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Acceptance value of MC

  1. Aug 14, 2009 #1
    Dear Experimental Particle Physicist,

    I have a simple question on how to calculate the Acc value from a MC dataset. I wish to know the Acc value for njet = 1 in some MC dataset which has 1000 events say. If there are 1 event that has one jet, would the following Acc value be 1/1000 ?

    Also if I was to introduce a second dataset, would the new Acc value be a sum of the two, or be similar to the above method, ie total output / total events ?

    Thank you for reading

  2. jcsd
  3. Aug 14, 2009 #2
    The acceptance is usually defined as [tex]\frac{\text{number of events reconstructed}}{\text{number of events generated}}[/tex]

    From the onset, you see that there is a drastic dependence on the generation. Depending on whether you have the same definition as me, you would need to have at least one jet in all of your generated events for instance. If you do not have one jet at least in every event generated, I would classify your strategy as including part of the luminosity in your MC (it's not necessarily bad, just different definition).

    If you were to introduce a second data set, you could treat the entire sample as just twice more statistics in a single MC sample. So the definition should remain the same, and provided you have small statistical fluctuations, the acceptance should not depend on how many events you generate.

    Hope that helps.
  4. Aug 14, 2009 #3
    Dear Humanino,

    Thank you for your shift reply. That has helped a lot. Indeed the reconstructed has at least one jet. The two data sets are for the e channel and the muon channel, these having four data sets themselves. I am grateful for your help.

    One last question has popped in my mind. I need to introduce njet = 2 into the Acc value. I have calculated the Acc for one dataset by (njet=1 events + njet=2 events) / total events . Would this be correct, or would I need to have twice the total events?

    Thank you once again
  5. Aug 14, 2009 #4
    If you want to calculate the acceptance for two jet events, my definition would be :
    [tex]\text{acceptance for two jets}=\frac{\text{number of events with at least two jets reconstructed}}{\text{number of events with at least two jets generated}}[/tex]
    Again, there are other possibilities depending on the conventions.

    If your purpose is to calculate the ratio of two jet events to one jet events, I think what you need is
    [tex]\frac{\text{two jet events reconstructed in data}}{\text{one jet events reconstructed in data}}\times\frac{\text{acceptance for one jet}}{\text{acceptance for two jets}}[/tex]
    where consistently and as before
    [tex]\text{acceptance for one jets}=\frac{\text{number of events with at least one jet reconstructed}}{\text{number of events with at least one jet generated}}[/tex]
    in this manner, you do not need to evaluate luminosity.
    Last edited: Aug 15, 2009
  6. Aug 15, 2009 #5
    Dear Humanino,

    You've been a great help! That has sorted most of my problems out, thank you. Another thing has come up, and hopefully this will be the last question, I've realised that my data sets are of different partons. I done the calculation like so:

    (Sum of weighted events reconstructed) / (sum of weighted events generated)

    Would this be the correct approach?

    Thanks again

  7. Aug 15, 2009 #6
    If I understand correctly, you generate your partons with a flat distribution in phase space and take care of the cross-sections by weighting your events with the known parton distribution functions. So instead of having more partons at low x, you just give them a higher weight than at higher x accordingly, and same story for different flavors. In that case, your formula is correct.
  8. Aug 15, 2009 #7
    Dear Humanino,

    Thank you once again. I'm really grateful for your help. Thanks

  9. Aug 15, 2009 #8
    Please do to take my advices for granted. I hope you cross check with a collaborator knowing better the details of your investigation.
  10. Aug 15, 2009 #9
    Dear Humanino

    Of course, I have a meeting with my advisor this monday. Thank you for your kind advise.

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