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Is there a way to verify that Bivariate Normal correlation?

  1. Oct 28, 2015 #1
    My main difficulty is finding E(XY), where if we find it by integrating, we get ∫xy f(x,y) dy dx where f(x,y) is the bivariate normal density function. This integral is a nightmare... are there any tricks to evaluating this integral, or finding E(XY) for that matter?
     
  2. jcsd
  3. Oct 28, 2015 #2

    Mark44

    Staff: Mentor

    I think that this would be your integral: ∫ ∫ xy f(x,y) dy dx; that is, an iterated integral. Because of the presence of ##e^{-x^2 - y^2}## (I'm really simplifying here), a standard trick is to change the integral to polar coordinates instead of Cartesian coordinates, although the presence of xy in the integrand might make the integral doable without changing to polar coordinates.
     
  4. Oct 29, 2015 #3
    I apologize, that was indeed what I meant by my integral. There is also an e^(-xy) term in the bivariate normal density, do you think it's still possible without transforming the integral to polar coordinates?

    And thank you for the reply.
     
  5. Oct 29, 2015 #4

    Mark44

    Staff: Mentor

    It's hard to say without seeing the integral. The trick I suggested was the standard one for evaluating ##\int_{-\infty}^{\infty} e^{-x^2}dx##. In Cartesian form, this integral can't be done, but by looking at the related iterated integral, and converting to polar form, the integration is straightforward.
     
  6. Oct 29, 2015 #5
    I actually tried that approach, but it's not quite clear cut.

    I am not sure how to put TeX into these forums, but the PDF of the bivariate normal density can be found in this link: http://mathworld.wolfram.com/BivariateNormalDistribution.html

    Do you have any advice on how to find E[XY] with that PDF? That is, double integrating over 2-dimensional reals xy f(x,y).
     
  7. Oct 29, 2015 #6

    Mark44

    Staff: Mentor

    What are the means and st. deviations of your distribution? If both variables are N(0, 1) the formula is a lot simpler. It would be helpful to actually see the integral you're trying to evaluate.
     
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