Why Effect Heterogeneity Matters
A rare-variant set may contain effects that differ in magnitude, and some may increase a quantitative phenotype while others decrease it. A burden score adds variant contributions, so opposing effects can cancel. SKAT instead uses a weighted sum of squared score contributions. Squaring prevents cancellation of opposite signs, but it does not guarantee high power: power also depends on the total signal, variant weights, genotype correlation, and the null distribution of the statistic.
The focus here is how those pieces combine under heterogeneous alternatives. The derivation uses a quantitative-trait score model. It is a clean setting for seeing the mechanism; binary-trait score tests have analogous large-sample behavior but require their own null-model variance calculations.
Model, Statistic, and Assumptions
Let \(y\) be an \(n\)-vector of quantitative phenotypes, \(C\) an \(n\times q\) covariate matrix, and \(G\) an \(n\times m\) matrix of genotypes for \(m\) variants. Consider the model
Here \(\gamma\) contains nuisance covariate effects and \(\beta\) contains the variant effects. For the exact finite-sample results below, treat \(C\) and \(G\) as fixed, assume \(C\) has full column rank, and assume independent Gaussian errors with common variance \(\sigma^2\). The null hypothesis is \(H_0:\beta=0\). Define the residual projection matrix
The score vector for the variant effects under the null is \(U=G'P_Cy\). With nonnegative variant weights \(w_j\), let \(W=\operatorname{diag}(w_1,\ldots,w_m)\). The SKAT quadratic statistic is \(Q=U'WU\). The weights can reflect prespecified variant properties; they must be chosen without using the observed phenotype in a way that invalidates the test’s calibration.
Under \(H_0\), \(U\) has mean zero and covariance matrix \(\sigma^2H\). Under the specified alternative, its mean is \(H\beta\). Thus the statistic’s behavior changes not only through the effect vector but also through the genotype information matrix \(H\). A zero or nearly zero eigenvalue of the weighted information matrix means that a corresponding genotype combination is unobserved or weakly informed by the data.
Why Squared Scores Avoid Sign Cancellation
A useful first case is orthogonal, equally informative variants: take \(H=I_m\), \(W=I_m\), and suppose \(\sigma^2\) is known. Then the score coordinates are independent, with \(U_j/\sigma\sim N(\beta_j/\sigma,1)\). Therefore,
The noncentrality parameter \(\lambda\) is the total squared standardized signal in this special case. The signs of the effects do not appear in it. A burden score, by contrast, adds the score coordinates before testing, so positive and negative contributions can offset each other. This contrast explains a key way SKAT can gain power: it can detect a set with real effects even when their directions are mixed.
Worked Example: Power for a Mixed-Effect Architecture
Suppose four orthogonal variants have standardized effects \((2,2,0,0)\), so \(\lambda=2^2+2^2=8\). Under the null, \(Q/\sigma^2\) has a central chi-square distribution with \(df=4\). At significance level \(\alpha=0.05\), its upper-tail critical value is approximately 9.488. Under this alternative, the power is the probability that a noncentral chi-square random variable exceeds that value.
A noncentral chi-square variable with \(df=4\) and noncentrality parameter 8 has a Poisson-mixture representation. With \(K\sim\operatorname{Pois}(8/2)=\operatorname{Pois}(4)\), its upper-tail probability is
For example, \(P(K=k)=e^{-4}4^k/k!\); evaluating the mixture through \(k=20\) gives approximately 0.6047, with the remaining terms negligible at the reported precision. Thus this hypothetical test has about 60.47% power at the 5% level under the stated model and architecture. This is not a claim about a real study; it illustrates how a specified noncentrality translates into power.
Worked Example: Opposing Effects and Burden Cancellation
Consider four orthogonal variants with standardized effects \((1,1,-1,-1)\). The SKAT noncentrality parameter is
Consequently, \(Q/\sigma^2\sim\chi^2_4(4)\), and its expected value is \(4+4=8\), compared with 4 under the null. The quadratic statistic retains signal from all four effects because each score contribution is squared.
For comparison, the standardized burden score in this orthogonal example is \(B=(U_1+U_2+U_3+U_4)/(2\sigma)\). Its variance is 1, and its mean under this alternative is \((1+1-1-1)/2=0\). Thus its distribution is the same as under the null; a two-sided level-0.05 burden test has only its 5% null rejection probability in this idealized case. This calculation shows the cancellation mechanism, not a universal ranking of the two tests.
General Alternative Distribution
With correlation among variants or unequal weights, the orthogonal case no longer describes the full distribution. Set \(A=W^{1/2}HW^{1/2}\) and \(z=W^{1/2}U/\sigma\). Then \(Q/\sigma^2=z'z\), and under the model \(z\) has covariance \(A\) and mean \(W^{1/2}H\beta/\sigma\). Let \(a_j\) be the positive eigenvalues of \(A\), with corresponding unit eigenvectors \(v_j\). Resolving \(z\) in these eigen-directions gives the alternative distribution
Here each \(\chi^2_1(\delta_j)\) is a noncentral chi-square variable with one degree of freedom and noncentrality \(\delta_j\), and the variables in the sum are independent. Under the null, all \(\delta_j=0\); the null distribution is therefore a weighted sum of central chi-square variables. This is the mixture chi-square calibration for SKAT.
