# bivariate marginal density function

• Apr 8th 2008, 12:08 AM
lllll
bivariate marginal density function
find the marginal density function for $X$ and $Y$.

$f(x,y) = \left\{ \begin{array}{rcl}
x+y & \mbox{for} & 0 \leq x \leq 1, 0 \leq y \leq 1 \\
0 & \mbox{for} & \mbox{other}
\end{array}\right.$

so far I have

$f(x) = \left\{ \begin{array}{rcl}
\int_{0}^{1} x+y \ dy & \mbox{for} & 0 \leq x \leq 1 \\
0 & \mbox{for} & \mbox{other}
\end{array}\right.$

= $f(x) = \left\{ \begin{array}{rcl}
xy+\frac{y^2}{2} \ \bigg{|}^{1}_{0} & \mbox{for} & 0 \leq x \leq 1 \\
0 & \mbox{for} & \mbox{other}
\end{array}\right.$
= $f(x) = \left\{ \begin{array}{rcl}
x+\frac{1}{2} & \mbox{for} & 0 \leq x \leq 1 \\
0 & \mbox{for} & \mbox{other} \end{array}\right.$

is this correct?

And how would you find:
$P \left( X \geq \frac{1}{2} \bigg{|}Y \geq \frac{1}{2} \right)$ ?
• Apr 8th 2008, 06:23 AM
falcald
Your density for $X$ is ok. The density for $Y$ will be the same.

Well, for $P\left(X\geq\frac{1}{2}\bigg{|}Y\geq\frac{1}{2}\ri ght)$.

You have to apply: $\frac{P\left(X\geq\frac{1}{2} , Y\geq\frac{1}{2} \right)}{P \left(Y \geq \frac{1}{2} \right)}$.

So, you have: $\frac{\int_{\frac{1}{2}}^{1} \int_{\frac{1}{2}}^{1} x+y \ dy \ dx}{\int_{\frac{1}{2}}^{1} y+\frac{1}{2} \ dy}$

Hope this helps.

Regards,

Federico.