Moment generating function (Probability)

[zetafunction]

given a probability distribution P(x) >0 on a given interval , if we define the moment generatign function

$$M(x)= \int_{a}^{b}dt e^{xt}P(t)dt$$

my question is , if the moment problem is determined, then could we say that ALL the zeros of M(x) are PURELY imaginary ? $$ia$$ or this is only for certain P(x) ??

 

[HallsofIvy]

given a probability distribution P(x) >0 on a given interval , if we define the moment generatign function

$$M(x)= \int_{a}^{b}dt e^{xt}P(t)dt$$

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

my question is , if the moment problem is determined, then could we say that ALL the zeros of M(x) are PURELY imaginary ? $$ia$$ 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

$$M(x)= \frac{1}{2}\int_{-1}^1 e^{xt}dt= \frac{e^x- e^{-1}}{2x}= \frac{sinh(x)}{x}$$

That is equal to 0 for x= 0.

 

[zetafunction]

but for $$x= 2\pi i m$$ 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 $$J_0 (ix)$$ that will have ONLY imaginary roots.

 

[bpet]

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?

$$M(x)= \frac{1}{2}\int_{-1}^1 e^{xt}dt= \frac{e^x- e^{-1}}{2x}= \frac{sinh(x)}{x}$$
That is equal to 0 for x= 0.

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