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Identifying Restrictions · Tutorial 253 of 499

Neighboring-Case Restrictions in a Three-Visit Trial

Learn to assign neighboring patterns to missing visits in a three-visit trial and compute the means implied by those restrictions.

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What You'll Learn

  • Represent three-visit monotone dropout patterns using observation indicators.
  • Assign a neighboring pattern to each missing visit, including chained assignments.
  • Calculate pattern-specific means under neighboring-case restrictions.
  • Combine pattern-specific means using pattern proportions to obtain visit means.
  • Recognize the assumptions, limitations, and diagnostic checks relevant to these restrictions.

Why Neighboring Patterns Matter

In a three-visit trial with dropout, each participant's observed visit pattern helps determine which outcomes are available to estimate directly. A restriction is needed to say how a missing outcome in one pattern relates to outcomes in other patterns. A neighboring-case restriction makes that link using patterns adjacent in the dropout sequence. The assignment must be done visit by visit: the nearest pattern in the sequence may itself lack the outcome, so a restriction can need to pass through more than one pattern.

The calculations below use a monotone pattern: once a participant misses a visit, later visits are also missing. The restrictions are assumptions about pattern-specific means, not facts established by the observed data. Their purpose is to supply values for means that the data alone do not reveal.

Definition: Write a three-visit observation pattern as \(r=(R_1,R_2,R_3)\), where \(R_j=1\) means visit \(j\) is observed and \(R_j=0\) means it is missing. Under monotone dropout, the patterns are \(111\) (all visits observed), \(110\) (visits 1 and 2 observed), and \(100\) (visit 1 observed). Let \(\mu_{j,r}=E(Y_j\mid R=r)\) be the mean outcome at visit \(j\) in pattern \(r\), whether or not that outcome is observed there. A neighboring-case restriction equates a missing-outcome mean in one pattern to the corresponding outcome mean in an adjacent pattern in the dropout sequence. If that adjacent pattern also lacks the outcome, the equality can be applied successively along the sequence.

For these three patterns, the adjacent steps are \(111\leftrightarrow110\leftrightarrow100\). Thus, the missing visit 3 mean in pattern \(110\) can be linked to the observed visit 3 mean in pattern \(111\). In pattern \(100\), visit 2 can be linked to the observed visit 2 mean in pattern \(110\). Visit 3 in pattern \(100\) can be linked to the restricted visit 3 mean in pattern \(110\), which is itself linked to the observed mean in pattern \(111\). This is a chain of identifying assumptions; it does not make the visit 3 outcomes for patterns \(110\) or \(100\) observed.

Set Up the Restrictions Before Calculating

For a simple marginal mean analysis, the neighboring-case equalities are

$$ \mu_{3,110}=\mu_{3,111},\qquad \mu_{2,100}=\mu_{2,110},\qquad \mu_{3,100}=\mu_{3,110}. $$

The first equality supplies the visit 3 mean for pattern \(110\); the second supplies the visit 2 mean for pattern \(100\); and the third supplies its visit 3 mean. Combining the first and third gives \(\mu_{3,100}=\mu_{3,111}\). The observed means that anchor the chain are the visit 3 mean among pattern \(111\) participants and the visit 2 mean among pattern \(110\) participants.

The mean equalities are marginal over the participants in each pattern. They do not assert that participants in different patterns have identical observed histories, nor do they state equality conditional on those histories. The restriction is most credible when the adjacent patterns are sufficiently comparable for the outcome and target population at issue. In a real analysis, that comparability requires subject-matter justification; the equalities cannot be verified for outcomes that are missing in the relevant patterns.

Worked Example: Assigning Each Missing Visit

Suppose a hypothetical three-visit trial has 100 participants: 40 with pattern \(111\), 35 with pattern \(110\), and 25 with pattern \(100\). The observed pattern-specific means are:

PatternObserved visitsObserved means
1111, 2, 3\(\bar{Y}_{1,111}=10,\ \bar{Y}_{2,111}=12,\ \bar{Y}_{3,111}=14\)
1101, 2\(\bar{Y}_{1,110}=9,\ \bar{Y}_{2,110}=11\)
1001\(\bar{Y}_{1,100}=8\)

The bars here denote sample means among participants in the stated pattern. For this illustration, assume the neighboring-case restrictions are acceptable for the target means and use these observed means as estimates of their corresponding pattern-specific means.

