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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?

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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 $\mu$.

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    $\begingroup$ This yields $X$ and $T$ such that the distributions of the sequences $(X\circ T^n)_n$ and $(X_n)_n$ coincide (which is the reason why the statement in the question should be modified). $\endgroup$ – Did Apr 19 '13 at 17:05
  • $\begingroup$ And of course: +1. $\endgroup$ – Did Apr 19 '13 at 17:59
  • $\begingroup$ thank you for your reply. But I want to know where the prerequisite "polish space" works. $\endgroup$ – lapuda Apr 20 '13 at 9:37
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    $\begingroup$ @lapuda One needs a theorem to ensure the existence of the measure $\mu$ in David's answer. For instance Kolmogorov's consistency theorem under the Polish assumption. $\endgroup$ – Stéphane Laurent Apr 23 '13 at 20:22
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    $\begingroup$ @lapuda That sounds right but they are some points to check: measurability of $f$ and the fact that the coordinates process has the same finite-dimensional distributions as $(X_n)$. I have not tried to do this exercise. $\endgroup$ – Stéphane Laurent Apr 24 '13 at 17:31

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