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Finding the pdf from generating functions?

  1. Feb 4, 2007 #1
    finding the pdf from generating functions??

    The generating function of a Poisson distribution is given by
    f(z) = Exp[-lamda(1-z)],
    where lambda is the mean and variance of the poisson process.

    Now suppose I have an embedded Poisson process, that is, f(f(z)), the new generating function would then be
    f(f(z)) = Exp[-lambda(1-Exp[-lambda(1-z)])].

    Now the question is how to I get the probability density of f(f(z))??
    I know I could differentiate and put z=0, but the problem is that I need values for fairly large numbers, that is P(X=100).., hence is not really practical to get the 100th derivative of the generating function.

    I have been told I could use a Fast Fourier Transform, but after googling FFT and probability densities I couldn't really find anything comprehensiable for a layperson like me!

    So any suggestions as to how I would get the density of f(f(z)), or even some sort of approximation, so I can get an idea of what the distribution looks like? Any help would be great!!

    (I actually need to know what the distribution of f(f(f(......))), looks like, but I presume if I can work out f(f(z)), then extrapolating to n cases is similar??)
  2. jcsd
  3. Feb 4, 2007 #2
    i don't know if it's much help but
    if G(w) is Fourier transform of f(z) then
    the fourier transform of the nth derivative of f(z) is equale to (iw)^n multiplied by G(w).
    hence u can calculate the fourier transformation of the nth derivative and continue the calculation from there..
  4. Feb 4, 2007 #3
    So for example, if I had the generting function of say, f(f(f(f(z)))), (as defined in the first post), then the 200th derivative of f(f(f(f(z)))), is simply (iw)^200 multiplied by G(w), where G(w) is the Fouirer Transform of f(f(f(f(z))))??

    Sorry, I'm just trying to get my head around all this stuff, as I'm kinda new to this stuff and not too sure what I'm doing!
  5. Feb 4, 2007 #4
    nope , the fourier tranformation of nth derivative of f(f(f(f(z)))) is (iw)^n multiplied by g(w).
    when u know the fourier transformation , of a certain function , u can apply the inverse fourier transformation to obtain the function..
    ((iw)^200)*g(w) does not equale the nth derivative of f(f(f(f(z)))).
    read a bit from this site about the transformation and the inverse transformation
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