VVUQ · Module 7 of 10
Uncertainty Propagation
Uncertain inputs make an uncertain output. Two methods carry the uncertainty through a model: first-order propagation when the model is close to linear over the uncertainty range, and Monte Carlo sampling when it is not. Both answer only the question you set up, so what you assume about the inputs matters as much as the arithmetic.
Readiness check
Learning objectives
- Calculate a first-order output standard uncertainty from local sensitivity coefficients and input standard uncertainties.
- Include covariance through the 2ρijdidj term and determine its direction from the effect and correlation signs.
- Distinguish the propagated output uncertainty u(y) from the Monte Carlo sampling error SE(Ȳ).
- Identify when first-order propagation is inadequate over the uncertainty range.
This module carries uncertainty through a model. Tick only what you can do closed-notes.
- Take a partial derivative of a function.
- Combine uncorrelated standard uncertainties as a root sum of squares.
- Recall relative uncertainty as a fraction.
- Distinguish a standard deviation from the standard error of a mean.
- Recall that a random sample estimates a mean.
The core idea
Uncertainty propagation turns input standard uncertainties into an output standard uncertainty. First-order propagation weights each input by its sensitivity coefficient and adds the variances, plus covariance terms whenever the inputs are correlated. Monte Carlo draws from an adopted joint input distribution and estimates the output distribution. Neither method checks whether the inputs, the dependence, or the model were right.
u2(y) ≈ Σ ci2u2(xi) + 2 Σi<j cicj cov(xi,xj)ci = ∂f/∂xi at the expansion pointu(y) ≈ sY, SE(Ȳ) = sY/√NA model maps uncertain inputs to an output, and UQ must carry that input uncertainty through to the result. First-order propagation linearises the model about a stated expansion point, usually the nominal inputs. Each input gets a sensitivity coefficient ci = ∂f/∂xi evaluated there, and the output standard uncertainty follows from u2(y) ≈ Σ ci2u2(xi) + 2 Σi<j cicj cov(xi,xj). The covariance sum is not optional decoration. It vanishes when the relevant input pairs are treated as uncorrelated. Independence is a stronger condition that is sufficient, but not necessary, for that simplification.
The method is fast and analytic. What it requires is that the model be close to linear over the range the input uncertainties actually span, that the input uncertainties be compatible standard uncertainties as Module 5 set out, and that the correlation structure be either quantified or explicitly assumed away. It does not require the inputs to be Gaussian. A distributional assumption enters only later, when an output standard uncertainty is turned into an interval through a coverage factor.
If dependence is physically plausible but insufficiently quantified, do not silently set the covariance to zero. State the zero-covariance assumption and examine plausible correlation scenarios when the result is decision-relevant.
Monte Carlo draws from an adopted joint input distribution, evaluates the model, and estimates the resulting output distribution and its summaries. It handles nonlinearity and any input distribution you can sample, at the cost of many model runs. What it does not remove is everything else: the dependence structure must be represented in how the samples are drawn, the model must be evaluable and adequate for the purpose, and increasing N does not correct wrong input distributions, wrong dependence, model-form error, or numerical error. Monte Carlo propagates the adopted input model faithfully, including wherever that model is wrong.
A sampling run carries two quantities that are easy to confuse. The underlying propagated output standard uncertainty u(y) follows from the model and the adopted input distribution, and it is what the propagation question asks for; the sample standard deviation sY is this run's estimate of it. The standard error SE(Ȳ) = sY/√N describes the sampling uncertainty of the Monte Carlo estimate of the expected output across repeated runs. It does not describe the spread of possible output values. Increasing N does not reduce the underlying propagated output uncertainty; it reduces Monte Carlo sampling error in estimated summaries such as the mean, the standard deviation, and quantiles.
The skills, taught in order
Seven skills build first-order propagation, its limits, Monte Carlo propagation, and the two different uncertainties a sampling run produces.
7.1 Sensitivity coefficients, effects, and components
Linearising the model about a stated expansion point, usually the nominal inputs, gives each input a sensitivity coefficient ci = ∂f/∂xi evaluated there. Three quantities follow from it, and they are worth keeping apart.
di = ciu(xi)ui(y) = |di| = |ci|u(xi)di2 = ci2u2(xi)The coefficient ci is signed and carries units of output per unit of input. The sensitivity-weighted first-order effect di is signed and carries the units of the output. The standard-uncertainty component associated with input i, ui(y), is the magnitude of that effect: it is non-negative, also in the units of the output. The diagonal contribution to output variance is di2, in the units of the output squared.
