From Log Deviations to Multiplicative Differences
In a nonlinear mixed-effects model, an exponential random effect gives individuals different parameter values by multiplying a typical value by a positive factor. This is useful for parameters such as clearance, volume, or rate constants: the resulting individual values remain positive, and the random effect has a direct ratio interpretation. The key is to distinguish the random effect’s scale from the parameter’s scale. A variance on the log scale is not itself a coefficient of variation on the parameter scale.
Consider a population parameter \(P\), with typical value \(P_{\mathrm{typ}}>0\), for individual \(i\). Let \(\eta_i\) be that individual’s random effect. In the standard exponential specification, \(\eta_i\) has mean zero and variance \(\omega^2\), and is normally distributed. The individual parameter is \(P_i=P_{\mathrm{typ}}\exp(\eta_i)\). The log-scale distributional assumption matters for the exact mean, variance, and geometric interpretations developed below.
Taking logarithms gives \(\log P_i=\log P_{\mathrm{typ}}+\eta_i\). Consequently, \(\log P_i\) is normal with mean \(\log P_{\mathrm{typ}}\) and variance \(\omega^2\). The distribution of \(P_i\) is lognormal. Because \(\eta_i\) has median zero, the multiplicative deviation has median one: half the population is below the typical value and half above it, under this model. The typical value is also the population geometric mean, since the geometric mean is obtained by exponentiating the mean log value. It is not generally the arithmetic mean.
Geometric Means and Multiplicative Deviations
A geometric mean summarizes positive values by averaging their logarithms and exponentiating the result. For a population of individual parameters, its model-based value is \(\exp\{E[\log P_i]\}\). Since \(E[\log P_i]=\log P_{\mathrm{typ}}+E[\eta_i]=\log P_{\mathrm{typ}}\), the population geometric mean is \(P_{\mathrm{typ}}\). The observed geometric mean in a finite sample need not equal \(P_{\mathrm{typ}}\) exactly; it varies with the sampled random effects.
The model is especially easy to interpret through ratios. A random effect of \(\eta_i=0.2\) does not mean that the parameter is 0.2 units higher. It means that the parameter is multiplied by \(\exp(0.2)\), or about 1.221. A negative random effect gives a factor below one. Equal-sized positive and negative log deviations are reciprocal factors: \(\exp(a)\exp(-a)=1\). This symmetry is on the log scale, not in the absolute parameter differences.
Worked Example: Reading One Individual’s Multiplicative Deviation
Suppose a hypothetical model has a typical clearance of \(5.00\) L/h, and one individual has \(\eta_i=-0.25\). Calculate that individual’s clearance and interpret the deviation.
The model gives \(CL_i=CL_{\mathrm{typ}}\exp(\eta_i)\). Substituting the values, \(CL_i=5.00\exp(-0.25)=5.00(0.7788)=3.894\) L/h.
The individual’s clearance is about \(0.7788\) times the typical clearance, or about \(22.1\%\) lower: \((0.7788-1)\times100\%=-22.12\%\). The random effect is not a decrease of \(0.25\) L/h. It is a log-scale value that implies a multiplicative decrease. The parameter remains positive.
What Log-Scale Variance Implies on the Parameter Scale
The random-effect variance \(\omega^2\) controls the spread of log parameters. Under the normal assumption, it also determines the arithmetic mean and relative variability of the parameters. The derivation follows by evaluating the exponential moment of a normal random variable. For \(t\) real, completing the square in the normal density gives \(E[\exp(t\eta_i)]=\exp(t^2\omega^2/2)\). In particular, using \(t=1\) yields the mean of the multiplicative factor, and using \(t=2\) yields its second moment.
The coefficient of variation (CV), defined as the standard deviation divided by the arithmetic mean, therefore satisfies \(\mathrm{CV}=\sqrt{\exp(\omega^2)-1}\). This is a relative measure and has no units. Conversely, if a target population CV is specified under this model, then \(\omega^2=\log(1+\mathrm{CV}^2)\). These relationships are exact under the stated lognormal model; treating \(\omega\) itself as the CV is only a small-variability approximation.
A second useful summary is the geometric standard deviation, \(\exp(\omega)\). Since \(\eta_i\) has standard deviation \(\omega\), values one standard deviation above and below the mean of the log parameter correspond to multiplying the typical value by \(\exp(\omega)\) and \(\exp(-\omega)\), respectively. If the normal assumption holds, about 68% of individual parameters lie between those two factors of \(P_{\mathrm{typ}}\). The interval is multiplicatively symmetric around the typical value, not additively symmetric.
Worked Example: Translating Log-Scale Variance Into Parameter Variability
Suppose a hypothetical concentration parameter has typical value \(8.00\) mg/L and random-effect variance \(\omega^2=0.1484\). Calculate the population geometric mean, arithmetic mean, CV, and one-log-standard-deviation multiplicative factors.
The population geometric mean is the typical value, \(8.00\) mg/L. For the arithmetic mean, \(E[P_i]=8.00\exp(0.1484/2)=8.00\exp(0.0742)\approx8.616\) mg/L. It exceeds the geometric mean because the lognormal distribution’s upper values raise the arithmetic average.
The coefficient of variation is \(\sqrt{\exp(0.1484)-1}=\sqrt{1.15998-1}=\sqrt{0.15998}\approx0.4000\), or about \(40.0\%\). It describes the population standard deviation as a proportion of the arithmetic mean, not a percentage difference from the typical value.
