1 / 9

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.

Download Presentation

Tarea de Modelos Ocultos de Markov

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  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

More Related