Talk:Quadratic form (statistics)

${\displaystyle \operatorname {cov} \left[\epsilon ^{T}\Lambda _{1}\epsilon ,\epsilon ^{T}\Lambda _{2}\epsilon \right]=2\operatorname {tr} \left[\Lambda _{1}\Sigma \Lambda _{2}\Sigma \right]+4\mu ^{T}\Lambda _{1}\Sigma \Lambda _{2}\mu }${\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 ${\displaystyle \Lambda _{1}^{T}=\Lambda _{2}}${\displaystyle \Lambda _{1}^{T}=\Lambda _{2}} and they are not symmetric, the above formula contradicts the variance formula, since in that case ${\displaystyle \epsilon ^{T}\Lambda _{1}\epsilon =\epsilon ^{T}\Lambda _{2}\epsilon }${\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 ${\displaystyle \Lambda }${\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 \Lambda }:

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

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

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

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

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

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