The Intercept Shift Under Case-Control Sampling

Why the Intercept Changes

Case-control studies deliberately select cases and controls in proportions that need not match those in the source population. This design can preserve the logistic relationship between covariates and outcome, but the observed sample is not a miniature version of the source population: the relative number of cases has been altered by sampling. The intercept records that altered baseline odds.

The distinction is important when interpreting predictions. A logistic model fitted to a case-control sample can describe how the odds vary with covariates while assigning probabilities that reflect the sample’s case-control selection, not the source population’s risks. The amount of the intercept shift is determined by the case and control sampling fractions.

Definition: Let \(Y\in\{0,1\}\) be the outcome, \(X\) the covariates, and \(S=1\) indicate inclusion in the analyzed sample. The case sampling fraction is \(s_1=P(S=1\mid Y=1)\), and the control sampling fraction is \(s_0=P(S=1\mid Y=0)\). For a source population, suppose \(P(Y=1\mid X=x)=p(x)\) follows the logistic model \(\operatorname{logit}p(x)=\alpha+x^\mathsf{T}\beta\). The key sampling condition is \(P(S=1\mid Y=y,X=x)=s_y\): within each outcome group, selection does not additionally depend on \(X\). Assume \(s_1,s_0>0\), so both outcomes can occur in the sampled data.

The condition on selection is stronger than merely saying that cases and controls are sampled at different rates. It says that, among cases, covariate values do not affect inclusion; the same is true among controls. This is the outcome-only selection setting considered here. It yields a shift that is constant across covariate values.

Deriving the Sampled-Data Intercept

Write \(q(x)=P(Y=1\mid X=x,S=1)\) for the outcome probability among sampled individuals with covariates \(x\). To derive it, compare the selected case and control probabilities at the same covariate value. Selection multiplies the source probability of a case by \(s_1\), and that of a control by \(s_0\).

Theorem: Under the definition and sampling condition above, the sampled conditional odds satisfy \(\operatorname{odds}\{q(x)\}=(s_1/s_0)\operatorname{odds}\{p(x)\}\). Consequently, \[ \operatorname{logit}q(x)=\left\{\alpha+\log(s_1/s_0)\right\}+x^\mathsf{T}\beta. \] Thus the sampled logistic model has the same slopes \(\beta\) as the source model and an intercept shifted by \(\log(s_1/s_0)\).

Proof. For any covariate value \(x\) for which the conditional probabilities are defined and \(P(S=1\mid X=x)>0\), Bayes’ rule and the sampling condition give

q(x)=P(Y=1\mid X=x,S=1) =\frac{P(S=1\mid Y=1,X=x)P(Y=1\mid X=x)} {P(S=1\mid X=x)} =\frac{s_1p(x)}{s_1p(x)+s_0\{1-p(x)\}}.

Because \(s_1,s_0>0\), the sampled probabilities for both outcomes are positive whenever the corresponding source probabilities are positive. Dividing the expression for \(q(x)\) by \(1-q(x)\) yields

\frac{q(x)}{1-q(x)} =\frac{s_1p(x)}{s_0\{1-p(x)\}} =\frac{s_1}{s_0}\exp\{\alpha+x^\mathsf{T}\beta\}.

Taking logarithms gives \(\operatorname{logit}q(x)=\alpha+\log(s_1/s_0)+x^\mathsf{T}\beta\). The added term contains no \(x\), so it changes the intercept and not any covariate slope. \(\square\)

The same calculation also gives the sampled probability directly:

q(x)=\frac{s_1p(x)}{s_1p(x)+s_0\{1-p(x)\}} =\operatorname{expit}\!\left(\alpha+\log(s_1/s_0)+x^\mathsf{T}\beta\right), \qquad \operatorname{expit}(t)=\frac{e^t}{1+e^t}.

These equations express two different but equivalent views of the effect. Sampling multiplies the odds by \(s_1/s_0\), while it adds \(\log(s_1/s_0)\) to the log-odds. It does not multiply the probability by that ratio.

Worked Examples

Worked Example: Oversampling Cases Eightfold Relative to Controls

Suppose the source model is \(\operatorname{logit}p(x)=-3+0.8x\). Cases are sampled with fraction \(s_1=0.8\), and controls with fraction \(s_0=0.1\). Find the sampled intercept and compare source and sampled probabilities at \(x=0\) and \(x=1\).

The sampling-fraction ratio is \(s_1/s_0=0.8/0.1=8\), so the intercept shift is \(\log 8\approx2.07944\). The sampled model is therefore

\operatorname{logit}q(x)=(-3+\log 8)+0.8x\approx-0.92056+0.8x.

At \(x=0\), the source probability is \(p(0)=e^{-3}/(1+e^{-3})\approx0.04743\). Its source odds are \(e^{-3}\approx0.04979\); after multiplying by 8, the sampled odds are about \(0.39830\). Hence \(q(0)=0.39830/(1+0.39830)\approx0.28485\).

At \(x=1\), the source linear predictor is \(-3+0.8=-2.2\), so \(p(1)=e^{-2.2}/(1+e^{-2.2})\approx0.09975\). The source odds \(e^{-2.2}\approx0.11080\) become sampled odds \(8e^{-2.2}\approx0.88642\), giving \(q(1)=0.88642/(1+0.88642)\approx0.46990\). The probabilities rise substantially in the sampled data, although the covariate slope remains \(0.8\).

