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Censoring and Informative Dropout · Tutorial 215 of 501

One-Period Derivation of Inverse Censoring Weights

See why weighting each observed outcome by the inverse probability of remaining observed corrects selection in a single follow-up period.

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What You'll Learn

  • Define the single-period observation probability from measured history.
  • Derive the inverse censoring weight from conditional independence and positivity.
  • Distinguish correcting selection across history strata from changing outcomes within a stratum.
  • Calculate weighted and complete-case means in explicit examples.
  • Diagnose near-violations of positivity and understand what weighting cannot repair.

Why a Single-Period Weight Is Needed

Suppose a study aims to describe an outcome \(Y\) for an eligible population, but \(Y\) is recorded only for people who remain uncensored through one follow-up period. If remaining observed is related to measured characteristics \(H\), the observed cases can have a different mix of those characteristics than the target population. When \(H\) also predicts \(Y\), the observed-case outcome distribution can therefore be misleading.

Inverse probability of censoring weighting corrects this selection by giving each observed person weight inversely proportional to their probability of being observed. The derivation below is for one follow-up period only. It requires that the measured history capture the relevant dependence between observation and outcome, and that every history value in the target population have a positive chance of being observed.

Definition: Let \(R=1\) indicate that a person’s outcome is observed at the follow-up, and \(R=0\) indicate censoring before that outcome is observed. Let \(H\) denote measured history available for all eligible people before the observation decision. Define the observation probability \(g(H)=P(R=1\mid H)\). The single-period inverse probability of censoring weight for an observed person is \(1/g(H)\); their weighted contribution is \(R/g(H)\), which is zero when \(R=0\).

The outcome \(Y\) is understood to be defined for the target population, whether or not it was recorded. For example, it could be the outcome that would have been measured at the scheduled follow-up had the person not been censored. This is a missing-outcome setting: if the outcome is not meaningfully defined for some people, the target estimand must be specified before weighting can address anything.

Assumptions and the Weighting Identity

Two assumptions give the weight its correction property. First, conditional independent censoring requires \(R\) and \(Y\) to be conditionally independent given \(H\). In words, among people with the same measured history, observation does not select people according to their outcome. Second, positivity requires \(g(H)>0\) with probability one in the target population. To make the weighted expectations below well-defined, also assume \(Y\) is integrable, meaning \(E[|Y|]<\infty\).

Theorem: Suppose \(R\) and \(Y\) are conditionally independent given \(H\), \(g(H)=P(R=1\mid H)>0\) almost surely, and \(Y\) is integrable. Then $$ E\!\left[\frac{R Y}{g(H)}\right]=E[Y]. $$ More generally, for any integrable function \(f(H,Y)\), $$ E\!\left[\frac{R f(H,Y)}{g(H)}\right]=E[f(H,Y)]. $$

Proof. Conditional independence gives $$ E[R f(H,Y)\mid H]=E[R\mid H]E[f(H,Y)\mid H] =g(H)E[f(H,Y)\mid H]. $$ Because \(g(H)>0\), division by \(g(H)\) yields $$ E\!\left[\frac{R f(H,Y)}{g(H)}\middle|H\right]=E[f(H,Y)\mid H]. $$ Taking expectations of both sides gives the claimed identity by the law of iterated expectations. The weighted quantity is integrable: applying the same argument to \(|f(H,Y)|\) gives \(E[R|f(H,Y)|/g(H)]=E[|f(H,Y)|]<\infty\). Thus the conditional expectation and subsequent expectation are well-defined. Setting \(f(H,Y)=Y\) gives the first identity. \(\square\)

This identity also establishes a distributional result, not just a mean correction. Define the weighted probability of an outcome event \(A\) by dividing its expected weighted observed count by the expected total weight. Since \(E[R/g(H)]=1\), applying the identity to \(f(H,Y)=\mathbf{1}(Y\in A)\) gives $$ \frac{E\!\left[R\mathbf{1}(Y\in A)/g(H)\right]} {E[R/g(H)]} =P(Y\in A). $$ Thus, at the population level, weighting the observed outcomes reconstructs the target outcome distribution. The same argument applies to functions of both \(H\) and \(Y\), so the weighted joint distribution is also the target joint distribution.

Worked Example: Correcting a Biased Mean Across Two History Strata

Let \(H\) be binary, with \(P(H=0)=P(H=1)=0.5\). Suppose \(Y\) is binary, with \(E[Y\mid H=0]=0.2\) and \(E[Y\mid H=1]=0.8\). The observation probabilities are \(g(0)=0.9\) and \(g(1)=0.3\), and assume conditional independent censoring.

