# 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 independent $$(E,\mathcal E)$$-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 $$0=n_0<\cdots and $$B\in\mathcal E^{\otimes k}$$, we obtain $$$$\begin{split}1_{\left\{\:\tau\:<\:\infty\:\right\}}\operatorname P\left[\left(\tilde Y_{\tau+n_1},\ldots,\tilde Y_{\tau+n_k}\right)\in B\mid\mathcal F_\tau\right]&=1_{\left\{\:\tau\:<\:\infty\:\right\}}\bigotimes_{i=1}^k\kappa^{n_i-n_{i-1}}(X_\tau,B)\\&=1_{\left\{\:\tau\:<\:\infty\:\right\}}\operatorname P\left[\left(Y_{\tau+n_1},\ldots,Y_{\tau+n_k}\right)\in B\mid\mathcal F_\tau\right]\end{split}\tag3$$$$ 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.

• Do you assume that $X$ and $Y$ are independent?
– saz
Jan 13 '19 at 12:02
• @saz Oops! Somehow forgotten to mention that. Yes, of course. Jan 14 '19 at 0:08

Since $$X_{\tau} = Y_{\tau}$$ on $$\{\tau<\infty\}$$ it holds that

$$\bar{Y}_n = 1_{\{\tau \geq n\}} Y_n + 1_{\{\tau \leq n-1\}} X_n.$$

Using that $$\{\tau \leq n-1\} \in \mathcal{F}_{n-1}^Z$$ we find from the pull out property of conditional expectation that

$$\mathbb{E}(f(\bar{Y}_n) \mid \mathcal{F}_{n-1}^Z) = 1_{\{\tau \geq n\}} \mathbb{E}(f(Y_n) \mid \mathcal{F}_{n-1}^Z) + 1_{\{\tau \leq n-1\}} \mathbb{E}(f(X_n) \mid \mathcal{F}_{n-1}^Z)$$

for any bounded measurable function $$f$$. Since $$X$$ and $$Y$$ are, by assumption, independent it follows (see the lemma below) that

$$\mathbb{E}(f(\bar{Y}_n) \mid \mathcal{F}_{n-1}^Z) = 1_{\{\tau \geq n\}} \mathbb{E}(f(Y_n) \mid \mathcal{F}_{n-1}^Y) + 1_{\{\tau \leq n-1\}} \mathbb{E}(f(X_n) \mid \mathcal{F}_{n-1}^X).$$

By assumption, $$X$$ and $$Y$$ are both Markov chains with transition kernel $$\kappa$$, and so

\begin{align*} \mathbb{E}(f(\bar{Y}_n) \mid \mathcal{F}_{n-1}^Z) &= 1_{\{\tau \geq n\}} \int f(y) \, \kappa(Y_{n-1},dy) + 1_{\{\tau \leq n-1\}} \int f(y) \, \kappa(X_{n-1},dy) \\ &= \int f(y) \, \kappa(\bar{Y}_{n-1},dy). \end{align*}

Since $$n \in \mathbb{N}$$ is arbitrary, this shows that $$(\bar{Y}_n)_{n \in \mathbb{N}}$$ is a Markov chain with transition kernel $$\kappa$$.

Lemma Let $$Z \in L^1(\mathbb{P})$$ be a random variable which is measurable with respect to a $$\sigma$$-algebra $$\mathcal{A}$$. If $$\mathcal{G},\mathcal{H}$$ are further $$\sigma$$-algebras such that $$\mathcal{H}$$ is independent from $$\sigma(\sigma(Z),\mathcal{G})$$, then $$\mathbb{E}(Z \mid \sigma(\mathcal{G},\mathcal{H})) = \mathbb{E}(Z \mid \mathcal{G}).$$

• Your lemma is just that if $\mathcal F$ is independent of $\mathcal G\vee\mathcal H$, then $\mathcal F$ is conditionally independet of $\mathcal H$ given $G$, right? That is clear since $\left\{G∩H:G∈\mathcal G\text{ and }H∈\mathcal H\right\}$ is a $\cap$-stable generator of $\mathcal G\vee\mathcal H$ and $\text P[F∩G∩H]=\text P[F]\text P[G∩H]=\text E[1_F]\text E[1_G\text P[H\mid\mathcal G]]=\text E[1_{F∩G}\text P[H\mid\mathcal G]]$ for all $F∈\mathcal F$, $G∈\mathcal G$ and $H∈\mathcal H$ (since $F$ is independent of $G∩H$ and $F$ is independent of $\mathcal G$). Jan 14 '19 at 22:32
• @0xbadf00d Note that $\mathcal{H}$ is assumed to be independent, not $\mathcal{F}$. It might well be that it is possible to formulate the lemma in terms of conditional independence but I don't see the point in it... for the proof I need the lemma the way I formulated it.
– saz
Jan 15 '19 at 7:38