SKAT-O as an Adaptive Burden-Variance Test

Why Adapt Between Burden and Variance-Component Evidence?

A rare-variant set can be associated with an outcome in different ways. If most variants have effects in the same direction, their score contributions can reinforce one another, making a burden test effective. If effects vary in direction or size, a burden score can cancel; a variance-component test such as SKAT can retain evidence because it accumulates squared contributions. The introductory material on choosing between burden and variance-component tests motivates this trade-off. SKAT-O addresses uncertainty about which pattern is present by adapting across a family that connects the two approaches.

Adaptation does not mean that SKAT-O first discovers the true genetic architecture and then tests it as if it had been specified in advance. It examines evidence across candidate tests, and its final p-value must account for that search. The shared null hypothesis and the calibration of the adaptive procedure are therefore as important as the individual statistics.

Model, Score Vector, and the SKAT-O Family

Consider a set of \(d\) variants and a phenotype measured on \(n\) individuals. Let \(G\) be the \(n \times d\) genotype matrix, and let the null model include the covariates and phenotype distribution appropriate to the analysis. Under that model, let \(U=(U_1,\ldots,U_d)'\) be the vector of variant score contributions, and let \(V\) be its covariance matrix under the null. Covariate adjustment is incorporated into the score and its covariance; linkage disequilibrium is reflected in the off-diagonal entries of \(V\).

Let \(W=\operatorname{diag}(w_1,\ldots,w_d)\) be the diagonal matrix of prespecified nonnegative variant weights. Define the weighted score vector \(Z=WU\). For a candidate mixing value \(0 \leq \rho \leq 1\), define \(R_\rho=(1-\rho)I+\rho\mathbf{1}\mathbf{1}'\), where \(I\) is the \(d \times d\) identity matrix and \(\mathbf{1}\) is the \(d\)-vector of ones. The SKAT-O quadratic statistic at that value is

Q_\rho=Z'R_\rho Z =U'W\left[(1-\rho)I+\rho\mathbf{1}\mathbf{1}'\right]WU.

Definition: SKAT-O is an adaptive test built from the family \(Q_\rho\). At \(\rho=0\), \(Q_0=U'W^2U\), the SKAT quadratic statistic. At \(\rho=1\), \(Q_1=(\mathbf{1}'WU)^2\), the squared weighted burden score. Values between the endpoints blend the two forms of evidence. The final SKAT-O p-value accounts for examining a prespecified collection of these candidate tests.

The burden endpoint is often written as a standardized one-degree-of-freedom score statistic, \(T_B=(\mathbf{1}'WU)^2/(\mathbf{1}'WVW\mathbf{1})\). This standardization changes the scale, not the burden direction being tested. The SKAT endpoint is instead calibrated using its quadratic-form null distribution. Thus raw values of \(Q_0\) and \(Q_1\) should not be compared as if they had the same reference distribution.

Hypotheses and Assumptions

The global null for the set is that every variant has zero effect in the specified association model. In a regression representation with variant-effect vector \(\beta\), this is \(H_0:\beta_1=\cdots=\beta_d=0\). Each fixed-\(\rho\) score test assesses this same null, but is sensitive to a different structure of departures. The burden direction emphasizes a nonzero weighted aggregate effect; the SKAT component emphasizes departures with variant effects that can differ across the set. The combined test is not a test of the claim that at least one particular \(\rho\) is “the true” architecture.

The usual score-test calibration relies on a correctly specified null model and a valid estimate of the score covariance \(V\). In large samples, the null score vector is approximated by a mean-zero multivariate normal distribution. Genotype coding, variant weights, covariate adjustment, and any adjustment for relatedness or population structure must be appropriate to the study. Weights and the variant set should be specified without using the outcome evidence being tested, unless the selection is explicitly accounted for. Sparse binary outcomes or highly unbalanced data may require more careful score calibration than a simple normal approximation.

