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Identifying Restrictions · Tutorial 244 of 499

Deriving Available-Case Mixtures Across Dropout Patterns

Learn to derive pattern-weighted available-case distributions and see exactly which unobserved pattern components require an identifying assumption.

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What You'll Learn

  • Define dropout patterns by the last visit observed and identify which patterns contribute data at a given visit.
  • Derive the available-case distribution using the law of total probability.
  • Calculate pattern weights and combine pattern-specific distributions with them.
  • Distinguish the observed available-case mixture from the full-population distribution.
  • Identify the unobserved pattern-specific components and state what the available-case restriction assumes.

Why the Available-Case Distribution Is a Mixture

At a particular visit, the people with observed outcomes may come from several dropout patterns. For example, someone observed at visit 2 might drop out after visit 2 or might remain in the study through visit 3. The distribution among everyone observed at visit 2 therefore combines the visit-2 distributions for each of those patterns. The combination is weighted by how common each pattern is among people observed at that visit, not by equal weights.

This distinction matters because a pattern-specific distribution for an outcome that was not observed cannot be read off from the available-case data. The available-case mixture can be derived and estimated from observed outcomes; using it to represent the missing pattern components requires an additional identifying restriction.

Definition: Consider \(K\) scheduled visits and monotone dropout. Let \(G=g\) mean that a participant was observed through visit \(g\), and not thereafter; \(G=K\) denotes observation through all visits. Let \(Y_t\) be the outcome at visit \(t\), and let \(R_t=1\) indicate that \(Y_t\) is observed. Then \(R_t=1\) exactly when \(G\geq t\). Write \(\pi_g=P(G=g)\) and \(F_{t\mid g}(y)=P(Y_t\leq y\mid G=g)\). The available-case distribution at visit \(t\) is \(F_t^{AC}(y)=P(Y_t\leq y\mid R_t=1)\).

The distribution \(F_{t\mid g}\) is a conceptual pattern-specific distribution even when \(Y_t\) is not observed in pattern \(g\). If \(g\geq t\), it is a distribution of outcomes observed at visit \(t\). If \(g<t\), it concerns outcomes after that pattern's dropout and cannot be directly estimated from the observed visit-\(t\) data.

Deriving the Pattern-Weighted Mixture

Among people observed at visit \(t\), only patterns \(g=t,\ldots,K\) are represented. Conditional on being observed, their pattern weights are \(P(G=g\mid R_t=1)=\pi_g/\sum_{h=t}^K\pi_h\). Applying the law of total probability across these mutually exclusive patterns gives the available-case mixture.

$$ F_t^{AC}(y) =\sum_{g=t}^{K}P(G=g\mid R_t=1)F_{t\mid g}(y) =\sum_{g=t}^{K}\frac{\pi_g}{\sum_{h=t}^{K}\pi_h}F_{t\mid g}(y). $$

The weights are nonnegative and sum to one. In a sample, if \(n_g\) participants have pattern \(g\), the corresponding empirical weights among the available cases are \(n_g/\sum_{h=t}^K n_h\). The mixture is not generally an unweighted average of the pattern-specific distributions.

The derivation does not require that dropout be independent of outcomes. It is an identity for the distribution among people observed at visit \(t\), given the pattern definition and observed data. Monotone dropout makes the set of contributing patterns especially clear. The identity does not, by itself, determine the outcome distributions in patterns with \(g<t\).

Worked Example: Combining Two Observed Patterns

Suppose a hypothetical three-visit study has 100 participants: 20 with \(G=1\), 30 with \(G=2\), and 50 with \(G=3\). At visit 2, the \(G=2\) pattern has outcome proportions 0.2, 0.6, and 0.2 at values 0, 1, and 2. The \(G=3\) pattern has proportions 0.2, 0.4, and 0.4 at those values. Find the available-case distribution at visit 2.

The available cases are the \(G=2\) and \(G=3\) participants, so there are \(30+50=80\). Their mixture weights are \(30/80=0.375\) and \(50/80=0.625\). The \(G=1\) participants do not contribute because their visit-2 outcomes are unobserved.