The distinction earns its keep as soon as covariance appears. A standard uncertainty is never negative, so ui(y) is always reported as a magnitude. The cross term for an input pair is 2ρijdidj. Whether it adds to or subtracts from the output variance depends jointly on the signs of the two sensitivity-weighted effects and the sign of their correlation. Discarding the effect signs too early makes the covariance term impossible to interpret correctly.
An input matters in proportion to both its own uncertainty and how strongly the output responds to it, so a large coefficient alone does not make an input important. Worked example 1 shows a case where one coefficient is twice the other and the two components ui(y) are equal. When inputs are correlated, the covariance terms mean the diagonal contributions alone do not give a complete ranking, because a cross term can add to or subtract from the total.
7.2 First-order propagation and covariance
The general first-order result carries a second sum:
u2(y) ≈ Σ ci2u2(xi) + 2 Σi<j cicj cov(xi,xj)= Σ ci2u2(xi) + 2 Σi<j cicj ρij u(xi)u(xj)where ρij is the correlation coefficient, between −1 and +1. The covariance sum vanishes when the relevant input pairs are treated as uncorrelated. Independence is a stronger condition that is sufficient, but not necessary, for this simplification. Inputs become correlated for ordinary reasons: two quantities measured on the same instrument, two properties from the same material batch, two parameters fitted from the same data set.
Written in terms of the first-order effects, the cross term for a pair is 2ρijdidj. Its direction therefore depends jointly on the signs of the two effects and on the sign of their correlation, which leaves four cases. Same-sign effects with positive correlation increase the output variance. Same-sign effects with negative correlation decrease it. Opposite-sign effects with positive correlation decrease it. Opposite-sign effects with negative correlation increase it.
An area L·W has two positive effects, so positive correlation inflates the result and negative correlation deflates it. A ratio a/b has a negative effect on the denominator, so that pair has opposite signs and both directions swap: positive correlation deflates the result and negative correlation inflates it. Reading the direction from the correlation alone is not safe.
If dependence is physically plausible but insufficiently quantified, do not silently set the covariance to zero. State the zero-covariance assumption and examine plausible correlation scenarios when the result is decision-relevant.
7.3 Products and relative uncertainty
For a product or quotient with zero relevant covariance terms, the formula reduces to combining relative uncertainties as a root sum of squares: (u(y)/y)2 = Σ (u(xi)/xi)2. This handy form covers many engineering formulas where quantities multiply. It inherits the zero-covariance assumption from the general formula, and it also hides the signs, since the coefficient on a denominator is negative and squaring removes the evidence. Where the inputs may be correlated, return to the general form, in which the signs survive.
7.4 When linearisation is not enough
First-order propagation is a local approximation, and its quality depends on how far the input uncertainties reach from the expansion point. It may be inadequate when the input uncertainty range is large relative to the curvature of the model, when the output distribution is skewed or multimodal, when thresholds or discontinuities lie inside the input range, when input distributions are bounded or asymmetric, when interactions between inputs are strong, or when the expansion point is not representative of the region the analysis cares about.
A second effect is easy to miss, because first-order propagation leaves the output central value at f evaluated at the expansion point. Take y = X2 at a nominal x = 4. For the exact nonlinear comparison in this illustration, assume X is Gaussian with mean μX = 4 and standard deviation σX equal to the stated u(x), so that
E[X2] = μX2 + σX2sd(X2) = √(4μX2σX2 + 2σX4)The first-order propagation formula does not require Gaussian inputs. Gaussianity is assumed here only to calculate the exact nonlinear reference mean and standard deviation.
At σX = 0.1 the first-order u(y) = 0.8 sits against an exact 0.800125, so the approximation is essentially exact. Raise σX to 2 and the first-order 16.0 understates the exact 16.9706 by about 6%, while the mean of y moves from 16 to 20, a shift of a quarter of the nominal value. When linearisation fails, the central value often moves before the spread is badly wrong.
7.5 Monte Carlo propagation
Monte Carlo draws from an adopted joint input distribution, evaluates the model, and estimates the resulting output distribution and its summaries. Drawing each input from its own marginal distribution represents the joint distribution only where those inputs carry no dependence the model would feel. Where dependence exists, it must be represented in how the samples are drawn, and no sample count reveals its absence.
The model must be evaluable and adequate for the purpose, and increasing N does not correct wrong input distributions, wrong dependence, model-form error, or numerical error. Any moment you report has to exist: a mean and a standard deviation are meaningful summaries only for outputs that have them. Skewed or multimodal outputs are better described with appropriate quantiles or other distributional summaries than with a mean and a standard deviation alone.
