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In order to estimate the length $L$ of an object the distance from its edges to the $0$ of a graded ruler are measured. Assume this object has its edges at $x$ and $y$ (mean values) with standard deviations $\sigma_x$ and $\sigma_y$ respectively, and it has a constant length, thus the two variables are not independent. It holds $y = x + L$. The function we then measure is $ L = f(x,y) = y - x$.

Using the error propagation formula

$$\sigma_L^2 =\left| \frac{\partial f}{\partial x}\right| ^2\sigma^2_x+\left| \frac{\partial f}{\partial y}\right|^2\sigma^2_y+2\frac{\partial f}{\partial x}\frac{\partial f}{\partial y}\sigma_{xy}$$

$$\sigma_L^2 = \left|-1 \right| ^2\sigma^2_x+\left| +1 \right|^2\sigma^2_y+2 (-1)(+1)\sigma_{xy}$$

$$\sigma_L^2 = 2 \sigma^2_x - 2 \sigma_{xy},$$

because $\sigma_x = \sigma_y$. Moreover $\sigma_{xy} = E[(x - E[x])(y - E[y])] = \sigma^2_x = \sigma^2_y$, using again $x = y - L$. This yields $\sigma_L^2 = 0$.

Intuitively I expect twice the error: $\sigma_L = 2 \sigma_x$. Why doesn't it work out like that?

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    $\begingroup$ I didn't think this needed the homework-and-exercises tag according to our policy, but even if it does, I think it takes only a trivial edit (which I've made) to make it conceptual and on topic. $\endgroup$ – David Z Jan 11 '16 at 10:07
  • $\begingroup$ the last time i used this formula, the third term also had a modulus sign on it. Intuitively it should also be so. Note that taking the modulus on the third term gives you the result that you expected. $\endgroup$ – Bruce Lee Jan 11 '16 at 10:18
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You're making two measurements, each with an error, and then subtracting. But, here's the key:

Assume... it has a constant length, thus the two variables are not independent.

This isn't true: the two variables are completely independent. That you made an error in one direction at one end of the object indicates nothing about your error at the other end of the object (unless you add further constraints to your description of the problem, e.g. exact same measuring technique at each end).

But, if you assume that the length is a constant, then there is no variability. Bingo: standard deviation of zero.

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  • $\begingroup$ yes, a constant length means zero uncertainty, it was in the assumptions $\endgroup$ – user46925 Jan 11 '16 at 12:58

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