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Characteristic function

  1. Oct 5, 2006 #1
    Hi all,

    Im currently researching into Multivariate distributions, in particular Im trying to derive the characteristic function of the bivariate distribution of a gaussian. While knowing that a gaussian density function cannot be integrated how is it possible to find the characteristic function. I have been working on it but I keep bumping in to a dead end. The following is what I did in the most recent dead end: (does anybody know why the latex thing aint working? I havnt been on here for a while)

    so I have a double integral of the

    exp{i*x*t_x+i*y*t_y}*(joint density function)

    then I did the substitutions
    u=x-m_x
    and
    v=y-m_y

    I then simplified it down such that I get the following in the exponential expression

    u^2*s_y^2-2*(si)*u*v*s_x*s_y+v^2*s_x^2

    where si is the correlation coefficient, the problem is I cant complete the square because of the existance of si, would anybody know what I would need to do?

    I had looked at Wolfram Mathworld website, its very interesting how it uses Eulers formula but isnt there any other way of doing it?

    Regards

    Steven
     
  2. jcsd
  3. Oct 6, 2006 #2

    mathman

    User Avatar
    Science Advisor
    Gold Member

    let w=au+bv and z=bu-av, where a^2+b^2=1. There will be some value of a where there will not be a cross product term for wz. Good luck!
     
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