cumulants and infinite divisibility 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.
 A: 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 
  $$
\kappa_1 = a + \int_{\mathbb{R} \setminus 0} \frac{x^3}{1+x^2} \, \mathrm{d} M(x), \quad
\kappa_2 = \sigma^2 + \alpha_2, \quad \kappa_n = \alpha_n \text{ if } n \geqslant 3.
$$
  where $\alpha_n$ is the $n$-th order moment of $M$, i.e. $\alpha_n = \int_{\mathbb{R} \setminus 0} x^n \, \mathrm{d} M(x)$.

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.
