_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right)d\tilde{N}\left(s\right)\right|}^2\hfill \\ \hfill & \le 3\mathbb{E}{\int }_{t_k}^t\left[\Delta |f_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right){|}^2+|g_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right){|}^2+\lambda {|u_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right)|}^2\right]ds\hfill \\ \hfill & \le 3{\Delta }^2{\left(l(\Delta )\right)}^2+3\Delta {\left(l(\Delta )\right)}^2+3\lambda \Delta {\left(l(\Delta )\right)}^2\hfill \\ \hfill & =3\Delta {\left(l(\Delta )\right)}^2(\Delta +1+\lambda ).\hfill \end{array}$$

Noting that $\Delta {\left(l(\Delta )\right)}^2\le {\Delta }^{{\displaystyle \frac{1}{2}}}$, we derive (19) immediately. Then, letting $\Delta \to 0$ yields the other assertion (20). This completes the proof. □

For later use, we will show the following theorem, which gives an upper bound for the second moment of $x_{\Delta }\left(t\right)$ for any $\Delta \in (0,{\Delta }^{*}]$.

Theorem 2. 

Let Assumptions 1 and 2 hold. For any $T>0$, we have

$$\underset{0<\Delta \le {\Delta }^{*}}{sup}\underset{0\le t\le T}{sup}\mathbb{E}{|x_{\Delta }\left(t\right)|}^2\le \left[|x_0{|}^2+4KT+2\sqrt{3}T{\left({\Delta }^{*}+1+\lambda \right)}^{{\displaystyle \frac{1}{2}}}\right]e^{4KT}.$$

(21)

Proof. 

Fix any $\Delta \in (0,{\Delta }^{*}]$ and $T>0$. Using the Itô–Doeblin formula, we derive from (15) that, for $0\le t\le T$,

$$\begin{array}{cc}\hfill |x_{\Delta }{\left(t\right)|}^2& = |x_0{|}^2+{\int }_0^t\left[2x_{\Delta }{\left(s\right)}^{\top }{f}_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right)-2\lambda x_{\Delta }{\left(s\right)}^{\top }{u}_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right)+{|g_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right)|}^2\right]ds\hfill \\ \hfill & + {\int }_0^{t}2x_{\Delta }{\left(s\right)}^{\top }{g}_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right)dB\left(s\right)+{\int }_0^t\left[{(x_{\Delta }\left(s^{-}\right)+u_{\Delta }\left({\overline{x}}_{\Delta }\left(s^{-}\right)\right))}^2-{\left(x_{\Delta }\left(s^{-}\right)\right)}^2\right]dN\left(s\right).\hfill \end{array}$$

Then, we have

$$\begin{array}{cc}\hfill \mathbb{E}|x_{\Delta }{\left(t\right)|}^2& = |x_0{|}^2+\mathbb{E}{\int }_0^t\left[2x_{\Delta }{\left(s\right)}^{\top }{f}_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right)+|g_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right){|}^2+\lambda {|u_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right)|}^2\right]ds\hfill \\ \hfill & = |x_0{|}^2+\mathbb{E}{\int }_0^t\left[2{\overline{x}}_{\Delta }{\left(s\right)}^{\top }{f}_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right)+|g_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right){|}^2+\lambda {|u_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right)|}^2\right]ds\hfill \\ \hfill & + \mathbb{E}{\int }_0^{t}2{\left[x_{\Delta }\left(s\right)-{\overline{x}}_{\Delta }\left(s\right)\right]}^{\top }{f}_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right)ds.\hfill \end{array}$$

By Lemma 1, (12), (19), and (10), we obtain

$$\begin{array}{cc}\hfill \mathbb{E}|x_{\Delta }{\left(t\right)|}^2& \le |x_0{|}^2+4KT+4K{\int }_0^t\mathbb{E}|{\overline{x}}_{\Delta }{\left(s\right)|}^{2}ds+2\mathbb{E}{\int }_0^t\left|x_{\Delta }\left(s\right)-{\overline{x}}_{\Delta }\left(s\right)\right||f_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right)|ds\hfill \\ \hfill & \le |x_0{|}^2+4KT+2\sqrt{3}T{\left(\Delta +1+\lambda \right)}^{{\displaystyle \frac{1}{2}}}+4K{\int }_0^t\left(\underset{0\le r\le s}{sup}\mathbb{E}{|x_{\Delta }\left(r\right)|}^2\right)ds.\hfill \end{array}$$

Since the last sum is non-decreasing in $t\in [0,T]$, we have

$$\underset{0\le r\le t}{sup}\mathbb{E}|x_{\Delta }{\left(r\right)|}^2\le {|x_0|}^2+4KT+2\sqrt{3}T{\left(\Delta +1+\lambda \right)}^{{\displaystyle \frac{1}{2}}}+4K{\int }_0^t\left(\underset{0\le r\le s}{sup}\mathbb{E}{|x_{\Delta }\left(r\right)|}^2\right)ds.$$

Applying the Gronwall inequality yields that

$$\underset{0\le r\le T}{sup}\mathbb{E}{|x_{\Delta }\left(r\right)|}^2\le \left[|x_0{|}^2+4KT+2\sqrt{3}T{\left(\Delta +1+\lambda \right)}^{{\displaystyle \frac{1}{2}}}\right]e^{4KT}.$$

As this holds for any $\Delta \in (0,{\Delta }^{*}]$, we obtain the required assertion (21). □

4. Convergence

In this section, we will prove that

$$\underset{\Delta \to 0}{lim}\mathbb{E}{\left|x_{\Delta }\left(T\right)-x\left(T\right)\right|}^2=0\mathrm{and}\underset{\Delta \to 0}{lim}\mathbb{E}{\left|{\overline{x}}_{\Delta }\left(T\right)-x\left(T\right)\right|}^2=0$$

for any $T>0$. This is sufficient for some applications. For example, we utilize the truncated EM method to approximate the European call or put option value. In the rest of the paper, we always fix $T>0$ arbitrarily.

Lemma 4. 

Let Assumptions 1 and 2 hold. For any real number $R>|x_0|$, define the stopping time $\tau =inf\{t\ge 0:|x\left(t\right)|\ge R\}.$ Then,

$$\mathbb{P}\left(\tau \le T\right)\le {\displaystyle \frac{1}{R^2}}\left(|x_0{|}^2+2KT\right)e^{2KT}.$$

(22)

Proof. 

For any $0\le t\le T$, using the Itô–Doeblin formula, we have

$$\begin{array}{cc}\hfill {|x(t\wedge \tau )|}^2& = |x_0{|}^2+{\int }_0^{t\wedge \tau }\left[2x{\left(s\right)}^{\top }f\left(x\left(s\right)\right)-2\lambda x{\left(s\right)}^{\top }u\left(x\left(s\right)\right)+{|g\left(x\left(s\right)\right)|}^2\right]ds\hfill \\ \hfill & + {\int }_0^{t\wedge \tau }2x{\left(s\right)}^{\top }g\left(x\left(s\right)\right)dB\left(s\right)+{\int }_0^{t\wedge \tau }\left[2x{\left(s^{-}\right)}^{\top }u\left(x\left(s^{-}\right)\right)+{\left(u\left(x\left(s^{-}\right)\right)\right)}^2\right]dN\left(s\right).\hfill \end{array}$$

By Assumption 2, we derive that

$$\begin{array}{cc}\hfill {\mathbb{E}|x(t\wedge \tau )|}^2& = |x_0{|}^2+\mathbb{E}{\int }_0^{t\wedge \tau }\left[2x{\left(s\right)}^{\top }f\left(x\left(s\right)\right)+{|g\left(x\left(s\right)\right)|}^2+\lambda {|u\left(x\left(s\right)\right)|}^2\right]ds\hfill \\ \hfill & \le |x_0{|}^2+2K\mathbb{E}{\int }_0^{t\wedge \tau }\left(1+{|x\left(s\right)|}^2\right)ds\le |x_0{|}^2+2KT+2K{\int }_0^t\mathbb{E}{|x(s\wedge \tau )|}^{2}ds.\hfill \end{array}$$

The Gronwall inequality yields that

$${\mathbb{E}|x(T\wedge \tau )|}^2\le \left(|x_0{|}^2+2KT\right)e^{2KT}.$$

Then, by the Chebyshev inequality, we have

$$\mathbb{P}\left(|x(T\wedge \tau )|\ge R\right)\le {\displaystyle \frac{1}{R^2}}\left(|x_0{|}^2+2KT\right)e^{2KT}.$$

This implies

$$\mathbb{P}\left(\tau \le T\right)\le {\displaystyle \frac{1}{R^2}}\left(|x_0{|}^2+2KT\right)e^{2KT}.$$

The proof is therefore complete. □

Lemma 5. 

