Manual for SOA Exam MLC.

Transcription

1 Chapter 5 Life annuities Extract from: Arcones Manual for the SOA Exam MLC Fall 2009 Edition available at 1/70

2 Due n year deferred annuity Definition 1 A due n year deferred annuity guarantees payments made at the beginning of the year while an individual is alive starting in n years The present value of a due n year deferred annuity is denoted by n Ÿ x Definition 2 The actuarial present value of a due n year deferred annuity for (x) with unit payment is denoted by n ä x We have that n ä x = E[ n Ÿx] 2/70

3 Theorem 1 n Ÿx = v n ä Kx n I (K x > n) = { 0 if Kx n, v n ä Kx n if K x > n, and n ä x = v n ä k n k 1 q x k=n+1 Proof: If K x n, then T x n and no payment is made If K x n + 1, then T x (K x 1, K x ] and unit payments at times n,, K x 1 are made The present value of a unit annuity paid at times n,, K x 1 is v n ä Kx n Hence, n Ÿ x = v n ä Kx n I (K x > n) and n ä x = v n ä k n P{K x = k} = v n ä k n k 1 q x k=n+1 k=n+1 3/70

4 Theorem 2 If i 0, and n ä x = A 1 x:n n A x d n Ÿ x = Z 1 x:n n Z x d = A x:n 1 (1 A x+n) d 4/70

5 Theorem 2 If i 0, and n ä x = A 1 x:n n A x d n Ÿ x = Z 1 x:n n Z x d = A x:n 1 (1 A x+n) d Proof: So, n Ÿx = v n Kx ä Kx n I (K n 1 v n x > n) = v I (K x > n) d = v n v Kx I (K x > n) = Z x:n 1 n Z x d d n ä x = A 1 x:n n A x d = A 1 x:n A 1 x:n A x+n d = A x:n 1 (1 A x+n) d 5/70

6 Example 1 Suppose that A 1 x:n = 03, A x+n = 06 and i = 005 Find n ä x 6/70

7 Example 1 Suppose that A 1 x:n = 03, A x+n = 06 and i = 005 Find n ä x Solution: We have that d = i 1+i = = 1 21 and n A x = A 1 x:n A x+n = (03)(06) = 018, n ä x = A 1 x:n n A x d = /21 = 252 7/70

8 Theorem 3 (current payment method) n Ÿ x = v k I (K x > k) = k=n k=n Z 1 x:k and n ä x = v k kp x = n E x ä x+n k=n 8/70

9 Theorem 3 (current payment method) n Ÿ x = v k I (K x > k) = and n ä x = k=n k=n Z 1 x:k v k kp x = n E x ä x+n k=n Proof: Given k n, a payment at time k is made if and only if the individual is alive at time k An individual is alive at time k if and only if T x > k Hence, n Ÿ x = v k I (T x > k) = k=n k=n Z 1 x:k 9/70

10 Theorem 3 (current payment method) n Ÿ x = v k I (K x > k) = k=n k=n Z 1 x:k and Proof: Thus, n ä x = v k kp x = n E x ä x+n k=n n ä x = v k kp x = v k+n k+n p x = v k+n np x kp x+n k=n k=0 k=0 =v n np x v k kp x+n = n E x ä x+n k=0 10/70

11 Theorem 4 If i = 0, n ä x = n p x (1 + e x+n ) Proof: n ä x = n E x ä x+n = n p x (1 + e x+n ) 11/70

12 Example 2 Suppose that A 1 x:n = 03, A x+n = 06 and i = 005 Find n ä x 12/70

13 Example 2 Suppose that A 1 x:n = 03, A x+n = 06 and i = 005 Find n ä x Solution: ä x+n = /21 = 84, n ä x = n E x ä x+n = (03)(84) = /70

14 Theorem 5 [ ( ) ] [ ] 2 2 E n Ÿ x = v n np x E (Ÿx+n) = v 2n np x (2ä x+n (2 d) 2ä x+n ) d Proof: Using that K x n K x > n and K x+n have the same distribution [ ( ) ] [ ( ] 2 2 E n Ÿ x = E v 2n ä Kx n ) I (Kx > n) [ ( ) 2 =v 2n np x E ä Kx n K x > n] [ ( ] 2 ] =v 2n np x E ä Kx+n ) = v 2n 2 np x E [Ÿ x+n = v 2n np x (2ä x+n (2 d) 2ä x+n ) d 14/70

