Strong convergence rate of the stochastic theta method for nonlinear hybrid stochastic differential equations with piecewise continuous arguments

Abstract

We consider the strong convergence of the stochastic theta (ST) method for highly nonlinear hybrid stochastic differential equations with piecewise continuous arguments (SDEPCAs). There are three major ingredients. The first is the pth moment boundedness of the ST method. Second, the mean square convergence rate of the ST method for hybrid SDEPCAs is given by means of the forward–backward Euler–Maruyama method. The third ingredient is a numerical simulation, which shows the agreement with the theoretical convergence rate.

1 Introduction

Stochastic differential equations with piecewise continuous arguments (SDEPCAs) play an important role in stochastic theory. Such models are applicable in a variety areas including biology, control science and neural networks (Li 2014; Mao et al. 2014; You et al. 2015; Xie and Zhang 2020). Component failures, changes in subsystem interconnections and sudden environmental disturbances can lead to abrupt changes of structures and parameters in many practical systems. To tackle these problems, hybrid systems driven by continuous-time Markov chains have become a powerful tool (Jobert and Rogers 2006; Smith 2002; Song and Mao 2018). Furthermore, it is well known that Markov chain can work as a stabilizing factor (Li and Mao 2020; Hu et al. 2020), that is, the whole system can be stable even though some subsystems are stable and others are unstable, such property is referred to as switching-dominated stability in Zhang et al. (2019).

Since explicit solutions are almost impossible to obtain for such systems, it becomes extremely important to solve them numerically. Finite time convergence analysis of an Euler type method for stochastic differential equations (SDEs) with Markovian switching was given in Mao et al. (2007) and Yuan and Mao (2004). It has been extended to stochastic differential delay equations (SDDEs) with Markovian switching (Li and Hou 2006; Milošević and Jovanović 2011; Zhang and Xie 2019), SDDEs with Markovian switching and Poisson jump (Li and Chang 2007; Wang and Xue 2007) and neutral SDDEs with Markovian switching (Yin and Ma 2011; Zhou and Wu 2009), etc. Numerical invariant measure of the backward Euler–Maruyama method for SDEs with Markovian switching was investigated in Li et al. (2018). It is worth mentioning that most of previous studies for hybrid systems is devoted to those equations driven by a continuous-time and homogeneous Markov chain independent of the Brownian motion, and the switching process r(t) was assumed to have a finite state space. Recently, Yin et al. extended the study to the Markov process (X(t), r(t)) by allowing the generator r(t) to depend on the current state X(t) and to have a countable state space (Yin and Zhu 2010; Nguyen and Yin 2016).

To our best knowledge, there are few works on SDEPCAs with Markovian switching. An SDEPCA belongs to the SDDEs, but the delay term is different from $t-\tau$ and may be a discontinuous function. Moreover, although the SDEPCAs are retarded, the solutions of these equations are determined by only a finite set of initial data, rather than a function, as in the case of general SDDEs (Wiener 1993; Mao 2007). Because of these characteristics, we cannot simply generalize the properties of hybrid SDDEs to hybrid SDEPCAs.

In this work, we concentrate on the numerical solutions for highly nonlinear hybrid SDEPCAs, the stochastic theta (ST) scheme, which is an extension of the Euler–Maruyama method and the backward Euler–Maruyama, is adopted. The rest of this paper is organized as follows. Some basic notations and assumptions are introduced in Sect. 2. The ST method for hybrid SDEPCAs is established in Sect. 3. Section 4 is devoted to the pth moment boundedness of the ST method. Then we go further to reveal the strong convergence rate of the numerical method in Sect. 5. Finally, a numerical experiment is given in Sect. 6 to verify our theoretical convergence order.

2 Notations and preliminaries

Throughout this paper, unless otherwise specified, we 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 (i.e., it is right continuous and ${\mathcal {F}}_0$ contains all ${\mathbb {P}}$-null sets). If A is a vector or matrix, its transpose is denoted by $A^\mathrm{{T}}$. Let $B(t)=(B_1(t),\dots ,B_d(t))^\mathrm{{T}}$ be a d-dimensional Brownian motion defined on the probability space. If x is a vector, $\Vert x\Vert$ denotes its Euclidean norm. $\langle x,y\rangle$ denotes the inner product of vectors x and y. If A is a matrix, its trace norm is denoted by $\Vert A\Vert =\sqrt{{{\,\mathrm{trace}\,}}(A^\mathrm{{T}}A)}$. For two real numbers a and b, we will use $a\vee b$ and $a\wedge b$ for the $\max \left\{ a,b\right\}$ and $\min \left\{ a,b\right\}$, respectively. Let ${\mathcal {L}}^p_{{\mathcal {F}}_t}(\Omega ;{\mathbb {R}}^n)$ denotes the family of ${\mathcal {F}}_t$-measurable ${\mathbb {R}}^n$-valued random variables $\xi$ with ${\mathbb {E}}\Vert \xi \Vert ^p<\infty$. Let ${\mathcal {L}}^p([a,b];{\mathbb {R}}^n)$ denotes the family of ${\mathbb {R}}^n$-valued ${\mathcal {F}}_t$-adapted processes $\{f(t)\}_{a\le t\le b}$ such that $\int _a^{b}\Vert f(t)\Vert ^p \mathrm{{d}}t<\infty$, a.s., ${\mathcal {L}}^p({\mathbb {R}}_+;{\mathbb {R}}^n)$ denotes the family of processes $\{f(t)\}_{t\ge 0}$ such that for every $T>0,$ $\{f(t)\}_{0\le t\le T}\in {\mathcal {L}}^p([0,T];{\mathbb {R}}^n)$. ${\mathcal {M}}^p([a,b];{\mathbb {R}}^n)$ denotes the family of processes $\{f(t)\}_{a\le t\le b}$ in ${\mathcal {L}}^p([a,b];{\mathbb {R}}^n)$ such that ${\mathbb {E}}\int _a^{b}\Vert f(t)\Vert ^p \mathrm{{d}}t<\infty$. $[\cdot ]$ denotes the greatest integer function.

Let $r(t),t\ge 0,$ be a right-continuous Markov chain on the probability space taking values in a finite state space $S=\left\{ 1,2,\dots ,N\right\}$ with generator $\Gamma =(\gamma _{ij})_{N\times N}$ given by

$$\begin{aligned} {\mathbb {P}}\left\{ r(t+\Delta )=j\vert r(t)=i\right\} = {\left\{ \begin{array}{ll} \gamma _{ij}\Delta +o(\Delta ),&{}\mathrm{{if}} ~~i\ne j,\\ 1+\gamma _{ii}\Delta +o(\Delta ),&{} \mathrm{{if}}~~ i=j, \end{array}\right. } \end{aligned}$$

where $\Delta >0$. Here $\gamma _{ij}\ge 0$ is the transition rate from i to j when $i\ne j$, and

$$\begin{aligned} \gamma _{ii}=-\sum _{j\ne i}\gamma _{ij}. \end{aligned}$$

We assume that the Markov chain $r(\cdot )$ is independent of the Brownian motion $B(\cdot )$.

Consider the following hybrid SDEPCAs

$$\begin{aligned} \mathrm{{d}}x(t)=f(x(t),x([t]),r(t))\mathrm{{d}}t+g(x(t),x([t]),r(t))\mathrm{{d}}B(t),\quad t\ge 0, \end{aligned}$$

(1)

with initial data $x(0)=x_0\in {\mathbb {R}}^n$ and $r(0)=i_0\in S,$ where

$$\begin{aligned} f:{\mathbb {R}}^n\times {\mathbb {R}}^n\times S\rightarrow {\mathbb {R}}^n\quad \mathrm{{and}} \quad g:{\mathbb {R}}^n\times {\mathbb {R}}^n\times S\rightarrow {\mathbb {R}}^{n\times d}. \end{aligned}$$

Let us give the definition of the solution.

Definition 1

An ${\mathbb {R}}^n$-valued stochastic process $\left\{ x(t)\right\} _{t\ge 0}$ is called a solution of (1) if it has the following properties:

• (1) $\left\{ x(t)\right\}$ is continuous and ${\mathcal {F}}_t$-adapted;

• (2) $\left\{ f(x(t),x([t]),r(t))\right\} \in {\mathcal {L}}^1({\mathbb {R}}_+;{\mathbb {R}}^n)$, $\left\{ g(x(t),x([t]),r(t))\right\} \in {\mathcal {L}}^2({\mathbb {R}}_+;{\mathbb {R}}^{n\times d})$;

• (3) Equation (1) is satisfied on each interval $[n,n+1)\subset [0,\infty )$ with integral end-points almost surely.

A solution $\left\{ x(t)\right\}$ is said to be unique if any other solution $\left\{ {\bar{x}}(t)\right\}$ is indistinguishable from $\left\{ x(t)\right\}$, that is

$$\begin{aligned} {\mathbb {P}}\left\{ x(t)={\bar{x}}(t)\ \mathrm{{for~all}}\ t\ge 0\right\} =1. \end{aligned}$$

We impose some assumptions:

Assumption 2.1

For every integer $R\ge 1$, there exists a constant $L(R)>0$ such that

$$\begin{aligned} \Vert f(x,y,i)-f({\bar{x}},{\bar{y}},i)\Vert \vee \Vert g(x,y,i)-g({\bar{x}},{\bar{y}},i)\Vert \le L(R)(\Vert x-{\bar{x}}\Vert +\Vert y-{\bar{y}}\Vert ), \end{aligned}$$

for all $i\in S$ and those $x, y,{\bar{x}},{\bar{y}}\in {\mathbb {R}}^n$ with $\Vert x\Vert \vee \Vert y\Vert \vee \Vert {\bar{x}}\Vert \vee \Vert {\bar{y}}\Vert \le R$.

Assumption 2.2

There exists a constant $\alpha >0$ such that

$$\begin{aligned} x^\mathrm{{T}}f(x,y,i)\le \alpha (1+\Vert x\Vert ^2+\Vert y\Vert ^{2}), \,\forall x, y\in {\mathbb {R}}^n, \,\forall i\in S. \end{aligned}$$

Assumption 2.3

There exist constants $L_1>0$ and $h_1\ge 1$ such that

$$\begin{aligned} \Vert f(x,y,i)\Vert \le L_1(1+\Vert x\Vert ^{h_1}+\Vert y\Vert ^{h_1}), \,\forall x, y\in {\mathbb {R}}^n, \,\forall i\in S. \end{aligned}$$

Assumption 2.4

There exist constants $L_2>0$ and $h_2\ge 1$ such that

$$\begin{aligned} \Vert g(x,y,i)\Vert \le L_2(1+\Vert x\Vert +\Vert y\Vert ^{h_2}), \,\forall x, y\in {\mathbb {R}}^n, \,\forall i\in S. \end{aligned}$$

Theorem 2.5

Under Assumptions 2.1–2.4, for any $T>0$, there exists a unique global solution x(t) to Eq. (1) on $t\in [0,T]$ with the initial data $x_0$. Moreover, the solution has the property that ${\mathbb {E}}\Vert x(t)\Vert ^p<\infty$ for all $t\in [0,T]$.

Proof

For any given $i\in S$, we first prove that there exists a unique global solution to the SDEPCA

$$\begin{aligned} \mathrm{{d}}x(t)=f(x(t),x([t]),i)\mathrm{{d}}t+g(x(t),x([t]),i)\mathrm{{d}}B(t) \end{aligned}$$

(2)

on $t\in [0,T]$ with the initial data $x_0$ and the solution has the property that ${\mathbb {E}}\Vert x(t)\Vert ^p<\infty$ for all $t\in [0,T]$. To distinguish between the solution of (1) and that of (2), we denote the solution of (2) by y(t).

In a similar way as the proof of Theorem 3.15 in Mao and Yuan (2006), there is a unique maximal local solution y(t) exists on $[0,\eta _{e})$ under the local Lipschitz condition, where $\eta _{e}$ is the explosion time. Then for each integer $R\ge \Vert x_0\Vert$, define the stopping time $\eta _R=\inf \{t\in [0,\eta _e):\Vert y(t)\Vert \ge R\}$. Clearly, $\eta _R$ is increasing as $R\rightarrow \infty$. We denote that $\eta _{\infty }=\lim _{R\rightarrow \infty }\eta _R$ and $\inf \emptyset =\infty$. Hence, $\eta _{\infty }\le \eta _e$ almost surely. If we can obtain $\eta _{\infty }=\infty$ almost surely, then $\eta _e=\infty$ almost surely. In what follows, we will prove $\eta _{\infty }=\infty$ almost surely and ${\mathbb {E}}\Vert y(t)\Vert ^p<\infty$.

