Tutorials › Marginal Structural Models › Product Weights for Longitudinal Follow-Up

Censoring and Informative Dropout · Tutorial 216 of 501

Product Weights for Longitudinal Follow-Up

Learn to factor the probability of complete follow-up into period-specific retention probabilities and interpret each reciprocal factor in the longitudinal weight.

Advanced 12 min read

What You'll Learn

  • Define period-specific conditional retention probabilities along an observed history.
  • Derive the product form for the probability of remaining uncensored through follow-up.
  • Construct the cumulative inverse censoring weight and interpret each factor.
  • Calculate how low retention in one period changes the total weight.
  • Recognize positivity, time-ordering, and history-specification problems that can undermine product weights.

Why Follow-Up Weights Accumulate Across Periods

In longitudinal studies, a participant must remain under observation at every period to contribute a complete history through the end of follow-up. The probability of being observed at the end is therefore built from a sequence of conditional retention probabilities. A single overall probability can conceal where follow-up is being lost; the product representation makes each period’s contribution explicit.

The earlier material on the one-period derivation of inverse censoring weights gives the basic idea: observations with a smaller probability of remaining observed receive a larger reciprocal weight. With repeated follow-up decisions, the same logic applies sequentially. The weight accumulates one inverse conditional probability at a time.

Definition: Let \(R_t=1\) indicate that a participant is observed through period \(t\), and let \(R_t=0\) indicate that the participant has been censored by then. Follow-up is monotone: \(R_t=1\) implies \(R_{t-1}=1\), and \(R_0=1\) for everyone at baseline. Let \(\mathcal H_t\) be the measured history available immediately before the period-\(t\) retention decision, for a participant still observed at that point. The conditional retention probability is $$ q_t(\mathcal H_t)=P(R_t=1\mid R_{t-1}=1,\mathcal H_t). $$ For a participant observed through period \(K\), the longitudinal inverse censoring weight is $$ W_K=\prod_{t=1}^{K}\frac{1}{q_t(\mathcal H_t)}. $$ For a participant not observed through \(K\), define the endpoint weight as zero; the product above is evaluated only for participants with \(R_K=1\).

The history \(\mathcal H_t\) must use the study’s actual time ordering. It contains information measured before the retention decision at \(t\), such as prior treatment and outcomes and any covariates collected before that decision. A variable measured only after the decision cannot be used as though it predicted that decision. Histories may differ between participants, so the same period can contribute different factors to different people’s weights.

Factorizing the Probability of Complete Follow-Up

For a fixed full-data history, the chain rule expresses the probability of remaining observed through \(K\) as a product of successive conditional probabilities. To use observed-history retention models for that factorization, assume sequential independent censoring: conditional on the history available before the period-\(t\) decision, retention at \(t\) is independent of the later full-data variables relevant to the analysis. Also assume that the retention probability is positive for histories under consideration.

Theorem: Under monotone follow-up and sequential independent censoring, for a full-data history \((\mathcal H_1,\ldots,\mathcal H_K)\) with positive conditional retention probabilities, $$ P(R_K=1\mid \mathcal H_1,\ldots,\mathcal H_K) =\prod_{t=1}^{K}q_t(\mathcal H_t). $$ Consequently, the reciprocal probability of remaining observed along that history is the product weight \(W_K\).

Proof. Write the event \(R_K=1\) as the intersection of the successive retention events \(R_1=1,\ldots,R_K=1\). By the chain rule, $$ P(R_K=1\mid \mathcal H_1,\ldots,\mathcal H_K) =\prod_{t=1}^{K}P(R_t=1\mid R_1=\cdots=R_{t-1}=1,\mathcal H_1,\ldots,\mathcal H_K). $$ Monotone follow-up means that \(R_1=\cdots=R_{t-1}=1\) is equivalent to \(R_{t-1}=1\). Sequential independent censoring permits the conditional probability for the period-\(t\) decision, given the full history, to be replaced by the probability conditional on being at risk and the history available at that decision: $$ P(R_t=1\mid R_{t-1}=1,\mathcal H_1,\ldots,\mathcal H_K) =P(R_t=1\mid R_{t-1}=1,\mathcal H_t)=q_t(\mathcal H_t). $$ Substitution into every factor gives the stated product. Positivity ensures its reciprocal is finite. \(\square\)

Each factor corresponds to one transition from being observed at \(t-1\) to remaining observed through \(t\). The product is not generally the product of unconditional retention rates: the conditional probability at each step is evaluated at that participant’s history on the path. If histories or retention probabilities vary, the period-specific factors vary too.

