Manual for SOA Exam MLC.

Transcription

1 Chapter 5. Life annuities. Section 5.7. Computing present values from a life table Extract from: Arcones Manual for the SOA Exam MLC. Spring 2010 Edition. available at 1/30

2 Recall that: Theorem 1 Assuming a uniform distribution of deaths, we have that: (i) A x = i A x. (ii) A 1 x:n = i A1 x:n. (iii) n A x = i n A x. (iv) A x:n = i A1 x:n + A 1 x:n. Theorem 2 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. 2/30

3 Whole life annuities Recall that: ä x = 1 A x, d a x = v A x, d a x = 1 A x, x a (m) x = 1 A(m) x, = v 1/m A (m) x. 3/30

4 Theorem 3 Under a uniform distribution of deaths within each year, x = 1 i a (m) i (m) A x = x = v 1/m i a x = 1 i A x i (m) A x = i (m) äx + i (m) i i (m), = 2 äx + i 2. i (m) a x + d i (m), 4/30

5 Theorem 3 Under a uniform distribution of deaths within each year, x = 1 i a (m) i (m) A x = x = v 1/m i a x = 1 i A x i (m) A x = i (m) äx + i (m) i i (m), = 2 äx + i 2. i (m) a x + d i (m), Proof: Using that ä x (m) we get that = 1 A(m) x, A (m) x = i A i (m) x and ä x = 1 Ax d, x = 1 A(m) x = 1 i A i (m) x = 1 i (1 dä i (m) x ) i i (m) = di i (m) äx + 1 = i (m) äx + i (m) i i (m). 5/30

6 Theorem 3 Under a uniform distribution of deaths within each year, x = 1 i a (m) i (m) A x = x = v 1/m i a x = 1 i A x Proof: Using that a x (m) we get that i (m) A x = i (m) äx + i (m) i i (m), = 2 äx + i 2. i (m) a x + d i (m), = v 1/m A (m) x, A (m) x = i A i (m) x and a x = v Ax d, a (m) x = v 1/m A (m) x = v 1/m i = v 1/m i i (m) (v da x ) = d i i (m) i (m) A x a x + v 1/m + v i i (m). 6/30

7 Theorem 3 Under a uniform distribution of deaths within each year, x = 1 i a (m) i (m) A x = x = v 1/m i a x = 1 i A x i (m) A x = i (m) äx + i (m) i i (m), = 2 äx + i 2. i (m) a x + d i (m), Proof: We have that So, v 1/m + v i i (m) a (m) x = = v 1/m i (m) vi i (m) = d i (m). i (m) a x + d i (m). 7/30

8 Theorem 3 Under a uniform distribution of deaths within each year, x = 1 i a (m) i (m) A x = x = v 1/m i a x = 1 i A x i (m) A x = i (m) äx + i (m) i i (m), = 2 äx + i 2. i (m) a x + d i (m), Proof: We know that a x = 1 Ax, A x = i A x and ä x = 1 Ax Hence, d. a x = 1 A x = 1 i A x = 1 i (1 dä x) = 2 äx + i 2. 8/30

9 Example 1 Conser the life table x l x Suppose that i = 6.5%. (i) Calculate ä (12) 80, a(12) 80 and a 80 using that A 80 = (ii) Calculate ä (12) 80, a(12) 80 and a 80 using that ä 80 = /30

10 Solution: (i) We have that i = 6.5%, d = = %, i (12) = 12(( ) ) = %, d (12) = 12(1 ( ) 1 12 ) = %. So, ä (12) 80 = (50000) 1 i = , A i (12) x d (12) = ( ) a (12) 80 = ä (12) 80 1 = = , a 80 = 1 i A x = ln(1.065) = ln(1.065) 10/30

11 Solution: (ii) ä (12) 80 = i (12) d (12) äx + i (12) i i (12) d (12) (0.065)( ) = ( )( ) ( ) ( )( ) = , a (12) 80 = ä (12) 80 1 = = , a 80 = 2 äx + i 2 = (0.065)( ) (ln(1.065)) 2 ( ) ln(1.065) (0.065) + (ln(1.065)) 2 = /30