Applying Itô’s formula to $\Vert y(t)\Vert ^p, p\ge 2$, we have

$$\begin{aligned} \begin{aligned} d(\Vert y(t)\Vert ^p)\le&p\Vert y(t)\Vert ^{p-2}\left( y^\mathrm{{T}}(t)f(y(t),y([t]),i)+\frac{p-1}{2}\left\| g(y(t),y([t]),i)\right\| ^2\right) \mathrm{{d}}t\\&+p\Vert y(t)\Vert ^{p-2}y^\mathrm{{T}}(t)g(y(t),y([t]),i)\mathrm{{d}}B(t). \end{aligned} \end{aligned}$$

(3)

• Take any $t\in [0,1)$, integrating both sides of (3) from 0 to $t\wedge \eta _R$, then

  $$\begin{aligned} \begin{aligned}&\Vert y(t\wedge \eta _R)\Vert ^p\\&\quad \le \Vert y(0)\Vert ^p+p\int _0^{t\wedge \eta _R}\Vert y(s)\Vert ^{p-2}\left( y^\mathrm{{T}}(s)f(y(s),y(0),i)+\frac{p-1}{2}\left\| g(y(s),y(0),i)\right\| ^2\right) \mathrm{{d}}s\\&\qquad +p\int _0^{t\wedge \eta _R}\Vert y(s)\Vert ^{p-2}y^\mathrm{{T}}(s)g(y(s),y(0),i)\mathrm{{d}}B(s). \end{aligned} \end{aligned}$$

  Hence,

  $$\begin{aligned} \begin{aligned}&{\mathbb {E}}\Vert y(t\wedge \eta _R)\Vert ^p\\&\quad \le \Vert x_0\Vert ^p+p{\mathbb {E}}\int _0^{t\wedge \eta _R}\Vert y(s)\Vert ^{p-2}\left( y^\mathrm{{T}}(s)f(y(s),x_0,i)+\frac{p-1}{2}\left\| g(y(s),x_0,i)\right\| ^2\right) \mathrm{{d}}s. \end{aligned} \end{aligned}$$

  By Assumptions 2.2 and 2.4, together with Young’s inequality, one has

  $$\begin{aligned} \begin{aligned} {\mathbb {E}}\Vert y(t\wedge \eta _R)\Vert ^p\le&\Vert x_0\Vert ^p+p{\mathbb {E}}\int _0^{t\wedge \eta _R}\Vert y(s)\Vert ^{p-2}\Bigg (\alpha (1+\Vert y(s)\Vert ^2+\Vert x_0\Vert ^2)\\&\qquad \qquad +\frac{(p-1)L_2^2}{2}(1+\Vert y(s)\Vert +\Vert x_0\Vert ^{h_2})^2\Bigg )\mathrm{{d}}s\\ \le&\Vert x_0\Vert ^p+p{\mathbb {E}}\int _0^{t\wedge \eta _R}\bigg \{\left( \gamma _0+\alpha \Vert x_0\Vert ^2+3(p-1)L_2^2/2\Vert x_0\Vert ^{2h_2}\right) \Vert y(s)\Vert ^{p-2}\\&\qquad \qquad \qquad \qquad +\gamma _0\Vert y(s)\Vert ^{p}\bigg \}\mathrm{{d}}s\\ \le&\Vert x_0\Vert ^p+p{\mathbb {E}}\int _0^{t\wedge \eta _R}\bigg \{\gamma _0^{\frac{p}{2}}+\alpha ^{\frac{p}{2}}\Vert x_0\Vert ^p+(3(p-1)L_2^2/2)^{\frac{p}{2}}\Vert x_0\Vert ^{ph_2}\\&\qquad \qquad \qquad \qquad +(3+\gamma _0)\Vert y(s)\Vert ^{p}\bigg \}\mathrm{{d}}s\\ \le&pt\gamma _0^{\frac{p}{2}}+(1+pt\alpha ^{\frac{p}{2}})\Vert x_0\Vert ^p+pt(3(p-1)/2)^{\frac{p}{2}}L_2^p\Vert x_0\Vert ^{ph_2}\\&\qquad \qquad \qquad \qquad +(3+\gamma _0)p\int _0^{t}{\mathbb {E}}\Vert y(s\wedge \eta _R)\Vert ^{p}\mathrm{{d}}s, \end{aligned} \end{aligned}$$

  where $\gamma _0=\alpha +3(p-1)L_2^2/2$. Now it can be obtained from Gronwall’s inequality that

  $$\begin{aligned} \begin{aligned} {\mathbb {E}}\Vert y(t\wedge \eta _R)\Vert ^p\le&\left( pt\gamma _0^{\frac{p}{2}}+(1+pt\alpha ^{\frac{p}{2}})\Vert x_0\Vert ^p+pt(3(p-1)/2)^{\frac{p}{2}}L_2^p\Vert x_0\Vert ^{ph_2}\right) e^{(3+\gamma _0)pt}\\ \le&\beta _0e^{(3+\gamma _0)p}, \end{aligned} \end{aligned}$$

  (4)

  where

  $$\begin{aligned} \beta _0=p\gamma _0^{\frac{p}{2}}+(1+p\alpha ^{\frac{p}{2}})\Vert x_0\Vert ^p+p(3(p-1)/2)^{\frac{p}{2}}L_2^p\Vert x_0\Vert ^{ph_2}<\infty . \end{aligned}$$

  Thus,

  $$\begin{aligned} {\mathbb {E}}\Vert y(1\wedge \eta _R)\Vert ^p=\lim _{t\rightarrow 1}{\mathbb {E}}\Vert y(t\wedge \eta _R)\Vert ^p\le \beta _0e^{(3+\gamma _0)p}. \end{aligned}$$

  (5)

  Let $I_G$ denote the indicator function of the set G, then

  $$\begin{aligned} \beta _0e^{(3+\gamma _0)p}\ge {\mathbb {E}}\Vert y(1\wedge \eta _R)\Vert ^p\ge {\mathbb {E}}\left( \Vert y(\eta _R)\Vert ^pI_{\{\eta _R\le 1\}}\right) \ge R^p{\mathbb {P}}(\eta _R\le 1); \end{aligned}$$

  hence,

  $$\begin{aligned} {\mathbb {P}}(\eta _R\le 1)\le \frac{\beta _0e^{(3+\gamma _0)p}}{R^p}. \end{aligned}$$

  Let $R\rightarrow \infty$, we have ${\mathbb {P}}(\eta _{\infty }\le 1)=0$, which gives

  $$\begin{aligned} {\mathbb {P}}(\eta _{\infty }> 1)=1, \end{aligned}$$

  combining (4) and (5), it can be obtained that

  $$\begin{aligned} {\mathbb {E}}\Vert y(t)\Vert ^p\le \beta _0e^{(3+\gamma _0)p},\quad t\in [0,1]. \end{aligned}$$

• For any $t\in [1,2)$, similar to the process above, integrating both sides of (3) from 1 to $t\wedge \eta _R$, and then taking the expectations, we can arrive at

  $$\begin{aligned} \begin{aligned}&{\mathbb {E}}\Vert y(t\wedge \eta _R)\Vert ^p\\&\quad \le \!{\mathbb {E}}\Vert y(1)\Vert ^p\!+\!p{\mathbb {E}}\int _1^{t\wedge \eta _R}\Vert y(s)\Vert ^{p-2}\left( y^\mathrm{{T}}(s)f(y(s),y(1),i)+\frac{p-1}{2}\left\| g(y(s),y(1),i)\right\| ^2\right) \mathrm{{d}}s\\&\quad \le {\mathbb {E}}\Vert y(1)\Vert ^p+p{\mathbb {E}}\int _1^{t\wedge \eta _R}\Vert y(s)\Vert ^{p-2}\Bigg (\alpha (1+\Vert y(s)\Vert ^2+\Vert y(1)\Vert ^2)\\&\qquad +\frac{(p-1)L_2^2}{2}(1+\Vert y(s)\Vert +\Vert y(1)\Vert ^{h_2})^2\Bigg )\mathrm{{d}}s\\&\quad \le {\mathbb {E}}\Vert y(1)\Vert ^p+p{\mathbb {E}}\int _1^{t\wedge \eta _R}\bigg \{\left( \gamma _0+\alpha \Vert y(1)\Vert ^2+3(p-1)L_2^2/2\Vert y(1)\Vert ^{2h_2}\right) \Vert y(s)\Vert ^{p-2}\\&\qquad +\gamma _0\Vert y(s)\Vert ^{p}\bigg \}\mathrm{{d}}s\\&\quad \le {\mathbb {E}}\Vert y(1)\Vert ^p+p{\mathbb {E}}\int _1^{t\wedge \eta _R}\bigg \{\gamma _0^{\frac{p}{2}}+\alpha ^{\frac{p}{2}}\Vert y(1)\Vert ^p+(3(p-1)L_2^2/2)^{\frac{p}{2}}\Vert y(1)\Vert ^{ph_2}\\&\qquad +(3+\gamma _0)\Vert y(s)\Vert ^{p}\bigg \}\mathrm{{d}}s\\&\quad \le p(t-1)\gamma _0^{\frac{p}{2}}+(1+p(t-1)\alpha ^{\frac{p}{2}}){\mathbb {E}}\Vert y(1)\Vert ^p+p(t-1)(3(p-1)/2)^{\frac{p}{2}}L_2^p{\mathbb {E}}\Vert y(1)\Vert ^{ph_2}\\&\qquad +(3+\gamma _0)p\int _1^{t}{\mathbb {E}}\Vert y(s\wedge \eta _R)\Vert ^{p}\mathrm{{d}}s. \end{aligned} \end{aligned}$$

  By Gronwall’s inequality, one has

  $$\begin{aligned} \begin{aligned} {\mathbb {E}}\Vert y(t\wedge \eta _R)\Vert ^p\le \beta _1e^{(3+\gamma _0)p}, \end{aligned} \end{aligned}$$

  (6)

  where

  $$\begin{aligned} \beta _1=\left( p\gamma _0^{\frac{p}{2}}+(1+p\alpha ^{\frac{p}{2}}){\mathbb {E}}\Vert y(1)\Vert ^p+p(3(p-1)/2)^{\frac{p}{2}}L_2^p{\mathbb {E}}\Vert y(1)\Vert ^{ph_2}\right) <\infty . \end{aligned}$$

  Hence,

  $$\begin{aligned} \begin{aligned} {\mathbb {E}}\Vert y(2\wedge \eta _R)\Vert ^p=\lim _{t\rightarrow 2}{\mathbb {E}}\Vert y(t\wedge \eta _R)\Vert ^p\le \beta _1e^{(3+\gamma _0)p}, \end{aligned} \end{aligned}$$

  (7)

  it gives

  $$\begin{aligned} \beta _1e^{(3+\gamma _0)p}\ge {\mathbb {E}}\Vert y(2\wedge \eta _R)\Vert ^p\ge {\mathbb {E}}\left( \Vert y(\eta _R)\Vert ^pI_{\{\eta _R\le 2\}}\right) \ge R^p{\mathbb {P}}(\eta _R\le 2). \end{aligned}$$

  Taking $R\rightarrow \infty$, yields

  $$\begin{aligned} {\mathbb {P}}(\eta _{\infty }\le 2)\le \lim _{R\rightarrow \infty }\frac{\beta _1e^{(3+\gamma _0)p}}{R^p}=0, \end{aligned}$$

  which implies

  $$\begin{aligned} {\mathbb {P}}(\eta _{\infty }> 2)=1, \end{aligned}$$

  then combining (6) and (7), it can be obtained that

  $$\begin{aligned} {\mathbb {E}}\Vert y(t)\Vert ^p\le \beta _1e^{(3+\gamma _0)p},\quad t\in [1,2]. \end{aligned}$$

• Repeating this procedure we can deduce that, for any integer $j\ge 1$,

  $$\begin{aligned} {\mathbb {P}}(\eta _{\infty }> j)=1, \end{aligned}$$

  and

  $$\begin{aligned} {\mathbb {E}}\Vert y(t)\Vert ^p\le \beta _je^{(3+\gamma _0)p},\quad t\in [j,j+1], \end{aligned}$$

  (8)

  where

  $$\begin{aligned} \beta _j=\left( p\gamma _0^{\frac{p}{2}}+(1+p\alpha ^{\frac{p}{2}}){\mathbb {E}}\Vert y(j)\Vert ^p+p(3(p-1)/2)^{\frac{p}{2}}L_2^p{\mathbb {E}}\Vert y(j)\Vert ^{ph_2}\right) <\infty . \end{aligned}$$

Since $j\ge 1$ is an arbitrary integer, we can conclude that $\eta _{\infty }=\infty$ almost surely and ${\mathbb {E}}\Vert y(t)\Vert ^p< \infty , \forall t\ge 0$.

Now we are in a position to prove Eq.(1) has a unique global solution x(t) and the solution has the property that ${\mathbb {E}}\Vert x(t)\Vert ^p<\infty$.

