A Three-Visit Workshop on Neighboring Patterns

Patterns and the Target Mean

With three scheduled visits and monotone dropout, a participant who misses a visit is assumed not to return at a later visit. Write a pattern as three indicators in visit order: 1 means the outcome is observed and 0 means it is missing. The possible patterns are $111$, $110$, and $100$. Thus, $111$ participants have all three outcomes, $110$ participants are observed at visits 1 and 2 but not visit 3, and $100$ participants are observed only at visit 1.

These patterns are ordered by how soon dropout occurs. The neighboring pairs are $111$ with $110$, and $110$ with $100$. The first pair shares observed history through visit 2; the second shares observed history through visit 1. A neighboring-pattern restriction borrows a mean for a missing visit from the adjacent pattern. It is an assumption about population means, not a fact established by the observed table.

Definition: Let $Y_j$ be the outcome at visit $j$, and let $R$ denote the observed pattern. Define $\mu_{j,r}=E(Y_j\mid R=r)$, the population mean at visit $j$ among members of pattern $r$. A neighboring-pattern mean restriction equates a missing-outcome mean in one pattern to the corresponding mean in an adjacent pattern. For the three-pattern chain considered here, the restrictions are $$ \mu_{3,110}=\mu_{3,111},\qquad \mu_{2,100}=\mu_{2,110},\qquad \mu_{3,100}=\mu_{3,110}. $$

The first restriction borrows the visit-3 mean for pattern $110$ from pattern $111$. The second borrows the visit-2 mean for pattern $100$ from pattern $110$. The third borrows the visit-3 mean for pattern $100$ from pattern $110$. Since the first and third restrictions connect the visit-3 means across adjacent patterns, they imply $\mu_{3,100}=\mu_{3,110}=\mu_{3,111}$. No restriction is needed for a visit already observed in a pattern.

To turn these assumptions into a numerical estimate, use sample means from the available donor patterns. Write $n_r$ for the number of participants in pattern $r$, and $\bar y_{j,r}$ for the observed sample mean at visit $j$ in that pattern. Use $\bar y^*_{j,r}$ for a completed-data sample mean after assigning values to the missing outcomes. These assignments are plug-in calculations based on the population restrictions; they do not make the population means equal to the observed donor sample means.

From Restrictions to a Weighted Mean

Suppose the target is the population mean at visit 3 in the population represented by these three patterns. Let $p_r=P(R=r)$. Since the patterns partition that population, the law of total expectation gives

E(Y_3)=\sum_{r\in\{111,110,100\}}p_r\mu_{3,r}.

Under the stated chain of visit-3 restrictions, this becomes $E(Y_3)=(p_{111}+p_{110}+p_{100})\mu_{3,111}=\mu_{3,111}$. This is a population implication of the restrictions. In a sample, estimate the overall mean by weighting each completed pattern mean by its sample proportion:

\bar y^*_{3,\mathrm{overall}} =\sum_r \frac{n_r}{n}\bar y^*_{3,r}, \qquad n=\sum_r n_r.

The weights are essential: a pattern containing many participants contributes more to the overall mean than a small pattern. The same calculation applies at visit 2, using the observed visit-2 means for patterns $111$ and $110$, and the assigned mean for pattern $100$. The weighted formula estimates the mean for the cohort represented by the pattern counts; it is not an unweighted average of the pattern means.

Worked Example: Set Up the Neighboring Restrictions

A hypothetical three-visit study has the patterns $111$, $110$, and $100$. The question is how to specify the missing means without treating sample estimates as population quantities.

For pattern $110$, visit 3 is missing. Its neighboring pattern $111$ has visit 3 observed, so the population restriction is $\mu_{3,110}=\mu_{3,111}$. For pattern $100$, visit 2 is missing and pattern $110$ is its neighbor with visit 2 observed; the restriction is $\mu_{2,100}=\mu_{2,110}$. Pattern $100$'s missing visit-3 mean is linked to pattern $110$ by $\mu_{3,100}=\mu_{3,110}$.

Together, the two visit-3 restrictions imply that all three pattern-specific visit-3 population means are equal. The implication follows by substitution: $\mu_{3,100}=\mu_{3,110}$ and $\mu_{3,110}=\mu_{3,111}$, hence $\mu_{3,100}=\mu_{3,111}$. The visit-2 restriction does not imply equality with pattern $111$, because no such visit-2 restriction was specified.

This setup relies on monotone dropout and on the assumed adjacent-pattern equalities. It does not follow merely from the fact that patterns are adjacent or have some observed visits in common.

Worked Calculation With Three Patterns

Consider the following hypothetical data, with the outcome measured in points. The numbers in the table are observed pattern-specific sample means; a dash indicates that the visit is missing for that entire pattern.

Pattern	Participants $n_r$	Visit 1 mean	Visit 2 mean	Visit 3 mean
$111$	40	8	12	18
$110$	30	7	10	—
$100$	30	6	—	—

Worked Example: Calculate the Implied Overall Means

First apply the restrictions as sample plug-in assignments. For pattern $110$, use the observed visit-3 sample mean in pattern $111$: $\bar y^*_{3,110}=\bar y_{3,111}=18$ points. For pattern $100$, assign the visit-2 sample mean from pattern $110$: $\bar y^*_{2,100}=\bar y_{2,110}=10$ points. The visit-3 chain gives $\bar y^*_{3,100}=\bar y^*_{3,110}=18$ points. These are sample assignments based on the restrictions, not claims that population means equal the observed values 18 or 10.

