Error propagation and symmetric errors

  • #1
Malamala
315
27
Hello! I am a bit confused about how to interpret symmetric error when doing error propagation. For example, if I have ##E = \frac{mv^2}{2}##, and I do error propagation I get ##\frac{dE}{E} = 2\frac{dv}{v}##. Which implies that if I have v being normally distributed, and hence having a symmetric error (by this I mean that the upper and lower limit of the uncertainty interval in the value is the same), the same is true for E, as that formula is predicting just one ##dE## and people would quote (at least in most of the papers I read) it as ##E \pm dE##. However if I have say, ##E = 1000## eV, ##dE = 100## eV, ##m = 100## amu and I calculate the associated speed, I get, for ##E = 1000## eV, ##v = 43926## m/s, for ##E = 1100## eV, ##v = 46070## m/s and for ##E = 900## eV, ##v = 41672## m/s and as you can see, the associated uncertainties on v are not symmetric. So how should I think of the symmetric uncertainty given when doing error propagation, as when plugging in the numbers this is not the case. Doesn't it mean that if I have a symmetric (gaussian) error on ##v## the error on ##E## will not be symmetric? And is this not the case most of the time, except when there is a linear relationship between the variables? Thank you!
 
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  • #2
First, you need to decide if you are going to analyze and report relative uncertainties or uncertainties. You are confusing things by mixing them. ##\sigma_v## is the uncertainty in ##v## and ##\frac{\sigma_v}{v}## is the relative uncertainty in ##v##. If one is symmetric then the other is usually not, so you need to be clear about what you are saying is symmetric.

By the way, I would not use ##dv## for an uncertainty, I would use ##\sigma_v##. Your notation risks lots of confusion with derivatives.
 

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