{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

final08sol

# final08sol - STAT455/855 Fall 2008 Applied Stochastic...

This preview shows pages 1–3. Sign up to view the full content.

STAT455/855 Fall 2008 Applied Stochastic Processes Final Exam, Solutions 1. (15 marks) (a) (9 marks) The classes and classifications are: (i) { 0 } , which is positive recurrent with period 1, and { 1 , 2 } , which is positive recurrent with period 2. (ii) { 0 } , which is positive recurrent with period 1, and { 1 , 2 } , which is positive recurrent with period 1. (iii) { 0 } , 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 P ( X n = j X 0 = 0) + P ( X n = j X 0 = 1) + P ( X n = j X 0 = 2) . (i) For j = 0, P ( X n = 0 X 0 = 0) = 1 and P ( X n = 0 X 0 = 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) = 0 for all n , P ( X n = 1 X 0 = 1) = 1 for n even and 0 for n odd, and P ( X n = 1 X 0 = 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) = 0 for all n as in part(i). If X 0 = k with k = 1 or 2 then lim n →∞ P ( X n = j X 0 = k ) is the stationary probability of state j for the Markov chain with state space { 1 , 2 } and one-step transition matrix . 4 . 6 . 7 . 3 .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
STAT 455/855 -- Final Exam Solutions, 2008 Page 2 of 7 Letting π 1 and π 2 denote the stationary probabilities of states 1 and 2, re- spectively, we have from the global balance equations for this chain that π 1 and π 2 satisfy π 1 = . 4 π 1 + . 7 π 2 and π 2 = . 6 π 1 + . 3 π 2 , together with the normalization constraint π 1 + π 2 = 1. Both balance equa- tions above give . 6 π 1 = . 7 π 2 or π 2 = (6 / 7) π 1 . Thus, π 1 satisfies π 1 (1+6 / 7) = 1, or π 1 = 7 / 13. This then gives π 2 = 6 / 13. Thus, we have lim n →∞ P ( X n = 1 X 0 = k ) = 7 13 lim n →∞ P ( X n = 2 X 0 = k ) = 6 13 , for k = 1 or 2. Therefore, lim n →∞ P 1 ( n ) = 1 3 7 13 + 7 13 = 14 39 , lim n →∞ P 2 ( n ) = 1 3 6 13 + 6 13 = 12 39 .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 7

final08sol - STAT455/855 Fall 2008 Applied Stochastic...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online