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 1/44
2 Properties of the APV for continuous insurance The following table shows the definition of all the variables in the previous section: type of insurance whole life insurance payment Z x = ν Tx n year term life insurance Z 1 x:n = ν Tx I (T x n) n year deferred life insurance n Z x = ν Tx I (n < T x ) n year pure endowment life insurance Z 1 x:n = ν n I (n < T x ) n year endowment life insurance min(tx Z x:n = ν,n) 2/44
3 Theorem 1 We have that Z x = Z 1 x:n + n Z x, A x = A 1 x:n + n A x, 2 A x = 2 A 1 x:n + 2 n A x, Var(Z x ) = Var(Zx:n 1 ) + Var( n Z x ) 2A 1 x:n n A x Cov(Z 1 x:n, n Z x ) = A 1 x:n n A x = 1 ( ) A 1 x:n 2 + n A 2 2 x A x. 2 3/44
4 Theorem 2 We have that Z x:n = Z 1 x:n + Z 1 x:n, A x:n = A 1 x:n + A 1 x:n, 2 A x:n = 2 A 1 x:n + 2 A 1 x:n, Var(Z x:n ) = Var(Z 1 x:n ) + Var(Z x:n ) 1 2A 1 x:n A 1 x:n Cov(Z 1 x:n, Z 1 x:n ) = A 1 x:n A 1 x:n = 1 2 ( A 1 x:n 2 + A 1 x:n 2 A x:n 2 ). 4/44
5 Theorem 3 We have that n A x = n E x A x+n. 5/44
6 Theorem 3 We have that n A x = n E x A x+n. Proof: We have that n A x = n ν t tp x µ x+t dt = =ν n np x ν t tp x+n µ x+n+t dt = n E x A x+n. 0 0 ν t+n n+tp x µ x+n+t dt 6/44
7 Theorem 3 We have that n A x = n E x A x+n. Proof: We have that n A x = n ν t tp x µ x+t dt = =ν n np x ν t tp x+n µ x+n+t dt = n E x A x+n. 0 0 ν t+n n+tp x µ x+n+t dt Corollary 1 A x = A 1 x:n + n E x A x+n. 7/44
8 Example 1 Suppose that δ = 0.04 and (x) has force of mortality µ = Calculate: 8/44
9 Example 1 Suppose that δ = 0.04 and (x) has force of mortality µ = Calculate: (i) A x and Var(Z x ). 9/44
10 Example 1 Suppose that δ = 0.04 and (x) has force of mortality µ = Calculate: (i) A x and Var(Z x ). Solution: (i) We have that A x = µ µ + δ = = , 2 A x = µ µ + 2δ = (2)(0.04) = , Var(Z x ) = 2 A x (A x ) 2 = ( ) 2 = /44
11 Example 1 Suppose that δ = 0.04 and (x) has force of mortality µ = Calculate: (ii) A 1 x:10 and Var(Z x:10 ). 1 11/44
12 Example 1 Suppose that δ = 0.04 and (x) has force of mortality µ = Calculate: (ii) A 1 x:10 and Var(Z x:10 ). 1 Solution: (ii) We have that A 1 x:10 = e 10δ 10p x = e (10)(0.04) e (10)(0.03) = e 0.7 = , 2 A 1 x:10 = e 10(2)δ 10p x = e (10)(2)(0.04) e (10)(0.03) = e 0.11 = , Var(Z x:10 ) 1 = 2 A 1 x:10 A 1 2 x:10 = ( ) 2 = /44
13 Example 1 Suppose that δ = 0.04 and (x) has force of mortality µ = Calculate: (iii) 10 A x and Var( 10 Z x ). 13/44
14 Example 1 Suppose that δ = 0.04 and (x) has force of mortality µ = Calculate: (iii) 10 A x and Var( 10 Z x ). Solution: (iii) We have that 10 A x = n E x A x+n = ( )( ) = , 2 10 A x = 2 ne x 2A x+n = ( )( ) = , Var( 10 Z x ) = ( ) 2 = /44
15 Example 1 Suppose that δ = 0.04 and (x) has force of mortality µ = Calculate: (iv) A 1 x:10 and Var(Z 1 x:10 ). 15/44
16 Example 1 Suppose that δ = 0.04 and (x) has force of mortality µ = Calculate: (iv) A 1 x:10 and Var(Z 1 x:10 ). Solution: (iv) We have that A 1 x:10 = A x 10 A x = = , 2 A 1 x:10 = 2 A x 2 10 A x = = , Var(Z 1 x:10 ) = 2 A 1 x:10 A 1 2 x:10 = ( ) 2 = /44
17 Example 1 Suppose that δ = 0.04 and (x) has force of mortality µ = Calculate: (v) A x:10 and Var(Z x:10 ). 17/44
