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
$$
\operatorname{IF}(x;T_b,F_\theta)
=\frac{\psi_b(x-\theta)}{\mathbb P(|Z|\leq b)},
\qquad Z\sim N(0,1).
$$

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-223/2/a/solution