Rare-Event Logit Under Case-Control Sampling

Why Case-Control Sampling Changes Logit Predictions

When an outcome is rare, a sample that reflects its population prevalence may contain very few cases. A case-control design addresses this by deliberately sampling more cases relative to controls. This can make data collection practical, but it changes the outcome frequency in the sample. Consequently, a logit model fitted to that sample generally has an intercept that does not directly give population probabilities.

Under a specific sampling mechanism, the change is especially structured: the logit intercept shifts, while the slope coefficients—and therefore the model's conditional odds ratios—are preserved. If the population prevalence and the sample case fraction are known, the intercept can be corrected and population probabilities recovered. These conclusions depend on selection being based on outcome status alone, not on covariates within outcome groups.

Definition: Let $Y$ be a binary outcome, with $Y=1$ denoting a case, and let $X$ denote observed covariates. In the target population, suppose $$ P(Y=1\mid X=x)=\operatorname{expit}(\alpha+x^\top\beta), \qquad \operatorname{expit}(z)=\frac{\exp(z)}{1+\exp(z)}. $$ Case-control sampling has the required outcome-only form when the probability of inclusion is $s_1$ for cases and $s_0$ for controls, with both probabilities constant across $x$. In particular, selection does not further depend on covariates once outcome status is fixed.

The sample's case fraction, written $q=P_s(Y=1)$, need not equal the target population prevalence $\tau=P(Y=1)$. Here $P_s$ refers to probabilities in the sampled data. The sampling mechanism changes the marginal outcome frequency, and it also changes the conditional outcome odds at each covariate value by the same multiplicative factor. This constant factor is why the logit intercept changes but the slopes do not.

Deriving the Intercept Shift

For covariate value $x$, write $p(x)=P(Y=1\mid X=x)$ for the population probability and $p_s(x)=P_s(Y=1\mid X=x)$ for the sampled conditional probability. Among sampled observations with covariates $x$, the relative representation of cases and controls is multiplied by $s_1$ and $s_0$, respectively. Thus the sampled conditional odds are

\frac{p_s(x)}{1-p_s(x)} = \frac{s_1}{s_0}\frac{p(x)}{1-p(x)}.

Taking logarithms and substituting the population logit model gives the key result:

\begin{aligned} \operatorname{logit}\{p_s(x)\} &=\log\!\left(\frac{p_s(x)}{1-p_s(x)}\right)\\ &=\log\!\left(\frac{s_1}{s_0}\right) +\log\!\left(\frac{p(x)}{1-p(x)}\right)\\ &=\left\{\alpha+\log\!\left(\frac{s_1}{s_0}\right)\right\}+x^\top\beta. \end{aligned}

Theorem: Under outcome-only sampling with inclusion probabilities $s_1$ and $s_0$, the sampled logit model has slopes $\beta$ and intercept $\alpha_s=\alpha+\log(s_1/s_0)$. Therefore, the population intercept is $\alpha=\alpha_s-\log(s_1/s_0)$. The conditional odds ratio for a one-unit increase in covariate $x_j$, holding other covariates fixed, remains $\exp(\beta_j)$ in the sampled and target populations.

The odds-ratio result follows because the added term $\log(s_1/s_0)$ is constant in $x$: it cancels when comparing the conditional odds at two covariate values. This is a result about conditional odds ratios, not a claim that probabilities or risk ratios are preserved. In particular, applying the sample intercept directly usually gives incorrect population probabilities.

Using the Sample Case Fraction and Population Prevalence

Often $s_1$ and $s_0$ are not individually known, but the sample case fraction $q$ and population prevalence $\tau$ are available. The odds of being a case in the sample equal the population case odds multiplied by the relative inclusion probability:

\frac{q}{1-q} = \frac{s_1}{s_0}\frac{\tau}{1-\tau}.

Consequently, the relative sampling factor and the corrected intercept can be calculated as follows. The value $q$ is the actual case fraction in the case-control sample; in a design fixing the numbers of cases and controls, it is set by design.

\frac{s_1}{s_0} = \frac{q/(1-q)}{\tau/(1-\tau)}, \qquad \alpha = \alpha_s+\log\!\left(\frac{\tau}{1-\tau}\right) -\log\!\left(\frac{q}{1-q}\right).

Worked Example: Correcting an Intercept With a Known Prevalence

Hypothetical data. Suppose cases make up 50% of a case-control sample, while the target population prevalence is 2%. A fitted sample logit has intercept $\hat{\alpha}_s=-1.00$ and slope $\hat{\beta}=0.70$ for a covariate $x$. Assume inclusion depends only on case status, the logit specification is appropriate, and the 2% prevalence applies to the target population.

First calculate the relative sampling factor:

\widehat{\frac{s_1}{s_0}} = \frac{0.50/(1-0.50)}{0.02/(1-0.02)} = \frac{1}{0.02/0.98} = 49.

The correction subtracts its logarithm from the sample intercept:

\hat{\alpha} = -1.00-\log(49) = -1.00-3.8918 = -4.8918.

At $x=0$, the population probability estimate is $\operatorname{expit}(-4.8918)\approx0.00745$, or about 0.745%. At $x=1$, the corrected linear predictor is $-4.8918+0.70=-4.1918$, giving $\operatorname{expit}(-4.1918)\approx0.0149$, or about 1.49%. The sample-model probabilities at those covariate values would instead be $\operatorname{expit}(-1.00)\approx0.269$ and $\operatorname{expit}(-0.30)\approx0.426$. The slope and its conditional odds ratio, $\exp(0.70)\approx2.01$, are unchanged; the large probability difference reflects the altered sample case fraction.

