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Censoring and Informative Dropout · Tutorial 218 of 501

Stabilized Censoring Weights and Their Expectation

Learn how baseline-conditioned numerators stabilize censoring weights and why their conditional mean is one among those still observed when the censoring probabilities are correctly specified.

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What You'll Learn

  • Define numerator and denominator probabilities for stabilized censoring weights over discrete follow-up.
  • Derive the cumulative weight and state the positivity conditions it requires.
  • Prove its unconditional and survivor-conditional expectation properties given baseline covariates.
  • Calculate stabilized weights and expectations in one-period and longitudinal examples.
  • Diagnose misspecified probabilities and distinguish stabilized weights from unstabilized weights.

Why Stabilize Censoring Weights?

Inverse-probability-of-censoring weights compensate for the fact that some participants leave observation before a time point of interest. A denominator conditional on the observed history can make those who remain represent similar histories that were lost. But the resulting weights can vary substantially. Stabilization adds a numerator probability conditional on baseline covariates, producing weights whose conditional mean has a useful, precise interpretation.

The key qualification is that this mean is one among participants still observed, conditional on baseline covariates. The weighted indicator that a participant is observed does not have conditional mean one: its mean is the probability of remaining observed. Keeping these two statements separate prevents a common interpretation error.

Definition and Assumptions

Consider discrete follow-up over intervals \(k=0,\ldots,t-1\). Let \(V\) denote baseline covariates, and let \(H_k\) denote the observed history just before the censoring decision in interval \(k\), including \(V\) and any relevant treatment and time-varying covariate history. Let \(R_0=1\), and define \(R_k=1\) when the participant remains uncensored through the start of interval \(k\); thus \(R_{k+1}=1\) means that the participant also remains uncensored through interval \(k\).

Definition: For participants with \(R_k=1\), define the conditional retention probabilities $$ p_k(H_k)=P(R_{k+1}=1\mid R_k=1,H_k),\qquad q_k(V)=P(R_{k+1}=1\mid R_k=1,V). $$ The stabilized censoring weight through time \(t\) is $$ W_t=R_t\prod_{k=0}^{t-1}\frac{q_k(V)}{p_k(H_k)}. $$ The product is evaluated for participants still observed through \(t\); set \(W_t=0\) for those with \(R_t=0\).

The numerator probability \(q_k(V)\) is the actual conditional probability of retention given baseline covariates and being uncensored at the start of interval \(k\). The denominator \(p_k(H_k)\) further conditions on the observed history. These are conditional probabilities, not arbitrary constants. In applications they are estimated from censoring models.

For the displayed weights to be defined, require \(p_k(H_k)>0\) wherever the corresponding histories occur among participants at risk, and require \(q_k(V)>0\) for the baseline values under consideration. If a history has zero probability of remaining observed, no finite inverse-probability weight can recover its contribution. For causal or full-history interpretations, the usual assumptions about the observation process, including sequential independent censoring and appropriate consistency, are also needed; the expectation identity below is a probability calculation and does not by itself establish causal identification.

Expectation Properties

The numerator probabilities are chosen so their product equals the probability of remaining observed through time \(t\), conditional on \(V\). This follows from the chain rule:

$$ P(R_t=1\mid V)=\prod_{k=0}^{t-1}P(R_{k+1}=1\mid R_k=1,V) =\prod_{k=0}^{t-1}q_k(V). $$

The weight has a corresponding expectation property. The indicator \(R_t\) matters: it excludes people censored before \(t\), while each denominator factor cancels the conditional probability of surviving its interval.

