What a Pairwise Eta Plot Shows
A pairwise eta plot places each subject’s empirical Bayes estimates for two random effects on perpendicular axes. It is a quick way to inspect whether subjects estimated to be high on one random effect also tend to be high or low on another. For example, a plot of the random effects on log clearance and log volume may show an upward trend, a downward trend, a roughly round cloud, or a small number of unusual points.
That picture is useful, but it is not a direct view of the population random effects. Each plotted coordinate is estimated using that subject’s data and the fitted population model. In particular, empirical Bayes estimates are typically shrunk toward the population mean, and the amount of shrinkage can differ by subject and parameter. The plot should therefore be read as a diagnostic view of the estimated subject effects, not as a stand-alone estimate or test of the covariance matrix.
The zero-centered normal random-effects distribution is a model assumption, not a guarantee that the finite set of empirical Bayes estimates will form a symmetric ellipse. The plot is most readily interpreted when the model uses a common population distribution for independent subjects, the random effects are meaningfully defined, and there is enough information in the observations to estimate both components for a useful fraction of subjects. Even then, the plotted estimates have uncertainty and are affected by the observations, residual model, and fitting procedure.
Reading Association Without Over-Interpreting It
A cloud elongated from lower left to upper right has positive apparent association: subjects with larger estimated values of one component tend also to have larger estimated values of the other. A lower-left to upper-right description is not appropriate if one axis has been defined in the opposite direction; interpret the signs using the actual parameterization. A cloud elongated from upper left to lower right has negative apparent association. A roughly round cloud suggests little visible linear association, but does not establish that the population covariance is zero.
The visual pattern is descriptive. It can be affected by finite sample variability, unequal information, influential points, and the estimation of the empirical Bayes values. Also, a pairwise plot only addresses one pair of components at a time. With several random effects, patterns across separate pairs do not by themselves summarize all features of the joint distribution.
Worked Example: Calculating the Descriptive Association
Suppose a hypothetical fit gives these five pairs of empirical Bayes estimates for two random effects: \((-1,-0.8),(-0.5,-0.4),(0,0),(0.5,0.4),(1,1.8)\). The coordinates are in the model’s random-effects units. Calculate their sample correlation and state what it describes.
The first-coordinate mean is \(\bar{x}=0\), and the second-coordinate mean is \(\bar{y}=0.2\). Thus the centered values are \(x_i-\bar{x}=(-1,-0.5,0,0.5,1)\) and \(y_i-\bar{y}=(-1,-0.6,-0.2,0.2,1.6)\).
The sum of cross-products is \(1+0.3+0+0.1+1.6=3.0\). The sums of squared deviations are \(1+0.25+0+0.25+1=2.5\) for the first coordinate and \(1+0.36+0.04+0.04+2.56=4.0\) for the second. Therefore,
The five estimates have a strong positive descriptive association. This value is the sample correlation of these plotted estimates; it is not an estimate with quantified uncertainty, nor proof that the population random-effects correlation equals 0.949. In particular, five subjects provide little basis for treating the visual association as a precise population feature.
Why Shrinkage Can Change the Picture
An empirical Bayes estimate combines a subject’s information with the fitted population distribution. When a subject has weak information about a component, that estimate tends to be pulled more strongly toward the population mean. If two components have different information for the same subject, or if subjects differ in their sampling designs, the plotted points need not all be contracted in the same way.
A simple Gaussian calculation makes the mechanism explicit. Suppose a subject has an approximate data-based estimate \(z_i\) of a random-effects vector, with \(z_i\mid\eta_i\sim N(\eta_i,R_i)\) and \(\eta_i\sim N(0,\Omega)\). Here \(R_i\) describes uncertainty in that data-based estimate. The posterior distribution is normal, and its mean is the empirical Bayes estimate in this Gaussian illustration.
The log posterior, apart from a constant, is \(-\frac{1}{2}\eta_i^\mathsf{T}\Omega^{-1}\eta_i-\frac{1}{2}(z_i-\eta_i)^\mathsf{T}R_i^{-1}(z_i-\eta_i)\).
The quadratic coefficient is \(\Omega^{-1}+R_i^{-1}\), and the linear term is \(R_i^{-1}z_i\). Completing the square gives posterior covariance \((\Omega^{-1}+R_i^{-1})^{-1}\) and posterior mean \((\Omega^{-1}+R_i^{-1})^{-1}R_i^{-1}z_i\).
Using the equivalent matrix identity, the posterior mean is \(\Omega(\Omega+R_i)^{-1}z_i\). Because \(R_i\) can vary across subjects, the transformation from \(z_i\) to \(\hat\eta_i\) can vary across subjects too.
This derivation is exact for the stated Gaussian model; nonlinear NLME models generally do not reduce to this simple calculation. The illustration nevertheless explains the diagnostic caution: unequal information can produce unequal shrinkage, and correlated components can be affected together by the information matrix. If every point were multiplied by the same positive constant on each axis, the sample correlation would be unchanged. Subject-specific shrinkage is different: it can change relative positions and therefore alter the apparent association, spread, or shape of the cloud.
Worked Example: Unequal Information for Two Components
Consider a hypothetical subject for whom an approximate data-based estimate is \(z_i=(2,2)^\mathsf{T}\). Assume, for illustration, independent prior components with variance 1 each, and independent data-based estimation errors with variances 0.25 for component 1 and 4 for component 2. In this special diagonal case, the posterior mean for component \(j\) is \(k_jz_{ij}\), where \(k_j=1/(1+v_j)\) and \(v_j\) is the corresponding error variance.
