Manual for SOA Exam MLC.

Transcription

1 Chapter 5. Life annuities. Extract from: Arcones Manual for the SOA Exam MLC. Spring 2010 Edition. available at 1/114

2 Whole life annuity A whole life annuity is a series of payments made while an individual is alive. The payments can be made either at the beginning of the year, either at the end of the year, or continuously. 2/114

3 Whole life due annuity Definition 1 A whole life due annuity is a series payments made at the beginning of the year while an individual is alive. Definition 2 The present value of a whole life due annuity for (x) with unit payment is denoted by Ÿ x. Ÿ x is a random variable that depends on T x. Recall that K x is the interval of death of (x), i.e. K x is the positive integer such that the death of (x) takes place in the interval (K x 1, K x ]. Definition 3 The actuarial present value of a whole life due annuity for (x) with unit payment is denoted by ä x. We have that ä x = E[Ÿ x ]. 3/114

4 Theorem 1 (i) Ÿ x = ä Kx = K x 1 k=0 v k. (ii) If i 0, Ÿ x = 1 v Kx = 1 Z x, d d 1 v k ä x = k 1 q x = 1 A x d d k=1 and Var(Ÿ x ) = Var(Z x) 2 A x A 2 x d 2 = d 2, where Z x is the present value of a unit life insurance paid at the end of the year of death. (iii) If i = 0, Ÿ x = K x = K(x) + 1, ä x = e x + 1 and Var(Ÿ x ) = Var(K(x)). 4/114

5 Proof: (i) Since death happens in the interval (K x 1, K x ], payments of one are made at times 0, 1, 2,..., K x 1, i.e. the cashflow of payments is a K x year annuity due. Hence, Ÿ x = ä Kx = K x 1 k=0 v k. (ii) If i > 0, Therefore, Ÿ x = ä Kx = 1 v Kx ä x = E[ä Kx ] = ä k P{K x = k} = k=1 [ ] 1 Zx ä x = E = 1 A x, d d ( ) 1 Zx Var(Y x ) = Var d d = 1 Z x. d 1 v k k 1 q x, d k=1 = Var(Z x) 2 A x A 2 x d 2 = d 2. (iii) If i = 0, Ÿ x = ä Kx = K x = K(x) + 1, ä x = e x + 1 and Var(Ÿ x ) = Var(K(x)). 5/114

6 Example 1 i = 5%. K x has probability mass function: k P{K x = k} Find ä x and Var(Ÿ x ). 6/114

7 Example 1 i = 5%. K x has probability mass function: k P{K x = k} Find ä x and Var(Ÿ x ). Solution: We have that ä 1 = 1, ä 2 = and ä 3 = The probability mass of Ÿ x is given by Hence, k P{Ÿ x = k} ä x = (1)(0.2) + ( )(0.3) + ( )(0.5) = E[Ÿ 2 x ] = (1) 2 (0.2) + ( ) 2 (0.3) + ( ) 2 (0.5) = , Var(Ÿx) = ( ) 2 = /114

8 Example 2 Suppose that p x+k = 0.97, for each integer k 0, and i = 6.5%. Find ä x and Var(Ÿx). 8/114

9 Example 2 Suppose that p x+k = 0.97, for each integer k 0, and i = 6.5%. Find ä x and Var(Ÿx). Solution: Recall that if p k = p, for each k 1, then A x = Hence, A x = q x q x + i ä x = 1 A x d 2 A x = Var(Ÿ x ) = = = , = /1.065 = , qx q x +i. q x q x + i(2 + i) = (0.065)( ) = , 2 A x A 2 x ( )2 d 2 = (0.065/1.065) 2 = /114

10 Example 3 Assuming i = 6% and the life table in the manual, find ä /114

11 Example 3 Assuming i = 6% and the life table in the manual, find ä 45. Solution: We have that ä 45 = 1 A 45 d = /(1.06) = /114

12 Example 4 John, age 65, has $750,000 in his retirement account. An insurance company offers a whole life due annuity to John which pays $P at the beginning of the year while (65) is alive for $750,000. The annuity is priced assuming that i = 6% and the life table in the manual. The insurance company charges John 30% more of the APV of the annuity. Calculate P. 12/114

13 Example 4 John, age 65, has $750,000 in his retirement account. An insurance company offers a whole life due annuity to John which pays $P at the beginning of the year while (65) is alive for $750,000. The annuity is priced assuming that i = 6% and the life table in the manual. The insurance company charges John 30% more of the APV of the annuity. Calculate P. Solution: We have that ä 65 = 1 A 65 d = /(1.06) = The APV of this is this annuity is Pä 65 = ( )P. We have that = (1.3)( )P and P = (1.3)( ) = /114

14 Theorem 2 If i 0, and Ÿ 2 x = 2Ÿx (2 d) 2Ÿ x d E[Ÿx 2 ] = 2ä x (2 d) 2ä x. d Proof. From Ÿx = 1 Zx d, we get that Z x = 1 dÿx. Hence, Ÿ 2 x = 1 2Z x + 2 Z x d 2 = 2Ÿ x (2 d) 2Ÿ x. d = 1 2(1 dÿ x ) + 1 d(2 d) 2Ÿ x d 2 14/114

