Variant-Level Score Contributions and Influence Diagnostics

Why Examine Variant-Level Contributions?

A gene-based test can provide evidence that a set of variants is associated with an outcome without identifying which variants account for the observed statistic. Variant-level score contributions offer a descriptive decomposition: they show how much each variant contributes to the observed quadratic statistic. Leave-one-variant-out analysis asks a related but distinct question: how much does the test change when that variant is removed?

These are diagnostics, not automatic procedures for selecting causal variants. The score contributions depend on the test, the chosen weights, and the fitted null model. In particular, a large contribution is not necessarily independent of contributions from correlated variants, and it does not establish a causal effect.

Score Statistic and Its Additive Decomposition

Consider \(n\) individuals and a set of \(m\) variants. Let \(Y\) be the outcome vector, \(X\) the matrix of covariates in the null model, and \(G\) the \(n \times m\) genotype matrix. For a binary outcome, suppose the null model is logistic regression with a correctly specified mean model and no variant effects. Write \(\hat{\mu}_i\) for the fitted outcome probability for individual \(i\), and let \(U_j\) be the score for variant \(j\), evaluated under this null model:

U_j = \sum_{i=1}^{n} G_{ij}(Y_i-\hat{\mu}_i), \qquad U=(U_1,\ldots,U_m)'.

For inference with covariates estimated under the null, the score covariance accounts for that estimation. In this logistic model, let \(A\) be diagonal with entries \(\hat{\mu}_i(1-\hat{\mu}_i)\). The efficient score covariance is estimated by

V = G'\left[A-AX(X'AX)^{-1}X'A\right]G.

This expression assumes that the null-model information matrix \(X'AX\) is invertible. It describes the score dependence as well as the marginal score variances. Other outcome models use their corresponding null-model score and covariance.

For the weighted quadratic score test used in SKAT, let \(W\) be a diagonal matrix of fixed, nonnegative variant weights. The statistic is \(Q=U'WU\). The weights may reflect a prespecified weighting scheme; they are part of the test definition and should not be confused with observed evidence.

Definition: The contribution of variant \(j\) to the observed weighted quadratic statistic is \(q_j=w_jU_j^2\), where \(w_j\) is the \(j\)th diagonal entry of \(W\). Thus \(Q=\sum_{j=1}^{m}q_j\). If \(Q>0\), the descriptive contribution share is \(q_j/Q\).

The decomposition follows directly from diagonal \(W\):

Q=U'WU =\sum_{j=1}^{m} w_j U_j^2 =\sum_{j=1}^{m}q_j.

Each \(q_j\) is nonnegative. This additivity applies to the observed quadratic statistic; it does not mean the variant scores are independent. Under the null, the approximate distribution of \(Q\) is a weighted sum of independent chi-square random variables, with weights given by the eigenvalues of \(W^{1/2}VW^{1/2}\). Consequently, the null distribution depends on the joint score covariance, not just on the list of observed \(q_j\).

Worked Example: Computing Contributions and Shares

Suppose a hypothetical four-variant set has score vector \(U=(2,-1,3,0.5)'\) and diagonal weights \(W=\mathrm{diag}(1,2,0.5,4)\). These are assumed to come from a fitted null model with fixed weights. Calculate each contribution, the total statistic, and the contribution shares.

For the four variants, the contributions are \(q_j=w_jU_j^2\):

q_1=1(2^2)=4,\quad q_2=2((-1)^2)=2,\quad q_3=0.5(3^2)=4.5,\quad q_4=4(0.5^2)=1.

Therefore \(Q=4+2+4.5+1=11.5\). The shares are \(4/11.5=0.3478\), \(2/11.5=0.1739\), \(4.5/11.5=0.3913\), and \(1/11.5=0.0870\). Variant 3 has the largest observed contribution, about 39.13% of \(Q\). This ranks contributions under the specified weights and score calculation; it does not show that variant 3 has the largest causal effect.

Leave-One-Variant-Out Influence

A leave-one-variant-out diagnostic removes variant \(j\), recomputes the set statistic, and compares the result with the full-set analysis. If the null fit, scores for the remaining variants, and their weights are held fixed, then the reduced statistic is \(Q_{-j}=Q-q_j\). Its absolute decrease is \(q_j\), and, for \(Q>0\), its relative decrease is \(q_j/Q\). In this fixed-score calculation, contribution share and relative decrease coincide.

Inference requires more than subtracting a contribution. After removing a variant, the score covariance is reduced to the corresponding submatrix and the null reference distribution must be recalculated. In particular, the mixture weights can change, so the p-value after deletion cannot generally be obtained by subtracting \(q_j\) from the statistic while retaining the original p-value. If weights or other analysis choices are recomputed after deletion, that change should be stated because the comparison then reflects more than removal of one score coordinate.

Worked Example: A Leave-One-Out Change in Statistic and P-Value

Suppose, hypothetically, that three score coordinates are independent and standardized under the null, so \(V=I\), and all weights equal 1. The observed scores are \(U=(2,1,0)'\). Under these assumptions, the full statistic has a chi-square distribution with 3 degrees of freedom under the null.

