# 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. Jan 11, 2016 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. Jan 11, 2016 at 10:18