15 Example 5 Suppose that i = 0.075, ä x = 8.6 and 2 ä x = /114

16 Example 5 Suppose that i = 0.075, ä x = 8.6 and 2 ä x = 5.6. (i) Calculate Var(Ÿx) using the previous theorem. 16/114

17 Example 5 Suppose that i = 0.075, ä x = 8.6 and 2 ä x = 5.6. (i) Calculate Var(Ÿx) using the previous theorem. Solution: (i) We have that E[Ÿ 2 x ] = 2ä x (2 d) 2ä x d =91.6, = Var(Y x ) = 91.6 (8.6) 2 = (8.6) (2 (0.075/1.075))(5.6) 0.075/ /114

18 Example 5 Suppose that i = 0.075, ä x = 8.6 and 2 ä x = 5.6. (ii) Calculate Var(Ÿx) using A x and 2 A x. 18/114

19 Example 5 Suppose that i = 0.075, ä x = 8.6 and 2 ä x = 5.6. (ii) Calculate Var(Ÿx) using A x and 2 A x. Solution: (ii) Using that ä x = 1 Ax d, we get that A x = 1 dä x = 1 (0.075/1.075)(8.6) = 0.4. Since 2 A x uses a discount factor of v 2, 2 A x = 1 (1 v 2 ) 2ä x = 1 (1 (1.075) 2 )(5.6) = Hence, Var(Ÿ x ) = 2 A x A 2 x (0.4)2 d 2 = (0.075/1.075) 2 = /114

20 Theorem 3 (current payment method) Ÿ x = k=0 Z 1 x:k, ä x = ke x = v k kp x. k=0 k=0 20/114

21 Theorem 3 (current payment method) Ÿ x = k=0 Z 1 x:k, ä x = ke x = v k kp x. k=0 k=0 Proof: The payment at time k is made if and only if k < T x. Ÿ x = K x 1 k=0 v k = v k I (k < T x ) = k=0 ä x = E[Ÿx] = ke x = v k kp x. k=0 k=0 k=0 Z 1 x:k, 21/114

22 Example 6 Consider the life table x l x An 80 year old buys a due life annuity which will pay $50000 at the end of the year. Suppose that i = 6.5%. Calculate the single benefit premium for this annuity. 22/114

23 Example 6 Consider the life table x l x An 80 year old buys a due life annuity which will pay $50000 at the end of the year. Suppose that i = 6.5%. Calculate the single benefit premium for this annuity. Solution: ä 80 = k=0 v k l 80+k = 1 + (1.065) + (1.065) 2 l (1.065) + (1.065) 4 + (1.065) = = , (50000)ä 80 = (50000)( ) = /114

24 Remember that a decreasing due annuity has payments of n, n 1,..., 1 at the beginning of the year, for n years. The present value of a decreasing due annuity is n 1 (Dä) n i = v k (n k) = n a n i. d k=0 Theorem 4 Suppose that the mortality of (x) follows De Moivre s model with integer terminal age ω, where x and ω are nonnegative integers. Then, (i) ä x = (Dä) ω x ω x. (ii) If i 0, ä x = ω x a ω x d(ω x). (iii) If i = 0, ä x = ω x /114

25 Proof: (i) ä x = v k kp x = k=0 ω x 1 k=0 v k ω x k ω x = (Dä) ω x ω x. (ii) We have that ä x = 1 A x d = 1 a ω x ω x d = ω x a ω x. d(ω x) (iii) ä x = e x + 1 = ω x /114

26 Example 7 Suppose that i = 6% and De Moivre s model with terminal age 100. (i) Find ä 30 using A 30. (ii) Find ä 30 using ä x = ω x a ω x d(ω x). 26/114

27 Example 7 Suppose that i = 6% and De Moivre s model with terminal age 100. (i) Find ä 30 using A 30. (ii) Find ä 30 using ä x = ω x a ω x d(ω x). Solution: (i) We have that Hence, A 30 = a ω x i ω x = a ä 30 = 1 A 30 d = = = /114

28 Example 7 Suppose that i = 6% and De Moivre s model with terminal age 100. (i) Find ä 30 using A 30. (ii) Find ä 30 using ä x = ω x a ω x d(ω x). Solution: (i) We have that Hence, (ii) A 30 = a ω x i ω x = a ä 30 = 1 A 30 d = = = ä 30 = 70 a = = /114

29 Theorem 5 (Iterative formula for the APV of a life annuity due) ä x = 1 + vp x ä x+1. 29/114

30 Theorem 5 (Iterative formula for the APV of a life annuity due) ä x = 1 + vp x ä x+1. Proof: We have that ä x = v k kp x = 1 + v k p x k 1 p x+1 k=0 =1 + vp x k=1 v k 1 k 1p x+1 k=1 =1 + vp x v k kp x+1 = 1 + vp x ä x+1. k=0 30/114

