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- \title[Jacobian for $\PBzero \to \PKstar \Pmuon \APmuon$]{Jacobian for $\PBzero \to \PKstar \Pmuon \APmuon$ \\ proposed solution}
- \author{$\PBzero \to \PKstar \Pmuon \APmuon$ team}
- %\institute{$^1$~University of Zurich}
- \date{\today}
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- \begin{document}
- % --------------------------- SLIDE --------------------------------------------
- \frame[plain]{\titlepage}
- \author{$\PBzero \to \PKstar \Pmuon \APmuon$ team}
- %\institute{(UZH)}
- % ------------------------------------------------------------------------------
- % --------------------------- SLIDE --------------------------------------------
- \begin{frame}\frametitle{Reminder}
- \begin{small}
-
- \begin{itemize}
- \item We wanted to calculate the $P_i$ from $S_i$.
- \item Both Toy MC error propagation (generating toy experiments based on the covariance matrix) and bootstrapping the data set produces distribution that has a most probable value that is different to the central value in the data (see plot below, most probable value from toys is different then the generated one (red line)).
- \item As discussed during the referee meeting we considered including the Jacobian the this picture.
- \end{itemize}
- \end{small}
- \begin{center}
- \includegraphics[width=0.45\textwidth]{images/P2.png}
- \end{center}
- \end{frame}
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- \begin{frame}\frametitle{Introduction}
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- \begin{itemize}
- \item Lets write 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.
- \item NB: This is a different question than what is the distribution of P measured by the experiments.
- \end{itemize}
-
- \end{frame}
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- \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}
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- \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}
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- \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 correct we need to take the Jacobian into account.
- \item Let's make a toy example calculating $P_2$. Values used (Gaussian distributed: mean $\pm$ error): $F_l= 0.7679 \pm 0.2$, $A_{FB} = -0.329 \pm 0.13$.
- \item The Jacobian: $J=\dfrac{2}{3} \dfrac{1}{1-F_L}$
- \item Generated $F_l$ and $A_{FB}$:
- \end{itemize}
-
- \includegraphics[width=0.85\textwidth]{images/Fl_AFb.png}
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-
- \end{frame}
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-
- \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.
- \item The white point is the value from which the toy was generated.
- \end{itemize}
- \begin{center}
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- \begin{columns}
- \column{2.5in}
- \begin{small}
- Scatter plot $F_L:P_2$, no Jacobian
- \end{small}
-
- \column{0.5in}
- {~}
- \column{2.2in}
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- \begin{small}
- Scatter plot $F_L:P_2$, with Jacobian
- \end{small}
-
- \end{columns}
- \includegraphics[width=1.1\textwidth]{images/2DZ2.png}
-
- \end{center}
- \end{frame}
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-
- \begin{frame}\frametitle{Re parametrization of pdf}
- \begin{itemize}
- \item Re parametrization of the pdf gives exactly the same answer as toys taking into account the jacobian:
- \end{itemize}
- {~}\\{~}\\
-
- \begin{columns}
- \column{2.5in}
- \begin{small}
- Profile likelihood from re-parametrised pdf.
- \end{small}
- \includegraphics[width=0.9\textwidth]{images/LL_pdf.png}
- \column{2.5in}
- \begin{small}
- Profile likelihood from toys with Jacobian
- \end{small}
- \includegraphics[width=0.9\textwidth]{images/LL_toys.png}
- \end{columns}
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-
- \end{frame}
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- \begin{frame}\frametitle{Toys Conclusions}
- \begin{itemize}
- \item We understand the source of the bias in the most probable value.
- \item Jacobian gives the same answer as does the parametrization of pdf.
- \item When we work out the interval on P2 (etc), should we use this Jacobian weighting?
- \item One should not look just at 1D projections as on them the most probable value is not the correct one:
- \item Coverage of $P_i$ is ensured by the coverage of $S_i$.
-
- \end{itemize}
- \begin{columns}
- \column{2.5in}
- \includegraphics[width=0.6\textwidth]{images/P2.png}
- \column{2.5in}
- \includegraphics[width=0.6\textwidth]{images/cast.png}
- \end{columns}
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- \end{frame}
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- \end{document}