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Identifying Restrictions · Tutorial 248 of 499

Available-Case Restrictions With Unequal Pattern Weights

Learn to form available-case mixtures using pattern proportions, apply the restriction to missing outcomes, and understand what the resulting distribution does—and does not—assume.

Advanced 11 min read

What You'll Learn

  • Define the available-case distribution at a visit from the patterns that observed that outcome.
  • Calculate mixture weights from unequal pattern proportions rather than averaging patterns equally.
  • Derive the full-population marginal distribution implied by the available-case restriction.
  • Reconstruct missing-outcome means and check the arithmetic against the pattern proportions.
  • Distinguish marginal restrictions from restrictions applied within covariate strata.
  • Recognize when an available-case restriction is an assumption rather than a conclusion from observed data.

Why Pattern Weights Matter

When dropout patterns have different sizes, the available cases at a visit are not an equally weighted collection of patterns. A pattern representing half the study contributes more individuals to the available-case distribution than a pattern representing one tenth. That distinction matters when the available-case distribution is used to specify the distribution of outcomes that were not observed.

The available-case restriction makes this borrowing explicit: for a particular visit, the distribution among patterns missing that visit is set equal to the distribution among all patterns that observed it. The latter is a mixture, and its weights come from the proportions of the observed patterns. This tutorial derives that mixture and follows its implications for a three-visit monotone-dropout setting.

Patterns and the Available-Case Mixture

Suppose each participant is scheduled for visits 1, 2, and 3, with monotone dropout: after a participant misses a visit, they do not return at a later visit. Write \(Y_j\) for the outcome at visit \(j\). Use pattern labels \(111\), \(110\), and \(100\): the \(j\)th digit is 1 if the outcome at visit \(j\) is observed and 0 otherwise. Let \(\pi_g=P(G=g)\) be the proportion of the study population in pattern \(g\), where \(G\) is the dropout pattern.

Definition: At visit \(j\), let \(\mathcal{A}_j\) be the set of patterns in which \(Y_j\) is observed. The available-case distribution at that visit is \(F_j^{AC}(y)=P(Y_j\leq y\mid Y_j\text{ is observed})\). If pattern membership determines whether \(Y_j\) is observed, then \(F_j^{AC}(y)=\sum_{g\in\mathcal{A}_j}w_{jg}F_{jg}(y)\), where \(F_{jg}(y)=P(Y_j\leq y\mid G=g)\) and \(w_{jg}=\pi_g/\sum_{h\in\mathcal{A}_j}\pi_h\).

The weights are conditional proportions among people who observed the visit. They are nonnegative and sum to 1. In the three-visit setting, \(\mathcal{A}_2=\{111,110\}\) and \(\mathcal{A}_3=\{111\}\). Thus visit 2's available cases mix the \(111\) and \(110\) distributions, while visit 3's available cases come only from pattern \(111\).

The available-case restriction is an identifying assumption for patterns missing a visit. At visit 2, it sets \(F_{2,100}=F_2^{AC}\). At visit 3, it sets \(F_{3,110}=F_3^{AC}\) and \(F_{3,100}=F_3^{AC}\). These equalities concern distributions, not merely means. They are assumptions about unobserved outcomes; the observed data alone do not verify them.

Assumptions: The visit outcomes and dropout-pattern labels are well-defined; the pattern proportions describe the population (or their sample analogues are used for reconstruction); and each visit has at least one pattern with an observed outcome. The available-case restriction is imposed for the outcome and conditioning set being analyzed. The monotone pattern labels above are used for the examples; the weighted-mixture formula itself applies whenever the observed patterns at a visit are correctly identified.

Deriving the Implied Marginal Distribution

For any fixed visit \(j\), the overall population distribution is a mixture across all dropout patterns. Write \(\mathcal{M}_j\) for patterns missing \(Y_j\), and let \(q_j=\sum_{g\in\mathcal{A}_j}\pi_g\) be the population proportion with \(Y_j\) observed. The available-case restriction assigns \(F_{jg}=F_j^{AC}\) for every \(g\in\mathcal{M}_j\). Substitution into the full mixture gives:

$$ \begin{aligned} F_j(y) &=\sum_{g\in\mathcal{A}_j}\pi_g F_{jg}(y) +\sum_{g\in\mathcal{M}_j}\pi_g F_{jg}(y)\\ &=q_jF_j^{AC}(y)+(1-q_j)F_j^{AC}(y)\\ &=F_j^{AC}(y). \end{aligned} $$

The first equality is the law of total probability over dropout patterns. In the second, the observed-pattern contribution is \(q_jF_j^{AC}\) by the definition of the available-case mixture, while the missing-pattern contribution is \((1-q_j)F_j^{AC}\) by the restriction. Their coefficients sum to 1, proving the result. Thus the restriction implies that the full-population marginal distribution at that visit equals the available-case distribution.

