Midterm Exam I, Actuarial Mathematical Models I - Exercise Solutions

Sep 27, 2026

Uploaded by Hazel

Academic midterm exam material from King Saud University for the Mathematics Department, Actuarial Mathematical Models I. The document provides multiple exercises and worked solutions on survival and mortality models, including converting survival functions to c.d.f. and p.d.f., deriving future-lifetime survival functions, and computing multi-year death or survival probabilities. It also includes calculations under Gompertz’s law, constant force of mortality, constant force of interest, and UDD assumptions, along with use of life table quantities and select-and-ultimate mortality excerpts.

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Midterm Exam I, Actuarial Mathematical Models I

Exercise 1

Exercise 2

Exercise 3

Exercise 4

1 King Saud University Academic Year (G) 2020{2021 College of Sciences Academic Year (H) 1442 Mathematics Department Bachelor AFM: M. Eddahbi October 4, 2020 from 9 to 11 AM, ACTU-362-372 (two pages + formula sheet) Midterm Exam I, Actuarial Mathematical Models I Exercise 1 The survival function S0 (t) of the age-at death T0 is given by 1 S0 (t) = for t 􀀕 0 (1 + t)2 ; 1. Find F0 (t) the c.d.f of T0 2. Find f0 (t) the p.d.f of T0 3. Find Sx (t) the survival function of the future life time Tx of an age x: 4. Calculate 2p 10 . 5. Calculate 10j5q20 . Solution: 1. The c.d.f. F0 (t) of T0 is given by F0 (t) = 1 􀀀 S0 (t) = 1 􀀀 (1t)2 for all t 􀀕 0 : 2. The p.d.f. f0 (t) of T0 is given by f0 (t) = F00(t) = 􀀐 1 􀀀 (1t)2 􀀑 0 = (1t)3 for all t 􀀕 0 : 3. The survival function Sx (t) of future life time Tx is given in terms of S0 (t) by: (1t)3 Sx (t) = S0x0(+x)~~ ~~t) = ~~ ~~(1+xt))2 = (1(+x~~ ~~)2t)2 : 4. We know that tpx = Sx (t), then 2p 10 = S10 (2) = (1(+110)22)2 = ((~~ ~~1113))22 = 0:71598: 5. We know that mjnqx = m+nqx 􀀀 m qx = mpx 􀀀 m+npx , then 10j5q20 = 10p20 􀀀 15p20 = S20 (10) 􀀀 S20 (15) (1 + 20)2 (1 + 20)2 = 􀀀 = 0:11862: (1 + 20 + 10)2 (1 + 20 + 15)2 2 Exercise 2 1. You are given the following mortality table: | x | 20 | 21 | 22 | 23 | 24 | | --- | --- | --- | --- | --- | --- | | qx | | | | 0:0700 | | | `x | 30 ; 000 | | | | 23 ; 900 | | dx | 1 ; 200 | | | | | Determine the probability that a life age 21 will die within two years. 2. CANCELED You are given kjqx = 0:01 + 1~~k~~00 for k = 0 ; 1 ; : : : ; 9. Calculate 4px+6 3. Find tpx when the Mortality follows Gompertz's law (􀀖x = Bcx ) . You are given that 5p60 = 0:95 and 10p60 = 0:87 . Determine 30p60 . Solution: 1. The required probability is 2 q21 = 1 􀀀 ~~`~~`~~23~~21 . We need `21 and `23 which will be backed up as follows `21 = `20 􀀀 d20 = 30000 􀀀 1200 = 28800 now we shall use qx = `~~ x~~`x~~x+1~~ or equivalently `x = `1~~+1~~q~~ ~~x , thus `23 = 1~~24~~q23 = 1~~2~~~~9~~0~~0~~:~~0~~07 = 25698:92: So we get 25698:92 2 q21 = 1 􀀀 = 0:1076764: 28800 2. CANCELED By de􀀌nition of tpx we can write 4px+6 = ``xx and kj qx = `x+k~~ ~~`xx+1+k () `x+k 􀀀 `x+1+k = kj qx `x hence 9 9 `x 􀀀 `x+10 = X (`x+k 􀀀 `x+1+k) = `x X k=0 k=0 5 5 `x 􀀀 `x+6 = X (`x+k 􀀀 `x+1+k) = `x X k=0 k=0 kj qx kj qx thus 9 `x+10 = `x 1 􀀀X k=0 therefore kj qx! and `x+6 = `x 51 􀀀X k=0 kj qx! 4px+6 = ``xx = 11~~ ~~􀀀~~ ~~PP9~~k~~5k~~0~~0~~ ~~kkq~~ ~~xq~~ ~~x = 11~~ ~~􀀀~~ ~~PP9~~k~~5k~~0~~0((00::0101~~ ~~~~ ~~00::01k01k)) = 00::5579 = 0:6962: 3 3. By de􀀌nition of tpx we can write tpx = e􀀀Rxx+t 􀀖 sds = e􀀀B Rxx+t csds = e􀀀B Rt0 cx+sds = e 􀀀 lBn~~c~~(xc) (ct􀀀1) Take the log for x = 60, t = 5 and t = 10 in the previous formula we get, Bc60 (c5 􀀀 1) Bc60 (c10 􀀀 1) ~~ ~~ = 􀀀 ln(0:95) and ~~ ~~ = 􀀀 ln(0:87) ln(c) ln(c) Dividing the 􀀌rst into the second, c5 + 1 = ~~ ~~lnln((00::8795)) = 2:715015 c = p5 1:715015 = 1:11392 c30 􀀀 1 = 24:4453 Bc60ln(c30(c)~~ ~~􀀀~~ ~~1) = 􀀒 Bc60ln(c(5c)􀀀~~ ~~1)􀀓 􀀒 c30c5~~ ~~􀀀􀀀~~ ~~11􀀓= 􀀀 ln(5p60 ) 􀀒 c30c5~~ ~~􀀀􀀀~~ ~~11􀀓 = (􀀀 ln 0:95) 􀀒 c30c5~~ ~~􀀀􀀀~~ ~~11􀀓 = 1:753641 30p60 = e􀀀1:753641 = 0:1731 Exercise 3 1. Assume that the life table function is given by `x = 10000e􀀀0:05x , x 􀀕 0. 􀀌nd 3j10q25 2. Assume that the life table function at some ages is given as follows: (i) `40 = 94 ; 000, (ii)`41 = 93 ; 000, (ii) `42 = 92 ; 000 Assuming Constant Force of Mortality between integral ages, 􀀌nd 1:3q40:5 . 3. The force of mortality at some ages of a certain mortality table are given: (i) 􀀖60:5 = 0:02, (ii) 􀀖61:5 = 0:04 and (iii) 􀀖62:5 = 0:06 . Under UDD assumption, calculate the probability that a person age 60:5 will die within two years. Solution: 1. We have 3j10q25 = 13 q25 􀀀 3 q25 = 3p25 􀀀 13p25 = `28 `25 􀀀 `38 `25 = `28 􀀀 `38 = e􀀀0:05􀀂28 􀀀 e􀀀0:05􀀂38 `25 e􀀀0:05􀀂25 = e􀀀0:05􀀂3 􀀀 e􀀀0:05􀀂13 = 0:3386 2. We can write under

FAQs

How is the c.d.f. derived from the given survival function in Exercise 1?

Use F0(t)=1−S0(t) for t≥0, where S0(t) is provided. Then differentiate to obtain the p.d.f. as the derivative of F0(t).

What assumptions are used to compute probabilities over an interval in Exercise 3?

The document applies Constant Force of Mortality (CFM) for parts that interpolate between integral ages, and it applies the UDD (uniform distribution of deaths) assumption for the probability of dying within two years when forces of mortality are given.

How does Exercise 2 handle a mortality table to find the probability of death within two years?

It uses life table relationships between survival and death probabilities (e.g., expressing two-year death probability as 2q21=1−2p21) and computes required intermediate quantities from the provided table values.

Sep 27, 2026

Uploaded by Hazel