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magdon · #10 10-07-2013, 09:19 AM

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

$\epsilon^TH\epsilon$

Quote:

Originally Posted by Sweater Monkey

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?