Theorem: Suppose \(p_k(H_k)\) and \(q_k(V)\) are the conditional probabilities defined above, with the required denominators positive. Then $$ E[W_t\mid V]=P(R_t=1\mid V)=\prod_{k=0}^{t-1}q_k(V). $$ Consequently, for baseline values with \(P(R_t=1\mid V)>0\), $$ E\left[\left.\prod_{k=0}^{t-1}\frac{q_k(V)}{p_k(H_k)}\right|R_t=1,V\right]=1. $$

Proof. Write \(B_k=\prod_{j=0}^{k-1}q_j(V)/p_j(H_j)\), with \(B_0=1\). On the event \(R_k=1\), \(B_k\) is determined by the observed history through the start of interval \(k\). Conditional on that history and \(R_k=1\),

$$ E\left[\left.\frac{R_{k+1}}{p_k(H_k)}\right|R_k=1,H_k\right] =\frac{P(R_{k+1}=1\mid R_k=1,H_k)}{p_k(H_k)} =1. $$

It follows by iterated expectation that \(E[R_{k+1}B_k/p_k(H_k)\mid V]=E[R_kB_k\mid V]\). Multiplying by \(q_k(V)\), which is fixed given \(V\), gives \(E[R_{k+1}B_{k+1}\mid V]=q_k(V)E[R_kB_k\mid V]\). Starting with \(R_0B_0=1\) and applying this recursion through \(k=t-1\) yields \(E[W_t\mid V]=\prod_{k=0}^{t-1}q_k(V)\). The chain rule above identifies this product with \(P(R_t=1\mid V)\). Finally, since \(W_t=0\) when \(R_t=0\),

$$ E[W_t\mid V] =P(R_t=1\mid V)\, E\left[\left.\prod_{k=0}^{t-1}\frac{q_k(V)}{p_k(H_k)}\right|R_t=1,V\right]. $$

Dividing by the positive retention probability proves the conditional-mean statement. \(\square\)

Thus the cumulative ratio has mean one in the baseline stratum among those retained. Across all participants in that stratum, the weighted retention indicator instead has mean \(P(R_t=1\mid V)\), which is generally less than one.

Worked Example: One Interval and Two Histories

Within one baseline stratum, suppose a binary history \(X\) is equally likely to be 0 or 1 among participants at risk. The retention probabilities are \(p(X=0)=0.8\) and \(p(X=1)=0.4\). The baseline-conditioned numerator is their marginal retention probability:

$$ q(V)=P(R_1=1\mid V)=0.5(0.8)+0.5(0.4)=0.6. $$

For retained participants with \(X=0\), the stabilized ratio is \(0.6/0.8=0.75\); with \(X=1\), it is \(0.6/0.4=1.5\). The unconditional weighted retention contributions from the two histories are \(0.5(0.8)(0.75)=0.30\) and \(0.5(0.4)(1.5)=0.30\). Their sum, \(0.60\), equals \(P(R_1=1\mid V)\).

Among retained participants, the probabilities of \(X=0\) and \(X=1\) are \(0.4/0.6=2/3\) and \(0.2/0.6=1/3\). Therefore the mean stabilized ratio among them is \((2/3)(0.75)+(1/3)(1.5)=0.5+0.5=1\), as the theorem states.

Worked Example: Two Intervals

Fix a baseline stratum. In interval 0, everyone has retention probability \(p_0=0.8\), so \(q_0=0.8\). Among those retained into interval 1, suppose a binary history \(X\) has probabilities \(1/2\) and \(1/2\), with \(p_1(X=0)=0.9\) and \(p_1(X=1)=0.5\). The second numerator is \(q_1=0.5(0.9)+0.5(0.5)=0.7\).

The cumulative ratio for someone retained through both intervals is \((0.8/0.8)(0.7/p_1(X))\). It is \(7/9\) when \(X=0\), and \(1.4\) when \(X=1\). The probability of retention through both intervals is \(0.8(0.7)=0.56\). The unconditional weighted contributions are \(0.8(0.5)(0.9)(7/9)=0.28\) for \(X=0\), and \(0.8(0.5)(0.5)(1.4)=0.28\) for \(X=1\). Their sum is \(0.56\); dividing by the retention probability gives a mean ratio of \(0.56/0.56=1\) among those retained.

