Is the product of two mixing random variables also mixing?

Question

Question: Assume $X_t$ and $Y_t$ are random variables from the same probability space adapted to the filtration $\mathcal{F}_{-\infty}, ..., \mathcal{F}_t, ..., \mathcal{F}_{\infty}$. If $X_t$ and $Y_t$ are both $\alpha$-mixing of size $-\gamma$, does it follow that $X_t Y_t$ is $\alpha$-mixing of size $-\gamma$?

Does it help if we assume that $X_t$ and $Y_{t+m+k}$ are independent $\forall k > 0$ where $m < \infty$.

Motivation: Intuitively I feel like the above must be true, since mixing implies a type of asymptotic independence, but I'm having a hard time proving it to myself. It is well known that $g(X_t, X_{t-1}, ..., X_{t-\tau})$ will be $\alpha$-mixing for any measurable function $g$ and finite $\tau$, but it isn't immediately obvious to me that this implies that $X_t Y_t$ will also be $\alpha$-mixing...

Answer 1

In Bradley, Introduction to Strong Mixing conditions, volume 1, we can find the following result:

Given $\mathcal A,\mathcal B,\mathcal C,\mathcal D$ sub-$\sigma$-algebras of $\mathcal F$ such that $\mathcal A\vee \mathcal C$ and $\mathcal B\vee \mathcal D$ are independent, we have $$\alpha(\mathcal A\vee\mathcal B,\mathcal C\vee \mathcal D)\leqslant \alpha(\mathcal A,\mathcal C)+\alpha(\mathcal B,\mathcal D),$$ where $\mathcal U\vee \mathcal V$ is the $\sigma$-algebra generated by $\mathcal U$ and $\mathcal V$.

Denoting $\alpha$, $\alpha'$ and $\alpha''$ the mixing coefficient associated respectively to $((X_t,Y_t),t\in\Bbb Z)$, $(X_t,t\in\Bbb Z)$ and $(Y_t,t\in\Bbb Z)$. Using the previous result and the assumption of independence, one can show with the notations in the OP that $$\alpha(m+n)\leqslant \alpha'(n)+\alpha''(n).$$ In particular, an exponential rate is preserved.

If we don't have such an assumption, the result may not hold, for example if we take $Y_t:=X_{t^2}$.