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Expectation of XY

  1. Apr 29, 2005 #1
    X~N(0,1), Y=X^2~[tex]\chi^2[/tex](1), find E(XY).

    My thoughts are in the following:
    To calculate E(XY), I need to know f(x,y), since [tex]E(XY)=\int{xyf(x,y)dxdy}[/tex]. To calculate f(x,y), I need to know F(x,y), since f(x,y)=d(F(x,y)/dxdy.

    [tex]F(x,y)=P(X\leq x, Y\leq y) \\
    =P(X\leq x, X^{2} \leq y)\\
    =P(X\leq x, -\sqrt{y} \leq X \leq \sqrt{y})[/tex]
    Thus,
    [tex]F(x,y) =P(-\sqrt{y} \leq X \leq x)P(x<\sqrt{y})+P(-\sqrt{y} \leq X \leq \sqrt{y})P(x > \sqrt{y})[/tex]
    Then I don't know how to calculate the four components of probabilities accordingly. Anyone gives a hand?

    Thanks!
    gim :bugeye:
     
    Last edited: Apr 29, 2005
  2. jcsd
  3. Apr 29, 2005 #2
    I think you're making this too hard for yourself. x and y have 100% correlation. I think you essentially want to calculate the third moment of a normal distribution, since x*y = x^3. So find the expected value of x^3 = the third moment.
     
  4. May 1, 2005 #3
    Thank you for your useful hint! The result following your method is E(XY)=0, then cor(X,Y)=0. In this sense X and Y are uncorrelated, but they are fully associated.
    gim
     
  5. May 2, 2005 #4
    I am still wondering the joint distribution of X and Y. There must be a solution to that. If it is not too difficult, please give me some hints.
    Thanks!
    gim
     
  6. May 2, 2005 #5
    f(x,y) = f(x) * f(y|x)

    So, you need f(y|x). However, once you know x, you know y exactly, so
    f(y|x) = delta function(y - x^2).

    So f(x,y) = f(x) * delta function(y-x^2).

    I'm not sure if I've seen delta functions outside of physics, actually. Here's a writeup I found:

    http://www.tutorfusion.com/eTutor/physics/e&m/1/5/1_5_dirac_delta_function.htm

    If you don't want to use delta functions, I guess you could just say:

    f(x,y) = f(x) when y= x^2
    = 0 otherwise
     
    Last edited: May 2, 2005
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