# SVM decision function

our decision function e.g. in SVMs for binary classification (where the response is labeld by $y_i \in \{-1,1\}$) has the form:

$f(\mathbf{x}) = \text{sgn}(\mathbf{w}^\top \mathbf{x} + b)$ where $\mathbf{w}^\top \mathbf{x} + b =0$ is the equation of the separating hyperplane.

But what happens if a new example $\mathbf{x}_{new}$ lies on the hyperplane $\mathbf{w}^\top \mathbf{x}_{new} + b = 0$ then $f(\mathbf{x}_{new})=0$ because $\text{sgn}(0)=0$. In which class $y_i \in \{-1,1\}$ do we than classify our new example? Do we randomize between -1 and 1?

• I don't think it is really considered for other classifiers either. For probabilistic classifiers, often the criterion $f(x) \geq 0.5$ is used, but as far as I am aware this is merely a convention and makes little difference in practice. Equivalently there would be no problem in assigning the pattern to the "positive" class if $f(x)=0$ for a support vector machine. – Dikran Marsupial Jan 9 '13 at 12:20