Null Distribution: Why Each Candidate Needs Its Own Calibration

Under the null approximation, \(Z=WU\) has mean zero and covariance \(V_W=WVW\). A quadratic form in a mean-zero normal vector has a weighted sum of independent chi-square variables as its null distribution. For this test, the weights are eigenvalues of a matrix that incorporates both the score covariance and the selected \(R_\rho\).

Q_\rho \ \overset{d}{\approx}\ \sum_{k=1}^{d}\lambda_k(\rho)\,\chi^2_{1,k}, \qquad \lambda_k(\rho)=\text{eigenvalues of }V_W^{1/2}R_\rho V_W^{1/2},

where the \(\chi^2_{1,k}\) variables are independent chi-square random variables with one degree of freedom. If \(V_W\) is singular, the same result applies on its nonzero-variance subspace; zero eigenvalues contribute nothing. The eigenvalues show why the candidate statistics cannot be calibrated by treating every \(Q_\rho\) as a standard chi-square statistic. They also show how linkage disequilibrium and the score covariance affect the reference distribution.

Justification: Write \(Z=V_W^{1/2}\varepsilon\), where \(\varepsilon\) has independent \(N(0,1)\) entries under the normal approximation. Then \(Q_\rho=\varepsilon'V_W^{1/2}R_\rho V_W^{1/2}\varepsilon\). Diagonalize the symmetric matrix as \(A_\rho=P\Lambda P'\), with \(P'P=I\) and diagonal \(\Lambda\) containing the eigenvalues. Because \(P'\varepsilon\) is also a vector of independent standard normal variables, \(Q_\rho\) has the distribution \(\sum_k\lambda_k(\rho)(P'\varepsilon)_k^2\), which is the stated weighted chi-square mixture.

For a fixed \(\rho\), this mixture yields a candidate p-value. SKAT-O then uses the candidate evidence across a prespecified range of \(\rho\) values. If the smallest candidate p-value is selected, its null distribution is not the null distribution of a single fixed-\(\rho\) test: the candidates are correlated because they use the same score vector. The adaptive p-value must account for their joint dependence, using an appropriate SKAT-O calibration. Simply reporting the smallest candidate p-value as though that candidate had been chosen in advance is anti-conservative.

Worked Example: Concordant Scores Favor the Burden Direction

Suppose a hypothetical two-variant analysis has equal weights, \(V=I\), and observed score vector \(U=(2,2)'\). The assumptions for this illustration are that the null score approximation is bivariate normal with identity covariance and that both weights equal one.

At the SKAT endpoint, \(Q_0=2^2+2^2=8\). Under these assumptions \(Q_0\sim\chi^2_2\), so its candidate p-value is \(P(\chi^2_2\geq 8)=e^{-8/2}=e^{-4}=0.0183\).

At the burden endpoint, \(Q_1=(2+2)^2=16\). Its null distribution is \(2\chi^2_1\), since \((U_1+U_2)/\sqrt{2}\) is standard normal. Thus the burden candidate p-value is \(P(\chi^2_1\geq 16/2)=P(\chi^2_1\geq 8)\), approximately \(0.00468\). The concordant scores reinforce in the aggregate, so the burden candidate is more compelling in this example. These are candidate p-values, not the final adaptive SKAT-O p-value; selection across candidates still needs calibration.

Worked Example: Opposing Scores Can Cancel in a Burden

Now suppose a hypothetical analysis has equal weights, \(V=I\), and \(U=(2,-2)'\), under the same normal-score assumptions. The two contributions have equal magnitude but opposite signs.

The SKAT statistic is \(Q_0=2^2+(-2)^2=8\), again with a \(\chi^2_2\) null distribution. Its candidate p-value is \(e^{-4}=0.0183\). The burden statistic is \(Q_1=(2-2)^2=0\); with a continuous, correctly calibrated one-degree-of-freedom test, the corresponding upper-tail p-value is \(1\).

The contrast illustrates the intended complementarity: a squared-score test retains contributions from both variants, while a burden sum cancels them. It does not establish that the variants have opposing causal effects. Score signs are evidence relative to the fitted model, and a significant set-level result alone does not identify individual effects.

