I have a strictly stationary process $(X_t)_{t\in \mathbb{Z}}$. Is it then true that the vector-valued process $\left(\begin{pmatrix}X_t\\X_t^2\end{pmatrix}\right)_{t\in \mathbb{Z}}$ is also strictly stationary?
My reasoning is that since the mapping $x \mapsto \begin{pmatrix}x\\x^2\end{pmatrix}$ is measurable, strict stationarity is preserved. I just wanted to have this sanity checked.
If you are not sure, then why do you not try to write down a formal proof?
Fix any $t_1,\ldots,t_n \in \mathbb{Z}$ and $r \in \mathbb{Z}$. By the stationarity of $(X_t)_{t \in \mathbb{Z}},$
$$\begin{align*} \mathbb{P}(h(X_{t_1+r}) \in B_1,\ldots,h(X_{t_n+r}) \in B_n) &= \int_{\mathbb{R}^n} \prod_{j=1}^n 1_{B_j}(h(x_j)) \, d\mathbb{P}_{X_{t_1+r},\ldots,X_{t_n+r}}(x_1,\ldots,x_n) \\ &=\int_{\mathbb{R}^n} \prod_{j=1}^n 1_{B_j}(h(x_j)) \, d\mathbb{P}_{X_{t_1},\ldots,X_{t_n}}(x_1,\ldots,x_n) \\ &= \mathbb{P}(h(X_{t_1}) \in B_1,\ldots,h(X_{t_n}) \in B_n) \end{align*}$$
for any measurable mapping $h:\mathbb{R} \to \mathbb{R}^k$ and any Borel sets $B_1,\ldots,B_n \in \mathcal{B}(\mathbb{R}^k)$. Hence,
$$(h(X_{t_1+r}),\ldots,h(X_{t_n+r})) \stackrel{d}{=} (h(X_{t_1}),\ldots,h(X_{t_n})),$$
i.e. $(h(X_t))_{t \in \mathbb{Z}}$ is strictly stationary.