Created
August 21, 2018 11:16
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\documentclass[a4paper,12pt]{scrartcl} | |
\usepackage[ngerman]{babel} | |
\usepackage{amsthm} | |
\usepackage{mathtools} | |
\usepackage{aligned-overset} | |
\newcommand{\Lagr}{\mathcal{L}} | |
\newcommand{\matr}[1]{\mathbf{#1}} | |
\newcommand{\X}{\matr{X}} %Matrix von X | |
\newcommand{\y}{\matr{y}} %y als voller Vektor | |
\newcommand{\yct}{\ubar{\y}^\mathbf{T}} %Zentriert | |
\newcommand{\betahat}{\hat{\beta}} %betahat | |
\DeclarePairedDelimiter{\abs}{\lvert}{\rvert} | |
\begin{document} | |
\begin{align*} | |
\Lagr(\beta_{0},\beta) | |
&=\min_{\beta_{0}, \beta} | |
\biggl\{ | |
\frac{1}{2} \sum_{i=1}^N \Bigl(y_{i}-\beta_{0} | |
- \sum_{j=1}^p\beta_{j}\tilde{x}_{ij}\Bigr)^{\!2} | |
+ \lambda\sum_{j=1}^p \abs{\beta_{j}} \biggr\}\\ | |
&= \min_{\beta_{0}, \beta} | |
\biggl\{ | |
\frac{1}{2} \sum_{i=1}^N | |
\Bigl(y_{i}-\beta_{0} - | |
\sum_{j=1}^p\beta_{j}(x_{ij}-\bar{x}_{j})\Bigr)^{\!2} | |
+\lambda\sum_{j=1}^p \abs{\beta_{j}} \biggr\} | |
\end{align*} | |
\begin{align*} | |
\frac{\partial \Lagr}{\partial \beta_{0}} | |
&=-\sum_{i=1}^N \Bigl(y_{i}-\betahat_{0} | |
- \sum_{j=1}^p\betahat_{j}(x_{ij}-\bar{x}_{j})\Bigr)\\ | |
&= -\sum_{i=1}^{N}y_i +N\betahat_{0} | |
+\sum_{j=1}^p \Bigl(\betahat_{j}\sum_{i=1}^N x_{ij} \Bigr) | |
- N\sum_{j=1}^p\betahat_{j}\bar{x}_{j}\\ | |
&= -N\bar{y}+N\betahat_{0}+\sum_{j=1}^p\betahat_{j}N\bar{x}_{j} | |
-N\sum_{j=1}^p\betahat_{j}\bar{x}_{j}\\ | |
&= -N\bar{y}+N\betahat_{0} | |
\end{align*} | |
Für die Nullstelle der Ableitung muss also | |
\[ \betahat_0 = \bar{y} \] | |
gelten. | |
Aus der Annahme, dass $\y$ zentriert ist, folgt für den kritischen Punkt | |
\[\betahat_{0}=0.\] | |
\end{document} |
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