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 $$\Lambda$$:




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

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)