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( ! ) Warning: Declaration of syntax_plugin_divalign2_common::handle($match, $state, $pos, &$handler) should be compatible with DokuWiki_Syntax_Plugin::handle($match, $state, $pos, Doku_Handler $handler) in /net/www/gmelanco/dokuwiki/lib/plugins/divalign2/common.php on line 28
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( ! ) Warning: Declaration of syntax_plugin_divalign2_common::render($mode, &$renderer, $data) should be compatible with DokuWiki_Syntax_Plugin::render($format, Doku_Renderer $renderer, $data) in /net/www/gmelanco/dokuwiki/lib/plugins/divalign2/common.php on line 28
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A PCRE internal error occured. This might be caused by a faulty plugin

( ! ) Warning: Declaration of renderer_plugin_latexit::listitem_open($level) should be compatible with Doku_Renderer::listitem_open($level, $node = false) in /net/www/gmelanco/dokuwiki/lib/plugins/latexit/renderer.php on line 38
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 \documentclass[a4paper, oneside, 10pt]{article} \usepackage[english]{babel} \usepackage[utf8x]{inputenc} \date{\today} \title{} \author{} \begin{document} ====== Master 4TYE814U/4TYE808U MIAGE \& e-MIAGE -- Processus stochastiques et simulation ====== ===== Séance de travaux dirigé ===== ==== Exercice ==== On revient à la chaîne de Markov à deux états de la section précédente. * Calculez: \$\$ P(X\_1 = 0 | X\_0 = 0, X\_2 = 0) = \textbackslash{}frac\{(1-p)\textasciicircum{}2\}\{(1-p)\textasciicircum{}2 + pq\} \$\$ ---- On reprend la définition de la probabilité conditionnelle: \$\$ P(X\_1 = 0 | X\_0 = 0, X\_2 = 0) = \textbackslash{}frac\{P(X\_1 = 0 \textbackslash{}cap X\_0 = 0 \textbackslash{}cap X\_2 = 0)\}\{P(X\_0 = 0 \textbackslash{}cap X\_2 = 0)\} \$\$ et on utilise une propriété du calcul des probabilité, à savoir que \$\$ P(X\_0 = 0 \textbackslash{}cap X\_2 = 0) = P(X\_0 = 0 \textbackslash{}cap X\_1 = 0 \textbackslash{}cap X\_2 = 0) + P(X\_0 = 0 \textbackslash{}cap X\_1 = 1 \textbackslash{}cap X\_2 = 0) \$\$ Du coup, l'expression à évaluer est égale à: \$\$ P(X\_1 = 0 | X\_0 = 0, X\_2 = 0) = \textbackslash{}frac\{P(X\_1 = 0 \textbackslash{}cap X\_0 = 0 \textbackslash{}cap X\_2 = 0)\}\{P(X\_0 = 0 \textbackslash{}cap X\_1 = 0 \textbackslash{}cap X\_2 = 0) + P(X\_0 = 0 \textbackslash{}cap X\_1 = 1 \textbackslash{}cap X\_2 = 0)\} \$\$ Or, de l'exercice précédent on trouve que: \$\$ P(X\_2 = 0 \textbackslash{}cap X\_1 = 0 \textbackslash{}cap X\_0 = 0) = P(X\_2 = 0 | X\_1 = 0) P(X\_1 = 0 | X\_0 = 0) P(X\_0 = 0) \$\$ \$\$ P(X\_2 = 0 \textbackslash{}cap X\_0 = 1 \textbackslash{}cap X\_0 = 0) = P(X\_2 = 0 | X\_1 = 1) P(X\_1 = 1 | X\_0 = 0) P(X\_0 = 0) \$\$ On trouve donc: \$\$ P(X\_1 = 0 | X\_0 = 0, X\_2 = 0) \$\$ \$\$ = \textbackslash{}frac\{P(X\_2 = 0 | X\_1 = 0) P(X\_1 = 0 | X\_0 = 0) P(X\_0 = 0)\}\{P(X\_2 = 0 | X\_1 = 0) P(X\_1 = 0 | X\_0 = 0) P(X\_0 = 0) + P(X\_2 = 0 | X\_1 = 1) P(X\_1 = 1 | X\_0 = 0) P(X\_0 = 0)\} \$\$ \$\$ = \textbackslash{}frac\{P(X\_2 = 0 | X\_1 = 0) P(X\_1 = 0 | X\_0 = 0)\}\{P(X\_2 = 0 | X\_1 = 0) P(X\_1 = 0 | X\_0 = 0) + P(X\_2 = 0 | X\_1 = 1) P(X\_1 = 1 | X\_0 = 0)\} \$\$ (on simplifie le terme \$P(X\_0 = 0)\$ présent partout). Ne reste plus qu'à substituer les valeurs \$P(X\_2 = 0 | X\_1 = 0) = P(X\_1 = 0 | X\_0 = 0) = (1-p)\$, \$P(X\_2 = 0 | X\_1 = 1) = p\$ et \$P(X\_1 = 1 | X\_0 = 0) = q\$. --- [[miage:td\_analytique|Retourner au TD]] \end{document}