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${\displaystyle \operatorname {cov} \left[\epsilon ^{T}\Lambda _{1}\epsilon ,\epsilon ^{T}\Lambda _{2}\epsilon \right]=2\operatorname {tr} \left[\Lambda _{1}\Sigma \Lambda _{2}\Sigma \right]+4\mu ^{T}\Lambda _{1}\Sigma \Lambda _{2}\mu }$

If ${\displaystyle \Lambda _{1}^{T}=\Lambda _{2}}$ and they are not symmetric, the above formula contradicts the variance formula, since in that case ${\displaystyle \epsilon ^{T}\Lambda _{1}\epsilon =\epsilon ^{T}\Lambda _{2}\epsilon }$. How to resolve this? Btyner 04:48, 3 April 2006 (UTC) reply

Yikes, the expression was wrong in the case of nonsymmetric ${\displaystyle \Lambda }$s. I have noted the symmetric requirement, and added a section showing how to derive the general expression. Btyner 18:24, 6 April 2006 (UTC) reply

Nothing in the usual proof of the result for expectation seems to require symmetry of ${\displaystyle \Lambda }$:

${\displaystyle \operatorname {E} (\epsilon ^{T}\Lambda \epsilon )}$	=	${\displaystyle \operatorname {E} [\operatorname {tr} (\epsilon ^{T}\Lambda \epsilon )]}$
	=	${\displaystyle \operatorname {E} [\operatorname {tr} (\Lambda \epsilon \epsilon ^{T})]}$
	=	${\displaystyle \operatorname {tr} (\Lambda \operatorname {E} [\epsilon \epsilon ^{T}])}$
	=	${\displaystyle \operatorname {tr} (\Lambda [\mu \mu ^{T}+\Sigma ])}$
	=	${\displaystyle \mu ^{T}\Lambda \mu +\operatorname {tr} (\Lambda \Sigma )}$

Geomon 23:30, 21 December 2006 (UTC) reply

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