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Uncertainty Propagation and Predictive Intervals · Tutorial 764 of 799

Decomposing Predictive Variance by Uncertainty Source

Learn to derive and estimate the two components of predictive variance in a nested PBPK simulation, and recognize when a decomposition is unreliable or misinterpreted.

Advanced 11 min read

What You'll Learn

  • State the law of total variance for a prediction conditional on uncertain PBPK parameters.
  • Distinguish conditional within-subject variability from variation across parameter draws.
  • Estimate both variance components from a balanced nested simulation.
  • Correct the variance of conditional sample means for finite inner-simulation noise.
  • Diagnose unstable or negative component estimates without treating them as physical variances.
  • Interpret variance fractions without assuming they identify causal or independent parameter effects.

Why Decompose Predictive Variance?

A predictive distribution can be wide for more than one reason. Even at a fixed set of PBPK parameter values, a prediction may vary because the defined within-subject process is stochastic. In addition, changing uncertain parameter values can shift the conditional average prediction. These sources matter differently: reducing uncertainty in a parameter distribution cannot remove genuine within-subject variability, while collecting better parameter information may reduce the spread of conditional predictions.

The distinction depends on what the simulation randomizes. Here, let \(\Theta\) denote a draw of the uncertain PBPK parameter vector, and let \(Y\) be the prediction for a specified subject and prediction target under that draw. The conditional distribution of \(Y\) given \(\Theta\) represents the defined within-subject variability at fixed parameters. The distribution of \(\Theta\) represents parameter uncertainty. The subject, dose, endpoint, and other conditions must be specified consistently across the nested simulation.

Definition: The within-subject variance component is \(E_\Theta[\operatorname{Var}(Y\mid\Theta)]\), the expected prediction variance when parameters are held fixed. The between-draw parameter component is \(\operatorname{Var}_\Theta[E(Y\mid\Theta)]\), the variance across parameter draws in the corresponding conditional mean predictions. Their sum is the predictive variance of \(Y\), provided the relevant second moments are finite.

“Between-draw” refers to differences among draws from the parameter uncertainty distribution, not to differences between subjects. If a PBPK model is deterministic at fixed parameters and no within-subject random input is included, then \(\operatorname{Var}(Y\mid\Theta)=0\); the predictive variance in this setup is entirely due to parameter uncertainty. Random variability should not be added merely to make a decomposition have two nonzero terms.

Deriving the Law of Total Variance

Write \(\mu(\Theta)=E(Y\mid\Theta)\) for the conditional mean prediction and \(\sigma^2(\Theta)=\operatorname{Var}(Y\mid\Theta)\) for the conditional variance. The overall mean is \(E(Y)=E_\Theta[\mu(\Theta)]\). Add and subtract \(\mu(\Theta)\) inside the squared deviation from the overall mean:

$$ \begin{aligned} \operatorname{Var}(Y) &=E\left[(Y-E(Y))^2\right]\\ &=E\left[\{Y-\mu(\Theta)+\mu(\Theta)-E(Y)\}^2\right]\\ &=E\left[(Y-\mu(\Theta))^2\right] +E\left[(\mu(\Theta)-E(Y))^2\right]\\ &\quad+2E\left[(Y-\mu(\Theta))(\mu(\Theta)-E(Y))\right]. \end{aligned} $$

The cross-product has expectation zero. Conditional on \(\Theta\), the factor \(\mu(\Theta)-E(Y)\) is fixed and \(E[Y-\mu(\Theta)\mid\Theta]=0\). Therefore,

$$ \begin{aligned} E\left[(Y-\mu(\Theta))(\mu(\Theta)-E(Y))\right] &=E_\Theta\left[(\mu(\Theta)-E(Y)) E\{Y-\mu(\Theta)\mid\Theta\}\right]\\ &=0. \end{aligned} $$

The first remaining term is \(E_\Theta[\operatorname{Var}(Y\mid\Theta)]\), and the second is \(\operatorname{Var}_\Theta[\mu(\Theta)]\). Thus,

$$ \operatorname{Var}(Y) =E_\Theta[\operatorname{Var}(Y\mid\Theta)] +\operatorname{Var}_\Theta[E(Y\mid\Theta)]. $$

This identity is exact for the specified predictive distribution. It does not require normality or a linear PBPK model. It does require finite second moments; it also requires that the conditioning variable \(\Theta\) and the meaning of \(Y\) be clearly defined. Dependence among uncertain model parameters is retained when \(\Theta\) is sampled jointly from the intended distribution.

