weight decay and data normalization *not a homework question*

[Haowen]

I have a general question regarding weight decay regularization.

Since 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 large and all other w small, but potentially a poor (overfitting) classifier could have 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!

[melipone]

That's a good question. I would like to know the answer myself. Andrew Ng says in his class that is not subject to regularization without explaining why of course. Here, in this class, we've applied regularization to everything (if I am not mistaken).

[yaser]

Originally Posted by Haowen

I have a general question regarding weight decay regularization.

Since 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 large and all other w small, but potentially a poor (overfitting) classifier could have 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!

Hi,

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 . However, if your criterion is that the zero hypothesis is the simplest hypothesis in linear regression, then 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 (), 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 ) from regularization.

[Haowen]

Ok, I understand now. Thank you.

Including captures the practitioner's heuristic "guess" that hypothesis that are closer to zero are "simpler". So the check is, very loosely, that the above assumption is true. Making all the features have zero mean is probably sufficient (in many applications) for the assumption to be "reasonable". However, it is not strictly related since the assumption could hold (or not) for other reasons depending on the application.

In either case, performing validation should allow us to narrow down the types of regularizers that make sense for a particular data set and application.