Why Effect Patterns Matter
A gene-level rare-variant test combines evidence across multiple variants. The choice of how to combine that evidence matters: a test that gains signal when effects point in the same direction can lose signal when they oppose one another. Another test can retain evidence from variants acting in different directions, but may pay a price when effects are uniform and many variants contribute little or no signal.
The burden test and the sequence kernel association test (SKAT) use different combinations of variant-level scores. A burden test first combines scores into one weighted sum and then tests that sum. SKAT instead forms a weighted sum of squared scores. This difference explains why their relative power depends on the pattern of variant effects. It does not mean that either test is always more powerful.
The subscript 0 denotes a variance or distribution under the null hypothesis. In an analysis with covariates or correlated variant scores, the burden variance and the null distribution of SKAT must reflect that structure. For example, if the null score vector \(U=(U_1,\ldots,U_m)'\) has covariance matrix \(V\), then \(\operatorname{Var}_0(B)=a'Va\). SKAT’s null distribution is determined by the eigenvalues of the appropriately weighted score covariance matrix; it is not generally an ordinary chi-square distribution with \(m\) degrees of freedom.
For a first comparison, use a deliberately simplified score model. Suppose the scores are independent and have equal information \(I>0\), so that under a local alternative \(U_j\sim N(I\beta_j,I)\). Here \(\beta_j\) is the effect for variant \(j\) on the model’s effect scale, and \(I\) is the common information scale. This approximation isolates the effect-pattern comparison; real analyses must account for covariates, linkage disequilibrium, and the appropriate trait model.
What Each Statistic Retains
Under the simplified model, the weighted burden score has mean \(I\sum_j a_j\beta_j\) and variance \(I\sum_j a_j^2\). Its standardized score therefore has a one-degree-of-freedom noncentral chi-square distribution after squaring:
The quantity \(\lambda_B\) is the noncentrality parameter in this simplified model. It depends on the squared weighted sum of the effects. Consequently, effects of opposite signs can cancel before the score is squared.
For SKAT, the same simplified model gives a different distribution. The component scores are independent, and \(U_j/\sqrt I\sim N(\sqrt I\beta_j,1)\). Thus each squared standardized score has a noncentral chi-square distribution:
When weights differ, this is a weighted sum of noncentral chi-square variables, not in general one noncentral chi-square distribution with a single noncentrality parameter. The quantity \(I\sum_j a_j^2\beta_j^2\) is a useful weighted signal measure, but it is not, in general, a noncentrality parameter for the whole SKAT distribution. If all weights equal a common value \(c\), then \(Q_S/(c^2I)\sim\chi^2_m(I\sum_j\beta_j^2)\). In particular, with all weights equal to 1, \(Q_S/I\sim\chi^2_m(I\sum_j\beta_j^2)\).
A Projection View of the Burden Comparison
With equal weights, the burden test compresses the vector of effects into its sum, while the SKAT weighted signal measure depends on the sum of squared effects. This can be made precise using the Cauchy–Schwarz inequality:
For equal weights, the burden noncentrality is \(I(\sum_j\beta_j)^2/m\), while the corresponding SKAT weighted signal measure is \(I\sum_j\beta_j^2\). The inequality shows that the burden signal is the part of the effect pattern captured by the all-ones direction; it is largest relative to the squared-effect signal when the effects are aligned and equal. Equality holds when all \(\beta_j\) are equal. This comparison helps explain signal retention, but it does not by itself establish which test has greater power: the tests have different null distributions and critical values.
Worked Example: Concordant Effects
Suppose a hypothetical gene has three included variants with equal weights, and the assumed effects are \(\beta=(0.20,0.20,0.20)\) on the model’s effect scale. Let the common information be \(I=100\). The effects are concordant: each has the same positive direction.
The burden noncentrality is
The corresponding equal-weight SKAT distribution is \(Q_S/I\sim\chi^2_3(12)\), because \(I\sum_j\beta_j^2=100(0.04+0.04+0.04)=12\). The burden statistic has a one-degree-of-freedom noncentral chi-square distribution with parameter 12; the scaled SKAT statistic has three degrees of freedom and the same noncentrality parameter. The different degrees of freedom matter for power, so equal noncentrality does not mean equal power.
For a concrete score calculation, suppose the observed standardized scores \(z_j=U_j/\sqrt I\) are \((3,3,2)\). The equal-weight burden statistic is \((3+3+2)^2/3=64/3=21.33\), while SKAT’s statistic on this standardized-score scale is \(3^2+3^2+2^2=22\). These are compared with different null reference distributions: \(\chi^2_1\) for the burden statistic and \(\chi^2_3\) for SKAT’s equal-weight statistic on this scale. The similar numerical values do not make them interchangeable.