Estimating Components from Nested Simulations

A balanced nested simulation uses \(M\) outer parameter draws and \(R\) inner predictions for each draw. Write these as \(\theta_m\) and \(y_{mr}\), where \(m=1,\ldots,M\) and \(r=1,\ldots,R\). The inner predictions must represent the stated conditional variability at \(\theta_m\); repeated model runs that are identical at fixed parameters do not estimate a nonzero conditional variance.

$$ \bar{y}_{m\cdot}=\frac{1}{R}\sum_{r=1}^{R}y_{mr}, \qquad s_m^2=\frac{1}{R-1}\sum_{r=1}^{R}(y_{mr}-\bar{y}_{m\cdot})^2, \qquad W=\frac{1}{M}\sum_{m=1}^{M}s_m^2. $$

When inner draws are conditionally independent with the same conditional distribution at a given \(\theta_m\), \(W\) estimates the expected within-subject variance. The sample variance of the \(M\) conditional sample means, however, is not by itself an estimate of the parameter component: each mean contains inner-simulation noise.

$$ S_{\bar{y}}^2= \frac{1}{M-1}\sum_{m=1}^{M}(\bar{y}_{m\cdot}-\bar{y}_{\cdot\cdot})^2, \qquad \bar{y}_{\cdot\cdot}=\frac{1}{M}\sum_{m=1}^{M}\bar{y}_{m\cdot}. $$

For equal \(R\), the conditional variance of each inner mean is approximately \(\sigma^2(\theta_m)/R\). Averaging over parameter draws, this contributes approximately \(E_\Theta[\sigma^2(\Theta)]/R\) to the variance of the conditional sample means. This leads to the balanced-design estimator

$$ \widehat{V}_{\mathrm{parameter}}=S_{\bar{y}}^2-\frac{W}{R}, \qquad \widehat{V}_{\mathrm{total}}=W+\widehat{V}_{\mathrm{parameter}}. $$

Under independent outer draws from the parameter distribution and conditionally independent inner draws, these are method-of-moments estimates of the two components. The subtraction corrects the observed spread in conditional sample means for expected inner noise. With finite \(M\), the parameter-component estimate can be negative because of Monte Carlo error. A negative estimate signals insufficient precision for resolving a small component; the true variance component is nonnegative. Replacing it by zero may be useful for a display, but it changes the estimator and should be reported as a truncation, not as a negative physical variance.

Worked Example: Exact Normal Decomposition

Suppose, hypothetically, that a predicted concentration \(Y\), in mg/L, has conditional distribution \(Y\mid\Theta=\theta\sim N(\theta,3^2)\), while the parameter-dependent conditional mean is \(\Theta\sim N(20,4^2)\). The first variance is within-subject variation at fixed \(\Theta\); the second describes uncertainty across parameter draws. Both normal distributions have finite variances, so the law applies.

The expected conditional variance is \(E_\Theta[\operatorname{Var}(Y\mid\Theta)]=3^2=9\;(\mathrm{mg/L})^2\). The variance of the conditional mean is \(\operatorname{Var}_\Theta[E(Y\mid\Theta)]=\operatorname{Var}(\Theta)=4^2=16\;(\mathrm{mg/L})^2\). Therefore,

$$ \operatorname{Var}(Y)=9+16=25\;(\mathrm{mg/L})^2, \qquad SD(Y)=\sqrt{25}=5\;\mathrm{mg/L}. $$

In this hypothetical example, the within-subject component is \(9/25=0.360\), or 36.0%, of the total variance, and the parameter component is \(16/25=0.640\), or 64.0%. These fractions describe this predictive distribution and target; they do not show that a particular parameter causes 64.0% of variability.

Worked Example: Estimating Components from Nested PBPK Predictions

Consider hypothetical nested predictions of AUC, measured in mg·h/L. Three parameter draws each receive four independent inner draws. The simulated AUC values are:

At \(\theta_1\): 8, 10, 12, 14; at \(\theta_2\): 14, 16, 18, 20; at \(\theta_3\): 20, 22, 24, 26 mg·h/L. The nested design is balanced, and the inner draws are assumed to represent the defined within-subject randomness at fixed parameter values.

The conditional means are \(11,17,23\) mg·h/L. For each group, the squared deviations from its mean sum to \(20\;(\mathrm{mg\cdot h/L})^2\), so each sample variance is \(20/(4-1)=6.667\;(\mathrm{mg\cdot h/L})^2\). Hence \(W=(6.667+6.667+6.667)/3=6.667\;(\mathrm{mg\cdot h/L})^2\).

The mean of the three conditional means is \(17\) mg·h/L. Their squared deviations sum to \((11-17)^2+(17-17)^2+(23-17)^2=72\;(\mathrm{mg\cdot h/L})^2\), giving \(S_{\bar{y}}^2=72/(3-1)=36.00\;(\mathrm{mg\cdot h/L})^2\). Correct for the inner-mean noise, \(W/R=6.667/4=1.667\;(\mathrm{mg\cdot h/L})^2\). Thus,

$$ \widehat{V}_{\mathrm{parameter}}=36.00-1.667=34.33\;(\mathrm{mg\cdot h/L})^2, \qquad \widehat{V}_{\mathrm{total}}=6.667+34.33=41.00\;(\mathrm{mg\cdot h/L})^2. $$

The estimates are based on only three outer draws, so they are illustrative, not precise. Their corresponding estimated standard deviations are \(\sqrt{6.667}=2.582\) mg·h/L within-subject and \(\sqrt{34.33}=5.859\) mg·h/L across parameter draws. The decomposition is about variance, so these standard deviations are not added to obtain the total standard deviation.

