The Mixture Chi-Square Null Distribution of SKAT

Why the Null Distribution Is a Mixture

The SKAT statistic is a quadratic form in variant-level score information. Even when each score is approximately normal under the null hypothesis, the quadratic form does not generally have the same null distribution as a single chi-square statistic. Its distribution depends on how variability is spread across directions in the weighted score vector. The eigenvalues of that vector’s covariance matrix describe this spread and determine the null law.

The distinction matters for inference: a chi-square distribution with a fixed number of degrees of freedom is not generally a valid substitute for the SKAT null distribution. This tutorial derives the mixture form and explains what its eigenvalues do. It does not address numerical methods for evaluating tail probabilities.

Score Quadratic Form and Assumptions

Let \(U\) be the \(m\)-component vector of variant-level scores under the null hypothesis \(H_0\), and let \(V\) be its covariance matrix under \(H_0\). Let \(W\) be the symmetric positive-semidefinite matrix that defines the SKAT quadratic form. In the usual diagonal-weight case, \(W\) is diagonal and its entries are the nonnegative quadratic-form weights. Define

The distributional argument below uses the standard score-normal approximation: under \(H_0\), \(U\) is approximately multivariate normal with mean zero and covariance \(V\). Equivalently, the weighted score vector \(W^{1/2}U\) is approximately multivariate normal with mean zero and covariance \(W^{1/2}VW^{1/2}\). This approximation is exact if the score vector is exactly multivariate normal; in applications it is typically an asymptotic approximation under the null model and its regularity conditions. The weights are treated as fixed for this derivation, and the covariance matrix is assumed to describe the null variability adequately.

The matrix \(W^{1/2}VW^{1/2}\) is positive semidefinite: \(V\) is a covariance matrix and hence positive semidefinite, and \(W\) is also positive semidefinite. Thus its eigenvalues are nonnegative. If some eigenvalues are zero, their directions contribute nothing to \(Q\) and are omitted from the sum. The number \(r\) of positive eigenvalues is the rank of this matrix.

Deriving the Mixture Form

Set \(Z=W^{1/2}U\) and \(A=W^{1/2}VW^{1/2}\). Then \(Q=Z^{\mathsf T}Z\), and under the stated score-normal approximation \(Z\) is approximately \(N(0,A)\). Since \(A\) is symmetric and positive semidefinite, it has an orthogonal eigendecomposition \(A=P\Lambda P^{\mathsf T}\), where \(P^{\mathsf T}P=I\) and \(\Lambda\) is diagonal with the eigenvalues of \(A\). Rotating the vector by \(P^{\mathsf T}\) preserves its squared length. Its coordinates in the eigenvector directions have variances given by the eigenvalues.

A multivariate normal vector with diagonal covariance has independent coordinates. For each positive eigenvalue \(\lambda_j\), the corresponding coordinate \(Y_j\) is therefore distributed as \(\sqrt{\lambda_j}\,N_j\), where the \(N_j\) are independent standard normal random variables. Squaring and summing gives the result:

Here \(\overset{d}{=}\) denotes equality in distribution under the normal model, or the corresponding approximate equality under the score-normal approximation. A zero eigenvalue gives a coordinate with zero variance, so that coordinate is zero with probability one and adds no term. This completes the derivation: each independent squared standard-normal coordinate contributes a \(\chi^2_1\) variable, scaled by its eigenvalue.

Worked Examples

Why the Degrees of Freedom Are Not Generally Fixed

For a standard \(\chi^2_k\) distribution, the statistic is a sum of \(k\) independent squared standard normals, each with coefficient \(1\). SKAT instead produces the weighted sum \(\sum_{j=1}^{r}\lambda_jX_j\). Its number of nonzero terms is the rank \(r\), but the coefficients can differ across terms and can depend on the covariance structure as well as on the chosen kernel weights. Thus “degrees of freedom equal the number of variants” is generally wrong, and “degrees of freedom equal the rank” does not by itself specify the mixture law.

If all \(r\) positive eigenvalues are equal to a common value \(c\), the distribution simplifies to \(c\chi^2_r\). If \(c=1\), this is an ordinary \(\chi^2_r\); if \(c\ne1\), it is scaled. When the eigenvalues are unequal, the law generally remains a weighted sum rather than a single chi-square distribution. The eigenvalues are consequently part of the null-distribution specification, not merely a way to count its terms.

Two useful checks follow directly from the mixture representation. Its mean and variance are

These formulas can verify calculations and show why a simple degree-of-freedom count loses information: the mean depends on the sum of the eigenvalues, while the variance also depends on their squared sizes. They do not, on their own, turn the mixture into an ordinary chi-square distribution.

Common Mistake and Practical Meaning

Incorrect reasoning: “There are \(m\) variants, so the SKAT statistic has a \(\chi^2_m\) null distribution.” This treats every score direction as if it contributed an independent squared standard normal with coefficient \(1\). But the correct null form has coefficients equal to the eigenvalues of \(W^{1/2}VW^{1/2}\); some can be zero, and the positive ones need not equal one.

Correction: First establish the null score covariance and the quadratic-form weights, then use the eigenvalues of the resulting weighted covariance matrix to characterize the mixture. The number of variants alone does not determine either the number of nonzero components or their weights. In particular, correlated score directions can change the eigenvalues, even when the variant count is unchanged.

The mixture representation provides the bridge between the SKAT score statistic and its null calibration. It explains why two analyses with the same number of variants need not have the same null distribution, and why the covariance structure matters. A computed SKAT statistic is interpreted relative to this distribution under the null model; the mixture form does not imply that any one variant is individually associated or establish a causal effect.