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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?

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I suspect that (at least for real valued inputs) the chance of the output being exactly zero is essentially negligible, so randomising, or going with the most probable class etc. will be fine. It isn't a problem I have ever run into using kernel machines.

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  • $\begingroup$ Okay, but it could happen and I am wondering why this case is not considered mathematically / in the definition of the decision function. $\endgroup$ – Giuseppe Jan 9 '13 at 11:13
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    $\begingroup$ 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. $\endgroup$ – Dikran Marsupial Jan 9 '13 at 12:20

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