STT 455-6: Actuarial Models

Transcription

1 STT 455-6: Actuarial Models Albert Cohen Actuarial Sciences Program Department of Mathematics Department of Statistics and Probability A336 Wells Hall Michigan State University East Lansing MI Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

2 Copyright Acknowledgement Many examples and theorem proofs in these slides, and on in class exam preparation slides, are taken from our textbook Actuarial Mathematics for Life Contingent Risks by Dickson,Hardy, and Waters. Please note that Cambridge owns the copyright for that material. No portion of the Cambridge textbook material may be reproduced in any part or by any means without the permission of the publisher. We are very thankful to the publisher for allowing posting of these notes on our class website. Also, we will from time-to-time look at problems from released previous Exams MLC by the SOA. All such questions belong in copyright to the Society of Actuaries, and we make no claim on them. It is of course an honor to be able to present analysis of such examples here. Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

3 Survival Models An insurance policy is a contract where the policyholder pays a premium to the insurer in return for a benefit or payment later. The contract specifies what event the payment is contingent on. This event may be random in nature Assume that interest rates are deterministic, for now Consider the case where an insurance company provides a benefit upon death of the policyholder. This time is unknown, and so the issuer requires, at least, a model of of human mortality Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

4 Survival Models Define (x) as a human at age x. Also, define that person s future lifetime as the continuous random variable T x. This means that x + T x represents that person s age at death. Define the lifetime distribution F x (t) = P[T x t] (1) the probabiliity that (x) does not survive beyond age x + t years, and it s complement, the survival function S x (t) = 1 F x (t). Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

5 Conditional Equivalence We have an important conditional relationship P[T x t] = P[T 0 x + t T 0 > x] Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

6 Conditional Equivalence We have an important conditional relationship and so P[T x t] = P[T 0 x + t T 0 > x] = P[x < T 0 x + t] P[T 0 > x] F x (t) = F 0(x + t) F 0 (x) 1 F 0 (x) (2) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

7 Conditional Equivalence We have an important conditional relationship and so P[T x t] = P[T 0 x + t T 0 > x] = P[x < T 0 x + t] P[T 0 > x] (2) In general we can extend this to F x (t) = F 0(x + t) F 0 (x) 1 F 0 (x) S x (t) = S 0(x + t) S 0 (x) (3) S x (t + u) = S x (t)s x+t (u) (4) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

8 Conditions and Assumptions Conditions on S x (t) S x (0) = 1 lim t S x (t) = 0 for all x 0 S x (t 1 ) S x (t 2 ) for all t 1 t 2 and x 0 Assumptions on S x (t) d dt S x(t) exists t R + lim t t S x (t) = 0 for all x 0 lim t t 2 S x (t) = 0 for all x 0 The last two conditions ensure that E[T x ] and E[T 2 x ] exist, respectively. Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

9 Example 2.1 Assume that F 0 (t) = 1 ( 1 t probability that (0) survives beyond age 30 (30) dies before age (40) survives beyond age 65 ) 1 6 for 0 t 120. Calculate the Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

10 Example 2.1 Assume that F 0 (t) = 1 ( 1 t probability that (0) survives beyond age 30 (30) dies before age (40) survives beyond age 65 ) 1 6 for 0 t 120. Calculate the P[(0) survives beyond age 30] = S 0 (30) = 1 F 0 (30) ( = 1 30 ) 1 6 = P[(30) dies before age 50] = F 30 (20) = F 0(50) F 0 (30) 1 F 0 (30) = P[(40) survives beyond age 65] = S 40 (25) = S 0(65) S 0 (40) = (5) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

11 The Force of Mortality Recall from basic probability that the density of F x (t) is defined as f x (t) := d dt F x(t). Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

12 The Force of Mortality Recall from basic probability that the density of F x (t) is defined as f x (t) := d dt F x(t). It follows that f 0 (x) := d dx F F 0 (x + dx) F 0 (x) 0(x) = lim dx 0 + dx P[x < T 0 x + dx] = lim dx 0 + dx (6) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

13 The Force of Mortality However, we can find the conditional density, also known as the Force of Mortality via P[x < T 0 x + dx T 0 > x] µ x = lim dx 0 + dx P[T x dx] 1 S x (dx) = lim = lim dx 0 + dx dx 0 + dx 1 S x (dx) = lim dx 0 + dx 1 = lim dx 0 + S 0 (x+dx) S 0 (x) dx = 1 d S 0 (x) dx S 0(x) = f 0(x) S 0 (x) = 1 S 0 (x) S 0 (x + dx) lim S 0 (x) dx 0 + dx (7) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

14 The Force of Mortality In general, we can show µ x+t = 1 d S x (t) dt S x(t) = f x(t) S x (t) and integration of this relation leads to S x (t) = S 0(x + t) S 0 (x) x+t = e 0 µ sds e x 0 µsds = e x+t x µ sds = e t 0 µx+sds (8) (9) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

15 Example 2.2 Assume that F 0 (t) = 1 ( 1 t ) for 0 t 120. Calculate µ x Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

16 Example 2.2 Assume that F 0 (t) = 1 ( 1 t ) for 0 t 120. Calculate µ x ) d dx S 0(x) = 1 ( 6 1 x ) µ x = = 1 ( 1 x x ) 1 6 ( ( 1 ( 6 1 x 120 ) 5 ( 6 1 )) 120 (10) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

17 Gompertz Law / Makehams s Law One model of human mortality, postulated by Gompertz, is µ x = Bc x, where (B, c) (0, 1) (1, ). This is based on the assumption that mortality is age dependent, and that the growth rate for mortality is proportional to it s own value. Makeham proposed that there should also be an age independent component, and so Makeham s Law is µ x = A + Bc x (11) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

18 Gompertz Law / Makehams s Law Of course, when A = 0, this reduces back to Gompertz Law. Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

19 Gompertz Law / Makehams s Law Of course, when A = 0, this reduces back to Gompertz Law. By definition, S x (t) = e x+t x = e At B ln c cx (c t 1) µ sds = e x+t x (A+Bc s )ds (12) Keep in mind that this is a multivariable function of (x, t) R 2 + Some online resources: CDC National Vital Statistics report, Dec Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