31 Example 8 Suppose that ä x = ä x+1 = 10 and q x = Find i. 31/114

32 Example 8 Suppose that ä x = ä x+1 = 10 and q x = Find i. Solution: Using that ä x = 1 + vp x ä x+1, we get that 10 = 1 + v(0.99)(10), v = 10 1 (0.99)(10) (0.99)(10) and i = = 10%. 32/114

33 Theorem 6 If the probability function of time interval of failure is P{K x = k} = p k 1 x (1 p x ), k = 1, 2,... then ä x = E[ Z x ] = 1 = 1 + i. 1 vp x i + q x 33/114

34 Theorem 6 If the probability function of time interval of failure is P{K x = k} = p k 1 x (1 p x ), k = 1, 2,... then ä x = E[ Z x ] = 1 = 1 + i. 1 vp x i + q x Proof 1. We have that kp x = P{K k + 1} = So, ä x = v k kp x = v k px k = k=0 k=0 j=k+1 pj 1 x (1 p x ) = (1 p x ) pk x 1 p x = px k. 1 = 1 + i = 1 + i. 1 vp x 1 + i p x i + q x 34/114

35 Theorem 6 If the probability function of time interval of failure is P{K x = k} = p k 1 x (1 p x ), k = 1, 2,... then ä x = E[ Z x ] = 1 = 1 + i. 1 vp x i + q x Proof 1. We have that kp x = P{K k + 1} = So, ä x = v k kp x = v k px k = k=0 k=0 j=k+1 pj 1 x (1 p x ) = (1 p x ) pk x 1 p x = px k. 1 = 1 + i = 1 + i. 1 vp x 1 + i p x i + q x Proof 2. Since K x has a geometric distribution, K x and K x+1 have the same distribution and ä x = ä x+1. So, ä x = 1 + vp x ä x and ä x = 1 1 vp x. 35/114

36 Recall that ä = 1 d = 1 1 v. For a whole life annuity, we need to discount for interest and mortality and we get ä x = 1 1 vp x. 36/114

37 Example 9 An insurance company issues 800 identical due annuities to independent lives aged 65. Each of of this annuities provides an annual payment of Suppose that p x+k = 0.95 for each integer k 0, and i = 7.5%. 37/114

38 Example 9 An insurance company issues 800 identical due annuities to independent lives aged 65. Each of of this annuities provides an annual payment of Suppose that p x+k = 0.95 for each integer k 0, and i = 7.5%. (i) Find ä x and Var(Ÿ x ). 38/114

39 Example 9 An insurance company issues 800 identical due annuities to independent lives aged 65. Each of of this annuities provides an annual payment of Suppose that p x+k = 0.95 for each integer k 0, and i = 7.5%. (i) Find ä x and Var(Ÿ x ). Solution: (i) We have that 1 1 ä x = = 1 vp x 1 (1.075) 1 (0.95) = = 8.6, A x = q x 0.05 = q x + i = 0.4, 2 q x A x = q x + i(2 + i) = (0.075)( ) = , Var(Ÿ x ) = 2 A x A 2 x (0.4)2 d 2 = (0.075/1.075) 2 = /114

40 Example 9 An insurance company issues 800 identical due annuities to independent lives aged 65. Each of of this annuities provides an annual payment of Suppose that p x+k = 0.95 for each integer k 0, and i = 7.5%. (ii) Using the central limit theorem, estimate the initial fund needed at time zero in order that the probability that the present value of the random loss for this block of policies exceeds this fund is 1%. 40/114

41 Example 9 An insurance company issues 800 identical due annuities to independent lives aged 65. Each of of this annuities provides an annual payment of Suppose that p x+k = 0.95 for each integer k 0, and i = 7.5%. (ii) Using the central limit theorem, estimate the initial fund needed at time zero in order that the probability that the present value of the random loss for this block of policies exceeds this fund is 1%. Solution: (ii) Let Ÿ x,1,..., Ÿ x,800 be the present value per unit face value for the 800 due annuities. Let Q be the fund needed Ÿx,j] = (30000)(800)(8.6) = , 800 E[ j=1 800 Var( 30000Ÿ x,j ) = (30000) 2 (800)( ) = , j=1 Q = (2.326) = /114

42 Theorem 7 For the constant force of mortality model, ä x = 1 = 1 + i = 1 vp x i + q x 1 1 e (δ+µ), where q x = 1 e µ. 42/114

43 Theorem 7 For the constant force of mortality model, ä x = where q x = 1 e µ. 1 = 1 + i = 1 vp x i + q x 1 1 e (δ+µ), Proof: We have that P{K x = k} = e µ(k 1) (1 e µ ), k = 1, 2,... Theorem 6 applies with p x = e µ. We have that ä x = vp x =. 1 e (δ+µ) 43/114