For component 1, \(k_1=1/(1+0.25)=0.8\), so its empirical Bayes estimate is \(0.8\times2=1.6\). For component 2, \(k_2=1/(1+4)=0.2\), so its estimate is \(0.2\times2=0.4\). The plotted point is therefore \((1.6,0.4)\), not \((2,2)\). Component 2 is pulled much closer to zero because its data-based estimate is less precise under these assumptions.
This single calculation does not establish how a whole plot’s correlation will change. Across subjects, different component-specific information can compress points by different amounts, so the overall shape and association may change. A useful reading of the plot asks whether unequal sampling or weak information for one component could explain part of the visible pattern.
Asymmetry and Extreme Subjects
Under a bivariate normal population model, the underlying random-effects distribution is symmetric about zero. A strongly asymmetric empirical Bayes cloud can therefore prompt questions about the model, but it does not diagnose the cause on its own. Nonlinearity, sparse subject-specific data, shrinkage, a few influential subjects, or genuine departures from the assumed random-effects distribution can all contribute. The shape of the estimates is not a direct picture of the shape of the latent random effects.
An extreme point deserves careful attention because it can dominate the apparent direction or strength of association. Check the subject’s data, design, and fitted individual predictions, and consider whether the point has unusually weak information or an unusual combination of observations. Do not remove it merely because it looks inconvenient: it may be a valid subject whose pattern reveals a model limitation. Conversely, one point should not be allowed to define the covariance interpretation without checking how much the visual summary depends on it.
Worked Example: An Extreme Subject Changes the Apparent Correlation
Suppose four hypothetical pairs of empirical Bayes estimates are \((-1,-1),(0,0),(1,1),(4,0)\). The last subject is extreme on the first component but not on the second. Calculate the sample correlation with and without that subject to see its influence on the descriptive pattern.
For all four points, the means are \(\bar{x}=1\) and \(\bar{y}=0\). The centered cross-product sum is \(2+0+0+0=2\); the sums of squared deviations are \(4+1+0+9=14\) for \(x\) and \(1+0+1+0=2\) for \(y\). Hence \(r=2/\sqrt{14\times2}=2/\sqrt{28}\approx0.378\).
For the first three points alone, the means are both zero. Their cross-product sum is \(1+0+1=2\), and each sum of squared deviations is 2. Thus \(r=2/\sqrt{2\times2}=1.00\). The extreme subject greatly reduces the observed sample correlation. This comparison identifies sensitivity in a small descriptive sample; it does not justify deleting that subject or establish the population covariance.
Practical Checks and Common Pitfalls
Interpret the pairwise plot alongside the subject-level information that generated its coordinates. Marginal eta distributions and shrinkage summaries can help reveal whether one component is more compressed than another; eta-versus-covariate displays can help identify remaining structure related to observed covariates. Residual and individual-fit diagnostics can help determine whether unusual observations or subject profiles accompany an extreme eta estimate. These checks complement the pairwise plot rather than turning it into a formal test.
- Check the scale and parameterization. Eta values are dimensionless deviations on the model’s random-effects scale, often a log scale. A positive association means larger values of both defined components tend to occur together; it does not automatically mean both original parameters increase in the same way if one parameter is defined inversely.
- Check how many subjects are informative. A dense cloud of highly shrunk estimates near zero can hide structure. Unequal sampling or parameter information can also make some subjects more influential in one coordinate.
- Check asymmetry and leverage. Look for one-sided tails, clusters, or isolated points, and assess whether a few subjects determine the apparent direction. Treat those features as prompts for investigation, not diagnoses.
- Compare the plot with the fitted covariance model. The plotted empirical Bayes estimates and the fitted \(\Omega\) are related but are not interchangeable. A visual pattern that seems inconsistent with the fitted covariance warrants investigation of information, model assumptions, and fitting, rather than an immediate conclusion that the estimate is wrong.
A common mistake is to calculate the correlation of empirical Bayes estimates and report it as the population random-effects correlation. The calculation describes the plotted points, but those points are posterior estimates with unequal uncertainty and shrinkage. Use the plot to identify patterns and possible failures; use the fitted model’s covariance estimates and their uncertainty for statements about the model’s population covariance.
Pairwise eta plots are most useful as one part of a diagnostic argument. They can make an apparent relationship, asymmetry, or influential subject visible quickly, but they cannot by themselves distinguish a true population correlation from estimation effects, establish normality, or determine which model change is appropriate. Their diagnostic value is greatest when the interpretation accounts for parameter definitions, subject-specific information, and corroborating displays.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- A pairwise eta plot has an upward-sloping cloud. What does this establish, and what does it not establish about the population covariance?
- In the Gaussian illustration, the prior variance is 2 and the data-based estimate error variance is 2 for one component. What is the shrinkage factor, and what is the empirical Bayes estimate if \(z=3\)?
- Why can subject-specific shrinkage alter a pairwise cloud even though multiplying every subject’s coordinate by one common positive constant would not change the sample correlation?
- A cloud is asymmetric and one subject lies far to the right. Give two appropriate diagnostic responses and one inappropriate conclusion.
Solutions. 1. It establishes a positive descriptive association among the plotted empirical Bayes estimates. It does not establish the size, statistical certainty, or even the exact sign of the population covariance without considering estimation and shrinkage. 2. The factor is \(k=2/(2+2)=0.5\), so the estimate is \(0.5\times3=1.5\). 3. A common positive scaling preserves relative positions and correlation, but subject-specific scaling changes different points by different amounts and can change their centered cross-products and variances. 4. Inspect that subject’s data and information, and compare the pattern with complementary diagnostics such as residuals or marginal eta distributions. It is inappropriate to conclude from the plot alone that the random-effects distribution is nonnormal or to delete the subject automatically.