Pattern \(110\), missing visit 3. Its adjacent pattern \(111\) has observed visit 3 outcomes, so the restriction gives \(\widehat{\mu}_{3,110}=\bar{Y}_{3,111}=14\).

Pattern \(100\), missing visit 2. The adjacent pattern \(110\) has observed visit 2 outcomes, giving \(\widehat{\mu}_{2,100}=\bar{Y}_{2,110}=11\).

Pattern \(100\), missing visit 3. The adjacent pattern \(110\) lacks observed visit 3 outcomes, but its visit 3 mean has just been restricted to the pattern \(111\) mean. Therefore \(\widehat{\mu}_{3,100}=\widehat{\mu}_{3,110}=14\).

The resulting pattern-specific mean estimates are \((10,12,14)\) for pattern \(111\), \((9,11,14)\) for pattern \(110\), and \((8,11,14)\) for pattern \(100\). The assigned value 14 for visit 3 in pattern \(100\) is a chained restriction, not a direct observed mean from that pattern.

From Pattern Means to Visit Means

Once each pattern has a mean for a visit, the overall mean at that visit is a pattern-weighted average. Let \(n_r\) be the number of participants in pattern \(r\), and let \(n=\sum_r n_r\). The pattern proportion is \(\hat{p}_r=n_r/n\). For visit \(j\), the restricted mean estimate is

$$ \widehat{\mu}_j=\sum_r \hat{p}_r\widehat{\mu}_{j,r}. $$

This formula follows by partitioning the trial population into its mutually exclusive patterns. If \(R\) denotes a participant's pattern, then \(E(Y_j)=\sum_r P(R=r)E(Y_j\mid R=r)\). Replacing the population pattern proportions and conditional means by their sample estimates gives the displayed estimator. The identity itself is a partitioning result; the values for missing outcomes still depend on the restrictions.

Worked Example: Calculating the Restricted Visit Means

Use the hypothetical data and restricted pattern means from the preceding example. The pattern proportions are \(\hat{p}_{111}=40/100=0.40\), \(\hat{p}_{110}=35/100=0.35\), and \(\hat{p}_{100}=25/100=0.25\).

At visit 1, all three patterns have observed means:

$$ \widehat{\mu}_1 =(0.40)(10)+(0.35)(9)+(0.25)(8) =4.00+3.15+2.00 =9.15. $$

At visit 2, pattern \(100\)'s missing mean is supplied by the restriction:

$$ \widehat{\mu}_2 =(0.40)(12)+(0.35)(11)+(0.25)(11) =4.80+3.85+2.75 =11.40. $$

At visit 3, the two dropout patterns receive the chained value 14:

$$ \widehat{\mu}_3 =(0.40)(14)+(0.35)(14)+(0.25)(14) =5.60+4.90+3.50 =14.00. $$

These are restricted estimates of the trial's visit-specific mean outcomes, in the same units as \(Y\). The visit 2 and visit 3 estimates are not simply observed-data averages: they incorporate the stated neighboring-case assumptions.

Worked Example: Why Pattern Weights Must Be Used

Consider a second hypothetical trial with 20 participants in pattern \(111\), 30 in pattern \(110\), and 50 in pattern \(100\). Suppose the observed mean at visit 2 is 16 in pattern \(111\), and 12 in pattern \(110\). Under the neighboring restriction, pattern \(100\)'s visit 2 mean is assigned 12. The restricted visit 2 mean is therefore

$$ \widehat{\mu}_2 =(20/100)(16)+(30/100)(12)+(50/100)(12) =3.20+3.60+6.00 =12.80. $$

An unweighted average of the three pattern means would be \((16+12+12)/3=13.33\), which is not the trial mean estimate: it gives each pattern equal weight despite their different sizes. The correct weighting uses each pattern's share of trial participants. Here the larger pattern \(100\), whose mean is assigned 12, contributes half of the total weight.

