\documentclass{article}
\usepackage{kotex}
\title{Restricted regression with a constraint $\bm{L}\bm{\beta} = \bm{c}$}
\author{심송용}
\newcommand{\bm}[1]{\mbox{\boldmath{$#1$}}}
\begin{document}
\maketitle
Lagrangian Method for least squares method
\begin{eqnarray}
 & &  \displaystyle \arg_{\bm{\beta}} \min (\bm{y} - \bm{X}\bm{\beta})'(\bm{y} - \bm{X}\bm{\beta}) \\ 
 && \mbox{ suject to }  \bm{L}\bm{\beta} = \bm{c} \nonumber
\end{eqnarray}
Let 
\begin{equation}
 Q = (\bm{y} - \bm{X}\bm{\beta})'((\bm{y} - \bm{X}\bm{\beta}) + 2{\lambda}' (\bm{L}\bm{\beta} - \bm{c} )
\end{equation}
Partial differentiation and setting them zero gives
\begin{equation} \label{eq:aa}
\left\{
\begin{array}{l}
\displaystyle \frac{\partial Q} {\partial \bm{\beta}} = -2\bm{X}'\bm{y} + 2 (\bm{X}'\bm{X})\bm{\beta} + 2\bm{L}'\bm{\lambda} = \bm{0} \\*[1.2em]
\displaystyle \frac{\partial Q} {\partial \bm{\lambda}}  = 2(\bm{L}\bm{\beta} -\bm{c}) = \bm{0}
\end{array}
\right.
\end{equation}
Denote the solution of $\bm{\beta}$ to the first equation (\ref{eq:aa}) as $\bm{b}^*$. Then,
\begin{eqnarray}
\bm{b}^* &=& (\bm{X}'\bm{X})^{-1} \bm{X}'\bm{y} - (\bm{X}'\bm{X})^{-1}\bm{L}' \bm{\lambda} \\ \nonumber
&=& \bm{b} -  (\bm{X}'\bm{X})^{-1}\bm{L}' \bm{\lambda},
\end{eqnarray}
where $\bm{b} =(\bm{X}'\bm{X})^{-1} \bm{X}'\bm{y} $ is the unrestricted least squares estimate of $\bm{\beta}$.

By substituting this $\bm{b}^*$ into $\bm{L}\bm{b}^*=\bm{c}$, we have
\end{document}
