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Domain probability problem

  1. Feb 22, 2008 #1
    If [tex] f(x,y) = e^{-x-2y} [/tex] find [tex] P(X<Y) [/tex].

    So is this equaled to [tex] 1 - P(X>Y) = 1 - \int\limits_{0}^{\infty} \int\limits_{0}^{x} e^{-x-2y} \ dy \ dx = \frac{1}{3}?[/tex]
     
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  3. Feb 22, 2008 #2

    EnumaElish

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    What is the domain? [itex]\int_0^\infty\int_0^\infty f(x,y)dydx[/itex]=1/2, not 1.
     
  4. Feb 22, 2008 #3
    the domain is [tex] 0 < x < \infty [/tex], [tex] 0 < y < \infty [/tex]. It is equaled to [tex] 1 [/tex], so its a valid pdf.
     
  5. Feb 22, 2008 #4

    EnumaElish

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    On that domain the probability doesn't add up to 1.
     
  6. Feb 22, 2008 #5
    whoops, I meant [tex] -\infty < x < \infty [/tex], [tex] -\infty < y< \infty [/tex].
     
  7. Feb 22, 2008 #6

    EnumaElish

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    It looks like the lower bound should be -Log[2]/3.

    For a "suitable" lower bound that equates the double integral to 1, your formula is correct.

    Alternatively, since integration over [0, +infinity) gives 1/2, you might define the cumulative distribution F*(x,y) = F(x,y) + 1/2, where F(x,y) is the double integral of f(s,t) over s from 0 to x and t from 0 to y. In this case f is a valid pdf.
     
    Last edited: Feb 22, 2008
  8. Feb 22, 2008 #7
    forgetting about the bounds for the moment, is the general set up correct?
     
  9. Feb 22, 2008 #8

    EnumaElish

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    See my previous post (edited).
     
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