Here \(\omega=\sqrt{0.1484}\approx0.3852\). The one-standard-deviation factors are \(\exp(0.3852)\approx1.470\) and \(\exp(-0.3852)\approx0.6803\). Thus, under the normal random-effect assumption, the corresponding interval is approximately \(8.00(0.6803)=5.442\) to \(8.00(1.470)=11.76\) mg/L. It is a model-based central interval of about 68%, not a confidence interval for the typical value.
Worked Example: Choosing Log-Scale Variance for a Target CV
A hypothetical model requires a population CV of \(30.0\%\) for a positive parameter. Find the log-scale variance and standard deviation, then interpret the geometric standard deviation.
Under the exponential random-effects model, \(\mathrm{CV}^2=\exp(\omega^2)-1\). Solving for the variance gives \(\omega^2=\log(1+\mathrm{CV}^2)=\log(1+0.300^2)=\log(1.09)=0.08618\). Therefore \(\omega=\sqrt{0.08618}=0.2936\).
The geometric standard deviation is \(\exp(0.2936)=1.341\). A one-log-standard-deviation increase corresponds to multiplying the typical value by 1.341; the matching decrease corresponds to multiplying by its reciprocal, about \(0.7456\). As a check, the arithmetic mean is \(P_{\mathrm{typ}}\exp(0.08618/2)\approx1.044P_{\mathrm{typ}}\), slightly above the geometric mean. The variance has been selected to give a 30.0% CV exactly under the assumed model.
When This Interpretation Helps—and When It Can Mislead
Exponential random effects are natural when heterogeneity is plausibly proportional: an absolute difference of 1 L/h has a different relative meaning at clearances of 2 and 20 L/h, while a ratio has the same interpretation at either level. The model also guarantees positive parameters and lets log-scale effects combine additively. For example, two log-scale contributions \(\eta_1\) and \(\eta_2\) produce a combined multiplier \(\exp(\eta_1+\eta_2)=\exp(\eta_1)\exp(\eta_2)\).
An additive natural-scale alternative, \(P_i=P_{\mathrm{typ}}+b_i\), describes absolute deviations in the units of \(P\). It may be appropriate when absolute differences are the scientific premise, but it can assign positive probability to impossible negative values unless constrained. The exponential model instead describes ratios, but can produce a strongly right-skewed distribution for large \(\omega^2\). The two models answer different variability questions; they are not interchangeable merely by changing units.
A common mistake is to report \(\omega^2\) as a percentage of variability, or to say that \(\omega=0.30\) means a 30% CV. The correction is to state the scale and, when needed, transform it: for \(\omega=0.30\), the exact CV is \(\sqrt{\exp(0.30^2)-1}\approx0.307\), or \(30.7\%\). The approximation CV \(\approx\omega\) is reasonable only when \(\omega\) is small.
Another pitfall is to confuse the typical value with the arithmetic mean. For \(\omega^2>0\), \(E[P_i]=P_{\mathrm{typ}}\exp(\omega^2/2)>P_{\mathrm{typ}}\). Reporting one as the other changes the interpretation, particularly for appreciable variability. Always label whether a reported value is a typical value/geometric mean, a median, or an arithmetic mean, and include the parameter’s units.
The exact conversions rely on the normal distribution for \(\eta_i\), not just on the equation \(P_i=P_{\mathrm{typ}}\exp(\eta_i)\). If the random-effect distribution is not normal, a zero mean on the log scale can still identify the geometric mean, but the moment formulas and the stated one-standard-deviation coverage need not hold. Diagnostics should therefore include checking whether the fitted model and data support the assumed distribution and whether the implied parameter values are scientifically plausible. Sparse data or weak information about between-individual variability can also make \(\omega^2\) uncertain; a precise-looking conversion does not remove uncertainty in its estimate.
Check Your Understanding
Use the model assumptions stated in this tutorial. Solutions follow each question.
- If \(\eta_i=0.10\), what is the multiplicative deviation, and what does it mean? Solution: \(\exp(0.10)=1.105\), so the parameter is about 1.105 times the typical value, or 10.5% higher.
- For \(\omega^2=0.04\), calculate the arithmetic mean as a multiple of the typical value. Solution: \(\exp(0.04/2)=\exp(0.02)=1.020\), so the arithmetic mean is about 1.020 times the typical value.
- For \(\omega=0.20\), is the CV exactly 20%? Solution: No. It is \(\sqrt{\exp(0.20^2)-1}=\sqrt{\exp(0.04)-1}\approx0.2020\), or 20.2%.
- What is the population geometric mean in the model, and why? Solution: It is \(P_{\mathrm{typ}}\), because \(\exp\{E[\log P_i]\}=\exp\{\log P_{\mathrm{typ}}+E[\eta_i]\}=P_{\mathrm{typ}}\).
- Why can the geometric one-standard-deviation interval be multiplicatively symmetric but not additively symmetric? Solution: Its endpoints are \(P_{\mathrm{typ}}\exp(-\omega)\) and \(P_{\mathrm{typ}}\exp(\omega)\), whose ratios to \(P_{\mathrm{typ}}\) are reciprocals. Their absolute distances from \(P_{\mathrm{typ}}\) are generally unequal.