Find a modified coupling $((X_n,\tilde Y_n))_{n∈ℕ_0}$ with the same coupling time $τ$ and $\tilde Y_n=X_n$ for $n≥τ$ in the coupling lemma

Let

  • $ (\Omega,\mathcal A,\operatorname P)$ be a probability space
  • $ (E,\mathcal E)$ be a measurable space
  • $ (X_n)_{n\in\mathbb N_0}$ and $ (Y_n)_{n\in\mathbb N_0}$ be $ (E,\mathcal E)$ -valued time-homogeneous Markov chains on $ (\Omega,\mathcal A,\operatorname P)$ with common transition kernel $ \kappa$ and $ $ Z_n:=(X_n,Y_n)\;\;\;\text{for }n\in\mathbb N_0$ $
  • $ \mathcal F^X$ , $ \mathcal F^Y$ and $ \mathcal F^Z$ denote the filtraiton generated by $ X$ , $ Y$ and $ Z$ , respectively

It’s easy to see that $ $ \tau:=\inf\left\{n\in\mathbb N_0:X_n=Y_n\right\}$ $ is an $ \mathcal F^Z$ -stopping time and hence $ $ \tilde Y_n:=1_{\left\{\:n\:<\:\tau\:\right\}}Y_n+1_{\left\{\:n\:\ge\:\tau\:\right\}}X_n\;\;\;\text{for }n\in\mathbb N_0$ $ is $ \mathcal F^Z$ -adapted. Moreover, $ \mathcal F^Z=\mathcal F^X\vee\mathcal F^Y$ .

How can we show that $ \tilde Y$ is a time-homogeneous Markov chain with the same distribution as $ Y$ ?

I guess the basic idea is that $ Z$ is clearly a time-homogeneous Markov chain with transition kernel $ \pi$ satisfying $ $ \pi((x,y),B_1\times B_2)=\kappa(x,B_1)\kappa(y,B_2)\;\;\;\text{for all }x,y\in E\text{ and }B_i\in\mathcal E\tag1.$ $ Since $ \mathbb N_0$ is countable, $ Z$ is strongly Markovian at $ \tau$ and hence $ $ 1_{\left\{\:\tau\:<\:\infty\:\right\}}\operatorname E\left[f\left(\left(Z_{\tau+n}\right)_{n\in\mathbb N_0}\right)\mid\mathcal F_\tau\right]=1_{\left\{\:\tau\:<\:\infty\:\right\}}(\pi f)(Z_\tau)\;\;\;\text{almost surely}\tag2,$ $ where $ \pi f:=\int\pi(\;\cdot\;,{\rm d}z)f(z)$ , for all bounded and $ (\mathcal E\otimes\mathcal E)^{\otimes\mathbb N_0}$ -measurable $ f:(E\times E)^{\mathbb N_0}\to\mathbb R$ . So, if $ k\in\mathbb N_0$ , $ n_0,\ldots,n_k\in\mathbb N_0$ with $ 0=n_0<\cdots<n_k$ and $ B\in\mathcal E^{\otimes k}$ , we obtain \begin{equation}\begin{split}1_{\left\{\:\tau\:<\:\infty\:\right\}}\operatorname P\left[\left(\tilde Y_{\tau+n_1},\ldots,\tilde Y_{\tau+n_k}\right)\in B\mid\mathcal F_\tau\right]&=1_{\left\{\:\tau\:<\:\infty\:\right\}}\bigotimes_{i=1}^k\kappa^{n_i-n_{i-1}}(X_\tau,B)\&=1_{\left\{\:\tau\:<\:\infty\:\right\}}\operatorname P\left[\left(Y_{\tau+n_1},\ldots,Y_{\tau+n_k}\right)\in B\mid\mathcal F_\tau\right]\end{split}\tag3\end{equation} almost surely.

However, it’s neither clear to me how we can conclude that $ \tilde Y$ is Markovian (with respect to its generated filtration) nor why it has the same distribution as $ Y$ .

Clearly, the distribution of $ Y$ is uniquely determined by the finite-dimensional distributions $ \operatorname P\left[\left(Y_{n_1},\ldots,Y_{n_k}\right)\;\cdot\;\right]$ (and the same applies to $ \tilde Y$ ). Moreover, we may write $ $ \operatorname P\left[\left(\tilde Y_{n_1},\ldots,\tilde Y_{n_k}\right)\;\cdot\;\right]=\operatorname P\left[n<\tau,\left(Y_{n_1},\ldots,Y_{n_k}\right)\;\cdot\;\right]+\operatorname P\left[n\ge\tau,\left(X_{n_1},\ldots,X_{n_k}\right)\;\cdot\;\right]\tag4.$ $ Many pieces, I’m not able to combine.

