Commit ca1c63c9 by nanored

### début dm 3

parent 86541fb0
dm3.tex 0 → 100644
 \documentclass[11pt]{article} \usepackage[english]{babel} \usepackage[utf8]{inputenc} \usepackage[T1]{fontenc} \usepackage[left=2.7cm,right=2.7cm,top=3cm,bottom=3cm]{geometry} \usepackage{amsmath,amssymb,amsfonts} \usepackage{kpfonts} \usepackage{tikz} \usepackage{bbm} \usepackage{hyperref} \title{ \noindent\rule{\linewidth}{0.4pt} \huge Convex Optimization --- DM 3 \noindent\rule{\linewidth}{1pt} } \author{Yoann Coudert-\,-Osmont} \newcommand{\trans}{\mathsf{T}} \newcommand{\syst}[2]{\left\{ \begin{array}{#1} #2 \end{array} \right.} \newcommand{\diag}{\text{\textbf{diag}}~} \newcommand{\pmat}[1]{\begin{pmatrix} #1 \end{pmatrix}} \newcommand{\R}{\mathbb{R}} \newcommand{\one}{\mathbbm{1}} \begin{document} \maketitle \section*{Question 1} We transform our primal by adding a new variable in order to add a constraint: $$\begin{array}{llr} \min_{w, z} & \dfrac{1}{2} \| z \|_2^2 + \lambda \| w \|_1 \\[2mm] \text{s.t.} & X w - y = z \end{array}$$ Then, we compute the dual function of this new formulation: $$\begin{array}{lll} g(v) & = & \inf_{z, w} \dfrac{1}{2} \| z \|_2^2 + \lambda \| w \|_1 + v^\trans \left( Xw - y - z \right) \\[2mm] & = & - v^\trans y + \inf_{z, w} - v^\trans z + \dfrac{1}{2} \| z \|_2^2 + \left( X^\trans v \right)^\trans w + \lambda \| w \|_1 \\[3mm] & = & - v^\trans y - \dfrac{1}{2} \| 2 v \|_2^{2 \, *} - \lambda \left\| - \dfrac{1}{\lambda} X^\trans v \right\|_1^* \end{array}$$ In the previous homework I compute the conjugate of the square of 2-norm and the conjugate of the 1-norm. We have : $$\| x \|_2^{2 \, *} = \frac{1}{4} \| x \|_2^2 \qquad \| x \|_1^* = \syst{ll}{ 0 & \text{If } \| x \|_{\infty} \leqslant 1 \\ + \infty & \text{Otherwise} }$$ We then simplify the dual function: $$g(v) = \syst{ll}{ - \dfrac{1}{2} \| v \|_2^2 - y^\trans v & \text{If } \left\| X^\trans v \right\|_\infty \leqslant \lambda \\ - \infty & \text{Otherwise} }$$ Furthermore we can rewrite the condition by : $$\forall i = 1, ..., d \; \left| \left( X^\trans v \right)_i \right| \leqslant \lambda \Leftrightarrow X^\trans v \preceq \lambda \one_d \text{ and } - X^\trans v \preceq \lambda \one_d$$ Then we let the matrix: $$A = \pmat{X^\trans \\ -X^\trans} \qquad \text{such that } A v \preceq \lambda \one_{2d}$$ We also remark that the following equality: $$\dfrac{1}{2} \| v \|_2^2 = \dfrac{1}{2} v^\trans I_n v = v^\trans Q v \qquad \text{with } Q = \dfrac{1}{2} I_n$$ The dual problem is obtained by maximizing $g$ which is equivalent to minimizing $-g$. Thus we obtain the following dual: $$(QP) \qquad \begin{array}{llr} \min_v & v^\trans Q v + y^\trans v \\[2mm] \text{s.t.} & A v \preceq \lambda \one_{2d} \end{array}$$ We identify with the notations of the statement : \begin{center} \fbox{$\displaystyle Q = \dfrac{1}{2} I_n \quad p = y \quad A = \pmat{X^\trans \\ -X^\trans} \quad b = \lambda \one_{2d}$} \end{center} \end{document}\texttt{} \ No newline at end of file
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