From Dropout to a Mean Trajectory
With repeated measurements, monotone dropout means that after a participant misses a visit, no later outcomes are observed. The observed sample can therefore change from one visit to the next. If dropout depends on recorded history, the mean among participants still observed may not represent the population mean at that visit. Inverse probability of censoring weighting (IPCW) addresses this by giving more weight to observed participants whose histories made continued observation less likely.
The calculation is visit-specific. A participant observed at visit \(k\) must have remained under observation through every earlier interval, so the relevant weight is the inverse of the cumulative probability of that continued observation. We use these weights to estimate the mean at each visit separately; the resulting sequence is the estimated mean trajectory.
Only participants with \(R_{ik}=1\) contribute an outcome at visit \(k\). Their weights use the product of conditional retention probabilities through that visit—not just the probability of being observed at the final interval. The denominator normalizes the weights among those observed, making this a ratio estimator of a mean.
Worked Example: Calculating a Three-Visit Trajectory
Worked Example: Calculating a Three-Visit Trajectory
Suppose eight participants belong to baseline groups A and B, with four participants in each. All eight have a baseline outcome. At visit 1, three in A and two in B remain observed. At visit 2, two of the three in A and one of the two in B remain observed. Assume the fitted conditional retention probabilities are \(p_1(A)=0.75\), \(p_1(B)=0.50\), \(p_2(A)=2/3\), and \(p_2(B)=0.50\). The observed outcomes are:
| Group | Baseline outcomes | Observed visit 1 outcomes | Observed visit 2 outcomes |
|---|---|---|---|
| A | 2, 4, 6, 8 | 4, 6, 8 | 5, 9 |
| B | 4, 6, 8, 10 | 10, 12 | 15 |
Baseline. Everyone is observed and has weight 1. The mean is
Visit 1. The weights are \(1/0.75=4/3\) for observed participants in A and \(1/0.50=2\) in B. The weighted numerator is \((4/3)(4+6+8)+2(10+12)=24+44=68\). The weighted denominator is \(3(4/3)+2(2)=4+4=8\). Thus
Visit 2. The cumulative retention probabilities are \((0.75)(2/3)=0.50\) for A and \((0.50)(0.50)=0.25\) for B. The cumulative weights are therefore 2 and 4, respectively. The weighted numerator is \(2(5+9)+4(15)=28+60=88\); the denominator is \(2(2)+1(4)=8\). Hence
The estimated trajectory is \(6,\ 8.5,\ 11\). These are separate weighted estimates at each visit. They are not the mean of a fixed set of complete cases: participants can contribute at earlier visits and drop out before later ones.
Why the Weight Recovers a Population Mean
The calculation relies on assumptions about why observation ends. The first is monotone observation: once dropout occurs, later outcomes are not observed. The second is sequential independent censoring: conditional on \(H_{j-1}\), remaining observed at visit \(j\) is independent of the future full-data outcomes relevant to the target. In particular, participants with the same recorded history must not have different retention probabilities because of an unrecorded factor that also predicts their future outcomes. The third is positivity: \(p_j(H_{j-1})>0\) for histories that occur in the target population. Otherwise, some histories have no chance of being observed at the next visit and their outcomes cannot be recovered by weighting.
For the following identity, express sequential independent censoring as the corresponding full-data condition: conditional on the full outcome and covariate data, the probability of remaining observed through visit \(k\) is \(G_k\), where the retention probabilities in \(G_k\) are evaluated at the histories that would be recorded under full follow-up. This condition, together with positivity, makes the inverse weight cancel the observation probability.
Proof. Conditional on the full data, \(f\) and \(G_k\) are fixed, and \(E[R_k\mid\text{full data}]=G_k\). Therefore
Taking expectations of both sides gives the identity. Setting \(f=Y_k\) gives the outcome identity, and setting \(f=1\) gives the denominator identity. Integrability and positivity ensure the displayed quantities are defined. \(\square\)
For estimation, suppose participants are independent and identically distributed, the censoring probabilities are known, \(E[|Y_k|]<\infty\), and \(E[R_k/G_k]=1\). The weak law of large numbers applied to the numerator and denominator then gives the following result.
Proof. Write the sample numerator and denominator as averages by dividing each sum by \(n\). By the weak law of large numbers and the weighted-mean identity,
Since the denominator converges in probability to 1, it is nonzero with probability approaching 1. The continuous mapping theorem for the ratio therefore yields \(\widehat{\mu}_k\xrightarrow{p}E[Y_k]\). \(\square\)
In practice, retention probabilities are usually estimated rather than known. The same large-sample argument can be applied when the fitted probabilities converge to the correct probabilities and the estimated weights behave regularly—for example, with finitely many histories, consistent probability estimates, and probabilities bounded away from zero. If fitted probabilities are wrong, the cancellation in the identity need not hold.