18 Example 1 Suppose that δ = 0.04 and (x) has force of mortality µ = Calculate: (v) A x:10 and Var(Z x:10 ). Solution: (v) We have that A x:10 = A 1 x:10 + A 1 x:10 = = , 2 A x:10 = 2 A 1 x: A 1 x:10 = = , Var(Z 1 x:10 ) = 2 A 1 x:10 A 1 2 x:10 = ( ) 2 = /44
19 Theorem 4 Suppose that the survival function is s(x) = e µx, 0 x. Then, (i) A x = µ µ+δ. (ii) A 1 x:n = e n(µ+δ). (iii) n A x = e n(µ+δ µ µ+δ. (iv) A 1 x:n = (1 e n(µ+δ) ) µ µ+δ. (v) A x:n = (1 e n(µ+δ) ) µ µ+δ + e n(µ+δ). 19/44
20 Theorem 4 Suppose that the survival function is s(x) = e µx, 0 x. Then, (i) A x = µ µ+δ. (ii) A 1 x:n = e n(µ+δ). (iii) n A x = e n(µ+δ µ µ+δ. (iv) A 1 x:n = (1 e n(µ+δ) ) µ µ+δ. (v) A x:n = (1 e n(µ+δ) ) µ µ+δ + e n(µ+δ). Proof: (i) follows from a previous theorem. 20/44
21 Theorem 4 Suppose that the survival function is s(x) = e µx, 0 x. Then, (i) A x = µ µ+δ. (ii) A 1 x:n = e n(µ+δ). (iii) n A x = e n(µ+δ µ µ+δ. (iv) A 1 x:n = (1 e n(µ+δ) ) µ µ+δ. (v) A x:n = (1 e n(µ+δ) ) µ µ+δ + e n(µ+δ). Proof: (ii) A 1 x:n = e nδ np x = e n(µ+δ). 21/44
22 Theorem 4 Suppose that the survival function is s(x) = e µx, 0 x. Then, (i) A x = µ µ+δ. (ii) A 1 x:n = e n(µ+δ). (iii) n A x = e n(µ+δ µ µ+δ. (iv) A 1 x:n = (1 e n(µ+δ) ) µ µ+δ. (v) A x:n = (1 e n(µ+δ) ) µ µ+δ + e n(µ+δ). Proof: (iii) n A x = n E x A x+n = e n(µ+δ) µ µ + δ. 22/44
23 Theorem 4 Suppose that the survival function is s(x) = e µx, 0 x. Then, (i) A x = µ µ+δ. (ii) A 1 x:n = e n(µ+δ). (iii) n A x = e n(µ+δ µ µ+δ. (iv) A 1 x:n = (1 e n(µ+δ) ) µ µ+δ. (v) A x:n = (1 e n(µ+δ) ) µ µ+δ + e n(µ+δ). Proof: (iv) A 1 x:n = A x n A x = µ µ + δ µ e n(µ+δ) µ + δ =(1 e n(µ+δ) µ ) µ + δ. 23/44
24 Theorem 4 Suppose that the survival function is s(x) = e µx, 0 x. Then, (i) A x = µ µ+δ. (ii) A 1 x:n = e n(µ+δ). (iii) n A x = e n(µ+δ µ µ+δ. (iv) A 1 x:n = (1 e n(µ+δ) ) µ µ+δ. (v) A x:n = (1 e n(µ+δ) ) µ µ+δ + e n(µ+δ). Proof: (v) A x:n = A 1 x:n + A 1 x:n = (1 e n(µ+δ) µ ) µ + δ + e n(µ+δ). 24/44
25 Example 2 Find the APV of a 15-year term life insurance to (x) with unity payment if δ = 0.06 and (x) has a constant force of mortality µ = /44
26 Example 2 Find the APV of a 15-year term life insurance to (x) with unity payment if δ = 0.06 and (x) has a constant force of mortality µ = Solution: We have that A 1 x:15 = (0.05)(1 e (15)( ) ) = = (0.05)(1 e (15)(0.11) ) /44
27 Theorem 5 Assume de Moivre s law with terminal age ω and that ω and x are positive integers. Then, (i) A x = a ω x i (ii) A 1 x:n = e nδ ω x n (iii) n A x = e nδ a ω x n i ω x. (iv) A 1 x:n = a n i (v) A x:n = a n i ω x + e nδ ω x n 27/44
28 Theorem 5 Assume de Moivre s law with terminal age ω and that ω and x are positive integers. Then, (i) A x = a ω x i (ii) A 1 x:n = e nδ ω x n (iii) n A x = e nδ a ω x n i ω x. (iv) A 1 x:n = a n i (v) A x:n = a n i ω x n ω x + e nδ Proof: (i) It follows from a previous theorem. 28/44
29 Theorem 5 Assume de Moivre s law with terminal age ω and that ω and x are positive integers. Then, (i) A x = a ω x i (ii) A 1 x:n = e nδ ω x n (iii) n A x = e nδ a ω x n i ω x. (iv) A 1 x:n = a n i (v) A x:n = a n i ω x Proof: (ii) + e nδ ω x n A 1 x:n = e nδ np x = e nδ ω x n. ω x 29/44
