Tutorials › Quantitative Systems Pharmacology (QSP) › Detecting Parameter Interactions with Second-Order Indices

Sensitivity Analysis and Model Reduction · Tutorial 670 of 799

Detecting Parameter Interactions with Second-Order Indices

Learn to estimate and interpret second-order Sobol indices for selected QSP parameter pairs, including the assumptions, computation, and diagnostics needed to separate pairwise interaction variance from other effects.

Advanced 12 min read

What You'll Learn

  • Define a second-order Sobol index as the variance contribution from a pure two-parameter interaction.
  • Distinguish a pair interaction from main effects and from higher-order terms that also involve that pair.
  • Construct a pick-and-freeze estimator using independent base samples and hybrid parameter vectors.
  • Work through exact and Monte Carlo calculations for hypothetical QSP outputs.
  • Diagnose noisy or misleading estimates, including negative finite-sample estimates.
  • Choose when pairwise indices add useful detail beyond first- and total-order summaries.

Why Isolate Pairwise Interactions?

A global sensitivity analysis can show that two QSP parameters are influential without revealing whether their effects combine independently or interact. For example, a receptor abundance parameter and a binding-affinity parameter may each affect a predicted response, while their joint effect may also include a distinct contribution that cannot be assigned to either parameter alone. A second-order Sobol index quantifies that pure pairwise contribution for a specified output and input distribution.

The distinction matters when deciding which mechanisms to investigate, which parameter combinations to measure more carefully, or which terms might be simplified during model reduction. A pairwise index is a variance attribution under a defined input distribution; it is not evidence that a biological interaction is causal, and it is not a universal property of the model independent of that distribution.

Definition and Assumptions

Let \(Y=f(X_1,\ldots,X_d)\) be a scalar model output, where \(X_1,\ldots,X_d\) are uncertain parameters. The standard Sobol decomposition assumes that the inputs are mutually independent and that \(Y\) has finite, nonzero variance \(V=\operatorname{Var}(Y)\). The variance decomposition then assigns orthogonal components to individual inputs, pairs, triples, and larger subsets.

Definition: The second-order Sobol variance component for distinct inputs \(X_i\) and \(X_j\) is \(V_{ij}=\operatorname{Var}(\mathbb{E}[Y\mid X_i,X_j]) - V_i - V_j\), where \(V_i=\operatorname{Var}(\mathbb{E}[Y\mid X_i])\) and \(V_j=\operatorname{Var}(\mathbb{E}[Y\mid X_j])\). The second-order Sobol index is \(S_{ij}=V_{ij}/V\). It measures the fraction of output variance attributable to the pure interaction between \(X_i\) and \(X_j\), excluding their main effects and excluding terms that also require additional inputs.

Equivalently, in the functional ANOVA decomposition, write \(f=f_0+\sum_i f_i(X_i)+\sum_{i<j} f_{ij}(X_i,X_j)+\cdots\), with components orthogonal under the independent input distributions. Then \(V_{ij}=\operatorname{Var}(f_{ij})\). The zero-mean component \(f_{ij}\) is what remains after removing the mean and both one-input effects from the conditional response given the pair.

$$ f_{ij}(X_i,X_j)=\mathbb{E}[Y\mid X_i,X_j]-\mathbb{E}[Y\mid X_i]-\mathbb{E}[Y\mid X_j]+\mathbb{E}[Y]. $$

The independence assumption is essential to this variance attribution. If parameters are correlated, the standard orthogonal decomposition does not apply as written: effects can be shared or assigned differently depending on the attribution method. For correlated inputs, the interpretation of \(S_{ij}\) given here should not be used without an appropriate alternative decomposition.

Deriving a Pick-and-Freeze Estimator

A second-order component can be isolated using two independent parameter samples. Let \(A\) and \(B\) be independent input vectors, each drawn from the joint input distribution. Define \(A_B^i\) as vector \(A\) with coordinate \(i\) replaced by \(B_i\), and \(A_B^{ij}\) as \(A\) with both coordinates \(i\) and \(j\) replaced by their values in \(B\). For a model output \(f(\cdot)\), form the contrast \(D_{ij}=f(A_B^{ij})-f(A_B^i)-f(A_B^j)+f(A)\).

$$ S_{ij}= \frac{\mathbb{E}\!\left[f(B)D_{ij}\right]}{V}. $$

Here is the justification. A functional ANOVA term that does not depend on \(X_i\) or does not depend on \(X_j\) cancels in the four-term contrast. The surviving terms must depend on both. In the expectation of \(f(B)D_{ij}\), orthogonality and independence make contributions involving additional coordinates vanish: those coordinates in the contrast are taken from \(A\), independently of the corresponding coordinates in \(B\), and the relevant ANOVA components have mean zero. The only surviving covariance is the pure \(f_{ij}\) component with itself, whose expectation is \(\operatorname{Var}(f_{ij})=V_{ij}\). Dividing by \(V\) gives the stated index.

With \(N\) independent base pairs \(A^{(r)},B^{(r)}\), estimate the numerator by the average of the products \(f(B^{(r)})D_{ij}^{(r)}\), and divide by an estimate \(\hat V\) of the output variance. This ratio is a consistent Monte Carlo estimator under the stated assumptions. It is not guaranteed to lie between zero and one in a finite run: sampling noise can make its numerator negative or larger than the estimated total variance. The true \(S_{ij}\), by contrast, is nonnegative.

Worked Example: Exact Pairwise Decomposition in a Three-Parameter Model

Consider a hypothetical QSP output \(Y=X_1+2X_2+X_3+3X_1X_2\), where the three inputs are mutually independent and each takes values \(-1\) and \(1\) with equal probability. Each input has mean zero and variance one. The products \(X_1X_2\), \(X_1\), \(X_2\), and \(X_3\) are mutually orthogonal, so the output variance is the sum of the component variances.