20 Comparison with US Gov t data Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

21 Actuarial Notation Actuaries make the notational conventions tp x = P[T x > t] = S x (t) tq x = P[T x t] = F x (t) u tq x = P[u < T x u + t] = S x (u) S x (u + t) u tq x, also known as the deferred mortality probability, is the probability that (x) survives u years, and then dies in the subsequent t years. Another convention is that p x := 1 p x and q x := 1 q x. (13) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

22 Actuarial Notation-Relationships Consequently, tp x + t q x = 1 Similarly, u tq x = u p x u+t p x t+up x = t p x up x+t µ x = 1 d xp 0 dx ( xp 0 ) µ x+t = 1 tp x d dt t p x d dt t p x = µ x+t tp x µ x+t = f x(t) S x (t) f x(t) = µ x+t tp x tp x = e t 0 µx+sds (14) (15) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

23 Actuarial Notation-Relationships Also, since F x (t) = t 0 f x(s)ds, we have as a linear approximation tq x = t 0 sp x µ x+s ds Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

24 Actuarial Notation-Relationships Also, since F x (t) = t 0 f x(s)ds, we have as a linear approximation tq x = q x = = t µ x+ 1 2 sp x µ x+s ds sp x µ x+s ds e s 0 µx+v dv µ x+s ds µ x+s ds (16) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

25 Mean and Standard Deviation of T x Actuaries make the notational definition e x := E[T x ], also known as the complete expectation of life. Recall f x (t) = t p x µ x+t = d dt t p x, and e x = = = t f x (t)dt t tp x µ x+t dt t d dt t p x dt Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

26 Mean and Standard Deviation of T x Actuaries make the notational definition e x := E[T x ], also known as the complete expectation of life. Recall f x (t) = t p x µ x+t = d dt t p x, and e x = = = = t f x (t)dt t tp x µ x+t dt t d dt t p x dt tp x dt Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

27 Mean and Standard Deviation of T x Actuaries make the notational definition e x := E[T x ], also known as the complete expectation of life. Recall f x (t) = t p x µ x+t = d dt t p x, and e x = = = = E[Tx 2 ] = t f x (t)dt t tp x µ x+t dt t d dt t p x dt tp x dt t 2 f x (t)dt = 0 2t tp x dt Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

28 Mean and Standard Deviation of T x Actuaries make the notational definition e x := E[T x ], also known as the complete expectation of life. Recall f x (t) = t p x µ x+t = d dt t p x, and e x = = = = E[Tx 2 ] = t f x (t)dt t tp x µ x+t dt t d dt t p x dt tp x dt t 2 f x (t)dt = V [T x ] := E[T 2 x ] ( e x ) 2 0 2t tp x dt (17) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

29 Example 2.6 Assume that F 0 (x) = 1 ( 1 x for a.)x = 30 and b.)x = ) 1 6 for 0 x 120. Calculate e x, V [T x ] Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

30 Example 2.6 Assume that F 0 (x) = 1 ( 1 x for a.)x = 30 and b.)x = 80. ) ) 1 6 for 0 x 120. Calculate e x, V [T x ] Since S 0 (x) = ( 1 x 6 120, it follows that in keeping with the model where survival is constrained to be les than 120, tp x = S 0(x + t) ( ) 1 = 1 t x : x + t 120 S 0 (x) 0 : x + t > 120 Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

31 Example 2.6 So, and e x = E[T 2 x ] = = 120 x x 0 ( ( 1 2t t 120 x ( 1 ) dt = 7 t 120 x ) 2(120 x) 2 ) 1 6 dt (120 x) (18) ( e 30, e 80 ) = (77.143, ) (V [T 30 ], V [T 80 ]) = ( (21.396) 2, (9.509) 2) (19) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

32 Exam MLC Spring 2007: Q8 Kevin and Kira excel at the newest video game at the local arcade, Reversion. The arcade has only one station for it. Kevin is playing. Kira is next in line. You are given: (i) Kevin will play until his parents call him to come home. (ii) Kira will leave when her parents call her. She will start playing as soon as Kevin leaves if he is called first. (iii) Each child is subject to a constant force of being called: 0.7 per hour for Kevin; 0.6 per hour for Kira. (iv) Calls are independent. (v) If Kira gets to play, she will score points at a rate of 100,000 per hour. Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

33 Exam MLC Spring 2007: Q8 Calculate the expected number of points Kira will score before she leaves. (A) 77,000 (B) 80,000 (C) 84,000 (D) 87,000 (E) 90,000 Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

34 Exam MLC Spring 2007: Q8 Define and so E [Kira s playing time] = tp x = P [Kevin still there] tp y = P [Kira still there] = = (1 t p x ) tp y dt ( 1 e 0.7t ) e 0.6t dt ( e 0.6t e 1.3t) dt = = hrs. 1.3 (20) (21) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

35 Exam MLC Spring 2007: Q8 It follows that E [Kira s winnings] = $ hr = $ Hence, we choose (E). E [Kira s playing time] (22) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

36 Numerical Considerations for T x In general, computations for the mean and SD for T x will require numerical integration. For example, Table: 2.1: Gompertz Model Statistics: (B, c) = (0.0003, 1.07) x e x SD[T x ] x + e x Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

37 Curtate Future Lifetime Define K x := T x (23) and so P [K x = k] = P [k T x < k + 1] = k q x = k p x k+1 p x (24) = k p x k p x p x+k = k p x q x+k Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

38 Curtate Future Lifetime E [K x ] := e x = = k P [K x = k] k=0 k ( k p x k+1 p x ) k=0 = kp x by telescoping series.. k=1 Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

39 Curtate Future Lifetime E [K x ] := e x = = k P [K x = k] k=0 k ( k p x k+1 p x ) k=0 = kp x by telescoping series.. k=1 E [ Kx 2 ] = k 2 P [K x = k] k=0 = 2 k kp x = 2 k=1 k kp x e x k=1 kp x k=1 (25) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

