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Therapeutic Drug Monitoring and Individual Predictions · Tutorial 545 of 599

Propagating Assay and Residual Error Into Forecasts

See how to carry assay and residual error through a patient-level concentration likelihood and into forecasts without counting the same source of variability twice.

Advanced 11 min read

What You'll Learn

  • Combine independent assay and residual variances on the scale specified by the observation model.
  • Calculate concentration likelihoods and compare candidate individual parameter values.
  • Explain how total observation error changes posterior parameter uncertainty and updating.
  • Separate uncertainty in a model-predicted concentration from variability in a future measured concentration.
  • Construct predictive intervals under stated normal-model assumptions.
  • Recognize double counting, scale mismatch, and other errors in error propagation.

Why Error Propagation Matters in Individual Forecasts

A measured concentration is not usually an exact readout of a patient’s underlying pharmacokinetic state. It may differ from the model-predicted concentration because of assay imprecision and because the observation model allows residual variability. Those sources affect how much a concentration should influence an individual fit. They also affect how wide a forecast should be.

The key distinction is between uncertainty about the patient’s parameters and variation in a concentration around the model prediction. A patient’s parameter posterior reflects the information in the observed concentrations, including their measurement error. A future concentration then has additional uncertainty from future observation error. The introductory material on posterior parameter uncertainty and predictive uncertainty distinguishes these two kinds of uncertainty; here the focus is how assay and residual error enter their calculations.

Definition: Let \(y_j\) be an observed concentration and \(f_j(\theta)\) the model-predicted typical concentration at that patient’s sampling time and dosing history, for individual parameter vector \(\theta\). In an additive normal observation model, $$ y_j=f_j(\theta)+\epsilon_{\mathrm{res},j}+\epsilon_{\mathrm{assay},j}, $$ where residual error and assay error are independent, mean-zero terms with variances \(\sigma_{\mathrm{res},j}^{2}\) and \(\sigma_{\mathrm{assay},j}^{2}\). The total observation-error variance is \(v_j=\sigma_{\mathrm{res},j}^{2}+\sigma_{\mathrm{assay},j}^{2}\).

This definition requires care. The variances add because the two error terms are assumed independent and expressed on the same scale. If they are correlated, their covariance must also be included: \(\operatorname{Var}(\epsilon_{\mathrm{res}}+\epsilon_{\mathrm{assay}})=\sigma_{\mathrm{res}}^2+\sigma_{\mathrm{assay}}^2+2\operatorname{Cov}(\epsilon_{\mathrm{res}},\epsilon_{\mathrm{assay}})\). If the observation model is proportional, combined error is not generally a constant variance in concentration units. The specified error model, rather than a convenient default, determines the likelihood.

From Error Variance to Concentration Likelihood

Under the additive normal model, conditional on \(\theta\), the observation has distribution \(y_j\mid\theta\sim N(f_j(\theta),v_j)\), where the second argument is the variance. Its likelihood contribution is the normal density evaluated at the observed concentration:

$$ L_j(\theta)=\frac{1}{\sqrt{2\pi v_j}} \exp\left[-\frac{\{y_j-f_j(\theta)\}^{2}}{2v_j}\right]. $$

The density has units inverse to the concentration unit. It is not a probability that the exact continuous value occurred. When observations are conditionally independent given the individual parameters and model, the full likelihood is the product of their density contributions. In practical Bayesian fitting, this likelihood is combined with the individual’s parameter prior to obtain a posterior distribution.

Worked Example: Combining Error and Comparing Two Predictions

Suppose, hypothetically, an assay reports \(y=10.8\) mg/L. The additive residual-error variance is \(0.80\;(\mathrm{mg/L})^2\), and independent assay error has standard deviation \(0.30\) mg/L. Compare typical predictions of \(10.0\) mg/L and \(11.5\) mg/L using the normal likelihood.

First convert assay standard deviation to variance: \(0.30^2=0.09\;(\mathrm{mg/L})^2\). Thus \(v=0.80+0.09=0.89\;(\mathrm{mg/L})^2\), and the total standard deviation is \(\sqrt{0.89}=0.9434\) mg/L.

