2.1. Construction of the coupling process

In the spirit of Skorokhod [Reference Skorokhod25] in the study of processes with rapid switching, [Reference Ghosh, Arapostathis and Marcus8] presented a representation of a Markov chain in terms of a Poisson random measure and studied the stability of RSPs. This kind of representation theorem has been widely applied in the study of various properties of RSPs.

Before giving out our precise representation theorem for establishing the comparison theorem, let us recall the construction in [Reference Ghosh, Arapostathis and Marcus8] for comparison. When $\mathcal S=\{1,2,\ldots,N\}$ is a finite state space, let $\Delta_{ij}(x)$ be consecutive, left-closed, right-open intervals on $[0,\infty)$ , each having length $q_{ij}(x)$ . More precisely,

\begin{align*} & \Delta_{12}(x) = [0,q_{12}(x)), \ \Delta_{13}(x) = [q_{12}(x),q_{12}(x)+q_{13}(x)), \ \ldots, \ \Delta_{1N}(x) = \big[\textstyle\sum_{j\neq 1} q_{1j}(x), q_1(x)\big), \\ & \Delta_{21}(x) = [q_1(x), q_1(x)+q_{21}(x)), \ \ldots, \\ & \quad \vdots\end{align*}

Define a function $h\,:\,\mathbb R^d\times \mathcal S\times \mathbb R\to \mathbb R$ by $h(x,i,z)=\sum_{j\in\mathcal S,j\neq i}(j-i) \textbf{1}_{\Delta_{ij}(x)}(z)$ . Then, $(\Lambda_t)$ is a jumping process satisfying (2.2) as a solution to the SDE $\mathrm{d}\Lambda_t = \int_{[0,\infty)}h(X_t,\Lambda_{t-},z)\,\mathcal{N}(\mathrm{d}t,\mathrm{d}z)$ , where $(X_t)$ satisfies (2.1), and $\mathcal{N}(\mathrm{d}t,\mathrm{d}z)$ is a Poisson random measure with intensity $\mathrm{d}t\times\mathrm{d}z$ .

We can now introduce our construction of the coupling process in three steps. We assume that conditions (Q1) and (Q2) hold in this section.

Step 1: Construction of intervals. For every fixed $x\in\mathbb R^d$ and $i,j\in \mathcal S$ , we define the intervals $\Gamma_{ij}(x)$ , $\Gamma_{ij}^\ast$ , and $\bar\Gamma_{ij}$ in the following way. Starting from 0, the intervals $\Gamma_{ij}(x)$ for $j<i$ are defined on the positive half-line, while $\Gamma_{ij}(x)$ for $j>i$ are defined on the negative half-line. More precisely,

(2.5) \begin{equation} \begin{aligned} \Gamma_{i1}(x) & = [0, q_{i1}(x)), \\ \Gamma_{i2}(x) & = [q_{i1}(x), q_{i1}(x) + q_{i2}(x)), \\ & \ \,\vdots \\ \Gamma_{i(i-1)}(x) & = \big[\textstyle\sum_{j=1}^{i-2}q_{ij}(x),\sum_{j=1}^{i-1} q_{ij}(x)\big), \end{aligned}\end{equation}

and

(2.6) \begin{equation} \begin{aligned} \Gamma_{i(i+c_i )}(x) & = [{-}q_{i(i+c_i )}(x), 0), \\ \Gamma_{i(i+c_i -1)}(x) & = [{-}q_{i(i+c_i -1)}(x)-q_{i(i+c_i )}(x), -q_{i(i+c_i )}(x)), \\ & \ \,\vdots \\ \Gamma_{i(i+1)}(x) & = \big[{-}\textstyle\sum_{j=i+1}^{i+c_i } q_{ij}(x), -\sum_{j=i+2}^{i+c_i } q_{ij}(x)\big), \end{aligned}\end{equation}

where $c_i$ is given in (Q2). Analogously, by replacing $q_{ij}(x)$ in (2.5) and (2.6) with $\bar q_{ij}$ and $q_{ij}^\ast$ respectively, we can define the intervals $\bar\Gamma_{ij}$ and $\Gamma_{ij}^\ast$ . Here and in what follows, we put $\Gamma_{ij}(x)=\emptyset$ if $q_{ij}(x)=0$ and $\Gamma_{ii}(x)=\emptyset$ for the convenience of notation. This convention also applies to the intervals $\bar \Gamma_{ij}$ and $\Gamma_{ij}^\ast$ .

The sort order of the intervals $\Gamma_{ij}(x)$ , $\bar \Gamma_{ij}$ , and $\Gamma_{ij}^\ast$ according to $j>i$ or $j<i$ will play an important role in the argument below. The assumption on the existence of $c_i$ is used here such that on the negative half-line, the first interval starting from 0 is associated with the state $j\in\mathcal S$ satisfying $j=\max\{k\in \mathcal S; k>i, q_{ik}(x)>0\}$ . The common starting point 0 of $\Gamma_{ij}(x)$ , $\bar\Gamma_{ij}$ , and $\Gamma_{ij}^\ast$ for different $i\in \mathcal S$ also plays an important role in our construction of the order-preservation coupling process.

Step 2: Explicit construction of Poisson random measure. Here we use the method of [Reference Ikeda and Watanabe11, Chapter I, p. 44] to present a concrete construction of the Poisson random measure. Denote by $\textbf{m}(\mathrm{d} x)$ the Lebesgue measure over $\mathbb R$ . Let $\xi_k$ , $k=1,2,\ldots$ , be random variables taking values in $[{-}K_0,K_0]$ with $\mathbb P(\xi_k \in \mathrm{d} x) = {\textbf{m}(\mathrm{d} x)}/{2K_0}$ , and $\tau_k$ , $k=1,2,\ldots$ , be non-negative random variables such that $\mathbb P(\tau_k>t)=\mathrm{e}^{-2tK_0}$ , $t\geq 0$ . Assume that all $\xi_k$ and $\tau_k$ , $k\geq 1$ , are mutually independent. Let $\zeta_n =\tau_1 +\tau_2 +\cdots+\tau_n $ , $ n=1,2,\ldots$ , and $\zeta_0 =0$ , $\mathcal D_{\textbf{p}} = \bigcup_{n\geq 1}\{\zeta_n\}$ , and $\textbf{p}(t)=\sum_{0\leq s<t}\Delta \textbf{p}(s)$ , with $\Delta \textbf{p}(s)=0$ for $s\not\in \mathcal D_{\textbf{p}}$ , and $\Delta\textbf{p}(\zeta_n )=\xi_n$ , $n\geq 1$ , where $\Delta\textbf{p}(s)\,:\!=\,\textbf{p}(s) -\textbf{p}(s{-})$ . The finiteness of $K_0$ means that $\lim_{n\to \infty}\zeta_n=\infty$ almost surely (a.s.); that is, during each finite time period, there exists a finite number of jumps for $(\textbf{p}(t))$ . Let

\begin{equation*} \mathcal{N}_{\textbf{p}}([0,t]\times A) = \#\{s\in \mathcal D_{\textbf{p}}; 0\leq s\leq t, \Delta \textbf{p}(s)\in A\},\qquad t>0,\,A\in \mathscr{B}(\mathbb R).\end{equation*}

As a consequence, $\textbf{p}(t)$ and $\mathcal{N}_{\textbf{p}}(\mathrm{d} t,\mathrm{d} z)$ are respectively a Poisson point process and a Poisson random measure with intensity measure $\mathrm{d} t\,\textbf{m}(\mathrm{d} x)$ . It is always assumed that $\textbf{p}(t)$ is independent of the Brownian motion $(B_t)$ in (2.1).

