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:
It's his proof that $Z_n$ is Markov I'm stuck on. It seems like there are some details left out here.
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 $Z'_n\sim\mathrm{markov}(\pi, P)$