For the \(10.0\) mg/L prediction, the residual is \(10.8-10.0=0.8\) mg/L, so

$$ L_{10.0}=\frac{1}{\sqrt{2\pi(0.89)}}\exp\left(-\frac{0.8^2}{2(0.89)}\right) \approx 0.2952\;(\mathrm{mg/L})^{-1}. $$

For the \(11.5\) mg/L prediction, the residual is \(10.8-11.5=-0.7\) mg/L, giving

$$ L_{11.5}=\frac{1}{\sqrt{2\pi(0.89)}}\exp\left(-\frac{(-0.7)^2}{2(0.89)}\right) \approx 0.3211\;(\mathrm{mg/L})^{-1}. $$

The likelihood ratio is

$$ \frac{L_{11.5}}{L_{10.0}} =\exp\left(\frac{0.8^2-0.7^2}{2(0.89)}\right) \approx 1.088. $$

Under this observation model, the observed concentration is about 1.09 times as likely under the \(11.5\) mg/L prediction as under the \(10.0\) mg/L prediction. This is modest evidence in their relative likelihoods, not a probability that either prediction is correct.

Increasing the total variance makes the likelihood less sensitive to a fixed residual: the exponential penalty for that residual is smaller. But the normalizing factor changes too, so likelihoods should be calculated from the full density when comparing parameter values under a model. With a common variance across candidate values, the normalizing factor cancels in a likelihood ratio; it does not necessarily cancel when variances depend on the prediction.

How Error Changes the Posterior

The likelihood’s influence on an individual posterior depends on its precision, the reciprocal of its variance. To see this directly, consider a single unknown parameter \(\theta\) with a normal prior and a measurement whose model prediction is \(\theta\). Assume a known total observation variance \(v\). The prior and likelihood are

$$ \theta\sim N(\mu_0,\tau^2),\qquad y\mid\theta\sim N(\theta,v). $$

Multiplying the prior density by the likelihood gives a posterior density proportional to

$$ \exp\left[-\frac{1}{2}\left\{\frac{(\theta-\mu_0)^2}{\tau^2} +\frac{(y-\theta)^2}{v}\right\}\right]. $$

Collect the terms involving \(\theta\). The coefficient of \(\theta^2\) is \(1/\tau^2+1/v\), and the linear term is \(-2\theta(\mu_0/\tau^2+y/v)\). Completing the square therefore gives a normal posterior with variance equal to the reciprocal of the combined precision and mean equal to that variance times the precision-weighted sum:

$$ V=\left(\frac{1}{\tau^2}+\frac{1}{v}\right)^{-1},\qquad \mu_{\mathrm{post}}=V\left(\frac{\mu_0}{\tau^2}+\frac{y}{v}\right). $$

This derivation shows why adding assay variance can weaken an update: it increases \(v\), decreases the observation precision \(1/v\), and reduces the measurement’s weight relative to the prior. In a nonlinear population PK model, the posterior generally is not available in this closed form, but the likelihood still reflects the observation model’s total error variance.

Worked Example: Assay Error and a Normal Posterior Update

Suppose, hypothetically, an individual parameter \(\theta\), measured in mg/L in this simplified model, has prior \(N(10,4)\). A concentration \(y=11\) mg/L is modeled as \(N(\theta,v)\). Residual variance is \(0.80\;(\mathrm{mg/L})^2\) and assay variance is \(0.09\;(\mathrm{mg/L})^2\), so \(v=0.89\;(\mathrm{mg/L})^2\).

The posterior variance and mean are

$$ V=\left(\frac{1}{4}+\frac{1}{0.89}\right)^{-1} =\frac{3.56}{4.89} \approx 0.7280\;(\mathrm{mg/L})^2, $$
$$ \mu_{\mathrm{post}} =10+\frac{4}{4+0.89}(11-10) \approx 10.818\;\mathrm{mg/L}. $$

Thus \(\theta\mid y\sim N(10.818,0.7280)\) in this simplified example. For comparison, if assay variance were incorrectly omitted, \(v=0.80\) would give \(V=(1/4+1/0.80)^{-1}=0.6667\;(\mathrm{mg/L})^2\) and posterior mean \(10+4/(4+0.80)=10.833\) mg/L. Omitting assay error makes the posterior narrower and gives the observation slightly more weight than the stated model supports.

