Why Weight Observed Outcomes at a Fixed Horizon?
Suppose a study enrolls a population at baseline and plans to measure an outcome at time \(K\). Some participants drop out before that measurement. The mean among participants who remain observed is generally not the population mean: remaining observed may depend on characteristics associated with the outcome. Inverse-probability-of-censoring weighting (IPCW) addresses this selection by giving more weight to observed outcomes from participants whose histories made continued observation less likely.
The target here is the marginal mean of the outcome at the fixed horizon in the baseline population, \(E[Y(K)]\). This is a mean under the population’s actual course of follow-up and covariate histories; IPCW for censoring alone does not define a treatment contrast or a treatment intervention.
A participant censored before \(K\) contributes zero to the numerator, not an imputed outcome. An observed participant contributes their outcome multiplied by the inverse of their estimated probability of being observed through the horizon. The denominator of this Horvitz–Thompson estimator is the baseline sample size \(n\).
Identification: Why the Weight Recovers the Marginal Mean
The argument requires a censoring model that conditions on enough history to account for informative dropout. Write the full data as the trajectory that would be available without censoring, including \(Y(K)\) and any covariates needed to describe the censoring process. Sequential independent censoring means that, among participants uncensored at \(t-1\), the censoring decision at \(t\) is independent of the relevant full-data future conditional on the history \(H(t-1)\). In particular, given the full trajectory, the probability of remaining uncensored at \(t\) is \(q_t\{H(t-1)\}\). This condition allows the censoring probability to vary with observed history; it does not require censoring to be marginally independent of the outcome.
The other central condition is positivity: each history with positive probability in the target population must have a positive probability of remaining uncensored through the horizon. For stable estimation, a stronger practical condition is often imposed: \(G(K)\ge\epsilon\) for some \(\epsilon>0\). The identification argument itself can allow probabilities arbitrarily close to zero if the expectation exists, but such cases can make estimation highly variable.
Proof. Let \(\mathcal F\) denote a participant’s full-data trajectory, including \(Y(K)\) and the histories determining the censoring probabilities. By the chain rule for conditional probabilities and sequential independent censoring, $$ P\{R(K)=1\mid\mathcal F\} =\prod_{t=1}^{K}P\{R(t)=1\mid R(t-1)=1,\mathcal F\} =\prod_{t=1}^{K}q_t\{H(t-1)\} =G(K). $$ The histories in this expression are the full-data values along the trajectory; for a participant observed through \(K\), they are the histories used to compute the weight. Since \(Y(K)\) and \(G(K)\) are determined by \(\mathcal F\), conditioning on \(\mathcal F\) gives $$ E\left[\left.\frac{R(K)Y(K)}{G(K)}\right|\mathcal F\right] =\frac{Y(K)}{G(K)}P\{R(K)=1\mid\mathcal F\} =Y(K). $$ Taking expectations on both sides proves the identity. Integrability follows from \(E[|Y(K)|]<\infty\), because the absolute conditional expectation of the weighted quantity is \(|Y(K)|\). \(\square\)
Worked Calculations
Worked Example: One Censoring Decision and Two History Groups
At a single follow-up time, a baseline history \(L\) has two equally frequent groups in a sample of eight. The estimated probability of remaining observed is \(0.8\) in group \(L=0\) and \(0.5\) in group \(L=1\). Two participants in each group are observed at the horizon. Their outcomes are \(2,4\) for \(L=0\) and \(6,10\) for \(L=1\). Calculate the IPCW estimate.
The inverse weights are \(1/0.8=1.25\) for \(L=0\) and \(1/0.5=2\) for \(L=1\). The weighted outcome total is $$ (2+4)(1.25)+(6+10)(2)=7.5+32=39.5. $$ The denominator is all eight baseline participants, so $$ \widehat{\mu}_{\mathrm{IPCW}}(K)=39.5/8=4.9375. $$ For comparison, the mean among the four observed participants is \((2+4+6+10)/4=5.5\). The difference illustrates that complete-case averaging and IPCW are different procedures. This one realized estimate need not equal the population mean exactly; the identification result is an expectation statement under its assumptions.
Worked Example: Multiplying Probabilities Across Follow-Up
There are four baseline participants and two follow-up decisions. For participants A and B, the conditional retention probabilities are \(0.9\) and \(0.8\), so \(G(K)=0.9(0.8)=0.72\). For C and D, they are \(0.5\) and \(0.6\), so \(G(K)=0.5(0.6)=0.30\). A, B, and C are observed through \(K\), with outcomes \(10,14,20\); D is censored before \(K\). Calculate the estimate.
The observed weighted contributions are $$ \frac{10}{0.72}=13.8889,\qquad \frac{14}{0.72}=19.4444,\qquad \frac{20}{0.30}=66.6667. $$ D contributes zero because \(R_D(K)=0\). Therefore $$ \widehat{\mu}_{\mathrm{IPCW}}(K) =\frac{13.8889+19.4444+66.6667}{4} =25.0000. $$ The complete-case mean is \((10+14+20)/3=14.6667\). C’s large contribution reflects the low estimated probability of remaining observed along C’s history. That contribution is mathematically prescribed by the weights, but it also signals potential sensitivity to the censoring model and to individual observations.
Worked Example: Horvitz–Thompson Versus Normalized IPCW
In a sample of five, three participants are observed at the horizon. Their estimated \(G(K)\) values are \(0.5,0.5,0.25\), and their outcomes are \(4,8,12\). The other two are censored before \(K\). Calculate the Horvitz–Thompson estimate and the normalized, or Hájek, version.
