final rapporteurs ?

[these_gilles.git] / paper_lniv_gpu / MV.tex~

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% correct bad hyphenation here
% \hyphenation{op-tical net-works semi-conduc-tor}


\begin{document}
%
% paper title
% can use linebreaks \\ within to get better formatting as desired
\title{Isolines}


% author names and affiliations
% use a multiple column layout for up to two different
% affiliations

\author{
}



% use for special paper notices
%\IEEEspecialpapernotice{(Invited Paper)}


% make the title area
\maketitle


\section{\label{isolines}Isolines} 
Soit $I$ l'image de reference non bruitee, $I'$ l'image bruitee par un bruit blanc gaussien additif de valeur moyenne nulle et d'ecart type $\sigma$. 

Soit $\widehat{I}$ l'image reconstruite. 
Le niveau de gris de chaque pixel de coordonnees $(i,j)$ est note $z(i,j)$.

Dans une fenetre rectangulaire $\omega$, soit $S^n$ la ligne de niveau partielle a laquelle appartient le pixel central de la fenetre.\\
$S^n$ est un ensemble de $n$ pixels dont les niveaux de gris possedent la meme valeur moyenne $\mu_{S^n}$ et dont la variance est $\sigma^2$.\\

Soit $\overline{S^n}$ definie par $\omega = S^n \cup \overline{S^n}$.\\
Les valeurs moyennes $\mu_{ij}$ des niveaux de gris des pixels de $\overline{S^n}$ sont inconnues et supposees differentes en tout point de $\overline{S^n$}. On note  $\left\{\mu_{ij}\right\}_{\overline{S^n}}$ l'ensemble de ces valeurs moyennes.

Soit $Z$ est l'ensemble des niveaux de gris des pixels de $\omega$.\\

La vraisemblance est donnee par :
\begin{eqnarray}
\displaystyle P \left[Z | S^n, \mu_{S^n}, \left\{\mu_{ij}\right\}_{\overline{S^n}}, \sigma \right]
\label{LL1}
\end{eqnarray}

\begin{eqnarray}
\displaystyle \prod_{(i,j)\in S^n}{P\left[z(i,j) | \mu_{S^n}, \sigma \right]} \times \displaystyle\prod_{(i,j)\in \overline{S^n}}{P\left[z(i,j) | \left\{\mu_{ij}\right\}_{\overline{S^n}}, \sigma \right]}
\label{LL2}
\end{eqnarray}


on recherche $\left\{\mu_{ij}\right\}_{\overline{S^n}}$ tel que 
\begin{eqnarray}
\displaystyle \widehat{\mu_{ij}} = z(i,j)
\end{eqnarray}
ET
\begin{eqnarray}
\displaystyle\prod_{(i,j)\in \overline{S^n}}{P\left[z(i,j) | \left\{\widehat{\mu_{ij}}\right\}_{\overline{S^n}}, \sigma \right]=1}
\end{eqnarray}

Il reste la vraisemblance generalisee 
\begin{eqnarray}
\displaystyle \prod_{(i,j)\in S^n}{P\left[z(i,j) | \mu_{S^n}, \sigma \right]}
\end{eqnarray}

% that's all folks
\end{document}