44 Whole life discrete immediate annuity Definition 4 A whole life discrete immediate annuity is a series payments made at the end of the year while an individual is alive. Definition 5 The present value of a whole life immediate annuity for (x) with unit payment is denoted by Y x. Notice that Y x is a random variable. Y x depends on T x. It is easy to see that Y x = Ÿx 1. Definition 6 The actuarial present value of a whole life immediate annuity for (x) with unit payment is denoted by a x. We have that a x = ä x 1. 44/114

45 Theorem (i) Y x = a Kx 1 and a x = k=2 a k 1 k 1 q x. (ii) If i 0, Y x = 1 (1 + i)z x i = v Z x, a x = v A x d d and Var(Y x ) = where Z x is the present value of a life insurance paid at the end of the year of death. (iii) If i = 0, Y x = K x 1 = K(x), a x = e x and Var(Y x ) = Var(K(x)). 2 A x A 2 x d 2, 45/114

46 Theorem 8 (i) Y x = a Kx 1 and a x = k=2 a k 1 k 1 q x. 46/114

47 Theorem 8 (i) Y x = a Kx 1 and a x = k=2 a k 1 k 1 q x. Proof: (i) Since death happens in the interval (K x 1, K x ], payments of one are made at times 1, 2,..., K x 1, payments of one are made at times 1, 2,..., K x 1, i.e. the cashflow of payments is an annuity immediate. Hence, Y x = a Kx 1. If K x = 1, Y x = a 0 = 0. Therefore, a x = E[a Kx 1 ] = a k 1 P{K x = k} = k=2 a k 1 k 1 q x. k=2 47/114

48 Theorem 8 (ii) If i 0, Y x = v Z x, a x = v A x d d and Var(Y x ) = 2 A x A 2 x d 2, where Z x is the present value of a life insurance paid at the end of the year of death. 48/114

49 Theorem 8 (ii) If i 0, Y x = v Z x, a x = v A x d d and Var(Y x ) = 2 A x A 2 x d 2, where Z x is the present value of a life insurance paid at the end of the year of death. 1 v k 1 (ii) If i 0, a k 1 = i. Hence, Kx 1 v 1 Y x = a Kx 1 = = 1 (1 + i)z x [ ] i i v Zx a x = E = v A x, d d ( ) v Zx Var(Y x ) = Var d = v Z x, d = Var(Z x) 2 A x A 2 x d 2 = d 2. 49/114

50 Theorem 8 (iii) If i = 0, Y x = K x 1 = K(x), a x = e x and Var(Y x ) = Var(K(x)). 50/114

51 Theorem 8 (iii) If i = 0, Y x = K x 1 = K(x), a x = e x and Var(Y x ) = Var(K(x)). (iii) If i 0, Y x = a Kx 1 = K x 1 = K(x). 51/114

52 Example 10 Suppose that p x+k = 0.97, for each integer k 0, and i = 6.5%. Find a x and Var(Y x ). 52/114

53 Example 10 Suppose that p x+k = 0.97, for each integer k 0, and i = 6.5%. Find a x and Var(Y x ). Solution: We have that A x = q x q x + i = = , a x = v A x = (1.065) = , d 0.065/ q x A x = = q x + (2 + i)i ( )(0.065) = , 2 A x A 2 x ( )2 Var(Y x ) = d 2 = (0.065/1.065) 2 = /114

54 Example 11 Suppose that i = 6% and the De Moivre model with terminal age 100. Find a /114

55 Example 11 Suppose that i = 6% and the De Moivre model with terminal age 100. Find a 30. Solution: We have that A 30 = a = Hence, a 30 = ν A 30 d = (1.06) (0.06)(1.06) 1 = /114

56 Theorem 9 If i 0, and Y 2 x = 2Y x (2 + i) 2Y x i E[Yx 2 ] = 2a x (2 + i) 2a x. i 56/114

57 Theorem 9 If i 0, and Y 2 x = 2Y x (2 + i) 2Y x i E[Yx 2 ] = 2a x (2 + i) 2a x. i Proof: We have that Y x = 1 (1+i)Zx i. Hence, (1 + i)z x = 1 iy x. Similarly, 2 Y x = 1 (1+i)2 Z x i(2+i) and (1 + i) 2 Z x = 1 i(2 + i) 2Y x. Therefore, Y 2 x = 1 2(1 + i)z x + (1 + i) 2 2Z x i 2 = 1 2(1 iy x) + 1 i(2 + i) 2Y x i 2 = 2Y x (2 + i) 2Y x. i 57/114

58 Example 12 Suppose that i = 0.075, a x = 7.6 and 2 a x = 4.6. (i) Calculate Var(Y x ) using the previous theorem. (ii) Calculate Var(Y x ) using A x and 2 A x. 58/114

59 Example 12 Suppose that i = 0.075, a x = 7.6 and 2 a x = 4.6. (i) Calculate Var(Y x ) using the previous theorem. (ii) Calculate Var(Y x ) using A x and 2 A x. Solution: (i) We have that E[Yx 2 ] = 2a x (2 + i) 2a x 2(7.6) (2.075)(4.6) = i Var(Y x ) = 75.4 (7.6) 2 = = 75.4, 59/114

