When Is a Gaussian Quadratic Form Chi-Square?

Why the Eigenvalues Decide the Distribution

A quadratic form in a Gaussian vector can look like a sum of squares without actually having a chi-square distribution. The key distinction is whether the directions being squared are weighted equally. For a standard normal vector, the eigenvalues of the matrix give exactly those weights. This tutorial establishes the precise criterion for an unscaled chi-square law, then separates it from the more general case of a scaled chi-square law.

Let \(Z\) be a \(p\)-dimensional vector whose entries are independent \(N(0,1)\) random variables, and let \(A\) be a real symmetric \(p\times p\) matrix. We study \(Q=Z^\mathsf{T}AZ\). If a quadratic form is initially written with a nonsymmetric matrix, replacing it by its symmetric part \((A+A^\mathsf{T})/2\) leaves the quadratic form unchanged.

By the spectral theorem, there is an orthogonal matrix \(U\) such that \(A=U\Lambda U^\mathsf{T}\), where \(\Lambda\) is diagonal with eigenvalues \(\lambda_1,\ldots,\lambda_p\). The standard normal density is unchanged by orthogonal transformations, so \(Y=U^\mathsf{T}Z\) again has independent \(N(0,1)\) entries. Consequently,

This weighted-sum representation is the starting point. A chi-square law has unit weights on the included squared normals and zero weights on the rest. The main issue is proving that no different set of eigenvalues can produce the same law.

The Exact Criterion

We prove both directions. The forward direction uses the moment-generating function (MGF), including when some eigenvalues might be negative; thus nonnegativity of the quadratic form is not assumed in advance.

For \(Q=\sum_i\lambda_iY_i^2\), independence and the one-dimensional normal integral give, for \(t\) in an open interval about zero where each \(1-2t\lambda_i>0\),

The interval exists because there are finitely many finite eigenvalues. The same formula for a chi-square variable is \(M_{\chi^2_k}(t)=(1-2t)^{-k/2}\), for \(t<1/2\).

Both sides are rational functions of \(t\). Because they agree on an interval, multiplying by their denominators gives a polynomial that vanishes at infinitely many points; that polynomial is identically zero. The rational functions therefore agree wherever both are defined. Each nonzero eigenvalue \(\lambda_i\) contributes a pole at \(t=1/(2\lambda_i)\) on the left, with residue \(-1/2\). Eigenvalues that are equal contribute the same pole, so their residues add; distinct eigenvalues have distinct poles and cannot cancel one another. The right side has a pole only at \(t=1/2\), with residue \(-k/2\). Thus every nonzero eigenvalue must be \(1\). If there are \(r\) such eigenvalues, the residue at \(1/2\) on the left is \(-r/2\); equality gives \(r=k\). All other eigenvalues are zero. The spectral theorem now gives \(A^2=A\), and its rank is the number of its nonzero eigenvalues, namely \(k\). \(\square\)

The criterion is stronger than checking that \(Q\) is nonnegative or that its mean matches a proposed chi-square mean. It requires the entire spectrum to consist only of zeros and ones.

Scaled Chi-Square Laws Are Different

Equal positive eigenvalues produce a scaled chi-square law, not necessarily an unscaled one. The distinction can be characterized just as sharply.

To see the necessity, the MGF of \(c\chi^2_k\) is \((1-2ct)^{-k/2}\). Equating it to the MGF of \(Q\), taking logarithms, and differentiating gives

The rational-function argument used in the preceding proof applies again. The right side has its only pole at \(1/(2c)\); each nonzero eigenvalue on the left contributes a pole at \(1/(2\lambda_i)\), with residue \(-1/2\). Equality forces every nonzero eigenvalue to be \(c\), and matching residues forces exactly \(k\) such eigenvalues. Conversely, those eigenvalues give \(Q=c\sum_{i=1}^{k}Y_i^2\), which has the stated law.

What the Criterion Rules Out

A negative eigenvalue is an immediate warning sign: along its eigendirection, the weighted sum contains a negative multiple of a squared normal. The exact theorem already rules out a chi-square law, whose values are nonnegative, because the required spectrum contains only zeros and ones. But even positive eigenvalues do not suffice. They must also all equal \(1\) for an unscaled chi-square, or all equal the same \(c>0\) for a scaled chi-square.

A common mistake is to see a sum of squared normal coordinates and conclude that it is chi-square. For example, \(3Y_1^2+3Y_2^2\) is indeed a constant times a sum of two squared standard normals, but its law is \(3\chi^2_2\), not \(\chi^2_2\). Another tempting shortcut is to match means: many different eigenvalue lists can have the same sum and hence the same mean. Matching a mean, or even a few moments, is not a substitute for the eigenvalue criterion.