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Therapeutic Drug Monitoring and Individual Predictions · Tutorial 531 of 599

Posterior Parameter Uncertainty and Predictive Uncertainty

Learn how patient-specific parameter uncertainty and residual concentration variability combine in a posterior predictive distribution.

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What You'll Learn

  • Distinguish uncertainty about an individual’s PK parameters from residual variability in a future concentration.
  • Derive the law-of-total-variance decomposition for a Bayesian forecast.
  • Calculate parameter and residual contributions using posterior parameter draws.
  • Understand why a predictive distribution may be a mixture rather than a normal distribution.
  • Choose between a parameter credible interval and a predictive interval for the question being asked.
  • Recognize plug-in forecasts that omit one or both sources of uncertainty.

Two Different Questions About a Forecast

A Bayesian forecast for a patient uses the observed concentrations and dosing history to update a distribution for that patient’s PK parameters. It can then predict a future concentration. These steps involve two distinct sources of uncertainty: uncertainty about the patient’s parameters given the available data, and residual variation in a future observed concentration even if those parameters were known.

The distinction matters because the questions are different. A posterior distribution for clearance describes what values of clearance remain plausible for this patient. A predictive distribution for a future concentration describes what concentrations may actually be observed, accounting both for uncertainty about clearance and for residual variation around the model-predicted concentration. A single forecast value, such as the prediction at a maximum a posteriori (MAP) estimate, does not by itself express either distribution.

Definition: Let \(\theta\) denote a patient’s PK parameters, \(y\) the patient’s observed data, and \(Y^*\) a future concentration. Posterior parameter uncertainty is uncertainty described by \(p(\theta\mid y)\). Predictive uncertainty is uncertainty described by \(p(Y^*\mid y)\), after allowing for both the posterior distribution of \(\theta\) and the model’s residual variability.

Write \(m(\theta)\) for the model-predicted concentration at the future time, conditional on the patient’s parameters and the recorded dosing history. A common observation model is \(Y^*=m(\theta)+\epsilon^*\), where \(\epsilon^*\) is a future residual error. The form and variance of that error depend on the chosen residual model; they need not be constant across predicted concentrations.

Decomposing Predictive Variance

For a forecast with finite variance, the law of total variance gives a direct decomposition. The first term measures how much the model-predicted concentration changes across the posterior distribution of the patient’s parameters. The second term averages the residual variance that remains conditional on those parameters.

$$ \operatorname{Var}(Y^*\mid y) = \operatorname{Var}_{\theta\mid y}\!\left\{E(Y^*\mid\theta,y)\right\} + E_{\theta\mid y}\!\left\{\operatorname{Var}(Y^*\mid\theta,y)\right\}. $$

If the conditional mean is \(m(\theta)\), this becomes:

$$ \operatorname{Var}(Y^*\mid y) = \operatorname{Var}_{\theta\mid y}\{m(\theta)\} + E_{\theta\mid y}\{\operatorname{Var}(Y^*\mid\theta,y)\}. $$

The first term is the parameter-uncertainty contribution to forecast variance. The second is the expected residual-variance contribution. They are variances on the concentration scale, so they can be added; their standard deviations cannot simply be added. The predictive standard deviation is the square root of the total variance.

Justification: Let \(\mu(\theta)=E(Y^*\mid\theta,y)\) and \(\mu=E(Y^*\mid y)=E_{\theta\mid y}\{\mu(\theta)\}\). Add and subtract \(\mu(\theta)\) from \(Y^*-\mu\), then square and take expectations. The cross-product has expectation zero because \(E\{Y^*-\mu(\theta)\mid\theta,y\}=0\). The remaining terms are \(E_{\theta\mid y}\{\operatorname{Var}(Y^*\mid\theta,y)\}\) and \(\operatorname{Var}_{\theta\mid y}\{\mu(\theta)\}\), respectively.

This decomposition assumes the posterior distribution represents uncertainty about \(\theta\) conditional on the observed patient data, and that the conditional predictive distribution correctly describes future observations given \(\theta\) and the relevant dosing history. It does not require residual variance to be constant: the second term averages it over the posterior. If the forecast model has an additional random effect or uncertainty source, it must be included in the conditioning structure and propagated rather than silently omitted.

Worked Example: Additive Residual Error and a Two-Point Posterior

Suppose a hypothetical patient’s posterior distribution gives a 25% probability to parameter values that predict a future concentration of 8 mg/L, and a 75% probability to values that predict 12 mg/L. Assume additive residual error with conditional mean zero and standard deviation 2 mg/L at either prediction. Calculate the posterior mean forecast and both variance contributions.

