Put $q_s=F^{-1}(s)$ and $k=-F^{-1}(\alpha)>0$. Symmetry gives $F^{-1}(1-\alpha)=k$. Differentiating the <trimmed mean> functional along $F_t=(1-t)F+t\Delta_x$ gives
$$
\operatorname{IF}(x;T,F)
=\frac1{1-2\alpha}
\int_\alpha^{1-\alpha}
\frac{s-\mathbf1_{\{x\leq q_s\}}}{f(q_s)}\,ds.
$$
Under the substitution $y=q_s$, symmetry implies
$$
\int_\alpha^{1-\alpha}\frac{s}{f(q_s)}\,ds
=\int_{-k}^kF(y)\,dy=k,
$$
while
$$
\int_\alpha^{1-\alpha}
\frac{\mathbf1_{\{x\leq q_s\}}}{f(q_s)}\,ds
=\int_{-k}^k\mathbf1_{\{x\leq y\}}\,dy.
$$
Evaluating the last integral in the three regions $x<-k$, $|x|\leq k$, and $x>k$ yields
$$
\boxed{
\operatorname{IF}(x;T,F)
=\frac1{1-2\alpha}
\begin{cases}
-k,&x<-k,\\
x,&|x|\leq k,\\
k,&x>k.
\end{cases}}
$$
This is the <influence function of a trimmed mean>, namely the <Huber score> with clipping parameter $k$, divided by $1-2\alpha$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 223 / 2 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-223/2/a/solution