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Let $X$ be a random variable. It is well-known that $X$ is normally distributed if and only if its last cumulants $\kappa_3 = \kappa_4 = ... = 0$ vanish. I was wondering if there are standard distributions satisfying $\kappa_m = ... = \kappa_n = 0$ for arbitrary $m<n\le\infty$, especially for $m=3$.

Marcel
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  • Any symmetric distribution will have $\kappa_3=0$. – A.S. Mar 08 '16 at 13:05
  • That's true. What about higher moments vanishing? – Marcel Mar 09 '16 at 09:02
  • By standard distributions do you mean to enumerate and discuss common distributions instead of constructing one? – Ѕᴀᴀᴅ Jun 12 '18 at 00:25
  • @AlexFrancisco Yes, a common distribution would be nice as those are the ones people care more about. Nevertheless, a constructed one would be a starting point. – Marcel Jun 12 '18 at 14:20
  • @Marcel Then Wiki already lists cumulants of quite some common distributions. – Ѕᴀᴀᴅ Jun 12 '18 at 14:37
  • @AlexFrancisco True, but those don't satisfy the above property that some of them are vanishing. – Marcel Jun 13 '18 at 12:19
  • @AlexFrancisco Well, then think of what I call common in a more general way. Wiki only mentions very few distributions; there are much more distributions out there which are studied and which I am probably not be aware of. Sorry for the confusion. And as I said, a constructed one would be a starting point at least. – Marcel Jun 13 '18 at 13:07

1 Answers1

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The answer is no: the Normal distribution is the only one with a finite polynomial as cumulant generating function. See http://www.scholarpedia.org/article/Cumulants#Uniqueness

The proof is credited to Marcinkiewicz (1939). Marcinkiewicz, Math Z (1939) 44: 612. https://doi.org/10.1007/BF01210677

Edit: I read the post a bit too quickly; this applies to $n=\infty$, not the case when $n$ is finite.

Christopher
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