View Single Post
Old 02-17-2013, 03:26 PM
yaser's Avatar
yaser yaser is offline
Join Date: Aug 2009
Location: Pasadena, California, USA
Posts: 1,478
Default Re: weight decay and data normalization *not a homework question*

Originally Posted by Haowen View Post
I have a general question regarding weight decay regularization.

Since w_0 is a component inside the regularization term, it looks like it is possible to trade off distance from the origin for model complexity, e.g., I can have more complex models closer to the origin.

For this to make intuitive sense so that the regularization correctly "charges" the hypothesis a cost for being more complex, it seems to me that all the features must be normalized to have zero mean. Otherwise for example if all the data points are in a ball far from the origin, regularization could fail in the sense that a "good" classifier would have w_0 large and all other w small, but potentially a poor (overfitting) classifier could have w_0 small and other w large and achieve the same regularization cost.

I'm not sure about this reasoning, is it correct? Is this a concern in practice? Thanks!

If one matches the regularization criterion to a given problem, the regularizer may be more specific than general weight decay. For instance, when we discuss SVM next week, the regularizer will indeed exclude w_0. However, if your criterion is that the zero hypothesis is the simplest hypothesis in linear regression, then w_0 should be included in the regularizer.

As emphasized in the lecture, the choice of a regularizer in a real situation is largely heuristic. If you have information in a particular situation that suggests that one form of regularizer is more plausible than the other, then that overrules the general choices that are developed for a different, idealized situation.

In all of these cases, the amount of regularization (\lambda), which is determined through validation (discussed in the next lecture), is key to making sure that we are getting the most benefit (or the least harm if we choose a bad \Omega) from regularization.
Where everyone thinks alike, no one thinks very much
Reply With Quote