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Tarea de Modelos Ocultos de Markov

Alberto Reyes B.

00377984

• Calcular la probabilidad de la secuencia AASS para el siguiente modelo por (a) Metodo directo (b) Metodo iterativo.

q1 q2

A S

0.50.5 0.5

0.5

M1: 0.8 A

M2: 0.8 S

q1: lanzar M1

q2: lanzar M2

= {0.5, 0.5}

A=0.5 0.5

0.5 0.5

B=0.8 0.2

0.2 0.8

Probabilidad de la observación dado el estado

Probabilidad de transición entre estados

Probabilidad inicial del estado

a. Por metodo directo

P(O)= q1 bq1(O1) aq1q2 bq2(O2) .. aq(T-1) T bqT(OT)Todos los Q

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

O={A,A,S,S} T=4

P(O,Q)



= (0.5)(0.8) x(0.5)(0.8)x (0.5)(0.2) x(0.5)(0.8)= 6.4e-3



= (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)

a. Por método iterativo

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)

O={A,A,S,S}

T=4

N=2Variable forward

t(i)= P(O1, O2 .. Ot, qt=Si)

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

Inducción

t+1(j)= [ t(i) aij ] bj(Ot+1) i

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

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

i=1 i=2

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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