Choosing Initial Values for Random-Effect Variances

Why Random-Effect Variance Starts Need Interpretation

A random-effect variance is not just a number supplied to an optimizer. In a typical pharmacokinetic model, it controls the spread of individual parameters around a population-typical value. A start that is too small can imply that individuals are almost identical; one that is too large can imply that many individuals have values far from the typical value. Either may make the initial fit implausible and can complicate estimation.

The goal is not to guess the final variance. It is to choose a finite, positive starting value that represents a defensible degree of variability and gives the estimation procedure a reasonable place to begin. The earlier material on constructing initial values from pharmacokinetic summaries can help establish plausible typical parameter values; here the focus is the variability around those values.

Definition: For a positive individual parameter \(P_i\), a common lognormal random-effects model is \(P_i=P_{\mathrm{typ}}\exp(\eta_i)\), where \(P_{\mathrm{typ}}>0\) is the population-typical value and \(\eta_i\sim N(0,\omega^2)\). The random effect has variance \(\omega^2\) and standard deviation \(\omega\). Under this model, \(P_{\mathrm{typ}}\) is the median of the individual-parameter distribution, not generally its arithmetic mean.

The variance \(\omega^2\) is measured on the log scale. It is not the squared coefficient of variation of \(P_i\), and the two quantities should not be treated as interchangeable. The coefficient of variation (CV) of a positive random variable is its standard deviation divided by its mean. Under the lognormal model, the CV of \(P_i\) is determined by \(\omega^2\), regardless of the value of \(P_{\mathrm{typ}}\).

Converting an Expected CV to a Log-Scale Variance

Suppose the intended initial distribution should have a particular CV. For \(\eta_i\sim N(0,\omega^2)\), the lognormal moments give \(E[\exp(\eta_i)]=\exp(\omega^2/2)\) and \(\operatorname{Var}[\exp(\eta_i)]=\exp(\omega^2)(\exp(\omega^2)-1)\). Multiplication by \(P_{\mathrm{typ}}\) scales the mean and standard deviation by the same factor, so it does not change the CV.

The following derivation shows the conversion. The mean of \(P_i\) is \(P_{\mathrm{typ}}\exp(\omega^2/2)\), and its variance is \(P_{\mathrm{typ}}^2\exp(\omega^2)(\exp(\omega^2)-1)\). Therefore:

\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}.

Solving for the variance gives the initial-value conversion:

\omega^2=\ln\!\left(1+\operatorname{CV}(P_i)^2\right).

Use the CV as a proportion in this formula: a CV of 30% is \(0.30\), not \(30\). For small CVs, \(\ln(1+\operatorname{CV}^2)\) is approximately \(\operatorname{CV}^2\), so \(\omega\) is approximately the CV. This is a useful rough check, not a replacement for the exact conversion when variability is moderate or large.

Worked Example: Starting From a 30% CV

Suppose a positive clearance parameter has a plausible typical value of \(10\ \mathrm{L/h}\), and the intended starting distribution should have a between-subject CV of 30%. Find the log-scale variance and interpret the implied central 95% range.

Convert the CV. Enter \(0.30\) in the conversion:

\omega^2=\ln(1+0.30^2)=\ln(1.09)=0.08618, \qquad \omega=\sqrt{0.08618}=0.2936.

Thus a reasonable initial variance is approximately \(0.0862\), with log-scale standard deviation approximately \(0.294\). These are two ways to describe the same start; the variance is the quantity used when the model parameter is specified as a variance.

Translate the start to individual values. Since \(\log(P_i/P_{\mathrm{typ}})=\eta_i\), the central 95% range under the assumed normal random effect is approximately the typical value multiplied by \(\exp(\pm1.96\omega)\):

\left[\exp(-1.96\times0.2936),\ \exp(1.96\times0.2936)\right] \approx [0.562,\ 1.778].

