Pairwise Compatibility Versus Global Compatibility

Why Pairwise Checks Can Miss a Global Conflict

In fully conditional specification, each variable is updated from a model conditional on the other variables. It is natural to check whether two such conditional models can come from a joint distribution. But passing every two-model check does not guarantee that all the conditional models come from one common joint distribution. The distinction is between local agreement and global coherence: separate pairs can each fit together while making demands that cannot all hold at once.

We will make that distinction precise using three binary variables. The example is deliberately simple, so the conflict can be demonstrated without relying on a complicated joint-density calculation. The same logical issue can arise in richer conditional models: compatibility must be assessed for the full collection, not inferred from separate pairwise checks.

The qualification “at each fixed value” matters. A conditional distribution for \(X\) given \(Y,Z\) and one for \(Y\) given \(X,Z\) must be compatible within each stratum of \(Z\) for the pairwise check to pass. The same requirement applies to the other two pairs. Even then, the pairwise distributions constructed for the three checks may impose mutually inconsistent restrictions on the full triple.

A Pair of Conditional Models That Fits Together

Let \(X,Y,Z\) each take values 0 or 1. Consider a conditional model in which \(X\) differs from \(Y\) with probability 0.9, regardless of \(Z\):

For the pairwise check involving \(X\) and \(Y\), specify the corresponding model for \(Y\) given \(X,Z\): \(Y\) differs from \(X\) with probability 0.9, regardless of \(Z\). At either fixed value of \(Z\), both conditionals come from the following bivariate distribution.

Each row of this table has nonnegative probability, and the entries sum to 1. Both marginal probabilities are 0.5. Given \(Y=0\), for example, the conditional probability that \(X=1\) is \(0.45/(0.05+0.45)=0.9\); given \(Y=1\), it is \(0.05/(0.45+0.05)=0.1\). By symmetry, the conditional model for \(Y\mid X,Z=z\) has the same form. Thus this pair of conditional models is compatible for either value of \(z\).

A Three-Variable System With No Common Joint Distribution

Now impose the analogous conditional models for all three pairs. For each variable, specify that it differs from each of the other variables with probability 0.9, regardless of the value of the third variable. For example, in addition to the \(X\mid Y,Z\) model already given, require

and impose the same “different with probability 0.9” rule for \(X\) versus \(Z\), \(Y\) versus \(Z\), and the reverse conditionals. Every specified probability is strictly between 0 and 1. For the \(X,Y\) pair at each fixed \(z\), the table above verifies compatibility. For the \(X,Z\) pair at each fixed \(y\), use the same table with \(Z\) in place of \(Y\). For the \(Y,Z\) pair at each fixed \(x\), use it with \(Y,Z\) as the two variables. Each pairwise check succeeds.

Nevertheless, there cannot be a joint distribution for \(X,Y,Z\) having all these conditionals. To see why, define an indicator for each pair that equals 1 when the pair differs and 0 otherwise:

The conditional models imply that each pair differs with marginal probability 0.9. For example, summing over \(Z\) gives

This calculation does not require \(Z\) to have probabilities 0.5 and 0.5; it only requires that the specified conditional model applies at every value of \(Z\) with positive probability. Applying the same argument to the other pairs gives \(E(D_{XY}+D_{XZ}+D_{YZ})=2.7\).

But for any fixed values of three binary variables, the number of unequal pairs is either 0 or 2. If all three values agree, there are no unequal pairs. Otherwise, one value differs from the other two, giving exactly two unequal pairs. In particular, the sum of the three indicators can never exceed 2. Taking expectations therefore gives a contradiction:

Thus no joint distribution can generate all the specified conditional models. The pairwise constructions are individually valid, but their combined requirements exceed what any triple of binary values can satisfy.

What a Successful Pairwise Check Does—and Does Not—Show

Pairwise checks are useful. If a pair of conditional models cannot come from a common bivariate conditional distribution, then the entire collection cannot come from one joint distribution either. A failed pairwise check is sufficient evidence of global incompatibility. The converse is false: passing all pairwise checks is only a necessary condition, not a sufficient one.

A different way to establish global compatibility is to exhibit a single joint distribution and derive every requested conditional from it. For example, let \(X,Y,Z\) be mutually independent Bernoulli variables with probabilities \(P(X=1)=0.3\), \(P(Y=1)=0.5\), and \(P(Z=1)=0.7\). The joint distribution is the product of these three Bernoulli distributions. Its conditionals satisfy

for all conditioning values. The same joint distribution produces all three models, so they are globally compatible. This example also passes every pairwise check, but it does so for a stronger reason: an explicit common joint distribution has been supplied.

Practical Checks and Common Pitfalls

A systematic assessment separates local checks from a global check. First, check whether each pair of conditional models is compatible on the relevant conditioning strata. Then ask whether the pairwise distributions can all coexist in a single joint distribution. For finite variables, this global question can sometimes be framed as finding nonnegative joint cell probabilities that sum to 1 and reproduce every specified conditional probability. The binary example instead uses an inequality that every joint distribution must satisfy. In more complex settings, finding a joint distribution or deriving necessary constraints may require algebraic or numerical methods; pairwise tables alone do not settle the question.

When conditional probabilities are defined only on events with positive probability, positivity must be handled explicitly. A conditional such as \(P(X\mid Y=y,Z=z)\) is not determined by an ordinary ratio if \(P(Y=y,Z=z)=0\). Pairwise tables that assign zero probability to conditioning strata can therefore hide undefined or unconstrained conditionals. In the example, all pairwise table entries are positive, and the stipulated conditional probabilities are defined at every binary conditioning value. This avoids a zero-probability loophole; the failure is genuinely global.

• Pitfall: treating pairwise success as a proof. “Every pair has a bivariate distribution, so the full system must have a joint distribution” is invalid. The 0.9 example passes each pairwise check but implies an impossible expected count of unequal pairs.
• Correction: state what each check establishes. Pairwise incompatibility rules out global compatibility, while pairwise compatibility does not rule out a higher-order conflict.
• Pitfall: assuming conditional models can be averaged into a joint. Each model may be internally valid, yet the constraints they impose together may contradict a property that every possible joint outcome must obey.
• Correction: seek a common joint or a global constraint. A constructed joint distribution verifies compatibility directly. To disprove compatibility, derive a necessary implication of the conditionals and show it violates a bound that holds for every joint distribution.

This distinction matters when interpreting a fully conditional specification. If a common joint distribution exists, the conditional models can be understood as different views of that distribution. If it does not, the fact that each pair looks coherent cannot justify that interpretation. Pairwise checking remains a useful diagnostic, but it should be described as a partial check rather than a certificate of global coherence.