Two Different Questions About Conditional Models
In multiple imputation by chained equations (MICE), a model is specified for each variable with missing values, conditional on the other variables. Two questions about those models are easy to confuse. Compatibility asks whether the specified conditional distributions can all be conditional distributions of one joint distribution. Correct conditional specification asks whether each specified conditional distribution matches the corresponding conditional distribution in the data-generating population.
These are not the same standard. A collection of conditionals may fit together perfectly while describing the wrong population. Conversely, if every specified conditional is the true conditional under one population distribution, then those conditionals necessarily fit together: that population distribution is a common joint distribution. This distinction matters because internal coherence of a set of imputation models is not, by itself, evidence that the models represent the data well.
The word “correct” here is a population statement about the model distributions, not a claim that estimated coefficients equal their population values in a finite sample. A fitted model can differ from its target because of sampling variation even when its model family is appropriate. The distinction in this tutorial concerns whether the conditional distributions being represented are the right ones, and whether they can arise from one joint distribution.
For continuous variables, conditional densities are understood up to the usual almost-everywhere qualifications. The examples below use distributions with well-defined conditionals and common support, so the compatibility comparisons can be made directly.
Why Correct Conditionals Must Be Compatible
Suppose a population has a joint distribution \(P\). Its conditionals are defined from that same distribution. For two variables, for example, \(P(X\mid Y)\) and \(P(Y\mid X)\) are both obtained from the joint distribution \(P(X,Y)\). Thus, if two proposed models equal these true conditionals, \(P(X,Y)\) itself witnesses their compatibility. The same reasoning applies to any number of variables: the data-generating joint distribution is a common distribution for all the correctly specified conditionals.
This implication does not reverse. Compatibility only says that some joint distribution has the proposed conditionals. It does not say that this distribution is the data-generating distribution. The joint distribution that makes the models compatible may differ from the truth in its marginal distributions, dependence structure, or both.
Worked Example: Compatible Gaussian Conditionals That Miss the Truth
Suppose, hypothetically, that the true population distribution is bivariate normal with means zero, variances one, and correlation \(0.5\). A proposed imputation system instead uses the two conditionals implied by a bivariate normal distribution with means zero, variances one, and correlation \(0.8\). Determine whether the proposed models are compatible and whether they are correct.
For a standardized bivariate normal pair with correlation \(\rho\), the conditional distribution is \(Y\mid X=x\sim N(\rho x,1-\rho^2)\), and symmetrically \(X\mid Y=y\sim N(\rho y,1-\rho^2)\). These formulas apply here because both the true and proposed joint distributions are bivariate normal with unit marginal variances.
Under the proposed models, \(\rho=0.8\), so \(Y\mid X=x\sim N(0.8x,1-0.8^2)=N(0.8x,0.36)\); the model for \(X\mid Y=y\) has the same form. Both conditionals come from the single proposed bivariate normal joint distribution with correlation \(0.8\). They are therefore compatible.
But the true conditional uses \(\rho=0.5\): \(Y\mid X=x\sim N(0.5x,1-0.5^2)=N(0.5x,0.75)\). At \(x=1\), the proposed conditional mean is \(0.8\), while the true conditional mean is \(0.5\); the proposed conditional variance is \(0.36\), while the true conditional variance is \(0.75\). The models are compatible but not correctly specified. They describe a stronger association and a narrower conditional distribution than the hypothetical truth.
A Compatible Model Can Be Wrong in a Simpler Way
The distinction does not depend on continuous variables or on complicated regression forms. A model of conditional independence can be compatible by construction and still omit an association present in the population. The next example makes the mismatch visible directly from a small probability table.
Worked Example: Independent Binary Conditionals Against an Associated Truth
Suppose the true joint probabilities for two binary variables \(X\) and \(Y\) are as follows. The table gives \(P(X=x,Y=y)\) in a hypothetical population.
| \(Y=0\) | \(Y=1\) | Total | |
|---|---|---|---|
| \(X=0\) | 0.40 | 0.10 | 0.50 |
| \(X=1\) | 0.10 | 0.40 | 0.50 |
| Total | 0.50 | 0.50 | 1.00 |
A proposed model uses \(P(Y=1\mid X=x)=0.5\) for either value of \(x\), and \(P(X=1\mid Y=y)=0.5\) for either value of \(y\). These conditionals are compatible: they are exactly the conditionals of the joint model in which \(X\) and \(Y\) are independent Bernoulli variables, each with probability \(0.5\) of being one.
Now calculate the true conditionals from the table. For \(X=0\), \(P(Y=1\mid X=0)=0.10/0.50=0.20\); for \(X=1\), \(P(Y=1\mid X=1)=0.40/0.50=0.80\). Likewise, \(P(X=1\mid Y=0)=0.10/0.50=0.20\), and \(P(X=1\mid Y=1)=0.40/0.50=0.80\).
The true conditionals show a substantial association, while the proposed conditionals assert independence. For example, when \(X=1\), the proposed probability that \(Y=1\) is \(0.50\), compared with the true probability \(0.80\). The proposed models are mutually compatible, but neither conditional represents the true conditional distribution.
Correct Specification Provides a Common Joint Distribution
The converse direction of the distinction is useful: if all the conditional models are correct for the same data-generating population, incompatibility is not possible. For two discrete variables this follows directly from the definition of a joint probability. If \(q(Y\mid X)=P(Y\mid X)\), then combining it with the true marginal distribution of \(X\) reconstructs the true joint distribution. The other true conditional is then also a conditional of that same joint distribution.
This identity is not a procedure for discovering an unknown joint distribution from arbitrary fitted conditionals. It is a justification of the implication: when the conditionals are all genuinely correct under one population law, that law is already the common joint distribution. With more than two variables, the same principle holds because each true conditional is obtained from the one full joint distribution \(P(X_1,\ldots,X_p)\).
