Meaning of the linear transformation in sigmoid output for Bernoulli parameter estimation This is found in 'Deep learning' book, chap.6, ~p.178 (http://www.deeplearningbook.org/contents/mlp.html), which discusses the problem of predicting the output class of a binary variable
In this section of the book $y$ is the value of a binary variable, $z:=w^Th + b$ seems to be an affine transformation of $h$, while $h$ is a function of $x$. So at this point basically there is no prescribed relation between $z$ (and by extension $x$) and $y$. We just have that based on the assumption $P(y) = \sigma((2y-1)z)$ (which is advertised as being a Bernoulli distribution), still no prescribed relation between the input and the output.
Then I came across the following puzzling sentence:

Saturation occurs only when the model already has the right answer, when $y=1$ and $z$ is very positive, or $y= 0$ and $z$ is very negative

which by plugging in the designed distribution yield $P(y=1) \approx 1$ and $P(y=0) \approx 0$... How is that saying anything about the model ? Also why the paragraph says nothing about the meaning of the transformation for $z$ ? 
Thanks
 A: Consider the diagram of a simple feed-forward neural network on page 170:

Section 6.2.2.2 on page 178 is describing using this network with a sigmoid activation to predict the parameter of a Bernoulli distribution given some samples $x_0, x_1, ...$.
Forward propagation (prediction) in this network will look like:
$h = W^Tx + b_0$
$z = w^Th + b_1$
$\hat{y} = sigmoid(z)$
If the model prediction $\hat{y}$ is close to $1$ then $z$, the input to $sigmoid$ was very large, and we are in the nearly-flat region of the sigmoid where the gradient is small therefore where it is hard to learn via gradient-descent i.e. "saturation". But if we ever get to this point then we must already have the right answer, so it is not a problem. 
This is because we will never start with an initial guess of $\hat{y}$ near $0$, because our data will be standardized and our random initial weights will be small, so z could never blow up to the saturation region of an incorrect prediction on the first training iteration. Only as we perform gradient updates via backpropagation $z$ will head in the right direction towards $\infty$ until $sigmoid(z)$ is close to the correct prediction and the output error is sufficiently small. So at any time we will either be outside of the saturation region or we will have gotten there by converging to the correct prediction.
A symmetric argument holds for $\hat{y}=0$, where $z$ must be very negative.
