# The covariance matrix of quadratic form, without normal assumption

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.

It's right to assume normality.

• Do you agree with my edits? Mar 25 at 15:30
• Yes! Thanks for your editing. This is my first time to ask questions here. You made this more formally and tidy. Mar 25 at 15:34

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 $$x$$.
• @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. Mar 25 at 18:17