Why the Pattern Weights Matter
A visit-specific mean describes outcomes at one particular time, but participants in a longitudinal study may have different dropout patterns. The population includes participants who stopped attending before that visit as well as those whose outcome was recorded there. Consequently, calculating a population mean requires asking which pattern-specific outcome means contribute to the target and what is known about each one.
The key distinction is between the mean for the full population at a visit and the mean among participants whose outcome was observed at that visit. These are different quantities when dropout occurs and outcomes vary by pattern. The pattern-mixture representation makes the distinction explicit: pattern weights describe how much of the population belongs to each pattern, while pattern-specific means describe outcomes within each pattern.
Under monotone dropout, pattern 1 includes the first visit only, pattern 2 includes visits 1 and 2, and pattern 3 includes all three visits. Thus, the outcome at visit \(j\) is observed in patterns with \(g\geq j\). This observation rule determines which pattern-specific means can be calculated directly from observed outcomes. It does not, by itself, determine which patterns belong in the full-population mean.
Deriving the Full-Population Mean
For any visit \(j\), the patterns \(G=1,2,3\) form a partition of the population. Applying the law of total expectation gives
This identity includes every pattern, even one that dropped out before visit \(j\). For those patterns, the corresponding mean concerns outcomes that were not recorded. The identity is a probability result, not a missing-data assumption: it holds for the full-data outcome \(Y_j\) if the conditional means exist. Whether those means can be identified from observed data is a separate question.
The observed-pattern mean has a different formula. Let \(O_j=1\) indicate that a participant’s outcome at visit \(j\) is observed. Under the monotone pattern definition, \(O_j=1\) exactly when \(G\geq j\). Conditioning on observation and applying the same partition argument gives
The denominator is the fraction of the population that reaches visit \(j\). This is a weighted mean among the patterns that include that visit; the weights are normalized to sum to one. In contrast, the full-population mean uses all pattern weights, which already sum to one, and includes the pattern-specific means for outcomes after dropout.
Worked Example: Writing the Three-Visit Mixture
Suppose participants have one of three monotone dropout patterns, with weights \(\pi_1,\pi_2,\pi_3\). Write the full-population mean at each visit, and identify which means are directly observed within each pattern.
At visit 1, every pattern includes the visit. Therefore, all three pattern-specific means are observed, and
At visit 2, pattern 1 has already dropped out. The full-population mean is still
Here \(\mu_{22}\) and \(\mu_{23}\) are observed-pattern means, but \(\mu_{21}\) is not observed in pattern 1. The mean among participants observed at visit 2 instead is
At visit 3, only pattern 3 includes the visit, so its observed-pattern mean is \(\mu_{33}\). The full-population mean remains \(\pi_1\mu_{31}+\pi_2\mu_{32}+\pi_3\mu_{33}\). Both \(\mu_{31}\) and \(\mu_{32}\) concern unobserved outcomes. This example shows why “average the patterns that include the visit” answers a conditional, observed-participant question rather than automatically giving the full-population mean.
Numerical Calculation and Interpretation
The next example uses hypothetical pattern weights and pattern-specific means. The means for unobserved outcomes are stipulated as part of a full-data model or sensitivity specification; they cannot be calculated from the observed outcomes in the corresponding dropout patterns alone.
Worked Example: Computing All Three Population Means
Suppose a hypothetical study has 100 participants: 20 in pattern 1, 30 in pattern 2, and 50 in pattern 3. Thus, \(\pi_1=0.20\), \(\pi_2=0.30\), and \(\pi_3=0.50\). Assume the following pattern-specific means, measured in outcome points:
| Pattern | Last observed visit | \(\mu_{1g}\) | \(\mu_{2g}\) | \(\mu_{3g}\) |
|---|---|---|---|---|
| 1 | 1 | 10 | 8 | 7 |
| 2 | 2 | 11 | 12 | 10 |
| 3 | 3 | 9 | 11 | 14 |
For patterns 1 and 2, the visit 2 and visit 3 means shown after their last observed visit are assumptions about unobserved outcomes, not means directly estimated from outcomes recorded in those patterns.
At visit 1, all pattern means are observed. The population mean is
At visit 2, include all three patterns in the full-population mixture:
The mean among participants observed at visit 2 uses only patterns 2 and 3, and normalizes their weights:
At visit 3, the full-population mean is
Only pattern 3 is observed at visit 3, so the observed-participant mean is \(14\) points. It differs from the full-population mean because the latter also includes the assumed visit 3 means for patterns 1 and 2. These calculations are statements about the specified pattern means and weights; they do not establish that the stipulated unobserved means are correct.
