# Hitting times in ergodic dynamical systems

For an ergodic Dynamical System, we know that the return has average equal to 1 over the measure of the set in question. What can we say about the hitting time? Also has finite average? Is it proportional to 1 over the measure?

• I don't understand the question, and I suspect other experts may not be able to either. What is the difference between "return (time)" and "hitting time" here? I would have defined either as the least positive $n$ so that $T^n x \in A$. Commented Dec 12, 2023 at 16:52
• Return time is when at the initial stage you are in the set you want to return. Hitting time is when you start from elsewhere. Commented Dec 12, 2023 at 17:12
• Got it, thanks! Answer below. Commented Dec 13, 2023 at 16:06

Undeterred by previous failures, here is one more attempt at an example.

I don't think that in general much can be said about this average, and it is not necessarily finite. Consider any set $$A$$ of positive measure inside an invertible ergodic measure-preserving system $$(X,T,\mu)$$. The usual way to deal with expected return times is to consider the so-called Kakutani-Rokhlin staircase, where you partition $$A$$ by return times. Namely, for $$n > 0$$, define $$A_n = \{x \in A \ : \ Tx, T^2 x, \ldots, T^{n-1}x \notin A, T^n x \in A\}$$. Then (up to sets of measure $$0$$) $$A = \bigcup_{n > 0} A_n$$. It's also clear that $$X = \bigcup_{n > 0} \bigcup_{0 \leq i < n} T^i A_n$$ (and the union is disjoint) by ergodicity; $$\mu$$-a.e. point of $$X$$ occurs between two visits to $$A$$.

Now you get the result you cited about the average value of the return time $$R$$; it is $$n$$ on $$A_n$$, so $$\int_A R = \sum_{n > 0} \int_{A_n} R = \sum_{n > 0} n \mu(A_n) = \sum_{n > 0} \mu(\bigcup_{0 \leq i < n} T^i A_n) = \mu(X) = 1$$. Then the average value of $$R$$ over $$A$$ is indeed $$1/\mu(A)$$.

The hitting time on $$T^i A_n$$ is $$n-i$$. So, if you want the average of the hitting time $$H$$ over all of $$X$$, it's $$\int H = \sum_{n > 0} \sum_{0 \leq i < n} \int_{T^i A_n} H = \sum_{n > 0} \sum_{0 \leq i < n} (n-i) \mu(T^i A_n) = \sum_{n > 0} \frac{n(n+1)}{2} \mu(A_n)$$.

There's no real reason for this sum to converge; you can tweak the decay rate of $$\mu(A_n)$$ as long as $$\sum_{n > 0} n\mu(A_n) = 1$$. Here's an attempt at an explicit example. Construct a countable-state Markov chain consisting of a root vertex $$v$$ and, for all $$n > 0$$, a single directed paths from $$v$$ to itself. The loops should be vertex-disjoint except for $$v$$. The probability, from $$v$$, of entering the loop of length $$n$$ is $$\frac{4}{n(n+1)(n+2)}$$; it's easily checked that these sum to $$1$$. For any vertex in a loop, the following vertices/states are forced until return to $$v$$.

This chain is irreducible, and the state $$v$$ is positively recurrent, meaning that the mean first return time to $$v$$ is finite. To check this, one computes $$\sum_{n > 0} n \frac{4}{n(n+1)(n+2)} = \sum_{n > 0} \frac{4}{(n+1)(n+2)} = 2$$. This means that the stationary distribution $$\mu$$ is ergodic, and that the cylinder set $$[v]$$ of paths starting at $$v$$ has measure $$\mu([v]) = 1/2$$.

Now, choose $$A = [v]$$. The set $$A_n$$ is just the set of paths starting at $$v$$, followed by the path of length $$n$$, with measure $$1/2 \frac{4}{n(n+1)(n+2)} = \frac{2}{n(n+1)(n+2)}$$. Then the expected hitting time, as computed above, is $$\sum_{n > 0} \frac{n(n+1)}{2} \mu(A_n) = \sum_{n > 0} \frac{1}{n+2}$$, which diverges.

• Leaving this for now, but I believe we now have an ergodic example with infinite expected hitting time. Commented Dec 13, 2023 at 19:58