Let Assumptions 1 and 2 hold. For any real number $R>|x_0|$ and $\Delta \in (0,{\Delta }^{*}]$, define the stopping time $\rho =inf\{t\ge 0:|x_{\Delta }\left(t\right)|\ge R\}.$ Then,

$$\mathbb{P}\left(\rho \le T\right)\le {\displaystyle \frac{1}{R^2}}\left[|x_0{|}^2+4KT+24KT{\Delta }^{{\displaystyle \frac{1}{2}}}\left(\Delta +1+\lambda \right)+2\sqrt{3}T{\left(\Delta +1+\lambda \right)}^{{\displaystyle \frac{1}{2}}}\right]e^{8KT}.$$

(23)

Proof. 

For any $0\le t\le T$, using the Itô–Doeblin formula, we have

$$\begin{array}{cc}\hfill \mathbb{E}|x_{\Delta }{(t\wedge \rho )|}^2& = |x_0{|}^2+\mathbb{E}{\int }_0^{t\wedge \rho }\left[2x_{\Delta }{\left(s\right)}^{\top }{f}_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right)+|g_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right){|}^2+\lambda {|u_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right)|}^2\right]ds\hfill \\ \hfill & = |x_0{|}^2+\mathbb{E}{\int }_0^{t\wedge \rho }\left[2{\overline{x}}_{\Delta }{\left(s\right)}^{\top }{f}_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right)+|g_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right){|}^2+\lambda {|u_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right)|}^2\right]ds\hfill \\ \hfill & + \mathbb{E}{\int }_0^{t\wedge \rho }2{\left[x_{\Delta }\left(s\right)-{\overline{x}}_{\Delta }\left(s\right)\right]}^{\top }{f}_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right)ds.\hfill \end{array}$$

By Lemma 1, we may obtain

$$\begin{array}{cc}\hfill \mathbb{E}|x_{\Delta }{(t\wedge \rho )|}^2& \le |x_0{|}^2+4K\mathbb{E}{\int }_0^{t\wedge \rho }\left(1+|{\overline{x}}_{\Delta }{\left(s\right)|}^2\right)ds+2\mathbb{E}{\int }_0^{t\wedge \rho }\left|x_{\Delta }\left(s\right)-{\overline{x}}_{\Delta }\left(s\right)\right|\left|f_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right)\right|ds\hfill \\ \hfill & \le |x_0{|}^2+4KT+8K{\int }_0^t\mathbb{E}{\left|x_{\Delta }(s\wedge \rho )\right|}^{2}ds+8K{\int }_0^T\mathbb{E}{\left|x_{\Delta }\left(s\right)-{\overline{x}}_{\Delta }\left(s\right)\right|}^{2}ds\hfill \\ \hfill & + 2\mathbb{E}{\int }_0^T\left|x_{\Delta }\left(s\right)-{\overline{x}}_{\Delta }\left(s\right)\right|\left|f_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right)\right|ds.\hfill \end{array}$$

Then, by Lemma 3, (12), and the Hölder inequality, we have

$$\begin{array}{cc}\hfill \mathbb{E}|x_{\Delta }{(t\wedge \rho )|}^2& \le |x_0{|}^2+4KT+8K{\int }_0^t\mathbb{E}{\left|x_{\Delta }(s\wedge \rho )\right|}^{2}ds+24KT{\Delta }^{{\displaystyle \frac{1}{2}}}\left(\Delta +1+\lambda \right)\hfill \\ \hfill & + 2l(\Delta ){\int }_0^T{\left(\mathbb{E}{\left|x_{\Delta }\left(s\right)-{\overline{x}}_{\Delta }\left(s\right)\right|}^2\right)}^{{\displaystyle \frac{1}{2}}}ds\hfill \\ \hfill & \le |x_0{|}^2+4KT+24KT{\Delta }^{{\displaystyle \frac{1}{2}}}\left(\Delta +1+\lambda \right)+2\sqrt{3}T{\left(\Delta +1+\lambda \right)}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill & + 8K{\int }_0^t\mathbb{E}{\left|x_{\Delta }(s\wedge \rho )\right|}^{2}ds.\hfill \end{array}$$

Hence, by the Gronwall inequality, we derive that

$$\mathbb{E}|x_{\Delta }{(T\wedge \rho )|}^2\le \left[|x_0{|}^2+4KT+24KT{\Delta }^{{\displaystyle \frac{1}{2}}}\left(\Delta +1+\lambda \right)+2\sqrt{3}T{\left(\Delta +1+\lambda \right)}^{{\displaystyle \frac{1}{2}}}\right]e^{8KT}.$$

Finally, by the Chebyshev inequality, we can show the assertion (23) in the same way as Lemma 4. □

Theorem 3.

Let Assumptions 1 and 2 hold. Then,

$$\underset{\Delta \to 0}{lim}\mathbb{E}|x_{\Delta }{\left(T\right)-x\left(T\right)|}^2=0and\underset{\Delta \to 0}{lim}\mathbb{E}{|{\overline{x}}_{\Delta }\left(T\right)-x\left(T\right)|}^2=0.$$

(24)

Proof. 

Let $\epsilon >0$ be arbitrary and $\theta =\tau \wedge \rho$. We hence have

$$\begin{array}{cc}\hfill \mathbb{E}|x_{\Delta }{\left(T\right)-x\left(T\right)|}^2& =\mathbb{E}\left(|x_{\Delta }{\left(T\right)-x\left(T\right)|}^2{\mathbf{1}}_{\{\theta >T\}}\right)+\mathbb{E}\left(|x_{\Delta }{\left(T\right)-x\left(T\right)|}^2{\mathbf{1}}_{\{\theta \le T\}}\right).\hfill \end{array}$$

(25)

By Lemmas 4 and 5, we obtain

$$\mathbb{P}(\theta \le T)\le \mathbb{P}(\tau \le T)+\mathbb{P}(\rho \le T)\le {\displaystyle \frac{A}{R^2}},$$

where

$$A=\left(|x_0{|}^2+2KT\right)e^{2KT}+\left[|x_0{|}^2+4KT+24KT{\Delta }^{{\displaystyle \frac{1}{2}}}\left(\Delta +1+\lambda \right)2\sqrt{3}T{\left(\Delta +1+\lambda \right)}^{{\displaystyle \frac{1}{2}}}\right]e^{8KT}.$$

By Theorems 1 and 2, we can easily find that

$$\mathbb{E}|x_{\Delta }{\left(T\right)-x\left(T\right)|}^2\le 2\mathbb{E}\left(|x_{\Delta }{\left(T\right)|}^2+{|x\left(T\right)|}^2\right)<\infty .$$

Furthermore, for any $\gamma >0$, we have

$$\mathbb{E}\left(|x_{\Delta }{\left(T\right)-x\left(T\right)|}^2{\mathbf{1}}_{\{\theta \le T\}}\right)\le \gamma \mathbb{P}(\theta \le T)+{\int }_{\{|x_{\Delta }\left(T\right)-x\left(T\right){|}^2\ge \gamma \}}{|x_{\Delta }\left(T\right)-x\left(T\right)|}^{2}d\mathbb{P}.$$

