# Proof $E[Z'TZ]^2=\operatorname{tr}^2(T)+\operatorname{tr}(T^2)$ [duplicate]

How to prove second moment of a quadratic form where $$Z$$ has normal distribution with mean zero and covariance matrix identical?

• Jun 12, 2019 at 2:42
• In the one-variable case with $T=(t_{11}),$ you are asking us to show $$t_{11}= t_{11}E[Z^2] =E[t_{11}Z^2] = E[Z^\prime T Z] =\operatorname{tr}^2(T) + \operatorname{tr}(T)^2 = t_{11}^2+t_{11}^2 = 2t_{11}^2.$$
– whuber
Jun 12, 2019 at 15:09

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