LFD Book Forum - Exercise 4.6

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- - Exercise 4.6 (http://book.caltech.edu/bookforum/showthread.php?t=4667)

Hello, I have this answer for the Exercise 4.6 but I'm not sure if it's right?

Because $sign(w^{T}x) = sign(\alpha w^{T}x)$ for any $\alpha > 0$ , very small weights are still as powerful as large weights (all that matters is the accuracy of the calculations that computer being able to perform): That also means a hyperplane can be represented by many hypotheses, constraining the weights can reduce the number of hypotheses represents the same hyperplane. Hence soft-order constraint will be able to reduce the

component while likely not compromising the

component.

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Edit: I have just remembered that the growth function has already taken care of the issue many hypotheses representing the same hyperplane (and this issue does not affect the

component anyway (?)). So in this case the answer should be the hard-order constraint...? I'm really confused right now. :clueless:

Re: Exercise 4.6

I have the same question. Can someone help here?

From my understanding having small weights is not perfect for sign(s), since this will lead to a signal that is often around 0 and thus a small change of just one input has a high chance to lead to a completely different output, if the sign changes.

So it would be better to have big weights, thus the signal is always pushed to the big number region and the sign is more stable.

But I maybe I'm just wrong here.

Re: Exercise 4.6

Yes, the soft order constraint does not impact classification. Better regularize with the hard order constraint, or use the soft order constraint with the "regression for classification" algorithm.

Quote:

Originally Posted by ntvy95 (Post 12322)

component while likely not compromising the

component anyway (?)). So in this case the answer should be the hard-order constraint...? I'm really confused right now. :clueless:

Re: Exercise 4.6

Correct again.

So let us differentiate between the theory of machine learning and its implementation on finite precision computers. In theory, if you have an infinite precision machine, then the size of the weights does not matter because it is a mathematical fact that, for positive $\alpha$ ,

$sign(\alpha w^T x)=sign(w^T x)$

In finite precision, you typically want the weights to be around 1 and the inputs rescaled to be around 1 too (this is called input preprocessing and you can read about it in e-Chapter 9).

Quote:

Originally Posted by CountVonCount (Post 12488)

Re: Exercise 4.6

Thanks for this clarification. It helps a lot for understanding.

Re: Exercise 4.6

Thank you very much for your reply!

Re: Exercise 4.6

Thanks for the helpful discussion. Follow up is why does the hard constraint imply that the weights will be larger?