It is well known (see Anderson 2012) that almost every sample path of the Markov chain $r(\cdot )$ is a right-continuous step function with a finite number of sample jumps in any finite subinterval of ${\mathbb {R}}_+$. Hence, there is a sequence of stopping times $0=\tau _0<\tau _1<\cdots<\tau _k<\cdots$ such that

$$\begin{aligned} r(t)=\sum _{k=0}^{\infty }r(\tau _k)I_{[\tau _k,\tau _{k+1})}(t), \quad t\ge 0. \end{aligned}$$

We first consider Eq. (1) on $t\in [\tau _0,\tau _1)$, which becomes

$$\begin{aligned} \mathrm{{d}}x(t)=f(x(t),x([t]),i_0)\mathrm{{d}}t+g(x(t),x([t]),i_0)\mathrm{{d}}B(t), \end{aligned}$$

(9)

with initial data $x_0$. By the existence-and-uniqueness proof for SDEPCA (2), we know that Eq. (9) has a unique continuous solution which belongs to ${\mathcal {M}}^2([\tau _0,\tau _1);{\mathbb {R}}^n)$ and has the property that ${\mathbb {E}}\Vert x(t)\Vert ^p<\infty$. In particular, $x(\tau _1)=\lim _{t\rightarrow \tau _1^-}x(t)\in L^2_{{\mathcal {F}}_{\tau _1}}(\Omega ;{\mathbb {R}}^n)$. We next consider Eq. (1) on $t\in [\tau _1,\tau _2)$, which becomes

$$\begin{aligned} \mathrm{{d}}x(t)=f(x(t),x([t]),r(\tau _1))\mathrm{{d}}t+g(x(t),x([t]),r(\tau _1))\mathrm{{d}}B(t), \end{aligned}$$

(10)

with initial data $x(\tau _1)$ given by the solution of Eq. (9). Again we know that Eq. (10) has a unique continuous solution which belongs to ${\mathcal {M}}^2([\tau _1,\tau _2);{\mathbb {R}}^n)$ and has the property that ${\mathbb {E}}\Vert x(t)\Vert ^p<\infty$. Repeating this procedure, we see that Eq. (1) has a unique solution x(t) on [0, T] and has the property that

$$\begin{aligned} {\mathbb {E}}\Vert x(t)\Vert ^p<\infty , \quad \forall t\in [0,T]. \end{aligned}$$

The proof is completed. $\square$

3 Stochastic theta method

To define the ST scheme, let us first explain how to simulate the discrete Markov chain $\left\{ r_k^{\Delta },k=0,1,2,\dots \right\}$. Recall the property of the embedded discrete Markov chain:

Given a step size $\Delta >0$, let $r_k^{\Delta }=r(k\Delta )$ for $k\ge 0$. Then $\left\{ r_k^{\Delta },k=0,1,2,\dots \right\}$ is a discrete Markov chain with the one-step transition probability matrix

$$\begin{aligned} {\mathbb {P}}(\Delta )=({\mathbb {P}}_{ij}(\Delta ))_{N\times N}=e^{\Gamma \Delta }. \end{aligned}$$

Hence, the discrete Markov chain $\left\{ r_k^{\Delta },k=0,1,2,\dots \right\}$ can be simulated as follows: Let $r_0^{\Delta }=i_0$ and compute a pseudo-random number $\zeta _1$ from the uniform [0, 1] distribution. Define

$$\begin{aligned} r_1^{\Delta }={\left\{ \begin{array}{ll}i_1,\,\,&{}\mathrm{{if\,\,}} i_1\in S-\{N\} \mathrm{{\,\,such\,\, that\,\,}} \sum _{j=1}^{i_1-1}{\mathbb {P}}_{i_0,j}(\Delta )\le \zeta _1<\sum _{j=1}^{i_1}{\mathbb {P}}_{i_0,j}(\Delta ),\\ N,\,\,&{}\mathrm{{if\,\,}} \sum _{j=1}^{N-1}{\mathbb {P}}_{i_0,j}(\Delta )\le \zeta _1, \end{array}\right. } \end{aligned}$$

where we set $\sum _{j=1}^{0}{\mathbb {P}}_{i_0,j}(\Delta )=0$ as usual. In other words, we ensure that the probability of state s being chosen is given by ${\mathbb {P}}(r_1^{\Delta }=s)={\mathbb {P}}_{i_0,s}(\Delta )$. Generally, having calculated $r_0^{\Delta },r_1^{\Delta },\dots ,r_k^{\Delta },$ we compute $r_{k+1}^{\Delta }$ by drawing a uniform [0, 1] pseudo-random number $\zeta _{k+1}$ and setting

$$\begin{aligned} r_{k+1}^{\Delta }={\left\{ \begin{array}{ll}i_{k+1},\,\,&{}\mathrm{{if\,\,}} i_{k+1}\in S-\{N\} \mathrm{{\,\,such\,\, that\,\,}} \\ &{}\quad \sum _{j=1}^{i_{k+1}-1}{\mathbb {P}}_{r_k^{\Delta },j}(\Delta )\le \zeta _{k+1}<\sum _{j=1}^{i_{k+1}}{\mathbb {P}}_{r_k^{\Delta },j}(\Delta ),\\ N,\,\,&{}\mathrm{{if\,\,}} \sum _{j=1}^{N-1}{\mathbb {P}}_{r_k^{\Delta },j}(\Delta )\le \zeta _{k+1}. \end{array}\right. } \end{aligned}$$

This procedure can be carried out independently to obtain more trajectories.

After explaining how to simulate the discrete Markov chain, we can now define the ST approximate solution to Eq. (1). Let $\Delta =1/m$ be a given step size with integer $m\ge 1$, and let the gridpoints $t_k$ be defined by $t_k=k\Delta (k\in {\mathbb {N}})$. Since for arbitrary $k\in {\mathbb {N}}$, there exist $s\in {\mathbb {N}}$ and $l=0,1,2,\dots ,m-1$ such that $k=sm+l$, the adaptation of the ST method to (1) leads to a numerical process of the following type by setting $X_0=x(0)=x_0, r_0^{\Delta }=r(0)=i_0$,

$$\begin{aligned} \begin{aligned} X_{sm+l+1}=&X_{sm+l}+(1-\theta )f\left( X_{sm+l},X_{sm},r_{sm+l}^{\Delta }\right) \Delta \\&+\theta f\left( X_{sm+l+1},X_{sm},r_{sm+l+1}^{\Delta }\right) \Delta +g\left( X_{sm+l},X_{sm},r_{sm+l}^{\Delta }\right) \Delta B_{sm+l}, \end{aligned} \end{aligned}$$

(11)

for $s\in {\mathbb {N}}$ and $l=0,1,2,\dots ,m-1$, where $\Delta B_{sm+l}=B(t_{sm+l+1})-B(t_{sm+l})$, $\theta \in [0,1]$ is a free parameter that is specified a priori. $X_{sm+l}$ is an approximation to the exact solution $x(t_{sm+l})$.

Since the ST scheme is semi-implicit when $\theta \ne 0$, the first item that need to be considered is the existence and uniqueness of solutions of these equations. In that sense, we will employ the one-sided Lipschitz condition in the first argument of the function f, which is given in the following.

Assumption 3.1

There exists a positive constant L such that

$$\begin{aligned} \langle x-{\bar{x}},f(x,y,i)-f({\bar{x}},y,i)\rangle \le L\Vert x-{\bar{x}}\Vert ^2, \end{aligned}$$

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

Remark 1

By Lemma 3.1 in Mao and Szpruch (2013), it is obvious to obtain that the ST method has a unique solution if $\Delta <\frac{1}{L\theta }$. In the rest of this paper, we always assume that $\Delta <\frac{1}{L\theta }$.

To implement numerical scheme (11), we define a map F, let

$$\begin{aligned} F(X_{\kappa m+j})=X_{\kappa m+j}-\theta f(X_{\kappa m+j},X_{\kappa m},r_{\kappa m+j}^{\Delta })\Delta ,~\kappa \in {\mathbb {N}}, j=0,1,\dots , m-1, \end{aligned}$$

(12)

then we can represent (11) as follows:

$$\begin{aligned} F(X_{sm+l+1})=F(X_{sm+l})+f(X_{sm+l},X_{sm},r_{sm+l}^{\Delta })\Delta +g(X_{sm+l},X_{sm},r_{sm+l}^{\Delta })\Delta B_{sm+l}, \end{aligned}$$

(13)

for $l=0,1,\dots , m-2,$

$$\begin{aligned} \begin{aligned} F(X_{(s+1)m})=&F(X_{(s+1)m-1})+f(X_{(s+1)m-1},X_{sm},r_{(s+1)m-1}^{\Delta })\Delta \\&+g(X_{(s+1)m-1},X_{sm},r_{(s+1)m-1}^{\Delta })\Delta B_{(s+1)m-1}\\&+\theta \left( f(X_{(s+1)m},X_{sm},r_{(s+1)m}^{\Delta })-f(X_{(s+1)m},X_{(s+1)m},r_{(s+1)m}^{\Delta })\right) \Delta , \end{aligned} \end{aligned}$$

(14)

for $l=m-1$.

According to Assumption 3.1, there exists an inverse mapping $F^{-1}$ and the solution to (11) can be represented in the following form:

$$\begin{aligned} X_{sm+l+1}=F^{-1}\Big (X_{sm+l}+(1-\theta )f(X_{sm+l},X_{sm},r_{sm+l}^{\Delta })\Delta +g(X_{sm+l},X_{sm},r_{sm+l}^{\Delta })\Delta B_{sm+l}\Big ), \end{aligned}$$

for $l=0,1,\dots , m-2,$

$$\begin{aligned} \begin{aligned} X_{sm+l+1}=F^{-1}\Bigg (X_{sm+l}+(1-\theta )f(X_{sm+l},X_{sm},r_{sm+l}^{\Delta })\Delta +g(X_{sm+l},X_{sm},r_{sm+l}^{\Delta })\Delta B_{sm+l}\\ \qquad \qquad +\theta \left( f(X_{(s+1)m},X_{sm},r_{(s+1)m}^{\Delta })-f(X_{(s+1)m},X_{(s+1)m},r_{(s+1)m}^{\Delta })\right) \Delta \Bigg ), \end{aligned} \end{aligned}$$

for $l=m-1$. Clearly, $X_{sm+l+1}$ is ${\mathcal {F}}_{t_{sm+l+1}}$-measurable.

4 pth moment boundedness of the ST method

Throughout this section, we fix $T>0$ be arbitrary and show that the pth moment of the ST method is bounded. The following lemma shows that to guarantee the boundedness of moments for $X_{sm+l}$ it is enough to bound the moments of $F(X_{sm+l})$, where $F(X_{sm+l})$ is defined by (12).

Lemma 4.1

Suppose that Assumption 2.2 holds. Let $\delta$ be any given constant with $1-4\alpha \theta \Delta \ge \delta >0$. Then for any $p\ge 2$,

$$\begin{aligned} \Vert X_{sm+l}\Vert ^p\le 3^{\frac{p}{2}-1}\delta ^{-\frac{p}{2}}\left\{ \Vert F(X_{sm+l})\Vert ^p+\left( 1-\delta \right) ^{\frac{p}{2}}\Vert X_{sm}\Vert ^p+\left( 1-\delta \right) ^{\frac{p}{2}}\right\} . \end{aligned}$$

Moreover,

$$\begin{aligned} \Vert X_{sm+l}\Vert ^p\le 3^{\frac{p}{2}-1}\delta ^{-\frac{p}{2}}\left\{ \Vert F(X_{sm+l})\Vert ^p+\left( 1/\delta -1\right) ^{\frac{p}{2}}\Vert F(X_{sm})\Vert ^p+\left( 1-\delta \right) ^{\frac{p}{2}}\right\} . \end{aligned}$$

Proof

Using Assumption 2.2, we can arrive at

$$\begin{aligned} \begin{aligned} \Vert F(X_{sm+l})\Vert ^2=&\Vert X_{sm+l}-\theta f(X_{sm+l},X_{sm},r_{sm+l}^{\Delta })\Delta \Vert ^2\\ \ge&\Vert X_{sm+l}\Vert ^2-2\theta \Delta X_{sm+l}^\mathrm{{T}}f(X_{sm+l},X_{sm},r_{sm+l}^{\Delta })\\ \ge&\Vert X_{sm+l}\Vert ^2-2\alpha \theta \Delta (1+\Vert X_{sm+l}\Vert ^2+\Vert X_{sm}\Vert ^2), \end{aligned} \end{aligned}$$

(15)

which implies

$$\begin{aligned} \Vert X_{sm+l}\Vert ^2\le (1-2\alpha \theta \Delta )^{-1}\Vert F(X_{sm+l})\Vert ^2+\frac{2\alpha \theta \Delta }{1-2\alpha \theta \Delta }\Vert X_{sm}\Vert ^2+\frac{2\alpha \theta \Delta }{1-2\alpha \theta \Delta }, \end{aligned}$$

(16)

then applying the inequality $(x+y+z)^{\frac{p}{2}}\le 3^{\frac{p}{2}-1}(x^{\frac{p}{2}}+y^{\frac{p}{2}}+z^{\frac{p}{2}}),\forall x,y,z>0$, and the fact that $1-4\alpha \theta \Delta \ge \delta$, we obtain

$$\begin{aligned} \begin{aligned} \Vert X_{sm+l}\Vert ^p\le&3^{\frac{p}{2}-1}\delta ^{-\frac{p}{2}}\left\{ \Vert F(X_{sm+l})\Vert ^p+\left( 1-\delta \right) ^{\frac{p}{2}}\Vert X_{sm}\Vert ^p+\left( 1-\delta \right) ^{\frac{p}{2}}\right\} . \end{aligned} \end{aligned}$$

(17)

Furthermore, if $l=0$, one can get that

$$\begin{aligned} \Vert X_{sm}\Vert ^2\le (1-4\alpha \theta \Delta )^{-1}\Vert F(X_{sm})\Vert ^2+\frac{2\alpha \theta \Delta }{1-4\alpha \theta \Delta }, \end{aligned}$$

(18)

from (15) directly, substituting (18) into (16), then

$$\begin{aligned} \Vert X_{sm+l}\Vert ^2&\le (1-2\alpha \theta \Delta )^{-1}\Vert F(X_{sm+l})\Vert ^2\nonumber \\&\quad +\frac{2\alpha \theta \Delta }{(1-2\alpha \theta \Delta )(1-4\alpha \theta \Delta )}\Vert F(X_{sm})\Vert ^2+\frac{2\alpha \theta \Delta }{1-4\alpha \theta \Delta }, \end{aligned}$$

(19)

which gives

$$\begin{aligned} \Vert X_{sm+l}\Vert ^p\le 3^{\frac{p}{2}-1}\delta ^{-\frac{p}{2}}\left\{ \Vert F(X_{sm+l})\Vert ^p+\left( 1/\delta -1\right) ^{\frac{p}{2}}\Vert F(X_{sm})\Vert ^p+\left( 1-\delta \right) ^{\frac{p}{2}}\right\} . \end{aligned}$$

(20)

In particular, it from (18) that

$$\begin{aligned} \begin{aligned} \Vert X_{sm}\Vert ^p\le 2^{\frac{p}{2}-1}\delta ^{-\frac{p}{2}}\left\{ \Vert F(X_{sm})\Vert ^p+\left( 1-\delta \right) ^{\frac{p}{2}}\right\} . \end{aligned} \end{aligned}$$

(21)

$\square$

In what follows, for notational simplicity, we use the convention that C represents a generic positive constant independent of $\Delta$, the value of which may vary with each appearance. For example, $C=C+C$ and $C=C\times C$ are understood in an appropriate sense. Moreover, we may give specific expressions of C when needed. Let us begin to establish the fundamental result of this paper that reveals the boundedness of the pth moment for the ST scheme.

