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Consider a random variable $x$ with pdf $f(x)$, and have $x \ge 0$. The moment generating function is defined as $M(t)=\int^{\infty}_{-\infty}e^{-tx}f(x)dx$ (noted that we change the sign of $t$ compared with common case ). For convenient we consider only when $t>0$. The problem is: When would a function be the moment generating function of a distribution? There are some trivial conditions like $M(0)=1$ and $M(t)$ is strictly decreasing. But obviously these are not enough.

An observation is that the $M(t)$ should lie in the convex hull of $e^{-tx}$. From inequality techniques we have $\log M(t)$ up convex. Is this condition sufficient? An related but a bit different problem is given $t_1,t_2,\dots,t_n$ and $M_1,M_2,\dots,M_n$ when there exist $M(t)$ such that $M(t_i)=M_i$?


Add Jul 12:

The reason that the up convex condition is not enough are as follows: Consider a function $\bar M(t)$ that it's logarithm is linear in an interval, using Cauchy inequality we can demonstrate that $\bar M(t)$ is composed of one $e^{-tx}$. But we can find other continuation.

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    $\begingroup$ Don't restrict yourself to real $t$. $M(it)$ is, up to some constants, the Fourier transform of $f$. $\endgroup$ – Thomas Andrews Jul 12 '13 at 13:02
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Take a look at this http://en.wikipedia.org/wiki/Bernstein's_theorem_on_monotone_functions

completely monotone functions for example have a "moment generating function representation"

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