VVUQ · Module 8 of 10

Sensitivity Analysis

Uncertainty tells you how much the output could vary; sensitivity asks which inputs that variation is attributable to, within the model and the ranges you adopted. Ranking the inputs shows where effort might pay off, but which ranking you compute decides which question you have answered.

01

Readiness check

Learning objectives

  • Calculate a local sensitivity coefficient and a normalized sensitivity, stating units and sign.
  • Distinguish a sensitivity from an uncertainty contribution.
  • Interpret first-order and total-effect Sobol indices and interaction effects under the standard mutually independent-input decomposition.
  • Determine whether an input may be considered for factor fixing using its total-effect index and a stated decision tolerance, not its first-order index alone.

This module ranks inputs by influence. Tick only what you can do closed-notes.

  • Take a partial derivative of a function.
  • Recall a normalized or relative quantity.
  • Recall variance as a measure of spread.
  • Add fractions that should sum to one.
  • Recall that a first-order uncertainty contribution is |ci|u(xi), not ci alone.
0 or 1 weak itemsContinue with this module.
2 weak itemsRevisit propagation and sensitivity in Module 7.
02

The core idea

Local sensitivity measures how the output responds to a small change in one input at one point. Normalising it gives a dimensionless percent-for-percent response. Neither becomes an uncertainty contribution until input uncertainty is included. Global variance-based analysis apportions output variance over a stated joint input distribution: the first-order index is an input's main effect, and the total-effect index adds everything it does through interactions.

local: ci = ∂y/∂xi at a stated pointnormalized: Sinorm = (xi/y)(∂y/∂xi)Sobol: Si = Var(E[Y | Xi])/Var(Y), STi ≥ Si

Sensitivity analysis asks which inputs the output's variation is attributable to. The local sensitivity is the partial derivative ci = ∂y/∂xi at a stated point, holding the others fixed: signed, carrying output-per-input units, and valid only near that point and only for one input at a time. Its magnitude depends on the units and on how the model is parameterized, and a derivative that is zero at one point can be large elsewhere.

Normalising, Sinorm = (xi/y)(∂y/∂xi), gives a dimensionless local percent-for-percent response that compares inputs of different units. This course uses Sinorm for normalized local sensitivity; notation varies across the literature. Normalising removes the units, not the locality, and it is not an uncertainty contribution: that needs the input uncertainty as well, as Module 7 set out.

Global variance-based analysis instead varies the inputs over a stated joint distribution and apportions the output variance. Under the standard Sobol decomposition, which assumes mutually independent inputs, the first-order index Si = Var(E[Y | Xi])/Var(Y) is the main effect of input i acting alone, and the total-effect index STi adds everything input i does through interactions with the others. Within that framework STi ≥ Si always.

A ranking is not a property of the model. It depends on the input ranges and distributions, the dependence assumptions, the output quantity chosen, the model region, and the parameterization. Change any of these and the ranking can change. Sensitivity analysis is a statement about the model under those choices, not a demonstration of physical cause.

The skill works when: you name the ranking criterion, the input ranges, and the dependence assumption before reading anything off the ranking.
The skill breaks down when: one index is used to answer two different questions, most often when a small first-order index is read as licence to fix an input.
The concept. Under the standard independent-input decomposition, global sensitivity splits the output variance into each input's main effect plus interaction variance. The first-order indices rank main effects; the interaction remainder is not attributed to any single input until total-effect indices are computed.
03

The skills, taught in order

Seven skills build local sensitivity, its relationship to uncertainty contribution, global variance-based analysis, and what a ranking does and does not support.

8.1 Local sensitivity coefficients

The local sensitivity of the output to an input is the partial derivative ci = ∂y/∂xi evaluated at a stated point, with the other inputs held fixed. It measures the output response to a small perturbation of one input.

