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Pattern-Mixture Models for Dropout · Tutorial 203 of 499

Worked Three-Visit Gaussian Example

Work through a three-visit Gaussian pattern-mixture model, separating quantities estimated from observed data from assumptions needed to calculate population means.

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What You'll Learn

  • Define three monotone dropout patterns and their pattern weights.
  • Calculate pattern-specific means from observed outcomes.
  • Identify which pattern-specific means cannot be estimated from observed data alone.
  • Apply explicit extrapolation assumptions for outcomes after dropout.
  • Combine pattern means and weights to calculate population visit-specific means.
  • Assess sensitivity to assumptions and diagnose sparse or unsupported patterns.

A Three-Visit Pattern-Mixture Calculation

A pattern-mixture model describes the outcome distribution separately for each missingness pattern, then combines those distributions using the proportions of people in the patterns. With three scheduled visits, this makes it possible to distinguish two different sources of information: observed outcomes within each pattern, and assumptions needed to describe outcomes after dropout. Keeping those sources separate is essential when calculating an overall mean.

Consider a hypothetical study with outcomes \(Y_1,Y_2,Y_3\) at three visits and monotone dropout: a participant who misses a visit does not return at a later visit. Let \(R=1\) denote dropout after visit 1, \(R=2\) dropout after visit 2, and \(R=3\) completion through visit 3. Thus, participants with \(R=r\) have outcomes observed through visit \(r\). This coding labels the patterns; it does not rank their outcomes or imply why dropout occurred.

Definition: The pattern weight is \(\pi_r=P(R=r)\), the proportion of the target population in pattern \(r\). The pattern-specific mean at visit \(j\) is \(\mu_{rj}=E(Y_j\mid R=r)\). The population mean at visit \(j\) is the weighted mixture $$ \mu_j=\sum_{r=1}^{3}\pi_r\mu_{rj}. $$ The weights are nonnegative and sum to one. A pattern-specific mean for an outcome not observed in that pattern requires a model or assumption; it is not directly calculated from that pattern’s observed outcomes.

For the Gaussian example, suppose the full outcome vector within pattern \(r\) follows \(Y\mid R=r\sim N_3(\mu_r,\Sigma_r)\), where \(Y=(Y_1,Y_2,Y_3)^\mathsf{T}\), \(\mu_r=(\mu_{r1},\mu_{r2},\mu_{r3})^\mathsf{T}\), and \(\Sigma_r\) is a positive-definite covariance matrix. Only the first \(r\) components are observed for pattern \(r\). The Gaussian specification describes the pattern-specific outcome distribution, but it does not by itself identify the unobserved components of \(\mu_r\) from observed data.

In the calculations below, observed components of each pattern mean are estimated by their pattern-specific sample means. For simplicity, suppose the covariance matrices are specified by a common positive-definite working covariance matrix; covariance estimation is not the target of this example. Any values assigned to mean components after dropout come from stated extrapolation restrictions, not from the unobserved outcomes. This distinction is substantive: the resulting population mean depends on those restrictions.

Worked Data and Pattern-Specific Means

The following hypothetical data contain 12 participants. The pattern counts are \(n_1=3\), \(n_2=4\), and \(n_3=5\). The observed outcome values are:

PatternObserved visit 1Observed visit 2Observed visit 3
Dropout after visit 1 (R = 1)8, 10, 12Not observedNot observed
Dropout after visit 2 (R = 2)8, 9, 10, 910, 11, 12, 11Not observed
Completers (R = 3)8, 9, 10, 11, 1210, 11, 12, 13, 1412, 13, 14, 15, 16

Worked Example: Estimating the Observed Pattern Means

Calculate every pattern-specific mean that can be estimated directly from the displayed outcomes. The observed-data assumptions for these calculations are simply that the listed values are the outcomes recorded for the participants in each pattern; no missing outcome is treated as observed.

