Tutorials › Econometrics (Graduate) › Sequential Inverse-Probability Weights for Dropout

Panel Designs and Unbalanced Data · Tutorial 279 of 952

Sequential Inverse-Probability Weights for Dropout

Learn to construct cumulative retention weights and use them to recover moments for the baseline panel cohort when dropout is conditionally ignorable and retention probabilities are positive.

Intermediate 10 min read

What You'll Learn

  • Define period-specific retention probabilities using the observed history of panel members still at risk.
  • Derive the cumulative inverse-probability weight for remaining observed through each wave.
  • State sequential exchangeability and positivity conditions for identifying full-panel moments.
  • Distinguish a cohort-scaled inverse-probability estimate from a normalized weighted estimate.
  • Calculate weighted means and cross-moments from hypothetical panel data.
  • Recognize when the assumptions fail and what the weights can—and cannot—correct.

Why Construct Weights Sequentially?

In a panel with monotone dropout, a participant who leaves the study is not observed at later waves. The chance of appearing in the final complete-case sample therefore depends on surviving a sequence of retention decisions. If those decisions depend on observed histories, the probability of remaining through the final wave is built from period-specific conditional retention probabilities, not from one common probability for everyone.

Sequential inverse-probability weighting uses that product structure. It assigns larger weights to observed panel members whose histories made continued participation less likely. Under explicit assumptions, this reweights the observed complete cases to identify moments for the original baseline cohort. It does not, by itself, make dropout ignorable: the assumptions about how retention relates to unobserved data are essential.

Notation and the Sequential Weight

Consider a cohort of \(N\) panel members measured at waves \(t=1,\ldots,T\). Assume everyone is observed at wave 1. Let \(R_{it}=1\) if member \(i\) is observed at wave \(t\), and \(R_{it}=0\) otherwise. Monotone dropout means \(R_{it}=1\) implies \(R_{i,t-1}=1\): once a member drops out, that member is not observed again. Let \(H_{i,t-1}\) denote the observed history available just before the retention decision for wave \(t\), including baseline variables and any earlier observed outcomes or covariates.

For a member still observed at wave \(t-1\), define the conditional retention probability \(q_{it}=P(R_{it}=1\mid R_{i,t-1}=1,H_{i,t-1})\). The probability of being observed through wave \(t\), conditional on the sequence of histories encountered, is the product of these period-specific probabilities.

Definition: The sequential inverse-probability weight through wave \(t\) for an observed member is $$ W_{it}=\frac{R_{it}}{\prod_{s=2}^{t}q_{is}}. $$ For a complete panel through wave \(T\), the weight is \(W_{iT}=1/\prod_{s=2}^{T}q_{is}\) when \(R_{iT}=1\), and zero otherwise. The empty product at \(t=1\) is 1, so \(W_{i1}=R_{i1}\).

Each factor in the denominator is conditional on still being at risk and on the history observed before that wave's retention decision. Thus the product is generally not the product of unconditional retention rates. If a retention model is estimated, it is fitted among members who are observed at the preceding wave, using the relevant observed histories. The resulting weights are then calculated for members observed through the wave of interest.

Worked Example: Multiplying Conditional Retention Probabilities

Suppose a three-wave panel has everyone observed at wave 1. For one history at wave 2, the conditional probability of retention is \(q_{i2}=0.80\), and for the resulting wave-2 history the probability of retention at wave 3 is \(q_{i3}=0.75\). For another member's history, these probabilities are \(0.50\) and \(0.80\). Calculate each complete-panel probability and weight.

For the first history, the probability of retention through wave 3 is \((0.80)(0.75)=0.60\). A complete case with this history receives weight \(1/0.60=1.667\).

For the second history, the probability is \((0.50)(0.80)=0.40\), and the complete case receives weight \(1/0.40=2.500\). The second history receives more weight because its retention sequence made complete observation less likely. These products are conditional sequence probabilities; they need not be the same across panel members.