The coefficient is signed: a positive ci means the output rises locally as the input rises, a negative one that it falls, and zero means no first-order effect at that point, which is not the same as no influence elsewhere. It carries output-per-input units, so raw coefficients from different inputs are not comparable, and its magnitude changes if the model is reparameterized.

Where no analytic derivative is available, finite differences are the usual estimate, and the step is a real choice. For a sufficiently smooth deterministic model, a central difference has second-order truncation error, whereas a forward difference has first-order truncation error. Central differences usually provide better truncation accuracy for the same step, but require two evaluations and may still be degraded by roundoff, solver noise, parameter bounds, or discontinuities. There is no universal perturbation size: a blanket rule such as changing every input by 1% is undefined at zero and may be physically meaningless for a bounded, signed, or categorical parameter.

8.2 Normalized sensitivity

Dividing by the output and multiplying by the input gives a dimensionless local coefficient that keeps its sign:

Sinorm = (xi/y)(∂y/∂xi)

read as the local percent output change per percent input change. It lets inputs of different units be compared on one scale. State the point at which it was evaluated, because normalising removes the units, not the locality.

The normalized coefficient requires a nonzero output y. Its percent-to-percent interpretation also requires a nonzero input xi, because a relative input change is undefined at zero. Near an output zero the coefficient may become singular; near an input zero the algebraic value may remain finite, but the percentage interpretation is not meaningful.

8.3 Sensitivity versus uncertainty contribution

A sensitivity says how strongly the output responds. It does not say how much uncertainty an input contributes, because that also depends on how uncertain the input is. Module 7 supplies the pieces: the signed first-order effect di = ciu(xi), the non-negative standard-uncertainty component ui(y) = |di|, and the diagonal variance contribution di2.

For nonzero xi and y these carry over to relative form. The signed relative first-order effect is

di/y = Sinorm · u(xi)/xi

and the non-negative relative standard-uncertainty component is

ui(y)/|y| = |Sinorm| · u(xi)/|xi|

Rank uncertainty contributions with the magnitude expression. Keep the signed form wherever covariance is in play, since Module 7's cross term 2ρijdidj needs the signs. A signed relative effect is not itself a standard uncertainty.

The two rankings can disagree. An input with twice the normalized sensitivity but a quarter of the relative uncertainty contributes half as much. Ranking by ci answers what the model responds to; ranking by ui(y) answers where the output uncertainty is coming from.

Normalising the diagonal terms gives a share of the first-order output variance:

Pi = di2 / Σj dj2

Provided Σjdj2 > 0, Pi is a normalized local first-order variance share and lies between zero and one. If all first-order effects vanish at the evaluation point, the shares are undefined and the local ranking provides no first-order discrimination. This does not establish that the inputs are globally non-influential.

This is valid only under the zero-covariance first-order approximation of Module 7. It is not a Sobol index. Where covariance is present the total also contains cross terms, which cannot generally be assigned to individual inputs without adopting an allocation convention.

8.4 Global sensitivity analysis

Global methods vary all inputs across a stated joint distribution and measure the effect on the output over the whole range, not just at a point. They capture nonlinearity and interactions that local derivatives miss, at the cost of many model evaluations.

A global result is conditional throughout. It depends on the input ranges and distributions adopted, the dependence assumptions, the output quantity analysed, the region of the model explored, and the parameterization, and it requires the output variance to exist. Report those choices with the ranking, because a ranking without them is not reproducible and is not a property of the model.

Variance-based Sobol indices require a probabilistic representation of the inputs and output. Aleatory and epistemic components may be analysed jointly, conditionally, or separately depending on the adopted uncertainty framework. The indices reported here are conditional on the chosen distributions and independence assumptions; they are not automatically appropriate for interval-only or otherwise non-probabilistic epistemic representations.

