Why Conditional Models Need to Agree
In fully conditional specification, each variable can be assigned a model conditional on the other variables. For two continuous variables, a common choice is a linear regression with a Gaussian residual distribution in each direction. But two individually plausible models need not be the conditional distributions of the same joint distribution. Their parameters must agree in a precise way.
Here we focus on a pair of variables \(X\) and \(Y\). The earlier material on deriving Gaussian conditionals from a bivariate normal shows how a joint normal model produces linear conditional means. The question here is the reverse one: given both conditional models, which parameter values allow a proper joint bivariate normal distribution?
Assume the conditional models have constant, positive variances and take the form
The first slope \(b_{XY}\) describes the change in the conditional mean of \(X\) for a one-unit increase in \(Y\); the second slope \(b_{YX}\) describes the change in the conditional mean of \(Y\) for a one-unit increase in \(X\). The intercepts are \(a_X\) and \(a_Y\), and the conditional variances are \(\sigma^2_{X\mid Y}\) and \(\sigma^2_{Y\mid X}\). These variances are assumed strictly greater than zero.
The Compatibility Restrictions
For compatibility, the parameters must satisfy two conditions. First, the slopes and variances must obey a cross-equation equality. Second, the resulting joint distribution must be proper, which imposes a strict bound on the product of the slopes.
The first equality can also be written as \(b_{XY}\sigma^2_{Y\mid X}=b_{YX}\sigma^2_{X\mid Y}\). It is the slopes divided by their respective conditional variances—not the slopes alone—that must match. The second condition rules out an implied joint distribution with a zero or negative variance in some direction. Given these conditions, there is no additional restriction on the two intercepts.
Deriving the Conditions
Suppose first that a positive joint density exists. The logarithm of that density can be written as the logarithm of \(f(x\mid y)\) plus a term depending only on \(y\). Expanding the part of the conditional log-density that involves both \(x\) and \(y\), the coefficient of \(xy\) is \(b_{XY}/\sigma^2_{X\mid Y}\). Writing the same joint log-density using \(f(y\mid x)\) gives the coefficient \(b_{YX}/\sigma^2_{Y\mid X}\). These coefficients must be equal, giving the first restriction. This argument is valid because the Gaussian densities are positive and their log-densities are smooth.
The joint precision matrix—the inverse of the joint covariance matrix—implied by the conditional quadratic terms is
The off-diagonal entries agree with the second conditional model precisely when the slope-to-variance equality holds. A proper bivariate normal requires \(Q\) to be positive definite: its first diagonal entry must be positive, and its determinant must be positive. The first requirement follows from \(\sigma^2_{X\mid Y}>0\). Using the slope equality, the determinant is
The denominator is positive, so \(\det(Q)>0\) exactly when \(b_{XY}b_{YX}<1\). This proves why equality of the slope-to-variance ratios is not sufficient on its own: the resulting precision matrix must also be positive definite.
The intercepts determine the joint mean. Writing the conditional means as equations for that mean gives
When \(b_{XY}b_{YX}<1\), the matrix is invertible. Thus every pair of finite intercepts determines a finite joint mean; no further intercept constraint is needed. The joint covariance is \(Q^{-1}\), which is finite and positive definite under the same strict product condition.
Worked Example: Verifying Compatible Conditionals
Suppose the hypothetical conditional models are \(X\mid Y=y\sim N(2+0.6y,4)\) and \(Y\mid X=x\sim N(-1+1.35x,9)\). Check compatibility and, if compatible, find the joint mean.
Check the slope-to-variance equality. The two ratios are \(0.6/4=0.15\) and \(1.35/9=0.15\). They agree.
Check propriety. The slope product is \(0.6(1.35)=0.81<1\). Both conditional variances are positive, so the implied precision matrix is positive definite and the models are compatible with a proper bivariate normal.
Find the joint mean. Solve \(\mu_X-0.6\mu_Y=2\) and \(\mu_Y-1.35\mu_X=-1\). Substitution gives \((1-0.81)\mu_X=2+0.6(-1)=1.4\), so \(\mu_X=1.4/0.19\approx7.368\). Then \(\mu_Y=-1+1.35(7.368)\approx8.947\). These are the joint means in the units of \(X\) and \(Y\), respectively.
As a further check, the implied precision matrix is \(\begin{pmatrix}0.25&-0.15\\-0.15&0.1111\end{pmatrix}\); its determinant is approximately \(0.005278>0\). The positive determinant agrees with the product check.
Examples of Failed Compatibility
Worked Example: Slopes That Do Not Match Their Variances
Consider the hypothetical models \(X\mid Y=y\sim N(1+0.5y,4)\) and \(Y\mid X=x\sim N(3+0.5x,9)\). Both variances are positive. Do the models define conditionals of one bivariate normal distribution?
The required ratios are \(0.5/4=0.125\) and \(0.5/9\approx0.05556\). They are unequal, so the conditional models are incompatible. Although the product of the slopes is \(0.25<1\), that fact only addresses the positive-definiteness restriction; it cannot repair the failed cross-equation equality. There is no single bivariate normal distribution having both of these exact conditionals.
