Available-Case Mean Reconstruction in a Three-Visit Study

Why Reconstruct Means Visit by Visit?

In a three-visit study with monotone dropout, participants may contribute outcomes at all three visits, only the first two, or only the first. Thus, the group with an observed outcome changes across visits. An available-case mean uses everyone whose outcome at that particular visit is observed. To reconstruct a population mean from that mean, we need an additional restriction about the outcomes missing at that visit.

The calculation is useful because it makes the restriction concrete: the available-case mean at a visit is first calculated from observed patterns, then assigned as the mean for each pattern missing that visit. The resulting population mean is a weighted sum over all patterns. This tutorial focuses on that calculation, not on estimating uncertainty around it.

Definition: Let \(Y_j\) denote the outcome at visit \(j\), for \(j=1,2,3\). Under monotone dropout, label each participant’s observation pattern by three indicators: \(111\) means observed at all visits, \(110\) means observed at visits 1 and 2 only, and \(100\) means observed at visit 1 only. Write \(n_r\) for the number of participants in pattern \(r\), \(N\) for the total sample size, and \(\bar{Y}_{j,r}\) for the observed mean at visit \(j\) within pattern \(r\), when that outcome is observed.

The Available-Case Restriction and the Reconstruction

For visit \(j\), the available patterns are those with an observed outcome at visit \(j\). The available-case mean, written here as \(\bar{Y}^{\,\mathrm{AC}}_j\), is the mean among all participants in those patterns. In a pattern table, calculate it by adding each available pattern’s outcome total and dividing by the total number of participants available at that visit.

The available-case restriction states that, at a given visit, the mean outcome among participants in any pattern missing that visit is equal to the available-case mean at that visit. This is a restriction on pattern-specific means; it does not say that missing outcomes are known or that each missing individual’s outcome equals the available-case mean.

Available-case restriction: For every pattern \(r\) missing visit \(j\), set the pattern-specific mean \(\mu_{j\mid r}=E(Y_j\mid R=r)\) equal to the mean among participants whose visit-\(j\) outcome is observed. In a sample reconstruction, use \(\bar{Y}^{\,\mathrm{AC}}_j\) for that missing pattern’s mean. The reconstructed population mean is the sum of pattern proportions multiplied by their observed or restricted pattern-specific means.

For example, at visit 2, patterns \(111\) and \(110\) are available, while \(100\) is missing visit 2. At visit 3, only \(111\) is available; both \(110\) and \(100\) are missing visit 3. At visit 1, all three patterns are available, so there are no missing visit-1 outcomes in this monotone-dropout setup.

\bar{Y}^{\,\mathrm{AC}}_j = \frac{\sum_{r\in A_j} n_r\bar{Y}_{j,r}} {\sum_{r\in A_j}n_r}, \qquad \widehat{\mu}^{\,\mathrm{AC}}_j = \sum_{r\in A_j}\frac{n_r}{N}\bar{Y}_{j,r} + \sum_{r\notin A_j}\frac{n_r}{N}\bar{Y}^{\,\mathrm{AC}}_j

Here \(A_j\) is the set of patterns with visit \(j\) observed. The first expression is the available-case mean. The second is the reconstruction: observed pattern means are retained, and the available-case mean is used for each pattern missing that visit. Counts \(n_r\) divided by \(N\) are the observed pattern proportions in the sample.

Why the Reconstructed Mean Equals the Available-Case Mean

The equality follows directly from the restriction and the weights. The total proportion in available patterns is multiplied by their weighted mean, \(\bar{Y}^{\,\mathrm{AC}}_j\). The total proportion in missing patterns is also multiplied by that same mean. Together, those proportions sum to one.

