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

The normal location score is $x-\theta$. Under a bound on <gross-error sensitivity>, the variance-minimizing influence curve clips this score. For $b>0$, define the <Huber score>
$$
\psi_b(u)=[u]_{-b}^{b}
=\max(-b,\min(u,b)).
$$
The optimal <B-robust estimator> is the <Huber location estimator> $\widehat\theta_b$ defined by
$$
\sum_{i=1}^n\psi_b(x_i-\widehat\theta_b)=0.
$$
Equivalently, it has the explicit optimization form
$$
\widehat\theta_b
=\underset{t\in\mathbb R}{\arg\min}
\sum_{i=1}^n\rho_b(x_i-t),
$$
where the <Huber loss> is
$$
\rho_b(u)=
\begin{cases}
u^2/2,&|u|\leq b,\\
b|u|-b^2/2,&|u|>b.
\end{cases}
$$
At $N(\theta,1)$ its normalized influence function is
$$
\operatorname{IF}(x;T_b,F_\theta)
=\frac{\psi_b(x-\theta)}{\mathbb P(|Z|\leq b)},
\qquad Z\sim N(0,1).
$$

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 223 / 2 / a / Solution

{c}

The Huber score clips a residual at a fixed threshold:
$$
\psi_k(u)=\max(-k,\min(u,k)).
$$


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 Incoming links: Huber score