# Expected value of joint distribution

1. Oct 23, 2012

### mrkb80

1. The problem statement, all variables and given/known data
Suppose that $f_{X,Y}(x,y)=\lambda^2e^{-\lambda(x+y)},0\le x,0\le y$
find $E[X+Y]$

2. Relevant equations

3. The attempt at a solution

I just want to double check I didn't make a mistake:
$E[X+Y]=E[X]+E[Y]=\int_0^{\infty} x{\lambda} e^{-\lambda x} dx + \int_0^{\infty} y{\lambda} e^{-\lambda y} dy = -2 - \dfrac{2}{\lambda}$

Last edited: Oct 24, 2012
2. Oct 24, 2012

### haruspex

Since $0\le x,0\le y$, it's hard to think that the expected value could be negative. Try writing out your last step in detail.

3. Oct 24, 2012

### mrkb80

good point. I think I see my mistake(s):

$E[X] + E[Y]=\int_0^{\infty} x \lambda e^{-\lambda x} dx + \int_0^{\infty} y \lambda e^{-\lambda y} dy = - x \dfrac{1}{\lambda} e^{-\lambda x} |_0^{\infty} - \int_0^{\infty} e^{- \lambda x } dx - y \dfrac{1}{\lambda} e^{-\lambda y} |_0^{\infty} - \int_0^{\infty} e^{- \lambda y } dy =\dfrac{1}{\lambda} + \dfrac{1}{\lambda} = \dfrac{2}{\lambda}$

Last edited: Oct 24, 2012
4. Oct 24, 2012

### haruspex

Looks right now.

5. Oct 26, 2012

### mrkb80

many thanks, by the way.