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Case-Control and Sampling Designs · Tutorial 978 of 981

When The Rare-Disease Approximation Applies

Learn how source-population risks determine the gap between odds ratios and risk ratios, and how to assess whether the rare-disease approximation is adequate.

Advanced 10 min read

What You'll Learn

  • Define source-population risks, risk ratios, and odds ratios for a binary exposure
  • Derive the exact relationship between an odds ratio and a risk ratio
  • Bound the approximation error when both risks are small
  • Calculate how the approximation behaves for rare and common outcomes
  • Explain why a case-control sample’s case fraction does not establish that disease is rare

Why the Approximation Needs a Condition

In case-control studies, the odds ratio is often the measure of association directly estimated by logistic regression. It is sometimes described as a risk ratio when disease is rare. That description has a precise condition: the source-population risks being compared must be small. The fraction of cases in the sampled data is not the relevant quantity, because case-control sampling sets that fraction by design.

The distinction matters in practice. An odds ratio and a risk ratio are different measures, and for common outcomes they can differ substantially even when estimated without bias. The exact relationship lets us check the approximation rather than rely on a rule of thumb.

Definitions and the Exact Relationship

Consider a binary exposure, with \(E=0\) denoting the unexposed group and \(E=1\) the exposed group, and a binary disease outcome measured over a specified follow-up period. Define the source-population risks by \(p_0=P(Y=1\mid E=0)\) and \(p_1=P(Y=1\mid E=1)\). These are risks in the population that generated the cases and controls, not proportions calculated from the case-control sample.

Definition: Provided \(0<p_0,p_1<1\), the risk ratio is \(RR=p_1/p_0\). The odds ratio is \(OR=\{p_1/(1-p_1)\}/\{p_0/(1-p_0)\}\). Risk is the probability of disease; odds are the probability of disease divided by the probability of no disease.

The odds of disease and the risk of disease are close when the risk is small: for example, a risk of \(0.01\) corresponds to odds \(0.01/0.99\), not exactly \(0.01\). Applying this distinction to both exposure groups gives the following exact identity.

Theorem: For \(0<p_0,p_1<1\),
$$ OR=RR\frac{1-p_0}{1-p_1}. $$

To prove it, substitute \(p_1/p_0\) for \(RR\) and simplify:

$$ OR =\frac{p_1}{1-p_1}\frac{1-p_0}{p_0} =\frac{p_1}{p_0}\frac{1-p_0}{1-p_1} =RR\frac{1-p_0}{1-p_1}. $$

No rare-disease assumption is used in this identity. It shows exactly why the measures can differ: the odds ratio includes the factor \((1-p_0)/(1-p_1)\), which is close to one only when both risks are small. \(\square\)

When the Rare-Disease Approximation Applies

The identity gives an exact measure of relative discrepancy:

$$ \frac{OR}{RR}-1 =\frac{1-p_0}{1-p_1}-1 =\frac{p_1-p_0}{1-p_1}. $$

Thus the signed relative discrepancy of \(OR\) from \(RR\) depends on the risk difference and on the exposed-group risk. If \(p_1>p_0\), then \(OR>RR\). If \(p_1<p_0\), then \(OR<RR\). If the two risks are equal, both ratios are one.

Result: If \(0<p_0,p_1\leq\delta<1\), then
$$ \left|\frac{OR}{RR}-1\right| =\frac{|p_1-p_0|}{1-p_1} \leq\frac{\delta}{1-\delta}. $$

For the inequality, \(|p_1-p_0|\leq\max(p_0,p_1)\leq\delta\), and \(1-p_1\geq1-\delta>0\). Dividing the first bound by the second gives the stated result. As \(\delta\) decreases toward zero, \(\delta/(1-\delta)\) also decreases toward zero. Therefore, along any sequence of pairs of positive risks for which both risks tend to zero, the relative discrepancy tends to zero. This establishes the approximation \(OR\approx RR\) under the rare-risk condition, rather than merely asserting it. \(\square\)

The bound is conservative: it describes a worst case over all pairs of risks at most \(\delta\). It is not a guarantee that the error equals that bound. For a particular study, the exact discrepancy can be calculated from the risks if they are known.

Worked Example: A Rare Outcome With a Small Discrepancy

Suppose the unexposed risk is \(p_0=0.01\) and the exposed risk is \(p_1=0.02\). The risk ratio is

$$ RR=\frac{0.02}{0.01}=2. $$

The odds ratio is

$$ OR=\frac{0.02/(1-0.02)}{0.01/(1-0.01)} =\frac{0.02/0.98}{0.01/0.99} =\frac{0.02\times0.99}{0.98\times0.01} =\frac{0.0198}{0.0098} \approx2.0204. $$

The relative discrepancy is \((2.0204/2)-1\approx0.0102\), or about \(1.02\%\). The exact identity gives the same check: \((p_1-p_0)/(1-p_1)=0.01/0.98\approx0.0102\). Here the rare-disease approximation is close.

Common Outcomes: Numerical Failures of the Approximation

When risks are common, the factors \(1-p_0\) and \(1-p_1\) need not be close to one. The odds ratio can then be substantially farther from one than the risk ratio for a harmful exposure, or closer to one than the risk ratio for a protective exposure.

