Ascertainment Correction for Sequenced Case-Control Samples

Why Ascertainment Changes What the Sample Represents

Sequencing studies often recruit people because they are affected by a disease, or deliberately recruit a chosen number of cases and controls. This design can make rare-variant association analyses feasible, but the resulting sample is not a miniature version of the source population. In particular, the proportion of cases in the sample generally does not estimate disease prevalence. The key distinction is between how participants entered the study and how disease status relates to genotype in the population.

Ascertainment correction makes that distinction explicit. It is needed when interpreting the sample case fraction as a population quantity or when estimating absolute disease risks from a case-control logistic model. Under specific conditions, however, the genetic odds-ratio association estimate can remain valid even though the sample prevalence is distorted.

Definition: Let \(D=1\) indicate a case and \(D=0\) a control in the source population. Let \(K=P(D=1)\) be the source-population disease prevalence. Write \(A=1\) if a person is included in the study, and define the case and control inclusion probabilities as \(a_1=P(A=1\mid D=1)\) and \(a_0=P(A=1\mid D=0)\). The sampled case fraction is \(q=P(D=1\mid A=1)\). In general, \(q\ne K\).

The inclusion probabilities describe the design, not disease biology. For example, if investigators enroll most eligible cases but only a small fraction of eligible controls, then \(a_1\) is much larger than \(a_0\), and the sample will have a far higher case fraction than the source population. The sample fraction \(q\) is still useful for describing the recruited sample; it simply answers a different question from \(K\).

Correcting the Case Fraction for Ascertainment

By the definition of conditional probability, the fraction of sampled people who are cases is the number of cases included divided by all included people, expressed in probabilities. Cases enter with probability \(a_1\) from a source-population fraction \(K\); controls enter with probability \(a_0\) from fraction \(1-K\). Therefore,

q=\frac{a_1K}{a_1K+a_0(1-K)}.

Solving this equation for \(K\) gives the prevalence correction. Multiplying both sides by the denominator, collecting the terms involving \(K\), and dividing yields

q\{a_1K+a_0(1-K)\}=a_1K

qa_0=K\{a_1(1-q)+qa_0\}, \qquad K=\frac{qa_0}{a_1(1-q)+qa_0}.

This calculation requires the case and control inclusion probabilities, or equivalent design information. Knowing only \(q\) does not identify \(K\): the same sampled case fraction could result from different source prevalences paired with different recruitment fractions.

Worked Example: Recovering Prevalence From an Ascertained Sample

Suppose a hypothetical sequencing study includes 80% of eligible cases and 2.00% of eligible controls. Its sampled case fraction is \(q=0.2878\). Estimate the source-population prevalence.

Here \(a_1=0.80\), \(a_0=0.0200\), and \(q=0.2878\). Substitute these values into the correction:

K=\frac{0.2878(0.0200)} {0.80(1-0.2878)+0.2878(0.0200)} =\frac{0.005756}{0.56976+0.005756} =\frac{0.005756}{0.575516} \approx 0.0100015.

The estimated source prevalence is approximately \(0.0100\), or 1.00%. The sample itself has 28.78% cases, but that proportion reflects the much higher inclusion probability for cases. It is not an estimate of population prevalence without correcting for the design.

How Sampling Changes the Logistic Model

Let \(G\) denote a genotype measure, such as the number of copies of an allele, and suppose the source-population disease model is logistic:

\operatorname{logit}\{P(D=1\mid G)\}=\alpha+\beta G, \qquad \operatorname{logit}(u)=\log\left(\frac{u}{1-u}\right).

The coefficient \(\beta\) is the change in the log odds of disease for a one-unit increase in \(G\); \(\exp(\beta)\) is the corresponding disease odds ratio. Now assume that selection depends on disease status, but not on genotype once disease status is fixed. In other words, \(P(A=1\mid D,G)=a_D\), where \(a_D\) is \(a_1\) for cases and \(a_0\) for controls. Among sampled people with genotype \(G\),

P(D=1\mid G,A=1) =\frac{a_1P(D=1\mid G)} {a_1P(D=1\mid G)+a_0P(D=0\mid G)}.

