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2 u t px (u px )(t px u )“The probability that (x) survives u t years is equivalent to (x) firstsurviving u years to age x u, and then surviving t additional years toage x u t.” Be careful:u t qx6 (u qx )(t qx u ).“The right-hand side implies that it is possible for (x) to die within uyears, then somehow come back to life at age x u in order to die againwithin t years. This, of course, is not possible and cannot be equal tothe left-hand side which is the probability (x) dies within u t years.” u t qx u t qx- u qx u px -u t px (u px )(t qx u )This is a u-year deferred probability of death, “the probability that (x)dies between ages x u and x u t.” Note: 0 t qx t qx . Note: 1 qx , 1 px , andu 1 qxare written as qx , px , and u qx , respectively.qx may be referred to as a mortality rate, and px may be referred to as asurvival rate.

3Force of Mortality: µx µ(x) force of mortality at age x, given survival to age x. This issometimes called the “hazard rate” or “failure rate.”d[S (x)]0dµx dxS0 (x) dx[ln S0 (x)]“This is the instantaneous death rate for a life at age x.” µx t µx (t) force of mortality at age x t, given survival to x t.d[p ]µx t dtt pt xx dtd [ln t px ]“This is the instantaneous death rate for a life at age x t. Here, thevariable is time after age x. You could also obtain µx t by replacing x inµx with x t.” t px exp[ R x txµs ds] exp[ Rt0µx s ds] If c 0, then µ x s cµx s t p x (t px )c .For constant k, then µ x s µx s k t p x (e kt )(t px ).The constant k should be such that µ x s 0. fx (t) t px µx t probability density function of Tx .This comes from the above formula for µx t , recognizing that fx (t) dddt [t qx ] dt [t px ]. t qx t px Rt0 s px µx s dsR u t qx ts px µx s dsR u tus px µx s ds

4Curtate Future Lifetime: Kx curtate future lifetime for (x), a discrete random variable.Kx bTx c integer part of Tx . That is, Kx represents the complete number of future years survived by (x), where any fractional time survivedin the year of death is ignored. Note: Kx 0, 1, 2.Kx may also be written as K(x) or K. k qx P r(Kx k) P r(k Tx k 1) for k 0, 1, 2.This is the probability mass function of Kx , “the probability that (x) diesin the (k 1)st year, between ages x k and x k 1.” k 1 qx 0 qx 1 qx . k qx“The probability that (x) dies within k 1 years is the sum of theprobabilities that (x) dies in the first year, the second year, ., the (k 1)st year.”

5Other Features of Tx and Kx Distributions: e̊x E(Tx ) R 0t(t px )(µx t )dt R 0t px dtThis is the complete expectation of life for (x), the average time-untildeath for (x). That is, (x) is expected to die at age x e̊x . V ar(Tx ) E(Tx2 ) - [E(Tx )]2 2 e̊x:n E[min(Tx , n)] R 0tt px dt - [e̊x ]2Rn0 t px dtThis is the n-year temporary complete life expectancy for (x), the average number of years out of the next n years that (x) survives.This expectation helps define the recursion: e̊x e̊x:n n px e̊x n .“The average number of future years that (x) survives is the averagenumber of years out of the first n years that (x) survives plus the averagenumber of years (x) survives after the first n years (accounting for theprobability that (x) survives the first n years).”Similarly: e̊x:m n e̊x:m m px e̊x m:n . The 100α-th percentile of the distribution of Tx , πα , is such that:πα q x α for 0 α 1.Special case: α 0.50; π.50 is called the median future lifetime for (x).

6 ex E(Kx ) P k 0 k(k qx ) P k 1 k pxThis is the curtate expectation of life for (x), the average curtate futurelifetime for (x). e̊x ex 12 V ar(Kx ) E(Kx2 ) - [E(Kx )]2 ex:n E[min(Kx , n)] P k 1 (2k 1)k px - [ex ]2Pnk 1 k pxThis is the n-year temporary curtate life expectancy for (x).This expectation helps define the recursions: ex ex:n n px ex n andex:m n ex:m m px ex m:n .

7Special Mortality Laws:de Moivre’s Law: Tx has a continuous uniform distribution.The limiting age is ω such that 0 x x t ω. µx 1ω x(Note: x 6 ω) S0 (x) ω xω F0 (x) xω t px ω x tω x t qx tω x u t qx tω x fx (t) t px µx t e̊x 1ω x(Note: x 6 ω)ω x2 e̊x:n nn px n2 n qx“(x) can either survive n years with probability n px , or die within n yearswith probability n qx . Surviving n years contributes n to the expectation.Dying within n years contributes n2 to the expectation as future lifetimehas a uniform distribution - (x), on average, would die halfway throughthe n-year period.” V ar(Tx ) ex (ω x)212ω x 12 V ar(Kx ) (ω x)212-112

8Modified/Generalized de Moivre’s Law: Tx has a beta distribution.The limiting age is ω such that 0 x x t ω. Also, there is aparameter α 0. µx αω x S0 (x) (Note: x 6 ω)¡ ω x αω¡ α F0 (x) 1 - ω xω¡ α t px ω x tω x¡ α t qx 1 - ω x tω x¡ t αNote: t qx 6 ω x. e̊x ω xα 1 V ar(Tx ) α(ω x)2(α 1)2 (α 2)Note: α 1 results in uniform distribution/de Moivre’s Law.

