Propagation of Uncertainty II: Correlated Errors
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You should be comfortable with the first-order error propagation formula for independent variables and understand the basic definition of covariance as a measure of joint variability.
Why this matters
Ignoring correlations can lead to significant underestimation of uncertainty in derived quantities.
In experimental physics, measurements are rarely perfectly independent. Systematic effects, shared calibration standards, or physical constraints often mean that errors in one measurement are linked to errors in another. When we combine such measurements to calculate a derived quantity, the standard "sum of squares" rule for independent errors fails. We must account for the covariance terms, which can either increase or decrease the total uncertainty depending on the sign of the correlation. This lecture generalizes the propagation rules to handle these correlated cases rigorously.
The Intuition Behind Correlated Errors
Imagine measuring the length and width of a rectangle to find its area. If your ruler is consistently too short, both the length and width measurements will be biased in the same direction. This is a positive correlation. If you calculate the area, the errors in length and width reinforce each other, making the area error larger than if the errors were random and independent. Conversely, if measuring one quantity tends to make another smaller (a negative correlation), the errors might partially cancel out in the derived quantity. The key insight is that the "cross-terms" in the variance calculation represent these interactions. When variables are independent, these cross-terms vanish. When they are correlated, they become crucial components of the total uncertainty budget.
Variance of Linear Combinations
Let us start with the simplest case: a linear combination of random variables. Suppose we have random variables \( X_1, \dots, X_n \) with finite variances and fixed real constants \( a_1, \dots, a_n \). We define the linear combination \( S = \sum_{i} a_i X_i \). The variance of this sum is not simply the sum of the variances unless the variables are uncorrelated.
According to the general rule for linear combinations, the variance is given by the double sum of the covariances weighted by the coefficients:
If the variables are pairwise uncorrelated, then \( \mathrm{Cov}(X_i, X_j) = 0 \) for \( i \neq j \), and the expression collapses to the familiar weighted sum of variances:
However, when correlations exist, the terms where \( i \neq j \) contribute significantly. Note that \( \mathrm{Cov}(X_i, X_j) = \mathrm{Cov}(X_j, X_i) \), so the double sum effectively counts each pair twice (once as \( i,j \) and once as \( j,i \)).
The Covariance Matrix Formalism
For general differentiable functions, we use the Jacobian matrix approach. Let \( \mathbf{x} \in \mathbb{R}^n \) be a random input vector with mean \( \boldsymbol{\mu} \) and covariance matrix \( \boldsymbol{\Sigma}_x \). Let \( \mathbf{y} = \mathbf{f}(\mathbf{x}) \) be a differentiable vector-valued map from \( \mathbb{R}^n \) to \( \mathbb{R}^m \). The covariance of the output vector \( \mathbf{y} \) to first order is determined by the Jacobian matrix \( \mathbf{J} \).
The Jacobian \( \mathbf{J} \) is an \( m \times n \) matrix with elements \( J_{ij} = \frac{\partial f_i}{\partial x_j} \) evaluated at the mean \( \boldsymbol{\mu} \). The propagation of uncertainty is then given by the matrix equation:
This compact form encapsulates all the cross-terms. The diagonal elements of \( \boldsymbol{\Sigma}_y \) give the variances of the output components, while the off-diagonal elements give the covariances between different output components. This method is powerful because it handles any number of inputs and outputs simultaneously, provided the function is differentiable at the mean.
Interpreting the Cross-Terms
Consider a simple case with two inputs \( x_1 \) and \( x_2 \) and one output \( y = f(x_1, x_2) \). The Jacobian is a row vector \( \mathbf{J} = \left[ \frac{\partial f}{\partial x_1}, \frac{\partial f}{\partial x_2} \right] \). The covariance matrix \( \boldsymbol{\Sigma}_x \) is a \( 2 \times 2 \) matrix:
where \( \rho \) is the correlation coefficient. The variance of \( y \) is:
The last term is the correlation term. If \( \rho > 0 \) and the partial derivatives have the same sign, this term increases the total variance. If \( \rho < 0 \) or the derivatives have opposite signs, it decreases the variance. This demonstrates that correlation can either amplify or dampen the uncertainty in the result.
