If $\alpha(X_{0}, X_{n}) \rightarrow 0$ as $n \rightarrow \infty$, is the process $\{X_n\}_n$ strongly mixing?

Question

Let $\{X_n\}_{n\in\mathbb Z}$ be a strictly stationary sequence of random variables. If $\alpha(X_{0}, X_{n}) \rightarrow 0$ as $n \rightarrow \infty$, is $\{X_n\}_n$ strongly mixing?

Intuitively: To establish that a stationary sequence of random variables is strongly mixing, is it sufficient to check any pair of variables, one from the past versus one from the future?

Notation

Following usual conventions, let $(\Omega, \mathcal{F}, P)$ be a probability space and $\sigma(\dots)$ denote the $\sigma$-field $\subset \mathcal{F}$ generated by $(\dots)$. For any two $\sigma$-fields $\mathcal{A}, \mathcal{B} \subset \mathcal{F}$, define the following measure of dependence $$ \alpha(\mathcal{A}, \mathcal{B}) = \sup\lvert P(A \cap B) - P(A)P(B) \rvert, A\in\mathcal{A}, B\in\mathcal{B} $$ We write $\alpha(X,Y) \equiv \alpha(\sigma(X),\sigma(Y))$ for random variables $X,Y$.

Let $\{X_n\}_{n\in\mathbb Z}$ be a strictly stationary sequence of random variables. The strong mixing coefficient (in this strictly stationary case) is $$ \alpha(n) = \alpha(\sigma(\dots, X_{-1}, X_{0}), \sigma(X_{n}, X_{n+1}, X_{n+2}, \dots)) $$ The sequence is strongly mixing ($\alpha$-mixing) if $\alpha(n) \rightarrow 0$ as $n \rightarrow \infty$.

Why I am puzzled

I would like to use a central limit theorem under weak dependence (strong mixing), and have established that $\alpha(n) \rightarrow 0$ as $n \rightarrow \infty$, but am worried about the jump from 'any pair of random variables tend to independent if they are far enough apart' to 'the future is independent from the past if the separation is large enough'.

Reference

Richard C. Bradley (2005), Basic Properties of Strong Mixing Conditions. A Survey and Some Open Questions, Probability Surveys 2, 107-144.

Answer 1

I quoted a simplified version of Theorem 1.4 from On possible mixing rates for some strong mixing conditions for $N$-tuplewise independent random fields by Richard C. Bradley, but stated for sequence.

Theorem Let $N\geqslant 2$ be an integer and let $\left(c_n\right)_{n\geqslant 1}$ be a non-increasing sequence of numbers in $[0,1]$. Then there exists a strictly stationary sequence $\left(X_i\right)_{i\in\mathbb Z}$ such that the following properties hold

1. The random variable $X_0$ is uniformly distributed on $[0,1]$.
2. If $i_1,\dots,i_N$ are distinct integers, then the collection of random variables $X_{i_1 },\dots,X_{i_N}$ is independent.
3. For any $n\geqslant 1$, $4\alpha\left(n\right)=c_n$.

We can choose $N=2$ and $c_n=1$ for each $n$ to get a counter-example to the question.

Answer 2

From $\alpha(X_0,X_n)\to 0$ it couldn't deduce that $\alpha(\sigma(X_i,i\le 0),\sigma(X_j,j\ge n))\to 0$. The example is following. Let $X=\{X_n,n\in\mathbb{Z}\}$ be a stationary Gaussian process with $\mathsf{E}[X_n]=0$ and spectral density $f(\lambda)=1_{[0,\pi]}(\lambda)$. For this process, $$ \mathbb{E}[X_n\overline{X}_0]=\int_{-\pi}^{\pi}e^{in\lambda}f(\lambda)\mathrm{d}\lambda \to0,\qquad \text{as}\quad n\to\infty. \tag{1}$$ Using a result of Kolmogorov and Rozanov, $\alpha(X_0,X_n)\to 0$. But form the Th. 7.1 in Bradley's paper, mentioned in your question, $X$ is not strongly mixing.

Answer 3

I realized that my intuitions were treating $\{X_n\}_{n \in Z}$ as being a Markov chain. Indeed, I subsequently learned that if $\{X_n\}_{n \in Z}$ is a Markov chain (in addition to being strictly stationary) then $$ \alpha(n) = \alpha(\sigma(X_o), \sigma(X_n)) $$ (Bradley, 2005, Section 3.1).