Two Different Questions About Imputation
A set of conditional imputation models can be internally coherent yet still be poorly aligned with the analysis that will be performed on the completed data. These are different concerns. Compatibility asks whether the conditional models for variables with missing values could all be conditionals of one joint distribution. Congeniality concerns whether the imputation procedure and the planned analysis can support coherent inference together. Confusing these questions can make a technically well-defined chained-equations procedure seem more suitable for an analysis than it is.
The distinction matters in MICE because a user specifies a conditional model for each variable to be imputed. Those specifications may or may not correspond to a common joint model. Separately, even if they do correspond to one, the joint model may impose restrictions that do not support the analysis model or its target. Compatibility is therefore an internal property of the conditional models; congeniality is a relationship between imputation and analysis.
Compatibility: Can the Conditionals Come From One Joint Model?
For two variables \(X\) and \(Y\), compatibility requires a joint distribution \(p(x,y)\) whose factorizations agree: \(p(x,y)=p(x\mid y)p_Y(y)=p(y\mid x)p_X(x)\). Here \(p_X\) and \(p_Y\) are the marginal distributions implied by the same joint distribution. It is not enough that each conditional model looks plausible on its own. Their probabilities and parameters must fit together so that both factorizations describe one joint distribution.
Worked Example: Two Compatible Gaussian Conditionals
Suppose the proposed conditional models are \(X\mid Y=y\sim N(0.5y,0.75)\) and \(Y\mid X=x\sim N(0.5x,0.75)\). Check whether one joint distribution generates both.
Consider a bivariate normal distribution with means zero, marginal variances 1, and correlation \(0.5\). Its covariance matrix is \(\begin{pmatrix}1&0.5\\0.5&1\end{pmatrix}\), which is positive definite because its determinant is \(1-0.5^2=0.75>0\).
For a bivariate normal pair, the conditional mean of \(X\) given \(Y=y\) is \(\operatorname{Cov}(X,Y)\operatorname{Var}(Y)^{-1}y\), and its conditional variance is \(\operatorname{Var}(X)-\operatorname{Cov}(X,Y)^2/\operatorname{Var}(Y)\). Substitution gives \(0.5(1)^{-1}y=0.5y\) and \(1-0.5^2/1=0.75\). By symmetry, the conditional mean of \(Y\) given \(X=x\) is \(0.5x\), and its conditional variance is also \(0.75\).
Thus both proposed conditionals follow from the same bivariate normal joint distribution. They are compatible. This argument establishes compatibility by constructing a joint model; it does not establish that a particular data set follows that model.
A useful warning is that a familiar family of conditional models does not automatically imply compatibility. In particular, normal conditionals with linear means need not be conditionals of one bivariate normal distribution. More importantly, ruling out a bivariate normal joint distribution alone would not rule out every possible joint distribution. To establish incompatibility, the argument must exclude any joint distribution, or explicitly restrict the question to a specified family.
Worked Example: A Direct Check That Rules Out Any Joint Density
Consider \(X\mid Y=y\sim N(0.75y,1)\) and \(Y\mid X=x\sim N(0.375x,1)\), each with support over the real line. Suppose a joint distribution generated both conditionals. Because \(Y\mid X=x\) has a density, mixing these conditional densities over the marginal distribution of \(X\) shows that \(Y\) has a density \(p_Y\). Then the specified density for \(X\mid Y=y\) gives the joint density \(p(x,y)=p(x\mid y)p_Y(y)\) almost everywhere, so its two factorizations must agree.
Taking logarithms of \(p(x\mid y)p_Y(y)=p(y\mid x)p_X(x)\) and rearranging requires \(\log p(x\mid y)-\log p(y\mid x)=\log p_X(x)-\log p_Y(y)\). The right-hand side is a sum of a function of \(x\) alone and a function of \(y\) alone. Expand the left-hand side, omitting constants that do not depend on \(x\) or \(y\):
If the required log-density difference were \(a(x)+b(y)\) almost everywhere, its rectangular difference \(F(x_1,y_1)-F(x_1,y_2)-F(x_2,y_1)+F(x_2,y_2)\) would be zero for almost every quadruple, where \(F\) is that difference. For the displayed polynomial, the rectangular difference is \(0.375(x_1-x_2)(y_1-y_2)\), which is nonzero for almost every quadruple. The required factorization is impossible. These particular full-support normal conditionals are therefore incompatible with any joint density, not merely with a bivariate normal joint distribution.
Congeniality: Do Imputation and Analysis Support the Same Inference?
Compatibility does not settle whether the imputation model is appropriate for an analysis. The imputation model might define a coherent joint distribution but restrict an association that the analysis aims to estimate. In that case, the completed data can be generated coherently under the imputation model, while the analysis asks questions that the imputation procedure does not adequately represent.
Congeniality is a relationship between procedures, not a property of a conditional distribution in isolation. In particular, it concerns both the model used to generate missing values and the model or inferential procedure used on each completed data set. A shared model must support the analysis estimand, and the imputation procedure must represent uncertainty relevant to that estimand. Merely including the analysis variables in an imputation model does not by itself establish this.
Worked Example: Compatible Imputation Models, Poor Alignment With an Interaction Analysis
Let \(G\) be a fully observed binary group indicator, \(X\) a predictor with missing values, and \(Y\) a continuous outcome. Suppose an imputer specifies, within each group, the joint model \(X\mid G=g\sim N(0,1)\) and \(Y\mid X,G=g\sim N(X,1)\), with the same slope in both groups. This is a coherent joint model within each group, so the conditional models derived from it are compatible.
