First-Order Uncertainty Propagation with the Delta Method

Why Local Sensitivity Can Approximate Output Uncertainty

A PBPK prediction is a function of inputs such as tissue partition coefficients, clearance, binding fractions, and physiological volumes. If those inputs are uncertain, the prediction is uncertain too. A direct way to approximate that uncertainty is to ask how much the model output changes near a representative parameter value when each input changes slightly. The delta method uses those local changes, together with the input variances, to approximate the output variance.

The method is especially useful when the model is differentiable and the plausible input values are concentrated near their means. It gives a compact calculation and makes clear which parameter uncertainties contribute most to output uncertainty. It is a local approximation, however, not a guarantee that the model behaves linearly across the full range of plausible inputs.

Definition: Let \(\boldsymbol{\theta}=(\theta_1,\ldots,\theta_k)^\mathsf{T}\) be a vector of uncertain model parameters, with mean vector \(\boldsymbol{\mu}\) and covariance matrix \(\boldsymbol{\Sigma}\). For a differentiable scalar output \(Y=g(\boldsymbol{\theta})\), the first-order delta-method approximation to its variance is

\operatorname{Var}(Y)\approx \nabla g(\boldsymbol{\mu})^\mathsf{T} \boldsymbol{\Sigma} \nabla g(\boldsymbol{\mu}).

Here \(\nabla g(\boldsymbol{\mu})\) is the vector of partial derivatives of the output with respect to the parameters, evaluated at their means. If the parameters are mutually independent, this reduces to

\operatorname{Var}(Y)\approx \sum_{j=1}^{k} \left( \frac{\partial g}{\partial \theta_j}\bigg|_{\boldsymbol{\theta}=\boldsymbol{\mu}} \right)^2 \operatorname{Var}(\theta_j).

Deriving the First-Order Formula

The approximation comes from replacing the model near \(\boldsymbol{\mu}\) with its tangent plane. Assume that \(g\) is differentiable in a neighborhood of \(\boldsymbol{\mu}\), that the parameter vector has finite variances, and that uncertainty is sufficiently concentrated for changes in curvature over that neighborhood to be small. The first-order Taylor expansion is

g(\boldsymbol{\theta}) \approx g(\boldsymbol{\mu})+ \nabla g(\boldsymbol{\mu})^\mathsf{T} (\boldsymbol{\theta}-\boldsymbol{\mu}).

The first term is constant, so it contributes no variance. For any fixed vector \(\mathbf{a}\), the variance of a linear combination \(\mathbf{a}^\mathsf{T}\boldsymbol{\theta}\) is \(\mathbf{a}^\mathsf{T}\boldsymbol{\Sigma}\mathbf{a}\). Substituting \(\mathbf{a}=\nabla g(\boldsymbol{\mu})\) gives

\operatorname{Var}\{g(\boldsymbol{\theta})\} \approx \operatorname{Var}\!\left\{ \nabla g(\boldsymbol{\mu})^\mathsf{T} (\boldsymbol{\theta}-\boldsymbol{\mu}) \right\} = \nabla g(\boldsymbol{\mu})^\mathsf{T} \boldsymbol{\Sigma} \nabla g(\boldsymbol{\mu}).

This also shows why covariance matters: the output depends on how parameter changes occur together, not only on the separate parameter variances. If the inputs are independent, the off-diagonal covariances are zero and only the sum of squared sensitivities times variances remains. The approximation to the output mean at this order is \(E(Y)\approx g(\boldsymbol{\mu})\); curvature can make the true mean differ from that value.

Worked Example: Tissue Amount From Partitioning and Plasma Concentration

Hypothetical values. Consider a tissue with fixed volume \(V_t=2.0\ \mathrm{L}\). Suppose its predicted amount is \(A_t=V_tK_pC_p\), where \(K_p\) is the tissue-to-plasma partition coefficient and \(C_p\) is plasma concentration. Let \(K_p\) have mean \(3.0\) and standard deviation \(0.30\), and let \(C_p\) have mean \(2.0\ \mathrm{mg/L}\) and standard deviation \(0.20\ \mathrm{mg/L}\). Assume the two uncertain inputs are independent and are concentrated enough around their means for a first-order approximation to be reasonable.

At the mean inputs, the predicted amount is

A_t=2.0\ \mathrm{L}\times3.0\times2.0\ \mathrm{mg/L} =12.0\ \mathrm{mg}.

