Manual for SOA Exam MLC.
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1 Chapter 5 Life annuities Extract from: Arcones Manual for the SOA Exam MLC Fall 2009 Edition available at 1/94
2 Due n year temporary annuity Definition 1 A due n year term annuity guarantees payments made at the beginning of the year while an individual is alive for at most n payments The actuarial present value of an n-year term annuity due is denoted by Ÿ x:n Definition 2 The actuarial present value of an n-year term annuity due for (x) with unit payment is denoted by ä x:n 2/94
3 Theorem 1 (i) Ÿ x:n = ä min(kx,n) = {äkx if K x n, ä n if K x > n, (ii) If i 0, Ÿ x:n = 1 Z x:n d (iii) If i = 0, Ÿx:n = min(k x, n) 3/94
4 Theorem 1 (i) Ÿ x:n = ä min(kx,n) = {äkx if K x n, ä n if K x > n, (ii) If i 0, Ÿ x:n = 1 Z x:n d (iii) If i = 0, Ÿx:n = min(k x, n) Proof: (i) If K x n, benefit payments are made at times 0, 1,, K x 1 and Ÿx:n = ä k If K x > n, benefit payments are made at times 0, 1,, n 1 and Ÿ x:n = ä n Therefore, Ÿ x:n = ä min(kx,n) 4/94
5 Theorem 1 (i) Ÿ x:n = ä min(kx,n) = {äkx if K x n, ä n if K x > n, (ii) If i 0, Ÿ x:n = 1 Z x:n d (iii) If i = 0, Ÿx:n = min(k x, n) Proof: (ii) If i 0, Ÿ x:n = ä min(kx,n) = 1 v min(kx,n) d = 1 Z x:n d 5/94
6 Theorem 1 (i) Ÿ x:n = ä min(kx,n) = {äkx if K x n, ä n if K x > n, (ii) If i 0, Ÿ x:n = 1 Z x:n d (iii) If i = 0, Ÿx:n = min(k x, n) Proof: (iii) If i = 0, Ÿx:n = ä min(kx,n) = min(k x, n) 6/94
7 Theorem 2 n n 1 ä x:n = ä k k 1 q x + ä n np x = ä k k 1 q x + ä n n 1 p x k=1 k=1 7/94
8 Theorem 2 ä x:n = n n 1 ä k k 1 q x + ä n np x = ä k k 1 q x + ä n n 1 p x k=1 {äkx Proof: Since Ÿ x:n = ä min(kx,n) = if K x n, ä n if K x > n, n n ä x:n = ä k P{K x = k}+ä n P{K x > n} = ä k k 1 q x +ä n np x k=1 k=1 k=1 Since Ÿ x:n = ä min(kx,n) = {äkx if K x < n, ä n if K x n, n 1 n 1 ä x:n = ä k P{K x = k}+ä n P{K x n} = ä k k 1 q x +ä n n 1 p x k=1 k=1 8/94
9 Theorem 3 If i 0, ä x:n = 1 A x:n d and Var(Ÿ x:n ) = 2 A x:n (A x:n ) 2 d 2 9/94
10 Example 1 Suppose that p x = 098, p x+1 = 097 and v = 092 (i) Find ä x:3 using Theorem 2 (ii) Find ä x:3 using Theorem 3 (iii) Find Var(Ÿ x:3 ) 10/94
11 Example 1 Suppose that p x = 098, p x+1 = 097 and v = 092 (i) Find ä x:3 using Theorem 2 (ii) Find ä x:3 using Theorem 3 (iii) Find Var(Ÿ x:3 ) Solution: (i) ä x:3 = (1)(002) + ( )(098)(003) + ( (092) 2 )(098)(097) = , 11/94
12 Example 1 Suppose that p x = 098, p x+1 = 097 and v = 092 (i) Find ä x:3 using Theorem 2 (ii) Find ä x:3 using Theorem 3 (iii) Find Var(Ÿ x:3 ) Solution: (ii) A x:3 = (092)(002) + (092) 2 (098)(003) + (092) 3 (098)(097) = ä x:3 = = /94
