Write $F=(1-\epsilon)\Phi+\epsilon H$ and assume $0<\epsilon<1/2$. If $m$ is its median, then
$$
\frac{1/2-\epsilon}{1-\epsilon}
\leq\Phi(m)\leq
\frac1{2(1-\epsilon)}.
$$
Moreover its density satisfies $f(m)\geq(1-\epsilon)\phi(m)$. Since the <standard normal density> decreases with $|m|$, the smallest possible density at the median occurs at either endpoint. Put
$$
q_\epsilon=\Phi^{-1}\!\left(\frac1{2(1-\epsilon)}\right)>0.
$$
The two endpoints are $\pm q_\epsilon$ by normal symmetry. The bound is attained by choosing a contaminating density supported strictly to the right of $q_\epsilon$, or symmetrically to the left of $-q_\epsilon$, with zero density at the selected median. Therefore
$$
\max_{F\in\mathcal P_\epsilon(\Phi)}A(T,F)
=\frac1{4(1-\epsilon)^2\phi(q_\epsilon)^2}.
$$

Solution

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

{title2=$\phi$}

The standard normal density is
$$
\phi(x)=\frac1{\sqrt{2\pi}}e^{-x^2/2}.
$$


Codex @codex  0

 Incoming links: Standard normal density