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Factorizations of the Joint Distribution · Tutorial 30 of 499

The Selection Factorization

Learn to write, interpret, and check the selection factorization, and distinguish its general probability identity from assumptions imposed by a particular model.

Intermediate 10 min read

What You'll Learn

  • Write the joint distribution of outcomes and response patterns using the selection factorization.
  • Interpret the outcome and response-pattern factors and their conditioning variables.
  • Verify normalization and calculate joint probabilities from both factors.
  • Explain why the factorization itself does not assume missing at random or parameter independence.
  • Recognize common specification and interpretation errors.

Why Factor the Joint Law?

In a longitudinal study, the outcomes of interest and the pattern of which outcomes are recorded are jointly distributed. A useful way to describe that joint law separates two questions: how are the full outcomes distributed given baseline covariates, and how likely is each response pattern given those covariates and the full outcome vector? This representation is called the selection factorization.

The distinction matters because the response-pattern distribution can depend on outcomes that are not observed for a participant. The factorization makes that dependence explicit in the probability model. It is a way to express a joint distribution, not by itself a claim that a particular dependence structure is present, absent, or identifiable from observed data.

Definition and Notation

Let \(X\) denote baseline covariates, \(Y=(Y_1,\ldots,Y_T)\) the vector of outcomes at \(T\) scheduled times, and \(R=(R_1,\ldots,R_T)\) the response-indicator vector, where \(R_j=1\) means \(Y_j\) is observed and \(R_j=0\) means it is missing. A response pattern is a particular vector \(r\) of zeros and ones. For example, when \(T=3\), \(r=(1,0,1)\) means the first and third outcomes are observed and the second is missing.

Definition: The selection factorization of the conditional joint law of \(Y\) and \(R\), given \(X=x\), is
$$ f(y,r\mid x)=f(y\mid x)\,\pi(r\mid y,x), $$
where \(f(y\mid x)\) is the conditional distribution of the full outcome vector and \(\pi(r\mid y,x)=P(R=r\mid Y=y,X=x)\) is the conditional probability of response pattern \(r\). If \(X\) is random, the full joint law also includes its distribution: \(f(x,y,r)=f(x)f(y\mid x)\pi(r\mid y,x)\).

For discrete outcomes, \(f\) denotes a probability mass function; for continuous outcomes, it denotes a density with respect to an appropriate measure. The response-pattern variable is discrete, so \(\pi\) is a probability. The formula applies to mixed outcome types as well, with the corresponding joint measure understood. For each relevant \(x\), \(f(y\mid x)\) must be a normalized outcome distribution. For each \(y,x\), the pattern probabilities must satisfy \(\sum_r\pi(r\mid y,x)=1\) and \(\pi(r\mid y,x)\geq 0\).

The first factor describes how the complete outcome vector varies in the population conditional on baseline covariates. The second describes selection into each response pattern, conditional on the complete outcome vector and those same covariates. In particular, it conditions on all of \(Y\), not merely the components observed under pattern \(r\). If a conditional distribution is specified at a covariate or outcome value having probability zero, its value there does not change the joint law.

Why the Factorization Holds

The selection factorization follows directly from the probability chain rule. For a discrete outcome vector, the joint conditional probability of any outcome value \(y\) and pattern \(r\) can be written as the probability of \(y\), multiplied by the conditional probability of \(r\) given \(y\). The same argument applies to densities and conditional distributions.

$$ \begin{aligned} f(y,r\mid x) &= f(y\mid x)P(R=r\mid Y=y,X=x)\\ &= f(y\mid x)\pi(r\mid y,x). \end{aligned} $$

This identity also shows why the two factors must be normalized in their respective roles. Summing or integrating over outcomes and summing over response patterns gives

$$ \sum_r\int f(y\mid x)\pi(r\mid y,x)\,dy = \int f(y\mid x)\left\{\sum_r\pi(r\mid y,x)\right\}\,dy = \int f(y\mid x)\,dy =1. $$

For discrete outcomes, replace the integral by a sum. Thus, normalized factors produce a normalized joint law. Conversely, the chain rule recovers these factors from a joint law wherever the relevant conditional distributions are defined. This is a general identity: it does not require independence between \(Y\) and \(R\), nor does it require a specific parametric model.

