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1 + \clearpage
2 +
3   \section{Results}
4   \label{sec:results}
5  
4 %\noindent {\color{red} In the 11 pb everything is very
5 %simple because there are a few zeros.  This text is written
6 %for the full dataset under the assumption that some of these
7 %numbers will not be zero anymore.}
8
6   \begin{figure}[tbh]
7   \begin{center}
8 < \includegraphics[width=0.75\linewidth]{abcdData.png}
8 > \includegraphics[width=0.75\linewidth]{abcd_35pb.png}
9   \caption{\label{fig:abcdData}\protect Distributions of SumJetPt
10   vs. MET$/\sqrt{\rm SumJetPt}$ for SM Monte Carlo and data.  Here we also
11   show our choice of ABCD regions.}
12   \end{center}
13   \end{figure}
14  
18
15   The data, together with SM expectations is presented
16 < in Figure~\ref{fig:abcdData}.  We see $\color{red} 0$
17 < events in the signal region (region $D$).  The Standard Model
18 < MC expectation is {\color{red} 0.4} events.
16 > in Figure~\ref{fig:abcdData}.  We see 1 event in the
17 > signal region (region $D$).  The Standard Model MC
18 > expectation is 1.4 events.
19  
20   \subsection{Background estimate from the ABCD method}
21   \label{sec:abcdres}
22  
23   The data yields in the
24   four regions are summarized in Table~\ref{tab:datayield}.
25 < The prediction of the ABCD method is is given by $AC/B$ and
26 < is 0.5 events.
27 < (see Table~\ref{tab:datayield}.  
25 > The prediction of the ABCD method is is given by $A\times C/B$ and
26 > is 1.5 $\pm$ 0.9 events (statistical uncertainty only, assuming
27 > Gaussian errors), as shown in Table~\ref{tab:datayield}.  
28  
29   \begin{table}[hbt]
30   \begin{center}
31   \caption{\label{tab:datayield} Data yields in the four
32 < regions of Figure~\ref{fig:abcdData}.  The quoted uncertainty
32 > regions of Figure~\ref{fig:abcdData}, as well as the predicted yield in region D given
33 > by A$\times$C / B.  The quoted uncertainty
34   on the prediction in data is statistical only, assuming Gaussian errors.
35 < We also show the SM Monte Carlo expectations.}
36 < \begin{tabular}{|l|c|c|c|c||c|}
35 > We also show the SM Monte Carlo expectations, scaled to 34.85~pb$^{-1}$.}
36 > \begin{tabular}{l||c|c|c|c||c}
37 > \hline
38 >         sample   &              A   &              B   &              C   &              D   & A$\times$C / B  \\
39   \hline
40 <      &A   & B    & C   & D   & AC/B \\ \hline
41 < Data  &3   & 6    & 1   & 0   & $0.5^{+0.6}_{-0.5}$ \\
42 < SM MC &2.5 &11.2  & 1.5 & 0.4 & 0.4 \\
40 > $t\bar{t}\rightarrow \ell^{+}\ell^{-}$   &           7.96   &          33.07   &           4.81   &           1.20   &           1.16  \\
41 > $t\bar{t}\rightarrow \mathrm{other}$   &           0.15   &           0.85   &           0.09   &           0.04   &           0.02  \\
42 >   $Z^0$ + jets   &           0.00   &           1.16   &           0.08   &           0.08   &           0.00  \\
43 > $W^{\pm}$ + jets   &           0.00   &           0.10   &           0.00   &           0.00   &           0.00  \\
44 >       $W^+W^-$   &           0.19   &           0.29   &           0.02   &           0.07   &           0.02  \\
45 >   $W^{\pm}Z^0$   &           0.03   &           0.04   &           0.01   &           0.01   &           0.00  \\
46 >       $Z^0Z^0$   &           0.00   &           0.03   &           0.00   &           0.00   &           0.00  \\
47 >     single top   &           0.28   &           1.00   &           0.04   &           0.01   &           0.01  \\
48 > \hline
49 >    total SM MC   &           8.61   &          36.54   &           5.05   &           1.41   &           1.19  \\
50 > \hline
51 >           data   &             11   &             36   &              5   &              1   &1.53 $\pm$ 0.86  \\
52   \hline
53   \end{tabular}
54   \end{center}
# Line 52 | Line 60 | SM MC &2.5 &11.2  & 1.5 & 0.4 & 0.4 \\
60   %estimate of the $t\bar{t}$ contribution.  The result
61   %of this exercise is {\color{red} xx} events.