12 Deferred annuities Recall that: n ä x = n E x ä x+n = n E x n A x, d n a x = n E x a x+n = n E x n A x, d n ä x (m) = n E x x+n = n E x n A (m) x, n a x (m) = n x 1 m n E x, 12/30

13 Theorem 4 Under a uniform distribution of deaths within each year, x = n E x i i n A (m) x = i (m) n ä x + i (m) i i (m) ne x, n n a x (m) = n x 1 m n E x = n a x = n E x i n A x i (m) n a x + d i (m) ne x, = 2 n ä x + i 2 n E x. 13/30

14 Proof: For the deferred life annuity due, using that n ä x (m) and n ä x = = n E x x+n, ä(m) n ä x (m) nex n Ax d x+n = 1 A(m) x+n, we get that, A (m) x = i i (m) A x, n E x A x+n = n A x = n E x x+n = 1 A (m) x+n 1 i ne x = n E x = n E x i i n A (m) x = n E x i ( i (m) n E x d n ä x ) = i (m) n ä x + i (m) i i (m) ne x. A i (m) x+n 14/30

15 Proof: For a deferred life annuity immediate, using that n a x (m) = n ä x (m) 1 m n E x, n ä x (m) = n E x x+n, ä(m) x A (m) x = i i (m) A x, we get that n a x (m) = n x = n E x 1 i i (m) A x+n = v 1/m ne x i i (m) n A x and n a x (m) = 1 A(m) x, 1 m n E x = n E x x+n 1 1 A (m) m n E x = n E x 1 m n E x 1 m n E x = n E x (1 d(m) m ) ne x x+n i A i (m) x+n ( ) = n E x a (m) x+n = ne x i (m) a x + d i (m) = i (m) n a x + d i (m) ne x 15/30

16 Proof: For a deferred continuous life annuity, using that n a x = n E x a x+n, a x = 1 Ax A 1 x:n = i A1 x:n, ne x A x+n = n A x and n ä x = nex n Ax d, we get that n a x = n E x a x+n = n E x 1 A x+n = n E x i ( ne x d n ä x ) = n E x 1 i A x+n = 2 n ä x + i 2 ne x. = n E x i n A x 16/30

17 Temporary annuities Recall that ä x:n = 1 A x:n, d a x:n = 1 A x:n, x:n = 1 A(m) x:n, a (m) x:n = ä(m) x:n 1 m + 1 m n E x, 17/30

18 Theorem 5 Under a uniform distribution of deaths within each year, i i (m) A 1 x:n x:n = 1 ne x = a (m) x:n = i (m) a x:n + d i (m) (1 ne x ), a x:n = 1 ne x i A1 x:n i (m) äx:n + i (m) i i (m) (1 ne x ), = 2 äx:n + i 2 (1 ne x ). 18/30

19 Proof: For a n year term life annuity due, using that x:n = 1 A (m) x:n we get that x:n = 1 A(m) x:n, A x:n = A 1 x:n + ne x, A (m) x = i i (m) A x, ä x:n = 1 A x:n d, = 1 A1 (m) x:n n E x = 1 ne x i i (m) ( 1 däx:n n E x ) = i (m) äx:n + i (m) i i (m) (1 ne x ) = 1 i ne x i (m) A 1 x:n 19/30

20 Proof: For a n year term life annuity immediate, using that a (m) x:n = ä(m) x:n 1 m + 1 m n E x, A x (m) = i A i (m) x, ä x:n = 1 A x:n d a (m) x:n = ä(m) x:n 1 m + 1 m n E x x:n = 1 A (m) x:n, A x:n = A 1 x:n + ne x,,, we get that = i (m) äx:n + i (m) i i (m) (1 ne x ) 1 m + 1 m n E x = i (m) (1 + a x:n n E x ) + i (m) i i (m) (1 ne x ) 1 m (1 ne x ) = i (m) a x:n + (1 n E x ) ( i (m) + i (m) i i (m) 1 m ). 20/30

