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benhoob |
1.1 |
%\appendix
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\section{Trigger Efficiency Model}
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\label{sec:appendix_trigger}
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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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claudioc |
1.3 |
The model assumptions are the following:
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benhoob |
1.1 |
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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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claudioc |
1.4 |
\item The single muon trigger has 50\% efficiency for
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$|\eta|>2.1$~\cite{ref:evans}.
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benhoob |
1.1 |
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claudioc |
1.5 |
\item If a muon in fails the single muon trigger, it
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will also fail the double muon trigger. This is actually
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a conservative assumption.
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benhoob |
1.1 |
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\item The double muon trigger has efficiency
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claudioc |
1.5 |
equal to the square of the single muon efficiency. This is
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also a conservative assumption.
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benhoob |
1.1 |
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\item The $e\mu$ triggers have no efficiency if the muon has $|\eta|>2.1$.
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claudioc |
1.5 |
Again, this is conservative.
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benhoob |
1.1 |
\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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claudioc |
1.4 |
\item $\epsilon_{\mu}$=93\%, the single muon trigger efficiency plateau
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for $|\eta|<2.1$~\cite{ref:evans};
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claudioc |
1.5 |
\item $\epsilon'_{\mu}$=40\%, the single muon trigger efficiency plateau
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claudioc |
1.4 |
for $|\eta|>2.1$~\cite{ref:evans};
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benhoob |
1.1 |
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claudioc |
1.2 |
\item $f9$=0.215: fraction of data with the Mu9 trigger unprescaled.
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benhoob |
1.1 |
(run$\le 147116$).
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claudioc |
1.4 |
\item $f11$=0.273 fraction of data with the Mu9 trigger prescaled and
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benhoob |
1.1 |
the Mu11 trigger unprescaled.
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(147196 $\leq$ run $\leq$ 148058).
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claudioc |
1.2 |
\item $e10$=0.002: fraction of data with the 10 GeV unprescaled electron triggers.
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benhoob |
1.1 |
(run$\le 139980$).
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claudioc |
1.2 |
\item $e15$=0.086: fraction of data with the 15 GeV unprescaled electron triggers.
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benhoob |
1.1 |
(139980 $<$ run $\leq$ 144114).
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claudioc |
1.2 |
\item $e17$=0.127: fraction of data with the 100\% efficient 17 GeV unprescaled electron triggers.
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benhoob |
1.1 |
(144114 $<$ run $\leq$ 147116).
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claudioc |
1.2 |
\item $e17b$=0.273: fraction of data with 17 GeV unprescaled electron triggers
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benhoob |
1.1 |
with efficiency $\epsilon_e^b=90\%$ (as measured by tag-and-probe).
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(147116 $<$ run $\leq$ 148058).
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claudioc |
1.3 |
\item $emess$=0.512: the remainder of the run with several different electron
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benhoob |
1.1 |
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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claudioc |
1.4 |
analysis thresholds. Since electron triggers are very close
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to 100\% efficient\cite{ref:evans},
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benhoob |
1.1 |
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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claudioc |
1.4 |
This is a very small fraction of events.
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%In our model they can only be triggered by the dimuon trigger.
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benhoob |
1.1 |
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\begin{center}
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claudioc |
1.4 |
$\epsilon = \epsilon_{\mu}^2 + \alpha (1-\epsilon_{\mu}) \epsilon'_{\mu}$
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benhoob |
1.1 |
\end{center}
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claudioc |
1.4 |
\noindent where $\alpha=2$ if both muons are above 15 GeV, $\alpha=(1+f9+f11)$ if
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one of the muons is between 11 and 15 GeV, and $\alpha=(1+f9)$ if one of the muon
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is below 11 GeV.
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benhoob |
1.1 |
\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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claudioc |
1.4 |
$\epsilon = \epsilon_{\mu} + (1-\epsilon_{\mu})\Delta_{\mu}$
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\end{center}
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\noindent where
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\begin{center}
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$\Delta_{\mu} = \epsilon'_{\mu}$ ~~~~(2nd muon with $P_T \geq 15$ GeV) \\
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$\Delta_{\mu} = (f9+f11)\epsilon'_{\mu}$ ~~~~(2nd muon with $11 \leq P_T < 15$ GeV) \\
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$\Delta_{\mu} = f9\epsilon'_{\mu}$ ~~~~(2nd muon with $9 \leq P_T < 11$ GeV) \\
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benhoob |
1.1 |
\end{center}
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claudioc |
1.4 |
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benhoob |
1.1 |
\subsubsection{First muon with $11<P_T<15$ and $|\eta|<2.1$; second muon
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claudioc |
1.4 |
with $|\eta|>2.1$ and $P_T>20$}
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The single muon trigger at low $\eta$ is fully operative only for a fraction of the run,
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this efficiency is captured by the first term below.
