Why a Derived Quantity Has Its Own Variability
In a one-compartment model with first-order elimination, the elimination half-life is determined by volume of distribution and clearance. It is not an independent parameter: a subject’s half-life is the ratio of that subject’s volume to that subject’s clearance, multiplied by \(\ln 2\). Consequently, its population distribution depends on the variability in both parameters and on how their random effects covary.
This distinction matters in population pharmacokinetics because neither the clearance variance nor the volume variance alone describes how widely half-lives vary. For example, when clearance and volume random effects move together, some of their variation can cancel in the ratio. When they move in opposite directions, variation in the ratio can increase. The derivation below makes that relationship explicit.
Random Effects on Clearance and Volume
Assume clearance and volume follow correlated lognormal random-effects models:
Here, \(CL_{\mathrm{pop}}\) and \(V_{\mathrm{pop}}\) are positive typical values, and the subject-level random effects have a joint bivariate normal distribution with mean zero:
Thus, \(\omega_{CL}^{2}\) and \(\omega_{V}^{2}\) are the variances of the log-scale effects, and \(\omega_{CL,V}\) is their covariance. The covariance matrix must be positive semidefinite; equivalently, \(|\omega_{CL,V}|\leq \omega_{CL}\omega_V\). The correlation is \(\rho_{CL,V}=\omega_{CL,V}/(\omega_{CL}\omega_V)\) when both variances are positive.
Substituting the subject-specific clearance and volume into the definition of half-life gives
Define \(t_{1/2,\mathrm{typ}}=\ln(2)V_{\mathrm{pop}}/CL_{\mathrm{pop}}\), the half-life calculated from the typical clearance and volume. The log of the individual half-life is therefore the log typical half-life plus a contrast of two random effects:
Deriving the Distribution and Summaries
A linear combination of jointly normal random variables is normal. Its mean is the same linear combination of their means, and its variance follows from the variance rule for a difference:
Write this variance as \(\sigma_{h}^{2}\). Then \(\log(t_{1/2})\) is normal with mean \(\log(t_{1/2,\mathrm{typ}})\) and variance \(\sigma_h^2\). The corresponding half-life summaries are:
The median equals the typical half-life because the random-effects contrast has mean zero on the log scale. The arithmetic mean is larger when there is nonzero variability; this is the usual right-skew of a lognormal distribution. The coefficient of variation (CV) is dimensionless, while the mean, median, and standard deviation of half-life have units of time. For a lognormal variable, its standard deviation is its mean multiplied by its CV.
Worked Example: Calculating Half-Life Variability
Suppose a hypothetical model has \(CL_{\mathrm{pop}}=6\) L/h and \(V_{\mathrm{pop}}=30\) L. Let \(\omega_{CL}^{2}=0.09\), \(\omega_{V}^{2}=0.16\), and \(\rho_{CL,V}=0.50\). Find the typical half-life, the log-scale variance, the arithmetic mean half-life, and its CV.
First calculate the covariance: \(\omega_{CL,V}=\rho_{CL,V}\omega_{CL}\omega_V=0.50\sqrt{0.09}\sqrt{0.16}=0.50(0.30)(0.40)=0.0600\). The typical half-life is
The log-scale variance is
Hence the median is \(3.466\) h, and the arithmetic mean is \(3.4657\exp(0.130/2)=3.4657\exp(0.065)=3.698\) h. The CV is \(\sqrt{\exp(0.130)-1}=\sqrt{0.1388}=0.3726\), or about \(37.3\%\). The standard deviation is approximately \(3.698(0.3726)=1.378\) h. These results use the assumed joint-normal log-scale effects; the numerical summaries do not follow from the two marginal variances without the covariance.
Dependence Between Half-Life and Its Components
Half-life is mathematically constructed from clearance and volume, so it is generally dependent on both. The direction and strength of that dependence are not determined merely by the ratio formula: they also depend on the joint random-effects structure. Under the assumptions above, let \(d=\eta_V-\eta_{CL}\). For a normally distributed pair of effects, the covariance between exponentiated linear combinations can be calculated using the lognormal covariance identity.
For example, the half-life–clearance covariance is negative when \(\omega_{CL,V}<\omega_{CL}^{2}\), zero when the two are equal, and positive when \(\omega_{CL,V}>\omega_{CL}^{2}\). A negative association is intuitive if volume is held fixed, but it is not guaranteed when volume and clearance vary jointly. Similarly, half-life can be positively or negatively associated with volume depending on \(\omega_V^{2}-\omega_{CL,V}\). These covariance results describe the population model; they do not establish a causal effect of changing one parameter while holding everything else fixed.
Worked Example: Dependence on Clearance and Volume
Use the random-effects variances and covariance from the preceding hypothetical example. Calculate the correlations of half-life with clearance and with volume. The log-scale variance of half-life is \(0.130\), while the log-scale variances for clearance and volume are \(0.09\) and \(0.16\).