This result does not mean that every pattern has the same outcome distribution. At visit 2, for example, \(F_{2,111}\) and \(F_{2,110}\) may differ; the restriction assigns their weighted mixture to pattern \(100\). The overall distribution equals that mixture because the missing-pattern contribution has been defined to match it.

Worked Example: Unequal Pattern Weights at Visit 2

Suppose the pattern proportions are \(\pi_{111}=0.50\), \(\pi_{110}=0.30\), and \(\pi_{100}=0.20\). At visit 2, \(Y_2\) takes values 0 or 10. Among pattern \(111\), \(P(Y_2=10\mid G=111)=0.20\); among pattern \(110\), \(P(Y_2=10\mid G=110)=0.60\). Find the available-case probability of 10 and the probability assigned to pattern \(100\) by the restriction.

The observed fraction at visit 2 is \(0.50+0.30=0.80\). The normalized weights among available cases are \(0.50/0.80=0.625\) for pattern \(111\) and \(0.30/0.80=0.375\) for pattern \(110\). Therefore,

$$ P(Y_2=10\mid Y_2\text{ observed}) =(0.625)(0.20)+(0.375)(0.60) =0.125+0.225=0.350. $$

The available-case restriction assigns this probability to pattern \(100\): \(P(Y_2=10\mid G=100)=0.350\). The implied overall probability is \((0.50)(0.20)+(0.30)(0.60)+(0.20)(0.35)=0.10+0.18+0.07=0.35\). It matches the available-case probability, as the derivation requires. The calculation uses the unequal pattern proportions; it does not give the two observed patterns equal weight.

Means and Pattern-Weighted Reconstruction

If the outcome has a finite mean, the distributional restriction implies the corresponding mean restriction. Let \(\mu_{jg}=E(Y_j\mid G=g)\) when this expectation exists. The available-case mean is \(\mu_j^{AC}=\sum_{g\in\mathcal{A}_j}w_{jg}\mu_{jg}\). For any pattern missing the visit, the restriction gives \(E(Y_j\mid G=g)=\mu_j^{AC}\). The overall mean is then \(\sum_g\pi_g\mu_{jg}=\mu_j^{AC}\). A mean-only restriction can also be imposed directly, but it is weaker: matching means does not establish matching distributions.

Worked Example: Reconstructing a Missing-Pattern Mean

At visit 2, suppose \(\pi_{111}=0.45\), \(\pi_{110}=0.35\), and \(\pi_{100}=0.20\). The observed pattern means are 8 minutes for pattern \(111\) and 12 minutes for pattern \(110\). Under the available-case restriction, calculate the mean assigned to \(Y_2\) in pattern \(100\), then verify the overall mean.

The proportion observed at visit 2 is \(0.45+0.35=0.80\), so the available-case weights are \(0.45/0.80=0.5625\) and \(0.35/0.80=0.4375\). The available-case mean, in minutes, is

$$ \mu_2^{AC} =(0.5625)(8)+(0.4375)(12) =4.5+5.25=9.75\text{ minutes}. $$

The restriction assigns a mean of 9.75 minutes to pattern \(100\). The full-population mean reconstructed across patterns is \((0.45)(8)+(0.35)(12)+(0.20)(9.75)=3.60+4.20+1.95=9.75\) minutes. The result follows because the restriction gives the missing pattern the same mean as the observed pool. Using \((8+12)/2=10\) minutes would incorrectly give the observed patterns equal weight despite their different proportions.

Conditioning Changes the Weights

The mixture must use the conditioning set specified by the restriction. If a restriction is imposed within baseline-covariate strata, compute the observed-pattern weights separately within each stratum. A marginal mixture generally is not the same as a mixture within strata: both pattern proportions and outcome means can vary with the covariate.

