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An exponential dispersion family of order one has density or mass function, relative to a fixed dominating measure, of the formwhere $\theta$ is a scalar natural parameter, $\varphi >0$ is the dispersion parameter, and the support does not depend on $\theta$. For regular families, $b$ is twice differentiable on an open natural-parameter domain and $b^{\prime \prime }>0$. Known observation weights $w$ replace $\varphi$ by $\varphi /w$. “Order one” refers to the scalar canonical statistic $y$, rather than an arbitrary vector of canonical statistics.

Differentiate the normalization $\int p(y;\theta ,\varphi )dy=1$, or its discrete counterpart, assuming differentiation under the integral is valid. The first derivative gives $\mathbb{E}(Y-b^{\prime }(\theta ))=0$. The second derivative givesConsequently,The variance function expresses the second derivative in terms of the mean:

A generalized linear model specifies independent responses from an exponential dispersion family, a linear predictor ${\eta }_i=x_i^T\beta$, and a differentiable monotone link function relating it to the response mean by $g({\mu }_i)={\eta }_i$. The family, variance function, covariate design, link function and dispersion specification together define the model. The canonical link function identifies the linear predictor with the natural parameter:One advantage is that the likelihood equations become particularly simple. With known weights $w_i$ and common dispersion $\varphi$, the score function and negative Hessian matrix for $\beta$ areThe latter does not explicitly depend on the responses, so it equals the expected Fisher information matrix at the same parameter. With full-rank design and positive variances the likelihood is strictly concave in $\beta$, simplifying optimization. These statements concern the coefficient block at a fixed dispersion value.

Three explicit examples, taking unit weights, are as follows.

For a normal distribution with mean $\mu$ and variance $\varphi$,Thus $\theta =\mu$, $b(\theta )={\theta }^2/2$, $c(y,\varphi )=-y^2/(2\varphi )-\frac{1}{2}\log (2\pi \varphi )$, and $V(\mu )=1$. Its canonical link function is $\boxed{g(\mu )=\mu }$, the identity link.

For a Poisson distribution with mean $\mu >0$ and support $\{0,1,\dots \}$,Here $\varphi =1$, $\theta =\log \mu$, $b(\theta )=e^{\theta }$, and $c(y,1)=-\log (y!)$. Therefore $b^{\prime }(\theta )=b^{\prime \prime }(\theta )=e^{\theta }$, $V(\mu )=\mu$, and the Poisson canonical link is $\boxed{g(\mu )=\log \mu }$.

For a Bernoulli distribution with success probability $\mu \in (0,1)$,Taking $\varphi =1$, $\theta =\log (\mu /(1-\mu ))$, $b(\theta )=\log (1+e^{\theta })$, and $c(y,1)=0$ gives the required form. Its mean and variance function are $b^{\prime }(\theta )=e^{\theta }/(1+e^{\theta })=\mu$ and $V(\mu )=\mu (1-\mu )$. Its canonical link function is $\boxed{g(\mu )=\log (\mu /(1-\mu ))}$, the logit link. The fixed dispersion values in the Poisson and Bernoulli cases are part of those models; an arbitrary continuous dispersion is not being asserted for them.

To test $H_0:{\beta }_j=0$, first fit the unrestricted model and then the restricted model with that component fixed at zero, estimating all nuisance parameters, including dispersion when applicable, in each fit. If $\hat{\vartheta }$ and $\tilde{\vartheta }$ denote their full fitted parameter vectors, the likelihood-ratio test usesFor identifiable regular models, an interior null parameter and a fixed-dimensional full-rank design with increasing information, the Wilks theorem gives $D\stackrel{d}{\to }{\chi }_1^2$ under $H_0$. A level-$\alpha$ test rejects for $D$ greater than the $(1-\alpha )$-quantile of the chi-squared distribution with one degree of freedom.

Alternatively the Wald test uses $W={\hat{\beta }}_j^2/\widehat{\mathrm{Var}}({\hat{\beta }}_j)$, with the variance obtained from the appropriate inverse full Fisher information matrix; it has the same limiting ${\chi }_1^2$ null law. The score test fits only the restricted model. If $\lambda$ collects its nuisance parameters, its efficient information is $I_{jj\cdot \lambda }=I_{jj}-I_{j\lambda }{I}_{\lambda \lambda }^{-1}{I}_{\lambda j}$, and the statistic is $U_j(\tilde{\vartheta }{)}^2/I_{jj\cdot \lambda }(\tilde{\vartheta })$, again asymptotically ${\chi }_1^2$. Accounting for the nuisance block is necessary rather than using the unadjusted diagonal information.