How does varying the regularization parameter in an SVM change the decision boundary for a non-separable dataset? A visual answer and/or some commentary on the limiting behaviors (for large and small regularization) would be very helpful.
The regularization parameter (lambda) serves as a degree of importance that is given to misclassifications. SVM pose a quadratic optimization problem that looks for maximizing the margin between both classes and minimizing the amount of misclassifications. However, for non-separable problems, in order to find a solution, the miclassification constraint must be relaxed, and this is done by setting the mentioned "regularization".
So, intuitively, as lambda grows larger the less the wrongly classified examples are allowed (or the highest the price the pay in the loss function). Then when lambda tends to infinite the solution tends to the hard-margin (allow no miss-classification). When lambda tends to 0 (without being 0) the more the miss-classifications are allowed.
There is definitely a tradeoff between these two and normally smaller lambdas, but not too small, generalize well. Below are three examples for linear SVM classification (binary).
For non-linear-kernel SVM the idea is the similar. Given this, for higher values of lambda there is a higher possibility of overfitting, while for lower values of lambda there is higher possibilities of underfitting.
The images below show the behavior for RBF Kernel, letting the sigma parameter fixed on 1 and trying lambda = 0.01 and lambda = 10
You can say the first figure where lambda is lower is more "relaxed" than the second figure where data is intended to be fitted more precisely.
(Slides from Prof. Oriol Pujol. Universitat de Barcelona)