First-Order Propagation of Uncertainty

Derivation

First-Order Propagation of Uncertainty

D-110 Home PU-106 Threads chance Depends on Variance of Linear Combinations and Covariance, taylor-series-expansion

Statement

For a differentiable function f(x₁,…,x_n) of measured quantities with means μ_i, variances σ_i² and covariances Cov(x_i,x_j), the variance of f to first order in the deviations is σ_f² = ∑_i (∂f/∂x_i)² σ_i² + ∑_i≠j (∂f/∂x_i)(∂f/∂x_j) Cov(x_i,x_j); for independent inputs this reduces to the familiar sum of squared partial contributions, with all derivatives evaluated at the means.

Why it matters

Almost no physical quantity of interest is measured directly. We measure lengths, times, voltages and masses, then combine them through a formula — a density from mass over volume, a rate constant from a fitted slope, a gravitational acceleration from a pendulum period. The error-propagation formula tells us how the scatter in the raw measurements flows through that formula into the derived result, so that a reported value can carry an honest uncertainty rather than a spurious number of significant figures.

It is also the engine behind experiment design. Because each input contributes an independent, additive term to σ_f², the formula tells you immediately which measurement dominates your error budget and is therefore worth improving, and which is already negligible.

Assumptions

Local linearity.The function must be well approximated by its first-order Taylor polynomial across the region of a few standard deviations around the means; drop this and the truncated higher-order terms bias both the mean and the variance of f.

Finite, well-defined first and second moments.Each input must have a finite mean and variance; for heavy-tailed distributions (e.g. Cauchy) σ_i² does not exist and the formula is meaningless.

Continuously differentiable f at the operating point.The partial derivatives must exist and be finite at μ; at a kink, cusp or pole the linear coefficient is undefined and propagation fails.

Known correlation structure.The general form needs the full covariance matrix. If inputs share a common measurement (a shared calibration, a shared clock) and you wrongly assume independence, the cross terms are silently omitted and the quoted uncertainty is biased — too small for positive correlation, too large for negative.

Derivation

f(x) = f(μ) + ∑_i (∂f/∂x_i)|_μ (x_i − μ_i) + O(Δx²)

Multivariate Taylor expansion of f about the mean vector μ = (μ₁,…,μ_n), keeping only the linear term. A

E[f] ≈ f(μ) + ∑_i (∂f/∂x_i) E[x_i − μ_i] = f(μ)

Take expectations, using linearity of E[·] and E[x_i − μ_i] = 0 by definition of the mean; the derivatives are constants evaluated at μ. A

f − E[f] ≈ ∑_i (∂f/∂x_i) (x_i − μ_i)

Subtract step 2 from step 1: to first order, the fluctuation of f is a linear combination of the input fluctuations. A

σ_f² = E[(f − E[f])²] = E{(∑_i f_i(x_i−μ_i))(∑_j f_j(x_j−μ_j))}

Apply the definition of variance to f and substitute step 3, writing the square as a product of two independent summation indices i, j and abbreviating f_i ≡ ∂f/∂x_i. B

σ_f² = ∑_i ∑_j f_i f_j E[(x_i−μ_i)(x_j−μ_j)] = ∑_i ∑_j f_i f_j Cov(x_i,x_j)

Interchange the finite double sum with the expectation (linearity) and pull the constant derivatives out; each surviving expectation is exactly the covariance of the pair. This is the variance-of-a-linear-combination result applied to coefficients f_i. B

σ_f² = ∑_i f_i² σ_i² + ∑_i≠j f_i f_j Cov(x_i,x_j)

Split the double sum into diagonal terms (i = j, where Cov(x_i,x_i) = σ_i²) and off-diagonal terms. This is the exact first-order result. B

σ_f² = ∑_i (∂f/∂x_i)² σ_i²

Impose independence: Cov(x_i,x_j) = 0 for i ≠ j, killing every cross term and leaving the sum of squared contributions. A

Result

σ_f² = ∑_i=1ⁿ (∂f/∂x_i)² σ_i² (independent inputs)

Reading. Each input contributes a term equal to its own variance amplified by the squared sensitivity ∂f/∂x_i — how strongly f responds to a nudge in x_i. Because the terms are squares they always add (uncertainties never cancel), which is why independent errors combine "in quadrature": the total is the root-sum-of-squares, dominated by the largest single contribution.

Units check. The partial derivative ∂f/∂x_i carries units [f]/[x_i], so (∂f/∂x_i)² σ_i² has units ([f]/[x_i])² · [x_i]² = [f]². Every term shares the units [f]², so their sum is a genuine variance and σ_f = √σ_f² has the units of f.

