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Tarea de Modelos Ocultos de Markov

Tarea de Modelos Ocultos de Markov. Alberto Reyes B. 00377984. 0.5. 0.5. 0.5. q1. q2. 0.5. A. S. Calcular la probabilidad de la secuencia AASS para el siguiente modelo por (a) Metodo directo (b) Metodo iterativo.  = {0.5, 0.5}. Probabilidad inicial del estado.

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Tarea de Modelos Ocultos de Markov

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  1. Tarea de Modelos Ocultos de Markov Alberto Reyes B. 00377984

  2. 0.5 0.5 0.5 q1 q2 0.5 A S • Calcular la probabilidad de la secuencia AASS para el siguiente modelo por (a) Metodo directo (b) Metodo iterativo. = {0.5, 0.5} Probabilidad inicial del estado Probabilidad de transición entre estados A= M1: 0.8 A M2: 0.8 S q1: lanzar M1 q2: lanzar M2 Probabilidad de la observación dado el estado B=

  3. Todos los Q P(O)= q1 bq1(O1) aq1q2 bq2(O2) .. aq(T-1) T bqT(OT) • Por metodo directo P(O,Q) O={A,A,S,S} T=4 Suponiendo Q = q1 q1 q1 q1 (secuencia 1/16) P(O,Q)= q1 bq1(O1) aq1q1 bq1(O2) aq1q1 bq1(O3) aq1q1 bq1(O4) = (0.5)(0.8) x(0.5)(0.8)x (0.5)(0.2) x(0.5)(0.2)= 1.6e-3

  4. Suponiendo Q = q1 q1 q1 q2 (secuencia 2/16) P(O,Q)= q1 bq1(O1) aq1q1 bq1(O2) aq1q1 bq1(O3) aq1q2 bq2(O4) = (0.5)(0.8) x(0.5)(0.8)x (0.5)(0.2) x(0.5)(0.8)= 6.4e-3 Suponiendo Q = q1 q1 q2 q2 (secuencia 3/16) P(O,Q)= q1 bq1(O1) aq1q1 bq1(O2) aq1q2 bq2(O3) aq2q2 bq2(O4) = (0.5)(0.8) x(0.5)(0.8)x (0.5)(0.8) x(0.5)(0.8)= 0.0256 P(O)= 1.6e-3 + 6.4e-3 + 0.0256 + … + P(O,Q16)

  5. Por método iterativo O={A,A,S,S} T=4 N=2 Variable forward t(i)= P(O1, O2 .. Ot, qt=Si) Inicio: 1(i)= i b i(O1) 1(q1)= q1 b q1(O1) = (0.5) (0.8)= 0.4 (i=1) 1(q2)= q2 b q2(O1) = (0.5) (0.2)= 0.1 (i=2)

  6. i Inducción t+1(j)= [ t(i) aij ] bj(Ot+1) Para t=1 2(q1)= [1(q1) a11+ 1(q2) a21] b1(O2) (j=1) = [(0.4)(0.5) + (0.1)(0.5)] x 0.8 = 0.2 2(q2)= [1(q1) a12+ 1(q2) a22] b2(O2) (j=2) = [(0.4)(0.5) + (0.1)(0.5)] x 0.2 = 0.05 i=1 i=2

  7. i=1 i=2 Para t=2 3(q1)= [2(q1) a11+ 2(q2) a21] b1(O3) (j=1) = [(0.2)(0.5) + (0.05)(0.5)] x 0.2 = 0.025 3(q2)= [2(q1) a12+ 2(q2) a22] b2(O3) (j=2) = [(0.2)(0.5) + (0.05)(0.5)] x 0.8 = 0.1

  8. i=1 i=2 Para t=3 4(q1)= [3(q1) a11+ 3(q2) a21] b1(O4) (j=1) = [(0.025)(0.5) + (0.1)(0.5)] x 0.2 = 0.0125 4(q2)= [3(q1) a12+ 3(q2) a22] b2(O4) (j=2) = [(0.025)(0.5) + (0.1)(0.5)] x 0.8 = 0.05

  9. i Finalización P(O)= T(i) dado que T=4, i={1,2} = 4(q1) + 4(q2) = 0.0125 + 0.05 =0.0625

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