# Connection between invariant measure and positive recurrence for continuum state space markov chain

Let $$\{ X_n(\omega,x)\}_{n \ge 0}$$ be a Markov chain with and underlying probability space $$(\Omega,\Sigma,\mathbb{P})$$ and state space $$X= \mathbb{S}^1$$. Suppose this markov chain admits unique ergodic measure which is full supported.

I would like to estimate $$\mu(C)$$, where $$C \subset \mathbb{S}^1$$, is a particular subset with positive measure. I would like to know if in any way $$\mu(C)$$ is related to the following quantity: $$\sup_{x \in \mathbb{S}^1}\mathbb{E}_x[\tau^C(\omega,x)]$$ where $$\tau^C(\omega,x):= \min\{ n \ge 1 \colon X_n(\omega,x) \in C\}$$ is defined for $$x \in C$$. Here $$X_n(\omega,x)$$ is the position of the markov chain at time $$n$$ with initial condition $$x$$ and randomness $$\omega$$. In particular, I would like to know if some inequality of the sort $$\mu(C) \ge\sup_{x \in \mathbb{S}^1} \frac{1}{\mathbb{E}_x[\tau^C(\omega,x)]}$$ holds or not.

The reason for this question is because it is well known that for a Markov Chain in a discrete state space it is possible to recover the invariant measure via the positive recurrent states. I was wandering if something like this was possible too in this case. Thanks in advance.

• I think the answer is yes, and that you can do it along the same lines as the discrete case, i.e., the number of visits, j, up to time n is the the probability that the time of of the jth visit is greater than n, which is attacked with the elementary renewal the in the discrete case, but should be comparable to the bounds you stated.
– mike
Aug 5 at 8:14
• can you be more specific? Aug 5 at 10:23
• @GiuseppeTenaglia In the last display in your post, did you mean the supremum on the right-hand side to be in the denominator? Aug 17 at 18:03

Yes, let's assume that sup (which I think should be in the denominator in the last expression) is finite, as otherwise it is always true. Suppose it is bounded by A, the each return time is also quite finite, as by chebyshev inequality $$P(T > 2A) < \frac 1 2$$, and using the markov property this makes each return time subgeometric, compared with the same geometric distribution. Now using the ergodic theorem calculate $$\mu(C)$$ by counting the number of time the process is in C. Let $$a > \mu(C)$$. If $$T_1, T_1 + T_2,$$ etc are those times, then $${\sum^n 1_C(X_i)} < na = P(T_1 + ... + T_{na} > n)$$ and so both are about 1. Therefore you expect that $$n < E(T_1 + ... + T_{na}) < na \sup(E(T)$$ and therefore $$a > \frac 1 {\sup(E(T)}$$. I had started out arguing that they returns times were subgeometric, which I though I was going to use, but seem not to have, but I'm going to leave it there in case it occurs to me why I wanted it.

Yes, there are such inequalities, which follow from the Kac Lemma in Ergodic Theory. (Although the form of the inequality at the end of the post is too optimistic- did you want the supremum to be taken in the denominator?)

The sequence of states in an ergodic time homogenous Markov chain, started according to an invariant measure, is an ergodic measure preserving system under the shift operator (Uniqueness of the invariant measure is not needed here.)

According to the Kac' lemma (see [1] and the references therein) the first return time function

$$\tau^C(\omega,x):= \min\{ n \ge 1 \colon X_n(\omega,x) \in C\}$$ will satisfy that its expectation (when averaged over the starting point $$x \in C$$) is the reciprocal of $$\mu(C)$$, that is

$$\frac{1}{\mu(C)}\int_C {\mathbb E}_x\Bigl(\tau^C(\omega,x)\Bigr) \,d\mu(x)=\frac{1}{\mu(C)}\,.$$ Consequently, $$\inf_{x \in C} \frac{1}{\mathbb{E}_x[\tau^C(\omega,x)]}\le \mu(C) \le \sup_{x \in C} \frac{1}{\mathbb{E}_x[\tau^C(\omega,x)]} \,.$$

• @NawafBou-Rabee Thanks, corrected, did you spot any more typos? Aug 17 at 18:38
• Looks correct but, why is ergodicity not needed for this identity to hold? Aug 17 at 19:05
• Ergodicity is needed, but not uniqueness of the invariant measure. Aug 17 at 22:17
• e.g. consider the Markov chain which transitions from $x \in [0,1]$to $2x \mod 1$ or $4x \mod 1.$ Aug 18 at 6:28