# Why do length measurements apparently have zero uncertainty?

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?

• 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. – David Z Jan 11 '16 at 10:07
• 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. – Bruce Lee Jan 11 '16 at 10:18