Just to give a context, define the mixing coefficient by $$\alpha(j)=\sup_T\sup_{1\leq k\leq T-j}\sup\left\{\lvert P(A\cap B)-P(A)P(B)\rvert: B\in\mathcal{F}_{T,1}^{k},A\in\mathcal{F}_{T,k+j}^T\right\}$$ 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 $j\to\infty$.
In order to show that an array $\{f(Y_{t,T}),g(X_{t,T})\}$ is $\alpha$-mixing given that $\{Y_{t,T},X_{t,T}\}$ is $\alpha$-mixing and $f,g$ are measurable functions, I concluded that it is sufficient that $\sigma(f(Y_{t,T}))\subseteq \sigma(Y_{t,T})$ and $\sigma(f(X_{t,T}))\subseteq \sigma(X_{t,T})$.
- Is my conclusion right?
- The inclusions hold?
My arguments
(1) Denote $\mathcal{G}_{T,i}^{k}=\sigma((f(Y_{l,T}),g(X_{l,T})):i\leq l\leq k)$. Since $\sigma((Y_{l,T},X_{l,T}))=\sigma(\sigma(Y_{l,T})\cup\sigma(X_{l,T}))$, we have that $\sigma(f(Y_{t,T}))\subseteq \sigma(Y_{t,T})$ and $\sigma(f(X_{t,T}))\subseteq \sigma(X_{t,T})$ imply $\mathcal{G}_{T,i}^{k}\subseteq \mathcal{F}_{T,i}^{k}$. This, in turn, implies that the mixing coefficient for the array $\{f(Y_{t,T}),g(X_{t,T})\}$ is less equal $\alpha(j),\forall j$.
(2) $\sigma(f(X_{l,T}))=\{(X^{-1}\circ f^{-1})(A):A\in\mathcal{B}_\mathbb{R}\}\subseteq \{X^{-1}(B):B\in\mathcal{B}_\mathbb{R}\}=\sigma(X_{l,T})$. The same for $Y_{l,T}$.
I will appreciate if you give me feedbacks!
I think that your argument is correct. For the part (2), it would be better to keep $X_{l,T}$ instead of $X$.