# Derivation of uncertainty propagation?

For a function $$y =f(x), x=\left(x_1, x_2, ..., x_N\right)$$ the law of propagation of uncertainty, see GUM sect 5$^{[1]}$ (pdf), is generally given as $$u_y^2 = \sum_{i=1}^N \left(\frac{\partial f}{\partial x_i}\right)^2 u_{x_i}^2$$ where $u_{x_i}$ is the uncertainty associated with each $x_i$.

How do we derive this propagation law? I understand that it can be achieved by Taylor expanding the function, but I don't see why this is a sensible thing to do, or what is the physical meaning of it.

(I've asked basically the same question at Signal Processing here but haven't got any useful answers. Hopefully you guys are more suited to this question).

[1] Evaluation of measurement data – Guide to the expression of uncertainty in measurement
JCGM 100:2008 (html) [Sec 5 here]

• Can you give a citation or two for "the law of uncertainty propagation?" Commented May 6, 2014 at 14:26
• @Alexis added a reference to the GUM. There is also a wikipedia page, en.wikipedia.org/wiki/Propagation_of_uncertainty. I also noticed I missed all the ^2 in the equation so I've corrected that too. Commented May 6, 2014 at 14:35
• If X and Y are independent, Var(X+Y) = Var(X) + Var(Y). The variance is more complicated for a non-linear relationship, e.g., Var(XY). Using the first terms of the Taylor Series expansion of something like XY gives a linear function that is assumed to be a reasonable approximation in the region of interest. There is no guarantee that this is true and the GUM discusses the use of a second order Taylor Series approximation. Commented May 6, 2014 at 15:52
• @Thomas So, if I understand correctly for anything other than a linear function expanding the variance becomes quite hard. By using the Taylor expansion we simplify the equation into something solvable. Generally we only take the first term as higher terms become small (although not always). It all begins to make sense, thanks :). Commented May 7, 2014 at 15:04
• Yes. I am trying to find a reference for you that does a better job of explaining this than what I offered. Commented May 7, 2014 at 17:41

The idea behind the differential calculus is to study potentially complicated functions $f:\mathbb{R}^n \to \mathbb{R}^m$ by means of linear approximations. Everything flows from this single idea.

For $x\in \mathbb{R}^n$ "the" linear approximation to $f$ near $x$ (if a unique one exists) is called the "derivative" or "gradient" $Df$. (It typically changes from one point to another and so is a function of $x$.)

By definition, then, $$Df: T_x\mathbb{R}^n \to T_{f(x)}\mathbb{R}^m$$ is a linear map from the space of all vectors in $\mathbb{R}^n$ originating at $x$ to the space of all vectors in $\mathbb{R}^m$ originating at $y=f(x)$.

It is a theorem (Spivak, Theorem 2-7) that when we use the directions determined by the coordinates $(x_1, x_2, \ldots, x_n)$ for $\mathbb{R}^n$ and $(y_1, y_2, \ldots, y_m)$ for $\mathbb{R}^m$ as bases for $T_x\mathbb{R}^n$ and $T_{f(x)}\mathbb{R}^m$, respectively, then the entries in the $m\times n$ matrix for $Df$ are the partial derivatives

$$(Df)_{ij} = \frac{\partial y_i}{\partial x_j}.\tag{1}$$

Suppose the "errors" $u_{x_i}$ in the $x_i$ are constant multiples of the standard deviations of random variables $U_i$ describing uncertainties in the $x_i$. One way to express this is in terms of the squared errors: assume there is some positive number $\lambda$ (often $1$, sometimes $2$, occasionally something else) for which

$$u_{x_i}^2 = \lambda^2 \operatorname{Var}(U_i)$$

for each $i$. Applying the linear approximation $Df(x)$, and taking $m=1$ for simplicity (although the general case is scarcely any more difficult), we know from $(1)$ that (at least approximately) the random variable governing uncertainties in $y$ is equal to

$$V = (Df(x)) (U_1, U_2, \ldots, U_n)^\prime = \frac{\partial y}{\partial x_i}U_1 + \frac{\partial y}{\partial x_2}U_2 + \cdots + \frac{\partial y}{\partial x_n}U_n.$$

The formula quoted in the question arises when it is assumed there is no correlation among the $U_i$, whence all the covariances in the calculation of $\operatorname{Var}(V)$ vanish, yielding

\eqalign{u_y^2 = \lambda^2\operatorname{Var}(V) &= \lambda^2\left(\operatorname{Var}\left(\frac{\partial y}{\partial x_1}U_1\right) + \cdots + \operatorname{Var}\left(\frac{\partial y}{\partial x_n}U_n\right)\right)\\ &= \left(\frac{\partial y}{\partial x_1}\right)^2 \lambda^2\operatorname{Var}(U_1)+ \cdots + \left(\frac{\partial y}{\partial x_n}\right)^2 \lambda^2\operatorname{Var}(U_n)\\ &= \left(\frac{\partial y}{\partial x_1}\right)^2 u_{x_1}^2 + \cdots + \left(\frac{\partial y}{\partial x_n}\right)^2 u_{x_n}^2 } QED.

The assumptions needed to derive this conclusion provide insight into its meaning, interpretation, and scope:

1. $f$ must be differentiable at $x$: that is, $Df$ must exist at $x$. This means that within a neighborhood of $x$, $f$ can be approximated to second order in $|x|$ by a linear transformation.

2. The "errors" $u_{x_i}$ must be multiples of a standard deviation.

3. The random deviations (that model the uncertainty in $x$) must be uncorrelated.

4. The typical size of the deviations $U_i = x^{*}_i - x_i$, as measured by the errors $u_{x_i}$, must be small enough that $(Df(x))(x^{*}-x)$ remains a good approximation to $f(x^{*})$; and large deviations (where the approximation no longer holds) must be improbable.

Note that Taylor's Theorem is not needed: the result combines the most basic properties of differentiation with a fundamental property of variances.

### Reference

Michael Spivak, Calculus on Manifolds, W. A. Benjamin (1965). Chapters 1 & 2.

• What a wonderful answer, and a fantastic reference too ! For anyone interested, Spivak (1965) is not an easy read (for normal people) but very well worth the effort. It is available for free here: strangebeautiful.com/other-texts/spivak-calc-manifolds.pdf Commented Sep 25, 2020 at 19:29