Find a modified coupling $((X_n,\tilde Y_n))_{n∈ℕ_0}$ with the same coupling time $τ$ and $\tilde Y_n=X_n$ for $n≥τ$ in the coupling lemma

Let

  • $ (\Omega,\mathcal A,\operatorname P)$ be a probability space
  • $ (E,\mathcal E)$ be a measurable space
  • $ (X_n)_{n\in\mathbb N_0}$ and $ (Y_n)_{n\in\mathbb N_0}$ be $ (E,\mathcal E)$ -valued time-homogeneous Markov chains on $ (\Omega,\mathcal A,\operatorname P)$ with common transition kernel $ \kappa$ and $ $ Z_n:=(X_n,Y_n)\;\;\;\text{for }n\in\mathbb N_0$ $
  • $ \mathcal F^X$ , $ \mathcal F^Y$ and $ \mathcal F^Z$ denote the filtraiton generated by $ X$ , $ Y$ and $ Z$ , respectively

It’s easy to see that $ $ \tau:=\inf\left\{n\in\mathbb N_0:X_n=Y_n\right\}$ $ is an $ \mathcal F^Z$ -stopping time and hence $ $ \tilde Y_n:=1_{\left\{\:n\:<\:\tau\:\right\}}Y_n+1_{\left\{\:n\:\ge\:\tau\:\right\}}X_n\;\;\;\text{for }n\in\mathbb N_0$ $ is $ \mathcal F^Z$ -adapted. Moreover, $ \mathcal F^Z=\mathcal F^X\vee\mathcal F^Y$ .

How can we show that $ \tilde Y$ is a time-homogeneous Markov chain with the same distribution as $ Y$ ?

I guess the basic idea is that $ Z$ is clearly a time-homogeneous Markov chain with transition kernel $ \pi$ satisfying $ $ \pi((x,y),B_1\times B_2)=\kappa(x,B_1)\kappa(y,B_2)\;\;\;\text{for all }x,y\in E\text{ and }B_i\in\mathcal E\tag1.$ $ Since $ \mathbb N_0$ is countable, $ Z$ is strongly Markovian at $ \tau$ and hence $ $ 1_{\left\{\:\tau\:<\:\infty\:\right\}}\operatorname E\left[f\left(\left(Z_{\tau+n}\right)_{n\in\mathbb N_0}\right)\mid\mathcal F_\tau\right]=1_{\left\{\:\tau\:<\:\infty\:\right\}}(\pi f)(Z_\tau)\;\;\;\text{almost surely}\tag2,$ $ where $ \pi f:=\int\pi(\;\cdot\;,{\rm d}z)f(z)$ , for all bounded and $ (\mathcal E\otimes\mathcal E)^{\otimes\mathbb N_0}$ -measurable $ f:(E\times E)^{\mathbb N_0}\to\mathbb R$ . So, if $ k\in\mathbb N_0$ , $ n_0,\ldots,n_k\in\mathbb N_0$ with $ 0=n_0<\cdots<n_k$ and $ B\in\mathcal E^{\otimes k}$ , we obtain \begin{equation}\begin{split}1_{\left\{\:\tau\:<\:\infty\:\right\}}\operatorname P\left[\left(\tilde Y_{\tau+n_1},\ldots,\tilde Y_{\tau+n_k}\right)\in B\mid\mathcal F_\tau\right]&=1_{\left\{\:\tau\:<\:\infty\:\right\}}\bigotimes_{i=1}^k\kappa^{n_i-n_{i-1}}(X_\tau,B)\&=1_{\left\{\:\tau\:<\:\infty\:\right\}}\operatorname P\left[\left(Y_{\tau+n_1},\ldots,Y_{\tau+n_k}\right)\in B\mid\mathcal F_\tau\right]\end{split}\tag3\end{equation} almost surely.

However, it’s neither clear to me how we can conclude that $ \tilde Y$ is Markovian (with respect to its generated filtration) nor why it has the same distribution as $ Y$ .

Clearly, the distribution of $ Y$ is uniquely determined by the finite-dimensional distributions $ \operatorname P\left[\left(Y_{n_1},\ldots,Y_{n_k}\right)\;\cdot\;\right]$ (and the same applies to $ \tilde Y$ ). Moreover, we may write $ $ \operatorname P\left[\left(\tilde Y_{n_1},\ldots,\tilde Y_{n_k}\right)\;\cdot\;\right]=\operatorname P\left[n<\tau,\left(Y_{n_1},\ldots,Y_{n_k}\right)\;\cdot\;\right]+\operatorname P\left[n\ge\tau,\left(X_{n_1},\ldots,X_{n_k}\right)\;\cdot\;\right]\tag4.$ $ Many pieces, I’m not able to combine.