Conflicting Neighboring Constraints and Coherence

When Neighboring-Case Assignments Collide

A neighboring-case restriction uses outcomes from a related observation pattern to specify an unobserved outcome in a target pattern. Such restrictions can be useful, but imposing more than one assignment on the same missing outcome raises a consistency question: could all the assignments describe one distribution for the target pattern? If not, the restrictions are incoherent as a joint set. The problem is not that one calculation must be wrong; it is that the assignments cannot all hold at once.

The important comparison is at the same target outcome and on a common conditioning scale. A marginal mean from one source cannot be compared directly with a mean conditional on a more detailed history from another source. Nor does agreement of two means establish agreement of two distributions.

Definition: Let \(R=r\) denote a target observation pattern, \(Z\) an outcome missing in that pattern, and \(H\) a history available for conditioning. Suppose two neighboring-case assignments specify conditional distributions \(Q_1(z\mid h)\) and \(Q_2(z\mid h)\) for \(Z\) in pattern \(r\). They are distributionally coherent on the target support if \(Q_1(\cdot\mid h)=Q_2(\cdot\mid h)\) for almost every history \(h\) occurring in pattern \(r\). If the assignments specify only conditional means, they are mean-coherent if those means agree for almost every such \(h\). Mean coherence is weaker than distributional coherence.

The phrase “on the target support” matters. A restriction is relevant to histories that occur in the target pattern. If a source pattern contains no observations for some target history, its conditional distribution there is not directly supplied by that source’s observed data; an assignment in that region requires an additional extrapolation. Also, restrictions are assumptions about unobserved outcomes, not facts that can generally be verified from observed data alone.

A Basic Compatibility Check

Write \(\mu_j(h)=E_j(Z\mid H=h)\) for the conditional mean assigned by source \(j\). If both restrictions describe one target distribution, the conditional means derived from that distribution must be equal wherever both restrictions apply. If both restrictions specify full conditional distributions, those distributions must agree there as well.

This follows directly from the definition of expectation: the mean of a given probability distribution is unique. Thus a difference in assigned means proves that two distributional assignments cannot both describe the same target conditional law. Conversely, equal means do not prove the laws are equal.

Q_1(\cdot\mid h)=Q_2(\cdot\mid h) \quad\Longrightarrow\quad \int z\,dQ_1(z\mid h)=\int z\,dQ_2(z\mid h)

The converse fails. For example, the distributions that place probability one-half at each of \(0\) and \(2\), and probability one-half at each of \(-1\) and \(3\), both have mean \(1\). They are not the same distribution: their possible values and probabilities differ. This distinction is central when a restriction specifies a mean rather than a full law.

Name the target.
Identify the pattern, the missing outcome, and the histories on which each neighboring-case assignment is meant to apply.

Put constraints on the same scale.
Express each assignment as a conditional distribution or conditional mean given a common history, or explicitly marginalize over the relevant history distribution.

Check equality on the target support.
Different assigned means establish a mean-level conflict. Different assigned distributions establish a distribution-level conflict even if their means agree.

Locate the mismatch.
Check whether it comes from genuinely different donor assignments, different conditioning histories, different weights, or unsupported extrapolation.

Worked Mean Checks

Worked Example: Two Donor Assignments Give Different Target Means

Consider a hypothetical study with three visits. In target pattern \(T\), participants have a recorded first-visit outcome \(H\), but the second-visit outcome \(Z\) is missing. Suppose two neighboring-case restrictions are proposed for \(Z\), each conditional on the same history \(H\). In target pattern \(T\), 40% of participants have \(H=0\) and 60% have \(H=1\). Source A assigns conditional means 5 and 7 for these two histories; source B assigns 6 and 8. Outcomes are measured in points.

The restrictions are comparable because they target the same missing outcome and condition on the same history. Under source A, the implied target mean is

E_T(Z)=0.40(5)+0.60(7)=2.00+4.20=6.20\text{ points}.

Under source B, it is

E_T(Z)=0.40(6)+0.60(8)=2.40+4.80=7.20\text{ points}.

The same target distribution cannot have both means 6.20 and 7.20 points. The conditional means also disagree within each history: source A assigns 5 rather than 6 points at \(H=0\), and 7 rather than 8 points at \(H=1\). Therefore the assignments are incompatible at the conditional-mean level, not merely because of a choice of marginal weights. This does not tell us which restriction is preferable; choosing one requires a substantive modeling decision.

Worked Example: Equal Means Do Not Resolve a Distribution Conflict

In another hypothetical setting, suppose two neighboring-case assignments specify the distribution of a missing outcome \(Z\), measured in points, at the same target history. Assignment A puts probability \(0.50\) at 0 and probability \(0.50\) at 2. Assignment B puts probability \(0.50\) at \(-1\) and probability \(0.50\) at 3.

Both means are

E_A(Z)=0.50(0)+0.50(2)=1\text{ point}, \qquad E_B(Z)=0.50(-1)+0.50(3)=1\text{ point}.

So the restrictions are compatible if the only imposed requirement is that the target mean equal 1 point. But if each assignment is intended to specify the full target distribution, they conflict: under A, \(Z\) can only be 0 or 2, whereas under B it can only be \(-1\) or 3. A single conditional distribution cannot have both sets of probabilities. The conclusion depends on what the restrictions claim: mean equality alone, or equality of the entire distribution.

