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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 $1-\varepsilon$ $$ \big|\mathbb{P}[A,X_i\in S] - \mathbb{P}[A] \cdot \mathbb{P}[X_i \in S]\big| < \varepsilon. $$

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    $\begingroup$ 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? $\endgroup$ – Anthony Quas Nov 25 '17 at 19:57
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    $\begingroup$ 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. $\endgroup$ – Vladimir Nov 25 '17 at 21:21
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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 \begin{equation} p(i):=P(A,X_i\in S). \end{equation} We need to show that \begin{equation} p(J_n)\to P(A)P(X_1\in S) \end{equation} 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
\begin{equation} d_n(S):=\frac1n\,\sum_{i=1}^n I\{X_i\in S\}-P(X_1\in S)\to0 \end{equation} 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 $|d_n(S)|\le1$, \begin{equation} 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. \end{equation} So, $d_n(S)I\{A\}\to0$ in $L^1$ uniformly in all $A$. So, \begin{equation} 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) \end{equation} 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 $A$.

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  • $\begingroup$ Very interesting! So we don't really need mixing for this? Ergodicity is enough? Or did you use weak mixing in the last step? It feels like this property should imply (strong) mixing... $\endgroup$ – Vladimir Nov 26 '17 at 6:55
  • $\begingroup$ Also: is there a name / reference for this result? How about for the iid case? $\endgroup$ – Vladimir Nov 26 '17 at 6:57
  • $\begingroup$ What I really needed was the ergodicity. As mentioned in the answer, the ergodicity is implied already by the weak mixing. So, it is enough to assume the weak mixing. Of course, the strong mixing would be (more than) enough, since it implies the weak mixing. Now, of course, the iid condition implies the strong mixing. Thus, of the four notions of mixing -- the iid condition, the strong mixing, the weak one, and the ergodicity -- the latter, the weakest of the four ones, suffices here. $\endgroup$ – Iosif Pinelis Nov 26 '17 at 14:57
  • $\begingroup$ 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. $\endgroup$ – Iosif Pinelis Nov 26 '17 at 15:10
  • $\begingroup$ Thanks! Isn't weak mixing needed for the last step? Aren't you assuming that the product of the sequence with itself is ergodic? This isn't true for every ergodic sequence. $\endgroup$ – Vladimir Nov 26 '17 at 19:38

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