Now, choose $R=R\left(\epsilon \right)>0$ so large that the first term on the right of this inequality is smaller than $\epsilon /4$. With this value of $R=R\left(\epsilon \right)$, the second term goes to zero as $\gamma \to \infty$ by Lebesgue’s dominated convergence theorem. Thus, there are numbers $R\left(\epsilon \right)$ and $\gamma \left(\epsilon \right)$, such that for all $\gamma \ge \gamma \left(\epsilon \right)$,

$$\begin{array}{cc}\hfill \mathbb{E}\left(|x_{\Delta }{\left(T\right)-x\left(T\right)|}^2{\mathbf{1}}_{\{\theta \le T\}}\right)& \le {\displaystyle \frac{\epsilon }{2}}.\hfill \end{array}$$

If we can show that for all sufficiently small $\Delta$,

$$\mathbb{E}\left(|x_{\Delta }{\left(T\right)-x\left(T\right)|}^2{\mathbf{1}}_{\{\theta >T\}}\right)\le {\displaystyle \frac{\epsilon }{2}},$$

(26)

we then have

$$\begin{array}{c}\hfill \underset{\Delta \to 0}{lim}\mathbb{E}{|x_{\Delta }\left(T\right)-x\left(T\right)|}^2=0.\end{array}$$

In order to prove (26), we define the truncated functions

$$F_R\left(x\right)=f\left((|x|\wedge R){\displaystyle \frac{x}{|x|}}\right),G_R\left(x\right)=g\left((|x|\wedge R){\displaystyle \frac{x}{|x|}}\right)\mathrm{and}{U}_R\left(x\right)=u\left((|x|\wedge R){\displaystyle \frac{x}{|x|}}\right)$$

for all $x\in {\mathbb{R}}^n$. Without loss of generality, we may assume that ${\Delta }^{*}$ is already sufficiently small for ${\varphi }^{-1}\left(l\left({\Delta }^{*}\right)\right)\ge R$. So, for all $\Delta \in (0,{\Delta }^{*}]$, we derive that

$$F_R\left(x\right)=f_{\Delta }\left(x\right)=f\left(x\right),G_R\left(x\right)=g_{\Delta }\left(x\right)=g\left(x\right)\mathrm{and}{U}_R\left(x\right)=u_{\Delta }\left(x\right)=u\left(x\right)$$

for all $x\in {\mathbb{R}}^n$ with $|x|\le R$. Consider the jump-diffusion SDE

$$dz\left(t\right)=F_R\left(z\left(t\right)\right)dt+G_R\left(z\left(t\right)\right)dB\left(t\right)+U_R\left(z\left(t^{-}\right)\right)d\tilde{N}\left(t\right)$$

(27)

on $t\ge 0$ with initial value $z\left(0\right)=x_0$. By Assumption 1, it is easy to observe that the coefficients of (27) are globally Lipschitz continuous with the Lipschitz constant $K_R$. Then, in the same way as Theorem 1, we can show that the jump-diffusion SDE (27) has a unique global solution $z\left(t\right)$ on $t\ge 0$. For each stepsize $\Delta \in (0,{\Delta }^{*}]$, we can apply the EM method to the jump-diffusion SDE (27) and we denote by $z_{\Delta }\left(t\right)$ the continuous-time EM solution. We also denote by ${\overline{z}}_{\Delta }\left(t\right)$ the step function. Hence, we have

$$z\left(t\right)=x_0+{\int }_0^t{F}_R\left(z\left(s\right)\right)ds+{\int }_0^t{G}_R\left(z\left(s\right)\right)dB\left(s\right)+{\int }_0^t{U}_R\left(z\left(s^{-}\right)\right)d\tilde{N}\left(s\right)$$

(28)

and

$$z_{\Delta }\left(t\right)=x_0+{\int }_0^t{F}_R\left({\overline{z}}_{\Delta }\left(s\right)\right)ds+{\int }_0^t{G}_R\left({\overline{z}}_{\Delta }\left(s\right)\right)dB\left(s\right)+{\int }_0^t{U}_R\left({\overline{z}}_{\Delta }\left(s^{-}\right)\right)d\tilde{N}\left(s\right).$$

(29)

Moreover, for $t\in [t_k,t_{k+1})$,

$$\begin{array}{cc}\hfill z_{\Delta }\left(t\right)& ={\overline{z}}_{\Delta }\left(t\right)+F_R\left({\overline{z}}_{\Delta }\left(t\right)\right)(t-t_k)+G_R\left({\overline{z}}_{\Delta }\left(t\right)\right)(B\left(t\right)-B\left(t_k\right))+U_R\left({\overline{z}}_{\Delta }\left(t\right)\right)(\tilde{N}\left(t\right)-\tilde{N}\left(t_k\right)).\hfill \end{array}$$

(30)

We then have

$$\begin{array}{cc}\hfill \mathbb{E}|z_{\Delta }\left(t\right)-{\overline{z}}_{\Delta }{\left(t\right)|}^2& \le {\left(l(\Delta )\right)}^2\mathbb{E}{\left[(t-t_k)+(B\left(t\right)-B\left(t_k\right))+(\tilde{N}\left(t\right)-\tilde{N}\left(t_k\right))\right]}^2\hfill \\ \hfill & \le 3{\left(l(\Delta )\right)}^2[{(t-t_k)}^2+(t-t_k)+\lambda (t-t_k)]\hfill \\ \hfill & \le 3{\left(l(\Delta )\right)}^2({\Delta }^2+\Delta +\lambda \Delta )=3{\left({\Delta }^{{\displaystyle \frac{1}{4}}}l(\Delta )\right)}^2\left({\Delta }^{{\displaystyle \frac{3}{2}}}+{\Delta }^{{\displaystyle \frac{1}{2}}}+\lambda {\Delta }^{{\displaystyle \frac{1}{2}}}\right)\hfill \\ \hfill & \le 3\left({\Delta }^{{\displaystyle \frac{3}{2}}}+{\Delta }^{{\displaystyle \frac{1}{2}}}+\lambda {\Delta }^{{\displaystyle \frac{1}{2}}}\right).\hfill \end{array}$$

(31)

By (28) and (29), we obtain

$$\begin{array}{cc}\hfill |z\left(t\right)-z_{\Delta }{\left(t\right)|}^2& \le 3{\left[{\int }_0^t(F_R\left(z\left(s\right)\right)-F_R\left({\overline{z}}_{\Delta }\left(s\right)\right))ds\right]}^2+3{\left[{\int }_0^t(G_R\left(z\left(s\right)\right)-G_R\left({\overline{z}}_{\Delta }\left(s\right)\right))dB\left(s\right)\right]}^2\hfill \\ \hfill & +3{\left[{\int }_0^t(U_R\left(z\left(s^{-}\right)\right)-U_R\left({\overline{z}}_{\Delta }\left(s^{-}\right)\right))d\tilde{N}\left(s\right)\right]}^2.\hfill \end{array}$$

For any $t_1\in [0,T]$, we derive that

$$\begin{array}{cc}\hfill \mathbb{E}\left[\underset{0\le t\le t_1}{sup}{|z\left(t\right)-z_{\Delta }\left(t\right)|}^2\right]& \le 3\mathbb{E}\left[\underset{0\le t\le t_1}{sup}{\left({\int }_0^t(F_R\left(z\left(s\right)\right)-F_R\left({\overline{z}}_{\Delta }\left(s\right)\right))ds\right)}^2\right]\hfill \\ \hfill & +3\mathbb{E}\left[\underset{0\le t\le t_1}{sup}{\left({\int }_0^t(G_R\left(z\left(s\right)\right)-G_R\left({\overline{z}}_{\Delta }\left(s\right)\right))dB\left(s\right)\right)}^2\right]\hfill \\ \hfill & +3\mathbb{E}\left[\underset{0\le t\le t_1}{sup}{\left({\int }_0^t(U_R\left(z\left(s^{-}\right)\right)-U_R\left({\overline{z}}_{\Delta }\left(s^{-}\right)\right))d\tilde{N}\left(s\right)\right)}^2\right].\hfill \end{array}$$