Step 3: Construction of coupling processes. Define three functions $\vartheta$ , $\vartheta^\ast$ , and $\bar\vartheta$ as follows:

\begin{align*} \vartheta(x,i,z)&=\sum_{j\in\mathcal S,j\neq i}(j-i)\textbf{1}_{\Gamma_{ij}(x)}(z),\\ \vartheta^\ast(i,z)&=\sum_{j\in\mathcal S,j\neq i}(j-i)\textbf{1}_{\Gamma_{ij}^\ast}(z),\\ \bar \vartheta(i,z)&=\sum_{j\in \mathcal S,j\neq i}(j-i)\textbf{1}_{\bar \Gamma_{ij}}(z).\end{align*}

Then, consider the following SDEs:

(2.7) \begin{align} \mathrm{d}\Lambda_t & = \int_{\mathbb R}\vartheta(X_t,\Lambda_{t-},z)\,\mathcal{N}_{\textbf{p}}(\mathrm{d}t,\mathrm{d}z), \end{align}

(2.8) \begin{align} \mathrm{d}\bar\Lambda_t & = \int_{\mathbb R}\bar\vartheta(\bar\Lambda_{t-},z)\,\mathcal{N}_{\textbf{p}}(\mathrm{d}t,\mathrm{d}z), \end{align}

(2.9) \begin{align} \mathrm{d}\Lambda_t^\ast & = \int_{\mathbb R}\vartheta^\ast\big(\Lambda_{t-}^\ast,z\big)\,\mathcal{N}_{\textbf{p}}(\mathrm{d}t,\mathrm{d}z), \end{align}

and $\Lambda_0=\bar \Lambda_0=\Lambda_0^\ast=i_0\in\mathcal S$ . Here, recall that $(X_t)$ satisfies (2.1).

The fact that the solution to (2.7) satisfies (2.2) can be checked directly using the property that $\mathbb P(\mathcal{N}_{\textbf{p}}((0,\delta]\times A)\geq 2)=o(\delta)$ , $\delta>0$ . Next, we verify that $\big(\bar \Lambda_t\big)$ and $\big(\Lambda_t^\ast\big)$ given by (2.8) and (2.9) are the jumping processes associated with $\big(\bar q_{ij}\big)$ and $\big(q_{ij}^\ast\big)$ respectively. According to Itô’s formula, for any bounded measurable function F on $\mathcal S$ ,

\begin{align*} \mathbb E F\big(\bar\Lambda_t\big) & = F\big(\bar \Lambda_0\big) + \mathbb E\int_0^t\int_{\mathbb R}\big(F\big(\bar\Lambda_{s-}+\bar\vartheta\big(\bar\Lambda_{s-},z\big)\big)-F\big(\bar\Lambda_{s-}\big)\big)\, \mathcal{N}_{\textbf{p}}(\mathrm{d}s,\mathrm{d}z) \\ & = F\big(\bar\Lambda_0\big) + \mathbb E\int_0^t\int_{\mathbb R}\sum_{j\in \mathcal S}\big(F(j)-F\big(\bar\Lambda_{s-}\big)\big) \textbf{1}_{\bar\Gamma_{\bar \Lambda_{s-}j}}(z)\,\mathrm{d}s\,\textbf{m}(\mathrm{d} z) \\ & = F\big(\bar\Lambda_0\big) + \mathbb E\int_0^t\sum_{j\in\mathcal S}\bar q_{\bar\Lambda_{s-}j}\big(F(j)-F\big(\bar \Lambda_{s-}\big)\big)\,\mathrm{d}s.\end{align*}

Writing $\bar P_t F(i_0)=\mathbb E F\big(\bar\Lambda_t\big)$ with $\bar\Lambda_0=i_0$ , we obtain from the above integral equation that $\bar P_tF(i_0) = F(i_0) + \int_0^t \bar P_s(\bar QF)\,\mathrm{d}s$ , where $\bar QF(i) = \sum_{j\in \mathcal S,j\neq i} \bar q_{ij}(F(j)-F(i))$ , and the corresponding differential form is

(2.10) \begin{equation} \frac{\mathrm{d} {\bar P}_t F}{\mathrm{d} t}= \bar P_t \bar QF.\end{equation}

Due to (Q1), the Kolmogorov forward equation (2.10) admits a unique solution, and hence $\big(\bar\Lambda_t\big)$ is a continuous-time Markov chain with transition rate matrix $\bar Q=\big(\bar q_{ij}\big)$ (cf. [Reference Chen3, Corollary 2.24]). The corresponding conclusion for $\big(\Lambda_t^\ast\big)$ can be proved by the same method.

Consequently, through the previous three steps, we have completed the construction of the desired Markov chains $\big(\Lambda_t^\ast\big)$ and $\big(\bar \Lambda_t\big)$ used in Theorem 2.1.

2.2. Proofs of the comparison theorem and its application

Let us first present the proof of the comparison theorem.

Proof of Theorem 2.1. We only provide the proof of $\Lambda_t\leq \bar\Lambda_t$ for all $t\geq 0$ almost surely; the corresponding solution for $\Lambda_t$ and $\Lambda_t^\ast$ can be proved in the same way.

According to the representation of (2.7) and (2.8), the processes $(\Lambda_t)$ and $\big(\bar \Lambda_t\big)$ have no jumps outside $\mathcal{D}_{\textbf{p}}$ . So, we only need to prove that

(2.11) \begin{equation} \mathbb P\big(\Lambda_t \leq \bar\Lambda_t,\ t\in \mathcal{D}_{\textbf{p}}\big) = 1. \end{equation}

By the definition of $\bar q_{ij}$ , for $j>i$ , $\bar q_{ij}\geq q_{lj}(x)$ for all $1\leq l\leq i$ and all $x\in \mathbb R^d$ ; for $j<i$ , $\bar q_{ij}\leq q_{lj}(x)$ for all $j< l\leq i$ and all $x\in \mathbb R^d$ . By the sort order of the intervals $\Gamma_{ij}(x)$ and $\bar\Gamma_{ij}$ , we have, for $i<k\in \mathcal S$ ,

(2.12) \begin{align} \bigcup_{r\geq m}\Gamma_{ir}(x) & \subset \bigcup_{r\geq m}\bar \Gamma_{kr} \qquad \text{for all } m>k \text{ and all } x\in \mathbb R^d, \end{align}

(2.13) \begin{align} \bigcup_{r\leq m}\Gamma_{ir}(x) & \supset \bigcup_{r\leq m}\bar \Gamma_{kr} \qquad \text{for all } m<i \text{ and all } x\in\mathbb R^d. \end{align}

Assuming $i=\Lambda_{\zeta_{n-1}}\leq \bar \Lambda_{\zeta_{n-1}}=k$ for some $n\geq 1$ , we are going to show that $\Lambda_{\zeta_{n}}\leq \bar \Lambda_{\zeta_n}$ , whose proof is divided into four cases.

