We first examine a simple hidden Markov model (HMM). We observe a sequence of ro

Question

We first examine a simple hidden Markov model (HMM). We observe a sequence of rolls of a four-sided die at an "occasionally dishonest casino", where at time t the observed outcome X_t element {1, 2, 3, 4). At each of these times, the casino can be in one of two states Z_t element (1, 2). When Z_t = 1 the casino uses a fair die, while when Z_t = 2 the die is biased so that rolling a 1 is more likely. In particular: p (x_t = 1|z_t = 1) = p (X_t = 2|Z_t = 1) = p (X_t = 3|Z_t = 1) = p (X_t= 4|Z_t = 1) = 0.25, p (X_t = 1|z_t = 1) = 0.7, p (X_t = 2|Z_t = 2) = p (X_t = 3|Z_t = 2) = p (X_t= 4|Z_t = 2) = 0.1. Assume that the casino has an equal probability of starting in either state at time t = 1, so that p (Z1 = 1) = p (z1 = 2) = 0.5. The casino usually uses the same die for multiple iterations, but occasionally switches states according to the following probabilities:p(Z_t + = 1|Z_t = 1) = 0.8, p (Z_t + 1 = 2|Z_t = 2) = 0.9. The other transition probabilities you will need are the complements of these. a. Under the HMM generative model, what is p (z_1 = z_2 = z_3), the probability that the same die is used for the first three rolls? b. Suppose that we observe the first two rolls. What is p (z_1 = 1 |x_1 = 2, x_2 = 4), the probability that the casino used the fair die in the first roll? c. Using the backward algorithm, compute the probability that we observe the sequence x_1 = 2, x_2 = 3, x_3 = 3 x_4 = 3 and x_5 = 1. Show your work (i.e., show each of your belief for based on time). Consider the final distribution at time t = 6 for both p (Z_t = 1) = p (Z_t = 2) = 1.

Explanation / Answer

Solution:

Suppose the first state of the die is state 1. The probability of this is p(z1=1)=0.5. The probability that the same die is used(i.e. casino would be in the same state) is p(z2=1|z1=1)=0.8.

Now suppose the first state of the die is state 2. So, p(z1=2)=0.5 and p(z2=2|z1=2)=0.9.

Other transition probabilities can be written as

p(zt+1=2|zt=1)=1-p(zt+1=1|zt=1)=.2

p(zt+1=1|zt=2)=1-p(zt+1=2|zt=2)=.1

p(z3=1|z1=1) = [p(z3=1|z2=2)*p(z2=2|z1=1)]+[p(z3=1|z2=1)*p(z2=1|z1=1)] = 0.1*0.2+0.8*0.8 = 0.66

p(z3=2|z1=2) = [p(z3=2|z2=2)*p(z2=2|z1=2)]+[p(z3=2|z2=1)*p(z2=1|z1=2)] = 0.9*0.9+0.2*0.1 = 0.83

Hence, the total probability that the same die is used for the first three rolls(i.e. casino would be in the same state) is = {p(z1=1)*p(z3=1|z1=1)}*{p(z1=2)*p(z3=2|z1=2)} = 0.5*0.66+0.5*0.83 = 0.745

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