The following seems to hold for $X$ distributed as 0-centered normal with covariance $\Sigma$, any suggestions how to show this analytically?
$$E[\|X\|^4] = \operatorname{tr}(\Sigma)^2+2\operatorname{tr}(\Sigma^2)$$
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Someone already worked out an answer, but the distribution of $X^T X$ is also given here – LinAlg Oct 29 '19 at 18:02
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The linked thread above does not concern itself with a general $\Sigma$. – StubbornAtom Oct 29 '19 at 18:14
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@StubbornAtom yes it does – LinAlg Oct 30 '19 at 14:07
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1@StubbornAtom that is not a contradiction with the answer I referred to. I have posted an elegant answer based on that. – LinAlg Oct 30 '19 at 15:39
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Since $\lVert X\rVert^4=(X^TX)^2$, we have for $X\sim N(0,\Sigma)$,
\begin{align} E\,[\lVert X\rVert^4]=E\,[(X^TX)^2]&=\operatorname{Var}[X^TX]+(E\,[X^TX])^2 \\&=2\operatorname{tr}(\Sigma^2)+(\operatorname{tr}(\Sigma))^2 \end{align}
The expectation of $X^T X$ for any vector $X$ is relatively straightforward, as shown here.
The variance of $X^TX$ is a bit more work, but for normal $X$ it can be done by generalizing this approach here.
For a reference regarding derivation of variance of quadratic forms in general, you can look up Linear Regression Analysis by Seber and Lee (second edition, section 1.5).
StubbornAtom
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Write $\Sigma = P^T \Lambda P$, then:
$$X^TX = (\Sigma^{-1/2}X)^T P^T \Lambda P (\Sigma^{-1/2}X) = \sum_i \lambda_i U_i^2,$$ where $U = P \Sigma^{-1/2}X \sim N(0,I)$. So $X^TX$ is a weighted sum of independent $\chi^2_1$ variables. Therefore, the expected value of $X^TX$ is $\sum_i \lambda_i = \text{tr}(\Sigma)$ and the variance is $2 \sum_i \lambda^2 = 2\text{tr}(\Sigma^2)$.
LinAlg
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