Why Compatibility Matters
In MICE, each incomplete variable is imputed from a model conditional on other variables. This allows the conditional model for each variable to be chosen to suit its type and predictors. But individually plausible conditional models need not automatically describe one coherent joint distribution. Compatibility names the precise property that resolves this question: do the specified conditionals fit together as the conditionals of a single joint distribution?
This is a property of the conditional distributions themselves, not a claim that they fit the observed data well. Nor is it a diagnostic for the missingness mechanism. A set of conditional models can be compatible and still be poorly specified for the data, and good-looking diagnostics do not by themselves establish compatibility.
The distinction between fixed distributions and model families matters. A pair of model families may be compatible for some parameter values but not for others. In an applied analysis, the fitted conditional distributions are the particular specifications to examine; the fact that their model families could be compatible under different parameter values does not establish that the fitted specifications are compatible.
For discrete variables, a joint distribution assigns probabilities to pairs of values. Its conditional probabilities must agree with the corresponding entries and margins of that same joint distribution. This gives a direct way to verify compatibility when a candidate joint distribution is available. A useful identity behind the definition is that, for any pair with positive probability, \(p(x,y)=p(y) p(x\mid y)=p(x) p(y\mid x)\). Both conditional specifications must be consistent with one set of joint probabilities and margins; it is not enough for each conditional model to be reasonable on its own.
Checking Compatibility With a Candidate Joint Distribution
For two binary variables, write each possible pair in a \(2\times2\) table. The table entries are joint probabilities, so they must be nonnegative and sum to one. To check that the table supports specified conditional models, divide each cell by the appropriate column total to obtain \(X\mid Y\), and by the appropriate row total to obtain \(Y\mid X\). If both resulting sets of probabilities match the specified conditionals, the candidate joint distribution verifies compatibility.
Worked Example: Compatible Dependent Conditionals
Suppose \(X\) and \(Y\) are binary, and a candidate joint distribution is given below. Rows correspond to \(X\), columns to \(Y\).
| \(Y=0\) | \(Y=1\) | Row total | |
|---|---|---|---|
| \(X=0\) | 0.40 | 0.10 | 0.50 |
| \(X=1\) | 0.10 | 0.40 | 0.50 |
| Column total | 0.50 | 0.50 | 1.00 |
All four entries are nonnegative and their sum is \(0.40+0.10+0.10+0.40=1.00\), so this is a joint probability distribution. Its column totals give \(P(X=1\mid Y=0)=0.10/0.50=0.20\) and \(P(X=1\mid Y=1)=0.40/0.50=0.80\).
Its row totals give \(P(Y=1\mid X=0)=0.10/0.50=0.20\) and \(P(Y=1\mid X=1)=0.40/0.50=0.80\). Thus, if the two specified models have exactly these conditional probabilities, this table is a witness that they are compatible. The conditionals are dependent: the probability of \(X=1\) changes with \(Y\), and the probability of \(Y=1\) changes with \(X\).
A candidate joint distribution is a sufficient demonstration of compatibility, because the definition asks whether at least one such distribution exists. It does not need to be the only possible joint distribution. Conversely, merely writing down two sets of conditional probabilities does not provide such a demonstration: their consistency must still be checked.
Compatibility Can Also Hold Under Independence
Independence is one special case, not the definition of compatibility. Under independence, the conditional distribution of either variable does not change with the value of the other. The next example verifies this case by showing one joint distribution that has both specified conditional models.
Worked Example: Compatible Independent Conditionals
Suppose the specified models are \(P(X=1\mid Y=y)=0.30\) for either value of \(y\), and \(P(Y=1\mid X=x)=0.60\) for either value of \(x\). Consider the following candidate joint distribution:
| \(Y=0\) | \(Y=1\) | Row total | |
|---|---|---|---|
| \(X=0\) | 0.28 | 0.42 | 0.70 |
| \(X=1\) | 0.12 | 0.18 | 0.30 |
| Column total | 0.40 | 0.60 | 1.00 |
The entries sum to \(0.28+0.42+0.12+0.18=1.00\). For \(Y=0\), the conditional probability of \(X=1\) is \(0.12/0.40=0.30\); for \(Y=1\), it is \(0.18/0.60=0.30\). For \(X=0\), the conditional probability of \(Y=1\) is \(0.42/0.70=0.60\); for \(X=1\), it is \(0.18/0.30=0.60\). Both conditionals match the stated models, so they are compatible.
The table entries also equal the products of the marginal probabilities. For example, \(P(X=0,Y=1)=0.70(0.60)=0.42\), and \(P(X=1,Y=0)=0.30(0.40)=0.12\). This verifies that the candidate represents independence. Compatibility does not require independence; the previous example was compatible with dependent conditionals.
A Conditional Pair That Cannot Share a Joint Distribution
The definition also clarifies what incompatibility means. If two conditional rules require mutually contradictory events to each have probability one, no joint distribution can satisfy both. The contradiction is enough to rule out compatibility; it is not necessary to estimate a joint distribution and discover a numerical failure.
