
#1




Exercise 3.4

#2




Re: Exercise 3.4
You can consider doublechecking your answer of 3.4(b). Hope this helps.
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#3




Re: Exercise 3.4
I am not sure how to approach part (a). Are we supposed to explain why that insample estimate intuitively makes sense, or (algebraically) manipulate expressions given earlier into it?

#4




Re: Exercise 3.4
Algebraically manipulate earlier expressions and you should get 3.4(a). It is essentially a restatement of .
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#5




Re: Exercise 3.4
I'm not sure if I'm going about part (e) correctly.
I'm under the impression that where as derived earlier and This lead me to I carried out the expansion of this expression and then simplified into the relevant terms but my final answer is 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. 
#6




Re: Exercise 3.4
1. I got $y^{\prime}=y\epsilon+\epsilon^{\prime}$.
and $\hat{y}y^{\prime}=H\epsilon +\epsilon^{\prime}$. 
#7




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:
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#8




Re: Exercise 3.4
Quote:
I think my problem is how I'm looking at the trace of the matrix. I'm under the impression that produces an NxN matrix with a diagonal of all values and 0 elsewhere. I come to this conclusion because the are all independent so when multiplied together the covariance of any two should be zero while the covariance of any should be the variance of . So then the trace of this matrix should have a sum along the diagonal of , shouldn't it? 
#9




Re: Exercise 3.4
Yes, that is right. You have to be more careful but use similar reasoning with
Quote:
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#10




Re: Exercise 3.4

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