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Biological Plausibility and Parameter Uncertainty · Tutorial 175 of 799

Propagating Uncertainty Under Parameter Dependence

Compare independent and correlated PBPK parameter sampling and calculate how their dependence changes the uncertainty in a model output.

Advanced 12 min read

What You'll Learn

  • Distinguish marginal parameter distributions from their joint dependence structure.
  • Derive output variance for an affine approximation with two uncertain parameters.
  • Compare independent, positively correlated, and negatively correlated parameter samples.
  • Generate paired log-parameter draws with a specified correlation.
  • Diagnose when an independence assumption or a correlation model is not credible.
  • Identify the limits of covariance-based uncertainty summaries.

Why Parameter Dependence Matters

A PBPK uncertainty analysis samples uncertain inputs and propagates each sampled parameter set through the model. The parameter distributions alone do not fully describe those sets: the joint distribution also specifies whether parameters tend to be high or low together. Sampling each marginal distribution independently therefore makes a substantive assumption. If the parameters are dependent, independent sampling can overstate or understate the spread of a predicted exposure.

This tutorial isolates that effect using a local affine approximation to a PBPK output. The approximation keeps the arithmetic transparent: for an affine output, the effect of dependence on variance is exact. A real PBPK model may respond nonlinearly, so this calculation is not a substitute for propagating joint draws through the full model.

Definition: Two uncertain parameters are independent if their joint distribution factors into the product of their marginal distributions. They are dependent if that factorization does not hold. Their linear dependence can be summarized by covariance, \(\operatorname{Cov}(U,V)=E[(U-E[U])(V-E[V])]\), and, when both standard deviations are positive, correlation, \(\rho=\operatorname{Cov}(U,V)/(\sigma_U\sigma_V)\). Correlation lies between \(-1\) and \(1\); zero correlation does not, in general, imply independence.

We will work with two positive PBPK inputs: a tissue-to-plasma partition coefficient \(K_p\) and an intrinsic-clearance parameter \(CL_{\mathrm{int}}\). Define their centered log deviations from reference values \(K_{p,0}\) and \(CL_{\mathrm{int},0}\) by \(U=\ln(K_p/K_{p,0})\) and \(V=\ln(CL_{\mathrm{int}}/CL_{\mathrm{int},0})\). Log-scale sampling represents positive parameters without assigning probability to negative values. Suppose \(U\) and \(V\) have means zero, standard deviations \(\sigma_U\) and \(\sigma_V\), and correlation \(\rho\).

For illustration, take the local output approximation

$$ Y=30+10U+10V \quad \text{mg}\cdot\text{h}/\text{L}. $$

Here \(Y\) is a hypothetical exposure endpoint, and the coefficients are local sensitivities on the log-parameter scale: a one-unit increase in either log deviation changes the approximated output by \(10\ \text{mg}\cdot\text{h}/\text{L}\). This affine relation is deliberately a simplified surrogate, not a universal PBPK equation.

How Dependence Enters Output Uncertainty

For any two random variables with finite variances and constants \(a\), \(b\), and \(c\), expanding the squared deviation of \(Y=c+aU+bV\) gives

$$ \begin{aligned} \operatorname{Var}(Y) &=E\left[\{a(U-E[U])+b(V-E[V])\}^2\right]\\ &=a^2\sigma_U^2+b^2\sigma_V^2+2ab\operatorname{Cov}(U,V)\\ &=a^2\sigma_U^2+b^2\sigma_V^2+2ab\rho\sigma_U\sigma_V. \end{aligned} $$

The middle line follows by multiplying out the square: the two squared terms have expectations \(a^2\sigma_U^2\) and \(b^2\sigma_V^2\), while the cross-product has expectation \(2ab\operatorname{Cov}(U,V)\). Independence implies zero covariance, so the cross-term vanishes. Dependence changes the variance through this term; it does not change the marginal variances in this calculation.

When \(a\) and \(b\) have the same sign, positive correlation increases the variance relative to independent sampling, while negative correlation decreases it. If the sensitivities have opposite signs, the direction reverses. Thus, “positive correlation increases uncertainty” is not a general rule without considering how each parameter affects the output.

