Talk:Quadratic form (statistics)

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

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

Yikes, the expression was wrong in the case of nonsymmetric $\Lambda$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)

Is symmetry really necessary for the expectation result?

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

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

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)