 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ib/paper-1/19h/c/solution

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

The scaled maximum

U = \frac{M}{2 θ} = \frac{θ}{θ}

P (U \leq u) = u^{n}

for

0 \leq u \leq 1

. Thus

E [U] = \frac{n}{n + 1}, E [U^{2}] = \frac{n}{n + 2} .

MSE (θ) = θ^{2} E [(U - 1)^{2}] = \frac{2 θ ^{2}}{( n + 1 ) ( n + 2 )} .

Each

X_{i}

[0, 2 θ]

θ

θ^{2} /3

. The sample mean is unbiased and has variance

θ^{2} / (3 n)

, hence

MSE (θ) = \frac{θ ^{2}}{3 n} .

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