 Solution

For an exponential dispersion family, differentiating the normalization integral with respect to $\theta$ gives an expected score of zero. The score is $(Y-b^{\prime }(\theta ))/\varphi$, hence $E(Y)=b^{\prime }(\theta )$. Differentiating again, the expected squared score is $b^{\prime \prime }(\theta )/\varphi$; it is also $\mathrm{Var}(Y)/{\varphi }^2$. ThusThe canonical link function sets the linear predictor equal to the natural parameter: $g(\mu )=\theta =(b^{\prime }{)}^{-1}(\mu )$. The variance function is $V(\mu )=b^{\prime \prime }((b^{\prime }{)}^{-1}(\mu ))$, so $\mathrm{Var}(Y_i)=\varphi V({\mu }_i)$; the dispersion parameter $\varphi$ is separated from the mean-dependent factor.

The saturated statistical model has a separate mean for each response and fits ${\tilde{\mu }}_i=y_i$, with boundary parameters interpreted by limits. At the same fixed dispersion, its log likelihood and the fitted model's log-likelihood define the scaled devianceThe unscaled deviance isKeeping dispersion fixed in this definition is essential; it is not a comparison with a separately estimated zero variance in a Gaussian saturated fit.

For the normal distribution, expanding its density exponent givesConsequentlyHere $V(\mu )=1$ and the canonical link is the identity, so ${\mu }_i={\beta }^T{x}_i$. The saturated natural parameter is ${\tilde{\theta }}_i=y_i$. Substitution into the deviance formula yields

For the cholesterol fits, all formulae include an intercept. Model 1 has age and BMI with three coefficients; models 2 and 3 have one of these covariates with two coefficients; model 4 has only the intercept. Their residual statistical degrees of freedom are respectively $27,28,28,29$. Their reported deviances are residual sums of squares, because these are Gaussian linear models.

The relevant hypothesis is $H_0:{\beta }_{\mathrm{BMI}}=0$ in the full model, against $H_1:{\beta }_{\mathrm{BMI}}\ne 0$, with age already present. Thus compare model 2 with model 1 using the partial F-test. Its extra sum of squares is $31.64-27.10=4.54$, and the common variance is estimated from model 1 by ${\hat{\sigma }}^2=27.10/27$. ThereforeSince $4.52325>4.210$, reject at $5\%$; the two-sided coefficient-test equivalent has $p\simeq 0.0427$. BMI is associated with cholesterol after adjustment for age, under the linear Gaussian model. The RSS values alone do not identify the sign of its coefficient or the size of the association.

For comparison, the marginal BMI test would compare model 4 with model 3, giving $F=13.88/(35.82/28)=10.8498$ on $(1,28)$ statistical degrees of freedom, with $p\simeq 0.00268$. This does not answer the adjusted question and makes the BMI evidence appear stronger than the conditional test. Age also contributes after BMI: comparing model 3 with model 1 gives $F=8.72/(27.10/27)=8.6878$ on $(1,27)$ statistical degrees of freedom, with $p\simeq 0.00653$. Thus retaining both predictors is supported at the stated level.

A raw deviance difference must not simply be compared with the supplied ${\chi }_1^2$ cutoff: Gaussian dispersion is unknown and the differences here are unscaled. If ${\sigma }^2$ were known, the scaled difference $4.54/{\sigma }^2$ would have a ${\chi }_1^2$ law. Estimating it from the full fit gives the exact finite-sample $F$ comparison above, with the appropriate denominator statistical degrees of freedom.