How can one derive the log-likelihood of the saturated model of an exponential family in general?
Differentiating the log likelihood w.r.t $\theta$ gives $y_i=\hat{\mu_i}$ but I don't think replacing $y_i$ with $\hat{\mu_i}$ in the equation for the general log-likelihood would be helpful? Also what about maximising w.r.t $\phi$?
for $\exp\Big(\frac{y_i\theta_i-b(\theta_i)}{\alpha(\varphi)}+c(y_i,\varphi)\Big)$
log-likelihood of the saturated model of an exponential family in general is
$\sum_i\Big(\frac{y_i\hat\theta_i-b(\hat\theta_i)}{\alpha(\varphi)}+c(y_i,\varphi)\Big)$ where $\hat\theta_i=g(\mu_i)$ and put $\mu_i=y_i$
we maximize the log-likelihood, if we need to estimate any of a parameters and maximize w.r.t. that parameter. Otherwise, not to get saturated model.
to get log-likelihood of the saturated model, we just see the perfect fit(i.e all $\mu_i=y_i$)..... log-likelihood of the saturated model is useful to compare scaled deviance from observed model. we don't maximize it.