13 Example 1 Suppose that p x = 098, p x+1 = 097 and v = 092 (i) Find ä x:3 using Theorem 2 (ii) Find ä x:3 using Theorem 3 (iii) Find Var(Ÿ x:3 ) Solution: (iii) 2 A x:3 = (092) 2 (002) + (092) 4 (098)(003) + (092) 6 (098)(097) = Var(Z x:3 ) = ( ) 2 = Var(Ÿ x:3 ) = (008) 2 = /94
14 Example 2 Suppose that v = 091 and De Moivre s model with terminal age 100 Find ä 40:20 using Theorem 3 14/94
15 Example 2 Suppose that v = 091 and De Moivre s model with terminal age 100 Find ä 40:20 using Theorem 3 Solution: With i = 9/91, we have that A 40:20 = A 1 40:20 + A 1 40:20 = a v 20 20p 40 = a (091)20 = , ä 40:20 = 1 A 40:20 d = = /94
16 Theorem 4 If i 0, and Ÿ 2 x:n = 2Ÿx:n (2 d) 2Ÿ x:n d E[Ÿx:n 2 ] = 2ä x:n (2 d) 2ä x:n d 16/94
![x:n d E[Ÿx:n 2 ] = 2ä](/docs-images/48/18476578/images/page_16.jpg "x:n (2 d) 2ä x:n d")
17 Theorem 4 If i 0, and Ÿ 2 x:n = 2Ÿx:n (2 d) 2Ÿ x:n d E[Ÿx:n 2 ] = 2ä x:n (2 d) 2ä x:n d Proof: From Ÿx:n = 1 Z x:n d, we get that Z x:n = 1 dÿx:n Hence, Ÿ 2 x:n = 1 2Z x:n + 2 Z x:n d 2 = 1 2(1 dÿ x:n ) + 1 d(2 d) 2Ÿ x:n d 2 = 2Ÿ x:n (2 d) 2Ÿ x:n d 17/94
18 Theorem 5 and n 1 Ÿ x:n = k=0 Z 1 x:k n 1 ä x:n = v k kp x k=0 18/94
19 Theorem 5 and n 1 Ÿ x:n = k=0 Z 1 x:k n 1 ä x:n = v k kp x k=0 Proof: The payment at time k, where 0 k n 1, is made if T x > k Hence, n 1 n 1 Ÿ x:n = v k I (T x > k) = Z 1 x:k k=0 k=0 19/94
20 Example 3 Suppose that p x = 098, p x+1 = 097 and v = 092 Find ä x:3 using the previous theorem 20/94
21 Example 3 Suppose that p x = 098, p x+1 = 097 and v = 092 Find ä x:3 using the previous theorem Solution: ä x:3 = (1) + (092)(098) + (092) 2 (098)(097) = /94
22 Theorem 6 If i = 0, ä x:n = 1 + e x:n 1 22/94
23 Theorem 6 If i = 0, ä x:n = 1 + e x:n 1 Proof: ä x:n = E[min(K x, n)] = E[min(K(x) + 1, n)] =1 + E[min(K(x), n 1)] = 1 + e x:n 1 23/94
24 Theorem 7 Under De Moivre s model, ä x:n = (ω x n)ä n + (Dä) n ω x 24/94
25 Theorem 7 Under De Moivre s model, ä x:n = (ω x n)ä n + (Dä) n ω x Proof: We have that n 1 n 1 ä x:n = v k kp x = n 1 = k=0 k=0 v k ω x n ω x k=0 n 1 v k ω x k ω x n 1 = v k ω x n + n k ω x k=0 + v k n k ω x = (ω x n)ä n + (Dä) n ω x k=0 25/94
26 Example 4 Suppose that v = 091 and De Moivre s model with terminal age 100 Find ä 40:20 using previous theorem 26/94
27 Example 4 Suppose that v = 091 and De Moivre s model with terminal age 100 Find ä 40:20 using previous theorem Solution: With i = 9/91, we have that ä 20 = , (Dä) 20 = , ä 40:20 = (40)ä 20 + (Dä) 20 = 60 = (40)( ) /94
28 Theorem 8 We have that Hence, Ÿ x = Ÿ x:n + n Ÿ x ä x = ä x:n + n ä x = ä x:n + n E x ä x+n 28/94
29 Theorem 8 We have that Hence, Ÿ x = Ÿ x:n + n Ÿ x ä x = ä x:n + n ä x = ä x:n + n E x ä x+n Proof: Using some previous theorems, n 1 Ÿ x:n + n Ÿ x = v k I (K x > k) + v k I (K x > k) = v k I (K x > k) = Ÿ x k=0 k=0 k=n 29/94