The full statistic is \(Q=2^2+1^2+0^2=5\). Its upper-tail p-value is \(P(\chi^2_3\geq 5)\approx 0.1718\). Removing variant 1 leaves \(Q_{-1}=1^2+0^2=1\). The two remaining independent standardized scores give a chi-square null distribution with 2 degrees of freedom, so the reduced p-value is \(P(\chi^2_2\geq1)=e^{-1/2}\approx0.6065\).

The statistic fell by \(4\), or \(4/5=80\%\), and the p-value increased under the reduced test. The p-values are computed from different null distributions, with 3 and 2 degrees of freedom. This example relies on independent standardized scores; in genomic data, score correlation often makes that simplification inappropriate.

Why Linkage Disequilibrium Complicates Interpretation

Variants in linkage disequilibrium can have correlated score statistics. The quadratic decomposition still allocates the observed value of \(Q\) exactly, but it does not allocate independent evidence. A variant with a large contribution may partly reflect a signal also represented by a correlated variant. Conversely, deleting one variant can alter the null distribution for the remaining set even when their observed scores are unchanged.

Worked Example: Equal Contributions, Correlated Scores

Consider two hypothetical variants with equal weights, observed scores \(U=(2,-2)'\), and null score covariance

V= \begin{pmatrix} 1 & 0.9\\ 0.9 & 1 \end{pmatrix}.

The observed statistic is \(Q=2^2+(-2)^2=8\), with contributions \(q_1=q_2=4\). Each variant therefore accounts for half of the observed statistic. But the scores are strongly correlated under the null: their covariance is 0.9. The eigenvalues of \(V\) are \(1.9\) and \(0.1\), so the null distribution of \(Q\) is \(1.9\chi^2_1+0.1\chi^2_1\), with independent chi-square variables in that mixture representation. The two contribution shares do not describe two independent pieces of evidence. This calculation uses unit weights; with other weights the relevant eigenvalues come from \(W^{1/2}VW^{1/2}\).

This distinction is central to interpretation: contribution shares describe the observed statistic, whereas the covariance and its eigenvalues determine its null calibration. A contribution table is not a substitute for a valid p-value calculation.

Quadratic Tests, Burden Tests, and Diagnostic Checks

For a burden test, a weighted aggregate score might be \(B=a'U\), where \(a\) is a prespecified vector of variant weights. The squared, standardized statistic is \(B^2/(a'Va)\), when \(a'Va>0\). Expanding the numerator gives

(a'U)^2=\sum_{j=1}^{m}a_j^2U_j^2+ 2\sum_{j<k}a_ja_kU_jU_k.

The cross-terms mean that the burden statistic does not have the same simple, nonnegative variant-by-variant decomposition as the SKAT quadratic statistic. A variant’s apparent role can depend on the signs and magnitudes of other scores. Thus, do not apply SKAT contribution shares as though they were a unique decomposition of a burden result. Omnibus tests likewise have their own construction and do not automatically inherit this additive interpretation.

A practical diagnostic starts by confirming that the genotype coding, allele orientation, null-model fit, and weights match the prespecified analysis. Then examine \(q_j\), \(q_j/Q\), and the leave-one-out changes. Check whether the largest contributions correspond to plausible genotype counts and reliable calls, and whether the conclusion is driven by one variant or remains similar across deletions. The null model should account for the covariates and outcome structure used in the analysis; the score-based approximation also requires a suitable null calibration. Rare allele counts, model misspecification, or other departures from the analysis assumptions can undermine interpretation even when the arithmetic is correct.

A common mistake is to say, “This variant contributes 40% of the statistic, so it explains 40% of the association,” or to treat its deletion p-value as a formal test that the variant is causal. The correction is that 40% is a share of one observed test statistic under one weighting and null-model specification. It is neither a fraction of biological effect nor a causal estimate. A deletion analysis is a sensitivity check; any follow-up claim about an individual variant requires an analysis designed for that question and appropriate control of the inferential process.

Key takeaway: For a weighted quadratic score test, \(Q=\sum_j w_jU_j^2\) gives an exact descriptive contribution for each variant. Dependence among scores still matters for the null distribution, and leave-one-out p-values must be recalibrated for the reduced set.

Check Your Understanding

Use the definitions and calculations in this tutorial to answer the questions. Each solution follows its question.

A two-variant quadratic test has \(U=(1,3)'\) and weights \(W=\mathrm{diag}(2,1)\). Find the contributions, total statistic, and shares.
In a fixed-score leave-one-out analysis, what happens to \(Q\) when variant \(j\) is removed? Does this alone determine the change in p-value?
Why can two equal SKAT contribution shares fail to represent equal, independent evidence?
For a burden statistic \(B=a'U\), why is there not generally an additive decomposition into nonnegative variant contributions?

Solutions. (1) \(q_1=2(1^2)=2\) and \(q_2=1(3^2)=9\), so \(Q=11\). The shares are \(2/11=0.1818\) and \(9/11=0.8182\). (2) With scores and weights for the remaining variants fixed, \(Q_{-j}=Q-q_j\). This does not determine the p-value change because the reduced score covariance, and hence the null reference distribution, may change. (3) Correlation between score coordinates means their contributions are not independent; the joint covariance affects the null distribution. (4) Squaring the aggregate score creates cross-terms \(2a_ja_kU_jU_k\), which can be positive or negative, so the statistic does not reduce to a sum of nonnegative variant-only terms.