 Solution

An exponential dispersion family has density or mass function, relative to a fixed measure,Here $\theta$ is the natural parameter, $b$ is the cumulant function, and $\varphi$ is the dispersion parameter. Assume the natural parameter lies in the interior of its domain and derivatives can pass through the normalizing integral. Differentiating normalization once and twice givesThe variance function is $V(\mu )=b^{\prime \prime }((b^{\prime }{)}^{-1}(\mu ))$, where the mean-to-natural-parameter inverse exists. The dispersion parameter may be fixed, as in a Poisson distribution, rather than estimated.