[an error occurred while processing this directive]

Stochastic Processes and

Advanced Mathematical Finance

Dunbar, Fall 2010

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 - q_{1}).

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 - q_{1}.
Therefore the joint probability of the two events in
succession is p(1 - q_{1}).

Consider a special case just to check the results. Consider
p = 1/2 = q. Then it
is easy to compute that p(1
- q_{1}) =
(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!

Back to main section with problem statement.

[an error occurred while processing this directive] [an error occurred while processing this directive]

Last modified: [an error occurred while processing this directive]