# asymmetric random walk

do asymmetric random walks also return to the origin infinitely?

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No they do not. –  Did May 3 '11 at 19:57
Why not? Presumably because rand({0,1}) doesn't converge? –  Joseph Weissman May 3 '11 at 20:28
I'm writing out a proof for this, and I"m getting that $\sum f_n$ doesn't converge. –  gentisse May 3 '11 at 20:38

This is a consequence of the law of large numbers. The position $S_n$ at time $n$ is the sum of $S_0$ and of $n$ i.i.d. displacements, each with expectation $m\ne0$, hence $S_n/n\to m$ almost surely. In particular, $|S_n|\ge |m|n/2$ for every $n\ge N$ where $N$ is random and almost surely finite, which implies $S_n\ne0$. Since $(S_n)$ does not visit zero after time $N$, the number of visits of zero is almost surely finite. The starting point $S_0$ is irrelevant.

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No. Heuristic: If the walk goes right with probability $1/2+\alpha/2>1/2$ then the expected position after $n$ steps is $\alpha n,$ while the expected variation is only $O(\sqrt n).$ Thus the walk crosses the origin only finitely often.
Proof sketch: let $P(x,y)$ be the generating function of all walks which end up at the origin for the first time, with $x$ meaning left and $y$ meaning right. You can write a recurrence relation for the walks and deduce an expression for $P$ by solving a quadratic. Now substitute $pt$ for $x$ and $1-p$ for $y$.