# 11.2 Logistic regression¶

Logistic regression is an example of a binary classifier, where the output takes one two values 0 or 1 for each data point. We call the two values classes.

Formulation as an optimization problem

Define the sigmoid function

$S(x)=\frac{1}{1+\exp(-x)}.$

Next, given an observation $$x\in\real^d$$ and a weights $$\theta\in\real^d$$ we set

$h_\theta(x)=S(\theta^Tx)=\frac{1}{1+\exp(-\theta^Tx)}.$

The weights vector $$\theta$$ is part of the setup of the classifier. The expression $$h_\theta(x)$$ is interpreted as the probability that $$x$$ belongs to class 1. When asked to classify $$x$$ the returned answer is

$\begin{split}x\mapsto \begin{cases}\begin{array}{ll}1 & h_\theta(x)\geq 1/2, \\ 0 & h_\theta(x)<1/2.\end{array}\end{cases}\end{split}$

When training a logistic regression algorithm we are given a sequence of training examples $$x_i$$, each labelled with its class $$y_i\in \{0,1\}$$ and we seek to find the weights $$\theta$$ which maximize the likelihood function

$\prod_i h_\theta(x_i)^{y_i}(1-h_\theta(x_i))^{1-y_i}.$

Of course every single $$y_i$$ equals 0 or 1, so just one factor appears in the product for each training data point. By taking logarithms we can define the logistic loss function:

$J(\theta) = -\sum_{i:y_i=1} \log(h_\theta(x_i))-\sum_{i:y_i=0}\log(1-h_\theta(x_i)).$

The training problem with regularization (a standard technique to prevent overfitting) is now equivalent to

$\min_\theta J(\theta) + \lambda\|\theta\|_2.$

This can equivalently be phrased as

(11.13)$\begin{split}\begin{array}{lrllr} \minimize & \sum_i t_i +\lambda r & & & \\ \st & t_i & \geq - \log(h_\theta(x)) & = \log(1+\exp(-\theta^Tx_i)) & \mathrm{if}\ y_i=1, \\ & t_i & \geq - \log(1-h_\theta(x)) & = \log(1+\exp(\theta^Tx_i)) & \mathrm{if}\ y_i=0, \\ & r & \geq \|\theta\|_2. & & \end{array}\end{split}$

Implementation

As can be seen from (11.13) the key point is to implement the softplus bound $$t\geq \log(1+e^u)$$, which is the simplest example of a log-sum-exp constraint for two terms. Here $$t$$ is a scalar variable and $$u$$ will be the affine expression of the form $$\pm \theta^Tx_i$$. This is equivalent to

$\exp(u-t) + \exp(-t)\leq 1$

and further to

(11.14)$\begin{split}\begin{array}{rclr} (z_1, 1, u-t) & \in & \EXP & (z_1\geq \exp(u-t)), \\ (z_2, 1, -t) & \in & \EXP & (z_2\geq \exp(-t)), \\ z_1+z_2 & \leq & 1. & \end{array}\end{split}$

This formulation can be entered using affine conic constraints (see Sec. 6.2 (From Linear to Conic Optimization)).

Listing 11.8 Implementation of $$t\geq \log(1+e^u)$$ as in (11.14). Click here to download.
# Adds ACCs for t_i >= log ( 1 + exp((1-2*y[i]) * theta' * X[i]) )
# Adds auxiliary variables, AFE rows and constraints
def softplus(task, d, n, theta, t, X, y):
task.appendcons(n)      # z1 + z2 = 1
task.appendafes(4*n)    #theta * X[i] - t[i], -t[i], z1[i], z2[i]
z1, z2 = nvar, nvar+n
zcon = ncon
thetaafe, tafe, z1afe, z2afe = nafe, nafe+n, nafe+2*n, nafe+3*n
for i in range(n):

# z1 + z2 = 1

# Affine conic expressions
afeidx, varidx, fval = [], [], []

## Thetas
for i in range(n):
for j in range(d):
afeidx.append(thetaafe + i)
varidx.append(theta + j)
fval.append(-X[i][j] if y[i]==1 else X[i][j])

# -t[i]
afeidx.extend([thetaafe + i for i in range(n)] + [tafe + i for i in range(n)])
varidx.extend([t + i for i in range(n)] + [t + i for i in range(n)])
fval.extend([-1.0]*(2*n))

# z1, z2
afeidx.extend([z1afe + i for i in range(n)] + [z2afe + i for i in range(n)])
varidx.extend([z1 + i for i in range(n)] + [z2 + i for i in range(n)])
fval.extend([1.0]*(2*n))

# Add a single row with the constant expression "1.0"

# Add an exponential cone domain

# Conic constraints
for i in range(n):
acci += 2


Once we have this subroutine, it is easy to implement a function that builds the regularized loss function model (11.13).

Listing 11.9 Implementation of (11.13). Click here to download.
# Model logistic regression (regularized with full 2-norm of theta)
# X - n x d matrix of data points
# y - length n vector classifying training points
# lamb - regularization parameter
def logisticRegression(env, X, y, lamb=1.0):
n, d = int(X.shape[0]), int(X.shape[1])         # num samples, dimension

# Variables [r; theta; t; u]
nvar = 1+d+2*n
r, theta, t = 0, 1, 1+d

# Objective lambda*r + sum(t)

# Softplus function constraints
softplus(task, d, n, theta, t, X, y);

# Regularization
# Append a sequence of linear expressions (r, theta) to F
for i in range(d):
task.putafefentry(numafe + i + 1, theta + i, 1.0)


In the next figure we apply logistic regression to the training set of 2D points taken from the example ex2data2.txt . The two-dimensional dataset was converted into a feature vector $$x\in\real^{28}$$ using monomial coordinates of degrees at most 6.