It may be a good thing to add to the documentation that the solutions of the SVDD and the OCSVM are identical when the Gaussian kernel is used for both methods with the same kernel bandwidth and C = 1/νN. This is mentioned in the original SVDD paper and in Tax thesis.

@albertthomas88 Is it because with Gaussian kernel all vectors have unit norms? I noticed the fact of equivalence when all vectors are normalized, but didn't think how it applies to Gaussian kernel.

Also it explains why the results were so similar in "Novelty and Outlier Detection" example. I need to change it.

In this light I think it's better to use linear kernel by default in SVDD, otherwise we have two similar (by default) classes.

Thanks for brining this point.

Added: My interpretation wasn't quite correct. Two methods are equivalent when K(x, y) = K(x - y) and parameters C and nu have proper relation.

@nmayorov Yes the two methods are equivalent if you have unit norm vectors, which is the case with the Gaussian kernel because in the feature space you have norm(Phi(x))^2 = k(x,x) = 1.

More generally, the equivalence holds true if all vectors have the same norm, i.e., when k(x,x) is constant, which is the case if k(x,y) only depends on x-y. This is mentioned in the OCSVM paper by Scholkopf et aL.

I don't know if this is a problem to have two similar classes by default but the OCSVM and the SVDD are known to perform best when the Gaussian kernel is used.

@albertthomas88

Yes the two methods are equivalent if you have unit norm vectors

Say we use linear kernel in 2-D space and all vectors have unit norms, but the decision boundary of SVDD is a circle and of OneClassSVM is a line. They can't be equivalent! Something is wrong here.

And the statement about K(x, y) = K(x - y) - I sort of took it on faith, but again this argument of the mismatch of the shapes of the boundaries troubles me. OK, we work in infinity-dimensional space, but can it make a plane become a sphere? Intuitive explanation might be that in the original space the decision boundaries become the same nevertheless, but it's a quite tricky thing.