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On wikipedia I read that the expected number of steps till a 1D simple random walk hits either $a$ or $-b$ is equal to $ab$. (I have seen this result also on other websites.) However, no proof or further reference is given. Could someone please explain how they arrived at this result?

Thanks in advance, Claus

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marked as duplicate by Did probability Oct 31 at 12:41

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Have a look at… – coffeemath Feb 18 '13 at 16:16

1 Answer 1

The expected time $t_x$ to hit either boundary starting at $x$ satisfies the recurrence

$$ t_x=1+\frac12\left(t_{x-1}+t_{x+1}\right)\;. $$

The general solution of this second-order linear recurrence is $t_x=-x^2+mx+n$, and you can easily check that $t_x=(a-x)(x+b)$ is of this form and satisfies the boundary conditions $t_a=t_{-b}=0$; thus $t_0=(a-0)(0+b)=ab$.

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The recurrence is second order. – vonbrand Oct 30 at 14:54
@vonbrand: Quite right, thanks, I've fixed it. – joriki Oct 31 at 12:38

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