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

The <gross-error sensitivity> is $\gamma^*(T,F)=\sup_x|\operatorname{IF}(x;T,F)|$. At $N(\theta,1)$, the <influence function of the sample median> has magnitude $1/(2\varphi(0))=\sqrt{\pi/2}$, so
$$
\gamma^*(\widehat\theta_{\mathrm{med}})=\sqrt{\frac\pi2}.
$$
For the <Huber location estimator>, $\psi_k(u)=\max(-k,\min(u,k))$ and $\mathbb E\psi_k'(Z)=\mathbb P(|Z|\leq k)=2\Phi(k)-1$, giving
$$
\gamma^*(\widehat\theta_{\mathrm{Hub},k})
=\frac{k}{2\Phi(k)-1}.
$$
For the symmetric normal law, the trimmed population mean is $\theta$. Writing $q_\gamma=\Phi^{-1}(1-\gamma)$ in the supplied influence function gives
$$
\gamma^*(\widehat\theta_{\mathrm{trim},\gamma})
=\frac{q_\gamma}{1-2\gamma}.
$$

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 223 / 1 / a / Solution

{c}

The Huber location estimator solves $\sum_i\psi_k(x_i-\theta)=0$, where $\psi_k$ clips its argument to $[-k,k]$.


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