40 Relationship between e x and e x Recall e x = 0 tp x dt = j=0 j+1 By trapezoid rule for numerical integration, we obtain tp x dt 1 2 ( jp x + j+1 p x ), and so j+1 j e x j=0 1 2 ( jp x + j+1 p x ) = jp x = e x j=1 j tp x dt (26) As with all numerical schemes, this approximation can be refined when necessary. (27) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

41 Comparison of e x and e x Approximation matches well for small values of x Table: 2.2: Gompertz Model Statistics: (B, c) = (0.0003, 1.07) x e x e x Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

42 Notes An extension of Gompertz - Makeham Laws is the GM(r, s) formula µ x = h 1 r (x) + e h2 s (x), where h 1 r (x), h 2 s (x) are polynomials of degree r and s, respectively. Hazard rate in survival analysis and failure rate in reliability theory is the same as what actuaries call force of mortality. Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

43 Homework Questions HW: 2.1, 2.2, 2.5, 2.6, 2.10, 2.13, 2.14, 2.15 Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

44 Life Tables Define for a model with maximum age ω and initial age x 0 the radix l x0, where l x0 +t = l x0 tp x0 (28) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

45 Life Tables It follows that l x+t = l x0 x+t x0 p x0 = l x0 x x0 p x0 tp x = l x tp x Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

46 Life Tables It follows that l x+t = l x0 x+t x0 p x0 = l x0 x x0 p x0 tp x = l x tp x (29) tp x = l x+t l x We assume a binomial model where L t is the number of survivors to age x + t. Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

47 Life Tables So, if there are l x independent individuals aged x with probability t p x of survival to age x + t, then we interpret l x+t as the expected number of survivors to age x + t out of l x independent individuals aged x. Symbolically, E[L t L 0 = l x ] = l x+t = l x tp x (30) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

48 Life Tables So, if there are l x independent individuals aged x with probability t p x of survival to age x + t, then we interpret l x+t as the expected number of survivors to age x + t out of l x independent individuals aged x. Symbolically, E[L t L 0 = l x ] = l x+t = l x tp x (30) Also, define the expected number of deaths from year x to year x + 1 as ( d x := l x l x+1 = l x 1 l ) x+1 = l x (1 p x ) = l x q x (31) l x Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

49 Example 3.1 Table: 3.1: Extract from a life table x l x d x Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

50 Example 3.1 Calculate: a.) l 40 b.) 10 p 30 c.) q 35 d.) 5 q 30 e.) P [(30) dies between age 35 and 36] Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

51 Example 3.1 Calculate: a.) l 40 = l 39 d 39 = Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

52 Example 3.1 Calculate: a.) l 40 = l 39 d 39 = b.) 10 p 30 = l 40 l 30 = Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

53 Example 3.1 Calculate: a.) l 40 = l 39 d 39 = b.) 10 p 30 = l 40 l 30 = c.) q 35 = d 35 l 35 = Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

54 Example 3.1 Calculate: a.) l 40 = l 39 d 39 = b.) 10 p 30 = l 40 l 30 = c.) q 35 = d 35 l 35 = d.) 5 q 30 = l 30 l 35 l 30 = Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

55 Example 3.1 Calculate: a.) l 40 = l 39 d 39 = b.) 10 p 30 = l 40 l 30 = c.) q 35 = d 35 l 35 = d.) 5 q 30 = l 30 l 35 l 30 = e.) P [(30) dies between age 35 and 36] = l 35 l 36 l 30 = Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

56 Fractional Age Assumptions So far, the life table approach has mirrored the survival distribution method we encountered in the previous lecture. However, in detailing the life table, no information is presented on the cohort in between whole years. To account for this, we must make some fractional age assumptions. The following are equivalent: Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

57 Fractional Age Assumptions So far, the life table approach has mirrored the survival distribution method we encountered in the previous lecture. However, in detailing the life table, no information is presented on the cohort in between whole years. To account for this, we must make some fractional age assumptions. The following are equivalent: UDD1 For all (x, s) N [0, 1), we assume that s q x = s q x UDD2 For all x N, we assume R x := T x K x U(0, 1) R x is independent of K x. Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

58 Proof of Equivalence Proof: UDD1 UDD2: Assume for all (x, s) N [0, 1), we assume that sq x = s q x. Then P [R x s] = and so R x U(0, 1). = P [R x s, K x = k] k=0 P [k T x k + s] k=0 = kp x sq x+k = kp x s q x+k k=0 k=0 = s kp x q x+k = s P [K x = k] = s k=0 k=0 (32) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

59 Proof of Equivalence To show independence of R x and K x, P [R x s, K x = k] = P [k T x k + s] = k p x sq x+k = s kp x q x+k = P[R x s] P[K x = k] (33) UDD2 UDD1: Assuming UDD2 is true, then for (x, s) N [0, 1) we have sq x = P [T x s] = P [K x = 0, R x s] = P[R x s] P[K x = 0] (34) = s q x QED Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

60 Corollary Recall that s q x = lx lx+s l x. It follows now that sq x = sq x = s d x l x = l x l x+s l x Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

61 Corollary Recall that s q x = lx lx+s l x. It follows now that sq x = sq x = s d x l x = l x l x+s l x l x+s = l x s d x which is a linear decreasing function of s [0, 1) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

62 Corollary Recall that s q x = lx lx+s l x. It follows now that sq x = sq x = s d x l x = l x l x+s l x l x+s = l x s d x which is a linear decreasing function of s [0, 1) (35) q x = d ds [ sq x ] = f x (s) = s p x µ x+s But, since q x is constant in s, we have f x (s) is constant for s [0, 1). Read over Examples Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

63 Constant Force of Mortality for Fractional Age For all (x, s) N [0, 1), we assume that µ x+s does not depend on s, and we denote µ x+s := µ x. It follows that p x = e 1 0 µx+sds = e µ x sp x = e s 0 µ x du = e µ x s = (p x ) s sp x+t = e s 0 µ x du = (p x ) s when t + s < 1 ( 1) k+1 (µ q x = 1 e µ x = x) k µ x k! k=1 sq x = 1 e µ x t µ x t, where the last two lines assume µ x 1. Read Examples 3.6, 3.7 and Sections 3.4, 3.5, 3.6. (36) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