Theorem 4.2

Let Assumptions 2.2–2.4 and 3.1 hold, and $\theta \ge 0.5$. Then for any $p\ge 2$, the ST scheme (11) has the following property:

$$\begin{aligned} {\mathbb {E}}\left\{ \sup _{0\le t_{sm+l}\le T}\Vert X_{sm+l}\Vert ^p\right\} \le C. \end{aligned}$$

Proof

For any $s\in {\mathbb {N}}, l=0,1,\dots ,m-2$, using Assumption 2.2 and $\theta \ge 0.5$, we have

$$\begin{aligned}&\Vert F(X_{sm+l+1})\Vert ^2 \nonumber \\&\quad =\Vert F(X_{sm+l})+f(X_{sm+l},X_{sm},r_{sm+l}^{\Delta })\Delta +g(X_{sm+l},X_{sm},r_{sm+l}^{\Delta })\Delta B_{sm+l}\Vert ^2\nonumber \\&\quad =\Vert F(X_{sm+l})\Vert ^2+2\langle X_{sm+l},f(X_{sm+l},X_{sm},r_{sm+l}^{\Delta })\rangle \Delta \nonumber \\&\qquad +(1-2\theta )\Vert f(X_{sm+l},X_{sm},r_{sm+l}^{\Delta })\Vert ^2\Delta ^2+\Vert g(X_{sm+l},X_{sm},r_{sm+l}^{\Delta })\Delta B_{sm+l} \Vert ^2\nonumber \\&\qquad +2\langle F(X_{sm+l})+f(X_{sm+l},X_{sm},r_{sm+l}^{\Delta })\Delta , g(X_{sm+l},X_{sm},r_{sm+l}^{\Delta })\Delta B_{sm+l}\rangle \nonumber \\&\quad \le \Vert F(X_{sm+l})\Vert ^2+2\alpha \Delta (1+\Vert X_{sm+l}\Vert ^2+\Vert X_{sm}\Vert ^2)+\Vert \Delta M_{sm+l}\Vert ^2+\Delta N_{sm+l}, \end{aligned}$$

(22)

where

$$\begin{aligned} \begin{aligned} \Delta M_{sm+l}=&g(X_{sm+l},X_{sm},r_{sm+l}^{\Delta })\Delta B_{sm+l},\\ \Delta N_{sm+l}=&2\langle F(X_{sm+l})+f(X_{sm+l},X_{sm},r_{sm+l}^{\Delta })\Delta ,\Delta M_{sm+l}\rangle . \end{aligned} \end{aligned}$$

Then we can infer that

$$\begin{aligned} \begin{aligned} \Vert F(X_{sm+l+1})\Vert ^2\le&\Vert F(X_{sm})\Vert ^2+2\alpha +2\alpha \Vert X_{sm}\Vert ^2+2\alpha \Delta \sum _{i=sm}^{sm+l}\Vert X_{i}\Vert ^2\\&+\sum _{i=sm}^{sm+l}\Vert \Delta M_{i}\Vert ^2+\sum _{i=sm}^{sm+l}\Delta N_{i}, \end{aligned} \end{aligned}$$

substituting (18) and (19) into the last equation and recalling that $1-4\alpha \theta \Delta \ge \delta$, we acquire

$$\begin{aligned} \begin{aligned} \Vert F(X_{sm+l+1})\Vert ^2\le&\frac{2\alpha \Delta }{\delta }\sum _{i=sm}^{sm+l}\Vert F(X_{i})\Vert ^2+\left( 1+\frac{2\alpha }{\delta ^2}\right) \Vert F(X_{sm})\Vert ^2+\frac{2\alpha }{\delta }\\&+\sum _{i=sm}^{sm+l}\Vert \Delta M_{i}\Vert ^2+\sum _{i=sm}^{sm+l}\Delta N_{i}. \end{aligned} \end{aligned}$$

Thus, according to the inequality $\left( \sum _{i=1}^k\vert a_i\vert \right) ^p\le k^{p-1}\sum _{i=1}^k\vert a_i\vert ^p$, $p\ge 1$, one can arrive at

$$\begin{aligned} \begin{aligned}&{\mathbb {E}}\left\{ \sup _{j=0,1,\dots ,l}\Vert F(X_{sm+j+1})\Vert ^p\right\} \\&\quad \le C_1\Delta \sum _{i=sm}^{sm+l}{\mathbb {E}}\Vert F(X_{i})\Vert ^p+C_2{\mathbb {E}}\Vert F(X_{sm})\Vert ^p+C_1\\&\qquad +C_3(l+1)^{\frac{p}{2}-1}\sum _{i=sm}^{sm+l}{\mathbb {E}}\Vert \Delta M_{i}\Vert ^{p}+C_3\underbrace{{\mathbb {E}}\left( \sup _{j=0,1,\dots ,l}\sum _{i=sm}^{sm+j}\Delta N_{i}\right) ^{\frac{p}{2}}}_A, \end{aligned} \end{aligned}$$

(23)

where $C_1=5^{\frac{p}{2}-1}\left( \frac{2\alpha }{\delta }\right) ^{\frac{p}{2}}, C_2=5^{\frac{p}{2}-1}\left( 1+\frac{2\alpha }{\delta ^2}\right) ^{\frac{p}{2}}, C_3=5^{\frac{p}{2}-1}$, and we use the fact that $l\Delta \le 1, l=0,1,\dots ,m-2.$

Since $\Delta B_{i}$ is ${\mathcal {F}}_{t_{i}}$-independent, with the help of Hölder’s inequality and Assumption 2.4, we have

$$\begin{aligned} \begin{aligned} {\mathbb {E}}\Vert \Delta M_{i}\Vert ^{p}\le&{\mathbb {E}}\Vert g(X_{i},X_{sm},r_{i}^{\Delta }) \Vert ^{p}({\mathbb {E}}\Vert \Delta B_{i} \Vert ^{2p})^{\frac{1}{2}}\\ \le&C\Delta ^{\frac{p}{2}}L_2^{p} 3^{p-1}{\mathbb {E}}\left( 1+\Vert X_{i}\Vert ^{p}+\Vert X_{sm}\Vert ^{ph_2}\right) , \end{aligned} \end{aligned}$$

(24)

combining (20) and (21), we can obtain that

$$\begin{aligned} \begin{aligned} {\mathbb {E}}\Vert \Delta M_{i}\Vert ^{p}\le&C\Delta ^{\frac{p}{2}}\Big (C_4{\mathbb {E}}\Vert F(X_{i})\Vert ^{p}+C_5{\mathbb {E}}\Vert F(X_{sm})\Vert ^{p}+C_6{\mathbb {E}}\Vert F(X_{sm})\Vert ^{ph_2}+C_7\Big ), \end{aligned} \end{aligned}$$

(25)

where

$$\begin{aligned}&C_4=L_2^{p} 3^{\frac{3p}{2}-2}\delta ^{-\frac{p}{2}},\quad C_5=L_2^{p} 3^{\frac{3p}{2}-2}\left( 1-\delta \right) ^{\frac{p}{2}}\delta ^{-p},\quad C_6=L_2^{p} 3^{p-1}2^{\frac{ph_2}{2}-1}\delta ^{-\frac{ph_2}{2}},\\&C_7=L_2^{p} 3^{p-1}\left( 1+3^{\frac{p}{2}-1}\left( 1-\delta \right) ^{\frac{p}{2}}\delta ^{-\frac{p}{2}}+2^{\frac{ph_2}{2}-1}\left( 1-\delta \right) ^{\frac{ph_2}{2}}\delta ^{-\frac{ph_2}{2}}\right) . \end{aligned}$$

According to the definition of $\Delta N_{sm+l}$ and $F(X_{sm+l})$, using the time discrete Burkholder–Davis–Gundy type inequality, we yield

$$\begin{aligned} A\le&\left( \frac{2}{\theta }\right) ^{\frac{p}{2}}{\mathbb {E}}\left( \sup _{j=0,1,\dots ,l}\sum _{i=sm}^{sm+j}\left( X_i+(\theta -1)F(X_i)\right) ^\mathrm{{T}}g(X_i,X_{sm},r_i^{\Delta })\Delta B_i\right) ^{\frac{p}{2}}\nonumber \\ \le&\frac{p}{2}\left( \frac{2}{\theta }\right) ^{\frac{p}{2}}\Delta ^{\frac{p}{4}}\left( \sum _{i=sm}^{sm+l}\left( {\mathbb {E}}\Vert (X_i+(\theta -1)F(X_i))^\mathrm{{T}}g(X_i,X_{sm},r_i^{\Delta })\Vert ^{\frac{p}{2}}\right) ^{\frac{4}{p}}\right) ^{\frac{p}{4}}\nonumber \\ \le&\frac{p}{2}\left( \frac{2}{\theta }\right) ^{\frac{p}{2}}\Delta ^{\frac{p}{4}}(l+1)^{\frac{p}{4}-1}\sum _{i=sm}^{sm+l}{\mathbb {E}}\Vert (X_i+(\theta -1)F(X_i))^\mathrm{{T}}g(X_i,X_{sm},r_i^{\Delta })\Vert ^{\frac{p}{2}}\nonumber \\ \le&\frac{p}{4}\left( \frac{2}{\theta }\right) ^{\frac{p}{2}}\Delta \sum _{i=sm}^{sm+l}\left( {\mathbb {E}}\Vert X_i+(\theta -1)F(X_i)\Vert ^p+{\mathbb {E}}\Vert g(X_i,X_{sm},r_i^{\Delta })\Vert ^p\right) \nonumber \\ \le&2^{\frac{3p}{2}-3}p\Delta \theta ^{-\frac{p}{2}}\sum _{i=sm}^{sm+l}\left\{ {\mathbb {E}}\Vert X_i\Vert ^p+{\mathbb {E}}\Vert F(X_i)\Vert ^p\right\} +2^{\frac{p}{2}-2}p\theta ^{-\frac{p}{2}}\Delta \sum _{i=sm}^{sm+l}{\mathbb {E}}\Vert g(X_i,X_{sm},r_i^{\Delta })\Vert ^p. \end{aligned}$$

(26)

Similar to (24) and (25), one can get that

$$\begin{aligned} {\mathbb {E}}\Vert g(X_i,X_{sm},r_i^{\Delta })\Vert ^p\le C_4{\mathbb {E}}\Vert F(X_{i})\Vert ^{p}+C_5{\mathbb {E}}\Vert F(X_{sm})\Vert ^{p}+C_6{\mathbb {E}}\Vert F(X_{sm})\Vert ^{ph_2}+C_7.\nonumber \\ \end{aligned}$$

(27)

By substituting (20) and (27) into (26), we have

$$\begin{aligned} A\le C_{8}\Delta \sum _{i=sm}^{sm+l}{\mathbb {E}}\Vert F(X_{i})\Vert ^p+C_{9}{\mathbb {E}}\Vert F(X_{sm})\Vert ^p+C_{10}{\mathbb {E}}\Vert F(X_{sm})\Vert ^{ph_2}+C_{11}, \end{aligned}$$

(28)

where

$$\begin{aligned} \begin{aligned} C_{8}&=2^{\frac{3p}{2}-3}p\theta ^{-\frac{p}{2}}\left( 1+3^{\frac{p}{2}-1}\delta ^{-\frac{p}{2}}\right) +2^{\frac{p}{2}-2}\theta ^{-\frac{p}{2}}pC_4,\\ C_{9}&=2^{\frac{3p}{2}-3}p\theta ^{-\frac{p}{2}}3^{\frac{p}{2}-1}\left( 1-\delta \right) ^{\frac{p}{2}}\delta ^{-p}+2^{\frac{p}{2}-2}\theta ^{-\frac{p}{2}}pC_5,\\ C_{10}&=2^{\frac{p}{2}-2}\theta ^{-\frac{p}{2}}pC_6,\quad C_{11}=2^{\frac{3p}{2}-3}p\theta ^{-\frac{p}{2}}3^{\frac{p}{2}-1}\left( 1-\delta \right) ^{\frac{p}{2}}\delta ^{-\frac{p}{2}}+2^{\frac{p}{2}-2}\theta ^{-\frac{p}{2}}pC_7. \end{aligned} \end{aligned}$$

It follows from (23), (25) and (28) that

$$\begin{aligned} \begin{aligned} {\mathbb {E}}\left\{ \sup _{j=0,1,\dots ,l}\Vert F(X_{sm+j+1})\Vert ^p\right\} \le&C_{12}\Delta \sum _{i=0}^{l-1}{\mathbb {E}}\left\{ \sup _{j=0,1,\dots ,i}\Vert F(X_{sm+j+1})\Vert ^p\right\} \\&+(C_{12}\Delta +C_{13}){\mathbb {E}}\Vert F(X_{sm})\Vert ^p+ C_{14}{\mathbb {E}}\Vert F(X_{sm})\Vert ^{ph_2}+ C_{15}, \end{aligned} \end{aligned}$$

where we set $\sum _{i=0}^{-1}{\mathbb {E}}\Vert F(X_{sm+i+1})\Vert ^p=0$ as usual. Here $C_{12}=C_1+C_3C_4C+C_3C_{8}$, $C_{13}=C_2+C_3C_5C+C_3C_{9}$, $C_{14}=C_3C_6C+C_3C_{10}$, $C_{15}=C_1+C_3C_7C+C_3C_{11}$. Using the discrete Gronwall inequality (Theorem 2.5 in Mao and Yuan 2006), we obtain

$$\begin{aligned}&{\mathbb {E}}\left\{ \sup _{j=0,1,\dots ,l}\Vert F(X_{sm+j+1})\Vert ^p\right\} \nonumber \\&\quad \le \left( (C_{12}\Delta +C_{13}){\mathbb {E}}\Vert F(X_{sm})\Vert ^p+ C_{14}{\mathbb {E}}\Vert F(X_{sm})\Vert ^{ph_2}+ C_{15}\right) e^{C_{12}}. \end{aligned}$$

