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What is (is there) the count process, which has its standard deviation proportional to its mean?

Note that I am not talking here about Poisson process, which has its variance proportional to mean. That is the process in question is overdispersed (in comparison to Poisson) for the average number of counts greater than 1, but underdispersed otherwise.

Background:
One often speaks of count processes in terms of how they modify the variance of a Poisson process: e.g., negative binomial is seen as overdispersed Poisson, while generalized Poisson process is claimed to represent the underdispersed Poisson. The question is thus whether there's a particular modification for modeling the situations where the standard deviation (rather than the variance) is proportional to the mean.

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Mean-variance relationships are defined for families of probability distributions, especially families indexed by a single parameter. As we vary the unknown parameter, both the mean and the variance of the distribution may change. Viewing the variance as a function of the mean defines the mean-variance relationship.

The answer to your question depends on how you define a "count process". One can easily define a family of probability distributions on the non-negative integers such that the expected value and standard deviation are always equal. However, such a family would be highly artificial and IMO unsuitable for any real-world data analysis.

Whether you want to call such a family a "count process" is up to you. Any distribution on the non-negative integers can be represented as a non-homogeneous Poisson process where the rate parameter depends on the number of events to that point.

The Poisson distribution defines a linear exponential family of probability distributions as the expected value varies. There is no linear exponential family on the non-negative integers with standard deviation equal to or proportional to the mean. So there is no "nice" distributional family with the mean-variance relationship that you specify.

If your counts are large, then the negative binomial family is what you are looking for. The negative binomial standard deviation is proportional to the expected value when the expected value $\mu$ is large. This relationship cannot hold however for small means. Indeed it can be proved that any linear exponential family on the non-negative integers converges to Poisson as $\mu$ converges to zero.

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  • $\begingroup$ +1 what you say makes sense. What I head in mind is that one often speaks of count processes in terms of hoy they modify variance of the Poisson process: e.g., negative binomial is seen as overdispersed Poisson, while generalized Poisson process is claimed to represent the underdispersed Poisson. The question is thus whether there's a particular modification for modeling the situations where standard deviation (rather than the variance) is proportional to the mean. $\endgroup$
    – Roger V.
    Feb 16, 2023 at 13:39
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    $\begingroup$ @RogerVadim I assume that by Generalized Poisson (GP) you mean the distribution proposed by Consul & Jain (Technometics 1973). You are using "counting process" to be synonymous with "count distribution" but they are not the same thing. Anyway, I understood your question. No one has ever proposed a count distribution family with SD proportional to the mean. I doubt that anyone ever will because counts just don't behave that way, either mathematically or in the real world. $\endgroup$ Feb 17, 2023 at 4:52
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    $\begingroup$ @RogerVadim Even if such a distributional family existed, your remark about the distribution being overdispersed for mu>1 and underdispersed for mu<1 would only be true if the proportionality constant (the dispersion parameter) was equal to 1. $\endgroup$ Feb 17, 2023 at 4:53

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