Worked Example: Sampling Fewer Cases Than Controls

Let the source model be \(\operatorname{logit}p(x)=-1.5+0.7x\). Suppose \(s_1=0.1\) and \(s_0=0.5\). Find the sampled model, then compare probabilities at \(x=0\) and \(x=2\).

Here \(s_1/s_0=0.1/0.5=0.2\), and \(\log(0.2)\approx-1.60944\). Thus the sampled intercept is \(-1.5-1.60944=-3.10944\), and

\operatorname{logit}q(x)\approx-3.10944+0.7x.

At \(x=0\), \(p(0)=e^{-1.5}/(1+e^{-1.5})\approx0.18243\). The source odds are \(e^{-1.5}\approx0.22313\); the sampled odds are \(0.2(0.22313)\approx0.04463\). Therefore \(q(0)=0.04463/(1+0.04463)\approx0.04272\).

At \(x=2\), the source linear predictor is \(-1.5+1.4=-0.1\), giving \(p(2)=e^{-0.1}/(1+e^{-0.1})\approx0.47502\). The source odds \(e^{-0.1}\approx0.90484\) become \(0.2e^{-0.1}\approx0.18097\), so \(q(2)=0.18097/(1+0.18097)\approx0.15324\). The shift is negative because cases were sampled at a lower fraction than controls.

Worked Example: Reading the Shift From Population Counts

A source population has 1,000 cases and 9,000 controls. A study selects 200 cases and 450 controls. With no covariates, calculate the two sampling fractions, the intercept shift, and the case probability in the selected sample.

The case fraction is \(s_1=200/1000=0.2\); the control fraction is \(s_0=450/9000=0.05\). Their ratio is \(s_1/s_0=4\), so the intercept shift is \(\log 4\approx1.38629\).

The source case probability is \(p=1000/(1000+9000)=0.1\), with odds \(0.1/0.9=1/9\). The source intercept is therefore \(\log(1/9)\approx-2.19722\). The sampled odds are \(4(1/9)=4/9\), and the sampled case probability is

q=\frac{4/9}{1+4/9}=\frac{4}{13}\approx0.30769.

This agrees with the selected counts: \(200/(200+450)=200/650\approx0.30769\). The sampled intercept is \(-2.19722+1.38629\approx-0.81093\), which is also \(\log(4/9)\).

What the Shift Does—and Does Not—Say

A constant intercept shift has a precise implication for comparisons across covariate values. For two covariate vectors \(x_a\) and \(x_b\), the sampled log-odds difference is

\operatorname{logit}q(x_a)-\operatorname{logit}q(x_b) =(x_a-x_b)^\mathsf{T}\beta =\operatorname{logit}p(x_a)-\operatorname{logit}p(x_b).

The shift cancels in the difference. In contrast, probabilities generally do not differ by a constant amount: the logistic transformation is nonlinear, and the same odds multiplier can produce different probability changes at different starting risks. Nor does slope preservation mean that the sampled model’s raw fitted probabilities can automatically be read as source-population risks.

The derivation concerns the conditional model implied by the sampling design. A finite-sample fitted intercept is an estimate and will vary from sample to sample; it need not equal the population intercept shift exactly. The relationship also relies on selection being independent of covariates within outcome groups. If control selection, for example, varies with \(X\), the ratio of selection probabilities can depend on \(x\), and the shift need not be constant.

Common Mistakes

Adding the wrong log-ratio. The shift is \(\log(s_1/s_0)\), not \(\log(s_0/s_1)\). Oversampling cases relative to controls must raise sampled case odds, so the shift is positive when \(s_1>s_0\).
Applying the ratio to probabilities. The ratio \(s_1/s_0\) multiplies odds. Compute \(q=s_1p/\{s_1p+s_0(1-p)\}\), or multiply the odds and convert back; do not set \(q=(s_1/s_0)p\).
Confusing the sample case proportion with the fraction ratio. In the count example, the case proportion in the selected sample is \(200/650\approx0.30769\), whereas the sampling-fraction ratio is \(0.2/0.05=4\). They are different quantities. The ratio of fractions multiplies source odds; it is not itself a probability.
Assuming the result holds under any sampling design. If selection depends on covariates even after conditioning on the outcome, the selection ratio may vary with \(x\), so a single intercept shift is not justified by this derivation.

Key takeaway: When inclusion depends on outcome but not on covariates within outcome groups, the sampled conditional logit equals the source logit plus \(\log(s_1/s_0)\). Unequal case and control sampling shifts the intercept, not the slopes, and generally changes the fitted probabilities.

Check Your Understanding

Work through these questions using the definitions and derivation above.

If \(s_1=0.3\) and \(s_0=0.1\), what is the intercept shift, and should it raise or lower sampled case odds?
For source probability \(p=0.2\), calculate the sampled probability when \(s_1/s_0=3\).
Why does the covariate slope remain unchanged under the stated sampling condition?
A study has 100 selected cases and 300 selected controls. Can their ratio alone determine \(s_1/s_0\)? Explain what additional information is needed.
What feature of the selection mechanism could make the intercept shift depend on \(x\)?