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\usetheme{Sybila} 

\title[Jacobian for $\PBzero \to \PKstar \Pmuon \APmuon$]{Jacobian for $\PBzero \to \PKstar \Pmuon \APmuon$ \\ proposed solution}
\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)}
% ------------------------------------------------------------------------------
% --------------------------- SLIDE --------------------------------------------
\begin{frame}\frametitle{Introduction}

\begin{itemize}
\item Lets down explicit on what we all agree ( I hope at least ;) ).
\begin{itemize}
\item Measurement of $\overrightarrow{S}=(F_l,~S_x)$ is unbiased.
\item Error is also correctly estimated ensuring the correct coverage.
\end{itemize}
\item The questions what I am answering: what is the corresponding confidence and probability distribution in a new space: $\overrightarrow{P}=(F_l,~P_x)$.
\item To put it a bit more simple: I want to map one space on the other one.
\end{itemize}

\end{frame}

\begin{frame}\frametitle{Some mathematical theorems assumptions 1}
\begin{itemize}
\item We have our standard transformation of ($\overrightarrow{S} \to \overrightarrow{P}$):
\begin{footnotesize}
\begin{align*}   
F_l      &\leftarrow  F_l\\                                                                     
P_1 &\leftarrow 2\frac{S_3}{1-F_{\rm L}}\\                                                          
P_2 &\leftarrow \frac{1}{2}\frac{S_6^s}{1-F_{\rm L}} = \frac{2}{3}\frac{A_{\rm FB}}{1-F_{\rm L}}\\  
P_3 &\leftarrow -\frac{S_9}{1-F_{\rm L}}\\                                                          
P_4^\prime &\leftarrow \frac{S_4}{\sqrt{F_{\rm L}(1-F_{\rm L})}}\\                                  
P_5^\prime &\leftarrow \frac{S_5}{\sqrt{F_{\rm L}(1-F_{\rm L})}}\\                                  
P_6^\prime &\leftarrow \frac{S_7}{\sqrt{F_{\rm L}(1-F_{\rm L})}}\\                                  
P_8^\prime &\leftarrow \frac{S_8}{\sqrt{F_{\rm L}(1-F_{\rm L})}}.                                   
\end{align*}                                                                               
\end{footnotesize}
\end{itemize}
\end{frame}

   
 \begin{frame}\frametitle{Some mathematical theorems assumptions 2}
\begin{itemize}
\item We know about this transformation:
\begin{itemize}
\item The parameter space is bounded domain ($D$) \checkmark
\item The angular PDF is smooth function in the domain \checkmark
\item There exists 1:1 transformation between $\overrightarrow{S}$ and $\overrightarrow{P}$ \checkmark
\item Inside the domain the Jacobian is non-zero. ($J \neq 0$) \checkmark
\end{itemize}
\item Next slide you will know why those assumptions are needed.
\end{itemize}
\end{frame}  
   
   
    
\begin{frame}\frametitle{Some mathematical theorems assumptions 3}
\begin{itemize}
\item Now since there is 1:1 correspondence the central point in the $\overrightarrow{P}$ should be derived from the central point of the $\overrightarrow{S}$ basis.
\item Now the confidence belt. In the $\overrightarrow{S}$  a $68\%$ confidence belt ($D$) is:
\begin{align*}
\int_D f(\overrightarrow{S}) d \overrightarrow{S} = 0.68
\end{align*}
\item In this equation our $D$ is effectively the errors that we quote.
\item Now form analysis thats to previous slide we can write :
\begin{align*}
\int_D \underbrace{f(\overrightarrow{S})}_{\rm{What~we~simulate/bootstrap}} d \overrightarrow{S} =  \int_{\Delta} \underbrace{f'(\overrightarrow{P})}_{\rm{What~we~get~in~P}}  \times \vert J \vert d\overrightarrow{P} 
\end{align*}

\end{itemize}
\end{frame}  
     
   \begin{frame}\frametitle{Toys}
\begin{itemize}
\item So to get the integral(coverage) correct we need to take the Jacobian into account.
\item Let's make a toy example calculating $P_2$. Values used (mean $\pm$ error): $F_l= 0.7679 \pm 0.2$, $A_{FB} = -0.329 \pm 0.13$.
\end{itemize}
\includegraphics[width=0.85\textwidth]{images/Fl_AFb.png}
\end{frame}  
   
      \begin{frame}\frametitle{Toys}
\begin{itemize}
\item I am rejecting the non-physical regions:

\end{itemize}
\includegraphics[width=0.85\textwidth]{images/Fl_boundry.png}
\end{frame}  
   
   
\begin{frame}\frametitle{Toys}
\begin{itemize}
\item Now how does the new space look like.
\item Important to take into account the boundary as without all my theorems fall down.
\end{itemize}
\includegraphics[width=0.85\textwidth]{images/2D.png}
\end{frame}    
              
              
    
\begin{frame}\frametitle{Toys ZOOM}
\begin{itemize}
\item Now how does the new space look like.
\item Important to take into account the boundary as without all my theorems fall down.
\end{itemize}
\includegraphics[width=0.85\textwidth]{images/2DZ.png}
\end{frame}                 
              
   
   \begin{frame}\frametitle{Toys Conclusions}
\begin{itemize}
\item One needs to take into account the Jacobian!
\item The most probable value agrees perfectly. 
\item The confidence belt can be transformed analytically even :)
\item There is one hack here: one cannot look at $P_2$ stand alone any more:

\end{itemize}
\begin{columns}
\column{2.5in}
\includegraphics[width=0.85\textwidth]{images/P2.png}
\column{2.5in}
\includegraphics[width=0.85\textwidth]{images/cast.png}
\end{columns}



\end{frame}  
   
   
   
              
              
              
              
              
              
              
\end{document}