Case-Control Sampling and Effective Sample Size

Why Case-Control Sampling Matters

Many genetic studies of binary traits do not sample people in proportion to how common the trait is in the population. Instead, researchers recruit a specified number of cases and controls. This can make a rare condition feasible to study, but the resulting sample has a case fraction that differs from population prevalence. That difference matters in two ways: it changes the scale of the observed binary phenotype, and it affects how much statistical information the sample provides.

These are related but distinct issues. A liability-scale conversion addresses the meaning of a heritability estimate when the outcome is binary and sampling is ascertained. Effective sample size summarizes, approximately, the precision loss from an imbalance between cases and controls. Neither quantity changes the actual number of people enrolled.

Definition: Let $p$ denote the population prevalence of a binary trait, and let $\hat{p}$ denote the case fraction in an analyzed case-control sample. The sample fraction is $\hat{p}=n_{\mathrm{case}}/(n_{\mathrm{case}}+n_{\mathrm{control}})$. Under case-control ascertainment, $\hat{p}$ is set by recruitment and need not estimate $p$.

Observed and Liability Scales

A binary phenotype coded as 0 for controls and 1 for cases has variance $\hat{p}(1-\hat{p})$ in the analyzed sample. An observed-scale SNP heritability, written $h^2_{O,\mathrm{SNP}}$, describes the proportion of that binary phenotype’s variance attributed to the modeled SNP set on the scale used for estimation. Because the sampled case fraction determines the binary variance, an observed-scale estimate can change with the sampling design, even when the underlying population trait and genetic effects have not changed.

A liability model represents disease status as a threshold on a continuous, normally distributed latent quantity. Let liability have variance 1, and suppose a person is a case when liability exceeds $t$. If prevalence is $p$, then $t=\Phi^{-1}(1-p)$, where $\Phi$ is the standard normal cumulative distribution function. Write $\phi(t)$ for the standard normal density evaluated at that threshold.

Result: Under a liability-threshold model, with an appropriate population prevalence $p$, the liability-scale estimate corresponding to an observed-scale estimate is $$ h^2_{L,\mathrm{SNP}} = h^2_{O,\mathrm{SNP}} \frac{p^2(1-p)^2}{\hat{p}(1-\hat{p})\phi(t)^2}, \qquad t=\Phi^{-1}(1-p). $$ This conversion assumes a normally distributed liability with unit variance, a threshold defining case status, and an estimate that is appropriate for the stated ascertainment design. The estimate is a rescaling under that model, not a new measurement of genetic effects.

The population prevalence and sample case fraction play different roles in this expression. Prevalence determines the liability threshold and the density at that threshold; the sample fraction describes the case-control composition that affected the observed-scale estimate. Substituting the sample fraction for population prevalence merely because it is known precisely would generally answer the wrong question.

Worked Example: Converting an Ascertained Estimate to the Liability Scale

Suppose a hypothetical study estimates observed-scale SNP heritability $h^2_{O,\mathrm{SNP}}=0.020$. The trait’s population prevalence is $p=0.010$, while cases make up $\hat{p}=0.20$ of the analyzed case-control sample. Assume the liability-threshold model is appropriate and that the estimate has been obtained in a way that supports this conversion.

For prevalence 0.010, the threshold is $t=\Phi^{-1}(0.990)\approx 2.326$, and $\phi(2.326)\approx 0.02665$. Substitution gives $$ h^2_{L,\mathrm{SNP}} = 0.020 \frac{(0.010)^2(0.990)^2} {(0.20)(0.80)(0.02665)^2}. $$

The numerator factor is $(0.010)^2(0.990)^2=0.00009801$. The denominator is $(0.20)(0.80)(0.02665)^2\approx 0.0001136$. Thus $$ h^2_{L,\mathrm{SNP}} \approx 0.020\left(\frac{0.00009801}{0.0001136}\right) \approx 0.0173. $$

Under the stated assumptions, the converted liability-scale SNP heritability is approximately 0.0173, or 1.73% of liability variance. This is not a claim that 1.73% of individual disease outcomes are genetically determined; it is a variance proportion for the specified SNP set on a modeled liability scale.

Effective Sample Size for Imbalanced Studies

A study with many more controls than cases does not usually provide as much association information as a balanced study with the same total number of participants. A useful approximation is the effective sample size, $n_{\mathrm{eff}}$, defined from the case and control counts by $$ n_{\mathrm{eff}} = \frac{4}{1/n_{\mathrm{case}}+1/n_{\mathrm{control}}} = \frac{4n_{\mathrm{case}}n_{\mathrm{control}}} {n_{\mathrm{case}}+n_{\mathrm{control}}}. $$

The factor 4 makes $n_{\mathrm{eff}}$ equal to the actual total sample size when cases and controls are balanced. For an imbalanced study, $n_{\mathrm{eff}}$ is smaller than that total. It is a precision-equivalent summary under approximations, not a count of the individuals actually analyzed and not a universal replacement for sample size in every heritability estimator.

