markov.py

Created by vovenastream

Created on September 30, 2025

1.53 KB


Objectif
# --------
# Modeliser des passages entre etats par une matrice de transition P,
# calculer l'etat suivant S1 = S0*P, etat n (S0*P^n) et l'etat stationnaire.
#
# Ce qu'on te demande typiquement
# -------------------------------
# - Construire le diagramme/matrice de transition P a partir d'un enonce.
# - Verifier que chaque LIGNE de P somme a 1.
# - Calculer S1, S2... a partir d'un S0 (vecteur-ligne).
# - Trouver l'etat stationnaire S (S = S*P et somme des composantes = 1).
#
# Formules importantes
# --------------------
# - S1 = S0 * P ;   Sn = S0 * P^n.
# - Stationnaire S : S = S*P avec s1 + s2 + ... = 1.
# - Pour 2 etats : si P = [ [p11, p12], [p21, p22] ],
#     poser S = [s, 1-s] et resoudre s = s*p11 + (1-s)*p21.
#
# Methode pas-a-pas
# -----------------
# 1) Lire l'enonce -> identifier les etats (A, non-A ; L, non-L ; etc.).
# 2) Construire P (les proba de rester/aller) par lignes (somme=1).
# 3) Si S0 donne, S1 = S0*P ; sinon passe directement a l'etat stationnaire.
# 4) Stationnaire (2 etats) : resoudre s = s*p11 + (1-s)*p21 -> s.
#    Le deuxieme composant est 1-s.
# 5) Interpretaion : pourcentage a long terme dans chaque etat.
#
# Petit exemple (2 etats)
# -----------------------
# P = [ [0.8, 0.2],
#       [0.6, 0.4] ]
# S0 = [0.1, 0.9]
# S1 = S0*P = [ 0.1*0.8 + 0.9*0.6 , 0.1*0.2 + 0.9*0.4 ]
#            = [ 0.08 + 0.54 , 0.02 + 0.36 ] = [ 0.62 , 0.38 ]
# Stationnaire : s = s*0.8 + (1-s)*0.6 = 0.8s + 0.6 - 0.6s
#  -> s - 0.2s = 0.6 -> 0.8s = 0.6 -> s = 0.75
# Donc S = [0.75, 0.25].  Verif rapide : S*P = S

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