Let $X(t)$ and $Y(t)$ be two stationary random processes and $$Z(t)=X(t)Y(t)$$By stationary I mean stationary in the strict sense which is $$F_{X(t_1),...,X(t_k)}(x_1,...,x_k)=F_{X(t_1+h),...,X(t_k+h)}(x_1,...,x_k)$$for all $h,t_1,...,t_k\in \mathbb{R}$ and $k\in\mathbb{N}$.
$1$. Is $Z(t)$ a stationary random process?
$2$. If $X(t)$ and $Y(t)$ are independent random processes, is $Z(t)$ a stationary random process?
My try:
$1$. Intuitively, the answer is no but I couldn't find a counterexample. For example if we set $Y(t) = aX(t)$ then $Z(t) = aX^2(t)$ and it's stationary.
$2$. In this case the answer seems to be yes. This post provides an answer but I don't understand why independence implies $$(X_{t_1 + h}, X_{t_2 + h}, \ldots, X_{t_n + h},Y_{t_1 + h}, Y_{t_2 + h}, \ldots, Y_{t_n + h}) \stackrel{d}{=} (X_{t_1 }, X_{t_2 }, \ldots, X_{t_n }, Y_{t_1 }, Y_{t_2 }, \ldots, Y_{t_n }) \tag{1}$$ Also, why $(1)$ implies $(2)$?$$f(X_{t_1 + h}, X_{t_2 + h}, \ldots, X_{t_n + h},Y_{t_1 + h}, Y_{t_2 + h}, \ldots, Y_{t_n + h})\stackrel{d}{=} f(X_{t_1 }, X_{t_2 }, \ldots, X_{t_n }, Y_{t_1 }, Y_{t_2 }, \ldots, Y_{t_n }) \tag{2}$$
Regarding part 1, here is a discrete-time counterexample:
Let $X_k$ be i.i.d. $\pm 1$ valued variables of mean zero, and let $\{W_k\}{k \in \mathbb Z}$ be an independent copy of $\{X_k\}_{k \in \mathbb Z}$. Define $Y_k=X_k$ for $k$ even and
$Y_k=W_k$ for $k$ odd. Then $\{X_k\}$ is a stationary sequence and so is $\{Y_k\}$. However, $\{Y_kX_k\}_{k \in \mathbb Z}$ is not stationary.
To get a continuous time example, pick $U$ uniform in $[0,1]$ and independent of the previous variables. For real $t$, let $X(t)=X_k$ iff $k \in \mathbb Z$ and $k-1+U < t \le k+U$. Define $Y(t)$ from $\{Y_k\}$ similarly, with the same $U$.
Regarding part 2:
Equation (1) holds because both the distributions of both sides have the form $\mu \times \nu$ with the same $\mu$ and $\nu$.
Equation (2) holds because we applied the same function to both sides of (1).