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Inverse Efficiency Matrix (error)

  1. Dec 17, 2014 #1


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    Hi, let's say in some experiment with ##Z^0## (eg LEP) you are able to determine the "misidentification" of your particles.
    Then you can find the efficiency matrix ##M_{eff}## which is given (for ##Z^0## decays to leptons or hadrons):

    [itex] \begin{pmatrix} N_e \\ N_\mu \\ N_\tau \\ N_{had} \end{pmatrix}_{detect} = M_{eff} \begin{pmatrix} N_e \\ N_\mu \\ N_\tau \\ N_{had} \end{pmatrix}_{real} \Rightarrow \vec{N}_{det} = M_{eff} \vec{N}_{real} [/itex]

    With some errors in its elements, so in general you measure ##M_{eff} \pm \delta M_{eff}##.

    My question is then, since I want to find what is actually my real measured events for the decay products:

    [itex] \vec{N}_{real} = M_{eff}^{-1} \vec{N}_{det}[/itex]

    How can you find the error in the elements of [itex]M_{eff}^{-1}[/itex]?

    One example I thought of is quite primitive, and in general very untrustworthy, for example get [itex]M_{eff}+\delta M_{eff}[/itex] find its inverse, and do the same for [itex]M_{eff} - \delta M_{eff}[/itex]. Then I know how much (in the worst case scenario the elements will diverge from the mean valued [itex]M_{eff}^{-1}[/itex].

    I find it more plausible to create different matrices [itex] A_i[/itex] which have as elements different values in the range allowed from [itex]M_{eff} \pm \delta M_{eff}[/itex], take their inverse [itex]A_i^{-1}[/itex] and then mean value it and find its statistical error.
    However with a ##4 \times 4## matrix (or 16 elements) I don't know how many [itex]A_i[/itex]'s can give a trustworthy result...any help?
  2. jcsd
  3. Dec 17, 2014 #2


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    Those are matrices with 16 elements, so you have 16 different uncertainties. To make it worse, they are correlated. Adding the positive value to all won't work.
    If you rely on Monte Carlo for your matrix, you can directly determine the inverse matrix so your statistics tool gives the uncertainties.

    Generating many random matrices looks like a possible approach. As many as you need for the precision you are interested in. The more correlations you take into account the harder it gets.
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