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]

Consider the standard linear regression model [math]Y_i = \mathbf{X}_{i,\ast} \bbeta + \varepsilon_i[/math] for [math]i=1, \ldots, n[/math] and with the [math]\varepsilon_i[/math] i.i.d. normally distributed with zero mean and a common variance. Let the first covariate correspond to the intercept. The model is fitted to data by means of the minimization of the sum-of-squares augmented with a lasso penalty in which the intercept is left unpenalized: [math]\lambda_1 \sum_{j=2}^p | \beta_j |[/math] with penalty parameter [math]\lambda_1 \gt 0[/math]. The penalty parameter is chosen through leave-one-out cross-validation (LOOCV). The predictive performance of the model is evaluated, again by means of LOOCV. Thus, creating a double cross-validation loop. At each inner loop the optimal [math]\lambda_1[/math] yields an empty intercept-only model, from which a prediction for the left-out sample is obtained. The vector of these prediction is compared to the corresponding observation vector through their Spearman correlation (which measures the monotonicity of a relatonship and -- as a correlation measure -- assumed values on the [math][-1,1][/math] interval with an analogous interpretation to the ‘ordinary’ correlation). The latter equals [math]-1[/math]. Why?