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Moment generating function (Probability)

  1. Dec 12, 2009 #1
    given a probability distribution P(x) >0 on a given interval , if we define the moment generatign function

    [tex] M(x)= \int_{a}^{b}dt e^{xt}P(t)dt [/tex]

    my question is , if the moment problem is determined, then could we say that ALL the zeros of M(x) are PURELY imaginary ? [tex] ia [/tex] or this is only for certain P(x) ??
  2. jcsd
  3. Dec 12, 2009 #2


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    Too many "dt"s! You mean
    [tex]M(x)= \int_a^b e^{xt}P(t)dt[/tex]

    Look at a very simple example: x is uniformly distributed between -1 and 1. That is, P(x)= 1/2 for all x between -1 and 1. Then

    [tex]M(x)= \frac{1}{2}\int_{-1}^1 e^{xt}dt= \frac{e^x- e^{-1}}{2x}= \frac{sinh(x)}{x}[/tex]

    That is equal to 0 for x= 0.
  4. Dec 12, 2009 #3
    but for [tex] x= 2\pi i m [/tex] is 0 for every integer 'm' hyperbolic sine has ALL pure imaginary roots, perhaps in order to have only imaginary roots you only need positive and even probability distribution P(x)= P(-x) for example surely you can try with a certain probability distribution to get a similar result for Bessel function [tex] J_0 (ix) [/tex] that will have ONLY imaginary roots.
  5. Dec 12, 2009 #4
    Perhaps you could elaborate a bit further on how you arrived at the conjecture?

    Doesn't M(0)=1 for all probability distributions?
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