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The normal distribution on $\mathbb{R}$, the exponential distribution on $\mathbb{R}_{\geq 0}$, and the geometric distribution on $\mathbb{N}$ are examples of distributions that are both infinitely divisible and entropy maximizers. On the other hand, the Poisson distribution is an infinitely divisible distribution on $\mathbb{N}$ without maximizing entropy, while the uniform distribution on the interval $[a,b]$ maximizes entropy but is not infinitely divisible.

Can anything be said about the relationship between these two classes of distributions?

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  • $\begingroup$ I don't see a-priori why there should be some intrinsic relationship between infinite divisibility and max-ent, but maybe i am missing some intuition here. $\endgroup$
    – Suvrit
    Jun 18, 2011 at 13:31
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    $\begingroup$ Doesn't the Poisson distribution maximize entropy with respect to the measure that assigns $1/n!$ to $\{n\}$ for $n \in \{0,1,2,\ldots\}$, among all measures on $\{0,1,2,\ldots\}$ having a given expected value? (BTW, a nice way to see istantly that the uniform distribution on $[0,1]$ is not infinitely divisible is that its fourth cumulant is negative. The even-degree cumulants of infinitely divisible distributions are non-negative.) $\endgroup$ Jun 19, 2011 at 16:01
  • $\begingroup$ So I see I posted a remark 11 years ago, and now I think there is a mistake there. It is true that among compound Poisson distributions, the fourth cumulant is never negative, but now I think that is probably not true of infinitely divisible distributions generally. $\endgroup$ Oct 14, 2022 at 20:34

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The Poisson distribution has maximum entropy under the condition of having a specific mean and being a sum of Bernoulli random variables. The condition of being a sum of Bernoulli random variables can be weakened to being ultra log-concave.

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  • $\begingroup$ Is this still true if one considers sums of integer multiples of Bernoulli random variables? $\endgroup$ Mar 28, 2015 at 20:40

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