Why Look for Excess Variation?
A burden test combines variant contributions into a single score. If some rare alleles increase a phenotype while others decrease it, their contributions can cancel. C-alpha takes a different view: it asks whether the observed case-carrier counts vary across variants more than expected under no association. A variant with more case carriers than expected and one with fewer can both add to that variation.
This makes C-alpha useful when effects may point in opposing directions. It is not a test that identifies which variants are causal, estimates their effect directions, or guarantees power for every heterogeneous architecture. Its central quantity is an aggregate of squared, null-centered deviations.
Model and Null Calibration
The conditional null distribution for one variant is hypergeometric: among \(N\) people, \(M_j\) are carriers, and \(n\) people are labelled cases. Under the null, every allocation of the \(n\) case labels is equally likely, conditional on the observed margins. Thus \(X_j\) counts how many carriers fall among those \(n\) cases. This model conditions away the overall case proportion and the variant’s observed carrier count.
The test assumes that, under the null, case labels are exchangeable under the chosen sampling or randomization scheme, possibly conditional on design strata or covariates. Population structure, relatedness, ascertainment, or covariate imbalance can make unrestricted exchangeability implausible. In such settings the null procedure must reflect the study design and adjustment strategy; simply permuting labels across everyone may not be valid. Genotype calls and the definition of a carrier must also be sufficiently reliable.
C-alpha is not generally calibrated by treating its variant-specific terms as independent. Linkage disequilibrium and shared carriers can make those terms dependent. A practical calibration is to generate null assignments that preserve the relevant design and genotype structure, recalculate the entire statistic each time, and compare the observed value with that empirical null distribution. This retains dependence among variants.
Why the Statistic Is Centered at Zero
For any random variable \(X_j\) with mean \(E_j\) and variance \(V_j\), the definition of variance gives \(E[(X_j-E_j)^2]=V_j\). Therefore, under the conditional null for variant \(j\),
Summing over variants gives \(E(C_\alpha)=0\) under the null. This expectation does not require independence across variants: linearity of expectation is enough. Independence, or an appropriate model for dependence, is needed if one wants to calculate the variance of the sum analytically. This is one reason whole-set permutation calibration is valuable.
The alternative interpretation is also informative. If the case-carrier count at a variant has alternative mean \(\mu_j\) and variance \(V_{j,\mathrm{alt}}\), then
The squared mean shift is nonnegative whether \(\mu_j\) is above or below its null expectation. If the alternative variance is close to the null variance, a shift in either direction tends to increase the expected contribution. This explains the test’s sensitivity to effects in opposing directions; it does not mean every alternative necessarily yields a positive contribution or strong power.
Worked Example: Calculating a Variant’s Centered Contribution
Suppose a hypothetical case-control sample contains \(N=100\) people, with \(n=40\) cases. For one variant, \(M_j=10\) people are carriers, of whom \(X_j=7\) are cases. Calculate its expected case-carrier count, null variance, and C-alpha contribution.
Under the conditional null, the expected count is \(E_j=nM_j/N=(40)(10)/100=4\) case carriers. The variance is
The observed squared deviation is \((7-4)^2=9\). The centered contribution is therefore \(9-2.1818\approx 6.8182\). The observed count is higher than expected, and this variant contributes positively to the statistic. That contribution alone is not a calibrated p-value: inference concerns the combined statistic and its null distribution.
Opposing Directions and a Worked Set Calculation
The burden-versus-variance-component distinction is especially clear when two variants have shifts in opposite directions. A sum of signed deviations can cancel, whereas C-alpha sums squared deviations after subtracting each null variance. The following example isolates that contrast; it does not establish that the variants are independent.
Worked Example: Enrichment and Depletion Both Add to C-alpha
In a hypothetical sample, \(N=100\) people include \(n=40\) cases. Two variants each have \(M_j=10\) carriers. Variant 1 has \(X_1=7\) case carriers; variant 2 has \(X_2=1\). Calculate the two contributions and their sum.
For each variant, \(E_j=(40)(10)/100=4\), and the conditional variance is \(V_j=(40)(60)(10)(90)/[100^2(99)]\approx2.1818\). For variant 1, the centered contribution is \((7-4)^2-2.1818=9-2.1818\approx6.8182\). For variant 2, it is \((1-4)^2-2.1818=9-2.1818\approx6.8182\).
Thus the two-variant statistic is \(C_\alpha\approx6.8182+6.8182=13.6364\). The first variant is enriched among cases relative to its null expectation, while the second is depleted. Both have the same squared deviation and both contribute positively. By contrast, their signed deviations from expectation are \(+3\) and \(-3\), which sum to zero. The numerical statistic alone does not determine significance; it must be compared with a null distribution that accounts for the set and its dependence structure.