Worked Example: A Negative Corrected Component

Suppose a hypothetical nested run has \(R=4\), an average within-group sample variance \(W=16\;(\mathrm{mg/L})^2\), and three conditional sample means of 9, 10, and 11 mg/L. These means have average 10 mg/L and sample variance \(S_{\bar{y}}^2=[(9-10)^2+(10-10)^2+(11-10)^2]/2=1.000\;(\mathrm{mg/L})^2\). The estimated inner-mean noise contribution is \(W/R=16/4=4.000\;(\mathrm{mg/L})^2\).

The corrected estimate is \(1.000-4.000=-3.000\;(\mathrm{mg/L})^2\). The parameter variance cannot truly be negative. The result means that, with this small outer sample, the observed spread among conditional sample means is smaller than the noise expected from the inner draws. It does not establish that parameter uncertainty has no effect. More independent outer draws are needed to estimate a small between-draw component; increasing \(R\) can also reduce noise in each conditional mean.

Diagnostics, Interpretation, and Alternatives

A decomposition is only as meaningful as its simulation design. Check that the outer draws represent the intended joint parameter uncertainty distribution and that any dependence among parameters has been preserved. Check that inner replicates genuinely vary according to the intended conditional process, use consistent units and prediction definitions, and are conditionally independent or have their dependence accounted for. A deterministic model run repeatedly at exactly the same inputs contributes no empirical within-subject variation.

The estimates also have different Monte Carlo demands. More inner draws stabilize each conditional mean and variance; more outer draws are essential for learning the distribution of conditional means. A large \(R\) cannot compensate for too few \(\theta_m\) draws when the parameter component is the main target. Examine component estimates across reproducible runs or increased simulation sizes, and report the numbers of outer and inner draws. A very small or negative corrected component, or a component estimate that changes substantially as \(M\) increases, is a warning that the decomposition is not yet resolved.

A direct pooled variance is an alternative summary when only total predictive spread is needed, but it does not identify sources. In a nested design, outcomes sharing an outer parameter draw are clustered; treating all \(MR\) values as independent observations can give misleading precision for the total or its components. The corrected nested estimator uses the design to separate conditional variation from variability in conditional means. Other uncertainty-propagation approaches, such as variance-based sensitivity analyses, ask different questions and may attribute variability to particular inputs; they are not substitutes for this two-component identity, especially when parameters are dependent.

The decomposition is distributional, not causal. A parameter component can reflect multiple correlated parameters, nonlinear responses, and interactions in the PBPK model. It cannot generally be divided into separate parameter contributions by simply examining marginal parameter variances. Likewise, “within-subject” here is defined by the conditional simulation. If the scientific target instead involves a population of subjects, the random quantities and conditioning structure must be specified for that target; this calculation alone does not separate between-subject spread from parameter uncertainty.

Key takeaway: Predictive variance equals expected variance conditional on parameters plus variance of conditional means across parameter draws. Estimate each from the matching level of a nested simulation, correct conditional-mean spread for inner Monte Carlo noise, and treat unstable components as a diagnostic signal rather than a biological conclusion.

Check Your Understanding

Work each problem before reading the solutions. Variances are in the square of the prediction's units.

  1. If \(E_\Theta[\operatorname{Var}(Y\mid\Theta)]=12\) and \(\operatorname{Var}_\Theta[E(Y\mid\Theta)]=7\), what is the total predictive variance and standard deviation?
  2. In a balanced design, \(R=5\), \(W=10\), and the sample variance of the conditional means is \(6.667\). Estimate the parameter component and total variance.
  3. Why is the sample variance of conditional sample means not, by itself, an estimate of parameter uncertainty?
  4. A deterministic PBPK model gives exactly the same prediction on every repeated run at fixed parameter values. What is the within-subject variance component for that simulation setup?

Solutions. (1) The total variance is \(12+7=19\) squared units, and the standard deviation is \(\sqrt{19}=4.359\) prediction units. (2) The inner-mean noise correction is \(W/R=10/5=2\); the parameter estimate is \(6.667-2=4.667\), and the total estimate is \(10+4.667=14.667\) squared units. (3) Each conditional sample mean has finite inner-simulation error, which adds approximately \(W/R\) to its across-draw sample variance; subtract that contribution to estimate variation in the conditional means. (4) It is zero, because \(Y\) has no variation conditional on the fixed parameter values in that setup.