Saturated model for a Bernoulli response I have a have a saturated GLM for a Bernoulli response.
Let $Y_i \backsim \text{ Ber}(\pi_i)$. The saturated model yields $\pi_i^{(s)}=y_i$, where $y_i$ is the observed value, which implies that 
$$\theta_i^{(s)}=\log\!\bigg(\frac{y_i}{1-y_i}\bigg)$$
However, since $y_i=0 \text{ or } y_i=1$ isn't this problematic with the log? 
Edit: It seems I've phrased myself quite poorly so I'll just say what I'm given about those models:
Saturated model - unlinks the $θ_i$
$\cdot$  Were bound by the constraint $g(\mu_i) = x_i^T\beta$ on the means
$\cdot$ Here we have one parameter $θ_i$
for each response $y_i$
$\cdot$ 
 MLE $θ^{(s)}_i$ for $θ_i$ in the saturated model is just $θ^{(s)}_i = \arg \max_θf(y_i|θ_i)$
In particular, if we use this information for $Y_i$ which has a $\text{Bin}(m_i, \pi_i)$ distribution, we get that that
$$θ^{(s)}_i=\log(\frac{y_i}{m_i-y_i})$$
In particular, taking $m_i=1$ (making it Bernoulli), we get that 
$$θ^{(s)}_i=\log(\frac{y_i}{1-y_i})$$
My issue is that $y_i=0 \text{ or } 1$, since $Y_i$ is Bernoulli, but then $θ^{(s)}_i$ seems to be undefined
I also don't understand what "were bound by the constraint..." means
 A: A saturated model means that you are fitting a separate parameter for every data point.  In general, the Bernoulli distribution has a parameter, $\pi$, that ranges over the interval $(0, 1)$.  That is, the parameter cannot really equal $0$ or $1$, only asymptotically approach it.  When you fit a separate parameter for each datum, as you do here, you fit either $\pi_i = 0$ or $\pi_i=1$.  This means you are asserting there is no possible variation for that observation.  You are no longer really fitting a Bernoulli, but instead a degenerate distribution.  The logit is not an appropriate link function for the degenerate distribution.  That's the issue.  
This isn't really a 'good' model, so I wouldn't worry too much about this.  The point of a saturated model is typically to provide an extreme case that can be used in a comparison.  For example, you could test a fitted model against the saturated model to see if your model seems to be leaving something out (although you wouldn't necessarily know what).  It doesn't really matter that the saturated model is unlikely to be a good model for other purposes.  You would use it for that test and then set it aside.  
A: That is not the way I interpret a saturated model. The saturated model is applicable to a regression model where the unique exposure levels have, as predicted values, the empirical frequency of outcomes for every level. That means, if $X$ takes values 1, 2, and 3, the predicted value of $Y$ at each of those levels (from the regression model) is $E[Y| X=1]$, $E[Y| X=2]$, and $E[Y| X=3]$. Hopefully there is a sufficient sample size so that the marginal frequencies of $X$ do not have small cell counts.
If there is only one observation per exposure level, you are correct that the predicted value is just the $Y$ value: 0 or 1. Characterizing those predictions with log odds ratios requires a value of $-\infty$ or $\infty$ which is the limit of $\log(p/(1-p))$ as $p \rightarrow 0$ or $1$ respectively. So your expression is correct. The solution to the MLE is said to lie on the boundary.