Worked Example: Matching Ratios but an Improper Joint Model
Now consider \(X\mid Y=y\sim N(0+0.8y,1)\) and \(Y\mid X=x\sim N(2+3.2x,4)\).
The ratios match: \(0.8/1=3.2/4=0.8\). But the product of the slopes is \(0.8(3.2)=2.56\), which exceeds 1. The determinant of the implied precision matrix is \((1-2.56)/(1\cdot4)=-0.39\), which is negative. The matrix is not positive definite, so these conditionals cannot arise from a proper bivariate normal distribution. Matching the cross-equation ratios is necessary but, by itself, not sufficient.
Interpreting the Restrictions and Checking Models
When the conditionals are compatible, their slope product has an additional interpretation. For a bivariate normal distribution with nonzero marginal variances, the two conditional slopes are \(b_{XY}=\operatorname{Cov}(X,Y)/\operatorname{Var}(Y)\) and \(b_{YX}=\operatorname{Cov}(X,Y)/\operatorname{Var}(X)\). Their product is therefore the squared correlation, so it lies between 0 and 1. This explains why compatible nonzero slopes have the same sign and why their product must be strictly less than 1 for a nondegenerate joint distribution.
This identity is a consequence of compatibility; it should not be used as a replacement for checking both restrictions. Slopes with product less than 1 can still fail the slope-to-variance equality, as the second worked example showed. Conversely, matching ratios do not prevent the product from exceeding 1, as the third example showed.
For a pair of fitted conditional models, a practical check is to record both conditional variances and both slopes, calculate the two ratios, and then calculate the slope product. Interpret these checks in light of how the models were specified and estimated. For estimated parameters, small numerical discrepancies may reflect rounding or estimation uncertainty; a substantial mismatch indicates that the fitted conditionals do not exactly represent a common bivariate normal model. The algebraic restrictions describe exact compatibility, not a statistical test for compatibility.
Each conditional must have a linear mean in the other variable, a Gaussian distribution, and a constant positive conditional variance.
Calculate \(b_{XY}/\sigma^2_{X\mid Y}\) and \(b_{YX}/\sigma^2_{Y\mid X}\). A difference violates the necessary cross-equation restriction.
Verify \(b_{XY}b_{YX}<1\). If it is not strictly less than 1, the implied joint precision matrix is not positive definite.
Passing both checks establishes compatibility for this pair under the specified Gaussian model. It does not show that the model describes the data well.
Common Pitfalls and Limitations
- Checking only whether the slopes are similar. The condition compares each slope with its own conditional variance. Equal slopes need not satisfy the required equality when the variances differ.
- Checking only the product. A product below 1 does not establish compatibility unless the slope-to-variance ratios also match.
- Treating equality at the boundary as acceptable. If \(b_{XY}b_{YX}=1\), the implied precision determinant is zero. The resulting covariance is singular, so there is no proper, nondegenerate bivariate normal distribution.
- Assuming compatibility means the model is correct. Compatibility concerns whether the two conditional distributions can belong to one joint distribution. It does not establish that either conditional form, its constant variance assumption, or its Gaussian errors are appropriate for a particular data set.
- Applying the test outside its assumptions. Nonlinear conditional means, non-Gaussian distributions, or variances that depend on the conditioning variable require different compatibility arguments. The restrictions derived here are specific to the stated linear Gaussian pair.
In MICE, these restrictions clarify what it means for two Gaussian update models to be consistent with a shared joint model. They also explain why specifying each regression separately does not automatically guarantee a common joint distribution. The comparison with an approach that starts from a joint distribution is direct: a joint model determines its conditionals, while separately specified conditionals must pass the compatibility checks to correspond to that same kind of joint model. Compatibility is a structural property, not a convergence diagnostic for an imputation chain.
Check Your Understanding
Use the two compatibility conditions, and state which condition is satisfied or violated in each case.
- For \(X\mid Y\sim N(0+0.4Y,2)\) and \(Y\mid X\sim N(1+0.8X,4)\), are the conditionals compatible with a proper bivariate normal? Show both checks.
- For slopes \(b_{XY}=0.5\), \(b_{YX}=0.5\), and conditional variances \(\sigma^2_{X\mid Y}=2\), \(\sigma^2_{Y\mid X}=2\), which restriction fails, if any?
- Can two conditional models with matching slope-to-variance ratios and slope product equal to 1 define a proper bivariate normal? Explain.
- Why is the condition \(b_{XY}b_{YX}<1\) not enough by itself to establish compatibility?
Solutions. (1) The ratios are \(0.4/2=0.2\) and \(0.8/4=0.2\), and the product is \(0.4(0.8)=0.32<1\); both conditions hold, so the models are compatible. (2) The ratios are both \(0.5/2=0.25\), and the product is \(0.25<1\); neither restriction fails, so the pair is compatible. (3) No. A product of 1 makes the precision determinant zero, so the implied distribution is singular rather than a proper nondegenerate bivariate normal. (4) The product check addresses positive definiteness only after the off-diagonal terms implied by the two conditionals agree. The slope-to-variance equality is independently necessary.