 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ib/paper-1/7h/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

F (x) = x^{2} / θ^{2}

[0, θ]

, so its median is

θ / 2

L (θ) = 2^{n} (\prod_{i} X_{i}) θ^{- 2 n} 1_{{θ \geq X_{(n)}}},

which decreases over its admissible range. Hence

θ = X_{(n)} = max_{i} X_{i}

m = \frac{X _{(n)}}{2} .

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