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miage_solution:solution_kodja_markov_2_etats

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miage_solution:solution_kodja_markov_2_etats [2017/01/08 19:41]
melancon created
miage_solution:solution_kodja_markov_2_etats [2018/01/21 20:13]
melancon
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-=== Exercice ===+ 
 +====== 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. On revient à la chaîne de Markov à deux états de la section précédente.
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 $$ $$
  
 +----
  
 On reprend la définition de la probabilité conditionnelle:​ 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)}+P(X_1 = 0 | X_0 = 0, X_2 = 0) = \frac{P(X_1 = 0 \cap X_0 = 0 \cap X_2 = 0)}{P(X_0 = 0 \cap X_2 = 0)}
 $$ $$
  
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 $$ $$
-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)+P(X_0 = 0 \cap X_2 = 0) = P(X_0 = 0 \cap X_1 = 0 \cap X_2 = 0) + P(X_0 = 0 \cap X_1 = 1 \cap X_2 = 0)
 $$ $$
  
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 $$ $$
-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)}+P(X_1 = 0 | X_0 = 0, X_2 = 0) = \frac{P(X_1 = 0 \cap X_0 = 0 \cap X_2 = 0)}{P(X_0 = 0 \cap X_1 = 0 \cap X_2 = 0) + P(X_0 = 0 \cap X_1 = 1 \cap X_2 = 0)}
 $$ $$
  
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 $$ $$
-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 \cap X_1 = 0 \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 \X_0 = 1 \X_0 = 0) = P(X_2 = 0 | X_1 = 1) P(X_1 = 1 | X_0 = 0) P(X_0 = 0)+P(X_2 = 0 \cap X_0 = 1 \cap X_0 = 0) = P(X_2 = 0 | X_1 = 1) P(X_1 = 1 | X_0 = 0) P(X_0 = 0)
 $$ $$
  
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 [[miage:​td_analytique|Retourner au TD]] [[miage:​td_analytique|Retourner au TD]]
 +
miage_solution/solution_kodja_markov_2_etats.txt · Last modified: 2018/01/21 20:13 by melancon