User:JabberWok/Sandbox

Essentially the two samples your looking at are highly correlated by the simple fact that they involve the same events. For instance given two cuts tight(T) and loose(L) then

number passing tight cut $$:= N_T $$

number passing loose cut $$:= N_L = N_T + N_{TL}$$

When comparing to see if there is a differences between quantities calculated with the T and L cuts you are only concerned about the error on $$N_{TL}$$.

Therefore for your difference in the ratio of cross sections for the eta opposite (population a) and eta same (population b), the error is proportional to the difference in number of events.




 * $$0 \,$$
 * $$= \frac{N_{T}(a)}{N_{T}(b)} - \frac{N_{L}(a)}{N_{L}(b)} \,$$
 * $$= \frac{N_{T}(a)}{N_{T}(b)} - \frac{N_{T}(a)+N_{TL}(a)}{N_{T}(b)+N_{TL}(b)} \,$$
 * }
 * $$= \frac{N_{T}(a)}{N_{T}(b)} - \frac{N_{T}(a)+N_{TL}(a)}{N_{T}(b)+N_{TL}(b)} \,$$
 * }

Where this difference should be equal to 0.

The exact way to treat this would be propagate the error through for this expression.

However, as long as $$N_{TL}(b)$$ is small the denominators can be taken as approximately the same, and in that case the above simplifies to


 * $$\frac{N_{TL}(a)}{N_T(b)} =~0 \,$$

with error bar $$~\frac{\sqrt{N_{TL}(a)}}{N_T(b)}$$ (the term due to the error on $$N_T(b)$$ is small).

that is the error bar that should be plotted on the plot.