Why the Hierarchy Matters
A population PK data set may include subjects recruited at several sites, with sites belonging to particular studies. If a PK parameter differs systematically across studies or sites, a model with only a subject-level random effect can attribute those group differences to individual variability. The model may then describe the observed data while misrepresenting how much variability belongs to each level.
A nested random-effects model represents the hierarchy directly: subjects are nested within sites, and sites are nested within studies. It separates variation associated with studies, variation among sites within a study, and variation among subjects within a site and study. The variance components answer different questions, so a reported “between-subject variability” needs a clear statement of whether it is conditional on a particular site and study or marginal across the modeled hierarchy.
Model for a PK Parameter
Consider a positive PK parameter, such as clearance \(CL\), modeled on the log scale. Let \(j\) index studies, \(k\) index sites within study \(j\), and \(i\) index subjects within site \(k\) of study \(j\). The subject’s parameter is modeled as follows:
Here \(CL_{\mathrm{typ}}\) is the typical value for the reference covariate pattern, \(u_j\) is the random study effect, \(v_{jk}\) is the random effect for site \(k\) within study \(j\), and \(\eta_{jki}\) is the subject-level random effect. For the scalar parameter considered here, a standard model assumes
The effects are assumed mutually independent across levels and independent between distinct groups at the same level. Thus, two subjects at one site share the study and site effects but have different subject effects. Subjects at different sites in the same study share only the study effect. Subjects in different studies share none of these random effects. The model also assumes that the variance components are common across their respective levels, unless a more elaborate model explicitly allows otherwise.
This equation specifies the distribution of the individual parameter, not the residual error in a measured concentration. A typical concentration model may use \(CL_{jki}\) to predict observations and separately model residual error. The variance components above describe variation in the latent subject parameter across the hierarchy; they are not residual variances.
How the Variance Components Combine
For a randomly selected subject from the modeled hierarchy, the deviation of log clearance from its typical value is \(u_j+v_{jk}+\eta_{jki}\). Each effect has mean zero, and the effects are independent. Therefore, the mean deviation is zero, and the variance of the sum is the sum of the variances:
The independence assumption matters: in general, the variance of a sum also includes twice each pairwise covariance. If the effects were correlated, simply adding their variances would not give the total. In this nested specification, independence makes the addition valid.
The same hierarchy gives a useful covariance result. Two distinct subjects in the same site share \(u_j\) and \(v_{jk}\). Their latent log-clearance deviations therefore have covariance \(\omega^2_{\mathrm{study}}+\omega^2_{\mathrm{site}}\). If they are in different sites of the same study, they share \(u_j\) only, so the covariance is \(\omega^2_{\mathrm{study}}\). Subjects in different studies have covariance zero under the stated model.
Worked Examples
Worked Example: Total and Conditional Variability
Suppose a hypothetical analysis estimates log-scale variance components of \(0.04\) for study, \(0.09\) for site within study, and \(0.16\) for subjects within site and study. Calculate total log-scale variance and the corresponding coefficients of variation on the original clearance scale.
Total across the hierarchy. By independence, add the three components:
The total log-scale standard deviation is \(\sqrt{0.29}=0.5385\). For a lognormal parameter whose log variance is \(\omega^2\), the coefficient of variation is \(\sqrt{\exp(\omega^2)-1}\). Thus, across studies and sites, the clearance CV is \(\sqrt{\exp(0.29)-1}=\sqrt{1.3364-1}\approx0.5800\), or about \(58.0\%\).
Conditional on a fixed study and site. Only the subject effect varies, so the relevant log-scale variance is \(0.16\), not \(0.29\). The conditional clearance CV is \(\sqrt{\exp(0.16)-1}\approx0.4165\), or about \(41.7\%\). These are different summaries: \(58.0\%\) includes between-study and between-site variation, while \(41.7\%\) describes subject variability within a fixed site and study.
Worked Example: Covariance for Subjects at Different Levels
Use the same hypothetical variance components: study variance \(0.04\), site variance \(0.09\), and subject variance \(0.16\). Consider two distinct subjects’ latent log-clearance deviations. Find their covariance if they are (a) at the same site, (b) at different sites in the same study, or (c) in different studies.
(a) At the same site, both subjects share the study and site effects. Their subject effects are independent. The covariance is therefore \(0.04+0.09=0.13\) log-clearance-squared units.
(b) At different sites in the same study, the subjects share only the study effect. The covariance is \(0.04\) log-clearance-squared units.
(c) In different studies, they share no random effects under this model. The covariance is \(0\).