64 Homework Questions HW: 3.1, 3.2, 3.4, 3.7, 3.8, 3.9, 3.10 Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

65 Contingent Events We have spent the previous two lectures on modeling human mortality. The need for such models in insurance pricing arises when designing contracts that are event-contingent. Such events include reaching retirement before the end of the underlying life (x). However, one can also write contracts that are dependent on a life (x) being admitted to college (planning for school), and also on (x) s external portfolios maintaining a minimal value over a time-interval (insuring external investments.) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

66 Contingent Events: General Case Consider a probability space (Ω, F, P) and an event A F. If we are working with a force of interest δ s (ω) and the time of event W as τ W, then we have under the stated probability measure P the Expected Present Value of a payoff K(ω) contingent upon W Actuarial Encounters of the Third Kind!! EPV = E[K(ω)e τ W 0 δ s(ω)ds ] (37) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

67 Some Initial Simplifying Assumptions K(ω) = 1 for all ω Ω (a.s.) δ s (ω) = δ for all ω Ω (a.s.) W := {event that (x) dies} τ W := T x P is obtained via historical observation and is thus a physical measure. Specifically, we use t p x obtained from life tables or via models of human mortality We do not assume now that a unique risk-neutral pricing measure P exists. Standard Ultimate Survival Model with assumes Makeham s law with (A, B, c) = ( , , 1.124) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

68 Recall... The equivalent interest rate i := e δ 1 per year The discount factor v := 1 1+i = e δ per year ( ) The nominal interest rate i (p) = p (1 + i) 1 p 1 compounded p times per year The effective rate of discount d := 1 v = i v = 1 e δ per year ( ) The nominal rate of discount d (p) := p 1 v 1 p compounded p times per year Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

69 Whole Life Insurance: Continuous Case Consider now the random variable Z = v Tx = e δtx (38) which represents the present value of a dollar upon death of (x). We are interested in statistical measures of this quantity: E[Z] = Āx := E[e δtx ] = E[Z 2 ] = 2 Ā x := E[e 2δTx ] = 0 e 2δt tp x µ x+t dt Var(Z) = 2 Ā x ( ) 2 Ā x [ ] ln (z) P[Z z] = P T x δ 0 e δt tp x µ x+t dt (39) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

70 Whole Life Insurance: Yearly Case Assuming payments are made at the end of the death year, our random variable is now Z = v Kx +1 = e δkx δ and so E[Z] = A x := E[v Kx +1 ] = = E[Z 2 ] = k=0 k=0 v k+1 k q x v 2k+2 k q x v k+1 P[K x = k] k=0 Var(Z) = 2 A x (A x ) 2 [ ] δ ln (z) P[Z z] = P K x δ (40) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

71 1 thly Whole Life Insurance: m Case Instead of only paying at the end of the last whole year lived, an insurance contract might specify payment upon the end of the last period lived. In this case, if we split a year into m periods, and define For example, if K x = 19.78, then K (m) x = K (m) x = 1 m mt x (41) 19 m = = 19.5 m = = m = = m = 12 Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

72 1 thly Whole Life Insurance: m Case It follows that we need r { 0, 1 m, 2 m [ ] P K x (m) = r = P,..., m 1 m [ r T x < r + 1 m to compute statistics for our random variable m+1, 1, m,...} ] = r 1 q x (42) m Z = v K (m) x + 1 m (43) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

73 1 thly Whole Life Insurance: m Case E[Z] = A (m) x E[Z 2 ] = 2 A (m) x = ( Var(Z) = 2 A (m) P[Z z] = P x := E[v K x (m) k=0 A (m) x + 1 m ] = v 2k+2 m k m 1 m q x ) 2 [ K x (m) ln (z) 1 δ m k=0 ] v k+1 m k m 1 m q x (44) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

74 Recursion Method One of the computational tools we share directly with quantitative finance is the method of backwards-pricing. In option pricing, we assume the contract has a finite term. Here, we assume a finite lifetime maximum of ω <. It follows that [ ] A ω 1 = E v K ω 1+1 = E [ v 1] = v (45) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

75 Recursion Method One of the computational tools we share directly with quantitative finance is the method of backwards-pricing. In option pricing, we assume the contract has a finite term. Here, we assume a finite lifetime maximum of ω <. It follows that [ ] A ω 1 = E v K ω 1+1 = E [ v 1] = v (45) At age ω 2, we have P[K ω 2 = 0] = q ω 2 and so A ω 2 = E [v ] K ω 2+1 = q ω 2 v + p ω 2 E [v ] (1+K ω 1)+1 = q ω 2 v + p ω 2 v E [v ] K ω 1+1 = q ω 2 v + p ω 2 v 2 (46) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

76 Recursion Method In general, we have the recursion equation for a life (x) that satisfies A x = vq x + vp x A x+1 A ω 1 = v (47) in the whole life case, and A (m) x A (m) ω 1 m = v 1 m 1 q x + v 1 m 1 p x A (m) m m x+ 1 m = v 1 m (48) in the 1 m thly case. Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

77 Recursion Method Recall that for Makeham s law we have, respectively 1 m ( p x = e A m Bc x c 1 ) m 1 ln (c) 1p x = e (49) Bc x A ln (c) (c 1) and for the power law of survival, S 0 (x) = ( 1 x ω ) a for some a, ω > 0 1 p x = m ( ω x 1 m ω x ) a ( ) (50) ω x 1 a 1p x = ω x Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

78 Recursion Method For any positive integer m, it follows that for Makeham s law: ( ( )) = v 1 m 1 e A m Bc x c m 1 1 ln (c) A (m) x A (m) ω 1 m ( + v 1 m e A m Bc x c 1 ) m 1 ln (c) A (m) x+ 1 m = v 1 m (51) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

79 Recursion Method For any positive integer m, it follows that for Power law: ( ( ) A (m) x = v 1 ω x 1 a ) m m 1 ω x ( ) + v 1 ω x 1 a m m A (m) ω x x+ 1 m A (m) ω 1 m = v 1 m (52) HW Project: For Power law with a = 3 5 and ω = 101 Generate a spreadsheet like Table 4.1 in the text, including values for 2 A (m) x Repeat Example 4.3 with the Power law model Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

