Tutorials › Nonlinear Mixed-Effects Models (NLME) › Random-Effect Variance on Natural and Log Scales

Between-Subject, Between-Occasion and Residual Variability · Tutorial 108 of 797

Random-Effect Variance on Natural and Log Scales

Learn to translate a log-scale random-effect variance into the natural-scale variability of a pharmacokinetic parameter.

Intermediate 9 min read

What You'll Learn

  • Distinguish log-scale random-effect variance from natural-scale parameter variance.
  • Derive the parameter median and arithmetic mean from the normal random-effect model.
  • Calculate natural-scale variance, standard deviation, and coefficient of variation.
  • Convert a target coefficient of variation to the required log-scale variance.
  • Recognize assumptions and common errors when interpreting these quantities.

Why the Scale of Variability Matters

In a nonlinear mixed-effects model, a random effect may be added to a parameter after taking its logarithm. The model’s variance is then specified on the log scale, while the parameter itself is reported on its natural scale—for example, clearance in L/h. These are different scales, so the variance parameter cannot be read directly as the variance of clearance. Translating between them is important for interpreting the size of between-subject variability and for reporting results in units meaningful to pharmacometrics.

This tutorial considers a single positive parameter with a normally distributed random effect. The central results are exact: the parameter’s median, arithmetic mean, variance, and coefficient of variation can all be derived from the normal distribution’s moment-generating function. The conclusions depend on the stated distributional assumption; they are not approximations based merely on the size of the variance.

Model and Assumptions

Definition: Let \(P_i\) be a positive parameter for individual \(i\), and let \(P_{\mathrm{typ}}>0\) be its typical value when the random effect is zero. The log-scale random-effect model is \(P_i=P_{\mathrm{typ}}\exp(\eta_i)\), where \(\eta_i\sim N(0,\omega^2)\). Thus, \(\omega^2\) is the variance of the random effect on the log scale, and \(\omega=\sqrt{\omega^2}\) is its standard deviation. The induced distribution of \(P_i\) is lognormal.

The zero-mean assumption for \(\eta_i\) means that \(P_{\mathrm{typ}}\) is the parameter value at \(\eta_i=0\). Because a normal distribution with mean zero has median zero, the median of \(P_i\) will equal \(P_{\mathrm{typ}}\). This does not imply that \(P_{\mathrm{typ}}\) equals the arithmetic mean of \(P_i\).

The formulas below assume \(\omega^2\geq 0\), a common \(P_{\mathrm{typ}}\) for the individuals being described, and the specified normal distribution for \(\eta_i\). They describe variation due to this random effect alone. If a fixed covariate multiplies \(P_{\mathrm{typ}}\), the resulting median and arithmetic mean are multiplied by that same positive factor, while the coefficient of variation is unchanged.

Deriving the Natural-Scale Moments

For a normal random variable \(\eta_i\sim N(0,\omega^2)\), its moment-generating function is \(E[\exp(t\eta_i)]=\exp(t^2\omega^2/2)\). Applying it at \(t=1\) gives the mean of \(P_i\). Applying it at \(t=2\) gives the second raw moment, from which the variance follows.

$$ \begin{aligned} \operatorname{Median}(P_i) &=P_{\mathrm{typ}}\exp\{\operatorname{Median}(\eta_i)\} =P_{\mathrm{typ}},\\ E[P_i] &=P_{\mathrm{typ}}E[\exp(\eta_i)] =P_{\mathrm{typ}}\exp(\omega^2/2),\\ E[P_i^2] &=P_{\mathrm{typ}}^2E[\exp(2\eta_i)] =P_{\mathrm{typ}}^2\exp(2\omega^2),\\ \operatorname{Var}(P_i) &=E[P_i^2]-\{E[P_i]\}^2\\ &=P_{\mathrm{typ}}^2\left\{\exp(2\omega^2)-\exp(\omega^2)\right\}\\ &=P_{\mathrm{typ}}^2\exp(\omega^2)\left\{\exp(\omega^2)-1\right\}. \end{aligned} $$

The coefficient of variation (CV) is the standard deviation divided by the arithmetic mean. It is dimensionless, so it can express relative variability without depending on the parameter’s units.

$$ \begin{aligned} \operatorname{CV}(P_i) &=\frac{\sqrt{\operatorname{Var}(P_i)}}{E[P_i]}\\ &=\frac{P_{\mathrm{typ}}\exp(\omega^2/2)\sqrt{\exp(\omega^2)-1}} {P_{\mathrm{typ}}\exp(\omega^2/2)}\\ &=\sqrt{\exp(\omega^2)-1}. \end{aligned} $$

These results separate the roles of the model quantities. The typical value sets the parameter’s scale and units. The log-scale variance determines the gap between median and arithmetic mean, the natural-scale variance, and the CV. In particular, \(\omega^2\) itself is not the natural-scale variance or the squared CV.

