Case-Control Sampling And Outcome Prevalence

Why Case-Control Sampling Changes the Event Proportion

Suppose an insurer wants to study a relatively uncommon event, such as a hypothetical high-cost claim within a specified coverage period. A simple random sample from the eligible population may contain too few events to support a useful analysis. A case-control design addresses this by selecting records according to outcome status: it samples cases, records with the event, and controls, records without it. This can make the dataset useful for studying differences between the groups, but it also changes the proportion of cases in the sample.

The distinction to keep in view is between prevalence in the source population and prevalence in the selected dataset. In a case-control dataset, the observed case proportion is often determined partly or entirely by the sampling plan. It should not automatically be read as the event prevalence in the population from which the records came.

Definition: Let \(Y=1\) indicate that an eligible record has the event during the defined outcome period, and let \(Y=0\) indicate that it does not. The source-population prevalence is \(p=P(Y=1)\). Let \(R=1\) indicate that a record is selected. Write \(s_1=P(R=1\mid Y=1)\) for the case inclusion probability and \(s_0=P(R=1\mid Y=0)\) for the control inclusion probability. The case probability among selected records is \(q=P(Y=1\mid R=1)\).

These definitions assume the outcome is defined consistently for the eligible population, including the same outcome horizon. The inclusion probabilities describe the sampling process, not the event itself. In case-control sampling, \(s_1\) and \(s_0\) are commonly different by design. The sample therefore has a different case mix from the source population.

Relating Source and Sample Prevalence

For a record in the selected dataset, the probability of being a case depends on both its chance of being a case in the source population and its chance of being selected if it is a case. Bayes' rule gives the relationship below.

q=P(Y=1\mid R=1) =\frac{P(R=1\mid Y=1)P(Y=1)}{P(R=1)} =\frac{p s_1}{p s_1+(1-p)s_0}.

Here \(p s_1\) is the joint probability that a record is a case and is selected. The denominator adds that probability to the joint probability of being a noncase and selected, \((1-p)s_0\). Dividing the selected-case probability by the probability of selection gives the case probability among selected records.

Important interpretation: The expression \(p s_1/[p s_1+(1-p)s_0]\) is the case probability for a randomly chosen selected record under the stated sampling model. It is also the ratio of expected selected case counts to expected total selected counts. It is not, in general, the exact expected realized sample proportion when records are independently selected and the selected group sizes are random. In a large sample it describes the limiting sample fraction; under fixed case and control quotas, the realized sample fraction is exactly the quota fraction.

The same relationship can be written in odds form. Provided the probabilities are strictly between zero and one, divide the sample case probability by the sample noncase probability:

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

Thus, outcome-dependent sampling multiplies the source-population odds by the ratio of case to control inclusion probabilities. If cases are sampled more heavily than controls, \(s_1/s_0>1\), and the selected records have higher case odds than the source population. This relation does not mean that the sample case proportion is a direct estimate of \(p\).

Worked Examples

Worked Example: Interpreting Random Outcome-Based Selection

A hypothetical eligible population has 12,000 records, of which 2.5% have a specified event. Records are independently selected with probability 0.80 if they are cases and 0.05 if they are controls. Find the case probability among selected records and interpret it carefully.

Method. Use the conditional-probability formula for the outcome distribution of selected records. First calculate the source counts:

Cases: \(12{,}000(0.025)=300\). Controls: \(12{,}000(0.975)=11{,}700\).

The expected selected case count is \(300(0.80)=240\), and the expected selected control count is \(11{,}700(0.05)=585\). The ratio of these expected counts is

\frac{240}{240+585} =\frac{240}{825} =0.2909.

Equivalently, using probabilities,

q=\frac{(0.025)(0.80)} {(0.025)(0.80)+(0.975)(0.05)} =\frac{0.020}{0.020+0.04875} =\frac{0.020}{0.06875} =0.2909.

The selected-record case probability is about 0.2909, even though source prevalence is 0.0250. The expected selected counts provide the ratio used in the calculation; because the actual counts are random, this ratio is not generally the exact expectation of the realized sample case proportion. Common mistake avoided: reporting 29.09% as the source-population event prevalence. The source prevalence remains 2.5%.

Worked Example: Reading a Fixed Case-Control Sample

A hypothetical study selects exactly 90 cases and 810 controls from an eligible population. The source population is known from a separate complete registry to have event prevalence 1.5%. What is the sample case proportion, and can it be used as the source prevalence?

Method. Under fixed group quotas, calculate the proportion directly from the selected group sizes:

\text{Sample case proportion} =\frac{90}{90+810} =\frac{90}{900} =0.10.

The sample contains 10% cases by construction. The source prevalence is 1.5%, as given by the registry, so the sample proportion is not an estimate of that prevalence. The quota design deliberately included cases at a much higher rate than they occur in the source population.

As a check, the source population would have 15 cases per 1,000 records at 1.5% prevalence, whereas the sample has 100 cases per 1,000 sampled records. The two figures describe different populations: all eligible records versus selected records. Common mistake avoided: substituting the observed 10% sample proportion for the known 1.5% population prevalence.

Worked Example: Recovering a Prevalence Estimate With Known Inclusion Probabilities

A hypothetical case-control dataset contains 100 cases and 300 controls. Records were independently selected with known inclusion probabilities of 0.60 for cases and 0.06 for controls. Use inverse-probability weighting to estimate source prevalence. Explain the role of the observed sample proportion.

