Opened 6 years ago
Last modified 6 years ago
#1339 new How to
B tagging efficiencies are independent?
| Reported by: | localname | Owned by: | |
|---|---|---|---|
| Priority: | major | Milestone: | |
| Component: | Delphes code | Version: | Delphes 3 |
| Keywords: | b tagging | Cc: |
Description
Now Delphes allow different b tagging working points. But different taggings are independent events. Which means a jet can be tightly tagged but not loosely tagged. This may cause some substantial deviation from reality?
For instance, if one has two WPs (T for the tight one and L for another, with b efficiency P_T and P_L), I expect the possibilty of a b jet get tagged as:
P_T: tightly tagged.
P_L-P_T: loosely tagged. The subtraction is because a tightly tagged b jet must also pass the loose tag.
1-P_L: not tagged as a b jet.
If two tags are independent, the chances become:
P_T: tightly tagged.
(1-P_T)*P_L: loosely tagged only.
(1-P_T)(1-P_L): not tagged.
In this toy model, I can find a solution to change the P's to match the first case for arbitrary number of WPs, but is it even necessary?