For lognormal variables formed from normal effects, the correlation of exponentials is \(\{\exp(c)-1\}/\sqrt{[\exp(a)-1][\exp(b)-1]}\), where \(a\) and \(b\) are the two log-scale variances and \(c\) is their log-scale covariance. For half-life and clearance, \(c=\omega_{CL,V}-\omega_{CL}^{2}=0.0600-0.0900=-0.0300\). Therefore,
For half-life and volume, \(c=\omega_V^{2}-\omega_{CL,V}=0.160-0.0600=0.100\). Thus,
Under this particular hypothetical model, half-life has a modest negative correlation with clearance and a positive correlation with volume. Those signs and magnitudes arise from the specified covariance structure, not from a universal rule that half-life must be negatively correlated with clearance and positively correlated with volume.
How the Clearance–Volume Correlation Changes Half-Life
The covariance contribution is especially clear when the marginal variances are held fixed. Since \(\omega_{CL,V}=\rho_{CL,V}\omega_{CL}\omega_V\), increasing the clearance–volume correlation decreases the variance of their difference \(\eta_V-\eta_{CL}\). Positive correlation tends to make the two effects move together, reducing their contrast; negative correlation tends to make them move in opposite directions, increasing it.
Worked Example: Same Marginal Variability, Different Correlation
Keep the hypothetical marginal variances \(\omega_{CL}^{2}=0.09\) and \(\omega_V^{2}=0.16\), and the typical half-life \(3.4657\) h, but change the random-effects correlation from \(0.50\) to \(-0.50\). The covariance becomes \((-0.50)(0.30)(0.40)=-0.0600\), so
The median remains \(3.4657\) h because the typical values and zero-mean log effects are unchanged. The arithmetic mean is \(3.4657\exp(0.370/2)=3.4657\exp(0.185)=4.170\) h, and the CV is \(\sqrt{\exp(0.370)-1}=\sqrt{0.4477}=0.6691\), or \(66.9\%\). Compared with the \(37.3\%\) CV when the correlation was \(0.50\), half-life variability is substantially greater even though the marginal clearance and volume variances are exactly the same.
Assumptions, Diagnostics, and Alternatives
The lognormal result requires a specific set of assumptions: half-life is defined by the one-compartment, first-order elimination relationship; clearance and volume are positive and have the stated exponential random-effects forms; and their random effects are jointly normal with finite covariance. The covariance matrix must be valid. Under those conditions, the distribution and moment formulas follow directly. If the random effects are not jointly normal, the identity \(\log(t_{1/2})=\log(t_{1/2,\mathrm{typ}})+\eta_V-\eta_{CL}\) still holds, but the contrast need not be normal, and the lognormal summaries need not apply.
Useful checks include verifying that the estimated covariance matrix is positive semidefinite, examining the estimated random-effects distributions, and computing the half-life distribution from simulated paired clearance and volume values. Pairing matters: independently simulating clearance and volume would erase their fitted covariance and could misstate the derived variability. For a fitted model, compare the distribution of simulated half-lives with the algebraic result when the joint-normal assumptions hold. Large disagreement can indicate an implementation or calculation error, or that the simulation includes features beyond the assumptions used in the formula.
The direct algebraic calculation is exact under the stated model and is preferable when its assumptions hold. Simulation is a flexible alternative for describing derived quantities under more complicated models or non-normal random effects, but its accuracy depends on the number of draws and on preserving the dependence structure. A first-order approximation such as the delta method can be useful for other derived functions, especially when an exact distribution is unavailable; for this ratio of lognormal parameters, it is unnecessary under the assumptions above.
Another possible model is to assign a random effect directly to half-life. That approach is not automatically equivalent to assigning random effects to clearance and volume. To preserve the implied half-life distribution, its log-scale variance must equal \(\omega_V^2+\omega_{CL}^2-2\omega_{CL,V}\), and the joint relationships with clearance and volume must also be compatible. Treating half-life as an unrelated random effect can therefore discard important dependence or create an incoherent parameterization.
Check Your Understanding
Try each question, then compare your reasoning with the solutions below.
- If \(\omega_{CL}^{2}=0.04\), \(\omega_V^{2}=0.09\), and \(\omega_{CL,V}=0.020\), what is the log-scale variance of half-life?
- For zero-mean jointly normal log-scale effects, is the typical half-life the median or the arithmetic mean of individual half-lives? Explain briefly.
- Suppose \(\omega_{CL}^{2}=0.04\), \(\omega_V^{2}=0.09\), and \(\omega_{CL,V}=0.060\). What is \(\operatorname{Cov}(\eta_V-\eta_{CL},\eta_{CL})\), and what does its sign imply about half-life–clearance covariance?
- Why should clearance and volume be simulated as paired values when calculating half-life variability from a fitted joint random-effects model?
Solutions. (1) \(\sigma_h^2=0.09+0.04-2(0.020)=0.090\). (2) It is the median: the log half-life is normal with mean \(\log(t_{1/2,\mathrm{typ}})\), so exponentiating that mean gives the median. The arithmetic mean is \(t_{1/2,\mathrm{typ}}\exp(\sigma_h^2/2)\). (3) The covariance is \(\omega_{CL,V}-\omega_{CL}^{2}=0.060-0.040=0.020\), which is positive. The lognormal covariance identity then gives a positive half-life–clearance covariance, provided the variances are nonzero. (4) The fitted effects can be correlated. Paired simulation preserves their covariance; independent simulation removes it and can consequently misstate the variance of \(\eta_V-\eta_{CL}\), and thus of half-life.