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Biological Plausibility and Parameter Uncertainty · Tutorial 176 of 799

Nonlinear Responses to Uncertain Inputs

Learn why nonlinear PBPK outputs can be skewed and why a prediction from mean inputs may differ from the mean prediction across uncertain inputs.

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What You'll Learn

  • Distinguish a prediction at mean inputs from the mean of predictions across uncertain inputs.
  • Use curvature and Jensen’s inequality to anticipate the direction of a mean shift.
  • Work through nonlinear hepatic clearance, saturable elimination, and inverse-clearance examples.
  • Recognize how nonlinear transformations can create asymmetric output distributions.
  • Choose between plug-in predictions, local approximations, and distribution propagation.
  • Diagnose when a mean prediction hides important features of the output distribution.

Why Nonlinear Responses Need Careful Interpretation

A PBPK model maps inputs, such as intrinsic clearance, tissue partition coefficients, or enzyme abundance, to outputs such as clearance or exposure. When that mapping is nonlinear, a prediction obtained by substituting mean input values need not equal the mean prediction across plausible input values. The discrepancy is not a computational error: it is a consequence of the model’s curvature.

The same nonlinearity can change the shape of an output distribution. Even if an input is symmetric, a nonlinear transformation may produce an asymmetric output; conversely, a skewed input distribution may be compressed or reshaped by the model. Consequently, reporting only a prediction at the mean inputs can conceal both a shift in the output mean and an important tail of plausible outcomes.

Definition: Let \(\boldsymbol{\theta}\) denote uncertain model inputs with a specified joint distribution, and let \(Y=g(\boldsymbol{\theta})\) be a deterministic model output. The prediction at mean inputs is \(g(E[\boldsymbol{\theta}])\), when the mean input vector is defined and biologically meaningful. The mean prediction is \(E[g(\boldsymbol{\theta})]\). These quantities generally differ when \(g\) is nonlinear.

Here, the distribution describes uncertainty about inputs and the output distribution induced by that uncertainty. The model is assumed deterministic for each input set; no additional random output variation is included. Input dependence must be represented in the joint distribution rather than silently discarded. The expectation \(E[g(\boldsymbol{\theta})]\) is defined only when the output is integrable under that distribution.

Curvature Explains the Mean Shift

For a single uncertain input \(X\), the size and direction of the difference between \(E[g(X)]\) and \(g(E[X])\) depend on the curvature of \(g\). If \(g\) is convex over the full range of possible inputs, Jensen’s inequality gives \(E[g(X)] \geq g(E[X])\). If \(g\) is concave over that range, the inequality reverses. These are global statements: if curvature changes sign over the range, neither direction is guaranteed.

A local Taylor expansion helps explain the effect when uncertainty is concentrated near the mean. Write \(\mu_X=E[X]\). Expanding \(g(X)\) around \(\mu_X\) gives a second-order approximation:

$$ g(X) \approx g(\mu_X) + g'(\mu_X)(X-\mu_X) + \frac{1}{2}g''(\mu_X)(X-\mu_X)^2. $$

Taking expectations eliminates the linear term because \(E[X-\mu_X]=0\), while \(E[(X-\mu_X)^2]=\operatorname{Var}(X)\). Thus, for sufficiently concentrated inputs and a smooth function,

$$ E[g(X)] \approx g(\mu_X) + \frac{1}{2}g''(\mu_X)\operatorname{Var}(X). $$

This is a local approximation, not an exact identity in general. Its leading correction has the sign of the local curvature and grows with input variance. If the input range is wide, curvature changes substantially, or the output has a long tail, this approximation may be poor; direct evaluation over the input distribution is then more informative.

Worked Example: Uncertain Intrinsic Clearance and Hepatic Clearance

Consider a hypothetical well-stirred liver model with hepatic blood flow \(Q=90.0\ \mathrm{L/h}\). For this example, assume the well-stirred model applies, flow is fixed, and the only uncertain input is intrinsic clearance \(CL_{\mathrm{int}}\). Let \(CL_{\mathrm{int}}\) take values \(5.00,\ 20.0,\) or \(80.0\ \mathrm{L/h}\) with probabilities \(0.603111111,\ 0.262777778,\) and \(0.134111111\), respectively. The probabilities sum to 1.

