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Conditional expectation (w/ transformation)

  1. Aug 29, 2006 #1
    Any hints on how to solve for E(Y|X) given the ff:

    Suppose U and V are independent with exponential distributions
    [tex]f(t) = \lambda \exp^{-\lambda t}, \mbox{ for } t\geq 0[/tex]

    Where X = U + V and Y = UV.

    I am having difficulty finding f(Y|X)...
    Also, solving for f(X,Y), I am also having difficulty transforming U and V to X and Y. I was able to define U and V to X and Y, but the terms are so complicated that its difficult to get the Jacobian.

    So maybe, there's no need for transformation?

    Help please. Thanks!
    Last edited: Aug 29, 2006
  2. jcsd
  3. Aug 29, 2006 #2
    here's what I was able to get so far.

    Since U and V are independent,
    then f(U,V) = f(U)f(V)
    [tex] f(U,V) = \lambda^{2} \exp^{-\lambda (u+v)}, \mbox{ for } u\geq 0 v\geq 0 [/tex]

    To solve for E(Y|X), I would need to find f(Y|X)
    [tex]f(Y|X) = \frac{f(X,Y)}{f(X)}[/tex]

    to get f(X,Y), I need to transform U and V to X and Y.
    [tex] f(X,Y) = \lambda^{2} \exp^{-\lambda (x)} |J| [/tex]
    where J is the jacobian.

    my rpoblem is in solving for the Jacobian.

    Since X = U+V
    and Y =UV

    then either
    [tex]U = \frac{2Y}{X+ \sqrt{X^{2} - 4Y}}[/tex]
    [tex]V = \frac{X+\sqrt{X^{2} - 4Y}}{2}[/tex]

    [tex]U = \frac{2Y}{X- \sqrt{X^{2} - 4Y}}[/tex]
    [tex]V = \frac{X-\sqrt{X^{2} - 4Y}}{2}[/tex]

    problem is, getting the Jacobian of U and V would result in a very complicated and long expresion.

    so I was thinking, maybe, I don't need to do the transformation. If not. What should I do?

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