 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2017/prelims/mathematics-iii/5/b/iv/solution

Starting from

X

n

, the markov property and the time-independent transition rules give

P (M_{2 n} = X ∣ M_{n} = X) = p_{n}

. Therefore

P (M_{n} = X ∣ M_{2 n} = X) = \frac{P ( M _{n} = X , M _{2 n} = X )}{p _{2 n}} = \frac{p _{n}^{2}}{p _{2 n}} .

Write

p_{n} = \frac{3}{4} (\frac{3}{4})^{n} (1 + \frac{1}{3} (- \frac{1}{3})^{n})

\frac{3}{4} \frac{( 1 + \frac{1}{3} ( - \frac{1}{3} ) ^{n} ) ^{2}}{1 + \frac{1}{3} ( \frac{1}{9} ) ^{n}} ⟶ \frac{3}{4} .

The conditioning selects paths which have avoided the absorbing state up to time

2 n

; it is therefore quite different from the unconditioned probability

p_{n}

, which tends to zero.

 New to topics? Read the docs here!