Answer:
Step-by-step explanation:
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 = 2) = p (x_t = 4 | z_t = 1) = 0.25, p (X_t = 1 | z_t = 2) = 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 = 1 | z_t = 1) = 0.8, p (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 (z1 = z2 = z3), 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 (z1 = 1 | x1 = 2, x2 = 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 x1 = 2, x2 = 3, x3 = 3, x4 = 3 and x5 = 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.
ANSWER:
Let say we have that the first state of the die is state 1. Therefore the probability of this is p(z1=1)=0.5.
Also 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.
Again, 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
With this, the total probability that the same die is used for the first three rolls (i.e. casino would be in the same state) is given thus;
{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
Prob = 0.745