The derivations
Propagation with the Covariance Matrix
Given a random input vector x ∈ ℝn with mean μ and covariance matrix Σx, and a differentiable vector-valued map y = f(x) with f: ℝn → ℝm, the covariance of the output to first order is Σy = J Σx JT, where J is the m×n Jacobian Jij = ∂fi/∂xj evaluated at μ.
Variance of Linear Combinations and Covariance
For random variables X1, …, Xn with finite variances and fixed real constants a1, …, an, the variance of the linear combination S = ∑i ai Xi is the double sum Var(S) = ∑i ∑j ai aj Cov(Xi, Xj), which collapses to the weighted sum of variances ∑i ai2 Var(Xi) when the variables are pairwise uncorrelated.
A common pitfall is assuming that if two variables are independent, their functions are also independent. This is not necessarily true. For example, if \( X \) and \( Y \) are independent, \( X^2 \) and \( Y^2 \) are independent, but \( X \) and \( X^2 \) are not. Always verify the independence of the specific variables entering the propagation formula, not just the original measurements. C
Worked example
Problem. A resistor's power dissipation is given by \( P = V^2 / R \). The voltage \( V \) is measured as \( 10.0 \pm 0.5 \ \mathrm{V} \) and the resistance \( R \) as \( 5.0 \pm 0.2 \ \Omega \). The correlation coefficient between \( V \) and \( R \) is \( \rho = 0.5 \). Calculate the uncertainty in \( P \).
- Identify the function: \( P(V, R) = V^2 R^{-1} \).
- Compute the partial derivatives at the mean values \( V=10.0 \), \( R=5.0 \): \( \frac{\partial P}{\partial V} = 2V/R = 2(10.0)/5.0 = 4.0 \ \mathrm{W/V} \). \( \frac{\partial P}{\partial R} = -V^2/R^2 = -(10.0)^2/(5.0)^2 = -4.0 \ \mathrm{W/\Omega} \).
- Identify variances: \( \sigma_V^2 = 0.5^2 = 0.25 \ \mathrm{V^2} \), \( \sigma_R^2 = 0.2^2 = 0.04 \ \Omega^2 \).
- Compute the covariance: \( \mathrm{Cov}(V, R) = \rho \sigma_V \sigma_R = 0.5 \times 0.5 \times 0.2 = 0.05 \ \mathrm{V \cdot \Omega} \).
- Apply the propagation formula: \( \sigma_P^2 = (4.0)^2 (0.25) + (-4.0)^2 (0.04) + 2(4.0)(-4.0)(0.05) \). \( \sigma_P^2 = 16(0.25) + 16(0.04) - 32(0.05) \). \( \sigma_P^2 = 4.0 + 0.64 - 1.60 = 3.04 \ \mathrm{W^2} \).
- Take the square root: \( \sigma_P = \sqrt{3.04} \approx 1.74 \ \mathrm{W} \).
Units check. The term \( (\partial P/\partial V)^2 \sigma_V^2 \) has units \( (\mathrm{W/V})^2 \mathrm{V^2} = \mathrm{W^2} \). The term \( 2 (\partial P/\partial V)(\partial P/\partial R) \mathrm{Cov}(V,R) \) has units \( (\mathrm{W/V})(\mathrm{W/\Omega})(\mathrm{V \cdot \Omega}) = \mathrm{W^2} \). All terms are consistent in \( \mathrm{W^2} \).
Three quick tests
- If two variables are uncorrelated, what happens to the cross-terms in the variance propagation formula?
Answer
They vanish, reducing the formula to the sum of weighted variances. - Can a positive correlation coefficient \( \rho \) ever decrease the total uncertainty?
Answer
Yes, if the partial derivatives with respect to the correlated variables have opposite signs. - In the matrix formula \( \boldsymbol{\Sigma}_y = \mathbf{J} \boldsymbol{\Sigma}_x \mathbf{J}^T \), what does the diagonal element \( (\boldsymbol{\Sigma}_y)_{ii} \) represent?
Answer
The variance of the \( i \)-th output variable \( y_i \).
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-111 · D-109 · Threads: chance