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Let $\{X_t, t\in\mathbb R\}$ be a well-behaved$^*$ stationary ergodic process.

I'm interested in the uniform convergence of averages: $$ \sup_{|x|\le R_n} \left|\frac1{2n}\int_{x-n}^{x+n} X_t dt - \mathbb{E}[X_0]\right|\to 0, n\to \infty, $$ for some $R_n\gg n$. Are there any results of this type?


$^*$Precisely, I'm looking at the exponent of the so-called shot-noise potential: $$ X_t = \exp\left\{\sum_{x\in \Pi} \phi(x-t)\right\}, $$ where $\Pi$ is a Poisson point process, and $\phi$ can be assumed as good as needed (e.g. continuous with finite support).

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    $\begingroup$ I'd look into concentration of measure. If the $X_t$ are close to being i.i.d. and the tails of $X_t$ are nice enough, you can get bounds such as $\mathbb{P} (|(2n)^ {-1}\int X_t - \mathbb{E} (X_0)| > \varepsilon) \leq C(\varepsilon, n)$, whence $\mathbb{P} ( \sup |(2n)^ {-1}\int X_t - \mathbb{E} (X_t)| > \varepsilon) \leq R_n C(\varepsilon, n)$. Then you only need to find $R_n$ such that $\lim R_n C(\varepsilon, n) = 0$ for all $\varepsilon$ to get convergence in distribution. $\endgroup$ – D. Thomine Jun 23 '19 at 21:19
  • $\begingroup$ @D.Thomine, the problem is that the only viable bound I can obtain so far is for the variance, which is of order $1/n$. This gives, through Chebyshev's inequality, the estimate of the same order for the probability, so $R_n\gg n$ is, unfortunately, impossible. $\endgroup$ – zhoraster Jun 24 '19 at 16:01
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Uniform convergence holds when $R_n$ is at most a power of $n$.

Using the tail of a Poisson variable, you can easily infer that $P(X_t>r) \le r^{-C\log \log r}$ for some $C$ that depends on the maximum and the finite support of $\phi$.
Thus when $\phi$ has finite support, $X_t$ has finite moments of all orders.

The $2k$'th moment of $$ S_x:= \left|\frac1{2n}\int_{x-n}^{x+n} X_t dt - \mathbb{E}[X_0]\right| $$ can then be bounded by $C_k n^{-k}$, where $C_k$ depends on $k$ and $\phi$. Therefore $P(S_x>\epsilon) \le C_k (n\epsilon)^{-k}$.
From this, one can apply chaining (see Talagrand's book springer.com/gp/book/9783642540745 ) to prove uniform convergence in the original formulation, provided $R_n/n^{k} \to 0$ as $n \to 0$.

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  • $\begingroup$ I have also obtained such estimate in slightly more general case already. Thank you very much anyway! $\endgroup$ – zhoraster Jun 28 '19 at 9:53
  • $\begingroup$ By the way, is there any good reference on the last sentence in your answer, i.e. getting a uniform (in $x$) estimate from the pointwise? I'm now doing this by hand... $\endgroup$ – zhoraster Jun 28 '19 at 10:34
  • $\begingroup$ There are general methods for this, like the Dudley integral (see e.g. his book "Uniform Central Limit Theorems") or Talagrand's chaining (see his book springer.com/gp/book/9783642540745 ). The setting you mention can be handled directly, I will modify my answer to clarify this. $\endgroup$ – Yuval Peres Jun 28 '19 at 11:39
  • $\begingroup$ Sorry, I don't see how does the containment in one of the $R_n/n$ intervals imply the desired property... Say, the average of $\sin$ over any interval of length $2\pi$ is zero, but the one over interval of length $\pi$ can be large... Anyway, I am indeed able to proceed with a kind of chaining argument, thanks! $\endgroup$ – zhoraster Jun 28 '19 at 13:16
  • $\begingroup$ You are right, I ignored the subtraction, will correct the answer. $\endgroup$ – Yuval Peres Jun 28 '19 at 13:22
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I found an elementary way to proceed through Chebyshev inequality.

Assume that $\phi\in C(\mathbb{R})$ and $|\phi(x)|\le \frac{C}{1+|x|^{1+\beta}}$ for some $\beta>0$. It is known (Carmona-Molchanov 1995) that for any $\delta>0$, $$X_t = o(|t|^\delta), t\to \infty,\tag{1}$$ a.s. It is also not hard to see that $$ \operatorname{var}\left(\frac1{2n}\int_{-n}^{n} X_t dt\right) = O\Bigl(\frac1n\Bigr), n\to\infty. $$ Now take some $r\in (1,2)$ and $a\in (r-1,1)$ and consider $A_n = \{k n^{r-a}, k=-[n^{a}],\dots,[n^a]+1\}$. Since $r-a<1$ and thanks to $(1)$, the average does not change a lot between the points of $A_n$, so for any $\varepsilon>0$, $$ \limsup_{n\to\infty}\mathrm{P}\left(\sup_{|x|\le n^r} \left|\frac1{2n}\int_{x-n}^{x+n} X_t dt - \mathbb{E}[X_0]\right|>\varepsilon\right) \\ = \limsup_{n\to\infty}\mathrm{P}\left(\sup_{|x|\in A_n} \left|\frac1{2n}\int_{x-n}^{x+n} X_t dt - \mathbb{E}[X_0]\right|>\varepsilon\right)\\ \le \limsup_{n\to\infty} \sum_{x\in A_n} \mathrm{P}\left(\sup_{|x|\in A_n} \left|\frac1{2n}\int_{x-n}^{x+n} X_t dt - \mathbb{E}[X_0]\right|>\varepsilon\right)\\ \le \limsup_{n\to\infty}\frac{cn^a}{\varepsilon^2} \operatorname{var}\left(\frac1{2n}\int_{-n}^{n} X_t dt\right)=0. $$ (With a little bit more effort an almost sure convergence can be shown for any $r>1$, but the above is enough for me.)

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