One Joint Law, Three Ways to Organize It
A selection model, a pattern-mixture model, and a shared-parameter model organize probabilities differently. Comparing their formulas is meaningful only when they refer to the same joint law. In particular, a conditional distribution in one factorization may average over a latent variable that is explicit in another.
Let \(Y=(Y_1,Y_2)\) denote two longitudinal outcomes, and let \(R\) denote the observation pattern. Use \(R=0\) for neither outcome observed, \(R=1\) for \(Y_1\) observed and \(Y_2\) missing, and \(R=2\) for both outcomes observed. The complete-data joint law concerns \(Y_1,Y_2,R\), including outcomes that may be unobserved in a realized record. The examples below use binary outcomes so that probabilities can be calculated by finite sums.
For discrete outcomes the first two forms are chain-rule identities wherever the relevant conditioning events have positive probability. The shared-parameter form is also an exact representation for a specified model if integrating out \(B\) gives the stated joint law. Its conditional-independence assumptions, when imposed, are modeling restrictions rather than consequences of the chain rule.
A Specified Shared-Parameter Law
Consider a hypothetical population model with \(B\in\{0,1\}\), \(P(B=0)=P(B=1)=0.5\). Conditional on \(B=b\), let \(Y_1\) and \(Y_2\) be independent Bernoulli outcomes with success probability \(q_b\), where \(q_0=0.2\) and \(q_1=0.8\). Also conditional on \(B\), let \(R\) be independent of \(Y\), with the following pattern probabilities:
| Value of \(B\) | \(P(R=0\mid B)\) | \(P(R=1\mid B)\) | \(P(R=2\mid B)\) |
|---|---|---|---|
| 0 | 0.2 | 0.3 | 0.5 |
| 1 | 0.8 | 0.1 | 0.1 |
Each row of pattern probabilities sums to one. Conditional on \(B=b\), the shared-parameter model therefore specifies
This model has two important assumptions: the outcomes are conditionally independent of each other given \(B\), and the pattern is conditionally independent of the outcomes given \(B\). Marginally, \(Y_1,Y_2,\) and \(R\) need not be independent, because they are associated through \(B\). The model is a concrete joint law to factor in the other two ways; the calculations below do not assume that the selection or pattern-mixture factors have the conditional-independence form used given \(B\).
Worked Example: Obtain the Marginal Pattern Probabilities
First compute the probabilities of the three patterns by averaging over \(B\). For \(R=0\),
Similarly, \(p(R=1)=0.5(0.3)+0.5(0.1)=0.20\), and \(p(R=2)=0.5(0.5)+0.5(0.1)=0.30\). These marginal probabilities are positive and sum to one. They are the pattern-probability factors in the pattern-mixture representation.
Deriving the Selection and Pattern-Mixture Factors
The selection factor \(p(R=r\mid Y=y)\) averages over the latent \(B\) conditional on the outcomes. Bayes’ rule gives the posterior weights \(p(B=b\mid Y=y)\), so
The second equality uses the specified conditional independence of \(R\) and \(Y\) given \(B\). In contrast, a pattern-mixture conditional averages over the distribution of \(B\) within a pattern:
Here the last equality uses the same conditional independence assumptions. Thus the averaging weights differ: selection uses \(p(B\mid Y)\), while pattern-mixture uses \(p(B\mid R)\). Neither conditional generally equals its counterpart given \(B\).
Worked Example: Calculate One Selection Conditional
For \(Y_1=Y_2=1\), the outcome probability is
The joint probability of these outcomes and \(R=2\) is
Consequently, the selection conditional is
Equivalently, the posterior probability of \(B=1\) given two successes is \(0.32/0.34=0.9412\), so averaging the pattern probabilities gives \(0.5(1-0.9412)+0.1(0.9412)=0.1235\). The result differs from either conditional pattern probability given \(B\), as it should.
Worked Example: Calculate Pattern-Mixture Outcome Distributions
For the complete-observation pattern, the probability of two successes conditional on \(R=2\) is
For comparison, the four outcome probabilities conditional on \(R=2\) are \(p(0,0\mid R=2)=0.162/0.30=0.54\), \(p(1,0\mid R=2)=0.048/0.30=0.16\), \(p(0,1\mid R=2)=0.048/0.30=0.16\), and \(p(1,1\mid R=2)=0.042/0.30=0.14\). They are nonnegative and sum to one.
For \(R=0\), the joint probability of two successes and that pattern is \(0.5(0.2^2)(0.2)+0.5(0.8^2)(0.8)=0.004+0.256=0.260\). Thus \(p(Y_1=1,Y_2=1\mid R=0)=0.260/0.50=0.52\). These pattern-specific outcome distributions differ because the patterns have different mixtures of \(B\).
Checking That the Three Forms Agree
The selection and pattern-mixture expressions must reconstruct the same joint probability as the shared-parameter expression. The following calculation verifies this for \(Y=(1,1)\) and \(R=2\):
Sum over \(B\): \(0.5(0.2^2)(0.5)+0.5(0.8^2)(0.1)=0.042\).
