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{\bf Question}

A pdf is defined by

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

\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}

$f(x,y)=C(x+2y)\ \ \ 0<x<2,\ 0<y<1$

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

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

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

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

$X$ 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) & = & \ds\frac{x+1}{4} \cdot
\left|\ds\frac{dx}{dz}\right|,\ \ \ 1<z<9\\ & = & \ds\frac{x+1}{4}
\cdot \ds\frac{(x+1)^3}{18}\\ & = & \ds\frac{(x+1)^4}{72}\\ & = &
\ds\frac{81}{z^2 \cdot 72} = \ds\frac{9}{8} \cdot
\ds\frac{1}{z^2},\ \ 1<z<9 \end{eqnarray*}

\end{description}

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