Non-integrable ergodic theory - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T17:04:40Z http://mathoverflow.net/feeds/question/63132 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/63132/non-integrable-ergodic-theory Non-integrable ergodic theory Anthony Quas 2011-04-27T08:56:06Z 2011-05-14T06:04:55Z <p>Can anyone help me out with proofs/counterexamples? I'm working on an operator-valued multiplicative ergodic theorem and need what may(?) be a well-known fact. This fact (if true) would help me get rid of an annoying asymmetry in the conclusion of a theorem.</p> <p>I'm assuming that $T$ is an invertible ergodic transformation of a probability space $(X,\mathcal B,\mu)$ and that $f$ is a measurable (but not necessarily integrable) function on $X$.</p> <blockquote> Is it true that $f(T^nx)/n \to 0$ a.e. if and only if $f(T^{-n}x)/n\to 0$ a.e.? </blockquote> <p>Comments:</p> <p>(1) In probability language if you define $X_n=f(T^nx)$ this is a stationary sequence of random variables. Borel-Cantelli 1 shows that if $\mathbb E |X_0|&lt;\infty$ (i.e. $f\in L^1$) then $X_n/n\to 0$ as $n\to\pm\infty$: The probability that $|X_n|/n > 1/k$ is $\mathbb P(|X_0| > n/k)$. The sum of this series is over-estimated by $k\mathbb E|X_0| &lt; \infty$. Hence almost surely $|X_n|/n &lt; 1/k$ for all large $n$. Since this is true for all $k$ you get $X_n/n\to 0$.</p> <p>If the $X_n$ are i.i.d. random variables, then the converse holds by Borel-Cantelli 2. This shows that for i.i.d. random variables, the boxed question has an affirmative answer.</p> <p>(2) In the case where the $X_n$ are not i.i.d. I believe there are examples where $X_n/n\to 0$ almost surely even though $\mathbb E|X_0|=\infty$.</p> <p>In the case that $f\in L^1$, both sides of the implication in the main question are true. The unresolved case is $f\not\in L^1$.</p> http://mathoverflow.net/questions/63132/non-integrable-ergodic-theory/63281#63281 Answer by camomille for Non-integrable ergodic theory camomille 2011-04-28T10:58:16Z 2011-04-28T11:07:31Z <p>This is not an answer to the question, but this is too long for a comment.</p> <p>I give what looks like an example where $f$ is not $L^1$ but where $f(T^n\cdot)/n \to 0$ a.e. I will state it in a probabilistic way (sorry about that).</p> <p>Let $\mu$ be a probability measure on ${1,2,\dots}$. I assume that $m=\int k\mu(dk)$ is finite. I define a new probability measure by $\hat{\mu}(dk)=k/m.\mu(dk)$. Let us consider a Markov chain $((A_n,B_n))$ on $\Omega=\{(a,b) : a \ge 1, 1 \le b \le a\} \subset N \times N$ with the following transitions probabilities :</p> <p>1) From $(a,b)$ with $b>1$ one goes to $(a,b-1)$.</p> <p>2) From $(a,1)$ one goes to $(c,c)$ where $c$ is chosen according to $\mu$.</p> <p>Now let $\nu$ be a probability measure on $\Omega$ defined by $\nu(\{(a,b)\})=\mu(\{a\})/m$. The above Markov chain admits $\nu$ as a stationary distribution and I consider it under this distribution. Note that $A_n$ is distributed according to $\hat{\mu}$. Now set $X_n=g(A_n)$ where $g$ is in $L^1(\mu)$ but not in $L^1(\hat{\mu})$. I think that $(X_n)_n$ is a counterexample (consider the Markov Chain seen at the times at which it jumps from $(a,\cdot)$ to $(c,c)$, then the first coordinates make a sequence of i.i.d.r.v. distributed according to $\mu$).</p> http://mathoverflow.net/questions/63132/non-integrable-ergodic-theory/64963#64963 Answer by Ori Gurel-Gurevich for Non-integrable ergodic theory Ori Gurel-Gurevich 2011-05-14T06:04:55Z 2011-05-14T06:04:55Z <p>$\newcommand{\R}{\mathbb R}$ $\newcommand{\P}{\mathbf P}$ $\newcommand{\Z}{\mathbb Z}$</p> <p>I found this question very interesting and gave it much thought this week. I believe I have a proof now. I think it would be interesting to see what generalizations one can get from this argument. Below I use probabilistic notation, which I'm more used to.</p> <p>Let <code>$\{X_n\}_{n\in \Z}$</code> be a stationary stochastic process, taking values in $\R$. Let $L=\limsup_{n\to -\infty} \frac{X_n}{|n|}$ and $R=\limsup_{n\to \infty} \frac{X_n}{|n|}$.</p> <p><strong>Theorem:</strong> $L=R$, almost surely.</p> <p><strong>Proof:</strong> It is enough to prove the theorem for ergodic processes. We may also assume, WLOG, that all the values are nonnegative integers.</p> <p>Let $A$ be the event that for some $n&lt;0$ we have $X_n\ge |n|$ and let $B$ be the event that for some $n>0$ we have $X_n\ge |n|$.</p> <p><strong>Lemma:</strong> $\P(A) \le 2 \P(B) .$</p> <p><strong>Proof of lemma:</strong> For a given realization of $X_n$, let $I$ be all indices $i$ for which $T^i X$ is in $A$ and let $J$ be all the indices for which $T^i X$ is in $B$. We claim that the density of $J$ is no more than twice the density of $I$, which then implies the conclusion.</p> <p>$I$ can be written as <code>$\cup_{n \in \Z} \{n,\ldots,n+X_n\}$</code>, while <code>$J=\cup_{n \in \Z} \{n-X_n,\ldots,n\}$</code>. In particular, $J$ is contained in <code>$\bar{J}=\cup_{n \in \Z} \{n-X_n,\ldots,n+X_n\}$</code>. But if we write $I$ as a union of <em>disjoint</em> intervals, then in $\bar{J}$ each of these intervals is extended to the left by at most the length of the interval. Hence, the density of $\bar{J}\supset J$ is at most twice that of $I$. $\blacksquare$</p> <p>Of course, by symmetry, we also have $\P(B) \le 2 \P(A)$. Let $A_K$ be the event that for some $n&lt;0$ we have $X_n \ge \max(|n|,K)$ and define $B_K$ analogously. By applying the lemma to the process <code>$\{Y_n\}_{n\in \Z}$</code> defined by $Y_n=X_n$ if $X_n\ge K$ and $Y_n=0$ otherwise, we get that $\P(A_K) \le 2 \P(B_K)$ and vice verse.</p> <p>In particular, $\lim_{K\to \infty} \P(A_K) =0$ if and only if $\lim_{K\to \infty} \P(B_K) = 0$. These limits necessary exists, since these are monotone decreasing events.</p> <p>For ergodic processes, we have that $L$ and $R$ are a.s. constant. Now, if $L&lt;1$ a.s. then we have $\P(A_K)\to 0$. In the other direction, if $\P(A_K)\to 0$ then a.s. $L\le 1$. Similar implications hold for $R$ and $B_K$. Hence, we get that if $L&lt;1$ then $R\le 1$ and vice verse. By applying these to a rescaled process $X_n / \alpha$, we get that for any $\alpha$ we have $L&lt;\alpha$ implies $R\le \alpha$ (and vice verse), so we must have $L=R$. $\blacksquare$</p>