For case (i), $\Lambda_{\zeta_n}=m\geq k$ , by (2.7) and the construction of the Poisson random measure $\mathcal{N}_{\textbf{p}}(\mathrm{d} t,\mathrm{d} z)$ , we get $\xi_n\in \Gamma_{im}\big(X_{\zeta_n}\big)$ . By (2.12), this yields that $\xi_n\in \bigcup_{r\geq m}\bar \Gamma_{kr}$ . Together with (2.8), this means that $\bar\Lambda_{\zeta_n}$ must jump into the set $\{l\in\mathcal S;\, l\geq m\}$ . Whence, $\Lambda_{\zeta_n}\leq \bar \Lambda_{\zeta_n}$ .

For case (ii), $\Lambda_{\zeta_n}=m$ with $i<m<k$ , (2.7) implies that $\xi_n\in \Gamma_{im}\big(X_{\zeta_n}\big)$ , and hence $\xi_n<0$ . So, $\xi_n\not\in \bigcup_{j\leq k}\bar \Gamma_{kj}\subset [0,\infty)$ , which means that $\bar\Lambda_{\zeta_n}$ cannot jump into the set $\{j\in\mathcal S; j\leq k\}$ , and hence $\Lambda_{\zeta_n}\leq \bar \Lambda_{\zeta_n}$ . But, if $\Lambda_{\zeta_n}=m$ with $m\leq i$ and $\bar\Lambda_{\zeta_n}\leq i$ , this situation is studied in case (iii).

For case (iii), $\Lambda_{\zeta_n}=m$ with $m\leq i$ and $\bar\Lambda_{\zeta_n}>i$ , it is obvious that $\Lambda_{\zeta_n}\leq \bar\Lambda_{\zeta_n}$ . If $\bar \Lambda_{\zeta_n}=m'\leq i$ , (2.8) and (2.13) yield $\xi_n\in \bar\Gamma_{km'}\subset \bigcup_{r\leq m'}\Gamma_{ir}\big(X_{\zeta_n}\big)$ . Whence, $\Lambda_{\zeta_n}$ jumps into $\{j\in \mathcal S;\, j\leq m'\}$ , and hence $\Lambda_{\zeta_n}\leq \bar \Lambda_{\zeta_n}$ still holds.

For case (iv), if $\bar \Lambda_{\zeta_n}=m$ with $i<m<k$ , then $\xi_n\in \bar \Gamma_{km}$ , $\xi_n>0$ , and $\xi_n\notin \bigcup_{j>i}\Gamma_{ij}\big(X_{\zeta_n}\big)$ . So, $\Lambda_{\zeta_n}\leq i<m=\bar \Lambda_{\zeta_n}$ .

Consequently, if $\Lambda_{\zeta_{n-1}}\leq \bar \Lambda_{\zeta_{n-1}}$ , we have $\Lambda_{\zeta_n}\leq \bar \Lambda_{\zeta_n}$ for $n\geq 1$ . By induction on n, we prove that (2.11) holds, and finally $\mathbb P(\Lambda_t\leq \bar \Lambda_t,\, t\geq 0) = 1$ . The proof of Theorem 2.1 is complete.

Proof of Theorem 2.2. For each $i\in\mathcal S$ , denote by $\big(X_t^{(i)}\big)$ the solution to the SDE

\begin{equation*}\mathrm{d} X_t^{(i)}=b\big(X_t^{(i)}, i\big)\,\mathrm{d}t+\sigma\big(X_t^{(i)},i\big)\,\mathrm{d}B_t, \qquad X_0^{(i)}=x,\end{equation*}

and by $\big(P_t^{(i)}\big)$ its associated semigroup. According to Itô’s formula and Gronwall’s inequality, it follows from (2.3) that $P_t^{(i)}\rho (x)=\mathbb E\rho \big(X_t^{(i)}\big)\leq \mathrm{e}^{ \beta_i t} \rho (x)$ . Let $\zeta_n$ , $\mathcal{N}_{\textbf{p}}(\mathrm{d} t,\mathrm{d} z)$ , and $\big(\bar \Lambda_t\big)$ be defined as at the beginning of this section. Let $\mathbb E^{\mathcal{N}_{\textbf{p}}}[{\cdot}]=\mathbb E[\,\cdot\mid\mathscr{F}^{\mathcal{N}_{\textbf{p}}}]$ be the conditional expectation with respect to the $\sigma$ -algebra $\mathscr{F}^{\mathcal{N}_{\textbf{p}}}=\sigma\{\textbf{p}(s);\, s\geq 0\}$ . The mutual independence of $\mathcal{N}_{\textbf{p}}$ and $(B_t)$ yields

\begin{equation*} \mathbb E^{\mathcal{N}_{\textbf{p}}}\big[\rho\big(X_{\zeta_n}\big)\big] \leq \mathbb E^{\mathcal{N}_{\textbf{p}}} \big[\rho\big(X_{\zeta_{n-1}}\big)\big] + \mathbb E^{\mathcal{N}_{\textbf{p}}} \bigg[\int_{\zeta_{n-1}}^{\zeta_n}\beta_{\Lambda_{\zeta_{n-1}}}\rho(X_s)\,\mathrm{d}s\bigg]. \end{equation*}

By Theorem 2.1, $\beta_{\Lambda_s}\leq \beta_{\bar \Lambda_s}$ a.s. Furthermore, since $(\bar\Lambda_s)$ depends only on $\mathcal{N}_{\textbf{p}}$ , we have

\begin{equation*} \mathbb E^{\mathcal{N}_{\textbf{p}}}\big[\rho\big(X_{\zeta_n}\big)\big] \leq \mathbb E^{\mathcal{N}_{\textbf{p}}}\big[\rho\big(X_{\zeta_{n-1}}\big)\big] + \beta_{\bar\Lambda_{\zeta_{n-1}}} \mathbb E^{\mathcal{N}_{\textbf{p}}} \bigg[\int_{\zeta_{n-1}}^{\zeta_n} \rho(X_s)\,\mathrm{d} s\bigg]. \end{equation*}

Then, as $\bar \Lambda_s\equiv \bar \Lambda_{\zeta_{n-1}}$ for $s\in [\zeta_{n-1},\zeta_n)$ by (2.8),

\begin{equation*} \mathbb E^{\mathcal{N}_{\textbf{p}}} \big[\rho\big(X_{\zeta_n}\big)\big] \leq \exp\bigg\{\int_{\zeta_{n-1}}^{\zeta_n} \beta_{\bar \Lambda_s}\,\mathrm{d}s\bigg\} \mathbb E^{\mathcal{N}_{\textbf{p}}}\big[\rho\big(X_{\zeta_{n-1}}\big)\big]. \end{equation*}