Worked Example: Three Retention Decisions

Suppose a participant’s conditional probabilities of remaining observed in three successive periods are \(0.90\), \(0.80\), and \(0.75\). Find the probability of remaining observed through period 3 and the corresponding endpoint weight.

The probability of complete follow-up along this history is $$ 0.90\times0.80\times0.75=0.54. $$ The exact reciprocal factors are \(1/0.90=10/9\), \(1/0.80=5/4\), and \(1/0.75=4/3\). Thus $$ W_3=\frac{10}{9}\times\frac{5}{4}\times\frac{4}{3} =\frac{50}{27}=1.851851\ldots, $$ which rounds to \(1.852\) at three decimal places. If instead the factors are first rounded to \(1.111\), \(1.25\), and \(1.333\), their displayed product is \(1.111\times1.25\times1.333=1.85120375\), which rounds to \(1.851\). The small discrepancy is due to rounding the factors before multiplication; the exact product is the appropriate calculation.

Each period contributes its own inverse retention probability. The weight is larger than one because the participant’s observed path had probability \(0.54\), below one, under the specified conditional retention probabilities.

How One Period Changes the Cumulative Weight

The product structure allows the contribution of any period to be isolated. Holding the earlier history fixed, a smaller \(q_t\) multiplies the accumulated weight by a larger amount, \(1/q_t\). The increment depends on the retention probability at that participant’s history, not just on the period number.

Result: For a participant retained through period \(t\), define the cumulative weight \(W_t=\prod_{j=1}^{t}1/q_j(\mathcal H_j)\), with \(W_0=1\). Then $$ W_t=\frac{W_{t-1}}{q_t(\mathcal H_t)} \quad\text{and}\quad \log W_t=\sum_{j=1}^{t}-\log q_j(\mathcal H_j). $$

Derivation. By definition, \(W_t\) is the product of the first \(t\) reciprocal probabilities. Separate the final factor: $$ W_t=\left(\prod_{j=1}^{t-1}\frac{1}{q_j(\mathcal H_j)}\right)\frac{1}{q_t(\mathcal H_t)} =W_{t-1}\frac{1}{q_t(\mathcal H_t)}. $$ Taking logarithms of the product gives \(\log W_t=\sum_{j=1}^{t}\log(1/q_j(\mathcal H_j))\), and \(\log(1/q_j)=-\log q_j\). This proves both identities. \(\square\)

Worked Example: A Low-Retention Period Has a Larger Increment

Two participants are both observed through period 3. Along the first participant’s history, the retention probabilities are \(0.95\), \(0.60\), and \(0.90\). Along the second participant’s history, they are \(0.95\), \(0.90\), and \(0.90\). Compare their endpoint weights.

For the first participant, the cumulative retention probability is $$ 0.95\times0.60\times0.90=0.513, $$ so the weight is \(1/0.513=1000/513=1.949317\ldots\), or \(1.949\) to three decimals.

For the second participant, the cumulative retention probability is $$ 0.95\times0.90\times0.90=0.7695, $$ so the weight is \(1/0.7695=2000/1539=1.299545\ldots\), or \(1.300\) to three decimals. The paths differ only in the period-2 retention probability. The first participant’s period-2 reciprocal factor is \(1/0.60=1.6667\ldots\), whereas the second’s is \(1/0.90=1.1111\ldots\); that difference accounts for the larger first weight.

Worked Example: Separating Earlier and Later Contributions

Suppose a participant’s retention probabilities are \(0.80\), \(0.80\), and \(0.50\). Calculate the weight through period 2 and then through period 3, assuming the participant remains observed at all three decisions.

