Craig's Criterion for Independent Quadratic Forms

Why a Matrix Condition Can Decide Independence

A quadratic form in a Gaussian vector can combine many coordinates, so independence is not always visible by inspection. Craig’s criterion gives an exact test: for a centered Gaussian vector with nonsingular covariance, two symmetric quadratic forms are independent precisely when a particular matrix product is zero. The criterion is stronger than a check of covariance alone; it captures the way the forms share Gaussian directions.

When the covariance is singular, some directions never vary. The criterion must then be applied to the parts of the matrices that act on the Gaussian vector’s support. This distinction matters: a raw matrix product can be nonzero even when the forms, as random variables, are independent.

Setup and Statement of the Criterion

A symmetric nonnegative definite matrix has a symmetric nonnegative definite square root \(\Sigma^{1/2}\). We can represent \(X\) in distribution as \(X=\Sigma^{1/2}Z\), where \(Z\sim N_p(0,I_p)\). This representation also applies when \(\Sigma\) is singular. Define the effective matrices

Then \(Q_A=Z'CZ\) and \(Q_B=Z'DZ\). Thus the forms depend on \(A\) and \(B\) only through their action on the varying Gaussian directions.

For positive definite \(\Sigma\), the equivalence follows because \(\Sigma^{1/2}\) is invertible: \(CD=\Sigma^{1/2}A\Sigma B\Sigma^{1/2}\), so \(CD=0\) exactly when \(A\Sigma B=0\). For singular \(\Sigma\), the criterion is the support-adjusted condition \(CD=0\), not necessarily the raw condition \(A\Sigma B=0\).

Proof of Craig’s Criterion

We first obtain the joint moment-generating function near the origin. For a symmetric matrix \(H\) and standard normal \(Z\), diagonalizing \(H\) by an orthogonal matrix reduces \(Z'HZ\) to a weighted sum of independent squared standard normal variables. The one-dimensional Gaussian integral gives \(E[\exp(t\lambda Z_1^2)]=(1-2t\lambda)^{-1/2}\) when \(1-2t\lambda>0\). Therefore, for \(s,t\) sufficiently close to zero,

The neighborhood can be chosen so that \(I_p-2sC-2tD\) is positive definite; the integral and determinant formula are then finite and the square root is unambiguous.

If \(Q_A\) and \(Q_B\) are independent, their joint moment-generating function factors into the product of their marginal moment-generating functions. The displayed formula consequently implies, for \(s,t\) near zero,

To extract a matrix consequence, use the expansion \(\log\det(I_p-H)=-\sum_{k\geq 1}\operatorname{tr}(H^k)/k\), valid when the eigenvalues of \(H\) have absolute value less than one. Taking logarithms of the determinant identity shows that the coefficient involving \(s^2t^2\) in \(\operatorname{tr}((2sC+2tD)^4)\) must be zero: the separate terms on the right have no mixed powers. Expanding the fourth power and using invariance of trace under cyclic rearrangement, that coefficient, apart from the nonzero common factor \(2^4\), is

Set \(a=\operatorname{tr}(C^2D^2)\) and \(M=CD\). Since \(C\) and \(D\) are symmetric, \(a=\operatorname{tr}(M'M)=\|M\|_F^2\), where \(\|M\|_F^2\) is the sum of the squares of the entries of \(M\). Also, Cauchy–Schwarz for the Frobenius inner product gives \(\lvert\operatorname{tr}(M^2)\rvert=\lvert\operatorname{tr}(M'M')\rvert\leq\|M\|_F^2=a\). Hence

The left side is zero, so \(a=0\), which forces every entry of \(M\) to be zero. Thus \(CD=0\). This proves necessity.

Conversely, suppose \(CD=0\). Since \(C\) and \(D\) are symmetric, \(DC=(CD)'=0\) as well. The range of \(C\) is therefore orthogonal to the range of \(D\). By the spectral theorem, \(Z'CZ\) depends only on the projection of \(Z\) onto the range of \(C\), and \(Z'DZ\) only on its projection onto the range of \(D\). Orthogonal projections of a standard Gaussian vector onto orthogonal subspaces are independent: in an orthonormal basis adapted to those subspaces, they use disjoint groups of independent standard normal coordinates. Functions of these independent projections are independent, so \(Q_A\) and \(Q_B\) are independent. This proves sufficiency and completes the theorem. \(\square\)

Worked Applications

Nonzero Means and the Scope of the Test

The theorem above is stated for a centered Gaussian vector. A useful sufficient condition also covers a nonzero mean when \(\Sigma\) is positive definite. Write \(X=\mu+\Sigma^{1/2}Z\), with \(Z\sim N_p(0,I_p)\), and use the effective matrices \(C=\Sigma^{1/2}A\Sigma^{1/2}\) and \(D=\Sigma^{1/2}B\Sigma^{1/2}\). Then

If \(CD=0\), the ranges of \(C\) and \(D\) are orthogonal. Because \(\Sigma\) is invertible, \(\Sigma^{1/2}A\mu=C\Sigma^{-1/2}\mu\) lies in the range of \(C\), and \(\Sigma^{1/2}B\mu=D\Sigma^{-1/2}\mu\) lies in the range of \(D\). Thus the first form depends, apart from a constant, only on the standard normal projection onto the range of \(C\), and the second only on the projection onto the orthogonal range of \(D\). They are independent. This establishes sufficiency for arbitrary \(\mu\); the centered theorem’s full equivalence is the claim proved above.

Common Pitfall and Why the Criterion Matters

A common error is to treat \(A\Sigma B=0\) as the criterion regardless of the covariance. For positive definite \(\Sigma\), this is equivalent to \(CD=0\). For singular \(\Sigma\), however, the random vector explores only its covariance support, and the correct test is \(CD=0\). Example 3 shows why imposing the raw product can reject independent forms.

Another error is to infer independence from a single scalar calculation such as zero covariance. Craig’s criterion is a matrix condition, and its proof uses the full joint Gaussian transform. A scalar summary does not generally encode whether the forms use overlapping Gaussian directions. The criterion is especially useful for checking proposed decompositions of Gaussian sums of squares: it provides a precise independence test before any distributional labels are assigned.