Setting $m_t\,:\!=\,\sup\{n\in\mathbb N;\, \zeta_n\leq t\}$ , deducing recursively, we obtain

\begin{align*} \mathbb E^{\mathcal{N}_{\textbf{p}}} [\rho(X_t)] & \leq \exp\bigg\{\int_{\zeta_{m_t}}^t \beta_{\bar \Lambda_s}\,\mathrm{d} s\bigg\} \mathbb E^{\mathcal{N}_{\textbf{p}}} \big[\rho\big(X_{\zeta_{m_t}}\big)\big] \\ & \leq \exp\bigg\{\int_{\zeta_{m_t}}^t\beta_{\bar \Lambda_s}\,\mathrm{d} s\bigg\} \exp\bigg\{\int_{\zeta_{m_t-1}}^{\zeta_{m_t}}\beta_{\bar \Lambda_s}\,\mathrm{d} s\bigg\} \mathbb E^{\mathcal{N}_{\textbf{p}}} \big[\rho\big(X_{\zeta_{m_t-1}}\big)\big] \\ & \leq \cdots \leq \exp\bigg\{\int_0^t\beta_{\bar \Lambda_s}\,\mathrm{d} s\bigg\}\rho(x). \end{align*}

Consequently, $\mathbb E[\rho(X_t)]\leq \mathbb E\big[\exp\big\{\int_0^t\beta_{\bar \Lambda_s}\,\mathrm{d}s\big\}\big]\rho(x)$ , and further, by Hölder’s inequality, for $p'\in (0,1]$ ,

(2.14) \begin{equation} \mathbb E[\rho^{p'}(X_t)] \leq \mathbb E\bigg[\exp\bigg\{\int_0^tp'\beta_{\bar \Lambda_s}\,\mathrm{d}s\bigg\}\bigg]\rho^{p'}(x). \end{equation}

According to [Reference Bardet, Guerin and Malrieu1, Propositions 4.1, 4.2], (2.4) yields that there exist $p'\in (0,1]$ , $C,\eta >0$ such that $\mathbb E\big[\exp\big\{\int_0^tp'\beta_{\bar \Lambda_s}\,\mathrm{d}s\big\}\big]\leq C\mathrm{e}^{-\eta t}$ . Combining this with (2.3) and (2.14), the desired conclusion follows immediately.

3.1. Construction of the coupling process

In this part, we introduce the coupling process $(\Lambda_t,\tilde \Lambda_t)$ on $\mathcal S\times \mathcal S$ that will be used in the proof of Theorem 3.1. Similarly to Section 2, it is also constructed using the Skorokhod representation theorem. However, there are several subtle differences to fit the current purpose.

Step 1: Construction of intervals Due to (H1) and (H2), we let

\begin{equation*} \Gamma_{1k}=[(k-2)K_0,(k-2)K_0+q_{1k}),\quad \widetilde \Gamma_{1k}=[(k-2)K_0,(k-2)K_0+\tilde q_{1k})\end{equation*}

for $2\leq k\leq c_0+1$ , and $U_1=[0,c_0K_0)$ . By (H1), $q_{1k}\leq K_0$ and $\tilde q_{1k}\leq K_0$ , and hence $\Gamma_{1k}\bigcap \Gamma_{1j}=\emptyset$ and $\widetilde \Gamma_{1k}\bigcap \widetilde \Gamma_{1j}=\emptyset$ , $k\neq j$ . Moreover, $\Gamma_{1k}\subset U_1$ and $\widetilde \Gamma_{1k}\subset U_1$ for all $2\leq k\leq c_0+1$ . For $n\geq 2$ , define

\begin{equation*} \Gamma_{nk} = \begin{cases} \big[2(n-1)c_0K_0-(n-k)K_0, 2(n-1)c_0K_0-(n-k)K_0+q_{nk}\big), & \!\!\text{$n-c_0\leq k<n$,} \\[5pt] \big[2(n-1)c_0K_0+(k-n-1)K_0, 2(n-1)c_0K_0+(k-n-1)K_0+q_{nk}\big), & \!\!\text{$n+c_0\geq k>n$}. \end{cases}\end{equation*}

Define, similarly, $\widetilde \Gamma_{nk}$ by replacing $q_{nk}$ above with $\tilde q_{nk}$ . Let $U_n=\big[(2n-3)c_0 K_0, (2n-1)c_0 K_0\big)$ , $n\geq 2$ . So $\Gamma_{nk}\subset U_n$ and $\widetilde \Gamma_{nk}\subset U_n$ for all $k,n\in\mathcal S$ with $|k-n|\leq c_0$ .

Compared with the intervals constructed in [Reference Ghosh, Arapostathis and Marcus8], the starting point of each interval $\Gamma_{nk}$ constructed above does not depend on any other intervals $\Gamma_{ij}$ , $i,j\in\mathcal S$ with $i\neq n$ , $j\neq k$ . We construct $\Gamma_{nk}$ in this way in order to remove the term $N^2$ that appears in (3.5).

Step 2: Construction of Poisson random measure Denote by $\textbf{m}(\mathrm{d} x)$ the Lebesgue measure over the real line $\mathbb R$ . Let $\xi_i^{(k)}$ , $k,i=1,2,\ldots$ , be $U_k$ -valued random variables with $\mathbb P\big(\xi_i^{(k)}\in \mathrm{d} x\big)={\textbf{m}(\mathrm{d} x)}/{\textbf{m}(U_k)}$ , and $\tau_i^{(k)}$ , $k,i=1,2,\ldots$ , be non-negative random variables such that $\mathbb P\big(\tau_i^{(k)}>t\big)=\exp[{-}t\textbf{m}(U_k)]$ , $t\geq 0$ . Assume all $\xi_i^{(k)}$ , $\tau_i^{(k)}$ to be mutually independent. Write $\zeta_n^{(k)}=\tau_1^{(k)}+\cdots+\tau_n^{(k)}$ for $k,n\ge 1$ and $\zeta_0^{(k)}=0$ for $k\ge 1$ . We define $\mathcal D_{\tilde {\textbf{p}}}=\bigcup_{k\geq 1}\bigcup_{n\geq 0}\big\{\zeta_n^{(k)} \big\}$ and $\tilde {\textbf{p}}(t)=\sum_{0\leq s<t} \Delta\tilde{\textbf{p}}(s)$ , $\Delta \tilde{\textbf{p}}(s)=0$ for $s\not\in \mathcal D_{\tilde{\textbf{p}}}$ , and $\Delta\tilde{\textbf{p}}\big(\zeta_n^{(k)}\big)=\xi_n^{(k)}$ , $k,n=1,2,\ldots$ Let

\begin{equation*} \mathcal{N}_{\tilde {\textbf{p}}}([0,t]\times A) = \#\big\{s\in \mathcal D_{\tilde{\textbf{p}}}; \, 0< s\leq t, \tilde {\textbf{p}}(s)\in A\big\}, \qquad t>0,\, A\in \mathscr{B}([0,\infty)).\end{equation*}

As a consequence, we get a Poisson random process $(\tilde{\textbf{p}}(t))$ and its associated Poisson random measure $\mathcal{N}_{\tilde {\textbf{p}}}(\mathrm{d} t,\mathrm{d} x)$ with intensity measure $\mathrm{d} t\,\textbf{m}(\mathrm{d} x)$ .

The construction of $({\tilde {\textbf{p}}}(t))$ is more complicated than that of $(\textbf{p}(t))$ in Section 2, which is caused by the fact that the union of $\Gamma_{nk}$ for $n,k=1,2,\ldots$ may be unbounded.

