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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 linear regression model [math]\mathbf{Y} = \mathbf{X} \bbeta + \vvarepsilon[/math] with [math]\vvarepsilon \sim \mathcal{N} ( \mathbf{0}_n, \sigma^2 \mathbf{I}_{nn})[/math]. This model (without intercept) is fitted to data using the lasso regression estimator [math]\hat{\bbeta}(\lambda_1) = \arg \min_{\bbeta} \| \mathbf{Y} - \mathbf{X} \bbeta \|_2^2 + \lambda_1 \| \bbeta \|_1[/math]. The relevant summary statistics of the data are:

[[math]] \begin{eqnarray*} \mathbf{X} = \left( \begin{array}{rr} 1 & -1 \\ -1 & 1 \end{array} \right), \, \mathbf{Y} = \left( \begin{array}{r} -5 \\ 4 \end{array} \right), \, \mathbf{X}^{\top} \mathbf{X} = \left( \begin{array}{rr} 2 & -2 \\ -2 & 2 \end{array} \right), \mbox{ and } \, \mathbf{X}^{\top} \mathbf{Y} = \left( \begin{array}{r} -9 \\ 9 \end{array} \right). \end{eqnarray*} [[/math]]

  • Specify the full set of lasso regression estimates with [math]\lambda_1 = 2[/math] that minimize the lasso loss function for these data.
  • Now consider fitting the linear regression model with the fused lasso estimator [math]\hat{\bbeta}(\lambda_f) = \arg \min_{\bbeta} \| \mathbf{Y} - \mathbf{X} \bbeta \|_2^2 + \lambda_f | \beta_1 - \beta_2|[/math]. Determine [math]\lambda_{f}^{(0)} \gt 0[/math] such that [math]\hat{\bbeta}(\lambda_f) = (0, 0)^{\top}[/math] for all [math]\lambda_f \gt \lambda_{f}^{(0)}[/math].