Expected number of steps till a random walk hits a or -b.

Question

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

Have a look at http://math.stackexchange.com/questions/288298/symmetric-random-walk-with-bounds?rq=1 — coffeemath, Feb 18 '13 at 16:16

joriki · Answer 1 · 2015-10-31T12:38:39.667

10

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 $t_0=(a-0)(0+b)=ab$.

edited Oct 31 '15 at 12:38

answered Feb 18 '13 at 17:12

joriki

238,052

The recurrence is second order. – vonbrand Oct 30 '15 at 14:54
1

@vonbrand: Quite right, thanks, I've fixed it. – joriki Oct 31 '15 at 12:38

Expected number of steps till a random walk hits a or -b.

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