Past exam of the probability, statistics and options I course of the University of Oxford / 2011 / M / 2 / d

Table of contents
- Solution d

Solution

With probability

2/3

, the chain starts at

+ 1

and

T = 1

. Conditional on starting at

- 1

, it takes a geometric number

G

of transitions to first reach

+ 1

, with success probability

1 - p

, so

T = 1 + G

. The probability generating function is therefore

E [s^{T}] = \frac{2}{3} s + \frac{1}{3} s \frac{( 1 - p ) s}{1 - p s} = \frac{2}{3} s + \frac{( 1 - p ) s ^{2}}{3 ( 1 - p s )} .

q

is absent because the process stops at its first visit to

+ 1

. At

s = 1

the expression equals one, confirming eventual hitting with probability one.

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