For pattern \(R=1\), only visit 1 is observed: $$ \hat{\mu}_{11}=\frac{8+10+12}{3}=\frac{30}{3}=10. $$

For pattern \(R=2\), the visit 1 and visit 2 means are $$ \hat{\mu}_{21}=\frac{8+9+10+9}{4}=\frac{36}{4}=9, \qquad \hat{\mu}_{22}=\frac{10+11+12+11}{4}=\frac{44}{4}=11. $$

For pattern \(R=3\), all visits are observed: $$ \hat{\mu}_{31}=\frac{8+9+10+11+12}{5}=\frac{50}{5}=10, $$ $$ \hat{\mu}_{32}=\frac{10+11+12+13+14}{5}=\frac{60}{5}=12, \qquad \hat{\mu}_{33}=\frac{12+13+14+15+16}{5}=\frac{70}{5}=14. $$

These arithmetic means estimate the observed components of the pattern means. The data provide no direct sample mean for \(\mu_{12}\), \(\mu_{13}\), or \(\mu_{23}\), because those outcomes are not observed in the corresponding patterns. A Gaussian model for the full vector does not turn those absent values into data.

Weights and the Population Mean

For a sample representing the target population, the empirical pattern weights are the pattern counts divided by the total sample size. Here \(n=3+4+5=12\), so the weights are

$$ \hat{\pi}_1=\frac{3}{12}=0.2500,\qquad \hat{\pi}_2=\frac{4}{12}=0.3333,\qquad \hat{\pi}_3=\frac{5}{12}=0.4167. $$

They sum to one: \(0.2500+0.3333+0.4167=1.0000\). To calculate a population visit mean, a mean is needed for every pattern at that visit. At visit 1, all patterns have observed outcomes, so the empirical mixture mean can be calculated without an extrapolation for later outcomes.

Worked Example: Calculating the Visit 1 Population Mean

Use the three observed visit 1 means and their empirical pattern weights. The mixture formula is \(\hat{\mu}_1=\sum_r\hat{\pi}_r\hat{\mu}_{r1}\): $$ \hat{\mu}_1=(0.2500)(10)+(0.3333)(9)+(0.4167)(10). $$

Using the exact fractions avoids rounding intermediate values: $$ \hat{\mu}_1=\left(\frac{3}{12}\right)(10)+\left(\frac{4}{12}\right)(9)+\left(\frac{5}{12}\right)(10) =\frac{30+36+50}{12} =\frac{116}{12}=9.6667. $$

The estimated population mean at visit 1 is 9.667 outcome units. The unit is the same unit in which \(Y_1\) was recorded. This is a weighted mean of the pattern-specific visit 1 means, not the unweighted average of the three pattern means.

Extrapolation Assumptions for Later Visits

At visits 2 and 3, not every pattern has observed outcomes. To show the full calculation, impose explicit mean restrictions for the missing components: in pattern \(R=1\), assume the mean increases by 1 outcome unit from visit 1 to visit 2 and by another 1 unit from visit 2 to visit 3; in pattern \(R=2\), assume the mean at visit 3 is 1 unit above the observed visit 2 mean. These are illustrative extrapolation assumptions, not findings established by the observed data. They give

$$ \hat{\mu}_1=(10,11,12)^\mathsf{T},\qquad \hat{\mu}_2=(9,11,12)^\mathsf{T},\qquad \hat{\mu}_3=(10,12,14)^\mathsf{T}. $$

The directly estimated components are the observed means calculated above. The components \(\hat{\mu}_{12}=11\), \(\hat{\mu}_{13}=12\), and \(\hat{\mu}_{23}=12\) are set by the stated restrictions. In an applied analysis, restrictions of this kind need substantive justification and should be examined for sensitivity.

Worked Example: Combining Pattern Means at Visits 2 and 3

Using the specified extrapolation assumptions, calculate the population mean at each later visit. At visit 2, the pattern means are 11, 11, and 12: $$ \hat{\mu}_2=\left(\frac{3}{12}\right)(11)+\left(\frac{4}{12}\right)(11)+\left(\frac{5}{12}\right)(12) =\frac{33+44+60}{12} =\frac{137}{12}=11.4167. $$

At visit 3, the pattern means are 12, 12, and 14: $$ \hat{\mu}_3=\left(\frac{3}{12}\right)(12)+\left(\frac{4}{12}\right)(12)+\left(\frac{5}{12}\right)(14) =\frac{36+48+70}{12} =\frac{154}{12}=12.8333. $$

Under the stated restrictions, the estimated population means are 11.42 outcome units at visit 2 and 12.83 outcome units at visit 3. The visit 2 and visit 3 estimates are conditional on the extrapolation assumptions for outcomes after dropout. They should not be reported as though all three pattern means at those visits were observed.

