If $X_t$ and $Y_t$ are both alpha mixing in the same sense for example decaying exponentially $\alpha(\ell) = O(\exp(-cn))$, $0 < c$.
Then can we say that the stacked random vector process $(X_t, Y_t)$ is also alpha mixing?
How about if we assume that $X_t$ and $Y_t$ are uncorrelated?
I am trying to extend this answer to hold not only for independent process but also when we just assume they are uncorrelated?
Therefore, my idea was to show that the stacked random vector is alpha mixing and then use that any measurable function of an alpha mixing process is also alpha mixing.
Take an i.i.d. sequence $\left(\varepsilon_t\right)_{t\in\mathbb Z}$ such that $\varepsilon_0$ takes the values $-1$, $0$ and $1$ with probability $1/3$. Let $X_t=\varepsilon_t$ and $Y_t=\varepsilon_{2t}^2$. Note that for each $s,t$, $\operatorname{Cov}(X_s,Y_t)=\mathbb E\left[X_sY_t\right]=0$ (this is clear if $s\neq t^2$ and if $s=t^2$, then $\mathbb E\left[X_sY_t\right]=\mathbb E\left[X_s^3\right]=0$.
Moreover, the sequence $\left((X_t,Y_t)\right)_{t\in\mathbb Z}$ is not alpha-mixing. Otherwise, $$ \alpha\left(\sigma\left((X_n,Y_n\right),\sigma\left((X_{2n},Y_{2n}\right) \right) $$ would converge to $0$, which is not the case since $$ \alpha\left(\sigma\left((X_n,Y_n\right),\sigma\left((X_{2n},Y_{2n}\right) \right)\geqslant \alpha\left(\sigma\left( Y_n\right),\sigma\left( X_{2n} \right) \right)=\alpha\left(\sigma(\varepsilon_{2n}),\sigma(\varepsilon_{2n})\right)\geqslant \mathbb P\left(\varepsilon_n=0\right)-\mathbb P\left(\varepsilon_n=0\right)^2=\frac 29. $$