Through period 2, the cumulative retention probability is \(0.80\times0.80=0.64\), so $$ W_2=\frac{1}{0.64}=1.5625. $$ The period-3 factor is \(1/0.50=2\). Therefore $$ W_3=W_2\times2=1.5625\times2=3.125. $$ The period-3 decision doubles the weight already accumulated through period 2. This illustrates why a single low conditional retention probability can have a substantial effect even if earlier retention was relatively high.

Assumptions, Diagnostics, and Failure Modes

The product is a mathematical factorization, but its use for longitudinal censoring adjustment depends on the assumptions behind the conditional probabilities. The main requirements are a correctly ordered and adequately specified observed history, sequential independent censoring given that history, and positivity: \(q_t(\mathcal H_t)>0\) for the histories represented in the target population. If a history has zero probability of retention at some period, no finite product weight can represent it among retained participants.

A very small estimated retention probability creates a very large reciprocal factor. Products can therefore become extreme even when no single factor seems dramatic. Useful warning signs include a small minimum estimated \(q_t\), a strongly right-skewed distribution of cumulative weights, and a few observations carrying a large fraction of the total weight. These patterns may reflect genuine weak positivity, sparse histories, model misspecification, or a combination. They are reasons to investigate the fitted conditional probabilities and the histories they represent; they do not, by themselves, identify which problem is responsible.

  • Check the time ordering. Confirm that each predictor in \(\mathcal H_t\) was available before the period-\(t\) retention decision.
  • Inspect period-specific probabilities. A tolerable endpoint average can conceal near-zero retention probabilities in a particular history and period.
  • Examine cumulative weights. Look at their range and concentration, and determine whether a small number of participants dominate.
  • Check support. Ask whether histories represented in the target population have observed participants who remain under follow-up.

The product construction differs from using one overall, unconditional probability of complete follow-up. An overall probability ignores how retention depends on evolving history; a conditional product preserves those period-by-period distinctions. It also differs from simply analyzing complete cases without weights, which does not account for differences in the probability of being retained. The product weight does not, on its own, guarantee valid adjustment: it cannot repair unmeasured predictors of dropout, violations of sequential independent censoring, or a lack of positivity. The later material on why inverse probability of censoring weighting recovers the uncensored history distribution addresses what the weights accomplish under the required assumptions.

Common Pitfall: Multiplying the Wrong Probabilities

A common error is to multiply period-specific marginal retention rates as though they were the conditional probabilities along a participant’s history. For example, if the overall retention rate in each of two periods is \(0.90\), multiplying those rates gives \(0.81\). That calculation is justified for the relevant history only if the period-specific conditional probabilities along that history are in fact \(0.90\) at both periods. When later retention depends on prior history, the conditional probability for period 2 can differ from its marginal rate.

The correction is to define \(q_t\) conditional on being at risk and on the history available before that period’s decision, then multiply those conditional probabilities along the history in question. A second error is to include a factor for a period after a participant has already been censored. Such a factor is not observed for that participant; an endpoint weight is zero for someone not followed through the endpoint, and the complete-follow-up product is calculated only for those retained through every decision.

Key takeaway: A longitudinal censoring weight is built sequentially: each period contributes the reciprocal of the conditional probability of remaining observed at that period, given the history then available. The endpoint weight is their product, so every factor and every time ordering matters.

Check Your Understanding

Answer each question, then compare with the brief solutions below.

  1. If the conditional retention probabilities over two periods are \(0.80\) and \(0.50\) along a history, what are the probability of remaining observed through period 2 and the endpoint weight?
  2. Why must the period-2 retention probability be conditioned on being observed through period 1?
  3. A participant has \(W_2=2\) and then remains observed at period 3, where \(q_3=0.40\). What is \(W_3\)?
  4. Name one assumption needed to interpret the product of conditional retention probabilities as the probability of complete follow-up along a full-data history.

Solutions. (1) The probability is \(0.80\times0.50=0.40\), and the weight is \(1/0.40=2.5\). (2) The period-2 decision occurs only for someone still at risk; conditioning on \(R_1=1\) defines the relevant transition probability. (3) \(W_3=W_2/q_3=2/0.40=5\). (4) Sequential independent censoring given the measured history is one required assumption; positivity of the conditional retention probabilities is also required for finite weights.

Clinical Biostats