62  
63 + \clearpage
64 +
65   \subsection{Background estimate from the $P_T(\ell\ell)$ method}
66   \label{sec:victoryres}
67  
68 + We first use the $P_T(\ell \ell)$ method to predict the number of events
69 + in control region A, defined in Sec.~\ref{sec:abcd} as
70 + $125<{\rm SumJetPt}>300$~GeV and $\met/\sqrt{\rm SumJetPt}>$8.5.
71 + We count the number of events in region
72 + $A'$, defined in Sec.~\ref{sec:othBG} by replacing the above $\met/\sqrt{\rm SumJetPt}$
73 + cut with the same cut on the quantity $P_T(\ell\ell)/\sqrt{\rm SumJetPt}$,
74 + and find $N_{A'}=6$. To predict the yield in region A we take
75 + $N_A = K \cdot K_C \cdot N_{A'} = 10.4 \pm 4.2$
76 + (statistical uncertainty only, assuming Gaussian errors),
77 + where we have taken $K = 1.73$ and $K_C = 1$. This yield is in good
78 + agreement with the observed yield of 11 events, as shown in
79 + Table~\ref{tab:victory_control} and displayed in Fig.~\ref{fig:victory} (left).
80 + {\color{red} \bf Perform DY estimate for this control region}.
81 +
82 + Encouraged by the good agreement between predicted and observed yields
83 + in the control region, we proceed to perform the $P_T(\ell \ell)$ method
84 + in the signal region ${\rm SumJetPt}>300$~GeV.
85   The number of data events in region $D'$, which is defined in
86   Section~\ref{sec:othBG} to be the same as region $D$ but with the
87   $\met/\sqrt{\rm SumJetPt}$ requirement
88 < replaced by a $P_T(\ell\ell)/\sqrt{\rm SumJetPt}$ requirement
89 < is $N_{D'}=1$.  Thus the BG prediction is
90 < $N_D = K \cdot K_C \cdot N_{D'} = 1.5$
91 < where $K=1.5 \pm xx$ as derived in Sec.~\ref{sec:victory} and
92 < $K_C = 1$.
93 < Note that if we were to subtract off from region $D'$
94 < the {\color{red} 0.4 $\pm$ 0.4} DY events estimated from
95 < Section~\ref{sec:othBG}, the background
96 < prediction would change to $N_D=0.9 \pm xx$ events.
97 <
98 < %%%TO BE REPLACED
99 < %{\color{red}As mentioned previously, for the 11/pb analysis
73 < %we use the $K$ factor from data and take $K=1$.
74 < %This will change for the full dataset.  We will also pay
75 < %more attention to the statistical errors.}
76 <
77 < %The number of data events in region $D'$, which is defined in
78 < %Section~\ref{sec:othBG} to be the same as region $D$ but with the
79 < %$\met/\sqrt{\rm SumJetPt}$ requirement
80 < %replaced by a $P_T(\ell\ell)/\sqrt{\rm SumJetPt}$ requirement
81 < %is $N_{D'}=1$.  Thus the BG prediction is
82 < %$N_D = K \cdot K_{\rm fudge} \cdot N_{D'} = 1.5$
83 < %where we used $K=1.5 \pm xx$ and $K_{\rm fudge}=1.0 \pm 0.0$.
84 < %Note that if we were to subtract off from region $D'$
85 < %the {\color{red} 0.4 $\pm$ 0.4} DY events estimated from
86 < %Section~\ref{sec:othBG}, the background
87 < %prediction would change to $N_D=0.9 \pm xx$ events.
