Thread: Exercise 3.4
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Old 10-07-2013, 05:53 AM
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magdon magdon is offline
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Default Re: Exercise 3.4

You got it mostly right. Your error is assuming both term, the H term and the one without the H give an N to cancel the N in the denominator. One term gives an N and the other gives a (d+1).


Quote:
Originally Posted by Sweater Monkey View Post
I'm not sure if I'm going about part (e) correctly.

I'm under the impression that E_{\text{test}}(\mathbf{w}_{\text{lin}})=\frac{1}{N}||X{\mathbf{w}}_{lin}-\mathbf{y'}||^2

where \hat{\mathbf{y}}=X{\mathbf{w}}_{lin}=X\mathbf{w}^*+H\mathbf{\epsilon} as derived earlier
and \mathbf{y'}=\mathbf{w}^{*T}\mathbf{x}+\mathbf{\epsilon'}=X\mathbf{w}^*+\mathbf{\epsilon'}

This lead me to \frac{1}{N}||H\mathbf{\epsilon}-\mathbf{\epsilon'}||^2

I carried out the expansion of this expression and then simplified into the relevant terms but my final answer is \sigma^2(1+(d+1)) because the N term cancels out.

Am I starting out correctly up until this expansion or is my thought process off from the start? And if I am heading in the right direction is there any obvious reason that I may be expanding the expression incorrectly? Any help would be greatly appreciated.
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