(29)

By induction, we divide the proof into several steps to show

$$\begin{aligned} {\mathbb {E}}\left\{ \sup _{0\le t_{sm+l}\le T}\Vert X_{sm+l}\Vert ^p\right\} \le C. \end{aligned}$$

Step 1. For $s=0, l=0,1,\dots , m-2$, (29) implies

$$\begin{aligned} {\mathbb {E}}\left\{ \sup _{j=0,1,\dots ,l}\Vert F(X_{j+1})\Vert ^p\right\} \le \left( (C_{12}\Delta +C_{13})\Vert F(X_{0})\Vert ^p+C_{14}\Vert F(X_{0})\Vert ^{ph_2}+ C_{15}\right) e^{C_{12}}.\nonumber \\ \end{aligned}$$

(30)

Noting that $F(X_0)=X_0-f(X_0,X_0,i_0)\Delta , X_0=x_0$, using Assumption 2.3, we can easily get that

$$\begin{aligned} \Vert F(X_{0})\Vert ^p\le C(p,\Vert x_0\Vert ,L_1,h_1)\qquad \mathrm{{and}}\qquad \Vert F(X_{0})\Vert ^{ph_2}\le C(p,\Vert x_0\Vert ,L_1,h_1,h_2). \end{aligned}$$

Substituting the last equations into (30) we have

$$\begin{aligned} \begin{aligned} {\mathbb {E}}\left\{ \sup _{j=0,\dots ,m-1}\Vert F(X_{j})\Vert ^p\right\} \le C, \end{aligned} \end{aligned}$$

(31)

and it follows from Lemma 4.1 that

$$\begin{aligned} {\mathbb {E}}\left\{ \sup _{j=0,1,\dots ,m-1}\Vert X_{j}\Vert ^p\right\} \le C. \end{aligned}$$

(32)

Repeating the procedures as discussed above, we can also get that

$$\begin{aligned} {\mathbb {E}}\left\{ \sup _{j=0,\dots ,m-1}\Vert F(X_{j})\Vert ^{ph_2}\right\} \le C \quad \mathrm{{and}}\quad {\mathbb {E}}\left\{ \sup _{j=0,1,\dots ,m-1}\Vert X_{j}\Vert ^{ph_1}\right\} \le C. \end{aligned}$$

(33)

Next we show that ${\mathbb {E}}\Vert X_m\Vert ^p\le C$ and ${\mathbb {E}}\Vert F(X_m)\Vert ^p\le C$. Applying (12), (13) and (14) again, by Assumptions 2.3 and 2.4, one has

$$\begin{aligned} \begin{aligned}&\Vert X_m-\theta f(X_m,X_0,r_m^{\Delta })\Delta \Vert ^2\\&\quad =\Vert F(X_{m-1})+f(X_{m-1},X_0,r_{m-1}^{\Delta })\Delta +g(X_{m-1},X_0,r_{m-1}^{\Delta })\Delta B_{m-1}\Vert ^2\\&\quad \le 3\Vert F(X_{m-1})\Vert ^2+3\Vert f(X_{m-1},X_0,r_{m-1}^{\Delta })\Delta \Vert ^2+3\Vert g(X_{m-1},X_0,r_{m-1}^{\Delta })\Delta B_{m-1}\Vert ^2,\\ \end{aligned} \end{aligned}$$

hence by (31)–(33), we infer that

$$\begin{aligned} \begin{aligned}&{\mathbb {E}}\Vert X_m-\theta f(X_m,X_0,r_m^{\Delta })\Delta \Vert ^p\\&\quad \le 3^{p-1}\bigg \{{\mathbb {E}}\Vert F(X_{m-1})\Vert ^p+{\mathbb {E}}\Vert f(X_{m-1},X_0,r_{m-1}^{\Delta })\Delta \Vert ^p+{\mathbb {E}}\Vert g(X_{m-1},X_0,r_{m-1}^{\Delta })\Delta B_{m-1}\Vert ^p\bigg \}\\&\quad \le 3^{p-1}\bigg \{{\mathbb {E}}\Vert F(X_{m-1})\Vert ^p+3^{p-1}L_1^{p}(1+{\mathbb {E}}\Vert X_{m-1}\Vert ^{ph_1}+\Vert X_0\Vert ^{ph_1})\Delta ^p\\&\qquad +3^{p-1}L_2^{p}(1+{\mathbb {E}}\Vert X_{m-1}\Vert ^{p}+\Vert X_0\Vert ^{ph_2})({\mathbb {E}}\Vert \Delta B_{m-1}\Vert ^{2p})^{\frac{1}{2}}\bigg \}\\&\quad \le C. \end{aligned} \end{aligned}$$

Moreover, repeating the process (15)–(17), one can get that

$$\begin{aligned} \begin{aligned} {\mathbb {E}}\Vert X_{m}\Vert ^p\le&3^{\frac{p}{2}-1}\Bigg \{\delta ^{-\frac{p}{2}}{\mathbb {E}}\Vert X_m-\theta f(X_m,X_0,r_m^{\Delta })\Delta \Vert ^p+\left( \frac{1-\delta }{\delta }\right) ^{\frac{p}{2}}\Vert X_{0}\Vert ^p+\left( \frac{1-\delta }{\delta }\right) ^{\frac{p}{2}}\Bigg \}\\ \le&C. \end{aligned} \end{aligned}$$

Using $F(X_{sm+l})=X_{sm+l}-\theta f(X_{sm+l},X_{sm},r_{sm+l}^{\Delta })\Delta$ again, we obtain

$$\begin{aligned} \begin{aligned} {\mathbb {E}}\Vert F(X_{m})\Vert ^p=&{\mathbb {E}}\Vert X_{m}-\theta f(X_{m},X_{m},r_{m}^{\Delta })\Delta \Vert ^p\\ \le&2^{p-1}{\mathbb {E}}\Vert X_{m}\Vert ^p+2^{p-1}\theta ^p\Delta ^p{\mathbb {E}} \Vert f(X_{m},X_{m},r_{m}^{\Delta })\Vert ^p\\ \le&2^{p-1}{\mathbb {E}}\Vert X_{m}\Vert ^p+6^{p-1}\theta ^p\Delta ^pL_1^p(1+2{\mathbb {E}}\Vert X_{m}\Vert ^p)\\ \le&C. \end{aligned} \end{aligned}$$

Step 2. For $s=1$, according to (29), we have

$$\begin{aligned} \begin{aligned}&{\mathbb {E}}\left\{ \sup _{j=0,1,\dots ,m-2}\Vert F(X_{m+j+1})\Vert ^p\right\} \\&\quad \le \left( (C_{12}\Delta +C_{13}){\mathbb {E}}\Vert F(X_{m})\Vert ^p+C_{14}{\mathbb {E}}\Vert F(X_{m})\Vert ^{ph_2}+ C_{15}\right) e^{C_{12}}\\&\quad \le \left( (C_{12}\Delta +C_{13})C+C_{14}C+ C_{15}\right) e^{C_{12}}:=C,\\ \end{aligned} \end{aligned}$$

which gives

$$\begin{aligned} {\mathbb {E}}\left\{ \sup _{j=1,2\dots ,m-1}\Vert F(X_{m+j})\Vert ^p\right\} \le C, \end{aligned}$$

then it follows from Lemma 4.1 that

$$\begin{aligned} {\mathbb {E}}\left\{ \sup _{j=1,2\dots ,m-1}\Vert X_{m+j}\Vert ^p\right\} \le C. \end{aligned}$$

Adopting the same procedures as in the Step 1, we can arrive at ${\mathbb {E}}\Vert X_{2m}\Vert ^p\le C$, then ${\mathbb {E}}\Vert F(X_{2m})\Vert ^p\le C$ follows from (12).

Step 3. For $s\in \{2,3,\dots ,[T]\}$, the following assertion can be proved in the same way as shown before, for any fixed T, there exists a constant C independent of $\Delta$ such that

$$\begin{aligned} {\mathbb {E}}\left\{ \sup _{j=1,\dots ,m}\Vert F(X_{sm+j})\Vert ^{p}\right\} \le C\qquad \mathrm{and }\qquad {\mathbb {E}}\left\{ \sup _{j=1,\dots ,m}\Vert X_{sm+j}\Vert ^{p}\right\} \le C. \end{aligned}$$

Combining Steps 1–3, we can get that

$$\begin{aligned} {\mathbb {E}}\left\{ \sup _{0\le t_{sm+l}\le T}\Vert X_{sm+l}\Vert ^p\right\} \le C, \end{aligned}$$

for any fixed T. The proof is completed. $\square$

5 Rate of strong convergence

It is convenient to work with a continuous extension of a numerical method here, because the continuous extension enables us to use the powerful continuous-time stochastic analysis to formulate theorems on numerical approximations. For this purpose, we introduce a new numerical scheme, which is called the forward–backward Euler–Maruyama (FBEM) scheme, to help us get a well-defined continuous-time numerical approximation.

First we compute the discrete values $X_{sm+l}$ from the ST method, then we define the discrete FBEM scheme on $[s,s+1)\subset [0,\infty ),s\in {\mathbb {N}}$ by

$$\begin{aligned} {\hat{X}}_{sm+l+1}={\hat{X}}_{sm+l}+f(X_{sm+l},X_{sm},r_{sm+l}^{\Delta })\Delta +g(X_{sm+l},X_{sm},r_{sm+l}^{\Delta })\Delta B_{sm+l}, \end{aligned}$$

(34)

where $l=0,1,\dots , m-1, {\hat{X}}_0=X_0=x(0)$.

Let $X(t)=X_{sm+l}$, ${\bar{X}}(t)={\hat{X}}_{sm+l}$, $r^{\Delta }(t)=r_{sm+l}^{\Delta }$ for $t\in [t_{sm+l},t_{sm+l+1})$, and the continuous FBEM scheme is defined by

$$\begin{aligned} {\hat{X}}(t)={\hat{X}}_{0}+\int _{0}^t f(X(u),X(s),r^{\Delta }(u))\mathrm{{d}}u+\int _{0}^t g(X(u),X(s),r^{\Delta }(u)) \mathrm{{d}}B(u), \end{aligned}$$

(35)

on each interval $[t_{sm+l},t_{sm+l+1})$. We would like to remark that the continuous and discrete FBEM schemes coincide at the grid points, that is, ${\hat{X}}(t_{sm+l})={\hat{X}}_{sm+l}$.

Now we impose some stronger versions of Assumptions 2.2–2.4 to get the convergence rate.

Assumption 5.1

For any constant $K>0$, there exists a positive constant $K_1$ such that

$$\begin{aligned}&(x-{\bar{x}})^\mathrm{{T}}(f(x,y,i)-f({\bar{x}},{\bar{y}},i))+K\Vert g(x,y,i)-g({\bar{x}},{\bar{y}},i)\Vert ^2\\&\quad \le K_1(\Vert x-{\bar{x}}\Vert ^2+\Vert y-{\bar{y}}\Vert ^2), \end{aligned}$$

for all $i\in S$ and $x, {\bar{x}},y, {\bar{y}}\in {\mathbb {R}}^n$.

Assumption 5.2

There exist constants $K_2>0$ and $\rho _1\ge 1$ such that

$$\begin{aligned} \begin{aligned} \Vert f(x,y,i)-f({\bar{x}},{\bar{y}},i)\Vert&\le K_2(1+\Vert x\Vert ^{\rho _1}+\Vert y\Vert ^{\rho _1}+\Vert {\bar{x}}\Vert ^{\rho _1}\\&\quad +\Vert {\bar{y}}\Vert ^{\rho _1})\left( \Vert x-{\bar{x}}\Vert +\Vert y-{\bar{y}}\Vert \right) , \end{aligned} \end{aligned}$$

for all $i\in S$ and $x,y,{\bar{x}},{\bar{y}}\in {\mathbb {R}}^n.$

Assumption 5.3

There exist constants $K_3>0,K_4>0$ and $\rho _2\ge 1$ such that

$$\begin{aligned}&\Vert g(x,y,i)-g({\bar{x}},y,i)\Vert \le K_3\Vert x-{\bar{x}}\Vert , \\&\Vert g(x,y,i)-g(x,{\bar{y}},i)\Vert \le K_4(1+\Vert y\Vert ^{\rho _2}+\Vert {\bar{y}}\Vert ^{\rho _2})\Vert y-{\bar{y}}\Vert , \end{aligned}$$

for all $i\in S$ and $x, y, {\bar{x}},{\bar{y}}\in {\mathbb {R}}^n$.

Assumption 5.4

There exists a positive constant $K_5$ such that

$$\begin{aligned} \Vert f(0,0,i)\Vert \vee \Vert g(0,0,i)\Vert \le K_5, \end{aligned}$$

for all $i\in S$.

