Why Reading a Factorization Carefully Matters
A joint factorization is more than a way to rearrange symbols. Each factor describes a particular probability distribution, conditional on particular variables and defined over a particular support. In missing-data problems, an especially consequential mistake is to read a distribution conditional on a missingness pattern as though it described only the outcomes observed in that pattern. Another is to drop the probability of the pattern when converting between factorizations. A third is to infer marginal independence from a conditional-independence statement.
The checks in this tutorial apply to discrete probabilities and, with sums replaced by integrals where appropriate, to densities. Let \(Y=(Y_1,\ldots,Y_K)\) be the vector of outcomes that would be represented in the full-data law, and let \(R\) identify the missingness pattern. For a particular pattern \(r\), write \(Y_{\mathrm{obs}}(r)\) for the components observed under that pattern and \(Y_{\mathrm{mis}}(r)\) for the remaining components.
The distinction is about the variable being distributed, not merely the conditioning notation. Under a dropout pattern, \(p(y\mid R=r)\) still assigns probabilities to values of outcomes that are not observed for people in that pattern. Those values are part of the full-data model; they are not directly recorded in that pattern.
Marginalizing to the Observed-Data Contribution
The joint law can be factored by the chain rule as \(p(y,r)=p(r)p(y\mid r)\). If a unit has pattern \(r\), the observed-data contribution is obtained by summing the joint probability over every compatible value of the missing outcomes. The pattern probability remains in the contribution.
The equality follows from the definition of a marginal probability and the chain rule. For continuous missing outcomes, replace the sum by an integral over their support. If the pattern itself is conditioned upon rather than included in a joint contribution, the relevant distribution is \(p(y_{\mathrm{obs}}(r)\mid r)\); it is not the joint contribution \(p(r)p(y_{\mathrm{obs}}(r)\mid r)\). Keeping track of which of these is being used prevents an otherwise easy-to-miss extra or missing factor.
Worked Example: A Dropout Pattern and Its Observed Marginal
Suppose \(Y_1\) and \(Y_2\) are binary outcomes. Pattern \(R=0\) means \(Y_1\) is observed and \(Y_2\) is missing. In a hypothetical model, \(P(R=0)=0.25\), and the full-outcome distribution conditional on this pattern is:
| \(Y_1\) | \(Y_2\) | \(P(Y_1,Y_2\mid R=0)\) |
|---|---|---|
| 0 | 0 | 0.10 |
| 0 | 1 | 0.30 |
| 1 | 0 | 0.20 |
| 1 | 1 | 0.40 |
The four conditional probabilities sum to \(0.10+0.30+0.20+0.40=1.00\), so this is a normalized distribution for the full pair \((Y_1,Y_2)\) among units with \(R=0\). To find the observed distribution when \(Y_1=1\), sum over the missing \(Y_2\):
Thus the observed-data joint contribution for \(Y_1=1,R=0\) is
The value \(0.40\) is the probability of one complete outcome pair conditional on the pattern; it is not the probability of the observed event \(Y_1=1\) conditional on the pattern. The latter includes both possible values of the missing \(Y_2\).
Do Not Drop the Normalizing Pattern Probability
A second common source of error arises when changing the order of conditioning. The chain rule gives \(p(y,r)=p(y)p(r\mid y)=p(r)p(y\mid r)\). Equating these expressions and dividing by \(p(r)\) gives Bayes' rule:
The denominator \(p(r)\) is the normalizing probability that makes the conditional distribution sum or integrate to one over \(y\). The formula requires \(p(r)>0\); if a pattern has probability zero, conditioning on that pattern is not defined by this ratio.
Worked Example: Checking a Bayes Conversion
Let \(Y\) be binary in a hypothetical population. Suppose \(P(Y=1)=0.40\), \(P(Y=0)=0.60\), \(P(R=1\mid Y=1)=0.75\), and \(P(R=1\mid Y=0)=0.25\). First find the pattern probability:
Therefore,
Without the denominator, the numerator \(0.30\) is the joint probability \(P(Y=1,R=1)\), not the conditional probability \(P(Y=1\mid R=1)\). A quick normalization check exposes the omission: the two numerators for \(Y=0\) and \(Y=1\) sum to \(0.15+0.30=0.45\), not one. Dividing both by \(P(R=1)=0.45\) gives \(P(Y=0\mid R=1)=0.333\) and \(P(Y=1\mid R=1)=0.667\), which sum to one.
Conditional Independence Is Not Marginal Independence
A factorization such as \(p(y_1,y_2\mid r)=p(y_1\mid r)p(y_2\mid r)\) states that \(Y_1\) and \(Y_2\) are conditionally independent given \(R\), on patterns with positive probability. It does not generally state that \(Y_1\) and \(Y_2\) are independent after \(R\) is no longer conditioned on. Marginalizing over \(R\) can create dependence because the pattern can associate with both outcomes.
