5
$\begingroup$

I have two questions, somehow related. The first one, has to do with the Birkhoff's ergodic theorem. In its classical formulation, it states that if we have a probability space $(X,\mathcal{B},\mu)$ and a measure preserving transformation $T:X\to X$, then for every $f\in L^1$ the averages $$ \dfrac{1}{n}\sum_{k=0}^{n-1}f\circ T^k (x) $$ converge to a function $\hat{f}\in L^1$, invariant by $T$. While the classical proofs (for example, the ones found in Walters, Ward-Einsiedler, etc) make use of the maximal inequality and the maximal ergodic theorem, it is not clear to me why the normalization by $n$ is the right one. Is it possible to normalize by other sequences $a_n\nearrow\infty$? In the sense that $$ (*)\qquad \dfrac{1}{a_n}\sum_{k=0}^{n-1}f\circ T^k (x) $$ could provide some additional information, or maybe not.

My second question is essentially the same, but in the infinite measure context. Jon Aaronson proved in On the ergodic theory of non-integrable functions and infinite measure spaces, 1977 that the averages $(*)$ do not provide any useful information regardless of the choice of the sequence $a_n$. While it is possible to follow the calculations, I do not get the idea behind this result. Any insight about this would be very appreciated. Thanks in advance.

$\endgroup$
  • 1
    $\begingroup$ You may be interested in an area called Weighted Ergodic Averages (that is indeed more general). In the infinite case, regarding your question, I suggest you to look at Melbourne-Terhesiu, Operator renewal theory and mixing rates for dynamical systems with infinite measure. Inventiones Mathematicae 189 (2012) 61-110. $\endgroup$ – user39115 Jun 22 '16 at 22:41
5
$\begingroup$

If you're dealing with a probability space, the constant function $\mathbf 1$ belongs to $L^1$, so that you have no choice but to normalize by $n$ if you want to take Cesàro averages of $L^1$ functions (if you normalize by something larger, it converges to 0 almost everywhere; if you normalize by something smaller, it diverges almost everywhere).

You might ask about functions whose integral is 0. Here, you might try dividing by something smaller than $n$. However, in this case you can use Rokhlin towers to build functions in $L^1_0$ ($L^1$ functions that integrate to 0) for which the limit superior of the averages is $\infty$ almost everywhere, so you certainly should not expect almost everywhere convergence in this setting.

In terms of positive results with smaller normalization, there is work of Forni and others on controlling the deviations of Birkhoff averages for smooth functions for certain classes of transformations, $T$. There is also a substantial body of work on 'dynamical central limit theorems', where if one takes a hyperbolic map $T$ and a smooth function $f$ with integral 0, then the averages converge in distribution to a normal distribution.

Finally, you ask about infinite measure-preserving transformations. The way that I believe that you should understand the positive results is that you can study the ratios of Birkhoff sums of two functions both belonging to $L^1$, the so-called ratio ergodic theorems. So, for example, it makes sense to ask for the almost sure limit of $$ \frac{\sum_{n=0}^{N-1} 1/(1+|X_n|^2)} {\#\{n<N\colon X_n=0\}}$$ where $(X_n)$ is a simple symmetric random walk on $\mathbb Z$, even though $(1/N)*\text{numerator}\to 0$ and $(1/N)*\text{denominator}\to 0$.

$\endgroup$
1
$\begingroup$

The non convergence result of Aaronson in infinite measure space is proved by reducing it (via inducing and using the ratio ergodic theorem which Anthony mentioned) to a question of whether given a probability measure preserving transformation and a positive nonintegrable function $f$ (the return time function), can one normalize the Birkhoff sums of $f$ by some sequence $b_n$ and get $$\frac{1}{b_n}\sum_{k=1}^nf\circ T^k\xrightarrow[n\to\infty]{} 1,\ \ a.s.?$$ (The function $f$ in the reduction is the first return function into a set of positive and finite measure). Aaronson generalises the Theorems of Chow and Robbins (the case $f\circ T,f\circ T^2,...$ are i.i.d.) and shows that if $T$ is ergodic and $f$ is not integrable then for any sequence $b_n\to\infty$ either $$ \limsup \frac{1}{b_n}\sum_{k=0}^nf\circ T^k=\infty, \ \ a.s. $$ or $$ \liminf \frac{1}{b_n}\sum_{k=0}^nf\circ T^k=0, \ \ a.s. $$

$\endgroup$
0
$\begingroup$

Let us restrict to the case in which the mean is zero (as A.Quas has pointed out, convergence a.s. is either trivial or ruled out for other normalizations otherwise).

In this case, the sequences $(a_{n})_{n}$ needed to obtain convergence in distribution to a non-degenerate (non-constant) random variable $Y$, this is, such that $$\frac{1}{a_{n}}\sum_{k=0}^{n-1}f\circ T^{k}\Rightarrow_{n} Y$$ for some non-degenerate $Y$, are unique in the following sense: if $(a'_{n})_{n}, Y'$ is any other such pair of objects, then there exists $$0<b=\lim_{n}\frac{a'_{n}}{a_{n}}$$ and, in particular. $Y'=bY$ (in distribution).

To see this note that, by convergence of types results (see p. 193 in Probability and Measure, 3rd edition or the appendix of this paper of mine for a more detailed explanation), if $(Z_{n})_{n}, Z,Z'$ are random variables with $Z,Z'$ not constant and $(a_{n})_{n}, (a'_{n})_{n}$ are sequences of positive numbers such that $a_{n}Z_{n}\Rightarrow_{n} Z$ and $a_{n}'Z_{n}\Rightarrow_{n} Z'$ then, since

$$\frac{a_{n}}{a'_{n}}a'_{n}Z_{n}\Rightarrow_{n} Z$$ and $Z'_{n}:=a'_{n}Z_{n}\Rightarrow Z'$, the sequence $a_{n}/a'_{n}$ has a (strictly positive) limit. Adapt this argument taking $Z_{n}:=\sum_{k=0}^{n-1}f\circ T^{k}$.

This may not answer your question, but at least it will give you a good feeling of why people usually re-normalize by, say, $\sqrt{n}$ (instead of some other constant) when trying to prove the Central Limit Theorem: if the CLT holds with this normalization it basically can't hold with any other, and the limit has to be normal (and similarly for a CLT under any other renormalization).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.