Worked Example: Contradictory Deterministic Conditionals
Let \(X\) and \(Y\) be binary, with each conditional specified for both possible conditioning values. Suppose \(X\mid Y\) is deterministic with \(X=Y\): when \(Y=0\), \(X=0\) with probability one, and when \(Y=1\), \(X=1\) with probability one. Suppose also that \(Y\mid X\) is deterministic with \(Y=1-X\): when \(X=0\), \(Y=1\) with probability one, and when \(X=1\), \(Y=0\) with probability one.
The first rule requires \(P(X=Y)=1\). The second requires \(P(Y=1-X)=1\). For binary values, the events \(X=Y\) and \(Y=1-X\) are disjoint: no pair \((x,y)\) can satisfy both. If a joint distribution satisfied both requirements, it would assign probability one to each of two disjoint events, which is impossible because their probabilities would sum to two. Therefore these specified conditionals are incompatible.
This conclusion uses the fact that both conditional rules are required at both conditioning values. If a conditioning value has zero probability under a proposed joint distribution, the corresponding conditional distribution is not defined there; it cannot be used to claim that the joint matches a conditional rule specified for that value.
What Compatibility Does and Does Not Establish
Compatibility is a coherence condition. It says that the conditionals could be the full conditional distributions of one joint distribution. It does not say that this joint distribution is the right description of the population, that its parameters are estimated precisely, or that an imputation procedure has converged.
In particular, compatibility should not be confused with the following:
- Good fit: compatible conditionals can still be misspecified or fit the observed data poorly.
- Convergence: stable traces or repeated draws do not demonstrate that all specified conditionals come from a common joint distribution.
- Compatibility with an analysis model: the question here concerns whether the imputation conditionals share a joint distribution. Whether imputation supports a particular analysis is a separate question.
- Identical marginal distributions: matching one-variable margins alone does not establish that the joint associations implied by two conditional models agree.
One alternative is to specify a joint model first and use its implied conditionals. This approach starts with a single joint distribution, so its conditionals are coherent by construction. Fully conditional specification instead chooses a model for each incomplete variable. That flexibility can be useful for mixed data types, but it makes compatibility a substantive question rather than an automatic consequence. The two approaches should not be treated as interchangeable: each has modeling choices and limitations.
For fitted MICE models, a practical review can compare the roles and functional forms of variables across the conditional specifications, then check whether a proposed joint model can reproduce them. For simple discrete examples, tables make the check explicit. In more complex settings, those checks may require algebra or specialized computation, and empirical diagnostics alone may not decide the question. A failure to find a joint model in a limited search is not, by itself, a proof that none exists.
Record the variable being imputed, its conditioning variables, and the specified conditional distribution at each relevant conditioning value.
For fixed fitted models, ask whether those particular distributions share a joint distribution. For model families, distinguish this from asking whether some parameter values could be compatible.
For small discrete examples, test a candidate joint table against both sets of conditionals. If the requirements imply a direct contradiction, no matching joint distribution exists.
Compatibility establishes a common-joint representation, not correct specification, good empirical fit, convergence, or suitability for a particular analysis.
Common Mistakes and Limitations
- Assuming every collection of conditionals is compatible. Each model may be a valid probability distribution on its own, yet their requirements can conflict. Check for a common joint distribution rather than judging the models separately.
- Calling independence the definition. Independence gives one compatible case, but dependent conditionals can also be compatible, as the first worked example showed.
- Inferring compatibility from a successful chain. A MICE algorithm may produce draws or apparently stable summaries. Those facts do not prove that the specified conditional distributions are all the conditionals of one joint law.
- Ignoring conditioning values with zero probability. Conditional distributions at such values are undefined under that joint distribution. Compatibility must be assessed on the support where the conditional distributions are required to agree.
In realistic applications, full conditional models can involve many predictors, continuous variables, interactions, nonlinear terms, and different outcome types. Establishing or disproving the existence of a common joint distribution can then be difficult. The definition remains useful even when a complete check is impractical: it identifies the coherence question, separates it from fit and convergence, and discourages treating each conditional model as though it were automatically part of one joint model.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions below.
- A candidate joint table has cell probabilities \(0.15, 0.35, 0.20,\) and \(0.30\), arranged as \(X=0\) row \((0.15,0.35)\) and \(X=1\) row \((0.20,0.30)\), with columns \(Y=0,1\). What is \(P(X=1\mid Y=1)\)?
- For that same table, what is \(P(Y=1\mid X=0)\)?
- In one sentence, what must be shown to establish compatibility of two fixed conditional distributions?
- Why do stable MICE trace plots not prove compatibility?
Solutions. (1) The \(Y=1\) column total is \(0.35+0.30=0.65\), so \(P(X=1\mid Y=1)=0.30/0.65\approx0.4615\). (2) The \(X=0\) row total is \(0.15+0.35=0.50\), so \(P(Y=1\mid X=0)=0.35/0.50=0.7000\). (3) One must show that there exists a single joint distribution whose conditional distributions equal both specified conditionals on the relevant support. (4) Stable traces concern the behavior of the iterative procedure; they do not establish that the specified conditionals are jointly coherent as full conditionals of one distribution.