core34 min studyLesson 106 in Physics

Propagate uncertainty with covariance

Use sensitivity coefficients and covariance to distinguish uncertainty amplification from cancellation in sums, differences, and ratios.

Before you begin

On this page

The question for today

What you’ll understand

Compute first-order combined uncertainty from a covariance matrix.
Explain why a common offset cancels in a difference under a shared-offset model.
Identify situations where linear propagation needs a more complete method.

Two uncertain inputs are often linked: the same reference calibrates both rulers, the same clock times both events, or the same fit determines both slope and intercept. Combining their standard uncertainties in quadrature while ignoring that connection can either overestimate or underestimate the output uncertainty. Covariance records how input deviations vary together, and sensitivity coefficients determine how those joint deviations enter the quantity being inferred.

Linearize the measurement model locally

For a smooth measurement function, a first-order Taylor expansion approximates a small output deviation by a weighted sum of input deviations. Each weight is the derivative with respect to one input, evaluated at the input estimates. These sensitivity coefficients carry units that convert each input contribution into the output unit. A large derivative means a modest input uncertainty may matter greatly; it does not by itself mean that the measurement has a large total uncertainty.

The approximation requires more than differentiability somewhere. Input uncertainty must be sufficiently concentrated that neglected curvature has little influence on the intended uncertainty summary. Ratios with a denominator near zero, thresholds, strong nonlinearities, or markedly asymmetric input distributions can defeat this approximation. In such cases propagate distributions or use another justified higher-order treatment with the physical model and domain respected. A symmetric plus-or-minus number is not automatically a faithful summary of a skewed output distribution.

u_y^2\approx\sum_i c_i^2u_i^2+2\sum_{i<j}c_ic_j\,u_{ij},\qquad c_i=\left.\frac{\partial f}{\partial x_i}\right|_{\widehat{\mathbf x}}

Here uᵢ are input standard uncertainties and uᵢⱼ are input covariances. Each term has squared output units. This is first-order propagation about the input estimates, rather than a universal rule for every nonlinear distribution.↔ Scroll wide equations horizontally.

Covariance follows the shared information

The covariance of two inputs has the product of their units. A dimensionless correlation coefficient divides it by their standard uncertainties and lies between minus one and one. Zero correlation is enough to remove the pair's covariance term from a first-order variance calculation, but does not generally imply statistical independence. Estimate covariance from suitable joint observations or derive it from shared input quantities. Do not invent a correlation coefficient merely to make an uncertainty budget look smaller.

For two readings affected by the same additive zero error, taking their difference cancels that error exactly within the shared-offset model. Treating the two zero contributions as independent would miss the cancellation. Their sum instead contains the shared error twice, so ignoring covariance would underestimate it. Cancellation fails if the offset drifts between readings or if the two inputs have different sensitivity to the supposed common effect. The physical error model determines the covariance, not the visual similarity of two numbers.

u^2(x_2-x_1)=u_2^2+u_1^2-2\rho u_1u_2,\qquad u_{12}=\rho u_1u_2

The minus sign arises from the sensitivities +1 and −1. Positive correlation reduces uncertainty of a difference but increases uncertainty of a sum. This expression is exact for a linear difference with correctly specified second moments.↔ Scroll wide equations horizontally.

Build the budget from underlying inputs

A reliable strategy is to express each measured quantity in terms of underlying independent and shared components before propagating. For example, write two positions as their separate reading terms plus one shared offset. Propagate the final displacement from those underlying inputs. This approach exposes common-mode cancellation without guessing correlations and avoids double counting an uncertainty already included in a fitted parameter covariance. Retain the covariance matrix with fit results when later calculations depend on several fitted parameters.

An uncertainty budget also guides experimental design. A difference measured with one stable instrument may reject a common zero offset better than a comparison between separately calibrated instruments, while a ratio may reject a common multiplicative scale factor. Neither strategy eliminates effects that are not truly shared. Test stability by reversing measurement order, repeating references, or monitoring time dependence. Correlation is useful when physically understood; unexplained correlation is a reason to investigate the experiment rather than a free precision bonus.

Work it through

A displacement with a common zero offset

Two position estimates are $x_1=10\,\mathrm{mm}$ and $x_2=15\,\mathrm{mm}$ . Each has independent reading standard uncertainty $0.30\,\mathrm{mm}$ and the same shared zero standard uncertainty $0.40\,\mathrm{mm}$ . Assume the zero is unchanged and independent of the reading terms. Find uncertainty of the displacement.

Each individual total variance is $0.30^2+0.40^2=0.25\,\mathrm{mm^2}$ , giving standard uncertainty $0.50\,\mathrm{mm}$ . Their covariance is the shared variance $0.16\,\mathrm{mm^2}$ .
The displacement estimate is $15-10=5\,\mathrm{mm}$ . Its variance is $0.25+0.25-2(0.16)=0.18\,\mathrm{mm^2}$ , leaving only the two independent reading contributions.
Take the square root to obtain $u(d)=\sqrt{0.18}\approx0.424\,\mathrm{mm}$ . Ignoring covariance would give $\sqrt{0.50}\approx0.707\,\mathrm{mm}$ , which misses the exact cancellation of the unchanged common zero.

The result

The displacement is approximately $(5.00\pm0.42)\,\mathrm{mm}$ with standard uncertainty. This result depends on the same offset entering both readings identically and remaining stable between them.

The living notebook

Explore the model

Interactive lab

Interactive lab · vectors

Projection signs explain sensitivity signs

Use the angle and magnitude controls for one vector and inspect the signed x and y components. At magnitude 1.5, predict which component changes sign when angle moves from 60 to 120 degrees, then check. This gives a geometric reminder that signed sensitivities matter in covariance terms. The laboratory draws one vector only; it does not display an uncertainty ellipse, a covariance matrix, or correlated random samples.

Vectorx / y projectionsTrace marker

x component: 1.393
y component: 0.975
Magnitude: 1.7

Angle35°Magnitude1.7

t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe sign of a sensitivity determines whether a shared deviation reinforces or cancels another contribution.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

In a difference, the identical offset appears with coefficients +1 and −1 and cancels. A sum doubles it and an average retains it. The cancellation requires a genuinely shared unchanged offset, rather than two independently uncertain zero errors.

Your turn

Try it yourself

Two quantities have standard uncertainties $u_1=u_2=2\,\mathrm V$ and correlation $\rho=0.5$ . Compute standard uncertainties for their sum and difference, then explain why a zero difference between their estimates would not make its uncertainty vanish.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use the same covariance with opposite signs for the two output models.

Compare with the solution

The covariance is $0.5(2)(2)=2\,\mathrm{V^2}$ . The sum variance is $4+4+2(2)=12\,\mathrm{V^2}$ , giving $\sqrt{12}\approx3.46\,\mathrm V$ . The difference variance is $4+4-2(2)=4\,\mathrm{V^2}$ , giving $2\,\mathrm V$ . Equal central estimates only make the estimated difference zero; they do not remove uncertainty about that difference. Perfect cancellation of all contributions would require a stronger physical relation than the specified partial correlation.

Keep the idea

Take it with you

First-order propagation uses derivatives and the complete input covariance matrix.
A shared effect can cancel in a difference and reinforce in a sum.
Strong nonlinearity or near-singular models require a justified treatment beyond local linearization.

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