Worked Example: Conditional Independence but Marginal Dependence
Consider a hypothetical binary pattern indicator with \(P(R=0)=P(R=1)=0.50\). Given \(R=0\), let \(Y_1\) and \(Y_2\) be independent Bernoulli variables, each with probability \(0.10\) of being 1. Given \(R=1\), let them again be independent, each with probability \(0.90\) of being 1. The product form holds within each pattern by construction.
By averaging over patterns,
Conditional independence gives \(P(Y_1=1,Y_2=1\mid R=0)=0.10(0.10)=0.01\) and \(P(Y_1=1,Y_2=1\mid R=1)=0.90(0.90)=0.81\). Hence
If the outcomes were marginally independent, this joint probability would equal \(P(Y_1=1)P(Y_2=1)=0.50(0.50)=0.25\). Since \(0.41\ne0.25\), they are marginally dependent. Conditional independence describes distributions within pattern strata; it does not justify replacing the marginal joint distribution by a product of marginal distributions.
A Practical Audit for Factorizations
When reading or writing a factorization, verify the following in order. These checks do not determine whether a substantive model is appropriate; they determine whether the probability statement has been represented correctly.
Decide whether the expression is a joint law, a full-outcome conditional law, an observed-outcome marginal, or a conditional distribution given a pattern.
Check that each factor conditions on the variables shown and that a changed conditioning order is justified by the chain rule or Bayes' rule.
If outcomes are missing in a pattern, sum or integrate over their possible values to obtain the observed-outcome distribution.
Conditional probabilities must sum or integrate to one on their support. An independence statement must retain its conditioning variables unless marginal independence is separately established.
A frequent wrong argument is: “The model factors as \(p(y_1\mid r)p(y_2\mid r)\), so \(Y_1\) and \(Y_2\) are independent.” The correction is that this proves independence only conditional on \(R\). To claim marginal independence, one must check directly that \(p(y_1,y_2)=p(y_1)p(y_2)\) for the marginal law, or establish an additional condition that implies it.
Another wrong argument is: “The factor \(p(y\mid r)\) uses the outcomes observed under pattern \(r\), so it is already an observed-data distribution.” That conclusion confuses the pattern with the vector being distributed. If \(p(y\mid r)\) is defined for the complete vector \(Y\), missing components still appear in its support and must be marginalized to obtain \(p(y_{\mathrm{obs}}(r)\mid r)\).
What These Checks Do—and Do Not—Establish
The three errors have distinct repairs: marginalize over missing outcomes, retain the normalizing pattern probability when applying Bayes' rule, and preserve the conditioning in independence statements. The chain rule and marginalization identities ensure algebraic consistency, but they do not by themselves establish that a chosen full-data model is scientifically plausible or that its unobserved components can be learned from observed data. A factorization can be correctly normalized and still encode assumptions that need separate justification.
The most useful diagnostic is to ask what values each factor ranges over. If a factor is conditional on a pattern but ranges over all of \(Y\), it is a full-outcome conditional distribution. If it ranges only over the observed components, it is an observed-outcome marginal. Then check that the probabilities sum to one and that every independence claim says exactly what its conditioning set supports.
Check Your Understanding
Answer each question, then compare with the solutions below.
- For a pattern \(r\) in which only \(Y_1\) is observed, what operation converts \(p(y_1,y_2\mid r)\) into the observed-outcome distribution?
- Suppose \(P(R=1,Y=1)=0.18\) and \(P(R=1)=0.30\). Find \(P(Y=1\mid R=1)\), and state what the numerator alone represents.
- If \(Y_1\) and \(Y_2\) are conditionally independent given \(R\), what additional calculation is needed before claiming they are marginally independent?
- A proposed conditional distribution assigns probabilities \(0.12\) and \(0.58\) to the two values of a binary outcome. What normalization problem should be investigated?
Solutions. (1) Sum over the missing \(Y_2\): \(p(y_1\mid r)=\sum_{y_2}p(y_1,y_2\mid r)\). For continuous \(Y_2\), integrate instead. (2) \(P(Y=1\mid R=1)=P(R=1,Y=1)/P(R=1)=0.18/0.30=0.60\). The numerator is a joint probability, not a conditional probability. (3) Compute the marginal joint law by summing over \(R\), and compare \(P(Y_1=y_1,Y_2=y_2)\) with \(P(Y_1=y_1)P(Y_2=y_2)\); conditional independence alone is insufficient. (4) The probabilities sum to \(0.12+0.58=0.70\), not one. Check whether a value was omitted or whether the numbers are unnormalized weights; a conditional probability distribution must be normalized on its support.