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I used to believe that the negative binomial distribution (for count data) and Gamma distribution (for continuous data) shared the property that the variance can take arbitrary values regardless of the mean, which makes them quite flexible as opposed to the Poisson distribution.

However I've recently seen in a variety of places that statisticians consider the variance to be (proportional to) the square of the mean (the Wikipedia Variance function page even plainly claims that $V=E^2$ by the way). To me, this seems to be vacuously true just like any two non-zero real numbers are proportional to each other.

The main point that I'm trying to make is that when you use a GLM based on the Gamma or negative binomial families, the coefficient of proportionality (the shape parameter) can vary with the IV, it doesn't have to be constant.

What am I missing?

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    $\begingroup$ This is a great question. As I was researching to propose an answer, I bumped into two additional points of confusion. First, while I knew the the shape parameter could vary with the IV, I assumed that this parameter was assumed constant and estimated in the GLM (not sure now if that is correct). Second, I interpreted the variance function to be a function that converted the mean to the variance...up to a constant of proportionality. I.e., I thought $V(\mu) = \mu^2$ meant $V(\mu) = k \cdot \mu^2$. I'm hoping the submitted answers might address these sub-queries. $\endgroup$ – Gregg H May 4 at 13:04
  • $\begingroup$ The question is not vacuous if the proportionality coefficient is not an extra parameter. $\endgroup$ – Xi'an May 4 at 13:17
  • $\begingroup$ @Xi'an I consider that when you know the mean, you don't know the two parameters yet. You can't derive the shape parameter from the only datum of the mean, therefore the coefficient is an extra parameter from this point of view. But then again maybe I'm missing something. Could you expand this comment into an answer? $\endgroup$ – Arnaud Mortier May 4 at 19:24
  • $\begingroup$ What I mean is the case of families where $\mu_\theta = c \sigma^2_\theta$ and $c$ remains the same for all $\theta$'s. $\endgroup$ – Xi'an May 4 at 19:28
  • $\begingroup$ @Xi'an this joins the second point raised by Gregg H. If this is indeed an assumption always made that $c$ is constant across the whole model in a GLM(M), then this would be worth an answer. I was not aware of that. $\endgroup$ – Arnaud Mortier May 4 at 19:33

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