How Effect Configurations Shape SKAT Power

Why Effect Configuration Matters

A gene can contain several associated variants without those variants producing the same statistical pattern. Their effects may point in the same direction, oppose one another, or differ substantially in magnitude. Their allele frequencies can also differ, so equal per-allele effects need not contribute equally to the observed score. These configurations matter because a burden test combines variant scores before squaring, whereas SKAT squares variant scores before combining them.

Power is the probability that a test rejects its null hypothesis under a specified alternative. It is not a fixed property of the test alone: it depends on the significance level, sample size, variant weights, allele frequencies, and the effects assumed under the alternative. The calculations below isolate the role of effect configuration using a deliberately simplified score model. They clarify the comparison without replacing power calculations for a particular study design.

Model for comparison: Suppose there are \(m\) variants and their covariate-adjusted score components have been standardized so that, under the null, \(U_j\sim N(0,1)\), independently for \(j=1,\ldots,m\). Under a specified alternative, let \(U_j\sim N(\delta_j,1)\), where \(\delta_j\) is the standardized mean shift for variant \(j\). For equal variant weights, use the squared standardized burden statistic \(T_B=(\sum_{j=1}^{m}U_j)^2/m\) and the SKAT statistic \(Q=\sum_{j=1}^{m}U_j^2\). In this model, \(T_B\) has a noncentral chi-square distribution with 1 degree of freedom and noncentrality \(\lambda_B=(\sum_j\delta_j)^2/m\); \(Q\) has a noncentral chi-square distribution with \(m\) degrees of freedom and noncentrality \(\lambda_S=\sum_j\delta_j^2\).

The independent, unit-variance score assumption makes the comparison transparent. In real data, correlated genotypes and unequal variant weights affect the score covariance and the null distribution; SKAT is generally calibrated using a weighted mixture of chi-square distributions. The simplified model is therefore an explanatory calculation, not a universal formula to substitute for a study-specific analysis.

Why the Two Noncentralities Differ

Under the model, the burden test first adds the score shifts. A positive shift at one variant can therefore be offset by a negative shift at another. By contrast, SKAT adds squared shifts, so the direction of a shift does not cancel another variant’s contribution. This difference appears directly in the noncentrality parameters.

\lambda_B=\frac{(\delta_1+\cdots+\delta_m)^2}{m}, \qquad \lambda_S=\delta_1^2+\cdots+\delta_m^2.

These parameters do not by themselves determine which test has greater power: the burden statistic has 1 degree of freedom in this model, while SKAT has \(m\). At a fixed significance level \(\alpha\), the relevant power is the probability that the alternative statistic exceeds its own null critical value. Thus, both the noncentrality and the null distribution matter.

Worked Configurations

Worked Example: Aligned Effects and a Burden Advantage

Suppose two hypothetical variants have standardized score shifts \(\delta_1=0.4565\) and \(\delta_2=1\). Both effects point in the same direction. Compare equal-weight burden testing and SKAT at \(\alpha=0.05\), using the simplified model.

Burden noncentrality. There are \(m=2\) variants, so

\lambda_B=\frac{(0.4565+1)^2}{2} =\frac{1.4565^2}{2} =\frac{2.12139225}{2} =1.060696125 \approx 1.061.

The burden statistic has a noncentral chi-square distribution with 1 degree of freedom. Equivalently, its signed standardized score has a normal distribution with mean \((0.4565+1)/\sqrt{2}\approx1.030\) and variance 1. The two-sided critical value for the signed score is approximately 1.96, giving power \(P(|Z+1.030|>1.96)\approx0.178\), where \(Z\sim N(0,1)\).

SKAT noncentrality. Here,

\lambda_S=0.4565^2+1^2 =0.20839225+1 =1.20839225 \approx1.208.

SKAT has 2 degrees of freedom in this model. Its 0.05 critical value is approximately 5.991, so its power is \(P\{\chi^2_2(1.2084)>5.991\}\approx0.1515\). For example, this tail probability can be evaluated from the noncentral chi-square Poisson mixture with mean \(\lambda_S/2\). In this configuration, the calculated burden power is about 0.178 and the SKAT power is about 0.1515. Both effects reinforce one another in the burden sum, and the burden test is more powerful in this particular comparison.

Worked Example: Opposing Effects and Burden Cancellation

Now suppose four hypothetical variants have standardized score shifts \((1,1,-1,-1)\). The two positive and two negative shifts have equal magnitudes. Use a 0.05 significance level and the same simplified model.

For the burden test, the signed shifts sum to zero:

\lambda_B=\frac{(1+1-1-1)^2}{4}=0.

Thus its statistic has the same central null distribution under this specified alternative; its rejection probability remains 0.05 in the model. This is the cancellation that can make a burden test insensitive to a genuinely nonzero but directionally mixed configuration.

For SKAT, all four squared shifts contribute:

\lambda_S=1^2+1^2+(-1)^2+(-1)^2=4.

SKAT has 4 degrees of freedom, so its 0.05 critical value is approximately 9.488. Its power is \(P\{\chi^2_4(4)>9.488\}\), approximately 0.320. The probability can be computed using the noncentral chi-square mixture: it is a weighted sum of central chi-square upper-tail probabilities with degrees of freedom \(4+2k\), where the weights are Poisson probabilities with mean \(4/2=2\). In this example SKAT has power above 0.05, while the burden test has power 0.05, because the opposing shifts cancel only in the signed sum.

