# Independent random variable expected value

1. Apr 7, 2009

### Proggy99

1. The problem statement, all variables and given/known data
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 $$E(X^{2}Y)$$

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

3. 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.

Last edited: Apr 7, 2009
2. Apr 7, 2009

### n!kofeyn

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.

3. Apr 7, 2009

### Proggy99

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.

4. Apr 7, 2009

### n!kofeyn

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