STA3007_0506_t03_ssol

STA3007_0506_t03_ssol - STA 3007 Applied Probability 2005...

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Unformatted text preview: STA 3007 Applied Probability 2005 Tutorial 3 Suggested Solution 1. First Step Analysis i. (a) Rearrage the matrix as: P =     1 2 3 1 . 6 . 1 . 1 . 2 2 . 3 . 4 . 2 . 1 . . 1 . . 3 . . . 1 .     T = min { n ≥ ,X n = 0 or X n = 3 } U i = Pr { X T = 0 | X = i } , i = 1 , 2 U = R [ ] + QU U 1 = (1)(0 . 1) + U 1 (0 . 6) + U 2 (0 . 1) + (0)(0 . 2) U 2 = (1)(0 . 2) + U 1 (0 . 3) + U 2 (0 . 4) + (0)(0 . 1) U 1 = 0 . 3810 ,U 2 = 0 . 5238 (b) v i = E [ T | X = i ] V = [ 1 ] + QV v 1 = 1 + v 1 (0 . 6) + v 2 (0 . 1) v 2 = 1 + v 1 (0 . 3) + v 2 (0 . 4) v 1 = 3 . 333 ,v 2 = 3 . 333 ii. Consider the patterns HHH, HHT, HTH, HTT, TTT, TTH, THT, THH as 8 different state space. Since we successivly flip the coin and stop until the patterns HHT or HTH appears, these patterns HHT,HTH are considered as absorption states. We can setup the following transition probability matrix: P =             HHH HHT HTH HTT TTT TTH THT THH HHH . 5 . 5 . . . . . . HHT . 1 . . . . . . . HTH . . 1 . . . . . . HTT . . . . . 5 . 5 . . TTT . . . . . 5 . 5 . . TTH . . . . . . . 5 . 5 THT . . . 5 . 5 . . . . THH . 5 . 5 . . . . . .             Rearrage the matrix as followed: P =             HHH HTT TTT TTH THT THH HHT HTH HHH . 5 . . . . . . 5 ....
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This note was uploaded on 05/21/2011 for the course STA 3007 taught by Professor Kb during the Spring '11 term at CUHK.

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STA3007_0506_t03_ssol - STA 3007 Applied Probability 2005...

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