Undergrad Why do systematic uncertainties disappear using ratios?

[Photonino]

Hello,

I often hear the phrase "Well, since you are taking a ratio bin-by-bin, you don't have to care about the luminosity syst. uncertainty and the trigger efficiency syst. uncertainty".

I think I understand qualitatively why this is the case (It cancels out in the ratio, since both quantities are affected by the same uncertainty), but I would like to double check whether this intuition is correct and to what extent this statement is correct when performing a physics analysis.

Thank you very much in advance!

 

[ChrisVer]

Why don't you try it out with toys?
Thing is that a common scale factor is not going to affect much your distribution. If the ratio for example is \frac{N_{pass}}{N_{tot}} an uncertainty that will affect their normalization by 10% is going to give you \frac{1.1N_{pass}}{1.1N_{tot}} = \frac{N_{pass}}{N_{tot}}.
I don't have a quantitive explanation for uncertainties that depend on the variable at which you are looking at, but I think the assumption at bin-by-bin ratio is supposed to assume that within a bin the uncertainty can be considered a constant.
Now to what extend this is true- well I'd say you cannot tell beforehand... the rigorous way would be to try it out and show that it's giving a negligible outcome [compared to other uncertainties].

 

[mfb]

For the trigger efficiency you'll have to verify that the same efficiency applies to both numerator and denominator. For the luminosity uncertainty this is basically always the case - unless the datasets you compare are from different years or something like that.