Manual for SOA Exam MLC.
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1 Chapter 4. Life Insurance. Section 4.9. Computing APV s from a life table. Extract from: Arcones Manual for the SOA Exam MLC. Fall 2009 Edition. available at 1/29
2 Computing APV s from a life table Usually, survival functions do not have an analytical form. In this section, we will consider the computation of some of the present values of life insurance products using a life table. Suppose that l x is known whenever x is a nonnegative integer. Then, we can find np x, whenever x and n are integers. Assuming a uniform distribution of deaths over each year of death, K(x) and S x = T x K(x) are independent r.v. s and S x has a distribution uniform on (0, 1). Hence, K x and S x are independent r.v. s and T x = K(x) + S x = K x + S x 1. 2/29
3 Theorem 1 Assume a uniform distribution of deaths over each year of death. Suppose that b t, t 0, is constant in each interval (k 1, k], k = 1, 2,.... Then, E[b Tx ν Tx ] = E[b Kx ν Kx ] i δ. 3/29
4 Theorem 1 Assume a uniform distribution of deaths over each year of death. Suppose that b t, t 0, is constant in each interval (k 1, k], k = 1, 2,.... Then, E[b Tx ν Tx ] = E[b Kx ν Kx ] i δ. Proof: Since S x has a distribution uniform on (0, 1), so has 1 S x. So, E[ν Sx 1 ] = E[e δ(1 Sx ) ] = 1 0 e δt dt = eδ 1 δ = i δ. Hence, E[b Tx ν Tx ] = E[b Kx ν Tx ] = E[b Kx ν Kx +Sx 1 ] =E[b Kx ν Kx ]E[ν Sx 1 ] = E[b Kx ν Kx ] i δ 4/29
5 Theorem 1 Assume a uniform distribution of deaths over each year of death. Suppose that b t, t 0, is constant in each interval (k 1, k], k = 1, 2,.... Then, E[b Tx ν Tx ] = E[b Kx ν Kx ] i δ. The factor i δ has a life table interpretation. Suppose that a death happens during the k th year. Under the discrete life insurance a benefit payment of one is made at time k. Under the continuous life insurance and an uniform distribution of deaths, this death can happen uniformly on the interval [k 1, k]. So, for each dollar paid at time k in the discrete case, we get a continuous cashflow of a unit rate over the interval [k 1, k] in the continuous case. The present value at time k of this continuous cashflow is s 1 i = (1 + i)1 1 = i δ δ. 5/29
6 Theorem 2 Assuming a uniform distribution of deaths, we have that: (i) A x = i δ A x. (ii) A 1 x:n = i δ A1 x:n. (iii) n A x = i δ n A x. (iv) A x:n = i δ A1 x:n + A x:n 1. Proof: (i) We apply Theorem 1 with b t = 1, t 0. We have that A x = E[ν Tx ] = E[b Tx ν Tx ] and A x = E[ν Kx ] = E[b Kx ν Kx ]. (ii) We apply Theorem 1 with b t = I (t n), t 0. We have that A 1 x:n = E[I (T x n)ν Tx ] = E[b Tx ν Tx ] and A 1 x:n = E[I (K x n)ν Kx ] = E[b Kx ν Kx ]. (iii) We apply Theorem 1 with b t = I (t > n), t 0. (iv) A x:n = A 1 x:n + A 1 x:n = i δ A1 x:n + A x:n 1. 6/29
