 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2017/discrete-mathematics-continuous-mathematics-and-probability/8/b/iv/solution

P (M_{n} = X, M_{2 n} = X) = p_{n} P (M_{2 n} = X ∣ M_{n} = X) = p_{n}^{2} .

Since

p_{2 n} > 0

P (M_{n} = X ∣ M_{2 n} = X) = \frac{p _{n}^{2}}{p _{2 n}} = \frac{( 3 ^{n + 1} + ( - 1 ) ^{n} ) ^{2}}{4 ( 3 ^{2 n + 1} + 1 )} .

The explicit formula gives

p_{n} \sim \frac{3}{4} (3/4)^{n}

, so

n \to \infty lim P (M_{n} = X ∣ M_{2 n} = X) = \frac{3}{4} .

The conditioning event has positive probability at each finite time even though those return probabilities tend to zero. Taking an undefined conditional on an infinite-time zero-probability event is not needed.

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