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claudioc |
1.1 |
\section{Trigger efficiency}
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\label{sec:trgEff}
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claudioc |
1.2 |
As described in Section~\ref{sec:trigSel} we rely on a
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mixture of single and double lepton triggers. The trigger
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efficiency is very high because for most of the phase space
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we have two leptons each of which can fire a single lepton
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trigger -- and the single lepton triggers are very efficient.
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We apply to MC events a simplified model of the trigger efficiency
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as a function of dilepton species ($ee$, $e\mu$, $\mu\mu$), the $P_T$
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of the individual leptons, and, in the case of muons, the $|\eta|$
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of the muons. We believe that this model is adequate for
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the trigger efficiency precision needed for this analysis.
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The model assumptions are the following
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\begin{itemize}
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\item Muon and electron trigger turn-ons as a function of $P_T$
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are infinitely sharp. {\color{red} Can we add references?}
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\item All electron triggers with no ID have 100\%
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efficiency for electrons passing our analysis cuts. {\color{red}
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Can we add a reference? Pehaps the top documentation?}
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\item Electron triggers with (Tight(er))CaloEleId have 100\%
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efficiency with respect to our offline selection. This we
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verified via tag-and-probe on $Z\to ee$.
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\item Electron triggers with EleId have somewhat lower
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efficiency. This was also measured by tag-and-probe.
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\item The single muon trigger has zero efficiency for $|\eta|>2.1$.
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This is conservative, the trigger efficiency here is of order
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$\approx 40\%$.
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\item If a muon in $|\eta|>2.1$ fails the single muon trigger, it
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will also fail the double muon trigger.
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\item The double muon trigger has efficiency
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equal to the square of the single muon efficiency.
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\item The $e\mu$ triggers have no efficiency if the muon has $|\eta|>2.1$.
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\end{itemize}
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The model also uses some luminosity fractions and some trigger
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efficiencies.
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\begin{itemize}
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\item $\epsilon_{\mu}$=xx, the single muon trigger efficiency plateau.
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\item $f9$=xx: fraction of data with the Mu9 trigger unprescaled.
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(run$\le 147116$).
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\item $f11$=xx fraction of data with the Mu9 trigger prescaled and
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the Mu11 trigger unprescaled.
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(147196 $\leq$ run $\leq$ 148058).
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\item $e10$=xx: fraction of data with the 10 GeV unprescaled electron triggers.
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(run$\le 139980$).
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\item $e15$=xx: fraction of data with the 15 GeV unprescaled electron triggers.
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(139980 $<$ run $\leq$ 144114).
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\item $e17$=xx: fraction of data with the 100\% efficient 17 GeV unprescaled electron triggers.
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(144114 $<$ run $\leq$ 147116).
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\item $e17b$=xx: fraction of data with 17 GeV unprescaled electron triggers
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with efficiency $\epsilon_e^b=90\%$ (as measured by tag-and-probe).
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(147116 $<$ run $\leq$ 148058).
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\item $emess$=xx: the remainder of the run with several different electron
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triggers, all of $P_T>17$ GeV. For this period we measure the
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luminosity-weighted
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trigger efficiency $\epsilon(P_T)$ via tag and probe to be 99\%
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($17<P_T<22$, 97\% ($22<P_T<27$), 98\% ($27<P_T<32$) and
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100\% ($P_T>32$).
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\end{itemize}
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The full trigger efficiency model is described separately for
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$ee$, $e\mu$, and $\mu\mu$.
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\subsection{$ee$ efficiency model}
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\label{sec:eemodel}
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This is the easiest. Throughout the 2010 run we have always
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had dielectron triggers with thresholds lower than our (20,10)
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analysis thresholds. Since electron triggers are 100\% efficient,
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the trigger efficiency for $ee$ is 100\%. We have verified that
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the efficiency of the dielectron trigger is 100\% with respect
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to the single electron trigger using $Z \to ee$ data.
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\subsection{$\mu\mu$ efficiency model}
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\label{sec:mmmodel}
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We consider different cases.
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\subsubsection{Both muons in $|\eta|<2.1$ and with $P_T>15$ GeV}
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This is the bulk of the $\mu\mu$.
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\begin{center}
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$\epsilon = 1 - (1-\epsilon_{\mu})^2$
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\end{center}
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\subsubsection{Both muons in $|\eta|<2.1$, one muon with $11<P_T<15$ GeV}
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In this case there must be a muon with $P_T>20$ GeV. The single muon
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trigger is operative for the full dataset on this muon. Some loss
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of efficiency can be recovered when the 2nd muon fires the trigger.
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But this can happen only for a fraction of the run. The dimuon trigger
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cannot fire in our model to recover any of the efficiency lost by
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the single muon trigger on the high $P_T$ muon.
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\begin{center}
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$\epsilon = \epsilon_{\mu} + (1-\epsilon_{\mu})\epsilon_{\mu}(f9+f11)$
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\end{center}
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\subsubsection{Both muons in $|\eta|<2.1$, one muon with $10<P_T<11$ GeV}
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Same basic idea as above.
