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Prove: In a random walk starting at the origin the probability to reach the position a > 0 before returning to the origin equals p(1 - q1).
This problem is taken from W. Feller, in Introduction to Probability Theory and Applications, Volume I, Chapter XIV, Section 9, problem 2 (a), page 367.
If the walker starts at the origin and goes to -1 at the first step, then the walk must return to the origin again before possibly reaching a > 0. Hence we need only consider the possibility of the walk starting from the point 1 at the first step, and then reaching the value a > 0 before returning to the origin. The probability of going to 1 is p, and then from 1 the subsequent independent probability of reaching a before returning to the origin is the same as the probability of the gambler achieving success, which is 1 - q1. Therefore the joint probability of the two events in succession is p(1 - q1).
Consider a special case just to check the results. Consider p = 1/2 = q. Then it is easy to compute that p(1 - q1) = (1/2) * (1 - (1 - 1/a)) = 1/(2a). Now take a = 2, and start from the origin. It is easy to see that the only way to go from the origin to the point a = 2 before returning to the origin is to go from 0 to 1, and then 1 to 2 in sequence, so from direct computation, the probability of such a path is 1/4. The formula gives the probability 1/(2a) = 1/4, so the formula agrees with direct calculation in this special case, and we are reassured!
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