Worked Example: Independent Log-Parameter Sampling

Suppose the log deviations have standard deviations \(\sigma_U=0.30\) and \(\sigma_V=0.20\), and are sampled independently. Calculate the output standard deviation for the affine approximation above.

Independence gives \(\rho=0\). Substituting \(a=b=10\) into the variance formula,

$$ \operatorname{Var}(Y) =10^2(0.30)^2+10^2(0.20)^2 =9+4=13\;(\text{mg}\cdot\text{h}/\text{L})^2. $$

Therefore, \(\operatorname{SD}(Y)=\sqrt{13}=3.606\ \text{mg}\cdot\text{h}/\text{L}\). Because both deviations have mean zero, \(E[Y]=30\ \text{mg}\cdot\text{h}/\text{L}\). If the joint distribution is bivariate normal, this affine output is normally distributed. Its central 90% interval is approximately \(30\pm1.645(3.606)\), or \([24.07,35.93]\ \text{mg}\cdot\text{h}/\text{L}\). This is a probability interval under the assumed parameter distribution and output model, not a confidence interval for an estimated population mean.

Worked Example: Positive and Negative Dependence

Keep the same marginal standard deviations and output approximation, but suppose the parameter correlation is \(\rho=0.75\). Compare the output variance with independent sampling.

The covariance is \(\operatorname{Cov}(U,V)=0.75(0.30)(0.20)=0.045\). The cross-term is \(2(10)(10)(0.045)=9\;(\text{mg}\cdot\text{h}/\text{L})^2\), so

$$ \operatorname{Var}(Y)=9+4+9=22\;(\text{mg}\cdot\text{h}/\text{L})^2, \qquad \operatorname{SD}(Y)=\sqrt{22}=4.690\ \text{mg}\cdot\text{h}/\text{L}. $$

The mean remains \(30\ \text{mg}\cdot\text{h}/\text{L}\), but the standard deviation is larger than the independent-sampling value of \(3.606\). Under the same bivariate-normal assumption, the central 90% interval becomes \(30\pm1.645(4.690)=[22.28,37.72]\ \text{mg}\cdot\text{h}/\text{L}\).

For contrast, if \(\rho=-0.75\), the cross-term is \(-9\), giving variance \(9+4-9=4\;(\text{mg}\cdot\text{h}/\text{L})^2\) and standard deviation \(2.000\ \text{mg}\cdot\text{h}/\text{L}\). The paired inputs then tend to offset each other in this output. The unchanged marginal distributions do not imply unchanged output uncertainty.

Generating Joint Parameter Draws

A simulation must preserve the intended dependence within each row of a parameter set. For two jointly normal log deviations, one convenient construction starts with independent standard normal random variables \(Z_1\) and \(Z_2\), each distributed as \(N(0,1)\). Set

$$ U=\sigma_U Z_1, \qquad V=\sigma_V\left(\rho Z_1+\sqrt{1-\rho^2}\,Z_2\right). $$

This construction gives \(E[U]=E[V]=0\), \(\operatorname{Var}(U)=\sigma_U^2\), and \(\operatorname{Var}(V)=\sigma_V^2\). Its covariance is \(\sigma_U\sigma_V\rho\): the \(Z_1\) contribution to \(V\) is shared with \(U\), while the independent \(Z_2\) contribution adds variation without adding covariance. The square root is real for \(-1\leq\rho\leq1\).

Worked Example: Constructing One Correlated Draw

Use \(\sigma_U=0.30\), \(\sigma_V=0.20\), and \(\rho=0.75\). For one hypothetical draw, let \(Z_1=1.00\) and \(Z_2=-0.50\). First,

$$ U=0.30(1.00)=0.300, \qquad V=0.20\left[0.75(1.00)+\sqrt{1-0.75^2}(-0.50)\right]. $$

Since \(\sqrt{1-0.75^2}=\sqrt{0.4375}=0.6614\), the second deviation is \(V=0.20(0.75-0.3307)=0.08386\). The corresponding parameter ratios are \(K_p/K_{p,0}=e^{0.300}=1.350\) and \(CL_{\mathrm{int}}/CL_{\mathrm{int},0}=e^{0.08386}=1.087\). The output approximation for this pair is

$$ Y=30+10(0.300)+10(0.08386) =33.84\ \text{mg}\cdot\text{h}/\text{L}. $$

Repeating this construction with new independent \(Z_1,Z_2\) values produces a collection of paired draws. The shared \(Z_1\) term creates the specified positive dependence; drawing \(U\) and \(V\) separately from their marginals would discard it.

