Is linear combination of strong mixing process still strong mixing? Suppose the process $\left\{X_{t}:t\in Z\right\}$ is absolutely continuously distributed and strong mixing with the coefficient $\alpha_{X}(s) \rightarrow 0$ defined as
\begin{equation}
\alpha_{X}(s) \equiv \sup \left\{ |P(A\cap B) - P(A)P(B)| : -\infty < t<\infty, A \in X^{t}_{-\infty}, B\in X^{\infty}_{t+s} \right\}.
\end{equation}
Meanwhile, suppose there is a absolutely continuously distributed process $\{Y_{t}:t\in Z\}$ with coefficient $\alpha_{Y}(s)$.
Then, is the process $\frac{1}{2}(X_{t}+Y_{t})$ necessarily strong mixing? If so, what is the mixing coefficient?
In general, suppose $\{X_{it}:t\in Z\}$ is strong mixing with $\alpha_{i}(s)$. Is $\underset{N \rightarrow \infty}{\lim} \displaystyle \frac{1}{N}\sum_{i=1}^{N}X_{it}$ still strong mixing? If so, what is the mixing coefficient?
 A: In general no: let $\left(X_t\right)_{t\in\mathbb Z}$ be an i.i.d. sequence, where $X_0$ has a standard normal law. Let $Y_t=X_{t^3}$; then both sequences $\left(X_t\right)_{t\in\mathbb Z}$ and $\left(Y_t\right)_{t\in\mathbb Z}$ are independent hence $\alpha$-mixing. But the sequence $\left(X_t+Y_t\right)_{t\in\mathbb Z}$ is not $\alpha$-mixing. Indeed, for all non-negative integer $n$, 
$$
\alpha\left(\sigma\left(X_n+Y_n\right),\sigma\left(X_{n^3}+Z_{n^3}\right) \right)
=
\alpha\left(\sigma\left(X_n+X_{n^3}\right),\sigma\left(X_{n^3}+X_{n^9}\right)\right)
$$
and since $\left(X_{n},X_{n^3},X_{n^9}\right)$ has the same distribution as 
$\left(U,V,W\right)$ (an i.i.d. vector with $U$ standard normal, we get that 
$$
\alpha\left(\sigma\left(X_n+Y_n\right),\sigma\left(X_{n^3}+Z_{n^3}\right) \right)
=
\alpha\left(\sigma\left(U+V\right),\sigma\left(V+W\right)\right)
$$
and the later term is positive an does not depend on $n$. 
We use the fact that the sequences $\left(X_t\right)_{t\in\mathbb Z}$ and $\left(Y_t\right)_{t\in\mathbb Z}$ are not independent (and actually really dependent). In general, if $\left(X_t\right)_{t\in\mathbb Z}$ and $\left(Y_t\right)_{t\in\mathbb Z}$ are two mixing sequences and $\left(X_t\right)_{t\in\mathbb Z}$ is independent of $\left(Y_t\right)_{t\in\mathbb Z}$, then the sequence $\left(X_t+Y_t\right)_{t\in\mathbb Z}$ is $\alpha$-mixing with rate $\alpha(n)\leqslant \alpha_X(n)+\alpha_Y(n)$.
