ABy Admin
Jun 25'23

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]

Show that the genalized ridge regression estimator, [math]\hat{\bbeta}(\mathbf{\Delta}) = (\mathbf{X}^{\top} \mathbf{X} + \mathbf{\Delta})^{-1} \mathbf{X}^{\top} \mathbf{Y}[/math], too (as in Question) can be obtained by ordinary least squares regression on an augmented data set. Hereto consider the Cholesky decomposition of the penalty matrix: [math]\mathbf{\Delta} = \mathbf{L}^{\top} \mathbf{L}[/math] Now augment the matrix [math]\mathbf{X}[/math] with [math]p[/math] additional rows comprising the matrix [math]\mathbf{L}[/math], and augment the response vector [math]\mathbf{Y}[/math] with [math]p[/math] zeros.