Pattern Weights and the Population Mixture

From Pattern-Specific Distributions to a Population Distribution

In a pattern-mixture model, outcomes are described conditionally on a missingness pattern. That description is not yet the population distribution. To recover the distribution for the target population, the conditional distributions must be combined in proportion to how common their patterns are in that population. The pattern proportions are the mixture weights.

This weighting step has a direct interpretation: a randomly selected member of the target population belongs to one dropout pattern, and that person’s outcome distribution is the distribution associated with that pattern. The population distribution averages over the pattern membership; it does not treat every pattern as equally common.

Definition: Let $G$ denote a person’s dropout pattern, with possible values $g=1,\ldots,K$, and let $Y_t$ be the outcome at visit $t$. The pattern weight is $\pi_g=P(G=g)$. For a specified visit, let $F_{g,t}(y)=P(Y_t\leq y\mid G=g)$ be the pattern-specific outcome distribution. If the patterns are mutually exclusive and exhaustive, the population distribution is $$ F_t(y)=P(Y_t\leq y)=\sum_{g=1}^{K}\pi_g F_{g,t}(y), \qquad \sum_{g=1}^{K}\pi_g=1. $$

The same identity applies to a full outcome vector: its population distribution is the weighted sum of the pattern-specific joint distributions, provided each describes the same target vector in the same target population. In practice, some components of that vector may be unobserved for people who dropped out. A model for those components is then needed to specify the relevant pattern-specific distribution. Pattern counts alone do not supply it.

Why the Weights Matter

A pattern-specific distribution answers a conditional question: what is the outcome distribution among people in pattern $g$? The population mixture answers a different question: what is the outcome distribution among people in the target population, allowing for the fact that different patterns have different prevalence?

For any event $A$ concerning the outcome, the same combination follows by partitioning the population into patterns:

P(A)=\sum_{g=1}^{K}P(A\mid G=g)P(G=g).

This is the law of total probability. For an outcome with a finite expectation in every pattern, the corresponding mean is the weighted average of the pattern-specific means:

E(Y_t)=\sum_{g=1}^{K}\pi_g E(Y_t\mid G=g).

The weights are not the number of visits observed in a pattern, nor do they represent how much information that pattern contributes to estimating its conditional distribution. They represent the population proportions in the patterns. A pattern with a smaller population proportion contributes less to the population mixture, even if its conditional mean is far from the others.

Worked Example: A Population Mean at the Final Visit

Suppose a hypothetical longitudinal cohort has three monotone dropout patterns across three visits: $G=1$ completes all visits, $G=2$ is observed through visit 2 and then drops out, and $G=3$ is observed only at visit 1. Consider the visit-3 outcome $Y_3$, measured in points. Assume the target population pattern proportions are $0.50$, $0.30$, and $0.20$, respectively. A specified pattern-mixture model gives conditional means of 8.0, 6.5, and 5.0 points.

For $G=2$ and $G=3$, visit-3 outcomes are not observed in this hypothetical setup. Their stated conditional means therefore represent model-based distributions for those unobserved outcomes, not sample averages directly calculated from visit-3 measurements.

The weights sum to one: $0.50+0.30+0.20=1.00$. Thus, the patterns partition the target population. The population mean is

E(Y_3) =(0.50)(8.0)+(0.30)(6.5)+(0.20)(5.0) =4.00+1.95+1.00 =6.95\text{ points}.

The estimated or model-implied population mean is 6.95 points under the stated pattern proportions and conditional means. It is not the unweighted average, $(8.0+6.5+5.0)/3=6.5$ points, because the three patterns are not equally prevalent.

Deriving and Calculating Mixture Summaries

The distribution formula can be derived by partitioning the event $Y_t\leq y$ according to pattern membership. Because the events $G=1,\ldots,G=K$ are disjoint and together cover the population,

\begin{aligned} P(Y_t\leq y) &=\sum_{g=1}^{K}P(Y_t\leq y,\ G=g)\\ &=\sum_{g=1}^{K}P(Y_t\leq y\mid G=g)P(G=g)\\ &=\sum_{g=1}^{K}\pi_gF_{g,t}(y). \end{aligned}

The second equality uses the definition of conditional probability. To obtain the mean, partition the expectation across the same disjoint patterns: each pattern contributes its conditional mean multiplied by the probability of being in that pattern. This gives $E(Y_t)=\sum_g\pi_gE(Y_t\mid G=g)$, assuming the expectations exist.

Worked Example: A Mixture Probability From Discrete Outcomes

Continue with the hypothetical visit-3 outcome, now specified by conditional distributions. In pattern $G=1$, let $Y_3$ be 6 or 10 points, each with probability $0.50$. In pattern $G=2$, let it be 4 or 8 points, each with probability $0.50$. In pattern $G=3$, let it be 2 or 6 points, each with probability $0.50$. Keep the pattern weights $0.50$, $0.30$, and $0.20$. Find the population probability that $Y_3\leq 6$.

Within pattern 1, the probability is $P(Y_3\leq6\mid G=1)=0.50$, because only the value 6 qualifies. Within pattern 2 it is $P(Y_3\leq6\mid G=2)=0.50$, because only 4 qualifies. Within pattern 3 it is $P(Y_3\leq6\mid G=3)=1.00$, because both values qualify. Therefore,

P(Y_3\leq6) =(0.50)(0.50)+(0.30)(0.50)+(0.20)(1.00) =0.25+0.15+0.20 =0.60.

