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.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-223/1/c/solution