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Unformatted text preview: STAT 455/855 Midterm Solutions Fall, 2008 1. (15 marks) (a) (5 marks) The equivalence classes are { 1 , 3 , 6 } , { 2 , 5 } , and { 4 } . Class { 1 , 3 , 6 } is transient and has period 1. Class { 2 , 5 } is recurrent and has period 2. Class { 4 } is recurrent and has period 1. (b) (10 marks) Define p 1 as in the problem and p 3 and p 6 as the probabilities of ultimately ending up in state 4, starting in state 3 and 6, respectively. If we start in state 1, then conditioning on the first move we obtain p 1 = 1 4 p 3 + 1 2 (1) + 1 4 p 6 . Similarly, if we start in state 3, then conditioning on the first move we obtain p 3 = 1 4 p 1 + 1 4 (0) + 1 4 (0) + 1 4 p 6 . Finally, starting in state 6 and conditioning on the first move we obtain p 6 = 1 2 p 1 + 1 2 p 3 . Thus, we have the three equations p 1 = 1 2 + 1 4 p 3 + 1 4 p 6 (1) p 3 = 1 4 p 1 + 1 4 p 6 (2) p 6 = 1 2 p 1 + 1 2 p 3 (3) in the three unknowns p 1 , p 3 , and p 6 . Plugging (3) into (2), we have p 3 = 1 4 p 1 + 1 4 1 2 p 1 + 1 2 p 3 = 3 8 p 1 + 1 8 p 3 , or 7 8 p 3 = 3 8 p 1 or p 3 = 3 7 p 1 . (4) Plugging (4) into (3) we get p 6 = 1 2 p 1 + 1 2 3 7 p 1 or p 6 = 5 7 p 1 . (5) Finally, plugging (4) and (5) into (1) we get p 1 = 1 2 + 1 4 3 7 p 1 + 1 4 5 7 p 1 = 1 2 + 2 7 p 1 or 5 7 p 1 = 1 2 or p 1 = 7 10 = 0 . 7 . STAT 455/855  Solutions: Midterm p. 2 2. (15 marks) (a) (5 marks) We solve for the smallest nonnegative solution of the equation s = G ( s )....
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This note was uploaded on 09/15/2010 for the course STAT 455/855 taught by Professor Glentakahara during the Fall '09 term at Queens University.
 Fall '09
 GLENTAKAHARA

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