\documentclass[11pt]{article} %\usepackage{amsbsy} % for \boldsymbol and \pmb %\usepackage{graphicx} % To include pdf files! \usepackage{amsmath} \usepackage{amsbsy} \usepackage{amsfonts} %\usepackage[colorlinks=true, pdfstartview=FitV, linkcolor=blue, citecolor=blue, urlcolor=blue]{hyperref} % For links \usepackage{fullpage} % Good for US Letter paper \topmargin=-0.75in \textheight=9.5in \usepackage{fancyhdr} \renewcommand{\headrulewidth}{0pt} % Otherwise there's a rule under the header \setlength{\headheight}{15.2pt} \fancyhf{} \pagestyle{fancy} \cfoot{Page \thepage {} of 2} % \pagestyle{empty} % No page numbers \begin{document} %\enlargethispage*{1000 pt} \begin{flushright} Name \underline{\hspace{60mm}} \\ $\,$ \\ Student Number \underline{\hspace{60mm}} \end{flushright} \vspace{2mm} \begin{center} {\Large \textbf{STA 431 Quiz 1}}\\ \vspace{1 mm} \end{center} \noindent % \emph{Calculators are allowed.} %\vspace{3mm} \begin{enumerate} \item (5 points) Let the random variable $x$ have expected value $\mu_x$, let the random variable $y$ have expected value $\mu_y$, and let $a$ be a non-zero constant. Circle one the of following statements and prove it, \emph{using the definition of covariance} from the formula sheet. %\begin{center} \begin{tabular}{cccc} $Cov(ax,y) = a^2Cov(x,y)$, & $Cov(ax,y) = aCov(x,y)$, & $Cov(ax,y) = Cov(x,y)$, & $Cov(ax,y) = 0$ \end{tabular} %\end{center} \newpage \item (5 points) Let \begin{eqnarray*} y_1 & = & \alpha_1 + \beta_1 x + \epsilon_1 \\ y_2 & = & \alpha_2 + \beta_2 x + \epsilon_2, \end{eqnarray*} where $E(x) = \mu$, $Var(x) = \sigma^2_x$, $E(\epsilon_1) = E(\epsilon_2) = 0$, $Var(\epsilon_1) = \sigma^2_1$ and $Var(\epsilon_2) = \sigma^2_2$. The random variables $x$, $\epsilon_1$ and $\epsilon_2$ are independent. Using anything you wish from the formula sheet, calculate $Cov(y_1,y_2)$. \textbf{Circle your final answer.} \end{enumerate} \end{document}