Remark 2

Assumptions 5.1–5.4 imply Assumptions 2.2–2.4. Suppose that Assumptions 5.1–5.4 hold, we can easily get that

$$\begin{aligned} \begin{aligned} x^\mathrm{{T}}f(x,y,i)=&(x-0)^\mathrm{{T}}(f(x,y,i)-f(0,0,i))+x^\mathrm{{T}}f(0,0,i)\\ \le&K_1(\Vert x\Vert ^2+\Vert y\Vert ^2)+\frac{1}{2}\Vert x\Vert ^2+\frac{1}{2}\Vert f(0,0,i)\Vert ^2\\ \le&\left( K_1+\frac{1}{2}+\frac{1}{2}K_5^2\right) (1+\Vert x\Vert ^2+\Vert y\Vert ^2), \end{aligned} \end{aligned}$$

which is Assumption 2.2, and

$$\begin{aligned} \begin{aligned} \Vert f(x,y,i)\Vert \le&\Vert f(x,y,i)-f(0,0,i)\Vert +\Vert f(0,0,i)\Vert \\ \le&K_2(1+\Vert x\Vert ^{\rho _1}+\Vert y\Vert ^{\rho _1})(\Vert x\Vert +\Vert y\Vert )+K_5\\ \le&(6K_2+K_5)\left( 1+\Vert x\Vert ^{2\rho _1}+\Vert y\Vert ^{2\rho _1}\right) , \end{aligned} \end{aligned}$$

which is Assumption 2.3, as well as

$$\begin{aligned} \begin{aligned} \Vert g(x,y,i)\Vert \le&\Vert g(x,y,i)-g(0,y,i)\Vert + \Vert g(0,y,i)-g(0,0,i)\Vert +\Vert g(0,0,i)\Vert \\ \le&K_3\Vert x\Vert +K_4(1+\Vert y\Vert ^{\rho _2})\Vert y\Vert +K_5\\ \le&K_6(1+\Vert x\Vert +\Vert y\Vert ^{\rho _2+1}), \end{aligned} \end{aligned}$$

which is Assumption 2.4, where $K_6=K_3+2K_4+K_5$.

Under the assumptions above, we can determine the rate of strong convergence for ST scheme (11) to the solution of (1). First, we need some lemmas.

Lemma 5.5

Let Assumptions 5.1–5.4 hold, and $\theta \ge 0.5$. Then for any $p\ge 2$ and sufficiently small step size $\Delta$, ${\hat{X}}_{sm+l}$ and $X_{sm+l}$ obey

$$\begin{aligned} {\mathbb {E}}\left\{ \sup _{0\le t_{sm+l}\le T}\Vert {\hat{X}}_{sm+l}-X_{sm+l}\Vert ^p\right\} \le C\Delta ^p. \end{aligned}$$

(36)

Proof

For any $s\in {\mathbb {N}}, l = 0,1,\dots ,m-1$, summing up both schemes of the discrete FBEM (34) and ST (11), respectively, we have

$$\begin{aligned} {\hat{X}}_{sm+l+1}-X_{sm+l+1}= & {} {\hat{X}}_{sm}-X_{sm}+\theta \big (f(X_{sm},X_{sm},r_{sm}^{\Delta })\nonumber \\&-f(X_{sm+l+1},X_{sm},r_{sm+l+1}^{\Delta })\big )\Delta , \end{aligned}$$

(37)

then we can infer that

$$\begin{aligned} \begin{aligned} {\hat{X}}_{sm+l+1}-X_{sm+l+1}=&\theta \Delta \big (f(X_{0},X_{0},r_{0}^{\Delta })-f(X_{sm+l+1},X_{sm},r_{sm+l+1}^{\Delta })\big )\\&+\theta \Delta \sum _{i=1}^s\big (f(X_{im},X_{im},r_{im}^{\Delta })-f(X_{im},X_{(i-1)m},r_{im}^{\Delta })\big ). \end{aligned} \end{aligned}$$

Thus, we obtain

$$\begin{aligned}&\Vert {\hat{X}}_{sm+l+1}-X_{sm+l+1}\Vert ^p \nonumber \\&\quad \le 4^{p-1}\theta ^p\Delta ^p\left( \Vert f(X_{0},X_{0},r_{0}^{\Delta })\Vert ^p+\Vert f(X_{sm+l+1},X_{sm},r_{sm+l+1}^{\Delta })\Vert ^p\right) \nonumber \\&\qquad +2^{p-1}\theta ^p\Delta ^ps^{p-1}\sum _{i=1}^s\Vert f(X_{im},X_{im},r_{im}^{\Delta })-f(X_{im},X_{(i-1)m},r_{im}^{\Delta })\Vert ^p. \end{aligned}$$

(38)

According to Assumptions 5.2, 5.4 and the inequality $(\vert a\vert +\vert b\vert )^p\le 2^{p-1}(\vert a\vert ^p+\vert b\vert ^p), p\ge 1$, we can acquire

$$\begin{aligned}&\Vert f(X_{sm+l+1},X_{sm},r_{sm+l+1}^{\Delta })\Vert ^p \nonumber \\&\quad \le 2^{p-1}\Vert f(X_{sm+l+1},X_{sm},r_{sm+l+1}^{\Delta })-f(0,0,r_{sm+l+1}^{\Delta })\Vert ^p+2^{p-1}\Vert f(0,0,r_{sm+l+1}^{\Delta })\Vert ^p\nonumber \\&\quad \le 2^{p-1}K_2^{p}(1+\Vert X_{sm+l+1}\Vert ^{\rho _1}+\Vert X_{sm}\Vert ^{\rho _1})^{p}(\Vert X_{sm+l+1}\Vert +\Vert X_{sm}\Vert )^{p}+2^{p-1}K_5^p\nonumber \\&\quad \le 2^{p-2}K_2^{p}\Big (3^{2p-1}\left( 1+\Vert X_{sm+l+1}\Vert ^{2p\rho _1}+\Vert X_{sm}\Vert ^{2p\rho _1}\right) \nonumber \\&\qquad +2^{2p-1}\left( \Vert X_{sm+l+1}\Vert ^{2p}+\Vert X_{sm}\Vert ^{2p}\right) \Big )+C. \end{aligned}$$

(39)

Similarly, one can also get that

$$\begin{aligned}&\Vert f(X_{im},X_{im},r_{im}^{\Delta })-f(X_{im},X_{(i-1)m},r_{im}^{\Delta })\Vert ^p \nonumber \\&\quad \le \frac{1}{2}K_2^{p}\Big (3^{2p-1}(1+3^{2p}\Vert X_{im}\Vert ^{2p\rho _1}+\Vert X_{(i-1)m}\Vert ^{2p\rho _1})\nonumber \\&\qquad +2^{2p-1}(\Vert X_{im}\Vert ^{2p}+\Vert X_{(i-1)m}\Vert ^{2p})\Big ). \end{aligned}$$

(40)

Substituting (39) and (40) into (38), with the help of Theorem 4.2, for $s\in [0,[T]]$, we yield

$$\begin{aligned} {\mathbb {E}}\left\{ \sup _{0\le t_{sm+l+1}\le T}\Vert {\hat{X}}_{sm+l+1}-X_{sm+l+1}\Vert ^p\right\} \le C\Delta ^p, \end{aligned}$$

and the assertion follows. $\square$

Lemma 5.6

Let Assumptions 5.1–5.4 hold, and $\theta \ge 0.5$. Then for any $p\ge 2$,

$$\begin{aligned} {\mathbb {E}}\left\{ \sup _{0\le t_{sm+l}\le T}\Vert {\hat{X}}_{sm+l}\Vert ^p\right\} \le C,\qquad {\mathbb {E}}\left\{ \sup _{0\le t\le T}\Vert {\hat{X}}(t)\Vert ^p\right\} \le C. \end{aligned}$$

Proof

It follows from Theorem 4.2 and Lemma 5.5 that

$$\begin{aligned} \begin{aligned} {\mathbb {E}}\left\{ \sup _{0\le t_{sm+l}\le T}\Vert {\hat{X}}_{sm+l}\Vert ^p\right\} \le&2^{p-1}{\mathbb {E}}\left\{ \sup _{0\le t_{sm+l}\le T}\Vert {\hat{X}}_{sm+l}-X_{sm+l}\Vert ^p\right\} \\&+2^{p-1}{\mathbb {E}}\left\{ \sup _{0\le t_{sm+l}\le T}\Vert X_{sm+l}\Vert ^p\right\} \le C. \end{aligned} \end{aligned}$$

(41)

Moreover, according to (35) and Hölder’s inequality,

$$\begin{aligned} \begin{aligned} \sup _{t_{sm+l}\le t< t_{sm+l+1}}\Vert {\hat{X}}(t)\Vert ^p\le&3^{p-1}\bigg (\Vert {\hat{X}}_{sm+l}\Vert ^p+\Delta ^{p}\Vert f(X_{sm+l},X_{sm},r_{sm+l}^{\Delta })\Vert ^p\\&+\sup _{t_{sm+l}\le t< t_{sm+l+1}}\left\| \int _{t_{sm+l}}^{t}g(X_{sm+l},X_{sm},r_{sm+l}^{\Delta })\mathrm{{d}}B(u)\right\| ^p\bigg ),\\ \end{aligned} \end{aligned}$$

it follows that

$$\begin{aligned} \begin{aligned}&{\mathbb {E}}\left\{ \sup _{0\le t\le T}\Vert {\hat{X}}(t)\Vert ^p\right\} \\&\quad \le {\mathbb {E}}\left\{ \sup _{0\le t_{sm+l}\le T}\sup _{t_{sm+l}\le t< t_{sm+l+1}}\Vert {\hat{X}}(t)\Vert ^p\right\} \\&\quad \le 3^{p-1}{\mathbb {E}}\left\{ \sup _{0\le t_{sm+l}\le T}\Vert {\hat{X}}_{sm+l}\Vert ^p\right\} +3^{p-1}\Delta ^{p}{\mathbb {E}}\left\{ \sup _{0\le t_{sm+l}\le T}\Vert f(X_{sm+l},X_{sm},r_{sm+l}^{\Delta })\Vert ^p\right\} \\&\qquad +3^{p-1}\underbrace{{\mathbb {E}}\left\{ \sup _{0\le t_{sm+l}\le T}\sup _{t_{sm+l}\le t< t_{sm+l+1}}\left\| \int _{t_{sm+l}}^{t}g(X_{sm+l},X_{sm},r_{sm+l}^{\Delta })\mathrm{{d}}B(u)\right\| ^p\right\} }_{I}. \end{aligned}\nonumber \\ \end{aligned}$$

(42)

Similar to (39), from Theorem 4.2 we can get that

$$\begin{aligned} {\mathbb {E}}\left\{ \sup _{0\le t_{sm+l}\le T}\Vert f(X_{sm+l},X_{sm},r_{sm+l}^{\Delta })\Vert ^p\right\} \le C. \end{aligned}$$

(43)

According to Remark 2 and Theorem 1.7.2 in Mao (2007), together with Theorem 4.2 again, we can infer that

$$\begin{aligned} I\le&\sum _{s=0}^{[T]}\sum _{l=0}^{m-1}{\mathbb {E}}\left\{ \sup _{t_{sm+l}\le t< t_{sm+l+1}}\left\| \int _{t_{sm+l}}^t g(X_{sm+l},X_{sm},r_{sm+l}^{\Delta })\mathrm{{d}}B(u)\right\| ^p\right\} \nonumber \\&\quad \le \sum _{s=0}^{[T]}\sum _{l=0}^{m-1}\left( \frac{p^3}{2(p-1)}\right) ^{\frac{p}{2}}\Delta ^{\frac{p}{2}}{\mathbb {E}}\Vert g(X_{sm+l},X_{sm},r_{sm+l}^{\Delta })\Vert ^p\nonumber \\&\quad \le 4^{p-1}\left( \frac{p^3}{2(p-1)}\right) ^{\frac{p}{2}}K_6^p\Delta ^{\frac{p}{2}}\sum _{s=0}^{[T]}\sum _{l=0}^{m-1}\left( 1+{\mathbb {E}}\Vert X_{sm+l}\Vert ^p+{\mathbb {E}}\Vert X_{sm}\Vert ^{p(\rho _2+1)}\right) \nonumber \\&\quad \le 4^{p-1}\left( \frac{p^3}{2(p-1)}\right) ^{\frac{p}{2}}K_6^p\Delta ^{\frac{p}{2}-1}([T]+1) \Bigg (1+{\mathbb {E}}\left\{ \sup _{0\le t_{sm+l}\le T}\Vert X_{sm+l}\Vert ^p\right\} \nonumber \\&\qquad +{\mathbb {E}}\left\{ \sup _{0\le t_{sm+l}\le T}\Vert X_{sm}\Vert ^{p(\rho _2+1)}\right\} \Bigg )\nonumber \\&\quad \le C\Delta ^{\frac{p}{2}-1}. \end{aligned}$$

(44)

Substituting (41), (43) and (44) into (42), one has ${\mathbb {E}}\left\{ \sup _{0\le t\le T}\Vert {\hat{X}}(t)\Vert ^p\right\} \le C$. $\square$

Lemma 5.7

Let Assumptions 5.1–5.4 hold, and $\theta \ge 0.5$, then for any $p\ge 2$,

$$\begin{aligned} \sup _{0\le t\le T}{\mathbb {E}}\Vert {\hat{X}}(t)-X(t)\Vert ^p\le C\Delta ^{\frac{p}{2}}. \end{aligned}$$

Proof

For any $t\in [0,T]$, there always exist $s\in {\mathbb {N}}$ and $l\in \{0,1,\dots , m-1\}$ such that $t\in [t_{sm+l}, t_{sm+l+1})$; hence,

$$\begin{aligned} {\mathbb {E}}\Vert {\hat{X}}(t)-X(t)\Vert ^p\le & {} 2^{p-1}{\mathbb {E}}\Vert {\hat{X}}(t)-{\bar{X}}(t)\Vert ^p+2^{p-1}{\mathbb {E}}\Vert {\bar{X}}(t)-X(t)\Vert ^p \nonumber \\= & {} 2^{p-1}{\mathbb {E}}\Vert {\hat{X}}(t)-{\bar{X}}(t)\Vert ^p+2^{p-1}{\mathbb {E}}\Vert {\hat{X}}_{sm+1}-X_{sm+l}\Vert ^p. \end{aligned}$$

(45)

