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Unformatted text preview: h (1) = 0 . 2 h (1) + 0 . 7 h (2) + 0 . 1 h (2) = 0 . 05 h (1) + 0 . 1 h (2) + 0 . 05 Solving gives h (1) = . 125 /. 685 and h (2) = . 045 /. 685. 9.7 (a) 1 → 2 but 2 6→ 1 so 1 is transient. 3 → 2 but 2 6→ 3 so 3 is transient. 5 → 2 but 2 6→ 5 so 5 is transient. { 2 , 4 } is a ﬁnite irreducible closed set so all these states are recurrent. (b) 3 → 6 but 6 6→ 3 so 3 is transient. 2 → 1 but 1 6→ 2 so 1 is transient. { 1 , 4 , 5 , 6 } is a ﬁnite irreducible closed set so all these states are recurrent 2...
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This note was uploaded on 09/29/2011 for the course MATH 4740 taught by Professor Durrett during the Spring '10 term at Cornell.
 Spring '10
 DURRETT
 Probability

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