\begin{aligned} \widehat{\mu}^{\,\mathrm{AC}}_j &= \sum_{r\in A_j}\frac{n_r}{N}\bar{Y}_{j,r} + \sum_{r\notin A_j}\frac{n_r}{N}\bar{Y}^{\,\mathrm{AC}}_j\\ &= \frac{\sum_{r\in A_j}n_r}{N}\bar{Y}^{\,\mathrm{AC}}_j + \frac{\sum_{r\notin A_j}n_r}{N}\bar{Y}^{\,\mathrm{AC}}_j\\ &= \frac{N}{N}\bar{Y}^{\,\mathrm{AC}}_j = \bar{Y}^{\,\mathrm{AC}}_j. \end{aligned}

The second line uses the definition of the available-case mean: the weighted sum of observed pattern means among available participants equals their total count times that mean. The last line follows because the available and missing patterns partition the sample. This identity is a useful arithmetic check. It does not establish that the restriction is true in the population.

Worked Three-Visit Calculation

Worked Example: Reconstructing Means at All Three Visits

Suppose a hypothetical study has \(N=100\) participants with monotone dropout. The table gives each pattern’s count and its observed means, in outcome units.

Pattern	Count	Visit 1 mean	Visit 2 mean	Visit 3 mean
111	40	10	12	14
110	30	11	13	Not observed
100	30	9	Not observed	Not observed

Visit 1. Everyone has an observed outcome, so the available-case mean is the overall observed mean:

\bar{Y}^{\,\mathrm{AC}}_1 = \frac{40(10)+30(11)+30(9)}{100} = \frac{400+330+270}{100} = 10.

There are no missing visit-1 outcomes to impute, so the reconstructed mean is also 10 outcome units.

Visit 2. Patterns \(111\) and \(110\) are available, with \(40+30=70\) participants. Their observed outcome total is \(40(12)+30(13)=480+390=870\). Therefore:

\bar{Y}^{\,\mathrm{AC}}_2 = \frac{40(12)+30(13)}{70} = \frac{870}{70} = 12.4286.

Pattern \(100\), with 30 participants, is missing visit 2. Under the restriction, its visit-2 mean is \(870/70\). The full-sample reconstruction is:

\widehat{\mu}^{\,\mathrm{AC}}_2 = \frac{40(12)+30(13)+30(870/70)}{100} = \frac{870+372.8571}{100} = 12.4286.

The numerator uses the unrounded available-case mean, \(870/70\), for the missing pattern. Its reconstructed mean is 12.4286 outcome units, equal to the visit-2 available-case mean, as the derivation predicts.

Visit 3. Only pattern \(111\) is available, so its mean of 14 is the available-case mean. Both other patterns are assigned that mean under the restriction:

\widehat{\mu}^{\,\mathrm{AC}}_3 = \frac{40(14)+30(14)+30(14)}{100} = \frac{1400}{100} = 14.

This is a restriction-based reconstruction, not a direct observation of the visit-3 population mean. Only 40 participants contributed observed visit-3 outcomes in this example.

Another Calculation and an Alternative Restriction

Worked Example: Using Counts Rather Than Rounded Proportions

In a second hypothetical study, \(N=80\), with 32 participants in pattern \(111\), 24 in \(110\), and 24 in \(100\). The observed visit-2 means are 18 outcome units for \(111\) and 15 for \(110\). First calculate the mean among the 56 available participants:

\bar{Y}^{\,\mathrm{AC}}_2 = \frac{32(18)+24(15)}{32+24} = \frac{576+360}{56} = \frac{936}{56} = \frac{117}{7} \approx 16.7143.

There are 24 participants in pattern \(100\), missing visit 2. Assigning that pattern the exact available-case mean \(117/7\), the reconstruction is:

\widehat{\mu}^{\,\mathrm{AC}}_2 = \frac{32(18)+24(15)+24(117/7)}{80} = \frac{936+401.1429}{80} = \frac{117}{7} \approx 16.7143.

Using counts directly avoids small discrepancies from rounding pattern proportions or the available-case mean too early. The result is in the same outcome units as the input means.

A useful comparison is the complete-case restriction, which uses completers rather than everyone available at the particular visit as the reference group. It can therefore produce a different mean when an intermediate dropout pattern has observed outcomes at that visit.