What Stabilization Changes—and What It Does Not

An unstabilized censoring weight uses numerator 1. Under the same conditions, its mean among participants retained through \(t\), conditional on \(V\), is \(1/P(R_t=1\mid V)\). Stabilization replaces that factor with a numerator that reflects baseline-specific retention. As a result, the stabilized ratio has mean one among retained participants within baseline strata, rather than forcing every baseline stratum to be represented as though retention were certain.

This is a property of the conditional mean, not a guarantee that each weight is close to one. Large ratios can remain when some histories have small retention probabilities, and weights can be variable even when the mean is exactly one. Stabilization also does not repair a positivity violation or incorrect denominator model.

The same distinction helps interpret a weighted sample. The observed participants, weighted by the stabilized ratio, have total expected weight proportional to \(P(R_t=1\mid V)\) within each baseline stratum. Their within-stratum mean weight is one, but the total weight still reflects how likely that stratum is to remain observed. Do not infer that the weighted sample has the original baseline distribution merely from the conditional-mean result.

Worked Example: A Misspecified Numerator

Suppose in a single baseline stratum the actual retention probability is \(0.6\), but the numerator model uses \(0.5\). If the denominator model is correct, the ratio for each retained participant is the chosen numerator divided by the actual retention probability in expectation. Its conditional mean among retained participants is therefore \(0.5/0.6=5/6\), not one. The unconditional mean of the weighted retention indicator is \(0.6(5/6)=0.5\), the misspecified numerator value.

This calculation shows why the theorem refers to the actual conditional probability \(q(V)\). Estimating it introduces sampling and model error; with a misspecified numerator, the exact conditional-mean-one property need not hold.

Diagnostics and Common Pitfalls

Useful checks focus on the probability models and on the distribution of the resulting weights, rather than on whether their overall average happens to be one.

  • Check retention probabilities. Inspect fitted denominator probabilities among participants at risk at each interval. Values near zero can create very large ratios; zeros indicate a positivity failure for the histories concerned.
  • Inspect weights by time and baseline stratum. Summaries of weight tails, maxima, and effective sample sizes can reveal instability that an overall mean conceals.
  • Compare numerator predictions with observed retention by baseline strata. A large discrepancy can signal an inadequate numerator model or sampling variation. The conditional-mean identity is exact for the true probabilities, not automatically for fitted values.
  • Do not confuse the two means. The theorem says the ratio has mean one among those retained, conditional on \(V\). It says the weighted indicator has mean \(P(R_t=1\mid V)\), not one.
  • Do not treat stabilization as a cure for extreme weights. If the denominator is small, a baseline numerator may still yield a large ratio. The underlying lack of follow-up support remains.

The main alternative is the unstabilized weight, whose numerator is one. It is simpler, but its survivor-conditional mean grows as the inverse of the retention probability. The stabilized form can improve scale while retaining history-specific inverse-probability adjustment. Neither choice removes the need for defensible censoring models, positivity, or the assumptions needed to interpret a weighted analysis.

Check Your Understanding

Work through each question, then compare your reasoning with the solutions.

  1. In one interval, \(P(R_1=1\mid V)=0.75\). What is the conditional mean of the stabilized ratio among participants with \(R_1=1\), and what is \(E[W_1\mid V]\)?
  2. Why is the denominator history-specific while the numerator conditions only on baseline covariates?
  3. If the true numerator retention probability is \(0.8\) but the numerator model uses \(0.6\), while the denominator is correct, what is the mean ratio among retained participants in that baseline stratum?
  4. Does a conditional mean of one among retained participants imply that all stabilized weights equal one? Explain.

Solutions. (1) The survivor-conditional mean of the ratio is one. Since \(W_1=R_1\) times that ratio, \(E[W_1\mid V]=P(R_1=1\mid V)=0.75\). (2) The denominator adjusts for the observed history that predicts retention; the numerator retains a baseline-conditioned probability rather than undoing all baseline-specific retention. (3) The mean ratio is \(0.6/0.8=0.75\), so the mean-one property fails under this misspecification. (4) No. A mean of one constrains only the average within the retained baseline stratum; individual weights may differ substantially, as in the one-interval example.

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