# Problem of cumulants

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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Are you asking if there exists a family of distributions, whose cumulants are closed under multiplication by a constant from a said set of non-negative reals closed under addition, and if so for which such sets besides non-negative reals and non-negative integers ? – Sasha Mar 6 '12 at 14:27
I've done some further editing to elaborate on the question a bit. – Michael Hardy Mar 6 '12 at 17:28
@DidierPiau : In Q3, obviously it would be absurd to accept either answer. I up-voted all three answers. But none actually answered the question. In Q2, I up-voted both answers. Both fully answer the question. There doesn't seem to be any particular reason to prefer one as the "accepted" answer. I'll look at the others that you cite. – Michael Hardy Mar 6 '12 at 20:27
@DidierPiau : ...and I wasn't "playing silly"; rather I was giving you the benefit of the doubt. Obviously my guesses as to your meaning have the possibility of being mistaken, whereas an explicit statement from you my resolve any doubts. – Michael Hardy Mar 6 '12 at 20:29
@DidierPiau : I've started a thread on meta about this: meta.math.stackexchange.com/questions/3772/… In response to a similar question, someone expressed and opinion that there should be no codified standards. In the question you pointed out about probability textbooks, I would never in a million years have guessed that someone would be offended by the lack of an "acceptance" when none of the three answers actually answers the question. I up-voted all of them because all of them are of some value. – Michael Hardy Mar 6 '12 at 21:44