Chapter_7_-_Question_Set_-_Solution

Chapter_7_-_Question_Set_-_Solution - ACTSC 331 - Life...

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Unformatted text preview: ACTSC 331 - Life contingencies 1 Exercises (Chapter 7) 1. A fully continuous 20-year payment, 30-year term insurance of $2000 is issued to (35) . We assume that the force of interest of 6%, and the force of mortality is such that t p x = exp & & : 0005246 1 : 1 x 1 : 1 t & 1 , t;x . Under these assumptions, n A 1 35+ n : 30 & n j a 35+ n : 20 & n j : 06873 11 : 39534 10 : 09710 7 : 35175 25 : 07952 & (a) Calculate the bene&t premium. (Answer: 12 : 063 ) Solution: The bene&t premium is given by P = 2000 A 1 35: 30 j a 35: 20 j = 2000(0 : 06873) 11 : 39534 = 12 : 063 . (b) De&ne the prospective loss rv j L for j = 10 ; 25 . Plot roughly the curves of j L in function of T for j = 10 ; 25 . Solution: The prospective loss rv is de&ned as 10 L = 8 > < > : 2000 v T & 10 & P a T & 10 j , 10 < T 20 , 2000 v T & 10 & P a 20 & 10 j , 20 < T 30 , & P a 20 & 10 j , T > 30 , and 25 L = 2000 v T & 25 , 25 < T 30 , , T > 30 . (c) Find the value of 10 L if T = 12 : 3 and the one of 25 L if T = 27 : 4 . (Answers: 1716 : 3 and 1731 : 8 ) Solution: We have ( 10 L j T = 12 : 3) = 2000 v 12 : 3 & 10 & P a 12 : 3 & 10 j = 2000 e & : 06(2 : 3) & (12 : 063) 1 & e & : 06(2 : 3) : 06 = 1716 : 3 , and ( 25 L j T = 27 : 4) = 2000 v 27 : 4 & 25 = 2000 e & : 06(2 : 4) = 1731 : 8 . (d) Compute the policy value j V = E [ j L j T > j ] for j = 10 and 25 . (Answers: 105 : 52 and 159 : 04 ) Solution : By de&nition, 10 V = 2000 A 1 45: 20 j & P a 45: 10 j = 2000(0 : 09710) & (12 : 063)(7 : 35175) = 105 : 52 , 1 and 25 V = 2000 A 1 60: 5 j = 2000(0 : 07952) = 159 : 04 . (e) Given that T > 10 , &nd the probability that the prospective loss 10 L is less than or equal to 1800 . (Answer: : 99377 ) Solution: From (b), we know that 10 L is a non-increasing function of the time-until-death rv T . Given that ( 10 L j T = 10) = 2000 , and ( 10 L j T = 20) = 2000 e & : 06(10) & P a 10 j = 2000 e & : 06(10) & (12 : 063) 1 & e & : 06(10) : 06 = 1006 : 9 , it follows that Pr( 10 L 1800 j T > 10) = Pr( T > t j T > 10) = t & & 10 p 45 , where t is the unique solution of 2000 v t & & 10 & P a t & & 10 j = 1800 . One concludes that t = 10 & 1 & ln 1800 + P & 2000 + P & = 10 & 1 : 06 ln 1800 + 12 : 063 : 06 2000 + 12 : 063 : 06 = 11 : 588 , which implies that Pr( 10 L 1800 j T > 10) = 1 : 588 p 45 = exp n & : 0005246(1 : 1) 45 & 1 : 1 1 : 588 & 1 o = 0 : 99377 . 2 2. A fully continuous 20-year deferred whole life insurance of $4000 is issued to (45) . The bene&t premium is payable as long as the insured is alive without exceeding the deferred period. We assume a force of interest of 6% , and the force of mortality is such that t p x = exp & & : 0005246 1 : 1 x 1 : 1 t & 1 , t;x ....
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Chapter_7_-_Question_Set_-_Solution - ACTSC 331 - Life...

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