30 Theorem 5 Assume de Moivre s law with terminal age ω and that ω and x are positive integers. Then, (i) A x = a ω x i (ii) A 1 x:n = e nδ ω x n (iii) n A x = e nδ a ω x n i ω x. (iv) A 1 x:n = a n i (v) A x:n = a n i ω x Proof: (iii) + e nδ ω x n n A x = n E x A x+n = e nδ ω x n ω x a ω x n i ω x n = e nδ a ω x n i ω x. 30/44
31 Theorem 5 Assume de Moivre s law with terminal age ω and that ω and x are positive integers. Then, (i) A x = a ω x i (ii) A 1 x:n = e nδ ω x n (iii) n A x = e nδ a ω x n i ω x. (iv) A 1 x:n = a n i (v) A x:n = a n i ω x n ω x + e nδ Proof: (iv) It follows from a previous theorem. 31/44
32 Theorem 5 Assume de Moivre s law with terminal age ω and that ω and x are positive integers. Then, (i) A x = a ω x i (ii) A 1 x:n = e nδ ω x n (iii) n A x = e nδ a ω x n i ω x. (iv) A 1 x:n = a n i (v) A x:n = a n i ω x Proof: (v) + e nδ ω x n A x:n = A 1 x:n + A 1 x:n = a n i ω x + ω x n e nδ ω x. 32/44
33 Example 3 Julia is 40 year old. She buys a 15-year term deferred life policy insurance which will pay $50,000 at the time of her death. Suppose that the survival function is s(x) = 1 x 100, 0 x 100. Suppose that the continuously compounded rate of interest is δ = /44
34 Example 3 Julia is 40 year old. She buys a 15-year term deferred life policy insurance which will pay $50,000 at the time of her death. Suppose that the survival function is s(x) = 1 x 100, 0 x 100. Suppose that the continuously compounded rate of interest is δ = (i) Find the actuarial present value of the benefit of this life insurance. 34/44
35 Example 3 Julia is 40 year old. She buys a 15-year term deferred life policy insurance which will pay $50,000 at the time of her death. Suppose that the survival function is s(x) = 1 x 100, 0 x 100. Suppose that the continuously compounded rate of interest is δ = (i) Find the actuarial present value of the benefit of this life insurance. Solution: (i) The density of T 40 is f T40 (t) = 1 60, 0 t 60. Hence, 15 A x = e (15)(0.05) a i (15)(0.05) 1 e (45)(0.05) = e 60 (0.05)(60) = e 0.75 e 3 = The actuarial present value is (50000)( ) = /44
36 Example 3 Julia is 40 year old. She buys a 15-year term deferred life policy insurance which will pay $50,000 at the time of her death. Suppose that the survival function is s(x) = 1 x 100, 0 x 100. Suppose that the continuously compounded rate of interest is δ = (ii) Find the standard deviation of the present value of the benefit of this life insurance. 36/44
37 Example 3 Julia is 40 year old. She buys a 15-year term deferred life policy insurance which will pay $50,000 at the time of her death. Suppose that the survival function is s(x) = 1 x 100, 0 x 100. Suppose that the continuously compounded rate of interest is δ = (ii) Find the standard deviation of the present value of the benefit of this life insurance. Solution: (ii) We have that 2 15 A x = e (15)(2)(0.05) a (1+i) = e 1.5 e 6 = , 6 (15)(0.05) 1 e (45)(2)(0.05) = e (2)(0.05)(60) Var( Z x ) = (50000) 2 ( ( ) 2 ) = /44