Step 3: Construction of coupling processes Define two functions $\vartheta$ , $\tilde \vartheta$ associated with $\Gamma_{ij}$ and $\widetilde \Gamma_{ij}$ , $i,j\in \mathcal S$ , by

(3.6) \begin{equation} \vartheta(i,z) = \sum_{j\in\mathcal S, j\neq i}(j-i)\textbf{1}_{\Gamma_{ij}}(z), \qquad \tilde \vartheta(i,z)=\sum_{j\in\mathcal S, j\neq i}(j-i)\textbf{1}_{\widetilde \Gamma_{ij}}(z).\end{equation}

Then, the desired coupling process $(\Lambda_t,\tilde \Lambda_t)$ is given as the solution of the following SDEs:

(3.7) \begin{align} \mathrm{d}\Lambda_t = \int_{[0,\infty)}\vartheta(\Lambda_{t-},z)\, \mathcal{N}_{\tilde{\textbf{p}}}(\mathrm{d}t,\mathrm{d} z), \qquad \Lambda_0=i_0\in\mathcal S, \end{align}

(3.8) \begin{align} \mathrm{d}\tilde\Lambda_t = \int_{[0,\infty)}\tilde\vartheta(\tilde\Lambda_{t-},z)\, \mathcal{N}_{\tilde{\textbf{p}}}(\mathrm{d}t,\mathrm{d}z), \qquad \tilde\Lambda_0=i_0\in\mathcal S. \end{align}

The transition rate matrix of the basic coupling is given by $q_{(ij)(kj)}=(q_{ik}-\tilde q_{jk})\vee 0$ , $q_{(ij)(ik)} =(\tilde q_{jk}-q_{ik})\vee 0$ , and $q_{(ij)(kk)}=q_{ik}\wedge \tilde q_{jk}$ , for all $i,j,k\in\mathcal S$ , and i,j not necessarily different.

By the previous construction of $\Gamma_{ij}$ , $\widetilde \Gamma_{ij}$ , and $\mathcal{N}_p(\mathrm{d} t,\mathrm{d} x)$ , the following properties hold:

1. (i) The processes $(\Lambda_t)$ and $\big(\tilde \Lambda_t\big)$ can jump only when the Poisson process $(\tilde{\textbf{p}}(t))$ jumps.

2. (ii) If $\Lambda_t=\tilde \Lambda_t=k$ , for $\delta>0$ , $\Lambda_{t+\delta}\neq \tilde \Lambda_{t+\delta}$ may happen only when $\zeta_n^{(k)}\in [t,t+\delta)$ and $\xi_n^{(k)}\in \Gamma_{kj}\Delta\widetilde \Gamma_{kj}$ for some $n\geq 1$ , where $A\Delta B\,:\!=\,(A\backslash B)\cup(B\backslash A)$ for Borel sets A, B.

3.2. Proof of Theorem 3.1 and its application

Proof of Theorem 3.1. To get the upper estimate, for $\delta>0$ , let

\begin{equation*} \alpha(\delta) = \sup_{i\in \mathcal S}\mathbb P(\Lambda_\delta\neq\tilde\Lambda_\delta \mid \Lambda_0=\tilde \Lambda_0=i), \qquad \beta(\delta) = 1-\alpha(\delta). \end{equation*}

Now we use the representations in (3.7) and (3.8) of $(\Lambda_t)$ and $\big(\tilde \Lambda_t\big)$ to estimate $\alpha(\delta)$ . Noting that $\Lambda_0=\tilde\Lambda_0=i_0$ for some $i_0\in\mathcal S$ , by (3.6), (3.7), and (3.8) we have

(3.9) \begin{align} \mathbb P\big(\Lambda_\delta\neq \tilde \Lambda_\delta\big) & = \mathbb P\big(\Lambda_\delta\neq\tilde\Lambda_{\delta}\mid\Lambda_0=\tilde\Lambda_0) = \mathbb P\big(\Lambda_\delta\neq\tilde\Lambda_\delta, \mathcal{N}_{\tilde{\textbf{p}}}([0,\delta]\times U_{i_0})\geq 1\big) \nonumber \\ & = \mathbb P\big(\Lambda_\delta\neq\tilde\Lambda_\delta,\mathcal{N}_{\tilde{\textbf{p}}}\big([0,\delta]\times U_{i_0}\big)=1\big) + \mathbb P\big(\Lambda_\delta\neq\tilde\Lambda_\delta,\mathcal{N}_{\tilde{\textbf{p}}}([0,\delta]\times U_{i_0})\geq 2\big). \end{align}

Since $H\,:\!=\,\textbf{m}(U_{i_0})=2c_{0}K_0<\infty$ , there exists a $C>0$ independent of the choice of $i_0\in\mathcal S$ such that

\begin{equation*}\mathbb P\big(\mathcal{N}_{\tilde{\textbf{p}}}([0,\delta]\times U_{i_0})\geq 2\big) = 1 - \mathrm{e}^{-H\delta}-H\delta\mathrm{e}^{-H\delta}\leq C\delta^2.\end{equation*}

Moreover,

(3.10) \begin{align} \mathbb P\big(\Lambda_\delta\neq\tilde\Lambda_\delta, \mathcal{N}_{\tilde{\textbf{p}}}\big([0,\delta]\times U_{i_0}\big)=1\big) & = \int_0^\delta\mathbb P\Big(\Lambda_{\delta}\neq\tilde\Lambda_\delta, \tau_1^{(i_0)}\in\mathrm{d}s,\tau_2^{(i_0)}>\delta-s\Big) \nonumber \\ & = \int_0^\delta\mathbb P\Big(\xi_1^{(i_0)}\in\cup_{j\in \mathcal S}\big(\Gamma_{i_0j}\Delta\widetilde\Gamma_{i_0j}\big), \tau_1^{(i_0)}\in\mathrm{d}s,\tau_2^{(i_0)}>\delta-s\Big) \nonumber \\ & = \delta\mathrm{e}^{-H\delta}\sum_{j\in\mathcal S,j\neq i_0}|q_{i_0j}-\tilde q_{i_0j}| \nonumber \\ & \leq \delta\mathrm{e}^{-H\delta}\big\|Q-\widetilde Q\big\|_{\ell_1}\leq \delta \big\|Q-\widetilde Q\big\|_{\ell_1}. \end{align}

In the light of (3.9) and (3.10), we get

(3.11) \begin{equation} \mathbb P\big(\Lambda_\delta\neq \tilde \Lambda_\delta\big) = \mathbb P(\Lambda_\delta\neq \tilde \Lambda_\delta\mid\Lambda_0=\tilde \Lambda_0=i_0) \leq \delta\big\|Q-\widetilde Q\big\|_{\ell_1}+C\delta^2, \end{equation}

and hence

(3.12) \begin{equation} \alpha(\delta)\leq \delta\big\|Q-\widetilde Q\big\|_{\ell_1}+C\delta^2,\qquad \beta(\delta)=1-\alpha(\delta) \geq 1-\delta\big\|Q-\widetilde Q\big\|_{\ell_1}-C\delta^2. \end{equation}

Next, let us consider the time $2\delta$ .