The formula follows by diagonalizing \(A\). In the eigenvector direction \(v_j\), the coordinate \(v_j'z\) has variance \(a_j\) and mean \(v_j'W^{1/2}H\beta/\sigma\). Dividing that coordinate by \(\sqrt{a_j}\) gives a unit-variance normal variable with squared mean \(\delta_j\). Its square is \(\chi^2_1(\delta_j)\), and its contribution to \(z'z\) is \(a_j\) times that square. Eigen-directions with \(a_j=0\) have no random score variation; under this model they also have zero alternative mean and contribute nothing.
Worked Example: The Same Effect Size in Different LD Directions
Suppose, in weighted eigen-directions, \(W=I\), \(\sigma^2=1\), and \(H=A=\operatorname{diag}(4,1)\). The eigenvalues are 4 and 1, so the null distribution of \(Q\) is \(4\chi^2_1+\chi^2_1\), with independent central terms. Consider first \(\beta=(1,0)\). The score mean is \(H\beta=(4,0)\), so
Now consider \(\beta=(0,1)\), which has the same Euclidean length but lies in the other eigen-direction. Its score mean is \((0,1)\), giving
The two distributions differ because the genotype information differs by direction. Their expected statistics are not themselves powers: power requires comparing each alternative distribution with the null critical region. The example establishes why effect magnitudes alone cannot determine power when LD and weights produce unequal eigenvalues.
Interpreting Heterogeneity, Weights, and Power
A random-effects description often used for SKAT represents variant effects as mean-zero random effects with a variance related to the variant weights. In that formulation, the variance component describes the scale of effect variation, while the score test assesses whether that component is zero. This connects SKAT’s quadratic form to alternatives with dispersed effects, including mixed signs. It does not mean that every variant must have a nonzero effect, or that the fitted model identifies each variant’s direction.
Weights matter because they change both the contributions to \(Q\) and the eigenvalues that determine its null reference distribution. A weight that emphasizes truly associated variants can improve sensitivity; a weight that emphasizes null or weakly associated variants can dilute it. Likewise, LD can combine variant signals into eigen-directions. Effects aligned with well-informed directions may contribute differently from effects aligned with poorly informed directions. The eigenvalue calculation, rather than a simple count of positive effects, captures this interaction.
Power is the probability that the test rejects \(H_0\) under a specified alternative. It is not determined by a \(p\)-value observed in one sample. For SKAT it depends on the effect vector, the genotype covariance, the weights, the residual variance, the sample size, and the significance threshold. The introductory material on random-effects models for variant-set association gives the variance-component perspective; here the key power implication is that the quadratic score accumulates squared signal, subject to the covariance structure.
Diagnostics, Alternatives, and Common Pitfalls
Before interpreting a power calculation, verify that its model reflects the analysis. Check that the tested variant set and weights were specified as intended, that the covariate adjustment is adequate, and that genotype coding and quality control are appropriate. Inspect whether the effective eigenvalue spectrum is concentrated in a few directions or spread across many; this helps explain what information the statistic can use. If the Gaussian linear-model assumptions are implausible, use a score variance and null calibration appropriate to the phenotype model rather than borrowing the quantitative-trait formula uncritically.
The main alternatives answer different questions. A burden test compresses variant scores into a signed aggregate and can be efficient when effects point in a common direction, but can lose signal through cancellation. SKAT retains squared contributions and is suited to alternatives with varying directions, though its power can be lower when the signal is highly aligned and a burden score captures it efficiently. Single-variant tests localize evidence variant by variant but do not pool weak signals across a set in the same way. These contrasts describe their mechanisms; they do not provide a universal selection rule.
- Mistake: “Mixed signs guarantee SKAT power.” Mixed signs prevent cancellation in the squared statistic, but weak effects, poor weights, limited sample size, or low-information genotype directions can still produce low power.
- Mistake: “More effect heterogeneity always raises SKAT power.” With orthogonal equal-information variants, equal sums of squared standardized effects give the same noncentrality, regardless of sign pattern or distribution of magnitudes.
- Mistake: “The noncentrality parameter is the power.” It indexes signal strength, but power also depends on the null critical value and the alternative distribution. In correlated settings, a weighted sum of noncentral chi-square variables may be needed.
- Mistake: “A large expected \(Q\) proves high power.” Power depends on the full distribution relative to the rejection threshold, not its mean alone. Two alternatives with different variance or eigenvalue structure can have different rejection probabilities even when their means are similar.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions below.
- In the orthogonal, equally weighted model, what quantity determines the noncentrality parameter, and does changing an effect’s sign change it?
- For standardized effects \((1,1,-1,-1)\), calculate the SKAT noncentrality parameter and the standardized burden mean.
- Why is the null distribution generally not an ordinary chi-square distribution when variants are correlated or have unequal weights?
- In the LD example with \(A=\operatorname{diag}(4,1)\), why can equal-length effect vectors produce different score distributions?
Solutions. (1) The noncentrality is \(\sum_j(\beta_j/\sigma)^2\); changing a sign leaves each squared term unchanged. (2) The SKAT noncentrality is \(1+1+1+1=4\), while the standardized burden mean is \((1+1-1-1)/2=0\). (3) The score covariance has unequal eigenvalues, so the quadratic form is a weighted sum of chi-square variables rather than a single unweighted chi-square. (4) The eigen-directions have different information: the score mean is \(H\beta\), and the variance contribution in each direction is set by its eigenvalue, so reallocating the same effect length changes the noncentral components and their weights.