Moment generating function (Probability)

  • #1
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Main Question or Discussion Point

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) ??
 

Answers and Replies

  • #2
HallsofIvy
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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]
Too many "dt"s! You mean
[tex]M(x)= \int_a^b 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) ??
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.
 
  • #3
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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.
 
  • #4
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my question is , if the moment problem is determined, then could we say that ALL the zeros of M(x) are PURELY imaginary ?
Perhaps you could elaborate a bit further on how you arrived at the conjecture?


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

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