Question: PLEASE ANSWER QUESTION 5 ONLY. I PUT Q4 AND ITS SOLUTION BELOW FOR REFERENCE. Q5. Hidden Markov Models: The Forward Algorithm (20 points) Consider the

 PLEASE ANSWER QUESTION 5 ONLY. I PUT Q4 AND ITS SOLUTION

PLEASE ANSWER QUESTION 5 ONLY. I PUT Q4 AND ITS SOLUTION BELOW FOR REFERENCE.BELOW FOR REFERENCE. Q5. Hidden Markov Models: The Forward Algorithm (20 points)Consider the same HMM as Q4. We have evidences E1 = E2

Q5. Hidden Markov Models: The Forward Algorithm (20 points) Consider the same HMM as Q4. We have evidences E1 = E2 = c. What is the updated belief B2(X) = P(X2 | E1 = c, E2 = c)? Consider the HMM shown below: (X1) (X2) EL E2 The prior probability P(Xo), dynamics model P(X4+1 | Xt), and sensor model P(Et | X+) are as follows. Provide the stationary distribution of X. Et X+ P(E|Xt) Xt+1 x P(X++1|X4) 0 0.3 X, P(X) 0 0 0.3 b 0 0.15 0 0.2 1 0 0.7 0 0.55 1 0.8 0 1 0.05 a 1 0.1 1 1 0.95 b 1 0.45 1 0.45 Solution Given P(x)= (0.3 0.7 } 10.05 0.95 un transition diagram Xat coco.. transition matrix 0.05 Which Wiran TP=0 where I let to be the stationary distribution satisting [110) rices [(0) Tou] [ 0.3 Tilo ) ILO) 0.7 [ 0.05 0.95 by solving above ean mo) 0.3 Tlo) + 0.05 Til) = 0.05 T1) = arro TiO) + Ti(1) = 1 tol + 14811(0) = 1 lusing To = /15 eqh (1) o ) TI) 5 14 Tilo) -(1) hence Ans Q5. Hidden Markov Models: The Forward Algorithm (20 points) Consider the same HMM as Q4. We have evidences E1 = E2 = c. What is the updated belief B2(X) = P(X2 | E1 = c, E2 = c)? Consider the HMM shown below: (X1) (X2) EL E2 The prior probability P(Xo), dynamics model P(X4+1 | Xt), and sensor model P(Et | X+) are as follows. Provide the stationary distribution of X. Et X+ P(E|Xt) Xt+1 x P(X++1|X4) 0 0.3 X, P(X) 0 0 0.3 b 0 0.15 0 0.2 1 0 0.7 0 0.55 1 0.8 0 1 0.05 a 1 0.1 1 1 0.95 b 1 0.45 1 0.45 Solution Given P(x)= (0.3 0.7 } 10.05 0.95 un transition diagram Xat coco.. transition matrix 0.05 Which Wiran TP=0 where I let to be the stationary distribution satisting [110) rices [(0) Tou] [ 0.3 Tilo ) ILO) 0.7 [ 0.05 0.95 by solving above ean mo) 0.3 Tlo) + 0.05 Til) = 0.05 T1) = arro TiO) + Ti(1) = 1 tol + 14811(0) = 1 lusing To = /15 eqh (1) o ) TI) 5 14 Tilo) -(1) hence Ans

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