\documentclass[12pt]{article} \usepackage{latexsym} \usepackage{graphicx} \usepackage{graphics} \usepackage[reqno]{amsmath} \usepackage{natbib} \usepackage{verbatim} \usepackage{url} \usepackage{lscape} %\usepackage{subfigure} \usepackage{subcaption} %\usepackage{fullpage} \usepackage{setspace} \usepackage{booktabs,array} \addtolength{\oddsidemargin}{-.45in} \addtolength{\evensidemargin}{-.45in} \addtolength{\textwidth}{1.in} %\addtolength{\textwidth}{1.45in} \addtolength{\topmargin}{-.225in} \addtolength{\textheight}{1.0in} % === dcolumnpackage=== \usepackage{dcolumn} \newcolumntype{.}{D{.}{.}{-1}} \newcolumntype{d}[1]{D{.}{.}{#1}} \usepackage{setspace} \usepackage{color,hyperref, longtable} \usepackage{epstopdf} \newcommand {\mabsingle} {\Large \renewcommand{\baselinestretch}{1} \normalsize} \newcommand {\mabdouble} {\renewcommand{\baselinestretch}{2} \Large \normalsize} \newcommand {\mabnote} {\Large \renewcommand{\baselinestretch}{1} \footnotesize \hspace{-.06in}} \setlength{\textfloatsep}{6pt} \setlength{\floatsep}{6pt} \setlength{\intextsep}{6pt} \usepackage[Glenn]{fncychap} % Glenn or Lenny are best styles \usepackage{fancybox} \newenvironment{shadowminipage}% {\begin{Sbox}\begin{minipage}} {\end{minipage}\end{Sbox}\shadowbox{\TheSbox} } \begin{document} \mabsingle %\begin{titlepage} \title{Answer Key for Ch14 Exercise1: Derive $\hat{\beta}$ \vspace{0.0in}} \maketitle \vspace{-0.2in} %\end{titlepage} \begin{enumerate} \item Apply the logic developed in this chapter to the model $Y_i = \beta_0 + \beta_1X_i + \epsilon_i$. Derive the OLS estimate for $\hat{\beta}_0$ and $\hat{\beta}_1$. \begin{enumerate} \item Write out the sum of squared residuals for the model \begin{eqnarray*} \sum \hat{\epsilon}_i^2 & = & \sum (Y_i - \hat{\beta}_0 - \hat{\beta}_1 X_{i})^2 \end{eqnarray*} \item We minimize the sum of squared residuals by taking the derivatives with respect to $\hat{\beta}_0$ (one equation) and $\hat{\beta}_1$ (a second equation) and setting them to zero.\footnote{\mabnote These are often referred to as ``normal equations'' (as if we haven't used the word \emph{normal} enough in statistics).} The estimates are the values which produce a derivative of zero. \begin{eqnarray*} \frac{\partial \sum \hat{\epsilon}_i^2}{\partial \hat{\beta}_0} & = & (-2)\sum (Y_i - \hat{\beta}_0 - \hat{\beta}_1 X_{i}) = 0\\ \frac{\partial \sum \hat{\epsilon}_i^2}{\partial \hat{\beta}_1} & = & (-2)\sum (Y_i - \hat{\beta}_0 - \hat{\beta}_1 X_{i})(X_i) = 0 \end{eqnarray*} \item Solving for $\hat{\beta}_0$ is relatively straightforward. Divide both sides by $(-2)$ and do the following steps: \begin{eqnarray*} \sum Y_i - \sum \hat{\beta}_0 - \sum \hat{\beta}_1 X_{i}) &=& 0 \mbox{ Separate the sum into components}\\ \sum Y_i - \sum \hat{\beta}_1 X_{i}) &=& \sum \hat{\beta}_0 \mbox{ Move $\hat{\beta}_0$ to RHS}\\ \sum Y_i - \hat{\beta}_1 \sum X_{i}) &=& N\hat{\beta}_0 \mbox{ Pull constants out of summations}\\ \frac{\sum Y_i}{N} - \hat{\beta}_1 \frac{\sum X_{i}}{N} &=& \hat{\beta}_0 \mbox{ Divide by $N$}\\ \overline{Y} - \hat{\beta}_1 \overline{X} &=& \hat{\beta}_0 \mbox{ Use definition of mean} \end{eqnarray*} \item To solve for $\hat{\beta}_1$ divide both sides of the derivative with respect to $\hat{\beta}_1$ by $(-2)$ and do the following steps: \begin{eqnarray*} \sum Y_iX_i - \sum \hat{\beta}_0 X_i - \sum \hat{\beta}_1 X_{i}^2 &=& 0 \mbox{ Separate the sum into components}\\ \sum Y_iX_i - \sum (\overline{Y} - \hat{\beta}_1 \overline{X})X_i - \sum \hat{\beta}_1 X_{i}^2 &=& 0 \mbox{ Substitute for $\hat{\beta}_0$}\\ \sum Y_iX_i - \sum \overline{Y}X_i &=& \hat{\beta}_1 (\sum X_{i}^2 -\overline{X}\sum X_i) \mbox{ Simplify}\\ \hat{\beta}_1 &=& \frac{\sum Y_iX_i - \sum \overline{Y}X_i}{\sum X_{i}^2 -\overline{X}\sum X_i} \mbox{ Solve for $\hat{\beta}_1$} \end{eqnarray*} \item We prefer the equation for $\hat{\beta}_1$ to be in the more intuitive mean-deviated form. \begin{enumerate} \item First substitute for $\sum X_i = N\overline{X}$. Be careful to note that $\sum Y_iX_i$ cannot be simplified; we can only use this for terms where the only thing in the summation is $X$ or $X$ times a constant. \begin{eqnarray*} \hat{\beta}_1 &=& \frac{\sum Y_iX_i - N\overline{Y}\overline{X}}{\sum X_{i}^2 -N\overline{X}^2} \end{eqnarray*} \item Next use facts that $\sum X_i = N\overline{X}$ and $\sum Y_i = N\overline{Y}$ to rewrite $(Y_i - \overline{Y})(X_i-\overline{X})$, which is the form we are seeking for the numerator of the $\hat{\beta}_1$ equation. \begin{eqnarray*} \sum (Y_i - \overline{Y})(X_i-\overline{X}) &=& \sum Y_iX_i - \overline{X}\sum{Y_i}- \overline{Y}\sum X_i + N\overline{Y}\overline{X} \\ &=& \sum Y_iX_i - N\overline{X}\overline{Y}- N\overline{Y}\overline{X} + N\overline{Y}\overline{X} \\ &=& \sum Y_iX_i - N\overline{X}\overline{Y} \end{eqnarray*} which is the value we had in the numerator of $\hat{\beta}_1$ in part (d) above, meaning we can use $\sum (Y_i - \overline{Y})(X_i-\overline{X})$ in the numerator. \item Do similar steps for the denominator of the $\hat{\beta}_1$ equation. \begin{eqnarray*} \sum (X_i - \overline{X})(X_i-\overline{X}) &=& \sum X_i^2 - 2\overline{X}\sum{X_i} + N\overline{X}^2 \\ &=& \sum X_i^2 - 2N\overline{X}^2 + N\overline{X}^2 \\ &=& \sum X_i^2 - N\overline{X}^2 \end{eqnarray*} which is the value we had in the denominator of $\hat{\beta}_1$ meaning we can use $\sum (X_i - \overline{X})^2$ in the denominator, meaning that \begin{eqnarray*} \hat{\beta}_1 &=& \frac{\sum (Y_i - \overline{Y})(X_i-\overline{X})}{\sum (X_i - \overline{X})^2} \end{eqnarray*} \end{enumerate} \end{enumerate} \end{enumerate} \end{document}