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I take the problem of cumulants to be this: given a sequence $(\kappa_1,\kappa_2,\kappa_3,\ldots)$, is it the sequence of cumulants of some probability distribution? In one sense, this is trivially equivalent to the problem of moments: the $n$th moment is a polynomial in the first $n$ cumulants and vice-versa. But cumulant sequences have a nice property that moment sequences don't have: the set of all such sequences is closed under addition. So draw a ray out from the origin $(0,0,0,\ldots)$. If the ray bumps into a cumulant sequence $(\kappa_1,\kappa_2,\kappa_3,\ldots)$, then $2(\kappa_1,\kappa_2,\kappa_3,\ldots),3(\kappa_1,\kappa_2,\kappa_3,\ldots),\ldots$ are also cumulant sequences.

For infinitely divisible distributions, for every real $t\ge 0$, the sequence $t(\kappa_1,\kappa_2,\kappa_3,\ldots)$ is a cumulant sequence.

Besides the nonnegative integers and the nonnegative reals, there are other sets of nonnegative reals closed under addition.

For which sets $T$ of nonnegative reals that are closed under addition is it the case that for some cumulant sequence $(\kappa_1,\kappa_2,\kappa_3,\ldots)$, for every real $t\ge 0$, $ t\in T \quad\text{iff}\quad t(\kappa_1,\kappa_2,\kappa_3,\ldots)\text{ is a cumulant sequence ?} $

(I'm guessing only closed sets, but there should be more than that to say about it, I would think.)

Later edit: Since there's no mad rush to answer this here, I've posted it to mathoverflow.

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    @zyx The part I elected as my favorite is "It is hard to read them as anything except..." So nasty, so pompous (and so wrong, of course). O well.2013-03-15

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