By the Cauchy–Schwarz inequality, the Doob martingale inequality, and the martingale isometry,

$$\begin{array}{cc}\hfill & \mathbb{E}\left[\underset{0\le t\le t_1}{sup}{|z\left(t\right)-z_{\Delta }\left(t\right)|}^2\right]\hfill \\ \hfill & \le 3T\mathbb{E}{\int }_0^{t_1}|F_R\left(z\left(s\right)\right)-F_R\left({\overline{z}}_{\Delta }\left(s\right)\right){|}^{2}ds+12\mathbb{E}{\int }_0^{t_1}{|G_R\left(z\left(s\right)\right)-G_R\left({\overline{z}}_{\Delta }\left(s\right)\right)|}^{2}ds\hfill \\ \hfill & +12\lambda \mathbb{E}{\int }_0^{t_1}{|U_R\left(z\left(s\right)\right)-U_R\left({\overline{z}}_{\Delta }\left(s\right)\right)|}^{2}ds\hfill \\ \hfill & \le 3K_R^2(T+4+4\lambda )\mathbb{E}{\int }_0^{t_1}{|z\left(s\right)-{\overline{z}}_{\Delta }\left(s\right)|}^{2}ds\hfill \\ \hfill & \le 6K_R^2(T+4+4\lambda )\mathbb{E}{\int }_0^{t_1}|z\left(s\right)-z_{\Delta }{\left(s\right)|}^{2}ds+6K_R^2(T+4+4\lambda )\mathbb{E}{\int }_0^{t_1}{|z_{\Delta }\left(s\right)-{\overline{z}}_{\Delta }\left(s\right)|}^{2}ds.\hfill \end{array}$$

Using (31), we get

$$\begin{array}{cc}\hfill \mathbb{E}\left[\underset{0\le t\le t_1}{sup}{|z\left(t\right)-z_{\Delta }\left(t\right)|}^2\right]& \le 6K_R^2(T+4+4\lambda ){\int }_0^{t_1}\mathbb{E}\left[\underset{0\le t\le s}{sup}{|z\left(t\right)-z_{\Delta }\left(t\right)|}^2\right]ds\hfill \\ \hfill & +18K_R^{2}T(T+4+4\lambda )\left({\Delta }^{{\displaystyle \frac{3}{2}}}+{\Delta }^{{\displaystyle \frac{1}{2}}}+\lambda {\Delta }^{{\displaystyle \frac{1}{2}}}\right).\hfill \end{array}$$

Applying the Gronwall inequality yields that

$$\begin{array}{cc}\hfill \mathbb{E}\left[\underset{0\le t\le T}{sup}{|z\left(t\right)-z_{\Delta }\left(t\right)|}^2\right]& \le 18K_R^{2}T(T+4+4\lambda ){\Delta }^{{\displaystyle \frac{1}{2}}}(\Delta +1+\lambda )e^{6K_R^2(T+4+4\lambda )}:=B{\Delta }^{{\displaystyle \frac{1}{2}}}.\hfill \end{array}$$

Consequently,

$$\mathbb{E}\left[\underset{0\le t\le T}{sup}{|z(t\wedge \theta )-z_{\Delta }(t\wedge \theta )|}^2\right]\le B{\Delta }^{{\displaystyle \frac{1}{2}}}.$$

Since

$$\mathbb{P}(x(t\wedge \tau )=z(t\wedge \tau ))=1\mathrm{and}\mathbb{P}(x_{\Delta }(t\wedge \rho )=z_{\Delta }(t\wedge \rho ))=1$$

for all $t\ge 0$, we have

$$\mathbb{E}\left[\underset{0\le t\le T}{sup}{|x(t\wedge \theta )-x_{\Delta }(t\wedge \theta )|}^2\right]\le B{\Delta }^{{\displaystyle \frac{1}{2}}},$$

(32)

which implies

$$\mathbb{E}|x(T\wedge \theta )-x_{\Delta }{(T\wedge \theta )|}^2\le B{\Delta }^{{\displaystyle \frac{1}{2}}}.$$

We therefore have

$$\begin{array}{cc}\hfill \mathbb{E}\left(|x_{\Delta }{\left(T\right)-x\left(T\right)|}^2{\mathbf{1}}_{\{\theta >T\}}\right)& =\mathbb{E}\left(|x_{\Delta }{(T\wedge \theta )-x(T\wedge \theta )|}^2{\mathbf{1}}_{\{\theta >T\}}\right)\hfill \\ \hfill & \le \mathbb{E}|x_{\Delta }{(T\wedge \theta )-x(T\wedge \theta )|}^2\hfill \\ \hfill & \le B{\Delta }^{{\displaystyle \frac{1}{2}}}.\hfill \end{array}$$

This implies (26) as desired. Finally, by Lemma 3, we can easily show that

$$\underset{\Delta \to 0}{lim}\mathbb{E}{|{\overline{x}}_{\Delta }\left(T\right)-x\left(T\right)|}^2=0.$$

The proof is therefore complete. □

5. Strong Convergence

The convergence of the previous section is not sufficient for approximating quantities that are path-dependent, for example, the European barrier option value. In these situations, we will need two stronger convergence results. In this section, we will show that

$$\underset{\Delta \to 0}{lim}\mathbb{E}\left(\underset{0\le t\le T}{sup}{|x_{\Delta }\left(t\right)-x\left(t\right)|}^2\right)=0\mathrm{and}\underset{\Delta \to 0}{lim}\mathbb{E}\left(\underset{0\le t\le T}{sup}{|{\overline{x}}_{\Delta }\left(t\right)-x\left(t\right)|}^2\right)=0$$

for any $T>0$. For this purpose, we should impose an additional condition.

Assumption 3.

Suppose that there is a $\overline{K}>0$, such that

$${|g\left(x\right)|}^2\vee {|u\left(x\right)|}^2\le \overline{K}{(1+|x|}^2),\forall x\in {\mathbb{R}}^n.$$

(33)

Lemma 6. 

Let Assumptions 1, 2, and 3 hold. Then

$$\begin{array}{cc}\hfill \mathbb{E}\left(\underset{0\le t\le T}{sup}{|x\left(t\right)|}^2\right)& \le 2(|x_0{|}^2+2KT)\left\{1+2T{e}^{2KT}\left[K+64\overline{K}+4\lambda \overline{K}\left(\overline{K}+2\right)\right]\right\}+{\displaystyle \frac{1}{2}}\hfill \\ \hfill & +16\overline{K}T\left(16+\lambda \overline{K}\right).\hfill \end{array}$$

(34)

Proof. 

For any $t\in [0,T]$, by the Itô–Doeblin formula and Assumption 2, we have

$$\begin{array}{cc}\hfill {|x\left(t\right)|}^2& =|x_0{|}^2+{\int }_0^t\left[2x{\left(s\right)}^{\top }f\left(x\left(s\right)\right)+{|g\left(x\left(s\right)\right)|}^2+\lambda {|u\left(x\left(s\right)\right)|}^2\right]ds\hfill \\ \hfill & +{\int }_0^{t}2x{\left(s\right)}^{\top }g\left(x\left(s\right)\right)dB\left(s\right)+{\int }_0^t\left[{|u\left(x\left(s\right)\right)|}^2+2x{\left(s^{-}\right)}^{\top }u\left(x\left(s^{-}\right)\right)\right]d\tilde{N}\left(s\right)\hfill \\ \hfill & \le |x_0{|}^2+2K{\int }_0^t\left(1+{|x\left(s\right)|}^2\right)ds+{\int }_0^{t}2x{\left(s\right)}^{\top }g\left(x\left(s\right)\right)dB\left(s\right)\hfill \\ \hfill & +{\int }_0^t\left[{|u\left(x\left(s\right)\right)|}^2+2x{\left(s^{-}\right)}^{\top }u\left(x\left(s^{-}\right)\right)\right]d\tilde{N}\left(s\right).\hfill \end{array}$$

