Let $m=F^{-1}(1/2)$ be the unique <median>. The <influence function of the sample median> is
$$
\operatorname{IF}(x;T,F)
=\frac{\mathbf1_{\{x>m\}}-\mathbf1_{\{x<m\}}}{2f(m)}
$$
away from $m$. Its second moment, equivalently the <asymptotic distribution of a sample median>, gives
$$
A(T,F)=\mathbb E_F[\operatorname{IF}(X;T,F)^2]
=\frac1{4f(F^{-1}(1/2))^2}.
$$
No symmetry assumption is used: the density is evaluated at the actual median of $F$.

Solution

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

 Solution

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