# Is the product of two mixing random variables also mixing?

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...

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It will work if the processes $(X_t)$ and $(Y_t)$ are independent. Otherwise, there should be counter examples as $Y_t:=X_{t^2}$. – Davide Giraudo Mar 6 '13 at 10:47
@DavideGiraudo $-\gamma$ defines the rate at which the mixing coefficients go to zero as $m$ (the time lag) goes to $\infty$. The most common usage in my experience is that "of size $-\gamma$" means that $\alpha_m = O(m^{-\gamma})$. Regarding the counter-examples, I'm afraid they are not obvious to me; but perhaps I am being dense. If it helps, I can assume that both $X_t$ and $Y_t$ are adapted to the same filtration $\mathcal{F}_{-\infty}, ..., \mathcal{F}_t, ..., \mathcal{F}_\infty$. – Colin T Bowers Mar 6 '13 at 11:05
@DavideGiraudo Also, thanks for your interest :-) I've updated the question. – Colin T Bowers Mar 6 '13 at 11:10
I saw your update. The problem of adaptedness is not an issue, I think (unless there are mixing conditions on the $\sigma$-algebras $\mathcal F_t$). How are $X_t$ and $Y_t$ linked in your context? Independence? – Davide Giraudo Mar 6 '13 at 11:57
@DavideGiraudo In my specific application, $X_t$ and $Y_t$ are both stationary ergodic and $\alpha$-mixing. I am trying not to place any restrictions on the possible dependence (or independence) between them other than that they are asymptotically independent, ie the relationship between $X_{t+m}$ and $Y_t$ approaches independence as $m \rightarrow \infty$. But if it is necessary to get traction, I'm happy to make stronger assumptions... Off to bed now. Cheers. ps if it helps, I could definitely assume $X_t$ is independent of $Y_{t+m}$ and $Y_{t-m}$ for some $m < \infty$... – Colin T Bowers Mar 6 '13 at 12:35

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 $Y_t:=X_{t^2}$.
Thanks Davide. Your answer, along with the reference, has proven useful. I understand your point about the $Y_t := X_{t^2}$ case now. – Colin T Bowers Apr 2 '13 at 3:56