A stochastic process $\{X_{t}:t\in \mathbb{T}\}$ is strictly stationary if its finite dimensional distribution is stable under any time shift: for any $r\in\mathbb{T}$, any $t_{1},\dots, t_{n}\in\mathbb{T}$ with any $n\in\mathbb{Z}_{\geq 1}$, we have $$(X_{t_{1}+r},\dots, X_{t_{n}+r})=_{d}(X_{t_{1}},\dots, X_{t_{n}}).$$ In other words, for any Borel sets $B_{1},\dots, B_{n}\in\mathcal{B}(\mathbb{R})$, we have $$\mathbb{P}(X_{t_{1}+r}\in B_{1},\dots, X_{t_{n}+r}\in B_{n})=\mathbb{P}(X_{t_{1}}\in B_{1},\dots,X_{t_{n}}\in B_{n}).$$
Let us assume that $\mathbb{T}:=\mathbb{Z}$. Fix $m\in\mathbb{Z}_{\geq 1}$ and let $g:\mathbb{R}^{m}\longrightarrow\mathbb{R}$ be a Borel function. Define $Y_{t}:=g(X_{t},X_{t+1},\dots, X_{t+m-1})$. I want to show that
The stochastic process $\{Y_{t}:t\in\mathbb{T}\}$ is also strictly stationary.
I can show it for $m=1$, but I don't know how to do it for $m\geq 1$ because of the following issue:
Fixing $n\in\mathbb{Z}_{\geq 1}$, let $t_{1},\dots, t_{n},r\in \mathbb{T}$. Let $B_{1},\dots, B_{n}\in\mathcal{B}(\mathbb{R})$, then since $g$ is Borel measurable, $g^{-1}(B_{1}),\dots, g^{-1}(B_{n})\in\mathcal{B}(\mathbb{R}^{m})$. So, we can write
\begin{align*} \mathbb{P}&(Y_{t_{1}}\in B_{1},\dots, Y_{t_{n}}\in B_{n})\\ &=\mathbb{P}((X_{t_{1}},\dots, X_{t_{1}+m-1})\in g^{-1}(B_{1}),\dots, (X_{t_{n}},\dots, X_{t_{n}+m-1})\in g^{-1}(B_{n})). \end{align*} The thing now is that I don't know how to further write it in the form of $$\mathbb{P}(X_{t_{1}}\in A_{1,1},\dots X_{t_{n}+m-1}\in A_{n,m})\ \text{for some Borel sets}\ A_{1,1},\dots, A_{n,m}\in\mathcal{B}(\mathbb{R}).$$ Only from here I can use the definition of the strict stationarity.
I've thought of using projection map, but we won't have $\pi^{-1}(g^{-1}(B_{1})$, but $\pi(g^{-1}(B_{1})$ instead, which is not necessarily a Borel set. I am happy to use other definitions of "$=_{d}$", such as
For any Borel set $B\in\mathcal{B}(\mathbb{R}^{n})$, we have $$\mathbb{P}((X_{t_{1}+r},\dots, X_{t_{n}+r})\in B)=\mathbb{P}((X_{t_{1}},\dots, X_{t_{n}})\in B).$$ But I don't think that this will help me either.
I think that we can simplify the matter to $m=2$, then the result follows by an induction.
The definition $$ \forall n,\forall B\in\mathcal{B}(\mathbb{R}^{n}),\forall r\in\mathbb Z,\quad \mathbb{P}((X_{1+r},\dots, X_{n+r})\in B)=\mathbb{P}((X_{1},\dots, X_{n})\in B) $$ is more convenient.
In order to check stationarity of $(Y_t)$, take a Borel set $C$ of $\mathbb R^n$ and $r\in\mathbb Z$. Then $$ \left\{ \left(Y_{1+r},\dots,Y_{n+r}\right)\in C\right\} =\left\{\left( X_{1+r},\dots,X_{n+r+m}\right)\in B\right\}, $$ where $$ B=\left\{\left(u_1,\dots,u_{m+n}\right)\in\mathbb R^{m+n}, \left(g\left(u_1,\dots,u_{m+1}\right),\dots,g\left(u_{n+1},\dots,u_{n+m}\right)\right)\in C\right\}. $$ Then we conclude by stationarity of $(X_t)$: $$ \mathbb P\left(\left(Y_{1+r},\dots,Y_{n+r}\right)\in C\right)= \mathbb P\left(\left( X_{1+r},\dots,X_{n+r+m}\right)\in B\right)=\mathbb P\left(\left( X_{1},\dots,X_{n+m}\right)\in B\right)\\ =\mathbb P\left(\left(Y_{1},\dots,Y_{n}\right)\in C\right). $$