80 Term Insurance: Continuous Case Consider now the case where payment is made in the continuous case, and death benefit is payable to the policyholder only if T x n. Then, we are interested in the random variable and so Z = e δtx 1 {Tx n} (53) Ā 1 x:n = E[Z] = 2 Ā 1 x:n = E [ Z 2] = n 0 n 0 e δt tp x µ x+t dt e 2δt tp x µ x+t dt (54) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

81 1 thly Term Insurance: m Case Consider again the case where the death benefit is payable at the end of the 1 m thly period in the death year to the policyholder only if K (m) x + 1 m n. Then, we are interested in the random variable and so Z = e (m) δ(k x + 1 m ) 1 { } (55) K x (m) + 1 m n A (m)1 x:n = E[Z] = 2 A (m)1 x:n = E [ Z 2] = mn 1 k=0 v k+1 m k m 1 m q x k=0 mn 1 v 2k+2 m (56) k m 1 q x m Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

82 Pure Endowment Pure endowment benefits depend on the survival policyholder (x) until at least age x + n. In such a contract, a fixed benefit of 1 is paid at time n. This is expressed via Z = e δn 1 {Tx n} A 1 x:n = E[Z] = v n np x = e δn np x (57) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

83 Endowment Insurance Endowment insurance is a combination of term insurance and pure endowment. In such a policy, the amount is paid upon death if it occurs with a fixed term n. However, if (x) survives beyond n years, the sum insured is payable at the end of the n th year. The corresponding present value random variable is δ min{tx,n} Z = e E[Z] = Āx:n = n 0 e δt tp x µ x+t dt + e δn np x = Ā1 x:n + A 1 x:n (58) This can be generalized once again to the 1 m thly case and for E[Z 2 ] Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

84 Deferred Insurance Benefits Suppose policyholder on a life (x) receives benefit 1 if u T x < u + n. Then Z = e δtx 1 {u Tx <u+n} E[Z] = u Ā 1 x:n = = u+n u n 0 e δt tp x µ x+t dt e δ(s+u) s+up x µ x+s+u ds n = e δu e δs up x sp x+u µ x+s+u ds 0 = e δu up x Ā 1 x+u:n (59) = Ā 1 x:u+n Ā1 x:u Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

85 Relationships By definition, we have A x = A 1 x:n + n A x = A 1 x:n + v n np x A x+n (60) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

86 Relationships By definition, we have What about relationship between Ā x and A x? A x = A 1 x:n + n A x = A 1 x:n + v n np x A x+n (60) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

87 Employing UDD: Āx vs A x If expected values are computed via information derived from life tables, then certainly Ā x must be approximated using techniques from previous lecture. Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

88 Employing UDD: Āx vs A x If expected values are computed via information derived from life tables, then certainly Ā x must be approximated using techniques from previous lecture. Recall that by the definition of s p x and the UDD, we have sp x µ x+s = f x (s) = d ds P[T x s] = d ds ( sq x ) = d ds (s q x) (61) = q x Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

89 Employing UDD: Āx vs A x It follows that using the UDD approximation leads to Ā x = 0 e δt tp x µ x+t dt = 1 k=0 k+1 k e δt tp x µ x+t dt = kp x v k+1 e δ e δs sp x+k µ x+k+s ds k=0 0 1 kp x v k+1 q x+k e δ e δs ds k=0 0 1 = v k+1 P[K x = k] e δ e δs ds = A x k=0 = A x i δ e δ e δs ds Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

90 Employing UDD: Āx vs A x It follows that using the UDD approximation leads to Ā x = 0 e δt tp x µ x+t dt = 1 k=0 k+1 k e δt tp x µ x+t dt = kp x v k+1 e δ e δs sp x+k µ x+k+s ds k=0 0 1 kp x v k+1 q x+k e δ e δs ds k=0 0 1 = v k+1 P[K x = k] e δ e δs ds = A x k=0 = A x i δ A x (m) i i (m) A x = 0 i ) A m ((1 + i) 1 x m e δ e δs ds (62) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

91 Claims Accelaration Approach : A (m) x vs A x Consider now a policy that pays the holder at the end of the 1 thly m period of death. In this case, the benefit is paid at one of the times r where { r K x + 1 m, K x + 2 m,..., K x + m } (63) m and so under the UDD, E [T payment K x = k] = m j=1 ( 1 m k + j ) = k + m + 1 m 2m (64) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

92 Claims Accelaration Approach : A (m) x vs A x Once again, it follows using the UDD aproximation A (m) x = E[v K x (m) = + 1 m ] = k=0 v k+1 m k m 1 m q x v E[Tpayment Kx =k] P [K x = k] k=0 k=0 = (1 + i) m 1 2m m+1 k+ v 2m k q x = (1 + i) m 1 2m Ax (1 + i) 1 2 Ax k=0 v k+1 k q x as m. Note that using the UDD approximation, both Āx A x are independent of x. and A(m) x A x (65) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

93 Variable Insurance Benefits Upon death, we have considered policies that pay the holder a fixed amount. What varied was the method and time of payment. If, however, the actual payoff amount depended on the time T x of death for (x), then we term such a contract a Variable Insurance Contract. Specifically, if the payoff amount dependent on T x is h(t x ), then Z = h(t x )e δtx E[Z] = 0 h(t)e δt tp x µ x+t dt (Ī Ā ) x := te δt tp x µ x+t dt 0 (Ī Ā ) n 1 x:n := te δt tp x µ x+t dt 0 (66) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

94 Example 4.8 Consider an n year term insurance issued to (x) under which the death benefit is paid at the end of the year of death. The death benefit if death occurs between ages x + k and x + k + 1 is valued at (1 + j) k. Hence, using the definition i := 1+i 1+j 1, Z = v Kx +1 (1 + j) Kx n 1 E[Z] = v k+1 (1 + j) k k q x k=0 = 1 n j v k+1 (1 + j) k+1 k q x k=0 = 1 n j k=0 k q x ( 1+i 1+j ) k+1 = j A1 x:n (67) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

