subprocess analysis with madanalysis

Asked by zheng on 2018-02-21

Hello,

I'm studying preliminary information of the p p > t h j process by analysing the 'events.lhe' file at parton level with madanalysis. Four subprocess folders are generated.

I'm wondering if the results from each subprocess is also stored separately somewhere so that I can analyse each subprocess and compare their separate contribution and also with the total at the same time. Or do I need to generate each subprocess separately to analyze and compare for this purpose? Thank you very much.

Best regards,
Ya-Juan

Question information

Language:
English Edit question
Status:
Solved
For:
MadAnalysis 5 Edit question
Assignee:
No assignee Edit question
Solved by:
Benjamin Fuks
Solved:
2018-03-19
Last query:
2018-03-19
Last reply:
2018-03-19

This question was reopened

Benjamin Fuks (fuks) said : #1

Hi,

You can't do that. MadAnalysis will analyse the full sample directly. For doing what you want, you would need one sample per subprocess.

Regards,

Benjamin

> On 20 Feb 2018, at 23:47 , zheng <email address hidden> wrote:
>
> New question #664727 on MadAnalysis 5:
> https://answers.launchpad.net/madanalysis5/+question/664727
>
> Hello,
>
> I'm studying preliminary information of the p p > t h j process by analysing the 'events.lhe' file at parton level with madanalysis. Four subprocess folders are generated.
>
> I'm wondering if the results from each subprocess is also stored separately somewhere so that I can analyse each subprocess and compare their separate contribution and also with the total at the same time. Or do I need to generate each subprocess separately to analyze and compare for this purpose? Thank you very much.
>
> Best regards,
> Ya-Juan
>
> --
> You received this question notification because you are an answer
> contact for MadAnalysis 5.
>

zheng (yjzheng218) said : #2

Thanks Benjamin Fuks, that solved my question.

zheng (yjzheng218) said : #3

Hello,

Following the same analysis, I have a question in using the 'reject' command for multiparticle rapidity.

I generated with Madgraph two process 1 ub>thd 2 cb>ths at the same time and analysed the lhe file. I wanted to see the effect of rapidity cut for the (thj). The folloing command is what I imposed.

ma5>define thj = t h1 j
ma5>plot Y (thj)
ma5>reject Y (thj) < 0
ma5>plot Y (thj)
ma5>reject Y (thj) < 0.05
ma5>plot Y (thj)
ma5>reject Y (thj) < 0.10
ma5>plot Y (thj)
ma5>reject Y (thj) < 0.15
ma5>plot Y (thj)
ma5>reject Y (thj) < 0.20
ma5>plot Y (thj)
ma5>submit ../Ythj_12ubcb

The number after cut is also copied as below,

Cuts
Signal (S)
Background (B)
S vs B
Initial (no cut)
130.159 +/- 0.251
Cut 1
38.49 +/- 5.21
Cut 2
36.75 +/- 5.14
Cut 3
35.06 +/- 5.06
Cut 4
33.33 +/- 4.98
Cut 5
31.61 +/- 4.89

From the Y (thj) distribution plot before any cut, I expect the command 'reject Y (thj) < 0' would keep more than half events (Hope there's some way that I can show you the plot). However, the cross section number is reduced from 13.02fb to 3.85fb, with only 1/3 of events left. How should I understand the inconsistency of exceptions from the plot? Do I have some misunderstanding in the command 'reject', or reading the output number after cut? Thank you very much.

Best regards,
Ya-Juan

Benjamin Fuks (fuks) said : #4

Hi,

Please provide me the working directory generated by MA5.

