From Conditional Statements to a Joint Law
In missing-data models, a verbal or mathematical statement often specifies only part of the joint distribution. The task is to translate it into factors that together describe the relevant variables, without silently dropping the missingness pattern, the outcome, a covariate, or a latent variable. The key is to read each conditional distribution literally: identify its outcome variable, its conditioning variables, and any independence restriction it encodes.
Let \(Y\) denote the complete outcome (possibly a vector of repeated outcomes), \(R\) the missingness pattern, and \(X\) observed covariates. When a shared-parameter representation is used, let \(U\) denote a latent variable. Write \(p(\cdot)\) for a probability mass function or density, as appropriate. A factorization conditional on \(X=x\) describes the joint law of the remaining variables within that covariate value; if \(X\) is random, include \(p(x)\) to obtain the full joint law.
The first two products are chain-rule factorizations of the same joint law, not different independence assumptions by themselves. The shared-parameter product makes a substantive conditional-independence claim: after conditioning on \(U\) and \(X\), the outcome and missingness pattern are independent. The latent variable is part of the joint law in that representation; to obtain a law for \(Y,R\) alone, sum or integrate over \(U\).
A Reliable Reading Procedure
Before multiplying terms, make a list of the variables the stated law must cover. Then match each conditional statement to a factor whose left-hand variable is the variable being distributed. A variable after the conditioning bar is a condition, not automatically a factor: for example, \(p(y\mid x)\) represents \(Y\) conditional on \(X\), but a full joint law for \(X,Y\) also needs \(p(x)\). Conversely, if the target is only the law conditional on \(X=x\), adding \(p(x)\) would change the target.
Decide whether the target is conditional on \(X\), a joint law including \(X\), or a joint law that also includes latent \(U\).
Record the variable on the left of the conditioning bar and the variables it is conditioned on. Treat an independence statement as a restriction on a conditional distribution.
Check that every target variable is represented, each factor has the stated conditioning variables, and the product is a normalized joint distribution on the stated support.
A product is not justified merely because its factors contain familiar variables. For example, \(p(y\mid x)p(r\mid x)\) is a joint law only if \(Y\perp R\mid X\). Without that independence statement, it generally imposes a restriction that was not given. A useful diagnostic is to ask whether the proposed product follows from a chain rule or from an explicitly stated conditional independence.
Worked Example: Reading a Selection Factorization
Suppose \(X=x\) is fixed, \(Y\) is binary, and \(R=1\) means the outcome is observed. The statements are \(P(Y=1\mid x)=0.40\), \(P(R=1\mid Y=1,x)=0.90\), and \(P(R=1\mid Y=0,x)=0.75\). Construct the joint distribution of \(Y,R\) given \(x\).
The statements supply an outcome distribution and a missingness distribution conditional on that outcome. They therefore give the selection product \(p(y\mid x)p(r\mid y,x)\). For the cell \(Y=1,R=1\), \(P(Y=1,R=1\mid x)=0.40(0.90)=0.36\).
For the other cells, use the complementary probabilities: \(P(Y=1,R=0\mid x)=0.40(1-0.90)=0.040\); \(P(Y=0,R=1\mid x)=0.60(0.75)=0.45\); and \(P(Y=0,R=0\mid x)=0.60(1-0.75)=0.15\).
The four values sum to \(0.36+0.040+0.45+0.15=1.00\), so they form a normalized conditional joint distribution. For instance, the response probability is \(P(R=1\mid x)=0.36+0.45=0.81\). The variables in the target conditional law, \(Y\) and \(R\), both appear; \(X\) is conditioned on rather than represented by a factor.
Reading Pattern-Mixture Statements
If the supplied statements instead describe the distribution of the pattern and then the outcome within each pattern, they match \(p(r\mid x)p(y\mid r,x)\). Every possible pattern with positive probability needs a conditional outcome distribution. If a pattern has probability zero, its conditional distribution is not determined by the joint law and is not needed to compute that law.
