Suppose $\{X_n\}_{n\geq 1}$ is a strict stationary process, that is for every $n$, $(X_0,\ldots,X_n)$ and $(X_1,\ldots,X_{n+1})$ have the same distribution.
Then there is a probability space $(\Omega,\mathcal{F},P)$, a r.v. $X$ and a measure-preserving transformation $T$ on $\Omega$ such that $X_n(\omega)= X(T^n(\omega))$ holds for all $n$.
I wonder why this is true when the $\{X_n\}$ have there values in a Polish space. can anyone explain this? Does the Kolmogorov extension theorem applies?
Let $(S,d)$ be the Polish space where the $X_n$ take their values. Let $f\colon \Omega\to S^{\Bbb N_0}$ be defined by $f(\omega):=(X_n(\omega),n\geqslant 1)$. There is a measure $\mu$ such that $\mu(A):=P\{\omega,f(\omega)\in A\}$ for all $A\in\mathcal B(S^{\Bbb N_0})$. Let $T\colon S^{\Bbb N_0}\to S^{\Bbb N_0}$ defined by $T((x_n,n\geqslant 1))=(x_{n+1},n\geqslant 1)$. Then $T$ preserves $\mu$.