A mean-matched count law can be overdispersed relative to Poisson

Given: The two-point count law in the Example and a comparison Poisson GLM with $a(\varphi )=1$, $b(\theta )=e^{\theta }$, and $\theta =\log 10$.

Generalized linear model with explicit conditioning: a GLM pairs a response family with a link $g({\mu }_i)={\eta }_i=x_i^{\mathsf{T}}\beta$ and a natural parameter map satisfying $b^{\mathrm{\prime }}({\theta }_i)={\mu }_i$.

Mean and variance from the cumulant function: for the stated family, mean is $b^{\mathrm{\prime }}(\theta )$ and variance is $a(\varphi )b^{\mathrm{\prime }\mathrm{\prime }}(\theta )$.

For the comparison Poisson response, the support is ${\mathbb{N}}_0$, the base measure is counting measure, and $f(k;\theta )=e^{-e^{\theta }}{e}^{k\theta }\mathrm{/}k!$. The exponential series and addition formula normalize these masses. Its first two parameter derivatives are ${\mathrm{\partial }}_{\theta }f(k;\theta )=f(k;\theta )(k-e^{\theta })$ and ${\mathrm{\partial }}_{\theta }^{2}f(k;\theta )=f(k;\theta )\{(k-e^{\theta }{)}^2-e^{\theta }\}$. On any compact interval $[u,v]$ they are dominated by the summable sequence $e^{vk}\{1+(k+e^v{)}^2+e^v\}\mathrm{/}k!$; the ratio test gives summability. Reindexing the same exponential series gives $\mathbb{E}[Y^2]={\lambda }^2+\lambda <\mathrm{\infty }$ for $\lambda =e^{\theta }$, so the family satisfies the fixed-support, moment, and derivative-envelope hypotheses. (The real exponential function and the number $e$ by a power series, The exponential addition formula $\exp (x+y)=\exp (x)\exp (y)$, The factorial $n!$ and the falling factorial $n^{\underset{‾}{k}}$, defined by recursion in $\mathbb{N}$, The exponential function is smooth and $(\exp {)}^{\mathrm{\prime }}=\exp$)