7
$\begingroup$

A couple of days after I posted this to stackexchange, no one's answered:

I take the problem of cumulants to be this: given a sequence $(\kappa_1,\kappa_2,\kappa_3,\ldots)$ of real numbers, 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. Even describing or classifying the (topologically) closed subsets of $\mathbb{R}^+$ that are closed under addition might be something of a chore, and I suspect many of them are not of this form.

That moderately complicated things might happen is at least hinted at by the case of $p\times p$ Wishart matrices, for which the number of degrees of freedom can be anywhere in the set $\lbrace 0,\dots,p-1\rbrace\cup (p-1,\infty)$. But that's for matrix-valued, rather than real-valued, random variables, so it's at most a hint.

$\endgroup$
5
  • 1
    $\begingroup$ Can you include a link to the stackexchange question? Have you left a link there to the MO version? $\endgroup$ Mar 8, 2012 at 22:10
  • $\begingroup$ Done. ${{{{{}}}}}$ $\endgroup$ Mar 8, 2012 at 22:29
  • 1
    $\begingroup$ One condition mentioned on Wikipedia (the source of all truth), is that the cumulant generating function can't be a polynomial of degree greater than 2. (The normal distribution manages degree 2.) So you can't have finite number of non-zero cumulants except in some special cases. $\endgroup$ Mar 9, 2012 at 0:46
  • 1
    $\begingroup$ ....and those special cases are the ones where all cumulants of degree $3$ or higher are $0$. If any cumulant of degree $\ge3$ is non-zero, then infinitely many are non-zero. However, this question is about the set of values of $t\ge 0$ for which $t\kappa$ is a cumulant sequence. $\endgroup$ Mar 9, 2012 at 18:39
  • 1
    $\begingroup$ The moment problem is solved and understood to a certain extent. See arxiv.org/pdf/2008.12698.pdf. I'm also looking for if there's a nice translation in terms of cumulants though.. $\endgroup$
    – Student
    Nov 16, 2023 at 16:09

1 Answer 1

7
$\begingroup$

This quote might explain why your question on MSE went unanswered:

There is one obstacle that stands in the way of dealing with cumulants. We do not know simple necessary and sufficient inequalities on cumulants that ensure that a given sequence of real numbers is the sequence of cumulants of a random variable. For characteristic functions, such a necessary and sufficient condition is Bochner's theorem. To be sure, such inequalities exist in abundance if one expresses moments in terms of cumulants. The problem is finding the simplest such inequalities. The explicit determination of fundamental inequalities satisfied by the cumulants of a random variable is surely the most important open problem in probability theory.

Gian-Carlo Rota, 1998: Twelve problems in probability no one likes to bring up - Problem eleven: cumulants.

A status update has been given by Di Nardo and Senato, 2009, unfortunately behind a pay wall.

$\endgroup$
10
  • 3
    $\begingroup$ Unfortunately G-C Rota passed taking away with him potential explanations of many mysterious statements like the one in the quote above...Somewhat related is this MO question: mathoverflow.net/questions/107526/… $\endgroup$ May 20, 2015 at 15:32
  • 1
    $\begingroup$ While Rota was writing that paper it grew from 10 problems to 14 and then contracted to 12. He published a number of top-10 lists not long before he died, including "Ten lessons I wish I had been taught" and some others. He wrote something called something like "Ten observations on teaching differential equations" but I don't know if he ever published it. $\endgroup$ May 20, 2015 at 18:00
  • 1
    $\begingroup$ @Michael Hardy: Published or not, here it is: web.williams.edu/Mathematics/lg5/Rota.pdf $\endgroup$ Apr 20, 2017 at 14:22
  • 1
    $\begingroup$ QUOTE FROM THE FIRST PARAGRAPH: One of many mistakes of my youth was writing a textbook in ordinary differential equations. It set me back several years in my career in mathematics. However, it had a redeeming feature: it led me to realize that I had no idea what a differential equation is. The more I teach differential equations, the less I understand the mystery of differential equations. END QUOTE $\qquad$ $\endgroup$ Apr 20, 2017 at 15:01
  • 1
    $\begingroup$ QUOTE FROM A LATER PARAGRAPH: The Administrative Director of the MIT mathematics department, who exercises supreme authority upon the faculty’s teaching, has only to wave a copy of my book at me, while staring at me in silence. At her prompting, I bow and fall into line; I will be the lecturer in the dreaded course for one more year, and I will repeat the mistakes I have been making every year since I first taught differential equations in 1958. END QUOTE The said administrative directore, Joanne Jonson (or Jonnson, or Jonsonn, or something else like that with no "h"), told me$\,\ldots\qquad$ $\endgroup$ Apr 20, 2017 at 15:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.