Why Factor a Longitudinal Distribution by Dropout Pattern?
With longitudinal data, participants may contribute different numbers of scheduled outcomes. A pattern-mixture representation describes the outcome trajectory separately within each missingness pattern, then expresses the population distribution in terms of those pattern-specific distributions and the probability of each pattern. The key idea is simple: partition the target population by the pattern each participant follows, and condition the outcome distribution on that pattern.
This is an exact probability identity, not by itself a model for why participants drop out. It does not require missingness to be independent of outcomes. Its important limitation is equally fundamental: although a pattern-specific distribution is well-defined for the full trajectory, some of its components may be unobserved for everyone in that pattern. The identity alone does not reveal those components.
Notation and Definition
Consider \(K\) scheduled visits and let \(Y=(Y_1,\ldots,Y_K)\) denote the full outcome trajectory for a participant in the target population. Let \(R\) denote the dropout pattern. For monotone dropout, index patterns by \(r=0,\ldots,K\), where \(R=r\) means that outcomes at visits \(1,\ldots,r\) are observed and outcomes at visits \(r+1,\ldots,K\) are missing. Thus \(R=K\) denotes a complete trajectory, while \(R=0\) denotes no observed outcomes. If the study design guarantees that an initial visit is observed for everyone, the pattern set can instead be restricted accordingly.
Here \(F_r\) and \(f_r\) concern the entire scheduled trajectory, including outcomes missing under pattern \(r\). A conditional distribution is mathematically defined even when some of its variables are not observed. This distinction is essential: defining \(f_r\) does not mean that the available data determine it.
The identity requires that the patterns form an exhaustive, mutually exclusive partition of the same target population, that the probabilities \(\pi_r\) refer to that population, and that each conditional trajectory distribution is defined for the same full outcome vector \(Y\). It does not require a missing-at-random assumption or any other restriction on the relationship between \(R\) and \(Y\). For outcomes without a joint density, the distribution-function version remains valid.
Deriving the Factorization
The decomposition follows by separating the event \(Y\in A\) according to the possible values of \(R\), for any measurable set of trajectories \(A\). Because exactly one dropout pattern applies to each participant, the events \(\{Y\in A,R=r\}\) are disjoint and their union is \(\{Y\in A\}\).
The second equality uses the definition of conditional probability, \(P(Y\in A,R=r)=P(R=r)P(Y\in A\mid R=r)\), whenever \(P(R=r)>0\). Patterns with probability zero contribute nothing; their conditional distribution need not be specified for this identity. Applying the result to the event \(\{Y\leq y\}\) gives the distribution-function factorization. If the conditional distributions have densities, differentiating the distribution-function identity, or applying the same partition argument to infinitesimal density contributions, gives \(f_Y(y)=\sum_r\pi_r f_r(y)\).
Worked Example: A Discrete Two-Visit Trajectory
Suppose a hypothetical study has two scheduled visits, with binary outcomes \(Y_1,Y_2\in\{0,1\}\). The three patterns are \(R=2\), \(R=1\), and \(R=0\). Assume the target-population pattern probabilities are \(0.40\), \(0.35\), and \(0.25\), respectively. The following hypothetical full-trajectory conditional distributions are specified. In the \(R=1\) and \(R=0\) rows, probabilities involving later outcomes describe unobserved outcomes as well as observed ones; they are posited distributions, not quantities directly available from those participants.
| Pattern | \(P(Y_1=1,Y_2=1\mid R)\) | \(P(Y_1=1,Y_2=0\mid R)\) | \(P(Y_1=0,Y_2=1\mid R)\) | \(P(Y_1=0,Y_2=0\mid R)\) |
|---|---|---|---|---|
| \(R=2\) | 0.30 | 0.20 | 0.25 | 0.25 |
| \(R=1\) | 0.20 | 0.30 | 0.20 | 0.30 |
| \(R=0\) | 0.12 | 0.18 | 0.28 | 0.42 |
Each row sums to 1, so each is a valid conditional distribution over the four possible two-visit trajectories. The pattern probabilities are nonnegative and sum to \(0.40+0.35+0.25=1\). For example, the population probability of the trajectory \((Y_1,Y_2)=(1,1)\) is
The same calculation gives the other trajectory probabilities: \(P(1,0)=0.23\), \(P(0,1)=0.24\), and \(P(0,0)=0.31\). They sum to \(0.22+0.23+0.24+0.31=1.00\). The calculation illustrates the factorization, but it does not establish that all entries in the table could be estimated from the observed data: later outcomes in incomplete patterns are missing.
Worked Example: A Continuous Trajectory Density
For a second hypothetical two-visit illustration, suppose \(Y_1\) and \(Y_2\) are measured in milligrams per litre, and the pattern probabilities are \(\pi_2=0.50\), \(\pi_1=0.30\), and \(\pi_0=0.20\). Assume each pattern-specific joint density is defined over the same full two-outcome space and integrates to 1. At the trajectory \((y_1,y_2)=(4,6)\) milligrams per litre, suppose the densities are \(f_2(4,6)=0.08\), \(f_1(4,6)=0.05\), and \(f_0(4,6)=0.02\), in inverse squared milligrams per litre. Then
This is a density value, not the probability of observing exactly \((4,6)\); for continuous outcomes, probabilities are assigned to regions by integrating the density. The calculation requires the densities to refer to the same outcome scale and to the same target population. It does not require pattern-specific densities to be equal.