Worked Example: Correct Binary Conditionals Reconstruct One Joint Distribution
Consider a hypothetical population with the following true joint probabilities: \(P(0,0)=0.30\), \(P(0,1)=0.20\), \(P(1,0)=0.10\), and \(P(1,1)=0.40\), where each ordered pair is \((X,Y)\). Verify that its two true conditional distributions are compatible.
First, \(P(X=0)=0.30+0.20=0.50\) and \(P(X=1)=0.10+0.40=0.50\). Therefore, \(P(Y=1\mid X=0)=0.20/0.50=0.40\), and \(P(Y=1\mid X=1)=0.40/0.50=0.80\).
The marginal probabilities for \(Y\) are \(P(Y=0)=0.30+0.10=0.40\) and \(P(Y=1)=0.20+0.40=0.60\). Thus, \(P(X=1\mid Y=0)=0.10/0.40=0.25\), and \(P(X=1\mid Y=1)=0.40/0.60=2/3\).
The proposed conditional models are correctly specified by assumption: they use exactly these four conditional probabilities. They are compatible because the original joint table produces both. For example, \(P(X=1,Y=1)=P(X=1)P(Y=1\mid X=1)=0.50(0.80)=0.40\); also \(P(X=1,Y=0)=0.50(0.20)=0.10\). The corresponding cells for \(X=0\) are \(0.50(0.40)=0.20\) and \(0.50(0.60)=0.30\). The reconstructed table is precisely the given joint table.
This example illustrates the logical implication, not a way to verify correctness from incomplete data alone. In practice, the population conditionals are unknown.
What to Check in Practice
Compatibility and conditional fit call for different questions and different evidence. Compatibility is a structural property: is there a joint distribution with all the specified conditionals? Correct specification is a modeling claim: do the fitted conditional forms approximate the actual conditional relationships in the population? Some conditional model families impose restrictions that can be investigated algebraically. Data-based checks, meanwhile, can reveal discrepancies between a conditional model and the observed information relevant to that model.
In a MICE analysis, useful checks for conditional specification include examining residual patterns for continuous outcomes, comparing observed and fitted outcome distributions, checking whether nonlinearities or interactions are needed, and assessing predictions in observed cases. For categorical outcomes, check whether fitted probabilities track observed category frequencies across relevant predictor patterns. These checks should reflect the model's outcome type and structure; a single overall fit statistic can conceal poor fit in important subgroups.
Such diagnostics have limits. When values are missing, the true full-data conditional relationships cannot generally be inspected directly for records with missing outcomes. Diagnostics based on observed cases also rely on which cases are observed and on the assumptions used to interpret them. A good fit to the observed data is not proof that a model is correct throughout the population. Nor does checking each conditional's predictive fit automatically establish that all the fitted conditionals share a joint distribution.
| Question | What it concerns | What it does not establish |
|---|---|---|
| Are the conditionals compatible? | Whether one joint distribution has all the proposed conditionals. | That this joint distribution is the true population distribution. |
| Is each conditional correctly specified? | Whether each proposed conditional matches its data-generating conditional. | That estimates from a finite sample equal the population quantities exactly. |
| Do observed-data diagnostics look adequate? | Whether particular discrepancies are visible in the information available for checking. | Either universal correctness or compatibility by itself. |
A joint-modeling approach starts by specifying one joint distribution and obtains its conditionals from that model; this guarantees compatibility within the assumed joint family. Its limitation is that the chosen joint distribution may be too restrictive or otherwise wrong. FCS specifies conditional models directly, which can make it easier to choose a model suited to each variable's type, but those conditionals still need scrutiny for compatibility and for fit to the population. Neither strategy removes the need to justify the modeling assumptions.
Common Mistake and Practical Interpretation
- Wrong reasoning: “The conditional models are compatible, so the imputations represent the true population relationships.” Compatibility only shows that the conditionals can arise from some common joint distribution. The Gaussian example showed compatible models that implied a different correlation and different conditional variances from the truth.
- Correction: Treat compatibility and conditional accuracy as separate checks. Ask which joint distribution the models could describe, and separately assess whether the conditional forms and fitted relationships are credible for the data-generating population.
- Another mistake: “If each conditional regression looks reasonable on its own, the set must be compatible.” A plausible individual model does not by itself show that there is one joint distribution generating the whole collection. Individual fit diagnostics and a compatibility argument answer different questions.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions below.
- In one sentence, distinguish compatibility from correct conditional specification.
- A proposed pair of binary conditionals both assign probability \(0.5\) to either outcome regardless of the other variable. What joint distribution makes them compatible?
- Suppose \(X,Y\) are standard bivariate normal with true correlation \(0.3\), but proposed conditionals are those from a standard bivariate normal with correlation \(0.6\). Are the proposed conditionals compatible? Are they correct?
- If every specified conditional equals the corresponding true conditional under one population joint distribution, what can be concluded about compatibility, and why?
- Why does a good observed-data fit for each conditional not prove that the entire FCS system is correct?
Solutions. (1) Compatibility asks whether some joint distribution generates all proposed conditionals; correctness asks whether each equals its population conditional. (2) The independent joint distribution with both variables Bernoulli with probability \(0.5\) of being one generates these conditionals. (3) They are compatible because both come from the bivariate normal joint distribution with correlation \(0.6\), but they are incorrect because the true correlation is \(0.3\); for example, the true conditional mean of \(Y\) at \(X=1\) is \(0.3\), not \(0.6\). (4) They are compatible: the data-generating joint distribution itself generates all the true conditionals. (5) The observed data may not reveal the full-data conditionals, and separate fit checks do not establish that one common joint distribution produces every conditional.