15 Example 3 Suppose that v = 091 and p x+k = 097 for each integer k 0 Find 40 ä x and Var( 40 Ÿ x ) 15/70

16 Example 3 Suppose that v = 091 and p x+k = 097 for each integer k 0 Find 40 ä x and Var( 40 Ÿ x ) Solution: We have that ne x = (097) 40 (091) 40 = , 1 ä x+n = 1 (097)(091) = , 40 ä x = n E x ä x+n = ( )( ) = /70

17 Example 3 Suppose that v = 091 and p x+k = 097 for each integer k 0 Find 40 ä x and Var( 40 Ÿ x ) Solution: 2ä 1 x+n = 1 (097)(091) 2 = , [ ( ) ] 2 E 40 Ÿ x = (091)80 (097) 40 ((2)( ) (2 009)( )) 009 = = (091)80 (097) 40 ((2)( ) (2 009)( )) 009 = Var( 40 Ÿ x ) = ( ) 2 = /70

18 Theorem 6 Under De Moivre s model and integers x and ω, (i) n ä x = v n (Dä) ω x n ω x (ii) If i 0, n ä x = v n (ω x n ä ω x n ) (ω x)d (ii) If i = 0, n ä x = (ω x n)(ω x n+1) 2(ω x) 18/70

19 Theorem 6 Under De Moivre s model and integers x and ω, (i) n ä x = v n (Dä) ω x n ω x (ii) If i 0, n ä x = v n (ω x n ä ω x n ) (ω x)d (ii) If i = 0, n ä x = (ω x n)(ω x n+1) 2(ω x) Proof: (i) We have that n ä x = n E x ä x+n = v n ω x n ω x (Dä) ω x n ω x n = v n (Dä) ω x n ω x 19/70

20 Theorem 6 Under De Moivre s model and integers x and ω, (i) n ä x = v n (Dä) ω x n ω x (ii) If i 0, n ä x = v n (ω x n ä ω x n ) (ω x)d (ii) If i = 0, n ä x = (ω x n)(ω x n+1) 2(ω x) Proof: (ii) If i 0, (Dä) n = n ä n d So, n ä x = v n (Dä) ω x n ω x = v n (ω x n ä ω x n ) (ω x)d = v n ω x n ä ω x n ) d ω x 20/70

21 Theorem 6 Under De Moivre s model and integers x and ω, (i) n ä x = v n (Dä) ω x n ω x (ii) If i 0, n ä x = v n (ω x n ä ω x n ) (ω x)d (ii) If i = 0, n ä x = (ω x n)(ω x n+1) 2(ω x) Proof: (iii) If i = 0, (Dä) n = n(n+1) 2 So, n ä x = v n (Dä) ω x n = ω x (ω x n)(ω x n + 1) 2(ω x) = (ω x n)(ω x n+1) 2 ω x 21/70

22 Example 4 Suppose that v = 091 and De Moivre s model with terminal age 100 Find 20 ä 40 22/70

23 Example 4 Suppose that v = 091 and De Moivre s model with terminal age 100 Find 20 ä 40 Solution 1: We have that i = 1 v v = 9 91 Hence, (Dä) 40 9/91 = 40 ä 40 9/ ä 40 = (091)20 (Dä) 40 9/91 60 = , = (091)20 ( ) 60 = /70

24 Example 4 Suppose that v = 091 and De Moivre s model with terminal age 100 Find 20 ä 40 Solution 1: We have that i = 1 v v = 9 91 Hence, (Dä) 40 9/91 = 40 ä 40 9/ ä 40 = (091)20 (Dä) 40 9/91 60 Solution 2: We have that = , = (091)20 ( ) 60 = E 40 = v p 40 = (091) = , 60 A 60 = a 40 9/91 = , 40 ä 60 = 1 A 60 = = , d ä 40 = 20 E 40 ä 60 = ( )( ) = /70

25 Theorem 7 Under constant force of mortality µ, n ä x = v n p n x 1 vp x Proof = e n(δ+µ) 1 e (δ+µ) n ä x = n E x ä x = v n p n x 1 1 vp x = v n p n x 1 vp x = e n(δ+µ) 1 e (δ+µ) 25/70