Worked Example: An Interior Candidate Blends the Evidence

For a third hypothetical two-variant analysis, let \(U=(1.5,1)'\), \(V=I\), and both weights equal one. The squared-score component is \(Q_0=1.5^2+1^2=3.25\); the burden endpoint is \(Q_1=(1.5+1)^2=6.25\). At \(\rho=0.5\),

Q_{0.5}=0.5(3.25)+0.5(2.5)^2 =1.625+3.125 =4.75.

For this candidate, the eigenvalues of \(R_{0.5}\) are \(1.5\) and \(0.5\). Therefore its null distribution is \(1.5\chi^2_{1,1}+0.5\chi^2_{1,2}\), not a chi-square distribution with a guessed number of degrees of freedom. The statistic \(4.75\) must be compared with that mixture to obtain its candidate p-value. This calculation shows how an interior candidate retains both the sum-of-squares component and the common-direction component; it does not license comparing \(4.75\) directly with \(3.25\) or \(6.25\) to decide which candidate is most significant.

Interpreting Adaptation, Diagnostics, and Limitations

A small calibrated SKAT-O p-value is evidence against the global no-association null for the analyzed set under the specified model. It does not prove that every variant is associated, identify causal variants, establish a direction of effect for the set, or reveal a unique genetic architecture. The candidate with the smallest p-value may be useful as a description of which pattern supplied stronger evidence, but the adaptation makes that selection data-dependent. It should not be treated as an independently confirmed biological conclusion.

Before interpreting a result, check whether the score covariance incorporates the study design and relevant dependence among variants. A mismatch between observed structure and the assumed \(V\), for example because population structure or relatedness is inadequately handled, can distort candidate and adaptive p-values. Check genotype coding, allele orientation, weight construction, variant inclusion, missingness, and the null-model fit. For binary phenotypes with sparse case counts, assess whether the chosen score calibration is reliable for the data; asymptotic approximations can be inaccurate in extreme settings.

The main alternatives answer narrower questions. A burden test focuses on one aggregate direction and can be powerful when effects align, but can lose power through cancellation. SKAT focuses on a variance-component alternative that allows heterogeneous effects and is less vulnerable to directional cancellation, but may be less efficient when effects are strongly aligned. SKAT-O trades some of the simplicity of either single test for adaptation, and pays for that search through its calibrated p-value. It does not remove limitations shared by the underlying score tests, such as poor null-model specification, low information for very rare variants, or sensitivity to the chosen variant set and weights.

Common mistake: “The candidate with the smallest p-value is the one SKAT-O selected, so I can report that p-value directly.” This ignores the search across correlated candidates and generally makes the evidence look stronger than it is. Report the properly calibrated adaptive p-value; describe a favorable candidate, if useful, as exploratory evidence about the pattern of scores rather than as a separately validated test.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

For \(d\) variants, what does the SKAT-O statistic reduce to at \(\rho=0\) and at \(\rho=1\)?
With two equal-weight variants, identity score covariance, and \(U=(3,-3)'\), calculate \(Q_0\) and \(Q_1\). Which candidate retains evidence that the burden cancels?
Why is the minimum candidate p-value not generally a valid final SKAT-O p-value?
Does a significant adaptive test prove that every variant has an effect? State what the test addresses instead.

Solutions. (1) At \(\rho=0\), \(Q_0=U'W^2U\), the SKAT quadratic statistic. At \(\rho=1\), \(Q_1=(\mathbf{1}'WU)^2\), the squared burden score. (2) \(Q_0=3^2+(-3)^2=18\), while \(Q_1=(3-3)^2=0\). The SKAT candidate retains the squared contributions; the burden cancels. (3) The candidates use the same score vector and are correlated, and choosing the minimum is a search. The final p-value must account for that joint selection. (4) No. The test addresses the global null that all variant effects in the specified set are zero under the model; it neither establishes an effect for every variant nor identifies causal variants.