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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.

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I am not sure I can put a finger on either my earlier claim or this fact, but I suspect that it has to do either with Bochner's theorem... which is to say that you cannot invert such defined ch.f. to obtain a function that would behave like a distribution function. – StasK Oct 29 '12 at 19:43
@StasK: Probably yes, but I am looking for the argument demonstrating this. – Arnold Neumaier Oct 30 '12 at 10:36
up vote 3 down vote accepted

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.)

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(+1) However, I believe the "ridge property" compares the absolute value of $f(z)$ to $f(\mathbb{Re}\ z)$ and that the comparison is enforced only for $\mathbb{Re}\ z \gt 0$. Otherwise, even quadratic polynomials would be ruled out. – whuber Nov 8 '12 at 16:15
@whuber: $f(z)$ is the log of the analytically continued characteristic function, hence the absolute value in the ridge property for the latter translates into that for the real part of $f$. If $f(z)=z^2$ then $\Re f(x+it)=x^2-t^2\le x^2=f(x)$, as stated. – Arnold Neumaier Nov 8 '12 at 17:32
I think something is missing, Arnold: your answer asserts that the CGF must satisfy the ridge property, not the CF. Your comment, on the contrary, states that the CF must satisfy the ridge property. Which is correct? – whuber Nov 8 '12 at 17:34
@whuber: in my first comment: latter = char.function. I didn't notice that my formulation was ambiguous. - In my answer, the ridge property as formulated there (with $\Re$ rather than absolute values) applies to the CGF, consistent with the quadratic case. Whereas the correct ridge property of the exponential must have an absolute value. – Arnold Neumaier Nov 8 '12 at 18:02
Perhaps. I think my concerns began when an earlier formulation of the property was edited; I was reacting to the change. What I would like to suggest is that if your characterization of the ridge property is not in the same form found in your reference, then it would help future readers to quote the reference verbatim and then restate it in the form you find suitable, so that they can see the connection. – whuber Nov 8 '12 at 18:07

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