8.5 Variance-based Sobol indices

This section and the next describe the standard Sobol, or functional-ANOVA, decomposition, which assumes mutually independent inputs. Every property stated below belongs to that framework. The first-order index is

Si = Vi/V, Vi = VarXi(E[Y | Xi]), V = Var(Y)

so Vi is the main-effect variance: the spread of the output's conditional mean as Xi varies over its distribution. Within this framework 0 ≤ Si ≤ 1. For an additive model the first-order indices sum to one; otherwise ΣiSi < 1, and the shortfall is interaction variance. Where the inputs are not mutually independent this decomposition is not unique in this form, and a specialised dependent-input formulation is required.

In practice these indices are estimated by sampling, so they carry sampling error. Report the estimator and the sample count, and check that the ranking has settled rather than only the point estimates. Two indices that are close may be unresolved at the sample size used. Estimators can also return slightly negative index estimates from noise alone, which is an estimation artefact rather than a negative share.

8.6 Total-effect indices and interactions

The first-order index deliberately excludes interactions. The total-effect index includes them:

STi = 1 − VarX~i(E[Y | X~i]) / Var(Y)

where X~i is every input except i. It captures everything input i does, alone and through interaction, so within this framework 0 ≤ STi ≤ 1 and STi ≥ Si always. The gap STi − Si is evidence of interactions involving input i. Summing those gaps double counts, because a shared interaction appears in the total effect of every input it involves; for the same reason the total-effect indices generally sum above one, with equality only for a purely additive model.

Interaction is not a property of the algebra. For y = x1x2 with mutually independent inputs, the interaction share is about 0.001 when each input has a 5% coefficient of variation and about 0.111 when each has 50%. The same product model, two very different decompositions, because the input distributions differ.

MeasureScopeIncludes interactions?Includes input uncertainty?
Local derivative cione stated pointnono
Normalized Sinormone stated point, dimensionlessnono
Component ui(y) = |ci|u(xi)one stated pointnoyes, to first order
Sobol first-order Siglobal, stated joint distributionno, main effect onlyyes, through the distribution
Sobol total-effect STiglobal, stated joint distributionyes, all involving iyes, through the distribution

Local measures are cheap and narrow. Only the last two answer a global question, and only the total-effect index accounts for what an input does through interaction. Both Sobol rows assume mutually independent inputs.

8.7 What a ranking supports, and choosing a method

Name the question before reading the ranking, because two different decisions call for two different indices.

Choosing what to learn. In the idealized factor-prioritization setting, suppose one input can be learned perfectly and fixed at its true value, with comparable investigation costs across inputs. The largest first-order index identifies the input expected to produce the greatest reduction in output variance, averaged over its possible true values. Real measurement or research programmes usually reduce uncertainty only partially and at different costs. The achievable uncertainty reduction, investigation cost, and decision value must therefore be considered alongside Si.

Choosing what to leave out. Under the standard independent-input Sobol decomposition, STi = 0 identifies a non-influential input. A sufficiently small estimated total-effect index may support fixing an input only relative to a stated tolerance for the resulting output-variance or decision impact. The first-order index cannot serve here: for y = x1 + x2x3 with independent inputs symmetric about zero, x2 and x3 each have a first-order index of exactly zero, yet together they carry a quarter of the output variance.

Reconsider before fixing anything when the decision concerns tails or rare events rather than variance, when the input acts on a constraint or a threshold, when the estimated ranking is itself uncertain, when the inputs are dependent, when the model region may shift, or when the decision is safety-critical.

Finally, separate the ranking from the decision. Uncertainty importance, design leverage, controllability, physical influence, and the cost of reducing an uncertainty are five different orderings. A high-sensitivity input you cannot control and a controllable input you already know well are different engineering situations, and no single index settles either. Sensitivity analysis reports that, within the adopted model and variation ranges, changes in an input are associated with the stated output response; causal conclusions need physical reasoning and experiment as well. In very high-dimensional problems, cheaper screening methods may be preferable for a first pass before variance-based indices are estimated.

Engineering connection: a sensitivity study turns a broad UQ result into a focused test plan, but only once you have said which decision it serves: which parameter to measure better, or which to leave out of the analysis.

