Why Case-Control Sampling Preserves Odds-Ratio Slopes

Why the Sampling Design Matters

In a case-control study, investigators choose how many people with the outcome (cases) and without it (controls) to include. The resulting proportion of cases is set by the design, not necessarily by the outcome frequency in the population that gave rise to the sample. This raises a natural question: can a logistic regression fitted to such a sample still estimate a genotype association?

Under a specific and important condition, it can. If inclusion depends on case-control status but not on genotype within either outcome group, sampling changes the logistic-regression intercept while preserving the genotype slopes. Since a slope represents a log odds ratio, the genotype odds ratios are preserved as well. The result does not mean that every quantity from the fitted model describes the source population: in particular, the sampled intercept generally does not.

Model and Sampling Conditions

Let \(Y\) be a binary outcome, with \(Y=1\) for a case and \(Y=0\) for a control. Let \(G\) denote a genotype predictor, such as allele dosage coded as 0, 1, or 2. In the source population, suppose the conditional probability of being a case follows the logistic model

P(Y=1\mid G)=\frac{\exp(\alpha+\beta G)}{1+\exp(\alpha+\beta G)}.

Here, \(\alpha\) is the source-population intercept and \(\beta\) is the change in log odds of the outcome for a one-unit increase in \(G\). This model assumes that the log odds are linear in the specified predictor. It can be extended to multiple genotype indicators or other predictors, provided the sampling condition below applies to all of them.

Definition: Outcome-dependent case-control sampling preserves logistic slopes when the probability of inclusion may differ between cases and controls, but is constant across genotypes within each outcome group. Write \(S=1\) if a person is selected. The condition is \(P(S=1\mid Y=1,G)=s_1\) and \(P(S=1\mid Y=0,G)=s_0\), where \(s_1\) and \(s_0\) are positive sampling probabilities that do not depend on \(G\).

This condition describes sampling within outcome groups: investigators may select a much larger fraction of cases than controls, but among cases the chance of selection must not vary with genotype, and among controls it must not vary with genotype. It is the genotype-independence of selection within outcome groups that matters. Merely calling a study “case-control” does not guarantee the condition.

Why the Intercept Shifts but the Slope Does Not

The key is to examine the odds of being a case among selected people with a given genotype. By the definition of conditional probability, those odds are the selected case probability divided by the selected control probability. The selection probabilities multiply the corresponding source-population probabilities:

\begin{aligned} \frac{P(Y=1\mid G,S=1)}{P(Y=0\mid G,S=1)} &= \frac{P(S=1\mid Y=1,G)P(Y=1\mid G)} {P(S=1\mid Y=0,G)P(Y=0\mid G)}\\ &= \frac{s_1}{s_0} \frac{P(Y=1\mid G)}{P(Y=0\mid G)}. \end{aligned}

The equality uses the sampling condition: within each outcome group, the selection probability is constant across genotypes. Under the source logistic model, the source-population odds are \(\exp(\alpha+\beta G)\). Therefore the selected-sample odds are

\frac{P(Y=1\mid G,S=1)}{P(Y=0\mid G,S=1)} = \frac{s_1}{s_0}\exp(\alpha+\beta G) = \exp\left(\alpha+\log\left(\frac{s_1}{s_0}\right)+\beta G\right).

Taking the logarithm gives a logistic model for the selected sample. Its intercept is \(\alpha+\log(s_1/s_0)\), and its genotype slope is still \(\beta\). The selection factor is the same for every genotype, so it adds a constant to the log odds rather than changing the amount by which the log odds vary with genotype.

Theorem: If the source-population conditional model is logistic and selection probabilities are constant across genotypes within each outcome group, then the selected-sample conditional model has the same genotype slope. Its intercept differs from the source intercept by \(\log(s_1/s_0)\). Consequently, the per-unit genotype odds ratio \(\exp(\beta)\) is unchanged.

The same reasoning works for a vector of predictors \(X\) in a source model with log odds \(\alpha+\boldsymbol{\beta}^{T}X\). If \(s_1\) and \(s_0\) are constant across all values of \(X\) within the respective outcome groups, selection adds \(\log(s_1/s_0)\) to the intercept and leaves every slope in \(\boldsymbol{\beta}\) unchanged. If the sampling probabilities vary with a predictor, however, their ratio can contribute a predictor-dependent term, and the simple intercept-only shift no longer follows.

Worked Example: A Per-Allele Slope Under Unequal Sampling

Suppose a hypothetical source population follows a logistic model with intercept \(\alpha=-2\) and per-allele slope \(\beta=\log(1.7)\). Researchers select 60% of cases and 3% of controls, with selection independent of genotype within each outcome group. Find the selected-sample intercept and per-allele odds ratio.

The case-to-control sampling-probability ratio is \(s_1/s_0=0.60/0.03=20\). The selected-sample intercept is therefore

\alpha^*=\alpha+\log\left(\frac{s_1}{s_0}\right) =-2+\log(20) =-2+2.9957 =0.9957.

The selected-sample slope remains \(\beta=\log(1.7)\), so the per-allele odds ratio is \(\exp(\beta)=1.7\). Thus, each additional allele is associated with 1.7 times the odds of case status in the selected sample, under this model and sampling design. The intercept has changed because cases were sampled at a higher rate; it should not be interpreted as the source-population intercept.

