The <Huber location estimator> uses
$$
\psi_k(u)=\operatorname{clip}(u,-k,k),
\qquad
\psi_k'(u)=\mathbf1_{\{|u|<k\}}
$$
away from the two corners. At $F=N(\theta,1)$, symmetry gives $T(F)=\theta$, and for $Z\sim N(0,1)$,
$$
\mathbb E\psi_k'(Z)=\mathbb P(|Z|<k)=2\Phi(k)-1.
$$
Hence
$$
\operatorname{IF}(x;T,F)
=\frac{\operatorname{clip}(x-\theta,-k,k)}{2\Phi(k)-1},
$$
which is bounded in $x$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 223 / 3 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-223/3/b/solution