ABy Admin
Jun 24'23

Exercise

[math] \require{textmacros} \def \bbeta {\bf \beta} \def\fat#1{\mbox{\boldmath$#1$}} \def\reminder#1{\marginpar{\rule[0pt]{1mm}{11pt}}\textbf{#1}} \def\SSigma{\bf \Sigma} \def\ttheta{\bf \theta} \def\aalpha{\bf \alpha} \def\ddelta{\bf \delta} \def\eeta{\bf \eta} \def\llambda{\bf \lambda} \def\ggamma{\bf \gamma} \def\nnu{\bf \nu} \def\vvarepsilon{\bf \varepsilon} \def\mmu{\bf \mu} \def\nnu{\bf \nu} \def\ttau{\bf \tau} \def\SSigma{\bf \Sigma} \def\TTheta{\bf \Theta} \def\XXi{\bf \Xi} \def\PPi{\bf \Pi} \def\GGamma{\bf \Gamma} \def\DDelta{\bf \Delta} \def\ssigma{\bf \sigma} \def\UUpsilon{\bf \Upsilon} \def\PPsi{\bf \Psi} \def\PPhi{\bf \Phi} \def\LLambda{\bf \Lambda} \def\OOmega{\bf \Omega} [/math]

The linear regression model, [math]\mathbf{Y} = \mathbf{X} \bbeta + \vvarepsilon[/math] with [math]\varepsilon \sim \mathcal{N}(\mathbf{0}_n, \sigma^2 \mathbf{I}_{nn})[/math] is fitted by means of the ridge regression estimator. The design matrix and response are:

[[math]] \begin{eqnarray*} \mathbf{X} = \left( \begin{array}{rr} 2 & -1 \\ 0 & 1 \end{array} \right) \quad \mbox{ and } \quad \mathbf{Y} = \left( \begin{array}{r} 1 \\ \tfrac{1}{2} \end{array} \right). \end{eqnarray*} [[/math]]

The penalty parameter is chosen as the minimizer of the leave-one-out cross-validated squared error of the prediction (i.e. Allen's PRESS statistic). Show that [math]\lambda = \infty[/math].