Worked Example: Natural-Scale Variability in Clearance

Suppose a hypothetical model has \(CL_i=5\exp(\eta_i)\) L/h, with \(\eta_i\sim N(0,0.09)\). Find the median, arithmetic mean, variance, standard deviation, and CV of clearance across individuals.

The typical value is \(CL_{\mathrm{typ}}=5\) L/h and \(\omega^2=0.09\). The median is the typical value: \[ \operatorname{Median}(CL_i)=5\text{ L/h}. \] The arithmetic mean is: \[ E[CL_i]=5\exp(0.09/2)=5\exp(0.045)\approx 5.230\text{ L/h}. \] The natural-scale variance is: \[ \operatorname{Var}(CL_i)=5^2\exp(0.09)\{\exp(0.09)-1\} =25(1.0942)(0.09417)\approx 2.576\;(\text{L/h})^2. \] Therefore, the standard deviation is \(\sqrt{2.576}\approx1.605\) L/h. The coefficient of variation is: \[ \operatorname{CV}(CL_i)=\sqrt{\exp(0.09)-1} =\sqrt{0.09417}\approx0.3069, \] or about 30.69%. The mean exceeds the median because the natural-scale distribution is right-skewed.

Reading Variance and CV in Context

The variance of the random effect, \(\omega^2\), is on the log scale. If \(P_i\) has units of L/h, then \(\operatorname{Var}(P_i)\) has units of \((\text{L/h})^2\), whereas \(\omega^2\) and the CV have no physical units. The log-scale variance is often convenient for specifying the model; the natural-scale variance and CV can be more immediately interpretable as descriptions of parameter spread.

A useful alternative expression is \(\omega^2=\ln(1+\operatorname{CV}^2)\). It follows by squaring the CV formula, adding one, and taking logarithms. This lets a modeler translate a desired relative variability into the log-scale variance used in the model. It does not make the CV equal to \(\omega\): that equality is only an approximation when variability is small.

Worked Example: Converting a Log-Scale Variance to Relative Spread

A hypothetical volume parameter has \(V_{\mathrm{typ}}=12\) L and \(\omega^2=0.25\). Calculate the median, arithmetic mean, natural-scale variance, and CV.

The median equals \(12\) L. The mean is: \[ E[V_i]=12\exp(0.25/2)=12\exp(0.125)\approx13.60\text{ L}. \] For the variance: \[ \operatorname{Var}(V_i)=12^2\exp(0.25)\{\exp(0.25)-1\} =144(1.2840)(0.2840)\approx52.52\text{ L}^2. \] The standard deviation is approximately \(\sqrt{52.52}=7.247\) L. Finally, \[ \operatorname{CV}(V_i)=\sqrt{\exp(0.25)-1} =\sqrt{0.2840}\approx0.5329, \] or 53.29%. This relatively large CV illustrates why the log-scale standard deviation, \(\omega=0.5\), should not simply be reported as a natural-scale CV of 50%.

Worked Example: Finding the Log-Scale Variance From a Target CV

Suppose a hypothetical parameter should have a CV of 30% under the lognormal random-effect model. Find \(\omega^2\) and, if \(P_{\mathrm{typ}}=4\) L/h, the arithmetic mean and natural-scale variance.

Rearrange \(\operatorname{CV}^2=\exp(\omega^2)-1\): \[ \omega^2=\ln(1+\operatorname{CV}^2)=\ln(1+0.30^2)=\ln(1.09)\approx0.08618. \] The corresponding log-scale standard deviation is \(\omega=\sqrt{0.08618}\approx0.2936\). The arithmetic mean is: \[ E[P_i]=4\exp(0.08618/2)=4\exp(0.04309)\approx4.176\text{ L/h}. \] The natural-scale variance is: \[ \operatorname{Var}(P_i)=4^2\exp(0.08618)\{\exp(0.08618)-1\} =16(1.09)(0.09)=1.570\;(\text{L/h})^2. \] Its standard deviation is approximately 1.253 L/h, and \(1.253/4.176\approx0.300\), confirming the target CV, up to rounding.