Worked Example: A Binary Outcome and One Response Indicator

Suppose a hypothetical population has a binary outcome \(Y\), with \(P(Y=1\mid X=x)=0.40\). Let \(R=1\) indicate that the outcome is observed. Suppose \(P(R=1\mid Y=1,X=x)=0.75\) and \(P(R=1\mid Y=0,X=x)=0.90\). These probabilities, together with their complements, specify the response factor for both patterns.

To find the joint probability of \(Y=1\) and an observed outcome, multiply the outcome probability by the conditional response probability:

$$ P(Y=1,R=1\mid x) = P(Y=1\mid x)P(R=1\mid Y=1,x) = 0.40(0.75) = 0.30. $$

The other three joint probabilities are

$$ \begin{aligned} P(Y=1,R=0\mid x)&=0.40(1-0.75)=0.10,\\ P(Y=0,R=1\mid x)&=0.60(0.90)=0.54,\\ P(Y=0,R=0\mid x)&=0.60(1-0.90)=0.06. \end{aligned} $$

They sum to \(0.30+0.10+0.54+0.06=1.00\), as required. The response probabilities differ by outcome value: \(0.75\) versus \(0.90\). That difference describes outcome-dependent response in this hypothetical model; the factorization itself permits it, but does not require it. The calculations assume the supplied probabilities define the conditional outcome and response distributions at the stated \(x\), and both distributions are normalized.

Interpreting the Factors in a Longitudinal Setting

With several outcomes, the same formula applies to the full vectors. The outcome factor is one joint distribution for all scheduled outcomes, while the response factor assigns probabilities to the possible response patterns conditional on those outcomes. This distinction is important: the factorization does not split the outcome distribution into visit-by-visit conditional distributions. Nor does it split the response mechanism into a sequence of indicator probabilities.

Worked Example: A Three-Pattern Distribution for Two Visits

Consider a hypothetical study with outcomes \(Y_1,Y_2\), and let \(R=(R_1,R_2)\). Suppose that at a particular outcome vector \(y=(y_1,y_2)\) and covariate value \(x\), the pattern probabilities are \(0.50\) for \(r=(1,1)\), \(0.30\) for \(r=(1,0)\), \(0.10\) for \(r=(0,1)\), and \(0.10\) for \(r=(0,0)\). They are nonnegative and sum to one. Suppose also that the outcome distribution assigns probability \(0.25\) to this \(y\) given \(x\).

The joint probability of this outcome vector and the pattern \(r=(1,0)\) is

$$ P(Y=y,R=(1,0)\mid x) = P(Y=y\mid x)P(R=(1,0)\mid Y=y,x) = 0.25(0.30) = 0.075. $$

The pattern means \(Y_1\) is recorded and \(Y_2\) is missing. The value \(0.075\) is a joint probability for the full outcome vector and that pattern, not just the probability of observing \(Y_1=y_1\). To check the response factor, add its four probabilities at this same \(y,x\): \(0.50+0.30+0.10+0.10=1.00\). The example assumes a valid conditional outcome probability and a valid response-pattern distribution; it makes no assumption that patterns have the same probabilities at other \(y,x\) values.

Baseline Covariates Belong in Both Conditional Factors

Conditioning on \(X\) allows the outcome distribution and response probabilities to vary with baseline characteristics. If \(X\) is omitted from a factor without justification, the resulting statement imposes a restriction: that factor no longer varies with \(X\) after conditioning on its other listed variables. Including \(X\) in the notation does not itself assert that every covariate has a nonzero effect; it makes clear which conditional laws are being described.

Worked Example: Calculating a Joint Probability at a Covariate Value

Suppose a hypothetical binary baseline covariate \(X\) has \(P(X=1)=0.50\). Let \(Y\) be a binary outcome with \(P(Y=1\mid X=1)=0.60\), and let \(R=1\) denote response, with \(P(R=1\mid Y=1,X=1)=0.50\). Then the joint probability of \(X=1,Y=1,R=1\) is

$$ \begin{aligned} P(X=1,Y=1,R=1) &=P(X=1)P(Y=1\mid X=1)P(R=1\mid Y=1,X=1)\\ &=0.50(0.60)(0.50)\\ &=0.15. \end{aligned} $$

The covariate probability is included because this calculation asks for a joint probability involving \(X\). If instead the question were conditional on \(X=1\), the answer would be \(0.60(0.50)=0.30\). These differ because \(0.15\) includes the probability of being in the \(X=1\) group, whereas \(0.30\) does not. The calculation assumes the three stated probabilities refer to compatible conditional distributions.