For \(P_{\mathrm{typ}}=10\ \mathrm{L/h}\), this corresponds to approximately \(5.62\) to \(17.8\ \mathrm{L/h}\). It is a model-implied distributional range, not a guaranteed range for a particular sample. Under this lognormal model, the median is \(10\ \mathrm{L/h}\), while the arithmetic mean is \(10\exp(0.08618/2)\approx10.44\ \mathrm{L/h}\).

Choose and Check a Starting Value

A practical choice begins with the scale of variability that is plausible for the parameter and model. If there is a defensible expected CV, convert it to \(\omega^2\). If variability is expressed instead as a plausible multiplicative spread, translate that spread into log units before choosing the variance. For example, a factor \(r>1\) above the typical value at a specified normal quantile \(z\) implies \(\omega=\ln(r)/z\), provided the intended upper quantile and lognormal model are clear.

After choosing a variance, check the distribution it implies. The central 95% range of the lognormal parameter, relative to its median, is \([\exp(-1.96\omega),\exp(1.96\omega)]\). This check often exposes a mistaken variance-versus-standard-deviation entry or a start that does not match the intended variability. The table shows some conversions for orientation; it does not define universal acceptable ranges.

Expected CV	Log-scale variance \(\omega^2\)	Log-scale SD \(\omega\)
10%	\(\ln(1.01)=0.00995\)	0.0998
20%	\(\ln(1.04)=0.03922\)	0.1980
30%	\(\ln(1.09)=0.08618\)	0.2936
50%	\(\ln(1.25)=0.2231\)	0.4724

Worked Example: Converting a 15% CV for Clearance

For a hypothetical model, take a typical clearance of \(4.0\ \mathrm{L/h}\) and an expected between-subject CV of 15%. Calculate a candidate variance start and check the implied distribution.

Calculate the variance and SD. The CV is \(0.15\):

\omega^2=\ln(1+0.15^2)=\ln(1.0225)=0.02225, \qquad \omega=\sqrt{0.02225}=0.1492.

A candidate variance start is therefore \(0.02225\). The approximate central 95% multipliers are \(\exp(\pm1.96\times0.1492)=\exp(\pm0.2924)\), or approximately \(0.7464\) and \(1.340\). Multiplying by \(4.0\ \mathrm{L/h}\) gives a range of approximately \(2.99\) to \(5.36\ \mathrm{L/h}\). The arithmetic is consistent with a relatively modest spread around the median of \(4.0\ \mathrm{L/h}\).

This is an initialization judgment, not evidence that the fitted CV will be 15%. The data and model determine the eventual estimate; this calculation only gives the optimizer a plausible starting variance.

Worked Example: Recognizing an Overly Broad Start

Suppose a hypothetical parameter has a typical value of \(8.0\) in its measurement units. A proposed log-scale variance start is \(\omega^2=0.50\), although the available scientific expectation is roughly a 20% CV. Determine what the proposed start implies.

Compute the implied CV.

\operatorname{CV} = \sqrt{\exp(0.50)-1} = \sqrt{0.6487} = 0.8054.

The start implies a CV of about 80.5%, far above the intended 20%. Its log-scale SD is \(\omega=\sqrt{0.50}=0.7071\), giving 95% multipliers \(\exp(\pm1.96\times0.7071)\approx[0.2501,3.999]\). At a median of 8.0, the implied central range is about 2.00 to 32.0 in the parameter's units. This start is not consistent with the stated expectation. A variance near \(\ln(1+0.20^2)=0.03922\) would better reflect that expectation as an initial value.

Multiple Random Effects and Practical Checks

For several positive individual parameters, apply the marginal conversion separately to each parameter with its own expected CV. If the random effects are correlated, their covariance entries are related to their marginal SDs by \(\operatorname{Cov}(\eta_j,\eta_k)=\rho_{jk}\omega_j\omega_k\). Choosing marginal variances does not by itself justify a particular correlation. Starts for correlations require separate scientific and numerical consideration; a variance start should not be inflated merely to compensate for an uncertain correlation.