The model’s hepatic clearance is \(CL_h=QCL_{\mathrm{int}}/(Q+CL_{\mathrm{int}})\). At the three possible input values, the predictions are:

$$ \begin{aligned} CL_h(5.00) &= \frac{90.0(5.00)}{90.0+5.00}=4.736842\ \mathrm{L/h},\\ CL_h(20.0) &= \frac{90.0(20.0)}{90.0+20.0}=16.363636\ \mathrm{L/h},\\ CL_h(80.0) &= \frac{90.0(80.0)}{90.0+80.0}=42.352941\ \mathrm{L/h}. \end{aligned} $$

The mean input is \(E[CL_{\mathrm{int}}]=19.000\ \mathrm{L/h}\). The mean prediction is the probability-weighted average of the three clearance predictions:

$$ \begin{aligned} E[CL_h] &=0.603111111(4.736842)+0.262777778(16.363636)+0.134111111(42.352941)\\ &=2.856842+4.300000+5.680000\\ &=12.836842\ \mathrm{L/h}. \end{aligned} $$

By contrast, the prediction at the mean intrinsic clearance is

$$ CL_h(E[CL_{\mathrm{int}}]) =\frac{90.0(19.000)}{90.0+19.000} =15.6881\ \mathrm{L/h}. $$

The mean prediction is lower because the well-stirred clearance function is concave in intrinsic clearance: its second derivative is negative for positive \(CL_{\mathrm{int}}\). The associated mean extraction ratio is \(12.836842\ldots/90.0=0.1426\), to four decimal places. The extraction ratio evaluated at the mean intrinsic clearance is \(15.688073\ldots/90.0=0.1743\). These are different summaries of the uncertain model output; dividing the rounded mean clearance \(12.84\) by \(90.0\) instead gives \(0.1427\) at four decimal places.

Nonlinearity Can Reshape the Distribution

A mean shift is only one consequence of a nonlinear response. The full distribution of predictions can be asymmetric, concentrated near a bound, or have a long tail. Those features matter when a decision depends on exceeding a threshold or when extreme exposure values are clinically important. A mean alone does not describe the probability of those outcomes.

Worked Example: Saturable Elimination Rate

Suppose a hypothetical elimination rate follows \(v(C)=V_{\max}C/(K_m+C)\), with fixed \(V_{\max}=12.0\ \mathrm{mg/h}\) and \(K_m=2.00\ \mathrm{mg/L}\). Assume the only uncertain quantity is concentration \(C\), which takes \(1.00,\ 2.00,\) and \(8.00\ \mathrm{mg/L}\) with probabilities \(0.50,\ 0.25,\) and \(0.25\). These inputs and the fixed kinetic parameters satisfy the model’s positive-value conditions.

The mean concentration is \(0.50(1.00)+0.25(2.00)+0.25(8.00)=3.00\ \mathrm{mg/L}\). The rate predictions are \(v(1.00)=4.00\ \mathrm{mg/h}\), \(v(2.00)=6.00\ \mathrm{mg/h}\), and \(v(8.00)=9.60\ \mathrm{mg/h}\). Therefore,

$$ E[v(C)]=0.50(4.00)+0.25(6.00)+0.25(9.60)=5.90\ \mathrm{mg/h}. $$

At the mean concentration, the prediction is \(v(3.00)=12.0(3.00)/(2.00+3.00)=7.20\ \mathrm{mg/h}\). The discrepancy has the expected direction: for positive \(C\), \(v''(C)=-2V_{\max}K_m/(K_m+C)^3<0\), so the response is concave. The output distribution has probability 0.50 at \(4.00\ \mathrm{mg/h}\), 0.25 at \(6.00\ \mathrm{mg/h}\), and 0.25 at \(9.60\ \mathrm{mg/h}\). It is not adequately described by its mean alone.

Worked Example: Inverse Clearance and a High Exposure Tail

For a hypothetical dose of \(100\ \mathrm{mg}\), assume complete systemic availability and linear, time-invariant pharmacokinetics, so that \(AUC=100/CL\). Suppose uncertain clearance \(CL\) is \(1.00,\ 2.00,\) or \(8.00\ \mathrm{L/h}\), with probabilities \(0.50,\ 0.25,\) and \(0.25\). Under these assumptions, the dose-to-clearance relationship applies for each input value.

