Understanding Logistic Regression II
Version 0.8
Consider a hyperplane $h$ that is used separate the domain $X$ into two categories (say {Good, Bad} or {1, 0} or {1, 1}) that we can collectively call to be the set $Y$. In other words, there is a true function $y = \text{sign}(h(x))$ that perfectly divides the data into the two categories of interest.
Another way of saying this is that learning the best hypothesis from the set of linear hyperplanes is a realizable learning problem. Actually, consider the above the definition of realizable learning.
Starting again, Consider a hyperplane $h$ that can map $x$ to $y$ as $y = \text{sign}(h(x))$ where $h(x) = \langle \theta, x \rangle$, where $\theta$ is the normal vector to the hyperplane.
We are only considering hyperplanes that pass through the origin here and not really addressing the general version with nonzero intercepts.
Note, that there is no probability distribution; Data is not generated stochastically. It isn’t the case that based on the value of $x$, or location of $x$ if we think of $x$ as designating a location in a vector space, $y$ is emitted stochastically. However, even this situation can be modeled stochastically, simply assume that $y$ is emitted stochastically, but the probability distribution over the possible values of $y$ is nonzero only at a single value given $x$.
Now logistic regression is a particular relaxation of the original problem. Instead of assuming that the probability distribution of $y$ given $x$ is nonzero for only value of $y$ we instead assume a parametric form of the distribution, the logistic function, which can approach the true lowentropy if we increase the magnitude of the value of $\theta$ used in the parameterization. This distribution is dependent basically on the signed distance of a point from the hyperplane defined by $\theta$. The logistic assumption says that $ p(y = 1  x) = \frac{1}{1 + \exp(h(x))} $.
Since $h(x) = \langle \theta, x \rangle$ therefore $p(y = 1  x) = \frac{1}{1 + \exp(+h(x))} = 1  p(y = 1  x) = p(y = 1  x)$
Here’s a different take on this, imagine we have a large number of different features $x_1, x_2, x_k$ on which $y$ is dependent. and we make the above assumption for each of these features. i.e. we say that
An improvement can be made to this assumption by incorporating the fact that certain features can have more influence than others in changing our beliefs. For example a single occurrence of the bigram “fantastic_movie” might be as potent as two occurrence of the bigram “good_acting” in a movie review for making us believe that a movie was truly “GOOD”. In general we can add many scalar multipliers, 1 for each feature
Now think of each distribution as an expert, and you want to incorporate the beliefs of all the experts into your final distribution $ p(y = 1  x_1, x_2, \ldots, x_k) $. The two standard ways to make a single distribution from a collecion of distributions is to either take their product “The Product of Experts” model or to make a mixture.
Consider what it would mean to take a product of the individual probability distributions:
Now essentially what we are saying is that each expert has a veto power, imagine how many terms there would be if a full expansion of the above product was carried out. The exact formula is $2^k$. Now logistic regression could be phrased as an approximation of this where we have kept only two terms so that
Actually the above way is not really a satisfying way to understand logistic regression and there there is a third and a better way to understand this that was proposed by Jaynes. Essentially the maxent principle finds the probability distribution that has the highest entropy after assuming that certain statistics derived from the data are good anough to be enforced as constraints for the predictive probability distribution.
Search for the entry “Why no maxent?” at David Mackay’s FAQ site to also get an understanding of why Maxent may not be a great idea.
Okay finally consider 4 scenarios

Data was completely linearly separable and generated by a distribution. Actually the probability of this event happening as the number of data points increase is lower. But even so with finite data such a situation can happen. The problem that can happen if we try to maximize the conditional log likelihod is that by increasing the magnitude of $\theta$ it is possible to keep making the loglikelihod higher and higher. After all when we increase the magnitude of $\theta$ the system keeps becoming more and more sure of itself and it never has to pay a penalty because well the data is linearly separable. Of course regularization is one way to solving this problem which can be interpreted as imposing a prior if one sticks to generative/statistical learning and do not use “learning theory” and its “regularized learning framework”. Of course if the data truly is linearly separable then even a simple perceptron such as the one below would do the job.

Data was linearly seperable and generated by a deterministic function. In this case instead of using loglikelihood based methods one can even use percetrons. They are easy to code and simple to execute. However, even percetrons can overfit and for such cases we have the margins in hinge loss minimization and slack and $C$ parameter in the dual problem solved in SVMs.
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import numpy as np
class BatchPerceptron(object):
def __init__(self, dim=2):
self.hyperplane = np.zeros((dim,), dtype='float32'))
def _distance(self, x):
return np.dot(self.hyperplane, x)
def predict(self, x):
y_hat = np.sign(self._distance(x))
return 1 if y_hat == 0 else y_hat
def prediction_is_incorrect(self, y, x):
return self.predict(x) != y
def update(self, data):
for (y, x) in data:
if self.prediction_is_incorrect(y, x):
self.hyperplane += (y * x)
return True
return False
bp = BatchPerceptron()
while bp.update(data):
pass
print(bp.hyperplane)
Perceptron Odyssey

Data not linearly seperable, generated by linear threshold. Impossible! Of course the ideas of slack and margin allow us to solve even this problem.

Data not linearly separable, generated stochastically. Even in such cases regularization may be needed. But this problem at least is guaranteed to have a solution because it is a convex minimization problem in which the minima actually exists.
Always remember that even the function $f(x) = x$ is a convex function but it’s doesn’t actually have a minima over the real line.