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

Let $m=F^{-1}(1/2)$ be the unique median. The influence function of the sample median isaway from $m$. Its second moment, equivalently the asymptotic distribution of a sample median, givesNo symmetry assumption is used: the density is evaluated at the actual median of $F$.

Write $F=(1-ϵ)\Phi +ϵH$ and assume $0<ϵ<1/2$. If $m$ is its median, thenMoreover its density satisfies $f(m)\ge (1-ϵ)\varphi (m)$. Since the standard normal density decreases with $|m|$, the smallest possible density at the median occurs at either endpoint. PutThe two endpoints are $\pm q_{ϵ}$ by normal symmetry. The bound is attained by choosing a contaminating density supported strictly to the right of $q_{ϵ}$, or symmetrically to the left of $-q_{ϵ}$, with zero density at the selected median. Therefore

Every symmetric contaminated distribution has median zero, soA symmetric contaminating density supported away from zero attains equality. HenceThis is strictly smaller than the unrestricted answer because $q_{ϵ}>0$ and $\varphi (q_{ϵ})<\varphi (0)$. Asymmetric contamination can move the median into a region of lower nominal density.