Propagating Posterior and Observation Uncertainty Into a Forecast

Let \(C^*=g(\theta)\) denote a future model-predicted typical concentration, conditional on the patient’s parameters, and let \(Y_{\mathrm{new}}\) be a future measured concentration. The posterior uncertainty in \(C^*\) comes from uncertainty in \(\theta\). A measured future concentration also contains its own residual and assay errors. If these future errors are independent of \(\theta\), have mean zero, and have constant additive variances, then the law of total variance gives

$$ \operatorname{Var}(Y_{\mathrm{new}}\mid y) =\operatorname{Var}\{g(\theta)\mid y\} +E\{\sigma_{\mathrm{res,new}}^2+\sigma_{\mathrm{assay,new}}^2\mid y\}. $$

The first term describes uncertainty in the patient-specific typical forecast. The second describes dispersion of a future measurement around that typical forecast. If the variance depends on \(\theta\) or on the predicted concentration, use its posterior expectation; if errors are correlated with parameter uncertainty or with other measurements, the covariance structure must also be accounted for.

For a linear forecast \(g(\theta)=a+b\theta\), with fixed \(a\) and \(b\), the posterior mean is \(a+b\mu_{\mathrm{post}}\) and the posterior variance is \(b^2V\). If the posterior for \(\theta\) is normal and future errors are normal with constant variance, the measured predictive distribution is also normal. This gives an exact interval in this special case. For nonlinear \(g\), a normal interval based only on a linear approximation may be inaccurate, particularly when posterior uncertainty is large or the transformation is strongly curved.

Worked Example: Separate Intervals for Typical and Measured Concentrations

Continue the hypothetical normal update above, with \(\theta\mid y\sim N(10.818,0.7280)\). At a future time, suppose the model’s typical concentration is \(C^*=2+0.6\theta\) mg/L. Future residual and assay variances are \(0.80\) and \(0.09\;(\mathrm{mg/L})^2\), respectively. Assume the linear model and all stated normal, independence, and constant-variance conditions hold.

The posterior mean and variance of the typical concentration are

$$ E(C^*\mid y)=2+0.6(10.818)=8.491\;\mathrm{mg/L}, \qquad \operatorname{Var}(C^*\mid y)=0.6^2(0.7280)=0.2621\;(\mathrm{mg/L})^2. $$

The posterior standard deviation is \(\sqrt{0.2621}=0.5119\) mg/L. A normal 95% interval for the patient’s typical concentration is therefore

$$ 2+0.6(10+4/4.89)\pm1.96\sqrt{0.6^2(3.56/4.89)}=[7.487,\;9.494]\;\mathrm{mg/L}. $$

For a future measured concentration, add the future observation-error variance \(0.80+0.09=0.89\;(\mathrm{mg/L})^2\). Its predictive variance is \(0.2621+0.89=1.1521\;(\mathrm{mg/L})^2\), with standard deviation \(1.073\) mg/L. The corresponding 95% predictive interval is

$$ 2+0.6(10+4/4.89)\pm1.96\sqrt{0.6^2(3.56/4.89)+0.89}=[6.387,\;10.595]\;\mathrm{mg/L}. $$

The first interval concerns the uncertain typical concentration for this patient. The wider second interval concerns a future measured value, which varies both because the patient’s parameters are uncertain and because the future observation has residual and assay error.

Concentration-Dependent Error and Model Choice

Assay precision is often described by a coefficient of variation, and residual variability may also increase with concentration. A simple proportional error model may specify a standard deviation proportional to \(f_j(\theta)\); an additive-plus-proportional model may use a variance such as \(\sigma_a^2+\sigma_p^2 f_j(\theta)^2\). These are model choices, not interchangeable descriptions. If independent assay and residual errors are both specified on the concentration scale, their variances can be added at that concentration. If the model is defined on a transformed scale, combine errors on that scale instead.

Worked Example: Combining Additive and Proportional Assay Error

Suppose, hypothetically, a predicted concentration is 12 mg/L. The residual-error standard deviation is an additive 0.50 mg/L, while assay imprecision is a 5% coefficient of variation applied to the predicted concentration. Assume independent, mean-zero normal errors on the concentration scale.