Taking the odds cancels the common denominator:

\frac{P(D=1\mid G,A=1)}{P(D=0\mid G,A=1)} =\frac{a_1}{a_0} \frac{P(D=1\mid G)}{P(D=0\mid G)}.

Taking logarithms then gives the sampled logistic model:

\operatorname{logit}\{P(D=1\mid G,A=1)\} =\alpha+\log\left(\frac{a_1}{a_0}\right)+\beta G.

Key result: If inclusion depends on case status alone, with the same case and control inclusion probabilities across genotypes, outcome-based sampling adds \(\log(a_1/a_0)\) to the logistic intercept and leaves the genotype slope \(\beta\) unchanged. Thus the sampled and source-population odds ratios \(\exp(\beta)\) agree under the stated model and sampling assumptions. The uncorrected sampled intercept does not estimate the source-population intercept.

This result explains why odds-ratio association estimates can be valid in a case-control study even when the case fraction is set by design. It does not imply that every statistic computed from an ascertained sample is unaffected. Absolute risks, prevalence, and the baseline odds require appropriate correction. The result also does not rescue a model whose assumptions fail, for example when selection within cases varies by genotype.

Worked Example: Checking an Odds Ratio Under Outcome-Based Sampling

Consider a hypothetical source population in which the disease risk is 0.0100 for people with genotype \(G=0\) and 0.0400 for people with \(G=1\). Cases are included with probability \(a_1=0.80\), and controls with probability \(a_0=0.0200\), with these probabilities independent of genotype within disease status. Compare the source and sampled odds ratios.

In the source population, the odds ratio comparing \(G=1\) with \(G=0\) is

\operatorname{OR}_{\mathrm{source}} =\frac{0.0400/(1-0.0400)}{0.0100/(1-0.0100)} =\frac{0.0400/0.9600}{0.0100/0.9900} =\frac{0.0416667}{0.0101010} =4.125.

For \(G=0\), the probability of being both a case and included is \(0.80(0.0100)=0.00800\); the probability of being both a control and included is \(0.0200(0.9900)=0.0198\). The sampled disease odds at \(G=0\) are therefore \(0.00800/0.0198=0.4040\). For \(G=1\), the corresponding probabilities are \(0.80(0.0400)=0.0320\) and \(0.0200(0.9600)=0.0192\), giving sampled disease odds \(0.0320/0.0192=1.6667\). Thus,

\operatorname{OR}_{\mathrm{sample}} =\frac{1.6667}{0.4040} \approx 4.125.

The odds ratio is preserved. Equivalently, both genotype-specific disease odds were multiplied by \(a_1/a_0=0.80/0.0200=40\), so their ratio did not change. The sampled risks themselves are not the source risks: for example, the sampled case fraction among \(G=0\) people is \(0.00800/(0.00800+0.0198)\approx0.2878\), not 0.0100.

Likelihoods and the Meaning of a Case-Control Analysis

The sampling mechanism also clarifies which likelihood is appropriate for which target. If genotypes in the source population have distribution \(f(G)\) and the source disease risk is \(p(G)=P(D=1\mid G)\), then the genotype distributions among cases and controls are

f(G\mid D=1)=\frac{f(G)p(G)}{K}, \qquad f(G\mid D=0)=\frac{f(G)\{1-p(G)\}}{1-K}.

A case-control sample typically contains a design-chosen number of cases and controls. Conditional on those numbers, its genotype likelihood is based on the genotype distributions within each disease group, rather than treating the observed case fraction as if it arose from random sampling of the population. The terms \(K\) and \(1-K\) normalize the case and control genotype distributions. In practice, the source genotype distribution may be a nuisance component, and the genetic association is often summarized through the disease odds ratio.

The prospective logistic likelihood models disease status given genotype. Under outcome-only sampling, it has the same slope parameter as the source model but a shifted intercept, as derived above. Thus a slope estimate from logistic regression can estimate the source log odds ratio under the stated assumptions, even though the sample’s disease fraction is artificial. The likelihood interpretation does not make the sample prevalence representative; it explains why a particular association parameter remains identifiable.

