# Propagation of large errors

Everything I've read about combining errors in quadrature when multiplying or dividing quantities with associated errors says that this works for "small error". What about large error? Say I want to compute A/B where A is +/- 1% and B is +/- 50%, can I still reasonably add the errors in quadrature?

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I assume that combining errors in quadrature means calculating the square root of the sum of the squares –  Henry Feb 14 '11 at 2:06
re-reading this question the answer feels like "ignore the tiny error in A" so let's go with the more general question: what if the errors in A and/or B are on the order of 10-30%... –  nicolaskruchten Feb 14 '11 at 2:07
@Henry yes. ... –  nicolaskruchten Feb 14 '11 at 2:09

For large error, the standard error of $A/B$ depends on the distributions of $A$ and $B$, not just on their standard errors. The distribution of $A/B$ is known as a ratio distribution, but which ratio distribution depends on the distributions of $A$ and $B$.

If we assume that $A$ and $B$ both have Gaussian (normal) distributions, then $A/B$ has a Gaussian ratio distribution, for which a closed form exists but is rather complicated. In general, this will be an asymmetric distribution, so it is not well characterised simply by its mean and standard deviation. However, it is possible to find a confidence interval for $A/B$ using Fieller's theorem.

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The formula for error propagation

$\sigma_f^2 = \Sigma (\frac{\delta f}{\delta x} \sigma_x)^2$

works exactly for normally distributed errors and linear functions $f(x_1,x_2,...)$

Since (most) functions can be linearly approximated, the above also works for small errors. For large errors, a symmetric distribution of $x$ leads to an asymmetric distribution of the error in $f$ (e.g. if $f(x)=x^{10}$, $f(0)=0$, $f(1)=1$, $f(2)=1024$, so the formula won't hold if $x=1$ and $\sigma_x=1$).

With large error, you may be able to calculate the transformation of the error distribution, otherwise you can perform a Monte Carlo simulation to estimate the distribution of $\sigma_f$.

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The first problem with large errors is that the expected value of the multiplication or division of the uncertain values will not be the multiplication or the division of the expected values. So while it is true that $E[X+Y]=E[X]+E[Y]$ and $E[X-Y]=E[X]-E[Y]$, it would usually not be true to say $E[XY]=E[X]E[Y]$ or $E[X/Y]=E[X]/E[Y]$, though they will be close for small errors. But for large errors, that effect will disrupt your propagation of error calculations.
Suppose for example you had $A$ being 270, 540 or 810 and $B$ being 3, 6 or 9. Then $A/B$ could be 30, 45, 60, 90 (three ways), 135, 180, or 270. While 90 may be the mode and median as well as 540/6, the mean is 110, and 30 is much closer to 90 (or to 110) than 270 is.