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The Poisson distribution mass function isIt is an exponential dispersion family of order one withIndeed, $b^{\prime }(\theta )=e^{\theta }=E(Y)$ and $\varphi b^{\prime \prime }(\theta )=e^{\theta }=\mathrm{Var}(Y)$. A generalized linear model specifies independent random variables from a response exponential family, together with a linear predictor and a link function relating it to the response mean. The Poisson canonical link is $g(\mu )=\log \mu$.

For the insurance counts, let class $1$ and merit $0$ be the reference levels in a regression factor. The fitted Poisson regression assumesThe insured exposure $n_{ij}$ is known; its logarithm is an offset, not a coefficient to estimate. The model has eight free mean parameters and fixes the dispersion parameter at one. It assumes additive class and merit effects on the log-rate scale, hence multiplicative effects on the rate scale without a class-by-merit interaction.

Apart from constants independent of the parameters, the log-likelihood isDifferentiating gives the Poisson regression margin-matching score equationsThus the fitted totals equal the observed totals in each class and merit category, including the reference categories by subtraction. In design matrix notation these equations are $Z^T(Y-\hat{\mu })=0$. Fisher scoring solves them iteratively; the reported three iterations describe the numerical fit, not an additional statistical result.

The regression intercept gives the baseline fitted claim rate $e^{-2.0357359}=0.130584$ per insured car year. The merit rate ratios, relative to merit $0$ at the same class, areThe class rate ratios, relative to class $1$ at the same merit, areEach predicted cell rate is the baseline rate times its class and merit multipliers. The output's coefficient-to-standard error ratios are asymptotic normal Wald test statistics under the fitted Poisson regression, despite the software label “t value”; there is no estimated Gaussian residual scale in this model. All nonreference effects have very large absolute ratios.

The null Poisson deviance is $33854.16$ on nineteen statistical degrees of freedom; adding the seven factor coefficients reduces it by $33274.6437$. But the final Poisson deviance is still $579.5163$ on twelve statistical degrees of freedom, and the deviance residuals include values near $-10.79$ and $11.63$. These are far beyond what a well-fitting unit-dispersion model would normally produce. The additive Poisson model fits substantially better than a common rate, but remains grossly inadequate. A class-by-merit interaction, overdispersion, or statistical dependence between claims could contribute; the summary alone does not distinguish these explanations. Its very small nominal standard errors and formal Wald tests rely on the rejected model, so improved mean structure and error assumptions are needed before treating them as reliable uncertainty assessments.

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.