95 Homework Questions HW: 4.1, 4.2, 4.3, 4.7, 4.9, 4.11, 4.12, 4.14, 4.15, 4.16, 4.17, 4.18 Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

96 Life Annuities A Life Annuity refers to a series of payments to or from an individual as long as that person is still alive. For a fixed rate i and term n, we recall the deterministic pricing theory: ä n i = 1 + v v n 1 = 1 v n d a n i = v v n = ä n i 1 + v n = 1 v n i n ā n i = v t dt = 1 v n 0 δ ä (m) n i = 1 ( ) m 1 + v 1 m v n 1 m = 1 v n d (m) a (m) n i = 1 (v ) m 1 m v n 1 m + v n = 1 v n i (m) (68) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

97 Whole Life Annuity Due Consider the case where 1 is paid out at the beginning of every period until death. Our present random variable is now and so Y := ä Kx +1 = 1 v Kx +1 d [ 1 v K x ] +1 ä x = E[Y ] = E d [ 1 v K x ] +1 V [Y ] = V d = A x A 2 x d 2 = 1 A x d = 1 d 2 V [1] + 1 d 2 V [v Kx +1 ] (69) (70) (71) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

98 Whole Life Annuity Due The present value random variable can also be represented as Y = v k 1 {Tx >k} (72) k=0 As P[T x > k] = t p x, we have the alternate expression for ä x [ ] ä x = E[Y ] = E v k 1 {Tx >k} = = k=0 E[v k 1 {Tx >k}] k=0 v k kp x = k=0 k q x ä k+1 k=0 (73) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

99 Term Annuity Due Define the present value random variable { äkx Y = +1 : K x {0, 1, 2,..., n 1} ä n : K x {n, n + 1, n + 2,...} Another expression is and so Y = ä min{kx +1,n} = 1 v min{kx +1,n} d (74) ä x:n = E[Y ] = 1 E [ min{kx v +1,n}] d = 1 A x:n d n 1 n 1 = v t tp x = k q x ä k+1 + n p x ä n t=0 k=0 (75) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

100 Whole Life and Term Immediate Annuity Define Y = k=1 v k 1 {Tx >k}. Then we have an annuity immediate that begins payment one unit of time from now. It follows that a x = ä x 1 V [Y ] = V [Y ] Also, if we define Y = a min{kx,n}, then (76) n a x:n = v t tp x = ä x:n 1 + v n np x (77) t=1 Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

101 Whole Life Continuous Annuity Define Y = ā Tx = 1 v Tx δ ā x = E[Y ] = 1 Āx δ = = 0 0 e δt 1 {Tx >t}dt e δt tp x dt (78) Note that if δ = 0, then ā x = e x Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

102 Term Continuous Annuity Define Y = ā min{tx,n}. Then min{tx 1 v,n} Y = δ ā x:n = E [Y ] = 1 Ā x:n δ = n 0 e δt tp x dt (79) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

103 Deferred Annuity Consider now the case of an annuity for (x) that will pay 1 at the end of each year, beginning at age x + u and will continue until death age x + T x. We define u ä x to be the Expected Present Value of this policy. It should be apparent that u ä x = ä x ä x:u = v t tp x t=u = v t tp x v t tp x = v t tp x ä x+u holds in the discrete case, and similarly in the continuous case, t=0 (80) u ā x = ā x ā x:u (81) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

104 Term Deferred Annuity For the cases of an annuity for (x) that will pay 1 at the end of each year, beginning at age x + u and will continue until death age x + T x up to a term of length n, or annuity-due payable 1 m thly. Then u a x:n = v u up x a x+u:n u ä x (m) = v u up x ä (m) x+u respectively. These combine with the previous slide to reveal the useful formulae: (82) ä x:n = ä x v n np x ä x+n ä (m) x:n = ä(m) x v n np x ä (m) x+n (83) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

105 Guaranteed Annuities There are instances where an age (x) wishes to buy a policy where payments are guaranteed to continue upon death to a beneficiary. In this case, define the present random variable as Y = ä n + Y 1, where { 0 : Kx {0, 1, 2,..., n 1} Y 1 = ä Kx +1 ä n : K x {n, n + 1, n + 2,...} and so [( ) ] E[Y 1 ] = E ä Kx +1 ä n 1 {Kx n} = n ä x = v n np x ä x+n Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

106 Guaranteed Annuities There are instances where an age (x) wishes to buy a policy where payments are guaranteed to continue upon death to a beneficiary. In this case, define the present random variable as Y = ä n + Y 1, where { 0 : Kx {0, 1, 2,..., n 1} Y 1 = ä Kx +1 ä n : K x {n, n + 1, n + 2,...} and so [( ) ] E[Y 1 ] = E ä Kx +1 ä n 1 {Kx n} = n ä x = v n np x ä x+n E[Y ] := ä x:n = ä n + v n np x ä x+n and E[Y (m) ] := ä (m) = x:n ä(m) n + v n np x ä (m) x+n (84) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

107 Example 5.4 A pension plan member is entitled to a benefit of 1000 per month, in advance, for life from age 65, with no guarantee. She can opt to take a lower benefit with a 10 year guarantee. The revised benefit is calculated to have equal EPV at age 65 to the original benefit. Calculate the revised benefit using the Standard Ultimate Survival Model, with interest at 5% per year. Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

108 Example 5.4 Let B denote the revised monthly benefit. Then the two options are per year, paid per month with Present Value Y 1 12B per year, paid per month with Present Value Y 2 Hence E[Y 1 Y 2 ] = 0 implies 12000ä (12) 65 = 12Bä (12) = 12B 65:10 ( ) ä (12) 10 + v 10 10p 65 ä (12) 75 Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

109 Example 5.4 Let B denote the revised monthly benefit. Then the two options are per year, paid per month with Present Value Y 1 12B per year, paid per month with Present Value Y 2 Hence E[Y 1 Y 2 ] = 0 implies 12000ä (12) 65 = 12Bä (12) = 12B B = :10 ( ) ä (12) 10 + v 10 10p 65 ä (12) 75 ä (12) 65 ä (12) 10 + v 10 10p 65 ä (12) 75 = = Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

