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Hi,

I'd appreciate some help on a Markov chain result I'm trying to show. I believe the following is sufficient for a continuous time Markov chain $(X_t)$ with a countable state space to be ergodic:

$(X_t)$ is irreducible.
There exists a finite subset $A$ of states such that, for all $a\in A$, conditional on $X_0 = a$, the expected return time back to $A$ is integrable, that is, $E_a[R_A]<\infty$ where $$ R_A := \inf ( t > 0 \mid X_t \in A ). $$
There exist $k,p>0$ such that for all $a,b \in A$, we have $$P (X_{t+k} = b \mid X_t = a) \geq p.$$

In other words, I'm under the impression that (2) and (3) implies positive recurrence. This is because if I start at any $a\in A$, then I take some finite time to return back to $A$. Then use the $k$ (finite no. of) steps to return back to $a$ to show postive recurrence. Whence (1) and (2)+(3) imply ergodicity.

Can some help show this claim or suggest why it's wrong?

Thanks Apus

edited Jan 29 2012 at 19:41

Didier Piau
4,193●1●10●15

asked Jan 23 2012 at 18:03

A Chuh
61●3

	I am not sure, but isn't it that ergodicity of ctMC is equivalent to the ergodicity of the underlying discrete-time Markov Chain? – Ilya Jan 27 2012 at 12:12
	I guess that in 2. you mean $\inf\{t>0:X_t\in A\}$. But if $X_0\in A$, this infimum is almost surely $0$ (hence finite), for every continuous time Markov chain, since $X_0\in A$ and $\inf\{t>0:X_t\ne X_0\}$ is almost surely positive. – Didier Piau Jan 28 2012 at 15:05
	@Ilya I don't think that's the case. Perhaps I can find a proof somewhere. @Didier Thanks for spotting the typo. Could you explain what you mean a little more please? – A Chuh Jan 29 2012 at 10:42
	What part of my previous comment is not clear to you? (Your revised 2. is not correct, I will modify it.) – Didier Piau Jan 29 2012 at 19:39

Ergodicity of a Markov chain

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