Simulating Typical and Individual Parameter Values

From Population Estimates to Individual Parameters

An NLME model connects population-level fixed effects and covariates to parameters for particular individuals. A simulation may need the typical clearance for a specified weight, or an individual clearance after accounting for a subject-specific random effect. These are related calculations, but they are not interchangeable: the first uses the typical random-effect value, while the second applies the model’s random-effects structure to a specified individual effect.

The essential workflow is to write down the fitted parameter relationship, insert the covariate values, and then apply the random effect on the scale and in the form specified by that relationship. A random effect should not simply be added to or multiplied by a parameter unless that is what the model says.

Definition: Write a model parameter for individual \(i\) as \(\theta_i=f(\theta,x_i,\eta_i)\), where \(\theta\) denotes the estimated fixed effects, \(x_i\) denotes that individual’s covariates, and \(\eta_i\) denotes the individual random effects. The typical parameter value at covariates \(x_i\) is \(f(\theta,x_i,0)\), obtained by setting the random-effect vector to zero. The individual parameter value is \(f(\theta,x_i,\eta_i)\), using the specified individual random effect.

The fixed effects and covariates are both involved in calculating a typical value. “Typical” does not mean “the value for a person with average covariates”; it means the model value at the covariates supplied, with random effects set to zero. If covariates are centered, their reference values follow the model’s centering convention. For example, if weight is divided by 70 kg, a weight of 70 kg gives a covariate ratio of one.

Write the Parameter Relationship Before Calculating

A model may represent a parameter using an additive random effect, an exponential random effect, or another specified function. Two common forms illustrate why the model equation matters:

\theta_i=\theta+\eta_i \qquad\text{and}\qquad \theta_i=\theta\exp(\eta_i).

In the first form, the random effect is added on the parameter scale. In the second, it is added on the log scale and then exponentiated, so the individual value is the typical value multiplied by \(\exp(\eta_i)\). These forms have different implications and cannot be substituted for one another. A covariate effect may also enter the model before the random effect is applied, as in:

\theta_i=\theta\,g(x_i)\exp(\eta_i), \qquad \theta_{\mathrm{typ}}(x_i)=\theta\,g(x_i).

Here \(g(x_i)\) is the covariate function. Setting \(\eta_i=0\) removes the random effect because \(\exp(0)=1\), but it does not remove the covariate effect. The estimated fixed effects are treated as fixed inputs for these calculations; this does not account for uncertainty in their estimates.

Assumptions and inputs: These calculations presume that the fitted model form is known and correctly transcribed; the fixed-effect estimates and the individual’s covariates are available; the covariates use the model’s original units, coding, and reference values; and the random effect is applied on the scale and through the transformation specified in the model. Any parameter constraints, such as positivity, must also be respected by the parameterization.

Worked Calculations

Worked Example: Typical Clearance at a Given Weight

Suppose a hypothetical fitted model gives typical clearance \(\theta_{CL}=5.800\) L/h at 70 kg and uses the allometric relationship \(CL_{\mathrm{typ}}(WT)=\theta_{CL}(WT/70)^{0.75}\). Calculate typical clearance for an individual weighing 84 kg.

No individual random effect is included in this calculation. Substitute the covariate and estimate:

CL_{\mathrm{typ}}(84) =5.800\left(\frac{84}{70}\right)^{0.75} =5.800(1.2)^{0.75}.

Since \(1.2^{0.75}\approx1.14653\), \(CL_{\mathrm{typ}}(84)\approx5.800(1.14653)=6.6499\) L/h. Thus the covariate-adjusted typical clearance at 84 kg is about 6.650 L/h. This is not yet an individual value incorporating a nonzero random effect.

Worked Example: Applying a Specified Exponential Random Effect

Suppose the covariate-adjusted typical clearance for a hypothetical individual is 6.4206 L/h, and the model specifies \(CL_i=CL_{\mathrm{typ},i}\exp(\eta_{CL,i})\). For a supplied random-effect value \(\eta_{CL,i}=0.15\), calculate the individual clearance.

The value 0.15 is supplied for this calculation; it is not being generated here. Apply the fitted exponential structure:

CL_i=6.4206\exp(0.15) =6.4206(1.161834\ldots) =7.4596729\ldots\ \mathrm{L/h}.

Rounded to four decimal places, \(CL_i\approx7.4597\) L/h; to three significant figures, it is 7.46 L/h. The positive random effect raises clearance above the covariate-adjusted typical value because \(\exp(0.15)>1\). The random effect is dimensionless in this parameterization, so the clearance retains units of L/h.

Worked Example: Combining a Covariate and a Random Effect

Consider a hypothetical absorption-rate model \(k_{a,i}=\theta_{ka}\exp\{\beta_{age}(AGE_i-40)+\eta_{ka,i}\}\), where \(\theta_{ka}=1.100\) h\(^{-1}\), \(\beta_{age}=-0.00800\) per year, and age is measured in years. For an individual aged 55 years with a supplied \(\eta_{ka,i}=0.100\), calculate both the typical and individual values.

