Life annuities. Actuarial mathematics 3280 Department of Mathematics and Statistics York University. Edward Furman.

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1 Edward Furman, Actuarial mathematics MATH3280 p. 1/53 Life annuities Actuarial mathematics 3280 Department of Mathematics and Statistics York University Edward Furman

2 Edward Furman, Actuarial mathematics MATH3280 p. 2/53 Discrete whole life annuity-due Consider a whole life annuity that pays 1 at the beginning of each year that the person age x survives. This is refereed to as the life annuity due. Then, for a n 1 + v + + v n 1 1 vn 1 v, we have that, Z.ȧ K+1. In our previous notations, we can think of b (1, 1,.,1) and v (1,v,.,v k ) Z b v 1 + v + + v K.ȧ K+1.

3 Figure 1: Annuity due for a person age x that dies between ages x + 5 and x + 6 The actuarial expected present value is then a x E[Z] E[.ȧ K+1 ] a k+1 P(K k) a k+1 kp x q x+k. Edward Furman, Actuarial mathematics MATH3280 p. 3/53

4 Edward Furman, Actuarial mathematics MATH3280 p. 4/53 The just derived expression is not convenient. It can be shown that a x v k kp x. Indeed: a x a k+1 kp x q x+k k v i P(K k) i0 ( ) P(K k) v i i0 ki P(K > i 1)v i i0 P(K i)v i i0 ip x v i. i0

5 Edward Furman, Actuarial mathematics MATH3280 p. 5/53 By summation by parts, i.e., by n f(k) g(k) [f(k)g(k)] n+1 m km where we obtain: k q x a k+1 leading to n km f(x) f(x + 1) f(x), g(k + 1) f(k). ( k p x k+1 p x ).ȧ k+1 ( k+1 p x k p x ).ȧ k+1, g(k) k p x and f(k).ȧ k+1.

6 Edward Furman, Actuarial mathematics MATH3280 p. 6/53 Now, we can rewrite a x a k+1 kp x q x+k (.ȧ 0 ( 1)) + noticing that k+1p x v k+1, g(k + 1) k+1 p x and f(k).ȧ k+1 vk+1. In view of the above, the following recursion holds: a x 1 + k+1p x v k vp x kp x+1 v k 1 + vp x a x+1. The initial value for the recursion is.ȧ 0.

7 Edward Furman, Actuarial mathematics MATH3280 p. 7/53 The variance for the whole life annuity is [ 1 v K+1] Var[Z] Var Var[vK+1 ] 1 v (1 v) 2 E[v2(K+1) ] E 2 [v K+1 ] (1 v) 2 In a similar manner, [ 1 v K+1] a x E 1 v 2 A x (A x ) 2 (1 v) 2. 1 A x d.ȧ.ȧ A x.

8 Edward Furman, Actuarial mathematics MATH3280 p. 8/53 Discrete n year term life annuity-due The present value rv of an n year term temporary life annuity due of 1 per year is {.ȧk+1, 0 K < n Z. a n, K n The expected actuarial present value is then a x:n It can be shown that: n 1 a k+1 kp x q x+k +.ȧ n np x. a x:n n 1 v k kp x.

9 Edward Furman, Actuarial mathematics MATH3280 p. 9/53 Indeed, a x:n n 1 n 1 k v i k q x + i0 n 1 i0 ki n 1 i0 n 1 i0 v i k q x + n 1 i0 n 1 i0 v i np x v i np x ( n 1 ) v i k q x + n p x ki ip x v i.

10 Edward Furman, Actuarial mathematics MATH3280 p. 10/53 We can again be interested in the following recursion n 1 i0 ip x v i 1 + n 1 i1 n vp x ip x v i 1 + i0 n 2 i0 i+1p x v i+1 ip x+1 v i 1 + vp x a x+1:n 1, where the initial value is.ȧ x+n:0 0. The variance can be found using the fact that a x:n 1 E[Z ] 1 v 1 A x:n 1 v, where Z is the present value rv of an endowment n years insurance.

