Why Combine Grouping With a Joint Test?
A rare-variant set may contain variants with different frequencies and effects. A single burden score combines variant information into one number per individual; that can be efficient when effects point in a similar direction, but effects that oppose one another may cancel. Testing every variant separately avoids that particular pooling step, but can be inefficient when individual variants are very rare. The combined multivariate and collapsing approach takes an intermediate route: it first collapses variants within prespecified groups, then tests the resulting group-level genotype indicators jointly.
The word multivariate here refers to having several genotype predictors in one model. The outcome in the basic example below is one continuous phenotype, not a vector of phenotypes. The joint test asks whether the group indicators, considered together, are associated with that outcome.
Groups are formed at the variant level; a person may carry variants in more than one group and therefore have multiple indicators equal to 1. The carrier indicator is a within-group collapse: it records whether there is any carrier genotype in the group, not which variant was carried or how many alleles were carried. The grouping rule, frequency cutoffs, and variant-set definition should be specified before examining the association result.
Model and Joint Test
For a single continuous phenotype, consider the linear model
Here \(Y_i\) is the phenotype for individual \(i\), \(\beta_0\) is the intercept, and \(\beta_j\) is the conditional mean difference associated with the \(j\)th group indicator, holding the other indicators fixed. The test is
For the usual exact partial F test, assume independent individuals, a correctly specified linear conditional mean, and errors that are normally distributed with mean zero and common variance \(\sigma^2\), independently of the indicators. Also require the design matrix to have full column rank: no group indicator can be an exact linear combination of the intercept and the other indicators. With \(K\) tested indicators and an intercept, the residual degrees of freedom are \(n-K-1\). If these assumptions hold, the partial F statistic has an \(F\) distribution with \(K\) and \(n-K-1\) degrees of freedom under \(H_0\).
Let \(SSE_0\) be the residual sum of squares from the intercept-only model and \(SSE_1\) the residual sum of squares from the model containing all \(K\) indicators. Then
The numerator measures the reduction in residual variation per added predictor; the denominator estimates the residual variance in the larger model. A large value is evidence against the joint null. This is an omnibus test: a small p-value indicates evidence that at least one group coefficient is nonzero, not that every group is associated or that any particular variant is causal.
Worked Examples
Worked Example: Forming Group-Level Carrier Indicators
Suppose a hypothetical variant set has four variants, assigned in advance to two frequency groups: group A contains variants 1 and 2, and group B contains variants 3 and 4. Genotypes are coded as alternate-allele dosages 0, 1, or 2. For each group, the carrier indicator is 1 if at least one dosage in that group is positive.
| Individual | Variant 1 | Variant 2 | Variant 3 | Variant 4 | \(C_{iA}\) | \(C_{iB}\) |
|---|---|---|---|---|---|---|
| 1 | 0 | 0 | 0 | 0 | 0 | 0 |
| 2 | 1 | 0 | 0 | 0 | 1 | 0 |
| 3 | 0 | 2 | 0 | 1 | 1 | 1 |
| 4 | 0 | 0 | 1 | 0 | 0 | 1 |
For individual 3, group A has dosages 0 and 2, so \(C_{3A}=1\); group B has dosages 0 and 1, so \(C_{3B}=1\). The indicators preserve whether each person carries something in each group, but discard the identity and dosage of the variants within that group. An individual carrying two variants in the same group still has an indicator of 1 for that group.
Worked Example: Calculating the Joint F Test
Consider 12 hypothetical individuals and two group indicators, A and B. There are three individuals in each of the four possible indicator patterns: (0,0), (1,0), (0,1), and (1,1). Their phenotype means in those patterns are respectively 0, 0.8165, 0.8165, and 1.6330 units. Suppose the within-pattern residual sum of squares is 9.000 units squared. These summaries are consistent with an additive pattern of group mean differences and within-pattern variation.
With three individuals per pattern, the grand mean is the average of the four pattern means. Each indicator is balanced, and the pattern means correspond to a difference of approximately 0.8165 units for each indicator. The intercept-only model has \(SSE_0=13.00\): 9.000 units squared of within-pattern residual variation plus 4.000 units squared explained by the differences between pattern means and the grand mean. The model including both indicators fits those pattern means, so \(SSE_1=9.000\) units squared.
There are \(K=2\) tested indicators and \(n-K-1=12-2-1=9\) residual degrees of freedom. Substitution gives
For an \(F\) distribution with degrees of freedom \((2,9)\), the upper-tail probability at \(F=2\) is
Thus the joint test gives \(F=2.000\) and \(p\approx0.1911\). At a significance level of 0.05, these data do not provide sufficient evidence to reject the joint null. This does not establish that all group effects are zero; it states that the observed test statistic is not unusual enough under the joint null to reject at that threshold.
