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Let $(X_i)_i, (Y_i)_i$ be two independent Markov chains on Polish state spaces $\mathcal{X}, \mathcal{Y}$ and with kernels $P, Q$. Suppose that they are both $\psi$-irreducible and aperiodic and have unique stationary distributions $\pi, \lambda$. Is it then true that the Markov chain $(X_i, Y_i)_i$ is also $\psi$-irreducible and aperiodic? It is straightforward to show that this chain has stationary distribution $\pi\times\lambda$, but $\psi$-irreducibility and aperiodicity aren't clear to me.

I have read that this is true for chains on countable state spaces, but for the general state space setting I run into some (product-)measure-theoretic problems. Any advice or perhaps a reference for this topic?

I am interested in the general case described above, but if it matters, I am especially interested in the case with $\mathcal{X}=\mathcal{Y}=[0,1]^m$ and $P=Q$. I hope it is true that this chain is also $\psi$-irreducible and aperiodic, but if not: what would perhaps be mild conditions under which it is?

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$\newcommand{\X}{\mathcal X}\newcommand{\Y}{\mathcal Y}\newcommand{\si}{\sigma}\newcommand{\B}{\mathscr B}$Since $\X$ and $\Y$ are Polish spaces, the corresponding Borel $\si$-algebras $\B(\X)$ and $\B(\Y)$ are countably generated, so that the theory of $\psi$-irreducibility expounded in the Meyn and Tweedie book is applicable.

Let us reproduce here relevant definitions and facts from that book:

P. 89: A chain $X$ on $\X$ is $\psi$-irreducible if it is $\phi$-irreducible for some non-trivial measure $\phi$ on $\B(\X)$ -- in the sense for each $x\in\X$ and each $A\in\B(\X)$ such that $\phi(A)>0$ there is some integer $n\ge1$ such that $P^n(x,A)>0$. (The condition that $\phi$ be nontrivial seems to be missing in the definition of the $\psi$-irreducibility in the book.)

By Proposition 4.2.2 (p. 90 in the book), if a chain $X$ on $\X$ is $\psi$-irreducible, then there is a probability measure (say $\psi_X$) which is in a certain sense maximal among all non-trivial measures $\phi$ on $\B(\X)$ such that the chain is $\phi$-irreducible. (This terminology in the book may be a bit confusing indeed.)

P. 91: $\B^+(\X):=\{A\in\B(\X)\colon\psi_X(A)>0\}$.

P. 109: A set $C\in\B(\X)$ is called a small set if for some integer $m\ge1$, some non-trivial measure $\nu_m$ on $\B(\X)$, all $x\in C$, and all $B\in\B(\X)$ we have $P^m(x,B)\ge\nu_m(B)$; then the set $C$ is called $\nu_m$-small.


Let now $X$ and $Y$ be two independent chains on $\X$ and $\Y$, which are $\psi_X$- and $\psi_\Y$-irreducible, with kernels $P$ and $Q$, respectively.

Fix any $(x,y)\in\X\times\Y$ and any $(A,B)\in\B^+(\X)\times\B^+(\Y)$. By Theorem 5.2.2 (p. 112 in the book), \begin{equation*} \exists k\ge1\ \exists C\subseteq A\ \text{s.t. } C\in\B^+(\X),\ C\text{ is $\nu_k$-small},\ \nu_k(C)>0. \end{equation*}

By Theorem 5.4.4 (p. 120 in the book), the period $d_X$ of the chain $X$ is the greatest common divisor (gcd) of the set \begin{equation*} E_C:=\{n\ge1\colon C\text{ is $(h_n\nu_k)$-small for some $h_n>0$}\}. \end{equation*} Also, it is easy to see and noted in the book that the set $E_C$ is closed w.r. to the addition. Since the chain $X$ is aperiodic, we have $d_X=1$, and hence eventually (that is, for all large enough $n$) we have $n\in E_C$, that is, eventually $C$ is $(h_n\nu_k)$-small, so that \begin{equation*} \forall u\in C\ P^n(u,A)\ge h_n\nu_k(A)\ge h_n\nu_k(C). \end{equation*} Also, the condition $C\in\B^+(\X)$ implies $a:=P^j(x,C)>0$ for some $j\ge1$. So, eventually \begin{equation*} P^n(x,A)\ge\int_C P^j(x,du)P^{n-j}(u,A)\ge\int_C P^j(x,du)h_n\nu_k(C) =ah_n\nu_k(C)>0. \end{equation*} Similarly, eventually $Q^n(y,B)>0$. Thus, eventually \begin{equation*} (P\otimes Q)^n((x,y),A\times B)=P^n(x,A)Q^n(y,B)>0. \tag{1}\label{1} \end{equation*}

Inequality \eqref{1} represents a kind of limited version of the $\psi$-irreducibility for the product chain. Also, again by Theorem 5.4.4 in the book, \eqref{1} implies that the product chain is aperiodic.


Here is an idea how to get the true $\psi$-irreducibility for the product chain. For the fixed $(x,y)\in\X\times\Y$ and any $F\in\B(\X\times\Y)=\B(\X)\otimes\B(\Y)$, let \begin{equation*} U(F):=U((x,y),F):=\sum_{n=1}^\infty (P\otimes Q)^n((x,y),F), \end{equation*} so that $U$ is a (possibly infinite) measure on $\B(\X\times\Y)$. Then \eqref{1} means that the measure $\psi_X\otimes\psi_Y$ is "absolutely continuous" w.r. to $U$ on the algebra generated by the product sets $A\times B\in\B(\X)\times\B(\Y)$.

So, it "only" remains to show that this (perhaps with some, hopefully mild, additional conditions) implies the true absolute continuity of $\psi_X\otimes\psi_Y$ w.r. to $U$ on the $\si$-algebra $\B(\X)\otimes\B(\Y)$, generated by the product sets $A\times B\in\B(\X)\times\B(\Y)$.

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  • $\begingroup$ Thanks for the detailed answer! Why is it sufficient to only consider rectangles of positive measure? It is not immediately clear to me why this is sufficient for irreducibility nor for aperiodicity. There may be sets of positive measure in the product space that don't contain any rectangles of positive measure. $\endgroup$
    – Dasherman
    Nov 18, 2022 at 0:56
  • $\begingroup$ @Dasherman : By Theorem 5.4.4 in the book, this answer does prove that the product chain is aperiodic -- because in that theorem one can use any $\nu_k$-small set $C\in\mathscr B^+(\mathcal X)$ with $\nu_k(C)>0$. As for the $\psi$-irreducibility, I have added an idea on how to extend its limited version to the complete one. $\endgroup$ Nov 18, 2022 at 4:26
  • $\begingroup$ I see, thank you. So to summarize: you've shown that the product chain is aperiodic, but the proof for $\psi$-irreducibility isn't complete yet. Do you think the claim is true without additional conditions, or do you perhaps have (a path towards) a counterexample in mind? $\endgroup$
    – Dasherman
    Nov 20, 2022 at 9:49
  • $\begingroup$ @Dasherman : Unfortunately, I don't know how to complete the proof of the $\psi$-irreducibility or to find a counterexample. I have posted a question about a special case of this at mathoverflow.net/questions/434854/… but have not received a valid answer yet. It appears that, if a counterexample exists, it should be rather exotic. $\endgroup$ Nov 20, 2022 at 14:28

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