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test2a-solution - ISyE 3232 Section A Stochastic...

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ISyE 3232 Stochastic Manufacturing and Service Systems Spring 2011 Section A J. G. Dai Test 2 (April 7, 2011) This is a closed book test. No calculator is allowed. 1. (30 points) Let X = { X n : n = 0 , 1 , 2 , . . . } be a discrete time Markov chain on state space S = { 1 , 2 , 3 , 4 } with transition matrix P = 0 1 2 0 1 2 1 4 0 3 4 0 0 1 2 0 1 2 1 4 0 3 4 0 . (a) Draw a transition diagram (b) Find P { X 2 = 4 | X 0 = 2 } . (c) Find P { X 2 = 2 , X 4 = 4 , X 5 = 1 | X 0 = 2 } . (d) What is the period of each state? (e) Let π = ( 1 8 , 1 4 , 3 8 , 1 4 ). Is π the unique stationary distribution of X ? Explain your answer. (f) Let P n be the n th power of P . Does lim n →∞ P n 1 , 4 = 1 4 hold? Explain your answer. (g) Let τ 1 be the first n 1 such that X n = 1. Compute E ( τ 1 | X 0 = 1). (If it takes you a long time to compute it, you are likely on a wrong track.) Solution. (a) 1 2 3 4 1/2 3/4 1/2 1/4 1/2 3/4 1/2 1/4 1
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(b) P { X 2 = 4 | X 0 = 2 } = P 21 P 14 + P 23 P 34 = 1 4 × 1 2 + 3 4 × 1 2 = 1 2 . (c) P { X 2 = 2 , X 4 = 4 , X 5 = 1 | X 0 = 2 } = P { X 5 = 1 | X 4 = 4 } P { X 4 = 4 | X 2 = 2 } P { X 2 = 2 | X 0 = 2 } = P 41 × 1 2 × ( 1 4 × 1 2 + 3 4 × 1 2 ) = 1 4 × 1 2 × 1 2 = 1 16 . (d) Each state has period 2. (e) YES. Since π = ( 1 8 , 1 4 , 3 8 , 1 4 ) satisfy π = πP and sum of the elements are 1, then π must be the stationary distribution. It is unique since this DTMC is irreducible. (f) No. The limit does not exist since the DTMC has period 2. (g) E ( τ 1 | X 0 = 1) = 1 π 1 = 8. 2. (20 points) Let P = . 2 . 8 0 0 0 . 5 . 5 0 0 0 0 . 25 0 . 75 0 0 0 . 5 0 . 5 0 0 0 0 1 Find approximately P 100 . Solution. Solving ( π 1 , π 2 ) . 2 . 8 . 5 . 5 = ( π 1 , π 2 ) we have ( π 1 , π 2 ) = ( 5 13 , 8 13 ). Let X be the DTMC with transition matrix P on state space S = { 1 , 2 , 3 , 4 , 5 } . Starting X 0 = 3, let f 3 be the probability that the DTMC
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