Show that a stochastic process is strictly stationary Let $X = (X_t)_{t \in \mathbb Z}$ and $Y = (Y_t)_{t \in \mathbb Z}$ be two independent and strictly stationary stochastic processes with mean zero, i.e., for any $(t_1,...,t_p)\in \mathbb Z^p$ and any $p\in \mathbb N$, we have
$$(X_{t_1},...., X_{t_p}) \overset d \sim (X_{t_1+h},...., X_{t_p+h})\quad \forall \, h >0$$
$$E[X_t]=0,\quad  \forall\, t \in \mathbb Z$$
The same for $Y = (Y_t)_{t \in \mathbb Z}$.
Now, consider a $U\sim \hbox{Bernoulli(p)}$ independent of $X$ and $Y$. Define $Z = (Z_t)_{t \in \mathbb Z}$ as:
\begin{equation}\label{abc}\tag{I}
Z=\begin{cases}
   X, & \text{ if $U =1$}\\
            Y, & \text{ if $U =0$}
   \end{cases}
\end{equation}
How to show that $Z = (Z_t)_{t \in \mathbb Z}$ is strictly stationary? i.e.:
$$(Z_{t_1},...., Z_{t_p}) \overset d \sim (Z_{t_1+h},...., Z_{t_p+h})\quad \forall \, h >0$$
I need express the distribution of  $(Z_{t_1},...., Z_{t_p})$ in function of the distribution $(X_{t_1},...., X_{t_p})$, $(Y_{t_1},...., Y_{t_p})$ and $U$. Denote the characteristic function, respectively as $\varphi_{Z_{t_1,...,t_p}}$, $\varphi_{X_{t_1,...,t_p}}$, $\varphi_{Y_{t_1,...,t_p}}$ and $\varphi_U$.
Are there any relations that can write $\varphi_{Z_{t_1,...,t_p}}$ in terms of $\varphi_{X_{t_1,...,t_p}}$, $\varphi_{Y_{t_1,...,t_p}}$ and $\varphi_U$?
 A: For $(t_1,\dots,t_p)\in\mathbb Z^p$, $h>0$ and $B$ a Borel subset of $\mathbb R^p$,
\begin{align}
\mathbb P\left( (Z_{t_1+h},...., Z_{t_p+h})\in B\right)&=\mathbb P\left( (Z_{t_1+h},...., Z_{t_p+h})\in B,U=0\right)+\mathbb P\left( (Z_{t_1+h},...., Z_{t_p+h})\in B,U=1\right)\\
&=\mathbb P\left( (Y_{t_1+h},...., Y_{t_p+h})\in B,U=0\right)+\mathbb P\left( (X_{t_1+h},...., X_{t_p+h})\in B,U=1\right)\\
&=\mathbb P\left( (Y_{t_1+h},...., Y_{t_p+h})\in B\right)\mathbb P\left(U=0\right)+\mathbb P\left( (X_{t_1+h},...., X_{t_p+h})\in B\right)\mathbb P\left(U=1\right)\\
&=\mathbb P\left( (Y_{t_1},...., Y_{t_p})\in B\right)\mathbb P\left(U=0\right)+\mathbb P\left( (X_{t_1},...., X_{t_p})\in B\right)\mathbb P\left(U=1\right)\\
&=\mathbb P\left( (Y_{t_1},...., Y_{t_p})\in B,U=0\right)+\mathbb P\left( (X_{t_1},...., X_{t_p})\in B,U=1\right)\\
&=\mathbb P\left( (Z_{t_1},...., Z_{t_p})\in B,U=0\right)+\mathbb P\left( (Z_{t_1},...., Z_{t_p})\in B,U=1\right)\\
&=\mathbb P\left( (Z_{t_1},...., Z_{t_p})\in B\right),
\end{align}
where the third and fifth equality is a consequence of independence of $U$ of $(X,Y)$.
