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Midterm_1_-_Solutions-_F08

# Midterm_1_-_Solutions-_F08 - ACTSC 331 Life Contingencies I...

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ACTSC 331 - Life Contingencies I Fall 2008 Test 1 1. ( 15 marks ) A fully continuous 20-year pure endowment of \$500 is issued to (45). The bene°t premium 500 P ( 20 E 45 ) is 13.94. Given that (1) 2 A 60: 1 5 j ° A 60: 1 5 j ! 2 = 0 : 034 (2) 2 A 60: 5 j ° ° A 60: 5 j ± 2 = 0 : 003 (3) A 60: 5 j = 0 : 79 (4) ° = 5% (5) 5 p 60 = 0 : 8 (a) (2 marks) de°ne the prospective loss r.v. 15 L Solution: De°nition of 15 L 15 L = ( 0 ° 500 P ( 20 E 45 ) a T ° 15 j , 15 < T < 20 500 v 5 ° 500 P ( 20 E 45 ) a 5 j , T > 20 = ( 0 ° 13 : 94 a T ° 15 j , 15 < T < 20 500 v 5 ° 13 : 94 a 5 j , T > 20 (b) ( 3 marks ) °nd the bene°t reserve at time 15 Solution: 500 15 V ( 20 E 45 ) = 500 5 E 60 ° 13 : 94 a 60: 5 j = 500 5 E 60 ° 13 : 94 1 ° A 60: 5 j 0 : 05 = 500 (0 : 8) e ° 0 : 05(5) ° 13 : 94 1 ° 0 : 79 0 : 05 = 252 : 97 (c) ( 8 marks ) calculate V ar ( 15 L j T > 15) 1

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Solution: The prospective loss 15 L can be written as 15 L = 500 15 Z ° 13 : 94 ² 1 ° 15 W ° ³ , where 15 Z = ´ 0 , 15 < T < 20 v 5 , T > 20 , and 15 W = ´ v T ° 15 , 15 < T < 20 v 5 , T > 20 . Thus, V ar ( 15 L j T > 15) = (500) 2 V ar ( 15 Z j T > 15) + ² 13 : 94 0 : 05 ³ 2 V ar ( 15 W j T > 15) +2 (500) ² 13 : 94 0 : 05 ³ Cov ( 15 Z 15 W j T > 15) , where V ar ( 15 Z j T > 15) = 0 : 034 , V ar ( 15 W j T > 15) = 0 : 003 , and Cov ( 15 Z 15 W j T > 15) = E [ 15 Z 15 W j T > 15] ° 5 E 60 A 60: 5 j = v 10 5 p 60 ° µ
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Midterm_1_-_Solutions-_F08 - ACTSC 331 Life Contingencies I...

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