26 Example 5 Suppose that v = 091 and the force of mortality is µ = 0005 Find 25 ä x 26/70

27 Example 5 Suppose that v = 091 and the force of mortality is µ = 0005 Find 25 ä x Solution: 25 ä x = (091)25 e (25)(0005) 1 (091)e 0005 = /70

28 Theorem 8 n ä x = vp x n 1 ä x+1 Proof: n ä x = v k kp x = vp x k=n =vp x n 1 ä x+1 k=n v k 1 k 1 p x+1 = vp x k=n 1 v k kp x+1 28/70

29 Example 6 Using i = 005 and a certain life table 10 ä 30 = 748 Suppose that an actuary revises this life table and changes p 30 from 095 to 096 Other values in the life table are unchanged Find 10 ä 30 using the revised life table 29/70

30 Example 6 Using i = 005 and a certain life table 10 ä 30 = 748 Suppose that an actuary revises this life table and changes p 30 from 095 to 096 Other values in the life table are unchanged Find 10 ä 30 using the revised life table Solution: We have that 10 ä 30 = vp x 9 ä 31 Hence, under the old table 9 ä 31 = (105)(748) = Since 9 ä 31 does not depend on p 30, using the revised life table 9 ä 31 = Hence, using the revised life table 10 ä 30 = (105) 1 (096)( ) = /70

31 n-year deferred annuity immediate Definition 3 An immediate n year deferred annuity guarantees payments made at the end of the year while an individual is alive starting n years from now The present value of an immediate n-year term annuity is denoted by n Y x Definition 4 The actuarial present value of an immediate n year deferred annuity for (x) with unit payment is denoted by n a x We have that n a x = E[ n Y x ] 31/70

32 Theorem 9 n Y x = v n a Kx n 1 I (K x > n + 1) = { 0 if Kx n + 1, v n a Kx n 1 if K x > n + 1, and n a x = v n a k n 1 k 1 q x k=n+2 32/70

33 Theorem 9 n Y x = v n a Kx n 1 I (K x > n + 1) = { 0 if Kx n + 1, v n a Kx n 1 if K x > n + 1, and n a x = v n a k n 1 k 1 q x k=n+2 Proof: If K x n + 1, then T x n and no payment is made If K x n + 2, then T x (K x 1, K x ] and unit payments at times n + 1,, K x 1 are made The present value of a unit annuity paid at times n + 1,, K x 1 is v n a Kx n 1 Hence, n Y x = v n a Kx n 1 I (K x > n + 1) = { 0 if Kx n + 1, v n a Kx n 1 if K x > n + 1, 33/70

34 Theorem 9 n Y x = v n a Kx n 1 I (K x > n + 1) = { 0 if Kx n + 1, v n a Kx n 1 if K x > n + 1, and n a x = v n a k n 1 k 1 q x k=n+2 Proof: and n a x = v n a k n 1 P{K x = k} = v n a k n 1 k 1 q x k=n+2 k=n+2 34/70

35 Theorem 10 n Y x = k=n+1 Z 1 x:k = n+1 Ÿx and n a x = k=n+1 v k kp x = n E x a x+n 35/70

36 Theorem 10 n Y x = k=n+1 Z 1 x:k = n+1 Ÿx and n a x = k=n+1 v k kp x = n E x a x+n Proof: Given k n + 1, a payment at time k is made if and only if the individual is alive at time k An individual is alive at time k if and only if T x > k Hence, n Y x = n a x = k=n+1 k=n+1 v k I (T x > k) = v k kp x = k=n+1 Z 1 x:k, v n+k n+k p x = k=1 =v n np x v k kp x+n = n E x a x+n k=1 v n v k np x kp x+n k=1 36/70

37 Theorem 11 If i = 0, n a x = n p x e x 37/70

38 Theorem 11 If i = 0, n a x = n p x e x Proof: We have that n a x = n E x a x+n = n p x e x 38/70

39 Theorem 12 Under constant force of mortality µ, n a x = e (n+1)(δ+µ) 1 e (δ+µ) 39/70

40 Theorem 12 Under constant force of mortality µ, n a x = e (n+1)(δ+µ) 1 e (δ+µ) Proof: We have that n a x = n+1 ä x = e (n+1)(δ+µ) 1 e (δ+µ) 40/70

41 Example 7 Suppose that v = 091 and p x = 097 for each x 0 Find 40 a x 41/70

42 Example 7 Suppose that v = 091 and p x = 097 for each x 0 Find 40 a x Solution: We have that 40 a x = (091)41 (097) 41 1 (091)(097) = /70

