Why Variability in Turnover Rates Matters
Two people can have the same biomarker concentration at baseline but respond differently over time because their biomarker is produced, converted, or degraded at different rates. A model with one shared set of turnover rates describes only one trajectory. A population model instead allows rates to vary between subjects and asks how that variation changes both individual predictions and the average trajectory.
This distinction is especially important for linked precursor-product systems. A change in precursor conversion can affect both the amount of precursor available and the rate at which biomarker enters the product pool. The population average of the resulting trajectories need not equal the trajectory calculated at average or typical rates.
Putting Random Effects on Turnover Rates
Consider a precursor pool \(P\) that is produced at rate \(k_{\mathrm{in}}\), degraded at rate constant \(k_{\mathrm{p}}\), and converted irreversibly to biomarker \(B\) at rate constant \(k_{\mathrm{c}}\). The biomarker is degraded at rate constant \(k_{\mathrm{b}}\). The rates \(k_{\mathrm{in}}\) have units of amount per time; the first-order rate constants have units of inverse time.
A common specification assigns a normally distributed random effect to the logarithm of each positive rate. For subject \(i\), write
The vector \(\boldsymbol{\eta}_i\) contains subject-specific deviations, and \(\Omega\) is their covariance matrix. Its diagonal elements are random-effect variances; its off-diagonal elements are covariances. Correlations are obtained by dividing a covariance by the product of the corresponding random-effect standard deviations. Because exponentiation is positive, this specification keeps all rates positive. With mean-zero normal random effects, \(k_{\mathrm{rate,pop}}\) is the median rate across subjects, not generally the arithmetic mean rate.
The lognormal specification assumes that random effects are adequately described by a multivariate normal distribution on the log-rate scale, that their distribution is stable across the modeled population, and that individual rates remain constant over the modeled time course unless the model explicitly makes them time-varying. It does not assume that the biomarker itself is normally distributed. Other distributions or structures may be appropriate if these assumptions are implausible.
Steady State Shows How Rate Variability Combines
For fixed individual rates, setting both derivatives to zero gives the subject-specific steady state. Solving the precursor equation first and substituting into the biomarker equation gives
Thus, increased production tends to increase both pools, while increased biomarker degradation tends to lower \(B_{0,i}\). The precursor degradation and conversion rates compete in the denominator for \(P_{0,i}\); conversion also appears in the numerator for \(B_{0,i}\). Consequently, the effect of changing \(k_{\mathrm{c},i}\) on the biomarker baseline is not necessarily proportional to the change in conversion rate.
Worked Example: Comparing Two Precursor-Product Baselines
Suppose two hypothetical subjects have the following constant rates, in units per day for production and per day for rate constants: \(k_{\mathrm{in}}=12\), \(k_{\mathrm{p}}=0.20\), and \(k_{\mathrm{b}}=0.50\) for both. Subject A has \(k_{\mathrm{c}}=0.30\), and subject B has \(k_{\mathrm{c}}=0.60\). Assume each has reached steady state.
For subject A, the precursor level is \(P_{0,A}=12/(0.20+0.30)=12/0.50=24\) amount units. The biomarker level is \(B_{0,A}=(0.30\times12)/(0.50\times0.50)=3.60/0.25=14.4\) amount units.
For subject B, \(P_{0,B}=12/(0.20+0.60)=12/0.80=15\) amount units. The biomarker level is \(B_{0,B}=(0.60\times12)/(0.50\times0.80)=7.20/0.40=18.0\) amount units.
The higher conversion rate lowers the precursor baseline from 24 to 15 amount units, because more precursor leaves that pool per unit time. At the same time, it raises the biomarker baseline from 14.4 to 18.0 amount units, because conversion supplies the biomarker. This example illustrates why variability in a linked-pool rate cannot be interpreted by looking at only one equation.
Population Means Are Not Typical-Subject Curves
For a model prediction \(y(t;\boldsymbol{\eta})\), the population mean is obtained by averaging the predictions over the random-effects distribution. In general, this is not the same as inserting the mean random effects into the model:
The inequality arises because turnover predictions are nonlinear functions of rates and rates are nonlinear functions of random effects. The typical-subject curve is useful as a reference, but it is not automatically representative of the average concentration or response in the population. In particular, a population average can combine subjects with different baselines and different response time courses.
A simple derivation makes this distinction explicit for a one-pool biomarker with constant production \(k_{\mathrm{in}}\) and first-order loss \(k_{\mathrm{out}}B\). Its steady state is \(B_0=k_{\mathrm{in}}/k_{\mathrm{out}}\). If independent normal random effects with variances \(\omega_{\mathrm{in}}^2\) and \(\omega_{\mathrm{out}}^2\) are placed on the log production and loss rates, then
The random part has variance \(\omega_{\mathrm{in}}^2+\omega_{\mathrm{out}}^2\) under independence. If the two random effects have covariance \(\omega_{\mathrm{in,out}}\), its variance is instead \(\omega_{\mathrm{in}}^2+\omega_{\mathrm{out}}^2-2\omega_{\mathrm{in,out}}\). For a normal random variable \(Z\), \(E[\exp(Z)]=\exp(E[Z]+\operatorname{Var}(Z)/2)\). Applying this identity gives
This result is specific to the one-pool steady-state ratio with normally distributed log-rate effects. The linked-pool steady-state expression contains a sum of rates in its denominator, so its expectation does not reduce to this same simple formula. For that model, population averages are usually calculated by numerical integration or by simulating random effects and averaging the resulting predictions.