30 Theorem 9 Under constant force of mortality, ä x:n = 1 v n p n x 1 vp x 30/94
31 Theorem 9 Under constant force of mortality, ä x:n = 1 v n p n x 1 vp x Proof: From ä x = ä x:n + n E x ä x+n = ä x:n + n E x ä x, we get ä x:n = (1 n E x )ä x = 1 v n p n x 1 vp x 31/94
32 Theorem 10 ä x:n = 1 + vp x ä x+1:n 1 32/94
33 Theorem 10 ä x:n = 1 + vp x ä x+1:n 1 Proof: n 1 n 1 n 1 ä x:n = v k kp x = 1 + v k kp x = 1 + vp x v k 1 k 1 p x+1 k=0 k=1 n 2 =1 + vp x v k kp x+1 = 1 + vp x ä x+1:n 1 k=0 k=1 33/94
34 For due annuities, we have that ä n = v n s n ä n is the present value of a due annuity s n is the accumulated value of a due annuity v n is the n year discount factor Definition 3 The actuarial accumulated value at time n of n year term temporary due annuity is defined by s x:n = äx:n ne x We have that ä x:n = n E x s x:n To take care that the number of living decreases over time, in actuarial computations, the n year discount factor is n E x = v n np x s x:n is the actuarial future value of a n year due life annuity to (x) 34/94
35 We have that s x:n = äx:n ne x n 1 = k=0 n 1 = äx:n k=0 v n = v k kp x np x v n np x 1 v n k n kp x+k = n 1 k=0 1 n ke x+k 1 Here, age x n ke x+k is the actuarial factor for time k to time n for a live 35/94
36 Immediate n year temporary annuity Definition 4 A immediate n year term annuity guarantees payments made at the end of the year while an individual is alive for n years The actuarial present value of an immediate n-year term annuity is denoted by Y x:n Definition 5 The actuarial present value of an n-year term life immediate annuity for (x) with unit payment is denoted by a x:n 36/94
37 Theorem 11 Y x:n = a min(kx 1,n) = { akx 1 if K x n, a n if K x > n, and n a x:n = a k 1 k 1 q x + a n np x k=2 37/94
38 Theorem 11 Y x:n = a min(kx 1,n) = { akx 1 if K x n, a n if K x > n, and n a x:n = a k 1 k 1 q x + a n np x k=2 Proof: If K x = 1, then no payment is made If 2 K x n + 1, then T x (K x 1, K x ] and payments are made at times 1,, K x 1 Hence, Y x:n = a k 1 = a Kx 1 If K x > n, then T x > n and the insurer makes payments at times 1,, n Hence, Y x:n = a n 38/94
39 Theorem 12 If i 0, and Y x:n = v Z x:n+1, a d x:n = v Z x:n+1 d Var(Y x:n ) = 2 A x:n+1 (A x:n+1 ) 2 d 2 39/94
40 Theorem 12 If i 0, and Y x:n = v Z x:n+1, a d x:n = v Z x:n+1 d Var(Y x:n ) = 2 A x:n+1 (A x:n+1 ) 2 d 2 Proof: Y x:n = a min(kx 1,n) min(kx v v,n+1) = d = 1 v min(kx 1,n) i = v Z x:n+1 d 40/94
41 Theorem 13 Y x:n = n k=1 Z 1 x:k and a x:n = n k=1 v k kp x 41/94