43 Theorem 13 Under De Moivre model and integers x and ω, n a x = v n (Da) ω x n 1 ω x Proof: We have that n a x = n ä x = v n+1 (Dä) ω x n 1 ω x = v n (Da) ω x n 1 ω x 43/70

44 Example 8 Suppose that v = 091 and the De Moivre model with terminal age 100 Find 20 a 40 44/70

45 Example 8 Suppose that v = 091 and the De Moivre model with terminal age 100 Find 20 a 40 Solution 1: We have that i = 1 v v = 9 91 Hence, 20 a 40 = (091)20 (Da) = /70

46 Example 8 Suppose that v = 091 and the De Moivre model with terminal age 100 Find 20 a 40 Solution 1: We have that i = 1 v v = 9 91 Hence, Solution 2: 20 a 40 = (091)20 (Da) = E 40 = v p 40 = (091) = , 60 A 60 = a 40 9/91 40 = , a 60 = v A 60 = = , d 20 a 40 = 20 E 40 a 60 = ( )( ) = /70

47 Theorem 14 E [ ( n Y x ) 2] = v n np x E [ Yx+n 2 ] v 2n np x (2a x+n (2 d) 2a x+n ) = d 47/70

48 Theorem 14 E [ ( n Y x ) 2] = v n np x E [ Yx+n 2 ] v 2n np x (2a x+n (2 d) 2a x+n ) = d Proof: Using that K x n K x > n has the distribution of K x+n, [ E ( n Y x ) 2] [ ( ] 2 = E v 2n a Kx n 1 ) I (Kx > n) [ ( ) 2 =v 2n np x E a Kx n 1 Kx > n] [ ( ] 2 =v 2n np x E a Kx+n 1 ) = v 2n np x E [ Yx+n 2 ] = v 2n np x (2a x+n (2 d) 2a x+n ) d 48/70

49 Theorem 15 n a x = vp x n 1 a x+1 49/70

50 Theorem 15 n a x = vp x n 1 a x+1 Proof: n a x = v k kp x = vp x =vp x k=n+1 k=n+1 k=n v k kp x+1 = vp x n 1 a x+1 v k 1 k 1 p x+1 50/70

51 n year deferred annuity continuous Definition 5 A n year deferred continuous annuity guarantees a continuous flow of payments while the individual is alive starting in n years Definition 6 The present value of an immediate n year term annuity is denoted by n Y x Definition 7 The actuarial present value of a whole life immediate annuity for (x) with unit payment is denoted by n a x We have that n a x = E[ n Y x ] 51/70

52 Theorem 16 = Tx n Y x = v s dsi (T x > n) = v n a Tx n I (T x > n) n { 0 if Tx n, v n a Tx n if T x > n 52/70

53 Theorem 16 = Tx n Y x = v s dsi (T x > n) = v n a Tx n I (T x > n) n { 0 if Tx n, v n a Tx n if T x > n Proof: If T x n, n Y x = 0 If T x > n, the unit continuous rate runs from n to T x and n Y x = Hence, n Y x = T x n Tx n v s ds = Tx n 0 v n+s ds = v n a Tx n v s dsi (T x > n) = v n a Tx n I (T x > n) 53/70

54 Theorem 17 n Y x = Z x:n 1 n Z x δ 54/70

55 Theorem 17 n Y x = Z x:n 1 n Z x δ Proof: We have that v n Tx a Tx n I (T n 1 v n x > n) = v I (T x > n) δ = e nδ e Tx δ I (T x > n) = Z x:n 1 n Z x δ δ 55/70

56 Theorem 18 n a x = n v n a t n tp x µ x+t dt Proof = n a x = E[v n a Tx n I (T x > n] = n v n a t n tp x µ x+t dt n v n a t n f Tx (t) dt 56/70

57 Theorem 19 (current payment method) Proof: Let n a x = h(t) = v t I (t > n) = n v t tp x dt { 0 if t n, v t if t > n, Then, H(t) = t 0 v s I (s > n) ds = t n v s dsi (t > n) = { 0 if t n, t n v s ds if t > n, By a previous theorem, n a x = E[H(T x )] = 0 h(t)s Tx (t) dt = n v t tp x dt 57/70