Worked Example: Averaging Individual Turnover Predictions
Consider two hypothetical subjects whose normalized biomarker predictions one day after a sustained 60% reduction in production are \(0.8032804\) and \(0.61155\). For a one-pool turnover model, the normalized prediction after a fractional input reduction \(I\), starting at baseline, is \(B(t)/B_0=1-I(1-\exp(-k_{\mathrm{out}}t))\). Here \(I=0.60\), \(t=1\) day, and the individual loss rates are defined as \(k_{\mathrm{out},1}=-\ln((0.8032804-0.40)/0.60)\) and \(k_{\mathrm{out},2}=-\ln((0.61155-0.40)/0.60)\), respectively. These are approximately \(0.3972\) and \(1.042\) per day.
Substitution gives \(B_1(1)/B_{0,1}=1-0.60(1-\exp(-k_{\mathrm{out},1}))=0.8032804\), and \(B_2(1)/B_{0,2}=1-0.60(1-\exp(-k_{\mathrm{out},2}))=0.61155\). The arithmetic population mean for these two equally weighted subjects is \((0.8032804+0.61155)/2=1.4148304/2=0.7074152\), which rounds to \(0.7074\), or 70.74% of baseline.
The individual predictions rounded to four decimal places are 0.8033 and 0.6116, but averaging those rounded values would give 0.70745. The reported population mean above is calculated from the unrounded predictions before rounding the final result. It describes the average of these two hypothetical subject predictions at one day, not the probability that an individual will have that value.
Computing and Checking Population Predictions
For a population model with several random effects, the mean prediction at a time \(t\) can be expressed as an integral over their distribution. In practice, simulation is often straightforward: draw random-effect vectors from the specified distribution, construct each subject’s rates, solve the model, and average the subject predictions. More draws improve numerical precision but do not correct a misspecified random-effects distribution or turnover structure.
State whether random effects apply to production, precursor degradation, conversion, biomarker degradation, or a subset. Define the log-rate parameterization and units.
Give the mean vector and covariance matrix, including any correlations that are estimated or fixed to zero. Check that the covariance matrix is positive semidefinite.
For every random-effect draw, transform effects into positive rates and solve the same turnover equations with the appropriate initial conditions.
Average unrounded predictions across draws for the mean. If describing spread, report a suitable prediction interval separately from the mean.
Useful diagnostics include checking whether estimated random-effect variances are near zero or poorly determined, whether correlations are unstable, and whether the model reproduces observed between-subject spread over time. Compare individual predictions with observations and inspect whether systematic patterns remain by baseline level or sampling time. A population mean curve alone can conceal poor predictions for subgroups or individuals.
The main alternative is a fixed-effects model with one shared rate per population. It is simpler and may be adequate when between-subject variation is negligible or the data cannot support its estimation, but it cannot represent a distribution of subject trajectories. A categorical subgroup model can represent specified group differences, but does not automatically describe continuous heterogeneity within groups. Random effects provide a distributional description, at the cost of stronger assumptions and greater data requirements.
Common Mistakes and Limitations
- Calling covariance a correlation. The off-diagonal entries of \(\Omega\) are covariances. A correlation must be standardized by the two random-effect standard deviations.
- Equating the typical subject with the population mean. The typical prediction sets random effects to zero; the population mean averages nonlinear predictions over their distribution.
- Averaging rounded predictions. Carry full precision through the population average and round only the final reported result.
- Assuming more random effects always improve a model. Sparse data may not identify several rate variances and covariances separately. Overparameterization can produce unstable estimates or boundary estimates.
Random effects describe between-subject parameter variation under a model; they do not by themselves establish why subjects differ or prove that a particular biological mechanism causes the variation. They also do not replace residual error, which represents unexplained observation-level variation conditional on the model. Estimates and population predictions remain conditional on the chosen pool structure, distributional assumptions, sampling design, and available information.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- In a lognormal rate model, what does setting a subject’s random effect to zero mean, and is that rate generally the arithmetic population mean?
- For the linked-pool model, why can increasing precursor conversion lower the precursor baseline while raising the biomarker baseline?
- A covariance matrix has diagonal entries \(\omega_1^2\) and \(\omega_2^2\), and off-diagonal entry \(\omega_{12}\). How is the correlation calculated?
- Two equally weighted subjects have unrounded normalized predictions 0.8032804 and 0.61155. What is their mean, rounded to four decimal places?
- Why might the population mean prediction differ from the typical-subject prediction?
Solutions. (1) It gives the population-typical rate, which is the median under a mean-zero normal effect on the log scale; it is generally not the arithmetic mean. (2) Conversion removes precursor and supplies product, so it acts in opposite directions in the two pool balances. (3) The correlation is \(\omega_{12}/(\sqrt{\omega_1^2}\sqrt{\omega_2^2})\). (4) \((0.8032804+0.61155)/2=0.7074152\), which rounds to \(0.7074\). (5) The prediction is nonlinear in rates and random effects, so averaging individual predictions generally does not equal prediction at the typical parameter values.