Dear experts,
After I normalize my histogram (to data luminosity for example), if I want to compute the integral btw [a, b], I wonder if I still have to use the “width option”. So can I do hist->Integral(a, b) or should it be hist->Integral(a, b, “width”)?
Regards
How did you normalize? What are the units or your y-axis afterwards?
“width” option will return to you the histogram area, i.e. in units of x*y (units of x times units of y axes)
if void “”, then it will return to you the sum of histogram entries, this means a value with just units of “y”.
Dear ferhue,
If I fill a 1D histogram with a BDT discriminant variable for example (on the x axis), for a signal sample, and I normalise to data luminosity, then I want to know the integral (number of events) btw [a, b], should I use “width” or not?
Regards
I can help you with your questions, but I will not replace your effort of thinking by yourself… That’s not the forum’s goal.
When normalizing, did you call histogram->Scale(1/someconstant); Or did you fill with weights?
Integral(a,b) --> returns number of events between bins a and b, (assuming that you did not call Scale or Fill with weights). If you scaled the histogram, then it returns the number of events times the scaling factor.
Integral(a,b,“width”) --> returns number of events times the binwidth. If you scaled, then it returns number of events times binwidth times scaling factor.
You should understand what is each option doing and conclude which option will give you what you want.
Dear ferhue,
thank you for your answer. Want I want is the “true” number of event, not just the histogram entries, so I understood that Integral(a, b, width), is the option that I need.
Regards