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We're looking for the $\operatorname{Cov}\left[x^T A x, ~x^T B x\right]$ where $x$ is random variable and mean-centered, but not independent and $A$ and $B$ are symmetric matrices. The fundamental term to compute is $E\left[x^T A x x^T B x\right]$, a quartic form. For the normal case, you can find this in the matrix cookbook. Has anyone figured this out in general?

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  • $\begingroup$ Not a whole lot of simplification is available for the general case you pose. But if you intend that the $x_i$ be iid, for instance, or exchangeable, or uncorrelated, then extensive simplification is possible. Could you clarify for us what assumptions you are making about the first four moments of $x$ (beyond "mean-centered," which implies all means are zero)? BTW, I'm curious about what you mean by "heterogeneous," because this expectation looks like a homogeneous quartic form in the components of $A$ and $B$ to me. $\endgroup$ – whuber Dec 15 '18 at 20:01
  • $\begingroup$ Maybe you don't need this anymore. I think you may find this video useful: youtube.com/watch?v=Z0jBgMDkfUg. $\endgroup$ – jwyao Apr 16 '20 at 23:12
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In the special case when $A$ and $B$ commute, they are simultaneously diagonalisable. If the spectrum of $A$ is made of the $\lambda_i$'s and the spectrum of $B$ of the $\xi_i$'s, then $$\text{Cov}[x^T A x, ~x^T B x]=\sum_i \lambda_i\xi \text{var}(X_i^2)$$ if $X_i$ is the $i$-th coordinate of $X$ in the orthonormal basis.

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    $\begingroup$ You must be assuming both $A$ and $B$ are symmetric (which is reasonable, but ought to be made explicit). $\endgroup$ – whuber Dec 13 '18 at 15:48
  • $\begingroup$ I think you must also be assuming normality (and independence), otherwise you can't ignore the off-diagonal terms. $\endgroup$ – Hasse1987 Oct 21 '20 at 17:46

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