Source. This module uses the variance-based sensitivity definitions and factor-prioritization/factor-fixing settings described by Saltelli et al. in Global Sensitivity Analysis: The Primer. It does not claim normative compliance.

04

Worked example 1: normalized sensitivity of drag, and what it does not settle

Aerodynamic drag scales with the square of speed, D ∝ v2. Find the normalized sensitivity of drag to speed, and say what it does and does not settle.

Figure 1. For a power law, the normalized sensitivity is just the exponent. Drag rising with the square of speed gives Snorm = 2 wherever v ≠ 0 and the quadratic relation holds: a 1% speed increase raises drag by about 2%. How much uncertainty speed contributes is a separate question.
  1. ProblemFind the normalized sensitivity of drag to speed in Figure 1, and say what it settles.
  2. Given / findD = c·v2. Find the normalized sensitivity Svnorm = (v/D)(∂D/∂v).
  3. AssumptionsDrag follows the quadratic law over the range of interest, and the operating point has v ≠ 0 and D ≠ 0, without which the normalized coefficient and its percent-to-percent reading do not apply.
  4. ModelSvnorm = (v/D)(∂D/∂v); for a power law D ∝ vn, Snorm = n.
  5. Equations∂D/∂v = 2c·vSvnorm = (2c·v)(v/(c·v2)) = 2
  6. SolveSvnorm = (2c·v)(v)/(c·v2) = 2. A 1% change in speed produces about a 2% change in drag.
  7. CheckFor any power law y ∝ xn the normalized sensitivity equals the exponent n, so the quadratic drag law gives exactly 2 wherever v ≠ 0, whatever the constant and whatever the operating speed. It is a local result that holds at every point where the power law holds, and says nothing about behaviour outside that range.
  8. ConclusionDrag responds twice as strongly to a relative change in speed as to a relative change in a linearly acting input. That is a statement about the model's response, not about uncertainty. The relative standard-uncertainty component is |Sinorm| × u(xi)/|xi|, so the ranking depends on the input uncertainties too. If speed is known to 0.5% and the linear input to 4%, the components are 2 × 0.5% = 1.0% and 1 × 4% = 4.0%: the less sensitive input contributes four times as much. Sensitivity alone does not decide where the output uncertainty comes from.
Result. Svnorm = 2 where v ≠ 0: 1% in speed gives about 2% in drag. Ranking uncertainty contributions needs u(xi)/|xi| as well.
05

Worked example 2: variance-based Sobol indices

A global sensitivity study over a stated joint input distribution, with the inputs treated as mutually independent, estimates a total output variance V = 100, a main-effect variance V1 = Var(E[Y | X1]) = 64, and V2 = 25. Find the first-order indices, the interaction share, and the total-effect indices, and say what each supports.

