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.

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!

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