Check the $\alpha$-mixing conditions: Stochastic process

Question

I want to determine if $\{Y_{t,T},X_{t,T}\}, t=1,\dotsc,T; T\geq 1$, is $\alpha$-mixing for some special cases. Let $X_{t,T}\sim i.i.d. U[0,1]$ and

1. $Y_{t,T}=1$ (degenerate random variable);
2. $Y_{t,T}=f(X_{t,T})$, $f$ measurable function;
3. $Y_{t,T}=g(\epsilon_{t,T})$, $g$ measurable (linear) function so that $\{\epsilon_{t,T}\}$ is a $\alpha$-mixing.

My attempt

My tools for this subject is rudimentary. I will write some ideas. Define the mixing coefficient by $$\alpha(j)=\sup_T\sup_{1\leq k\leq T-j}\sup\{\lvert P(A\cap B)-P(A)P(B)\rvert: B\in\mathcal{F}_{T,1}^{k},A\in\mathcal{F}_{T,k+j}^T\}$$ where $\mathcal{F}_{T,i}^{k}=\sigma((Y_{l,T},X_{l,T}):i\leq l\leq k)$ which is known to be equal $\sigma(\bigcup_{l=i}^k \sigma(Y_{l,T},X_{l,T}))$. So the array $\{Y_{t,T},X_{t,T}\}$ is called $\alpha$-mixing if $\alpha(j)\to0$ as $j\to\infty$. I will denote independence by $\perp$. I believe that if a sequence $\{X_{t}\}$ is independent, then $\mathcal{F}_{1}^k\perp \mathcal{F}_{k+j}^\infty$, implying that $\alpha(j)=0$. The same should hold for the triangular array $X_{t,T}$. As far as I see, it trivially means $\alpha$-mixing. Now I start to answer the questions for each case.

1. As $Y_{t,T} \perp X_{i,T},$ we have $\sigma(Y_{i,T},X_{i,T})\perp \sigma(Y_{i,T},X_{i,T})$ for $ t,i=1,\dotsc,T, T\geq 1$. It means $\mathcal{F}_{T,1}^k\perp \mathcal{F}_{T,k+j}^\infty$ for any $j>0$. So $\alpha(j)=0$ for all positive $j$. (Here, my arguments lack formality)
2. $\sigma(Y_{i,T},X_{i,T})$ is it the product sigma algebra? Or just $\sigma(\sigma(Y_{i,T})\cup \sigma(X_{i,T}))$? In both cases, $\sigma(f(X_{t,T}))=\{X^{-1}\circ f^{-1}(B):B\in\mathcal{B}_{\mathbb{R}}\}\subseteq \{X^{-1}(B):B\in\mathcal{B}_{\mathbb{R}}\}=\sigma(X_{t,T})$. This should imply $\sigma(f(X_{i,T}),X_{i,T})\subseteq \sigma(X_{i,T},X_{i,T})$. Using similar arguments of item 1 and the independence of $\{X_{t,T}\}$,$\alpha(j)=0$.
3. Using the argument of item 2, $g(\epsilon_{t,T})$ is $\alpha$-mixing and we can proceed considering $Y_{t,T}=\epsilon_{t,T}$. I see that $\sigma(\epsilon_{t,T},X_{t,T})$ is not necessarily independent of $\sigma(\epsilon_{i,T},X_{i,T})$ anymore due to $\epsilon$. I think that as $\epsilon_{i,T}$ is $\alpha$-mixing, the presence of independent $X_{t,T}$ doesn't affect the limit $\lim_{j\to\infty}\alpha(j)=0$.

As you can see, my ideas are rudimentary and based on intuition. Can you help me to prove these results in a properly way?

I wonder if there is a general result for $\alpha$-mixing sequences $\{Y_t\},\{X_t\}$ such that $Y_t$ and $X_t$ are independent that says $\{(Y_t,X_t)\}$ is $\alpha$-mixing.

Answer 1

If you want more formality, for 1. and 2., the inclusion $\mathcal F^k_{T,i}\subset \sigma(X_{\ell,T},i\leqslant \ell\leqslant k)$ holds. By independence, it follows that for all $T\geqslant 1$ and all $1\leqslant k\leqslant T-j$, $\sup\{\lvert P(A\cap B)-P(A)P(B)\rvert: B\in\mathcal{F}_{T,1}^{k},A\in\mathcal{F}_{T,k+j}^T\}=0$.

For 3. it seems that we have to assume that $\left(\varepsilon_{t,T}\right)$ is independent of $\left(X_{t,T}\right)$. Then we can use the inequality mentioned in this thread.