\documentclass[12pt]{article} \addtolength{\textheight}{2.7in} \addtolength{\topmargin}{-1.15in} \addtolength{\textwidth}{1.0in} \addtolength{\evensidemargin}{-0.5in} \addtolength{\oddsidemargin}{-0.65in} \setlength{\parskip}{0.1in} \setlength{\parindent}{0.0in} \newcommand{\given}{\, | \,} \pagestyle{empty} \raggedbottom \begin{document} \vspace*{-0.3in} \begin{flushleft} Prof.~David Draper \\ Department of Applied Mathematics and Statistics \\ University of California, Santa Cruz \end{flushleft} \begin{center} \textbf{\large AMS 131: Quiz 5} \end{center} \bigskip \begin{flushleft} Name: \underline{\hspace*{5.85in}} \end{flushleft} You're working on a problem involving two continuous random variables $X$ and $Y$, and you figure out that their joint PDF has the following form: \begin{equation} \label{joint-pdf} f_{ X, Y } ( x, y ) = \left\{ \begin{array}{cc} c \, x^2 & \textrm{for } 0 \le y \le 1 - x^2 \\ 0 & \textrm{otherwise} \end{array} \right\} \, . \end{equation} \begin{itemize} \item[(a)] Sketch the support $S$ of this bivariate distribution. \vspace*{0.9in} \item[(b)] Compute the normalizing constant $c$. \vspace*{0.9in} \item[(c)] It can be shown that the marginal PDFs of $X$ and $Y$ with this joint PDF are \begin{equation} \label{X-marginal} f_X ( x ) = \left\{ \begin{array}{cc} \frac{ 15 }{ 4 } x^2 ( 1 - x^2 ) & \textrm{for } -1 \le x \le 1 \\ 0 & \textrm{otherwise} \end{array} \right\} \end{equation} and \begin{equation} \label{Y-marginal} f_Y ( y ) = \left\{ \begin{array}{cc} \frac{ 5 }{ 2 } ( 1 - y )^{ \frac{ 3 }{ 2 } } & \textrm{for } 0 \le y \le 1 \\ 0 & \textrm{otherwise} \end{array} \right\} \, . \end{equation} Verify that both of these marginals are correct. \vspace*{0.9in} \item[(d)] Are $X$ and $Y$ independent in this joint distribution? Explain briefly. \end{itemize} \end{document}