Measurable function applied on stationary sequence Let $(X_t)_{t\in\mathbb{N}}$ be a strictly stationary sequence of real random variables and $f:\mathbb{R}\rightarrow\mathbb{R}$ a measurable function.
My simple question: Is $$ Y_t:=f(X_t)$$ also strictly stationary?
 A: Since $f$ is measurable, $f^{-1}(B)$ is Borel measurable for all Borel measurable $B\subseteq\mathbb R$. Thus, by definition of stationarity,
\begin{equation}
\begin{split}
\operatorname P\left[\left(Y_{t_0},\ldots,Y_{t_k}\right)\in B_0\times\cdots\times B_k\right]&=\operatorname P\left[\left(X_{t_0},\ldots,X_{t_k}\right)\in f^{-1}(B_0)\times\cdots\times f^{-1}(B_k)\right]\\&=\operatorname P\left[\left(X_{s+t_0},\ldots,X_{s+t_k}\right)\in f^{-1}(B_0)\times\cdots\times f^{-1}(B_k)\right]\\&=\operatorname P\left[\left(Y_{s+t_0},\ldots,Y_{s+t_k}\right)\in B_0\times\cdots\times B_k\right]
\end{split}
\end{equation}
for all $s\ge0$, $k\in\mathbb N_0$, $0\le t_0<\cdots<t_k$ and Borel measurable $B_0,\ldots,B_k\subseteq\mathbb R$.
A: Yes:

(WP) In mathematics, a stationary process (or strict(ly) stationary process or strong(ly) stationary process) is a stochastic process whose joint probability distribution does not change when shifted in time. Consequently, parameters such as the mean and variance, if they are present, also do not change over time and do not follow any trends...

...Consequently again, the distribution of every function of $(X_t,X_{t+1},\ldots,X_{t+n})$ does not depend on $t$, for every $n$, hence $Z_t=f(X_t,X_{t+1},\ldots,X_{t+n})$ defines a strictly stationary process $(Z_t)$, for every measurable function $f$.
