The error we obtain by using error bars or histogram.Sumw2() is basically the sqrt(N) or sqrt(sum of weights^2).
Shouldn’t the error be 1/sqrt(N) or 1/sqrt(sum of weights^2) ?
Any help would be highly appreciated.
Thank you.
No. And it is obvious.
For me it usually helps to look at extreme cases. Consider a very simple case: a histogram with 2 bins and 2e8+1 entries filled randomly into bin1 and bin2. Let the result be: bin1 has 1e8 entries, bin2 has 1e8+1 entries. Question: does the random number generator favor one bin? Answer: obviously not. The difference between 1e8 and 1e8+1 is negligible. Yet when you try your error formula the uncertainty would be 1/sqrt(1e8)=0.0001, so you are claiming that 1 entry difference is a relevant thing.
I don’ think I am very clear about it.
What I think is that the larger the entries in the bin, the smaller is the error? Isn’t that true?
If this is the case then the error should be 1/sqrt(N) rather than sqrt(N).
Am I wrong somewhere?
Could it be that you are mixing up relative and absolute values?
\frac{\sqrt{N}}{N} = \frac{1}{\sqrt{N}}
Thank you so much! You are right! 
This topic was automatically closed 14 days after the last reply. New replies are no longer allowed.