Independent random variable expected value

[Proggy99]

Homework Statement

Let the join probability density function of ZX and Y be given by
f(x,y)=\left\{\stackrel{2e^{-(x+2y)}\ \ \ \ \ if\ x\ \geq,\ \ \ y\ \geq\ 0}{0\ \ \ \ \ \ \ otherwise}
Find E(X^{2}Y)

Homework Equations

I approached this problem using a theorem from the book that states
E[h(X,Y)] =\int^{\infty}_{-\infty}\int^{\infty}_{-\infty}h(x,y)f(x,y)dxdy

The Attempt at a Solution

Using the above formula, I did the following:
\int^{\infty}_{0}\int^{\infty}_{0}x^{2}y2e^{-(x+2y)}dxdy=
\int^{\infty}_{0}\int^{\infty}_{0}2ye^{-2y}x^{2}e^{-x}dxdy=
\int^{\infty}_{0}2ye^{-2y}[-x^{2}e^{-x}\ +\ \int^{\infty}_{0}2x(-e^{-x})dx]dy=
\int^{\infty}_{0}2ye^{-2y}[-x^{2}e^{-x}\ +\ 2xe^{-x}\ +\ \int^{\infty}_{0}2e^{-x}dx]dy=
\int^{\infty}_{0}2ye^{-2y}[-x^{2}e^{-x}\ +\ 2xe^{-x}\ -\ 2e^{-x}]^{\infty}_{0}dy=
\int^{\infty}_{0}2ye^{-2y}[(0+0-0)-(0+0-2)]dy=
\int^{\infty}_{0}2y2e^{-2y}dy=
-2ye^{-2y}\ +\ \int^{\infty}_{0}-2e^{-2y}dy=
[-2ye^{-2y}\ +\ e^{-2y}]^{\infty}_{0}=
[(0+0)-(0+1)\ =\ -1

I know the answer is supposed to be positive one, not negative one. I have gone over my calculations several times and can not find where I am making a mistake. I also accept that real possibility that I am approaching this in the wrong way in the first place. can someone help me out on this problem? Thanks for any help.

 

[n!kofeyn]

The Attempt at a Solution

Using the above formula, I did the following:
\int^{\infty}_{0}\int^{\infty}_{0}x^{2}y2e^{-(x+2y)}dxdy=
\int^{\infty}_{0}\int^{\infty}_{0}2ye^{-2y}x^{2}e^{-x}dxdy=
\int^{\infty}_{0}2ye^{-2y}[-x^{2}e^{-x}\ +\ \int^{\infty}_{0}2x(-e^{-x})dx]dy=

I found that in this last step, you have a negative sign inside the integral which should not be there. Remember integration by parts is uv-\int v\,du, so you should have a negative in front of the integration sign, which will then cancel with the negative inside.

 

[Proggy99]

I found that in this last step, you have a negative sign inside the integral which should not be there. Remember integration by parts is uv-\int v\,du, so you should have a negative in front of the integration sign, which will then cancel with the negative inside.

gah, thanks for the catch. Looking past that, I realized I had consistently added rather than subtracted when doing my integration by parts, so I did the sign wrong three times resulting in a change in the final answer from negative to positive. Thanks for pointing it out.

 

[n!kofeyn]

No problem. Negative signs are always killers on those type of integrals.