Uniform convergence of averages for stationary ergodic process

Question

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).

Answer 1

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 $n \to 0$.

Answer 2

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.)