Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 223 / 2

For residual $r=y_i-x_i^{\top }\theta$, minimizeover $\gamma$. Soft thresholding gives $\hat{\gamma }=\mathrm{sgn}(r)(|r|-k\sigma {)}_{+}$, and the minimized value isThis is $\sigma {\rho }_k(r/\sigma )$ for the usual Huber loss. Multiplication by the positive constant $\sigma$ does not change the minimizing $\theta$, proving equivalence.

Solved by gpt-5.6-sol high.

For fixed residual $r=y_i-x_i^{\top }\theta$, choosing ${\gamma }_i=0$ costs $r^2$, while choosing ${\gamma }_i=r$ costs $\ell$. No other nonzero choice improves on ${\gamma }_i=r$. Thuswith $L=\sqrt{\ell }$. Summing over observations proves the equivalence.

Solved by gpt-5.6-sol high.

Away from the two nondifferentiable cutoffs, the skipped-mean score is ${\psi }_L(u)=2u{1}_{\{|u|\le L\}}$. The estimating equation is therefore

Solved by gpt-5.6-sol high.

Choose $v\in {\mathbb{R}}^p$ outside the finitely many hyperplanes $\{v:x_i^{\top }v=0\}$. Then $x_i^{\top }v\ne 0$ for every $i$. For ${\theta }_t=tv$, every residual $y_i-t{x}_i^{\top }v$ eventually has absolute value greater than $L$. Every indicator in the estimating equation is then zero, so the equation is satisfied for every sufficiently large $t$.

Thus arbitrarily large solutions already exist without contamination. Under the definition in the question, the skipped-mean regression estimator has breakdown point zero. This is a standard pathology of an exactly redescending score when every root is admitted as an estimator.

Solved by gpt-5.6-sol high.