11 Edward Furman, Actuarial mathematics MATH3280 p. 11/53 n year deferred whole life annuity-due Then Var[Z] Var[Z ] 2 (1 v) 2 A x:n (A x:n ) 2 (1 v) 2. The n year deferred whole life annuity-due of 1 payable at the beginning of each year while (x) survives from age x + n onward is described by the following rv { 0, 0 K < n Z a K+1 n, K n. n The actuarial expected present value is then n a x.ȧ x.ȧ n 1 x:n kp x v k v k kp x kn v k kp x

12 Edward Furman, Actuarial mathematics MATH3280 p. 12/53 n year certain and life annuity-due Show that Var[Z] 1 ( d v2n np x 2 a x+n (1 + v) 2.ȧ ) x+n (n a x ) 2. The n year certain and life annuity-due pays guaranteed payments for at least n years and changes to the regular life annuity-due thereafter. Thus, the present value rv is {.ȧn, 0 K < n Z a K+1, K n. The expected present value rv is a x:n E[Z] n 1 a n P(K k) + kn a k+1 P(K k)

13 Edward Furman, Actuarial mathematics MATH3280 p. 13/53 Hence, a x:n.ȧ np(k n 1) +.ȧ n nq x + kn kn a k+1 k q x. Show by summation by parts that a x:n.ȧ n + v k kp x. kn a k+1 P(K k)

14 Edward Furman, Actuarial mathematics MATH3280 p. 14/53 Whole life annuity-immediate This life annuity is similar to the whole life annuity-due but this one is payable at the end of the payments periods. The present value rv is Z a K. The actuarial expected present value is given by a x E[Z] Also, as far as a k kp x q x+k i1 k1 v i kp x q x+k ki k v i kp x q x+k i1 v i ip x. i1 a k 1 vk i 1 (1 + i)vk+1 i,

15 Edward Furman, Actuarial mathematics MATH3280 p. 15/53 Continuously payable annuities which leads to a x E [ 1 (1 + i)v K+1] i 1 (1 + i)a x i. The actuarial present value for a continuous whole life annuity is denoted by a x. The price of this annuity is then: a x E[Z] 0 a t tp x µ(x + t)dt, which after using integration by parts for u a t, dv t p x µ(x + t)dt, v t p x and du v t dt, we have that a x a t t p x v t tp x dt 0 v t tp x dt 0 te x dt.

16 Edward Furman, Actuarial mathematics MATH3280 p. 16/53 Whole life continuous annuity The following recursion is easily obtained: a x 1 0 v t tp x dt + a x:1 + vp x 0 1 v t tp x dt v t tp x+1 dt a x:1 + vp x a x+1, a 0. We know that for every t a t 1 vt 1 δa δ t + v t. This is true for the rv T as well, thus a T 1 vt δ 1 δa T + v T 1 δa x + A x.

17 Edward Furman, Actuarial mathematics MATH3280 p. 17/53 The variance is easily calculated as [ 1 v T ] [ v T ] Var[Z] Var Var δ δ 2 A x A 2 x δ 2 Example 1 Calculate the pdf and the cdf of Z. ( 1 v T ) ( ) F Z (z) P(Z z) P z P 1 v T δz δ ( ) ( ) P v T 1 δz P e δt 1 δz ( ) log(1 δz) P ( δt log(1 δz)) P T δ ( ) log(1 δz) F T for z < 1/δ. δ

18 Edward Furman, Actuarial mathematics MATH3280 p. 18/53 n-year life continuous annuity The pdf is then f Z (z) d dz F Z (z) d dz F T f T ( log(1 δz) δ ( ) log(1 δz) δ ) (1 δy) 1 for z < 1/δ. The present value rv of an n year term temporary life continuous annuity of 1 per year is { a Z T, 0 T < n a n, T n. The present value of Z is a x:n E[Z] n 0 a t tp x µ(x + t)dt + a n n p x.

19 Edward Furman, Actuarial mathematics MATH3280 p. 19/53 Integration by parts gives a x:n n 0 e δt tp x dt. Note that { a Z T 1 vt δ, 0 T < n a n 1 vn δ, T n. Thus, for Z being the n-year term endowment insurance, and [ 1 Z ] E[Z] a x:n E δ Var[Z] Var[Z] δ 2 1 A x:n δ 2 A x:n A 2 x:n δ 2.

20 Edward Furman, Actuarial mathematics MATH3280 p. 20/53 Deferred n year term continuous annuity Show at home that Var[Z] 2 δ (a x:n 2 a x:n ) a 2 x:n. We now consider the so called deferred n-year term annuity. The present value is { 0, 0 T < n Z n a T n a T a n, T n. We next show that Proof: Indeed n a x n a x n E x a x+n. n v n a t n tp x µ(x + t)dt,

21 Edward Furman, Actuarial mathematics MATH3280 p. 21/53 Further, for s t n, we have that n a x 0 v n np x v n a s s+n p x µ(x + s + n)ds 0 n E x a x+n Also, certainly n a x a x a x:n a s s p x+n µ(x + s + n)ds n v t tp x dt Calculate the variance of this annuity. n te x dt.