42 Theorem 13 Y x:n = n k=1 Z 1 and a x:k x:n = n k=1 v k kp x Proof: For 1 k n, a payment at the end of the k th year is received if and only if K x > k Hence, the present value of this payment is v k I (K x > k) = Z 1 x:k The total present value of the payments in an n-year term annuity is n k=1 Z 1 x:k 42/94
43 Theorem 14 Y x:n = Ÿ x:n+1 1, a x:n = ä x:n+1 1 and Var(Y x:n ) = Var(Ÿ x:n+1 ) 43/94
44 Theorem 14 Y x:n = Ÿ x:n+1 1, a x:n = ä x:n+1 1 and Var(Y x:n ) = Var(Ÿ x:n+1 ) Proof: We have that Ÿ x:n+1 1 = n k=0 Z 1 x:k 1 = n k=1 Z 1 x:k = Y x:n 44/94
45 Example 5 Suppose that p x = 098, p x+1 = 097, p x+2 = 096 and v = 092 (i) Find a x:3 using each of the three previous theorems (ii) Find Var(Y x:3 ) 45/94
46 Example 5 Suppose that p x = 098, p x+1 = 097, p x+2 = 096 and v = 092 (i) Find a x:3 using each of the three previous theorems (ii) Find Var(Y x:3 ) Solution: (i) Using the first theorem, a x:3 = 3 a Kx 1 P{K x = k} + a 3 P{K x > 3} k=2 =a 1 p x q x+1 + a 2 p x p x+1 q x+2 + a 3 p x p x+1 p x+2 =(092)(098)(003) + (092 + (092) 2 )(098)(097)(004) + (092 + (092) 2 + (092) 3 )(098)(097)(096) = /94
47 Example 5 Suppose that p x = 098, p x+1 = 097, p x+2 = 096 and v = 092 (i) Find a x:3 using each of the three previous theorems (ii) Find Var(Y x:3 ) Solution: Using the second theorem, A x:4 = (092)(002) + (092) 2 (098)(003) + (092) 3 (098)(097)(004) + (092) 4 (098)(097)(096) = , a x:3 = = /94
48 Example 5 Suppose that p x = 098, p x+1 = 097, p x+2 = 096 and v = 092 (i) Find a x:3 using each of the three previous theorems (ii) Find Var(Y x:3 ) Solution: Using the third theorem, a x:3 = 3 v k kp x k=1 =(092)(098) + (092) 2 (098)(097) + (092) 3 (098)(097)(096) = /94
49 Example 5 Suppose that p x = 098, p x+1 = 097, p x+2 = 096 and v = 092 (i) Find a x:3 using each of the three previous theorems (ii) Find Var(Y x:3 ) Solution: (ii) 2 A x:4 = (092) 2 (002) + (092) 4 (098)(003) + (092) 6 (098)(097)(004) + (092) 8 (098)(097)(096) = , Var(Y x:3 ) = ( )2 (008) 2 = /94
50 Theorem 15 and a x:n = ä x:n 1 + A 1 x:n Y x:n = Ÿ x:n 1 + Z 1 x:n 50/94
51 Theorem 15 and a x:n = ä x:n 1 + A 1 x:n Proof: We have that Ÿ x:n 1 + Z 1 Y x:n = Ÿ x:n 1 + Z 1 x:n x:n = n 1 k=0 Z 1 x:k 1 + Z 1 x:n = n k=1 Z 1 x:k = Y x:n 51/94
52 Theorem 16 We have that and Y x = n Y x + Y x:n a x = n a x + a x:n = n a x + n E x a x+n 52/94
53 Theorem 16 We have that and Y x = n Y x + Y x:n a x = n a x + a x:n = n a x + n E x a x+n Proof: We have that n Y x:n + n Y x = v k I (K x > k) + v k I (K x > k) = v k I (K x > k) = Y x k=1 k=1 k=n+1 53/94
54 Theorem 17 If i = 0, a x:n = e x:n 54/94
55 Theorem 17 If i = 0, a x:n = e x:n Proof: We have that a min(kx 1,n) = min(k x 1, n) = min(k(x), n) and a x:n = E[min(K(x), n)] = e x:n 55/94
56 Theorem 18 Under constant force of mortality, a x:n = (1 v n p n x )vp x 1 vp x 56/94