Worked Example: Comparing Available Cases With Completers

Return to the first hypothetical table and consider visit 2. The available-case mean is \(870/70=12.4286\), because both \(111\) and \(110\) have observed visit-2 outcomes. If, instead, the missing visit-2 mean for pattern \(100\) is restricted to the completers’ visit-2 mean of 12, the reconstruction is:

\widehat{\mu}^{\,\mathrm{CC}}_2 = \frac{40(12)+30(13)+30(12)}{100} = \frac{480+390+360}{100} = \frac{1230}{100} = 12.3.

The two reconstructions differ by \(12.4286-12.3=0.1286\) outcome units. The difference arises because the available-case donor mean includes the 30 participants in pattern \(110\), whose observed visit-2 mean is 13, while the complete-case donor mean uses only pattern \(111\). Neither calculation is automatically correct: each depends on its own restriction about the missing outcomes.

Assumptions, Checks, and Limitations

The reconstruction is interpretable only when its setup and restriction are clear. At minimum, check the following conditions:

Pattern structure is correctly represented. The labels \(111\), \(110\), and \(100\) describe monotone dropout: once a participant misses a visit, later visits are also missing. If the study has intermittent missingness, these three patterns do not describe all observation patterns.
Each available-case mean is estimable. At least one participant must have an observed outcome at the visit. If there are no available cases, the donor mean cannot be calculated from the table.
Observed means and counts refer to the same target sample. Pattern counts must sum to \(N\), and each pattern-specific mean must use the participants counted in that pattern who actually have that visit observed.
The restriction is an assumption, not a consequence of the table. Outcomes missing within a pattern cannot be compared directly with their unobserved values. The observed data alone do not verify that missing pattern-specific means equal the available-case mean.

A basic arithmetic diagnostic is to check that the pattern counts add to \(N\), identify the available patterns separately for each visit, and verify that the reconstructed mean equals the available-case mean under this restriction. A failure of that equality usually signals a denominator, weight, or arithmetic error. Passing the check confirms the calculation, not the substantive plausibility of the restriction.

The calculated means are point reconstructions. They do not by themselves provide standard errors, confidence intervals, or a measure of uncertainty due to the restriction. Small available groups can make their means unstable, and a point estimate can look precise even when it rests on very few observed outcomes. Nor does agreement between observed means across patterns prove the missing-outcome restriction: those observed comparisons concern outcomes that were observed, not the outcomes missing in the patterns being reconstructed.

Key takeaway: At each visit, calculate the mean among all patterns observed at that visit. Under the available-case restriction, use that mean for the missing patterns and weight all pattern means by their sample proportions. The reconstructed mean then equals the available-case mean; the equality is a consequence of the assumption and weighting, not evidence that the assumption holds.

Check Your Understanding

Answer each question, then compare with the solutions below.

In monotone three-visit dropout, which patterns contribute to the available-case mean at visit 2?
Suppose \(N=50\), with 20 participants in pattern \(111\) whose observed visit-2 mean is 8 and 10 in pattern \(110\) whose observed visit-2 mean is 11. Calculate the visit-2 available-case mean.
Using the data in question 2, suppose the remaining 20 participants are in pattern \(100\). Under the available-case restriction, calculate the reconstructed visit-2 mean.
Does equality between the reconstructed mean and the available-case mean demonstrate that the missing-outcome restriction is true? Explain.

Solutions. (1) Patterns \(111\) and \(110\) are observed at visit 2; pattern \(100\) is not. (2) The observed total is \(20(8)+10(11)=160+110=270\), and the available count is 30, so the mean is \(270/30=9\) outcome units. (3) Assign mean 9 to pattern \(100\): \(\{20(8)+10(11)+20(9)\}/50=(160+110+180)/50=450/50=9\) outcome units. (4) No. The equality follows algebraically from assigning the available-case mean to missing patterns and weighting by their proportions. Missing outcomes are unobserved, so the table cannot establish that this assumption is correct.