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The dispersion parameter scales the variance of an exponential dispersion family while leaving its mean function unchanged.

Writing $\varphi =1/a$, $\theta =-r/a<0$, and $b(\theta )=-\log (-\theta )$ puts the gamma density into exponential dispersion family formIts mean is $a/r$ and its variance is $a/r^2$.

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.

For $y=0,1,2,\dots$ and $\lambda >0$, the Poisson distribution has massMatch this to the exponential dispersion family withThen $\mu =b^{\prime }(\theta )=e^{\theta }=\lambda$ and $V(\mu )=b^{\prime \prime }(\theta )=\mu$. The canonical link function expresses the natural parameter in terms of the mean, so the Poisson canonical link is $g(\mu )=\log \mu$. Its conditional variance equals its conditional mean.

A generalized linear model specifies independent responses in an exponential dispersion family, their means ${\mu }_i=E(Y_i)$, and a link function connecting those means to a linear predictor. In the common notation,where $a_i$ are known positive weights, $\varphi$ is the common dispersion parameter, and $V({\mu }_i)=b^{\prime \prime }({\theta }_i)$ is the variance function. The systematic part isHere $x_i$ is a row of the known design matrix and $\beta$ contains the unknown regression coefficients. The link function is invertible on the permitted mean domain; the canonical link function takes the mean to the natural parameter. This separates distributional, predictor, and link assumptions: it does not require the response itself to be normally distributed or the mean itself to be linear in the covariates.

For $\lambda >0$, write the Poisson distribution mass function in exponential dispersion family form:Thus the natural parameter of an exponential family is $\theta =\log \lambda$ and the dispersion parameter is fixed at $1$. The counting-measure support is $0,1,\dots$; there is no independent free dispersion parameter in an ordinary Poisson model. The degenerate boundary $\lambda =0$ is obtained as a limit, rather than a finite natural parameter.

The exponential dispersion family identities give $\mathbb{E}Y=b^{\prime }(\theta )$ and $\mathrm{Var}(Y)=a(\varphi )b^{\prime \prime }(\theta )$. Since both derivatives of $e^{\theta }$ equal $e^{\theta }$,

For $0<p<1$, the Bernoulli distribution has massIn exponential dispersion family notation $\exp \{[y\theta -b(\theta )]/\varphi +c(y,\varphi )\}$, takeThe support is $\{0,1\}$; outside it the probability is zero. The natural parameter of an exponential family is the log odds, and the dispersion parameter is fixed rather than freely estimated. Differentiating the cumulant givesAt $p=0$ or $1$ the degenerate distributions are limiting boundary cases with natural parameter $-\infty$ or $+\infty$.

Write the exponential dispersion family in the conventionAssume that the support does not depend on the parameters, that $\theta$ is in the interior of the natural parameter space, and that differentiation can pass under the normalizing integral or sum. Differentiating $\int f=1$ at fixed dispersion parameter givesso the expectation is $b^{\prime }(\theta )$. A second differentiation givesConsequentlyHere $b$ is the cumulant function of an exponential family; the same identities follow by differentiating $\log \mathbb{E}{e}^{tZ}=[b(\theta +\varphi t)-b(\theta )]/\varphi$. For a known observation weight $w$, replace $\varphi$ by $\varphi /w$. These regularity assumptions for differentiation under the integral sign matter: differentiation of a parameter-dependent support can introduce additional boundary terms.

Let $H_i$ be the number healthy among $m_i=25$ patients and $Y_i=H_i/m_i$. The fitted grouped-binomial logistic regression isFor $y=h/m$ its mass has exponential dispersion family formwhereThus $\mathbb{E}Y=p=b^{\prime }(\theta )$ and

An exponential dispersion family has densities of the formwhere $\theta$ is the canonical parameter, $\varphi >0$ is the dispersion parameter, and the support does not depend on $\theta$. Its mean and variance are

For the given Gamma distribution, setThengivesThus the gamma laws form the Gamma exponential dispersion family.

Directly integrating, or using the cumulant-generating function of a gamma distribution,soDifferentiation at zero yields

For independent observations from an exponential dispersion family with common known dispersion parameter ${\sigma }^2$, the generalized linear model log-likelihood iswhere ${\mu }_i=b^{\prime }({\theta }_i)$ and the link function satisfies $g({\mu }_i)={\eta }_i=x_i^T\beta$.