For example, under this joint model \(X\) and the residual in \(Y=X+\varepsilon\) are independent with variances 1. Thus \(\operatorname{Var}(Y\mid G=g)=2\) and \(\operatorname{Cov}(X,Y\mid G=g)=1\). The normal conditional formula gives \(X\mid Y=y,G=g\sim N((1/2)y,1/2)\), since the conditional mean is \((1/2)y\) and the variance is \(1-1^2/2=1/2\). Together with \(Y\mid X,G=g\sim N(X,1)\), this gives conditionals derived from one joint model in each group.
Now suppose the planned analysis estimates a separate linear slope of \(Y\) on \(X\) in each group. A hypothetical complete-data pattern has \(X=-1,1\) in group 0 and corresponding \(Y=-1,1\); in group 1, let \(X=-1,1\) and corresponding \(Y=-2,2\). The least-squares slope through the origin in group 0 is \(\sum XY/\sum X^2=[(-1)(-1)+(1)(1)]/[(-1)^2+1^2]=2/2=1\). In group 1 it is \([(-1)(-2)+(1)(2)]/[(-1)^2+1^2]=4/2=2\). The estimated slope difference is \(2-1=1\) outcome unit per predictor unit.
The imputation model, however, imposes a common slope of 1 in both groups. Its conditional distribution for missing \(X\) is derived under that restriction, not under a model allowing a group-specific slope. Thus, although the conditional imputation models are compatible with each other, that imputation procedure is not generally congenial for inference on an unrestricted group-by-\(X\) interaction. If the imputer instead allowed the group-specific association and its uncertainty, alignment could be improved. This example illustrates the distinction; it is not a general recipe for specifying interaction imputations.
A Practical Comparison
| Question | Compatibility | Congeniality |
|---|---|---|
| What is being compared? | The conditional imputation models with one another. | The imputation procedure with the planned analysis. |
| Core criterion | One joint distribution generates all specified conditionals. | A common inferential framework supports the imputation and the analysis estimand. |
| What does success establish? | Internal coherence of the conditional distributions. | Potentially coherent inference for the intended analysis, under the shared assumptions. |
| What does success not establish? | That the joint model is scientifically adequate or matches the analysis. | That the conditionals are mutually compatible, or that the assumptions fit the data. |
How to Check, and What the Checks Cannot Tell You
For compatibility, a strong theoretical check is to write down a candidate joint model and derive every conditional from it. If that construction fails, compare the factorizations directly, as in the Gaussian example. When a candidate joint density is available, verify that it is nonnegative, integrates to one, and produces the stated conditionals. For more than two variables, pairwise checks alone need not establish that all the conditionals are compatible with one joint distribution.
For congeniality, first state the analysis estimand and identify the assumptions that let the analysis estimate it. Then ask whether the imputation model represents the relevant associations and their uncertainty. For example, compare restrictions on group-specific associations in imputation with the parameters being estimated in the analysis. This is a model-based review, not a test with a universal cutoff.
Observed-data and imputed-data diagnostics can reveal conspicuous distributional mismatches, implausible values, or unstable behavior. They cannot prove that a joint distribution exists or that imputation and analysis are congenial. The same observed data can be consistent with different assumptions about missing values, and a good-looking diagnostic plot does not show that all relevant analysis parameters are represented.
Alternatives address different concerns. A joint-model approach specifies a joint distribution directly, which can make its conditional implications explicit; it still requires suitable assumptions and may be difficult to specify for mixed or complex data. Fully conditional specification permits variable-specific models, but those models may not be compatible. A carefully designed chained-equations procedure can be useful even when no simple joint model is available, but its practical adequacy must not be confused with a proof of compatibility or congeniality.
Common Mistakes and Limitations
- Equating compatibility with congeniality. Compatible conditional models can jointly impose a restriction that conflicts with the analysis target. Check internal coherence and analysis alignment separately.
- Declaring incompatibility from a normal-theory contradiction alone. That rules out a joint distribution only if the argument establishes that any possible joint distribution must be normal. A direct factorization argument can rule out more broadly, as shown above.
- Assuming that excellent diagnostics prove either property. Diagnostics can expose problems but cannot verify a common probability model or guarantee valid inference under unverifiable missing-data assumptions.
- Treating a shared model as proof of correctness. A common model can be mathematically coherent and still be substantively wrong. Compatibility and congeniality do not establish that the model is true.
These concepts have limits as practical checklists. Congeniality has formal formulations that depend on the imputation and analysis procedures, so an informal claim of “alignment” should be tied to a stated estimand and assumptions. Compatibility can be hard to establish for many variables and mixed distributions. In either case, a conclusion is only as useful as the model assumptions used to reach it.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions below.
- Two conditional distributions are both normal. Does that fact establish compatibility? Explain.
- Why does a nonzero \(xy\) term in the factorization comparison rule out the two Gaussian conditionals in the second worked example?
- An imputation model is compatible with a joint distribution but constrains two group-specific slopes to be equal. The analysis estimates the slopes separately. Which property is established, and what concern remains?
- Can plots comparing observed and imputed values prove congeniality? Give one reason for your answer.
Solutions. (1) No. Normality of each conditional does not ensure that their parameters can arise from one joint distribution; the factorization constraints must also hold. (2) The log-density equality requires the difference to equal a function of \(x\) alone minus a function of \(y\) alone. Such a representation has zero rectangular difference for almost every quadruple, whereas the displayed \(xy\) term gives a nonzero rectangular difference for almost every quadruple, so the equality cannot hold. (3) Compatibility within the imputation model is established, but congeniality for inference on unrestricted separate slopes is in doubt because the imputation restriction excludes their difference. (4) No. Plots can reveal distributional discrepancies, but they do not establish a shared inferential model or verify assumptions about missing values.