Cheers,

Benjamin

> On 9 Mar 2018, at 14:22 , zheng <email address hidden> wrote:
>
> Question #664727 on MadAnalysis 5 changed:
> https://answers.launchpad.net/madanalysis5/+question/664727
>
> Status: Solved => Open
>
> zheng is still having a problem:
> Hello,
>
> Following the same analysis, I have a question in using the 'reject'
> command for multiparticle rapidity.
>
> I generated with Madgraph two process 1 ub>thd 2 cb>ths at the same time
> and analysed the lhe file. I wanted to see the effect of rapidity cut
> for the (thj). The folloing command is what I imposed.
>
> ma5>define thj = t h1 j
> ma5>plot Y (thj)
> ma5>reject Y (thj) < 0
> ma5>plot Y (thj)
> ma5>reject Y (thj) < 0.05
> ma5>plot Y (thj)
> ma5>reject Y (thj) < 0.10
> ma5>plot Y (thj)
> ma5>reject Y (thj) < 0.15
> ma5>plot Y (thj)
> ma5>reject Y (thj) < 0.20
> ma5>plot Y (thj)
> ma5>submit ../Ythj_12ubcb
>
> The number after cut is also copied as below,
>
> Cuts
> Signal (S)
> Background (B)
> S vs B
> Initial (no cut)
> 130.159 +/- 0.251
> Cut 1
> 38.49 +/- 5.21
> Cut 2
> 36.75 +/- 5.14
> Cut 3
> 35.06 +/- 5.06
> Cut 4
> 33.33 +/- 4.98
> Cut 5
> 31.61 +/- 4.89
>
>> From the Y (thj) distribution plot before any cut, I expect the command
> 'reject Y (thj) < 0' would keep more than half events (Hope there's some
> way that I can show you the plot). However, the cross section number is
> reduced from 13.02fb to 3.85fb, with only 1/3 of events left. How
> should I understand the inconsistency of exceptions from the plot? Do I
> have some misunderstanding in the command 'reject', or reading the
> output number after cut? Thank you very much.
>
> Best regards,
> Ya-Juan
>
> --
> You received this question notification because you are an answer
> contact for MadAnalysis 5.
>

zheng (yjzheng218) said : #5

Hi, I shared the files created by MA5 in dropbox, please see if you can read it. Thank you very much.

https://www.dropbox.com/sh/hu4iull4g5x32x4/AABFWg2Z50kFa4g6eXgcNREza?dl=0

Best regards,
Ya-Juan

Benjamin Fuks (fuks) said : #6

Hi Ya-Juan,

I have checked the code and it does well what it is supposed to do. As soon as one of the considered particle has a negative rapidity, the event is rejected. Remember that for each event, you have three entries in the histogram and that a single one of them being negative is sufficient to reject the event.

Cheers,

Benjamin

> On 12 Mar 2018, at 08:58 , zheng <email address hidden> wrote:
>
> Question #664727 on MadAnalysis 5 changed:
> https://answers.launchpad.net/madanalysis5/+question/664727
>
> Status: Answered => Open
>
> zheng is still having a problem:
>
> Hi, I shared the files created by MA5 in dropbox, please see if you can read it. Thank you very much.
>
> https://www.dropbox.com/sh/hu4iull4g5x32x4/AABFWg2Z50kFa4g6eXgcNREza?dl=0
>
> Best regards,
> Ya-Juan
>
> --
> You received this question notification because you are an answer
> contact for MadAnalysis 5.
>

zheng (yjzheng218) said : #7

Hi Benjamin.

thank you very much for your explanation. It seems MA5 are rejecting events by counting the rapidity of each particle independently even though it is defined as a multiparticle.

I assumed that defining multi particle thj would result in a( thj) system and keeps the four momentum of this multiparticle from the sum of the three. Therefore, I thought Y(thj) would calculate the rapidity of this multiparticle system from the four momentum. I wanted to reject the negative rapidity events of this system. In this case, do I have to make my own program or is there function in MA5 defined for this purpose?

Best regards,
Ya-Juan

Best Benjamin Fuks (fuks) said : #8

Hi Ya-Juan,

If you want to sum the three momenta, you need to encode the cut as
  select eta (t h j) > xxx
with spaces. Note that this will loop over all t h j combinations. You may want to focus in t[1], h[1], etc… (the leading top, the leading h).

Cheers,

Benjamin

> On 17 Mar 2018, at 21:38 , zheng <email address hidden> wrote:
>
> Question #664727 on MadAnalysis 5 changed:
> https://answers.launchpad.net/madanalysis5/+question/664727
>
> Status: Answered => Open
>
> zheng is still having a problem:
> Hi Benjamin.
>
> thank you very much for your explanation. It seems MA5 are rejecting
> events by counting the rapidity of each particle independently even
> though it is defined as a multiparticle.
>
> I assumed that defining multi particle thj would result in a( thj)
> system and keeps the four momentum of this multiparticle from the sum of
> the three. Therefore, I thought Y(thj) would calculate the rapidity of
> this multiparticle system from the four momentum. I wanted to reject the
> negative rapidity events of this system. In this case, do I have to make
> my own program or is there function in MA5 defined for this purpose?
>
> Best regards,
> Ya-Juan

zheng (yjzheng218) said : #9

Thanks Benjamin Fuks, that solved my question.