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We all know that a mgf of a random variable $m_X(t)$ is positive and $m(0)=1$. My question is: if a positive real function $f(t)$ satisfies $f(0)=1$ and the function is smooth enough (around 0), does there always exist a random variable $X$ whose mgf is $f(t)$? Or is there any other conditions that have to be met?

The origin of this problem is that I want to construct two independent random variables, $X_1$ and $X_2$, where $X_1$ follows uniform distribution between $[0,k]$, where $0<k<1$ and $X_1+X_2$ follows uniform distribution between $[0,1]$. $X_2$ apparently exists if $1/k$ is a positive integer, but what if $k$ is any real number?

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    $\begingroup$ There are more conditions. But anyway it is better to look at characteristic functions: the characteristic function of $X_1+X_2$ is the product of characteristic functions of $X_1$ and $X_2$, and if if $1/k$ is not an integer, then the zeroes of these functions mismatch. $\endgroup$ Nov 28, 2019 at 22:11
  • $\begingroup$ Thank you! This is very helpful! $\endgroup$ Nov 29, 2019 at 3:38

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Your question is equivalent to the existence of a distribution with the given moments, and that question is the so-called moment problem. The answer is "usually not".

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Just to give an idea of why the answer should be no: suppose $X$ is a random variable. To make life easy, I'll only talk about even moments, so no cancellation.

Jensen's inequality says $\mathbb{E} X^4\ge(\mathbb{E} X^2)^2$, with equality if and only if $X^2$ is almost surely constant. So not only is this an extra constraint (the second and fourth derivatives of your MGF at zero satisfy an inequality) but also if the constraint holds with equality it tells you all the even moments.

Similarly, $\mathbb{E} X^8\ge (\mathbb{E}X^4)^2$, and similarly equality if and only if $X^2$ is a.s. constant. What if this constraint doesn't actually hold with equality, but it's close to equality? Then $X^2$ has to be quite close to constant, and in particular $\mathbb{E} X^4$ has to be close to $(\mathbb{E} X^2)^2$ (one can quantify this). The higher moments in general tend to control the lower moments. (The other way round is false; putting a bit of probability mass on a very large number doesn't change lower moments much but does push up higher moments).

So even trying to characterise what are possible moments is not eay (and most things are not); and that's simply describing the derivatives at zero of an MGF. Now of course the point of an MGF is that it is its Taylor series at zero (i.e. it's analytic), but if you don't put that in as a condition, I can add an arbitrary function which is smooth, passes through zero and all of whose derivatives at zero are zero (non-constant such functions are well known to exist, e.g. $e^{-1/x^2}$).

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