Diagnostics, Alternatives, and Limits

Before interpreting a propagated distribution, check that the dependence model is supported and that the sampled pairs preserve it. Useful checks include:

  • Inspect paired draws. A scatterplot of \(U\) against \(V\) can reveal the direction, strength, curvature, or clusters in the simulated relationship. A single correlation value will not describe every possible dependence pattern.
  • Check the joint model. Confirm that the correlation or covariance comes from evidence or a defensible elicitation, rather than a convenient assumption. For multiple parameters, the proposed covariance matrix must be positive semidefinite; otherwise no joint distribution can have that matrix.
  • Verify the sampling implementation. Compare simulated marginal means, standard deviations, and correlations with their intended values. Check that each row passed to the PBPK model is a joint draw, rather than a mixture of values drawn from different rows.
  • Check plausibility and support. The log-scale construction ensures positive parameter values. It does not ensure that every combination is biologically plausible; an assumed joint distribution can still assign weight to implausible combinations.

Independent Monte Carlo sampling is appropriate when independence is justified, or when used explicitly as a comparison scenario. Joint sampling is preferable when evidence or scientific knowledge supports dependence. A covariance-based calculation such as the one derived here is an efficient check for an affine output, but it summarizes only second moments. Directly propagating joint samples through a PBPK model also reflects its actual output calculation, though its results still depend on the chosen joint input distribution. Marginal ranges alone cannot determine dependence: many different joint distributions share the same marginals.

The worked variance formula is exact for an affine output with finite input variances. The normal intervals and the specific sampling construction additionally assume a bivariate-normal distribution on the log scale. Correlation alone does not establish that distribution, and equal correlations can coexist with different joint shapes. For a nonlinear PBPK response, dependence can affect more than output variance; that requires a separate analysis of the response to uncertain inputs. Finally, epistemic uncertainty in parameters should not be confused with interindividual variability: a joint parameter distribution used to express uncertainty does not, by itself, represent a population of individuals.

Key takeaway: When uncertain parameters are dependent, preserve their joint structure in every sampled parameter set. For an affine output, the dependence contribution is \(2ab\rho\sigma_U\sigma_V\); its sign and size depend on both the correlation and the output sensitivities.

Check Your Understanding

Use \(Y=30+10U+10V\), \(\sigma_U=0.30\), and \(\sigma_V=0.20\) for the questions below.

  1. If \(\rho=0.50\), calculate \(\operatorname{Var}(Y)\) and \(\operatorname{SD}(Y)\), with units.
  2. What correlation gives \(\operatorname{Var}(Y)=16\;(\text{mg}\cdot\text{h}/\text{L})^2\)?
  3. Why can independent sampling be misleading when the marginal distributions are correct?
  4. In the joint-normal sampling construction, what does setting \(\rho=0\) do to the formula for \(V\)?

Solutions. (1) The covariance is \(0.50(0.30)(0.20)=0.030\). Thus the variance is \(9+4+2(10)(10)(0.030)=19\;(\text{mg}\cdot\text{h}/\text{L})^2\), and the standard deviation is \(\sqrt{19}=4.359\ \text{mg}\cdot\text{h}/\text{L}\). (2) The variance is \(13+200(0.30)(0.20)\rho=13+12\rho\). Solving \(16=13+12\rho\) gives \(\rho=0.25\). (3) Correct marginals do not preserve the joint distribution. Independent draws omit the covariance contribution and can therefore change the output uncertainty. (4) The formula reduces to \(V=0.20Z_2\). Because \(Z_1\) and \(Z_2\) are independent, \(U=0.30Z_1\) and \(V\) are independent under this bivariate-normal construction.