Figure 2. Under the standard independent-input decomposition, the output variance splits into main effects from each input and a remainder due to their interaction. Input x1 has the largest main effect at 64%. The remainder belongs to both inputs, which is why the total effects sum to more than one.
  1. ProblemFind the first-order indices, interaction share, and total-effect indices for the study in Figure 2, and say what each supports.
  2. Given / findV = 100, V1 = 64, V2 = 25. Find S1, S2, the interaction share S12, and ST1, ST2.
  3. AssumptionsThe inputs are treated as mutually independent, which the standard decomposition requires. V1 and V2 are main-effect variances, and with two inputs the remainder is the single pairwise interaction. The indices are estimates from a finite sample and carry sampling error that this problem does not quantify.
  4. ModelSi = Vi/V; with two inputs S12 = 1 − (S1 + S2), ST1 = S1 + S12 and ST2 = S2 + S12.
  5. EquationsS1 = V1/V, S2 = V2/VS12 = 1 − (S1 + S2)STi = Si + S12
  6. SolveS1 = 64/100 = 0.64, S2 = 25/100 = 0.25, S12 = 1 − (0.64 + 0.25) = 0.11. The total effects follow: ST1 = 0.64 + 0.11 = 0.75 and ST2 = 0.25 + 0.11 = 0.36.
  7. CheckThe first-order shares and the interaction sum to 1 by construction, since the interaction was obtained as the remainder; that is bookkeeping rather than a test. What the numbers do tell you is that the interaction is nonzero, so the model is not purely additive, and that the total effects sum to 1.11, above one, because the shared interaction is counted in both. Each index lies in [0,1] and each STi exceeds its Si, as they must within this framework.
  8. ConclusionInput x1 has the largest main effect. In the idealized factor-prioritization setting, where one input can be learned perfectly and fixed at its true value and investigation costs are comparable, x1 is the input expected to produce the greatest reduction in output variance, averaged over its possible true values. A real programme would also weigh how much the uncertainty could actually be reduced, at what cost, and what the decision is worth. If the question were instead which inputs could be fixed and dropped, the total effects are the relevant numbers, and ST2 = 0.36 says x2 cannot be dropped even though its main effect is only a quarter. All of this holds for this input distribution, this independence assumption, and this output quantity.
Result. S1 = 0.64, S2 = 0.25, S12 = 0.11; ST1 = 0.75, ST2 = 0.36, summing to 1.11.
06

Misconceptions and diagnostics

MistakeSymptomDiagnostic questionCorrection
Local derivative over a wide rangeRanking wrong for large variations"Is the input range small?"Use global sensitivity for wide ranges.
Comparing raw derivativesDifferent units confuse the ranking"Did I normalise?"Use normalized sensitivity to compare inputs on one scale.
Ignoring interactionsFirst-order indices do not sum to 1"Where did the rest of the variance go?"Under the independent-input decomposition the shortfall is interaction variance, not error, and it belongs to no single input.
Dropping an input because its first-order index is smallAn interacting input discarded"Is STi small, or only Si?"Fixing an input needs a small total-effect index against a stated tolerance. Si can be zero while STi is large.
Reading a sensitivity as an uncertainty contributionThe wrong input targeted"Have I included u(xi)?"The component is |ci|u(xi). A strong response to a well-known input contributes little.
Calling a local variance share a Sobol indexA local result reported as global"Over what distribution was this computed?"Pi = di2/Σdj2 is a first-order local share under zero covariance. A Sobol index needs a stated joint distribution.
Reading an estimated index as exactA rank order asserted from noise"What sample count, and has the ranking settled?"Estimated indices carry sampling error. Two close indices may be unresolved.
07

Practice ladder

Level 1 · Direct skill

For a power law y ∝ x3, what is the normalized sensitivity to x?

Show answer

The normalized sensitivity equals the exponent, so Snorm = 3: 1% in x gives about 3% in y. This holds wherever x ≠ 0 and y ≠ 0.

Level 2 · Mixed concept

An output variance of 200 has main-effect variances V1 = 150 and V2 = 40, for two inputs treated as mutually independent. Find the first-order indices and the interaction share, then say what you can conclude about the total effects.

Show answer

S1 = 150/200 = 0.75, S2 = 40/200 = 0.20, so S12 = 1 − 0.95 = 0.05. With two inputs the total effects follow: ST1 = 0.75 + 0.05 = 0.80 and ST2 = 0.20 + 0.05 = 0.25, summing to 1.05. Neither input is non-influential on these numbers, so neither can be fixed without accepting the corresponding loss of variance.

Level 3 · Independent problem

(a) For y = x1·x2 at x1 = 4, x2 = 5, find the normalized sensitivity to x1. (b) The same model is analysed globally with X1 and X2 mutually independent. Does the interaction share follow from the algebra?

