Propagation of uncertainties for a function of observables

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This is a personal derivation of a familiar result! Alex Read 14.05.2011

We want to estimate the uncertainty of a function [math]f[/math] of a set of [math]n[/math] uncorrelated observables [math]\vec{x}[/math] with associated uncertainties [math]\vec{\sigma}[/math]. We assume that the observables [math]\vec{x}[/math] are sampled from normal distributions which are narrow relative to the form of the function and thus approximate the function with its Taylor-series expansion, keeping only the lowest-order derivatives [math]f(\vec{x}+\vec{\delta_x}) \simeq f(\vec{x}) + {df(\vec{x}) \over d{\vec{x}}}\cdot \vec{\delta_x} + h.o.[/math], where [math]\delta_{x_i}[/math] represents the Gaussian-distributed deviation from the true mean (which is unknown, but estimated by the observed value) [math]x_i[/math]. In other words [math]\lt \delta_{x_i}\gt =0[/math] and [math]\lt \delta_{x_i}^2\gt =\sigma_{x_i}^2[/math].

The uncertainty in [math]f[/math] is estimated as the square root of its mean variance [math]\sqrt{\sigma^2_f}[/math] where [math]\sigma^2_f=\lt (f-\lt f\gt )^2\gt [/math]. We replace [math]f[/math] with the series approximation above and take advantage of that in this approximation [math]\lt f\gt = f(\vec{x})[/math]. In addition we write the dot product as an explicit sum, taking care to have separate indices for the terms in the square of the dot product, obtaining [math]\sigma^2_f = \lt (\Sigma_{i=1}^n {df(\vec{x}) \over dx_i} \delta_{x_i}) (\Sigma_{j=1}^n {df(\vec{x}) \over dx_j} \delta_{x_j}) \gt [/math]. The product of these two sums can be re-written as the sum of 2 terms: [math] \sigma^2_f = \lt \Sigma_{i=1}^n ({df(\vec{x}) \over dx_i} \delta_{x_i})^2 + \Sigma_{i=1}^n \Sigma_{j=1}^n (1-\delta_{ij}){df(\vec{x}) \over dx_i} {df(\vec{x}) \over dx_j}\delta_{x_i}\delta_{x_j} \gt [/math]

Since the sub-terms in the second term above are only non-zero for [math]i\neq j[/math], and since [math]x_i[/math] and [math]x_{j\neq i}[/math] are uncorrelated, and since the average of a product of an uncorrelated pair of Gaussian distributed variables is the product of their averages, the second term vanishes due to the fact that [math]\lt \delta_{x_i}\gt =0[/math]. Since the variance of a [math]\delta_{x_i}[/math] is [math]\sigma_{x_i}^2[/math], the final and familiar result is [math]\sigma^2_f = \Sigma_{i=1}^n ({df(\vec{x}) \over dx_i} \sigma_{x_i})^2[/math]. In other words, uncorrelated contributions to the total uncertainty of a function of observables are to be added in quadrature.