Manual for SOA Exam MLC.

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1 Chapter 4 Life Insurance Extract from: Arcones Manual for the SOA Exam MLC Fall 2009 Edition available at 1/13

2 Non-level payments paid at the end of the year Suppose that a life insurance provides a benefit of b k paid at the end of the k th year if death happens in this year The present value of this benefit is B x = b K ν K The actuarial present value of B x is E[B x ] = We have that E[B 2 x ] = b k ν k P{K x = k} = bk 2 ν2k P{K x = k} = b k ν k k 1 q x bk 2 ν2k k 1 q x 2/13

value of this benefit is B x = b K ν K The actuarial present value of B x is E[B x ] = We

3 Example 1 A whole life insurance on (50) pays plus the return of the net single premium with interest at δ = 003 at the end of the year of time of death The survival function for (50) follows the de Moivre s law with ω = 110 Calculate the net single premium for δ = 007 3/13

year of time of death The survival function for (50) follows the de

4 Example 1 A whole life insurance on (50) pays plus the return of the net single premium with interest at δ = 003 at the end of the year of time of death The survival function for (50) follows the de Moivre s law with ω = 110 Calculate the net single premium for δ = 007 Solution: We have that b k = Pe (003)k and 60 P = b k ν k k 1 q x = ( Pe (003)k )e (007)k 1 60 = e (007)k + P e (004)k = (50000)(e 007 e (007)61 ) 60(1 e 007 ) = P + (e 004 e (004)61 )P 60(1 e 004 ) Hence, P = = /13

have that b k = 50000 + Pe (003)k and 60 P = b k ν k k 1 q x = (50000 + Pe (003)k )e (007)k 1 60 = 50000 60 60 e (007)k + P 60 60 e (004)k =

5 Definition 1 An increasing by one whole life insurance pays k at time k, for each k 1, if the failure happens in the k th interval Definition 2 The actuarial present value of a unit increasing whole life insurance is denoted by (IA) x We have that (IA) x = kν k P{K x = k} = kν k k 1 q x 5/13

The actuarial present value of a unit increasing whole life insurance is

6 Theorem 1 For each x 0, (IA) x = A x + 1 E x (IA) x+1 6/13

7 Theorem 1 For each x 0, Proof: We have that = = s(x + 1) 1E x (IA) x+1 = ν s(x) (IA) x = A x + 1 E x (IA) x+1 k+1 s(x k 1) s(x k) kν s(x) k s(x + k 1) s(x + k) (k 1)ν, s(x) k=2 k s(x k 1) s(x k) kν s(x + 1) 7/13

s(x + 1 + k 1) s(x + 1 + k) kν s(x) k s(x + k 1) s(x + k)

8 Theorem 1 For each x 0, (IA) x = A x + 1 E x (IA) x+1 Proof: A x + 1 E x (IA) x+1 k s(x + k 1) s(x + k) = ν s(x) k s(x + k 1) s(x + k) + (k 1)ν s(x) k=2 k s(x + k 1) s(x + k) = kν = (IA) s(x) x 8/13

9 Example 2 Suppose that A 30 = 013, (IA) 30 = 045, ν = 094 and p 30 = 099 Find (IA) 31 9/13

10 Example 2 Suppose that A 30 = 013, (IA) 30 = 045, ν = 094 and p 30 = 099 Find (IA) 31 Solution: Since (IA) x = A x + 1 E x (IA) x+1, 045 = (094)(099) (IA) 31, and (IA) 31 = (094)(099) = /13

= A x + 1 E x (IA) x+1, 045 = 013 + (094)(099) (IA)

11 Definition 3 An increasing by one n th year term life insurance pays k at time k, where k 1, if the failure happens in the k th interval and k n Definition 4 The actuarial present value of a unit increasing n th year term life insurance is denoted by (IA) 1 x:n We have that (IA) 1 x:n = kν k P{K x = k} = kν k k 1 q x 11/13

The actuarial present value of a unit increasing n th year term life insurance

12 Definition 5 An decreasing by one n th year term life insurance pays n + 1 k at time k if the failure happens in the k th interval, where 1 k n Definition 6 The actuarial present value of a unit decreasing n th year term life insurance is denoted by (DA) 1 x:n We have that (DA) 1 x:n = (n +1 k)ν k P{K x = k} = (n +1 k)ν k k 1 q x 12/13

actuarial present value of a unit decreasing n th year term life insurance is

13 Theorem 2 (IA) x:n + (DA) x:n = (n + 1) (A) x:n Proof: (IA) 1 x:n + (DA)1 x:n = kν k k 1 q x + (n + 1 k)ν k k 1 q x = (n + 1)ν k k 1 q x = (n + 1) (A) 1 x:n 13/13

Manual for SOA Exam MLC.

Manual for SOA Exam MLC. Chapter 4. Life Insurance. c 29. Miguel A. Arcones. All rights reserved. Extract from: Arcones Manual for the SOA Exam MLC. Fall 29 Edition. available at http://www.actexmadriver.com/ c 29. Miguel A. Arcones.