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## Homework Statement

Let the join probability density function of ZX and Y be given by

[tex]f(x,y)=\left\{\stackrel{2e^{-(x+2y)}\ \ \ \ \ if\ x\ \geq,\ \ \ y\ \geq\ 0}{0\ \ \ \ \ \ \ otherwise}[/tex]

Find [tex]E(X^{2}Y)[/tex]

## Homework Equations

I approached this problem using a theorem from the book that states

E[h(X,Y)] =[tex]\int^{\infty}_{-\infty}\int^{\infty}_{-\infty}h(x,y)f(x,y)dxdy[/tex]

## The Attempt at a Solution

Using the above formula, I did the following:

[tex]\int^{\infty}_{0}\int^{\infty}_{0}x^{2}y2e^{-(x+2y)}dxdy=[/tex]

[tex]\int^{\infty}_{0}\int^{\infty}_{0}2ye^{-2y}x^{2}e^{-x}dxdy=[/tex]

[tex]\int^{\infty}_{0}2ye^{-2y}[-x^{2}e^{-x}\ +\ \int^{\infty}_{0}2x(-e^{-x})dx]dy=[/tex]

[tex]\int^{\infty}_{0}2ye^{-2y}[-x^{2}e^{-x}\ +\ 2xe^{-x}\ +\ \int^{\infty}_{0}2e^{-x}dx]dy=[/tex]

[tex]\int^{\infty}_{0}2ye^{-2y}[-x^{2}e^{-x}\ +\ 2xe^{-x}\ -\ 2e^{-x}]^{\infty}_{0}dy=[/tex]

[tex]\int^{\infty}_{0}2ye^{-2y}[(0+0-0)-(0+0-2)]dy=[/tex]

[tex]\int^{\infty}_{0}2y2e^{-2y}dy=[/tex]

[tex]-2ye^{-2y}\ +\ \int^{\infty}_{0}-2e^{-2y}dy=[/tex]

[tex][-2ye^{-2y}\ +\ e^{-2y}]^{\infty}_{0}=[/tex]

[tex][(0+0)-(0+1)\ =\ -1[/tex]

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.

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