Mixed Directions and Cancellation
When some variants increase the modeled trait and others decrease it, a burden sum may conceal their separate contributions. The sign convention for each effect must be consistent—for example, the same allele coding and effect scale across variants—before describing effects as concordant or mixed. SKAT squares each variant score, so a negative score does not cancel a positive one in its quadratic statistic. That feature can be useful when directions genuinely vary, although it does not guarantee high power in every mixed-effect setting.
Worked Example: Mixed-Direction Effects
Consider a hypothetical three-variant gene with equal weights, \(I=100\), and effects \(\beta=(0.30,-0.30,0.10)\). The first two effects oppose one another, so the weighted burden sum is small relative to their individual magnitudes.
The burden noncentrality is
For SKAT, the component noncentrality parameters are \(100(0.30)^2=9\), \(100(-0.30)^2=9\), and \(100(0.10)^2=1\). Therefore \(Q_S/I\) has the distribution \(\chi^2_1(9)+\chi^2_1(9)+\chi^2_1(1)\), where the three variables in the sum are independent. This is not a single noncentral chi-square distribution. Here the weighted signal measure is \(9+9+1=19\), much larger than the burden noncentrality of \(0.3333\); the comparison illustrates cancellation in the burden, not a universal ranking of test power.
For observed standardized scores \((4,-4,1)\), the burden statistic is \((4-4+1)^2/3=1/3=0.3333\), whereas the SKAT statistic is \(4^2+(-4)^2+1^2=33\). Again, each value requires its own null reference distribution. A large SKAT statistic can reflect evidence in opposing directions, while the burden score can be small because those directions cancel.
Sparse Effects and Dilution
A different mismatch occurs when only a small subset of the included variants affects the trait. With equal weights, a burden score averages its evidence across all \(m\) variants through its standardization. If many included variants have no effect, the combined score can be less concentrated than it would be if the active variants were known in advance. SKAT can retain squared evidence from the active component scores without requiring their signs to align.
Worked Example: One Active Variant
Suppose a hypothetical gene contains four equally weighted variants, but only the first has a nonzero effect: \(\beta=(0.40,0,0,0)\). Let \(I=50\). Under the simplified model, the burden noncentrality is
For SKAT, the first component has noncentrality \(50(0.40)^2=8\), and the other three are central. Thus \(Q_S/I\) is a sum of one \(\chi^2_1(8)\) variable and three independent \(\chi^2_1\) variables. Its weighted signal measure is 8, whereas the burden noncentrality is 2. The distinction reflects dilution of a single active effect in the equal-weight burden; it does not imply that SKAT always has greater power when effects are sparse.
If the observed standardized scores are \((5,0,0,0)\), the burden statistic is \(5^2/4=6.25\), and SKAT’s equal-weight statistic is \(5^2=25\). The burden statistic has a one-degree-of-freedom null reference distribution; SKAT’s has a four-degree-of-freedom reference distribution. A larger raw statistic is not, on its own, evidence of a smaller p-value.
Conditions Behind the Comparison
These examples use a simplified independent, equal-information score model to make effect-pattern consequences transparent. In practice, several features can change the comparison:
- Effect alignment and weights. A burden test is most naturally suited to effects that align with the chosen allele coding and weights. If the weights emphasize variants with larger effects and the signs are aligned, the burden projection can be efficient. Weights that poorly match the effects can weaken that advantage.
- Correlation among variants. Linkage disequilibrium can make score components correlated. Then the burden variance uses the full covariance \(a'Va\), and SKAT’s null distribution depends on the weighted covariance eigenvalues. Treating correlated scores as independent can give incorrect calibration.
- Trait model and covariates. The score vector and its covariance must come from the specified trait model, with covariates handled appropriately. The binary- and quantitative-trait formulations differ in their score construction and variance; the effect-pattern comparison does not remove the need for correct model specification.
- Variant set and weighting. Both tests depend on which variants are included and how they are weighted. If many variants are uninformative or the chosen weights are poorly matched to the effects, performance can change. A test’s theoretical advantage under one effect pattern may not carry over to another analysis.
The tests also answer the same broad gene-level association question through different alternatives. A significant result from either test is evidence against its gene-level null under the analysis assumptions; it does not, by itself, identify which variant is causal or establish the direction of every variant’s effect.
Common Mistake
Check Your Understanding
Use the definitions and simplified model above to answer the following.
- Why can two large variant effects contribute little to the burden noncentrality if they have opposite signs?
- With equal weights and \(m\) variants, write the burden noncentrality and the SKAT distribution under the simplified independent-score model.
- Why is \(I\sum_j a_j^2\beta_j^2\) not generally a single noncentrality parameter for SKAT when the weights differ?
- If only one of many included variants has an effect, what feature of an equal-weight burden can reduce its signal relative to the active component?
- Why should the numerical values of a burden statistic and a SKAT statistic not be compared without considering their null distributions?