Limiting cases

Single variable: σ_f = |df/dx| σ_x — the absolute error scales by the local slope.
Sum or difference f = x ± y: σ_f² = σ_x² + σ_y². Absolute errors add in quadrature (note: still + for a difference).
Product or quotient f = xy or x/y: (σ_f/f)² = (σ_x/x)² + (σ_y/y)². Relative errors add in quadrature.
Power law f = x^p: σ_f/|f| = |p| σ_x/|x| — the exponent multiplies the fractional error.
Zero sensitivity ∂f/∂x_i = 0: that input drops out to first order; its influence, if any, is second order.

Breaks when

Strong nonlinearity over the error scale. When the second derivative term ½ f″(Δx)² is not negligible against the linear term — large fractional uncertainties, or curved functions like exp/log evaluated far into their bend — the truncated Taylor series both biases E[f] and misestimates σ_f.
At or near a stationary point. Where ∂f/∂x_i = 0 (e.g. f = sinθ at θ = 90°), the first-order formula predicts zero uncertainty; the real spread is governed by the quadratic term and the distribution of f becomes one-sided.
Correlated inputs treated as independent. If Cov(x_i,x_j) ≠ 0 (shared calibration, common systematic) the omitted cross terms make the quoted σ_f wrong in a definite direction — understated for positive correlation.
Undefined or infinite moments. Heavy-tailed inputs (Cauchy-like) have no finite variance, and singular points (poles, kinks) have no finite derivative — the formula's ingredients simply do not exist.

Failure modes

Linear addition of errors. Reporting σ_f = σ_x + σ_y instead of √(σ_x² + σ_y²); overstates the uncertainty whenever inputs are independent.
Mixing absolute and relative rules. Adding relative errors for a sum, or absolute errors for a product. The rule depends on the operation.
Dropping the square on the derivative. Writing ∑ (∂f/∂x_i) σ_i — dimensionally wrong and typically underestimates.
"Errors cancel" for a difference. Believing x − y has smaller error than x + y; the variances add either way, and subtraction of near-equal numbers actually inflates the relative error.
Ignoring a shared systematic. Propagating two quantities from the same instrument as if independent, silently discarding a real covariance term.
Sign-flipping the covariance term. When correlations are included, forgetting that f = x − y carries −2Cov, not +2Cov.

Discussion

The formula is nothing more than the variance of a linear combination, with the combination coefficients supplied by calculus. Step 3 replaces the possibly complicated function f by a plane tangent to it at the operating point; from that moment the problem is purely one about a weighted sum of random variables, and all the machinery of second moments applies unchanged. This is why the derivation leans on exactly two priors — the Taylor expansion to build the linearization, and the variance of a linear combination to evaluate its spread.

Geometrically, the covariance form σ_f² = g^TΣg, with gradient g = ∇f and covariance matrix Σ, is a quadratic form. The independent case is the special case where Σ is diagonal. Rotating to the eigenbasis of Σ (the principal axes of the error ellipsoid) always restores the diagonal, quadrature-sum picture — correlation is just the statement that your chosen variables are not aligned with those principal axes.

There is a subtlety about where the derivatives are evaluated. Strictly, the coefficients in step 1 are constants only if fixed at the true means μ_i, which are unknown; in practice one evaluates them at the measured values, introducing a further error that is itself second order in the deviations and hence consistent with the first-order truncation already made. The same order-counting shows that the leading correction to σ_f² involves the second derivatives and the third and fourth moments of the inputs, so for a symmetric, mildly nonlinear problem the first-order estimate is typically good to a few percent even when a single fractional uncertainty reaches 10%.

Common misconceptions. The formula does not require the inputs to be Gaussian — only that their variances exist; Gaussianity is an extra assumption needed to turn σ_f into a 68% confidence interval, not to derive the propagation itself. Nor does a "0.05 cm" uncertainty mean a hard bound; it is a standard deviation, and quadrature addition is the correct combination precisely because it tracks variances, not worst-case ranges.

Worked examples

Example 1 — area of a rectangle (product rule). A plate measures L = 5.00 ± 0.05 cm and W = 3.00 ± 0.02 cm, the two measurements independent. Find the area and its uncertainty.

A = LW; ∂A/∂L = W, ∂A/∂W = L

Model and its two partial derivatives, symbols first. A

σ_A² = W²σ_L² + L²σ_W²

Insert the derivatives into the independent-input result. A

σ_A² = (3.00)²(0.05)² + (5.00)²(0.02)² = 0.0225 + 0.0100 = 0.0325 cm⁴

Substitute numbers with units; the two contributions are comparable, neither negligible. A

A = 15.00 cm², σ_A = √0.0325 = 0.180 cm²

Central value and square root of the variance. A

A = 15.00 ± 0.18 cm²

Reading. A 1% error in each edge yields about a 1.2% error in the area, the two combining in quadrature; the length term dominates slightly because 0.05/5.00 > 0.02/3.00.