If population prevalence \(K\) is known or estimated externally, the logistic intercept can be corrected. The sampling odds ratio satisfies

\frac{a_1}{a_0} =\frac{q/(1-q)}{K/(1-K)}, \qquad \alpha=\alpha_{\mathrm{sample}}-\log\left(\frac{a_1}{a_0}\right) =\alpha_{\mathrm{sample}}+\operatorname{logit}(K)-\operatorname{logit}(q).

The last equality follows because the marginal sampling odds \(q/(1-q)\) equal the source prevalence odds \(K/(1-K)\) multiplied by \(a_1/a_0\). The correction restores the source-model intercept; it does not change the slope under the outcome-only sampling conditions. An externally supplied prevalence must refer to the target population and the same disease definition and relevant time period.

Worked Example: Correcting a Logistic Intercept

Suppose a hypothetical study samples cases at probability 0.80 and controls at probability 0.0200. A logistic model fitted in the ascertained sample has intercept \(\alpha_{\mathrm{sample}}=-0.7000\) and genotype slope \(\beta=0.8000\). Correct the intercept to the source-population scale.

The sampling odds ratio is \(a_1/a_0=0.80/0.0200=40\), so the intercept shift is \(\log(40)\approx3.6889\). Since the sample intercept equals the population intercept plus this shift,

\alpha=-0.7000-\log(40) =-0.7000-3.6889 =-4.3889.

The corrected source model is \(\operatorname{logit}\{P(D=1\mid G)\}=-4.3889+0.8000G\). At \(G=0\), the model-implied source risk is \(1/\{1+\exp(4.3889)\}\approx0.0123\). The genotype odds ratio remains \(\exp(0.8000)\approx2.23\) per one-unit increase in \(G\); the intercept correction changes the risk level, not that odds ratio.

Conditions, Limits, and Common Mistakes

The preservation of a genetic odds ratio is conditional, not automatic. Before interpreting a case-control estimate as a source-population association, check the following:

Selection within disease status: inclusion must not depend on genotype after conditioning on case or control status. If carriers are more likely to be recruited among cases than noncarriers, the selection factor is genotype-dependent and can alter the estimated association.
Consistent sampling across relevant covariates: if case and control inclusion probabilities vary across covariate strata, a single intercept shift may not suffice. The sampling process and analysis must account for that variation.
Correct model and outcome: the stated slope result applies to the specified logistic disease model and its odds ratio. It does not turn an odds ratio into a risk ratio or establish a causal effect.
Target-population alignment: a prevalence correction is meaningful only when the external prevalence and the study’s source population refer to the same target and disease definition.

A frequent mistake is to read a sample with 30% cases as evidence that disease prevalence in the source population is about 30%. That reasoning treats the recruited people as if they had been sampled with equal probability regardless of disease status. The correction requires the design probabilities: the sampled case fraction is generated jointly by source prevalence and the relative inclusion of cases and controls.

A second mistake is to conclude that because the odds ratio can be preserved, every result from a case-control analysis needs no correction. The odds-ratio slope and absolute-risk scale are distinct targets. Correcting an intercept or interpreting risks requires ascertainment information or an appropriate external prevalence; the sample case fraction alone cannot supply it.

Key takeaway: Outcome-dependent sampling generally distorts disease prevalence and the logistic intercept. If selection depends only on case status and is genotype-independent within status, it shifts the intercept by \(\log(a_1/a_0)\) while preserving the logistic genotype slope and its odds ratio. State the sampling assumptions and correct the intercept or prevalence before making population-risk interpretations.

Check Your Understanding

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

A study samples 70% of eligible cases and 5% of eligible controls. Why is the observed case fraction not, by itself, an estimate of source-population prevalence?
Write the equation relating sampled case fraction \(q\), prevalence \(K\), and inclusion probabilities \(a_1\) and \(a_0\). What additional information is needed to solve for \(K\)?
Under what condition does outcome-based sampling leave a logistic genotype slope unchanged? What happens to the intercept?
Explain why multiplying the disease odds at every genotype value by the same sampling odds ratio preserves the odds ratio between two genotype values.
Give an example of genotype-dependent recruitment within case status and explain why it could invalidate the odds-ratio preservation result.