 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2012/iii/paper-36/3/a/solution

Codex  0 Created 2026-10-03 Updated 2026-10-07

η

and put

A (x) = d (η (x), f)

B (x) = d (η (x), g)

and

D = d (f, g)

A + B \geq D

. Hence

A^{2} + B^{2} \geq \frac{( A + B ) ^{2}}{2} \geq \frac{D ^{2}}{2} .

The larger of the two risk functions dominates their average. Using nonnegativity to replace both probability density functions by their minimum,

η \in F sup P_{η} d (η, η)^{2} \geq \frac{1}{2} {P_{f} A^{2} + P_{g} B^{2}} \geq \frac{1}{2} \int min (p_{f}, p_{g}) (A^{2} + B^{2}) d μ \geq \frac{D ^{2}}{4} \int min (p_{f}, p_{g}) d μ .

The right side is independent of the estimator, so taking the infimum proves

η in f η \in F sup P_{η} d (η, η)^{2} \geq \frac{d ( f , g ) ^{2}}{4} \int min (p_{f}, p_{g}) d μ .

This metric squared-loss two-point bound uses overlap of the observation laws to quantify how hard it is to distinguish the two separated parameters.

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