The assay standard deviation is \(0.05(12)=0.60\) mg/L, so its variance is \(0.36\;(\mathrm{mg/L})^2\). Residual variance is \(0.50^2=0.25\;(\mathrm{mg/L})^2\). The total variance and standard deviation at this prediction are

$$ v=0.25+0.36=0.61\;(\mathrm{mg/L})^2,\qquad \sqrt{v}=0.7810\;\mathrm{mg/L}. $$

If the hypothetical observed concentration is 13 mg/L, the residual is 1 mg/L and its likelihood density is

$$ L=\frac{1}{\sqrt{2\pi(0.61)}}\exp\left(-\frac{1^2}{2(0.61)}\right) \approx 0.225\;(\mathrm{mg/L})^{-1}. $$

Because assay variance depends on the predicted concentration in this setup, it must be recalculated for each candidate prediction. Treating it as a fixed variance would specify a different likelihood.

Diagnostics, Alternatives, and Common Pitfalls

Before using an error model for patient forecasting, check that it matches the units and scale of the concentration data. Review assay validation information and the fitted residual-error specification. Compare standardized residuals, \(\{y_j-f_j(\theta)\}/\sqrt{v_j}\), across concentrations and sampling times: systematic changes in their spread can indicate that a constant additive variance is unsuitable. A pattern in residuals may instead reflect model misspecification, so simply inflating the variance is not a general remedy.

A direct variance sum is appropriate for independent additive errors on the same scale. An alternative is to specify a joint error distribution with a covariance matrix when measurements share a calibration effect or other correlated error. A log-scale model is another alternative when multiplicative error is more appropriate; it requires a likelihood for the transformed data and a careful distinction between the mean on the log scale and the mean on the concentration scale. These approaches encode different assumptions and should not be mixed casually.

  • Double counting assay error. If the fitted residual variance already includes assay imprecision, adding the assay variance again exaggerates uncertainty. Establish what the residual model represents before combining terms.
  • Adding standard deviations instead of variances. For independent errors, add variances and then take the square root: \(s_{\mathrm{total}}=\sqrt{s_1^2+s_2^2}\), not \(s_1+s_2\).
  • Mixing scales. A variance in log concentration cannot be added directly to a variance in mg/L squared. Express the components on the same modeled scale.
  • Using only parameter uncertainty for a measured forecast. This omits future observation variability. Conversely, adding observation error to an interval intended only for the typical concentration changes the target of the interval.
  • Assuming all uncertainty is captured. The calculations rely on a specified structural model, dosing history, and error distribution. Uncertainty in residual-error parameters or model structure is not included unless it is explicitly represented and propagated.
Key takeaway: Use the observation model to combine error on the correct scale, let the resulting likelihood determine how strongly concentrations update individual parameters, and add future observation variance only when forecasting a future measured concentration.

Check Your Understanding

Answer each question, then compare with the solutions below.

  1. Independent assay and residual standard deviations are 0.20 and 0.40 mg/L. What is the combined standard deviation under an additive model?
  2. Why does increasing assay variance generally make one observed concentration less influential in an individual Bayesian update?
  3. A posterior typical forecast has mean 8 mg/L and variance \(0.25\;(\mathrm{mg/L})^2\). Future residual and assay variances are 0.36 and \(0.09\;(\mathrm{mg/L})^2\). Give the standard deviation for a future measured concentration.
  4. An assay has 4% proportional error at a predicted concentration of 15 mg/L. What assay variance should be combined with independent additive residual variance \(0.16\;(\mathrm{mg/L})^2\)?

Solutions. (1) The total variance is \(0.20^2+0.40^2=0.20\;(\mathrm{mg/L})^2\), so the combined standard deviation is \(\sqrt{0.20}=0.4472\) mg/L. (2) Greater assay variance increases total observation variance and decreases likelihood precision, so the data receive less weight relative to the prior. (3) The predictive variance is \(0.25+0.36+0.09=0.70\;(\mathrm{mg/L})^2\), giving a standard deviation of \(\sqrt{0.70}=0.8367\) mg/L. (4) The assay standard deviation is \(0.04(15)=0.60\) mg/L, so assay variance is \(0.36\;(\mathrm{mg/L})^2\). The total observation variance is \(0.36+0.16=0.52\;(\mathrm{mg/L})^2\), with standard deviation \(\sqrt{0.52}=0.7211\) mg/L.

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