Conditions for Identifying Full-Panel Moments

Let \(Z_i\) denote the member's full data across all waves, including values that may be unobserved after dropout, and let \(g(Z_i)\) be an integrable function of those data. Examples include a wave-specific outcome, a product of outcomes from two waves, or another moment needed to describe the full-panel distribution. The target is the baseline-cohort moment \(E[g(Z_i)]\), not merely the corresponding average among people who completed the panel.

Sequential exchangeability: At each wave \(t\geq2\), among members observed at wave \(t-1\), retention at wave \(t\) is independent of the full data not yet observed, conditional on the observed history \(H_{i,t-1}\). Equivalently, the conditional retention probability given the full data and that history equals \(q_{it}\).

Positivity: For every observed history with positive probability in the target cohort, the conditional probability of remaining observed at the next wave is strictly positive. For identification by complete cases, the product retention probability through \(T\) must therefore be positive for histories represented in the target population.

Sequential exchangeability is a missingness assumption, not a consequence of using weights. It says that, after conditioning on the recorded history, the retention decision does not further select members according to their unobserved data relevant to \(g\). Positivity rules out histories for which continued observation is impossible. If a subgroup has zero retention probability, no reweighting of observed members from that subgroup can recover its unobserved outcomes without additional assumptions.

Why the Product Weight Identifies the Moment

Under the stated conditions, the complete-case weighted moment satisfies

$$ E\left[\frac{R_{iT}}{\prod_{t=2}^{T}q_{it}}g(Z_i)\right]=E[g(Z_i)]. $$

The reason is that each inverse factor cancels one conditional retention decision. To see the cancellation, condition first on the full data and on the history available before wave \(T\). Among members observed at \(T-1\), sequential exchangeability gives \(E[R_{iT}\mid Z_i,H_{i,T-1},R_{i,T-1}=1]=q_{iT}\). Thus the expected factor \(R_{iT}/q_{iT}\) is 1 for those still at risk. Repeating this argument for wave \(T-1\), then wave \(T-2\), and so on, removes every retention indicator and its corresponding probability factor.

$$ E\left[ \frac{R_{iT}}{\prod_{t=2}^{T}q_{it}}g(Z_i) \right] = E\left[ \frac{R_{i,T-1}}{\prod_{t=2}^{T-1}q_{it}}g(Z_i) \right] =\cdots=E[g(Z_i)]. $$

This argument establishes identification of the population moment. In a finite sample, if the probabilities are known, the cohort-scaled inverse-probability estimator uses the baseline cohort size \(N\) in its denominator. A normalized, or Hájek, version divides by the sum of observed weights instead. The normalized version is often less sensitive to an overall mismatch in weight totals, but it is a ratio estimator and does not have the same exact finite-sample unbiasedness property as the cohort-scaled estimator when the probabilities are known.

Worked Example: Estimating a Full-Cohort Mean

A hypothetical baseline cohort contains \(N=5\) members. Three are observed at all three waves, with final-wave outcomes 10, 6, and 8 units. Their respective products of conditional retention probabilities are 0.60, 0.60, and 0.40. Assume the sequential exchangeability and positivity conditions hold, and treat these probabilities as known. Calculate the cohort-scaled and normalized weighted estimates of the full-cohort mean outcome.

The cohort-scaled estimate is the weighted sum divided by the baseline cohort size:

$$ \widehat{\mu}_{HT} = \frac{10/0.60+6/0.60+8/0.40}{5} = \frac{50/3+10+20}{5} = \frac{140/3}{5} = 9.333. $$

The estimate is 9.333 outcome units. Its numerator is based on the three observed complete cases, while the denominator remains the five baseline members.

The normalized estimate divides the same weighted sum by the sum of the three weights:

$$ \widehat{\mu}_{H} = \frac{10/0.60+6/0.60+8/0.40} {1/0.60+1/0.60+1/0.40} = \frac{140/3}{35/6} = 8.000. $$

This estimate is also in outcome units. The two estimates differ because one scales by the known baseline cohort size and the other normalizes by the observed weight total. The displayed calculations use unrounded probabilities and weights throughout; rounding intermediate values before the final division could change the last reported digit.

