= Solution
A <generalized linear model> assumes independent responses from one <exponential dispersion family>,
$$
f_i(y_i;\theta_i,\sigma_i^2)
=\exp\!\left\{
\frac{y_i\theta_i-b(\theta_i)}{\sigma_i^2}
+c(y_i,\sigma_i^2)
\right\},
$$
where $\sigma_i^2=a_i\sigma^2$, $\mu_i=\mathbb E[Y_i]=b'(\theta_i)$, and a <link function> relates the mean to the <linear predictor> by $g(\mu)=X\beta$. It follows from normalization of the density after shifting the <natural parameter of an exponential family>[natural parameter] that
$$
\mathbb E[e^{t_iY_i}]
=\exp\!\left\{
\frac{b(\theta_i+\sigma_i^2t_i)-b(\theta_i)}{\sigma_i^2}
\right\}.
$$
Independence therefore gives the joint <moment-generating function>
$$
\mathbb E[e^{t^TY}]
=\exp\!\left\{
\sum_{i=1}^n
\frac{b(\theta_i+a_i\sigma^2t_i)-b(\theta_i)}{a_i\sigma^2}
\right\},
$$
whenever every shifted natural parameter lies in its natural-parameter domain.
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