Propagation of Uncertainty I: The First-Order Formula
~7 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)
You should be comfortable calculating partial derivatives of multivariable functions and understand the definitions of mean, variance, and covariance for random variables.
Why this matters
In experimental physics, we rarely measure the quantity of interest directly; we measure its ingredients.
When you calculate a derived quantity, such as the speed of a particle from its position and time, the uncertainties in the position and time measurements do not simply add up. They combine in a specific, mathematically rigorous way. Understanding this combination is essential for reporting honest error bars. If you get the propagation wrong, your final result may appear more precise than it actually is, or you may discard valid data because your error bars are too large. This lecture provides the standard tool for this task: the first-order propagation of uncertainty.
The Core Idea
Imagine you are standing on a hill that is slightly tilted. If you take a small step in any direction, your change in height is approximately proportional to the size of your step and the steepness of the slope in that direction. This is the essence of linearization. In our context, the "height" is the derived quantity \( f \), and the "steps" are the small deviations of the measured variables \( x_i \) from their mean values. Because the deviations are small, we can ignore the curvature of the function and treat it as a flat plane locally. The total uncertainty in \( f \) is then determined by how sensitive \( f \) is to each \( x_i \) (the slope) and how uncertain each \( x_i \) is (the step size). If the variables are independent, the uncertainties add in quadrature (like the sides of a right-angled triangle). If they are correlated, we must also account for the fact that when one variable is high, the other tends to be high or low, which adds or subtracts from the total uncertainty.
The General Formula
Let \( f(x_1, \dots, x_n) \) be a differentiable function of \( n \) measured quantities. Let \( \mu_i \) be the mean of \( x_i \), \( \sigma_i^2 \) be the variance of \( x_i \), and \( \mathrm{Cov}(x_i, x_j) \) be the covariance between \( x_i \) and \( x_j \). The variance of \( f \), denoted \( \sigma_f^2 \), to first order in the deviations is given by the sum of the variances weighted by the squared partial derivatives, plus the sum of the covariances weighted by the product of the partial derivatives.
This formula is the most general form. It accounts for all possible correlations between the input variables. The partial derivatives are evaluated at the mean values \( (\mu_1, \dots, \mu_n) \). Note that the double sum \( \sum_{i \neq j} \) includes both \( i < j \) and \( i > j \). Since \( \mathrm{Cov}(x_i, x_j) = \mathrm{Cov}(x_j, x_i) \), this is often written as \( 2 \sum_{i < j} \dots \) for clarity, but the summation notation above is standard.
The Independent Case
In many introductory experiments, we assume that the measurement errors in the different variables are independent. This means that the error in measuring \( x_i \) gives no information about the error in measuring \( x_j \). Mathematically, this implies that the covariance is zero for all \( i \neq j \): \( \mathrm{Cov}(x_i, x_j) = 0 \). The general formula simplifies significantly.
This is the form you will use most often. It states that the variance of the result is the sum of the variances of the inputs, each scaled by the square of the sensitivity of the function to that input. The standard uncertainty \( \sigma_f \) is the square root of this sum. This "root-sum-square" rule is the standard for independent errors.
Interpreting the Terms
Each term in the sum \( \left( \frac{\partial f}{\partial x_i} \right)^2 \sigma_i^2 \) represents the contribution to the total variance from the uncertainty in \( x_i \). The factor \( \left| \frac{\partial f}{\partial x_i} \right| \) is the sensitivity coefficient. If \( f \) changes rapidly with \( x_i \), a small error in \( x_i \) leads to a large error in \( f \). If \( f \) is insensitive to \( x_i \), the error in \( x_i \) contributes little to the total uncertainty. It is good practice to calculate each term separately to identify which measurement dominates the uncertainty. This tells you where to focus your efforts to improve the precision of the final result.
The derivation
First-Order Propagation of Uncertainty
For a differentiable function f(x1,…,xn) of measured quantities with means μi, variances σi2 and covariances Cov(xi,xj), the variance of f to first order in the deviations is σf2 = ∑i (∂f/∂xi)2 σi2 + ∑i≠j (∂f/∂xi)(∂f/∂xj) Cov(xi,xj); for independent inputs this reduces to the familiar sum of squared partial contributions, with all derivatives evaluated at the means.
