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I have seen the answer $AB = 0$ is the condition, but I couldn't find an efficient proof.

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    $\begingroup$ You need also to account for the covariance of $X.$ As an extreme example, suppose components $2,3,\ldots, p$ of $X$ are identically zero: then the entries $a_{ij}$ and $b_{kl}$ could be anything provided $i,j,k,l\ne 1.$ $\endgroup$
    – whuber
    Commented Feb 11, 2021 at 18:43
  • $\begingroup$ Thank you @whuber for your response, X has mean vector "mu" and covariance matrix "sigma". $\endgroup$ Commented Feb 12, 2021 at 6:24

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For efficiency, use characteristic functions.

Let $|\cdot|$ denote the determinant. The following demonstration repeatedly evaluates the multinormal ($n$ dimensional) integral in the form

$$(2\pi)^{n/2}|T|^{n/2} = \int\cdots\int \exp\left(-\mathbf{x}\,T\,\mathbf{x}^\prime/2\right)\,\mathrm{d}^n\mathrm{x}$$

where $T$ is any symmetric matrix (with real or complex coefficients) for which the integral is finite.

The question of independence of functions of $X$ comes down to independence of the same functions of $X$ minus its mean (because the mean is constant), so we may assume the mean is zero.

Let's compute the characteristic functions of the variables $XAX^\prime$ and $XBX^\prime$ where the density of $X$ is proportional to $\exp(-\mathbf{x}V\mathbf{x}^\prime/2).$ (When the covariance $\Sigma$ is invertible, $V^{-1}=\Sigma.$ When $\Sigma$ is noninvertible all calculations are performed in a subspace where $\Sigma$ is supported and $A,B$ are understood as representing quadratic forms on this subspace. Thus we may suppose $V$ is invertible.)

By definition, for any real number $s$ the characteristic function of $XAX^\prime$ is

$$\begin{aligned} \phi_{XAX^\prime}(s) &=E\left[\exp(is\,XAX^\prime\right] \\ &=(2\pi|V|)^{-n/2}\int\cdots\int \exp\left(-\mathbf{x}\,\left(-2isA + V\right)\,\mathbf{x}^\prime/2\right)\,\mathrm{d}^n \mathrm{x} \\ &=|1_n - 2isAV^{-1}|^{n/2}. \end{aligned}$$

Since random variables $(U,V)$ are independent if and only if $\phi_U(s)\phi_V(t) = \phi_{(U,V)}(s,t),$ the question of independence comes down to whether

$$|1_n - 2isAV^{-1}|^{n/2}\ |1_n - 2itBV^{-1}|^{n/2} = |1_n - 2isAV^{-1} - 2itBV^{-1}|^{n/2}.$$

Taking roots and using the multiplicative property of determinants shows this equation is equivalent to

$$|1_n - 2isAV^{-1} -2itBV^{-1} - 4stAV^{-1}BV^{-1}| =|1_n - 2isAV^{-1} - 2itBV^{-1}|$$

for all $(s,t).$ That clearly is the case if and only if the extra term on the left hand side vanishes; that is,

$$AV^{-1}\,BV^{-1} = 0.$$

(Equivalently, $A\Sigma B = AV^{-1}B = 0.$)

When $\Sigma=1_n$ is the identity matrix, that indeed is the condition $AB=0$ quoted in the question, QED.


BTW, notice that when $A$ is any real square matrix (symmetric or not), the quadratic form can be written

$$X A X^\prime = \frac{1}{2}X\left(A + A^\prime\right) X^\prime,$$

so we may replace $A$ by $(A+A^\prime)/2,$ which is symmetric, and proceed as before. Therefore the requirement that $A$ and $B$ be symmetric is unnecessary. The condition that the two forms be independent becomes

$$(A+A^\prime)\,\Sigma\,(B+B^\prime)= 0.$$

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  • $\begingroup$ Is this really correct? $\endgroup$
    – Hunaphu
    Commented Feb 23, 2021 at 12:59
  • $\begingroup$ @Hunaphu As your answer indicates, it's correct, but I think it's incomplete: at the end I make a hand-waving assertion about the "only if" condition, without proving it; and it has bothered me that I could not offer an accessible proof. $\endgroup$
    – whuber
    Commented Feb 23, 2021 at 13:02
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The result is called Craig's theorem. Named after the 1943 note Note on the Independence of Certain Quadratic Forms by Craig.

See this paper for details and history.

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