57 Theorem 18 Under constant force of mortality, a x:n = (1 v n p n x )vp x 1 vp x Proof: We have that a x:n = a x n E x a x+n = (1 n E x )a x = (1 v n p n x )vp x 1 vp x 57/94
58 Theorem 19 Under De Moivre s model, a x:n = (ω x n 1)a n + (Da) n ω x 58/94
59 Theorem 19 Under De Moivre s model, Proof: We have that = = a x:n = (ω x n 1)a n + (Da) n ω x a x:n = n k=1 n k=1 n v k kp x = k=1 n k=1 v k ω x k ω x v k ω x n 1 + n + 1 k ω x v k ω x n 1 ω x + n k=1 v k n + 1 k ω x = (ω x n 1)a n + (Da) n ω x 59/94
60 Example 6 Suppose that v = 091 and the De Moivre s model with terminal age 100 (i) Find a 40:20 using the previous theorem (ii) Find a 40:20 using that a 40:20 = v A 40:21 d 60/94
61 Example 6 Suppose that v = 091 and the De Moivre s model with terminal age 100 (i) Find a 40:20 using the previous theorem (ii) Find a 40:20 using that a 40:20 = v A 40:21 d Solution: Since i = (1 v)/v = 9/91, a 20 9/91 = , (Da) 20 9/91 = , (39)( ) a 40:20 = = (ii) We have that A 40:21 = a 21 9/91 60 a 40:20 = (091) = , = /94
62 Example 7 Suppose that p x = 098, p x+1 = 097 and v = 092 Find Var(Y x:3 ) 62/94
63 Example 7 Suppose that p x = 098, p x+1 = 097 and v = 092 Find Var(Y x:3 ) Solution: We have that A x:4 =(092)(002) + (092) 2 (098)(003) + (092) 3 (098)(097)(004) + (092) 4 (098)(097)(096) = , 2 A x:4 =(092) 2 (002) + (092) 4 (098)(003) + (092) 6 (098)(097)(004) + (092) 8 (098)(097)(096) = , Var(Y x:4 ) = ( )2 (008) 2 = /94
64 Definition 6 The actuarial accumulated value at time n of an n year term immediate annuity is s x:n = a x:n ne x s x:n is the actuarial future value of a n year immediate life insurance policy to (x) We have that s x:n = a x:n ne x = a n x:n k=1 v n = v k kp x np x v n = np x n k=1 1 v n k n kp x+k 64/94
65 Theorem 20 a x:n = vp x ä x+1:n = vp x (1 + a x+1:n 1 ) 65/94
66 Theorem 20 a x:n = vp x ä x+1:n = vp x (1 + a x+1:n 1 ) Proof: We have that n n n 1 a x:n = v k kp x = vp x v k 1 k 1p x+1 = vp x v k kp x+1 =vp x ä x+1:n k=1 k=1 k=0 66/94
67 Continuous n year temporary annuity Definition 7 A n year term continuous annuity guarantees a continuous flow of payments at a constant rate for n years while the individual is alive Definition 8 The present value of a n year term continuous annuity is denoted by Y x:n Definition 9 The actuarial present value of an n year term continuous annuity for (x) with unit payment is denoted by a x:n We have that a x:n = E[Y x:n ] 67/94
68 Recall that the the present value of the continuous annuity is a n i = n 0 v s ds = 1 v n δ Theorem 21 Y x:n = min(tx,n) 0 v t dt = a min(tx,n) = { atx if T x n, a n if T x > n, Hence, a x:n = n 0 a t tp x µ x+t dt + a n P{T x > n} 68/94