The target mean is $$ E[Y]=0.5(0.2)+0.5(0.8)=0.1+0.4=0.5. $$ Among observed people, the probability masses from the two strata are \(P(H=0,R=1)=0.5(0.9)=0.45\) and \(P(H=1,R=1)=0.5(0.3)=0.15\). The total observed probability is \(0.60\), so the observed-case stratum proportions are \(0.45/0.60=0.75\) and \(0.15/0.60=0.25\). The unweighted observed-case mean is therefore $$ E[Y\mid R=1]=0.75(0.2)+0.25(0.8)=0.15+0.20=0.35. $$ It is below the target mean because the high-outcome stratum is less likely to be observed.

Now calculate the expected weighted numerator: $$ E\!\left[\frac{RY}{g(H)}\right] =0.5\frac{0.9(0.2)}{0.9} +0.5\frac{0.3(0.8)}{0.3} =0.1+0.4=0.5. $$ The expected total weight is $$ E\!\left[\frac{R}{g(H)}\right] =0.5\frac{0.9}{0.9}+0.5\frac{0.3}{0.3}=1. $$ The ratio is \(0.5/1=0.5\), matching the target mean. The inverse weights, \(1/0.9\) and \(1/0.3\), compensate for the unequal observation probabilities.

Why the Algebra Works Within and Across Strata

The weight does not alter the outcome distribution within a history stratum under conditional independent censoring. Its role is to restore each stratum’s representation in the target population. This can be seen directly in a discrete-history case. Write \(p_h=P(H=h)\) and \(g_h=P(R=1\mid H=h)\). Conditional independence implies that the observed outcome mean within stratum \(h\) is \(E[Y\mid H=h,R=1]=E[Y\mid H=h]\), whenever \(p_hg_h>0\). The expected weighted outcome contribution from that stratum is $$ E\!\left[\frac{RY\mathbf{1}(H=h)}{g(H)}\right] =\frac{p_hg_hE[Y\mid H=h,R=1]}{g_h} =p_hE[Y\mid H=h]. $$ Summing over strata recovers \(\sum_h p_hE[Y\mid H=h]=E[Y]\). In contrast, the complete-case mean combines stratum means using observed-case proportions \(p_hg_h/\sum_jp_jg_j\), which generally differ from the target proportions \(p_h\).

Worked Example: A Different Stratum Mix Produces a Different Mean

Suppose \(P(H=0)=0.25\), \(P(H=1)=0.75\), \(E[Y\mid H=0]=10\), and \(E[Y\mid H=1]=2\). Let \(g(0)=0.8\) and \(g(1)=0.4\), with conditional independent censoring.

The target mean is $$ E[Y]=0.25(10)+0.75(2)=2.5+1.5=4. $$ The observed stratum masses are \(0.25(0.8)=0.20\) and \(0.75(0.4)=0.30\), for a total observed probability of \(0.50\). Thus, observed-case proportions are \(0.20/0.50=0.40\) and \(0.30/0.50=0.60\), giving the unweighted observed mean $$ 0.40(10)+0.60(2)=4+1.2=5.2. $$ For the weighted mean, the stratum contributions to the numerator are $$ 0.25\frac{0.8(10)}{0.8}=2.5,\qquad 0.75\frac{0.4(2)}{0.4}=1.5. $$ They sum to \(4\). The corresponding denominator contributions are \(0.25(0.8/0.8)=0.25\) and \(0.75(0.4/0.4)=0.75\), which sum to \(1\). The weighted ratio is \(4/1=4\), the target mean.

The observed-case mean is not wrong for the population of observed people. It is simply not the mean for the original target population. The inverse weights change the relative contribution of the two history strata from \(0.40,0.60\) back to \(0.25,0.75\).

When the Weight Cannot Repair Selection

Positivity is essential, not a technical detail that can be ignored because a formula has been written down. If a history stratum has \(g(h)=0\), no one in that stratum contributes an observed outcome. The reciprocal \(1/g(h)\) is undefined, and the data contain no observed outcomes from which to recover that stratum’s outcome distribution without additional assumptions.

