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Setup: Suppose $X = \{X_n\}_{n\in\mathbb{Z}}$ is a stationary ergodic proces on the real line and let $A = \prod_{n\in\mathbb{Z}}A_n$ be a Borel measurable set such that $$ P(X \in A) = P\left(X_n\in A_n \quad\forall n\in\mathbb{Z}\right) > 0. $$ Finally, let $\tau:\mathbb{R}^\mathbb{Z}\rightarrow \mathbb{R}^\mathbb{Z}$ denote the shift.

Question: I need the result that, almost surely, there are infinitely many $k\in\mathbb{Z}$ such that $\tau^k(X) \in A$. My question is two-fold

  1. Is my short proof below correct.
  2. Can the result also be shown by using more basis results on ergodic sequences, like Poincarre's recurrence theorem.

Thank you in advance!

My proof: This uses Birkhoff's ergodic theorem and the fact that measurable functions of stationary ergodic sequences are also stationary ergodic. Define the sequence $\{Y_n\}_{n\in\mathbb{Z}}$ as $Y_n = 1 $ if $\tau^n(X) \in A$ and zero otherwise. Then $\{Y_n\}_{n\in\mathbb{Z}}$ is stationary ergodic and thus $$ \lim_{k\rightarrow\infty}\frac{1}{k}\sum_{n=0}^{k-1}Y_n = E(Y) = P(A) > 0, $$ which implies infinitely many of the $Y_n$ have to be equal to one.

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Yes it works.

No, this can't be deduced from the Poincare recurrence theorem. If it was possible, the ergodic hypothesis would not be needed. But without the ergodic hypothesis, it is easy to give a counterexample. Just take a constant sequence of random variables $X_n = X_0$ and constant targets $A_n =A_0$. Then you can get in $A$ only if you are in $A$ from the start.

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