# Existence of an almost surely convergent subsequence.

Edit: I slightly relaxed the question to positive probability instead of almost surely.

Let $(\Omega,\mathcal{F},P)$ be a probability space. I have a sequence of random variables $(X_n)_{n\in\mathbb{N}}$ and a random variable $X$ such that \begin{align} \sup_{n}|X_n(\omega)| \le |X(\omega)| \qquad\forall\omega\in\Omega. \end{align} I also have the property \begin{align} P(X_{n+1} = X_n) > \frac{n}{n+1}. \end{align} I am interested whether the probability of the existence of a converging subsequence of $(X_n)_{n\in\mathbb{N}}$ is nonzero. That is, whether there exists a subsequence $(n_k)_{k\in\mathbb{N}}$ such that \begin{align} P\left(\lim_{k\rightarrow\infty}X_{n_k}\text{ exists}\right)>0. \end{align}

Thoughts:

• It is enough to show that there is a subsequence that converges in probability since that implies the existence of a further subsequence that converges almost surely.
• By the first property and the fact that $\mathbb{R}$ is separable and complete we can use Prohorov's theorem to ensure the existence of a subsequence that converges in distribution.
• I could potentially assume $E(|X|)<\infty$, so that the family $(X_n)_{n\in\mathbb{N}}$ is uniformly integrable, but I am not sure whether that helps.

Thank you in advance!

## 1 Answer

Not necessarily.

Let $\xi_i$ be independent Rademacher where $P(\xi_i = 1) = (2i-1)/2i$, $P(\xi_i = -1) = 1/2i$. Then set $X_n = \prod_{i=1}^n \xi_i$, so $X_n = X_{n+1}$ iff $\xi_{n+1} = 1$. Think of $X_n$ as the state of a light switch at time $n$, and $\xi_i$ as indicating whether the switch was toggled at time $i$.

Given a subsequence $X_{n_k}$, let $Y_{k} = X_{n_{k+1}}/X_{n_k} = \prod_{i=n_{k}+1}^{n_{k+1}} \xi_i$ indicate whether the switch was toggled between times $n_k$ and $n_{k+1}$, so $X_{n_k} = X_{n_{k+1}}$ iff $Y_{k} = 1$. Note that the $Y_k$ are independent.

Now note that $E[\xi_i] = 1 - 1/(4i)$, so we have \begin{align*}E[X_m / X_n] &= \prod_{i=n+1}^m 1 - \frac{1}{4i} \\ &= \exp\left(\sum_{i=n+1}^m \ln\left(1-\frac{1}{4i}\right)\right) \\ &\le \exp\left(\sum_{i=n+1}^m -\frac{1}{4i} \right)\end{align*} using the fact that $\ln(1+x) \le x$. By divergence of the harmonic series, for each fixed $n$ this approaches 0 as $m \to \infty$. And we also have $E[X_m/X_n] \ge 0$, so this means $E[X_m / X_n] \to 0$ and therefore $P(X_m / X_n = \pm 1) \to 1/2$.

So if we pass to a subsequence such that each $n_{k+1}$ is sufficiently larger than $n_k$, we can ensure that $P(Y_k = -1) \ge 1/4$ for all $k$. By the second Borel–Cantelli lemma, this implies that $Y_k =-1$ infinitely often, almost surely. In other words, the sequence $X_{n_k}$ oscillates between $\pm 1$, and therefore diverges, almost surely.

Note you are right at the edge of success here: if you had $P(X_{n+1} = X_n) > 1 - 1/n^{1+\epsilon}$, the first Borel–Cantelli lemma would give you that $X_n$ is eventually constant, almost surely, so you have a.s. convergence of the whole sequence.

• Thank you for the great answer! – Marc May 23 '17 at 8:29