The inverse weights are \(2,2,4\), and the weighted outcome total is $$ 2(4)+2(8)+4(12)=8+16+48=72. $$ Dividing by the baseline sample size gives the Horvitz–Thompson estimate \(72/5=14.4\). The normalized estimate divides by the sum of the observed weights: $$ \widehat{\mu}_{\mathrm{H}}(K) =\frac{72}{2+2+4} =\frac{72}{8} =9. $$ Normalization can reduce sensitivity to the total weight in a particular sample, but it changes the finite-sample estimator: it is not the same estimator as \(n^{-1}\sum_i R_i(K)Y_i(K)/\widehat G_i(K)\). Under suitable conditions both can be consistent for the mean, but the normalization does not repair confounding by unmeasured predictors of dropout, a wrong censoring model, or severe positivity violations.
Estimated Weights and Consistency
In applications, the \(q_t\) values are usually estimated from observed follow-up data, for example by modeling the discrete-time censoring hazards. The resulting \(\widehat G_i(K)\) is the product of the estimated conditional retention probabilities along participant \(i\)’s observed history. Using estimated rather than known probabilities adds an estimation step to the argument.
Proof. First consider the estimator using the true probabilities. The summands \(R_i(K)Y_i(K)/G_i(K)\) are independent and identically distributed, and their expectation is \(E[Y(K)]\) by the identification theorem. Since \(Y(K)\) is bounded and \(G(K)\ge\epsilon\), their absolute values are bounded by \(\sup|Y(K)|/\epsilon\). The law of large numbers therefore gives $$ \frac{1}{n}\sum_{i=1}^{n}\frac{R_i(K)Y_i(K)}{G_i(K)} \xrightarrow{p}E[Y(K)]. $$ It remains to show that estimating \(G\) does not change the limit. On the event that both \(G_i(K)\) and \(\widehat G_i(K)\) are at least \(\epsilon/2\), $$ \left|\frac{R_iY_i}{\widehat G_i}-\frac{R_iY_i}{G_i}\right| \le |Y_i|\frac{|\widehat G_i-G_i|}{\widehat G_iG_i} \le \sup|Y(K)|\frac{2}{\epsilon^2}|\widehat G_i-G_i|. $$ Averaging over \(i\), the absolute difference between the estimated-weight and true-weight averages is at most \(\sup|Y(K)|(2/\epsilon^2)\sup_i|\widehat G_i(K)-G_i(K)|\), which converges in probability to zero. The event used in the bound has probability tending to one by the stated lower-bound condition. Combining this result with the convergence of the true-weight average proves the claim. \(\square\)
This is a sufficient consistency result, not a guarantee that any fitted censoring model works. In practice, consistency requires the estimated retention probabilities to represent the relevant conditional probabilities well enough. The uniform convergence condition stated here is one transparent way to make the plug-in argument; more specialized results can use other regularity conditions.
Diagnostics, Alternatives, and Limits
Inspect the estimated retention probabilities and the resulting weights before interpreting the estimate. Useful checks include the distribution and maximum of \(1/\widehat G_i(K)\), the number of participants observed at each time, and whether fitted retention probabilities approach zero in particular histories. Compare weighted and unweighted summaries of measured baseline or follow-up history among observed participants. Large weights concentrated in a few people, very small fitted probabilities, or little overlap between observed histories can indicate practical positivity problems. These checks can reveal instability; they cannot establish that all predictors of dropout have been measured or that the censoring model is correct.
The main alternative is a complete-case mean, which averages only outcomes observed through \(K\). It is generally not a population mean when observation depends on outcome-related history. Outcome regression offers a different route: model the horizon outcome using observed history and use that model to predict outcomes for the target population. IPCW instead models observation probabilities and reweights the observed outcomes. Each approach relies on assumptions about its modeled process; neither is automatically protected from misspecification. Combining models may be useful, but is outside this fixed-horizon derivation.
IPCW also has clear limits. If some target histories have zero probability of remaining observed, the target mean is not identified by these weights. If probabilities are positive but tiny, the estimator can be highly variable. Weight truncation can reduce extreme contributions, but generally changes the estimator and may introduce bias; it does not restore the original identification argument. Finally, IPCW for censoring alone does not adjust for treatment-confounding, intermittent missingness, or a competing event that changes the estimand. Those situations require an analysis matched to their own structure.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- At a fixed horizon, why does the Horvitz–Thompson IPCW estimator divide by the baseline sample size rather than the number of observed outcomes?
- A participant’s retention probabilities at two follow-up decisions are \(0.8\) and \(0.5\). What is the probability of remaining observed through both decisions, and what is the inverse weight?
- Name two assumptions needed for the identification identity and explain what each contributes.
- Suppose estimated retention probabilities are close to zero for a few histories. What problem can this create, and can truncation prove the original assumptions hold?
Solutions. (1) The Horvitz–Thompson form represents the baseline target population: each baseline participant contributes an inverse-probability weighted outcome, with zero contribution if censored. Its denominator is therefore \(n\); dividing by the number observed would instead normalize among observed participants. (2) The product is \(0.8(0.5)=0.4\), so the inverse weight is \(1/0.4=2.5\). (3) Sequential independent censoring makes the probability of observation through the horizon equal to the modeled product conditional on history, allowing weighting to recover the full-data mean. Positivity ensures that histories in the target population can be represented among those observed. (4) Small probabilities produce large weights and potentially high variance and sensitivity to model errors. Truncation may reduce instability but does not prove positivity or correct specification, and can introduce bias.