11.5 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

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

Implementation

As can be seen from (11.20) 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

(11.21)\[\begin{split}\begin{array}{rclr} (z_1, 1, u-t) & \in & \EXP & (z_1\geq \exp(u-t)), \\ (z_2, 1, -t) & \in & \EXP & (z_2\geq \exp(-t)), \\ z_1+z_2 & \leq & 1. & \end{array}\end{split}\]
Listing 11.10 Implementation of \(t\geq \log(1+e^u)\) as in (11.21). Click here to download.
# t >= log( 1 + exp(u) ) coordinatewise
def softplus(M, t, u):
    n = t.getShape()[0]
    z1 = M.variable(n)
    z2 = M.variable(n)
    M.constraint(Expr.add(z1, z2), Domain.equalsTo(1))
    M.constraint(Expr.hstack(z1, Expr.constTerm(n, 1.0), Expr.sub(u,t)), Domain.inPExpCone())
    M.constraint(Expr.hstack(z2, Expr.constTerm(n, 1.0), Expr.neg(t)), Domain.inPExpCone())

Once we have this subroutine, it is easy to implement a function that builds the regularized loss function model (11.20).

Listing 11.11 Implementation of (11.20). 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(X, y, lamb=1.0):
    n, d = int(X.shape[0]), int(X.shape[1])         # num samples, dimension
    M = Model()
    theta = M.variable(d)
    t     = M.variable(n)
    reg   = M.variable()

    M.objective(ObjectiveSense.Minimize, Expr.add(Expr.sum(t), Expr.mul(lamb,reg)))
    M.constraint(Var.vstack(reg, theta), Domain.inQCone())

    signs = list(map(lambda y: -1.0 if y==1 else 1.0, y))
    softplus(M, t, Expr.mulElm(Expr.mul(X, theta), signs))

    return M, theta

Example: 2D dataset fitting

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.

_images/logistic-regression.png

Fig. 11.6 Logistic regression example with none, medium and strong regularization (small, medium, large \(\lambda\)). Without regularization we get obvious overfitting.