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I've been reading Norris's book on Markov Chains, and there's one part of his proof of convergence of a Markov chains to equilibrium that leaves me a bit confused. He starts with $X_n$ and $Y_n$, with $X_0 \sim \lambda$ and $Y_0\sim \pi$, the invariant measure. He then introduces the coupling time $T$, at which both chains first reach a given state $b$, and $W_n = (X_n, Y_n)$, the joint process. Next, he introduces a piecewise defined $Z_n$ process, and proceeds to show that it is Markov:enter image description here

It's his proof that $Z_n$ is Markov I'm stuck on. It seems like there are some details left out here.

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  • $\begingroup$ What part did you understand up until? $\endgroup$ – spaceisdarkgreen Mar 22 '17 at 1:22
  • $\begingroup$ I found the "By symmetry, we can replace..." line a bit confusing, but I kind of buy it. I was mostly troubled by the last sentence, that $Z_n$ is Markov, but now see that, if $W_n'$ really is Markov on $(\mu, \tilde P)$, then its components must also be Markov because this is just a decoupled Markov Chain on the product space. $\endgroup$ – user2379888 Mar 22 '17 at 12:40
  • $\begingroup$ Yeah, that's right on the last point. $\endgroup$ – spaceisdarkgreen Mar 22 '17 at 23:43
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You have the Markov process $(X_n,Y_n)\sim \mathrm{markov}(\mu,\tilde P).$ $(Z_n,Z'_n)$ must be statistically identical to it by the symmetry argument given. Basically, you wait till the time when $(X_n,Y_n)=(b,b)$ and then pull a switcheroo. Since the transition matrix is symmetrical and the state is symmetrical, you've done nothing. Thus $(Z_n,Z'_n)\sim \mathrm{markov}(\mu,\tilde P)$ which means $Z_n\sim\mathrm{markov}(\lambda, P)$ and $Z'_n\sim\mathrm{markov}(\pi, P)$

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  • $\begingroup$ Just to be clear, you can make the switch at the stopping time because, at $T$, you know $X_{T} = Y_{T} = b$, and they then propagate, independently, with the same transition matrix, $P$, is that right? $\endgroup$ – user2379888 Mar 22 '17 at 12:41
  • $\begingroup$ @user2379888 intuitively, yes. The reason for constructing the pair is that, from the perspective of the markov chain on $I\times I$ with initial distribution $(\lambda,\pi)$ and transition matrix $\tilde P = P\otimes P$, you are waiting till it's in state $(b,b),$ doing nothing, and then carrying on. Thus $(Z_n,Z_n')$ is the same markov chain as $(X_n,Y_n).$ $\endgroup$ – spaceisdarkgreen Mar 22 '17 at 23:34

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