 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ib/paper-4/9h/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

E Z = n = 0 \sum \infty P (X_{n} = Y_{n}) = n = 0 \sum \infty x \in Z \sum p_{0 x} (n)^{2} = n = 0 \sum \infty x \in Z \sum p_{0 x} (n) p_{x 0} (n) .

p_{00} (2 n)

, so

E Z = n = 0 \sum \infty p_{00} (2 n) .

0

only at an even time. Since it is recurrent at

0

, part (a) gives

\sum_{m = 0}^{\infty} p_{00} (m) = \sum_{n = 0}^{\infty} p_{00} (2 n) = \infty.

Therefore

E Z = \infty

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