Ratio of two Tprofile histograms

[couet]

See the possible options here: ROOT: TRatioPlot Class Reference
(diff, diffsig …)

[tanmaya_shree]

Yes now it looks fine I guess. Yes, I will check. Thank you very much!

[Wile_E_Coyote]

@couet

See the possible options here: ROOT: TRatioPlot Class Reference
(diff, diffsig …)

The question is … which options make sense for TProfile histograms (e.g., which options will return “properly” calculated errors)?

[tanmaya_shree]

I tried these.
Ratio2.C (848 Bytes)

[tanmaya_shree]

A->ProjectionX();
What this does?
Also, I want to know what A->Sumw2(); does?

[couet]

@Wile_E_Coyote

The question is … which options make sense for TProfile histograms (e.g., which options will return “properly” calculated errors)?

Not sure about it. @moneta ?

[couet]

@tanmaya_shree

A->ProjectionX();
What this does?

https://root.cern/doc/master/classTProfile.html#a4d3fa02873a069643c90918d6c5a29bd

@tanmaya_shree

Also, I want to know what A->Sumw2(); does?

https://root.cern/doc/master/classTProfile.html#a9591898784d69e20b76a1841a58bac07

[moneta]

Hi,
You cannot perform a direct ratio of two TProfile’s histogram using TRatioPlot or TProfile::Divide. The resulting ratio of two TProfile’s is not anymore a TProfile, so the correct way of computing the ratio is to convert first the TProfile in a TH1 object using TProfile::ProjectionX()and then use TH1::Divide or TRatioPlot.
I will add an error message in TProfile::Divide explaining this.
However, note also that computing the correct error on a ratio of means is tricky, using a normal approximation as it is done in TH1::Divide can give also incorrect results.
If one is really interest in the correct confidence interval in the ratio, it should be estimated using simulation (i.e with bootstrapping)

Thank you for raising this issue.

Lorenzo