The All-Scatter-Is-Assay-Error Mistake

Why the Source of Scatter Matters

A concentration that differs from a deterministic pharmacokinetic reference can differ for at least two reasons: the latent concentration may have moved away from that reference, or the assay may have measured it imperfectly. These sources are not interchangeable. An assay error changes an observation; a process perturbation changes the state and can therefore affect later concentrations.

The all-scatter-is-assay-error mistake assigns every observed deviation to measurement error and leaves the latent path on its deterministic reference. This can fit the marginal spread of measurements while producing forecasts of future latent concentrations that are much too certain. The distinction is especially important when state perturbations persist over time.

A Model That Separates State and Observation

Let \(m(t)\) be a deterministic PK reference concentration, and write the latent concentration as \(X_t=m(t)+U_t\), where \(U_t\) is a state perturbation. The observation \(Y_t\) includes both this latent deviation and measurement error. For a concrete calculation, consider a mean-reverting perturbation:

Assume \(\lambda>0\), \(\sigma>0\), and that \(U_t\) is in its stationary distribution, \(N(0,V)\), where \(V=\sigma^2/(2\lambda)\). Assume measurement errors \(\epsilon_t\) are independent across observation times, independent of the state process, and distributed as \(N(0,R)\), with \(R>0\). These assumptions describe a simple model for local, additive deviations around a reference; they are not a claim that all PK concentrations are Gaussian or can take any value without constraint.

Over a time interval \(\Delta>0\), the state has the exact transition

where the transition innovation \(\eta_t\) is independent of \(U_t\) and of measurement errors. This follows by multiplying the SDE by \(e^{\lambda t}\) and integrating: the resulting stochastic integral has variance \(\sigma^2\int_0^\Delta e^{-2\lambda(\Delta-s)}\,ds=\sigma^2(1-e^{-2\lambda\Delta})/(2\lambda)=V(1-\rho^2)\). Thus a state deviation is partly retained at the next time, while a new innovation also enters.

Two Consequences of Persistence

The stationary state variance is \(V\), while the variance of a single observed deviation \(Y_t-m(t)\) is \(V+R\). That total variance alone cannot reveal how much is process noise and how much is measurement error. Dependence across time carries additional information.

To verify the result, use the transition over \(h\): \(U_{t+h}=e^{-\lambda h}U_t+\eta\), with \(\eta\) independent of \(U_t\). Therefore \(\operatorname{Cov}(U_t,U_{t+h})=e^{-\lambda h}\operatorname{Var}(U_t)=V e^{-\lambda h}\). For \(h>0\), the two measurement errors are independent of each other and of the state, so every covariance term involving an error is zero. At one time, independence instead gives \(\operatorname{Var}(U_t+\epsilon_t)=V+R\). This proves the result. \(\square\)

A second consequence concerns forecasting. Suppose the observed deviation at time \(t\) is \(y=Y_t-m(t)\), and no other observations are being conditioned on. Since \(U_{t+h}\) and \(Y_t-m(t)\) are jointly normal, the conditional mean and variance are

For completeness, the covariance of the future state with the observed deviation is \(V e^{-\lambda h}\), and the observed deviation has variance \(V+R\). For jointly normal variables, conditioning on a scalar variable with variance \(s^2\) gives conditional mean equal to the covariance divided by \(s^2\), times its observed deviation, and conditional variance equal to the original variance minus the squared covariance divided by \(s^2\). Here the future-state variance is \(V\), so substitution gives both expressions. In particular, the forecast variance can also be written as

The first term is new state variation accumulated after time \(t\); the second is the remaining uncertainty about the state at time \(t\), carried forward. Both are nonnegative. An assay-only model that fixes the latent state at \(m(t)\) has neither contribution and cannot represent this uncertainty.

Worked Comparisons and Diagnostics

In practice, inspect time-ordered deviations from the reference rather than only their overall spread. Persistent runs of positive or negative deviations, or positive covariance at nonzero lags, are evidence that independent assay error alone may be inadequate. Compare candidate models using residual time patterns and predictive checks, and check whether estimated process and assay variances are identifiable from the available sampling times. The covariance calculation above is a model implication, not a guarantee that a small data set will estimate it precisely.

Model Choice, Limitations, and a Common Mistake

Common mistake: “The observations vary around the deterministic PK curve, so the variance of those residuals is the assay-error variance; the latent future concentration is therefore known.” This reasoning equates a marginal observation variance with a measurement-error variance and ignores temporal dependence. In the example, the marginal variance \(0.05\) is the sum \(V+R\), not \(R\). The correction is to distinguish the state and observation components, then assess whether the observed time pattern supports persistence before forecasting latent concentrations.

The OU model is a useful calculation, not a universal description of PK variability. It assumes a stationary, mean-reverting Gaussian perturbation with constant parameters, independent measurement errors, and an additive observation scale. Concentrations near zero, proportional errors, dose-dependent dynamics, parameter uncertainty, or correlated assay errors may require different models. If the data do not contain enough temporal information, \(V\), \(R\), and \(\lambda\) can be difficult to distinguish; a fitted decomposition should then be reported with that uncertainty rather than treated as definitive.