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I am tying to analyze a random walk on an integer lattice $\mathbb{Z}^k$. For $k=1$, what is the probability that after $n$ steps the drunkard's distance from the origin is lower than $\sqrt{n}$?

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1 Answer 1

The exact values $$ 2^{-n}\sum_{k=\frac12(n-\sqrt{n})}^{k=\frac12(n+\sqrt{n})}{n\choose k} $$ are not easily computed except for small values of $n$ but their limit when $n\to\infty$ is known and given by the gaussian approximation $$ \sqrt{2/\pi}\int_0^1\mathrm e^{-x^2/2}\mathrm dx=\mathrm{erf}(1/\sqrt2)=0.682689... $$

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since this is a "base" question, can you please elaborate or provide a reference for your solution? (i am interested in this simple case, and for k>1) –  MrRoth Feb 26 '12 at 11:13
I did. Please check the link. –  Did Feb 26 '12 at 11:17
not talking about the approximation. about the first formula. doesn't the formula refer to the probability of a distance greater than $\sqrt{n}$ –  MrRoth Feb 26 '12 at 11:19
Some halves were missing, sorry about that. The argument is that, if $k$ steps are to the right and $n-k$ to the left, the position is $2k-n$, and one wants this to be between $-\sqrt{n}$ and $+\sqrt{n}$. –  Did Feb 26 '12 at 11:29
You asked for the probability to be at distance at most $\sqrt{n}$ after $n$ steps. The formula in my post provides that. –  Did Feb 27 '12 at 17:31

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