Worked Example: Why Testing Groups Jointly Can Matter
Suppose two groups have estimated conditional phenotype differences of \(+1.2\) and \(-1.2\) units for their carrier indicators. A simple combined carrier score that adds the two indicators has an average effect that can be close to zero when the groups contribute equally: the positive and negative components offset. The combined multivariate model keeps separate coefficients, so its joint null remains \(H_0:\beta_A=\beta_B=0\), rather than testing only a single pooled coefficient.
For illustration, suppose the estimated coefficient vector is \((1.2,-1.2)\) units and its estimated covariance matrix is diagonal with variances \(0.36\) units squared. The squared standardized contributions are
The joint Wald quadratic is therefore \(4.000+4.000=8.000\). Under a large-sample chi-square calibration with two degrees of freedom, its upper-tail probability is \(e^{-8/2}\approx0.01832\). This illustrative calculation shows how separate group coefficients can retain joint evidence when a one-dimensional sum cancels. It is not the same finite-sample F test as in the previous example; its chi-square calibration is an approximation whose adequacy depends on the model and sample size.
The joint approach does not eliminate all cancellation. Variants collapsed into the same carrier indicator are no longer distinguished. If variants within one group have opposing effects, the indicator may conceal that pattern before the joint test is performed.
Justification of the Partial F Statistic
The partial F test compares the fit of two nested models: the null model, which has only an intercept, and the full model, which adds \(K\) group indicators. Under the stated normal linear-model assumptions, the residual sum of squares divided by its degrees of freedom estimates the common error variance. Under \(H_0\), adding the \(K\) indicators should reduce residual sum of squares only through random variation. The reduction per added parameter and the full-model residual mean square therefore estimate the same \(\sigma^2\).
More formally, under \(H_0\), the scaled reduction \((SSE_0-SSE_1)/\sigma^2\) has a chi-square distribution with \(K\) degrees of freedom, and \(SSE_1/\sigma^2\) has a chi-square distribution with \(n-K-1\) degrees of freedom. These quantities are independent for nested linear models with normally distributed errors. Dividing each by its degrees of freedom gives independent estimates of the same variance; their ratio has an \(F_{K,n-K-1}\) distribution. This yields the displayed statistic and its reference distribution.
Design Choices, Diagnostics, and Limits
The result depends on choices made before fitting the model. Frequency groups should be defined using a stated frequency source and cutoff rule. A variant assigned to one group should not also be counted in another unless the modeling plan explicitly defines overlapping groups. Very fine grouping can leave indicators with few carriers; very broad grouping can combine variants whose effects differ. These choices affect both the predictors and the scientific interpretation of the test.
Check the design matrix before interpreting a p-value. If a group has no carriers, its indicator is all zero and its coefficient cannot be estimated. If two indicators are identical, their separate coefficients are not identifiable. High correlation between indicators does not necessarily make the model mathematically unidentified, but it can make coefficient estimates unstable and reduce power. Review carrier counts by group, the number of distinct indicator patterns, the rank of the design matrix, residual plots, and whether residual spread changes across fitted values. Outliers or strong departures from normality can make the exact small-sample F calibration unreliable; dependence between individuals also violates the stated model.
The method's main comparison is with one-dimensional burden tests and separate single-variant tests. A burden test can be more efficient when included effects are sufficiently aligned, but it compresses information across its included variants into one score. The combined approach preserves distinctions between groups and tests them jointly, while still reducing the number of predictors relative to testing every variant separately. Separate variant tests retain variant-level detail but can have little information for very rare variants and require attention to multiple testing. No approach is uniformly best: the suitable choice depends on the expected effect pattern and the available carrier counts.
The coefficients in the linear model describe conditional mean differences for the collapsed indicators, in phenotype units. They are not automatically variant-specific effects: each indicator represents a set of variants, and a carrier may carry more than one variant within a group. A significant omnibus result does not identify which group or variant is responsible; group-specific follow-up requires its own estimates and appropriate inferential care. The Gaussian linear-model test described here also should not be transferred unchanged to binary outcomes. The outcome model and calibration must match the phenotype and study design.
Check Your Understanding
Answer each question, then compare with the brief solutions.
- In a group containing three variants, an individual has dosages 0, 1, and 0. What is the group carrier indicator, and what information does it discard?
- A test includes \(K=3\) group indicators and \(n=20\) individuals. What are the numerator and residual degrees of freedom for the partial F test with an intercept?
- Why can a joint test of two group indicators detect evidence that a single pooled burden score misses?
- Name two conditions that can make a group-indicator regression rank deficient or its estimates unstable.
Solutions. (1) The indicator is 1 because at least one dosage is positive; it discards which variant was carried and the dosage details within the group. (2) The numerator degrees of freedom are \(K=3\), and the residual degrees of freedom are \(n-K-1=20-3-1=16\). (3) The joint model retains separate coefficients, so effects pointing in opposite directions across groups need not cancel into one pooled score. (4) An all-zero indicator or duplicate indicators cause rank deficiency; very low carrier counts or highly correlated indicators can make estimates unstable.