What the Factorization Does—and Does Not—Establish
The pattern-mixture representation rewrites a marginal outcome distribution in terms of dropout-pattern probabilities and conditional trajectory distributions. It is useful because it makes explicit that different dropout groups can have different outcome trajectories. It also makes explicit the components needed to reconstruct a population distribution: how common each pattern is and what full-trajectory distribution applies within that pattern.
However, a full trajectory distribution conditional on an incomplete pattern includes outcomes that were not observed for that pattern. For example, when \(R=1\) in the two-visit setting, \(Y_2\) is missing by definition. The data can describe the observed \(Y_1\) values in that pattern, but they do not directly reveal the conditional distribution of \(Y_2\) there. Additional assumptions, restrictions, or sensitivity specifications are needed to determine such unobserved components. The factorization is an identity, not an identification result.
The pattern-mixture approach is often contrasted with a selection model. A pattern-mixture model expresses the joint outcome-and-pattern distribution by first specifying pattern probabilities and outcome distributions conditional on pattern. A selection model instead expresses it through an outcome distribution and a pattern distribution conditional on outcomes. These are alternative factorizations of the same joint distribution; neither factorization alone removes the need to address what the observed data do not identify. For broader background on missingness terminology, see the standard subject of Overlaps Missing Data.
Worked Example: Checking a Factorization Before Using It
A hypothetical analyst proposes two patterns for a two-visit study, with \(\pi_1=0.60\) and \(\pi_2=0.50\), and gives a probability of 0.70 for \((Y_1,Y_2)=(1,1)\) in each pattern. Before computing a population probability, check the requirements.
Pattern probabilities. They sum to \(0.60+0.50=1.10\), which exceeds 1. If these are meant to be exhaustive pattern probabilities in one target population, they cannot both be correct. The proposed mixture is therefore invalid as stated.
Conditional event probabilities. The two values 0.70 are individually within the probability range \([0,1]\), but that fact does not fix the invalid pattern weights. If corrected weights were, for example, 0.60 and 0.40, and both conditional event probabilities remained 0.70, the factorization would give
This corrected calculation assumes the two patterns are exhaustive and mutually exclusive and that the conditional probabilities concern the same full trajectory in the same target population. A numerical result should not be reported until those structural checks are satisfied.
Diagnostics, Alternatives, and Limitations
Before applying the decomposition, check that the pattern labels describe disjoint possibilities and cover the target population. For monotone dropout, verify that each pattern corresponds to the stated sequence of observed visits; a participant observed at visit 2 but not visit 1 would not fit the simple number-of-visits coding used here. Check that the \(\pi_r\) values are nonnegative and sum to 1, and that every \(f_r\) is a valid distribution on the same full trajectory space. In a discrete example, each conditional probability table must sum to 1. In a density model, each conditional density must integrate to 1.
A further diagnostic is conceptual: ask which entries of each pattern-specific trajectory distribution are actually observed. If an argument treats an unobserved outcome distribution as directly known merely because it appears in the factorization, the representation has been mistaken for evidence. Pattern-specific distributions can also be poorly supported when a pattern is rare, and a zero-probability pattern has no empirically defined conditional distribution in the target population.
The pattern-mixture factorization is not the only way to write the joint law of outcomes and missingness. The selection-model factorization \(f(Y)P(R\mid Y)\) is the main alternative. The pattern-mixture form \(P(R)f(Y\mid R)\) is especially direct when the scientific description is organized by dropout groups; the selection form is direct when dropout is described conditional on outcomes. The two forms are not competing probability identities. Their practical differences arise from the modeling restrictions or assumptions imposed on the factors.
The principal limitation is that observed data generally do not identify the full conditional trajectories for incomplete patterns without additional assumptions. Consequently, the factorization alone does not determine a unique full-data distribution. It also does not imply that dropout patterns have the same outcome distribution, that pattern membership causes outcome differences, or that observed pattern proportions automatically represent the intended target population. Those questions depend on design, sampling, and modeling choices beyond the identity itself.
Check Your Understanding
For each question, decide what the factorization permits and check the arithmetic or assumptions. Full solutions follow.
- There are three exhaustive patterns with probabilities \(0.50\), \(0.30\), and \(0.20\). For an event \(A\) in the full trajectory, the conditional probabilities are \(0.40\), \(0.60\), and \(0.10\). Find \(P(A)\).
- Two proposed pattern probabilities are \(0.65\) and \(0.45\), and the patterns are claimed to be exhaustive and mutually exclusive. Is this a valid mixture? Explain.
- In a two-visit study, \(R=1\) means only \(Y_1\) is observed. Does the factorization by itself identify \(P(Y_2=1\mid R=1)\)? Why or why not?
- State the condition needed to use the density factorization \(f_Y(y)=\sum_r\pi_r f_r(y)\).
Solutions. (1) Apply the identity: \(P(A)=(0.50)(0.40)+(0.30)(0.60)+(0.20)(0.10)=0.20+0.18+0.02=0.40\). (2) No. The probabilities sum to \(0.65+0.45=1.10\), not 1, so they cannot describe exhaustive, mutually exclusive patterns in one population. (3) No. \(Y_2\) is unobserved under \(R=1\); the factorization defines the relevant conditional probability but does not identify its value from the observed outcomes alone. (4) The full outcome distribution and each pattern-specific conditional distribution must have densities with respect to a common measure on the same trajectory space; then the density mixture follows from the distributional identity.