7 Example 1 Assuming a uniform distribution of deaths, i = 6% and the life table in the manual, find: A 50, 15 E 50, 15 A 50, A 1 50:15 and A 50:15. 7/29
8 Example 1 Assuming a uniform distribution of deaths, i = 6% and the life table in the manual, find: A 50, 15 E 50, 15 A 50, A 1 50:15 and A 50:15. Solution: We have that A 50 = i δ A 50 = = , ln(1.06) 15E 50 = ν 15 15p 50 = (1.06) 15 l = (1.06) l = , A 65 = i δ A 65 = = , ln(1.06) 15 A 50 = 15 E 50 A 65 = ( )( ) = , A 1 50:15 = A A 50 = = , A 50:15 = A 1 50: E 50 = = /29
9 Example 2 Suppose that i = 0.05, q x = 0.03 and q x+1 = Assuming a uniform distribution of deaths, find A 1 x:2 and Var(Z 1 x:2 ). 9/29
10 Example 2 Suppose that i = 0.05, q x = 0.03 and q x+1 = Assuming a uniform distribution of deaths, find A 1 x:2 and Var(Z 1 x:2 ). Solution: We have that A 1 x:2 = νq x + ν 2 p x q x+1 = (1.05) 1 (0.03) + (1.05) 2 (0.97)(0.04) = , A 1 x:2 = i δ A1 x:2 = = , ln(1.05) 2 A 1 x:2 = ν2 q x + ν 4 p x q x+1 = (1.05) 2 (0.03) + (1.05) 4 (0.97)(0.04) = , 2 A 1 ((1 + i) 2 1) 2 A 1 x:2 x:2 = 2δ = ((1.05)2 1)( ) 2 ln(1.05) = Var(Z 1 x:2 ) = 2 A 1 x:2 (A 1 x:2 )2 = ( ) 2 = /29
11 Example 3 (i) A 80. (ii) A 1 80:3. (iii) 3 A 80. (iv) A 1 80:3. (v) A 80:3. (vi) A 80. (vii) A 1 80:3. (viii) 3 A 80. (ix) A 1 80:3. (x) A 80:3. 11/29
12 Example 3 (i) A 80. (ii) A 1 80:3. (iii) 3 A 80. (iv) A 1 80:3. (v) A 80:3. (vi) A 80. (vii) A 1 80:3. (viii) 3 A 80. (ix) A 1 80:3. (x) A 80:3. Solution: (i) A 80 = k=1 νk l 80+k 1 l 80+k l (1.065) 250 is (1.065) (1.065) (1.065) + (1.065) + (1.065) = = /29
13 Example 3 (i) A 80. (ii) A 1 80:3. (iii) 3 A 80. (iv) A 1 80:3. (v) A 80:3. (vi) A 80. (vii) A 1 80:3. (viii) 3 A 80. (ix) A 1 80:3. (x) A 80:3. Solution: (ii) A 1 80:3 = l ν = (1.065) l = /29
14 Example 3 (i) A 80. (ii) A 1 80:3. (iii) 3 A 80. (iv) A 1 80:3. (v) A 80:3. (vi) A 80. (vii) A 1 80:3. (viii) 3 A 80. (ix) A 1 80:3. (x) A 80:3. Solution: (iii) 3 A 80 = k=4 ν k l 80+k 1 l 80+k l =(1.065) + (50000)(1.065) + (1.065) = = /29
15 Example 3 (i) A 80. (ii) A 1 80:3. (iii) 3 A 80. (iv) A 1 80:3. (v) A 80:3. (vi) A 80. (vii) A 1 80:3. (viii) 3 A 80. (ix) A 1 80:3. (x) A 80:3. Solution: (iv) A 1 80:3 = A 80 3 A 80 = = /29
16 Example 3 (i) A 80. (ii) A 1 80:3. (iii) 3 A 80. (iv) A 1 80:3. (v) A 80:3. (vi) A 80. (vii) A 1 80:3. (viii) 3 A 80. (ix) A 1 80:3. (x) A 80:3. Solution: (v) A 80:3 = A 1 80:3 + A 1 = :3 = /29
17 Example 3 (i) A 80. (ii) A 1 80:3. (iii) 3 A 80. (iv) A 1 80:3. (v) A 80:3. (vi) A 80. (vii) A 1 80:3. (viii) 3 A 80. (ix) A 1 80:3. (x) A 80:3. Solution: (vi) A 80 = i δ A 80 = ( ) = ln(1.065) 17/29
18 Example 3 (i) A 80. (ii) A 1 80:3. (iii) 3 A 80. (iv) A 1 80:3. (v) A 80:3. (vi) A 80. (vii) A 1 80:3. (viii) 3 A 80. (ix) A 1 80:3. (x) A 80:3. Solution: (vii) A 1 80:3 = A 1 80:3 = /29
