Exercises_-_Chapter_8_-_Solutions

Exercises_-_Chapter_8_-_Solutions - ACTSCI 331 Life...

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ACTSCI 331 - Life Contingencies I Solutions (Chapter 8) j L for j = 0 ; :::; 14 , j L = 8 > > < > > : 1000 v K +1 j 3 a K +1 j j , K = j; j + 1 ; :::; 14 1000 v K +1 j 3 a 15 j j 2 15 j a K +1 15 j , K = 15 ; :::; 24 500 v K +1 j 3 a 15 j j 2 15 j a K +1 15 j , K = 25 ; :::; 29 500 v K +1 j 3 a 15 j j 2 15 j a 15 j , K = 30 ; 31 ; ::: for j = 15 ; :::; 24 , j L = 8 < : 1000 v K +1 j 2 a K +1 j j , K = j; j + 1 ; :::; 24 500 v K +1 j 2 a K +1 j j , K = 25 ; :::; 29 500 v K +1 j 2 a 30 j j , K = 30 ; 31 ; ::: for j = 25 ; :::; 29 j L = 500 v K +1 j 2 a K +1 j j , K = j; j + 1 ; :::; 29 500 v K +1 j 2 a 30 j j , K = 30 ; 31 ; ::: for j = 30 ; 31 ; ::: j L = 500 v K +1 j , K = j; j + 1 ; ::: (b) Under the equivalent principle, E [ Z ] = E [ Y ] 500 A 40 + 500 A 1 40: 25 j = ± a 40: 30 j a 40: 15 j ² . It follows that = 500 A 40 + 500 A 1 40: 25 j a 40: 30 j a 40: 15 j = 500 (0 : 202) + 500 (0 : 114) 2 (13 : 414) + (10 : 005) = 4 : 29 . 10 V = 500 A 50 + 500 A 1 50: 15 j ± 2 a 50: 20 j + a 50: 5 j ² = 500 (0 : 316) + 500 (0 : 149) (2 (10 : 806) + 4 : 386) = 500 (0 : 316) + 500 (0 : 149) 4 : 29 (2 (10 : 806) + 4 : 386) = 120 : 97 , 20 V = 500 A 60 + 500 A 1 60: 5 j 2 a 60: 10 j = 500 (0 : 458) + 500 (0 : 112) 2 (4 : 29) (6 : 929) = 225 : 55 1
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and 40 V = 500 A 80 = 500 (0 : 744) = 372 . 50 25 P (( IA ) 35 ) = 50 ( IA ) 35 a 35: 25 j = 50 (3 : 352) a 35: 25 j . Under the prospective approach, the reserve at time 10 can be expressed as 10 V = 50 ( IA ) 45 + 500 A 45 50 25 P (( IA ) 35 a 45: 15 j = 50 ( IA ) 45 + 500 A 45 50 (3 : 352) a 35: 25 j a 45: 15 j = 50 (4 : 191) + 500 A 45 50 (3 : 352) a 45: 15 j a 35: 25 j , where a 45: 15 j a 35: 25 j = 1 10 V 35: 25 j . while 1 10 V 35 = 1 A 45 1 A 35 , which allows to identify A 45 = 1 (1 10 V 35 ) (1 A 35 ) = 1 (1 & 0 : 111) (1 & 0 : 142) = 0 : 23724 . It follows that 10 V = 50 (4 : 191) + 500 (0 : 23724) 50 (3 : 352) (1 & 0 : 227) = 198 : 62 . L L = e 0 : 06( K +1) e & 0 : 08( K +1) P a K +1 j ; K = 0 ; 1 ; 2 ::: = e & 0 : 02( K +1) P 1 e & 0 : 08( K +1) 1 e & 0 : 08 ± , K = 0 ; 1 ; 2 ::: E ( Z ) = E ( Y ) A @2% 30 = P a @8% 30 . which implies P = A @2% 30 a @8% 30 . 10 L 10 L = e 0 : 06( K +1) e & 0 : 08( K +1 10) P a K +1 10 j ; K = 10 ; 11 ; 12 ::: = e 0 : 06(10) e & 0 : 02( K +1 10) P 1 e & 0 : 08( K +1 10) 1 e & 0 : 08 ± , K = 10 ; 11 ; 12 ::: 2
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10 V = E [ 10 L j T > 10] = e 0 : 6 A @2% 40 P a @8% 40 = e 0 : 6 A @2% 40 A @2% 30 a @8% 30 a @8% 40 = e 0 : 6 A @2% 40 A @2% 30 a @8% 40 a @8% 30 , where a @8% 40 a @8% 30 = 1 10 V @8% 30 = 1 & 0 : 033 , and 1
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This note was uploaded on 10/30/2009 for the course ACTSC 331 taught by Professor David during the Fall '09 term at Waterloo.

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Exercises_-_Chapter_8_-_Solutions - ACTSCI 331 Life...

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