# Thread: Expected value of Bivariate distribution

1. ## Expected value of Bivariate distribution

I'm a little stuck on this question:
$f(x) = \left\{ \begin{array}{rcl}
e^{-(y_1+y_2)} & \mbox{for} & y_1 > 0, \ \ y_2 >0 \\
0 & \mbox{for} & \mbox{otherwise}
\end{array}\right.$

a) What are $E(Y_1+Y_2)$ and $V(Y_1+Y_2)$ ?

b) What is $P(Y_1-Y_2>3)$ ?

so far I have:

a) $E(Y_1+Y_2) = E(Y_1) +E(Y_2)$

$E(Y_1) = \int_0^{\infty} y_1e^{-(y_1+y_2)} \ dy_1$

$= y_1e^{-(y_1+y_2)} - (-e^{-(y_1+y_2)}) \bigg{|}_0^{\infty}$

$= (0 + 0) - (0 -e^{-y_2})= e^{-y_2}$

$E(Y_2) = e^{-{y_1}}$

$E({Y_1}^2) = \int_0^{\infty} {y_1}^2 e^{-(y_1+y_2)} \ dy_1$

integration by parts gives me $=-{y_1}^2e^{-(y_1+y_2)} +2y_1 e^{-(y_1+y_2)}- 2e^{-(y_1+y_2)} \bigg{|}_{0}^{\infty}$

$= e^{-(y_1+y_2)}(-{y_1}^2+2y_1-2)\bigg{|}_{0}^{\infty}$

$= 0 - e^{-y_2}(2) = 2e^{-y_2}$

$E({Y_2}^2) = 2e^{-y_1}$

now to get the variance, it's $[E({Y_2}^2) +E({Y_1}^2)] - [(E({Y_1+Y_2}))^2]$

$= 2e^{-y_1} + 2e^{-y_2} - (e^{-y_1}+e^{-y_2})^2$

At this point I know I made a mistake since the answer in the back of the book is 2 and I got nowhere near that.

b) $y_1 >3 \ \ \mbox{and} \ \ 0 < y_2 < y_1-3$

$= \int_3^{\infty} \int_0^{y_1-3} e^{-(y_1+y_2)} \ dy_2\ dy_1$

$= \int_3^{\infty} \left( -e^{-(y_1+y_2)} \right) \bigg{|}_0^{y_1-3} \ dy_1$

$= \int_3^{\infty} \left( -e^{-(y_1+(y_1-3))} - (-e^{-(y_1+0)}) \right) \ dy_1$

$= \int_3^{\infty} \left( -e^{-(2y_1-3)} + e^{-y_1} \right) \ dy_1$

$= \int_3^{\infty} -e^{-(2y_1-3)} \ dy_1 + \int_3^{\infty} e^{-y_1} \ dy_1$

$= \frac{1}{2}e^{-(2y_1-3)}\bigg{|}_3^{\infty} - e^{-y_1} \bigg{|}_3^{\infty}$

= $(0 -\frac{1}{2}e^{-3}) + (0+e^{-3}) = \frac{1}{2}e^{-3}$

2. Originally Posted by lllll
I'm a little stuck on this question:
$f(x) = \left\{ \begin{array}{rcl}
e^{-(y_1+y_2)} & \mbox{for} & y_1 > 0, \ \ y_2 >0 \\
0 & \mbox{for} & \mbox{otherwise}
\end{array}\right.$
$Y_1$ and $Y_2$ are independent, with densities $e^{-y_1}$ and $e^{-y_2}$ , so:

$E(Y_1)=\int_0^{\infty}y_1 e^{-y_1} \ dy_1$

and so on

RonL