Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 223 / 1 / c / Solution

Every symmetric contaminated distribution has median zero, so

f (0) = (1 - ϵ) ϕ (0) + ϵ h (0) \geq (1 - ϵ) ϕ (0) .

A symmetric contaminating density supported away from zero attains equality. Hence

max_{F \in P_{ϵ} (Φ) F symmetric} A (T, F) = \frac{1}{4 ( 1 - ϵ ) ^{2} ϕ ( 0 ) ^{2}} = \frac{π}{2 ( 1 - ϵ ) ^{2}} .

This is strictly smaller than the unrestricted answer because

q_{ϵ} > 0

and

ϕ (q_{ϵ}) < ϕ (0)

. Asymmetric contamination can move the median into a region of lower nominal density.

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