The average recurrence time - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T05:14:47Z http://mathoverflow.net/feeds/question/83028 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/83028/the-average-recurrence-time The average recurrence time Daniel Mansfield 2011-12-09T04:24:54Z 2011-12-09T06:24:09Z <p>As seen on <a href="http://en.wikipedia.org/wiki/Ergodic_theory#Sojourn_time" rel="nofollow">wikipedia</a>, given a measure space $(X,\Sigma,\mu)$ with $\mu(X) &lt; \infty$ and a measure preserving transformation $T: X \mapsto X$. Let $A \subset X$ be a set of positive measure. Define $k_i$ as the power of $T$ such that $T^{k_i}x \in A$ for the $i$th time: that is to say $k_i$ is the "$i$'th return time to $A$". The difference between recurrence times is $R_i = k_i - k_{i-1}$ (assume for simplicity that $k_0 = 0$, that is $x \in A$)</p> <p>I would like know how to prove the following:</p> <p>$$\lim\limits_{n\mapsto\infty} \frac{R_1 + \cdots + R_n}{n} = \frac{\mu(X)}{\mu(A)}$$</p> <p>The <a href="http://en.wikipedia.org/wiki/Ergodic_theory#Sojourn_time" rel="nofollow">wikipedia article</a> indicates that this is a consequence of the ergodic theorem.</p> <p>Note that my definition of the $k_i$ above differs slightly from that of wikipedia, in as much as I have omitted to say that the $k_i$s are sorted in increasing order.</p> http://mathoverflow.net/questions/83028/the-average-recurrence-time/83033#83033 Answer by Robert Israel for The average recurrence time Robert Israel 2011-12-09T06:24:09Z 2011-12-09T06:24:09Z <p>As you might guess from the reference to the ergodic theorem, you do need to assume that $T$ is ergodic. For a counterexample where $T$ is not ergodic, consider the identity map and suppose $\mu(A) &lt; \mu(X)$. </p> <p>Note that $R_1 + \ldots + R_n = k_n$. Let $I_A$ be the indicator function of $A$, which of course is in $L^1(\mu)$.<br> The Birkhoff ergodic theorem says if $T$ is ergodic, $\lim_{k \to \infty} \frac{1}{k} \sum_{j=1}^k I_A(T^i x) = \frac{1}{\mu(X)} \int I_A d\mu$ almost everywhere, i.e. $S_k(x)/k \to \mu(A)/\mu(X)$ where $S_k(x)$ is the number of $i \in {1,2,\ldots, k}$ such that $T^i(x) \in A$. Since $\mu(A) > 0$, we also have $k/S_k(x) \to \mu(X)/\mu(A)$ almost everywhere. But if $k = k_n(x)$, $S_k = n$.</p>