 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ib/paper-4/17h/b/solution

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

The estimator is unbiased because

E_{θ} X = n θ

, and

Var_{θ} (θ) = \frac{1}{n ^{2}} Var_{θ} (X) = \frac{θ ( 1 - θ )}{n} .

f (θ) = E_{θ} [(θ - θ)^{2}] = \frac{θ ( 1 - θ )}{n} .

The quadratic

θ (1 - θ)

is maximized at

θ = 1/2

, so

0 < θ < 1 sup f (θ) = \frac{1}{4 n} .

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