9Constant Force of Mortality: Tx has an exponential distribution, x 0. There is another parameter that denotes the force of mortality: µ 0. µx µ S0 (x) e µx F0 (x) 1 - e µx t px e µt (px )t t qx 1 - e µt e̊x 1µ e̊x:n 1 e µnµ V ar(Tx ) ex 1µ2pxqx V ar(Kx ) px(qx )2Note1: A constant force of mortality implies that “age does not matter.”This can easily be seen from t px e µt ; x does not appear on the righthand side.Note2: Tx has an exponential distribution implies that Kx has a geometricdistribution.

10Gompertz’s Law: µx Bcx for x 0, B 0, c 1 S0 (x) exp[ lnBc (cx 1)] F0 (x) 1 - exp[ lnBc (cx 1)] t px exp[ lnBc cx (ct 1)] t qx 1 - exp[ lnBc cx (ct 1)]Makeham’s Law: µx A Bcx for x 0, A -B, B 0, c 1 S0 (x) exp[ Ax Bxln c (c F0 (x) 1 - exp[ Ax t px exp[ At Bxln c (cB x tln c c (c t qx 1 - exp[ At 1)] 1)] 1)]B x tln c c (c 1)]Note: A 0 results in Gompertz’s Law.Weibull’s Law: Tx has a Weibull distribution. µx kxn for x 0, k 0, n 0kxn 1 ] S0 (x) exp[ n 1k F0 (x) 1 - exp[ n 1xn 1 ]k t px exp[ n 1((x t)n 1 xn 1 )]k((x t)n 1 xn 1 )] t qx 1 - exp[ n 1

11Life Tables:Given a survival model with survival probabilities t px , one can construct alife table, also called a mortality table, from some initial age x0 (usually age0) to a maximum age ω (a limiting age). Let lx0 , the radix of the life table, represent the number of lives age x0 .lx0 is an arbitrary positive number. lω 0. lx t (lx )(t px ) for x0 x x t ω.lx t represents the expected number of survivors to age x t out of lxindividuals aged x. t dx lx - lx t (lx )(t qx ) for x0 x x t ω.represents the expected number of deaths between ages x and x tout of lx lives aged x.t dxNote 1: 1 dx is written as dx .Note 2: If n 1, 2., then n dx dx dx 1 . dx n 1 . t dx u lx u - lx u t (lx )(u t qx ).The Illustrative Life Table is the life table that is provided to the candidatetaking Exam MLC or Exam 3L. Some questions from either exam will involveIllustrative Life Table calculations. A web link to this table (and ALL examtables) is provided for each exam in Appendix A of this study supplement.

12Fractional Age Assumptions:Life Tables are usually defined for integer ages x and integer times t. Fora quantity that involves fractional ages and/or fractional times, one has tomake an assumption about the survival distribution between integer ages;that is, one has to interpolate the value of the quantity within each year ofage. Two common interpolation assumptions follow.Uniform Distribution of Deaths (UDD):One linearly interpolates within each year of age. For integer age x and 0 s s t 1: lx s lx - sdx (1 s)lx (s)lx 1 . This is a linear function for s. s qx sqx s px 1 - sqx µx s qx1 sqx(does not hold at s 1) fx (s) s px µx s qx (does not hold at s 1) s qx t sqx1 tqx e̊x ex 12 V ar(Tx ) V ar(Kx ) e̊x:n ex:n 11212 n qx Note: uniform distribution/de Moivre’s Law has the property of UDDacross all ages up to the limiting age ω.Furthermore, uniform distribution/de Moivre’s Law may be expressed aslx k(ω x) for 0 x ω where k 0.

