Talk:Quadratic form (statistics)

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\operatorname{cov}\left[\epsilon'\Lambda_1\epsilon,\epsilon'\Lambda_2\epsilon\right]=2\operatorname{tr}\left[\Lambda _1\Sigma\Lambda_2 \Sigma\right] + 4\mu'\Lambda_1\Sigma\Lambda_2\mu

If Λ1' = Λ2 and they are not symmetric, the above formula contradicts the variance formula, since in that case ε'Λ1ε = ε'Λ2ε. How to resolve this? Btyner 04:48, 3 April 2006 (UTC)

Yikes, the expression was wrong in the case of nonsymmetric Λs. I have noted the symmetric requirement, and added a section showing how to derive the general expression. Btyner 18:24, 6 April 2006 (UTC)

[edit] Is symmetry really necessary for the expectation result?

Nothing in the usual proof of the result for expectation seems to require symmetry of Λ:

\operatorname{E}(\epsilon' \Lambda \epsilon) = \operatorname{E}[ \operatorname{tr} ( \epsilon' \Lambda \epsilon) ]
= \operatorname{E}[ \operatorname{tr} ( \Lambda \epsilon \epsilon') ]
= \operatorname{tr}( \Lambda \operatorname{E}[ \epsilon \epsilon' ])
= \operatorname{tr}( \Lambda [ \mu \mu ' + \Sigma ])
= \mu '\Lambda \mu + \operatorname{tr}(\Lambda \Sigma )

Geomon 23:30, 21 December 2006 (UTC)

Good call. I must have been thinking about bilinear forms when I wrote that. Btyner 00:16, 23 January 2007 (UTC)