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Assume $\mathbf{x}$ is a random vector with mean $\mathbf{\mu}$ and covariance matrix $\mathbf{\Sigma}$. Symmetric matrices $\mathbf{A}$ and $\mathbf{B}$ are given.

Without assuming normality, how to prove that the covariance matrix of the quadratic form is the following?

$$ \mbox{Cov} \left( \mathbf{x'Ax,x'Bx} \right) = 2 \mbox{Tr} \left( \mathbf{A\Sigma B\Sigma} \right) + 4 \mathbf{\mu^T A\Sigma B \mu} $$

I found a proof but assuming normality. Can anyone please give me some hint or reference to prove this? Thanks a lot!

I found Robert's suggestion pretty reasonable. I found this question is taken from the book 《Modern multivariate analysis》, Casella. enter image description here

It's right to assume normality.

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  • $\begingroup$ Do you agree with my edits? $\endgroup$
    – Rodrigo de Azevedo
    Commented Mar 25, 2021 at 15:30
  • $\begingroup$ Yes! Thanks for your editing. This is my first time to ask questions here. You made this more formally and tidy. $\endgroup$
    – Regan
    Commented Mar 25, 2021 at 15:34

1 Answer 1

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Why would you think normality is not needed? Consider the $1$-dimensional case: the left side is a constant times the variance of $x^2$, which depends on the $4$'th moment; it is not just a function of the mean and variance of $x$.

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  • $\begingroup$ @Regan, when you get a moment you might follow up on Robert's suggestion as a counterexample. Work through the simple situation of a single dimension with $X$ being exponentially distributed with mean 1. $\endgroup$
    – A rural reader
    Commented Mar 25, 2021 at 18:17
  • $\begingroup$ Thanks a lot! I found the normality assumption is definitely needed. $\endgroup$
    – Regan
    Commented Mar 26, 2021 at 2:05

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