When the Random Effect Has a Nonzero Mean

The standard parameterization uses a mean-zero random effect, but the algebra also shows what changes if \(\eta_i\sim N(m,\omega^2)\). In that case, the median becomes \(P_{\mathrm{typ}}\exp(m)\), and the arithmetic mean becomes \(P_{\mathrm{typ}}\exp(m+\omega^2/2)\). The natural-scale variance is \(P_{\mathrm{typ}}^2\exp(2m+\omega^2)\{\exp(\omega^2)-1\}\), while the CV remains \(\sqrt{\exp(\omega^2)-1}\). The mean shift changes the location and scale of the parameter distribution but not its relative spread.

This distinction matters when interpreting model output. A model that assumes a zero-mean random effect gives the stated mean-zero formulas; one should not silently apply them after introducing a nonzero mean. In the usual mean-zero formulation, \(P_{\mathrm{typ}}\) is the median rather than the arithmetic mean, and the latter is larger whenever \(\omega^2>0\).

Assumptions, Diagnostics, and Alternatives

The moment formulas are exact if the random effect is normally distributed and enters the parameter through the stated exponential transformation. They do not establish that these assumptions describe a particular population well. Assessment should therefore consider whether the fitted model reproduces relevant features of the observed data and whether the implied parameter distribution is scientifically plausible. Simulations from the fitted model can reveal implausible tails or an overly broad range of positive parameter values.

Estimated individual random effects require care in such checks. They are informed by each individual’s observations and are generally shrunken toward zero, especially when an individual contributes little information. Consequently, the sample variance of estimated individual effects need not equal the fitted \(\omega^2\). It should not be used as a direct substitute for the model-based variance without accounting for estimation and shrinkage.

A log-scale random effect guarantees a positive parameter, which is useful for quantities such as clearance and volume. An additive random effect on the parameter scale is a different model: it can allow negative values unless its distribution or parameter range prevents them, and its variance has the parameter’s squared units. The choice is not determined by algebra alone; it should be compatible with the parameter’s physical constraints and the scientific model. The lognormal model also imposes a particular right-skewed distribution and can imply a long upper tail, which may be implausible for some applications.

Common mistake: “The estimated \(\omega^2\) is 0.09, so clearance has a variance of 0.09 and a CV of 30%.” The first claim confuses log-scale and natural-scale variance, and the second is only approximately right here. For the model with \(CL_{\mathrm{typ}}=5\) L/h and \(\omega^2=0.09\), the natural-scale variance is about \(2.576\;(\text{L/h})^2\), and the CV is \(\sqrt{\exp(0.09)-1}\approx30.69\%\). Calculate each quantity from its defining formula and report its units.

Check Your Understanding

Use the mean-zero lognormal random-effect model unless a question says otherwise. Solutions are included so you can check the scale, units, and arithmetic.

  1. If \(P_{\mathrm{typ}}=8\) mg/L and \(\omega^2=0\), what are the median, arithmetic mean, and CV?
  2. For a positive parameter with \(\omega^2=\ln(1.25)\), calculate its CV.
  3. With \(P_{\mathrm{typ}}=3\) L/h and \(\omega^2=0.16\), is the arithmetic mean larger or smaller than the median? Calculate both.
  4. Explain why a reported log-scale variance of \(0.16\) is not a natural-scale variance in \((\text{L/h})^2\).

Solutions. (1) When \(\omega^2=0\), \(\eta_i=0\) with probability one. The median and mean are both \(8\) mg/L, and \(\operatorname{CV}=\sqrt{\exp(0)-1}=0\). (2) \(\operatorname{CV}=\sqrt{\exp\{\ln(1.25)\}-1}=\sqrt{0.25}=0.5\), or 50%. (3) The median is \(3\) L/h. The mean is \(3\exp(0.16/2)=3\exp(0.08)\approx3.250\) L/h, so it is larger. (4) The value \(0.16\) is the variance of the dimensionless log-scale random effect \(\eta_i\). The natural-scale variance depends on both \(P_{\mathrm{typ}}\) and \(\omega^2\), and has squared parameter units; here it is \(3^2\exp(0.16)\{\exp(0.16)-1\}\) in \((\text{L/h})^2\), not \(0.16\;(\text{L/h})^2\).