Let $x_i^T$ be row $i$ of the design matrix, $\eta_i=x_i^T\beta=1/\mu_i$, and $\gamma=1/\phi$. The full <log-likelihood> is
$$
\ell(\beta,\gamma)
=\sum_{i=1}^{61}
\left\{
\gamma\log\gamma+\gamma\log\eta_i-\log\Gamma(\gamma)
+(\gamma-1)\log y_i-\gamma y_i\eta_i
\right\}.
$$
Differentiation gives the <score function>
$$
\nabla_\beta\ell
=\gamma X^T(\mu-Y),
\qquad
\mu_i=\eta_i^{-1},
$$
and
$$
-\nabla_\beta^2\ell
=\gamma X^T\operatorname{diag}(\mu_i^2)X.
$$
The Hessian does not depend on $Y$, so the <Fisher information matrix> is
$$
\mathcal I_\beta
=\frac1\phi X^TWX,
\qquad
W=\operatorname{diag}(\mu_i^2).
$$

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-218/1/b/solution