Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 218 / 1

Put $\varphi =1/\gamma$ and $\theta =-1/\alpha$. The density can be writtenwhereThus this is an exponential dispersion family. Its derivative identities giveThe variance function is $V(\mu )={\mu }^2$. In the usual Gamma-GLM convention the canonical inverse link isit differs by a minus sign from the natural parameter $\theta =-1/\mu$.

Let $x_i^T$ be row $i$ of the design matrix, ${\eta }_i=x_i^T\beta =1/{\mu }_i$, and $\gamma =1/\varphi$. The full log-likelihood isDifferentiation gives the score functionandThe Hessian does not depend on $Y$, so the Fisher information matrix is

The score equation $X^T(\mu -Y)=0$ is independent of $\gamma$, because $\gamma >0$ is only a common factor. Fixing $\varphi =1$ or estimating it therefore gives the same maximum-likelihood estimator $\hat{\beta }$.

Under the usual full-rank and regularity conditions,with $W$ evaluated consistently at the fitted means. Consequently every standard error from "mod2" is $\sqrt{\hat{\varphi }}$ times the corresponding standard error computed with dispersion one in "mod1". In particular,

The null hypothesis isagainst the alternative that at least one coefficient is nonzero. The code uses the scaled reduction in exponential-family deviance,and compares it with ${\chi }_3^2$.

That chi-squared calibration is appropriate when the dispersion is known, and is asymptotically valid after consistent dispersion estimation. With unknown Gamma dispersion, the standard finite-sample GLM comparison instead usesagainst an $F_{3,57}$ distribution. Its p-value is approximately $0.034$, so the correctly calibrated test still rejects at the 5% level and gives evidence that component type or position contributes to mean failure time.