Thus,

$$\begin{array}{cc}\hfill \mathbb{E}\left(\underset{0\le t\le T}{sup}{|x\left(t\right)|}^2\right)& \le |x_0{|}^2+2KT+2K\mathbb{E}\left(\underset{0\le t\le T}{sup}{\int }_0^t{|x\left(s\right)|}^{2}ds\right)\hfill \\ \hfill & +2\mathbb{E}\left[\underset{0\le t\le T}{sup}\left|{\int }_0^{t}x{\left(s\right)}^{\top }g\left(x\left(s\right)\right)dB\left(s\right)\right|\right]\hfill \\ \hfill & +\mathbb{E}\left(\underset{0\le t\le T}{sup}\left|{\int }_0^t\left[{|u\left(x\left(s\right)\right)|}^2+2x{\left(s^{-}\right)}^{\top }u\left(x\left(s^{-}\right)\right)\right]d\tilde{N}\left(s\right)\right|\right).\hfill \end{array}$$

By the Burkholder–Davis–Gundy inequality, the Hölder inequality, the Doob martingale inequality, and the martingale isometry, we derive that

$$\begin{array}{cc}\hfill \mathbb{E}\left(\underset{0\le t\le T}{sup}{|x\left(t\right)|}^2\right)& \le |x_0{|}^2+2KT+2K\mathbb{E}{\int }_0^T{|x\left(s\right)|}^{2}ds+8\sqrt{2}\mathbb{E}{\left({\int }_0^T{|x\left(s\right)|}^2{|g\left(x\left(s\right)\right)|}^{2}ds\right)}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill & +{\left[4\mathbb{E}{\left({\int }_0^T\left(|u\left(x\left(s^{-}\right)\right){|}^2+2x{\left(s^{-}\right)}^{\top }u\left(x\left(s^{-}\right)\right)\right)d\tilde{N}\left(s\right)\right)}^2\right]}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill & \le |x_0{|}^2+2KT+2K\mathbb{E}{\int }_0^T{|x\left(s\right)|}^{2}ds+8\sqrt{2}\mathbb{E}{\left({\int }_0^T{|x\left(s\right)|}^2{|g\left(x\left(s\right)\right)|}^{2}ds\right)}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill & +2{\lambda }^{{\displaystyle \frac{1}{2}}}{\left[\mathbb{E}{\int }_0^T{\left({|u\left(x\left(s\right)\right)|}^2+2x{\left(s\right)}^{\top }u\left(x\left(s\right)\right)\right)}^{2}ds\right]}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill & \le |x_0{|}^2+2KT+2K{\int }_0^T\mathbb{E}{|x\left(s\right)|}^{2}ds+8\sqrt{2}\mathbb{E}{\left[\left(\underset{0\le t\le T}{sup}{|x\left(t\right)|}^2\right){\int }_0^T{|g\left(x\left(s\right)\right)|}^{2}ds\right]}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill & +2{\lambda }^{{\displaystyle \frac{1}{2}}}\mathbb{E}{\left[\left(\underset{0\le t\le T}{sup}{|u\left(x\left(t\right)\right)|}^2\right){\int }_0^T{\left(|u(x\left(s\right))|+2|x(s)|\right)}^{2}ds\right]}^{{\displaystyle \frac{1}{2}}}.\hfill \end{array}$$

By Theorem 1 and Assumption 3, we may obtain

$$\begin{array}{cc}\hfill \mathbb{E}\left(\underset{0\le t\le T}{sup}{|x\left(t\right)|}^2\right)& \le (|x_0{|}^2+2KT)(1+2KT{e}^{2KT})+{\displaystyle \frac{1}{4}}\mathbb{E}\left(\underset{0\le t\le T}{sup}{|x\left(t\right)|}^2\right)+128\mathbb{E}{\int }_0^T{|g\left(x\left(s\right)\right)|}^{2}ds\hfill \\ \hfill & +{\displaystyle \frac{1}{4\overline{K}}}\mathbb{E}\left(\underset{0\le t\le T}{sup}{|u\left(x\left(t\right)\right)|}^2\right)+4\lambda \overline{K}\mathbb{E}{\int }_0^T{\left(|u(x\left(s\right))|+2|x(s)|\right)}^{2}ds\hfill \\ \hfill & \le (|x_0{|}^2+2KT)(1+2KT{e}^{2KT})+{\displaystyle \frac{1}{4}}\mathbb{E}\left(\underset{0\le t\le T}{sup}{|x\left(t\right)|}^2\right)+128\mathbb{E}{\int }_0^T\overline{K}(1+|x\left(s\right){|}^2)ds+{\displaystyle \frac{1}{4}}\hfill \\ \hfill & +{\displaystyle \frac{1}{4}}\mathbb{E}\left(\underset{0\le t\le T}{sup}{|x\left(t\right)|}^2\right)+8\lambda {\overline{K}}^{2}T+8\lambda \overline{K}\left(\overline{K}+2\right){\int }_0^T\mathbb{E}{|x\left(s\right)|}^{2}ds\hfill \\ \hfill & \le (|x_0{|}^2+2KT)\left\{1+2T{e}^{2KT}\left[K+64\overline{K}+4\lambda \overline{K}\left(\overline{K}+2\right)\right]\right\}+{\displaystyle \frac{1}{2}}\mathbb{E}\left(\underset{0\le t\le T}{sup}{|x\left(t\right)|}^2\right)+{\displaystyle \frac{1}{4}}\hfill \\ \hfill & +8\overline{K}T\left(16+\lambda \overline{K}\right).\hfill \end{array}$$

This implies the required assertion (34) easily. □

Lemma 7. 

Let Assumptions 1, 2, and 3 hold. Then

$$\begin{array}{c}\hfill \underset{0<\Delta \le {\Delta }^{*}}{sup}\mathbb{E}\left(\underset{0\le t\le T}{sup}{|x_{\Delta }\left(t\right)|}^2\right)\le 2(Q_1+Q_2)+256\overline{K}T(1+HT)+{\displaystyle \frac{1}{2}}+16\lambda {\overline{K}}^{2}T(1+H)+32\lambda \overline{K}HT,\end{array}$$

(35)

where

$$H:=\left[|x_0{|}^2+4KT+2\sqrt{3}T{(\Delta +1+\lambda )}^{{\displaystyle \frac{1}{2}}}\right]e^{4KT},Q_1:={|x_0|}^2+4KT+4KTHand{Q}_2:=6T{\Delta }^{{\displaystyle \frac{1}{4}}}(\Delta +1+\lambda ).$$

Proof. 

Fix any $\Delta \in (0,{\Delta }^{*}]$. Using the Itô–Doeblin formula, Lemma 1, and (12), we can derive that

