How to prove second moment of a quadratic form where $Z$ has normal distribution with mean zero and covariance matrix identical?
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$\begingroup$ Also, see stats.stackexchange.com/questions/48066/… $\endgroup$– StatsStudentCommented Jun 12, 2019 at 2:42
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$\begingroup$ In the one-variable case with $T=(t_{11}),$ you are asking us to show $$t_{11}= t_{11}E[Z^2] =E[t_{11}Z^2] = E[Z^\prime T Z] =\operatorname{tr}^2(T) + \operatorname{tr}(T)^2 = t_{11}^2+t_{11}^2 = 2t_{11}^2.$$ $\endgroup$– whuber ♦Commented Jun 12, 2019 at 15:09
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Suppose $A\in\mathbb R^{m\times n}$ and $B\in\mathbb R^{n\times m}.$ Then $\operatorname{tr}(AB) = \operatorname{tr}(BA).$ The proof of that is routine.
So we have $Z\sim N_n(0, I_n)$ and $T\in\mathbb R^{n\times n}.$ Then \begin{align} & \operatorname E(Z'TZ) = \operatorname E(\operatorname{tr}(Z'TZ)) = \operatorname E(\operatorname{tr}(TZZ')) \\[8pt] = {} & \operatorname{tr}(\operatorname E(TZZ')) \quad \text{since tr is linear} \\[8pt] = {} & \operatorname{tr}(T\operatorname E(ZZ')) \quad \text{since $T$ is constant} \\[8pt] = {} & \operatorname{tr}(T I_n) = \operatorname{tr}(T). \end{align} So it appears that your proposed identity is mistaken
answered Jun 12, 2019 at 5:12