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miage_solution:solution_kodja_markov_2_etats [Guy Melançon - Enseignement]

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miage_solution:solution_kodja_markov_2_etats

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=== Exercice === On revient à la chaîne de Markov à deux états de la section précédente. * Calculez: $$ P(X_1 = 0 | X_0 = 0, X_2 = 0) = \frac{(1-p)^2}{(1-p)^2 + pq} $$ On reprend la définition de la probabilité conditionnelle: $$ P(X_1 = 0 | X_0 = 0, X_2 = 0) = \frac{P(X_1 = 0 \& X_0 = 0 \& X_2 = 0)}{P(X_0 = 0 \& X_2 = 0)} $$ et on utilise une propriété du calcul des probabilité, à savoir que $$ P(X_0 = 0 \& X_2 = 0) = P(X_0 = 0 \& X_1 = 0 \& X_2 = 0) + P(X_0 = 0 \& X_1 = 1 \& X_2 = 0) $$ Du coup, l'expression à évaluer est égale à: $$ P(X_1 = 0 | X_0 = 0, X_2 = 0) = \frac{P(X_1 = 0 \& X_0 = 0 \& X_2 = 0)}{P(X_0 = 0 \& X_1 = 0 \& X_2 = 0) + P(X_0 = 0 \& X_1 = 1 \& X_2 = 0)} $$ Or, de l'exercice précédent on trouve que: $$ P(X_2 = 0 \& X_1 = 0 \& X_0 = 0) = P(X_2 = 0 | X_1 = 0) P(X_1 = 0 | X_0 = 0) P(X_0 = 0) $$ $$ P(X_2 = 0 \& X_0 = 1 \& X_0 = 0) = P(X_2 = 0 | X_1 = 1) P(X_1 = 1 | X_0 = 0) P(X_0 = 0) $$ On trouve donc: $$ P(X_1 = 0 | X_0 = 0, X_2 = 0) $$ $$ = \frac{P(X_2 = 0 | X_1 = 0) P(X_1 = 0 | X_0 = 0) P(X_0 = 0)}{P(X_2 = 0 | X_1 = 0) P(X_1 = 0 | X_0 = 0) P(X_0 = 0) + P(X_2 = 0 | X_1 = 1) P(X_1 = 1 | X_0 = 0) P(X_0 = 0)} $$ $$ = \frac{P(X_2 = 0 | X_1 = 0) P(X_1 = 0 | X_0 = 0)}{P(X_2 = 0 | X_1 = 0) P(X_1 = 0 | X_0 = 0) + P(X_2 = 0 | X_1 = 1) P(X_1 = 1 | X_0 = 0)} $$ (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]]

miage_solution/solution_kodja_markov_2_etats.1483900912.txt.gz · Last modified: 2017/01/08 19:41 by melancon