Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 3 / 5J / c

For the canonical link function, ${\theta }_i={\eta }_i=x_i^T\beta$. Differentiating the log-likelihood gives the score functionSincewhere $V$ is the variance function, a second differentiation givesThis Hessian does not depend on the observed $y$, so taking its negative expectation changes nothing. Hence