21 Proof: We have that Hence, i (m) + i (m) i ( i (m) 1 d(m) m = i (m) 1 m = + i (m) i + i (m)d(m) m i (m) i (m) = i (m) v 1/m iv i (m) a (m) x:n = ) i(1 d) = d i (m). i (m) a x:n + d i (m) (1 ne x ). 21/30

22 For a n year term life continuous annuity, using that a x:n = 1 A x:n, A x:n = A 1 x:n + n E x, A 1 x:n = i A1 x:n, ä x:n = 1 A x:n d, we get that a x:n = 1 A x:n = 1 A1 x:n n E x ( ) 1 däx:n n E x = 1 ne x i = 1 ne x i A1 x:n = 2 äx:n + i 2 (1 ne x ). 22/30

23 Linear interpolation of the actuarial discount factor. Another way to interpolate is to assume that actuarial discount factor is linear. If we assume k+ j E x = v k+ j m m k+ j p x is linear in j, m then k+ j m E x = k E x j m ( k+1e x k E x ), j = 0, 1,..., m 1. The actuarial discount factor appears in annuities computations. We have that x = 1 m k=0 v k m k p x = 1 m m m 1 k=0 j=0 v k+ j m k+ j m p x = 1 m m 1 k=0 j=0 k+ j m E x. 23/30

24 Theorem 6 Assuming that k+ j E x is linear in j, then m x = ä x m 1 2m, x:n = ä x:n m 1 2m (1 ne x ), n x = n ä x m 1 2m ne x. 24/30

25 Proof: Using that m 1 j=0 j = (m 1)m 2, we have that x = 1 m = 1 m ( k=0 = m + 1 2m = m + 1 2m = m + 1 2m m 1 k=0 j=0 ke x m 1 2 ke x + m 1 2m k=0 ke x + m 1 2m k=0 ke x + m 1 2m k=0 m 1 k+ j E x = 1 m m ) ( k+1 E x k E x ) k+1e x k=0 ke x k=1 k=0 j=0 ( ke x j ) m ( k+1e x k E x ) ke x m 1 2m = ä x m 1 2m. k=0 25/30

26 We have that x:n = 1 m = 1 m n 1 = m + 1 2m = m + 1 2m k=0 n 1 n 1 k=0 j=0 Chapter 5. Life annuities. m 1 k+ j m E x = 1 m n 1 m 1 k=0 j=0 ( v k ke x m 1 ( k+1 E x k E x ) 2 k=0 n 1 + m 1 2m k=0 n 1 ke x + m 1 2m ke x + m 1 2m k=0 =ä x:n m 1 2m (1 ne x ). n 1 k+1e x k=0 ( v k ke x j ) m ( k+1e x k E x ) ) n ke x = m + 1 2m k=1 ke x m 1 2m + m 1 2m n E x n 1 ke x k=0 26/30

27 Using that ä x (m) = x:n + n ä x (m) the last formula. and ä x = ä x:n + n ä x, we can get 27/30

28 Example 2 Conser the life table x l x Suppose that i = 6.5%. Calculate ä (12) 80 assuming that the actuarial discount factor is linear. 28/30

29 Example 2 Conser the life table x l x Suppose that i = 6.5%. Calculate ä (12) 80 assuming that the actuarial discount factor is linear. Solution: Using that ä 80 = , we get that ä (12) 80 = (2)(12) = /30

30 In the continuous case, we assume k+t E x = v k+t k+t p x is linear in t, 0 t 1. In this case k+te x = k E x t( k+1 E x k E x ), 0 t 1. Letting m in the previous theorem, we get that: Theorem 7 Assuming that k+t E x is linear in t, 0 t 1, a x = ä x 1 2, a x:n = ä x:n 1 2 (1 ne x ), n a x = n ä x 1 2 ne x. 30/30