The main-effect variances are \(V_1=1\), \(V_2=2^2=4\), and \(V_3=1\). The pairwise term is \(3X_1X_2\), with variance \(3^2\operatorname{Var}(X_1X_2)=9\), because \(X_1X_2\) also has mean zero and variance one. Thus \(V=1+4+1+9=15\), and

$$ S_{12}=\frac{V_{12}}{V}=\frac{9}{15}=0.600. $$

There are no other pairwise components in this model, so \(S_{13}=S_{23}=0\). The value \(0.600\) means that 60.0% of the output variance under these specified input distributions is due to the pure \(X_1\)-by-\(X_2\) interaction. It does not mean that either parameter alone explains 60.0%; their main effects contribute \(1/15\) and \(4/15\), respectively.

Worked Example: Estimating a Pair Index from Pick-and-Freeze Samples

Suppose a hypothetical Monte Carlo run evaluates a model for two selected parameters \(X_1\) and \(X_2\), alongside any other independent inputs. For each row, the analyst calculates \(D_{12}=f(A_B^{12})-f(A_B^1)-f(A_B^2)+f(A)\), then multiplies it by \(f(B)\). In a run with \(N=1{,}000\) base pairs, suppose the mean of those products is \(8.94\), and the estimated output variance from the base evaluations is \(14.9\) response-units squared.

The estimated index is

$$ \hat{S}_{12}=\frac{8.94}{14.9}=0.600. $$

The estimate attributes about 60.0% of output variance to the pure interaction of the selected pair, under the input distributions used to generate the samples. This is an estimate, not an exact fraction: Monte Carlo error affects both the numerator and the variance estimate. It should therefore be reported with an uncertainty assessment or stability check, especially if it will drive a consequential modeling decision.

Separating Pair Effects from Higher-Order Effects

A second-order index does not capture every interaction in which both selected parameters participate. For instance, a three-input component \(f_{ijk}(X_i,X_j,X_k)\) belongs to the third-order variance contribution, not to \(V_{ij}\). The total-order index for \(X_i\) includes all variance components involving \(X_i\), including pairwise and higher-order terms, but \(S_{ij}\) isolates only the pure two-input component.

Worked Example: A Strong Three-Way Effect with No Pairwise Indices

For a hypothetical output \(Y=X_1X_2X_3\), let the three inputs again be independent, equally likely \(-1\) or \(1\). The output has mean zero and variance one. Conditioning on any single input or any pair leaves the remaining independent input with mean zero, so \(\mathbb{E}[Y\mid X_i,X_j]=0\) for every pair. Therefore \(V_{ij}=0\) and \(S_{ij}=0\) for all three pairs.

The output is nevertheless entirely determined by a three-way interaction: its third-order variance component is one. This example shows why a small \(S_{ij}\) does not prove that the two parameters never participate together in the model response. Their joint influence may occur only as part of a higher-order term.

Choosing Pairs, Diagnosing Estimates, and Comparing Approaches

Computing every pair requires \(d(d-1)/2\) pair-specific contrasts, each formed from model evaluations on hybrid inputs. For a large QSP parameter set, this can be expensive, so selected pairs are often chosen using mechanistic hypotheses or earlier screening results. Pair indices are most informative when the scientific question concerns a specific combination, such as two parameters controlling a shared pathway or feedback loop.

Check the estimator before interpreting a ranking. Confirm that the inputs were sampled independently from the intended distributions, that every hybrid vector changes only the specified coordinates, and that all runs use the same deterministic model settings or a consistent treatment of simulation noise. Compare estimates across independent batches or increasing sample sizes. A negative estimate is a warning about finite-sample noise, not evidence of a negative true variance component. A value above one can also arise from noisy numerator or denominator estimates and should prompt more computation and a check of the implementation.

First- and total-order Sobol indices answer broader questions: the main effect of one parameter, and the total variance contribution involving that parameter, respectively. Second-order indices add detail by isolating selected pure pair terms. The Fourier Amplitude Sensitivity Test offers a different variance-decomposition strategy, while pick-and-freeze estimates are convenient for targeted pairs when the model can be evaluated at hybrid inputs. Neither approach removes the need to define the output and input distributions carefully.

A practical limitation is that a pairwise table is incomplete when higher-order interactions matter. Also, indices depend on the parameter distributions and on the chosen output; changing either can change the variance allocation. Treat a large pair index as evidence of a substantial interaction contribution to variance under the specified analysis, not as a causal effect or a direct measure of biological mechanism strength.

Key takeaway: \(S_{ij}\) is the fraction of output variance assigned to the pure interaction component for a selected pair of independent inputs. It excludes both main effects and interactions that also involve other inputs.

Check Your Understanding

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

  1. If \(V=20\) response-units squared and \(V_{ij}=5\) response-units squared, what is \(S_{ij}\), and what does it mean?
  2. Why can a finite-sample estimate of \(S_{ij}\) be negative even though the true index cannot be negative?
  3. In the model \(Y=X_1X_2X_3\) with independent centered inputs, what do zero pairwise indices imply about the three-way contribution?
  4. What assumption makes the standard orthogonal Sobol decomposition appropriate, and what issue arises if the parameters are correlated?

Solutions. (1) \(S_{ij}=5/20=0.25\), so the pure pair component accounts for 25% of output variance under the specified distributions. (2) Monte Carlo error can make the estimated numerator negative; the true component is a variance and is nonnegative. (3) Nothing about the three-way contribution: it can be large or even account for all the variance, as in the worked example. (4) The inputs must be mutually independent; with correlated parameters, the standard orthogonal attribution does not apply as written, so effects may be shared or assigned differently.

Clinical Biostats