# A continuous weighted ergodic average

Let $(X_t)_{t\ge 0}$ be a continuous stationary ergodic process (w.r.t. the shift operator), and $f$ be a bounded Borel function. Prove or disprove:

$$\lim_{n\to\infty} \frac{1}{\ln n}\sum_{k=1}^n (\ln(k+1)-\ln k)f(X_{\ln k}) = \mathbb{E}[f(X_0)] \qquad \text{a.s.}$$

There are several results on the pointwise convergence of weighted ergodic average in the discrete case, while I did not find the counterpart in the continuous case.

I can prove that this result is true under an additional assumption, which is closely related to strong mixing and seems to hold in the cases I am interested in (see below for more details).

Assumption: for any bounded Borel function $f$, there exists constants $c_0>0, t_0>0$ and exponent $\beta>0$ (which may depend on $f$) such that $|\text{cov}(f(X_0),f(X_t))|\le c_0t^{-\beta}$ for any $t>t_0$. In other words, the correlation enjoys a polynomial decay with an arbitrarily slow speed.

Proof based on this assumption:

Since $\ln(k+1)-\ln k=k^{-1}+O(k^{-2})$, it is equivalent to prove that $$\lim_{n\to\infty} \frac{1}{\sum_{k=1}^n k^{-1}}\sum_{k=1}^n \frac{f(X_{\ln k})}{k} = \mathbb{E}[f(X_0)] \qquad \text{a.s.}$$ W.l.o.g. we assume that $f\ge 0$. Denote the normalized partial sum on the LHS by $S_n$, we have \begin{align} \text{var}(S_n) &\le \frac{c_1}{(\ln n)^2}\sum_{i,j=1}^n \frac{|\text{cov}(f(X_{\ln i}), f(X_{\ln j}))|}{ij}\\ &= \frac{c_1}{(\ln n)^2}\left(\sum_{|\ln i-\ln j|\le t_0} \frac{|\text{cov}(f(X_{\ln i}), f(X_{\ln j}))|}{ij} + \sum_{|\ln i-\ln j|> t_0} \frac{|\text{cov}(f(X_{\ln i}), f(X_{\ln j}))|}{ij}\right)\\ &\le \frac{c_1}{(\ln n)^2}\left(\sum_{|\ln i-\ln j|\le t_0} \frac{\|f\|_\infty^2}{ij} + \sum_{|\ln i-\ln j|> t_0} \frac{c_0}{ij|\ln i-\ln j|^\beta}\right)\\ &\le \frac{c_2}{(\ln n)^2}\sum_{i=1}^n \frac{1}{i}\left(\sum_{j=ie^{-t_0}}^{ie^{t_0}} \frac{\|f\|_\infty^2}{j} + \sum_{|\ln i-\ln j|> t_0} \frac{c_0}{j|\ln i-\ln j|^\beta}\right)\\ &\le \frac{c_3}{(\ln n)^2}\sum_{i=1}^n \frac{1}{i}\left(1 + \int_1^{ie^{-t_0}} \frac{dx}{x(\ln i-\ln x)^\beta}+\int_{ie^{t_0}}^n \frac{dx}{x(\ln x-\ln i)^\beta}\right)\\ &\le \frac{c_4}{(\ln n)^2}\sum_{i=1}^n \frac{(\ln n)^{1-\beta}}{i} \le \frac{c_5}{(\ln n)^\beta}. \end{align}

Now we fix some $\alpha>1$ and consider the sequence $k_m=[e^{\alpha^m}]$. Along the subsequence $S_{k_m}$, for any $\epsilon>0$, we have \begin{align} \sum_{m=1}^\infty P(|S_{k_m}-\mathbb{E}[f(X_0)]|>\epsilon) \le \frac{1}{\epsilon^2} \sum_{m=1}^\infty \text{var}(S_{k_m}) \le \frac{1}{\epsilon^2} \sum_{m=1}^\infty \frac{c_6}{\alpha^{\beta m}} < \infty \end{align} then by Borel--Cantelli $S_{k_m}\to \mathbb{E}[f(X_0)]$ a.s.. For the original sequence $S_n$ with $k_m\le n<k_{m+1}$, by $f\ge 0$ we have $\frac{\ln k_m}{\ln k_{m+1}}S_{k_m}\le \frac{\ln k_m}{\ln n}S_{k_m} \le S_n\le \frac{\ln k_{m+1}}{\ln n}S_{k_{m+1}}\le \frac{\ln k_{m+1}}{\ln k_m}S_{k_{m+1}}$, and thus \begin{align} \alpha^{-1}\cdot \mathbb{E}[f(X_0)]\le \liminf_{n\to\infty} S_n \le \limsup_{n\to\infty} S_n \le \alpha\cdot \mathbb{E}[f(X_0)] \qquad \text{a.s.} \end{align} Finally choosing $\alpha\to 1^+$ completes the proof.