Why the Effective-Size Formula Works

For a simple justification, consider the approximate sampling variance of a log-odds-ratio estimate for a genetic association near the null. If genotype variation is comparable in cases and controls and the other relevant conditions are similar, this variance contains a term proportional to $1/n_{\mathrm{case}}+1/n_{\mathrm{control}}$. For a balanced study with $n_{\mathrm{eff}}$ participants, the corresponding term is $4/n_{\mathrm{eff}}$, because each group has $n_{\mathrm{eff}}/2$ participants: $$ \frac{1}{n_{\mathrm{eff}}/2}+\frac{1}{n_{\mathrm{eff}}/2} = \frac{4}{n_{\mathrm{eff}}}. $$

Equating the two terms gives $4/n_{\mathrm{eff}}=1/n_{\mathrm{case}}+1/n_{\mathrm{control}}$, which rearranges to the effective-size formula. This derivation explains the precision comparison: effective size is the balanced-study size with the same approximate case-control information. It relies on comparable genotype variation and similar analysis conditions; it does not account automatically for every source of uncertainty in SNP heritability estimation.

Worked Example: Effective Size With Many More Controls

A hypothetical study includes 1,200 cases and 4,800 controls. Its actual total sample size is 6,000, but the group sizes are unequal.

Using the formula, $$ n_{\mathrm{eff}} = \frac{4(1{,}200)(4{,}800)}{1{,}200+4{,}800} = \frac{23{,}040{,}000}{6{,}000} = 3{,}840. $$

The study therefore has approximately the association precision of a balanced study with 3,840 participants under the stated approximation, rather than that of a balanced study with 6,000. As a check, $4/(1/1{,}200+1/4{,}800)=3{,}840$ as well. The effective size is less than the actual total because the cases are the limiting group.

Worked Example: Comparing Two Designs With the Same Total

Consider two hypothetical studies, each with 10,000 participants. Study A has 5,000 cases and 5,000 controls. Study B has 500 cases and 9,500 controls. Compare their effective sample sizes and approximate standard errors under the same association model and comparable genotype variation.

For Study A, $$ n_{\mathrm{eff,A}} = \frac{4(5{,}000)(5{,}000)}{10{,}000} = 10{,}000. $$ For Study B, $$ n_{\mathrm{eff,B}} = \frac{4(500)(9{,}500)}{10{,}000} = 1{,}900. $$

When standard error is approximately proportional to $1/\sqrt{n_{\mathrm{eff}}}$, the ratio of Study B’s standard error to Study A’s is $$ \frac{SE_B}{SE_A} \approx \sqrt{\frac{10{,}000}{1{,}900}} \approx 2.294. $$

Thus the severe imbalance in Study B corresponds to substantially less association precision than the balanced design, despite equal total sample sizes. This comparison is approximate: it concerns the information summarized by effective size and does not assert that every heritability method’s standard error is determined by this ratio alone.

Using Effective Size in Heritability Work

Effective sample size is useful when describing the information in binary-trait GWAS results and when a method or downstream analysis calls for an effective rather than a raw sample size. For example, in summary-statistic analyses, reporting the case and control counts alongside $n_{\mathrm{eff}}$ makes the imbalance transparent. The effective size can also help explain why two studies with similar total enrollment may have different precision.

For SNP heritability, keep three quantities conceptually separate: the observed-scale estimate, its conversion to the liability scale, and the precision of the estimate. Effective size describes an approximate precision consequence of case-control imbalance. It does not itself perform the liability conversion, determine population prevalence, or correct an estimate for an unsuitable sampling model. Conversely, converting scales does not recover information that was lost through having relatively few cases.

Quantity	What it describes	What it does not do
Observed-scale SNP heritability	Genetic variance relative to variance of the analyzed binary phenotype	Provide a prevalence-independent comparison across different case fractions
Liability-scale SNP heritability	Genetic variance relative to unit liability variance under a threshold model	Remove the assumptions needed for the liability model or ascertainment correction
Effective sample size	Approximate precision-equivalent size for an imbalanced case-control design	Equal the actual enrollment or encode every source of uncertainty in a heritability estimate

Common Pitfalls

Using the recruited case fraction as population prevalence. Wrong reasoning: “Cases are 20% of my sample, so the trait prevalence is 20%.” Correction: case-control recruitment can set the sample fraction; use a justified population prevalence for the liability threshold.
Reporting an observed-scale estimate as if it were on the liability scale. The numerical value depends on the analyzed binary scale and sampling fraction. State the scale and the assumptions behind any conversion.
Treating effective size as the number enrolled. In an imbalanced study, $n_{\mathrm{eff}}$ summarizes approximate information; it does not change the actual case and control counts.
Assuming the effective-size formula captures all precision differences. It is an approximation whose comparison presumes similar genotype variation and analysis conditions. Differences in other features can also affect uncertainty.

Key takeaway: Case-control ascertainment affects both the scale on which binary-trait heritability is expressed and the precision available for estimation. Use population prevalence and sample case fraction in their distinct roles for liability conversion, and use effective sample size as an approximate summary of imbalance—not as a substitute for the study’s actual counts.

Check Your Understanding

Try these questions using the definitions and assumptions above.

A study has 800 cases and 3,200 controls. Calculate its effective sample size and compare it with its actual total sample size.
Explain why the case fraction in an ascertained sample need not equal population prevalence.
In the liability-scale conversion, what determines the threshold $t$, and what role does the sample fraction play?
Two studies have equal total sample sizes, but one is balanced and the other has many more controls than cases. Which is expected to have the greater effective sample size, and why?
Does a liability-scale conversion restore precision lost through case-control imbalance? Explain the difference between scale and precision.