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Presentations / Zurich_group / 07_07_2015 / mchrzasz.tex
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\usetheme{Sybila} 

\title[DY templates determination]{DY templates determination}
\author{Marcin Chrz\k{a}szcz$^{1}$}
\institute{$^1$~University of Zurich}
\date{\today}

\begin{document}
% --------------------------- SLIDE --------------------------------------------
\frame[plain]{\titlepage}
\author{Marcin Chrz\k{a}szcz{~}}
\institute{(UZH)}
% ------------------------------------------------------------------------------
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\begin{frame}\frametitle{Fast reminder}


\begin{itemize}
\item On the previous meetings I have shown that I can perform fits to data, using two background templates and a signal template.
\item For that fits I used just the $\PZzero$ data template.
\item Since we are fitting the isolation variable:
\end{itemize}
\begin{equation}
\mu\mu_{\rm{iso}} = \log (\max (p_T^{\rm{cone}}(\mu^i,0.5)-p_T^{\rm{cone}}(\mu^i,0.1)), \mu^i \in \lbrace\mu^=,\mu^-\rbrace 
\end{equation}
\begin{itemize}
\item The isolation is mass dependent, so for each dimuon mass we need to have a new signal template.
\item Today on how to derive them.
\end{itemize}
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\end{frame}
\begin{frame}\frametitle{Nicola template}
\begin{columns}
\column{3in}
\begin{itemize}
\item Take DY MC, calculate the scaling factor between given mass bin and the $\PZzero$ mass bin.
\item Has problem later on for $4<y<4.5$ as there is very little $\PZzero$ in data.
\item Example of the scan.

\end{itemize}
\includegraphics[width=0.9\textwidth]{{images/DY_NICOLA/3.0_3.25_15000.0_20000.0Nicola}.png}\\

\column{2in}
\includegraphics[width=0.95\textwidth]{{images/DY_NICOLA/2.0_2.25_10500.0_12000.0_0.988}.png}\\
\includegraphics[width=0.95\textwidth]{{images/DY_NICOLA/2.0_2.25_10500.0_12000.0_1.006}.png}
\end{columns}
\end{frame}

\begin{frame}\frametitle{My template simple version}
\begin{columns}
\column{3in}
\begin{itemize}

\item Take DY MC, calculate the scaling factor between given bin and the template derived from the $\PZzero$ and $\PUpsilon(1S)$
\item We have lots of $\PUpsilon(1S)$ in data in $4<y<4.5$
\item Template before scaling is a linear combination of $\PUpsilon(1S)$ and $\PZzero$  templates. Considered using a $\log$ dependence but didn't find significance difference. 

\end{itemize}
\includegraphics[width=0.6\textwidth]{{images/DY_ME1/3.0_3.25_15000.0_20000.0Me}.png}\\

\column{2in}
\only<2>{
\includegraphics[width=0.95\textwidth]{{images/DY_ME1/2.0_2.25_12000.0_15000.0Me}.png}\\
\includegraphics[width=0.95\textwidth]{{images/DY_NICOLA/2.0_2.25_12000.0_15000.0Nicola}.png}
}
\end{columns}
\end{frame}



\begin{frame}\frametitle{My template hard version}
\begin{columns}
\column{3in}
\begin{itemize}

\item Take DY MC, calculate the scaling factor between given bin and the template derived from the $\PZzero$ and $\PUpsilon(1S)$
\item but instead of just scaling the distribution, scale also the ration between bin 0 and the rest.
\item Now we fit two parameters.

\end{itemize}
\includegraphics[width=0.6\textwidth]{{images/DY_ME2/3.0_3.25_15000.0_20000.0me2}.png}\\

\column{2in}
\includegraphics[width=0.6\textwidth]{{images/DY_ME2/3.25_3.5_15000.0_20000.0me2}.png}\\
\includegraphics[width=0.6\textwidth]{{images/DY_ME2/3.5_3.75_15000.0_20000.0me2}.png}\\
\includegraphics[width=0.6\textwidth]{{images/DY_ME2/3.75_4.0_15000.0_20000.0me2}.png}\\
\end{columns}
\end{frame}



\begin{frame}\frametitle{Conclusions}
\begin{itemize}
\item In the fit we see that the second parameter is need to better describe the template.
\item This would be my default option after fixing the plotting bug ;)
\item After that do the fit and write down the note.
\end{itemize}
\end{frame}


              
\end{document}