Let $m=F^{-1}(1/2)$. Since $F$ has an everywhere positive density, it is continuous and strictly increasing, so $F(m)=1/2$. Membership in the <Kolmogorov neighborhood of a distribution> gives
$$
\Phi(m)-\varepsilon\leq\frac12\leq\Phi(m)+\varepsilon.
$$
By the symmetry of the <standard normal distribution>,
$$
-\Phi^{-1}\left(\frac12+\varepsilon\right)
\leq m\leq
\Phi^{-1}\left(\frac12+\varepsilon\right).
$$
The stated asymptotic-bias formula for the <sample median> therefore yields
$$
\boxed{\sup_{F\in\mathcal P_\varepsilon^K(\Phi)\cap\mathcal M}
b(\{T_n\},F)\leq b_1},
\qquad
b_1=\Phi^{-1}\left(\frac12+\varepsilon\right).
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 223 / 1 / a / Solution

{c}
{title2=$\mathcal P_\varepsilon^K(F)$}

The Kolmogorov $\varepsilon$-neighborhood of a cumulative distribution function $F$ consists of all distribution functions $G$ satisfying
$$
\sup_t|G(t)-F(t)|\leq\varepsilon.
$$


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 Incoming links: Kolmogorov neighborhood of a distribution