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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 $S_t$?

Jong
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1 Answers1

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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.

nullUser
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  • Thanks! So is it possible to conclude that $P(S_n \rightarrow \infty)>0$? – Jong Aug 16 '15 at 04:30
  • That is correct, $P(S_n \to \infty) = 0$ because case 2 and case 4 are disjoint, and case 4 happens with probability $1$. – nullUser Aug 16 '15 at 04:32
  • To clarify, case 4 means that, given a probability space $\Omega$, the event ${ \omega|S_n(\omega) \text{ does not converge } }$ occurs with probability 1. In other words, $S_n$ has a limit on a measure 0 set? – Jong Aug 16 '15 at 04:47
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    That is correct. recall that $S_n(\omega)$ converges if and only if $\limsup S_n(\omega) = \liminf S_n(\omega)$. Since 4. occurs with probability $1$, this implies that with probability $1$, $S_n$ does not converge. – nullUser Aug 16 '15 at 04:50