Why Compatibility Matters
In fully conditional specification (FCS), an imputation procedure updates each variable using a model for that variable conditional on the others. The resulting collection of conditional models is not automatically a single coherent joint model. Compatibility gives a precise way to ask whether such a joint model exists: could all the specified conditional distributions be obtained from one probability distribution for the variables?
This is a question about the conditional distributions themselves, not about how quickly an algorithm runs. A chain can have a stationary distribution without the update models being conditionals of a proposed common joint distribution. Conversely, compatibility alone does not establish that a chain will converge quickly, or that the specified models describe the data well.
For continuous variables, the same definition applies to conditional distributions or densities, with agreement understood almost everywhere under the relevant marginal distribution. A conditional distribution at a conditioning value that has probability zero is not identified by the joint distribution: its value can be assigned as a version of the conditional without changing the joint law. Compatibility therefore concerns agreement on the support that the joint distribution actually gives positive probability (or, in continuous settings, on the relevant sets up to probability zero).
The definition concerns fixed conditional distributions. In an FCS procedure, conditional models may include parameters that are drawn or updated during the algorithm. Compatibility can be asked at fixed parameter values; asking whether entire model families can be conditionals of one joint model is a stronger, family-level question. Neither question is answered merely by observing that an implementation produces imputations.
What a Joint Distribution Requires
A joint distribution does more than provide a separate conditional rule for each variable. The conditionals must agree with one another wherever their conditioning events overlap. For a finite discrete example, write \(p_{xy}=P(X=x,Y=y)\). If the relevant cell probabilities are positive, a conditional probability is a ratio of joint-cell probabilities. Thus several conditional specifications can imply different values for the same joint-cell ratio; if they conflict, no joint distribution can generate all of them.
The following calculation is a direct consequence of the definition. For two binary variables, let \(a=P(X=0,Y=0)\), \(b=P(X=1,Y=0)\), \(c=P(X=0,Y=1)\), and \(d=P(X=1,Y=1)\). If the cells involved are positive, then
The two routes from \(a\) to \(d\) must give the same ratio: \(d/a=(d/c)(c/a)=(d/b)(b/a)\). This is not a general compatibility test for every setting, but it illustrates the central requirement: all conditionals that come from one joint distribution must yield mutually consistent joint probabilities.
Worked Example: Verifying Compatibility From a Joint Table
Suppose hypothetical binary variables \(X\) and \(Y\) have the following proposed joint distribution. The entries are probabilities and sum to one.
| \(Y=0\) | \(Y=1\) | Row total | |
|---|---|---|---|
| \(X=0\) | 0.20 | 0.10 | 0.30 |
| \(X=1\) | 0.30 | 0.40 | 0.70 |
| Column total | 0.50 | 0.50 | 1.00 |
All four cells are positive, so each conditional below is defined. The conditionals for \(X\) are
The conditionals for \(Y\) are
These four conditional distributions are compatible: the displayed joint table is a witness, meaning a joint distribution that induces all four. This example verifies compatibility by construction; it does not imply that every set of conditionals can be assembled into such a table.
Worked Example: Detecting Incompatible Binary Conditionals
Consider hypothetical specifications \(P(X=1\mid Y=0)=0.20\), \(P(X=1\mid Y=1)=0.80\), \(P(Y=1\mid X=0)=0.70\), and \(P(Y=1\mid X=1)=0.60\). Suppose a joint distribution generated all four. Because all four full conditional probabilities are strictly between zero and one, the joint distribution must have full support: choose a \(y\) with \(P(Y=y)>0\); the interior \(X\mid Y=y\) conditional gives positive probability to both \(X\) values, and the interior \(Y\mid X\) conditionals then give positive probability to both \(Y\) values for each \(X\). Thus all four cells are positive. Use the cell notation \(a,b,c,d\) above.
The first conditional implies \(b/(a+b)=0.20\), hence \(b/a=0.20/0.80=0.25\). The second implies \(d/(c+d)=0.80\), hence \(d/c=0.80/0.20=4\). From the third, \(c/(a+c)=0.70\), so \(c/a=0.70/0.30=7/3\). From the fourth, \(d/(b+d)=0.60\), so \(d/b=0.60/0.40=1.5\).
Combining the first and fourth ratios gives \(d/a=(d/b)(b/a)=1.5(0.25)=0.375\). Combining the second and third gives \(d/a=(d/c)(c/a)=4(7/3)=28/3\), approximately \(9.333\). These cannot both be the same ratio. Therefore no joint distribution generates the four specifications; they are incompatible.
Support and Zero-Probability Conditioning Events
Compatibility does not generally require a joint distribution to assign positive probability to every possible combination of values. In a discrete setting, the support is the set of combinations with positive joint probability. Conditional distributions are constrained by the joint distribution on its support, while assignments at zero-probability conditioning events do not by themselves make those events have positive probability.
Worked Example: An Interior Conditional at a Null Event
Let a hypothetical joint distribution assign probability \(0.5\) to \((X=0,Y=0)\), probability \(0.5\) to \((X=1,Y=0)\), and zero to both cells with \(Y=1\). Then \(P(Y=1)=0\). The conditional distribution \(P(X\mid Y=1)\) is not determined by this joint distribution because \(Y=1\) never occurs. A version of the conditional can nevertheless be assigned, for example \(P(X=1\mid Y=1)=0.40\), without changing any joint probability.
