Let $f\colon\Bbb{R}^n\to\Bbb{R}$ be the decision function of an SVM using the radial basis function (RBF), $$ k(\mathbf{x},\mathbf{x}')=\exp\Big(-\gamma\|\mathbf{x}-\mathbf{x}'\|^2\Big). $$ That is, $$ f(\mathbf{x})=\sum_{i=1}^{N} a_i k(\mathbf{x},\mathbf{x}_i) + \rho, $$ where $a_i\in\Bbb{R}$, $\forall i=1,\ldots,N$, and $\mathbf{x}_i\in\Bbb{R}^n$ is the $i$-th support vector (SV).

The label of an unseen, testing datum, $\mathbf{x}_t$, is determined by the sign of the decision function $f$ when evaluated on $\mathbf{x}_t$, i.e., $$ y_t = \operatorname{sgn}\big(f(\mathbf{x}_t)\big). $$

What would be an appropriate and meaningful strategy in order to additionally compute a probabilistic "degree of confidence", $p_t\in[0.1]$, expressing the confidence for assigning the label $y_t$ to the testing sample $\mathbf{x}_t$?

Concerning my attempts, I tried the sigmoid function, but it does not work. Any help?

Thanks in advance!

  • $\begingroup$ Platt scaling is the standard method. It's supported natively in libsvm. $\endgroup$ – Dougal Apr 3 '15 at 17:30
  • $\begingroup$ Thanks @Dougal, I saw that in the libsvm paper, but didn't really get it. I implement my own svm-like classifier, so I need to implement this too. I wonder how difficult would be to find $A,B$? Have you ever tried to do it by yourself? $\endgroup$ – nullgeppetto Apr 3 '15 at 17:33
  • $\begingroup$ The libsvm authors wrote a paper about implementing it. $\endgroup$ – Dougal Apr 3 '15 at 17:36
  • $\begingroup$ Yes, that's what I said right above ("I saw that in the libsvm paper"), but I find it a bit vague... Thanks anyway! You actually answered my question. $\endgroup$ – nullgeppetto Apr 3 '15 at 17:37
  • 1
    $\begingroup$ My bad @Dougal, I never clicked on the link, I just thought you were talking about the well-known and general implementation paper. So, yes, this is much more informative, I will study it and I will return to ask some question, if necessary! Thanks again! $\endgroup$ – nullgeppetto Apr 3 '15 at 17:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.