Let Xi = lunch condition on day i (rice/noodles) P(Xi+1 = rice| Xi-1 = rice, Xi...

Let Xi = lunch condition on day i (rice/noodles)

P(Xi+1 = rice| Xi-1 = rice, Xi = rice) = 0.7

P(Xi+1 = rice| Xi-1 = noodles, Xi = rice) = 0.6

P(Xi+1 = rice| Xi-1 = rice, Xi = noodles) = 0.3

P(Xi+1 = rice| Xi-1 = noodles, Xi = noodles) = 0.55

Q1. Is {Xn} a Markov Chains? Why?

Q2. How to transform the process into a M.C. ?

Solutions

Expert Solution

Since the transition probabilities to the state "rice" or "noodles" depends on last two states, {Xn} is not a Markov Chains.

Let there would be four states - (rice, rice) , (noodles, rice), (rice, noodles) , (noodles, noodles) which denotes the lunch condition on last two days (i, i-1).

The transition probability from state (rice, rice) to (rice, rice) is,

P(Yi+1 = (rice, rice) | Yi = (rice, rice) ) = (P(Xi+1 = rice | Xi-1 = rice, Xi = rice) = 0.7

P(Yi+1 = (rice, noodles) | Yi = (rice, rice) ) = 0

P(Yi+1 = (noodles, rice) | Yi = (rice, rice) ) = (P(Xi+1 = noodles | Xi-1 = rice, Xi = rice) = 1 - (P(Xi+1 = rice | Xi-1 = rice, Xi = rice) = 1 - 0.7 = 0.3

P(Yi+1 = (noodles, noodles) | Yi = (rice, rice) ) = 0

P(Yi+1 = (rice, rice) | Yi = (rice, noodles) ) = P(Xi+1 = rice| Xi-1 = noodles, Xi = rice) = 0.6

P(Yi+1 = (rice, noodles) | Yi = (rice, noodles) ) = 0

P(Yi+1 = (noodles, rice) | Yi = (rice, noodles) ) = P(Xi+1 = noodles| Xi-1 = noodles, Xi = rice) = 1 - P(Xi+1 = rice| Xi-1 = noodles, Xi = rice) = 1 - 0.6 = 0.4

P(Yi+1 = (noodles, noodles) | Yi = (rice, noodles) ) = 0

P(Yi+1 = (rice, rice) | Yi = (noodles, rice) ) = 0

P(Yi+1 = (rice, noodles) | Yi = (noodles, rice) ) = P(Xi+1 = rice| Xi-1 = rice, Xi = noodles) = 0.3

P(Yi+1 = (noodles, rice) | Yi = (noodles, rice) ) = 0

P(Yi+1 = (noodles, noodles) | Yi = (noodles, rice) ) = P(Xi+1 = noodles| Xi-1 = rice, Xi = noodles) = 1 - P(Xi+1 = rice| Xi-1 = rice, Xi = noodles) = 1 - 0.3 = 0.7

P(Yi+1 = (rice, rice) | Yi = (noodles, noodles) ) = 0

P(Yi+1 = (rice, noodles) | Yi = (noodles, noodles) ) = P(Xi+1 = rice| Xi-1 = noodles, Xi = noodles) = 0.55

P(Yi+1 = (noodles, rice) | Yi = (noodles, noodles) ) = 0

P(Yi+1 = (noodles, noodles) | Yi = (noodles, noodles) ) = P(Xi+1 = noodles| Xi-1 = noodles, Xi = noodles) = 1 - P(Xi+1 = rice| Xi-1 = noodles, Xi = noodles) = 1 - 0.55 = 0.45

For all the states, the transition probabilities to the other states only depends on last states, and hence {Yn} is a Markov chain.

orchestra answered 1 year ago

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