Why the Mixture Formula Holds

The weighting step follows from conditioning on the mutually exclusive and exhaustive dropout patterns. For a visit \(j\), the law of total expectation gives

$$ E(Y_j)=\sum_{r=1}^{3}E(Y_j\mid R=r)P(R=r) =\sum_{r=1}^{3}\mu_{rj}\pi_r. $$

The patterns partition the population, so \(\sum_rP(R=r)=1\). The formula therefore averages the conditional means in proportion to the population sizes of their patterns. Replacing the population weights with \(n_r/n\), and the pattern means with estimates, gives the empirical plug-in estimate. This justification requires each pattern to be defined consistently and the weights to refer to the same target population as the means.

The formula does not say that observed participants in one pattern represent the missing outcomes in another. Instead, it says how to combine pattern-specific means once those means have been specified or estimated. When all components at a visit are observed within their patterns, the calculation uses observed data throughout. When some components are unobserved, the mixture formula remains valid, but the corresponding pattern means depend on modeling assumptions.

Sensitivity, Diagnostics, and Alternatives

The weights and observed means can be checked directly: counts must add to the total, weights must sum to one, and each displayed sample mean must use the observed values in its own pattern. For a Gaussian specification, the covariance matrix must be positive definite. Also examine whether fitted means and implied trajectories are plausible on the outcome scale. In very small patterns, sample means are imprecise and an unrestricted covariance matrix may be poorly estimated or not estimable; a parsimonious or externally specified covariance structure may be needed.

A useful sensitivity check changes an unobserved pattern mean while holding observed quantities and weights fixed. For example, increase the assumed visit 3 mean in pattern \(R=1\) from 12 to 16, leaving all other values unchanged. The visit 3 population mean changes by the pattern 1 weight times the four-unit increase: $$ \hat{\mu}_3^{\mathrm{new}}=12.8333+(0.2500)(16-12)=12.8333+1.0000=13.8333. $$ This shows exactly how much the population estimate depends on that extrapolation. It does not establish which assumed value is correct.

A common mistake is to calculate the visit 3 population mean using only the completers’ mean of 14, or to treat that value as the mean for everyone. That reasoning silently assumes that the other patterns have the same visit 3 mean as completers. The correction is to include every pattern with its weight and make explicit any assumptions used for unobserved pattern means. A second mistake is to call the imputed or restricted means observed-data estimates; label those components as assumption-dependent.

The main alternatives answer different questions. A complete-case mean uses only pattern \(R=3\) and does not, by itself, represent the full population. An available-case mean at a visit uses everyone observed at that visit, but its contributing pattern mix can differ across visits. A selection model instead represents the outcome distribution together with a model for missingness; it is a different factorization, not a justification for omitting pattern-specific assumptions. Pattern-mixture calculations make the pattern weights and the pattern-specific outcome assumptions visible, but cannot identify unobserved outcomes without restrictions or additional information.

Results are especially fragile when a dropout pattern is empty or very small, when pattern membership is incorrectly coded, or when extrapolated means are implausible. If a pattern has zero weight in the target population, its mean does not affect that population mixture, although it may still matter for other targets. If all patterns have positive weight but a visit-specific mean is unidentified in one or more patterns, the overall mean is not identified from the observed data alone. Report the observed estimates, the extrapolation assumptions, and sensitivity results separately.

Key takeaway: First estimate pattern weights and the pattern means supported by observed outcomes. Then identify every unobserved pattern-visit mean and state its assumption before applying the weighted mixture formula. The mixture arithmetic is straightforward; its interpretation depends on the quality and transparency of the extrapolation.

Check Your Understanding

Answer each question, then compare with the solutions below.

  1. In the example, what is the estimated weight for dropout after visit 2?
  2. Which visit 2 pattern-specific means are directly estimated from observed data, and which is assumption-dependent?
  3. If the assumed visit 3 mean for pattern \(R=2\) were 15 rather than 12, what would the visit 3 mixture mean be, with all other quantities unchanged?
  4. Why is the completers’ visit 3 mean not automatically the population visit 3 mean?

Solutions. (1) The weight is \(4/12=0.3333\). (2) The pattern \(R=2\) visit 2 mean, 11, and the pattern \(R=3\) visit 2 mean, 12, are directly estimated; the pattern \(R=1\) visit 2 mean, 11, comes from the specified extrapolation restriction. (3) The increase is \(0.3333(15-12)=1.000\), so the mixture mean becomes \(12.8333+1.000=13.8333\). (4) Completers represent only pattern \(R=3\), with weight \(5/12\); the population mean also depends on the visit 3 means in the other patterns and their weights.

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