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?

3$\begingroup$ 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 $T^n x \in A$. $\endgroup$– Ronnie PavlovCommented Dec 12, 2023 at 16:52

2$\begingroup$ 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. $\endgroup$– AutovetorCommented Dec 12, 2023 at 17:12

$\begingroup$ Got it, thanks! Answer below. $\endgroup$– Ronnie PavlovCommented Dec 13, 2023 at 16:06
1 Answer
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 measurepreserving system $(X,T,\mu)$. The usual way to deal with expected return times is to consider the socalled KakutaniRokhlin staircase, where you partition $A$ by return times. Namely, for $n > 0$, define $A_n = \{x \in A \ : \ Tx, T^2 x, \ldots, T^{n1}x \notin A, T^n x \in A\}$. 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 $ni$. So, if you want the average of the hitting time $H$ over all of $X$, it's $\int H = \sum_{n > 0} \sum_{0 \leq i < n} \int_{T^i A_n} H = \sum_{n > 0} \sum_{0 \leq i < n} (ni) \mu(T^i A_n) = \sum_{n > 0} \frac{n(n+1)}{2} \mu(A_n)$.
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 countablestate 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 vertexdisjoint 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.

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