Propagating Measurement Uncertainty

Why Measurement Errors Propagate

A computed quantity often depends on several measured inputs. For instance, an area depends on measured lengths, and a density depends on measured mass and volume. Since each measurement may differ from its true value, the output may differ from the value computed using the recorded measurements.

The total differential gives a local, first-order way to estimate this effect. When input errors are bounded, the absolute contributions from the individual inputs combine to estimate the largest first-order output change. The absolute values matter: if the errors can vary independently, their effects might reinforce rather than cancel.

Measurement Errors and the Total Differential

Suppose a quantity is modeled by a differentiable function \(f:\mathbb{R}^n\to\mathbb{R}\). Let \(\mathbf{a}=(a_1,\ldots,a_n)\) be the recorded input values, and let the actual inputs be \(a_i+\Delta x_i\). Assume the measurement errors satisfy \(|\Delta x_i|\leq\varepsilon_i\), where each \(\varepsilon_i\geq 0\) is a stated error bound. The recorded output is \(f(\mathbf{a})\); the actual modeled output is \(f(\mathbf{a}+\Delta\mathbf{x})\).

Using the given error bounds, the absolute contribution from input \(x_i\) is estimated by \(\left|\frac{\partial f}{\partial x_i}(\mathbf{a})\right|\varepsilon_i\). Adding these contributions gives the standard first-order estimate of the largest output uncertainty:

Each partial derivative is evaluated at the recorded inputs. Its absolute value measures the local sensitivity of the output to that input, while \(\varepsilon_i\) measures the size of the allowed input error. Their product has the units of the output.

Why the Absolute Contributions Add

Differentiability at \(\mathbf{a}\) means the output change can be written as a linear part plus a remainder that becomes negligible compared with the size of the input change as that change approaches zero. In precise terms, with the Euclidean norm \(\|\cdot\|\), there is a remainder \(r(\Delta\mathbf{x})\) such that

For the linear part, the triangle inequality and the error bounds give

This explains the sum of absolute contributions. If the errors are allowed to vary independently, the bound for the linear part is attained by choosing each \(\Delta x_i\) at an endpoint of its allowed interval, with sign chosen to make its term have the same sign as the other terms. Thus the sum gives the worst case for the linearized change, not a prediction that every error will reach its maximum.

A Reliable Calculation Procedure

Worked Examples

Common Pitfall: Letting Signs Cancel

Suppose \(q(x,y)=x-y\), with measured values \(x=10\), \(y=4\), and error bounds \(0.1\) for both inputs. The partial derivatives are \(q_x=1\) and \(q_y=-1\). A signed calculation that assumes both errors are positive gives \(dq=(1)(0.1)+(-1)(0.1)=0\). That does not estimate the worst-case effect: the errors might have opposite signs. The correct first-order uncertainty estimate uses absolute contributions:

Indeed, choosing \(\Delta x=0.1\) and \(\Delta y=-0.1\) gives \(\Delta q=\Delta x-\Delta y=0.2\). The signed differential is useful when particular input changes are known; for worst-case bounded errors that can vary independently, take the absolute value of each contribution before adding.

How to Interpret the Estimate

The method estimates the effect of small, bounded input errors on a smooth local model. It is useful when a computed quantity depends on several measurements and one needs to identify which inputs contribute most to the output uncertainty. A large partial derivative indicates that small changes in that input can have a relatively strong local effect; a small error bound can still make its total contribution small.

Keep the assumptions and limits in view. Evaluate partial derivatives at the recorded inputs, use consistent units, and use absolute contributions when estimating the worst-case first-order change. If errors are not small, or if the function changes rapidly over the allowed input range, the linear estimate may be less accurate. The differential alone does not establish a rigorous bound for finite errors; that would require an additional estimate of the nonlinear remainder.