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I encountered a Lemma:

For any irreducible aperiodic Markov chain $(X_0, X_1, \ldots)$with state space $S =\{s_1,\ldots, s_k \}$ and transition matrix $P$, we have for any two states $s_i,s_j \in  S$ that if the chain starts in state $s_i$ , then $$P(T_{i, j} < \infty ) = 1$$

The hitting time $T_{i, j}$ is defined as $$T_{i, j} = \min\{ n \geq 1: X_n = s_j\}$$

Is aperiodicity indispensable for this lemma? Or is there an example of a irreducible but periodic Markov chain with finite state space that admits two states $s_m,s_n \in  S$ such that $$P(T_{m, n} < \infty ) < 1$$

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No, aperiodicity is not needed. To see this, let $d$ be the period of the chain and consider the chain observed at every $d$th time point : $Y_n:=X_{dn}$. Then $Y_n$ is aperiodic and visits every state with probability one, hence so does $X_n$.

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  • $\begingroup$ Thank you. Hope you don't mind, I'm just beginning to self-study this topic. Is there a sharpened version of this lemma I can refer to? $\endgroup$ – Metta World Peace Sep 28 '13 at 2:08
  • $\begingroup$ What do you mean by "sharpened"? $\endgroup$ – user940 Sep 28 '13 at 2:09
  • $\begingroup$ Sorry for confusion, I'm not a native speaker of English. I mean , where can I find a proof for a theorem in which aperiodicity is not needed, or even better, some other conditons are not needed either? $\endgroup$ – Metta World Peace Sep 28 '13 at 2:13
  • $\begingroup$ I've added some more explanation. $\endgroup$ – user940 Sep 28 '13 at 2:18
  • $\begingroup$ Tons of thanks! $\endgroup$ – Metta World Peace Sep 28 '13 at 2:21

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