Moments Beyond the Mean

The same identity applies to any integrable function \(g\) of the full panel data. For example, if a target is the average product of outcomes at waves 1 and 3, take \(g(Z_i)=Y_{i1}Y_{i3}\). The complete-case weighted estimate then targets that cross-moment. Such moments can be useful for describing the joint distribution across waves or forming other quantities; the weighting identity does not require \(g\) to be a single outcome.

Worked Example: A Weighted Cross-Moment

In a hypothetical baseline cohort of \(N=6\), three members are observed at all three waves. Their wave-1 and wave-3 outcomes, along with their retention probabilities, are: member A has \(Y_{i1}=2\), \(Y_{i3}=5\), \(q_{i2}=0.80\), \(q_{i3}=0.75\); member B has \(Y_{i1}=1\), \(Y_{i3}=4\), \(q_{i2}=0.50\), \(q_{i3}=0.80\); member C has \(Y_{i1}=3\), \(Y_{i3}=2\), \(q_{i2}=0.50\), \(q_{i3}=0.60\). Calculate the cohort-scaled estimate of \(E[Y_{i1}Y_{i3}]\).

The products of outcomes are 10, 4, and 6. The corresponding retention probability products are 0.60, 0.40, and 0.30, so the inverse weights are \(1/0.60\), \(1/0.40\), and \(1/0.30\). Therefore,

$$ \widehat{E[Y_{i1}Y_{i3}]} = \frac{10/0.60+4/0.40+6/0.30}{6} = \frac{50/3+10+20}{6} = \frac{140/3}{6} = 7.778. $$

The estimated cross-moment is 7.778 outcome-units-squared. As with a mean, identification relies on the sequential assumptions applying to the full-data function being estimated, here the product of the two outcomes.

Practical Interpretation and Common Pitfalls

Sequential weights clarify which histories are represented by each observed complete case: a person with a low modeled chance of remaining in the panel represents more members of the baseline cohort than a person with a high chance. The method can therefore address selection induced by observed histories under sequential exchangeability. It cannot correct selection that still depends on unobserved outcomes after conditioning on those histories, nor can it recover information from histories with zero probability of retention.

The probability models must also respect the order of information. A retention probability for wave \(t\) is conditional on information available before the wave-\(t\) retention decision. If a current-wave outcome is only measured when the person is retained, it cannot be included as an observed predictor of that same retention decision. A different timing convention, in which a measurement occurs before dropout, would require defining the history and decision order accordingly.

  • Using one retention rate for everyone. This ignores the dependence of retention on observed history. The sequential product uses a conditional probability at each wave for the history actually encountered.
  • Multiplying unconditional rates. Conditional retention probabilities generally vary with earlier histories, so a product of population-average rates need not equal an individual's probability of completing the panel.
  • Treating weighting as proof of exchangeability. Weights do not establish that missingness is conditionally independent of unobserved data. That is an identifying assumption requiring substantive justification.
  • Ignoring positivity or extreme weights. Very small retention probabilities produce large weights and can make estimates unstable. Positivity is an identification condition; diagnostics can reveal practical problems but cannot create support where none exists.
  • Mixing the two denominators. The cohort-scaled estimator divides by \(N\); the normalized estimator divides by the sum of observed weights. They are distinct estimates and should be labeled and interpreted accordingly.
Key takeaway: For monotone dropout, the complete-panel inverse-probability weight is the reciprocal of the product of period-specific conditional retention probabilities. Sequential exchangeability and positivity make this weighting identify full-panel moments for the baseline cohort.

Check Your Understanding

Answer each question using the definitions and assumptions in this tutorial.

  1. A member's conditional retention probabilities at waves 2, 3, and 4 are 0.80, 0.50, and 0.75. What is the probability of remaining observed through wave 4, and what complete-panel weight corresponds to it?
  2. State sequential exchangeability in words. Why does it concern unobserved data as well as the observed history?
  3. What does positivity rule out, and why can a weight not recover outcomes for a history with zero retention probability?
  4. For a cohort-scaled inverse-probability estimate of a mean, what belongs in the denominator: the baseline cohort size or the sum of observed weights? How does the normalized estimate differ?
  5. Explain why using a single unconditional retention rate at every wave can fail to represent a member's probability of completing the panel.
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