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Expected value of joint distribution

  1. Oct 23, 2012 #1
    1. The problem statement, all variables and given/known data
    Suppose that [itex]f_{X,Y}(x,y)=\lambda^2e^{-\lambda(x+y)},0\le x,0\le y [/itex]
    find [itex]E[X+Y][/itex]


    2. Relevant equations



    3. The attempt at a solution

    I just want to double check I didn't make a mistake:
    [itex]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} [/itex]
     
    Last edited: Oct 24, 2012
  2. jcsd
  3. Oct 24, 2012 #2

    haruspex

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    Since [itex]0\le x,0\le y [/itex], it's hard to think that the expected value could be negative. Try writing out your last step in detail.
     
  4. Oct 24, 2012 #3
    good point. I think I see my mistake(s):


    [itex] 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}[/itex]
     
    Last edited: Oct 24, 2012
  5. Oct 24, 2012 #4

    haruspex

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    Looks right now.
     
  6. Oct 26, 2012 #5
    many thanks, by the way.
     
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