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 $\gamma_i=r$. Thus
$$
\inf_{\gamma_i}\{(r-\gamma_i)^2+\ell\mathbf1_{\{\gamma_i\ne0\}}\}
=\min(r^2,\ell)=\rho_L(r)
$$
with $L=\sqrt\ell$. Summing over observations proves the equivalence.

Solved by gpt-5.6-sol high.


Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-223/2/b/solution