Worked Example: Building a Pattern-Mixture Product
Again take a fixed \(X=x\), binary \(Y\), and \(R\in\{0,1\}\). Suppose \(P(R=1\mid x)=0.80\), \(P(Y=1\mid R=1,x)=0.50\), and \(P(Y=1\mid R=0,x)=0.25\). Find the joint probabilities.
These statements match the pattern-mixture product \(p(r\mid x)p(y\mid r,x)\). For \(R=1,Y=1\), the joint probability is \(0.80(0.50)=0.40\); for \(R=1,Y=0\), it is \(0.80(0.50)=0.40\). For \(R=0,Y=1\), it is \(0.20(0.25)=0.050\); for \(R=0,Y=0\), it is \(0.20(0.75)=0.15\).
The values sum to \(1.00\), and both \(Y\) and \(R\) are represented. As a check, \(P(Y=1\mid x)=0.40+0.050=0.45\). The corresponding selection conditional is \(P(R=1\mid Y=1,x)=0.40/0.45=0.8889\), illustrating that the product can also be rearranged into the other chain-rule order when the relevant denominators are positive.
When the Statement Includes a Shared Parameter
A statement that \(Y\perp R\mid U,X\) means that for each supported \(u,x\), the conditional joint distribution factors as \(p(y,r\mid u,x)=p(y\mid u,x)p(r\mid u,x)\). Combining that restriction with the distribution of \(U\) gives a product for the joint law of \(Y,R,U\). To obtain the marginal law of \(Y,R\), \(U\) must then be removed by summation for discrete \(U\) or integration for continuous \(U\).
Worked Example: Constructing and Marginalizing a Shared-Parameter Product
At a fixed \(X=x\), let \(U\) be binary with \(P(U=0\mid x)=P(U=1\mid x)=0.50\). Suppose \(Y\perp R\mid U,x\). When \(U=0\), let \(P(Y=1\mid U,x)=0.20\) and \(P(R=1\mid U,x)=0.60\). When \(U=1\), let those probabilities be \(0.80\) and \(0.90\), respectively. Find the marginal joint distribution of \(Y,R\) given \(x\).
The joint law including \(U\) is \(p(u\mid x)p(y\mid u,x)p(r\mid u,x)\). For \(U=0\), conditional independence gives the cell probabilities \((Y=1,R=1):0.20(0.60)=0.12\), \((1,0):0.20(0.40)=0.080\), \((0,1):0.80(0.60)=0.48\), and \((0,0):0.80(0.40)=0.32\).
For \(U=1\), the corresponding conditional cells are \(0.80(0.90)=0.72\), \(0.80(0.10)=0.080\), \(0.20(0.90)=0.18\), and \(0.20(0.10)=0.020\). Each set of four conditional probabilities sums to one. Weighting each set by \(P(U=u\mid x)=0.50\) and adding gives \(P(Y=1,R=1\mid x)=0.5(0.12)+0.5(0.72)=0.42\); \(P(1,0\mid x)=0.080\); \(P(0,1\mid x)=0.33\); and \(P(0,0\mid x)=0.17\).
The marginal cells sum to one. The product includes \(U\) in the joint law, while the weighted sum correctly removes it from the marginal law. If \(X\) were random and the target were the full law of \(X,Y,R,U\), the additional factor \(p(x)\) would be required.
Justification, Checks, and Comparisons
The chain rule justifies the two ordinary factorizations. For any supported \(y,r,x\), factor the joint conditional probability first by \(Y\), or first by \(R\):
For example, the first equality follows from the definition of conditional probability: \(p(y\mid x)p(r\mid y,x)=p(y\mid x)p(y,r\mid x)/p(y\mid x)=p(y,r\mid x)\) whenever \(p(y\mid x)>0\). The second equality follows by the same argument conditioning first on \(R\). In discrete cases, summing either product over \(y,r\) gives one because each conditional distribution sums to one. The shared-parameter product follows by applying the same rule to \(U\) and then using the stated conditional independence; marginalization sums or integrates that normalized joint law over \(u\).