22 Edward Furman, Actuarial mathematics MATH3280 p. 22/53 n-year certain and life annuity Pays 1 unit each year for n years and becomes a usual life annuity thereafter. { a n, 0 T < n Z. a T, T n The actuarial present value is a x:n n 0 a n n q x + a n t p x µ(x + t)dt + Integration by parts would lead to n a x:n a n + n a t tp x µ(x + t)dt. n v t tp x dt. a t tp x µ(x + t)dt

23 Edward Furman, Actuarial mathematics MATH3280 p. 23/53 We also have that a x:n a n + n a x a n + n E x a x+n a n + (a x a x:n ). Prove that the variance of the n-year certain and whole life annuity is equal to the variance of the n-year differed annuity. What is Var[a x:n ] Var[ n a x ]? Lastly, recall the so called accumulation function n n S n (1 + i) n t 0 dt vt dt v n. 0

24 Edward Furman, Actuarial mathematics MATH3280 p. 24/53 Intuitively, we have that s x:n n 0 vt tp x dt v n np x a x:n ne x. The latter equation represents the actuarial accumulated value at the end of the term of an n-year temporary life annuity of 1 per year payable continuously while (x) survives. The benefit is available at age x + n if x survives till then. Example 2 Consider the timescale measured in intervals of length 1/m where a unit is a year. Let a whole life insurance for a unit amount be payable at the end of the m-thly interval in which death occurs. 1. What is the discount function for this insurance? What is the expected present value?

25 2. Show that A (m) x i i (m) A x. Solution 1. Let K 0, 1,., be the number of complete insurance years lived prior to death and let J 0, 1,.,m 1 be the number of complete m-ths of the year lived in the year of death. For a death occurring in the m-th of the year following age x + k + j/m and the benefit is payable at the end of the m-th period, we have that E[v K+(J+1)/m ] m 1 j0 m 1 j0 v k+(j+1)/m P(K k,j j) v k+(j+1)/m kp x j/m 1/m q x+k A (m) x. Edward Furman, Actuarial mathematics MATH3280 p. 25/53

26 Edward Furman, Actuarial mathematics MATH3280 p. 26/53 2. Recall that under UDD, j/m 1/mq x+k (j+1)/m q x+k (j)/m q x+k UDD j + 1 m q x+k j m q x+k 1 m q x+k. A (m) x m 1 v k j0 m 1 v k+1 j0 v (j+1)/m kp x j/m 1/m q x+k v (j+1)/m 1 kp x 1 m q x+k m 1 v k+1 kp x q x+k (1 + i) 1 (j+1)/m 1 m j0

27 Edward Furman, Actuarial mathematics MATH3280 p. 27/53 A (m) x v k+1 kp x q x+k (1 + i) v k+1 kp x q x+k (1 + i) m 1 j0 m 1 v (j+1)/m 1 m (v (1/m) ) j+1 1 m j0 v k+1 kp x q x+k (1 + i) 1 v(m 1+1)/m v1/m1 m 1 v 1/m v k+1 kp x q x+k (1 + i) 1 m v k+1 kp x q x+k (1 + i) 1 m 1 v v 1/m (1 v 1/m ) 1 v (1 + i) 1/m 1,

28 Edward Furman, Actuarial mathematics MATH3280 p. 28/53 m-thly payable annuities which after recalling that i (m) m((1 + i) 1/m 1) becomes A (m) x v k+1 (1 v)(1 + i) kp x q x+k i (m) v k+1 kp x q x+k i i (m) i i (m)a x. We denote by.ȧ(m) x the actuarial present value of a life annuity of 1 per year payed in installments of 1/m at the beginning of each m-th of the year while (x) survives. Let J be the number of complete mths of a year lived in the year of death. Then a (m) x E[Z] E[.ȧ(m) K+(J+1)/m ],

29 where a (m) n mn 1 j0 Show at home that 1 m (v1/m ) j 1 v n m(1 (1 d) 1/m ) 1 vn d (m). a (m) x 1 A(m) x d (m). Another representation of the expected present value is The variance is Var[Z] Var a (m) x 1 m (v 1/m ) h h/mp x. h0 [ 1 v K+(j+1)/m d (m) ] 2 A (m) x (A (m) x ) 2 (d (m) ) 2 Edward Furman, Actuarial mathematics MATH3280 p. 29/53

30 Edward Furman, Actuarial mathematics MATH3280 p. 30/53 We know that for every t Thus a t 1 vt 1 v 1 d.ȧ t + v t. a T 1 vt 1 v 1 d.ȧ T + vt 1 d.ȧ x + A x. We then have the following relationship Moreover, 1 d.ȧ x + A x d (m).ȧ(m) x a (m) x d d (m) a x 1.ȧ(m) 1 a x.ȧ(m) d (m)(a(m) (A (m) + A (m) x. x A x ) x A x ).