7.6 Output uncertainty versus Monte Carlo sampling error
A sampling run carries two quantities that answer different questions. The underlying propagated output standard uncertainty u(y) follows from the model and the adopted input distribution; the sample standard deviation sY is this run's estimate of it. The standard error SE(Ȳ) = sY/√N describes the sampling uncertainty of the Monte Carlo estimate of the expected output across repeated runs. It does not describe the spread of possible output values.
Increasing N does not reduce the underlying propagated output uncertainty. It reduces Monte Carlo sampling error in estimated summaries such as the mean, the standard deviation, and quantiles. The sample standard deviation and the output quantiles carry Monte Carlo sampling error of their own, so convergence should be checked for the actual quantity being reported, not only for the mean. Quantiles far into a tail need many more samples than a mean does. Record the sample count and the random seed so a run can be reproduced.
7.7 Choosing a method
Use first-order propagation where the model is close to linear over the range the input uncertainties span, where the coefficients are available or cheap, and where speed matters. Use Monte Carlo where the model is strongly nonlinear over that range, where the output distribution itself is needed rather than a single standard uncertainty, or where tail behaviour matters. Running both is a good habit, but reading agreement between them as confirmation is a mistake: both consume the same input distributions and the same correlation assumptions, so agreement indicates that the model is close to linear over the range and says nothing about whether those inputs were right.
| Method | Assumes | Cost | Handles |
|---|---|---|---|
| First-order propagation | local linearity over the uncertainty range; compatible standard uncertainties; a stated correlation assumption | one derivative per input | mild nonlinearity |
| Monte Carlo | an adopted joint input distribution; an evaluable, adequate model | many model runs | strong nonlinearity; input distributions you can sample; the full output distribution |
First-order propagation is cheap but local; Monte Carlo is general but costly. Neither removes the need for the input distributions, the correlation structure, and the model to be right.
Module 6 raised a case this module cannot settle by formula. Where aleatory and epistemic components are represented differently, combining them into one propagated standard uncertainty is a modelling choice to be stated rather than a step to be taken automatically. The examples here combine components because they are supplied on compatible standard-uncertainty bases, which is a simplification and not a general rule.
Engineering connection: the input uncertainty term in the validation uncertainty of Module 5 is produced by this propagation, which is why Module 5 insists on compatible standard-uncertainty bases before anything is combined.
Source. This module uses the first-order law of propagation of uncertainty and Monte Carlo propagation framework described in JCGM 100:2008 and JCGM 101:2008. It does not claim normative compliance beyond what it implements here.
Worked example 1: first-order propagation for a product
An area is computed as y = L·W with L = 10.00 ± 0.10 m and W = 5.00 ± 0.05 m, both standard uncertainties. Find the area, its standard uncertainty, and each input's standard-uncertainty component.
- ProblemFind the area, its standard uncertainty, and each input's standard-uncertainty component for the product in Figure 1.
- Given / findy = L·W, L = 10.00 ± 0.10 m and W = 5.00 ± 0.05 m as standard uncertainties. Find y, the signed first-order effects dL and dW, the components uL(y) and uW(y), and u(y).
- AssumptionsL and W are supplied as compatible standard uncertainties in metres, and they are small enough relative to the nominal values for the first-order form to apply. The pair is treated as uncorrelated for the main result, so cov(L,W) = 0. That is an assumption rather than a finding, and the conclusion shows what it is worth.
- Modelu2(y) = cL2u2(L) + cW2u2(W) + 2cLcWρ u(L)u(W), with cL = ∂y/∂L = W and cW = ∂y/∂W = L. With ρ = 0 the last term drops and, for a product, the relative shortcut (u(y)/y)2 = (uL/L)2 + (uW/W)2 gives the same answer.
- Equationsy = L·WcL = W, cW = Lu(y) = √(cL2u2(L) + cW2u2(W)) at ρ = 0
- Solvey = 10.00 × 5.00 = 50.00 m². Coefficients: cL = W = 5.00 m and cW = L = 10.00 m. First-order effects: dL = cLuL = 5.00 × 0.10 = +0.500 m² and dW = cWuW = 10.00 × 0.05 = +0.500 m², both in the units of the output. Both are positive here, so the components are uL(y) = uW(y) = 0.500 m². At ρ = 0, u(y) = √(0.5002 + 0.5002) = 0.707 m². The relative shortcut agrees: √(0.0102 + 0.0102) = 0.01414, and 50.00 × 0.01414 = 0.707 m².