110 Example 5.4 Let B denote the revised monthly benefit. Then the two options are per year, paid per month with Present Value Y 1 12B per year, paid per month with Present Value Y 2 Hence E[Y 1 Y 2 ] = 0 implies 12000ä (12) 65 = 12Bä (12) = 12B 65:10 ( ) ä (12) 10 + v 10 10p 65 ä (12) 75 ä (12) 65 B = 1000 ä (12) 10 + v 10 10p 65 ä (12) 75 = = V [Y 1 Y 2 ] = 0? (85) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

111 Linearly Increasing Annuities Define an annuity where the payments increase linearly at times t = 0, 1, 2,.. provided that (x) is alive at time t (I ä) x = (t + 1) v t tp x t=0 n 1 (86) (I ä) x:n = (t + 1) v t tp x t=0 Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

112 Linearly Increasing Annuities If then the annuity is payable continuously, with payments increasing by 1 at each year end and the rate of payment in the t th year constant and equal to t for t {1, 2,..m,.., n}, then h(t) = m1 {m t<m+1}, and the EPV is If h(t) = t, then n 1 (I ā) x:n = (m + 1) m ā x:1 (87) m=0 (Ī ā) x:n = n 0 te δt tp x dt (88) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

113 Evaluating Annuities Using Recursion By recursion, we observe ä x = 1 + vp x + v 2 2p x + v 3 3p x +... = 1 + vp x ( 1 + vpx+1 + v 2 2p x+1 + v 3 3p x ) = 1 + vp x ä x+1 Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

114 Evaluating Annuities Using Recursion By recursion, we observe ä x = 1 + vp x + v 2 2p x + v 3 3p x +... = 1 + vp x ( 1 + vpx+1 + v 2 2p x+1 + v 3 3p x ) = 1 + vp x ä x+1 (89) ä (m) x = 1 m + v 1 m 1 p x ä (m) m x+ 1 m Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

115 Evaluating Annuities Using Recursion By recursion, we observe ä x = 1 + vp x + v 2 2p x + v 3 3p x +... = 1 + vp x ( 1 + vpx+1 + v 2 2p x+1 + v 3 3p x ) = 1 + vp x ä x+1 (89) ä (m) x = 1 m + v 1 m 1 p x ä (m) m x+ 1 m Consider the case where there is a maximum age in the model, and so ä ω 1 = 1 ä (m) ω 1 m = 1 m (90) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

116 Evaluating Annuities Using UDD Recall that under the UDD assumption, A (m) x = i i (m) A x Ā x = i δ A x (91) and by definition, ä x = 1 A x d ä (m) x = 1 A(m) x d (m) ā x = 1 Āx δ (92) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

117 Evaluating Annuities Using UDD It follows that ä (m) x = 1 A(m) x d (m) = 1 i i (m) A x d (m) = i (m) ia x i (m) d (m) = i (m) i(1 dä x ) i (m) d (m) id = i (m) d (m) äx i i (m) i (m) d (m) := α(m)ä x β(m) ā x = id δ 2 äx i δ δ 2 (93) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

118 Evaluating Annuities Using UDD For term annuities, we have ä (m) x:n = ä(m) x v n np x ä (m) x+n = α(m)ä x β(m) v n np x (α(m)ä x+n β(m)) = α(m) ( ä x v n np x ä (m) x+n = α(m) ä x:n β(m) (1 v n np x ) ä x:n m 1 2m (1 v n np x ) ) β(m) (1 v n np x ) (94) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

119 Woolhouse s Formula Consider a function g : R + R such that lim t g(t) = 0, then 0 g(t)dt = h k=0 g(kh) h h2 g(0) g (0) h4 720 g (0) +... (95) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

120 Woolhouse s Formula Define and so for h = 1, g(t) = v t tp x g (t) = t p x δe δt v t tp x µ x+t (96) g (0) = δ µ x ā x = g(k) g (0) k=0 v k kp x (δ + µ x) k=0 = ä x (δ + µ x) (97) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

121 Woolhouse s Formula Correspondingly, for h = 1 m, ā x 1 m = k=0 g k=0 ( ) k 1 m 2m m 2 g (0) v k m k p x 1 m 2m 1 12m 2 (δ + µ x) = ä (m) x 1 2m 1 12m 2 (δ + µ x) (98) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

122 Woolhouse s Formula Equating the previous two approximations for ā x, we obtain ä (m) x ä x m 1 2m m2 1 12m 2 (δ + µ x) (99) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

123 Woolhouse s Formula For term annuities, we obtain the approximation ä (m) x:n ä x:n m 1 2m (1 v n np x ) m2 1 12m 2 (δ+µ x v n np x (δ+µ x+n )) (100) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

124 Woolhouse s Formula Letting m, we get ā x ä x (δ + µ x) ā x:n ä x:n 1 2 (1 v n np x ) 1 12 (δ + µ x v n np x (δ + µ x+n )) (101) For ā x with δ = 0, the approximation above reduces further to e x (e x + 1) µ x (102) NB: For life tables, we can compute these quantities using the approximation µ x 1 2 [ln (p x) + ln (p x+1 )] Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

125 Select and Ultimate Survival Models Notation: Aggregate Survival Models: Models for a large population, where tp x depends only on the current age x. Select (and Ultimate) Survival Models: Models for a select group of individuals that depend on the current age x and Future survival probabilities for an individual in the group depend on the individual s current age and on the age at which the individual joined the group d > 0 such that if an individual joined the group more than d years ago, future survival probabilities depend only on current age. So, after d years, the person is considered to be back in the aggregate population. Ultimately, a select survival model includes another event upon which probabilities are conditional on. Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

126 Select and Ultimate Survival Models Notation: d is the select period The mortality applicable to lives after the select period is over is known as the ultimate mortality. A select group should have a different mortality rate, as they have been offered (selected for) life insurance. A question, of course, is the effect on mortality by maintaining proper health insurance. Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