Calibrating and Interpreting the Test
For a valid permutation test, the permitted rearrangements must represent the null distribution under the design. With unrestricted exchangeable case labels, each rearrangement keeps the observed genotypes fixed and assigns exactly \(n\) people as cases. If covariates or strata constrain exchangeability, the randomization scheme must respect those constraints. Each rearrangement produces a complete set of counts \(X_j\) and one recalculated \(C_\alpha\).
If \(B\) random null rearrangements are generated and \(b\) yield a statistic at least as large as the observed statistic, a common one-sided Monte Carlo estimate is \(p=(b+1)/(B+1)\). The added one prevents a reported zero p-value when none of the sampled rearrangements is as extreme as the observed result. The test is upper-tailed because its target is excess variation. A p-value is the probability, under the specified null calibration, of a statistic at least this large; it is not the probability that the null hypothesis is true.
Worked Example: Empirical Upper-Tail Calibration
Suppose the hypothetical two-variant example above gives \(C_\alpha=13.6364\). A valid conditional permutation procedure generates \(B=999\) null statistics, of which \(b=28\) are at least 13.6364. The Monte Carlo p-value is
At significance level \(\alpha=0.05\), this result is unusual under the stated permutation null, so the test rejects that null at the 5% level. The conclusion is evidence of excess variation in case-carrier counts across the tested variants, not proof that every included variant is associated or that the effects have particular directions. The calculation is meaningful only if the permutation scheme is valid for the study.
Diagnostics, Comparisons, and Limitations
A useful diagnostic is to inspect the observed and expected carrier counts, as well as the individual centered contributions. This helps determine whether the result is driven by one sparse variant or by a pattern spread across the set. It does not replace formal calibration: selecting or excluding variants after examining their contributions can change the test’s error rate. Also check whether the null procedure preserves the case count, relevant strata, and dependence among genotypes.
Sparse data are common in rare-variant analysis. If a variant has very few carriers, its count distribution is discrete and may be far from a convenient continuous approximation. A normal approximation to the aggregate statistic can then be inaccurate, particularly for small samples or strongly dependent variants. Permutation or another justified null calibration is preferable when feasible. Too few permutations also limit p-value resolution: with \(B\) random permutations, the smallest value from the plus-one formula is \(1/(B+1)\).
| Approach | Primary signal | Strength and limitation |
|---|---|---|
| Burden test | A signed aggregate effect across variants | Can be powerful when effects align; opposing directions can cancel. |
| C-alpha | Excess dispersion of observed counts around null expectations | Can detect heterogeneous directions; does not by itself identify direction or causal variants. |
| Variance-component test | A quadratic score for heterogeneous variant effects | Also accommodates differing effect directions; its formulation and calibration depend on the chosen model and genotype structure. |
C-alpha and variance-component tests share a broad sensitivity to heterogeneity, but their statistics and null calibrations are not interchangeable. The earlier material on variance-component tests develops that family more generally. C-alpha’s defining construction is the sum of per-variant squared count deviations minus their conditional null variances.
A common mistake is to conclude that a depleted variant must make a negative contribution because it is less common in cases. Its signed deviation is negative, but C-alpha squares that deviation. For example, a count one carrier below expectation by three contributes \(9-V_j\), just as a count three carriers above expectation does. The correction is to distinguish the direction of a deviation from its squared magnitude, and then subtract the null variance before interpreting the contribution. A negative centered contribution is possible when the observed squared deviation is smaller than its null expectation; it does not, by itself, indicate a protective effect.
Worked Example: A Negative Centered Contribution
Use \(N=100\), \(n=40\), and \(M_j=10\), so \(E_j=4\) and \(V_j\approx2.1818\), as calculated earlier. Suppose a hypothetical variant has \(X_j=4\) case carriers. Its deviation is zero, so its contribution is \((4-4)^2-2.1818=-2.1818\). The contribution is negative because the observed squared deviation is below its null expectation. It is not evidence that the variant decreases risk; that conclusion would require a direction-sensitive effect estimate and an appropriate inferential analysis.
Check Your Understanding
Answer each question, then compare with the brief solutions below.
- In a sample with \(N=120\), \(n=48\), and \(M_j=5\), what is the expected number of case carriers for variant \(j\)?
- Why can a variant depleted among cases contribute positively to C-alpha?
- For a single variant, why does subtracting \(V_j\) center its contribution at zero under the conditional null?
- A study has related participants and population subgroups. Why might unrestricted case-label permutation be invalid?
- In 499 valid null permutations, 9 statistics are at least as large as the observed statistic. Calculate the plus-one Monte Carlo p-value.
Solutions. (1) \(E_j=nM_j/N=(48)(5)/120=2\) case carriers. (2) C-alpha squares the deviation, so a negative signed deviation can have a large positive square; the contribution is positive when that square exceeds the null variance. (3) Under the null, \(E[(X_j-E_j)^2]=V_j\), so the expected centered contribution is \(V_j-V_j=0\). (4) Relatedness or subgroup structure can make case labels nonexchangeable across all participants; permutations must preserve the relevant design or model structure. (5) \(p=(9+1)/(499+1)=10/500=0.020\).