60 Example 12 Suppose that i = 0.075, a x = 7.6 and 2 a x = 4.6. (i) Calculate Var(Y x ) using the previous theorem. (ii) Calculate Var(Y x ) using A x and 2 A x. Solution: (ii) Using that a x = 1 (1+i)Ax i, we get that A x = 1 iax 1+i = 1 (0.075)(7.6) = 0.4. Since 2 A x uses a interest factor of (1 + i) 2, 2 A x = 1 i(2 + i) 2a x 1 (0.075)( )(4.6) (1 + i) 2 = (1.075) 2 = Hence, Var(Y x ) = 2 A x A 2 x (0.4)2 d 2 = (0.075/1.075) 2 = /114

61 Theorem 10 (current payment method) Y x = k=1 Z 1 x:k and a x = k=1 A 1 x:k = ke x = k=1 ν k kp x, k=1 where Z 1 x:k = νn I (n < K x ) is the present value of an n-year pure endowment life insurance 61/114

62 Theorem 10 (current payment method) Y x = k=1 Z 1 x:k and a x = k=1 A 1 x:k = ke x = k=1 ν k kp x, k=1 where Z 1 x:k = νn I (n < K x ) is the present value of an n-year pure endowment life insurance Proof: The payment at time k is made if and only if k < T x. Hence, Y x = ν k I (k < T x ) = and a x = k=1 k=1 A 1 x:k = ke x = k=1 k=1 Z 1 x:k ν k kp x. k=1 62/114

63 Theorem 11 If the probability function of time interval of failure is is P{K = k} = p k 1 x (1 p x ), k = 1, 2,... then a x = E[Z x ] = vp x = 1 q 1 vp x q + i. 63/114

64 Theorem 11 If the probability function of time interval of failure is is P{K = k} = p k 1 x (1 p x ), k = 1, 2,... then a x = E[Z x ] = vp x = 1 q 1 vp x q + i. Proof: We have that k p x = P{K k + 1} = p k x. So, a = v k kp x = k=1 v k px k = k=1 vp x 1 vp x = p 1 + i p x = 1 q x q x + i. 64/114

65 Example 13 Suppose that p x+k = 0.95, for each integer k 0, and i = 7.5%. Find a x. 65/114

66 Example 13 Suppose that p x+k = 0.95, for each integer k 0, and i = 7.5%. Find a x. Solution: We have that a x = p 1 + i p x = = /114

67 Theorem 12 For the constant force of mortality model, a x = vp x = 1 q x 1 vp x q x + i = e (δ+µ) 1 e (δ+µ). 67/114

68 Theorem 12 For the constant force of mortality model, a x = vp x = 1 q x 1 vp x q x + i = e (δ+µ) 1 e (δ+µ). Proof: Previous theorem applies with p x = e µ. So, a x = vp x = 1 q x 1 vp x q x + i = e (δ+µ) 1 e (δ+µ). 68/114

69 Theorem 13 a x = νp x (1 + a x+1 ). 69/114

70 Theorem 13 a x = νp x (1 + a x+1 ). Proof: We have that a x = ν k kp x = ν k p x k 1 p x+1 k=1 =νp x (1 + =νp x (1 + k=1 ) ν k 1 k 1p x+1 k=2 ν k kp x+1 ) = νp x (1 + a x+1 ). k=1 70/114

71 Example 14 Using i = 0.05 and a certain life table a 30 = Suppose that an actuary revises this life table and changes p 30 from 0.95 to Other values in the life table are unchanged. Find a 30 using the revised life table. 71/114

72 Example 14 Using i = 0.05 and a certain life table a 30 = Suppose that an actuary revises this life table and changes p 30 from 0.95 to Other values in the life table are unchanged. Find a 30 using the revised life table. Solution: Using that 4.52 = a 30 = vp 30 (1 + a 31 ) = (1.05) 1 (0.95)(1 + a 31 ), we get that a 31 = (4.52)(1.05) = The revised value of a 30 is (1.05) 1 (0.96)( ) = /114

73 Whole life continuous annuity Definition 7 A whole life continuous annuity is a continuous flow of payments with constant rate made while an individual is alive. Definition 8 The present value random variable of a whole life continuous annuity for (x) with unit rate is denoted by Y x. Definition 9 The actuarial present value of a whole life continuous annuity for (x) with unit rate is denoted by a x. We have that a x = E[Y x ]. 73/114

74 Theorem 14 (i) Y x = a Tx. (ii) If δ 0, Y x = 1 Z x, a x = 1 A x δ δ and Var(Y x ) = 2 A x A 2 x δ 2. (iii) If δ = 0, Y x = T x and a x = e x. 74/114

75 Theorem 14 (i) Y x = a Tx. (ii) If δ 0, Y x = 1 Z x, a x = 1 A x δ δ and Var(Y x ) = 2 A x A 2 x δ 2. (iii) If δ = 0, Y x = T x and a x = e x. Proof: (i) Since payments are received at rate one until time T x, Y x = ā Tx. 75/114

