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The reversibility of a Markov chain is defined in the following way with some basic propositions.

Unfortunately all examples of reversible Markov chains shown in my textbook so far are irreducible, giving me an impression (I think it is false) that all reversible Markov chains are irreducible.

I am curious about is there any example of reversible yet reducible Markov chains, so the initial distribution satisfying the detailed balance is not the only stationary distribution. Thank you!

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If the chain has an absorbing state, then the answer is no, because then $P(X_0=x_0,...X_n=x_n)>0$ but $P(X_n=x_n,...,X_0=x_0)=0$, where $x_n$ is the absorbing state. On the other hand, suppose your markov chain can be broken down into two disjoint communicating classes, i.e. two disconnected Markov chains. Then if the individual markov chains are reversable, then so is the full chain, since it's impossible to cross from one to the other.

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  • $\begingroup$ Can we claim that A time-reversible Markov chain must be irreducible ? @alex-r $\endgroup$
    – John Stone
    Jan 14 at 12:10
  • $\begingroup$ I try my best to understand your explanation, but I'm still a little confused @Alex-R. $\endgroup$
    – John Stone
    Jan 14 at 12:14
  • $\begingroup$ Suppose the chain has an absorbing state $k$, then we have $p_{kk}=1$ and $p_{kj}=0$ for $\forall j\ne k$. Thus, your first part explanation means if the chain has an absorbing state then there doesn't exist examples of reversible yet reducible Markov chains, since An reducible Markov chain always have some absorbing states while the time-reversibility is not satisfied as you shown. Is my understanding right? @Alex-R $\endgroup$
    – John Stone
    Jan 14 at 12:41
  • $\begingroup$ As for your second part explanation, which means, if the chain is reducible (two disjoint communicating classes), then the individual Markov chains can not be reversible, otherwise it will lead to the full chain is reversible, which contradicts the chain is reducible. Is my understanding right? @AlexR $\endgroup$
    – John Stone
    Jan 14 at 12:58

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