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$\def\P{\mathsf{P}}$

Let $(X_n)_{n\in\mathbb{Z}_+}$ be a Markov chain with a transition kernel $P(x,dy)$. Consider now a product Markov chain $(X^1_n,X^2_n)_{n\in\mathbb{Z}_+}$ with the transition kernel $P(x_1,dy_1)P(x_2,dy_2)$.

Recall that an invariant probability measure $\pi$ is called ergodic, if for any measurable set $A$ such that $P(x,A)=1$ for any $x\in A$ we have $\pi(A)$ is either $0$ or $1$.

Now let $\mu$ be an ergodic measure for $(X_n)_{n\in\mathbb{Z}_+}$. Is it true that the measure $\mu\otimes\mu$ would be ergodic for the product Markov chain $(X^1_n,X^2_n)_{n\in\mathbb{Z}_+}$?

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    $\begingroup$ Haven't checked the details, but it seems like this should be a straightforward exercise in the monotone class or $\pi$-$\lambda$ lemma. Where did you get stuck? $\endgroup$
    – Nate Eldredge
    Commented Oct 17, 2016 at 18:45
  • $\begingroup$ @NateEldredge It's not so clear to me how the monotone class lemma would work here. Let's say we proved that if an invariant set $A$ from $E\times E$ (I denoted by $E$ the state space of the Markov process) is of the form $A_1\times A_2$ then $\mu\otimes\mu (A)$ is either $0$ or $1$. Such sets definetely form a $\pi$-system, let's call it $\mathcal{A}$. It's clear that all invariant sets form a $\sigma$-algebra (and hence a $\lambda$-system), let's call it $\mathcal I$. By the monotone class theorem we have $\sigma(\mathcal{A})\subset \mathcal I$. But why $\sigma(\mathcal{A})=\mathcal I$? $\endgroup$
    – John
    Commented Oct 17, 2016 at 22:29
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    $\begingroup$ @NateEldredge Do you see now why I got stuck? Because this statement is not true, see a very suprising counterexample by R W below! Do you still think that it was "a straightforward exercise in the monotone class"? $\endgroup$
    – John
    Commented Oct 18, 2016 at 7:24
  • $\begingroup$ My experience is that Nate generally means to be helpful and didn't intend to be dismissive by writing 'straightforward'. If he guessed wrong, then he guessed wrong. $\endgroup$
    – Todd Trimble
    Commented Oct 18, 2016 at 11:53
  • $\begingroup$ @John: Obviously I was wrong. Sorry for the misdirection. $\endgroup$
    – Nate Eldredge
    Commented Oct 18, 2016 at 12:25

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No - this is not true. The minimal example is provided by the deterministic Markov chain with two states (so that the state space is $\mathbb Z_2=\{0,1\}$) with the deterministic transitions $x\mapsto x+1\, (\!\!\!\! \mod 2)$. This chain is obviously ergodic in your sense with respect to the stationary measure $\mu$ with $\mu(0)=\mu(1)=1/2$. [I emphasize that your definition of ergodicity differs from the usual probabilistic definition, and rather coincides with what is called ergodicity in the theory of dynamical systems, see Different uses of the word "ergodic" .] Then the product Markov chain is also deterministic and has two orbits $\{(0,0),(1,1)\}$ and $\{(0,1),(1,0)\}$, so that it is not ergodic.

EDIT By the way, what is still true, is that if the tail $\sigma$-algebra of either chain is trivial (which is sometimes called ergodicity by probabilists), then the tail $\sigma$-algebra of the product is also trivial. The reason why this property does not imply the one you were asking about is precisely the same why the product of two ergodic (in the usual "dynamical" sense) transformations is not necessarily ergodic, and is related to mixing. Namely, in the Markov case the space of ergodic components of the chain is the quotient of its tail boundary (the one determined by the tail $\sigma$-algebra) by the time shift.

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  • $\begingroup$ Wow! Great and very surprising counterexample, R W, thank you very much! $\endgroup$
    – John
    Commented Oct 18, 2016 at 7:25

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