Dtyjlui

Let

• 
  $(Omega,mathcal A,operatorname P)$ be a probability space


• 
  $(E,mathcal E)$ be a measurable space


• 
  $(X_n)_{ninmathbb N_0}$ and $(Y_n)_{ninmathbb N_0}$ be independent $(E,mathcal E)$-valued time-homogeneous Markov chains on $(Omega,mathcal A,operatorname P)$ with common transition kernel $kappa$ and $$Z_n:=(X_n,Y_n);;;text{for }ninmathbb N_0$$
  


• 
  $mathcal F^X$, $mathcal F^Y$ and $mathcal F^Z$ denote the filtraiton generated by $X$, $Y$ and $Z$, respectively

It's easy to see that $$tau:=infleft{ninmathbb N_0:X_n=Y_nright}$$ is an $mathcal F^Z$-stopping time and hence $$tilde Y_n:=1_{left{:n:<:tau:right}}Y_n+1_{left{:n:ge:tau:right}}X_n;;;text{for }ninmathbb N_0$$ is $mathcal F^Z$-adapted. Moreover, $mathcal F^Z=mathcal F^Xveemathcal F^Y$.

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 $$pi((x,y),B_1times B_2)=kappa(x,B_1)kappa(y,B_2);;;text{for all }x,yin Etext{ and }B_iinmathcal Etag1.$$ Since $mathbb N_0$ is countable, $Z$ is strongly Markovian at $tau$ and hence $$1_{left{:tau:<:infty:right}}operatorname Eleft[fleft(left(Z_{tau+n}right)_{ninmathbb N_0}right)midmathcal F_tauright]=1_{left{:tau:<:infty:right}}(pi f)(Z_tau);;;text{almost surely}tag2,$$ where $pi f:=intpi(;cdot;,{rm d}z)f(z)$, for all bounded and $(mathcal Eotimesmathcal E)^{otimesmathbb N_0}$-measurable $f:(Etimes E)^{mathbb N_0}tomathbb R$. So, if $kinmathbb N_0$, $n_0,ldots,n_kinmathbb N_0$ with $0=n_0<cdots<n_k$ and $Binmathcal E^{otimes k}$, we obtain begin{equation}begin{split}1_{left{:tau:<:infty:right}}operatorname Pleft[left(tilde Y_{tau+n_1},ldots,tilde Y_{tau+n_k}right)in Bmidmathcal F_tauright]&=1_{left{:tau:<:infty:right}}bigotimes_{i=1}^kkappa^{n_i-n_{i-1}}(X_tau,B)\&=1_{left{:tau:<:infty:right}}operatorname Pleft[left(Y_{tau+n_1},ldots,Y_{tau+n_k}right)in Bmidmathcal F_tauright]end{split}tag3end{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 $operatorname Pleft[left(Y_{n_1},ldots,Y_{n_k}right);cdot;right]$ (and the same applies to $tilde Y$). Moreover, we may write $$operatorname Pleft[left(tilde Y_{n_1},ldots,tilde Y_{n_k}right);cdot;right]=operatorname Pleft[n<tau,left(Y_{n_1},ldots,Y_{n_k}right);cdot;right]+operatorname Pleft[ngetau,left(X_{n_1},ldots,X_{n_k}right);cdot;right]tag4.$$ Many pieces, I'm not able to combine.

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