38 Example 3 Julia is 40 year old. She buys a 15-year term deferred life policy insurance which will pay $50,000 at the time of her death. Suppose that the survival function is s(x) = 1 x 100, 0 x 100. Suppose that the continuously compounded rate of interest is δ = (ii) Find the standard deviation of the present value of the benefit of this life insurance. Solution: (ii)the standard deviation of Z x is = /44
39 Example 4 Julia is 40 year old. She buys a 15 year endowment policy insurance which will pay $50,000 at the time of her death, or in 15 years whatever comes first. Suppose that the survival function is s(x) = 1 x 100, 0 x 100. Suppose that the continuously compounded rate of interest is δ = /44
40 Example 4 Julia is 40 year old. She buys a 15 year endowment policy insurance which will pay $50,000 at the time of her death, or in 15 years whatever comes first. Suppose that the survival function is s(x) = 1 x 100, 0 x 100. Suppose that the continuously compounded rate of interest is δ = (i) Find the actuarial present value of the benefit of this life insurance. 40/44
41 Example 4 Julia is 40 year old. She buys a 15 year endowment policy insurance which will pay $50,000 at the time of her death, or in 15 years whatever comes first. Suppose that the survival function is s(x) = 1 x 100, 0 x 100. Suppose that the continuously compounded rate of interest is δ = (i) Find the actuarial present value of the benefit of this life insurance. Solution: (i) The density of T 40 is f T40 (t) = 1 60, 0 t 60. Hence, A 40:15 = a 15 i e (15)(0.05) = 1 e (15)(0.05) (0.05)(60) e 60 = 1 e e 0.75 (0.75) = The actuarial present value is (50000)( ) = /44
42 Example 4 Julia is 40 year old. She buys a 15 year endowment policy insurance which will pay $50,000 at the time of her death, or in 15 years whatever comes first. Suppose that the survival function is s(x) = 1 x 100, 0 x 100. Suppose that the continuously compounded rate of interest is δ = (ii) Find the standard deviation of the present value of the benefit of this life insurance. 42/44
43 Example 4 Julia is 40 year old. She buys a 15 year endowment policy insurance which will pay $50,000 at the time of her death, or in 15 years whatever comes first. Suppose that the survival function is s(x) = 1 x 100, 0 x 100. Suppose that the continuously compounded rate of interest is δ = (ii) Find the standard deviation of the present value of the benefit of this life insurance. Solution: We have that 2 A 40:15 = a 15 (1+i) = 1 e (15)(2)(0.05) (2)(0.05)(60) = , + e (15)(2)(0.05) e 60 = 1 e e 1.5 (0.75) 6 Var(50000Z 40:15 ) = (50000) 2 ( ( ) 2 ) = /44
44 Example 4 Julia is 40 year old. She buys a 15 year endowment policy insurance which will pay $50,000 at the time of her death, or in 15 years whatever comes first. Suppose that the survival function is s(x) = 1 x 100, 0 x 100. Suppose that the continuously compounded rate of interest is δ = (ii) Find the standard deviation of the present value of the benefit of this life insurance. Solution: The standard deviation of 50000Z 40:15 is = /44
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