Deriving Power for Case-Control GWAS

Why Power in a Case-Control GWAS Has Its Own Calculation

For a quantitative trait, power is often described through the expected change in the trait per allele relative to its variability. In a case-control study, the outcome is binary, and association is commonly represented by a change in the log odds of disease per additional effect allele. The planned numbers of cases and controls both matter: adding participants to the smaller group often contributes more information than adding the same number to the larger group.

This tutorial derives a useful approximation for the power of a single additive logistic association test. It connects the effect-allele frequency and disease odds ratio to the expected test statistic, then compares that statistic with a significance threshold chosen in advance. The calculation is a planning approximation, not a guarantee of the result in a realized study.

Model and Quantities Needed

Definition: Let $G$ be an individual's effect-allele count, coded as 0, 1, or 2, and let $Y=1$ indicate case status and $Y=0$ control status. The additive logistic model is $$ \operatorname{logit}\{P(Y=1\mid G)\}=\beta_0+\beta G. $$ The per-allele disease odds ratio is $\mathrm{OR}=\exp(\beta)$, so $\beta=\log(\mathrm{OR})$. The model assumes that each additional effect allele changes the log odds by the same amount $\beta$.

The odds ratio compares disease odds for individuals differing by one effect allele under this model. For example, $\mathrm{OR}=1.15$ means that the disease odds are multiplied by 1.15 per additional allele; it does not mean that disease risk rises by 15 percentage points. An odds ratio below 1 corresponds to a negative $\beta$. For a two-sided test, power depends on the magnitude of the effect, not its direction.

Let $p$ be the effect-allele frequency in the source population, $n_{\mathrm{case}}$ the number of cases, and $n_{\mathrm{control}}$ the number of controls. Under Hardy-Weinberg equilibrium, the genotype count has variance

\operatorname{Var}(G)=2p(1-p).

This variance is largest when $p=0.5$ and smaller when the allele is uncommon or nearly fixed. Thus, for a fixed sample size and per-allele effect, a less variable genotype generally provides less information for association.

Information and the Noncentrality Parameter

Under the null hypothesis of no association, the logistic model has a common case probability in the ascertained sample. With the intercept estimated, the approximate information for the slope is the genotype variance multiplied by an effective case-control sample size:

I_\beta \approx 2p(1-p)\frac{n_{\mathrm{case}}n_{\mathrm{control}}} {n_{\mathrm{case}}+n_{\mathrm{control}}}.

The fraction $n_{\mathrm{case}}n_{\mathrm{control}}/(n_{\mathrm{case}}+n_{\mathrm{control}})$ is small if either group is small. In particular, when the group sizes are equal to $m$, this fraction is $m/2$, not $m$. The approximate standard error of the estimated log odds ratio is $SE(\hat{\beta})\approx 1/\sqrt{I_\beta}$.

A short justification explains the case-control factor. Under the null, the fitted case probability is approximately $\pi=n_{\mathrm{case}}/(n_{\mathrm{case}}+n_{\mathrm{control}})$. The logistic information for the slope after accounting for the intercept is approximately the weighted sum of squared deviations of $G$ from its mean. The logistic weight is $\pi(1-\pi)$, and the sum of genotype squared deviations is approximately $(n_{\mathrm{case}}+n_{\mathrm{control}})\operatorname{Var}(G)$. Therefore,

\pi(1-\pi)(n_{\mathrm{case}}+n_{\mathrm{control}})\operatorname{Var}(G) = \frac{n_{\mathrm{case}}n_{\mathrm{control}}} {n_{\mathrm{case}}+n_{\mathrm{control}}}\operatorname{Var}(G).

Substituting the Hardy-Weinberg genotype variance gives the information formula above. This uses the null genotype variance as a planning approximation; it is not an exact expression for every alternative or sampling scheme.