$$\begin{array}{cc}\hfill |x_{\Delta }{\left(t\right)|}^2& =|x_0{|}^2+{\int }_0^t\left[2x_{\Delta }{\left(s\right)}^{\top }{f}_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right)+|g_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right){|}^2+\lambda {|u_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right)|}^2\right]ds\hfill \\ \hfill & +{\int }_0^{t}2x_{\Delta }{\left(s\right)}^{\top }{g}_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right)dB\left(s\right)+{\int }_0^t\left[|u_{\Delta }\left({\overline{x}}_{\Delta }\left(s^{-}\right)\right){|}^2+2x_{\Delta }{\left(s^{-}\right)}^{\top }{u}_{\Delta }\left({\overline{x}}_{\Delta }\left(s^{-}\right)\right)\right]d\tilde{N}\left(s\right)\hfill \\ \hfill & =|x_0{|}^2+{\int }_0^t\left[2{\overline{x}}_{\Delta }{\left(s\right)}^{\top }{f}_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right)+|g_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right){|}^2+\lambda {|u_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right)|}^2\right]ds\hfill \\ \hfill & +{\int }_0^{t}2{(x_{\Delta }\left(s\right)-{\overline{x}}_{\Delta }\left(s\right))}^{\top }{f}_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right)ds+{\int }_0^{t}2x_{\Delta }{\left(s\right)}^{\top }{g}_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right)dB\left(s\right)\hfill \\ \hfill & +{\int }_0^t\left[|u_{\Delta }\left({\overline{x}}_{\Delta }\left(s^{-}\right)\right){|}^2+2x_{\Delta }{\left(s^{-}\right)}^{\top }{u}_{\Delta }\left({\overline{x}}_{\Delta }\left(s^{-}\right)\right)\right]d\tilde{N}\left(s\right)\hfill \\ \hfill & \le |x_0{|}^2+{\int }_0^{t}4K(1+|{\overline{x}}_{\Delta }{\left(s\right)|}^2)ds+2l(\Delta ){\int }_0^t|x_{\Delta }\left(s\right)-{\overline{x}}_{\Delta }\left(s\right)|ds\hfill \\ \hfill & +{\int }_0^{t}2x_{\Delta }{\left(s\right)}^{\top }{g}_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right)dB\left(s\right)+{\int }_0^t\left[|u_{\Delta }\left({\overline{x}}_{\Delta }\left(s^{-}\right)\right){|}^2+2x_{\Delta }{\left(s^{-}\right)}^{\top }{u}_{\Delta }\left({\overline{x}}_{\Delta }\left(s^{-}\right)\right)\right]d\tilde{N}\left(s\right).\hfill \end{array}$$

By Lemma 3, Theorem 2, the Burkholder–Davis–Gundy inequality, the Hölder inequality, the Doob martingale inequality, and the martingale isometry, we have

$$\begin{array}{cc}\hfill \mathbb{E}\left(\underset{0\le t\le T}{sup}{|x_{\Delta }\left(t\right)|}^2\right)& \le |x_0{|}^2+4KT+4K{\int }_0^T\mathbb{E}|{\overline{x}}_{\Delta }{\left(s\right)|}^{2}ds+2l(\Delta ){\int }_0^T\mathbb{E}|x_{\Delta }\left(s\right)-{\overline{x}}_{\Delta }\left(s\right)|ds\hfill \\ \hfill & +2\mathbb{E}\left[\underset{0\le t\le T}{sup}\left|{\int }_0^t{x}_{\Delta }{\left(s\right)}^{\top }{g}_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right)dB\left(s\right)\right|\right]\hfill \\ \hfill & +\mathbb{E}\left\{\underset{0\le t\le T}{sup}\left|{\int }_0^t\left[|u_{\Delta }\left({\overline{x}}_{\Delta }\left(s^{-}\right)\right){|}^2+2x_{\Delta }{\left(s^{-}\right)}^{\top }{u}_{\Delta }\left({\overline{x}}_{\Delta }\left(s^{-}\right)\right)\right]d\tilde{N}\left(s\right)\right|\right\}\hfill \\ \hfill & \le Q_1+Q_2+8\sqrt{2}\mathbb{E}{\left[\left(\underset{0\le t\le T}{sup}{|x_{\Delta }\left(t\right)|}^2\right){\int }_0^T|g_{\Delta }{({\overline{x}}_{\Delta }\left(s\right)|}^{2}ds\right]}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill & +2{\lambda }^{{\displaystyle \frac{1}{2}}}\mathbb{E}{\left[\left(\underset{0\le t\le T}{sup}{|u_{\Delta }\left({\overline{x}}_{\Delta }\left(t\right)\right)|}^2\right){\int }_0^T{\left(|u_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right)|+2|x_{\Delta }\left(s\right)|\right)}^{2}ds\right]}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill & \le Q_1+Q_2+{\displaystyle \frac{1}{4}}\mathbb{E}\left(\underset{0\le t\le T}{sup}{|x_{\Delta }\left(t\right)|}^2\right)+128\mathbb{E}{\int }_0^T{|g_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right)|}^{2}ds\hfill \\ \hfill & +{\displaystyle \frac{1}{4\overline{K}}}\mathbb{E}\left(\underset{0\le t\le T}{sup}{|u_{\Delta }\left({\overline{x}}_{\Delta }\left(t\right)\right)|}^2\right)+4\lambda \overline{K}\mathbb{E}{\int }_0^T{\left(|u_{\Delta }\left({\overline{x}}_{\Delta }\left(s\right)\right)|+2|x_{\Delta }\left(s\right)|\right)}^{2}ds.\hfill \end{array}$$

However, we may easily observe that $|g_{\Delta }{\left(x\right)|}^2\vee |u_{\Delta }{\left(x\right)|}^2\le \overline{K}{(1+|x|}^2)$. Therefore, we have

$$\begin{array}{cc}\hfill \mathbb{E}\left(\underset{0\le t\le T}{sup}{|x_{\Delta }\left(t\right)|}^2\right)& \le Q_1+Q_2+{\displaystyle \frac{1}{4}}\mathbb{E}\left(\underset{0\le t\le T}{sup}{|x_{\Delta }\left(t\right)|}^2\right)+128\overline{K}T\left(1+{\int }_0^T\mathbb{E}{|{\overline{x}}_{\Delta }\left(s\right)|}^{2}ds\right)\hfill \\ \hfill & +{\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{4}}\mathbb{E}\left(\underset{0\le t\le T}{sup}{|{\overline{x}}_{\Delta }\left(t\right)|}^2\right)+8\lambda {\overline{K}}^{2}T+8\lambda {\overline{K}}^2{\int }_0^T\mathbb{E}{|{\overline{x}}_{\Delta }\left(s\right)|}^{2}ds\hfill \\ \hfill & +16\lambda \overline{K}{\int }_0^T\mathbb{E}{|x_{\Delta }\left(s\right)|}^{2}ds\hfill \\ \hfill & \le Q_1+Q_2+{\displaystyle \frac{1}{2}}\mathbb{E}\left(\underset{0\le t\le T}{sup}{|x_{\Delta }\left(t\right)|}^2\right)+128\overline{K}T(1+HT)+{\displaystyle \frac{1}{4}}\hfill \\ \hfill & +8\lambda {\overline{K}}^{2}T(1+H)+16\lambda \overline{K}HT,\hfill \end{array}$$

which means

$$\mathbb{E}\left(\underset{0\le t\le T}{sup}{|x_{\Delta }\left(t\right)|}^2\right)\le 2(Q_1+Q_2)+256\overline{K}T(1+HT)+{\displaystyle \frac{1}{2}}+16\lambda {\overline{K}}^{2}T(1+H)+32\lambda \overline{K}HT.$$

This is the desired assertion. □

Theorem 4. 

Let Assumptions 1, 2, and 3 hold. Then

$$\underset{\Delta \to 0}{lim}\mathbb{E}\left(\underset{0\le t\le T}{sup}{|x_{\Delta }\left(t\right)-x\left(t\right)|}^2\right)=0.$$

(36)

Proof. 

Since

$$\begin{array}{cc}\hfill \mathbb{E}\left(\underset{0\le t\le T}{sup}{|x_{\Delta }\left(t\right)-x\left(t\right)|}^2\right)& =\mathbb{E}\left({\mathbf{1}}_{\{\theta >T\}}\underset{0\le t\le T}{sup}{|x_{\Delta }\left(t\right)-x\left(t\right)|}^2\right)+\mathbb{E}\left({\mathbf{1}}_{\{\theta \le T\}}\underset{0\le t\le T}{sup}{|x_{\Delta }\left(t\right)-x\left(t\right)|}^2\right)\hfill \end{array}$$

we can show the assertion (36) in the same way as in the proof of Theorem 3 by Lemmas 4, 5, 6, 7, and (32). □

We can easily verify that it is much easier to estimate ${\overline{x}}_{\Delta }\left(t\right)$ than $x_{\Delta }\left(t\right)$ in practice. It is therefore more desirable to obtain the following theorem.

Theorem 5. 