The posterior mean of the conditional prediction is:

$$ E_{\theta\mid y}\{m(\theta)\} =0.25(8)+0.75(12) =2+9 =11\ \text{mg/L}. $$

The parameter-uncertainty contribution is the posterior variance of the two possible conditional means:

$$ \operatorname{Var}_{\theta\mid y}\{m(\theta)\} =0.25(8-11)^2+0.75(12-11)^2 =0.25(9)+0.75(1) =3\ (\text{mg/L})^2. $$

The conditional residual variance is \(2^2=4\ (\text{mg/L})^2\) at either prediction, so its posterior expectation is also \(4\ (\text{mg/L})^2\). Therefore:

$$ \operatorname{Var}(Y^*\mid y)=3+4=7\ (\text{mg/L})^2, \qquad SD(Y^*\mid y)=\sqrt{7}=2.646\ \text{mg/L}. $$

The 3 \((\text{mg/L})^2\) component reflects uncertainty in the patient-specific model prediction; the 4 \((\text{mg/L})^2\) component reflects residual variability at a given parameter value. The forecast mean is 11 mg/L, but this mean alone does not describe the predictive distribution. In this example the distribution is a mixture of two residual distributions, not necessarily a single normal distribution.

Posterior Prediction in Practice

When posterior calculations are represented by draws, the decomposition can be computed directly. For each draw \(\theta^{(s)}\), calculate the conditional prediction \(m^{(s)}=m(\theta^{(s)})\) and the conditional residual variance \(v^{(s)}=\operatorname{Var}(Y^*\mid\theta^{(s)},y)\). The variation among \(m^{(s)}\) estimates the parameter component; the average of \(v^{(s)}\) estimates the residual component. Preserve the model’s joint posterior draws when calculating predictions: using unrelated marginal parameter summaries can fail to represent combinations supported by the posterior.

1
Draw patient parameters.
Obtain draws from the posterior \(p(\theta\mid y)\), conditional on the patient’s available data.
2
Calculate conditional predictions.
For each draw, evaluate the PK model at the intended future time and dosing history to obtain \(m^{(s)}\).
3
Represent residual variability.
For each draw, calculate the appropriate conditional residual variance or simulate a future residual from the specified observation model.
4
Summarize the target.
Use the \(m^{(s)}\) values for parameter-driven uncertainty in the conditional prediction. Use simulated future observations, or the variance decomposition, for predictive uncertainty.

Worked Example: Residual Variance That Depends on the Prediction

In a second hypothetical forecast, posterior draws imply conditional mean concentrations of 10 mg/L and 14 mg/L, each with probability 0.5. Suppose the conditional residual variances are 1 and 4 \((\text{mg/L})^2\), respectively, and each conditional residual has mean zero. Calculate the total predictive variance.

The posterior mean of the conditional predictions is \(0.5(10)+0.5(14)=12\) mg/L. Thus the parameter-uncertainty contribution is:

$$ 0.5(10-12)^2+0.5(14-12)^2 =0.5(4)+0.5(4) =4\ (\text{mg/L})^2. $$

The expected conditional residual variance is:

$$ 0.5(1)+0.5(4)=2.5\ (\text{mg/L})^2. $$

The total predictive variance and standard deviation are therefore:

$$ \operatorname{Var}(Y^*\mid y)=4+2.5=6.5\ (\text{mg/L})^2, \qquad SD(Y^*\mid y)=\sqrt{6.5}=2.550\ \text{mg/L}. $$

Using only the residual variance at the posterior mean prediction would not generally give the correct expected residual contribution. The calculation instead averages the conditional variances over the posterior draws, as required when residual variability changes with the predicted concentration.

Intervals, Alternatives, and Diagnostics

A credible interval for a parameter, or for a function of parameters such as \(m(\theta)\), summarizes posterior uncertainty about that quantity. It does not include future residual variation unless that variation is specifically added. A predictive interval for \(Y^*\) includes both parameter uncertainty and residual variation under the assumed model. Its endpoints are usually obtained from quantiles of the posterior predictive distribution; a variance decomposition alone does not determine those quantiles.

The predictive distribution is generally a mixture over parameter values. Even if each conditional residual distribution is normal, the mixture need not be normal. Consequently, a normal interval formed as the predictive mean plus or minus 1.96 predictive standard deviations is an approximation, not a universal exact 95% predictive interval. Posterior simulation is useful when the prediction is nonlinear, the posterior is asymmetric or multimodal, or the conditional residual distribution changes across parameter values.