19 Example 3 (i) A 80. (ii) A 1 80:3. (iii) 3 A 80. (iv) A 1 80:3. (v) A 80:3. (vi) A 80. (vii) A 1 80:3. (viii) 3 A 80. (ix) A 1 80:3. (x) A 80:3. Solution: (viii) 3 A 80 = i δ 3 A 80 = ( ) = ln(1.065) 19/29
20 Example 3 (i) A 80. (ii) A 1 80:3. (iii) 3 A 80. (iv) A 1 80:3. (v) A 80:3. (vi) A 80. (vii) A 1 80:3. (viii) 3 A 80. (ix) A 1 80:3. (x) A 80:3. Solution: (ix) A 1 80:3 = i δ A1 80:3 = ( ) = ln(1.065) 20/29
21 Example 3 (i) A 80. (ii) A 1 80:3. (iii) 3 A 80. (iv) A 1 80:3. (v) A 80:3. (vi) A 80. (vii) A 1 80:3. (viii) 3 A 80. (ix) A 1 80:3. (x) A 80:3. Solution: (x) A 80:3 = A 1 80:3 +A 1 80:3 = = /29
22 Theorem 3 Assuming a uniform distribution of deaths, we have that: (i) (I A) x = i δ (IA) x. (ii) (IA) x = i δ (IA) x + i ( 1 δ δ 1 ) d Ax. 22/29
23 To calculate actuarial present values for life insurance paid at the end of the m thly interval of interval, we also need to interpolate life tables. Theorem 4 Assuming a uniform distribution of deaths, we have that: (i) A (m) x = i A i (m) x. (m) (ii) A1 x:n = (iii) n A (m) x = i (iv) A (m) x:n = i i (m) A 1 x:n. i n A (m) x. i A 1 i (m) x:n + A x:n 1. 23/29
24 i The factor has a life table interpretation. Suppose that i (m) l x+k l x+k+1 deaths happens during the k th year and (l x+k l x+k+1 ) is multiple of m. Under the discrete annual life insurance the total benefit payment made at time k is l x+k l x+k+1. Under a uniform distribution of deaths, l x+k l x+k+1 [ m ] deaths happen in each of the intervals k 1 + j 1 m, k 1 + j m, j = 1,..., m. The cashflow of payments under the m th ly plan is l x+k l x+k+1 l x+k l x+k+1 l x+k l x+k+1 l x+k l x+k+1 m Payments m m m Time k m k m k m k 1 + m m The present value at time k of this cashflow is (l x+k l x+k+1 )s (m) 1 i = (l x+k l x+k+1 ) (1 + i)1 1 i (m) m We get that the present value under the m th ly plan is s (m) 1 i times the present value under the annual plan. = (l x+k l x+k+1 ) i i (m) = i i (m) 24/29
25 Example 4 (i) A (12) 80. (ii) 3 A (12) (12) 80. (iii) A 1. (iv) A(12) x:n 80:3. 25/29
26 Example 4 (i) A (12) 80. (ii) 3 A (12) (12) 80. (iii) A 1 x:n. (iv) A(12) Solution: (i) A (12) 80 = i i (12) A 80 = = : ((1.065) 1/12 1) ( ) 26/29
27 Example 4 (i) A (12) 80. (ii) 3 A (12) (12) 80. (iii) A 1 x:n. (iv) A(12) Solution: (ii) 3 A (12) 80 = i i (12) 3 A 80 = = : ((1.065) 1/12 1) ( ) 27/29
28 Example 4 (i) A (12) 80. (ii) 3 A (12) (12) 80. (iii) A 1 x:n. (iv) A(12) Solution: (iii) 80:3. (12) A1 x:n = A(12) 80 3 A (12) 80 = = /29
29 Example 4 (i) A (12) 80. (ii) 3 A (12) (12) 80. (iii) A 1 x:n. (iv) A(12) Solution: (iv) A (12) 80:3. (12) 80:3 = A 1 x:n + A 1 = :3 = /29
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