 Solution

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

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

θ_{0} = 1/2

and

z = (θ - 1/2)^{2}

. Since

θ (1 - θ) = \frac{1}{4} - z, 0 \leq z < \frac{1}{4},

the risk is affine in

z

g_{w, 1/2} (θ) = \frac{w ^{2}}{4 n} + z ((1 - w)^{2} - \frac{w ^{2}}{n}) .

Its supremum is therefore controlled by the midpoint

z = 0

and the endpoint limit

z \to 1/4

. Requiring both endpoint values to be at most the MLE's maximal risk

1/ (4 n)

gives

\frac{w ^{2}}{4 n} \leq \frac{1}{4 n}, \frac{( 1 - w ) ^{2}}{4} \leq \frac{1}{4 n} .

Equivalently,

∣ w ∣ \leq 1, ∣1 - w ∣ \leq \frac{1}{n} .

Their intersection is

1 - \frac{1}{n} \leq w \leq 1 .

For precisely this range, the shrinkage estimator has maximal mean squared error no greater than that of

θ

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