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Factorizations of the Joint Distribution · Tutorial 36 of 499

Factorizing Nonmonotone Missingness Patterns

Learn to enumerate nonmonotone response patterns and express their joint distribution with the complete outcome vector as a pattern mixture.

Intermediate 10 min read

What You'll Learn

  • Encode each visit’s response status in a binary pattern vector.
  • Distinguish nonmonotone patterns from monotone dropout histories.
  • Factor the joint law into pattern probabilities and pattern-conditional outcome distributions.
  • Check that pattern probabilities and conditional distributions are valid.
  • Interpret pattern-specific outcome distributions and their limitations.
  • Identify common errors when coding or reading nonmonotone patterns.

Why Nonmonotone Patterns Need Their Own Representation

In a longitudinal study, a participant may miss one assessment and return for a later one. Such a history is not monotone dropout: the observed visits do not simply stop at a particular time. To describe the joint distribution of outcomes and response histories, we therefore need to represent each distinct history as its own pattern, rather than treating every history as a dropout time.

The central factorization is simple, but its interpretation matters. It separates the probability of a response pattern from the distribution of the full outcome vector among individuals with that pattern. That conditional distribution includes outcomes that were not observed for some individuals. The factorization is a representation of a joint law; it does not, by itself, make those missing outcomes observable or specify how they should be estimated.

Definition: For \(T\) scheduled visits, let \(Y=(Y_1,\ldots,Y_T)\) be the full outcome vector and \(R=(R_1,\ldots,R_T)\) the response-pattern vector, where \(R_t=1\) if \(Y_t\) is observed and \(R_t=0\) otherwise. A pattern is a particular vector \(r\in\{0,1\}^T\). It is nonmonotone if the observed visits do not form an initial uninterrupted sequence followed only by missing visits; for example, \(r=(1,0,1)\) is nonmonotone.

The set of possible patterns is \(\mathcal R=\{0,1\}^T\), so there are \(2^T\) patterns before any are ruled out as impossible by the study design. For a pattern \(r\), define its probability as \(\pi_r=P(R=r)\). Define \(f_r(y)\) as the conditional distribution of the complete outcome vector \(Y\) given \(R=r\). When \(Y\) is continuous, \(f_r\) denotes a density; when \(Y\) is discrete, it denotes a probability mass function.

Pattern-mixture factorization: For every pattern \(r\) with \(\pi_r>0\), the joint law factors as $$ P(R=r,Y\in A)=\pi_r\,P(Y\in A\mid R=r) $$ for any event \(A\) in the outcome space. In density or mass-function notation, this is \(p(r,y)=\pi_r f_r(y)\). The pattern probabilities satisfy \(\sum_{r\in\mathcal R}\pi_r=1\), and each \(f_r\) is a normalized distribution over the full outcome vector.

Encoding and Factorizing the Patterns

The ordering of visits must be fixed and used consistently. A one in position \(t\) means that visit \(t\) was observed; a zero means it was not. Patterns \(r=(1,0,1)\) and \(r=(1,1,0)\) are distinct, even though both have two observed visits, because they identify different sets of observed outcomes. In particular, a count of observed visits is not enough to identify a pattern.

For a nonmonotone pattern, the observed components can be separated from the missing components using the positions indicated by \(r\). For example, with three visits and \(r=(1,0,1)\), the observed components are \(Y_1\) and \(Y_3\), while \(Y_2\) is missing. The conditional law \(f_{101}(y_1,y_2,y_3)\) is still a law for all three outcomes; it is not merely a distribution of the two observed values. Its marginal distribution for \((Y_1,Y_3)\) describes the observed outcomes in that pattern, while its distribution for \(Y_2\) involves an unobserved outcome.

Worked Example: Listing Three-Visit Patterns

Consider a hypothetical study with three scheduled visits. A participant is assessed at visits 1 and 3 but misses visit 2. Write the response pattern and identify the observed and missing outcome components.

In the fixed visit order \(1,2,3\), the response indicators are \(R_1=1\), \(R_2=0\), and \(R_3=1\). Thus the pattern is \(r=(1,0,1)\). The observed outcomes are \(Y_1\) and \(Y_3\); \(Y_2\) is missing.

With three binary response indicators, the full pattern set has \(2^3=8\) members:

\(000, 001, 010, 011, 100, 101, 110, 111\).