76 Theorem 14 (i) Y x = a Tx. (ii) If δ 0, Y x = 1 Z x, a x = 1 A x δ δ and Var(Y x ) = 2 A x A 2 x δ 2. (iii) If δ = 0, Y x = T x and a x = e x. Proof: (ii) If δ 0, a n i = 1 e nδ δ, Y x = a Tx = 1 v Tx δ = 1 Z x, a x = E δ [ ] 1 Z x δ = 1 A x δ and ( ) 1 Z x Var(Y x ) = Var δ = Var(Z x) 2 A x A 2 x δ 2 = δ 2. 76/114

77 Theorem 14 (i) Y x = a Tx. (ii) If δ 0, Y x = 1 Z x, a x = 1 A x δ δ and Var(Y x ) = 2 A x A 2 x δ 2. (iii) If δ = 0, Y x = T x and a x = e x. Proof: (iii) If δ = 0, a t i = t. Hence, Y x = T x and a x = e x. 77/114

78 Example 15 Suppose that v = 0.92, and the force of mortality is µ x+t = 0.02, for t 0. Find a x and Var(Y x ). 78/114

79 Example 15 Suppose that v = 0.92, and the force of mortality is µ x+t = 0.02, for t 0. Find a x and Var(Y x ). Solution: Using previous theorem, A x = 0.02 ln(0.92) = , a x = 1 A x = = , δ ln(0.92) A x = (2)( ln(0.92)) = , Var(Z x ) 2 A x A 2 x ( )2 δ 2 = δ 2 = ( ln(0.92)) 2 = /114

80 We define m Y x as the present value of whole life continuous annuity with unit rate at a force of interest m times the original force, i.e. m 1 e Tx mδ Y x = mδ. We define m a x = E [m ] Y x. It is not true that m a x = E [ (Y x ) m]. 80/114

81 Theorem 15 If δ 0, E[Y 2 x] = 2(a x 2 a x ). δ 81/114

82 Theorem 15 If δ 0, E[Y 2 x] = 2(a x 2 a x ). δ Proof: From m a x = 1 m A x mδ, we get that m A x = 1 mδ ma x. Hence, [ (1 ) ] E[Y 2 2 [ Zx 1 2Zx + 2 ] Z x x] = E = E δ δ 2 [ 1 2(1 δax ) + 1 2δ 2a ] x =E δ 2 = 2(a x 2 a x ). δ 82/114

83 Example 16 Suppose that a x = 12, 2 a x = 7 and δ = (i) Find Var(Y x ) using that Var(Y x ) = 2 A x A 2 x δ 2. (ii) Find Var(Y x ) using previous theorem. 83/114

84 Example 16 Suppose that a x = 12, 2 a x = 7 and δ = (i) Find Var(Y x ) using that Var(Y x ) = 2 A x A 2 x δ 2. (ii) Find Var(Y x ) using previous theorem. Solution: (i) We have that A x = 1 a x δ = 1 (12)(0.05) = 0.4, 2 A x = 1 2 a x 2δ = 1 (7)(2)(0.05) = 0.3. Hence, (ii) We have that Var(Y x ) = 2 A x A 2 x 0.3 (0.4)2 δ 2 = (0.05) 2 = 56. E[Y 2 x] = 2(a x 2 a x ) 2(12 7) = = 200, δ 0.05 Var(Y x ) = 200 (12) 2 = /114

85 Theorem 16 (current payment method) a x = 0 v t tp x dt = 0 te x dt. Proof: Let h(x) = v x, x 0. Let H(x) = x 0 h(t) dt = x 0 v t dt = ā x, x 0. By a previous theorem, E[Y x ] = E[ā Tx ] = E[H(T x)] = 0 h(t)s Tx (t) dt = 0 v t tp x dt. 85/114

86 Example 17 Suppose that δ = 0.05 and t p x = 0.01te 01.t, t 0. Calculate a x. 86/114

87 Example 17 Suppose that δ = 0.05 and t p x = 0.01te 01.t, t 0. Calculate a x. Solution: Using that 0 t n e t/β dt = βn+1 n!, = a x = 0 0 v t tp x dt = 0 e (0.05)t 0.01te 0.1t dt 0.01te 0.15t dt = 0.01 (0.15) 2 = /114

88 Theorem 17 The cumulative distribution function of Y x = ā Tx is 0 ( ) if y < 0, F Y x (y) = F Tx ln(1 δy) δ if 0 < y a ω x, 1 if a ω x y. Proof. The function y = H(x) = ā x = 1 e δx δ y = H(x) = 1 e δx δ is increasing. If, then δy = 1 e δx, e δx = 1 δy and x = ln(1 δy) δ. Hence, the inverse function of H is H 1 (y) = ln(1 δy) δ, y 0. Hence, if 0 < y, F Y x (y) = P{Y x y} = P {H(T x ) y} = P { T x H 1 (y) } { } ( ) ln(1 δy) ln(1 δy) =P T x = F Tx. δ δ 88/114