For outcome value 0, the available-case probability is \(0.375(0.2)+0.625(0.2)=0.200\). For value 1, it is \(0.375(0.6)+0.625(0.4)=0.475\). For value 2, it is \(0.375(0.2)+0.625(0.4)=0.325\). These probabilities sum to \(0.200+0.475+0.325=1.000\), as required.

Equivalently, the available-case cumulative distribution is 0.200 at \(y=0\), 0.675 at \(y=1\), and 1 at \(y=2\). The same result follows by weighting the pattern-specific cumulative probabilities: at \(y=1\), \(0.375(0.8)+0.625(0.6)=0.675\). Every quantity used here is observed within a pattern represented at visit 2.

What the Mixture Does Not Identify

The full-population distribution of \(Y_t\), if outcomes at visit \(t\) were defined for everyone, would include all dropout patterns:

$$ F_t(y)=\sum_{g=1}^{K}\pi_g F_{t\mid g}(y). $$

For \(g\geq t\), the pattern-specific outcomes at visit \(t\) are observed, subject to ordinary sampling uncertainty. For \(g<t\), they are not observed. Thus the available-case mixture identifies the distribution conditional on being observed at visit \(t\), but does not identify the full-population distribution \(F_t\). The missing components are the pattern-specific distributions \(F_{t\mid g}\) for \(g<t\), and their population weights are \(\pi_g\).

The available-case restriction is an assumption that fills this gap by setting each missing pattern-specific distribution equal to the observed available-case distribution at that visit: \(F_{t\mid g}(y)=F_t^{AC}(y)\) for every \(g<t\). Under this restriction, the full-population distribution equals the available-case distribution. The observed-data mixture does not prove the restriction; its validity cannot be checked directly using outcomes that are missing in those patterns.

$$ \begin{aligned} F_t(y) &=\sum_{g<t}\pi_gF_{t\mid g}(y)+\sum_{g=t}^K\pi_gF_{t\mid g}(y)\\ &=\sum_{g<t}\pi_gF_t^{AC}(y) +\left(\sum_{g=t}^K\pi_g\right)F_t^{AC}(y)\\ &=F_t^{AC}(y)\sum_{g=1}^K\pi_g =F_t^{AC}(y). \end{aligned} $$

This short derivation relies on two facts: the restriction equates every early-dropout component to the available-case distribution, and the pattern probabilities sum to one. Without that restriction, the first line contains unobserved distributions and the full-population distribution is not determined by the available cases alone.

Worked Example: A Visit-2 Mean and an Unidentified Component

In a hypothetical study, 30 participants have \(G=1\), 30 have \(G=2\), and 40 have \(G=3\). At visit 2, the observed mean is 6 points in each of patterns \(G=2\) and \(G=3\). Find the available-case mean and show what is not identified about the population mean.

Among visit-2 available cases, the pattern weights are \(30/(30+40)=3/7\) and \(40/(30+40)=4/7\). Therefore \(\bar{Y}_2^{AC}=(3/7)(6)+(4/7)(6)=6\) points. The observed available-case mean is 6 points.

Let \(\mu_{2\mid 1}\) denote the unknown mean at visit 2 among the \(G=1\) pattern. The full-population mean is \((30/100)\mu_{2\mid 1}+(30/100)(6)+(40/100)(6)=0.3\mu_{2\mid 1}+4.2\) points. For instance, if \(\mu_{2\mid 1}=4\), this expression is 5.4 points; if \(\mu_{2\mid 1}=8\), it is 6.6 points. Both values are compatible with the stated observed means, so the full-population mean is not identified by them.

Under the available-case restriction, the \(G=1\) visit-2 distribution is set equal to the visit-2 available-case distribution. In particular, its mean is 6 points, giving \(0.3(6)+0.3(6)+0.4(6)=6\) points for the population mean. This conclusion follows from the restriction, not from observing the \(G=1\) visit-2 outcomes.

Later Visits and Boundary Patterns

As the visit index increases, fewer dropout patterns remain observable. At the final visit, only \(G=K\) contributes to the available-case distribution. Earlier dropout patterns then all have unobserved final-visit outcomes. This is a boundary case of the same mixture formula: its observed mixture has one component, while the full-population distribution still has components for every pattern.