Let Assumptions 1, 2, and 3 hold. Then

$$\underset{\Delta \to 0}{lim}\mathbb{E}\left(\underset{0\le t\le T}{sup}{|{\overline{x}}_{\Delta }\left(t\right)-x\left(t\right)|}^2\right)=0.$$

(37)

To prove the theorem, we need the following lemma.

Lemma 8. 

Let $\Delta \in (0,{\Delta }^{*}]$. Then

$$\mathbb{E}\left(\underset{0\le t\le T}{sup}{|x_{\Delta }\left(t\right)-{\overline{x}}_{\Delta }\left(t\right)|}^2\right)\le 3{\Delta }^{{\displaystyle \frac{3}{2}}}+{\displaystyle \frac{108}{25}}{\left[15(T+1)\right]}^{{\displaystyle \frac{1}{3}}}{\Delta }^{{\displaystyle \frac{1}{6}}}+48\lambda {\Delta }^{{\displaystyle \frac{1}{2}}}.$$

(38)

Consequently

$$\underset{\Delta \to 0}{lim}\mathbb{E}\left(\underset{0\le t\le T}{sup}{|x_{\Delta }\left(t\right)-{\overline{x}}_{\Delta }\left(t\right)|}^2\right)=0.$$

(39)

Proof. 

Let $M=[T/\Delta ]$, $n=0,1,\cdots ,M$. For any $t\in [0,T]$, we have

$$\begin{array}{cc}\hfill x_{\Delta }\left(t\right)-{\overline{x}}_{\Delta }\left(t\right)& =(t-[t/\Delta ]\Delta )f_{\Delta }\left({\overline{x}}_{\Delta }([t/\Delta ]\Delta )\right)+g_{\Delta }\left({\overline{x}}_{\Delta }([t/\Delta ]\Delta )\right)(B\left(t\right)-B([t/\Delta ]\Delta ))\hfill \\ \hfill & +u_{\Delta }\left({\overline{x}}_{\Delta }([t/\Delta ]\Delta )\right)(\tilde{N}\left(t\right)-\tilde{N}([t/\Delta ]\Delta )).\hfill \end{array}$$

Thus, we derive that

$$\begin{array}{cc}\hfill \mathbb{E}\left(\underset{0\le t\le T}{sup}{|x_{\Delta }\left(t\right)-{\overline{x}}_{\Delta }\left(t\right)|}^2\right)& \le 3{\Delta }^2{\left(l(\Delta )\right)}^2+3{\left(l(\Delta )\right)}^2\mathbb{E}\left(\underset{0\le t\le T}{sup}{|B\left(t\right)-B([t/\Delta ]\Delta )|}^2\right)\hfill \\ \hfill & +3{\left(l(\Delta )\right)}^2\mathbb{E}\left(\underset{0\le t\le T}{sup}{|\tilde{N}\left(t\right)-\tilde{N}([t/\Delta ]\Delta )|}^2\right).\hfill \end{array}$$

(40)

By the Hölder inequality and the Doob martingale inequality, we obtain

$$\begin{array}{cc}\hfill \mathbb{E}\left(\underset{0\le t\le T}{sup}{|B\left(t\right)-B([t/\Delta ]\Delta )|}^2\right)& \le \mathbb{E}\left(\underset{0\le n\le M}{max}\underset{n\Delta \le t\le (n+1)\Delta }{sup}{|B\left(t\right)-B(n\Delta )|}^2\right)\hfill \\ \hfill & \le {\left[\mathbb{E}\left(\underset{0\le n\le M}{max}\underset{n\Delta \le t\le (n+1)\Delta }{sup}{|B\left(t\right)-B(n\Delta )|}^6\right)\right]}^{{\displaystyle \frac{1}{3}}}\hfill \\ \hfill & \le {\left[\sum _{n=0}^M\mathbb{E}\left(\underset{n\Delta \le t\le (n+1)\Delta }{sup}{|B\left(t\right)-B(n\Delta )|}^6\right)\right]}^{{\displaystyle \frac{1}{3}}}\hfill \\ \hfill & \le {\left[\sum _{n=0}^M{\left({\displaystyle \frac{6}{5}}\right)}^6\mathbb{E}{|B((n+1)\Delta )-B(n\Delta )|}^6\right]}^{{\displaystyle \frac{1}{3}}}\hfill \\ \hfill & ={\left[{\left({\displaystyle \frac{6}{5}}\right)}^6\sum _{n=0}^{M}15{\Delta }^3\right]}^{{\displaystyle \frac{1}{3}}}\hfill \\ \hfill & \le {\left({\displaystyle \frac{6}{5}}\right)}^2{\left[15(T+1)\right]}^{{\displaystyle \frac{1}{3}}}{\Delta }^{{\displaystyle \frac{2}{3}}}.\hfill \end{array}$$

(41)

By the Doob maximal inequality and the Doob martingale inequality, we have

$$\begin{array}{cc}\hfill \mathbb{E}\left(\underset{0\le t\le T}{sup}{|\tilde{N}\left(t\right)-\tilde{N}([t/\Delta ]\Delta )|}^2\right)& \le \mathbb{E}\left(\underset{0\le n\le M}{max}\underset{n\Delta \le t\le (n+1)\Delta }{sup}{|\tilde{N}\left(t\right)-\tilde{N}(n\Delta )|}^2\right)\hfill \\ \hfill & \le 4\mathbb{E}\left(\underset{M\Delta \le t\le (M+1)\Delta }{sup}{|\tilde{N}\left(t\right)-\tilde{N}(M\Delta )|}^2\right)\hfill \\ \hfill & \le 16\mathbb{E}|\tilde{N}((M+1)\Delta )-\tilde{N}{(M\Delta )|}^2=16\lambda \Delta .\hfill \end{array}$$

(42)

Substituting (41) and (42) into (40) yields

$$\mathbb{E}\left(\underset{0\le t\le T}{sup}{|x_{\Delta }\left(t\right)-{\overline{x}}_{\Delta }\left(t\right)|}^2\right)\le 3{\Delta }^2{\left(l(\Delta )\right)}^2+{\displaystyle \frac{108}{25}}{\Delta }^{{\displaystyle \frac{2}{3}}}{\left(l(\Delta )\right)}^2{\left[15(T+1)\right]}^{{\displaystyle \frac{1}{3}}}+48\lambda \Delta {\left(l(\Delta )\right)}^2.$$

Consequently, the required assertion (38) follows by applying (10). Letting $\Delta \to 0$, (39) is obtained. The proof is complete. □

Using the elementary inequality ${\left|a+b\right|}^2\le {2|a|}^2+2{|b|}^2$, we can show ${\left|{\overline{x}}_{\Delta }\left(t\right)-x\left(t\right)\right|}^2\le 2{\left|{\overline{x}}_{\Delta }\left(t\right)-x_{\Delta }\left(t\right)\right|}^2+2{\left|x_{\Delta }\left(t\right)-x\left(t\right)\right|}^2$. Therefore, Theorem 5 follows from Theorem 4 and Lemma 8 immediately.

6. Examples

In this section, two examples are considered, and their simulation results are given to to illustrate our theoretical results of Theorems 4 and 5. Due to the strong convergence result, which can be used to justify the method within Monte Carlo simulations that compute the expected payoff of financial products (see [24]), we can discuss the Monte Carlo simulation of European put option prices.

Example 1. 

