Why can a polynomial of degree $>2$ not be a cumulant generating function? Why can a polynomial of degree $>2$ not be a cumulant generating function?
I read somewhere that this is impossible but can't retrieve the source. 
The answer by StasK to Higher order generalization of the multivariate normal distribution mentions a related 
statement ''Once you depart from zero third cumulant, all higher order cumulants have to be non-zero, as well: there is no distribution for which $\kappa_4=0$ if $\kappa_3\ne 0$," but also gives no source.
 A: In the mean time, I found out that the result (rephrased in terms of characteristic functions) was first described in the paper
J. Marcinkiewicz, Sur une propriete de la loi de Gauss,
Mathematische Zeitschrift 44 (1939) 612-618.
The result is also proved on p.213 of
E. Lukacs,
Characteristic Functions, 2nd ed., 
Griffin, London 1970.
The proof is quite lengthy.
The remarks on p.224 there imply that the nonexistence of polynomial cumulant generating functions of degree $>2$ is a consequence of the fact that every entire cumulant generating functions $f(x)$ must satisfy a ridge property of the form $\Re f(x+it) \le f(x)$ for all real $x$ and $t$. (This is equivalent to the traditional ridge property $|c(t+iy)|\le c(iy)$ for the characteristic function $c(x)$, and can be obtained from the latter by taking the logarithm.)
A: For future reference, this set of lecture notes provides a rather succinct proof of the Marcinkiewicz theorem:
https://math.uc.edu/~brycw/probab/charakt/charakt.pdf
(and excels by being neither paywalled nor in French)
Seems the key is to reframe the problem as showing that the characteristic function for the difference between two variables with the polynomial characteristic must be a Gaussian, which can only be true if the original characteristic was that of a Gaussian itself. They then only have to work with even ordered polynomials as the difference variable induces cancellation. They then bound the characteristic function from below by showing the highest order must have a negative coefficient (for the whole function to be bounded) and from above by using Jensens inequality. This then leads to a contradiction unless the polynomial is a quadratic.