The local sensitivities are

\frac{\partial A_t}{\partial K_p}=V_tC_p=2.0\times2.0=4.0\ \mathrm{mg}, \qquad \frac{\partial A_t}{\partial C_p}=V_tK_p=2.0\times3.0=6.0\ \mathrm{L}.

The units of the first derivative are milligrams per unit of \(K_p\); those of the second are liters, or milligrams per milligram-per-liter. Applying the independent-input formula,

\operatorname{Var}(A_t)\approx (4.0\ \mathrm{mg})^2(0.30)^2+ (6.0\ \mathrm{L})^2(0.20\ \mathrm{mg/L})^2 =1.44+1.44=2.88\ \mathrm{mg}^2.

Thus the first-order output standard deviation is \(\sqrt{2.88}=1.697\ \mathrm{mg}\), approximately \(1.70\ \mathrm{mg}\). Relative to the \(12.0\ \mathrm{mg}\) prediction, this is about \(14.1\%\). The result is a local standard-deviation approximation; by itself it is not a confidence interval or a probability statement about where the output lies.

Applying the Calculation to PBPK Outputs

For a PBPK model, the output \(g(\boldsymbol{\theta})\) might be a tissue amount, a concentration at a specified time, or an exposure summary. The calculation follows the same sequence: specify the output as a function of uncertain inputs, evaluate its partial derivatives at the input means, and combine those derivatives with the input variances and any relevant covariances. Each derivative should be interpreted in the units of output per unit of parameter.

For independent inputs, a parameter contributes more to the approximate output variance when its variance is large, its local sensitivity is large, or both. This is a statement about its contribution at the chosen parameter values. It is not a general ranking of parameter importance at other doses, times, or physiological conditions, where the model derivatives may differ.

Worked Example: Uncertainty in AUC From Uncertain Clearance

Hypothetical values. For a simple linear exposure prediction, take \(\mathrm{AUC}=D/CL\), where dose \(D=100\ \mathrm{mg}\) is fixed and clearance \(CL\) has mean \(5.0\ \mathrm{L/h}\) and standard deviation \(0.50\ \mathrm{L/h}\). Assume clearance remains positive over its plausible range, its uncertainty is modest relative to its mean, and the function is locally well approximated by its tangent near \(5.0\ \mathrm{L/h}\).

The output at the mean clearance is

\mathrm{AUC}=\frac{100\ \mathrm{mg}}{5.0\ \mathrm{L/h}} =20.0\ \mathrm{mg\cdot h/L}.

The derivative with respect to clearance, evaluated at its mean, is

\frac{\partial \mathrm{AUC}}{\partial CL} =-\frac{D}{CL^2} =-\frac{100}{(5.0)^2} =-4.00\ \mathrm{mg\cdot h/L\ per\ (L/h)}.

Therefore,

\operatorname{Var}(\mathrm{AUC}) \approx(-4.00)^2(0.50\ \mathrm{L/h})^2 =4.00\ (\mathrm{mg\cdot h/L})^2, \qquad SD(\mathrm{AUC})\approx2.00\ \mathrm{mg\cdot h/L}.

The negative derivative means that a local increase in clearance decreases AUC. The variance uses the squared derivative, so its sign is positive. The calculation indicates an approximate standard deviation of \(2.00\ \mathrm{mg\cdot h/L}\) around the local prediction; it does not imply that the exact output distribution is symmetric.

Worked Example: Hepatic Extraction With Two Uncertain Inputs

Hypothetical values. In a simplified well-stirred liver relationship, let hepatic extraction be \(E=x/(Q+x)\), where \(x=f_uCL_{\mathrm{int}}\), \(Q\) is hepatic blood flow, \(f_u\) is the unbound fraction, and \(CL_{\mathrm{int}}\) is intrinsic clearance. Fix \(Q=90\ \mathrm{L/h}\). Let \(f_u\) have mean \(0.10\) and standard deviation \(0.010\), and let \(CL_{\mathrm{int}}\) have mean \(20\ \mathrm{L/h}\) and standard deviation \(2.0\ \mathrm{L/h}\). Assume the two inputs are independent, \(Q\) is fixed, and their uncertainty is local enough for the first-order approximation.

At the mean inputs, \(x=0.10\times20=2.0\ \mathrm{L/h}\), so

E=\frac{2.0}{90+2.0}=0.02174.

