3
$\begingroup$

Let $(\Omega,\mathcal A,\operatorname P)$ be a probability space, $\tau:\Omega\to\Omega$ be a measurable map on $(\Omega,\mathcal A)$ with $\operatorname P\circ\:\tau^{-1}=\operatorname P$, $Y_n:\Omega\to[-\infty,\infty)$ be $\mathcal E$-measurable for $n\in\mathbb N$ with $\operatorname E\left[Y_1^+\right]<\infty$ and $$Y_{m+n}\le Y_m+Y_n\circ\tau^m\;\;\;\text{for all }m,n\in\mathbb N\tag1$$ and $$M_n:=\max(Y_1,\ldots,Y_n)\;\;\;\text{for }n\in\mathbb N.$$

It's easy to show the following extension of the maximal ergodic theorem: $$\operatorname E[Y_1;M_n\ge0]\ge0\;\;\;\text{for all }n\in\mathbb N.\tag2$$

The ordinary maximal ergodic theorem is given by the special case, where $$Y_n=\sum_{i=0}^{n-1}X\circ\tau^i\;\;\;\text{for all }n\in\mathbb N$$ for some integrable real-valued random variable on $(\Omega,\mathcal A,\operatorname P)$. In that special case, it can be deduced from $(2)$ that $$\operatorname P\left[\sup_{n\in\mathbb N}\left|\frac{Y_n}n\right|\ge c\right]\le\frac1c\operatorname E[|Y_1|]\;\;\;\text{for all }c>0\tag3.$$

Can we extend this result to the general case?

Improve this question
$\endgroup$
6
  • $\begingroup$ After a bit of searching, I found the answer to my question in comment. So it is true that (2) holds in general and it is proved in literature that (3) holds for the special case where $Y_n=\sum_k X\circ \tau ^k$ indeed, except that there is a mistake in your formulation : you have to remove the absolute values (and this is very important). This is actually written without the absolute values on the wikipedia page you're refering too. Also, if you edit the question, you should add the tag ergodic theory. I answer below your question (positively) with the correct formulation. $\endgroup$
    – M. Dus
    Commented Apr 30, 2020 at 14:06
  • $\begingroup$ @M.Dus I'm sorry for my late response. What I mean is that $(2)$ and $(3)$ (with the absolute value) are known to be true when $(Y_n)$ is additive. It's easy to see (by more or less precisely the same proof as in the additive case) that $(2)$ remains to hold true when $(Y_n)$ is subadditive, but it's not clear to me how $(3)$ generalizes and how this generalization can be proven. $\endgroup$
    – 0xbadf00d
    Commented Apr 30, 2020 at 14:11
  • $\begingroup$ Could you provide a proof for (3) with the absolute values then ? I'm currently writing a proof as an answer for (3) without them in the general case. $\endgroup$
    – M. Dus
    Commented Apr 30, 2020 at 14:15
  • $\begingroup$ see, for example, Lemma 2.6 (2) on p. 34 in wt.iam.uni-bonn.de/fileadmin/WT/Inhalt/people/Andreas_Eberle/…. (Note that you need to take $Y_n=nA_nF$.) $\endgroup$
    – 0xbadf00d
    Commented Apr 30, 2020 at 14:17
  • $\begingroup$ Right thank you ! I have a (simple) proof for the statement without the absolute values, which seems to be the important result in literature, but I don't know about yours. Do you still want me to write it up ? $\endgroup$
    – M. Dus
    Commented Apr 30, 2020 at 14:24

1 Answer 1

Reset to default
0
$\begingroup$

The answer is no in general, but yes if the sequence $Y_n$ is non-negative.

First, let us focus on the case where $Y_n$ is non-positive. Then, $\sup \frac{1}{n}|Y_n|=-\inf \frac{1}{n}Y_n$. If you assume moreover that all the $Y_n$ are $L^1$, then by Kingman's subadditive theorem, $\frac{1}{n}Y_n$ converges to $Y=\inf \frac{1}{n}Y_n$ almost surely. Notice then that $\mathbb{P}\left (\sup\frac{1}{n} |Y_n|\geq c\right )=\mathbb{P}\left (Y\leq -c\right )$. Whenever $Y$ has positive probability of taking the value $-\infty$, you cannot bound $\mathbb{P}\left (Y\leq -c\right )$ by something converging to 0 as $c$ goes to infinity.

Here is a concrete counter-example. Let $Y_n$ be the constant function $Y_n=-n^2$. Then $Y_n$ is subadditive and satisifes all your assumptions. You have $\sup \frac{1}{n}|Y_n|=+\infty$ and so for any $c$, $\mathbb{P}\left (\sup\frac{1}{n} |Y_n|\geq c\right )=1$, so you don't have $\mathbb{P}\left (\sup\frac{1}{n} |Y_n|\geq c\right )\leq \frac{1}{c}\mathbb{E}(|Y_1|)=\frac{1}{c}$.

