Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook