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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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    $\begingroup$ 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. $\endgroup$
    – StasK
    Commented Oct 29, 2012 at 19:43
  • $\begingroup$ @StasK: Probably yes, but I am looking for the argument demonstrating this. $\endgroup$
    – Arnold Neumaier
    Commented Oct 30, 2012 at 10:36

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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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  • $\begingroup$ (+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. $\endgroup$
    – whuber ♦
    Commented Nov 8, 2012 at 16:15
  • $\begingroup$ @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. $\endgroup$
    – Arnold Neumaier
    Commented Nov 8, 2012 at 17:32
  • $\begingroup$ 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? $\endgroup$
    – whuber ♦
    Commented Nov 8, 2012 at 17:34
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    $\begingroup$ @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. $\endgroup$
    – Arnold Neumaier
    Commented Nov 8, 2012 at 18:02
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    $\begingroup$ 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. $\endgroup$
    – whuber ♦
    Commented Nov 8, 2012 at 18:07
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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.

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