Background: the original problem is as follows: for i.i.d. stable r.v.s $X_1,X_2,\cdots$ with parameter $\alpha\in(0,2]$, the partial sum $S_n=\sum_{k=1}^n X_k$ satisfies $$\lim_{n\to\infty} \frac{1}{\ln n}\sum_{k=1}^n \frac{1}{k}f(k^{-\frac{1}{\alpha}}S_k) = \mathbb{E}[f(X_1)]\qquad \text{a.s.}$$ for any bounded Borel function $f$.

Together with CLT, this is a generalization of P. Levy's result: for i.i.d. $X_n$ with continuous symmetric distribution, $$\lim_{n\to\infty} \frac{1}{\ln n}\sum_{k=1}^n \frac{1}{k}\cdot 1_{S_k>0} = \frac{1}{2}\qquad \text{a.s.}$$ Also, this is closely related to the almost sure central limit theorem: for i.i.d. $X_n$ with $\mathbb{E}(X_1)=0, \mathbb{E}(X_1^2)=1$, then $$\lim_{n\to\infty} \frac{1}{\ln n}\sum_{k=1}^n \frac{1}{k}\cdot 1_{\frac{S_k}{\sqrt{k}}<x} = \Phi(x)\qquad \text{a.s.}$$ where $\Phi(\cdot)$ is the CDF of the standard normal distribution.

One way to prove the original problem is to construct a continuous-time Levy process $(S_t)_{t\ge 0}$ with $S_1\overset{d}{=} X_1$, and then consider its Lamperti transform $(Z_t)=(e^{-t/\alpha}S_{e^t})$. It is straightforward to show that $(Z_t)$ is stationary. Ergodicity follows from the Blumenthal's zero–one law for cadlag stochastic processes with independent increments. Now we arrive at the problem stated here.

In fact, more is true for this special example: $(Z_t)$ is strongly mixing since we can verify from $(S_t)$ that $\ln\mathbb{E}[\exp(i(\theta_1 Z_0+\theta_2 Z_t))]- \ln\mathbb{E}[\exp(i\theta_1 Z_0)]-\ln\mathbb{E}[\exp(i\theta_2 Z_t)]\to 0$ as $t\to\infty$ (For this point, see Remark 3.7.3 in P. Embrechts and M. Maejima: Self-similar Processes. Princeton University Press, Princeton, NJ, 2002). Moreover, for Brownian motion ($\alpha=2$), by Renyi maximal correlation for joint Gaussian variables we know that $$|\text{cov}(f(Z_0),f(Z_t))|\le \text{var}(f)\cdot |\text{corr}(Z_0,Z_t)|= \text{var}(f)\cdot e^{-t/2}$$ decays exponentially, which is much stronger than what we have assumed!

Remaining question:

1. Is there any way to prove this result using the results/proofs in ergodic theory (e.g., some maximal inequality)? In fact, from pointwise ergodic theorem we know that $$\frac{1}{\ln t}\int_0^t \frac{1}{u}f(X_{\ln u})du \to \mathbb{E}[f(X_0)] \qquad \text{a.s.}$$ as $t\to\infty$, which is very close to the question here. Moreover, this problem looks very similar to the weighted ergodic average problem $\sum_{k=0}^{n-1}a_{n,k}f\circ T^k(X)$ and the subsequence problem $\frac{1}{n}\sum_{k=0}^{n-1}f\circ T^{m_k}(X)$.

2. For general stable processes other than Brownian motion, I highly believe that our additional assumption still holds, but I did not find proper references on maximal correlation/strongly mixing coefficient. Can anyone help me with that?