88 < %{\color{red} When we do this with a real
89 < %$K_{\rm fudge}$, the fudge factor will be different
90 < %after the DY subtraction.}
91 <
92 < As a cross-check, we use the $P_T(\ell \ell)$
93 < method to also predict the number of events in the
94 < control region $120<{\rm SumJetPt}<300$ GeV and
95 < \met/$\sqrt{\rm SumJetPt} > 8.5$.  We predict
96 < $5.6^{+x}_{-y}$ events and we observe 4.
97 < The results of the $P_T(\ell\ell)$ method are
98 < summarized in Figure~\ref{fig:victory}.
88 > replaced by a $P_T(\ell\ell)/\sqrt{\rm SumJetPt}$ requirement,
89 > is $N_{D'}=2$.  
90 > We next subtract off the expected DY contribution of
91 > {\color{red} \bf $N_{DY}$ = 0.8 $\pm$ 0.8 (update DY estimate)} events, as calculated
92 > in Sec.~\ref{sec:othBG}. The BG prediction is
93 > $N_D = K \cdot K_C \cdot (N_{D'}-N_{DY}) = 1.8^{+2.5}_{-1.8}$ (statistical
94 > uncertainty only, assuming Gaussian errors), where $K=1.54 \pm xx$
95 > as derived in Sec.~\ref{sec:victory} and $K_C = 1$.
96 > This prediction is consistent with the observed yield of
97 > 1 event, as summarized in Table~\ref{tab:victory_signal} and Fig.~\ref{fig:victory}
98 > (right).
99 >
100  
101   \begin{figure}[hbt]
102   \begin{center}
103 < \includegraphics[width=0.48\linewidth]{victory_control.png}
104 < \includegraphics[width=0.48\linewidth]{victory_sig.png}
103 > \includegraphics[width=0.48\linewidth]{victory_control_35pb.png}
104 > \includegraphics[width=0.48\linewidth]{victory_signal_35pb.png}
105   \caption{\label{fig:victory}\protect Distributions of
106   tcMet/$\sqrt{\rm SumJetPt}$ for the control and signal region.
107   We show the oberved distributions in both Monte Carlo and data.
# Line 110 | Line 111 | ${P_T(\ell\ell)}/\sqrt{\rm SumJetPt}$ in
111   \end{figure}
112  
113  
114 + \begin{table}[hbt]
115 + \begin{center}
116 + \caption{\label{tab:victory_control}Results of the dilepton $p_{T}$ template method in the control region
117 + $125 < \mathrm{sumJetPt} < 300$~GeV. The predicted and observed yields for
118 + the region $\mathrm{tcmet}/\sqrt{\mathrm{sumJetPt}}>$~8.5 are shown for data
119 + and MC. The error on the prediction for data is statistical only, assuming
120 + Gaussian errors.}
121 + \begin{tabular}{lccc}
122 + \hline
123 +              & Predicted           &   Observed &  Obs/Pred \\
124 + \hline
125 + total SM   MC &      7.10           &       8.61 &      1.21 \\
126 +         data &    10.38 $\pm$ 4.24 &         11 &      1.06 \\
127 + \hline
128 + \end{tabular}
129 + \end{center}
130 + \end{table}
131 +
132 + \begin{table}[hbt]
133 + \begin{center}
134 + \caption{\label{tab:victory_signal}Results of the dilepton $p_{T}$ template method in the signal region
135 + $\mathrm{sumJetPt} > 300$~GeV. The predicted and observed yields for
136 + the region $\mathrm{tcmet}/\sqrt{\mathrm{sumJetPt}}>$~8.5 are shown for data
137 + and MC. The error on the prediction for data is statistical only, assuming
138 + Gaussian errors.}
139 + \begin{tabular}{lccc}
140 + \hline
141 +              & Predicted                &   Observed &  Obs/Pred \\
142 + \hline
143 + total SM   MC &      0.96                &       1.41 &      1.46 \\
144 +         data &  $1.8^{+2.5}_{-1.8}$     &          1 &      0.56 \\
145 + \hline
146 + \end{tabular}
147 + \end{center}
148 + \end{table}
149 +
150 +
151   \subsection{Summary of results}
152   To summarize: we see no evidence for an anomalous
153   rate of opposite sign isolated dilepton events

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