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Math 489/Math 889
Stochastic Processes and
Advanced Mathematical Finance
Dunbar, Fall 2010


Problem Statement

If q > p, conclude from the preceding problem: In a random walk starting at the origin, the number of visits to the point a > 0, that take place before the first return to the geometric distribution with ratio 1 - qqa-1. (Why is the condition q > p necessary?)

Solution

This problem is adapted from W. Feller, in Introduction to Probability Theory and Applications, Volume I, Chapter XIV, Section 9, problem 3, page 367.

This problem is conceptually and computationally simpler if p = 1/2 = q. It is convenient to first calculate and record some probabilities, then solve the problem.

Starting from the origin, the probability to reach a > 0 before returning to the origin is p(1 - q1) = p(1 - (1 - 1/a)) = p(1/a) = 1/(2a).

Starting at a, the probability to reach the origin before returning to the starting point is qqa-1 = q(1 - (a - 1)/a) = q(1/a) = 1/(2a). This makes sense, since this situation is symmetric with the previous situation and so should have the same probability.

The ratio for the geometric probability is supposed to be 1 - qqa-1 = 1 - q(1 - (a - 1)/a) = 1 - q(1/a) = 1 - 1/(2a) = (2a - 1)/(2a).

The probability of 0 visits to a before hitting the origin is the complement of the probability of one (or more) visits to a before hitting the origin. Since the random walk is recurrent (that is, will hit every point eventually from any starting point, this is where the necessity of p = q = 1/2 comes in!) the probability of hitting the origin again from a (perhaps after more visits to a) is certain, so the probability of one (or more) visits to a before hitting the origin is the same as the probability of a first visit to a, namely p(1 - q1) = 1/(2a). Then the desired probability of 0 visits is (1 - 1/(2a)) = (2a - 1)/(2a).

The probability of exactly one visit to the origin is the probability of passing from the origin to a without first returning to the origin, followed by passing from a to the origin without first returning to a. This is

 2 p(1 - q1)qqa-1 = (1/(2a))(1/(2a)) = 1/(4a ).

The probability of exactly two visits to a before returning to the origin is the probability of passing from the origin to a without first returning to the origin, followed by a bridge from to a to a without touching the origin, followed by passing from a to the origin without first returning to a. The middle probability is the same as the already computed probability of passing exactly 0 visits to a from the origin, by the symmetry of the situation. This is

p(1 - q1)((2a- 1)/(2a))qqa-1 = (1/(2a))(1/(2a)) = 1/(4a2)(2a - 1)/(2a).

Now the pattern is clear, and we see that the number of visits to a with out first returning to the origin is a geometric random variable with ratio (2a - 1)/(2a).


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