In statistics and probability literature, a strictly stationary stochastic process $\{X_t\}\in\mathbb{R}$ is called $\alpha$-mixing if

$\alpha(n)=\sup_{A\in\mathcal{F}_{-\infty}^{0}, B\in\mathcal{F}_{n}^{\infty}}|P(A)P(B)-P(AB)|\longrightarrow 0$

as $n\rightarrow 0$ where $\mathcal{F}_{a}^{b}$ denotes the $\sigma$-algebra generated by $\{X_t; i\leq t\leq j\}$.

Let us now consider an $\mathbb{R}^d$-valued (stationary) stochastic process $\{Z_t\}$ defined as $\{Z_t\}:=\{Z_{1t},Z_{2t},...,Z_{dt}\}^{T}$. I was wondering if the process $\{Z_t\}$ being mixing would imply that its marginal processes $\{Z_{it}\}$ $\forall i$ being mixing or vice versa..

I understand this should be rather straightforward, but could not rigorously provide a proof with this.. This is not a homework, just personal curiosity.


Let us denote by $\alpha(n)$ the $n$th $\alpha$-mixing coefficient of $\{X_t\}$.

If $f\colon\mathbb R^d\to\mathbb R$ is a Borel-measurable function then $\alpha'(n)\leqslant \alpha(n)$, where $\alpha'(n)$ is the $n$th $\alpha$-mixing coefficient of $\{f(X_t)\}$. This because $\sigma\{f(X_i),a\leqslant i\leqslant b\}\subset\sigma\{X_i,a\leqslant i\leqslant b\}$.

Here is a question about a partial converse.

  • $\begingroup$ Thanks very much Davide! Could you please kindly provide me with a sketch proof for this? (or perhaps recommend a book that covers this?) Also, I was wondering if the converse of the statement would hold. I.e. If $\{Z_{it}\}$ is mixing for all $i$, would $\{Z_t\}:=\{Z_{1t},Z_{2t},...,Z_{dt}\}$ be mixing as well? Many thanks in advance.. $\endgroup$
    – John
    Oct 13 '14 at 22:16
  • $\begingroup$ @JohnK Braley's book is a good reference, there is also Doukhan's book. See edit for the other questions. $\endgroup$ Oct 14 '14 at 19:02
  • $\begingroup$ I am very sorry for the delayed response, thanks so much for your help! $\endgroup$
    – John
    Oct 21 '14 at 22:22
  • $\begingroup$ You are welcome. $\endgroup$ Oct 22 '14 at 8:45

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