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Expectation of random variable

  1. Apr 18, 2007 #1
    I have two random variables X and Y, and I need to calculate E(XY). The expectation of X, E(X) = aZ, and the expectation of Y, E(Y) = bZ, where a and b are known constants and Z is a random variable.

    So the question is how would I calculate E(XY)?
    I was thinking that I could do the following:
    E(XY) = E(aZ,bZ)
    => E(XY) = ab*E(ZZ)
    => E(XY) = ab*E(Z^2)

    Is it correct to do this?? or how would I do it?
     
  2. jcsd
  3. Apr 18, 2007 #2

    Hurkyl

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    E(X) can't be equal to aZ: E(X) is a number, and aZ is a random variable. Are you sure you stated the problem right?


    Anyways, there generally aren't short-cuts to computing the expectation of a product of random variables.
     
  4. Apr 19, 2007 #3
    Yes, you're right I've stated the problem wrong! I can restate in another, much easier way.

    So basically I need to calculate E(XY), where E(X) = aE(Y), where the constant a is less than 1.

    So any ideas on how to go about calculating E(XY)??
    Any help or directions would be great!
     
    Last edited: Apr 19, 2007
  5. Apr 19, 2007 #4

    Hurkyl

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    Can't be done with the information given.
     
  6. Apr 19, 2007 #5
    What more information would I need to calculate this?
    For example, I know what the E(Y) and Var(Y) is going to be, I also know what the constant a is going to be.
    So I know what the mean and variance of X and Y are going to be and the constant a, so what more information do I need to get E(XY)?

    Overall I'm trying to calculate the Cov(XY) = E(XY) - E(X)E(Y), and seeming as X and Y are dependent, shouldn't I be able to work out the covariance between them??
    I think I have all the information neccessary to get this expression, I'm probably just not supplying it to you here?
     
  7. Apr 19, 2007 #6

    Hurkyl

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    Do you have their distribution? You can compute E(XY) directly, rather than looking for a shortcut involving other things you can compute.
     
  8. Apr 19, 2007 #7
    No unfortunately I'm unable to get the distribution of XY (if that's what you were talking about).
    I just have the mean and variance of X and Y to play with and the constant a.

    So when you say
    Is there a general formulae for calculating E(XY) for dependent variables??
    I could only find a formulae for independent variables.
     
  9. Apr 19, 2007 #8

    Hurkyl

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    [tex]E[XY] = \sum_{a, b} a b \mathop{\mathrm{P}}(X = a \mathrm{\ and\ } Y = b)[/tex]
    (Or an integral, if appropriate)
     
    Last edited: Apr 19, 2007
  10. Apr 19, 2007 #9
    Ok thanks for that, I'll have a look into it.

    Also I was thinking maybe I could do it the following way, but I'm not sure my
    "random variable algebra" is correct:

    So again suppose I need to calculate E(XY), where E(X) = aE(Y), where the constant a <= 1.
    We have E(X|Y) = aY
    => E(YX|Y) = aYY = aY^2
    => E(XY) = E(E(YX|Y)) = E(aY^2)
    => E(XY) = aE(Y^2)

    Would this be correct??
     
  11. Apr 29, 2007 #10
    You say you are given that [tex]E[X]=aE[Y][/tex]. This does not imply that [tex]E[X|Y]=aY[/tex]. As an example, suppose [tex]X\sim N(a,1)[/tex], [tex]Y\sim N(1,1)[/tex], and they are independent. Then [tex]E[X]=a=a\cdot 1=aE[Y][/tex], but [tex]E[X|Y]=E[X]=a[/tex]. Clearly, [tex]a\ne aY[/tex].

    If you are given that [tex]E[X|Y]=aY[/tex], then your calculations are correct.
     
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