The pattern \(101\) is nonmonotone because observation resumes at visit 3 after visit 2 was missed. The pattern \(110\), by contrast, has observations at visits 1 and 2 followed by a missed visit 3; it is a monotone dropout history. Both are distinct patterns, and their conditional outcome distributions need not be the same.

This coding assumes each scheduled visit has a well-defined observed-or-missing status and that the visit order is known. It does not assume that the outcomes or response patterns are independent across visits.

Why the Factorization Holds

The factorization follows directly from the definition of conditional probability, not from a missingness assumption. For any event \(A\) involving the complete outcome vector and any pattern \(r\) with positive probability,

$$ P(R=r,Y\in A) =P(R=r)P(Y\in A\mid R=r) =\pi_r P(Y\in A\mid R=r). $$

If \(Y\) has a density or mass function, the same identity applies to infinitesimal or pointwise outcome events, giving \(p(r,y)=\pi_r f_r(y)\). Summing the joint law over all patterns recovers the marginal outcome distribution:

$$ f_Y(y)=\sum_{r\in\mathcal R}\pi_r f_r(y). $$

This is a mixture representation: the full outcome distribution is a weighted average of the pattern-specific distributions, with weights equal to the pattern probabilities. The identity requires only that the joint law exists. If \(\pi_r=0\), the conditional distribution given \(R=r\) is not determined by the observed joint law; the corresponding joint contribution is zero, so no value for \(f_r\) is needed to represent that contribution.

Worked Example: Computing Joint Probabilities From Pattern Components

Suppose a hypothetical two-visit study has three response patterns with positive probability: \(11\), \(10\), and \(01\). Assume their probabilities are \(\pi_{11}=0.50\), \(\pi_{10}=0.30\), and \(\pi_{01}=0.20\), which sum to 1. For pattern \(10\), suppose the conditional distribution of the complete binary outcome vector is specified by the following probabilities.

Complete outcome vector \(y=(y_1,y_2)\)\(f_{10}(y)\)
\((0,0)\)0.20
\((0,1)\)0.30
\((1,0)\)0.10
\((1,1)\)0.40

The four conditional probabilities sum to \(0.20+0.30+0.10+0.40=1.00\), so they form a valid distribution for \(Y\) given pattern \(10\). The joint probability of pattern \(10\) and complete outcome vector \((1,1)\) is

$$ P(R=10,Y=(1,1))=\pi_{10}f_{10}(1,1) =0.30(0.40)=0.12. $$

Similarly, \(P(R=10,Y=(0,1))=0.30(0.30)=0.09\). These joint probabilities concern the full vector, including \(Y_2\), which is unobserved under pattern \(10\). The calculation is valid as a statement about the specified joint law; it does not mean that \(Y_2\) was recorded for those participants.

Pattern-Specific Distributions and the Overall Distribution

The pattern-mixture representation makes clear that outcome distributions may differ across response histories. Such differences are part of the joint law and are not ruled out by the factorization. Conversely, the factorization does not claim that response history causes outcome differences. It is a way to organize the joint distribution, not a causal interpretation.

For an outcome component with a finite conditional mean in each pattern, the overall mean is a weighted average of the pattern-specific means. For example, writing \(\mu_{t,r}=E(Y_t\mid R=r)\), the law of total expectation gives

$$ E(Y_t)=\sum_{r\in\mathcal R}\pi_r\mu_{t,r}. $$

This identity includes patterns in which \(Y_t\) is missing. In those patterns, \(\mu_{t,r}\) is a feature of the full conditional distribution, but it cannot generally be calculated from that pattern’s observed values alone. Any estimate of such a quantity requires information or assumptions beyond the algebraic factorization.

Worked Example: A Mixture Mean Across Patterns

Suppose a hypothetical two-visit study has patterns \(11\), \(10\), and \(01\) with probabilities \(0.50\), \(0.30\), and \(0.20\). Assume the conditional means of \(Y_1\) are \(10\), \(8\), and \(12\) units, respectively. These are stipulated features of the pattern-specific full outcome distributions; in pattern \(01\), \(Y_1\) is missing.

The marginal mean of \(Y_1\) is

$$ E(Y_1) =(0.50)(10)+(0.30)(8)+(0.20)(12) =5+2.4+2.4=9.8\text{ units}. $$

The arithmetic checks because the weights sum to \(0.50+0.30+0.20=1.00\). The result is the mean under the specified full joint law. It is not an estimate obtained simply by averaging observed \(Y_1\) values: the mean of 12 units in pattern \(01\) concerns outcomes missing in that pattern and would need to be justified or estimated using additional modeling choices or information.