Definition: The approximate noncentrality parameter for the additive logistic association statistic is $$ \delta=\beta\sqrt{I_\beta} =\log(\mathrm{OR})\sqrt{2p(1-p)\frac{n_{\mathrm{case}}n_{\mathrm{control}}} {n_{\mathrm{case}}+n_{\mathrm{control}}}}. $$ Under the alternative, the standardized Wald statistic is approximated by a normal random variable with variance 1 and mean $\delta$. For a two-sided test, its power is approximately $P(|Z|>z_{1-\alpha/2})$, where $Z\sim N(\delta,1)$ and $\alpha$ is the prespecified significance threshold.

Writing $\Phi$ for the standard normal cumulative distribution function and $c=z_{1-\alpha/2}$, the two-sided power is

\operatorname{Power}\approx P(Z<-c)+P(Z>c) =\Phi(-c-\delta)+1-\Phi(c-\delta).

If the effect is protective, $\delta$ is negative. The two-sided power is unchanged if $\delta$ is replaced by $|\delta|$, because the two tails exchange roles. The threshold must be supplied for the intended test. A genome-wide analysis may use a much smaller $\alpha$ than a single-test calculation; the formula does not select that threshold.

Worked Calculations

Worked Example: A Modest Risk-Increasing Effect at a Stringent Threshold

Suppose a hypothetical study plans 8,000 cases and 8,000 controls. Assume an effect-allele frequency $p=0.40$, a per-allele disease odds ratio of 1.15, and a two-sided prespecified threshold $\alpha=5\times10^{-8}$. Use the approximation to calculate power.

First, calculate the effective case-control factor and genotype variance:

\frac{n_{\mathrm{case}}n_{\mathrm{control}}}{n_{\mathrm{case}}+n_{\mathrm{control}}} =\frac{8{,}000(8{,}000)}{16{,}000}=4{,}000, \qquad 2p(1-p)=2(0.40)(0.60)=0.48.

Thus $I_\beta\approx 0.48(4{,}000)=1{,}920$, and $\beta=\log(1.15)$. Using unrounded values in the multiplication,

\delta=\log(1.15)\sqrt{1{,}920} \approx 0.13976194(43.817805) \approx 6.1241.

The two-sided normal critical value at this threshold is approximately $c=z_{1-2.5\times10^{-8}}\approx5.4513$. Therefore,

\operatorname{Power}\approx \Phi(-5.4513-6.1241)+1-\Phi(5.4513-6.1241) \approx \Phi(-11.5754)+1-\Phi(-0.6728) \approx 0.749.

Under the stated approximation and assumptions, this design has about 74.9% power to detect the specified effect at the prespecified threshold. It does not have 74.9% probability that the association is true.

Worked Example: A Protective Effect and Unequal Group Sizes

Consider a hypothetical design with 1,900 cases and 5,700 controls, effect-allele frequency $p=0.20$, per-allele odds ratio $\mathrm{OR}=0.80$, and a two-sided threshold $\alpha=0.01$. Calculate the approximate power.

The effective sample-size factor is

\frac{1{,}900(5{,}700)}{1{,}900+5{,}700} =\frac{10{,}830{,}000}{7{,}600} =1{,}425.

The genotype variance is $2(0.20)(0.80)=0.32$, so $I_\beta\approx0.32(1{,}425)=456$. Since the odds ratio is below 1, $\beta=\log(0.80)<0$, and the noncentrality is negative:

\delta=\log(0.80)\sqrt{456} \approx -0.22314355(21.3541565) \approx -4.7650.

For the two-sided threshold, $c=z_{0.995}\approx2.5758$. Using the symmetry of the two-sided power formula, calculate with $|\delta|=4.7650$:

\operatorname{Power}\approx \Phi(-2.5758-4.7650)+1-\Phi(2.5758-4.7650) \approx \Phi(-7.3408)+\Phi(2.1892) \approx 0.9857.

The approximate power is 98.57% for detecting this protective per-allele association with a two-sided test at $\alpha=0.01$. The unequal allocation is reflected in the factor 1,425; it is not equivalent to treating all 7,600 participants as if they provided the same information.