Method. Give each sampled record a weight equal to the reciprocal of its inclusion probability. A case receives weight \(1/0.60=1.6667\); a control receives weight \(1/0.06=16.6667\). The weighted case total is

100\left(\frac{1}{0.60}\right)=166.6667.

The weighted control total is

300\left(\frac{1}{0.06}\right)=5{,}000.

The inverse-probability-weighted prevalence estimate is the weighted case total divided by the total weighted count:

\widehat{p} =\frac{166.6667}{166.6667+5{,}000} =\frac{166.6667}{5{,}166.6667} \approx 0.0323.

To check the direction, cases were ten times as likely to be selected as controls, so the unweighted sample case proportion \(100/400=0.25\) should be much higher than the weighted estimate. The weighting adjusts for the known difference in inclusion probabilities; it does not make this finite-sample estimate exact. Common mistake avoided: treating the raw 25% as source prevalence, or weighting cases and controls in the wrong direction. The less frequently sampled controls receive the larger weight.

Why the Formula Holds, and When the Sample Fraction Is Fixed

The formula follows directly from conditional probability. Among all records in the source population, the joint probability of being both a case and selected is \(P(Y=1,R=1)=p s_1\). The joint probability of being a selected control is \(P(Y=0,R=1)=(1-p)s_0\). These are disjoint ways to be selected, so

$$ P(R=1)=p s_1+(1-p)s_0. $$

The conditional probability that a selected record is a case is the joint probability of being a selected case divided by the probability of being selected. Substitution gives \(q=p s_1/[p s_1+(1-p)s_0]\). This derivation concerns the distribution of a selected record; it does not assert that a random realized sample ratio equals a ratio of expected counts.

There are two common designs to distinguish:

Fixed case and control quotas. If the design takes exactly \(n_1\) cases and \(n_0\) controls, the observed sample proportion is exactly \(n_1/(n_1+n_0)\). It is fixed by design, not discovered from the source population.
Random selection within outcome groups. If each record is selected with a probability depending on outcome, the selected case and control counts can vary. The formula gives the case probability for a randomly chosen selected record and the ratio of expected selected counts. For large samples, the realized proportion will typically be close to that value, but it need not equal it in every sample.

The formula assumes the inclusion probabilities used are appropriate for the target population and are specified by outcome status as described. If selection also depends on other characteristics, a single pair of probabilities \(s_1,s_0\) may not adequately represent the design. The analyst must understand the actual sampling process before interpreting a sample proportion or applying weights.

Practical Interpretation and Exam Traps

Case-control sampling can be useful when events are uncommon because it ensures that the dataset contains enough cases for analysis. Its cost is that the case fraction no longer reflects the source prevalence. A reported sample proportion may describe the study design rather than the event frequency in the eligible population.

Before interpreting any case proportion, identify what was fixed or randomized:

If the study fixed the number of cases and controls, calculate the sample proportion from those quotas and do not call it source prevalence.
If outcome-specific inclusion probabilities are known, use them to relate the source and selected distributions; use appropriate weighting when estimating source quantities.
If the sampling design or inclusion probabilities are unknown, the observed sample proportion alone does not determine source prevalence.
Keep the outcome definition and eligible population aligned. A prevalence for one outcome horizon or eligibility rule is not automatically the prevalence for another.

A frequent exam error is to compute the fraction of cases in a case-control dataset and report it as the probability of the event in the population. Another is to call the ratio of expected selected counts the exact expected sample fraction under random inclusion. The first confuses the selected and source populations; the second confuses a ratio of expectations with an expectation of a ratio. State the design, define the target quantity, and use the appropriate interpretation.

Key takeaway: Outcome-based sampling changes the selected records' event prevalence. A case-control sample proportion is a description of the sample or its design, not automatically a source-population prevalence estimate. The relation \(q=p s_1/[p s_1+(1-p)s_0]\) makes the effect of differing inclusion probabilities explicit.

Check Your Understanding

For each problem, distinguish the source prevalence from the selected-record case probability or the fixed sample proportion.

A source population has event prevalence 4%. Cases are selected with probability 0.50 and controls with probability 0.10. Find the case probability among selected records under the stated outcome-based sampling model. (Answer: \(q=0.1724\), or 17.24%.)
A study fixes its sample at 75 cases and 225 controls. What is the exact sample case proportion? Does that number by itself estimate source prevalence? (Answer: 0.25, or 25%; no, the proportion is fixed by the quotas and does not by itself estimate source prevalence.)
Source prevalence is 8%; case inclusion probability is 0.25 and control inclusion probability is 0.05. Calculate the case probability among selected records. (Answer: \(q=0.3030\), or 30.30%.)
The case probability among selected records is 0.40. Case inclusion probability is 0.80 and control inclusion probability is 0.20. Use the odds relationship to recover source prevalence. (Answer: \(p=1/7\approx0.1429\), or 14.29%.)
A random sample contains 40 cases and 160 controls. Their inclusion probabilities are 0.50 and 0.02, respectively. Calculate the inverse-probability-weighted prevalence estimate. (Answer: \(\frac{40/0.50}{40/0.50+160/0.02}=\frac{80}{8{,}080}\approx0.0099\), or 0.99%.)

What You'll Learn