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I was reading error propagation from "Introduction to Statistics and Data Analysis for Physicists" by G. Bohm and G. Zech, and I stumbled upon this example,

Given are the sides a, b with a reading error $\delta_1$ and a relative scaling error $\delta_2$, caused by a possible extension or shrinkage of the measuring tape. We want to calculate the error $\delta$F of the area F = ab. We find, $$(\delta_a)^{2} = (\delta_1)^{2} + (a\delta_2)^{2}\\(\delta_b)^{2} = (\delta_1)^{2} + (b\delta_2)^{2}\\C_{ab} = ab(\delta_2)^{2}$$ where $C_{ab}$ is covariance

I'm not able to understand how this covariance comes about.

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  • $\begingroup$ I'll answer this in the next days, meanwhile have a look at this answer to better understand the framework. $\endgroup$ – Massimo Ortolano Oct 28 '16 at 11:09
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Covariance is associated with the variance, which is not the same as the second moment. To compute the variance on needs to subtract the averages and the same applies for the convariance. Easiest way would be simply to assume that the averages are zero, because the the second moment gives the variance and the expectation value of the product would be the covariance.

So, while the variance is given by $$ \sigma^2 = E\{(x-\mu)^2\} = E\{x^2\} - \mu^2 , $$ the covariance (in your notation) is given by $$ C_{ab} = E\{(a-\delta_a)(b-\delta_b)\} = E\{ab\} - \delta_a\delta_b . $$ So what the covariance is telling us is that the quantities are not statistically independent, because statistical independence would have meant that $$ E\{ab\} = E\{a\}E\{b\} = \delta_a\delta_b . $$ So the reason why you are given a nonzero value for the covariance is to take into account the fact that $a$ and $b$ are not statistically independent.

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