The first-order approximation assumes that the deviations \( (x_i - \mu_i) \) are small compared to the scale over which the function changes significantly. If the function is highly non-linear (e.g., \( f(x) = e^x \) with large \( \sigma_x \)), the first-order formula can be a poor estimate of the true uncertainty. In such cases, higher-order terms in the Taylor expansion become significant, or one must use Monte Carlo methods to propagate the uncertainty. Always check if the relative uncertainty is small enough for the linear approximation to be valid. C
Worked example
Problem. The density \( \rho \) of a solid cylinder is calculated from its mass \( m \), diameter \( d \), and height \( h \) using the formula \( \rho = \frac{4m}{\pi d^2 h} \). The measured values are \( m = 10.0 \pm 0.1 \ \mathrm{g} \), \( d = 2.0 \pm 0.1 \ \mathrm{cm} \), and \( h = 5.0 \pm 0.1 \ \mathrm{cm} \). Assume the errors are independent. Calculate the density and its standard uncertainty.
- Identify the function: \( f(m, d, h) = \frac{4m}{\pi d^2 h} \).
- Calculate the partial derivatives at the mean values. \( \frac{\partial f}{\partial m} = \frac{4}{\pi d^2 h} \) \( \frac{\partial f}{\partial d} = \frac{4m}{\pi} (-2) d^{-3} h^{-1} = -\frac{8m}{\pi d^3 h} \) \( \frac{\partial f}{\partial h} = \frac{4m}{\pi d^2} (-1) h^{-2} = -\frac{4m}{\pi d^2 h^2} \)
- Substitute the mean values \( m=10, d=2, h=5 \). \( \frac{\partial f}{\partial m} = \frac{4}{\pi (4)(5)} = \frac{1}{5\pi} \approx 0.0637 \ \mathrm{g/cm^3/g} \) \( \frac{\partial f}{\partial d} = -\frac{8(10)}{\pi (8)(5)} = -\frac{80}{40\pi} = -\frac{2}{\pi} \approx -0.6366 \ \mathrm{g/cm^3/cm} \) \( \frac{\partial f}{\partial h} = -\frac{4(10)}{\pi (4)(25)} = -\frac{40}{100\pi} = -\frac{2}{5\pi} \approx -0.1273 \ \mathrm{g/cm^3/cm} \)
- Calculate the variance contribution from each variable. \( \sigma_m^2 = (0.1)^2 = 0.01 \) \( \sigma_d^2 = (0.1)^2 = 0.01 \) \( \sigma_h^2 = (0.1)^2 = 0.01 \) Term \( m \): \( (0.0637)^2 (0.01) \approx 0.0000406 \) Term \( d \): \( (-0.6366)^2 (0.01) \approx 0.004053 \) Term \( h \): \( (-0.1273)^2 (0.01) \approx 0.000162 \)
- Sum the variances: \( \sigma_\rho^2 \approx 0.0000405 + 0.004053 + 0.000162 = 0.004256 \ \mathrm{g^2/cm^6} \).
- Take the square root: \( \sigma_\rho = \sqrt{0.004256} \approx 0.065 \ \mathrm{g/cm^3} \).
- Calculate the central value: \( \rho = \frac{4(10)}{\pi (4)(5)} = \frac{40}{20\pi} = \frac{2}{\pi} \approx 0.6366 \ \mathrm{g/cm^3} \).
Units check. The derivative \( \partial f / \partial m \) has units \( \mathrm{g/cm^3 / g} = \mathrm{cm^{-3}} \). Squared, it is \( \mathrm{cm^{-6}} \). Multiplied by \( \sigma_m^2 \) in \( \mathrm{g^2} \), the term is \( \mathrm{g^2 cm^{-6}} \). The square root is \( \mathrm{g/cm^3} \), which matches the units of density. The dominant term is from the diameter, which makes sense because \( d \) is in the denominator squared.
Three quick tests
- If \( f = x + y \) and \( x, y \) are independent with \( \sigma_x = \sigma_y = \sigma \), what is \( \sigma_f \)?
Answer
\( \sqrt{2}\sigma \). The derivatives are both 1, so \( \sigma_f^2 = 1^2\sigma^2 + 1^2\sigma^2 = 2\sigma^2 \). - If \( f = xy \) and \( x, y \) are independent, what is the relative uncertainty \( \sigma_f / f \)?
Answer
\( \sqrt{(\sigma_x/x)^2 + (\sigma_y/y)^2} \). This is a standard result for products of independent variables. - Why does the covariance term vanish for independent variables?
Answer
By definition, independent random variables have zero covariance. The error in one variable provides no information about the error in the other.
Problem set for PU-106 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.
Derivations cited: D-110 · Threads: chance