 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/iii/paper-36/2/d/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

Write

H_{k} = \sum_{j = 1}^{k} 1/ j

, with

H_{0} = 0

. The truncated uniform prior gives the posterior distribution

p (N ∣ y, M) = \frac{N ^{- 1}}{H _{M} - H _{y - 1}}, y \leq N \leq M .

E [N ∣ y, M] = \frac{M - y + 1}{H _{M} - H _{y - 1}} .

Each term in the numerator of the expectation is

N / N = 1

, which explains the exact count

M - y + 1

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