In SVM, we minimise the Lagrangian (primal form):

$$L_p = \frac{1}{2}\Vert w \Vert^2 - \sum_{i=1}^m \alpha_i[y_i(w \cdot x_i + b)-1]$$

by taking partial derivatives with respect to $w$ and $b$, obtaining the two equations:

$$w = \sum_i \alpha_i y_i x_i$$ and $$\sum \alpha_i y_i = 0,$$ with the coefficients $\alpha_i \geq 0$. The weight vector is expressed as a linear combination of the support vectors, i.e. the vectors that lie on the margin, for which the coefficients $ \alpha_i $ are non-zero.

Using the formulation that includes slack variables, we obtain this new constraint: $$0 \leq \alpha_i \leq C,$$ where $C$ determines how much points on the other side of the margin will be penalised.

My question is the following: What would be the intuitive and geometrical meaning of constraining the sum of the support vector coefficients, say like this:

$$\sum_i \alpha_i = 1?$$


1 Answer 1


It's basically a harder constraint. Note that $\sum_i\alpha_i=C$ is equivalent to $\sum_i \alpha_i=1$ since we can redefine $\hat{\alpha}_i=\alpha_i/C$. So the original constraint becomes $0\leq \alpha_i\leq 1$, whereas now you have $\sum_i\alpha_i=1$. I'm not sure if there's a geometric meaning but it would in some sense be a form of L1 regularization on your slack variables, causing your classifier to be extra sensitive to certain classes.


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