Binomial Option on TH1::Divide Command


I am dividing two different distributions, and one of them is a subset of the other (numerator is subset of denominator). These are kinematic distributions (e.g. event count/cross section vs. jet Pt), and their shape is not necessarily Gaussian. The y-axis is not necessarily integral-valued either. This ratio resembles an “efficiency” quantity in the sense that the it is of the form “events that passed the cut/total number of events”. The errors of the numerator are obviously correlated to those of the denominator, so the correct error of the ratio may be done in terms of (1) error propagation or of (2) an estimation of the confidence interval of the ratio’s distribution. If you’re able to use distributions with an integral number of counts/events, then you can use the TEfficiency class, which has well defined options for how to handle the binomial proportion’s confidence interval. For fractional entries (e.g. TH1D), you cannot use TEfficiency due to the nature of the binomial distribution being integral-valued.

My question concerns working with it in terms of error propagation. If I am working with TH1D’s, and I use the TH1D::Divide command, I’ve been told to check the binomial option “B” to correctly propagate errors. I’ve looked in the TH1 code, and I don’t understand where this binomial error derivation comes from. The link listed to give further information is broken:
Could someone explain to me where this formula comes from? Is there an advantage of using this method as opposed to some of the others in TEfficiency?



TEfficiency with the kNormal stat error option implements exactly the same algorithm as TH1::Divide with the option “B”.
Both they works in case of integer content or weighted histograms.

The formula for weighted histogram is derived from the weighted maximum likelihood function. This is discussed in this post (this is the correct link, I will fix also in the TH1.cxx code)

Best Regards


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