= Solution
Every symmetric contaminated distribution has median zero, so
$$
f(0)=(1-\epsilon)\phi(0)+\epsilon h(0)
\geq(1-\epsilon)\phi(0).
$$
A symmetric contaminating density supported away from zero attains equality. Hence
$$
\max_{\substack{F\in\mathcal P_\epsilon(\Phi)\\F\text{ symmetric}}}
A(T,F)
=\frac1{4(1-\epsilon)^2\phi(0)^2}
=\frac{\pi}{2(1-\epsilon)^2}.
$$
This is strictly smaller than the unrestricted answer because $q_\epsilon>0$ and $\phi(q_\epsilon)<\phi(0)$. Asymmetric contamination can move the median into a region of lower nominal density.
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