Thread: Exercise 3.4
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Old 10-07-2013, 09:19 AM
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magdon magdon is offline
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Default Re: Exercise 3.4

Yes, that is right. You have to be more careful but use similar reasoning with

\epsilon^TH\epsilon

Quote:
Originally Posted by Sweater Monkey View Post
Yes I realize that only one term should have the N so the issue must be in how I'm expanding the expression.

I think my problem is how I'm looking at the trace of the \mathbf{\epsilon}^T\mathbf{\epsilon} matrix.

I'm under the impression that \mathbf{\epsilon}^T\mathbf{\epsilon} produces an NxN matrix with a diagonal of all \sigma^2 values and 0 elsewhere. I come to this conclusion because the \epsilon are all independent so when multiplied together the covariance of any two should be zero while the covariance of any \epsilon_i\epsilon_i should be the variance of \sigma^2. So then the trace of this matrix should have a sum along the diagonal of N\sigma^2, shouldn't it?
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