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Independent random variable expected value

  1. Apr 7, 2009 #1
    1. The problem statement, all variables and given/known data
    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]

    2. Relevant 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]


    3. 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.
     
    Last edited: Apr 7, 2009
  2. jcsd
  3. Apr 7, 2009 #2
    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 [itex]uv-\int v\,du[/itex], so you should have a negative in front of the integration sign, which will then cancel with the negative inside.
     
  4. Apr 7, 2009 #3
    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.
     
  5. Apr 7, 2009 #4
    No problem. Negative signs are always killers on those type of integrals.
     
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