Applying Hölder’s inequality and Theorem 1.7.1 in Mao (2007), for $t\in [t_{sm+l}, t_{sm+l+1})$,

$$\begin{aligned} \begin{aligned}&{\mathbb {E}}\Vert {\hat{X}}(t)-{\bar{X}}(t)\Vert ^p\\&\quad ={\mathbb {E}}\Big \Vert \int _{t_{sm+l}}^t f(X_{sm+l},X_{sm},r^{\Delta }_{sm+l})\mathrm{{d}}u+\int _{t_{sm+l}}^t g(X_{sm+l},X_{sm},r^{\Delta }_{sm+l})\mathrm{{d}}B(u)\Big \Vert ^p\\&\quad \le 2^{p-1}\Delta ^{p}{\mathbb {E}}\Vert f(X_{sm+l},X_{sm},r^{\Delta }_{sm+l})\Vert ^p\\&\qquad +2^{p-1}\left( \frac{p(p-1)}{2}\right) ^{\frac{p}{2}}\Delta ^{\frac{p}{2}}{\mathbb {E}}\left\| g(X_{sm+l},X_{sm},r^{\Delta }_{sm+l})\right\| ^p, \end{aligned} \end{aligned}$$

similar to (39), by Assumptions 5.2–5.4 and Theorem 4.2, we can easily get

$$\begin{aligned} {\mathbb {E}}\left\| f(X_{sm+l},X_{sm},r^{\Delta }_{sm+l})\right\| ^p \le C \qquad \mathrm{and}\qquad {\mathbb {E}}\left\| g(X_{sm+l},X_{sm},r^{\Delta }_{sm+l})\right\| ^p \le C, \end{aligned}$$

which means

$$\begin{aligned} \begin{aligned} {\mathbb {E}}\Vert {\hat{X}}(t)-{\bar{X}}(t)\Vert ^p\le C\Delta ^{p}+C\Delta ^{\frac{p}{2}}\le C\Delta ^{\frac{p}{2}}. \end{aligned} \end{aligned}$$

By substituting the last equation and (36) into (45), the desired assertion follows. $\square$

Lemma 5.8

Let Assumptions 5.1–5.4 hold, and $\theta \ge 0.5$, then for any $s\in {\mathbb {N}}$,

$$\begin{aligned} \begin{aligned} {\mathbb {E}}\int _s^{s+1}\Vert f(X(u),X(s),r^{\Delta }(u))-f(X(u),X(s),r(u))\Vert ^2\mathrm{{d}}u\le C\Delta , \end{aligned} \\ \begin{aligned} {\mathbb {E}}\int _s^{s+1}\Vert g(X(u),X(s),r^{\Delta }(u))-g(X(u),X(s),r(u))\Vert ^2\mathrm{{d}}u\le C\Delta . \end{aligned} \end{aligned}$$

Proof

Using Assumption 5.2, we have

$$\begin{aligned} \begin{aligned}&{\mathbb {E}}\int _{t_{sm+l}}^{t_{sm+l+1}}\Vert f(X(u),X(s),r^{\Delta }(u))-f(X(u),X(s),r(u))\Vert ^2 \mathrm{{d}}u\\&\quad ={\mathbb {E}}\int _{t_{sm+l}}^{t_{sm+l+1}}\Vert f(X(u),X(s),r^{\Delta }(u))-f(X(u),X(s),r(u))\Vert ^2 I_{\left\{ r(u)\ne r(t_{sm+l})\right\} } \mathrm{{d}}u\\&\quad \le 2{\mathbb {E}}\int _{t_{sm+l}}^{t_{sm+l+1}}\Vert f(X(u),X(s),r^{\Delta }(u))-f(0,0,r^{\Delta }(u))\Vert ^2 I_{\left\{ r(u)\ne r(t_{sm+l})\right\} } \mathrm{{d}}u\\&\qquad +2{\mathbb {E}}\int _{t_{sm+l}}^{t_{sm+l+1}}\Vert f(X(u),X(s),r(u))-f(0,0,r(u))\Vert ^2 I_{\left\{ r(u)\ne r(t_{sm+l})\right\} } \mathrm{{d}}u\\&\quad \le 24K_2^2\int _{t_{sm+l}}^{t_{sm+l+1}}{\mathbb {E}}\left\{ \left( 1+\Vert X(u)\Vert ^{2\rho _1}+\Vert X(s)\Vert ^{2\rho _1}\right) \left( \Vert X(u)\Vert ^2+\Vert X(s)\Vert ^2\right) I_{\left\{ r(u)\ne r(t_{sm+l})\right\} }\right\} \mathrm{{d}}u\\&\quad =24K_2^2\int _{t_{sm+l}}^{t_{sm+l+1}}{\mathbb {E}}\Big ({\mathbb {E}}\big \{\left( 1+\Vert X_{sm+l}\Vert ^{2\rho _1}+\Vert X_{sm}\Vert ^{2\rho _1}\right) \left( \Vert X_{sm+l}\Vert ^2+\Vert X_{sm}\Vert ^2 \right) \vert r(t_{sm+l})\big \}\\&\qquad \times {\mathbb {E}}\big \{I_{\left\{ r(u)\ne r(t_{sm+l})\right\} }\vert r(t_{sm+l})\big \}\Big )\mathrm{{d}}u, \end{aligned} \end{aligned}$$

where in the last step we use the fact that $X(u)=X_{sm+l},X(s)=X_{sm}$ and $I_{\left\{ r(u)\ne r(t_{sm+l})\right\} }$ when $t_{sm+l}< u< t_{sm+l+1}$ are conditionally independent with respect to the $\sigma$-algebra generated by $r(t_{sm+l})$. By the property of Markov chain, one has

$$\begin{aligned} \begin{aligned} {\mathbb {E}}\left\{ I_{\left\{ r(u)\ne r(t_{sm+l})\right\} }\vert r(t_{sm+l})\right\} =&\sum _{i\in S}I_{\left\{ r(t_{sm+l})=i\right\} }{\mathbb {P}}(r(u)\ne i\vert r(t_{sm+l})=i)\\ =&\sum _{i\in S}I_{\left\{ r(t_{sm+l})=i\right\} }\sum _{j\ne i}(\gamma _{ij}(u-t_{sm+l})+o(u-t_{sm+l}))\\ \le&\Big (\max _{1\le i\le N}(-\gamma _{ii})\Delta +o(\Delta )\Big )\sum _{i\in S}I_{\left\{ r(t_{sm+l})=i\right\} }\\ \le&{\bar{C}}(\Delta +o(\Delta )), \end{aligned} \end{aligned}$$

where ${\bar{C}}=\max _{1\le i\le N}(-\gamma _{ii})$. Hence,

$$\begin{aligned} \begin{aligned}&{\mathbb {E}}\int _{t_{sm+l}}^{t_{sm+l+1}}\Vert f(X(u),X(s),r^{\Delta }(u))-f(X(u),X(s),r(u))\Vert ^2 \mathrm{{d}}u\\&\quad \le C(\Delta +o(\Delta ))\int _{t_{sm+l}}^{t_{sm+l+1}}{\mathbb {E}}\left\{ \left( 1+\Vert X_{sm+l}\Vert ^{2\rho _1}+\Vert X_{sm}\Vert ^{2\rho _1}\right) \left( \Vert X_{sm+l}\Vert ^2+\Vert X_{sm}\Vert ^2\right) \right\} \mathrm{{d}}u\\&\quad \le C\Delta (\Delta +o(\Delta ))\left( 1+{\mathbb {E}}\Vert X_{sm+l}\Vert ^{4\rho _1}+{\mathbb {E}}\Vert X_{sm}\Vert ^{4\rho _1}+{\mathbb {E}}\Vert X_{sm+l}\Vert ^4+{\mathbb {E}}\Vert X_{sm}\Vert ^4\right) . \end{aligned} \end{aligned}$$

Then using Theorem 4.2, one can get that

$$\begin{aligned}&{\mathbb {E}}\int _s^{s+1}\Vert f(X(u),X(s),r^{\Delta }(u))-f(X(u),X(s),r(u))\Vert ^2\mathrm{{d}}u\nonumber \\&\quad =\sum _{l=0}^{m-1}{\mathbb {E}}\int _{t_{sm+l}}^{t_{sm+l+1}}\Vert f(X(u),X(s),r^{\Delta }(u))-f(X(u),X(s),r(u))\Vert ^2 \mathrm{{d}}u\nonumber \\&\quad \le C(\Delta +o(\Delta ))\left( 1+{\mathbb {E}}\Vert X_{sm+l}\Vert ^{4\rho _1}+{\mathbb {E}}\Vert X_{sm}\Vert ^{4\rho _1}+{\mathbb {E}}\Vert X_{sm+l}\Vert ^4+{\mathbb {E}}\Vert X_{sm}\Vert ^4\right) \nonumber \\&\quad \le C(\Delta +o(\Delta )). \end{aligned}$$

(46)

The second inequality can also be proved similarly. $\square$

Theorem 5.9

Under Assumptions 5.1–5.4, and $\theta \ge 0.5$, the continuous FBEM method (35) strongly converges to the solution of hybrid SDEPCAs (1), that is

$$\begin{aligned} {\mathbb {E}}\left\{ \sup _{0\le t\le T}\Vert {\hat{X}}(t)-x(t)\Vert ^2\right\} \le C\Delta . \end{aligned}$$

Proof

Let $e(t)={\hat{X}}(t)-x(t)$, $e_{\Delta }(t)={\hat{X}}(t)-X(t)$. For any $t\in [0,T]$, there always exists $s\in {\mathbb {N}}$ such that $t\in [s,s+1)$, hence

$$\begin{aligned} \begin{aligned} {\mathbb {E}}\left\{ \sup _{0\le t\le T}\Vert e(t)\Vert ^2\right\} \le {\mathbb {E}}\left\{ \sup _{0\le s\le [T]}\sup _{s\le t<s+1}\Vert e(t)\Vert ^2\right\} \le \sum _{s=0}^{[T]}{\mathbb {E}}\left\{ \sup _{s\le t< s+1}\Vert e(t)\Vert ^2\right\} . \end{aligned} \end{aligned}$$

(47)

For any $t\in [s,s+1)$, it follows from (34) and (35) that

$$\begin{aligned} \begin{aligned} e(t)=&e(s)+\int _{s}^{t}(f(X(u),X(s),r^{\Delta }(u))-f(x(u),x(s),r(u)))\mathrm{{d}}u\\&+\int _{s}^{t}(g(X(u),X(s),r^{\Delta }(u))-g(x(u),x(s),r(u)))\mathrm{{d}}B(u), \end{aligned} \end{aligned}$$

then according to the generalised Itô formula (Mao and Yuan 2006), one has

$$\begin{aligned} \begin{aligned} \Vert e(t)\Vert ^2&=\Vert e(s)\Vert ^2+\int _{s}^{t}2e(u)^\mathrm{{T}}(f(X(u),X(s),r^{\Delta }(u))-f(x(u),x(s),r(u)))\mathrm{{d}}u\\&\quad +\int _{s}^{t}\Vert g(X(u),X(s),r^{\Delta }(u))-g(x(u),x(s),r(u))\Vert ^2\mathrm{{d}}u\\&\quad +\int _{s}^{t}2e(u)^\mathrm{{T}}(g(X(u),X(s),r^{\Delta }(u))-g(x(u),x(s),r(u)))\mathrm{{d}}B(u). \end{aligned} \end{aligned}$$

Hence, it is easy to see that

$$\begin{aligned} \begin{aligned} {\mathbb {E}}\left\{ \sup _{s\le u\le t}\Vert e(u)\Vert ^2\right\}&\le {\mathbb {E}}\Vert e(s)\Vert ^2+2{\mathbb {E}}\left\{ \sup _{s\le u\le t}M(u)\right\} \\&\quad +2{\mathbb {E}}\int _{s}^{t} e(u)^\mathrm{{T}}\left( f(X(u),X(s),r^{\Delta }(u))-f(x(u),x(s),r(u))\right) \mathrm{{d}}u\\&\quad +{\mathbb {E}}\int _{s}^{t} \left\| g(X(u),X(s),r^{\Delta }(u))-g(x(u),x(s),r(u))\right\| ^2\mathrm{{d}}u, \end{aligned}\nonumber \\ \end{aligned}$$

(48)

where

$$\begin{aligned} M(u)=\int _{s}^ue(v)^\mathrm{{T}}\left( g(X(v),X(s),r^{\Delta }(v))-g(x(v),x(s),r(v))\right) \mathrm{{d}}B(v). \end{aligned}$$

Applying the Burkholder–Davis–Gundy inequality and $2ab\le a^2+b^2$, we obtain

$$\begin{aligned} \begin{aligned} {\mathbb {E}}\left\{ \sup _{s\le u\le t}M(u)\right\}&\le 4\sqrt{2}{\mathbb {E}}\left( \int _{s}^t\left\| e(u)^\mathrm{{T}}\left( g(X(u),X(s),r^{\Delta }(u))-g(x(u),x(s),r(u))\right) \right\| ^2\mathrm{{d}}u\right) ^{\frac{1}{2}}\\&\quad \le \frac{1}{4}{\mathbb {E}}\left( \sup _{s\le u\le t}\Vert e(u)\Vert ^2\right) \\&\qquad +32{\mathbb {E}}\int _s^t \Vert g(X(u),X(s),r^{\Delta }(u))-g(x(u),x(s),r(u))\Vert ^2\mathrm{{d}}u. \end{aligned} \end{aligned}$$

Substituting the last equation into (48), we get

$$\begin{aligned} \begin{aligned}&\frac{1}{2}{\mathbb {E}}\left\{ \sup _{s\le u\le t}\Vert e(u)\Vert ^2\right\} \\&\quad \le {\mathbb {E}}\Vert e(s)\Vert ^2+2\underbrace{{\mathbb {E}}\int _{s}^{t} e(u)^\mathrm{{T}}\left( f(X(u),X(s),r^{\Delta }(u))-f(x(u),x(s),r(u))\right) \mathrm{{d}}u}_{J_1}\\&\qquad +65\underbrace{{\mathbb {E}}\int _{s}^{t} \left\| g(X(u),X(s),r^{\Delta }(u))-g(x(u),x(s),r(u))\right\| ^2\mathrm{{d}}u}_{J_2}. \end{aligned} \end{aligned}$$