\begin{align*} \mathbb P\big(\Lambda_{2\delta}\neq \tilde \Lambda_{2\delta}\big) & = \mathbb P\big(\Lambda_{2\delta}\neq \tilde\Lambda_{2\delta},\Lambda_\delta= \tilde \Lambda_{\delta}\big) + \mathbb P\big(\Lambda_{2\delta}\neq \tilde\Lambda_{2\delta},\Lambda_\delta\neq \tilde \Lambda_\delta\big) \\ & = \mathbb P\big(\Lambda_{2\delta}\neq \tilde \Lambda_{2\delta}\mid\Lambda_\delta=\tilde \Lambda_\delta\big) \mathbb P\big(\Lambda_\delta=\tilde \Lambda_\delta\big) + \mathbb P\big(\Lambda_{2\delta}\neq \tilde \Lambda_{2\delta}\mid\Lambda_\delta\neq \tilde \Lambda_{\delta}\big) \mathbb P\big(\Lambda_\delta\neq \tilde \Lambda_{\delta}\big) \\ & \leq \alpha(\delta)\mathbb P\big(\Lambda_\delta=\tilde \Lambda_\delta\big) + \mathbb P\big(\Lambda_\delta\neq \tilde \Lambda_\delta\big) \\ & = \alpha(\delta) + (1-\alpha(\delta))\mathbb P\big(\Lambda_\delta\neq \tilde \Lambda_\delta\big) \\ & \leq \alpha(\delta)(1+\beta(\delta)), \end{align*}

where we have used the time homogeneity of Markov chain $(\Lambda_t,\tilde \Lambda_t)$ to get the estimate $\mathbb P\big(\Lambda_{2\delta}\neq\tilde\Lambda_{2\delta}\mid\Lambda_{\delta}=\tilde\Lambda_{\delta}\big) \leq \alpha(\delta)$ . Analogously, using the time homogeneity of $(\Lambda_t)$ and $\big(\tilde \Lambda_t\big)$ ,

\begin{align*} \mathbb P\big(\Lambda_{3\delta}\neq \tilde \Lambda_{3\delta}\big) & \leq \mathbb P\big(\Lambda_{3\delta}\neq \tilde\Lambda_{3\delta}\mid\Lambda_{2\delta}=\tilde \Lambda_{2\delta}\big) \mathbb P\big(\Lambda_{2\delta}= \tilde \Lambda_{2\delta}\big) + \mathbb P\big(\Lambda_{2\delta}\neq \tilde \Lambda_{2\delta}\big) \\ & \leq \alpha(\delta) + \beta(\delta) \mathbb P\big(\Lambda_{2\delta} \neq \tilde\Lambda_{2\delta}\big) \\ & \leq \alpha(\delta)\big(1+\beta(\delta)+\beta(\delta)^2\big) = 1 - \beta(\delta)^3. \end{align*}

Deducing inductively, we get

\begin{equation*} \mathbb P\big(\Lambda_{k\delta}\neq \tilde \Lambda_{k\delta}\big)\leq \alpha(\delta)\sum_{m=1}^k\beta(\delta)^{m-1} = 1 - \beta(\delta)^{k} \qquad \text{for}\ k\geq 4. \end{equation*}

Therefore, for $t>0$ , setting $K(t)=[{t}/{\delta}]$ , $t_k=k\delta$ for $k\leq K$ , and $t_{K+1}=t$ , by (3.12),

\begin{align*} \int_0^t\mathbb P\big(\Lambda_s\neq \tilde \Lambda_s\big)\mathrm{d} s & \leq \sum_{k=0}^{K(t)}\int_{t_k}^{t_{k+1}} \big(\mathbb P\big(\Lambda_s\neq\tilde\Lambda_s,\Lambda_{k\delta}=\tilde\Lambda_{k\delta}\big) + \mathbb P\big(\Lambda_s\neq\tilde\Lambda_s,\Lambda_{k\delta}\neq\tilde\Lambda_{k\delta}\big)\big)\,\mathrm{d}s \\ & \leq \sum_{k=0}^{K(t)}\int_{t_k}^{t_{k+1}} \mathbb P\big(\Lambda_s\neq\tilde\Lambda_s\mid\Lambda_{k\delta}\neq\tilde\Lambda_{k\delta}\big)\,\mathrm{d}s + \sum_{k=0}^{K(t)}\int_{t_k}^{t_{k+1}}\mathbb P \big(\Lambda_{k\delta}\neq\tilde\Lambda_{k\delta}\big)\,\mathrm{d}s \\ & \leq \alpha(\delta)t + \delta\Bigg(K(t)+1 - \sum_{k=0}^{K(t)} \beta(\delta)^k\Bigg) \\ & \leq \alpha(\delta)t+\delta(K(t)+1)-\delta\sum_{k=0}^{K(t)}\big(1-\delta\big\|Q-\widetilde Q\big\|_{\ell_1}-C\delta^2\big)^k \\ & \leq \alpha(\delta)t + \delta(K(t)+1) - \frac{1-\big(1-\delta\big\|Q-\widetilde Q\big\|_{\ell_1}-C\delta^2\big)^{K(t)+1}}{\big\|Q-\widetilde Q\big\|_{\ell_1}+C\delta}. \end{align*}

Letting $\delta\downarrow 0$ and then dividing both sides by t, we get the upper estimate

\begin{equation*}\frac{1}{t}\int_0^t\!\mathbb P\big(\Lambda_s\neq \tilde \Lambda_s\big)\,\mathrm{d}s \leq 1-\frac{1}{t\big\|Q-\widetilde Q\big\|_{\ell_1}}\Big(1-\mathrm{e}^{-\big\|Q-\widetilde Q\big\|_{\ell_1}t}\Big)\leq 1.\end{equation*}

Using the inequality $\mathrm{e}^{-x}\leq 1-x+\frac 12 x^2$ for $x\geq 0$ , we further get

\begin{equation*}\frac{1}{t}\int_0^t\!\mathbb P\big(\Lambda_s\neq \tilde \Lambda_s\big)\,\mathrm{d}s \leq \min\bigg\{\frac 12 \big\|Q-\widetilde Q\big\|_{\ell_1} t, 1\bigg\}.\end{equation*}

Therefore, the upper estimates (3.3) and (3.4) have been proved.

