In Proposition 5.4.5 of Meyn and Tweedie's Markov Chains and Stochastic Stability, it is said that if a chain $\Phi$ is $\psi$-irreducible and aperiodic, then every $m$-skeleton of it is also $\psi$-irreducible and aperiodic. The authors did not give an explicit proof of this.

It is easy to show aperiodicity using induction, but I had a hard time to figure out the proof for $\psi$-irreducibility. Is induction and contradiction a right direction to go? In general, if there is no aperiodicity of the chain, is it also true that all the $m$-skeletons are $\psi$-irreducible?


I think it can be proved as follows:

For any $C \in B^{+}(X)$, w.o.l.g. (see Theorem 5.2.2), we can assume that $C$ is a $\nu_M$ small set with $\nu_M(C) > 0$ . Since, $\Phi$ is aperiodic (and because this property does not depend on the small set as shown by the proof of Theorem 5.4.4), $g.c.d.(E_C)=1$. By Lemma D.7.4, there exists $k_0$ s.t. for all $k \geq k_0$, $C$ is $\nu_k$ small with $\nu_k = \delta_k \nu_M$ and with $\delta_k >0$.

For $x \in X$, since $\Phi$ is $\psi$-irreducible, there is some $r$ s.t. $P^r(x, C) >0$. Hence, by Chapman-Kolmogorov, for any $k \geq k_0$ we get: $$P^{r+k}(x, C) \geq \int_C P^r(x, dy) P^k(y, C) \geq P^r(x, C) \delta_k \nu_M(C) >0 \quad .$$ Therefore, for any $m$, there exist $i$ s.t. $P^{i\cdot m}(x, C) >0$, which yields the result.


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