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Model to be chosen if Poisson distribution mean and variance are not the same, say If mean is greater than variance or variance is greater than mean?

KARThik
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  • You're correct that if the mean and variance aren't the same, the distribution is not Poisson. Beyond that, there's no general answer to your question. It's as if you asked "I have an animal that is not a cow. What animal is it?" – pjs Oct 07 '17 at 17:38
  • My question was which model or models can be analyzed if Poisson is not the right one for the count distribution? – KARThik Oct 08 '17 at 00:19
  • Yes, I understood what you were asking. I was trying to use humor to point out that your question is totally open-ended. Observing some fact that shows your distribution is not Poisson gives no hint as to the vast number of things it could be instead. – pjs Oct 08 '17 at 00:47
  • Your title implies that you can have a Poisson distribution with mean and variance that differ. If they actually differ, it won't be Poisson; it seems odd to suggest that it is Poisson. Do you mean to say "count data" or "count variable" instead? Do you have any predictor variables? (e.g. where you might have been considering a GLM/Poisson regression) – Glen_b Oct 08 '17 at 05:25
  • The term to search for is 'over-dispersion'. https://en.wikipedia.org/wiki/Overdispersion – Paul McGee Apr 08 '22 at 16:47

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If you only have mean and variance, and they are not equal, obviously you have to try two-parametric discrete distribution. From the top of my head:

  • Binomial
  • Negative binomial
  • Hypergeometric distribution
  • Negative Hypergeometric
  • Compound distributions like Gamma-Poisson mix
Severin Pappadeux
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  • Only knowing two of the moments does not limit the distribution to two-parameter distributions. – pjs Oct 12 '17 at 13:39
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    @pjs that's true, but Occam tells us to start with two-parametric distributions first before digging deeper hole – Severin Pappadeux Oct 12 '17 at 21:19
  • Have to disagree with that, Occam tells us to pursue likely outcomes before pursuing unlikely ones. There's no reason to believe any of the ones you listed is any more likely than the good old multinomial, for instance. – pjs Oct 12 '17 at 23:41