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E(x / Y)

  1. Jun 22, 2007 #1

    DLS

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    If x and y are independent, does the following relation hold: E(X / Y) =
    E(X) * E(1 / Y) ?
    :uhh:
     
  2. jcsd
  3. Jun 23, 2007 #2

    CompuChip

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    I'll assume that E is the expectation value and X, Y are stochastic variables.
    Certainly, E(X Y) = E(X) E(Y) if (and only if) X and Y are independent (Meester, A Natural Introduction to Probability Theory). So you would be done if you could prove X, Y independent => X, 1/Y independent. Can you do that?
     
  4. Jun 23, 2007 #3

    DLS

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    Thanks. I don't really know how to prove that for the moment. Working on it...
     
  5. Jun 23, 2007 #4

    EnumaElish

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  6. Jun 23, 2007 #5

    CompuChip

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    I did find this result in the book cited before:
    (Theorem 5.6.12) Let X1, ..., Xn be independent continuous random variables and let g1, ..., gn be regular functions. Then g1(X1), ..., gn(Xn) are independent random variables, under a certain definition of regularity, which I think [tex]y \mapsto 1/y[/tex] satisfies.
    If you want I can replicate the proof.
     
  7. Jun 23, 2007 #6

    CompuChip

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    I typed it out anyway, it's attached as PDF.

    I just put in the theorem and the proof. You should check for yourself that it indeed proves the theorem (e.g., that what is proved is equivalent to your definition independence of variables) and that what you have satisfies the conditions of the theorem (that is, [tex]g_1: X \mapsto X[/tex] and [tex]g_2: Y \mapsto 1/Y[/tex] are regular according to the given definition). Finally, you should then probably show that for independent variables the identity E(XY) = E(X)E(Y) from my first post holds, by writing out some definition.

    Will that help you?
     

    Attached Files:

  8. Jun 23, 2007 #7
    Assuming E(1/Y) exists. I'd imagine you can find cases where it doesn't exist.
     
  9. Jun 24, 2007 #8

    DLS

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    That is a nice proof. Thx. It really helps.
     
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