Writing \(x=f_uCL_{\mathrm{int}}\), the derivatives are

\frac{\partial E}{\partial f_u} =\frac{QCL_{\mathrm{int}}}{(Q+f_uCL_{\mathrm{int}})^2} =\frac{90\times20}{92^2} =0.2127,

\frac{\partial E}{\partial CL_{\mathrm{int}}} =\frac{Qf_u}{(Q+f_uCL_{\mathrm{int}})^2} =\frac{90\times0.10}{92^2} =0.001063\ \mathrm{h/L}.

The first derivative is per unit change in the dimensionless fraction \(f_u\); the second is per \(\mathrm{L/h}\) change in intrinsic clearance. With independent inputs,

\operatorname{Var}(E)\approx (0.2127)^2(0.010)^2+ (0.001063\ \mathrm{h/L})^2(2.0\ \mathrm{L/h})^2 =9.04\times10^{-6}.

The approximate standard deviation is \(0.00301\) in extraction fraction, equivalent to about \(0.301\) percentage points. This gives a local measure of uncertainty in \(E\), not an interval constrained to the range from zero to one.

Checking Whether the Approximation Is Trustworthy

A useful diagnostic is to compare the input uncertainty with the scale over which the model’s slope changes. For example, the clearance example has a coefficient of variation of \(0.50/5.0=10\%\), so the uncertain values are relatively close to the expansion point. That supports—but does not prove—the local approximation. If a parameter has a broad or strongly skewed distribution, the tangent may represent only a small part of its plausible range.

Check the following before relying on a delta-method result:

Differentiability: the output must have well-defined local derivatives at the parameter means. A threshold, discontinuity, or switching rule can invalidate a single local linear approximation.
Scale and support: a local linear calculation can imply values outside a parameter’s physical bounds, such as a negative fraction, even when the underlying parameter cannot be negative.
Curvature: if the slope changes substantially across plausible inputs, second-order effects may matter for the output mean and variance.
Input dependence: if uncertain parameters are correlated, omitting their covariance terms can overstate or understate the output variance. Use the full covariance-matrix expression when those covariances are known and relevant.
Units and evaluation point: derivatives must use the stated parameter units and be evaluated at the same parameter values used for the output prediction.

For a linear output function, the first-order variance formula is exact. For a nonlinear function, it is an approximation whose quality depends on local curvature and the spread of the uncertain inputs. When those conditions are doubtful, an exact transformation may be available for a simple model; otherwise, a more complete uncertainty analysis is preferable. The delta method is particularly valuable as a transparent local calculation and as a check on whether the approximate uncertainty is of a plausible magnitude.

A Common Misinterpretation

Common pitfall: “The derivative is negative, so the output variance should be negative.” This confuses the direction of local change with the magnitude of uncertainty. The derivative’s sign describes whether the output rises or falls as an input increases. In the independent-input variance formula, each derivative is squared, so each variance contribution is nonnegative. With correlated inputs, cross-terms can be negative, but the full covariance-matrix expression is the variance of the linearized output and cannot be negative.

Another error is to call the approximate output mean \(g(\boldsymbol{\mu})\) the exact mean of a nonlinear prediction. The first-order expansion omits curvature, so the true \(E\{g(\boldsymbol{\theta})\}\) need not equal \(g(\boldsymbol{\mu})\). The approximation should be reported as local, particularly when inputs are skewed, bounded, or widely dispersed.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions below.

For \(Y=4\theta\), where \(\theta\) has variance \(0.25\), what is the delta-method variance of \(Y\)?
If two independent inputs contribute variance terms of \(0.36\) and \(0.64\), what is the approximate output standard deviation?
For \(\mathrm{AUC}=D/CL\), is the local derivative with respect to \(CL\) positive or negative, and what does its sign mean?
Why can \(g(\boldsymbol{\mu})\) differ from the true mean output when \(g\) is nonlinear?

Solutions. (1) The derivative is \(4\), so \(\operatorname{Var}(Y)\approx4^2(0.25)=4\). (2) The total approximate variance is \(0.36+0.64=1.00\), giving standard deviation \(1.00\) in output units. (3) The derivative is \(-D/CL^2\), which is negative for positive dose and clearance; locally, increasing clearance decreases AUC. (4) A nonlinear function has curvature, so averaging its transformed values need not equal transforming the input means; the first-order calculation omits that curvature.