= Solution
Write $\mu=\alpha/\gamma$ and $\phi=1/\alpha$. With $\theta=-\gamma/\alpha=-1/\mu<0$, the <Gamma exponential dispersion family> representation is
$$
\log f(y)=\frac{y\theta-b(\theta)}{\phi}+c(y,\phi),\qquad b(\theta)=-\log(-\theta),
$$
where
$$
c(y,\phi)=\left(\frac1\phi-1\right)\log y-\frac1\phi\log\phi-\log\Gamma(1/\phi).
$$
Substitution gives the original gamma density, so both unknown parameters are included. The <exponential-family derivative identities> yield $\mu=b'(\theta)=-1/\theta$ and $b''(\theta)=1/\theta^2=\mu^2$. Therefore
$$
\boxed{V(\mu)=\mu^2,\quad\phi=\alpha^{-1},\quad\operatorname{Var}(Y)=\phi\mu^2.}
$$
The exact natural-parameter <canonical link function> is $g_{\rm can}(\mu)=-1/\mu$. \b[The usual inverse-link convention is $g(\mu)=1/\mu$], as used by the printed R fits; multiplying the link and coefficients by $-1$ gives the natural-parameter convention. The sign convention changes neither the fitted means nor the weight matrix below.
Back to article page
