 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-1/29l/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Put

M_{n} = max_{1 \leq i \leq n} ∣ X_{i} ∣

. The likelihood is

L (θ) = \frac{1}{( 2 θ ) ^{n}} 1_{{θ \geq M_{n}}} .

It decreases on its nonzero range, so

θ = M_{n} .

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