Assumptions, Checks, and Failure Modes

No missing-at-random, missing-completely-at-random, or missing-not-at-random condition is needed for the identity \(p(r,y)=\pi_r f_r(y)\). It is a factorization of a joint probability distribution. The practical requirements are instead that the response pattern be coded unambiguously, the pattern probabilities be valid, and each conditional distribution be defined on the full outcome space. If one imposes a parametric model for \(\pi_r\) or \(f_r\), that model adds assumptions that must be assessed separately.

Several checks help catch errors before the factorization is used:

  • Check visit order and coding. Verify that every position corresponds to the same scheduled visit for every participant and that 1 always means observed.
  • Check pattern distinctions. Preserve the location of missing visits. Patterns with the same number of observed outcomes are not interchangeable.
  • Check normalization. The \(\pi_r\) values must be nonnegative and sum to 1. For each positive-probability pattern, \(f_r\) must integrate or sum to 1 over the full outcome space.
  • Check the outcome space. A conditional distribution for \(Y\mid R=r\) must account for every component of \(Y\), not only the components observed under \(r\).
  • Check empty patterns. A pattern absent from a finite sample can still have positive probability in the population. Do not conclude its population probability is exactly zero just because no sampled person has that pattern.

A particularly common mistake is to write \(f(y_{\mathrm{obs}}\mid r)\) and treat it as the full pattern-specific outcome distribution. That notation describes only the observed components. The pattern-mixture factorization uses a conditional law for the complete \(Y\); the observed-component distribution is a marginal of that law. Another mistake is to treat a nonmonotone history such as \(101\) as if it were equivalent to dropout after visit 1. That loses the information that visit 3 was observed and assigns outcomes to the wrong observed set.

Key takeaway: A nonmonotone response history is a distinct binary pattern. Its contribution to the joint law is its pattern probability multiplied by the conditional distribution of the full outcome vector for that pattern. This identity organizes the law, but it does not identify missing outcomes from observed data alone.

How This Factorization Relates to Alternatives

The pattern-mixture factorization groups the joint law by response pattern: first specify how probable each pattern is, then specify the outcome distribution within that pattern. The selection factorization instead writes the joint law in terms of an outcome distribution and a response distribution conditional on outcomes. These are alternative factorizations of the same joint law, not automatically different laws. Converting between their components requires care and is a separate calculation.

Sequential response-indicator factorizations can describe response histories through a sequence of conditional probabilities. They may be convenient for modeling visit-by-visit response, but the pattern-mixture representation keeps the full response history explicit and associates each history with its own outcome distribution. For nonmonotone data, this makes it especially important not to collapse histories into a single dropout time.

The representation has limitations. With many visits there are \(2^T\) possible patterns, and some may be rare or absent in the sample. Pattern-specific distributions can then be difficult to estimate. Moreover, the conditional distributions of outcomes missing under a pattern are not directly observed in that pattern. The factorization alone supplies neither a unique estimation procedure nor assumptions that resolve this information gap.

Check Your Understanding

Answer each question, then compare with the solutions below.

  1. For four visits, what pattern records visits 1, 2, and 4 as observed and visit 3 as missing? Is it monotone?
  2. If a study has pattern probabilities \(0.45\), \(0.35\), and \(0.20\), and the conditional probability of \(Y=y\) in the second pattern is \(0.60\), what is the joint probability of that pattern and \(Y=y\)?
  3. Does the pattern-mixture factorization require a missing-at-random assumption? Explain briefly.
  4. Why is a distribution of observed outcomes under pattern \(101\) not the same object as \(f_{101}(y_1,y_2,y_3)\)?

Solutions. (1) The pattern is \(r=(1,1,0,1)\). It is nonmonotone because observation resumes at visit 4 after visit 3 is missed. (2) The joint probability is \(0.35(0.60)=0.21\). (3) No. The factorization follows from the definition of conditional probability; missingness assumptions would be additional modeling restrictions, not requirements for the identity. (4) The observed-outcome distribution concerns only \(Y_1\) and \(Y_3\), whereas \(f_{101}\) is a distribution for the complete vector, including the missing component \(Y_2\).

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