Case-Control Sample Allocation and Rare-Variant Power

Why Case-Control Allocation Matters

A fixed sequencing budget forces a design choice: sequence more cases, more controls, or a balance of the two? Equal numbers can be convenient, but they do not necessarily provide the most information. The useful allocation depends on the contrast being studied, the variability of the gene-level signal in each group, and the cost of sequencing each participant.

The calculations here use a simple gene-level endpoint: whether a participant carries at least one qualifying rare variant in the gene. This provides a transparent approximation for comparing designs. Actual gene-level tests may use richer variant-level information, and their power can depend on the test, variant effects, and calibration. The allocation rule is therefore a planning guide under stated assumptions, not a universal guarantee of optimality for every gene-based test.

Definition: Let $p_1$ be the probability that a case carries at least one qualifying variant in the gene, and let $p_0$ be the corresponding probability for a control. The carrier-prevalence contrast is $\Delta=p_1-p_0$. For $n_1$ independent cases and $n_0$ independent controls, the approximate variance of the estimated contrast is $$ V=\frac{p_1(1-p_1)}{n_1}+\frac{p_0(1-p_0)}{n_0}. $$ Its standard error is $SE=\sqrt{V}$, and the signal-to-noise ratio under the alternative is approximately $Z=\Delta/SE$.

This variance formula treats each participant’s gene-level carrier status as a binary observation. Under independent sampling within each group, the sample carrier proportion has variance $p(1-p)/n$; adding the case and control contributions gives the variance of their difference. The approximation assumes the carrier probabilities used for design are appropriate for the target population and that case and control observations are independent. It also assumes comparable variant definition and callability across groups, so a difference in detection is not mistaken for a difference in carrier prevalence.

In an ascertained case-control sample, $p_1$ and $p_0$ describe the planned case and control groups, not the proportion of cases in the general population. For design calculations, use prevalence assumptions that correspond to the intended analysis and sampling scheme. The contrast is an association measure; it does not by itself establish a causal effect.

Power and a Fixed Sequencing Budget

Suppose the planned test is two-sided at a gene-level significance threshold $\alpha_g$, and the normal approximation is adequate. Write $z_{1-\alpha_g/2}$ for the corresponding standard-normal quantile. The approximate power is

\operatorname{Power}\approx \Phi\!\left(\frac{|\Delta|}{\sqrt{V}}-z_{1-\alpha_g/2}\right) +\Phi\!\left(-\frac{|\Delta|}{\sqrt{V}}-z_{1-\alpha_g/2}\right),

where $\Phi$ is the standard-normal cumulative distribution function. This expression makes the allocation objective clear: when $\Delta$ and $\alpha_g$ are fixed, lowering $V$ increases the signal-to-noise ratio and therefore increases approximate power.

Let $c_1$ and $c_0$ be the per-person costs for sequencing a case and a control, respectively, and let $B$ be the available budget. Ignoring fixed costs for the moment, a feasible allocation satisfies $c_1n_1+c_0n_0=B$. Define $a=p_1(1-p_1)$ and $b=p_0(1-p_0)$. The variance to minimize is $a/n_1+b/n_0$.

Deriving the Cost-Aware Allocation

The continuous optimum follows by minimizing the variance subject to the budget. Form the Lagrangian $L=a/n_1+b/n_0+\lambda(c_1n_1+c_0n_0-B)$. Setting its derivatives with respect to $n_1$ and $n_0$ to zero gives

-\frac{a}{n_1^2}+\lambda c_1=0, \qquad -\frac{b}{n_0^2}+\lambda c_0=0.

Thus $a/(c_1n_1^2)=b/(c_0n_0^2)=\lambda$. Taking the ratio and solving for the allocation gives

\frac{n_1}{n_0} = \sqrt{\frac{a c_0}{b c_1}} = \sqrt{\frac{p_1(1-p_1)c_0}{p_0(1-p_0)c_1}}.

The solution is a minimum: $a/n_1+b/n_0$ is convex for positive sample sizes, and the budget constraint is linear. For equal per-person costs, the group with the larger carrier-status variance receives more participants. A group that is more expensive to sequence receives relatively fewer, all else equal. The rule optimizes precision for the specified contrast; it does not simply assign participants in proportion to the prevalence difference.

