Initial Values for Between-Subject Variability

Why the Scale of Variability Matters

An initial value for between-subject variability is not simply the observed standard deviation of clearance or volume entered into an estimation program. In a common population model, each subject’s positive parameter is the typical population parameter multiplied by an exponential random effect. The model estimates the variance of that random effect on the log scale, whereas summaries of individual parameters are often reported on the original scale. Translating between the scales correctly gives a defensible starting value.

The translation also depends on what the observed spread represents. Differences between subjects in their underlying clearances or volumes are between-subject variability. Differences between an observation and the model prediction for that subject are residual unexplained variability. These components can both contribute to the spread of measured concentrations, but they are not interchangeable. The earlier material on residual error models concerns the observation-level component; this tutorial focuses on initial values for the parameter-level random effects.

Lognormal Random Effects and the Variance Conversion

Definition: For subject $i$, a lognormal model for a positive parameter $P_i$ is $$ P_i=TVP\exp(\eta_{P,i}), \qquad \eta_{P,i}\sim N(0,\omega_P^2). $$ Here $TVP$ is the typical parameter, $\eta_{P,i}$ is the subject-specific random effect, and $\omega_P^2$ is its variance on the log scale. The random effects are assumed to have mean zero, to follow the stated normal distribution, and to be independent across subjects. The model implies that $TVP$ is the median and geometric mean of the population distribution of $P_i$, not generally its arithmetic mean.

The variance conversion follows from the moments of a lognormal variable. If $Z\sim N(0,\omega_P^2)$, then $E[\exp(Z)]=\exp(\omega_P^2/2)$ and $E[\exp(2Z)]=\exp(2\omega_P^2)$. Therefore,

E[P_i]=TVP\exp(\omega_P^2/2), \qquad \operatorname{Var}(P_i)=TVP^2\exp(\omega_P^2)\left[\exp(\omega_P^2)-1\right].

The coefficient of variation (CV) is the standard deviation divided by the arithmetic mean. Substituting the two moments gives a relationship that does not depend on $TVP$:

CV^2=\frac{\operatorname{Var}(P_i)}{E[P_i]^2} =\exp(\omega_P^2)-1, \qquad \omega_P^2=\ln(1+CV^2).

Thus, a sample SD and mean can provide a rough candidate for the log-scale variance: calculate $CV=s/\bar{x}$, then use $\ln(1+CV^2)$. This is a moment-based translation, not a claim that the sample summaries are precise estimates of the population distribution. It is most interpretable when the parameters are positive and plausibly lognormal, the subjects are reasonably representative of the modeled population, and major systematic differences between subjects have not been mistaken for random variation.

Worked Example: Converting Clearance Spread to a Variance

Suppose a hypothetical preliminary summary of subject clearances has arithmetic mean $5.00$ L/h and SD $1.50$ L/h. Treat these as a rough description of a positive, approximately lognormal between-subject distribution. Find a candidate initial value for $\omega_{CL}^2$, and determine the $TVCL$ that makes the model’s arithmetic mean equal to $5.00$ L/h.

First, the observed CV is $1.50/5.00=0.3000$, or 30.00%. The log-scale variance is

\omega_{CL}^2=\ln(1+0.3000^2)=\ln(1.09)=0.08618.

The corresponding log-scale SD is $\sqrt{0.08618}=0.2936$. To match the arithmetic mean, use $E[CL_i]=TVCL\exp(\omega_{CL}^2/2)$. Equivalently, since $\exp(\omega_{CL}^2)=1+CV^2=1.09$,

TVCL=\frac{5.00\ \text{L/h}}{\exp(0.08618/2)} =\frac{5.00\ \text{L/h}}{\sqrt{1.09}} =4.789\ \text{L/h}.

The candidate variance is $0.08618$, a dimensionless log-scale quantity. The typical clearance is $4.789$ L/h under the stated mean-matching choice. If a separately chosen typical clearance is instead retained as the median or geometric mean, it need not equal the sample arithmetic mean.

Choosing a Typical Parameter Consistently

A variance starting value and a typical-parameter starting value should be interpreted together. For a lognormal distribution with fixed $TVP$, increasing $\omega_P^2$ increases the arithmetic mean above $TVP$, even though the median remains $TVP$. If preliminary summaries provide an arithmetic mean $m$ and CV, then a mean-matching typical value is

TVP=\frac{m}{\exp(\omega_P^2/2)} =\frac{m}{\sqrt{1+CV^2}}.

