Convergence of series of expectations of indicator functions.

Question

Let $\{X_n\}_{n\in\mathbb{N}}$ be a sequence of stationary ergodic random variables. Let $A$ be an event such that $p = P(X_1\in A)>0$.

Question: Does the following series converge: \begin{align} \sum_{k=1}^{\infty}E\left(\prod_{n=1}^{k}1_{\{X_n\in A\}}\right). \end{align}

Notes: The potential difficulty lies in the stationary ergodicness instead of independence. If $\{X_n\}_{n\in\mathbb{N}}$ is an iid sequence, then \begin{align} \sum_{k=1}^{\infty}E\left(\prod_{n=1}^{k}1_{\{X_n\in A\}}\right) = \sum_{k=1}^{\infty}\prod_{n=1}^{k}E\left(1_{\{X_n\in A\}}\right) = \sum_{k=1}^{\infty}\prod_{n=1}^{k}p = \sum_{k=1}^{\infty}p^k<\infty. \end{align}

Answer 1

Using the representation of a stationary sequence by dynamical systems, the question is equivalent to the following one.

Let $\left(\Omega,\mathcal A,\mu,T\right)$ be an ergodic dynamical system, let $A\in\mathcal A$ such that $\mu\left(A\right)\lt 1$ and let $p_n:=\mu\left(\bigcap_{i=1}^nT^{-i}A\right)$. Is the sequence $\left(p_n\right)_{n\geqslant 1}$ summable?

What is true is that $p_n\to 0$. Indeed, let $B:=\bigcap_{i\geqslant 1}T^{-i}A$. Then one can check that the sets $ T^{-l}\left(B\Delta T^{-1}B\right)$, $l\geqslant 1$ are pairwise disjoint, hence $\mu\left(B\Delta T^{-1}B\right)=0$. By ergodicity, $\mu\left(B\right)=0$.

However, it may be not true that the sequence $\left(p_n\right)_{n\geqslant 1}$ is summable. We shall use a variation of Lemma 3. 9 in Lesigne, Emmanuel; Volný, Dalibor Large deviations for martingales. (English summary) Stochastic Process. Appl. 96 (2001), no. 1, 143–159.

Lemma. Let $\left(\Omega,\mathcal A,\mu,T\right)$ be an aperiodic dynamical system. Let $\left(n_k\right)_{k\geqslant 1}$ be an increasing sequence of integers and $\left(\delta_k\right)_{k\geqslant 1}$ a sequence of numbers in $(0,1)$ such that for any $k\geqslant 1$, $$\tag{*}n_k\sum_{l\gt k}\delta_l\leqslant \frac 1{16}\delta_k.$$ Then there exists pairwise disjoint set $A_k$, $k\geqslant 1$ such that for any $k\geqslant 1$, $\mu\left(A_k\right)\leqslant \delta_k$ and $\mu\left(\bigcap_{i=1}^{n_k}T^{-i}A_k\right)\geqslant \delta_k/2$.

Sketch of proof. For each $k$, by Rokhlin's lemma, there exists a set $B_k\in\mathcal A$ such that $$\frac{3}{16}\frac{\delta_k}{n_k}\leqslant \mu\left(B_k\right)\leqslant \frac{1}{4}\frac{\delta_k}{n_k}$$ and the sets $T^{-i}B_k$, $1\leqslant i\leqslant 4n_k$ are pairwise disjoint. Then define $C_k:=\bigcup_{j=1}^{4n_k}T^{-j}B_k$ and finally let $A_k:=C_k\setminus \bigcup_{l\gt k}C_l$.

• Since $\mu\left(C_k\right)\leqslant \delta_k$, we have $\mu\left(A_k\right)\leqslant\delta_k$.
• Pairwise disjointness follows by definition.
• We have $$\mu\left(\bigcap_{i=1}^{n_k}T^{-i}A_k\right) \geqslant \mu\left(\bigcap_{i=1}^{n_k}T^{-i}C_k\right)-n_k\sum_{l\gt k}\mu\left(C_l\right)$$ and using $(*)$ and the inclusion $\bigcup_{j=n_k+1}^{4n_k}T^{-j}B_k\subset\bigcap_{j=1}^{n_k}T^{-j}C_k$, we get the third property.

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Now, let $A:=\bigcup_{k=1}^{\infty}A_k$ and let $\delta_k:=1/n_k$. Then we have to find an increasing sequence of integers $\left(n_k\right)_{k\geqslant 1}$ such that $16n_k^2\sum_{i\gt k}n_i^{-1}\leqslant 1$ and the series $\sum_{n=1}^{\infty}p_n$ diverges. Observe that \begin{align} \sum_{n=n_{k}+1}^{n_{k+1}}\mu\left(\bigcup_{i=1}^nT^{-i}A\right) &\geqslant \sum_{n=n_{k}+1}^{n_{k+1}}\mu\left(\bigcup_{i=1}^nT^{-i}A_{k+1}\right)\\ &\geqslant \sum_{n=n_{k}+1}^{n_{k+1}}\mu\left(\bigcup_{i=1}^{n_{k+1}}T^{-i}A_{k+1}\right)\\ &=\left(n_{k+1}-n_k\right)\mu\left(\bigcup_{i=1}^{n_{k+1}}T^{-i}A_{k+1}\right)\\ &\geqslant \left(n_{k+1}-n_k\right)\frac{1}{2n_{k+1}}. \end{align}

Now, one can choose $n_k=2^km_k$, where $m_k$ is such that $16\cdot 2^{k+1}m_k^2\leqslant m_{k+1}$. In this way, $$16n_k^2\sum_{i\gt k}n_i^{-1} \leqslant 16\cdot 2^{2k}m_k^2\sum_{i\gt k}2^{-i}m_{k+1}^{-1}\leqslant 1.$$ Since $n_k/n_{k+1}\to 0$, the quantity $\sum_{n=n_{k}+1}^{n_{k+1}}p_n$ can be made bigger than $1/4$ for $k$ large enough.