However, small remark : the answer is yes for non-positive $Y_n$ if you have the property that $\mathbb{E}(\frac{1}{n}|Y_n|)\leq \mathbb{E}(|Y_1|)$. Indeed, using Markov inequality, you get $\mathbb{P}\left (\frac{1}{n} |Y_n|\geq c\right )\leq \frac{1}{c}\mathbb{E}(\frac{1}{n}|Y_n|)\leq \frac{1}{c}\mathbb{E}(|Y_1|)$ and this is true for all $n$, so this is true for the almost sure limit, using dominated convergence.

About the non-negative case now. Inequality (3) is usually stated without the absolute values in literature : $$\mathbb{P}\left (\sup\frac{1}{n} Y_n\geq c\right )\leq \frac{1}{c}\mathbb{E}(|Y_1|).$$

This statement is true in general and so in particular, the answer to your question is yes whenever $Y_n$ is non-negative.

Indeed, consider a subbaditive sequence $Y_n$, that is satisfying your condition $Y_{n+m}\leq Y_m+Y_n\circ \tau^m$. Let $Z_n=\sup_{k=1,...,n}\frac{1}{k}Y_k$. Also let $\widetilde{Y}_n=\sum_{j=0}^{n-1}Y_1\circ \tau^j$ and finally, let $\widetilde{Z}_n=\sup_{k=1,...,n}\frac{1}{k}\widetilde{Y}_k$. As you claim, the result is true for the sequence $\widetilde{Y}_n$.

Note that since $Z_n$ in non-decreasing, you have $\mathbb{P}\left (\sup\frac{1}{n} Y_n\geq c\right )=\lim_n\mathbb{P}(Z_n\geq c)$ so we just need to prove that $\mathbb{P}(Z_n\geq c)\leq \frac{1}{c}\mathbb{E}(Y_1)$.

Now for fixed $n$, for every $x$, there exists $1\leq k(x)\leq n$ such that $Z_n=\frac{1}{k(x)}Y_{k(x)}$. Because of subadditivity, you have $Y_{k(x)}\leq \sum_{j=0}^{k(x)-1}Y_1\circ \tau^j(x)=\widetilde{Y}_{k(x)}(x)$. So $\frac{1}{k(x)}Y_{k(x)}\leq \frac{1}{k(x)}\widetilde{Y}_{k(x)}(x)\leq \widetilde{Z}_n(x)$. This proves that for any $x$, $Z_n(x)\leq \widetilde{Z}_n(x)$ so $\mathbb{P}(Z_n\geq c)\leq \mathbb{P}(\widetilde{Z}_n\geq c)$. Using that $\widetilde{Z}_n$ also is non-decreasing, you get $\mathbb{P}(\widetilde{Z}_n\geq c)\leq \mathbb{P}\left (\sup\frac{1}{n} \widetilde{Y}_n\geq c\right )$ and so you can use the result for $\widetilde{Y}_n$.

Improve this answer
$\endgroup$
4
  • $\begingroup$ @0xbadf00d I added an answer for the non-positive case : unfortunately the answer is no. However, I also added some comments and the right assumption might be that $\mathbb{E}(\frac{1}{n}|Y_n|)\leq \mathbb{E}(|Y_1|)$. $\endgroup$
    – M. Dus
    Commented May 1, 2020 at 11:43
  • $\begingroup$ @0xbadf00d Oh but I realized that for non-postive $Y_n$, then $|Y_n|\geq \sum_{k=0}^{n-1}Y_1\circ \tau ^k$. So if $\mathbb{E}(\frac{1}{n} |Y_n|)\leq \mathbb{E}(|Y_1|)$, then we have $Y_n=\sum_{k=0}^{n-1}Y_1\circ \tau^k$ almost surely. In my opnion, this leads to the following question : Assume that $Y_n$ is non-positive and integrable. Let $Y$ be the almost sure limit of $Y_n$ and $Y'$ the almost sure limit of $\sum_{k=0}^{n-1}Y_1\circ \tau^k$. Do we have $Y=Y'$ almost surely ? BTW, sorry for the many messages, but I find this question very interesting ! $\endgroup$
    – M. Dus
    Commented May 1, 2020 at 15:30
  • $\begingroup$ I'm a bit out of time recently, but regarding your counterexample: Please note that it is only a counterexample for $c>1$, but usually one is interested in infinitesimal small $c>0$. So, it wouldn't be a restriction to assume $c\in(0,1)$. $\endgroup$
    – 0xbadf00d
    Commented May 12, 2020 at 8:59
  • $\begingroup$ @0xbadf00d No worries ! You're right, but also note that you can change $Y_n$ to be $-\frac{c}{2}n^2$. In this situation, you still have that $\inf \frac{|Y_n|}{n}$ is $+\infty$ so $\mathbb{P}(\sup \frac{1}{n}|Y_n|\geq c)=1$. On the other hand, $\frac{1}{c}\mathbb{E}(|Y_1|)=1/2$. This is not a counter-example for any $c$, but for any $c$, you get a counter-example. Now the question becomes "does (3) holds for any small enough $c$ ?" Also, in my last comment, I obviously meant "assume that $Y_n$ is non-positive and integrable AND THAT (3) HOLDS, then do we have $Y=Y'$ a.s. ? $\endgroup$
    – M. Dus
    Commented May 12, 2020 at 16:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .