Why do length measurements apparently have zero uncertainty?

Question

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?

Answer 1

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.