From: Benjamin Auder <benjamin.a@mailoo.org>
Date: Mon, 2 Feb 2015 12:16:29 +0000 (+0100)
Subject: remove source latex file from doc
X-Git-Url: https://git.auder.net/app_dev.php/current/rpsls.js?a=commitdiff_plain;h=ef3c3e248b53f93329e0482e616f1b92281bdd9e;p=synclust.git

remove source latex file from doc
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-\documentclass{article}
-\usepackage{a4wide}
-\usepackage{graphicx}
-\def\pen{\textrm{pen}}
-\def\L{\mathcal{L}}
-\title{\bf Detecting areas with synchronous temporal dynamics}
-\author{Christophe Giraud}
-
-\begin{document}
-\maketitle
-
-\noindent This document summarizes the algorithm used when function \textit{findSyncVarRegions()} is called with first argument method=``convex''. Reading first the article \emph{Delimiting synchronous populations from monitoring data} by Giraud et al. is recommanded, since we use here the same notations.
-
-\section{Model and estimation procedure}
-
-\subsection{Goal}
-
-We write $Z_{stk}$ for the $k$th observations, year $t$, site $s$ and $z_{st}=\sum_{k}Z_{stk}$.
-Our goal is to estimate regions $R$ such that
-\begin{equation}\label{model}
-Z_{stk}\sim \textrm{Poisson}(\exp(\theta_{s}+f(x_{s},t)))\quad\textrm{with } f(x,t)\approx \sum_{R}\rho_{R}(t){\bf 1}_{x\in R}.
-\end{equation}
- In other words, we try to estimate $f$ with the a priori that
-\begin{itemize}
-\item for each year $t$ the map $x \to f(x,t)$ is piecewise constant
-\item the boundary of the regions where $x \to f(x,t)$ is constant are the same for all year $t$.
-\end{itemize}
-The main difficulty is to detect the regions $R$.
-
-\subsection{Estimation procedure}
-Let $G$ be a graph and write $V(s)$ for the set of the neighbors of $s$ in G.
-The estimators $\widehat \theta$ and $\widehat f$ are defined as minimizers of 
-$$\mathcal{L}(\theta,f)+\alpha \pen(f):=\sum_{s,t}[e^{\theta_{s}+f_{st}}-z_{st}(\theta_{s}+f_{st})]+\alpha
-\sum_{s\stackrel{G}{\sim}u}\|f_{s.}-f_{u.}\|/D_{su}$$
-with boundary conditions: $f_{s1}=0$ for all $s$. We typically choose $D_{su}=1/|V(s)|+1/|V(u)|$.
- 
-\section{Optimization algorithm}
-
-The following quantity is to be minimized
-$$\mathcal{L}(\theta,f)+\alpha \pen(f):=\sum_{s,t}[e^{\theta_{s}+f_{st}}-z_{st}(\theta_{s}+f_{st})]+\alpha\sum_{s\stackrel{G}{\sim}u}\|f_{s.}-f_{u.}\|/D_{su}$$
-with boundary conditions: $f_{s1}=0$ for all $s$.
-This last expression can be rewritten into
-$$\mathcal{L}(\theta,f)+\alpha \pen(f)=\sum_{s,t}[e^{\theta_{s}+f_{st}}-z_{st}(\theta_{s}+f_{st})]+\alpha
-\sum_{s\stackrel{G}{\sim}u}\max_{\|\phi_{su}\|\leq 1}\langle\phi_{su},f_{s.}-f_{u.}\rangle/D_{su}$$
-with $\phi_{su}\in\mathbf R^T$.
-
-\newpage
-\noindent Let us introduce
-$$F(\theta,f,\phi)=\sum_{s,t}[e^{\theta_{s}+f_{st}}-z_{st}(\theta_{s}+f_{st})]+\alpha
-\sum_{s<u}{\bf 1}_{s\stackrel{G}{\sim}u}\ \langle\phi_{su},f_{s.}-f_{u.}\rangle/D_{su}.$$
-We can reformulate the quantity to be optimized using $F$ as follows.
-$$\mathcal{L}(\theta,f)+\alpha \pen(f)=\max_{\max_{s< u}\|\phi_{su}\|\leq 1}F(\theta,f,\phi).$$
-The penalized log-likelihood can now be minimized with the following steps.
-
-\subsection*{Application}
-
-Iterate until convergence:
-\begin{enumerate}
-\item gradient descent in $\theta$:\\* $\theta\leftarrow \theta - h \nabla_{\theta}F$
-\item gradient descent in $f$ with condition $f[\ ,1]=0$\\*
-$f[\ ,-1]\leftarrow f[\ ,-1]-h'\nabla_{f[\ ,-1]}F$ 
-\item gradient ascent in $\phi$\\*
-$\phi_{su}\leftarrow \phi_{su}+h''\nabla_{\phi_{su}}F$
-\item $\phi_{su}\leftarrow \phi_{su}/\max(1,\|\phi_{su}\|)$
-\end{enumerate}
-Return($\theta,f$)
-
-%\subsection*{Gradient en $\theta$:}
-%On a
-%$$\mathcal{L}(\theta,f)=\sum_{s}\left[e^{\theta_{s}}\sum_{t}e^{f_{st}}-\theta_{s}\sum_{t}z_{st}\right]+\ldots$$
-%Donc 
-%$$\partial_{\theta_{s}}F=e^{\theta_{s}}\sum_{t}e^{f_{st}}-\sum_{t}z_{st}$$
-%
-%
-%\subsection*{Gradient en $f$:} on note $\phi_{su}=-\phi_{us}$ pour $s>u$
-%$$\partial_{f_{st}}F=e^{\theta_{s}}e^{f_{st}}-z_{st}+\alpha\sum_{u\in V(s)}\phi_{su}/D_{su}$$
-%
-%
-%\subsection*{Gradient en $\lambda$:}
-%pour $s<u$ avec $s\sim u$
-%$$\nabla_{\phi_{su}}F=\alpha(f_{s.}-f_{u.})/D_{su}.$$
-
-\end{document}