Worked Example: A Common Outcome With a Harmful Exposure

Let \(p_0=0.20\) and \(p_1=0.40\). The risk ratio is still \(2\), but the odds ratio is

$$ OR=\frac{0.40/0.60}{0.20/0.80} =\frac{2/3}{1/4} =\frac{8}{3} \approx2.667. $$

Thus the odds ratio is about \(33.3\%\) larger than the risk ratio: \((2.667/2)-1\approx0.333\). The exact discrepancy formula verifies this: \((0.40-0.20)/(1-0.40)=0.20/0.60=1/3\). Calling this odds ratio “approximately a risk ratio” would conceal a substantial difference.

Worked Example: A Common Outcome With a Protective Exposure

Suppose \(p_0=0.30\) and \(p_1=0.15\). Then

$$ RR=\frac{0.15}{0.30}=0.50, \qquad OR=\frac{0.15/0.85}{0.30/0.70} =\frac{0.15\times0.70}{0.85\times0.30} =\frac{0.105}{0.255} \approx0.412. $$

Here \(OR/RR\approx0.824\), so the odds ratio is about \(17.6\%\) below the risk ratio. The exact relative discrepancy is \((0.15-0.30)/(1-0.15)=-0.15/0.85\approx-0.176\). The direction agrees with the general result: for a protective exposure, \(p_1<p_0\), so \(OR<RR\).

Why the Baseline Risk Matters

An odds ratio alone does not determine a risk ratio. To see why, solve the exact relationship for \(p_1\), given \(p_0\) and an odds ratio \(\theta\):

$$ \theta=\frac{p_1/(1-p_1)}{p_0/(1-p_0)} \quad\Longrightarrow\quad p_1=\frac{\theta p_0}{1-p_0+\theta p_0}. $$

The resulting risk ratio \(p_1/p_0\) therefore depends on the baseline risk \(p_0\), not just on \(\theta\). This is why an odds ratio of a given size cannot automatically be reported as the same-sized risk ratio.

Worked Example: The Same Odds Ratio, Different Risk Ratios

Take an odds ratio of \(2\). When the unexposed risk is \(p_0=0.10\), the exposed risk is

$$ p_1=\frac{2(0.10)}{1-0.10+2(0.10)} =\frac{0.20}{1.10} \approx0.1818, \qquad RR=\frac{0.1818}{0.10}\approx1.818. $$

When \(p_0=0.50\), the same calculation gives

$$ p_1=\frac{2(0.50)}{1-0.50+2(0.50)} =\frac{1.00}{1.50} =\frac{2}{3}, \qquad RR=\frac{2/3}{1/2}=\frac{4}{3}\approx1.333. $$

In both cases the odds ratio is exactly \(2\), but the risk ratios are about \(1.818\) and \(1.333\). The approximation is closer in the first case because the risks are smaller.

Using the Approximation in Case-Control Analysis

Under the usual case-control sampling setup, cases and controls are sampled at different rates from their source-population outcome groups. The earlier material on retrospective logistic likelihoods and outcome-dependent sampling establishes why the logistic slope can identify an odds ratio under the relevant sampling assumptions. It does not turn that odds ratio into a risk ratio. The rare-disease approximation is a separate step that concerns the source-population risks.

In particular, a case-control sample might contain half cases and half controls even when disease is rare in the source population. That sample fraction reflects the sampling design; it is not an estimate of \(p_0\) or \(p_1\). Assessing the approximation requires information about the risks in the source population, or a justified basis for treating them as small.

For covariate-specific comparisons, apply the identity within the covariate pattern being compared: use the corresponding exposed and unexposed source-population risks. A statement that risks are small in one stratum does not establish that they are small in every stratum. Nor should an odds ratio be relabeled a risk ratio merely because a fitted logistic model reports an odds ratio.

Common Mistake and Practical Check

  • Wrong reasoning: “Cases are rare in my case-control data, so the odds ratio is approximately a risk ratio.”
  • Correction: Case-control sampling determines the sample case fraction. The approximation depends on the disease risks in the source population, not the sampled fraction.
  • Wrong reasoning: “The outcome is binary, so odds and risks are interchangeable.”
  • Correction: They are different quantities. Their ratio is \((1-p_0)/(1-p_1)\); check the risks before treating that factor as close to one.

A useful practice is to report the measure actually estimated as an odds ratio and explain the rare-risk condition if interpreting it as a risk ratio. When source-population risks are not small, report the odds ratio as such; do not silently substitute a risk-ratio interpretation.

Key takeaway: The exact relation is \(OR=RR(1-p_0)/(1-p_1)\). The odds ratio approximates the risk ratio when both source-population risks are small, not merely when the case-control sample contains few cases.

Check Your Understanding

Use the exact relationship to assess each claim.

  1. If \(p_0=0.03\) and \(p_1=0.06\), calculate the risk ratio and odds ratio. How large is the relative discrepancy?
  2. For a harmful exposure with \(p_1>p_0\), which is larger, the odds ratio or the risk ratio? Explain using the exact identity.
  3. Why does a sample with equal numbers of cases and controls provide no evidence by itself that the source-population disease risk is one half?
  4. An odds ratio equals \(2\), and the unexposed risk is \(0.25\). Use the formula for \(p_1\) to find the exposed risk and the corresponding risk ratio.
  5. State the condition under which the rare-disease approximation is justified, and explain why “the outcome is uncommon in the sample” is not an equivalent condition.
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