Multiple Censoring Processes and Joint Follow-Up Weights

Why Joint Follow-Up Weights Need an Order

Follow-up can end for more than one reason. A participant might withdraw, become unreachable, or be artificially censored after a protocol deviation. If these mechanisms occur in sequence, the probability of remaining observed is a probability of passing every censoring decision—not a collection of unrelated marginal probabilities. The relevant factors must reflect the history available at each decision and the order in which the decisions are made.

This construction extends the usual inverse probability of censoring weight. It is useful when different processes have distinct predictors or occur at different points in follow-up. The introductory material on inverse probability of censoring weighting establishes the basic role of reweighting observed records; here the focus is how to build the joint retention probability when censoring has several components.

Sequential Definition

Suppose follow-up has $K$ intervals. In interval $k$, there are $R_k$ ordered censoring decisions. Let $A_{kr}=1$ mean that a participant passes decision $r$ in interval $k$, and let $A_{kr}=0$ mean that follow-up ends at that decision. Order the decisions chronologically, first by interval and then by $r$. Define $O_{k0}=1$ for someone observed at the start of interval $k$, and

O_{kr}=O_{k,r-1}A_{kr}, \qquad r=1,\ldots,R_k, \qquad O_{k+1,0}=O_{k,R_k}.

Thus $O_{kr}=1$ means the participant has passed all decisions through decision $r$ in interval $k$. Let $H_{kr}$ denote the measured history available immediately before decision $r$, including earlier outcomes and covariates and, when relevant, the outcomes of earlier censoring decisions in the same interval.

Definition: The conditional retention probability at decision $(k,r)$ is $q_{kr}(H_{kr})=P(A_{kr}=1\mid O_{k,r-1}=1,H_{kr})$. The joint probability of remaining observed through interval $K$, along a participant's history, is the product of these ordered conditional probabilities: $$ Q_K=\prod_{k=1}^{K}\prod_{r=1}^{R_k}q_{kr}(H_{kr}). $$ For a participant observed through that endpoint, the unstabilized joint follow-up weight is $W_K^C=1/Q_K$.

Each factor is conditional on being at risk for that decision. In general, $q_{kr}$ can depend on the outcome of an earlier decision in the same interval; a decision after follow-up has ended is never reached and has no applicable retention probability for that participant. The product is a chain-rule factorization. It does not assert that the censoring mechanisms are independent.

Worked Calculations

Worked Example: Two Sequential Mechanisms at One Visit

At a scheduled visit, participants first need to remain reachable and then need to remain eligible under a protocol rule. Among participants who reach the first decision, the estimated probability of remaining reachable is $0.90$. Among those who pass it, the conditional probability of remaining protocol-eligible is $0.80$. Find the probability of passing both decisions and the corresponding weight.

The first factor is $q_1=0.90$. The second is conditional on passing the first, so $q_2=0.80$. Therefore

Q=0.90(0.80)=0.72,\qquad W^C=\frac{1}{0.72}=1.388\overline{8}.

An observed participant who passes both decisions represents about $1.389$ people in the reweighted target population at this visit. The multiplication uses the second mechanism's conditional probability, not a probability calculated in a different risk set.

Worked Example: Accumulating Weight Across Two Intervals

In interval 1, the conditional retention probabilities for two ordered mechanisms are $0.95$ and $0.80$. In interval 2, for a participant with the observed history, they are $0.90$ and $0.70$. Calculate the probability of remaining observed through interval 2 and the endpoint weight.

The within-interval products are $0.95(0.80)=0.76$ and $0.90(0.70)=0.63$. Passing all four decisions has probability

$$ Q_2=0.95(0.80)(0.90)(0.70) =0.76(0.63)=0.4788. $$

The corresponding weight is

W_2^C=\frac{1}{0.4788}\approx 2.0886.

This endpoint weight includes all four decisions. Using only the interval-2 factors would produce $1/0.63\approx1.5873$, which does not account for selection earlier in follow-up.

Worked Example: Why Marginal Retention Probabilities Can Mislead

Suppose the probability of passing mechanism 1 is $0.90$, and the probability of passing mechanism 2 among participants who passed mechanism 1 is $0.80$. The marginal probability of passing mechanism 2 in the original group is $0.85$. What is the joint retention probability?

The chain rule gives

P(A_1=1,A_2=1)=P(A_1=1)P(A_2=1\mid A_1=1) =0.90(0.80)=0.72.

Multiplying the two marginal probabilities instead gives $0.90(0.85)=0.765$, which is not the joint probability. The value $0.85$ averages over people who did and did not pass mechanism 1; it is not the retention probability in the second mechanism's risk set. No independence assumption justifies replacing the conditional probability by that marginal value.

Identification: When Does the Weight Recover the Full-Data Mean?

A large weight alone does not make an observed outcome representative. Identification requires that the measured histories capture the information needed to account for censoring, and that every relevant history has a positive chance of remaining observed.

Assumptions for a joint weighting identity: Consider the ordered censoring decisions through the target endpoint, and let $\mathcal F$ denote the complete-data trajectory relevant to a full-data quantity $Y$. Assume (1) consistency of the recorded histories and outcomes with their full-data values when observed; (2) sequential conditional independent censoring, so that at each decision the conditional probability of passing it, given the complete trajectory and prior passage, equals $q_{kr}(H_{kr})$; (3) positivity, $q_{kr}(H_{kr})>0$ for histories with positive probability in the target population; and (4) integrability, $E[|Y|]<\infty$. Then the inverse joint retention weight recovers the full-data mean.