Worked Example: Equal Costs and a Fixed Number of Participants

Suppose a hypothetical study can sequence 2,000 participants at equal per-person costs. The planning assumptions are $p_1=0.040$ for cases and $p_0=0.010$ for controls. The intended two-sided gene-level threshold is $\alpha_g=0.05$. Find the cost-aware allocation and compare its approximate power with equal enrollment.

Allocation. Here $a=0.040(0.960)=0.0384$, $b=0.010(0.990)=0.00990$, and the costs are equal. Therefore, $$ \frac{n_1}{n_0}=\sqrt{\frac{0.0384}{0.00990}}=1.969. $$ With $n_1+n_0=2{,}000$, the continuous solution is approximately $n_1=1{,}326.5$ cases and $n_0=673.5$ controls. Use the feasible rounded allocation of 1,327 cases and 673 controls.

Approximate power for the rounded allocation. The contrast is $\Delta=0.040-0.010=0.0300$. Its estimated variance for planning is $$ V=\frac{0.0384}{1{,}327}+\frac{0.00990}{673} =0.00002894+0.00001471 \approx 0.00004365. $$ Thus $SE=\sqrt{0.00004365}\approx0.006607$ and $Z=0.0300/0.006607\approx4.541$. Since $z_{0.975}\approx1.960$, the approximate power is $\Phi(4.541-1.960)+\Phi(-4.541-1.960)$, or about $0.995$.

With equal enrollment, $n_1=n_0=1{,}000$, the variance is $0.0384/1{,}000+0.00990/1{,}000=0.00004830$, so $SE\approx0.006950$ and $Z\approx4.317$. The corresponding approximate power is about $\Phi(4.317-1.960)+\Phi(-4.317-1.960)=0.991$. Under these assumptions, the unequal allocation gives slightly higher power for the same total number and equal costs.

Unequal Sequencing Costs

The cost-aware ratio can shift substantially when cases and controls have different costs. Costs may include recruitment and phenotyping as well as sequencing, provided the cost assigned to each participant is a reasonable marginal cost for the design. Fixed setup costs do not alter the ratio formula if they are the same for all candidate allocations, but they reduce the budget remaining for participants.

Worked Example: Twice the Cost per Case

Consider a hypothetical design in which a case costs 2 budget units and a control costs 1. The planning variances are $a=0.02855$ for cases and $b=0.009900$ for controls. A carrier probability near $0.02942$ gives the stated case variance, since $0.02942(1-0.02942)\approx0.02855$; the control probability is $0.0100$. Let the participant budget be 3,000 units.

The optimal continuous ratio is $$ \frac{n_1}{n_0} =\sqrt{\frac{0.02855(1)}{0.009900(2)}} =\sqrt{1.4419} \approx1.201. $$ The cost constraint is $2n_1+n_0=3{,}000$. Substituting $n_1\approx1.201n_0$ gives $n_0\approx3{,}000/(2(1.201)+1)\approx882$ controls and approximately 1,059 cases. The rounded allocation uses $2(1{,}059)+882=3{,}000$ budget units.

The contrast under these assumptions is $\Delta=0.02942-0.01000=0.01942$. Its variance is $$ V=\frac{0.02855}{1{,}059}+\frac{0.009900}{882} =0.0000269594+0.0000112245 =0.0000381839. $$ Therefore $SE=\sqrt{0.0000381839}\approx0.006179$, and $Z=0.01942/0.006179\approx3.143$. At a two-sided threshold of $\alpha_g=0.01$, $z_{0.995}\approx2.576$, giving approximate power $\Phi(3.143-2.576)+\Phi(-3.143-2.576)\approx0.715$. This is a design-based approximation, conditional on the stated carrier rates and cost assumptions.

Rare Carriers and a Stringent Gene-Level Threshold

A larger relative effect does not guarantee high power when carriers are uncommon. If both groups have low carrier prevalence, the number of informative participants can be small even in a substantial sample. In addition, a stringent gene-level threshold raises the signal-to-noise ratio required for rejection. The next calculation illustrates why the allocation rule should be paired with a power calculation rather than treated as a complete design answer.

