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Exercises_-_Chapter_8_-_Solutions

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

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ACTSCI 331 - Life Contingencies I Solutions (Chapter 8) 1. (a) De°nition of 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 °v 15 ° j ° a K +1 ° 15 j , K = 15 ; :::; 24 500 v K +1 ° j ° 3 ° ° a 15 ° j j ° 2 °v 15 ° j ° a K +1 ° 15 j , K = 25 ; :::; 29 500 v K +1 ° j ° 3 ° ° a 15 ° j j ° 2 °v 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 . For the bene°t reserve, we have 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 . 2. First, let us identify the form of the bene°t premium 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 . 3. (a) De°nition of 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 ³ 1 ° e ° 0 : 08( K +1) 1 ° e ° 0 : 08 ´ , K = 0 ; 1 ; 2 ::: It follows that the bene°t premium is given by E ( Z ) = E ( Y ) A @2% 30 = P ° a @8% 30 . which implies P = A @2% 30 ° a @8% 30 . De°nition of 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 ³ 1 ° e ° 0 : 08( K +1 ° 10) 1 ° e ° 0 : 08 ´ , K = 10 ; 11 ; 12 ::: 2
It follows that the bene°t reserve is given by 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 ° 10 V @2% 30 = ° a @2% 40 ° a @2% 30 = 1 ° A @2% 40 1 ° A @2% 30 , which implies A @2% 30 = 1 ° 1 ° A @2% 40 1 ° 10 V @2% 30 = 1 ° 1 ° 0 : 688 1 ° 0 : 094 = 0 : 65563 . One concludes that 10 V = e 0 : 6 (0 : 688) ° (0 : 65563) (1 ° 0 : 033) = 0 : 6196 . (b) De°nition of L L = ( e 0 : 06( K +1) e ° 0 : 08( K +1) ° P ° a K +1 j , K = 0 ; 1 ; :::; 19 e 0 : 06( K +1) e ° 0 : 08( K +1) ° P ° a 20 j , K = 20 ; 21 ; ::: It follows that the bene°t premium is given by E ( Z ) = E ( Y ) A @2% 30 = P ° a @8% 30: 20 j . which implies P = A @2% 30 ° a @8% 30: 20 j . 3

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De°nition of 10 L 10 L = ( e 0 : 06( K +1) e ° 0 : 08( K +1 ° 10) ° P ° a K +1 ° 10 j , K = 10 ; 11 ; :::; 19 e 0 : 06( K +1) e ° 0 : 08( K +1 ° 10) ° P ° a 10 j , K = 20 ; 21 ; ::: = ( e 0 : 06(10) e ° 0 : 02( K +1 ° 10) ° P ° a K +1 ° 10 j , K = 10 ; 11 ; :::; 19 e 0 : 06(10) e ° 0 : 02( K +1 ° 10) ° P ° a 10 j , K = 20 ; 21 ; ::: It follows that the bene°t reserve is given by 10 V = E [ 10 L j T > 10] = e 0 : 6 A @2% 40 ° P ° a @8% 40: 10 j = e 0 : 6 A @2% 40 ° A @2% 30 ° a @8% 30: 20 j ° a @8% 40: 10 j = e 0 : 6 A @2% 40 ° A @2% 30 ° a @8% 40: 10 j ° a
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