| Statements supplied | Product that matches them | Extra check |
|---|---|---|
| \(p(y\mid x)\) and \(p(r\mid y,x)\) | \(p(y\mid x)p(r\mid y,x)\) | Confirm the outcome distribution and pattern probabilities cover their supports. |
| \(p(r\mid x)\) and \(p(y\mid r,x)\) | \(p(r\mid x)p(y\mid r,x)\) | Supply an outcome distribution for each pattern with positive probability. |
| \(p(u\mid x)\), \(p(y\mid u,x)\), \(p(r\mid u,x)\), and \(Y\perp R\mid U,X\) | \(p(u\mid x)p(y\mid u,x)p(r\mid u,x)\) | Include \(U\) in the joint law, or sum/integrate it out for the marginal law. |
These forms differ in what is specified directly. A selection product starts with the outcome distribution and models the pattern conditional on the outcome; a pattern-mixture product starts with pattern probabilities and models outcomes within patterns. Both are valid chain-rule representations of a joint law, and changing the order alone does not add an independence assumption. A shared-parameter product is different in that it explicitly invokes conditional independence given \(U,X\). The earlier material on Comparing Three Factorizations Of One Joint Law gives the broader comparison; here the practical point is to use the conditionals actually supplied rather than infer a preferred model from their appearance.
Three checks catch many errors. First, check coverage: every variable in the target must be in a factor or, if latent, be included and then explicitly marginalized. Second, check normalization: the factors must define probability distributions on their supports. Third, check compatibility: a factor must use the conditioning variables in the given statement, and a product of marginal conditionals must not be substituted for a conditional joint distribution unless independence is stated.
Common Pitfalls and Limitations
- Dropping a variable. Writing only \(p(y\mid x)\) does not give the joint law of \(Y,R\mid X=x\). Add a factor for \(R\) in the stated chain-rule form.
- Treating conditioning as a factor. \(X\) appearing after a vertical bar does not mean \(p(x)\) is included. Add \(p(x)\) only when the target is a joint law including \(X\).
- Assuming independence from separate marginals. The product \(p(y\mid x)p(r\mid x)\) requires \(Y\perp R\mid X\); it does not follow from having two conditional marginal distributions.
- Leaving a latent variable in a marginal answer. A product involving \(U\) describes a joint law that includes \(U\). Sum or integrate over it if the requested law is only for \(Y,R\).
A zero-probability conditioning event needs particular care: a conditional distribution given that event is not defined by the joint law in the usual way. Likewise, conditional factors need only be specified on supported conditioning values. Factorizations organize assumptions and distributions; writing a valid product does not by itself show that its ingredients can be learned from observed data or that its assumptions are empirically testable.
Check Your Understanding
For each item, identify the appropriate product and check what variables the target law contains. Solutions follow.
- Given \(p(r\mid x)\) and \(p(y\mid r,x)\), what product gives the law of \(Y,R\mid X=x\)?
- Given \(p(y\mid x)\) and \(p(r\mid x)\), what additional statement is needed to multiply these into a joint law?
- If \(p(u\mid x)\), \(p(y\mid u,x)\), \(p(r\mid u,x)\), and \(Y\perp R\mid U,X\) are given, what operation gives the law of \(Y,R\mid X=x\)?
- A researcher writes \(p(x)p(y\mid x)p(r\mid y,x)\) but says the target is \(Y,R\mid X=x\). What is the issue?
Solutions. (1) Use \(p(r\mid x)p(y\mid r,x)\); it represents both \(R\) and \(Y\) conditional on \(x\). (2) The needed statement is \(Y\perp R\mid X\), so that \(p(y,r\mid x)=p(y\mid x)p(r\mid x)\). (3) Sum or integrate \(p(u\mid x)p(y\mid u,x)p(r\mid u,x)\) over \(u\), according to whether \(U\) is discrete or continuous. (4) The factor \(p(x)\) makes this a joint law including \(X\), rather than the requested law conditional on \(X=x\); omit it for that target.