# Uniform convergence of averages for stationary ergodic process

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).

• 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. – D. Thomine Jun 23 '19 at 21:19
• @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. – zhoraster Jun 24 '19 at 16:01

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 $$n \to 0$$.

• I have also obtained such estimate in slightly more general case already. Thank you very much anyway! – zhoraster Jun 28 '19 at 9:53
• 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... – zhoraster Jun 28 '19 at 10:34
• 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. – Yuval Peres Jun 28 '19 at 11:39
• 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! – zhoraster Jun 28 '19 at 13:16
• You are right, I ignored the subtraction, will correct the answer. – Yuval Peres Jun 28 '19 at 13:22

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.)