(49)

Using Assumption 5.2, Lemma 5.8, Hölder’s inequality, Theorem 4.2, Lemmas 5.6, and  5.7, one can acquire that

$$\begin{aligned} \begin{aligned} J_1&\le 2{\mathbb {E}}\int _{s}^{t} \Vert e(u)\Vert ^2\mathrm{{d}}u+{\mathbb {E}}\int _{s}^{t}\Vert f(X(u),X(s),r^{\Delta }(u))-f(X(u),X(s),r(u))\Vert ^2\mathrm{{d}}u\\&\quad +{\mathbb {E}}\int _{s}^{t}\Vert f(X(u),X(s),r(u))-f({\hat{X}}(u),{\hat{X}}(s),r(u))\Vert ^2\mathrm{{d}}u\\&\quad +2{\mathbb {E}}\int _{s}^{t} e(u)^\mathrm{{T}}\left( f({\hat{X}}(u),{\hat{X}}(s),r(u))-f(x(u),x(s),r(u))\right) \mathrm{{d}}u\\&\le 2{\mathbb {E}}\int _{s}^{t} \sup _{s\le v\le u}\Vert e(v)\Vert ^2\mathrm{{d}}u+C\Delta \\&\quad +2{\mathbb {E}}\int _{s}^{t} e(u)^\mathrm{{T}}\left( f({\hat{X}}(u),{\hat{X}}(s),r(u))-f(x(u),x(s),r(u))\right) \mathrm{{d}}u. \end{aligned} \end{aligned}$$

According to Lemma 5.8, Assumption 5.3, Theorem 4.2, Lemmas 5.6 and 5.7, we yield

$$\begin{aligned} \begin{aligned} J_2&\le 3{\mathbb {E}}\int _s^t \Vert g(X(u),X(s),r^{\Delta }(u))-g(X(u),X(s),r(u))\Vert ^2\mathrm{{d}}u\\&\quad +3{\mathbb {E}}\int _s^t \Vert g(X(u),X(s),r(u))-g({\hat{X}}(u),{\hat{X}}(s),r(u))\Vert ^2\mathrm{{d}}u\\&\quad +3{\mathbb {E}}\int _s^t \Vert g({\hat{X}}(u),{\hat{X}}(s),r(u))-g(x(u),x(s),r(u))\Vert ^2\mathrm{{d}}u\\&\le C\Delta +6{\mathbb {E}}\int _s^t \left( K_3^2\Vert e_{\Delta }(u)\Vert ^2+3K_4^2\left( 1+\Vert X(s)\Vert ^{2\rho _2}+\Vert {\hat{X}}(s)\Vert ^{2\rho _2}\right) \Vert e_{\Delta }(s)\Vert ^2\right) \mathrm{{d}}u\\&\quad +3{\mathbb {E}}\int _s^t \Vert g({\hat{X}}(u),{\hat{X}}(s),r(u))-g(x(u),x(s),r(u))\Vert ^2\mathrm{{d}}u\\&\le C\Delta +18K_4^2\int _s^t\left\{ {\mathbb {E}}\left( 1+\Vert X(s)\Vert ^{2\rho _2}+\Vert {\hat{X}}(s)\Vert ^{2\rho _2}\right) ^2\right\} ^{\frac{1}{2}}\left\{ {\mathbb {E}}\Vert e_{\Delta }(s)\Vert ^4\right\} ^{\frac{1}{2}}\mathrm{{d}}u\\&\quad +3{\mathbb {E}}\int _s^t \Vert g({\hat{X}}(u),{\hat{X}}(s),r(u))-g(x(u),x(s),r(u))\Vert ^2\mathrm{{d}}u\\&\le C\Delta +3{\mathbb {E}}\int _s^t \Vert g({\hat{X}}(u),{\hat{X}}(s),r(u))-g(x(u),x(s),r(u))\Vert ^2\mathrm{{d}}u. \end{aligned} \end{aligned}$$

Substituting $J_1$ and $J_2$ into (49), using Assumption 5.1, then

$$\begin{aligned} \begin{aligned} {\mathbb {E}}\left\{ \sup _{s\le u\le t}\Vert e(u)\Vert ^2\right\} \le&2{\mathbb {E}}\Vert e(s)\Vert ^2+8{\mathbb {E}}\int _{s}^{t} \sup _{s\le v\le u}\Vert e(v)\Vert ^2\mathrm{{d}}u+C\Delta \\&+8K_1{\mathbb {E}}\int _{s}^{t} (\Vert e(u)\Vert ^2+\Vert e(s)\Vert ^2)\mathrm{{d}}u\\ \le&2{\mathbb {E}}\Vert e(s)\Vert ^2+8(2K_1+1){\mathbb {E}}\int _{s}^{t} \sup _{s\le v\le u}\Vert e(v)\Vert ^2\mathrm{{d}}u+C\Delta .\\ \end{aligned} \end{aligned}$$

(50)

For $s=0, t\in [0,1)$, (50) implies

$$\begin{aligned} {\mathbb {E}}\left\{ \sup _{0\le u\le t}\Vert e(u)\Vert ^2\right\} \le 8(2K_1+1){\mathbb {E}}\int _{0}^{t} \sup _{0\le v\le u}\Vert e(v)\Vert ^2\mathrm{{d}}u+C\Delta , \end{aligned}$$

then according to the Gronwall inequality and the continuity of $\Vert e(u)\Vert ^2$, we have

$$\begin{aligned} {\mathbb {E}}\left\{ \sup _{0\le u\le 1}\Vert e(u)\Vert ^2\right\} \le C\Delta e^{8(2K_1+1)}=C\Delta . \end{aligned}$$

In particular, we know that ${\mathbb {E}}\Vert e(1)\Vert ^2\le C\Delta$.

For $s=1, t\in [1,2)$, (50) implies

$$\begin{aligned} {\mathbb {E}}\left\{ \sup _{1\le u\le t}\Vert e(u)\Vert ^2\right\} \le 2{\mathbb {E}}\Vert e(1)\Vert ^2+8(2K_1+1){\mathbb {E}}\int _{1}^{t} \sup _{1\le v\le u}\Vert e(v)\Vert ^2\mathrm{{d}}u+C\Delta , \end{aligned}$$

using the Gronwall inequality and the continuity of $\Vert e(u)\Vert ^2$ once more, we can also get

$$\begin{aligned} {\mathbb {E}}\left\{ \sup _{1\le u\le 2}\Vert e(u)\Vert ^2\right\} \le C\Delta , \end{aligned}$$

in particular, ${\mathbb {E}}\Vert e(2)\Vert ^2\le C\Delta$.

Repeating the same procedures, for any $s\in [0,[T]]$, we always have

$$\begin{aligned} {\mathbb {E}}\left\{ \sup _{s\le u\le s+1}\Vert e(u)\Vert ^2\right\} \le C\Delta , \end{aligned}$$

substituting this inequality into (47), which gives

$$\begin{aligned} {\mathbb {E}}\left\{ \sup _{0\le t\le T}\Vert e(t)\Vert ^2\right\} \le C([T]+1)\Delta =C\Delta . \end{aligned}$$

The proof is completed. $\square$

We are now ready to formulate the main theorem of this paper.

Theorem 5.10

Let Assumptions 5.1–5.4 hold, and $\theta \ge 0.5$, there exists a positive constant C, independent of $\Delta$, such that the ST method (11) strongly converges to the solution of hybrid SDEPCAs (1), that is

$$\begin{aligned} \sup _{0\le t\le T}{\mathbb {E}}\Vert X(t)-x(t)\Vert ^2\le C\Delta . \end{aligned}$$

Proof

It is apparent from the triangle inequality that

$$\begin{aligned} \sup _{0\le t\le T}{\mathbb {E}}\Vert X(t)-x(t)\Vert ^2\le 2\sup _{0\le t\le T}{\mathbb {E}}\Vert X(t)-{\hat{X}}(t)\Vert ^2+2\sup _{0\le t\le T}{\mathbb {E}}\Vert {\hat{X}}(t)-x(t)\Vert ^2, \end{aligned}$$

then the assertion follows from Lemma 5.7 and Theorem 5.9. $\square$

6 Numerical simulation

In this section, we consider the following scalar nonlinear hybrid SDEPCA

$$\begin{aligned} \mathrm{{d}}x(t)=f(x(t),x([t]),r(t))\mathrm{{d}}t+g(x(t),x([t]),r(t))\mathrm{{d}}B(t),\quad t\ge 0, \end{aligned}$$

(51)

where $f:{\mathbb {R}}\times {\mathbb {R}}\times S\rightarrow {\mathbb {R}}$,

$$\begin{aligned} f(x(t),x([t]),i)= {\left\{ \begin{array}{ll} -x^3(t)+x(t), &{} \mathrm{{if}}~ i=1, \\ -x^3(t)+x([t]), &{} \mathrm{{if}}~ i=2, \end{array}\right. } \end{aligned}$$

and $g:{\mathbb {R}}\times {\mathbb {R}}\times S\rightarrow {\mathbb {R}}$,

$$\begin{aligned} g(x(t),x([t]),i)= {\left\{ \begin{array}{ll} x(t)+\sin (x([t])), &{} \mathrm{{if}}~ i=1, \\ \sin (x(t))+\cos (x([t])), &{} \mathrm{{if}} ~i=2, \end{array}\right. } \end{aligned}$$

with the initial conditions $x_0=1$ and $i_0=1\in S=\{1,2\}$. Here B(t) is a scalar Brownian motion on $(\Omega ,{\mathcal {F}},\{{\mathcal {F}}_t\}_{t\ge 0},{\mathbb {P}})$, and r(t) is a right-continuous Markov chain taking values in S with the generator $\Gamma =(\gamma _{ij})_{2\times 2}=\begin{bmatrix}-1 &{} 1 \\ 2 &{} -2 \end{bmatrix}.$

By a straight calculation, one has

$$\begin{aligned} \begin{aligned}&(x-{\bar{x}})(f(x,y,1)-f({\bar{x}},{\bar{y}},1))+K\Vert g(x,y,1)-g({\bar{x}},{\bar{y}},1)\Vert ^2\\&\quad =(x-{\bar{x}})(-x^3+x+{\bar{x}}^3-{\bar{x}})+K\vert x+\sin y-{\bar{x}}-\sin {\bar{y}}\vert ^2\\&\quad \le -(x^2+{\bar{x}}^2)\vert x-{\bar{x}}\vert ^2-x{\bar{x}}\vert x-{\bar{x}}\vert ^2+\vert x-{\bar{x}}\vert ^2+2K\vert x-{\bar{x}}\vert ^2+2K\vert \sin y-\sin {\bar{y}}\vert ^2\\&\quad \le (2K+1)(\vert x-{\bar{x}}\vert ^2+\vert y-{\bar{y}}\vert ^2), \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned}&(x-{\bar{x}})(f(x,y,2)-f({\bar{x}},{\bar{y}},2))+K\Vert g(x,y,2)-g({\bar{x}},{\bar{y}},2)\Vert ^2\\&\quad =(x-{\bar{x}})(-x^3+y+{\bar{x}}^3-{\bar{y}})+K\vert \sin x-\sin {\bar{x}}+\cos y-\cos {\bar{y}}\vert ^2\\&\quad =-(x^2+{\bar{x}}^2+x{\bar{x}})\vert x-{\bar{x}}\vert ^2+(x-{\bar{x}})(y-{\bar{y}})+2K\vert x-{\bar{x}}\vert ^2+2K\vert y-{\bar{y}}\vert ^2\\&\quad \le \left( 2K+\frac{1}{2}\right) (\vert x-{\bar{x}}\vert ^2+\vert y-{\bar{y}}\vert ^2), \end{aligned} \end{aligned}$$

which means the coefficients satisfy Assumption 5.1. Similarly we can also verify that the coefficients satisfy other conditions of Theorem 5.10. We generate 2000 different discretized Brownian paths and use the numerical solution of the backward EM method with step-size $\Delta =2^{-15}$ as the “exact solution”.

Let $\epsilon$ and $\eta$ denote the errors in mean square,

$$\begin{aligned} \begin{aligned} \epsilon (T)&={\mathbb {E}}\vert x(T)-X_{Tm}\vert ^2=\frac{1}{2000}\sum _{i=1}^{2000}\vert x(T,\omega _i)-X_{Tm}(\omega _i)\vert ^2,\\ \eta (T)&={\mathbb {E}}\left\{ \sup _{0\le t_{sm+l}\le T}\vert x(t_{sm+l})-X_{sm+l}\vert ^2\right\} \\&=\frac{1}{2000}\sum _{i=1}^{2000}\left( \max _{0\le t_{sm+l}\le T}\vert x(t_{sm+l},\omega _i)-X_{sm+l}(\omega _i)\vert ^2\right) . \end{aligned} \end{aligned}$$

We calculate the errors in mean square $\epsilon (1),\epsilon (2), \epsilon (3)$ and $\eta (2), \eta (3)$ with step sizes $2^{-6},2^{-7},2^{-8},2^{-9},2^{-10}$, respectively. The log–log mean square error plots corresponding to those chosen values of $\Delta$ and $\theta$ are given in Figs. 1 and 2. It is well known that the slope of a line in the log–log error plot implies the order of convergence for the numerical method. Graphically, the mean square error lines’ slopes are close to the reference lines’ slope. Therefore, it can be seen from Figs. 1 and  2 that the order of convergence in mean square for the ST method is close to 0.5.

Fig. 1

Convergence rate of the ST method for Eq. (51)

Fig. 2

Convergence rate (uniform) of the ST method for Eq. (51)