# $X$ is a $p$-variate normal vector, and $A$ and $B$ are two symmetric matrices. When will $X^\prime AX$ and $X^\prime BX$ be independent?

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

• 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 $i,j,k,l\ne 1.$
– whuber
Commented Feb 11, 2021 at 18:43
• Thank you @whuber for your response, X has mean vector "mu" and covariance matrix "sigma". Commented Feb 12, 2021 at 6:24

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

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

• Is this really correct? Commented Feb 23, 2021 at 12:59
• @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.
– whuber
Commented Feb 23, 2021 at 13:02

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.