answer all parts for exercise 9.2.4 and show all work. se 92.3 The tranobability
ID: 3364305 • Letter: A
Question
answer all parts for exercise 9.2.4 and show all work.
se 92.3 The tranobability matrix of a discre obability matrix of a discrete-time Markov chain is given by 'e random ote by Y, and buted, and are s an alternating r interest with at the process associating a 0 0 0 1.0 0 0 1.0 0 0 P=10001.0 0 0 0.8 0.2 00 0.4 0 0.6 0 0 w all sample paths of length 4 that begin in state 1. What is the probability of being in each of the states1 hrough 5 after four steps beginning in state 1? Evercise ransition probability matrix is given by 9.2.4 Consider a discrete-time Markov chain consisting of four states a, b, c, and d and whose own. As soon tioning state s the random orresponding ribed above, 0.0 0.0 1.0 0.0 0.0 0.4 0.6 0.0 0.8 0.0 0.2 0.0 0.2 0.3 0.0 0.5 ime that the emachine is Compute the following probabilities 9.2.5 at time step n, n A Markov chain with two states a and b has the following conditional probabilities: If it is in . 1,2, ..., then it stays in state a with probability 0.5(0.5)". If it is in state b at 0Explanation / Answer
Question 9.2.4:
The required probabilities here are computed as:
a) The path here for which the probability is to be computed is:
a --> c --> c --> c --> c
The probability here is computed as:
pac *pcc*pcc*pcc = 1*0.2*0.2*0.2 = 0.008
Therefore 0.008 is the required probability here.
b) The path here is given as:
a --> b --> c --> d
The required probability computed here as:
pab*pbc*pcd = 0
because we cannot go from state a to state b in one step
Therefore 0 is the required probability here.
c) The path here is:
b --> c --> a --> c --> c
The probability is computed as:
pbc*pca*pac*pcc = 0.6*0.8*1*0.2 = 0.096
Therefore 0.096 is the required probability here.
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