Assumptions, Identification, and Diagnostics
The mixture identity rests on a small set of precise conditions: pattern membership is defined for each participant; the patterns are mutually exclusive and exhaustive; the weights are population proportions summing to one; and the conditional mean \(E(Y_j\mid G=g)\) exists for every pattern included in the full-data target. The outcome at a visit after dropout need not be observed for the mathematical identity to hold, but its conditional mean must be supplied or identified to evaluate the identity numerically.
Observed data directly support \(\mu_{jg}\) when pattern \(g\) includes visit \(j\). For \(g<j\), the outcome is missing by design of the pattern. A pattern-mixture analysis therefore needs assumptions or a model for those means. Different choices can yield different full-population marginal means while fitting the same observed data. The earlier material on the population mixture establishes the role of pattern weights; here the crucial additional step is to align each visit-specific mean with every pattern in the target population.
Several checks help catch errors in a calculation or implementation:
- Check the weights. Confirm that each pattern is counted once, weights are nonnegative, and \(\sum_g\pi_g=1\). If using estimated weights, check that they correspond to the target population.
- Check the target. For the full-population mean, include patterns that dropped out before the visit. For the mean among observed participants, restrict to \(g\geq j\) and divide by \(\sum_{g=j}^{3}\pi_g\).
- Check the means available in the data. A pattern-specific mean for a visit after that pattern’s last observed visit cannot be described as empirically observed in that pattern. Label the assumption or model that supplies it.
- Check the visit ordering. With monotone dropout, the set of patterns observed at visit \(j\) is \(g\geq j\). A different dropout structure requires redefining the pattern-specific inclusion sets.
Worked Example: Sensitivity to an Unobserved Pattern Mean
Return to the hypothetical weights and means above. Suppose a sensitivity analysis lowers the assumed visit 2 mean for pattern 1 from 8 to 6, leaving the other means unchanged. The full-population visit 2 mean changes only through the pattern 1 term:
The change is \(10.30-10.70=-0.40\) points. Equivalently, the assumed pattern-specific mean changed by \(-2\) points, and its contribution to the population mean changed by \((0.20)(-2)=-0.40\) points. The observed-participant mean at visit 2 remains \(11.375\) points because pattern 1 does not contribute to that conditional mean. This isolates the effect of changing an assumption about an outcome that was not observed in that pattern.
Common Pitfall and Alternatives
A common mistake is to compute the mean at visit 3 using only pattern 3 and call it the population mean. In the example, that calculation gives 14 points, which is the mean among participants observed at visit 3. The full-population mean under the stipulated pattern-specific means is 11.40 points. The correction is to state the estimand first: if it is the full-population mean, use every pattern weight and specify the unobserved pattern-specific means; if it is the observed-participant mean, use only patterns that include the visit and normalize their weights.
The two calculations answer different questions, so neither is inherently an incorrect summary. An observed-participant mean is directly supported by outcomes recorded at that visit, but the composition of the observed group changes across visits. A full-population mean maintains the same target population across visits, but after dropout it depends on assumptions for unobserved outcomes. Complete-case summaries restrict attention to participants observed at every visit and thus target yet another selected group. These alternatives should not be treated as interchangeable estimates of the same quantity.
The mixture identity itself does not choose or validate assumptions for the missing means, nor does it provide uncertainty intervals. Sparse patterns can make pattern-specific estimates unstable, and pattern weights or means may be estimated rather than known. Those limitations affect how confidently a calculated marginal mean can be reported; they do not alter the weighting identity. Always report which target mean is being calculated and how unobserved pattern-specific means were obtained.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions below.
- In a three-visit study, which patterns contribute to the full-population mean at visit 2? Which contribute to the mean among participants observed at visit 2?
- Suppose \((\pi_1,\pi_2,\pi_3)=(0.25,0.25,0.50)\), and the visit 2 pattern means are \((6,8,10)\). Calculate the full-population visit 2 mean.
- Using the weights and means in question 2, calculate the mean among participants observed at visit 2.
- Why can the visit 3 mean among observed participants differ from the full-population visit 3 mean?
- If an assumed pattern 1 mean at visit 3 increases by 3 outcome points and \(\pi_1=0.20\), by how much does the full-population visit 3 mean change, with all else fixed?
Solutions. (1) The full-population mean includes patterns 1, 2, and 3. The observed-participant mean includes patterns 2 and 3, with their weights normalized by \(\pi_2+\pi_3\). (2) The full-population mean is \((0.25)(6)+(0.25)(8)+(0.50)(10)=1.50+2.00+5.00=8.50\) outcome points. (3) The observed-participant mean is \(\{(0.25)(8)+(0.50)(10)\}/(0.25+0.50)=7.00/0.75=9.333\) outcome points. (4) The observed-participant mean uses only pattern 3 at visit 3, whereas the full-population mean also includes patterns 1 and 2 through their assumed visit 3 means. (5) The full-population mean increases by \((0.20)(3)=0.60\) outcome points.