Why Nested Case-Control Sampling Is Event-Specific
A prospective cohort may contain many people whose covariates must be measured repeatedly or expensively. A nested case-control design reduces that burden: when a cohort member experiences the event of interest, controls are sampled from the people still at risk at that event time. The resulting data are therefore organized around event-specific risk sets, not around one permanent division of the cohort into “cases” and “controls.”
This distinction matters for both analysis and interpretation. A person sampled as a control at one event time may be sampled again later, and may subsequently become a case. The relevant probability for design-based weighting is the probability that the person is included as a control in a particular risk set. It is generally not the probability that the person appears anywhere in the study sample.
This convention makes the sampling unit a subject-time pair: subject \(i\) as a potential control at event time \(t_j\). Write \(A_j\) for the eligible control pool and \(R_j=|A_j|\) for its size. Let \(S_j\) be the sampled controls from that pool. If a risk-set definition or sampling protocol instead permits the case to be sampled as a control for another event at the same time, that protocol requires its own explicit eligibility rules and inclusion probabilities.
Inclusion Probabilities Within a Risk Set
Suppose \(m_j\) controls are selected by simple random sampling without replacement from the \(R_j\) eligible controls. Each eligible person then has the same probability of inclusion in that event-specific sample.
To verify the probability, there are \(\binom{R_j}{m_j}\) equally likely samples. Of these, \(\binom{R_j-1}{m_j-1}\) contain a specified subject \(i\). Thus
The formula applies to the eligible control pool, not automatically to a risk set that includes the case. For example, if a risk set contains 20 people including the case and 4 controls are sampled from the other 19 people, the control inclusion probability is \(4/19\), not \(4/20\).
Worked Example: A Uniform Risk-Set Sample
At an event time, 12 cohort members are at risk, including the person who experiences the event. The study samples 3 controls from the other 11 people by simple random sampling without replacement. Calculate the control inclusion probability, weight, and an estimated total for a covariate.
There are \(R_j=11\) eligible controls, and \(m_j=3\). For each eligible subject, \(\pi_{ij}=3/11\), so the inverse-probability weight is \(w_{ij}=11/3\).
Suppose the covariate is an indicator for prior exposure, and the three sampled controls have values \(1,0,1\). Their weighted total is \((11/3)(1+0+1)=22/3\). This is an estimate of the number exposed among all 11 eligible controls, not a claim that the actual finite-population count is \(22/3\). The estimate may be fractional because it is a weighted estimate.
If the case’s exposure value is needed in a total over the full risk set, add that known case contribution separately. It is not part of the sampled-control total in this example.
Why Inverse-Probability Weights Recover Risk-Set Totals
Let \(g(X_{ij})\) be a fixed quantity for subject \(i\) at time \(t_j\), such as an exposure indicator or a covariate value. The finite-population total of this quantity over the eligible control pool is \(T_j=\sum_{i\in A_j}g(X_{ij})\). If the inclusion probability \(\pi_{ij}\) is known and positive, the Horvitz–Thompson estimator of that total is \(\widehat T_j=\sum_{i\in S_j}g(X_{ij})/\pi_{ij}\).
For a proof, let \(I_{ij}\) equal 1 if subject \(i\) is sampled as a control for event \(j\), and 0 otherwise. Then \(E[I_{ij}]=\pi_{ij}\), and
By linearity of expectation, with the covariate values fixed under the design,
This proof does not require the inclusion indicators to be independent. In simple random sampling without replacement, for example, they are dependent because the sample size is fixed; the expectation argument still holds.
Worked Example: Unequal Probabilities and Weighted Totals
An event-specific control pool contains six people: two exposed and four unexposed. A design samples controls independently with known inclusion probabilities of \(0.5\) for each exposed person and \(0.25\) for each unexposed person. In one realized sample, one exposed and one unexposed person are selected. Estimate the total number exposed in the pool.
For the exposure indicator \(g\), the selected exposed person contributes \(1/0.5=2\) to the Horvitz–Thompson estimate. The selected unexposed person contributes \(0/0.25=0\). Therefore \(\widehat T_j=2\), the estimate of the total number exposed.
