# hw7sols - ISyE 3232 H Ayhan Stochastic Manufacturing and...

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Solutions to Homework 7 October 24, 2011 1. According to the fact that the row summation of transition matrix P equals one, we can ﬁll the missing values in P: 1 5 4 5 0 2 5 1 2 1 10 0 1 10 9 10 . And α is the initial distribution, so the summation of its entries equals one too. So we have α = ( 1 5 , 2 5 , 2 5 ) . a) Pr( x 1 = 0 | x 0 = 1 ) = P 1 , 0 = 2 5 . b) We let α i denote the distribution of X i (so α 0 = α ). Then we have α i = α 0 P i . So X 1 ’s distribution α 1 = αP = ( 1 5 , 2 5 , 2 5 ) 1 5 4 5 0 2 5 1 2 1 10 0 1 10 9 10 = ( 1 5 , 2 5 , 2 5 ) . c) α 15 = αP 15 . We ﬁnd from b) α 1 = αP = ( 1 5 , 2 5 , 2 5 ) = α . So αP i +1 = αP i = ... = α . Let i = 14 ; we have α 15 = αP 15 = α . d) It’s irreducible. Every state can communicate with each other. e) State 0 can reach itself in one step. So the Markov Chain is aperiodic. f) Yes, it’s postively recurrent. Because it’s a irreducible and ﬁnite Markov Chain. 1 ISyE 3232 H. Ayhan Stochastic Manufacturing and Service Systems Fall 2011

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2. (a) In order to show that { X n ,n
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hw7sols - ISyE 3232 H Ayhan Stochastic Manufacturing and...

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