69 Recall that the the present value of the continuous annuity is a n i = n 0 v s ds = 1 v n δ Theorem 21 Y x:n = min(tx,n) 0 v t dt = a min(tx,n) = { atx if T x n, a n if T x > n, Hence, a x:n = n 0 a t tp x µ x+t dt + a n P{T x > n} Proof: Notice that the continuous payments are paid until min(t x, n) 69/94
70 Theorem 22 n a x:n = E[a min(tx,n) ] = v t tp x dt 0 70/94
71 Theorem 22 n a x:n = E[a min(tx,n) ] = v t tp x dt Proof: Let h(t) = v t I (t n) and let H(t) = t 0 h(s) ds = t 0 0 v s I (s n) ds = min(t,n) By a previous theorem, [ ] min(tx,n) a x:n = E[Y x:n ] = E v s ds = E[H(T x )] = 0 h(t) t p x dt = 0 0 v t I (t n) t p x dt = 0 n 0 v s ds v t tp x dt 71/94
72 Theorem 23 If δ 0, and Y x:n = 1 Z x:n, a δ x:n = 1 A x:n δ Var(Y x:n ) = 2 A x:n (A x:n ) 2 δ 2 72/94
73 Theorem 23 If δ 0, and Y x:n = 1 Z x:n, a δ x:n = 1 A x:n δ Var(Y x:n ) = 2 A x:n (A x:n ) 2 δ 2 Proof: Using that a t = 1 v t δ, we get that Y x:n = a min(tx,n) = 1 v min(tx,n) δ = 1 Z x:n δ 73/94
74 Theorem 24 If δ = 0, Y x:n = min(t x, n) and Y x:n = e x:n 74/94
75 Theorem 24 If δ = 0, Y x:n = min(t x, n) and Y x:n = e x:n Proof: If δ = 0, Y x:n = a min(tx,n) = min(t x, n) and Y x:n = E[min(T x, n)] = e x:n 75/94
76 Theorem 25 Under De Moivre s model, a x:n = (ω x n)a n + ( Da ) n ω x 76/94
77 Theorem 25 Under De Moivre s model, Proof: We have that = = a x:n = n 0 n 0 a x:n = (ω x n)a n + ( Da ) n ω x n 0 v t tp x dt = n v t ω x n + n t dt ω x v t ω x n n dt + ω x = (ω x n)a n + ( Da ) n ω x 0 0 v t ω x t ω x v t n t ω x dt dt 77/94
78 Example 8 Suppose that δ = 6% and deaths are uniformly distributed with terminal age 105 (i) Calculate a 65:20 using that a x:n = (ω x n)a n +(Da) n ω x (ii) Calculate a 65:20 using that a x:n = 1 A x:n δ 78/94
79 Example 8 Suppose that δ = 6% and deaths are uniformly distributed with terminal age 105 (i) Calculate a 65:20 using that a x:n = (ω x n)a n +(Da) n ω x (ii) Calculate a 65:20 using that a x:n = 1 A x:n δ Solution: (i) We have that a 20 = 1 e (20)(006) = , 006 ( Da )20 = 20 a 20 = , 006 a 65:20 = (ω x n)a n + ( Da ) n ω x (20)( ) = = /94
80 Example 8 Suppose that δ = 6% and deaths are uniformly distributed with terminal age 105 (i) Calculate a 65:20 using that a x:n = (ω x n)a n +(Da) n ω x (ii) Calculate a 65:20 using that a x:n = 1 A x:n δ Solution: (ii) We have that A 65:20 = a v p 65 = 1 e (20)(006) (20)(006) 20 + e (40)(006) 40 = , a 65:20 = 1 A 65:20 δ = (006) = /94
81 Theorem 26 E[Y 2 x:n ] = 2(a x:n 2 a x:n ) δ 81/94
82 Theorem 26 E[Y 2 x:n ] = 2(a x:n 2 a x:n ) δ Proof: From m a x:n = 1 m A x:n mδ, we get that m A x:n = 1 mδ ma x:n Hence, ) 2 [ E[Y 2 1 Z x:n ] = E ( x:n 1 2Z x:n + 2 Z x:n = E δ δ 2 [ 1 2(1 δax:n ) + 1 2δ 2a ] x:n =E δ 2 = 2(a x:n 2 a x:n ) δ ] 82/94