- CheckThe coefficient on W is twice the coefficient on L, yet the two components are equal at 0.500 m², so each input supplies half the diagonal variance. A large coefficient does not make an input important on its own. The root sum of squares, 0.707 m², falls below the linear sum of 1.000 m² because variances add rather than magnitudes.
- ConclusionThe area is 50.00 m² with a standard uncertainty of 0.707 m², conditional on the zero-covariance assumption. The two first-order effects share a sign here, so the cross term 2ρdLdW takes the sign of ρ: at ρ = +0.5 u(y) rises to 0.866 m², and at ρ = −0.5 it falls to 0.500 m². With opposite-sign effects both directions would swap. Where L and W could plausibly be correlated, for instance because one instrument measured both, the assumption belongs in the report beside the number rather than inside it.
Worked example 2: output uncertainty and Monte Carlo sampling error
A Monte Carlo propagation with N = 400 samples gives an output sample standard deviation sY = 1.5. Report the propagated output uncertainty and the standard error of the estimated output mean, say what each one describes, and say what raising N would change.
- ProblemSeparate the two uncertainties carried by the run in Figure 2.
- Given / findN = 400 and sY = 1.5 in the units of the output. Find u(y), SE(Ȳ), and the sample count that halves SE(Ȳ).
- AssumptionsThe 400 draws come from the adopted joint input distribution, and the output has a finite mean and variance so that these summaries exist. For the interval below, the sampling distribution of the mean is taken as approximately normal at this sample count. Whether the adopted input distribution is right lies outside what this calculation can address.
- Modelu(y) ≈ sY is the estimated propagated output standard uncertainty. SE(Ȳ) = sY/√N is the sampling uncertainty of the Monte Carlo estimate of the expected output. They answer different questions, and SE(Ȳ) scales as 1/√N.
- Equationsu(y) ≈ sYSE(Ȳ) = sY/√Nhalve SE(Ȳ) ⇒ 4× N
- SolveThe output sample standard deviation is sY = 1.5, so the estimated propagated output standard uncertainty is u(y) ≈ 1.5. The standard error of the estimated mean is SE(Ȳ) = 1.5/√400 = 1.5/20 = 0.075. An approximate 95% confidence interval for the Monte Carlo estimate of the expected output is ȳ ± 1.96(0.075) = ȳ ± 0.147, under the large-sample normality assumption above. To halve SE(Ȳ) to 0.0375, N must be 4 × 400 = 1600.
- CheckThe two intervals differ by a factor of √N, since 1.96 sY divided by 1.96 sY/√N is √400 = 20. The ±0.147 is not wrong arithmetic; it is the wrong quantity if it is reported as the output uncertainty. A 95% coverage interval for the output quantity itself is ȳ ± 1.96(1.5) only where the output distribution is explicitly assumed approximately Gaussian. Otherwise, obtain output coverage intervals from the empirical output quantiles.
- ConclusionThe estimated propagated output uncertainty is about 1.5, and that answers the propagation question. The 0.075 says how precisely this run located the expected output across repeated runs, and cutting it in half costs four times the runs. Raising N does not reduce the underlying propagated output uncertainty; it reduces Monte Carlo sampling error in the estimated mean, standard deviation, and quantiles alike. The sample standard deviation and the quantiles carry sampling error of their own, so convergence should be checked for the quantity actually being reported, not only for the mean. No sample count corrects a wrong input distribution, an unrepresented dependence, model-form error, or numerical error.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| First-order propagation on a strongly curved model | Both the spread and the central value are off | "Is the model close to linear over the uncertainty range?" | Judge curvature over the range, not at a point. Move to Monte Carlo where it fails. |
| Ranking inputs by coefficient | The wrong input targeted | "What is ui(y) = |ci|u(xi), not just ci?" | The component is the magnitude of the sensitivity-weighted effect, and correlated inputs add cross terms 2ρijdidj whose direction depends on both effect signs together with the correlation sign, which the diagonal misses. |
| Setting covariance to zero because nobody measured it | An assumption absorbed into the answer | "Do these inputs share an instrument, a batch, or a fit?" | State the zero-covariance assumption and examine plausible correlation scenarios when the decision depends on it. |
| Reporting SE(Ȳ) as the output uncertainty | Prediction uncertainty too small by a factor of √N | "Am I describing the output, or my estimate of its expected value?" | u(y) ≈ sY is the output uncertainty; sY/√N is the sampling error of the estimated mean. |
| Raising N to fix a wrong model | A precise answer that stays wrong | "Is this error Monte Carlo sampling error?" | N reduces sampling error in the estimated summaries. Input distributions, dependence, model form, and numerical error are untouched. |
| Reading agreement between the two methods as confirmation | Misplaced confidence | "What did both methods assume?" | Agreement indicates local linearity. Both used the same inputs and the same correlation assumption. |
Practice ladder
For y = x2 at x = 4 with u(x) = 0.1, find u(y) by first-order propagation, and state the units of the coefficient, of the signed first-order effect, and of the standard-uncertainty component.