This assignment is strictly between zero and one, but it does not imply \(P(Y=1)>0\). The other full conditional values on events with positive probability are \(P(X=1\mid Y=0)=0.50\) and \(P(Y=1\mid X=0)=P(Y=1\mid X=1)=0\). They agree with the displayed joint distribution. Thus a conditional value at a zero-probability conditioning event alone does not require that event to occur.
There is an important contrasting fact for finite binary variables. If every full conditional probability is strictly between zero and one at every conditioning value, compatibility forces full support. To see this, start with any \(y\) for which \(P(Y=y)>0\); at least one exists. The interior conditional \(P(X=x\mid Y=y)\) gives positive joint probability to both \(x\) values. Consequently each \(X=x\) has positive marginal probability. The interior conditionals \(P(Y=y\mid X=x)\) then give positive probability to both \(Y\) values for each \(x\). All four joint cells are therefore positive. This conclusion depends on all the stated interior conditions, not on an interior assignment at a single null conditioning event.
Using Compatibility in FCS
For MICE, compatibility is a structural property of the conditional update models. If they are compatible, there is a joint distribution whose conditionals match them (at the parameter values under consideration). This provides a coherent joint-distribution interpretation of those conditionals. If they are incompatible, the update rules cannot all be interpreted as conditionals of one common joint distribution. The algorithm may still define a transition process, but that fact does not supply the missing joint model.
Compatibility is also distinct from correct specification. A compatible set of conditionals can arise from a joint distribution that is a poor representation of the data-generating process. Conversely, an algorithm's observed stability is not evidence that its specified conditionals have a common joint distribution. The earlier material on stationarity and convergence addresses behavior of the chain; compatibility asks whether a particular joint distribution can induce the specified conditional laws. These are related questions, not interchangeable answers.
Practical checks depend on the model and its support. In a small finite table, one can often calculate implied cell ratios or solve for joint probabilities, as in the examples. In richer continuous or mixed-variable settings, direct checks may be difficult. Diagnostics of imputed values and chain behavior can reveal problems such as implausible values or poor mixing, but they do not generally prove compatibility. Nor does passing a compatibility check establish adequate fit, suitable missingness assumptions, or validity of a substantive analysis.
Identify the variables, the specified conditional distributions, and—if parameters are present—the fixed values or model families under consideration.
Determine which conditioning configurations have positive probability under a proposed joint distribution. Do not treat a value assigned at a null event as evidence that the event has positive probability.
In a small discrete example, compare the joint-cell probabilities implied by different conditionals. A contradiction rules out compatibility; a joint distribution that induces every conditional establishes it.
Compatibility establishes the existence of a joint distribution for the specified conditionals. It does not establish model adequacy, chain convergence, or inferential validity by itself.
Common Mistakes and Limitations
- Assuming any collection of conditional models has a joint model. Separate conditional specifications may imply conflicting joint probabilities. Check their consistency rather than presuming a common source.
- Equating compatibility with convergence. Compatibility is about the existence of a joint distribution inducing the conditionals. Convergence concerns the behavior of an algorithm over iterations; one does not establish the other.
- Inferring support from an assignment at a null event. A conditional version at a zero-probability conditioning value does not force that event to occur. In the finite binary case, full support follows only when all full conditional probabilities are interior at every conditioning value.
- Treating compatibility as proof that the models are correct. A common joint distribution can exist and still be unsuitable for the scientific problem or data.
The main alternative is to specify a joint distribution first and derive its conditional distributions, rather than specifying each conditional separately. That approach makes a common joint law explicit, but requires choosing and fitting a joint model. FCS specifies conditional models directly, which can be convenient when different variable types call for different conditional model forms; its flexibility does not guarantee those forms are compatible. Neither approach removes the need to assess model assumptions and the intended analysis.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions below.
- In one sentence, define compatibility of a collection of conditional distributions.
- For the joint table in the first worked example, calculate \(P(X=1\mid Y=0)\) and \(P(Y=1\mid X=1)\). What establishes compatibility?
- Suppose \(b/a=0.5\), \(d/b=2\), \(c/a=3\), and \(d/c=1\) are all implied by specified conditionals for a binary pair, with positive cells required. Are the conditionals compatible? Show the check.
- Does specifying \(P(X=1\mid Y=1)=0.4\) force \(P(Y=1)>0\)? Explain.
- For finite binary \(X,Y\), explain why compatibility and strictly interior full conditional probabilities at every conditioning value imply full support.
Solutions. (1) The conditionals are compatible if some joint distribution induces each one wherever its conditioning values have positive probability. (2) \(P(X=1\mid Y=0)=0.30/(0.20+0.30)=0.60\), and \(P(Y=1\mid X=1)=0.40/(0.30+0.40)=4/7\), approximately \(0.5714\). The joint table itself induces all the listed conditionals, so it is a witness to compatibility. (3) The route through \(b\) gives \(d/a=(d/b)(b/a)=2(0.5)=1\), while the route through \(c\) gives \(d/a=(d/c)(c/a)=1(3)=3\). The contradiction means the specifications are incompatible. (4) No. If \(P(Y=1)=0\), that conditional is at a null conditioning event and its assigned version does not alter the joint distribution. (5) At least one \(Y\) value has positive probability. The interior conditional of \(X\) given that value makes both \(X\) values have positive joint probability. Both \(X\) values therefore have positive marginal probability; the interior conditionals of \(Y\) given each \(X\) value then give positive probability to both \(Y\) values for each \(X\). Hence all four cells have positive probability.