= Solution
Use the <exponential dispersion family> convention $f(y)=\exp\{[y\theta-b(\theta)]/\phi+c(y,\phi)\}$. For the <Gamma distribution>, take
$$
\phi=\alpha^{-1},\qquad \theta=-\lambda/\alpha<0,
\qquad b(\theta)=-\log(-\theta),
$$
$$
c(y,\phi)=(\alpha-1)\log y+\alpha\log\alpha-\log\Gamma(\alpha).
$$
Substitution reproduces the gamma density exactly. Differentiating its normalization with respect to $\theta$ yields $\mathbb EY=b'(\theta)$ and $\operatorname{Var}Y=\phi b''(\theta)$: the first identity follows from zero mean score, and differentiation again gives the variance. Therefore
$$
\boxed{\mathbb EY=\alpha/\lambda,\qquad\operatorname{Var}Y=\alpha/\lambda^2.}
$$
The <canonical link function> is $g(\mu)=\theta=-1/\mu$. Some conventions reverse the sign and call $1/\mu$ the reciprocal link; this only reverses the regression coefficients. We retain the genuine natural parameter $\theta=-1/\mu$.
In the <Gamma regression with canonical link>, $\eta_i=x_i^T\beta<0$ and $\mu_i=-1/\eta_i$. Differentiating the independent-observation log-likelihood gives
$$
\boxed{U(\beta)=\phi^{-1}X^T(y-\mu),\qquad
I(\beta)=\phi^{-1}X^T\operatorname{diag}(\mu_i^2)X.}
$$
Here $d\mu_i/d\eta_i=\mu_i^2$, so the negative Hessian equals this information matrix. <Fisher scoring> or Newton iteration is
$$
\beta_{\rm new}=\beta+I(\beta)^{-1}U(\beta),
$$
with the means and information recalculated at each iteration; damping can preserve $X\beta<0$ and increase the likelihood. Full column rank makes the information positive definite for positive means.
The saturated model has mean $y_i$. Subtracting fitted from saturated log-likelihood yields the unscaled <deviance>
$$
\boxed{D=2\sum_i\left[\frac{y_i-\widehat\mu_i}{\widehat\mu_i}
-\log\frac{y_i}{\widehat\mu_i}\right].}
$$
Twice the actual log-likelihood ratio is $D/\phi$. In particular, the printed parenthetical $\widehat\mu=X\widehat\beta$ is incorrect for this nonidentity link: \b[$X\widehat\beta$ estimates $\eta$, and $\widehat\mu_i=-1/(x_i^T\widehat\beta)$].
Back to article page
