FinalexamACTL2003-Yr2009-Sol

FinalexamACTL2003-Yr2009-Sol - ACTL 2003/5103 Final...

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ACTL 2003/5103 Final Examination-Sample Solutions-year 2009 Question 1 (a) . Sol. { 1 , 5 } transient { 3 } recurrent { 2 , 4 } recurrent (b) Sol. P T = ± 1 5 4 5 1 2 0 S = ( I - P T ) ( - 1) = ± 4 5 - 4 5 - 1 2 1 - 1 = 1 4 5 1 2 4 5 4 5 - 4 5 - 1 2 1 = ± 5 2 2 5 4 2 (c) Sol. Pr ( X 2 = 3) = 5 X i =1 Pr ( X 2 = 3 | X 0 = i ) Pr ( X 0 = i ) = 5 X i =1 P (2) i 3 i 15 = 1 15 ( 1 5 0 0 0 4 5 )(0 0 1 0 1 2 ) T + 2 15 (0 1 3 0 2 3 0)(0 0 1 0 1 2 ) T + 3 15 (0 0 1 0 0)(0 0 1 0 1 2 ) T + 4 15 (0 1 2 0 1 2 0)(0 0 1 0 1 2 ) T + 5 15 ( 1 2 0 1 2 0 0)(0 0 1 0 1 2 ) T = 1 15 × 2 5 + 0 + 3 15 + 0 + 5 15 × 1 2 = 59 150 = 0 . 39 An easier way is to use the fact that 3 is absorbing ( so p (2) 33 = 1), the tran- sition probability from a recurrent state to a transient state is always 0, (so 1
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p (2) 23 = 0 = p (2) 43 ). (d) Sol. The chain is not ergodic due to any of the following reasons: it is reducible and it is not positive recurrent. (e) Sol. Since when the chain enters in class { 2 , 4 } it will always stay in a state within the class. Go starting from state 2, the Markov chain is same as a Markov chain starting from state 2 with state space { 2 , 4 } and transition probability matrix ± 1 3 2 3 1 2 1 2 . Note the new Markov chain is ergodic and therefore the limiting proba- bilities lim n →∞ p 24 and lim n →∞ p 42 exist. Let x = lim n →∞ p 42 and y = lim n →∞ p 24 . Then x = 1 3 x + 1 2 y y = 2 3 x + 1 2 y x + y = 1 Solving the equations gives x = 3 7 and y = 4 7 . So lim n →∞ p 24 = 4 7 . Question 2 (a) (i) Sol. Let X n denote the state of a policyholder in year n . { X n } is a Markov chain with state space { 1 , 2 , 3 } . 0 . 9 0 . 1 0 0 . 7 0 . 2 0 . 1 0 0 . 7 0 . 3 2
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(ii) Sol. Solving π 1 = 0 . 9 π 1 + 0 . 7 π 2 π 2 = 0 . 1 π 1 + 0 . 2 π 2 + 0 . 7 π 3 π 3 = 0 . 1 π 2 + 0 . 3 π 3 π 1 + π 2 + π 3 = 1 gives π 1 = 49 57 , π 2 = 7 57 , and π 3 = 1 57 . The long-run average premium a policyholder pays per year=100 π 1 + 200 π 2 + 300 π 3 = 115 . 79 (b) Sol. Method 1: Let the state equal to the capital at the end of a year. The state space S = { 0 , 1 , 2 } . The one-step transition probability matrix is 1 0 0 1 3 1 3 1 3 1 4 1 4 1 2 Note the statement “after receiving the capital injection at the beginning of a year, the capital is 3 million” is equivalent to“ at the end of last year, the capital is 2 million”. Since 0 is absorbing, so whenever the chain reaches state 0, it will always stay there. So the desired probability is
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FinalexamACTL2003-Yr2009-Sol - ACTL 2003/5103 Final...

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