89 Theorem 18 The probability density function of Y x = ā Tx is f Tx ln(1 δy) δ f Y x (z) = 1 δy if 0 < y < a ω x, 0 else. Proof: For 0 < y < a ω x, ( f Y x (y) = d dy F Y x (y) = d ( ) dy F ln(1 δy) f Tx δ T x = δ 1 δy ln(1 δy) ). 89/114

90 Corollary 1 Under the De Moivre model with terminal age ω, Y x is a continuous r.v. with cumulative distribution function ln(1 δy) F Y x (y) = δ(ω x), if 0 < y a ω x. and density function f Y x (y) = 1 (1 δy)(ω x), if 0 < y a ω x. Proof: The claims follows noticing that F Tx (t) = f Tx (t) = 1 ω x, if 0 t ω x. t ω x and 90/114

91 Corollary 2 Under constant force of mortality µ, Y x is a continuous r.v. with cumulative distribution function and density function F Y x (y) = 1 (1 δy) µ δ, 0 < y 1 δ. f Y x (y) = µ(1 δy) µ δ 1, 0 < y < 1 δ. 91/114

92 Proof: Since F Tx (t) = 1 e µt, if 0 t < ; we have that for 0 < y 1 δ, F Y x (y) = F Tx ( =1 (1 δy) µ δ. ) ( ( ln(1 δy) = 1 exp µ δ )) ln(1 δy) δ Y x has density f Y x (y) = F Y x (y) = µ(1 δy) µ δ 1, 0 < y < 1 δ. 92/114

93 The present value of a continuously decreasing annuity is ( Da ) n = n i 0 (n t)v t dt = n a n i. δ 93/114

94 Theorem 19 Under De Moivre s model with terminal age ω, ( ) Da a x = ω x a ω x δ(ω x) = ω x i. ω x 94/114

95 Theorem 19 Under De Moivre s model with terminal age ω, ( ) Da Proof: We have that a x = ω x a ω x δ(ω x) = ω x i. ω x a x = 1 A x δ = 1 a ω x ω x δ = ω x a ω x δ(ω x) and a x = 0 v t tp x dt = ω x 0 v t ω x t ω x dt = ( Da ) ω x i ω x 95/114

96 Example 18 Suppose that i = 6.5% and deaths are uniformly distributed with terminal age 100. (i) Calculate the density of Y 40. (ii) Calculate a 40 using that a x = 1 Ax δ. (iii) Calculate a 40 using that a x = ω x a ω x δ(ω x). 96/114

97 Example 18 Suppose that i = 6.5% and deaths are uniformly distributed with terminal age 100. (i) Calculate the density of Y 40. (ii) Calculate a 40 using that a x = 1 Ax δ. (iii) Calculate a 40 using that a x = ω x a ω x δ(ω x). Solution: (i) We have that f Y 40 (y) = 1 (1 δy)(ω x) = 1 (60)(1 y ln(1.065)), if 0 < y < 1 (1.065) 60 ln(1.065) = /114

98 Example 18 Suppose that i = 6.5% and deaths are uniformly distributed with terminal age 100. (i) Calculate the density of Y 40. (ii) Calculate a 40 using that a x = 1 Ax δ. (iii) Calculate a 40 using that a x = ω x a ω x δ(ω x). Solution: (ii) We have that A 40 = a = 1 (1.065) 60 (60) ln(1.065) = Hence, a 40 = 1 A 40 δ = ln(1.065) = /114

99 Example 18 Suppose that i = 6.5% and deaths are uniformly distributed with terminal age 100. (i) Calculate the density of Y 40. (ii) Calculate a 40 using that a x = 1 Ax δ. (iii) Calculate a 40 using that a x = ω x a ω x δ(ω x). Solution: (iii) We have that a 60 = 1 (1.065) 60 ln(1.065) = Hence, a 40 = a (ln(1.065))(60) = (ln(1.065))(60) = /114

100 Theorem 20 Under constant force of mortality µ, a x = 1 µ+δ. Proof. We have that a x = 1 A x δ = 1 µ µ+δ δ = 1 µ + δ. 100/114

101 Example 19 Suppose that δ = 0.07 and the force of mortality is (i) Calculate the density of Y 40. (ii) Calculate a 40. (iii) Calculate the variance of Y /114

102 Example 19 Suppose that δ = 0.07 and the force of mortality is (i) Calculate the density of Y 40. (ii) Calculate a 40. (iii) Calculate the variance of Y 40. Solution: (i) The density of Y 40 is f Y 40 (t) = (0.02)(1 0.07y) 5 7, 0 < y < /114

103 Example 19 Suppose that δ = 0.07 and the force of mortality is (i) Calculate the density of Y 40. (ii) Calculate a 40. (iii) Calculate the variance of Y 40. Solution: (i) The density of Y 40 is f Y 40 (t) = (0.02)(1 0.07y) 5 7, 0 < y < (ii) a 40 = = /114