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For the remaining fraction, we rely on the double muon trigger as well as the
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single muon trigger at high $\eta$ (2nd term in the equation).
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\begin{center}
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$\epsilon = (f9+f11)(\epsilon_{\mu} + (1-\epsilon_{\mu})\epsilon'_{\mu})
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+ (1-f9-f11)(\epsilon_{\mu}^2 + (1-\epsilon_{\mu})\epsilon'_{\mu})$
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\end{center}
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\noindent which reduces to
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benhoob |
1.1 |
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\begin{center}
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claudioc |
1.4 |
$\epsilon = (f9+f11)\epsilon_{\mu} + (1-f9-f11)\epsilon_{\mu}^2
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+ (1-\epsilon_{\mu})\epsilon'_{\mu}$
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benhoob |
1.1 |
\end{center}
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\subsubsection{First muon with $10<P_T<11$ and $|\eta|<2.1$; second muon
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claudioc |
1.4 |
with $|\eta|>2.1$ and $P_T>20$}
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benhoob |
1.1 |
Same basic idea as above.
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\begin{center}
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claudioc |
1.4 |
% $\epsilon = f9~\epsilon_{\mu} + (1-f9)\epsilon_{\mu}^2$
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$\epsilon = f9\epsilon_{\mu} + (1-f9)\epsilon_{\mu}^2
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+ (1-\epsilon_{\mu})\epsilon'_{\mu}$
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benhoob |
1.1 |
\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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claudioc |
1.4 |
This is a 10\% effect to start with.
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The first term is from the electron efficiency. The 2nd term is the correction
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due to the single muon efficieny.
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benhoob |
1.1 |
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\begin{center}
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claudioc |
1.4 |
$\epsilon = \Delta_1 + (1-\Delta_1)\Delta_{\mu}$
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benhoob |
1.1 |
\end{center}
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\subsection{Summary of the trigger efficiency model}
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\label{sec:trgeffsum}
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We take the trigger efficiency for $ee$ as 100\%. The trigger efficiency
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claudioc |
1.3 |
for the $e\mu$ and $\mu\mu$ final states is summarized in
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claudioc |
1.5 |
Figures~\ref{fig:emuModel} and~\ref{fig:mumuModel}.
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benhoob |
1.1 |
We estimate the systematic uncertainties on the trigger modeling
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to be at the few percent level.
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claudioc |
1.5 |
\begin{figure}[htb]
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\begin{center}
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| 284 |
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\includegraphics[width=0.99\linewidth]{emuModel.png}
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\caption{\label{fig:emuModel}\protect Trigger efficiency for the
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$e\mu$ pair as a function of the $P_T$ of the two leptons.
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The top table corresponds to $|\eta(\mu)| < 2.1$, the bottom
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table to $|\eta(\mu)| > 2.1$.}
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\end{center}
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\end{figure}
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\clearpage
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claudioc |
1.3 |
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\begin{figure}[tbh]
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| 295 |
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\begin{center}
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| 296 |
claudioc |
1.5 |
\includegraphics[width=0.99\linewidth]{mumuModel.png}
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| 297 |
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\includegraphics[width=0.99\linewidth]{mumu24Model.png}
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| 298 |
claudioc |
1.3 |
\caption{\label{fig:mumuModel}\protect Trigger efficiency for the
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$\mu\mu$ pair as a function of the $P_T$ of the two muons.
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The top table corresponds to both muons having $|\eta| < 2.1$;
|
| 301 |
claudioc |
1.5 |
the middle table has one of the muon with $|\eta|<2.1$ and the
|
| 302 |
|
|
other muon with $|\eta|>2.1$; the bottom table has both muons with
|
| 303 |
|
|
have $|\eta|>2.1$.}
|
| 304 |
claudioc |
1.3 |
\end{center}
|
| 305 |
|
|
\end{figure}
|
| 306 |
|
|
|
| 307 |
|
|
\clearpage |