For the typical value, set \(\eta_{ka,i}=0\) but retain the age effect:

k_{a,\mathrm{typ}}(55) =1.100\exp\{-0.00800(55-40)\} =1.100\exp(-0.120) \approx1.100(0.886920) =0.9756\ \mathrm{h}^{-1}.

For the individual value, include the supplied random effect in the exponent:

k_{a,i} =1.100\exp\{-0.00800(15)+0.100\} =1.100\exp(-0.0200) \approx1.100(0.980199) =1.0782\ \mathrm{h}^{-1}.

The age effect lowers the typical value relative to \(\theta_{ka}\), while the positive random effect partly offsets that decrease. Both results have units of h\(^{-1}\). The exponent is dimensionless: the age coefficient has units of inverse years, and it multiplies an age difference in years.

Why Setting the Random Effect to Zero Defines “Typical”

The definition follows directly from the model’s individual-parameter function. For a given covariate vector, evaluate that function at zero random effect. If the model has an exponential random effect, then \(\exp(0)=1\); if it has an additive random effect, adding zero leaves the covariate-adjusted fixed-effect value unchanged.

\theta_{\mathrm{typ}}(x_i)=f(\theta,x_i,0), \qquad \theta_i=f(\theta,x_i,\eta_i).

This definition does not generally say that the typical parameter equals the arithmetic mean of individual parameters. For an additive model with a zero-mean random effect, the two may coincide under the model’s assumptions. With a nonlinear transformation, such as exponentiation, transforming a zero random effect gives the typical value on the model’s parameter scale, but it does not by itself establish the arithmetic average across individuals. The distinction matters whenever a simulation or report labels a value “typical,” “mean,” or “average.”

The calculation also separates two sources of variation. Covariate adjustment changes the typical value according to measured characteristics. The random effect then represents the model’s subject-specific departure from that covariate-adjusted value. Changing a covariate is not a substitute for applying a random effect, and a random effect should not be used to compensate for incorrectly coded covariates.

Checks, Alternatives, and Common Pitfalls

Before using calculated values, verify that the formula matches the fitted model and that its inputs are on the required scales. A practical check is to evaluate the model first at \(\eta_i=0\), then at the specified \(\eta_i\), and compare the results. For an exponential random effect, a positive value should multiply the covariate-adjusted typical value by a factor greater than one; a negative value should multiply it by a factor between zero and one. For an additive effect, check the change in the parameter’s own units.

Ignoring covariates: Setting \(\eta_i=0\) does not set covariates to zero or to their reference values. Use the individual’s covariates in the model.
Using the wrong scale: In \(\theta_i=\theta\exp(\eta_i)\), adding \(\eta_i\) directly to \(\theta\) is incorrect. Follow the model’s transformation.
Mixing parameterizations: An additive random effect and an exponential random effect imply different calculations. Do not infer one from the other.
Confusing typical with average: A zero random effect defines the typical value in this convention; nonlinear transformations can make that differ from an arithmetic mean.
Overstating what was simulated: Calculating values for supplied random effects does not generate those effects or account for uncertainty in fixed-effect estimates.

Compared with plug-in simulation at the population parameter estimates, this calculation can retain covariate adjustment and specified individual departures instead of using only a single population-level parameter set. A population typical value is useful as a reference, while individual values are useful when a simulation requires heterogeneity across individuals. The calculations remain conditional on the chosen fixed-effect estimates, covariates, and random-effect values. They do not, by themselves, represent uncertainty about those inputs or define how individual random effects should be generated.

Key takeaway: First calculate the covariate-adjusted typical value by setting the random effect to zero. Then obtain an individual value by applying the supplied random effect through the model’s exact structure. Keep the covariate function, transformation, units, and interpretation explicit at both steps.

Check Your Understanding

Answer each question, then compare with the solutions below.

A model is \(CL_{\mathrm{typ}}(WT)=4.500(WT/70)^{0.75}\) L/h. What is typical clearance at 70 kg?
A model specifies \(V_i=35.00\exp(\eta_{V,i})\) L. For supplied \(\eta_{V,i}=-0.100\), calculate \(V_i\) to three significant figures.
In \(k_{a,i}=1.20\exp\{0.010(AGE_i-50)+\eta_{ka,i}\}\) h\(^{-1}\), calculate the typical value at age 60 years.
Why is setting \(\eta_i=0\) not equivalent to setting every covariate to its reference value?
Does calculating an individual value for a supplied \(\eta_i\) specify how to generate random effects for a population simulation?

Solutions. (1) At 70 kg, \(WT/70=1\), so \(CL_{\mathrm{typ}}=4.500(1)^{0.75}=4.500\) L/h. (2) \(V_i=35.00\exp(-0.100)=35.00(0.904837)=31.6693\) L, or 31.7 L to three significant figures. (3) For the typical value set \(\eta_{ka,i}=0\): \(k_{a,\mathrm{typ}}=1.20\exp\{0.010(60-50)\}=1.20\exp(0.100)=1.3262\) h\(^{-1}\). (4) Zeroing the random effect changes only that model component; the covariate terms remain and must be evaluated at the specified covariate values. (5) No. It applies a supplied random effect to the parameter equation but does not specify a procedure or distribution for generating random effects.