Conditioning and Weights Can Create Apparent Conflicts

A comparison is valid only after both constraints refer to a common target quantity. Suppose one restriction gives a mean conditional on a coarse history \(H\), while another gives means conditional on a more detailed history \((H,K)\). To obtain a target mean from the more detailed restriction, average its conditional means using the distribution of \(K\) in the target pattern, conditional on \(H\). Using donor-pattern proportions instead can answer a different question.

Worked Example: Reweighting a Conditional Assignment

Suppose in a hypothetical target pattern the outcome \(Z\) is missing, \(H\) is fixed at a particular history, and a recorded covariate \(K\) takes values 0 and 1. In the target pattern, \(P_T(K=0\mid H)=0.25\) and \(P_T(K=1\mid H)=0.75\). One restriction assigns the target mean \(E_T(Z\mid H)=10\) points. A second assigns \(E_T(Z\mid H,K=0)=8\) and \(E_T(Z\mid H,K=1)=12\) points.

The second restriction implies the target mean after averaging over the target pattern’s \(K\) distribution:

E_T(Z\mid H) =0.25(8)+0.75(12) =2+9 =11\text{ points}.

This is one point higher than the first restriction’s 10-point assignment. If both claims are intended to determine \(E_T(Z\mid H)\), they are genuinely mean-incoherent under the stated target weights. Had the weights been \(0.50\) and \(0.50\), the second assignment would instead imply \(0.50(8)+0.50(12)=10\) points, agreeing at this marginal mean.

The calculation depends on using the target weights: the outcome mean sought is for the target pattern. Averaging the conditional means using another pattern’s \(K\) proportions would not check compatibility for this target mean. Also, agreement after averaging would establish agreement only for this marginal mean; it would not erase disagreement in the conditional means at \(K=0\) and \(K=1\) if both restrictions also claim those conditional means.

Derivation: Why Reweighting Must Use the Target Pattern

For discrete \(K\), partition the target pattern according to its possible \(K\) values. The conditional mean of \(Z\) at a fixed \(H=h\) is the weighted average of the conditional means within those groups, with weights given by the target conditional probabilities. This is the law of total expectation applied within pattern \(T\):

E_T(Z\mid H=h) =\sum_k E_T(Z\mid H=h,K=k)\,P_T(K=k\mid H=h).

The weights sum to one. Replacing \(P_T(K=k\mid H=h)\) with probabilities from a donor pattern generally changes the average unless the relevant distributions of \(K\) agree or the conditional means happen to make the change immaterial. Thus two reported marginal means can appear inconsistent simply because they were calculated for different populations. First standardize both assignments to the target pattern; only then decide whether they conflict.

Diagnostics, Alternatives, and Limits

A practical diagnostic is to write each restriction as a separate statement before combining them: which target pattern, which outcome, which conditioning variables, whether it specifies a distribution or only a mean, and which pattern supplies the borrowed information. Then check common support and target weights. A table can expose a conflict by placing the assigned conditional means side by side for every target history. For distributional restrictions, compare the probability laws, not only their means.

Common warning signs include two different numerical values for the same target conditional mean; statements that silently change from conditional to marginal quantities; donor-pattern weights used for a target-pattern average; and assignments for histories absent from a donor pattern. The last case is not automatically a contradiction, but the assignment there is not identified by that donor’s observed outcomes alone.

When assignments conflict, the main alternatives are to retain one restriction, revise one or more assignments, or formulate a different restriction that explicitly combines the sources. These choices change the assumptions and can change the resulting target distribution. Averaging two incompatible assignments is not a neutral repair: it creates a new assignment that must itself be justified. The introductory material on pattern-mixture models provides the setting for these restrictions; the compatibility check here concerns whether several of them can jointly describe one target law.

A key limitation is that observed data typically cannot decide which of two conflicting assumptions about a missing outcome is correct. The calculations identify logical incompatibility given the assumptions and supplied weights; they do not establish the truth of either assignment. Likewise, a lack of detected conflict in a limited set of means does not establish distributional coherence.

Common mistake: “The two neighboring patterns have the same mean, so their distributional restrictions are coherent.” This is incorrect: distinct distributions can share a mean. The correction is to match the level of the claim. Equal means support mean coherence only; full distributional coherence requires equality of the assigned laws on the target support.

Check Your Understanding

Answer each question, then compare with the solutions below.

Two assignments for the same target history give conditional means 4 and 4.5 points. Can they both describe one target conditional distribution? What does this establish?
Two assigned distributions have the same mean but different probabilities at possible outcome values. Are they distributionally coherent?
A restriction gives means 2 and 6 for \(K=0\) and \(K=1\). In the target pattern, the probabilities of these values are 0.75 and 0.25. What target mean does this restriction imply?
Why is it generally wrong to use a donor pattern’s \(K\) proportions to marginalize a restriction for the target pattern?

Solutions. (1) No. A single conditional distribution has a unique mean, so the assignments conflict at the mean level on that history. (2) No. Different probability laws are distributionally incompatible even when their means agree. (3) The implied mean is \(0.75(2)+0.25(6)=1.50+1.50=3\) points. (4) The target mean averages over the target pattern’s distribution of \(K\); donor weights generally describe a different population and can produce a different marginal quantity.