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Unformatted text preview: STAT455/855 Fall 2008 Applied Stochastic Processes Final Exam, Solutions 1. (15 marks) (a) (9 marks) The classes and classifications are: (i) { } , which is positive recurrent with period 1, and { 1 , 2 } , which is positive recurrent with period 2. (ii) { } , which is positive recurrent with period 1, and { 1 , 2 } , which is positive recurrent with period 1. (iii) { } , which is positive recurrent with period 1, and { 1 , 2 } , which is transient with period 1. (b) (6 marks) We find lim n P j ( n ), for j = 0 , 1 , 2, by conditioning on the initial state: P j ( n ) = P ( X n = j ) = 1 3 h P ( X n = j X = 0) + P ( X n = j X = 1) + P ( X n = j X = 2) i . (i) For j = 0, P ( X n = 0 X = 0) = 1 and P ( X n = 0 X = k ) = 0 for k = 1 or k = 2. Therefore, P ( X n = 0) = 1 / 3 for all n so the limiting value is 1 / 3. For j = 1, P ( X n = 1 X = 0) = 0 for all n , P ( X n = 1 X = 1) = 1 for n even and 0 for n odd, and P ( X n = 1 X = 2) = 0 for n even and 1 for n odd. Therefore, for all n (either n even or n odd), we obtain P ( X n = 1) = 1 / 3, and so the limiting value is 1 / 3. By subtraction, we obtain P ( X n = 2) = 1 / 3 for all n and in the limit. (ii) For j = 0, we obtain P ( X n = 0) = 1 / 3 for all n as in part(i). For j = 1 or j = 2, we have P ( X n = j X = 0) = 0 for all n as in part(i). If X = k with k = 1 or 2 then lim n P ( X n = j X = k ) is the stationary probability of state j for the Markov chain with state space { 1 , 2 } and onestep transition matrix " . 4 . 6 . 7 . 3 # . STAT 455/855  Final Exam Solutions, 2008 Page 2 of 7 Letting 1 and 2 denote the stationary probabilities of states 1 and 2, re spectively, we have from the global balance equations for this chain that 1 and 2 satisfy 1 = . 4 1 + . 7 2 and 2 = . 6 1 + . 3 2 , together with the normalization constraint 1 + 2 = 1. Both balance equa tions above give . 6 1 = . 7 2 or 2 = (6 / 7) 1 . Thus, 1 satisfies 1 (1+6 / 7) = 1, or 1 = 7 / 13. This then gives 2 = 6 /...
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 Fall '09
 GLENTAKAHARA

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