<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}

\newcommand{\ds}{\displaystyle}
\begin{document}
\parindent=0pt

{\bf Question}

A pdf is defined by

$$f(x,y)=\left\{\begin{array}{ll}C(x+2y) &amp; {\rm if}\ 0 &lt; x &lt; 2\
{\rm and}\ 0 &lt; y &lt; 1\\ 0 &amp; {\rm otherwise} \end{array} \right.$$

\begin{description}
\item[(a)]
Find the value of $C$.

\item[(b)]
Find the marginal pdf's of $X$ and $Y$.  Are $X$ and $Y$
independent?

\item[(c)]
Find the pdf of the random variable $Z=\ds\frac{9}{(X+1)^2}$.
\end{description}

\medskip

{\bf Answer}

$f(x,y)=C(x+2y)\ \ \ 0&lt;x&lt;2,\ 0&lt;y&lt;1$

\begin{description}
\item[(a)]
\begin{eqnarray*} &amp;&amp;\ds\int_0^2 \! \ds\int_0^1 (x+2y) \,dy\,dx\\
&amp; = &amp; \ds\int_0^2 \,dx \ds\int_0^1 (x+2y) \,dy\\ &amp; = &amp; \ds\int_0^2
\,dx \left[xy+y^2\right]_0^1\\ &amp; = &amp; \ds\int_0^2 (x+1) \,dx\\ &amp; =
&amp; \frac{x^2}{2}+\left.x\right|_0^2 = 2+2 = 4. \end{eqnarray*}

Therefore $C=\ds\frac{1}{4}.$

\item[(b)]
$f_X(x)=\ds\int_0^1 f(x,y) \,dy = \ds\frac{1}{4} \ds\int_0^1
(x+2y) \,dy = \ds\frac{x+1}{4},\ \ 0&lt;x&lt;2$

$f_Y(y)=\ds\int_0^2 f(x,y) \,dx = \ds\frac{1}{4} \ds\int_0^2
(x+2y) \,dx = \ds\frac{1}{2}+y,\ \ 0&lt;y&lt;2$

$X$ and $Y$ are not independent since $f(x,y) \ne f_X(x)f_Y(y).$


\item[(c)]
$$z=\ds\frac{9}{(x+1)^2}$$

$x=0 \Rightarrow z=9$ and $x=2 \Rightarrow z=1$

$\ds\frac{dz}{dx}=-\ds\frac{9}{(x+1)^4} \cdot 2(x+1)$

Therefore \begin{eqnarray*} g(z) &amp; = &amp; \ds\frac{x+1}{4} \cdot
\left|\ds\frac{dx}{dz}\right|,\ \ \ 1&lt;z&lt;9\\ &amp; = &amp; \ds\frac{x+1}{4}
\cdot \ds\frac{(x+1)^3}{18}\\ &amp; = &amp; \ds\frac{(x+1)^4}{72}\\ &amp; = &amp;
\ds\frac{81}{z^2 \cdot 72} = \ds\frac{9}{8} \cdot
\ds\frac{1}{z^2},\ \ 1&lt;z&lt;9 \end{eqnarray*}

\end{description}

\end{document}
</pre></body></html>