# On a finitary version of mixing

Let $(X_1,X_2,\ldots)$ be a stationary, mixing sequence of real random variables. Then it holds (for example) for any event $A$ that is measurable in $\sigma(X_1,X_2,\ldots)$ and any $S \subseteq \mathbb{R}$ that $$\lim_{i \to \infty} \big|\mathbb{P}[A,X_i\in S] - \mathbb{P}[A] \cdot \mathbb{P}[X_i \in S]\big|=0.$$

I am looking for a finitary version of this statement. That is, something of this sort: for every $\varepsilon>0$ there is an $n$ large enough such that, if $A$ is measurable in $\sigma(X_1,\ldots,X_n)$, and if $i$ is chosen uniformly in $\{1,\ldots,n\}$, then with probability $1-\varepsilon$ $$\big|\mathbb{P}[A,X_i\in S] - \mathbb{P}[A] \cdot \mathbb{P}[X_i \in S]\big| < \varepsilon.$$

• Are you asking to deduce the finitary statement from the limiting statement? Or are you asking if the finitary statement is already studied in the literature? – Anthony Quas Nov 25 '17 at 19:57
• I am asking whether this finitary statement, or something like it, has been studied in the literature. I don't know if it follows from the one above it, but I suspect something similar should be true and should be known. – Vladimir Nov 25 '17 at 21:21

Let $J_n$ be a random variable independent of $X:=(X_1,X_2,\ldots)$ and uniformly distributed in the set $\{1,\ldots,n\}$. For each $i\in\{1,\ldots,n\}$, let $$p(i):=P(A,X_i\in S).$$ We need to show that $$p(J_n)\to P(A)P(X_1\in S)$$ in probability uniformly in all $A\in\sigma(X_1,\ldots,X_n)$; the convergence here everywhere is as $n\to\infty$.
Let us show a bit more -- that this convergence is uniform over all $A$ in the underlying sigma-algebra (say $\Sigma$). Suppose then that the weak mixing condition holds (which of course will hold if the strong mixing holds). Then the sequence $X$ is ergodic; see e.g. the Remark on page 13 in Ergodic Theory. Now the ergodic theorem implies that
$$d_n(S):=\frac1n\,\sum_{i=1}^n I\{X_i\in S\}-P(X_1\in S)\to0$$ almost surely and hence in probability, where $I\{\cdot\}$ denotes the indicator function. So, for every $\varepsilon>0$ there is a natural $n_\varepsilon$ such that for all natural $n>n_\varepsilon$ we have $P(|d_n(S)|>\varepsilon)<\varepsilon$; therefore and because $|d_n(S)|\le1$, $$E|d_n(S)I\{A\}|\le E|d_n(S)| \le\varepsilon P(|d_n(S)|\le\varepsilon)+P(|d_n(S)|>\varepsilon)\le2\varepsilon.$$ So, $d_n(S)I\{A\}\to0$ in $L^1$ uniformly in all $A$. So, $$Ep(J_n)= \frac1n\,\sum_{i=1}^n P(A,X_i\in S) =Ed_n(S)I\{A\}+P(A)P(X_1\in S)\to P(A)P(X_1\in S)$$ uniformly in all $A$. Similarly, letting $(\tilde A,\tilde X)$ denote an independent copy of $(A,X)$, we have \begin{multline*} Ep(J_n)^2= \frac1n\,\sum_{i=1}^n P(A,X_i\in S)^2 =\frac1n\,\sum_{i=1}^n P(A,\tilde A,X_i\in S,\tilde X_i\in S) \\ \to P(A,\tilde A)P(X_1\in S,\tilde X_1\in S)=P(A)^2P(X_1\in S)^2 \end{multline*} uniformly in all $A$;
here we use the fact that the weak mixing property is preserved under the direct product (see e.g. page 5 in http://arxiv.org/abs/math/0603575v1 ) and hence the sequence of pairs $((X_1,\tilde X_1),(X_2,\tilde X_2),\dots)$ is ergodic. Thus, $Ep(J_n)\to P(A)P(X_1\in S)$ and $Var\,p(J_n)\to0$ uniformly in all $A$. So, by Chebyshev's inequality, indeed $p(J_n)\to P(A)P(X_1\in S)$ in probability uniformly in all $A$.
• I haven't done work in ergodic theory or mixing, nor have I followed developments there. That may be a reason why I haven't seen such a result in the literature. I don't think it could appear in a separate paper devoted to it, since it follows quickly from the ergodic theorem, as shown in the answer. Your construction involving a random index uniformly distributed in $\{1,\dots,n\}$ appears in papers on Stein's method in Berry--Esseen-type inequalities. So, you may want to search for such Stein-method papers that also involve stationary and/or mixing processes. – Iosif Pinelis Nov 26 '17 at 15:10