For a check, each subject’s total log-scale variance is \(0.29\). The variance of the difference between the two subjects’ log-clearance deviations is \(2(0.29-\operatorname{Cov})\). It is \(2(0.29-0.13)=0.32\) for subjects at the same site, \(2(0.29-0.04)=0.50\) for different sites in the same study, and \(2(0.29-0)=0.58\) for different studies. Subjects with fewer shared levels have greater variance in their difference.
Worked Example: Partitioning Variability Across Levels
A hypothetical model reports log-scale variance components of \(0.02\) for study, \(0.06\) for site, and \(0.12\) for subject. Calculate each component’s share of total variance, and compare the conditional and marginal clearance CVs.
The total variance is \(0.02+0.06+0.12=0.20\). The shares are \(0.02/0.20=10\%\) for study, \(0.06/0.20=30\%\) for site, and \(0.12/0.20=60\%\) for subject. These percentages partition the log-scale variance; they are not percentages of clearance itself.
Conditional on one study and site, the clearance CV is \(\sqrt{\exp(0.12)-1}=\sqrt{1.1275-1}\approx0.3571\), or about \(35.7\%\). Marginal across the modeled studies and sites, it is \(\sqrt{\exp(0.20)-1}=\sqrt{1.2214-1}\approx0.4705\), or about \(47.1\%\). The marginal CV is larger because it includes the additional study- and site-level variance.
Original-Scale Interpretation
The coefficient-of-variation conversion follows from the lognormal distribution. If \(Y=\exp(Z)\) and \(Z\sim N(m,\omega^2)\), then \(E(Y)=\exp(m+\omega^2/2)\) and \(\operatorname{Var}(Y)=\exp(2m+\omega^2)\{\exp(\omega^2)-1\}\). Dividing the variance by the squared mean and taking the square root gives
For the nested model, use \(\omega^2_{\mathrm{subject}}\) for a conditional, within-site-and-study CV, and use \(\omega^2_{\mathrm{study}}+\omega^2_{\mathrm{site}}+\omega^2_{\mathrm{subject}}\) for the marginal CV across the hierarchy. The CV is dimensionless; when reported as a percentage, multiply it by \(100\%\). The corresponding log-scale standard deviation is not itself the original-scale CV.
Assumptions, Diagnostics, and Limitations
Interpreting fitted components as separate sources of variability requires a design that contains information at each level. Estimating site variation requires multiple sites in at least some studies; estimating study variation requires multiple studies. Subject variation requires multiple subjects within groups. If, for example, each study has only one site, study and site effects cannot generally be separated from the observed grouping structure. Sparse or highly unbalanced recruitment can also make components imprecise, even when they are theoretically identifiable.
Check that the recorded identifiers represent the actual nesting: a site identifier reused in two studies must not accidentally be treated as one shared site if those are distinct study-specific sites. Review estimates and uncertainty for each component, convergence, and whether a component is close to the zero boundary. A near-zero estimate may indicate little support for that level in these data, but it can also reflect limited information or confounding between levels. Compare the implied within-site and between-site patterns with the observed data, and assess whether conclusions change under reasonable alternative specifications.
The model also imposes distributional assumptions: random effects are normally distributed on the log-parameter scale, centered at zero, independent across levels, and described by common variances within a level. A strongly influential group, systematic shifts not captured by the model, or marked departures from these assumptions can make a simple variance partition inadequate. Random effects represent modeled heterogeneity; they do not by themselves explain why studies or sites differ.
A subject-only random-effects model is simpler, but it pools all variation into the subject component and cannot describe the shared deviations among subjects at the same site or study. Fixed study or site effects can instead represent the specific groups observed, but they do not estimate a population variance for a broader set of studies or sites. The nested random-effects formulation is useful when the groups are treated as levels from a wider population and the goal includes estimating their variability. Which formulation is appropriate depends on the scientific target, design, and amount of information available.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions below.
- Write the random-effects expression for log clearance when subjects are nested within sites and sites within studies.
- Two distinct subjects are at different sites in the same study. Which random effect do they share?
- If study, site, and subject log-scale variances are \(0.01\), \(0.04\), and \(0.09\), what is the total log-scale variance?
- For that same model, which variance is used to calculate the within-site, within-study CV?
- Why can one site per study prevent separate estimation of study and site variance?
Solutions. (1) \(\log(CL_{jki})=\log(CL_{\mathrm{typ}})+u_j+v_{jk}+\eta_{jki}\), with study index \(j\), site-within-study index \(k\), and subject-within-site index \(i\). (2) They share the study effect \(u_j\), but not the site or subject effects. (3) Add the independent components: \(0.01+0.04+0.09=0.14\). (4) Use the subject variance \(0.09\), because study and site are fixed for this conditional comparison. (5) With only one site in each study, the data do not provide within-study contrasts among sites; a study shift and a site-within-study shift are confounded in the grouping structure.