58 Theorem 20 n a x = v n np x a x+n = n E x a x+n Proof Using that T x t T x > t and T x+t have the same distribution, n a x = E[v n a Tx n I (T x > n)] = v n np x E[a Tx n T x > n] =v n np x E[a Tx+n ] = v n np x a x+n = n E x a x+n 58/70

59 Theorem 21 If i = 0, n a x = n p x ex+n Proof n a x = n E x a x+n = n p x ex+n 59/70

60 Theorem 22 [ ( ) ] 2 E n Y x = v 2n np x E[ ( ) 2] Y x+n = v 2n 2 ( a x+n 2 ) a x+n np x δ Proof: Using that T x t T x > t and T x+t have the same distribution, [ ( ) ] [ ( ] 2 2 E n Y x = E v 2n a Tx n ) I (Tx > n) [ ( ) 2 =v 2n np x E a Tx n T x > n)] [ ( ] 2 =Ev 2n np x E a Tx+n ) = v 2n np x E[ ( ) 2] Y x+n =v 2n 2 ( a x+n 2 ) a x+n np x δ 60/70

61 Theorem 23 Under De Moivre s model, (i) n a x = v n (Da) ω x n ω x (ii) If i 0, n a x = v n (ω x n a ω x n ) (ω x)δ (iii) If i = 0, n a x = (ω x n)2 2(ω x) 61/70

62 Theorem 23 Under De Moivre s model, (i) n a x = v n (Da) ω x n ω x (ii) If i 0, n a x = v n (ω x n a ω x n ) (ω x)δ (iii) If i = 0, n a x = (ω x n)2 2(ω x) Proof: (i) We have that n a x = n E x a x+n = v n ω x n ω x ( Da ) ω x n ω x n = v ( ) n Da ω x n ω x 62/70

63 Theorem 23 Under De Moivre s model, (i) n a x = v n (Da) ω x n ω x (ii) If i 0, n a x = v n (ω x n a ω x n ) (ω x)δ (iii) If i = 0, n a x = (ω x n)2 2(ω x) Proof: (ii) If i 0, ( Da ) n = n an δ So, n a x = v n ω x n a ω x n δ ω x = v n ( ω x n a ω x n ) (ω x)δ 63/70

64 Theorem 23 Under De Moivre s model, (i) n a x = v n (Da) ω x n ω x (ii) If i 0, n a x = v n (ω x n a ω x n ) (ω x)δ (iii) If i = 0, n a x = (ω x n)2 2(ω x) Proof: (iii) If i = 0, ( Da ) n = n2 2 So, n a x = v n (ω x n)2 2 ω x = (ω x n)2 2(ω x) 64/70

65 Example 9 Suppose that v = 091 and De Moivre s model with terminal age 100 Find 20 a 40 65/70

66 Example 9 Suppose that v = 091 and De Moivre s model with terminal age 100 Find 20 a 40 Solution 1: We have that a 40 = 1 (091)40 = , ln(091) ( ) v n ω x n a ω x n 20 a 40 = = (ω x)δ = (091)20 ( ) (60)( ln(091)) 66/70

67 Example 9 Suppose that v = 091 and De Moivre s model with terminal age 100 Find 20 a 40 Solution 2: We have that 20E 40 = v p 40 = (091) = , 60 A 60 = a = 1 (091)40 (40)( log(091)) = , a 60 = 1 A 60 δ = log(091) = , 20 a 40 = 20 E 40 a 60 = ( )( ) = /70

68 Theorem 24 Under constant force of mortality, n a x = e n(µ+δ) µ+δ Proof We have that n a x = n E x a x+n = e n(µ+δ) µ+δ 68/70

69 Example 10 Suppose that v = 092, and the force of mortality is µ x+t = 002, for t 0 Find 20 a x and Var( 20 Y x ) Solution: We have that a x = 1 log(092) = , 20 a x = (092) 20 e (20)(002) ( ) = , 2 1 a x = 2 log(092) = , E[ ( ) 2] n Y x = v 2n 2 ( a x+n 2 ) a x+n np x δ =(092) 40 (20)(002) 2 ( ) e = , log(092) Var( 20 Y x ) = ( ) 2 = /70

70 Theorem 25 n a x = vp x n 1 a x+1 Proof n a x = n v t tp x dt = vp x v t 1 t 1 p x+1 dt =vp x v t tp x+1 dt = vp x n 1 a x+1 n 1 n 70/70