31 If we assume that J is uniformly distributed on the integers 0, 1,.,m 1, then A (m) x i i (m)a x S (m) A 1 x, which thus leads to a (m) x.ȧ(m) 1 a x S(m) 1 1 d (m) A x d(m).ȧ(m) a 1 x A x (S (m) 1) 1 d (m) d.ȧ x (1 d.ȧ x)(s (m) 1 1) d (m) 1 S(m) (1 d.ȧ 1 x) d (m). Edward Furman, Actuarial mathematics MATH3280 p. 31/53

32 Edward Furman, Actuarial mathematics MATH3280 p. 32/53 After all, we have that Thus: a (m) x a (m) x α(m) S (m) 1 S(m) d.ȧ 1 x (S (m) 1 S (m) 1 d (m) 1) 1 1 a x a (m) S(m) 1 d (m) α(m).ȧ x β(m), where a (m),β(m) S(m) d (m). Even simpler expression can be obtained if we assume that v k+j/m k+j/mp x is linear in j.

33 Edward Furman, Actuarial mathematics MATH3280 p. 33/53 Proof Indeed, as the linearity in e.g. s (0, 1) implies we have that f(x + s) f(x)(1 s) + f(x + 1)s, m 1 j0 m 1 j0 1 m vk+j/m k+j/mp x 1 m (( 1 j )v kkp x + jm ) m vk+1k+1p x v k kp x (v k kp x v k+1 k+1p x ) v k kp x (v k kp x v k+1 k+1p x ) m 1 j0 j m 2 (m 1) 2m.

34 Edward Furman, Actuarial mathematics MATH3280 p. 34/53 Thus, a (m) x 1 m (v 1/m ) h h/mp x h0 m 1 j0 v k kp x.ȧ x 1 m vk+j/m k+j/mp x (m 1) 2m. (m 1) 2m (v k kp x v k+1 k+1p x )

35 Edward Furman, Actuarial mathematics MATH3280 p. 35/53 We can use the same techniques to developing formulas for deferred and n-year term mthly payable annuities. However, it seems easier to say that e.g., a (m) x:n.ȧ(m).ȧ(m) 1 ending up with x v n np x a (m) x+n.ȧ(m) x n E x a (m) x+n a x β(m)a x n E x (.ȧ(m) a x+n β(m)a x+n ) a (m) x:n.ȧ(m) 1 a x:n β(m)(a 1x:n ). 1 Alternatively, as a (m) x α(m).ȧ x β(m)

36 Edward Furman, Actuarial mathematics MATH3280 p. 36/53 and we have that a (m) x+n α(m).ȧ x+n β(m), a (m) x:n α(m).ȧ x:n β(m)(1 n E x ).

37 Edward Furman, Actuarial mathematics MATH3280 p. 37/53 Similarly, for a deferred life annuity, we have that n a (m) x thus yielding.ȧ(m) x.ȧ(m) 1 n a (m) x.ȧ(m) x:n a x β(m)a x.ȧ(m) a x:n + β(m)a 1x:n,.ȧ(m) 1 n Alternatively, prove at home, n a (m) 1 a x β(m) n A x. x α(m) n a x β(m) n E x.

38 Edward Furman, Actuarial mathematics MATH3280 p. 38/53 m-thly payable immediate annuities We denote by a (m) x the actuarial present value of a life annuity of 1 per year payed in installments of 1/m at the end of each m-th of the year while (x) survives. Let J be the number of complete mths of a year lived in the year of death. Then a (m) x E[Z] E [ ] a (m) K+J/m E [ 1 v K+J/m i (m) ], Recall that a (m) n nm j1 1 m (v1/m ) j 1 vn i (m), i (m) m[(1 + i) 1/m 1].

39 Edward Furman, Actuarial mathematics MATH3280 p. 39/53 We showed that [ 1 (1 + i)v K+1] a x E i Similarly, we obtain that 1 (1 + i)a x i. leading to a (m) x 1 (1 + i)1/m A (m) x i (m), 1 ia x + (1 + i)a x i (m) a (m) x + ( 1 + i(m) m ) A (m) x. Meaningful?

