Error propagation and uncertainty budgets

Start with the measurement model

The map's title uses the familiar phrase 'error propagation', but this lesson propagates uncertainty information. A standard uncertainty is a standard-deviation measure, not a known signed error, a maximum error or a tolerance. A model Y = f(X₁, X₂) relates a specified scalar output Y to two inputs; a real measurement model must also include significant corrections and other relevant influences.

Uppercase symbols below denote random variables representing supplied information about quantities. They do not assert that a fixed object's dimensions physically change. For each original finite teaching model, E means a probability-weighted mean: multiply each possible value by its supplied probability and add. The probabilities are exact stipulations, sum to 1 and are not frequencies estimated from measurements. Input estimates are xᵢ = E(Xᵢ).

Let u²(Xᵢ) = E[(Xᵢ − xᵢ)²], with u(Xᵢ) its nonnegative square root. Variance has squared quantity units; standard uncertainty has the input's unit. We use u(Y) for the exact standard deviation of a supplied model and u_lin(Y) for its first-order approximation. In a real budget these quantities are evaluated from justified measurement information; none of the numerical models here assesses an actual instrument.

Scale inputs with signed sensitivities

A sensitivity coefficient cᵢ is the partial derivative ∂f/∂Xᵢ evaluated at the input estimates, holding the other input fixed. It describes the first-order output change caused by a small input change. A two-input linearization is Y_lin = f(x₁, x₂) + c₁(X₁ − x₁) + c₂(X₂ − x₂). For this approximation, the function must be differentiable near the relevant estimates and neglected terms must be small enough for the intended use. Differentiability alone does not guarantee that.

The units of cᵢ are output units divided by that input's units. Thus cᵢ times an input change has output units. A coefficient can be negative: for f(X, Z) = X − Z, c_X = +1 and c_Z = −1. The minus sign matters when combining dependent inputs. The magnitude |cᵢ|u(Xᵢ) is nonnegative; its square is that input's diagonal variance contribution.

For a rectangle's area A = LW, with positive side lengths in metres, c_L = w and c_W = l at means l and w. These coefficients have units m because area units m² divided by length units m gives m. Even when one input has a larger standard uncertainty, its scaled contribution need not be larger: the sensitivity also matters.

Retain covariance and its units

For two inputs with means x₁ and x₂, define C₁₂ = E[(X₁ − x₁)(X₂ − x₂)], their covariance. It has the product of the two input units and can be positive, negative or zero. When both standard uncertainties are nonzero, the correlation coefficient ρ = C₁₂/[u(X₁)u(X₂)] is dimensionless and must satisfy −1 ≤ ρ ≤ +1. Independence implies zero covariance when the second moments exist; zero covariance alone does not imply independence.

The first-order two-input variance is u_lin²(Y) = c₁²u²(X₁) + c₂²u²(X₂) + 2c₁c₂C₁₂. Each term has output units squared, even when the input units differ. The first two terms are nonnegative. The covariance term can have either sign, but a valid covariance model makes the total nonnegative. This is the exact variance of Y_lin; its adequacy for the original nonlinear Y still needs checking.

When the output is an affine function, meaning a constant plus fixed multiples of the inputs, this linearization is exact. For a sum with coefficients +1, +1, positive covariance increases variance; for a difference with +1, −1, it decreases variance. A shared reference or influence can create dependence, so separate table rows do not establish independence. Significant dependencies require evidence or a justified model.

Check a correlated model by enumeration

Supply length inputs X = 100 mm + (2 mm)S and Z = 60 mm + (3 mm)T. The dimensionless signs S and T take the four pairs and probabilities in the table. Each sign has mean zero, variance 1 and equal marginal probabilities for +1 and −1, but the joint probabilities are not products of the marginals. This is an exact finite information model, not an observed experiment.

The means are E(X) = 100 mm and E(Z) = 60 mm; u(X) = 2 mm and u(Z) = 3 mm. E(ST) = (3 − 1 − 1 + 3)/8 = 1/2, so C_XZ = (2 mm)(3 mm)/2 = 3 mm² and ρ = 1/2. For Y₊ = X + Z, the mean is 160 mm and the variance is 4 mm² + 9 mm² + 6 mm² = 19 mm². For Y₋ = X − Z, the mean is 40 mm and the variance is 4 mm² + 9 mm² − 6 mm² = 7 mm².