Worked Example: A Less Common Allele and Moderate Power

Suppose a hypothetical study has 500 cases and 1,000 controls, with $p=0.10$, $\mathrm{OR}=1.30$, and a two-sided threshold $\alpha=0.05$. Calculate approximate power.

Here the effective sample-size factor is

\frac{500(1{,}000)}{500+1{,}000} =\frac{500{,}000}{1{,}500} \approx333.3333, \qquad 2p(1-p)=2(0.10)(0.90)=0.18.

Thus $I_\beta\approx0.18(333.3333)=60$. With $\beta=\log(1.30)\approx0.2623643$,

\delta=\log(1.30)\sqrt{60} \approx0.2623643(7.745967) \approx2.0319.

At $\alpha=0.05$, $c=z_{0.975}\approx1.9600$. Therefore,

\operatorname{Power}\approx \Phi(-1.9600-2.0319)+1-\Phi(1.9600-2.0319) \approx\Phi(-3.9919)+\Phi(0.0719) \approx0.5287.

The approximate power is 52.87%. The result illustrates how a moderate effect can still yield limited power when the allele is less common and one study group is relatively small.

Assumptions and Interpretation

The calculation is most appropriate as a first-order design approximation when the following conditions are reasonably met:

The outcome is analyzed with an additive logistic model, and the stated odds ratio is per additional effect allele.
Individuals contribute independent observations, and the planned case and control counts are the numbers with usable genotype data.
The effect-allele frequency and Hardy-Weinberg genotype variance are appropriate for the target sample under the null approximation.
The effect is not so large, and the allele not so rare, that the normal approximation or the null-based information approximation becomes unreliable.
The selected significance threshold is the one intended for this particular test, and the planned test is two-sided as used in the formula.

Case-control sampling fixes the numbers recruited as cases and controls, so the sample case fraction need not equal disease prevalence in the source population. The logistic slope and its odds-ratio interpretation can be modeled without inserting the population prevalence into this power formula. However, the approximation does not account automatically for covariates, relatedness, genotype quality, deviations from the assumed effect model, or uncertainty in the effect and allele frequency. Such features can change the information available in a real analysis.

The formula also makes the role of allocation explicit. With a fixed total sample size, the effective factor $n_{\mathrm{case}}n_{\mathrm{control}}/(n_{\mathrm{case}}+n_{\mathrm{control}})$ is largest when the groups are balanced. If recruitment costs or availability require an unequal design, the smaller group can become the limiting component. This is a feature of the case-control information calculation, not a claim that every study must have equal group sizes.

Common Mistake

A frequent error is to use the total sample size in place of the effective case-control factor. For example, with equal groups of 1,000 cases and 1,000 controls, it is incorrect to use 2,000 as the factor multiplying $2p(1-p)$. The correct factor is

\frac{1{,}000(1{,}000)}{1{,}000+1{,}000}=500.

Using the total would overstate the information and the noncentrality, and therefore overstate power. A second calculation pitfall is rounding $\log(\mathrm{OR})$, the square root of the information, or their product too early. Keep full precision through $\delta$ and the normal-tail calculation, then report the final result to a sensible number of significant figures.

Key takeaway: For this approximate additive logistic calculation, allele frequency sets genotype variance, case and control counts set the effective sample size, and the per-allele odds ratio sets the log-effect. Their combination determines the noncentrality; the prespecified two-sided threshold then determines approximate power.

Check Your Understanding

Use the definitions and formulas in this tutorial to answer the following questions.

In the additive logistic model, what does the per-allele odds ratio describe, and how is it related to $\beta$?
For 600 cases and 1,400 controls, calculate the effective case-control factor $n_{\mathrm{case}}n_{\mathrm{control}}/(n_{\mathrm{case}}+n_{\mathrm{control}})$.
Why does $2p(1-p)$ become smaller as the effect-allele frequency approaches 0 or 1?
For a two-sided test, why does a protective odds ratio produce the same approximate power as a risk-increasing effect with the same absolute log odds ratio?
State one assumption or limitation that should be checked before using this approximation for a real GWAS design.