40 Edward Furman, Actuarial mathematics MATH3280 p. 40/53 We can also adjust the annuities due to obtain the necessary actuarial present values for the immediate annuities: a (m) x.ȧ(m) x 1/m since the only difference between the two is the payment at time 0. And why? a (m) x:n.ȧ(m) x:n 1/m + 1/m n E x,

41 Edward Furman, Actuarial mathematics MATH3280 p. 41/53 To conclude Summary of discrete life annuities of 1 per annum payable at the beginning of each year (due) or at the end of each year (immediate)

42 Useful relations between specific discrete annuities and insurances. Edward Furman, Actuarial mathematics MATH3280 p. 42/53

43 Summary of continuous life annuities of 1 per annum payable continuously. Edward Furman, Actuarial mathematics MATH3280 p. 43/53

44 Typical distributions for the present value rv T. Edward Furman, Actuarial mathematics MATH3280 p. 44/53

45 Edward Furman, Actuarial mathematics MATH3280 p. 45/53 Example 3 Using the assumption of a uniform distribution of deaths in each year of age and the Illustrative Life Table with interest at the effective annual rate of 6%, calculate a. a 20,a 50,a 80 b.v ar(a T ) for x20,50,80. Solution a. We showed that a x 1 A x UDD 1 δ ia x. δ δ Also, i/δ 0.06/ ln( ) , which leads to a 20 a 50 UDD A 20 ln(1.06) UDD A 50 ln(1.06) ln(1.06) ln(1.06)

46 Edward Furman, Actuarial mathematics MATH3280 p. 46/53 a 80 UDD A 80 ln(1.06) ln(1.06) b. Recall that [ v T ] Var[a T ] Var δ To calculate 2 A x, we note that 2 A x i δ 2 A x where δ 2δ 2 ln(1.06) and i e δ A x A 2 x δ 2

47 Edward Furman, Actuarial mathematics MATH3280 p. 47/53 Finally, 2 A x A x A x, which leads to 2 A Var[a T ] ln(1.06) A Var[a T ] ln(1.06) A Var[a T ] ln(1.06)

48 Edward Furman, Actuarial mathematics MATH3280 p. 48/53 Example 4 Using the values obtained in Exercise 3, calculate the standard deviation and the coefficient of variation,σ/µ, of the following present-value random variables. a. Individual annuities issued at ages 20,50,80 with life incomes of 1,000 per year payable continuously. b. A group of 100 annuities, each issued at age 50, with life income of 1,000 per year payable continuously. Solution a. We find directly from σ µ Var0.5 [a T ] E[a T ] that for x 20,σ/µ 0.111, x 50,σ/µ and x 80,σ/µ

49 Edward Furman, Actuarial mathematics MATH3280 p. 49/53 b. µ 100E[1000 a T(50) ] a 50 σ 2 100Var[1000 a T(50) ] Var[a T(50) ] σ Var 1/2 [a T(50) ] Thus σ µ

50 Edward Furman, Actuarial mathematics MATH3280 p. 50/53 Example 5 Assume that µ(x + t) µ and the force of interest is δ for all t 0. a. If Y a T, 0 T, display the formula for the distribution function of Y. b. If Y { 0 0 T < n a T a n T n, display the formula for the distribution function of Y. Solution a. We have already shown that in general ( ) log(1 δy) F Y (y) F T for y 1/δ. δ Under the CFM assumption, T is exponential with µ, and thus F T (t) 1 e µt.

51 Edward Furman, Actuarial mathematics MATH3280 p. 51/53 Hence, F Y (y) 1 exp ( ) log(1 δy) µ δ 1 (1 δy) µ/δ, y 1/δ b. Note again that Y denotes the present value of a continuously payable deferred annuity. Thus Y has a positive probability in zero F Y (0) P(Y 0) P(0 T n) F T (n) CFM 1 e µn. As to the case y > 0, we will use the fact that Y a T a n vn v T, δ from which follows that the largest value Y can take is v n /δ.

52 Edward Furman, Actuarial mathematics MATH3280 p. 52/53 ( v n v T ) F Y (y) P(Y y) P y P ( v T δy v n) δ ( ) ( ) P v T > v n δy P e δt > v n δy ( ) P ( δt > log(v n δy)) P T log(vn δy) δ ( ) F T log(vn δy) δ 1 (v n δy) µ/δ for 0 < y v n /δ.

53 Figure 2: The CDF of a differed n years continuously payable annuity under CFM. Edward Furman, Actuarial mathematics MATH3280 p. 53/53