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Where might I find a clear exposition of how to prove that a real-valued probability distribution for which all moments exist is infinitely divisible if and only if all of its cumulants of even order are non-negative? I know how to do that for compound Poisson processes, by using the law of total cumulance. I suspect you need to get into something more delicate for the more general case.

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up vote 2 down vote accepted

The book "Infinite Divisibility of Probability Distributions on the Real Line" by Fred W. Steutel and Klaas van Harn states (Theorem 7.4, see Google books):

Let $X$ be an infinitely divisible random variable with canonical triple $(a, \sigma^2, M)$, let $n\in \mathbb{N}$ and let $X$ have finite moment $\mu_n$ of order $n$. Then $n$-th order cumulant $\kappa_n$ of $X$ is given by $$ \kappa_1 = a + \int_{\mathbb{R} \backslash 0} \frac{x^3}{1+x^2} \mathrm{d} M(x), \quad \kappa_2 = \sigma^2 + \alpha_2, \quad \kappa_n = \alpha_n \text{ if } n \geqslant 3. $$ where $\alpha_n$ is the $n$-th order moment of $M$, i.e. $\alpha_n = \int_{\mathbb{R} \backslash 0} x^n \mathrm{d} M(x)$.

As a corollary to this theorem, even order cumulants, if they exists, are non-negative $\kappa_{2n} \geqslant 0$.

This takes care of one direction.

The counterexample to the other direction has been provided by Herman Rubin. Let $Z$ be a symmetric continuous random variable with the density $$ f_Z(z) = \frac{|z|}{2} \exp\left( -|z| \right) $$ Its moment generating function is $\mathcal{M}_Z(t) = \frac{1+t^2}{(1-t^2)^2}$, which means $\kappa_{2r+1}=0$ and $\kappa_{2r} = (r-1)! \left(2-(-1)^r\right) > 0$. The distribution is not infinitely divisible, as a symmetric infinitely divisible distribution is necessarily unimodal with the mode at the point of symmetry.

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Thank you. I now remember that this was done in another forum about five years ago---I'd forgotten. – Michael Hardy Jan 14 '12 at 0:38
@A.S. Thank you. I have applied the correction. – Sasha Jan 27 at 21:32

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