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The uncertainty in any particular measurement is $\sigma_E$. Resolution for these devices is almost always stated in relative terms as here, but take it like this because it depends on the energy measured. So just multiply by the energy. That is, express your signal in $\mathrm{GeV}$ and then find $$ \begin{align} \sigma_E = \left(\frac{0.1}{\sqrt{E}} ...


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This is mentioned, for example, in "Statistical Mechanics: Algorithms and Computations" by Krauth section 2.1.2. (But the author just uses computational/empirical details and you've gone into it more). So the result you get is correct and echoed elsewhere. The errors will blow up rapidly.


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While I am not familiar with the details of your experiment, I am quite familiar with the methods of maximum likelihood estimation. In the field in which I work, we model the system response as part of our model - so that when we compare the predicted measurements with the actual ones, we find our how close our model is to the underlying truth. It goes ...


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@Tom's answer appears absolutely correct, but omits to mention the assumptions behind these formulae: (i) The measured values of $A$ and $B$ follow a normal distribution. (ii) The measurement of $A$ does not depend on $B$ and vice-versa. i.e. the variables are independent. (iii) Over the range $\pm \sigma_A$ and $\pm \sigma_B$; $Z(A)$ and $Z(B)$ are ...


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One of the formulae in your question is not quite correct (as pointed out by Nikos M) but it should be ${\sigma_Z}^2 = ({\delta Z \over \delta A})^2 {\sigma_A}^2 + ({\delta Z \over \delta B})^2 {\sigma_B}^2$ This is the basic equation for propagating uncertainties. The version you have in your question without the squared terms is a simplification of ...


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If the data are normally distributed, then the variance of the variance is given by $$ Var(s^2) = \frac{2 \sigma^4}{n-1},$$ where $\sigma$ is the standard deviation. $$\sigma^2 = \frac{1}{n} \Sigma_i (x_i - \bar{x})^2 $$ And $s^2$ is the unbiased sample variance, calculated from the data, where $s^2 = n \sigma^2/(n-1)$. Formulae found here Hence the ...



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