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Old 04-15-2016, 02:50 AM
ntvy95 ntvy95 is offline
Join Date: Jan 2016
Posts: 37
Default Exercise 4.6

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 var component while likely not compromising the bias component.


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