2 ms·
Shouldn't an RBF kernel be expected to perform better than a 2nd degree poly kernel since, technically, it is infinite-dimensional? (I.e., the radial basis func
by eob 17y ago
Shouldn't an RBF kernel be expected to perform better than a 2nd degree poly kernel since, technically, it is infinite-dimensional? (I.e., the radial basis function can be expanded into an infinite-degree taylor series)
I think (not a ML expert by any means) that RBF should probably outperform most fixed-integer polynomial kernels, but it does so at the cost of introducing more structural risk into the model.
- alextp 17y agoIt does not really go like that. The higher the dimension (actually VC dimension, but polynomial dimension is a similar concept) the more points you would theoretically need to guarantee that the classifier is not overfitting (ie, representing the noise in the data set).