# Schölkopf's One-class SVM: Role of $\rho$ in the cost function

I have read and re-read the original paper and articles on Schölkopf's One-class SVM but elements of it still baffles me. The paper defines the cost function as: $$L = \frac{1}{2}||w^{2}|| + \frac{1}{\nu}\sum_{i}{(\xi_{i} - \rho)}$$

EDIT: The cost function above is wrong, the correct one reads: $$L = \frac{1}{2}||w^{2}|| - \rho + \frac{1}{\nu}\sum_{i}{\xi_{i}}$$ The rest of the argument below follows the premise of the wrong loss function.

such that $$w^{T}\cdot\Phi(x_{i}) \geq \rho - \xi_{i}$$ $$\xi_{i} \geq 0.$$

It seems to me there there is nothing in these equations that stop variable $\rho$ from increasing without bounds in trying to minimize $L$.

Here's my argument: rewriting the inequalities, we obtain: $$\xi_{i} - \rho \geq -w^{T}\cdot\Phi(x_{i})$$ $$\xi_{i} - \rho \geq -\rho$$ which are graphically represented below. The inequalities means that the offset slack $\xi_{i} - \rho$ lies above the blue line.

From this plot we can see that if a projected data point $w^{T}\cdot\Phi(x_{in})$ already lies inside the margin ($\leq \rho$), then increasing $\rho$ does nothing to change its offset slack $\xi_{in} - \rho$.

Conversely, if a point lies outside the margin, then increasing $\rho$ will have an effect of decreasing $\xi_{out} - \rho$.

Therefore, it seems like to minimize $L$, we are allowed to increase $\rho$ until $\rho = max(w^{T}\cdot\Phi(x_{i}))$ i.e. all the points lie on the 'wrong' side of the margin. Surely, this cannot be right. Am I missing anything in my logic?

• If you rearrange the inqualities you have $\xi_i - \rho \geq -w^T \phi(x_i)$, note the minus sign on the right. – MotiN May 15 '18 at 11:49
• @MotiN you are right! The mistaken has been corrected. – networker May 15 '18 at 16:47

$$L = \frac{1}{2}||w^{2}|| + (\frac{1}{\nu}\sum_{i}{\xi_{i})} - \rho.$$
Therefore there is tension between increasing $\rho$ for the last term and the increase in the sum of slacks in the middle term as $\rho$ increases.