Forecast approachWhat it representsMain limitation
Plug-in prediction at a MAP estimateOne conditional model predictionDoes not itself represent posterior parameter uncertainty or residual variation.
Posterior distribution of \(m(\theta)\)Variation in the conditional prediction across patient parameter valuesDoes not include future residual variation.
Posterior predictive distribution of \(Y^*\)Parameter uncertainty combined with conditional future-observation variationDepends on the adequacy of both the PK model and the residual model.
Variance decompositionSeparate contributions to predictive varianceDoes not by itself describe distribution shape or provide interval quantiles.

Useful checks include confirming that simulated parameter draws and future observations follow the stated conditioning, verifying the residual variance at the relevant predictions, and comparing the simulated predictive distribution with its variance decomposition. If a simulated predictive variance disagrees substantially with the sum of the two components, check the implementation for omitted variability, inconsistent weighting, or an incorrect conditional mean. Also inspect whether the posterior predictions are skewed or multimodal: one mean and standard deviation can conceal clinically relevant structure.

A common mistake is to call the posterior spread of \(m(\theta)\) a predictive interval for a future concentration. That spread answers how uncertain the model-predicted mean is across plausible patient parameters. It omits residual deviation of an actual future observation around that mean. The correction is to add residual uncertainty through the conditional predictive model, then obtain predictive summaries from the resulting distribution.

The decomposition quantifies uncertainty under the specified model; it does not guarantee that the model captures all clinically relevant uncertainty. Incorrect dosing history, misspecified residual behavior, or a PK model that does not represent the patient can make a narrow interval misleading. In addition, forecast uncertainty depends on the future time and dosing scenario: changing either can change both the parameter-driven spread and residual variance.

Worked Example: Lognormal Residual Error

For a third hypothetical patient, suppose the conditional mean future concentration \(m(\theta)\) is 8 mg/L or 12 mg/L with equal posterior probability. Use a lognormal residual model:

$$ Y^*=m(\theta)\exp(\epsilon^*-\sigma^2/2), \qquad \epsilon^*\sim N(0,\sigma^2), \qquad \sigma=0.2. $$

The subtraction of \(\sigma^2/2\) makes \(E(Y^*\mid\theta,y)=m(\theta)\). Under this model, the conditional variance is \(m(\theta)^2\{\exp(\sigma^2)-1\}\). First, the mean of the conditional predictions is \(0.5(8)+0.5(12)=10\) mg/L, and their posterior variance is:

$$ 0.5(8-10)^2+0.5(12-10)^2=4\ (\text{mg/L})^2. $$

Since \(\sigma^2=0.04\), the expected residual variance is:

$$ \frac{8^2+12^2}{2}\{\exp(0.04)-1\} =104(0.04081) =4.244\ (\text{mg/L})^2. $$

Thus the predictive variance is \(4+4.244=8.244\ (\text{mg/L})^2\), and the predictive standard deviation is \(\sqrt{8.244}=2.871\) mg/L. The parameter component and residual component have both been calculated on the concentration scale, even though the residual model was specified on a log scale.

Check Your Understanding

Answer each question, then compare with the solutions below.

  1. A forecast has parameter-uncertainty variance 5 \((\text{mg/L})^2\) and expected residual variance 4 \((\text{mg/L})^2\). What is its predictive standard deviation?
  2. Why is the posterior distribution of \(m(\theta)\) not generally the predictive distribution of a future observed concentration?
  3. Two equally probable posterior draws give conditional means of 6 and 10 mg/L. Conditional residual variance is 1 \((\text{mg/L})^2\) for each draw. Calculate the posterior mean forecast, parameter-uncertainty variance, and total predictive variance.
  4. Does a predictive variance decomposition establish that the predictive distribution is normal? Explain.

Solutions. (1) Add the variance components: \(5+4=9\ (\text{mg/L})^2\); the predictive standard deviation is \(\sqrt{9}=3\) mg/L. (2) \(m(\theta)\) varies across posterior parameter values but describes the conditional mean. A future observed concentration also varies because of residual error conditional on those parameters. (3) The mean is \(0.5(6)+0.5(10)=8\) mg/L. The parameter component is \(0.5(6-8)^2+0.5(10-8)^2=4\ (\text{mg/L})^2\). The expected residual variance is 1 \((\text{mg/L})^2\), so total predictive variance is \(4+1=5\ (\text{mg/L})^2\). (4) No. The decomposition determines variance, not distribution shape. A mixture of conditional distributions can be nonnormal even if each conditional residual distribution is normal.

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