104 Example 19 Suppose that δ = 0.07 and the force of mortality is (i) Calculate the density of Y 40. (ii) Calculate a 40. (iii) Calculate the variance of Y 40. Solution: (i) The density of Y 40 is f Y 40 (t) = (0.02)(1 0.07y) 5 1 7, 0 < y < (ii) a 40 = = (iii) 0.02 A 40 = = , A 40 = (2)0.07 = 0.125, Var(Y 40 ) = ( )2 (0.07) 2 = /114

105 Theorem 21 Suppose that T x has a p th quantile ξ p such that P{T x < ξ p } = p = P{T x ξ p }. Given b, δ > 0, the p th quantile of b 1 e δtx δ is b 1 e δξp δ. 105/114

106 Example 20 Suppose that i = 6% and De Moivre model with terminal age 100. (i) Calculate the 30% percentile and the 70% percentiles of Y 30. (ii) Calculate a 30. (iii) Calculate the variance of Y /114

107 Example 20 Suppose that i = 6% and De Moivre model with terminal age 100. (i) Calculate the 30% percentile and the 70% percentiles of Y 30. (ii) Calculate a 30. (iii) Calculate the variance of Y 30. Solution: (i) Let ξ 0.30 the 30% percentile of T 30. We have that 0.3 = F T30 (ξ 0.30 ) = ξ So, ξ 0.30 = 21. The 30% percentile of Y 30 = 1 (1.06) T 30 ln(1.06) is 1 (1.06) 21 ln(1.06) = Let ξ 0.70 the 70% percentile of T 30. We have that 0.7 = F T30 (ξ 0.70 ) = ξ So, ξ 0.70 = 49. The 70% percentile of Y 30 = 1 (1.06) T 30 ln(1.06) is 1 (1.06) 49 ln(1.06) = /114

108 Example 20 Suppose that i = 6% and De Moivre model with terminal age 100. (i) Calculate the 30% percentile and the 70% percentiles of Y 30. (ii) Calculate a 30. (iii) Calculate the variance of Y 30. Solution: (ii) We have that A 30 = a 70 i ω x = 1 (1.06) 70 (70) ln(1.06) = and a 30 = 1 A 30 δ = ln(1.06) = /114

109 Example 20 Suppose that i = 6% and De Moivre model with terminal age 100. (i) Calculate the 30% percentile and the 70% percentiles of Y 30. (ii) Calculate a 30. (iii) Calculate the variance of Y 30. Solution: (iii) We have that 2 A 30 = a 70 (1+i) 2 1 ω x = 1 (1.06) (2)(70) (70)(2) ln(1.06) = and Var(Y 30 ) = = A 30 (A 30 ) ( )2 δ 2 = (ln(1.06)) 2 109/114

110 Example 21 Suppose that v = 0.92 and that the force of mortality is µ x+t = 0.02, for t 0. (i) Calculate the density of Y x. (ii) Calculate the first, second and third quartiles of Y x. (iii) Calculate a x. 110/114

111 Example 21 Suppose that v = 0.92 and that the force of mortality is µ x+t = 0.02, for t 0. (i) Calculate the density of Y x. (ii) Calculate the first, second and third quartiles of Y x. (iii) Calculate a x. Solution: (i) We have that f Y x (y) = µ(1 δy) µ δ 1 = (0.02)(1 + ln(0.92)y) 0.02 ln(0.92) 1, 1 if 0 < y ln(0.92). 111/114

112 Example 21 Suppose that v = 0.92 and that the force of mortality is µ x+t = 0.02, for t 0. (i) Calculate the density of Y x. (ii) Calculate the first, second and third quartiles of Y x. (iii) Calculate a x. Solution: (ii) Let ξ p the 100(p)% percentile of Y x. We have that p = F Yx (ξ p ) = 1 (1 δξ p ) µ δ = (1 + ln(0.92)ξp ) So, ξ p = 1 (1 p) ln(0.92) 0.02 ln(0.92). The first quartile of Y x is 0.02 ln(0.92). ξ 0.25 = 1 (1 0.25) ln(0.92) 0.02 ln(0.92) = /114

113 Example 21 Suppose that v = 0.92 and that the force of mortality is µ x+t = 0.02, for t 0. (i) Calculate the density of Y x. (ii) Calculate the first, second and third quartiles of Y x. (iii) Calculate a x. Solution: (ii) The median of Y x is ξ 0.5 = 1 (1 0.5) ln(0.92) 0.02 ln(0.92) = The third quartile of Y x is ξ 0.75 = 1 (1 0.75) ln(0.92) 0.02 ln(0.92) = /114

114 Example 21 Suppose that v = 0.92 and that the force of mortality is µ x+t = 0.02, for t 0. (i) Calculate the density of Y x. (ii) Calculate the first, second and third quartiles of Y x. (iii) Calculate a x. Solution: (iii) We have that a x = 1 µ + δ = ln(0.92) = /114