For a given mean $\mu$, what is the entropy maximizing probability distribution on the nonnegative integers?

Different sources indicated either the geometric or the Poisson distribution for this. As I am new to the topic, it would be great if someone would give me a hint or point me to a source with a thorough explanation of these issues.


maximizing $S=\sum_{n=0}^\infty p_n \log p_n$ with the constraint $\sum_n n p_n=\mu$ and $\sum_n p_n =1$ gives $p_n = a b^n$, with Lagrange multipliers $a=1/(\mu+1)$ and $b=\mu/(\mu+1)$ determined by the constraints, so this is indeed a geometric distribution.

what is the source you are referring to that says the distribution is Poisson?

  • $\begingroup$ Just from the net. Probably because Poisson processes are often used to model the situation described above. $\endgroup$ Jan 24 '15 at 13:36
  • $\begingroup$ Hi, Actually there is a source for this "Binomial and Poisson Distributions as Maximum Entropy Distributions" which gives a proof that the maxent distribution for n going to infinity is poisson $\endgroup$
    – Pushpendre
    Jul 20 '15 at 19:47
  • 1
    $\begingroup$ EDIT : the above paper gives a proof that the binomial/poisson are maxent distribution amongst those distributions that are "n-generalized binomial distributions". In other words the binomial and poisson distribution are discrete maxent distributions with a given mean but only among the set of distributions which can be modeled as "n-generalized binomial". See the paper for the meaning of the term. $\endgroup$
    – Pushpendre
    Jul 20 '15 at 19:55
  • $\begingroup$ @Carlo Beenakker I ve been trying to do the calculation on my own. But I am not having much success in getting a=1/u+1 Using lagrange multiplier I’ve gotten up till Summation (pk) -1 =0 Summation (k * pk)- u=0 And -1-log(pk)=a*(1)+b*(k) $\endgroup$ Feb 13 '18 at 17:39
  • $\begingroup$ @Math.love.. --- These are the equations to solve for $a,b$ with $p_n=ab^n$: $\sum_n p_n = a/(1-b)=1$, $\sum_n np_n=ab/(1-b)^2=\mu$; the solution is $a=1/(\mu+1)$ and $b=\mu/(\mu+1)$. $\endgroup$ Feb 13 '18 at 20:32

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