The mean clearance is \(0.50(1.00)+0.25(2.00)+0.25(8.00)=3.00\ \mathrm{L/h}\). Predictions at the three values are \(100.0,\ 50.0,\) and \(12.50\ \mathrm{mg\cdot h/L}\). The mean predicted AUC is

$$ E[AUC]=0.50(100.0)+0.25(50.0)+0.25(12.50)=65.625\ \mathrm{mg\cdot h/L}. $$

The prediction at mean clearance is \(100/3.00=33.33\ \mathrm{mg\cdot h/L}\), much lower than the mean predicted AUC. The function \(100/CL\) is convex for positive clearance, so Jensen’s inequality predicts the direction of this difference. In this example, the low-clearance input produces a high-exposure value, illustrating how a nonlinear transformation can create an upper output tail.

Propagation, Diagnostics, and Alternatives

When an output is nonlinear, a plug-in prediction at mean inputs is a useful reference but not a substitute for the output distribution. The main alternative is to propagate plausible input sets through the model and summarize the resulting predictions. If input sets are sampled from an appropriate joint distribution, each output is calculated as \(g(\boldsymbol{\theta})\); the resulting values approximate the output distribution. The sample mean estimates \(E[g(\boldsymbol{\theta})]\), while quantiles describe its spread and asymmetry. The input dependence and model constraints must be respected in the sampled sets.

An analytical transformation or numerical integration can provide the output distribution or its moments when the model and input distribution are simple enough. First-order or second-order local approximations can be computationally efficient, but they may miss skewness and tail behavior when uncertainty is broad or curvature is strong. These approaches answer different practical needs: local approximations summarize behavior near a reference input, whereas propagation evaluates the model across its input distribution.

1
Inspect the response function.
Identify the output of interest, its units, and whether the function is convex, concave, or changes curvature over the plausible input range.
2
Check the input distribution.
Confirm that the distribution represents the intended uncertainty, has finite moments needed for the chosen summary, and preserves relevant dependence among inputs.
3
Evaluate predictions across inputs.
Use an analytical calculation, numerical integration, or input-set propagation. Compare the mean prediction with the prediction at mean inputs, but also inspect quantiles and the distribution’s shape.
4
Check stability and interpretation.
For numerical propagation, increase the number of input sets and check that means, quantiles, and tail summaries stabilize. Report the output quantity, units, assumptions, and the summary being interpreted.

Warning signs include a large gap between \(E[g(\boldsymbol{\theta})]\) and \(g(E[\boldsymbol{\theta}])\), output values clustering near a bound, a few extreme predictions dominating the mean, or quantiles that change materially as more input sets are evaluated. These signals do not by themselves prove an implementation error. They may reflect real curvature or a long-tailed input distribution, but they warrant checks of units, parameter ranges, model equations, and sampling stability.

A common mistake is to say, “The parameters average to these values, so the model prediction at those values is the average prediction.” That reasoning is valid for a linear function, not for a general nonlinear one. The correction is to distinguish \(g(E[\boldsymbol{\theta}])\) from \(E[g(\boldsymbol{\theta})]\) and calculate the latter by propagation or a justified approximation. Another pitfall is to infer output symmetry from input symmetry; a nonlinear transformation does not preserve symmetry in general.

Key takeaway: Curvature shifts the mean prediction away from the prediction at mean inputs, and nonlinear transformations can reshape the entire output distribution. When the difference or tail behavior matters, evaluate and report predictions across the uncertain input distribution rather than relying only on a mean-input calculation.

Check Your Understanding

Use the definitions and examples above to answer these questions; brief solutions follow.

  1. If \(g\) is convex across the full range of an uncertain input \(X\), which is at least as large: \(E[g(X)]\) or \(g(E[X])\)?
  2. For a concave response, what is the direction of the mean shift, and what condition on the range is needed to apply Jensen’s inequality in that direction?
  3. In the inverse-clearance example, why does a low-clearance input contribute disproportionately to the high-exposure tail?
  4. Name two checks that can help diagnose unstable or implausible output summaries from input-set propagation.

Solutions. (1) Jensen’s inequality gives \(E[g(X)]\geq g(E[X])\). (2) \(E[g(X)]\leq g(E[X])\) if the response is concave over the full support of \(X\). (3) AUC is inversely proportional to clearance, so smaller positive clearance produces a larger AUC. (4) For example, check units and parameter ranges, confirm that the joint input distribution preserves dependence, and increase the number of propagated input sets to assess stability of means and quantiles.