For the lower estimate, We will estimate the difference by induction. Due to (3.9),

\begin{align*} \mathbb P\big(\Lambda_\delta\neq \tilde \Lambda_\delta\big) & = \mathbb P\big(\Lambda_\delta\neq \tilde \Lambda_\delta\mid\Lambda_0=\tilde \Lambda_0\big) \\ & \geq \mathbb P\big(\Lambda_{\delta}\neq \tilde \Lambda_{\delta}, N_{\tilde{\textbf{p}}}\big([0,\delta]\times U_{i_0}\big)=1\big) \\ & = \delta\mathrm{e}^{-H\delta}\sum_{j\neq i}|q_{i_0j}-\tilde q_{i_0j}| \geq \delta\mathrm{e}^{-H\delta}\inf_{i\in \mathcal S}\sum_{j\neq i}|q_{ij}-\tilde q_{ij}| \,=\!:\,\tilde \alpha(\delta). \end{align*}

Then,

\begin{align*} \mathbb P\big(\Lambda_{2\delta}\neq \tilde \Lambda_{2\delta}\big) & = \mathbb P\big(\Lambda_{2\delta}\neq \tilde \Lambda_{2\delta},\Lambda_{\delta}=\tilde\Lambda_{\delta}\big) + \mathbb P\big(\Lambda_{2\delta}\neq \tilde \Lambda_{2\delta},\Lambda_\delta\neq \tilde \Lambda_{\delta}\big) \\ & = \mathbb P\big(\Lambda_{2\delta}\neq \tilde \Lambda_{2\delta}\mid\Lambda_{\delta}=\tilde \Lambda_{\delta}\big) \big(1-\mathbb P\big(\Lambda_{\delta}\neq \tilde \Lambda_{\delta}\big)\big) \\ & \quad +\mathbb P\big(\Lambda_\delta\neq \tilde \Lambda_{\delta}\big) - \mathbb P\big(\Lambda_{2\delta}=\tilde \Lambda_{2\delta}\mid\Lambda_{\delta}\neq \tilde \Lambda_{\delta}\big) \mathbb P\big(\Lambda_\delta\neq \tilde \Lambda_{\delta}\big) \\ & \geq \mathbb P\big(\Lambda_{2\delta}\neq \Lambda_{2\delta}\mid\Lambda_\delta=\tilde \Lambda_\delta\big) \big(1-\mathbb P\big(\Lambda_{\delta}\neq \tilde \Lambda_{\delta}\big)\big) + \mathbb P\big(\Lambda_\delta\neq \tilde \Lambda_{\delta}\big) \\ & \quad - \mathbb P\Big(\text{there exist jumps for $(\tilde{\textbf{p}}(t))$ during $[\delta,2\delta)$ localizing in $U_{\Lambda_\delta}\cup U_{\tilde \Lambda_\delta}$}\Big) \\& \mathbb P\big(\Lambda_\delta\neq \tilde \Lambda_{\delta}\big) \\ & \geq \mathbb P\big(\Lambda_{2\delta}\neq \Lambda_{2\delta}\mid\Lambda_\delta=\tilde \Lambda_\delta\big) \big(1-\mathbb P\big(\Lambda_{\delta}\neq \tilde \Lambda_{\delta}\big)\big) \\ & \quad + \mathbb P\big(\Lambda_\delta\neq \tilde \Lambda_{\delta}\big) - \big(1-\mathrm{e}^{-M\delta}\big)\mathbb P\big(\Lambda_\delta\neq \tilde \Lambda_\delta\big) \\ & \geq \tilde \alpha(\delta)(1-\alpha(\delta))+\mathrm{e}^{-M\delta}\tilde \alpha(\delta) \\ & \geq \tilde\alpha(\delta)+\big(\mathrm{e}^{-M\delta}-\delta\big\|Q-\widetilde Q\big\|_{\ell_1}-C\delta^2\big) \tilde\alpha(\delta), \end{align*}

where we have used (3.11) and $\mathbb P\big(N_{\tilde{\textbf{p}}}\big([\delta, 2\delta]\times \big(U_{\Lambda_\delta}\cup U_{\tilde \Lambda_{\delta}}\big)\big)\geq 1\big) \leq 1- \mathrm{e}^{-M\delta}$ , since $\textbf{m}(U_{i}\cup U_{j})\leq M=4c_0K_0$ for all $i,j\in\mathcal S$ . Setting $\gamma(\delta)=\mathrm{e}^{-M\delta}-\delta\big\|Q-\widetilde Q\big\|_{\ell_1}-C\delta^2$ , which is positive when $\delta$ is small enough, we rewrite the previous estimate in the form $\mathbb P\big(\Lambda_{2\delta}\neq \tilde \Lambda_{2\delta}\big) \geq \tilde \alpha(\delta)+\gamma(\delta)\tilde \alpha(\delta)$ . Repeating this procedure, we obtain

\begin{equation*} \mathbb P(\Lambda_{k\delta}\neq \tilde \Lambda_{k\delta}) \geq \tilde \alpha(\delta)\sum_{m=1}^{k}\gamma(\delta)^{m-1}, \qquad k\geq 3. \end{equation*}

Then, from this, for $K\in \mathbb N$ ,

\begin{align*} \int_0^{K\delta}\mathbb P\big(\Lambda_s\neq\tilde\Lambda_s\big)\,\mathrm{d}s & = \sum_{k=0}^{K }\int_{k\delta}^{(k+1)\delta} \big(\mathbb P\big(\Lambda_s\neq\tilde\Lambda_s,\Lambda_{k\delta}\neq\tilde\Lambda_{ k\delta}\big) + \mathbb P\big(\Lambda_s\neq\tilde\Lambda_s,\Lambda_{k\delta}=\tilde\Lambda_{k\delta}\big)\big)\,\mathrm{d}s \\ & \geq \sum_{k=0}^{K }\int_{k\delta}^{(k+1)\delta} \big(\mathbb P\big(\Lambda_{k\delta}\neq\tilde\Lambda_{k\delta}\big) - \mathbb P\big(\Lambda_{k\delta}\neq\tilde\Lambda_{k\delta},\Lambda_s=\tilde\Lambda_s\big)\big)\,\mathrm{d}s \\ & \geq \delta\sum_{k=1}^{K}\mathbb P\big(\Lambda_{k\delta}\neq\tilde\Lambda_{k\delta}\big) - \delta\big(1-\mathrm{e}^{-M\delta}\big)\sum_{k=1}^{K}\mathbb P\big(\Lambda_{k\delta}\neq\tilde\Lambda_{k\delta}\big) \\ & \geq \delta\mathrm{e}^{-M\delta}\sum_{k=1}^K\tilde\alpha(\delta)\frac{1-\gamma(\delta)^k}{1-\gamma(\delta)} \\ & = \mathrm{e}^{-M\delta}\frac{\delta\tilde\alpha(\delta)}{1-\gamma(\delta)} \bigg(K-\frac{\gamma(\delta)\big(1-\gamma(\delta)^K\big)}{1-\gamma(\delta)}\bigg). \end{align*}

Since

\begin{equation*}\lim_{\delta\downarrow 0}\frac{\tilde \alpha(\delta)}{1-\gamma(\delta)} = \frac{\inf\limits_{i\in\mathcal S}\sum\limits_{j\neq i}|q_{ij}-\tilde q_{ij}|}{M+\big\|Q-\widetilde Q\big\|_{\ell_1}}, \qquad \lim_{\delta\downarrow 0} \gamma(\delta)^{\frac{t}{\delta}}=\mathrm{e}^{-\big(M+\big\|Q-\widetilde Q\big\|_{\ell_1}\big) t},\end{equation*}

by taking $K=[{t}/{\delta}]$ in the previous estimation and letting $\delta$ tend downward to 0, we finally get

\begin{equation*} \int_0^t\mathbb P\big(\Lambda_s\neq\tilde\Lambda_s\big)\,\mathrm{d}s \geq \frac{\inf\limits_{i\in\mathcal S}\sum\limits_{j\neq i}|q_{ij}-\tilde q_{ij}|}{M+\big\|Q-\widetilde Q\big\|_{\ell_1}} \left(t-\frac{1}{M+\big\|Q-\widetilde Q\big\|_{\ell_1}}\left(1-\mathrm{e}^{-\Big(M+\big\|Q-\widetilde Q\big\|_{\ell_1}\Big)t}\right)\right), \end{equation*}

which is the desired lower estimate. The proof is complete.

