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\newcommand{\ds}{\displaystyle}
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{\bf Question}

Suppose that the joint pdf of $X$ and $Y$ is as follows:

$$f(x,y)=\left\{\begin{array}{ll} \frac{15}{4}x^2 & {\rm if}\ 0
\leq y \leq 1-x^2\\ 0 & {\rm otherwise}
\end{array} \right.$$

\begin{description}
\item[(a)]
Determine the marginal pdf's of $X$ and $Y$.  Are $X$ and $Y$
independent?

\item[(b)]
Find the conditional pdf of $Y$ given that $X=x$.

\item[(c)]
Find $E(Y|X=x)$.

\end{description}

\medskip

{\bf Answer}

$f(x,y)=\ds\frac{15}{4}x^2,\ \ 0 \leq y \leq 1-x^2$

\begin{description}
\item[(a)]
First check that its a pdf

\begin{eqnarray*} & & \ds\int \! \ds\int_{0 \leq y \leq 1-x^2}
\ds\frac{15}{4} x^2 \,dy \,dx\\ & = & \ds\frac{15}{4}
\ds\int_{-1}^1 x^2 \,dx \ds\int_0^{1-x^2} \,dy\\ & = &
\ds\frac{15}{4} \ds\int_{-1}^1 x^2(1-x^2) \,dx = 1 \end{eqnarray*}


$f_X(x) = \ds\frac{15}{4} x^2(1-x^2),\ \ -1\leq x \leq 1$

$f_Y(y) = \ds\frac{15}{4} \ds\int_{-\sqrt{1-y}}^{\sqrt{1-y}} x^2
\,dx = \ds\frac{5}{2}(1-y)^{\frac{3}{2}},\ \ 0<y<1$

Since
\begin{eqnarray*} & & 1-x^2 \geq y\\ & \Rightarrow & x^2 \leq 1-y\\
& \Rightarrow & -\sqrt{1-y} \leq x \leq \sqrt{1-y} \end{eqnarray*}

$X$ and $Y$ are not independent.

\item[(b)]
\begin{eqnarray*} f(y|X=x) = \ds\frac{f(x,y)}{f_X(x)} & = &
\ds\frac{15}{4} x^2 \cdot \ds\frac{1}{\frac{15}{4} x^2 (1-x^2)}\\
& = & \ds\frac{1}{1-x^2},\ \ \ 0 \leq y \leq 1-x^2.
\end{eqnarray*}

$Y|X=x\sim {\rm Uniform}(0,1-x^2)\ \ {\rm where}\ \ -1\le x\le 1$

\item[(c)]
$E(Y|X=x)=\ds\frac{1-x^2}{2}$ where $-1 \leq x \leq 1.$
\end{description}
\end{document}