Under these specified distributions and weights, 60% of the target population has a visit-3 outcome at or below 6 points. This calculation uses the whole distribution within each pattern; the pattern means alone would not determine this probability.

Worked Example: Estimating Pattern Weights From Counts

Suppose a hypothetical study enrolls $N=100$ people, classified into the same three dropout patterns. The observed counts are $n_1=52$, $n_2=28$, and $n_3=20$. Estimate the pattern proportions using the sample fractions.

The counts are exhaustive and non-overlapping because $52+28+20=100$. The empirical weights are

\hat{\pi}_1=\frac{52}{100}=0.52,\qquad \hat{\pi}_2=\frac{28}{100}=0.28,\qquad \hat{\pi}_3=\frac{20}{100}=0.20.

They sum to $0.52+0.28+0.20=1.00$. If the pattern-specific visit-3 means remain 8.0, 6.5, and 5.0 points, the plug-in mixture mean is

\widehat{E(Y_3)} =(0.52)(8.0)+(0.28)(6.5)+(0.20)(5.0) =4.16+1.82+1.00 =6.98\text{ points}.

The change from 6.95 to 6.98 points arises from the changed weights, with the conditional means held fixed. This calculation treats the observed pattern fractions as estimates of the target-population proportions. It does not establish that the pattern-specific mean for an unobserved visit is correct.

Conditions, Diagnostics, and Limits

The mixture identity is exact when the patterns form a partition and the conditional distributions and weights refer to the same population. Its use as an analysis also depends on how those quantities were obtained. Check the following conditions before interpreting a computed mixture:

Patterns are defined consistently. Each person must belong to exactly one pattern, and the set of patterns must cover the target population. Unclassified people or overlapping definitions make the weights incomplete or double-counted.
Weights are on the intended population scale. Sample fractions estimate population proportions when the sample is representative for pattern membership, or when the design and analysis appropriately account for unequal selection. Raw counts need not be population weights in a stratified or otherwise unequal-probability sample.
Every conditional distribution concerns the same outcome. For example, all $F_{g,3}$ must refer to visit 3 on the same measurement scale, rather than mixing visits or outcome definitions.
Unobserved outcomes are handled explicitly. For a dropout pattern that does not have observed measurements at the target visit, its conditional distribution must be supplied by assumptions or a model. The pattern weight does not identify that distribution.

Useful checks include verifying that $\hat{\pi}_g\geq0$ for every pattern and that $\sum_g\hat{\pi}_g=1$; checking that all pattern-specific probabilities lie between 0 and 1 and each distribution is normalized; and reproducing the mixture probability or mean by summing each pattern’s weighted contribution. A weight sum above or below one often signals a missing category, double counting, or a denominator inconsistent with the target population.

A population mixture is not generally the same as the conditional distribution among completers. It is also not the equally weighted average of the pattern-specific distributions unless the pattern proportions actually are equal. Replacing the population weights with equal weights changes the target distribution; restricting the calculation to observed completers changes the target group. Neither is a harmless computational shortcut.

Compared with a single pooled outcome distribution, the pattern-mixture representation makes pattern membership visible and allows pattern-specific distributions to differ. Compared with reporting each pattern separately, the weighted mixture answers a population-level question. The mixture does not, by itself, determine how unobserved outcomes behave, resolve uncertainty in estimated weights and conditional distributions, or establish a causal interpretation. Those require additional modeling, design, and inferential considerations.

Common Pitfall

Common mistake: “There are three dropout patterns, so average their three means equally to obtain the overall mean.” This is wrong unless the patterns each make up one-third of the target population. The correction is to multiply every pattern-specific mean by its pattern proportion and add the contributions: $E(Y_t)=\sum_g\pi_gE(Y_t\mid G=g)$. Also check whether any conditional mean for an unobserved outcome is model-based before treating the result as data-determined.

Check Your Understanding

Answer each question, then compare with the solutions below.

Two patterns have weights 0.70 and 0.30 and conditional means 4 and 9 units. What is the mixture mean?
Why do pattern counts not, by themselves, determine the population distribution at a visit that some patterns miss?
For weights 0.25 and 0.75, the conditional probabilities of an event are 0.20 and 0.60. What is the population probability?
A mixture calculation uses three pattern weights that sum to 0.92. Name one diagnostic concern.
When does an equally weighted average of pattern-specific means equal the population mixture mean?

Solutions. (1) $0.70(4)+0.30(9)=2.8+2.7=5.5$ units. (2) Counts determine pattern proportions, but not the conditional distribution of the unobserved outcome in patterns without measurements at that visit; a model or additional assumptions are needed. (3) $0.25(0.20)+0.75(0.60)=0.05+0.45=0.50$. (4) The patterns may fail to exhaust the target population, a category may be missing, or the weights may have been calculated with inconsistent denominators; verify definitions and normalize only when justified. (5) It equals the mixture mean when all patterns have equal weights, or in special cases where unequal weights happen to yield the same average because of the particular conditional means.