This adjustment is not always the appropriate way to initialize a model. For example, a typical value may already have been obtained by back-calculating a parameter from a representative concentration profile. In that case, changing it solely to force agreement with an unadjusted sample mean could be inconsistent with the source of the starting value. The analyst should record whether the chosen typical value represents a median, a geometric mean, or a value intended to match an arithmetic mean.

Worked Example: Converting Volume Variability and Matching the Mean

Consider hypothetical individual volume summaries with arithmetic mean $40.0$ L and SD $20.0$ L. Assume for this initialization exercise that volume is positive and its population distribution is reasonably represented by a lognormal model. Calculate the CV, $\omega_V^2$, and the $TVV$ that matches the arithmetic mean.

The CV is $20.0/40.0=0.5000$, or 50.00%. Therefore,

\omega_V^2=\ln(1+0.5000^2)=\ln(1.25)=0.2231.

The random-effect SD is $\sqrt{0.2231}=0.4724$. The mean-matching typical volume is

TVV=\frac{40.0\ \text{L}}{\sqrt{1.25}} =\frac{40.0\ \text{L}}{1.1180} =35.78\ \text{L}.

Under this model, the distribution’s median is $35.78$ L while its arithmetic mean is $40.0$ L. The original-scale SD is $20.0$ L by construction of the moment conversion. These are candidate initial values, not evidence that the population distribution is exactly lognormal or that the variance is well determined.

Why Residual Error Must Stay Separate

A concentration observation varies for more than one reason. Subjects can differ in their underlying clearance and volume, and repeated observations within a subject can deviate from the individual model prediction because of residual error. In general, an SD calculated by pooling raw concentrations across subjects and times does not isolate either component. It depends on the sampling times, dose, structural model, covariate distribution, and both kinds of variability.

A simple log-scale decomposition illustrates the distinction. Suppose, just for illustration, a log concentration has one unit of sensitivity to a subject random effect and an independent residual term:

\log C_{ij}=a+\eta_i+\epsilon_{ij}, \qquad \operatorname{Var}(\eta_i)=\omega^2,\quad \operatorname{Var}(\epsilon_{ij})=\sigma^2.

With independence, the variance of a single log concentration around $a$ is $\omega^2+\sigma^2$. This does not make the two components the same: $\eta_i$ is shared across observations from subject $i$, while $\epsilon_{ij}$ is an observation-level deviation. In a pharmacokinetic model, sensitivity to a parameter random effect can also vary with time and model structure, so the simple sum is only an illustration of the separation, not a general formula for converting pooled concentration spread into a parameter variance.

Worked Example: Do Not Use Residual CV as Clearance CV

In a hypothetical analysis, a preliminary clearance summary gives mean $5.00$ L/h and SD $1.50$ L/h. Separately, observed concentrations have a residual proportional-error CV of 20%. Find the clearance variance starting value using the clearance summary, and explain why substituting the residual CV would be wrong.

The between-subject clearance CV is $1.50/5.00=0.3000$. Its candidate variance is

\omega_{CL}^2=\ln(1+0.3000^2)=\ln(1.09)=0.08618.

If the 20% residual CV were incorrectly substituted, the calculation would instead give

\ln(1+0.2000^2)=\ln(1.04)=0.03922.

That value describes a conversion of a 20% CV under a lognormal assumption; it is not a clearance random-effect variance merely because it is expressed as a variance. The 20% figure describes observation-level residual error, while the clearance summary describes differences between subject parameters. They have different sources and roles in the model, so the candidate clearance variance from the stated clearance spread is $0.08618$, not $0.03922$.

Derivation, Assumptions, and Practical Checks

The moment conversion can be justified directly. For a normal random variable $Z\sim N(0,\omega^2)$, the normal moment-generating function gives $E[\exp(tZ)]=\exp(t^2\omega^2/2)$. Taking $t=1$ and $t=2$ yields $E[\exp(Z)]=\exp(\omega^2/2)$ and $E[\exp(2Z)]=\exp(2\omega^2)$. Since $P=TVP\exp(Z)$,

\operatorname{Var}(P)=E[P^2]-E[P]^2 =TVP^2\left[\exp(2\omega^2)-\exp(\omega^2)\right] =TVP^2\exp(\omega^2)\left[\exp(\omega^2)-1\right].

