Survival Models Solution Chapter 2

Suggested solutions to DHW textbook exercises Exercise 2.1 (a) The probability that a newborn life dies before age 60 is given by Pr[T0 ≤ 60] = F0 (60) = 1 − (1 − 60/105)1/5 = 1 − (45/105)1/5 = 1 − (3/7)1/5 = 0.1558791. (b) The probability that (30) survives to at least age 70 is Pr[T30

1 − F0 (70) 351/5 Pr[T0 > 70] = = 1/5 = > 40] = Pr[T0 > 30] 1 − F0 (30) 75



7 15

1/5 = 0.8586207.

(c) The probability that (20) dies between 90 and 100 is Pr[70 < T20 ≤ 80] =

F0 (100) − F0 (90) 151/5 − 51/5 Pr[90 < T0 ≤ 100] = = = 0.1394344. Pr[T0 > 20] 1 − F0 (20) 851/5

(d) First, derive the form of the force of mortality: dF0 (x)/dx f0 (x) = = µx = 1 − F0 (x) 1 − F0 (x) Thus, µ50 =

1 5

1− 1



1 x −4/5 105 105
x 1/5 − 105

=

1 . 5(105 − x)

1 = 0.003636364. 5(55)

(e) The median future lifetime of (50) is the solution m to 1  m 1/5 Pr[T50 > m] = = 1 − . 2 55 This leads us to m = 55[1 − (1/2)5 ] = 53.28125. (f) For a person currently age 50, his survival function is  1/5  1/5 55 − t Pr[T0 > 50 + t] t = = 1− , tp50 = Pr[T50 > t] = Pr[T0 > 50] 55 55 for 0 ≤ t ≤ 55. His complete expectation of life is therefore 1/5 Z 1 Z 55 Z 55  t dt = 55 u1/5 du = 55(5/6) = 45.83333. ˚ e50 = 1− tp50 dt = 55 0 0 0 (g) The curtate expectation of life at age 50 is 1/5  1/5  1/5  1/5 55 55  X X k 54 53 1 e50 = 1− = + + ··· + = 45.17675. kp50 = 55 55 55 55 k=1 k=1 The sum above can be done in an R program as follows: > k e sum(e) [1] 45.17675

Prepared by E.A. Valdez

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Suggested solutions to DHW textbook exercises Exercise 2.2 (a) The implied limiting age ω is the solution to G(ω) = 0 which leads us to 18000 − 110ω − ω 2 = −(ω − 90)(ω + 200) = 0. Thus, ω = 90 since the limiting age cannot be negative. (b) For G to be a legitimate survival function, it must satisfy 3 conditions: (i) G(0) = 1: trivial (ii) G(ω) = 0: verified in (a) above. (iii) G must be non-increasing. We check whether dG(x)/dx ≤ 0. dG(x) −2(55 + x) = dx 18000 which clearly is non-positive for all 0 ≤ x ≤ 90. (c) Now that we have verified G(x) is a legitimate survival function, we can write it as S0 (x) so that 15400 77 18000 − 110(20) − 202 = = = 0.8555556. 20p0 = Pr[T0 > 20] = S0 (20) = 18000 18000 90 This gives the probability that a newborn will survive to age 20. (d) The survival function for a life age 20 can be expressed as S0 (20 + t) Pr[T0 > 20 + t] = Pr[T0 > t] S0 (20) 2 [18000 − 110(20 + t) − (20 + t) ]/18000 = [18000 − 110(20) − 202 ]/18000 150t + t2 [18000 − 110(20) − 110t − 202 − 40t − t2 ]/18000 = =1− . [18000 − 110(20) − 202 ]/18000 15400

S20 (t) = Pr[T20 > t] =

(e) The probability that (20) will die between the ages of 30 and 40 is Pr[10 < T20 < 20] = S20 (10) − S20 (20) = =

150(20) + 202 150(10) + 102 − 15400 15400

1800 9 = = 0.1168831. 15400 77

(f) The force of mortality at age x is given by µx = so that µ50 =

−dS0 (x)/dx [110 + 2x]/18000 110 + 2x = = , S0 (x) [18000 − 110x − x2 ]/18000 18000 − 110x − x2 110 + 2(50) 21 = = 0.021. 2 18000 − 110(50) − 50 1000

Prepared by E.A. Valdez

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Suggested solutions to DHW textbook exercises Exercise 2.3 We are given (100 − x)1/2 1√ 100 − x = , for 0 ≤ x ≤ 100. 10 10 The probability that a newborn will die between ages 19 and 36 is given by S0 (x) =

19|17 q0

= Pr[19 < T0 ≤ 36] = S0 (19) − S0 (36) =

Prepared by E.A. Valdez

811/2 − 641/2 1 = = 0.10. 10 10

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Suggested solutions to DHW textbook exercises Exercise 2.4 (a) To show S0 is a legitimate survival function, we show 3 conditions: (i) S0 (0) = 1: trivial (ii) lim S0 (x) = 0: Since all parameters A, B, C and D are all positive, then the term x→∞

Ax + 12 Bx2 +

C Dx log D

−

C log D

→ ∞ as x → ∞ so that lim S0 (x) = e−∞ = 0. x→∞

(iii) S0 must be non-increasing: Define the term H(x) = Ax + 12 Bx2 + logCD Dx − logCD so dH(x) that = A + Bx + CDx and that dx dS0 (x) dH(x) = −e−H(x) , dx dx which is clearly strictly negative for all x. (b) We have n h io C C 2 x+t exp − A(x + t) + B(x + t) + D − log D log D S0 (x + t) n h io = Sx (t) = S0 (x) exp − Ax + Bx2 + logCD Dx − logCD    C 2 x t = exp − At + B(2xt + t ) + D (D − 1) . log D (c) The force of mortality at age x can be expressed as e−H(x) dH(x) −dS0 (x)/dx dx µx = = = A + Bx + CDx . −H(x) S0 (x) e The force of mortality has a similar form to that of Makeham’s except for the addition of a linear term on age x. (d) Solving all of part (d) requires use of a computer software. Here we give our solution coded in R. [slightly differ from textbook answers] (i) Note that we can express tp30

  = S30 (t) = exp − At + B(60t + t2 ) +

 C 30 t D (D − 1) . log D

The R code to compute this for different values of t is given by A