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Does a symmetric random walk in- one dimension- converge almost surely? Can we prove or disprove it by martingales?

peanut
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No, it diverges almost surely, in the sense that for all $n\in\mathbb N$ and all $M\in\mathbb N$ there is almost surely an $n_+\gt n$ such that $X_{n_+}\gt M$ and an $n_-\gt n$ such that $x_{n_-}\lt-M$.

joriki
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  • What if step sizes form a conditionally convergent series? I don't think this is OP's question, but it made me wonder. (Maybe I should post another question.) – 2'5 9'2 Nov 29 '12 at 05:20
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    @alex: See Byron's paper on the random harmonic series, and references therein. – joriki Nov 29 '12 at 05:40
  • Step sizes are by definition nonnegative: how could they form a conditionally convergent series? – Robert Israel Nov 29 '12 at 06:41
  • @Robert: I think the idea was that the sum of the step sizes doesn't converge, so the sum of the signed steps isn't absolutely convergent but might be conditionally convergent with probability $1$, as in the case of the random harmonic series. – joriki Nov 29 '12 at 06:59