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I Effect of errors due to branching and acceptances

  1. Oct 13, 2018 at 10:51 AM #1

    CAF123

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    Say I have say 10 measurements of an experimental observable e.g differential cross section with respect to rapidity , one measurement in each bin and corresponding statistical and systematic uncertainties given in % per bin.

    Now, suppose I want the value of this dsigma/dy after correcting for e.g acceptance and certain branching fraction.

    How are the statistical and systematic errors changed? The acceptance and branching typically come with an error so in what way will they affect the statistical and systematic errors?
     
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  3. Oct 13, 2018 at 12:34 PM #2

    mfb

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    Standard error propagation for the multiplication of two numbers.
    It gets more interesting if you have significant migration between the bins or other sources of correlations.
     
  4. Oct 13, 2018 at 12:45 PM #3

    CAF123

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    But does the acceptance and branching enter into the statistical or the systematic uncertainty? I think the acceptance is dependent on the rapidity so cannot be systematic (therefore statistical) and the branching would enter in the systematics?
     
  5. Oct 13, 2018 at 9:20 PM #4

    mfb

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    What exactly is a statistic uncertainty about that? If you repeat the measurement, do you expect a different result?

    In general both will come with a statistic and a systematic uncertainty. Treat them separately. Combine all sources of statistical uncertainty, combine all sources of systematic uncertainty.
     
  6. Oct 14, 2018 at 5:20 AM #5

    CAF123

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    Thanks, in the article I am studying the acceptance and branching are given each in the form ##A \pm \delta A## only, so how should I interpret the ##\delta A##? Added in quadrature of the statistical and systematic errors? If that might be the case, then it seems without further information on the decomposition of this error it is impossible for me to split into statistical and systematic error contributions?
     
  7. Oct 14, 2018 at 5:43 AM #6

    mfb

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    I don't know, I would have to see the article. If it is both combined then you can't split it of course.
    If you have external sources of uncertainty it is common to treat them as separate class. Something like that: ##52 \pm 2 \text{(stat)} \pm 3 \text{(syst)} \pm 1 \text{(BF)}##.
     
  8. Oct 15, 2018 at 3:07 PM #7

    CAF123

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    Indeed, if I had the errors for the branching and acceptance spelled out as ##A \pm \delta A_{\text{stat}} \pm \delta A_{\text{sys}} ## for acceptance and ##B\pm \delta B_{\text{stat}} \pm \delta B_{\text{sys}} ## for the branching then assuming independent sources of statistical and systematic errors, my net statistical on d(sigma)/dy would be $$\delta _{\text{stat}} = \sqrt{\delta A_{\text{stat}}^2 + \delta B_{\text{stat}}^2 + \dots}$$ and similarly for the systematic. I think that should be correct.

    It's common to see experimental formulae for differential cross sections expressed in terms of parameters that are estimated by experimentalists, e.g the acceptance appears in the denominator. Could you also get the effect of the error on the differential cross section due to the acceptance through partial derivatives? And this would coincide with simply adding in quadrature?
     
  9. Oct 16, 2018 at 5:16 AM #8

    mfb

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    I guess so - it works if I understand your description correctly.
     
  10. Oct 18, 2018 at 3:54 AM #9

    CAF123

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    Can we do this explicitly? Given d(sigma)/dyi = f(Ai,...), where Ai is the acceptance in bin i and dots denote other experimental quantities such as track efficiencies, purities etc..., we have in the notation used in previous posts $$\delta_{\text{stat}}^2 \left( \frac{d \sigma}{d y_i} \right) = \left( \frac{\delta f}{\delta A_i} \delta A_{i,\text{stat}} \right)^2 + \dots$$ so that $$\delta_{\text{stat}} \left( \frac{d \sigma}{d y_i} \right) = \sqrt{\left( \frac{\delta f}{\delta A_i} \delta A_{i,\text{stat}} \right)^2 + \dots}$$ But comparing this to the centered equation in #7, $$\delta_{\text{stat}} = \sqrt{\delta A_{\text{stat}}^2 + \dots}$$ there is an extra multiplicative factor ##\left( \frac{\delta f}{\delta A}\right)^2## in the case where I used partial derivatives. What's the reconciliation?
     
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