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I know every state in an irreducible aperiodic finite-state Markov chain is positive recurrent, which means their recurrent times are all finite.

Do we have a conclusion that the mean recurrent times of any two states are the same, by irreducibility? Thanks for clarifying.

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Irreducibility means that from every state you can reach any other state. Clearly, this does not imply that the (mean) time to return to any state is the same across all states.

Consider, for instance, a simple Markov chain with 2 states, A and B. Let $a$ and $b$ the self-loop probabilities of remaining at state A and B, respectively. Then, the mean time to return to states A and B, once you leave it, are $1/(1-b)$ and $1/(1-a)$, respectively. They are not the same, except if $a=b$.

Note that in this example the Markov chain is aperiodic, as the two states have self-transitions. Therefore, its period is 1.

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  • $\begingroup$ Thanks for the example. If we have an extra condition that the Markov chain is aperiodic, do we still have the conclusion that the mean recurrent times of any two states are different? (Your example is periodic with period 2). Thanks. $\endgroup$ Commented Nov 20, 2022 at 4:56
  • $\begingroup$ the chain is ergodic (aperiodic and with 1 recurrent class) $\endgroup$
    – Daniel S.
    Commented Nov 20, 2022 at 19:36
  • $\begingroup$ I see. I thought $a=0$ and $b=0$ but I was wrong. Thanks. $\endgroup$ Commented Nov 20, 2022 at 23:31

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