Example 2 — g from a pendulum (nonlinear, relative-error route). A simple pendulum gives g = 4π²L/T² with L = 1.000 ± 0.005 m and period T = 2.006 ± 0.004 s.

g = 4π² L T⁻² ⇒ (σ_g/g)² = (σ_L/L)² + (2σ_T/T)²

For a product of powers the fractional variances add, each weighted by the squared exponent (1 for L, 2 for T). B

σ_L/L = 0.00500; 2σ_T/T = 2(0.004)/2.006 = 0.00399

Evaluate the two fractional terms; note the period enters twice as strongly. A

(σ_g/g)² = (0.00500)² + (0.00399)² = 4.09×10⁻⁵ ⇒ σ_g/g = 0.00640

Root-sum-of-squares of the fractional contributions. A

g = 4π²(1.000)/(2.006)² = 9.811 m s⁻²; σ_g = 9.811 × 0.00640 = 0.063 m s⁻²

Central value, then convert the fractional uncertainty back to absolute. A

g = 9.811 ± 0.063 m s⁻²

Reading. Despite the timing being nominally more precise, the factor-of-2 exponent makes the period and length contribute almost equally; halving the length error alone would barely help. Units check: 4π² is dimensionless, m/s² follows directly, and σ_g shares those units.

Problems

Sum in quadrature. Two independent lengths x = 12.0 ± 0.3 cm and y = 7.0 ± 0.4 cm are joined end to end. Find L = x + y and σ_L.
Solution
∂L/∂x = ∂L/∂y = 1, so σ_L² = σ_x² + σ_y² = 0.3² + 0.4² = 0.09 + 0.16 = 0.25 cm². σ_L = 0.50 cm. Result: L = 19.0 ± 0.5 cm.
Quotient. A density is ρ = m/V with m = 25.0 ± 0.1 g and V = 10.0 ± 0.2 cm³, independent. Find ρ ± σ_ρ.
Solution
ρ = 25.0/10.0 = 2.500 g cm⁻³. Relative: (σ_ρ/ρ)² = (0.1/25.0)² + (0.2/10.0)² = (0.00400)² + (0.0200)² = 1.6×10⁻⁵ + 4.0×10⁻⁴ = 4.16×10⁻⁴. σ_ρ/ρ = 0.0204, so σ_ρ = 2.500 × 0.0204 = 0.051. The volume dominates entirely. Result: ρ = 2.500 ± 0.051 g cm⁻³.
Power-law weighting. Kinetic energy E = ½mv² with m = 2.00 ± 0.02 kg and v = 20.0 ± 0.5 m s⁻¹. Find E ± σ_E.
Solution
E = ½(2.00)(20.0)² = 400 J. Since E ∝ mv²: (σ_E/E)² = (σ_m/m)² + (2σ_v/v)² = (0.0100)² + (2×0.5/20.0)² = (0.0100)² + (0.0500)² = 1.0×10⁻⁴ + 2.5×10⁻³ = 2.6×10⁻³. σ_E/E = 0.0510, σ_E = 20.4 J. The velocity, doubly weighted, dominates. Result: E = 400 ± 20 J.
Correlated inputs. Two readings from one drifting instrument have σ_x = σ_y = 0.50 units and Cov(x,y) = +0.20 units². Compare σ of f = x − y against the value you would wrongly get assuming independence.
Solution
For a difference, ∂f/∂x = +1, ∂f/∂y = −1, so σ_f² = σ_x² + σ_y² − 2 Cov(x,y) = 0.25 + 0.25 − 2(0.20) = 0.10, so σ_f = 0.316 units. Assuming independence gives √0.50 = 0.707 units. The positive correlation makes the difference more precise (common drift cancels), and ignoring it would overstate the error by more than a factor of 2. Result: σ_f = 0.32 vs a naive 0.71.
Breakdown at a stationary point. A quantity is f = sinθ with θ = 90.0° ± 2.0°. Apply the first-order formula, then explain why it fails and estimate the true leading effect.
Solution
Convert: 2.0° = 0.0349 rad. First order: σ_f = |cosθ| σ_θ = |cos 90°| × 0.0349 = 0 — the formula predicts zero uncertainty because θ sits exactly at a maximum of sin. This is a genuine breakdown: the sensitivity vanishes, so the leading behaviour is quadratic. Second order: Δf ≈ ½ f″ (Δθ)² = ½(−sin 90°)(0.0349)² = −6.1×10⁻⁴. So f is not scattered symmetrically about 1 but pushed one-sidedly below it by about 6×10⁻⁴; the true spread and bias are both set by the second derivative, exactly the regime the linear formula cannot describe. Result: first order gives 0; real effect ≈ −6×10⁻⁴ (one-sided, second order).