Therefore u(Y₊) = √19 mm ≈ 4.359 mm and u(Y₋) = √7 mm ≈ 2.646 mm. The square root is nonnegative. Both results are exact model standard deviations before the displayed decimal approximation, because both outputs are affine. Omitting covariance would give 13 mm² for both variances and would misrepresent both outputs. These standard deviations are not guaranteed error bounds or normal-distribution coverage intervals.

Build a rectangular-area budget

For a separate rectangle, supply independent inputs L = 3 m + ε and W = 2 m + η. Here ε takes −0.03 m and +0.03 m with equal probabilities, and η independently takes −0.02 m and +0.02 m with equal probabilities. Each of the four paired states therefore has probability 1/4. The positive sides have means l = 3 m, w = 2 m, standard uncertainties u(L) = 0.03 m and u(W) = 0.02 m, and zero covariance.

At these means, the area estimate f(l, w) is 6 m². The sensitivity coefficients are c_L = 2 m and c_W = 3 m. Each scaled standard contribution is 0.06 m²; each diagonal variance contribution is 0.0036 m⁴. The first-order budget gives u_lin²(A) = 0.0072 m⁴ and u_lin(A) = 0.06√2 m² ≈ 0.08485 m². Equal variance contributions are 50% each of this first-order variance, despite the unequal input uncertainties.

A budget records more than an arithmetic total. Give the input estimate, uncertainty measure and information source, the coefficient and its units, the resulting output contribution, and any dependencies. Here the exact supplied distributions are the information sources; no observations, calibration certificates, sample counts or evaluated degrees of freedom are supplied. This small table is complete only for the stipulated two-input model.

Compare the product with its approximation

The full area is A = (l + ε)(w + η) = lw + wε + lη + εη. The first-order model omits εη. Because the supplied ε and η are independent and have zero means, E(εη) = 0 and E(A) = lw = 6 m². This equality follows from the model's dependence structure; transforming input means is not a universal way to obtain a nonlinear output mean.

In this model the three varying terms wε, lη and εη have zero pairwise covariances. For example, E[(wε)(εη)] = wE(ε²)E(η) = 0. Their variances add, giving Var(A) = w²u²(L) + l²u²(W) + u²(L)u²(W). The extra term is (0.03 m)²(0.02 m)² = 0.00000036 m⁴. Hence the exact variance is 0.00720036 m⁴ and the exact standard deviation is √(0.00720036 m⁴).

The extra variance is 0.005% of the first-order variance in this particular case. The displayed first-order uncertainty is consequently a close approximation here, but the exact variance is different. That percentage is not an error bound for other models. Direct enumeration provides a separate check: the four areas are 5.8806, 5.9994, 5.9994 and 6.1206 m², each with probability 1/4.

Notice a failure at a zero gradient

For a third model, X is a signed displacement along one fixed spatial axis, measured in metres relative to a fixed origin. Supply X = −a, 0 or +a with probabilities 1/4, 1/2 and 1/4, where a = 0.02 m. The output Y = X² is a squared-displacement quantity with units m²; it is not automatically a physical area. Its input mean is E(X) = 0 m.

Evaluating f(x) = x² at the mean gives f(0) = 0 m², and its sensitivity c = 2x is 0 m there. First-order propagation therefore gives u_lin(Y) = 0 m². Yet the table's two nonzero X states both give Y = a², whose total probability is 1/2. Thus E(Y) = a²/2 = 0.0002 m², E(Y²) = a⁴/2, and Var(Y) = a⁴/4 = 0.00000004 m⁴. The exact u(Y) is a²/2 = 0.0002 m². First order missed both the output mean and its dispersion.

The GUM's 2026 amendment explicitly addresses nonlinearity in the output estimate as well as the older variance warning. It describes higher-order treatment or propagation of distributions using a Monte Carlo method; the particular next-order mean formula it supplies assumes independent normally distributed inputs. Our finite three-point model is not normal. We instead obtain its exact moments from its stated probabilities. A zero derivative by itself does not establish zero output uncertainty.

Audit and report the budget

Check the model, signs, units and dependence information before trusting a total. The covariance contribution belongs to a pair, so adding the same pair twice counts it twice. Also investigate whether an input uncertainty already includes an effect before adding that effect again as a separate contribution. The GUM specifically warns about overlap between a component evaluated from other information and variability already represented by statistical analysis.

A standard uncertainty does not by itself specify the probability of a symmetric interval. An expanded uncertainty U = k u_c multiplies a combined standard uncertainty u_c by a stated coverage factor k; interpreting its interval needs the relevant distribution and evaluation conditions. Neither the finite models here nor a budget table automatically justify a normal 95% coverage claim. Report whether the uncertainty is first-order or evaluated from a fuller model, and retain the assumptions and information needed to reproduce it.

