Alpha mixing property of a $\mathbb{R}^d$ valued Stochastic Process

Question

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.

Answer 1

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.