# Tag Info

0

Figured it out. Didn't see the obvious before that $\phi(x)$ is constant: $$cov[\phi(x)^Tw,w^T\phi(x')]=\mathbb{E}[(\phi(x)^Tw-\mathbb{E}[\phi(x)^Tw])(w^T\phi(x')-\mathbb{E}[w^T\phi(x')])]=\mathbb{E}[\phi(x)^T(w-\mathbb{E}[w])(w^T-\mathbb{E}[w^T])\phi(x')]=\phi(x)^T \mathbb{E}[(w-\mathbb{E}[w])(w^T-\mathbb{E}[w^T])] \phi(x')= \phi(x)^T S_N \phi(x')$$

0

Try this package: https://CRAN.R-project.org/package=WeightSVM It uses a modified version of 'libsvm' and is able to deal with instance weighting. You can assign lower weights to some subjects.

1

One way to construct kernels over discrete spaces is by considering a graph representation of the data and defining a smooth function over nodes of graphs via graph fourier transform. This class of kernels (introduced by Kondor and Lafferty) is called as diffusion kernels. Concretely, the kernel over the graph nodes $V$ is given by: \begin{align} K(V,V) = \...

0

As usεr11852 says, cross validation does make sense for optimizing linear SVMs. According to Hastie et al.: The Entire Regularization Path for the Support Vector Machine, Journal of Machine Learning Research 5 (2004) 1391–1415, inside the cross validation you don't need to compute one SVM per C value as it is possible to compute the complete path of SVMs ...

0

Yes, it makes full sense to use cross-validation to find optimal hyper-parameters values in the case of SVM with a linear kernel. If anything, the choosing the regularisation parameter $C$ in the Lagrange formulation is analogous to choosing the ridge regularisation parameter in ridge regression. Therefore it is necessary for our training data to be scaled ...

Top 50 recent answers are included