What the Factorization Does—and Does Not—Assume

A key feature of the selection factorization is that it places the full outcome vector among the conditions for the response-pattern distribution. Consequently, it can represent response probabilities that depend on unobserved outcomes as well as observed ones. The formula alone does not assert that this dependence occurs. A substantive restriction on \(\pi(r\mid y,x)\) is an additional modeling assumption.

Likewise, writing separate factors does not automatically make their model parameters distinct or statistically independent. A model might use separate parameter sets for \(f(y\mid x)\) and \(\pi(r\mid y,x)\), or might link them through shared parameters. Such choices concern the model built on the factorization, not the probability identity. The earlier material on ignorability conditions provides the relevant context for when response behavior can be left out of likelihood-based outcome analysis; the selection factorization itself does not establish ignorability.

The pattern-mixture factorization is a different way to express the joint law, organized around response patterns and outcome distributions within patterns. It is a useful contrast in perspective: the selection form starts with the distribution of the full outcomes and then describes response conditional on them. Conversion between the two forms requires additional probability calculations and is not needed to interpret the selection factorization.

Diagnostics, Failure Modes, and Limitations

A proposed specification should first pass basic probability checks. For each \(x\), the outcome distribution must integrate or sum to one. For each \(y,x\), the probabilities over all possible response patterns must be nonnegative and sum to one. If a response model assigns probabilities outside zero to one, or probabilities that do not sum to one across patterns, it is not a valid conditional distribution.

A second diagnostic is conceptual: check that the factor conditions on the variables the model intends to use. A response probability written only as a function of \(x\) asserts no dependence on \(y\) given \(x\); a probability written as a function of \(y,x\) allows dependence on the full outcome vector. The data may not reveal how response depends on outcomes that are missing. Therefore, a scientifically plausible full-data dependence cannot be validated simply by inspecting observed values.

The factorization is a representation, not a guarantee that every component is estimable from observed data. In particular, when response can depend on unobserved outcomes, multiple choices for the outcome and response factors may be compatible with the same observed information. Additional assumptions, restrictions, or sensitivity analyses may be needed for substantive conclusions. These limitations are separate from the algebraic correctness of the factorization.

  • Pitfall: treating the factorization as an independence statement. The reasoning “the joint law is a product, so outcomes and response are independent” is wrong. The second factor is conditional on the full outcomes; independence would require a further restriction such as \(\pi(r\mid y,x)=\pi(r\mid x)\).
  • Pitfall: conditioning only on observed outcomes by definition. The selection factor is defined conditional on the full outcome vector. Replacing that vector by observed components changes the conditioning statement and may no longer describe the same joint law.
  • Pitfall: assuming a valid factorization is empirically identified. Normalized factors define a joint law, but that does not mean the observed data uniquely determine both factors. Identification depends on what is observed and on additional assumptions.
Key takeaway: The selection factorization writes the joint law as \(f(y\mid x)\pi(r\mid y,x)\). The first term describes full outcomes; the second describes response patterns conditional on those full outcomes and baseline covariates. The formula is a chain-rule identity, while restrictions on either factor are substantive modeling choices.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

  1. In \(f(y\mid x)\pi(r\mid y,x)\), what does each factor describe, and which factor conditions on the full outcome vector?
  2. Suppose \(P(Y=1\mid x)=0.30\) and \(P(R=1\mid Y=1,x)=0.80\). Calculate \(P(Y=1,R=1\mid x)\).
  3. For a study with two visits, what does \(R=(0,1)\) mean if 1 denotes observed and 0 denotes missing?
  4. Does the selection factorization alone imply that response is independent of the outcomes given \(X\)? Explain.

Solutions. (1) \(f(y\mid x)\) describes the distribution of the full outcome vector given baseline covariates. \(\pi(r\mid y,x)\) describes the conditional probability of a response pattern given the full outcome vector and covariates, so it is the factor conditioning on \(Y\). (2) By multiplication, \(P(Y=1,R=1\mid x)=0.30(0.80)=0.24\). (3) The first outcome is missing and the second is observed. (4) No. The response factor is allowed to depend on \(y\); conditional independence would require the additional restriction \(\pi(r\mid y,x)=\pi(r\mid x)\).

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