Checking the Result with Counts

For a binary genotype contrast, the odds ratio can also be computed directly from a two-by-two table. If sampling multiplies the number of cases by one common factor and the number of controls by another common factor, those factors cancel in the cross-product odds ratio. The following example makes that cancellation explicit.

Worked Example: Source and Sample Odds Ratios

Consider a hypothetical source population with the following counts. Let \(G=0\) denote the reference genotype group and \(G=1\) the comparison group. Suppose 50% of cases and 10% of controls are selected, with the same sampling fraction for both genotype groups within each outcome.

Genotype group	Source cases	Source controls	Selected cases	Selected controls
\(G=0\)	100	900	50	90
\(G=1\)	150	850	75	85

In the source population, the odds ratio comparing \(G=1\) with \(G=0\) is the case-to-control odds for \(G=1\), divided by the case-to-control odds for \(G=0\):

OR_{\text{source}} = \frac{150/850}{100/900} = \frac{150\times900}{850\times100} = 1.588.

In the selected sample, the corresponding calculation is

OR_{\text{sample}} = \frac{75/85}{50/90} = \frac{75\times90}{85\times50} = 1.588.

The odds ratios agree. The case sampling fraction multiplies both genotype groups’ case counts by 0.50, and the control sampling fraction multiplies both groups’ control counts by 0.10. Neither changes the relative case-to-control odds between genotype groups. The counts are hypothetical and illustrate the sampling identity; actual estimates also have sampling uncertainty.

What the Result Does—and Does Not—Justify

For the specified logistic model, fitting the case-control sample estimates the same conditional genotype slope as the source-population model, subject to the usual sampling uncertainty and model assumptions. The slope can be exponentiated to obtain a genotype odds ratio. For an additive dosage \(G\), \(\exp(\beta)\) is the odds ratio for a one-allele increase; for a binary genotype contrast, it compares the odds in the coded group with those in the reference group.

The intercept is different. It absorbs the ratio of the case and control sampling probabilities, so it reflects the constructed sample as well as the source population. A case-control sample’s fraction of participants who are cases is generally not the source population’s case prevalence. Consequently, the fitted intercept cannot ordinarily be used as if it were the source-population intercept to obtain population probabilities. Recovering population absolute risks requires additional information about the source population and the sampling design; the slope-preservation result alone does not provide it.

This distinction also clarifies the scope of the conclusion. It concerns odds ratios and logistic slopes, not every measure of association. In particular, preserving an odds ratio does not establish that a risk difference or an absolute risk is preserved. Nor does it establish a causal effect of genotype: it is a statement about a conditional statistical association under a model and sampling scheme.

Worked Example: When Genotype-Dependent Sampling Breaks Preservation

Use the hypothetical source counts from the previous example, where the source odds ratio is 1.588. Now suppose controls are still selected at 10% in both genotype groups, but cases are selected at 50% for \(G=0\) and 80% for \(G=1\). Find the resulting sampled odds ratio.

The selected counts are 50 cases and 90 controls for \(G=0\), and \(150(0.80)=120\) cases and \(850(0.10)=85\) controls for \(G=1\). Therefore,

OR_{\text{sample}} = \frac{120/85}{50/90} = \frac{120\times90}{85\times50} = 2.541.

The sampled odds ratio, 2.541, is larger than the source odds ratio, 1.588. The reason is that cases with \(G=1\) were more likely to be selected than cases with \(G=0\), so selection no longer applies a common multiplier to case counts across genotype groups. In odds-ratio terms, the sample odds ratio is multiplied by \(0.80/0.50=1.6\): \(1.588\times1.6=2.541\). The preservation result does not apply to this sampling scheme.

Common Mistakes and Practical Checks

Claiming that case-control sampling leaves every coefficient unchanged. The guaranteed shift is in the intercept; genotype slopes are preserved only under the stated within-outcome sampling condition.
Assuming that any case-control design meets the condition. Check whether selection within cases and within controls is independent of genotype. Genotype-dependent recruitment or participation can change the slope.
Interpreting the fitted intercept as a population baseline. The intercept includes the case-to-control sampling ratio. A sample with an investigator-chosen case fraction does not generally reveal population prevalence.
Equating a preserved odds ratio with a preserved risk difference or causal effect. The result is specifically about conditional odds and logistic slopes; it does not by itself supply population risks or establish causation.

A useful check is to write the selected odds as the source odds multiplied by \(s_1/s_0\). If that ratio is constant across genotype values, its logarithm is an intercept shift. If it varies with genotype, it can change the slope, so the simple preservation result cannot be invoked.

Check Your Understanding

Answer each question using the sampling assumptions and model described above.

If cases are selected at 40% and controls at 5%, what is the shift from the source intercept to the selected-sample intercept?
Under the preservation condition, a per-allele slope is \(\log(1.4)\). What per-allele odds ratio does the selected-sample model imply?
Why does selecting more cases than controls not, by itself, change a genotype odds ratio?
What feature of the sampling design fails if cases with one genotype are more likely to be recruited than cases with another genotype?
Why should the intercept from a case-control sample not automatically be used to report source-population absolute risks?