Worked Example: History-Specific Retention at the Final Visit
Worked Example: History-Specific Retention at the Final Visit
At a final visit, suppose the four observed participants have two possible prior histories. For history A, the conditional retention probabilities at visits 1 and 2 are \(0.80\) and \(0.50\); the observed outcomes are 2, 4, and 6. For history B, they are \(0.50\) and \(0.20\); the observed outcome is 14. Calculate the weighted mean and compare it with the unweighted observed mean.
For A, \(G_2=(0.80)(0.50)=0.40\), so \(W_2=1/0.40=2.5\). For B, \(G_2=(0.50)(0.20)=0.10\), so \(W_2=1/0.10=10\). The weighted numerator is \(2.5(2+4+6)+10(14)=30+140=170\), and the denominator is \(3(2.5)+10=17.5\). Thus
The unweighted observed mean is \((2+4+6+14)/4=6.5\). The difference arises because the sole observed participant with history B represents a history with a lower estimated probability of remaining observed. Whether the weighted value is accurate still depends on the censoring assumptions and the probability model; the calculation alone does not verify them.
Common Pitfalls, Diagnostics, and Alternatives
- Using only the last-interval probability. The visit-2 weight must include the probability of being observed at visit 1 as well. For example, if the interval probabilities are 0.8 and 0.5, the cumulative probability is 0.4 and the weight is 2.5, not 2.
- Using a single unweighted observed mean. This targets the mean among those observed at that visit, which need not equal the population mean when observation depends on outcome-related history.
- Assuming a large weight proves bias correction. A large weight may reflect low estimated retention, but it does not establish that the model is correct or that all relevant predictors of dropout were measured.
- Ignoring positivity. If a history has zero retention probability, its contribution cannot be reconstructed from observed outcomes. A finite weight in the software does not fix a true zero probability.
As a numerical check, confirm that the cumulative probability for each participant is the product of the interval probabilities along that participant’s observed history, and that the denominator includes the weights of the same observed participants as the numerator. Examine whether fitted retention probabilities are implausibly close to zero and whether the resulting estimates are driven by very few observations. Such checks can reveal problems, but they do not prove the assumptions hold; they are not substitutes for the more detailed weight diagnostics considered separately.
Complete-case analysis uses only participants observed at every visit. It is simple, but can discard substantial information and is generally biased for population means if complete observation depends on outcome-related history. Outcome regression instead models the outcome among observed participants and predicts or averages over the target population; its validity depends on that outcome model and the assumptions needed for prediction. IPCW models observation rather than the outcome and reweights observed outcomes. These approaches rely on different models, and neither weighting nor regression automatically protects against unmeasured predictors of dropout. Their relative performance depends on the data and model assumptions.
Worked Example: A Near-Zero Retention Probability
At one visit, three participants are observed with outcomes 2, 4, and 20. Their estimated cumulative retention probabilities are 0.80, 0.80, and 0.10. Calculate the weighted mean, then show what happens if the last weight is capped at 4.
The uncapped weights are \(1.25,\ 1.25,\ 10\). The weighted numerator is \(1.25(2)+1.25(4)+10(20)=2.5+5+200=207.5\), and the denominator is \(1.25+1.25+10=12.5\). The weighted mean is \(207.5/12.5=16.6\).
If the weight 10 is capped at 4, the numerator becomes \(2.5+5+4(20)=87.5\), and the denominator becomes \(1.25+1.25+4=6.5\). The capped estimate is \(87.5/6.5\approx13.46\). Capping reduces the influence of the participant with low estimated retention, but changes the weighting scheme and, in general, changes the target or introduces bias. It is not a remedy for structural positivity failure.
Check Your Understanding
Answer each question, then check the solutions below.
- A participant’s conditional retention probabilities through visit 3 are 0.9, 0.8, and 0.5. What is the cumulative retention probability and the inverse censoring weight at visit 3?
- At a visit, two observed outcomes are 3 and 9, with weights 2 and 4. Calculate the Hájek weighted mean.
- Why does monotone dropout require multiplying retention probabilities across intervals?
- What does a zero retention probability for a target-population history imply for IPCW?
Solutions. (1) The cumulative probability is \((0.9)(0.8)(0.5)=0.36\), so the weight is \(1/0.36=25/9\approx2.78\). (2) The weighted numerator is \(2(3)+4(9)=42\); the denominator is \(2+4=6\), giving \(42/6=7\). (3) Observation at the visit requires remaining observed through every prior interval, so the probability of continued observation is the product of the conditional probabilities along that history; its inverse corrects for that cumulative selection. (4) No participant with that history can be observed at the visit, so the corresponding outcome information is absent. The inverse probability is undefined, and the target mean for that history is not recoverable by IPCW without additional assumptions or a changed target population.