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Hw4-ORIE3510_S12_solns

Hw4-ORIE3510_S12_solns - ORIE 3510 Homework 4 Solutions...

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ORIE 3510 – Homework 4 Solutions Instructor: Mark E. Lewis due 2PM, Wednesday February 22, 2012 (ORIE Hallway drop box) 1. If p ij were (strictly) positive, then p ( n ) ji would be 0 for all n (otherwise, i and j would communicate). But then the process, starting in i , has a positive probability of at least p ij of never returning to i . This contradicts the recurrence of i . Hence p ij = 0. 2. a) The classes are { 1 , 2 , 6 , 9 } , { 3 , 4 , 7 } and { 5 , 8 } b) { 1 , 2 , 6 , 9 } and { 3 , 4 , 7 } are recurrent. { 5 , 8 } is transient. 1/2 1/2 3 1 1/2 1 2 7 1 4 1 1 2/3 3/4 1/2 6 9 8 1 1/3 5 1/4 c) Let d ( i ) denote the period of state i . We have that d (1) = d (2) = d (6) = d (9) = 4 d (3) = d (4) = d (7) = 1 d (5) = d (8) = 2 3. If M i =0 p ij = 1 for all i , then π j = 1 / ( M + 1) satisfies π j = M X i =0 π i p ij and M X j =0 π j = 1 . 1
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Since the chain is irreducible and aperiodic, there is a unique nonnegative solution to the steady state equation. That is, π j given as above is the limiting probabilities and π is the limiting distribution.
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