Worked Example: The Final-Visit Mixture

Return to the hypothetical counts of 20 participants with \(G=1\), 30 with \(G=2\), and 50 with \(G=3\). At visit 3, only the \(G=3\) pattern is observed. Suppose its 50 outcomes consist of 10 scores of 40, 25 scores of 60, and 15 scores of 80.

The available-case weight for \(G=3\) is \(50/50=1\), so \(F_3^{AC}=F_{3\mid 3}\). Its observed mean is \(\{10(40)+25(60)+15(80)\}/50=(400+1500+1200)/50=62\) points.

The full-population mean would be \(0.2\mu_{3\mid 1}+0.3\mu_{3\mid 2}+0.5(62)\) points. The first two means are unobserved. For example, if they were 40 and 80 points, the full mean would be \(0.2(40)+0.3(80)+0.5(62)=63\) points. The available-case restriction instead sets both missing-pattern distributions equal to \(F_{3\mid 3}\), yielding a population mean of 62 points. The difference between these results illustrates the identifying role of the assumption.

Assumptions, Checks, and Alternatives

Before using this construction, verify that the pattern labels match the study's observation process. The simple indexing \(G=g\) assumes monotone dropout: after a participant misses a visit, later outcomes are not observed. If the data allow intermittent missingness, the pattern definition must represent the actual observation histories rather than silently treating them as monotone.

Useful checks concern the observable calculation, not the unobserved restriction. Confirm that pattern counts among available cases give nonnegative weights summing to one; that each included pattern actually has outcomes at visit \(t\); and that the weighted probabilities form a valid distribution. If a calculated mixture probability is negative or the probabilities do not sum to one, the weighting or pattern-specific summaries have been mishandled. Sparse patterns can also make empirical pattern distributions unstable.

No diagnostic based only on the observed outcomes can establish that the missing-pattern distributions equal \(F_t^{AC}\). This restriction is not a claim that dropout is unrelated to outcomes in every sense; it is a specific equality of outcome distributions across the observed and unobserved patterns at a given visit. Its plausibility must be assessed using subject-matter knowledge and sensitivity to alternative assumptions.

The main contrast is with a complete-case restriction, which uses only participants observed through all visits as the reference group. At an intermediate visit, the available-case distribution also includes participants who are observed at that visit but drop out later. It can therefore differ from the complete-case distribution. Pattern-specific extrapolation is another alternative: it assigns missing-pattern distributions using a model or sensitivity parameters rather than setting them equal to the available-case distribution. These approaches answer different assumption-based questions; none is made correct by the mixture identity itself.

Common pitfall: “The available-case distribution is a weighted mixture of observed pattern distributions, so it identifies the population distribution.” The first clause is correct, but the conclusion is not. The available-case mixture includes only patterns with outcomes observed at that visit. The full-population mixture also contains the unobserved components for patterns that dropped out earlier; equating those components to the available-case distribution is an additional, untestable restriction.

Check Your Understanding

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

  1. At visit 2 of a three-visit monotone-dropout study, which patterns enter the available-case mixture?
  2. At visit 2, pattern weights among available cases are 0.25 for \(G=2\) and 0.75 for \(G=3\). At \(y=5\), their pattern-specific cumulative probabilities are 0.4 and 0.8. What is \(F_2^{AC}(5)\)?
  3. At visit 2, which pattern-specific distributions remain unobserved, and why?
  4. State the available-case restriction for patterns missing the visit-\(t\) outcome. Does the observed mixture verify it?

Solutions. (1) Patterns \(G=2\) and \(G=3\), because both are observed through visit 2. (2) \(F_2^{AC}(5)=0.25(0.4)+0.75(0.8)=0.10+0.60=0.70\). (3) \(F_{2\mid 1}\) is unobserved because the \(G=1\) pattern drops out before visit 2; its visit-2 outcomes are missing. (4) The restriction is \(F_{t\mid g}(y)=F_t^{AC}(y)\) for each \(g<t\). No: those pattern-specific outcomes are unobserved, so equality cannot be checked from the observed mixture alone.

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