3
$\begingroup$

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 $0=n_0<\cdots<n_k$ 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.

$\endgroup$
2
  • $\begingroup$ Do you assume that $X$ and $Y$ are independent? $\endgroup$
    – saz
    Jan 13, 2019 at 12:02
  • $\begingroup$ @saz Oops! Somehow forgotten to mention that. Yes, of course. $\endgroup$
    – 0xbadf00d
    Jan 14, 2019 at 0:08

1 Answer 1

3
$\begingroup$

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 $$\mathbb{E}(Z \mid \sigma(\mathcal{G},\mathcal{H})) = \mathbb{E}(Z \mid \mathcal{G}).$$

$\endgroup$
2
  • $\begingroup$ 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 $\mathcal G$). $\endgroup$
    – 0xbadf00d
    Jan 14, 2019 at 22:32
  • $\begingroup$ @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. $\endgroup$
    – saz
    Jan 15, 2019 at 7:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .