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Let $\{X_k\}_{k\in\mathbb{N}}$ be a stationary ergodic sequence on a probability space $(\Omega,\mathcal{F},P)$ with shift $T$. Also, let $\{v_k\}_{k\in\mathbb{N}}$ be a sequence of random variables in $\mathbb{N}$ such that $v_k$ is a function of $X_1,\ldots,X_{v_k-1}$ and $v_k>v_{k-1}$. Suppose $A\in\mathcal{F}$ is an event with $P(A)>0$.

I am trying to investigate the quantity $$ \limsup_{k\rightarrow\infty}P(T^{v_k}A), $$ and specifically whether extra assumptions are needed to ensure this quantity is nonzero.

In the case that $\{X_k\}_{k\in\mathbb{N}}$ is iid one can use \begin{align} P(T^{v_k}A) = \sum_{\ell=1}^{k}P(T^{v_k}A\mid v_k = \ell)P(v_k=\ell) \stackrel{!}{=} \sum_{\ell=1}^{k}P(T^{\ell}A)P(v_k=\ell) = P(A) > 0, \end{align} however with only the SE assumption I am stuck, since the conditioning on $v_k=\ell$ cannot be ignored at !. An alternative approach I am thinking of is using some form of LLN to get $$ \lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=1}^{n} 1_{\{T^{v_k}A\}} = P(A) > 0, $$ which would imply that the limsup is nonzero. However, this does not necessarily seem to hold. I am quite unexperienced at working with stationary ergodic sequences, any help is highly apreciated!

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Mixing may be sufficient (not sure).

Recall that a transformation is mixing if $\mu(T^{-k}A \cap B) \rightarrow \mu(A)\mu(B)$ as $k \rightarrow \infty $ for all $A$, $B$.

If $T$ is just weak-mixing, we still have $\mu(T^{-k}A \cap B) \rightarrow \mu(A)\mu(B)$ but only along a subsequence of density one, so I guess it does not hold.

In the ergodic case, I would try to build a counterexample by using an irrational rotation $R_\theta$ of the circle (identify the circle with $[0,1[$ and take $X_k(x) = R^k_\theta(x)$. Note that $X_1$ is just the identity). Take for $A$ some interval containing $0$ and cut it into subintervals of size $1/k$. Define $v_k$ on each of these subintervals so that under $T^{v_k}$, they end close to $[0,1/k[$. So they stack at time $v_k$, $A$ is sent close to $[0,1/k[$ by $T^{v_k}$ and the limsup should be 0. Note that $v_k$ depends only on $x=X_1(x)$.

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  • $\begingroup$ Thank you for your response. That counterexample idea will hopefully help me understand possible difficulties better. $\endgroup$
    – Marc
    Jul 28, 2016 at 12:44
  • $\begingroup$ In the full generality of your question one can find also counterexamples in some mixing systems (as that in general mixing does not imply mixing after one step). If your sequence is continued fraction mixing then your proof will work. Sequences that are known to be cf mixing are $\phi,\phi\circ T,....$ where $\phi$ is a nice function and $T$ is a Gibbs Markov map. If you restrict your question to more specific $v_k$ like stopping times then you might have a different picture. $\endgroup$
    – user78465
    Jul 31, 2016 at 10:52
  • $\begingroup$ Thank you for your response. The $v_k$ are a function of $X_1,\ldots,X_{v_{k-1}}$, so doesn't this imply they are stopping times, at least in the iid case? Does that make the problem easier? $\endgroup$
    – Marc
    Aug 1, 2016 at 12:46

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