The conditional independence condition is a substantive assumption: after conditioning on the stated history, the decision to remain observed carries no further information about the full-data quantity. It is not guaranteed by fitting a well-calibrated prediction model.

Theorem: Under the stated assumptions, let $O_K$ indicate that the participant passes every censoring decision through the target endpoint, and let $Q_K$ be the product of the corresponding conditional retention probabilities. Then $$ E\!\left[\frac{O_KY}{Q_K}\right]=E[Y]. $$ Here $O_KY$ is evaluated from observed data only when $O_K=1$; when $O_K=0$, its weighted contribution is defined to be zero.

Proof. Index the ordered decisions by $j=1,\ldots,J$, where $J=\sum_{k=1}^{K}R_k$. Write $A_j$ for passage at decision $j$, $O_J=\prod_{j=1}^{J}A_j$, and $q_j$ for its conditional retention probability. Conditional on the complete trajectory, the sequential assumption gives

P(O_J=1\mid\mathcal F) =\prod_{j=1}^{J}P(A_j=1\mid A_1=\cdots=A_{j-1}=1,\mathcal F) =\prod_{j=1}^{J}q_j =Q_K.

Because $Y$ and $Q_K$ are functions of the complete trajectory, conditional expectation therefore yields

E\!\left[\left.\frac{O_JY}{Q_K}\right|\mathcal F\right] =\frac{Y}{Q_K}P(O_J=1\mid\mathcal F) =Y.

Positivity makes the ratio defined on the target support, and integrability permits taking expectations. Applying iterated expectation gives $E[O_JY/Q_K]=E[Y]$. $\square$

The theorem explains why the denominator is a product: it is the conditional probability of passing the entire ordered sequence. For an estimated weight, the $q_j$ values are replaced by estimates from the at-risk data. The identity describes the population target under the assumptions; it does not make estimated weights exact in a finite sample.

Stabilization, Modeling, and Failure Modes

A stabilized joint weight replaces some denominator factors with numerator probabilities based on a reduced history. For a chosen numerator $Q_K^{\mathrm{num}}$, the form is $W_K^{C,\mathrm{stab}}=Q_K^{\mathrm{num}}/Q_K^{\mathrm{den}}$. Numerator factors must refer to a coherent ordered retention process as well. Stabilization can reduce weight variability, but it does not repair a denominator model that omits important predictors of censoring or violates positivity. When treatment weights are also used, the treatment and censoring components are multiplied as in the combined treatment-and-censoring weighting framework.

Component probabilities are commonly estimated among those at risk at each decision, using models appropriate to the decision and its measured history. Pooled logistic regression is one option for discrete-time hazards. The fitted probability at each step must match the event coding and risk set used to define that step; multiplying predictions from models with incompatible risk sets does not produce the desired joint probability.

Several checks help reveal construction problems. Confirm the stated within-interval order against the data collection and censoring rules. For each factor, verify which participants were at risk and which history variables were available before that decision. Trace a few individual products by hand. Very small fitted retention probabilities create large inverse weights; probabilities equal to zero in a target-relevant history violate positivity rather than merely creating a computational inconvenience. The weight distribution and its sensitivity to model choices can flag problems, but diagnostics cannot establish conditional independent censoring.

Approach	What it models	Main caution
Sequential component models	Conditional passage at each ordered mechanism	Requires correct risk sets, histories, and ordering at every step
One model for overall retention	Passing the entire interval or follow-up endpoint	Can obscure which histories predict which component and may discard useful timing information
Separate marginal probabilities multiplied together	Each mechanism without conditioning on earlier passage	Generally wrong unless the required conditional probabilities or independence conditions actually hold

The sequential construction is especially useful when components have different timing or predictors. A single overall-retention model may still be reasonable if it directly models the relevant joint event conditional on an adequate history. Neither representation solves unmeasured causes of censoring, lack of support, or misspecification. The estimated weight is an adjustment for selection under assumptions, not a substitute for explaining why follow-up ended.

Check Your Understanding

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

At two ordered decisions, the conditional passage probabilities are $0.92$ and $0.75$. Calculate the joint passage probability and the inverse weight.
Why is multiplying two marginal passage probabilities generally not a valid way to calculate joint retention?
At each of two intervals, a participant must pass two mechanisms with conditional probabilities $0.90$, $0.80$, $0.95$, and $0.60$, in that order. Calculate the endpoint weight.
Name two assumptions needed for the joint weighting identity, and explain what a zero retention probability means.

Solutions. (1) $Q=0.92(0.75)=0.69$, so $W^C=1/0.69\approx1.4493$. (2) The second factor in the chain rule is conditional on passing the first decision. A marginal probability averages over a different group and need not equal that conditional probability. (3) $Q=0.90(0.80)(0.95)(0.60)=0.4104$, so $W^C=1/0.4104\approx2.4366$. (4) Sequential conditional independent censoring and positivity are two required assumptions; consistency and integrability are also part of the theorem's conditions. A zero retention probability on a target-relevant history is a positivity violation: no observed participant from that history can represent its full-data outcomes through inverse weighting.