Random Walk Limit Behavior

Question

Suppose $\{ X_t \}$ is a sequence of i.i.d. random variables, with support $\{-1,1\}$ and distribution $P(1)=P(-1)=1/2$. Thus, $S_t = \sum_{s=1}^{t} X_s$ is a zero mean random walk. Also, $S_t$ is a martingale, but the conditions for Doob's martingale convergence theorem do not apply. What is it possible to say about the limiting behavior of $S_t$?

Answer 1

For any random walk on $\mathbb{R}$ there are only four possibilities. Exactly one of the following happens with probability one.

1. $S_n = 0$ for all $n$
2. $S_n \to \infty$
3. $S_n \to -\infty$
4. $-\infty = \liminf S_n < \limsup S_n = \infty$

This is because $\limsup S_n$ is an exchangeable random variable, meaning reordering finitely many of the $X_i$ doesn't change it's value.

In your case we end up in option 4. since clearly we are not in 1., and by symmetry if we were in 2. then we should also be in 3., so it must be that we are in 4.