Support Vector Machines for Binary Classification

Understanding Support Vector Machines

• Separable Data

• Nonseparable Data

• Nonlinear Transformation with Kernels

Separable Data

You can use a support vector machine (SVM) when your data has exactly two classes. An SVM classifies data by finding the best hyperplane that separates all data points of one class from those of the other class. The best hyperplane for an SVM means the one with the largest margin between the two classes. Margin means the maximal width of the slab parallel to the hyperplane that has no interior data points.

The support vectors are the data points that are closest to the separating hyperplane; these points are on the boundary of the slab. The following figure illustrates these definitions, with + indicating data points of type 1, and – indicating data points of type –1.

Mathematical Formulation: Primal.  This discussion follows Hastie, Tibshirani, and Friedman [1] and Christianini and Shawe-Taylor [2].

The data for training is a set of points (vectors) x_j along with their categories y_j. For some dimension d, the x_j ? R^d, and the y_j = ±1. The equation of a hyperplane is

f(x)=x′β+b=0

where β ? R^d and b is a real number.

The following problem defines the best separating hyperplane (i.e., the decision boundary). Find β and b that minimize ||β|| such that for all data points (x_j,y_j),

yjf(xj)≥1.

The support vectors are the x_j on the boundary, those for which yjf(xj)=1.

For mathematical convenience, the problem is usually given as the equivalent problem of minimizing ?β?. This is a quadratic programming problem. The optimal solution (ˆβ,ˆb) enables classification of a vector z as follows:

class(z)=sign(z′ˆβ+ˆb)=sign(ˆf(z)).

ˆf(z) is the classification score and represents the distance z is from the decision boundary.

Mathematical Formulation: Dual.  It is computationally simpler to solve the dual quadratic programming problem. To obtain the dual, take positive Lagrange multipliers α_j multiplied by each constraint, and subtract from the objective function:

LP=12β′β−?jαj(yj(xj′β+b)−1),

where you look for a stationary point of L_P over β and b. Setting the gradient of L_P to 0, you get

Substituting into L_P, you get the dual L_D:

LD=?jαj−12?j?kαjαkyjykxj′xk,

which you maximize over α_j ≥ 0. In general, many α_j are 0 at the maximum. The nonzero α_j in the solution to the dual problem define the hyperplane, as seen in Equation 1, which gives β as the sum of α_jy_jx_j. The data points x_j corresponding to nonzero α_j are the support vectors.

The derivative of L_D with respect to a nonzero α_j is 0 at an optimum. This gives