There are $n=40+30+30=100$ participants. At visit 2, the completed-data overall mean is $$ \bar y^*_{2,\mathrm{overall}} =\frac{40(12)+30(10)+30(10)}{100} =\frac{480+300+300}{100} =10.8\text{ points}. $$

At visit 3, the observed mean for pattern $111$ is 18 points, and the two missing pattern means are both assigned 18 points. Therefore, $$ \bar y^*_{3,\mathrm{overall}} =\frac{40(18)+30(18)+30(18)}{100} =\frac{720+540+540}{100} =18.0\text{ points}. $$

The visit-3 result is also a direct consequence of the sample assignments: every pattern-specific completed mean is 18 points, so any weighted average using pattern proportions summing to one is 18 points. This numerical equality depends on the chain of restrictions; it is not an observed-data result alone.

Checks, Comparisons, and Sensitivity

Before calculating, check that the table really represents monotone dropout and that each donor pattern has an observed mean for the visit it supplies. Here, $111$ supplies visit 3 to $110$, while $110$ supplies visit 2 and visit 3 to $100$. If a donor pattern is empty, its sample mean is unavailable; if the donor visit is not observed, the proposed plug-in calculation cannot be made as written. A sparse donor pattern may also produce an unstable sample mean even when it is technically available.

Observed means can be compared across patterns at visits where both patterns have data, such as visits 1 and 2 for $111$ and $110$. Large differences can flag concern that patterns differ in their outcome histories. But similarity at observed visits does not establish equality of their missing-outcome means. Those missing means are not observed for direct comparison. Thus, a table can reveal a computational or compatibility problem, but it cannot verify the identifying restriction for outcomes that were never recorded.

Worked Example: Quantify a Departure From the Chain

Return to the hypothetical table above. As a sensitivity check, suppose the visit-3 mean for pattern $100$ were 2 points higher than the value borrowed through the neighboring chain, while the other pattern means remained as assigned. This is not the original restriction; it asks how the estimate changes under a stated departure.

The revised pattern $100$ visit-3 mean is $18+2=20$ points. The revised overall mean is $$ \frac{40(18)+30(18)+30(20)}{100} =\frac{720+540+600}{100} =18.6\text{ points}. $$

The change from 18.0 to 18.6 points equals the pattern $100$ proportion, $30/100=0.30$, multiplied by the 2-point departure: $0.30(2)=0.60$ points. This calculation shows how the overall estimate responds to a chosen deviation; it does not determine which deviation is scientifically appropriate.

Neighboring-pattern mean restrictions are not the same as complete-case analysis, which uses only pattern $111$ for a visit-3 calculation and does not use the other patterns' assumed means. They are also more specific than asserting marginal equality across all patterns without regard to dropout order. A distributional restriction is stronger than a restriction on means: equality of means alone does not assert equal variances or equal outcome distributions. These approaches answer different modeling questions and should not be treated as interchangeable.

A key limitation is that the overall mean is identified by the observed data plus the restrictions, not by the observed data alone. Different restrictions on the missing means can yield different overall means while leaving the observed table unchanged. Neighboring restrictions can be useful as a transparent, structured assumption, but their plausibility must be assessed in the substantive setting. The calculation also presumes that the three patterns cover the target cohort and that the observed pattern counts are suitable weights for that target.

Common Pitfall

A frequent error is to write $\mu_{3,110}=18$ points because the observed sample mean for pattern $111$ is 18 points. This mixes a population quantity, $\mu_{3,110}$, with a finite-sample estimate, $\bar y_{3,111}$. The restriction says $\mu_{3,110}=\mu_{3,111}$; it does not make either population mean equal to the observed value 18. For a plug-in calculation, the correct statement is $\bar y^*_{3,110}=\bar y_{3,111}=18$ points. Apply the same distinction to every missing mean: state the population restriction with $\mu$, then separately show the sample assignment with $\bar y^*$ and observed $\bar y$.

Key takeaway: In a three-visit monotone-dropout chain, specify each neighboring restriction for the population means first. Then assign missing sample means from the observed donor means, weight the completed pattern means by their pattern proportions, and report the resulting overall mean as conditional on those restrictions.

Check Your Understanding

Answer each question, then compare with the solutions below.

Which patterns are neighbors in the three-visit monotone-dropout chain, and what observed history does each neighboring pair share?
Suppose $\bar y_{3,111}=14$ points and $\bar y_{2,110}=9$ points. Give the corresponding plug-in assignments for $\bar y^*_{3,110}$ and $\bar y^*_{2,100}$. What do the population restrictions say?
If the pattern proportions are 0.5, 0.3, and 0.2, and completed visit-3 means are 16, 16, and 16 points, calculate the overall mean.
In the preceding question, suppose the completed visit-3 mean for pattern $100$ is 21 points instead. Calculate the overall mean and its increase from 16 points.

Solutions. (1) The neighbors are $111$ and $110$, which share observed visits 1 and 2, and $110$ and $100$, which share observed visit 1. (2) The plug-in assignments are $\bar y^*_{3,110}=\bar y_{3,111}=14$ points and $\bar y^*_{2,100}=\bar y_{2,110}=9$ points. The population restrictions are $\mu_{3,110}=\mu_{3,111}$ and $\mu_{2,100}=\mu_{2,110}$; they do not set those population means to the sample values. (3) The weighted mean is $0.5(16)+0.3(16)+0.2(16)=16$ points. (4) The weighted mean is $0.5(16)+0.3(16)+0.2(21)=17$ points, an increase of $17-16=1$ point.