Worked Example: A Stratum With No Chance of Observation

Let \(P(H=0)=P(H=1)=0.5\). Suppose \(g(0)=1\) and \(g(1)=0\), and the outcome means are \(E[Y\mid H=0]=0\) and \(E[Y\mid H=1]=10\). The target mean is $$ E[Y]=0.5(0)+0.5(10)=5. $$ But only people with \(H=0\) are observed. Their outcomes are all that the observed data reveal, so an observed-case mean is \(0\). The weight for \(H=1\) would require \(1/0\), which is undefined.

To see the identification problem, consider a second possible population with exactly the same distribution of \(H\), the same observation rule, and the same outcome distribution for \(H=0\), but with \(E[Y\mid H=1]=100\). It produces the same observed data, yet its target mean is \(0.5(0)+0.5(100)=50\), rather than \(5\). The observed data cannot distinguish these target means. No finite reweighting of observed cases can supply the missing outcomes.

Near-violations also matter in practice. If \(g(H)\) is positive but very small for some people, their weights are very large. A few observations may then dominate a sample estimate, making it unstable even though the population identity holds. Inspect estimated observation probabilities and the distribution of the resulting weights; unusually large weights or sparse history strata warn that the result may be driven by limited support. Weighting also does not correct conditional dependence between \(R\) and \(Y\) left over after conditioning on \(H\), or errors in estimating \(g(H)\).

How This Approach Compares With Alternatives

A complete-case analysis uses only \(R=1\) observations without correcting their altered history distribution. It is valid for the target outcome mean under stronger conditions—for example, if observation is independent of \(Y\) without conditioning on \(H\)—but can be biased when observation and outcome are associated through \(H\). Inverse probability weighting instead keeps the target population as its reference and adjusts the contributions of observed people using \(H\).

Outcome-model approaches, including regression adjustment or multiple imputation, use a model for the outcome given measured variables to address missing outcomes. They differ from inverse probability weighting in the model they rely on: weighting models observation, whereas those approaches model outcomes or missing values. Neither strategy makes unmeasured selection disappear. In this one-period derivation, the essential comparison is that weighting reconstructs the target distribution by correcting the observed sample’s selection probabilities, under conditional independent censoring and positivity.

The identity is a population statement. In data, \(g(H)\) is usually unknown and must be estimated; substituting an estimated probability does not make the sample weighted mean equal the target mean exactly. Its performance depends on the adequacy of the observation-probability model and on having enough observations across the relevant histories. This tutorial’s result concerns one period; it does not derive the cumulative weights needed when observation must continue through multiple follow-up periods.

1
Specify the target outcome.
State the population and follow-up outcome whose distribution is sought, including how the outcome is defined for people who were censored.
2
State the observation assumptions.
Identify the measured history \(H\), require conditional independent censoring given \(H\), and check that every target history has positive observation probability.
3
Apply the single-period contribution.
For each person, use \(R/g(H)\): zero if the outcome is unobserved, and the reciprocal observation probability if it is observed.
4
Assess support and stability.
Look for very small observation probabilities, large weights, or history strata with few or no observed people; these can undermine practical reliability or identification.
Common pitfall: “The observed cases have the same outcome mean as the target population within every history stratum, so the complete-case mean is unbiased.” The first claim can follow from conditional independent censoring, but it does not imply the second: the complete cases generally have different proportions across strata. The inverse weight restores those proportions by dividing each observed contribution by its stratum’s observation probability.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

  1. In a one-period study, what are \(R\) and \(g(H)\), and which people receive a nonzero inverse censoring contribution?
  2. State the conditional independent censoring and positivity assumptions used in the weighting identity.
  3. Suppose \(P(H=0)=0.6\), \(P(H=1)=0.4\), and \(g(0)=g(1)=0.5\). If the stratum means are \(2\) and \(8\), respectively, what are the target mean and the complete-case mean?
  4. Why can inverse weighting fail to identify the target mean if \(g(h)=0\) for a history value in the target population?

Solutions. (1) \(R=1\) means the follow-up outcome is observed and \(R=0\) means it is censored before observation; \(g(H)=P(R=1\mid H)\). Only \(R=1\) people have a nonzero contribution \(R/g(H)\). (2) Require \(R\) and \(Y\) to be conditionally independent given \(H\), and require \(g(H)>0\) almost surely in the target population. (3) The target mean is \(0.6(2)+0.4(8)=1.2+3.2=4.4\). Since both strata have the same observation probability, the observed-case stratum proportions remain \(0.6\) and \(0.4\), so the complete-case mean is also \(4.4\). (4) No one with that history is observed, and \(1/g(h)\) is undefined. Different outcome distributions in that unobserved stratum can produce identical observed data but different target means.

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