Multiply the outcome marginal by the pattern conditional: \(0.34(0.1235)\), which equals \(0.042\) before rounding.
Multiply the pattern probability by its outcome conditional: \(0.30(0.14)=0.042\).
The general justification is the product rule and marginalization. For any \(y,r\) with positive denominators,
where the final equality holds for the specified shared-parameter model. The equality does not mean that \(p(r\mid y)=p(r\mid b)\), or that \(p(y\mid r)=p(y\mid b)\). It means that each factorization reconstructs the same joint probability using its own conditioning variables.
What the Comparison Does—and Does Not—Establish
The selection factorization conditions the pattern distribution on the complete outcome vector. The pattern-mixture factorization conditions the outcome distribution on the pattern. The shared-parameter factorization conditions both on a latent variable and then averages it out. Their practical differences are summarized below.
| Representation | Factors | Conditioning distribution to obtain | Modeling feature |
|---|---|---|---|
| Selection | \(p(Y)p(R\mid Y)\) | \(R\) given the complete \(Y\) | Directly describes pattern probabilities across outcome values |
| Pattern-mixture | \(p(R)p(Y\mid R)\) | Complete \(Y\) within each pattern | Directly describes pattern-specific outcome distributions |
| Shared parameter | \(\sum_b p(B)p(Y\mid B)p(R\mid B)\) here | Conditional outcome and pattern laws given \(B\) | Uses a latent variable and, in this example, conditional independence |
A factorization identity should not be confused with a claim that its factors are empirically determined or easy to estimate. For example, \(p(Y\mid R=0)\) refers to outcomes when neither visit was observed; those outcomes are not directly recorded for that pattern. Likewise, \(p(R\mid Y)\) conditions on the complete outcome vector, including values missing under some patterns. The formulas define the complete-data law, but observed records do not automatically supply every factor.
A shared-parameter formulation can make dependence more structured: here the same \(B\) affects both outcome probabilities and the observation pattern. That is useful when the conditional model is scientifically or statistically appropriate, but it adds assumptions. A different distribution for \(B\), different conditional outcome models, or dependence between \(R\) and \(Y\) even after conditioning on \(B\) would define a different shared-parameter model. The selection and pattern-mixture forms can still be written for that resulting joint law, but their conditional factors must be recalculated.
Diagnostics and Common Pitfalls
For a specified finite joint law, basic checks are concrete: conditional distributions must be nonnegative and sum to one; marginal pattern probabilities must sum to one; and multiplying either pair of factors must recover each joint cell. In continuous settings, sums are replaced by integrals and densities must normalize. A zero-probability pattern has no defined conditional distribution \(p(Y\mid R=r)\) under the usual ratio definition. Similarly, \(p(R\mid Y=y)\) is undefined at outcome values with zero probability. Do not divide by zero or assign a conditional distribution without stating an additional convention or model extension.
A common mistake is to take a conditional factor from one representation and reuse it in another—for example, setting \(p(R\mid Y)=p(R\mid B)\) because both describe observation patterns. The correction is to identify the variables being conditioned on and average over the omitted variable. In this example, \(p(R\mid Y)\) averages \(p(R\mid B)\) using \(p(B\mid Y)\), whereas \(p(Y\mid R)\) averages \(p(Y\mid B)\) using \(p(B\mid R)\).
Another mistake is to treat all three representations as equally restrictive models simply because they factor the same joint law. The first two are chain-rule decompositions of a joint distribution. A chosen parametric specification for their factors can impose restrictions, and the shared-parameter model above imposes explicit conditional independence. If fitted models do not reproduce the same joint probabilities, they are not merely different writings of one law; at least one modeling specification or calculation differs.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- In the example, what is \(p(R=1)\)? Show how it follows from the shared-parameter pattern probabilities.
- Why does calculating \(p(R=2\mid Y_1=Y_2=1)\) require weights conditional on the outcomes rather than the marginal probabilities of \(B\)?
- Using the reported values, verify the pattern-mixture reconstruction of \(p(Y_1=Y_2=1,R=2)\).
- What prevents \(p(Y\mid R=0)\) from being defined by the usual conditional-probability ratio if \(p(R=0)=0\)?
- Does conditional independence of \(R\) and \(Y\) given \(B\) imply their marginal independence? Explain briefly.
Solutions. (1) \(p(R=1)=0.5(0.3)+0.5(0.1)=0.20\). (2) Given two successes, \(B=1\) is more likely: \(p(B=1\mid Y_1=Y_2=1)=0.32/0.34=0.9412\). The selection conditional therefore averages using these outcome-specific posterior weights, not the prior weights \(0.5,0.5\). (3) \(p(R=2)p(Y_1=1,Y_2=1\mid R=2)=0.30(0.14)=0.042\), matching the shared-parameter joint probability. (4) The ratio would divide \(p(Y,R=0)\) by \(p(R=0)=0\), so it is undefined; there are no observations in that pattern under the law. (5) No. Both \(Y\) and \(R\) depend on \(B\), so averaging over \(B\) can associate them even when they are independent conditional on \(B\).