Before fitting, check the following:

Units and parameterization: confirm that the model expects a variance, not an SD, CV, or standard error.
Scale: confirm that the random effect is on the log scale if using the lognormal CV conversion.
Distributional implication: translate the proposed variance into a CV or a range of individual values and assess its plausibility in context.
Parameter bounds: use a finite, permissible starting value. A strictly positive start is generally sensible when a nonzero variance is being estimated, but it does not rule out a final estimate at the boundary.
Model meaning: check that the assumed random-effects structure and residual model are appropriate; a plausible start cannot repair a misspecified model.

If estimation behaves differently after a modest, defensible change in the variance start, record that sensitivity and investigate whether the model is weakly identified or the optimization is unstable. Do not select a start simply because it produces a preferred final estimate. Conversely, convergence from one start does not establish that the variance is well estimated or that the model is adequate.

Common Mistakes and Limitations

Entering the CV as the variance. A 30% CV is \(0.30\), but the corresponding log-scale variance is \(\ln(1.09)=0.08618\), not \(0.30\).
Confusing variance and SD. For a 30% CV, the log-scale SD is \(0.2936\), while the variance is \(0.08618\). Squaring or taking a square root changes the quantity.
Assuming a universal acceptable variance. The appropriate start depends on the parameter, population, model, and available knowledge. The conversions are guides to coherent initialization, not fixed standards.
Treating the start as a prior or result. An optimizer's initial value is a computational input. It is not a Bayesian prior, a fitted estimate, or proof that the chosen variability is true.
Ignoring the model assumption. The CV conversion above relies on a normal random effect applied exponentially. It does not automatically apply to additive random effects or to other parameter distributions.

These checks have limits. A plausible marginal CV cannot establish that the data contain enough information to estimate the variance, and a reasonable initial distribution cannot resolve confounding between model components. The distributional range is a diagnostic of the start under the assumed model, not a diagnostic that the model itself is correct.

Key takeaway: For a positive parameter with \(P_i=P_{\mathrm{typ}}\exp(\eta_i)\) and \(\eta_i\sim N(0,\omega^2)\), convert an expected CV using \(\omega^2=\ln(1+\operatorname{CV}^2)\). Then check the implied individual-parameter spread and verify that the software expects a log-scale variance. Treat the result as a defensible starting value, not a claim about the fitted variability.

Check Your Understanding

Answer the questions, then compare your work with the brief solutions below.

A positive parameter is modeled lognormally with an intended CV of 20%. Calculate the initial \(\omega^2\) and \(\omega\).
Explain why a variance start of \(0.04\) is not a 4% CV under the lognormal random-effects model.
For \(\omega^2=0.01\), calculate the implied CV to approximately one decimal place as a percentage.
If the model uses \(P_i=P_{\mathrm{typ}}\exp(\eta_i)\), is \(P_{\mathrm{typ}}\) the arithmetic mean of \(P_i\)? State the mean in terms of \(P_{\mathrm{typ}}\) and \(\omega^2\).
A colleague proposes a very broad variance start because a fit from a smaller start failed. Give two checks to make before accepting the broader start.

Solutions. (1) \(\omega^2=\ln(1+0.20^2)=\ln(1.04)=0.03922\), and \(\omega=\sqrt{0.03922}=0.1980\). (2) The implied CV is \(\sqrt{\exp(0.04)-1}\approx0.2020\), or about 20.2%, not 4%. (3) \(\sqrt{\exp(0.01)-1}\approx0.1003\), or about 10.0%. (4) No. The median is \(P_{\mathrm{typ}}\); the arithmetic mean is \(P_{\mathrm{typ}}\exp(\omega^2/2)\). (5) Translate the proposed variance into an implied CV or individual-value range and compare it with scientific expectations; also verify that the model and software parameterization use the intended log-scale variance. A failed fit alone does not justify an implausible starting distribution.