Next, we consider the application of Theorem 3.1. First, let us consider its application to perturbation theory on the invariant probability measures of continuous-time Markov chains on infinite state spaces. There have been many works on perturbation theory of Markov chains, such as [Reference Mitrophanov16, Reference Zeifman and Isaacson29] and references therein. We refer readers to the recent review paper [Reference Zeifman, Korolev and Satin30] for more discussions on this topic.

Let $P_t$ and $\widetilde P_t$ denote the semigroups with respect to the transition rate matrices Q and $\widetilde Q$ respectively. Assume that there exist invariant probability measures $\pi=(\pi_i)_{i\in\mathcal S}$ and $\tilde \pi=(\tilde \pi_i)_{i\in\mathcal S}$ associated respectively with $P_t$ and $\widetilde P_t$ , i.e. $\pi P_t=\pi$ , $\tilde \pi \widetilde P_t=\tilde \pi$ .

Corollary 3.1. Assume (H1) and (H2) hold. Suppose that there exists a function $\eta\,:\,[0,\infty)\to[0,2]$ satisfying $\int_0^\infty\eta_s\,\mathrm{d}s<\infty$ such that, for some $i_0\in\mathcal S$ , $\|P_t(i_0,\cdot)-\pi\|_{\mathrm{var}}\leq \eta_t$ and $\|\widetilde P_t(i_0,\cdot)-\tilde \pi\|_{\mathrm{var}}\leq \eta_t$ for $t\geq 0$ . Then,

\begin{equation*} \|\pi-\tilde \pi\|_{\mathrm{var}} \leq 2\sqrt 2\bigg(\int_0^\infty \eta_s\,\mathrm{d}s\bigg)^{1/2}\sqrt{\big\|Q-\widetilde Q\big\|_{\ell_1}}. \end{equation*}

Proof. For any bounded function h on $\mathcal S$ with $|h|_\infty\,:\!=\,\sup_{i\in\mathcal S} |h_i|\leq 1$ ,

\begin{align*} |\pi(h)-\tilde \pi(h)| & = \Bigg|\sum_{i\in\mathcal S} \pi_i h_i-\sum_{i\in\mathcal S}\tilde \pi_i h_i\Bigg| \\ & \leq \bigg|\pi(h)-\frac 1t\int_0^t P_sh(i_0)\,\mathrm{d}s\bigg| + \bigg|\tilde\pi(h)-\frac 1t\int_0^t\widetilde P_s h(i_0)\,\mathrm{d}s\bigg|\\ & + \bigg|\frac 1t\int_0^t P_s h(i_0)-\widetilde P_s h(i_0)\,\mathrm{d}s\bigg| \\ & \leq \frac 1t\int_0^t|P_s h(i_0)-\pi(h)|\,\mathrm{d}s + \frac 1t\int_0^t|\widetilde P_s h(i_0)-\tilde \pi(h)|\,\mathrm{d}s + \big\|Q-\widetilde Q\big\|_{\ell_1} t \\ & \leq \frac 2 t\int_0^\infty\eta_s\,\mathrm{d}s + \big\|Q-\widetilde Q\big\|_{\ell_1} t \qquad \text{for all } t>0, \end{align*}

where we have used (3.4) in Theorem 3.1. By taking $t=\Big(2\int_0^\infty\eta_s\,\mathrm{d}s/\big\|Q-\widetilde Q\big\|_{\ell_1}\Big)^{1/2}$ , we arrive at

\begin{equation*}\|\pi-\tilde \pi\|_{\mathrm{var}} = \sup_{|h|\leq 1}|\pi(h)-\tilde\pi(h)| \leq 2\sqrt 2\bigg(\int_0^\infty\eta_s\,\mathrm{d}s\bigg)^{\frac 12}\sqrt{\big\|Q-\widetilde Q\big\|_{\ell_1}}, \end{equation*}

which is the desired conclusion.

Second, we can apply Theorem 3.1 to improve all the main results [Reference Shao and Yuan23, Theorems 1.1--1.4]. Here we only state the improvement of [Reference Shao and Yuan23, Theorem 1.1] to save space.

Consider the regime-switching system $(X_t,\Lambda_t)$ satisfying

\begin{equation*} \mathrm{d} X_t=b(X_t,\Lambda_t)\,\mathrm{d} t+\sigma(X_t,\Lambda_t)\,\mathrm{d} B_t, \qquad X_0=x_0\in\mathbb R^d,\ \Lambda_0=i_0\in\mathcal S,\end{equation*}

with $(\Lambda_t)$ a Markov chain on $\mathcal S=\{1,2,\ldots,N\}$ , $2\leq N\leq \infty$ . In realistic applications, we can sometimes only get an estimation $\widetilde Q$ of the original transition rate matrix Q of $(\Lambda_t)$ . $\widetilde Q$ determines another Markov chain $\big(\tilde \Lambda_t\big)$ . Correspondingly, the studied system $(X_t)$ turns into $(\widetilde X_t)$ satisfying

\begin{equation*}\mathrm{d} \widetilde X_t=b\big(\widetilde X_t,\tilde \Lambda_t\big)\,\mathrm{d} t + \sigma\big(\widetilde X_t,\tilde \Lambda_t\big)\,\mathrm{d} B_t, \qquad \widetilde X_0=x_0,\ \tilde \Lambda_0=i_0.\end{equation*}

It is necessary to measure the difference between $X_t$ and $\widetilde X_t$ caused by the difference between Q and $\widetilde Q$ . We shall characterize the difference between $X_t$ and $\widetilde X_t$ via the Wasserstein distance between their distributions.

For any two probability measures $\mu$ , $\nu$ on $\mathbb R^d$ , define the $L_2$ -Wasserstein distance between them by

\begin{equation*}W_2(\mu,\nu)^2=\inf_{\pi\in\mathscr{C}(\mu,\nu)}\bigg\{\int_{\mathbb R^d\!\times\!\mathbb R^d} |x-y|^2\,\pi(\mathrm{d} x,\mathrm{d} y)\bigg\},\end{equation*}

where $\mathscr{C}(\mu,\nu)$ denotes the set of all the couplings of $\mu$ , $\nu$ on $\mathbb R^d\times \mathbb R^d$ .

Corollary 3.2. Assume that (Q1), (Q2), (H1), and (H2) hold. Denote by $\mathcal{L}(X_t)$ and $\mathcal{L}(\widetilde X_t)$ the distributions of $X_t$ and $\widetilde X_t$ respectively. Then

\begin{equation*} W_2\big(\mathcal{L}(X_t),\mathcal{L}\big(\widetilde X_t\big)\big)^2 \leq C(p,t)\Big(\big\|Q-\widetilde Q\big\|_{\ell_1}\Big)^{(p-1)/p}, \qquad t>0, \end{equation*}

where $p>1$ and C(p,t) is a positive constant depending only on p and t.