A useful budget helps decide what additional information or improved measurement would matter. Compare contributions after converting them to output units, while retaining covariance and nonlinear effects; do not rank inputs solely by their unscaled numerical uncertainties. The next subject, Calibration, standards, and traceability, examines the reference relationships that support actual measurement results.

Worked examples

Try it yourself

1. In the original area budget, u(L) = 0.03 m exceeds u(W) = 0.02 m. Does L therefore contribute more first-order variance? Calculate both scaled magnitudes and both variance contributions.

   Show the answer and reasoning

   No. Both scaled magnitudes are 0.06 m² and both variance contributions are 0.0036 m⁴.

   The respective sensitivities are 2 m and 3 m. Compare |c_L|u(L) and |c_W|u(W) in output units, then their squares. Ranking unscaled input uncertainties ignores the model's sensitivities.

2. Someone gives 13 mm² as the variance of both outputs in the original correlated length case. Which term is missing, and what are the corrected sum and difference variances?

   Show the answer and reasoning

   The missing covariance term is +6 mm² for the sum and −6 mm² for the difference. Their corrected variances are 19 mm² and 7 mm².

   Use 2c_Xc_ZC_XZ with C_XZ = 3 mm². The coefficient of Z changes sign between the two outputs; its squared diagonal contribution remains 9 mm². These are valid nonnegative variances of the supplied affine models.

3. Change only the joint sign probabilities to 1/8, 3/8, 3/8 and 1/8 in the same pair order. Keep X = 100 mm + (2 mm)S and Z = 60 mm + (3 mm)T. Find C_XZ, ρ and the sum and difference variances.

   Show the answer and reasoning

   C_XZ = −3 mm², ρ = −1/2, Var(X + Z) = 7 mm² and Var(X − Z) = 19 mm².

   The marginals remain balanced, so the input means and variances stay unchanged. E(ST) becomes (1 − 3 − 3 + 1)/8 = −1/2. The covariance contribution reverses sign, reversing which output has the larger variance.

4. For a new independent two-point rectangle model with the same means, reduce the L deviations to ±0.015 m and retain W deviations ±0.02 m. Find the first-order variance and the exact product variance.

   Show the answer and reasoning

   The first-order variance is 0.0045 m⁴; the exact product variance is 0.00450009 m⁴.

   The first diagonal term is (2 m × 0.015 m)² = 0.0009 m⁴; the second remains 0.0036 m⁴. The exact independent product adds (0.015 m)²(0.02 m)² = 0.00000009 m⁴. Improving one model component leaves the other contribution present.

5. Use the same three-point displacement probabilities but set a = 0.04 m. For Y = X², find f(E(X)), the exact E(Y), Var(Y) and u(Y). Does first-order propagation at the input mean detect them?

   Show the answer and reasoning

   f(E(X)) = 0 m², E(Y) = 0.0008 m², Var(Y) = 0.00000064 m⁴ and u(Y) = 0.0008 m². First order at the zero input mean gives zero uncertainty and misses the output mean.

   a² = 0.0016 m², so the output mean and standard deviation are each a²/2 = 0.0008 m². Squaring that standard deviation gives the stated variance. Doubling a multiplies the squared-displacement mean and standard deviation by 4 and its variance by 16.

6. In the original symmetric three-point model, view X and Y = X² as a pair. Find Cov(X, Y) and its unit. Does that value prove the pair is independent?

   Show the answer and reasoning

   Cov(X, Y) = 0 m³, but the pair is not independent for a = 0.02 m.

   Cov(X, Y) = E(X³) − E(X)E(X²) = 0 by symmetry, and m times m² gives m³. Yet Y is determined by X: when X = 0, Y = 0 with certainty, whereas its unconditional probability is only 1/2. Zero covariance need not remove nonlinear dependence.

What to carry forward

• A propagation calculation needs an explicit measurement model and justified input uncertainty and dependence information.
• Sensitivity coefficients retain signs and convert input variation into output units.
• Covariance can increase or decrease an output variance; separate input rows do not establish independence.
• The area case has first-order variance 0.0072 m⁴ and exact model variance 0.00720036 m⁴.
• A nonlinear output can change both the mean and dispersion; a zero derivative does not certify zero uncertainty.
• Record source information, units, covariance, approximation limits and the uncertainty measure; a budget alone supplies no normal coverage claim.