Consider the scalar SDEwPJs

$$dx\left(t\right)=(\alpha x\left(t\right)-\beta x^5\left(t\right))dt+\delta x\left(t\right)dB\left(t\right)+\gamma x\left(t^{-}\right)d\tilde{N}\left(t\right)$$

(43)

on $t\ge 0$ with the initial value $x\left(0\right)\in \mathbb{R}$, where α, β, δ, and γ are four positive numbers; $B\left(t\right)$ is a scalar Brownian motion; and $\tilde{N}\left(t\right)$ is a compensated Poisson process, which means that $\tilde{N}\left(t\right)=N\left(t\right)-\lambda t$, in which $N\left(t\right)$ is a scalar Poisson process with intensity λ. Clearly, its coefficients $f\left(x\right)=\alpha x-\beta x^5$, $g\left(x\right)=\delta x$, and $u\left(x\right)=\gamma x$ are locally Lipschitz-continuous for $x\in \mathbb{R}$. Also, for any $x\in \mathbb{R}$, we have

$$xf\left(x\right)+{\displaystyle \frac{1}{2}}{|g\left(x\right)|}^2+{\displaystyle \frac{\lambda }{2}}{|u\left(x\right)|}^2=\alpha x^2-\beta x^6+\left({\displaystyle \frac{{\delta }^2}{2}}+{\displaystyle \frac{\lambda }{2}}{\gamma }^2\right)x^2,$$

which is bounded above by a positive constant. This implies that Assumption 2 is satisfied. Moreover, Assumption 3 is satisfied with $\overline{K}={\delta }^2\vee {\gamma }^2$. Hence, we can conclude that the truncated EM solutions of the SDEwPJs (43) satisfy Theorems 4 and 5.

Let us start the simulation. We set $T=\alpha =\beta =\delta =\gamma =1$, $\lambda =0.5$ and $x\left(0\right)=2$. Hence, for $t\in [0,1]$, the SDEwPJs (43) can be expressed as

$$dx\left(t\right)=(x\left(t\right)-x^5\left(t\right))dt+x\left(t\right)dB\left(t\right)+x\left(t^{-}\right)d\tilde{N}\left(t\right).$$

(44)

To apply the truncated EM method, we need to find functions ϕ and l. Noting that

$$\underset{|x|\le r}{sup}(|x-x^5|\vee x\vee x)\le 4r^5,\forall r\ge 1,$$

we choose $\varphi \left(r\right)=4r^5$ and its inverse function ${\varphi }^{-1}\left(r\right)={(r/4)}^{{\displaystyle \frac{1}{5}}}$ for $r\ge 4$. We let $l(\Delta )={\Delta }^{-{\displaystyle \frac{1}{4}}}$ for $\Delta >0$. Therefore, $(|x|\wedge {\varphi }^{-1}\left(l(\Delta )\right)){\displaystyle \frac{x}{|x|}}=(|x|\wedge {({\displaystyle \frac{{\Delta }^{-\frac{1}{4}}}{4}})}^{{\displaystyle \frac{1}{5}}}){\displaystyle \frac{x}{|x|}}$. To simplify our analysis, we only consider a European put option whose exercise price is $K=5$. The expected payoff at expiry time T is given by $\mathbb{E}\left[\right(5-x\left(1\right))\vee 0]$. By Theorem 3, it is not hard to prove that $\mathbb{E}[(5-x_{\Delta }\left(1\right))\vee 0]$ converges to $\mathbb{E}\left[\right(5-x\left(1\right))\vee 0]$ as $\Delta \to 0$. The proof is standard and hence, we omit it (see [24,27]). Therefore, we can use Monte Carlo simulation to estimate the value of European put option.

Table 1. Simulation results under the SDEwPJs (44) for a European put option.

No. of Simulation Trials	No. of Time Steps	Standard Error	Computing Time (s)
5000	40	0.0081	1.5
20,000	100	0.0070	16
80,000	320	0.0061	212.4
120,000	480	0.0056	493.1

shows the corresponding simulation results. The simulation results clearly verify the effectiveness of theoretical results. We remark that the expected payoff from the truncated EM method for a path-dependent option is also convergent to the real expected payoff by Theorems 4 and 5. So Monte Carlo simulation is still effective for a path-dependent option (see also [27]).

Example 2. 

Consider the mean-reverting square root process with Poisson jumps

$$dx\left(t\right)=\kappa (\theta -x\left(t\right))dt+\sigma \sqrt{x\left(t\right)+\epsilon }dB\left(t\right)+\tau x\left(t^{-}\right)d\tilde{N}\left(t\right)$$

(45)

on $t\ge 0$, with the initial value $x\left(0\right)>0$, where $B\left(t\right)$ and $\tilde{N}\left(t\right)$ are the same as that in Example 1; κ, θ, σ, and τ are non-negative constants; and $\epsilon \in (0,1)$ is a sufficiently small positive constant. By Theorem 2.1 in [27], we know that for any initial value $x\left(0\right)>0$, there is a unique solution $x\left(t\right)$ to Equation (45) and the solution will never become negative with probability one. Its coefficients $f\left(x\right)=\kappa (\theta -x)$, $g\left(x\right)=\sigma \sqrt{x+\epsilon }$, and $u\left(x\right)=\tau x$ are clearly locally Lipschitz-continuous for $x\in {\mathbb{R}}_{+}$. For $\lambda {\tau }^2<2\kappa$, we have

$$\begin{array}{cc}\hfill xf\left(x\right)+{\displaystyle \frac{1}{2}}{|g\left(x\right)|}^2+{\displaystyle \frac{\lambda }{2}}{|u\left(x\right)|}^2& =x\kappa (\theta -x)+{\displaystyle \frac{{\sigma }^2}{2}}(x+\epsilon )+{\displaystyle \frac{\lambda }{2}}{\tau }^2{x}^2\hfill \\ \hfill & =\left(\kappa \theta +{\displaystyle \frac{{\sigma }^2}{2}}\right)x+\left({\displaystyle \frac{\lambda }{2}}{\tau }^2-\kappa \right)x^2+{\displaystyle \frac{{\sigma }^2}{2}}\epsilon \hfill \\ \hfill & \le {\displaystyle \frac{{\left(2\kappa \theta +{\sigma }^2\right)}^2}{16\kappa -8\lambda {\tau }^2}}+{\displaystyle \frac{{\sigma }^2}{2}}\epsilon .\hfill \end{array}$$

Thus, Assumption 2 holds. Furthermore, it is easy to show that

$${|g\left(x\right)|}^2\vee {|u\left(x\right)|}^2=\left[{\sigma }^2(x+\epsilon )\right]\vee \left({\tau }^2{x}^2\right)\le \left({\sigma }^2\vee 2{\tau }^2\right){(1+|x|}^2).$$

This means that Assumption 3 is satisfied with $\overline{K}={\sigma }^2\vee 2{\tau }^2$. Therefore, we can conclude by Theorems 4 and 5 that the assertions (36) and (37) hold.

Now let us start the simulation. We set $\kappa =2$, $T=\theta =\sigma =\tau =1$, $\lambda =0.5$ and $x\left(0\right)=1$. Hence, for $t\in [0,1]$, the SDEwPJs (45) can be written as

$$dx\left(t\right)=2(1-x\left(t\right))dt+\sqrt{x\left(t\right)+\epsilon }dB\left(t\right)+x\left(t^{-}\right)d\tilde{N}\left(t\right).$$

(46)

Noting that

$$\underset{|x|\le r}{sup}(2|1-x|\vee \sqrt{x+\epsilon }\vee |x|)\le 6r,\forall r\ge 1,$$

we choose $\varphi \left(r\right)=6r$ and its inverse function ${\varphi }^{-1}\left(r\right)=r/6$ for $r\ge 6$. We let $l(\Delta )={\Delta }^{-{\displaystyle \frac{1}{4}}}$ for $\Delta >0$. Thus, $(|x|\wedge {\varphi }^{-1}\left(l(\Delta )\right)){\displaystyle \frac{x}{|x|}}=(|x|\wedge ({\displaystyle \frac{{\Delta }^{-\frac{1}{4}}}{6}})){\displaystyle \frac{x}{|x|}}$. Similarly, we consider a European put option whose exercise price is $K=8$. The expected payoff at expiry time T is given by $\mathbb{E}\left[\right(8-x\left(1\right))\vee 0]$. The corresponding simulation results are listed in

Table 2. Simulation results under the SDEwPJs (46) for a European put option.

No. of Simulation Trials	No. of Time Steps	Standard Error	Computing Time (s)
5000	40	0.0459	1.4
20,000	100	0.0364	14.3
80,000	320	0.0333	185.8
120,000	480	0.0245	455.5

, from which we see that the effectiveness of theoretical results.