The design’s expected sample size is \(2(0.5)+4(0.25)=2\). Its probabilities are not those of a fixed-size simple random sample; using one common weight for both exposure groups would ignore the stated design. The inverse probability for each selected subject must correspond to that subject’s own probability of inclusion in this event-specific pool.
Repeated Selection and Subject-Time Weights
A control’s weight is attached to a particular risk-set contribution. Suppose a person is eligible at two event times, and at each time a separate simple random sample is drawn. If that person’s inclusion probabilities are \(1/4\) and \(1/2\), the corresponding event-specific weights are 4 and 2. They are not replaced by one weight based on the probability of appearing at least once.
For example, if the two samples are independent, the probability of appearing at least once is \(1-(1-1/4)(1-1/2)=5/8\). Its reciprocal, \(8/5\), is not the correct weight for either event-specific total: the two totals concern different risk sets and have inclusion probabilities \(1/4\) and \(1/2\), respectively. A person selected as a control earlier and later observed as a case likewise contributes as a sampled control to the earlier risk set and as the case to the later event.
Worked Example: A Person Selected at More Than One Event Time
A subject is eligible as a control for two events. At the first event, 2 controls are sampled from 8 eligible people; at the second, 2 are sampled from 4 eligible people. Assume separate samples are drawn independently and the subject remains eligible at both times. Find the subject’s event-specific weights and probability of being selected at least once.
At the first event the inclusion probability is \(2/8=1/4\), giving weight 4. At the second it is \(2/4=1/2\), giving weight 2.
The probability of missing the first sample is \(3/4\); the probability of missing the second is \(1/2\). By independence, the probability of missing both is \((3/4)(1/2)=3/8\). Therefore the probability of selection at least once is \(1-3/8=5/8\).
The reciprocal \(8/5\) of this last probability is not a substitute for the two event-specific weights. It describes inclusion in the union of two samples, whereas each risk-set total has its own pool, inclusion event, and weight.
Using Weights Without Overclaiming
Inverse-probability weights provide a principled way to estimate finite-cohort quantities that are sums over risk-set members. They can also be used in estimating procedures designed for the sampling structure, provided the relevant inclusion probabilities are known and the estimating procedure accounts for how the cases and controls were selected. When sampling probabilities depend on covariates, the weights must reflect that dependence.
A weighted total is design-unbiased under the conditions stated above. A ratio of two estimated totals, a nonlinear function of estimated risk-set totals, or a regression coefficient does not automatically inherit that exact unbiasedness. In particular, plugging estimated weighted totals into a nonlinear likelihood does not by itself prove that the resulting estimator is unbiased or consistent. The model, sampling design, and estimating procedure must be considered together.
Nested case-control analyses often use the matched risk sets directly to estimate relative effects. Under uniform risk-set sampling, the conditional comparison of each case with its sampled controls can preserve information about relative rates. That does not mean that a conditional comparison estimates every cohort quantity, such as an absolute risk or a full-population covariate total. Design weights are useful when the target involves the underlying cohort, but they do not turn every weighted regression into a valid design-based analysis automatically.
Common Mistake
A frequent error is to assign a control a single weight equal to the inverse probability that the person was ever selected in the study. This confuses a subject’s inclusion in the union of all sampled sets with inclusion in a particular event-specific risk set. The correction is to define the target quantity first, identify the risk set to which it refers, and use the inclusion probability for that subject in that risk set. A second error is to use \(m_j\) divided by the risk-set size when the case is excluded from control sampling; the denominator must be the eligible control-pool size \(R_j\), not a count that includes the case.
Check Your Understanding
Use the event-specific sampling pool and the target quantity to guide each answer.
- A risk set has 15 people including the case, and 4 controls are sampled from the remaining people by simple random sampling without replacement. What are the control inclusion probability and inverse-probability weight?
- State the conditions under which the Horvitz–Thompson estimator of a risk-set total is design-unbiased.
- A person has inclusion probability \(0.2\) at one event time and \(0.5\) at another. What weight applies to that person’s contribution to each event-specific total?
- Why is the reciprocal of the probability of being selected at least once generally not the correct weight for each event-specific risk-set total?
- Does design-unbiased estimation of risk-set totals imply that every weighted nonlinear regression estimator is unbiased? Explain.