127 Example 3.8 Consider men who need to undergo surgery because they are suffering from a particular disease. The surgery is complicated and P[survive one year after surgery] = 0.5 (d, l 60, l 61, l 70 ) = (1, 89777, 89015, 77946) Calculate P[A], P[B], P[C], where A = {(60),about to have surgery, will be alive at age 70} B = {(60),had surgery at age 59, will be alive at age 70} C = {(60),had surgery at age 58, will be alive at age 70} (103) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

128 Example 3.8 P[A] = P[(60),about to have surgery, alive at age 61] l70 l 61 = = P[B] = l 70 l 60 = P[C] = l 70 l 60 = (104) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

129 Select Survival Models S [x]+s (t) = P[(x + s) selected at (x), survives to(x + s + t)] tq [x]+s = P[(x + s) selected at (x)dies before(x + s + t)] µ [x]+s = force of mortality at (x + s) for select at (x) ( ) 1 S[x]+s (h) = lim h 0 + h tp [x]+s = 1 t q [x]+s = S [x]+s (t) = e t 0 µ [x]+s+udu (105) For t < d, we refer to to the above as part of the select model. For t d, they are part of the ultimate model. Please read through section on Select Life Tables. Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

130 Select Life Tables Sometimes, we wish to compute values from life tables. Consider again a model where x x 0, where x 0 is the initial age, and 0 t d. Then l x+d = d t p [x]+t l [x]+t (106) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

131 Example 3.9 Theorem Consider y x + d > x + s > x + t x x 0. Then y x tp [x]+t = l y l [x]+t s tp [x]+t = l [x]+s l [x]+t (107) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

132 Example 3.9 Proof. y x tp [x]+t = y x d p [x]+d d t p [x]+t = y x d p x+d d t p [x]+t = l y l x+d l x+d l [x]+t = l y l [x]+t s tp [x]+t = d t p [x]+t d sp [x]+s (108) = l x+d l [x]+t l [x]+s l x+d = l [x]+s l [x]+t Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

133 Example 3.11 A select survival model has a select period of three years. Its ultimate mortality is equivalent to the US Life Tables, 2002 Females of which an extract is shown below. Information given is that for all x 65, ( p[x], p [x 1]+1, p [x 2]+2 ) = (0.999, 0.998, 0.997). (109) Table: 3.5: Extract from US LIfe Tables, 2002 Females x l x Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

134 Example 3.11 Calculate the probability that a woman currently aged 70 will survive to age 75 given that 1 she was select at age 67: 2 she was select at age 68 3 she was select at age 69 4 she was select at age 70 Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

135 Example p [70 3]+3 = 5 p 70 = l 75 l 70 = p [70 2]+2 = l [68]+2+5 = l 75 l [68]+2 l [68]+2 = l 75 l 75 = l [68]+3 1p l [68] p [68]+2 = 4 p 71 1p [68]+2 = = (110) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

136 Example p [70 1]+1 = l [69]+1+5 l [69]+1 = l 75 l [69]+1 = l 75 l [69]+3 ( 1 p [69]+1 ) ( 1 p [69]+2 ) = l 75 l 72 ( 1 p [69]+1 ) ( 1 p [69]+2 ) = = p [70] = l 75 l 73 ( 1 p [70] ) ( 1 p [70]+1 ) ( 1 p [70]+2 ) = = (111) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

137 Example 3.12 Given a table of values for q [x], q [x 1]+1, q x and the knowledge that the model incorporates a 2 year selct period, compute 4p [70] 3q [60]+1 Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

138 Example 3.12 Given a table of values for q [x], q [x 1]+1, q x and the knowledge that the model incorporates a 2 year selct period, compute 4p [70] 3q [60]+1 4p [70] = p [70] p [70]+1 p [70]+2 p [70]+3 = p [70] p [70]+1 p 72 p 73 = ( 1 q [70] ) ( 1 q[70]+1 ) (1 q72 ) (1 q 73 ) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

139 Example 3.12 Given a table of values for q [x], q [x 1]+1, q x and the knowledge that the model incorporates a 2 year selct period, compute 4p [70] 3q [60]+1 4p [70] = p [70] p [70]+1 p [70]+2 p [70]+3 = p [70] p [70]+1 p 72 p 73 = ( 1 q [70] ) ( 1 q[70]+1 ) (1 q72 ) (1 q 73 ) 3q [60]+1 = q [60]+1 + p [60]+1 q 62 + p [60]+1 p 62 q 63 (112) = q [60]+1 + ( 1 q [60]+1 ) q62 + ( 1 q [60]+1 ) (1 q62 ) q 63 Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

140 Example 3.13 A select survival model has a two-year select period and is specified as follows. The ultimate part of the model follows Makeham s law, where (A, B, c) = ( , , 1.124): µ x = ( ) (1.124) x (113) The select part of the model is such that for 0 s 2, and so for 0 t 2, µ [x]+s = s µ x+s (114) ( tp [x] = e t 0 µ [x]+sds = exp [0.9 2 t t ln (0.9) + ct 0.9 t ln ( ) 0.9 c )] (115) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

141 Example 3.13 It follows that given an initial cohort at age x 0, that is given l x0, we can compute the entries of a select life table via l x = p x 1 l x 1 l [x]+1 = l x+2 p [x]+1 l [x] = l x+2 2p [x] (116) Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

142 Homework Questions HW: 3.1, 3.2, 3.4, 3.7, 3.8, 3.9, 3.10, 5.1, 5.3, 5.5, 5.6, 5.11, 5.14 Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

143 What is a Premium? When entering into a contract, the financial obligations of all parties must be specified. In an insurance contract, the insurance company agrees to pay the policyholder benefits in return for premium payments. The premiums secure the benefits as well as pay the company for expenses attached to the administation of the policy Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283

144 Premium Types A Net Premium does not explicitly allow for company s expenses, while a Office or Gross Premium does. There may be a Single Premium or or a series of payments that could even match with the policyholder s salary freequency. It is important to note that premiums are paid as soon as the contract is signed, otherwise the policyholder would attain coverage before paying for it with the first premium. This could be seen as an arbitrage opportunity - non-zero probability of gain with no money up front. Albert Cohen (MSU) STT 455-6: Actuarial Models MSU / 283