Worked Example: Variant Count With Aligned Shifts

Consider hypothetical configurations in which every included variant has the same positive standardized shift, \(\delta_j=0.5\). First include two such variants, then four. This isolates how adding consistently aligned effects changes the noncentralities.

With two variants, the burden and SKAT noncentralities are

\lambda_B=\frac{(0.5+0.5)^2}{2}=0.5, \qquad \lambda_S=0.5^2+0.5^2=0.5.

With four variants, they become

\lambda_B=\frac{(0.5+0.5+0.5+0.5)^2}{4}=1, \qquad \lambda_S=4(0.5^2)=1.

In this equal-shift configuration, adding aligned variants increases the noncentrality for both tests. But it also changes SKAT’s null degrees of freedom from 2 to 4. Consequently, one cannot infer the exact change in SKAT power from the noncentrality alone: its alternative tail probability must be compared with the critical value for the updated degrees of freedom. In actual applications, newly included variants may also have different weights, frequencies, and effects rather than identical shifts.

Worked Example: Allele Frequency and Effect Magnitude

Consider a hypothetical quantitative-trait study with \(N=5{,}000\) independent participants, residual variance 1, and two uncorrelated variants in Hardy–Weinberg equilibrium. In a simple additive model with an intercept and no other covariates, the genotype dosage variance for allele frequency \(p\) is \(2p(1-p)\). The standardized score shift is approximately \(\delta=\sqrt{N\,2p(1-p)}\,\beta\), where \(\beta\) is the per-allele effect in trait units.

Let variant 1 have frequency \(p_1=0.001\) and effect \(\beta_1=1\) trait unit per allele. Its shift is

\delta_1=\sqrt{5{,}000(2)(0.001)(0.999)}(1) =\sqrt{9.99} \approx3.161.

Let variant 2 have frequency \(p_2=0.01\) and effect \(\beta_2=-0.2\) trait units per allele. Its shift is

\delta_2=\sqrt{5{,}000(2)(0.01)(0.99)}(-0.2) =\sqrt{99}(-0.2) \approx-1.990.

Under the equal-weight score comparison, the resulting noncentralities are

\lambda_S=3.161^2+(-1.990)^2\approx9.990+3.960=13.95,

and

\lambda_B=\frac{(3.161-1.990)^2}{2} \approx\frac{1.171^2}{2} \approx0.685.

The rarer variant has the larger per-allele effect, but its score shift reflects both that effect and the genotype variance. The signs also matter: the shifts partly cancel in the burden sum, while their squares both contribute to SKAT. These calculations assume the stated simple model; additional covariates, genotype correlation, variant weights, and the discreteness of rare-variant counts can change the relevant power calculation.

What the Configurations Teach

In the simplified equal-weight model, equal shifts in the same direction make the signed sum coherent. When effect magnitudes are balanced but signs differ, SKAT’s squared contributions avoid the cancellation that can reduce burden power. Unequal magnitudes produce intermediate cases: a large effect in one direction may dominate several smaller effects in the other direction, so the burden sum need not vanish. A burden test can therefore perform well under some mixtures, but its behavior depends on the signed, weighted sum of the effects.

Variant count has no universal effect independent of the accompanying configuration. Adding aligned effects can strengthen the aggregate signal, as in the worked example. Adding variants with nearly zero effects can instead add noise or degrees of freedom, particularly for an unweighted SKAT statistic. Likewise, a rare allele’s frequency affects the amount of genotype information, but variant weighting may deliberately alter how much each variant contributes. Power comparisons must use the same variant set, weights, sample size, and significance level to be interpretable.

For a specified configuration, power is evaluated by taking the relevant null critical value and computing the alternative probability of exceeding it. In the model used here, these are noncentral chi-square tail probabilities. In a practical SKAT analysis with unequal kernel eigenvalues, the null statistic is generally a weighted mixture rather than an ordinary chi-square with \(m\) degrees of freedom; its calibration and power calculation must respect that distribution.

Common Mistake: Comparing Effect Sums Alone

A tempting but incorrect argument is: “The total effect is zero, so there is no association signal.” A zero signed sum only says that the effects cancel in the particular burden combination. In the four-variant example, the sum is zero but the sum of squared shifts is 4, so SKAT has a noncentral alternative distribution. Conversely, a large sum of squared effects does not guarantee that SKAT is more powerful than burden testing: the tests have different null distributions, and the result also depends on weights, variant frequencies, and the study design.

The correct approach is to state the effect configuration and the test model, calculate the test-specific alternative distribution, and compare its rejection probability with the test’s critical value at the chosen \(\alpha\). Do not treat a noncentrality parameter as a power value or assume one test is uniformly more powerful.

Key takeaway: Burden tests are sensitive to the signed, weighted combination of variant effects; SKAT accumulates squared score contributions and is less vulnerable to cancellation. Variant count, effect size, direction, and allele frequency jointly determine power, so conclusions must be tied to an explicit configuration and test calibration.

Check Your Understanding

Use the simplified independent, unit-variance score model unless a question states otherwise.

For three variants with shifts \((0.8,0.8,0.8)\), calculate \(\lambda_B\) and \(\lambda_S\).
For shifts \((1,-1)\), calculate both noncentralities and explain why they differ.
Why does an increase in SKAT noncentrality not by itself establish that SKAT power increased when additional variants are included?
In the quantitative-trait model, what two quantities determine the approximate standardized shift \(\delta\) besides sample size?
Explain why a zero burden shift does not establish that every variant has zero effect.