Hitting time

The following applies to both binomial (drunken man) and Wiener process:

Suppose we are at x = 0 and there are two absorbing barriers, a and -b (a,b>0). Then

p(absorbed at a) = b/(a+b)
p(absorbed at -b) = a/(a+b)
E[time until absorption] = ab

For a proof, refer to Zhou. Outline of the proof:

Let S_n be the Wiener process. Both S_n and (S_n)^2-N are martingales (N being the hitting time). Hence we have a set of 2 eqns with 2 unknowns p_a and E[N]:
E[S_n]=p_a*a+(1-p_a)*b=0
E[(S_n)^2-N]=p_a*(a^2)+(1-p_a)*(b^2)-E[N]=0