Expected value of joint distribution

[mrkb80]

Homework Statement

Suppose that $f_{X,Y}(x,y)=\lambda^2e^{-\lambda(x+y)},0\le x,0\le y$
find $E[X+Y]$

Homework Equations

The Attempt at a Solution

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

 

[haruspex]

Homework Statement

Suppose that $f_{X,Y}(x,y)=\lambda^2e^{-\lambda(x+y)},0\le x,0\le y$
find $E[X+Y]$

Homework Equations

The Attempt at a Solution

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

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.

 

[mrkb80]

good point. I think I see my mistake(s):$E[X] + E[Y]=\int_0^{\infty} x \lambda e^{-\lambda x} dx + \int_0^{\infty} y \lambda e^{-\lambda y} dy = - x \dfrac{1}{\lambda} e^{-\lambda x} |_0^{\infty} - \int_0^{\infty} e^{- \lambda x } dx - y \dfrac{1}{\lambda} e^{-\lambda y} |_0^{\infty} - \int_0^{\infty} e^{- \lambda y } dy =\dfrac{1}{\lambda} + \dfrac{1}{\lambda} = \dfrac{2}{\lambda}$

 

[haruspex]

Looks right now.

 

[mrkb80]

many thanks, by the way.