Scores in logistic regression I understand well that logistic regression is a well defined functions to fit the input variables to the probability bias of a Bernouili trial. However, I do not understand the origin of the linear assumption for modelling the log odd-ratio :
$$
\ell = \log_b \frac{p}{1-p} = \beta_0 + \beta_1 x_1 + \beta_2 x_2
$$
Are there any theoretical reasons behind the use of this heuristic to define the link function? Are there canonical ways to transform the input to justify that assumption?
 A: Nothing other than that the linear function is the simplest function. What else do you need for a heuristic? In practice, it is not that limiting, you can use polynomials, logarithms, splines, etc. in this framework.
A: The assumption of linearity is the simplest non-vacuous assumption about how the covariates can effect the expectation.  That the linearity assumption is made as on the log-odds scale is a bit strange.  But considering that a) the log odds scale is an unbounded quantity (i.e. we can have log odds from $-\infty$ to $\infty$ and have them correspond to probabilities on the unit interval), and b) the logit function is the canonical link for logistic regression, then the assumption of linearity on the log odds scale starts to not seem so crazy.
A: Simplified case:
Logistic regression uses the form:
$$
p(X)=\frac{e^{\beta_{0}+\beta_{1} X}}{1+e^{\beta_{0}+\beta_{1} X}}
$$
It is easy to see that no matter what values $\beta_{0}, \beta_{1}$ or $X$ take, $p(X)$ will have values between 0 and 1 .
A bit of rearrangement gives
$$
\log \left(\frac{p(X)}{1-p(X)}\right)=\beta_{0}+\beta_{1} X
$$
This monotone transformation is called the log odds or logit transformation of $p(X)$.
For multivariate case:
\begin{array}{c}
\log \left(\frac{p(X)}{1-p(X)}\right)=\beta_{0}+\beta_{1} X_{1}+\cdots+\beta_{p} X_{p} \\
p(X)=\frac{e^{\beta_{0}+\beta_{1} X_{1}+\cdots+\beta_{p} X_{p}}}{1+e^{\beta_{0}+\beta_{1} X_{1}+\cdots+\beta_{p} X_{p}}}
\end{array}
Why we say the logistic regression is a linear model?
Short answer: because of the logit function.
Long answer: A classifier is linear if its decision boundary on the feature space is a linear function: positive and negative examples are separated by a hyperplane.