Worked Example: Comparing Allocations at a Stringent Threshold

Suppose a hypothetical study has 2,000 budget units, cases cost 1.5 units each, controls cost 1 unit each, and the assumed carrier prevalences are $p_1=0.020$ and $p_0=0.005$. The planned two-sided gene-level threshold is $\alpha_g=0.001$.

Here $a=0.020(0.980)=0.0196$ and $b=0.005(0.995)=0.004975$. The optimal ratio is $$ \frac{n_1}{n_0} =\sqrt{\frac{0.0196(1)}{0.004975(1.5)}} =\sqrt{2.6261} \approx1.621. $$ The continuous budget solution is about 945 cases and 583 controls. To keep exactly within the budget after rounding, use 944 cases and 584 controls, for a total cost of $1.5(944)+584=2{,}000$ units.

The contrast is $\Delta=0.020-0.005=0.0150$. At this rounded allocation, $$ V=\frac{0.0196}{944}+\frac{0.004975}{584} =0.00002076+0.000008519 \approx0.00002928. $$ Thus $SE\approx0.005411$ and $Z\approx0.0150/0.005411=2.771$. For $\alpha_g=0.001$, $z_{0.9995}\approx3.291$, so approximate power is $\Phi(2.771-3.291)+\Phi(-2.771-3.291)\approx0.302$.

For comparison, 800 cases and 800 controls also cost 2,000 units. Their variance is $0.0196/800+0.004975/800=0.00003072$, giving $SE\approx0.005542$, $Z\approx2.706$, and power about $\Phi(2.706-3.291)+\Phi(-2.706-3.291)=0.279$. The cost-aware allocation improves approximate power, but power remains modest; the sample size, effect assumptions, or budget may need reconsideration.

Practical Interpretation and Common Pitfalls

The allocation depends on assumptions about both groups. If preliminary data give uncertain carrier prevalences, calculate power over a plausible range rather than treating one estimate as certain. Because the formula uses $p(1-p)$, a change in either prevalence changes the variance and potentially the recommended ratio. Also calculate with the actual gene-level threshold intended for the analysis; using a less stringent threshold can substantially overstate power when many genes are tested.

The simple calculation is most defensible when a carrier indicator is a useful summary of the gene-level signal and when the case and control groups have comparable measurement quality. More complex gene-based tests may use variant-specific information, and their power need not follow this binary-carrier approximation exactly. For those settings, the formula is a useful benchmark, not a replacement for test-specific power assessment.

Pitfall: assuming equal enrollment is always optimal. Equal counts ignore both group-specific variance and sequencing cost. Use the cost-aware ratio, then compare feasible integer allocations.
Pitfall: allocating in proportion to the effect size. The ratio is determined by $p_1(1-p_1)$, $p_0(1-p_0)$, and the relative costs, not by $|p_1-p_0|$ alone. The effect contrast affects the resulting power, even though it cancels from this variance-minimization rule.
Pitfall: treating estimated power as a guarantee. Power is conditional on the assumed prevalences, costs, test threshold, and approximation. Differences in callability or other systematic measurement features can invalidate the comparison.

Key takeaway: Under a fixed budget, the continuous allocation that minimizes the variance of a case-control carrier-prevalence contrast satisfies $n_1/n_0=\sqrt{p_1(1-p_1)c_0/[p_0(1-p_0)c_1]}$. Round to feasible sample sizes, recompute variance and power, and treat the result as conditional on the design assumptions and the gene-level test.

Check Your Understanding

Answer each question using the allocation and power principles above.

For fixed case and control costs, which quantities determine the variance-minimizing ratio $n_1/n_0$?
If the per-case sequencing cost increases while all other inputs remain unchanged, how does the optimal ratio change?
Why can an optimized case-control allocation still have low power for a rare-variant gene-level test?
For $p_1=0.03$, $p_0=0.01$, and equal per-person costs, write the formula for the optimal ratio and identify which group receives more participants.
Why should a power calculation use the planned gene-level significance threshold rather than automatically using 0.05?