What Is Preserved—and What Must Be Recalibrated

For a one-unit increase in $x_j$, holding the other regressors fixed, the population conditional odds are multiplied by $\exp(\beta_j)$. Since the case-control sampling factor is the same at both values of $x_j$, this conditional odds ratio is also the one represented by the sampled slope. But the probability change associated with that covariate depends on the intercept and on the covariate values. It is therefore not generally preserved by case-control sampling. A probability prediction, marginal effect, or classification threshold calculated from the uncorrected sample intercept should not be read as a population quantity.

This distinction is useful in practice. A case-control sample can support estimation of conditional odds-ratio relationships under the stated assumptions, while estimation of absolute population risk needs additional information, such as the target prevalence. The prevalence must describe the population to which predictions will apply; substituting the sample case fraction for it simply leaves the sampling distortion in place.

Worked Example: Corrected Probabilities and a Preserved Odds Ratio

Hypothetical data. A sample has a case fraction of $q=0.40$, and the target population prevalence is $\tau=0.01$. A sample logit model is $\operatorname{logit}\{p_s(x)\}=-1.50+1.20x$. Assume outcome-only sampling, an appropriate logit specification, and a prevalence applicable to the target population.

The sample case odds are $0.40/0.60=2/3$, and the population case odds are $0.01/0.99=1/99$. Their ratio is

\frac{s_1}{s_0} = \frac{2/3}{1/99} = 66, \qquad \log(66)=4.1897.

Thus the corrected population intercept is $\hat{\alpha}=-1.50-4.1897=-5.6897$, while the estimated slope remains $1.20$. At $x=1$, the sampled predictor is $-1.50+1.20=-0.30$, so the sampled probability is $\operatorname{expit}(-0.30)\approx0.4256$. The corrected population predictor is $-5.6897+1.20=-4.4897$, giving $\operatorname{expit}(-4.4897)\approx0.0111$, or about 1.11%.

The conditional odds ratio for a one-unit increase in $x$ is $\exp(1.20)\approx3.320$ in both models. That means the conditional odds are multiplied by about 3.320, not that the probability is multiplied by 3.320. The probability at $x=1$ is much lower after correction because the population prevalence is 1%, not 40%.

Assumptions and Limits of the Correction

The intercept correction is exact for the specified sampling mechanism and model, not a general repair for any unrepresentative sample. Its main assumptions are:

Selection depends on outcome status alone. Within cases and within controls, inclusion must not vary with the covariates in a way that changes their conditional distributions.
The conditional logit form is suitable. The model must represent the relationship of interest, including any needed nonlinear terms or interactions.
The prevalence and sample fraction are defined consistently. The target prevalence must refer to the population and period for which probabilities are sought, and $q$ must describe the case-control sample used to fit the model.
The covariates are measured comparably. Selection or measurement procedures should not introduce additional distortions to the covariate-outcome relationship.

If selection among cases depends on $x$, for example, the relative representation of cases versus controls can vary across $x$. The sampling factor is then no longer a constant multiplier of the odds, so a single intercept adjustment need not suffice and slope estimates may also be affected. Likewise, knowing prevalence alone does not fix misspecification or selection bias.

Worked Example: A Selection Rule That Breaks the Simple Correction

Hypothetical design. Suppose investigators sample every eligible control, but among cases they sample with probability 0.80 when $x=1$ and 0.20 when $x=0$. Assume the population logit model is correctly specified. This design does not satisfy outcome-only sampling, because case inclusion varies with $x$.

At $x=1$, the sampling multiplier on the population odds is $0.80/1=0.80$. At $x=0$, it is $0.20/1=0.20$. The sampled log-odds contrast between $x=1$ and $x=0$ therefore differs from the population contrast by

\log(0.80)-\log(0.20)=\log(4)\approx1.386.

Thus a fitted sampled slope for a binary $x$ is shifted by about 1.386 on the log-odds scale, even though the population model itself is correct. A single intercept correction cannot remove this difference: the sample odds multiplier is not constant across $x$. The correction formula should not be used for this design without a model that accounts for the covariate-dependent selection.

Inference and a Common Pitfall

If the sampling factor is known and treated as fixed, subtracting its log changes the intercept estimate but not the slope estimates. The uncertainty in a corrected intercept still comes from estimating the model. If the prevalence is itself estimated rather than known, uncertainty in that prevalence also contributes to uncertainty in the corrected intercept and probabilities; treating an estimated prevalence as fixed can understate uncertainty. Any reported population probability should identify the target population and the prevalence information used for calibration.

Common mistake: “The sample logit gives a 42.6% probability at $x=1$, so about 42.6% of comparable people in the population will be cases.” This treats the case-control sample as if its case fraction reflected population prevalence. In the second worked example, the sample probability at $x=1$ is about 42.6%, but the prevalence-corrected population probability is about 1.11%. The correction is justified only under the stated sampling assumptions and with suitable prevalence information.

Key takeaway: Under outcome-only case-control sampling, the sample logit intercept equals the population intercept plus the log relative inclusion probability for cases versus controls. Population prevalence and sample case fraction identify that shift; the slope odds ratios are preserved, but probabilities require the corrected intercept.

Check Your Understanding

Use the outcome-only sampling assumptions unless a question states otherwise.

A target population has prevalence $\tau=0.02$, and cases are 40% of the sample. What is the implied relative inclusion factor $s_1/s_0$?
Explain why the conditional odds ratio associated with a logit slope is preserved under outcome-only case-control sampling.
A sample logit has intercept $-2.00$, and $s_1/s_0=10$. What is the corrected population intercept?
Why can a correct sample slope still produce incorrect population probability predictions?
Give an example of a selection rule that would invalidate a single intercept correction, and explain which assumption it violates.