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\begin{center}
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$\epsilon = \epsilon_{\mu} + (1-\epsilon_{\mu})\epsilon_{\mu}f9$
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\end{center}
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\subsubsection{Both muons with $|\eta|>2.1$}
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This is a very small fraction of events. In our model they can only be
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triggered by the dimuon trigger.
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\begin{center}
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$\epsilon = \epsilon_{\mu}^2$
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\end{center}
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\subsubsection{First muon with $P_T>15$ and $|\eta|<2.1$; second muon
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with $|\eta|>2.1$}
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The single muon trigger is always operative. If it fails the double muon
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trigger also fails.
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\begin{center}
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$\epsilon = \epsilon_{\mu}$
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\end{center}
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\subsubsection{First muon with $11<P_T<15$ and $|\eta|<2.1$; second muon
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with $|\eta|>2.1$}
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The single muon trigger is operative only for a fraction of the run.
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For the remaining fraction, we must rely on the double muon trigger.
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\begin{center}
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$\epsilon = (f9+f11)\epsilon_{\mu} + (1-f9-f11)\epsilon_{\mu}^2$
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\end{center}
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\subsubsection{First muon with $10<P_T<11$ and $|\eta|<2.1$; second muon
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with $|\eta|>2.1$}
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Same basic idea as above.
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\begin{center}
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$\epsilon = f9~\epsilon_{\mu} + (1-f9)\epsilon_{\mu}^2$
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\end{center}
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\subsection{$e\mu$ efficiency model}
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\label{sec:emumodel}
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This is the most complicated case. The idea is that the muon trigger
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is used to get the bulk of the efficiency. Then the single electron
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trigger(s) and the $e\mu$ triggers are used to get back dome of the
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efficiency loss. The various cases are listed below.
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\subsubsection{Muon with $|\eta|<2.1$ and $P_T>15$}
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This is the bulk of the acceptance.
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\begin{center}
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$\epsilon = \epsilon_{\mu} + (1-\epsilon_{\mu})\Delta_1$
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\end{center}
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where $\Delta_1$ is the efficiency from the electron trigger:
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\begin{itemize}
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\item $P_T(ele)<15 \to \Delta_1=e10$
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\item $15<P_T(ele)<17 \to \Delta_1=e10+e15$
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\item $P_T(ele)>15 \to \Delta_1=e10+e15+e17+\epsilon_e^b~e17b+\epsilon(P_T)~emess$
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\end{itemize}
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\subsubsection{Muon with $|\eta|<2.1$ and $11<P_T>15$}
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This is the similar to the previous case, except that the muon
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trigger is operative only for a subset of the data taking period.
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\begin{center}
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$\epsilon = (f11+f9)\epsilon_{\mu} + \Delta_2 + \Delta_3$
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\end{center}
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Here $\Delta_2$ is associated with the period where the muon
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trigger was at 15 GeV, in which case we use electron triggers or
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$e\mu$ triggers. Note that the electron in this case must be
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above 20 GeV. This can happen only in the latter part of the run,
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thus we write
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\begin{center}
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$\Delta_2 = (1-f11-f9)~(\epsilon_{\mu}~+~
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(1-\epsilon_{\mu})\epsilon(P_T))$
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\end{center}
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\noindent where the first term is for the $e\mu$ trigger and the
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second term corresponds to $e\mu$ trigger failures, in which case we have
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to rely on the electron trigger.
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Then, $\Delta_3$ is associated with muon trigger failures in early runs,
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{\em i.e.}, run $<148819$. In this case the electron trigger picks it
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up and the $e\mu$ trigger does not help.
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\begin{center}
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$\Delta_3 = (f11+f9)(1-\epsilon_{\mu}) \cdot \epsilon_e$
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\end{center}
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\noindent where $\epsilon_e$ is the efficiency of the electron
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trigger for $P_T>20$. This is 100\% up to run 147716 (fraction
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$(e10_e15+e17)/(f11+f9)$; then it is somewhat lower up to
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run 148058, then it becomes very close to 100\% again.
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For this latter part of the run we approximate it as $\epsilon_e^b$.
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Thus:
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\begin{center}
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$\epsilon_e = (e10_e15+e17)/(f11+f9) +
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\epsilon_e^b(f11+f9-e10-e15-e17)/(f11+f9)$
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\end{center}
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\subsubsection{Muon with $|\eta|<2.1$ and $9<P_T>11$}
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Identical to the previous case, but replace $(f11+f9)$ with $f9$ everywhere.
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\subsubsection{Muon with $|\eta|>2.1$}
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This is a 10\% effect to start with. We assume no single muon efficiency,
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no $e\mu$ efficiency. Then we can only ise the single electron trigger.
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\begin{center}
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$\epsilon = \Delta_1$
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\end{center} |
