Chapter_7_-_Question_Set_-_Solution

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

This preview shows pages 1–4. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ¢ ....
View Full Document

## This note was uploaded on 01/04/2012 for the course ACTSC 331 taught by Professor David during the Fall '09 term at Waterloo.

### Page1 / 51

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online