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Rephrased question: Is it ever possible for a reducible Markov chain to be reversible?

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up vote 3 down vote accepted

Yes. Consider the Markov chain with just two states and no transitions between them. This is reducible, and any distribution is stationary and satisfies the condition for reversibility, since both sides of the detailed balance condition are zero.

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Reversibility implies that if the Markov chain can go from $x$ to $y$ in finite time with positive probability, the same holds for $y$ and $x$. Hence the only way to be reducible is that there exists some states $x$ and $y$ such that the chain cannot go from $x$ to $y$ neither from $y$ to $x$. In other words there exists a partition of the state space such that the Markov chain starting from a state in a given class stays in this class forever with full probability and such that the Markov chain restricted to this class is irreducible.

In other words, one considers a collection $(Q_i)_i$ of reversible irreducible transition kernels on disjoint non empty state spaces $(S_i)_i$ with stationary measures $\mu_i$. The state space of the Markov chain is the union of the spaces $S_i$ and while in $S_i$, the chain uses the kernel $Q_i$. Every barycenter of the stationary measures $(\mu_i)_i$ is stationary and the chain is reversible but reducible as soon as there is more than one state space $S_i$.

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