Show answer

(a) ∂y/∂x1 = x2 = 5, so S1norm = 5 × (4/20) = 1. A product is locally linear in each factor, so the normalized sensitivity is 1. (b) No. For y = X1X2 with mutually independent inputs, Var(Y) = μ12σ22 + μ22σ12 + σ12σ22, and the interaction share is σ12σ22/Var(Y). With a 5% coefficient of variation on each input that is about 0.001; with 50% it is about 0.111. The same product model gives very different decompositions because the input distributions differ, so a product term is not automatically all interaction.

Transfer task | Real engineering

A model has ten uncertain inputs but a limited test budget. Explain how sensitivity analysis focuses the effort, which index answers which question, and what the result would not establish.

What good work looks like

Run a global analysis over an explicitly stated joint input distribution and report both first-order and total-effect indices. Use the first-order indices to choose which two or three inputs to measure better: in the idealized factor-prioritization setting they identify the inputs whose perfect knowledge would give the greatest expected variance reduction, though a real programme must also weigh how far the uncertainty can actually be reduced, at what cost, and what the decision is worth. Use the total-effect indices, not the first-order ones, before fixing anything at nominal, and say what loss of output variance the chosen tolerance accepts: an input with a near-zero main effect can still carry substantial variance through interaction. Report the estimator and sample count and check that the ranking has settled, not just the point estimates. Close down by saying what the result does not establish: it holds for the ranges and distributions adopted, assumes the dependence structure specified, and does not show that changing an input in the real system would produce the modelled effect.

08

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Differentiate this model yourself and check my sensitivity coefficients, including their signs and units."
"Here is my finite-difference step and the resulting estimate. Is the step in a stable range for this model?"
"Give me three models; I will compute the normalized sensitivity of each and say where its percent-to-percent reading stops being meaningful."
"Rank the inputs for me." Computing the sensitivities is the skill.
"Which input matters most?" Matters for which decision, under which distribution, is the question you are meant to answer.
"Pick the input distributions and correlations." These are modelling judgements needing evidence, and they set the ranking.
"Tell me which inputs are negligible." Fixing an input needs a total-effect index and a stated tolerance, not a plausible-sounding answer.

Portfolio task

For a real model, compute both a local ranking and a global one, and compare them. State the point at which the local coefficients were evaluated, the joint input distribution and dependence assumption behind the global indices, and the output quantity. Report first-order and total-effect indices, say which decision each supports, and note where the two rankings disagree and why.

Must include: a local ranking with each input's component ui(y), a global ranking with first-order and total-effect indices, the input distributions and dependence assumption, the sample count, and a statement of what the ranking does not establish.
09

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. What is a local sensitivity coefficient, and what are its units and sign?

ci = ∂y/∂xi at a stated point. It is signed and carries output-per-input units, so raw coefficients from different inputs are not comparable.

2. Why is a normalized sensitivity not an uncertainty contribution?

Because it contains no input uncertainty. The relative component is |Sinorm| × u(xi)/|xi|, so an input with the larger normalized sensitivity can contribute less if it is better known.

3. What does a first-order Sobol index give, and what does it assume?

Si = Var(E[Y | Xi])/Var(Y), the main effect of input i acting alone, under the standard decomposition that assumes mutually independent inputs and a stated joint distribution.

4. What does the total-effect index add, and why do the STi sum above one?

It adds everything input i does through interactions, so STi ≥ Si. A shared interaction is counted in the total effect of every input it involves, so within this framework the sum exceeds one unless the model is additive.

5. Which index bears on leaving an input out, and on what terms?

The total-effect index: STi = 0 identifies a non-influential input, and a small estimated STi may support fixing one only against a stated tolerance for the output-variance or decision impact. A small first-order index does not settle it, since an input can have Si = 0 and still carry variance through interaction.

TodayFinish this quiz and Levels 1 and 2 of the ladder.
+1 dayRe-derive a normalized sensitivity and Sobol indices from a blank page.
+3 daysRank the inputs of three new models.
+7 daysCalibrate a model and predict, Module 9.
+30 daysReuse sensitivity ranking to focus a test plan.