Timeline for Discrete Maximum Entropy Distribution with given mean
Current License: CC BY-SA 3.0
8 events
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| Feb 13, 2018 at 20:32 | comment | added | Carlo Beenakker | @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 $b=\mu/(\mu+1)$. | |
| Feb 13, 2018 at 17:39 | comment | added | Math.love.. | @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) | |
| Jul 20, 2015 at 19:55 | comment | added | Pushpendre | 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. | |
| Jul 20, 2015 at 19:47 | comment | added | Pushpendre | 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 | |
| Jan 24, 2015 at 13:36 | comment | added | J Fabian Meier | Just from the net. Probably because Poisson processes are often used to model the situation described above. | |
| Jan 24, 2015 at 13:35 | vote | accept | J Fabian Meier | ||
| Jan 24, 2015 at 12:49 | history | edited | Carlo Beenakker | CC BY-SA 3.0 |
added 2 characters in body
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| Jan 24, 2015 at 12:44 | history | answered | Carlo Beenakker | CC BY-SA 3.0 |