Dividing by $E[P]^2=TVP^2\exp(\omega^2)$ gives $CV^2=\exp(\omega^2)-1$, and taking the natural logarithm gives the stated conversion. The derivation requires a lognormal parameter distribution; it does not say that every empirical collection of estimates is lognormal or that the sample CV is a stable estimate of its population counterpart.

Before using observed spread to set an initial value, check the following:

Are the summaries on the parameter scale? Clearance or volume summaries can be considered for this conversion; concentration spread pooled across observations cannot be treated as their parameter CV.
Are the observations comparable? Strong covariate or design differences can create apparent spread that should not be attributed wholly to unexplained random effects.
Are the summaries informative? A small number of subjects, extreme values, or uncertain individual parameter estimates can make the sample CV unstable.
Is the lognormal shape plausible? A few large values can inflate the arithmetic SD, and a lognormal assumption may be poor if the population is a mixture or contains influential outliers.
Is the parameter positive and the calculation valid? The method assumes a positive parameter and a nonnegative CV. A zero observed SD gives a zero moment-based candidate, but does not establish that the true population variance is zero.

Individual parameter estimates from sparse data are not direct measurements of each subject’s true parameter. Their apparent spread may be compressed by shrinkage or affected by estimation error. Consequently, an empirical CV from such estimates is a practical starting-value clue, not an unqualified estimate of the random-effect variance. Compare it with scientific plausibility and whether the proposed starting value allows stable estimation; the fitted model and its diagnostics must ultimately be assessed on their own merits.

Alternatives and Common Pitfalls

The moment conversion is convenient when summaries are reported as an arithmetic mean and SD. If suitable positive individual values are available, another descriptive route is to calculate the SD of their logarithms; under the assumed model, that estimates the log-scale SD directly, whose square is a candidate $\omega_P^2$. The two routes need not agree in a finite sample, especially with skewness or outliers. A third option is to choose a scientifically plausible variance and examine whether nearby starting values lead to materially different estimation behavior. None of these alternatives removes the need to distinguish population variability from residual error or systematic covariate differences.

Common mistake: “The clearance SD is $1.50$ L/h, so I will enter $1.50$ as the random-effect variance.” This is wrong twice: the SD is on the original clearance scale and has units, whereas $\omega_{CL}^2$ is a dimensionless variance on the log scale. Convert the SD to a CV by dividing by the mean, then use $\ln(1+CV^2)$. Also, do not use a concentration residual CV in that conversion as if it described between-subject clearance differences.

These calculations initialize diagonal random-effect variances for clearance or volume. They do not determine how clearance and volume random effects co-vary; that is a separate modeling choice. A candidate value can also be a poor initialization if the observed summaries are biased, the assumed distribution is unsuitable, or the model has limited information to estimate the variability. It should therefore be documented as an initial value and its source stated clearly, rather than reported as a final population estimate.

Check Your Understanding

For each question, work through the conversion and interpretation before reading the solution.

A hypothetical positive volume distribution has mean $30.0$ L and SD $9.00$ L. What are its CV and moment-based candidate $\omega_V^2$?
For a lognormal parameter with $CV=0.400$, calculate the variance $\omega^2$ and state whether it has parameter units.
A model’s observed concentrations have a 25% residual CV, while individual clearance summaries have CV 35%. Which CV is relevant to a clearance random-effect variance, and why?
Explain why the typical parameter is not necessarily the arithmetic mean under a lognormal random-effects model.

Solutions. (1) $CV=9.00/30.0=0.3000$. Thus $\omega_V^2=\ln(1+0.3000^2)=\ln(1.09)=0.08618$, dimensionless. (2) $\omega^2=\ln(1+0.400^2)=\ln(1.16)=0.1484$, also dimensionless because it is a variance on the log scale. (3) Use the 35% clearance CV: $\omega_{CL}^2=\ln(1+0.350^2)=\ln(1.1225)=0.1156$. The residual CV concerns observation-level deviations, not between-subject clearance. (4) Since $E[P]=TVP\exp(\omega^2/2)$, a positive variance makes the arithmetic mean larger than $TVP$, which is the median and geometric mean under the stated model.