83 Theorem 27 Y x = Y x:n + n Y x and a x = a x:n + n a x = a x:n + n E x a x+n 83/94
84 Theorem 27 Y x = Y x:n + n Y x and a x = a x:n + n a x = a x:n + n E x a x+n Proof: Under a n year term annuity consists of continuously payments in the interval (0, n) until time of death Under a n year deferred annuity payments consists of continuously payments in the interval (n, ) until time of death Hence, Y x = Y x:n + n Y x 84/94
85 Taking n = 1, in previous theorem, we get the recurrence relation for a x : Theorem 28 a x = a x:1 + vp x a x+1 85/94
86 Example 9 Suppose that a x = 10, q x = 002 and δ = 007 Deaths are uniformly distributed within each year of age Find a x+1 86/94
87 Example 9 Suppose that a x = 10, q x = 002 and δ = 007 Deaths are uniformly distributed within each year of age Find a x+1 Solution: If deaths are uniformly distributed within each year of age, then l x+t = l x td x and t p x = 1 tq x Hence, 10 = 1 0 = 1 e = 1 e e 007t (1 (002)t) dt + e 007 (098)a x (002) 0 (007) 2 e t t dt + e 0007 (098)a x+1 (002)(1 e 007 ( )) (007) 2 + e 0007 (098)a x+1 = a x+1 and a x+1 = = /94
88 Theorem 29 Under constant force of mortality, a x:n = 1 e n(µ+δ) µ+δ 88/94
89 Theorem 29 Under constant force of mortality, a x:n = 1 e n(µ+δ) µ+δ Proof: a x:n = a x n a x = 1 µ + δ e n(µ+δ) µ + δ = 1 e n(µ+δ) µ + δ 89/94
90 Example 10 Suppose that δ = 008, and the force of mortality is µ x+t = 001, for t 0 Find a x:10 and Var(Y x:10 ) 90/94
91 Example 10 Suppose that δ = 008, and the force of mortality is µ x+t = 001, for t 0 Find a x:10 and Var(Y x:10 ) Solution: We have that a x:10 = 1 e (10)( ) We have that = 1 e A x:10 = (1 e (10)( ) )(001) = 1 e e 09 = , 2 A x:10 = (1 e (10)(001+(2)008) )(001) (001 + (2)008) = 1 e e 17 = , = e (10)( ) + e (10)(001+(2)008) 91/94
92 Var(Z x:10 ) 2 A x:10 A 2 x: ( )2 δ 2 = δ 2 = (008) 2 = /94
93 Theorem 30 a x:m+n = a x:n + m E x a x+m:n Proof: a x:m+n = m+n 0 v t tp x dt = m 0 v t tp x dt + m+n =a x:m + v m mp x v t m t m p x+m dt n =a x:m + m E x v t tp x+m dt m =a x:m + m E x a x+m:n 0 m+n m v t tp x dt 93/94
94 The actuarial accumulated value at time n of an n year term temporary continuous annuity is s x:n = a x:n = a n x:n 0 ne x v n = v t tp x dt np x v n np x n 1 n 1 = 0 v n t dt = dt n t p x+t 0 n te x+t 1 n te x+t is the actuarial factor from time t to time n for a live age x 94/94
Manual for SOA Exam MLC.
Chapter 5 Life annuities Extract from: Arcones Manual for the SOA Exam MLC Fall 2009 Edition available at http://wwwactexmadrivercom/ 1/70 Due n year deferred annuity Definition 1 A due n year deferred
Manual for SOA Exam MLC.
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