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

A pdf is defined by

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

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

\item[(b)]
Find $P(X+Y>2),\ P(Y<\frac{1}{2}),\ P(X \leq 1),\ P(X=3Y)$.

\item[(c)]
Are $X$ and $Y$ independent?

\item[(d)]
Are the events $\{X<1\}$ and $\{Y \geq \frac{1}{2}\}$ independent?
\end{description}

\medskip

{\bf Answer}

\begin{description}
\item[(a)] $f(x,y)=Cy^2,\ \ \ 0 \leq x \leq 2.$

$\ds\int_0^2 \! \int_0^1 y^2 \,dy\,dx = \ds\int_0^2 \,dx
\left.\ds\frac{y^3}{3}\right|_0^1 = \ds\frac{2}{3} \Rightarrow
C=\ds\frac{3}{2}$

\item[(b)]
$f_X(x)=\ds\frac{1}{2},\ \ \ 0<x<2$

$f_Y(y)=3y^2,\ \ \ 0<y<1$

$X$ and $Y$ are independent. \setlength{\unitlength}{.5in}
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\begin{eqnarray*} P(X+Y>2) & = & \ds\frac{3}{2} \ds\int_1^2 \,dx \ds\int_{2-x}^1 y^2
\,dy\\ & = & \ds\frac{3}{2} \ds\int_1^2 \,dx
\left.\ds\frac{y^2}{3}\right|_{2-x}^1\\ & = & \ds\frac{1}{2}
\ds\int_1^2 \left\{1-(2-x)^3\right\} \,dx\\ & = & \ds\frac{3}{8}
\end{eqnarray*}



$P\left\{Y<\ds\frac{1}{2}\right\}=\ds\int_0^{\frac{1}{2}} 3y^2
\,dy =\ds\frac{1}{8}$

$P\left\{X\le 1\right\}=\ds\int_0^1 \ds\frac{1}{2} \,dx
=\ds\frac{1}{2}$

$P(X=3Y)=0$ because the distribution of $X$ and $Y$ is continuous.


\item[(c)]
$P\left(X \leq 1,Y \geq
\ds\frac{1}{2}\right)=\ds\frac{3}{2}\ds\int_0^1 \,dx
\ds\int_{\frac{1}{2}}^1 y^2 \,dy = \ds\frac{7}{16}$

$P(X \leq 1) \cdot P(Y \geq \frac{1}{2}) = \ds\frac{1}{2} \cdot
\ds\frac{7}{8} = \ds\frac{7}{16}$

Hence the events are independent.

\item[(d)]
Also follows from the general result that: if $X$ and $Y$ are
independent then

$$P\left\{X \in A,Y \in B\right\}=P(X \in A,Y \in B).$$
\end{description}

\end{document}