Assumptions, Diagnostics, and Alternatives

The arithmetic is simple, but its interpretation relies on several conditions. First, the dropout patterns must be defined consistently, and the monotone sequence must be appropriate for the data being analyzed. Second, every observed anchor mean must be based on participants with the stated pattern and visit. Third, the neighboring equalities must be defended as identifying restrictions for the target population and outcome scale. Finally, pattern proportions should reflect the population whose overall mean is being estimated; if the trial has sampling or analysis weights, the corresponding weighted proportions and means may be needed.

Useful checks include confirming that the pattern counts sum to the analysis sample size, verifying that every assigned value has a stated source, and recomputing each weighted mean from the pattern contributions. Compare the observed histories and baseline characteristics across adjacent patterns: large differences do not disprove a marginal equality, but they can make it less plausible. Also examine sensitivity to the anchor means. If a small change in an assumed equality substantially changes the overall estimate, the result is assumption-sensitive and should be reported as such.

A complete-case analysis uses only pattern \(111\), so it does not use the observed outcomes from dropout patterns to estimate the visit means. Available-case summaries use the outcomes observed at each visit but do not, by themselves, specify the missing pattern-specific means needed to describe a common target population. Neighboring-case restrictions instead supply those pattern-specific means through explicit cross-pattern equalities. These approaches answer different questions and rely on different assumptions; none makes the missing outcomes empirically observed.

A common mistake is to assign pattern \(100\)'s visit 3 mean directly from the visit 3 observations in pattern \(111\) and then describe that as a direct neighboring comparison. In the dropout sequence, pattern \(110\) lies between them and itself lacks observed visit 3 values. The correct account is the chain \(\mu_{3,100}=\mu_{3,110}=\mu_{3,111}\), with each equality stated as an assumption. In this particular three-pattern setup, the chain gives the same numerical value as direct equality to \(111\), but the distinction matters: it makes the required restrictions visible and prevents the borrowed value from being mistaken for observed evidence.

The method is limited by the strength and scope of its restrictions. Marginal mean equality does not imply equality of full outcome distributions, and a mean restriction does not automatically justify conclusions about variances, quantiles, or individual trajectories. Nor does a good fit to observed data validate an equality involving an unobserved mean. The resulting estimates should therefore be described as conditional on the selected neighboring-case restrictions, rather than as estimates made assumption-free by the pattern structure.

Key takeaway: Assign each missing visit to an adjacent pattern with an available or already restricted mean for that visit, show any chain explicitly, and then combine pattern-specific means using pattern proportions. The calculation is transparent; the identifying equalities remain assumptions.

Check Your Understanding

Answer each question using the three-pattern sequence \(111\leftrightarrow110\leftrightarrow100\). Solutions follow the questions.

  1. Under the neighboring restrictions, which observed mean anchors the missing visit 3 mean in pattern \(110\)?
  2. If the observed visit 2 mean in pattern \(110\) is 7.5, what value is assigned to the visit 2 mean in pattern \(100\)?
  3. If the visit 3 mean in pattern \(111\) is 18, what visit 3 mean is assigned to pattern \(100\), and through what chain?
  4. There are 10 participants in pattern \(111\), 20 in pattern \(110\), and 70 in pattern \(100\). Restricted visit 2 means are 9, 6, and 6, respectively. Calculate the restricted overall visit 2 mean.
  5. Does agreement between the observed visit 1 means across patterns establish the neighboring restriction for missing visit 3 means? Explain.

Solutions. (1) The observed visit 3 mean in pattern \(111\), because \(\mu_{3,110}=\mu_{3,111}\). (2) The assigned mean is 7.5, since \(\mu_{2,100}=\mu_{2,110}\). (3) The assigned mean is 18: \(\mu_{3,100}=\mu_{3,110}=\mu_{3,111}=18\). (4) The pattern proportions are 0.10, 0.20, and 0.70, so \((0.10)(9)+(0.20)(6)+(0.70)(6)=0.90+1.20+4.20=6.30\). (5) No. Agreement at visit 1 concerns a different outcome mean; it cannot establish an equality for unobserved visit 3 outcomes.

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