Show answer
c = ∂y/∂x = 2x = 8, in units of y per unit of x. The signed first-order effect is d = c u(x) = +8 × 0.1 = +0.8, and the standard-uncertainty component is u(y) = |d| = |8| × 0.1 = 0.8, both in the units of y. Only one input is uncertain here, so that component is the whole of u(y). The coefficient on its own does not carry the output units.
A Monte Carlo run with N = 100 gives sY = 2.0. Report the estimated propagated output uncertainty and the standard error of the estimated mean, and say what changes if N rises to 400.
Show answer
u(y) ≈ sY = 2.0. SE(Ȳ) = 2.0/√100 = 0.2. At N = 400, SE(Ȳ) = 2.0/20 = 0.1, halved. The estimate of u(y) also becomes more precise with the larger sample, but the underlying u(y) it is estimating does not fall: that is a property of the model and the adopted inputs, not of the run.
For a quotient y = a/b with a = 20 ± 0.4 and b = 4 ± 0.04, find the relative uncertainty of y with zero covariance. Then say which way the answer moves if a and b are positively correlated, and why.
Show answer
(ua/a)2 + (ub/b)2 = 0.022 + 0.012 = 0.0005, so the relative uncertainty is √0.0005 = 0.0224, or 2.24%, and u(y) = 5 × 0.0224 = 0.112. The two first-order effects have opposite signs: ∂y/∂a = 1/b = +0.25 gives da = +0.100, and ∂y/∂b = −a/b2 = −1.25 gives db = −0.050, so the components are ua(y) = 0.100 and ub(y) = 0.050, both non-negative. The cross term is 2ρdadb, and dadb is negative, so positive correlation reduces the first-order output variance while negative correlation would raise it. At ρ = +0.5 the relative uncertainty is about 1.73%. For a product, where both effects share a sign, those two directions swap.
You must propagate uncertainty through a strongly nonlinear simulation whose inputs come partly from a shared calibration. Argue for a method, say how you would decide the sample count, and state what your result would not establish.
What good work looks like
Monte Carlo, because first-order propagation is local and the model is strongly nonlinear over the input range, and because curvature moves the output central value as well as widening the spread. Sampling must come from the joint input distribution, since the shared calibration correlates those inputs and marginal sampling would represent a different joint distribution. Choose N from the precision needed on the quantity actually being reported, using a pilot run to estimate it, and note that tail quantiles need far more samples than a mean does. Record N and the seed so the run can be reproduced. A good answer closes by saying what the result does not establish: it does not show that the adopted input distributions were right, that the dependence was correctly represented, or that the model was adequate, since Monte Carlo propagates all three faithfully whether or not they are correct.
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Propagate real input uncertainties through a model two ways, first-order and Monte Carlo, and compare the results. State the expansion point, the uncertainty basis of each input, and your correlation assumption. Report the estimated output standard uncertainty and, separately, the standard error of the estimated mean. Where the two methods differ, explain why; where they agree, say what that agreement does and does not establish.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. Write the first-order propagation formula, including the term that is often dropped.
u2(y) ≈ Σ ci2u2(xi) + 2 Σi<j cicj cov(xi,xj). The second sum vanishes when the relevant pairs are treated as uncorrelated; independence is sufficient for that, but not necessary.
2. Why does a large sensitivity coefficient not make an input important?
Because the standard-uncertainty component is ui(y) = |ci|u(xi), a magnitude. A large coefficient on a well-known input can contribute less than a small coefficient on a poorly known one. With correlated inputs the ranking also needs the signed effects di = ciu(xi), because a cross term is 2ρijdidj and its direction depends on both effect signs together with the correlation sign.
3. What does Monte Carlo assume?
An adopted joint input distribution and an evaluable, adequate model. It drops the linearity requirement and little else.
4. A run gives sY and sY/√N. Which is the propagated output uncertainty?
u(y) ≈ sY. The quantity sY/√N is the sampling error of the estimated expected output across repeated runs, not the spread of possible output values.
5. What does raising N change, and what does it not?
It reduces Monte Carlo sampling error in estimated summaries such as the mean, standard deviation, and quantiles. It does not reduce the underlying propagated u(y), and it does not correct wrong input distributions, wrong dependence, model-form error, or numerical error.