Put $\phi=1/\gamma$ and $\theta=-1/\alpha$. The density can be written
$$
f(y;\theta,\phi)
=\exp\left[
\frac{y\theta-b(\theta)}{\phi}+c(y,\phi)
\right],
\qquad
b(\theta)=-\log(-\theta),
$$
where
$$
c(y,\phi)
=-\log\Gamma(1/\phi)-\phi^{-1}\log\phi
+(\phi^{-1}-1)\log y.
$$
Thus this is an <exponential dispersion family>. Its derivative identities give
$$
\mathbb EY=b'(\theta)=-\frac1\theta=\alpha,
\qquad
\operatorname{Var}(Y)=\phi b''(\theta)
=\phi\alpha^2=\frac{\alpha^2}{\gamma}.
$$
The <variance function> is $V(\mu)=\mu^2$. In the usual Gamma-GLM convention the canonical inverse link is
$$
g(\mu)=\frac1\mu;
$$
it differs by a minus sign from the natural parameter $\theta=-1/\mu$.

Solution

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

 Solution

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