Worked Example: Applying the Mean Restriction Within Strata

Suppose a baseline covariate \(X\) has two equally common strata, \(X=0\) and \(X=1\). At visit 2, the pattern proportions conditional on \(X=0\) are \(0.20,0.30,0.50\) for \(111,110,100\), respectively; for \(X=1\), they are \(0.60,0.20,0.20\). The observed \(Y_2\) means for patterns \(111,110\) are 4 and 10 when \(X=0\), and 6 and 14 when \(X=1\). Find the restricted mean for pattern \(100\) in each stratum and the resulting overall mean.

For \(X=0\), the observed proportion is \(0.20+0.30=0.50\), so the available-case weights are \(0.20/0.50=0.40\) and \(0.30/0.50=0.60\). The restricted mean for pattern \(100\) is \((0.40)(4)+(0.60)(10)=7.6\). For \(X=1\), the observed proportion is \(0.60+0.20=0.80\), giving weights \(0.60/0.80=0.75\) and \(0.20/0.80=0.25\). The restricted mean there is \((0.75)(6)+(0.25)(14)=8.0\).

Within each stratum, the full-population mean equals that stratum's available-case mean. Because the strata are equally common, the overall mean is \((0.50)(7.6)+(0.50)(8.0)=7.8\) outcome units. This is a conditional restriction; replacing it with a single marginal mixture would use different weights and would encode a different assumption.

Interpretation, Diagnostics, and Alternatives

The pattern weights determine which observed distributions contribute to the borrowed distribution. A small observed pattern receives a small mixture weight, even if its mean or distribution differs sharply from that of a larger pattern. At a later visit in this three-visit monotone setting, only pattern \(111\) observes \(Y_3\); the available-case distribution there is simply \(F_{3,111}\). The restriction assigns that distribution to both patterns missing visit 3. At visit 2, by contrast, the borrowed distribution is a mixture of two observed patterns.

The restriction is not a consequence of dropout being monotone, and it is not established by the observed outcomes. It is an identifying choice. Its implications can be checked for coherence with observed pattern summaries and the restrictions being imposed, but agreement of observed summaries cannot prove the assumption about outcomes that were not observed. A later analysis may reveal conflicts among restrictions or other incompatibilities.

The complete-case restriction is a different borrowing choice: it uses the complete-pattern distribution rather than the available-case pool. For visit 2, that replaces the mixture of patterns \(111\) and \(110\) with pattern \(111\) alone. An unweighted average of observed-pattern distributions is another possible calculation, but it is not the available-case mixture unless the observed pattern proportions are equal. Neighboring-case restrictions make a different assignment based on selected adjacent patterns; they should not be mistaken for the pooled available-case rule.

A common mistake is to say, “There are two observed patterns, so average their distributions 50–50.” That reasoning counts pattern labels rather than participants. The correction is to normalize each observed pattern's population proportion by the total proportion observed at that visit. Another mistake is to conclude that the restriction says all pattern-specific distributions are equal. It says that each missing pattern is assigned the available-case mixture; observed patterns may retain different distributions.

Key takeaway: For a visit with unequal dropout-pattern proportions, the available-case distribution is a pattern-weighted mixture with weights proportional to the observed pattern sizes. The available-case restriction assigns this mixture to patterns missing the visit, so the implied full-population marginal distribution at that visit equals the available-case distribution.

Check Your Understanding

Answer each question using pattern proportions among the population, and remember to normalize the observed-pattern weights.

  1. At a visit, two observed patterns have proportions 0.40 and 0.10. What are their available-case weights?
  2. The available-case mixture gives probability 0.25 to an event, and 30% of participants are missing that visit. Under the restriction, what is the event probability in the missing group and in the full population?
  3. Two observed patterns have proportions 0.20 and 0.30 and means 5 and 9. What available-case mean is assigned to a missing pattern?
  4. Why can a marginal available-case mixture differ from a mixture calculated separately within covariate strata?

Solutions. (1) The observed proportion is \(0.40+0.10=0.50\), so the weights are \(0.40/0.50=0.80\) and \(0.10/0.50=0.20\). (2) The restriction assigns probability 0.25 to the missing group. The full-population probability is \((0.70)(0.25)+(0.30)(0.25)=0.25\). (3) The available-case weights are \(0.20/0.50=0.40\) and \(0.30/0.50=0.60\), giving \((0.40)(5)+(0.60)(9)=7.4\) outcome units. (4) Pattern proportions and outcome distributions can differ across strata, so conditional mixtures use stratum-specific weights and need not equal the marginal mixture.

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