13Constant Force of Mortality (Exam MLC Only):One exponentially interpolates within each year of age. For integer age xand 0 s s t 1: lx s lx psx ln[lx s ] (1 s) ln[lx ] s ln[lx 1 ] s px psx s qx 1 - psx µx s -ln px µx (does not hold at s 1) fx (s) s px µx s -ln px (psx ) (does not hold at s 1) s qx t 1 - psx

141.2Exercises1.1 Suppose: F0 (t) 1 - (1 0.00026t2 ) 1 for t 0.Calculate the probability that (30) dies between ages 35 and 40.(A) 0.056(B) 0.058(C) 0.0601.2. You are given: s(x) (A) 0.009(B) 0.01110,000 x210,000(C) 0.013(D) 0.062(E) 0.064for 0 x 100. Calculate: q49 .(D) 0.015(E) 0.0172t1.3. Suppose: S0 (t) exp[ 2500] for t 0.Calculate the force of mortality at age 45.(A) 0.036(B) 0.039(C) 0.042(D) 0.045(E) 0.0481.4. The probability density function of the future lifetime of a brand new3machine is: f (x) 4x27c for 0 x c.Calculate: µ(1.1).(A) 0.06(B) 0.07(C) 0.08(D) 0.09(E) 0.101.5. You are given:(i) The probability that (30) will die within 30 years is 0.10.(ii) The probability that (40) will survive to at least age 45 and that another(45) will die by age 60 is 0.077638.(iii) The probability that two lives age 30 will both die within 10 years is0.000096.(iv) All lives are independent and have the same expected mortality.Calculate the probability that (45) will survive 15 years.(A) 0.90(B) 0.91(C) 0.92(D) 0.93(E) 0.94

151.6. You are given:(i) e50 20 and e52 19.33(ii) q51 0.035Calculate: q50 .(A) 0.028(B) 0.030(C) 0.032(D) 0.034(E) 0.0361.7. For a population of smokers and non-smokers:(i) Non-smokers have a force of mortality that is equal to one-half the forceof mortality for smokers at each age.(ii) For non-smokers, mortality follows a uniform distribution with ω 90.Calculate the difference between the probability that a 55 year old smokerdies within 10 years and the probability that a 55 year old non-smoker dieswithin 10 years.(A) 0.20(B) 0.22(C) 0.24(D) 0.26(E) 0.281.8. You are given:(i) The standard probability that (40) will die prior to age 41 is 0.01.(ii) (40) is now subject to an extra risk during the year between ages 40and 41.(iii) To account for the extra risk, a revised force of mortality is definedfor the year between ages 40 and 41.(iv) The revised force of mortality is equal to the standard force of mortalityplus a term that decreases linearly from 0.05 at age 40 to 0 at age 41.Calculate the revised probability that (40) will die prior to age 41.(A) 0.030(B) 0.032(C) 0.034(D) 0.036(E) 0.038

161.9. You are given:(i) Tx denotes the time-until-death random variable for (x).(ii) Mortality follows de Moivre’s Law with limiting age ω.(iii) The variance of T25 is equal to 352.0833.Calculate: e̊40:10 .(A) 7.5(B) 8.0(C) 8.51.10. You are given: µx (D) 9.0 180 x(E) 9.5for 0 x 80.Calculate the median future lifetime for (40).(A) 4.0(B) 4.3(C) 4.6(D) 4.9(E) 5.21.11. You are given:½µx 0.040.05for 0 x 40for 40 xCalculate: e̊25 .(A) 22(B) 231.12. You are given:(C) 24k q0(D) 25(E) 26 0.10 for k 0, 1, ., 9.Calculate: 2 p5 .(A) 0.50(B) 0.55(C) 0.60(D) 0.65(E) 0.70

171.13. For the current model of Zingbot:(i) s(x) ω xωfor 0 x ω(ii) var[T (5)] 102.083333.For the proposed model of Zingbot, with the same ω as the current model:α(1) s (x) ( ω xω ) for 0 x ω, α 0(2) µ 10 0.0166667.Calculate the difference between the complete expectation of life for abrand new proposed model of Zingbot and the complete expectation of lifefor a brand new current model of Zingbot.(A) 5.9(B) 6.1(C) 6.3(D) 6.5(E) 6.71.14. Mortality for Frodo, age 33, is usually such that:¡ 110 x t 2for 0 t 110 x.t px 110 xHowever, Frodo has decided to embark on a dangerous quest that willlast for the next three years (starting today). During these three years only,Frodo’s mortality will be revised so that he will have a constant force ofmortality of 0.2 for each year. After the quest, Frodo’s mortality will onceagain follow the above expression for t px .Calculate Frodo’s revised complete expectation of life.(A) 15.2(B) 15.4(C) 15.6(D) 15.8(E) 16.0

181.15. You are given:(i) µ(x) B(1.05)x for x 0, B 0.(ii) p51 0.9877Calculate: B.(A) 0.001(B) 0.002(C) 0.003(D) 0.004(E) 0.0051.16. You are given:(i) The force of mortality for Vivian is µVx µ for x 0, µ 0.(ii) The force of mortality for Augustine is µAx Calculate µ so that(A) 0.01610 p30190 xfor 0 x 90.is the same for Vivian and Augustine.(B) 0.018(C) 0.020(D) 0.022(E) 0.0241.17. Consider the following life table, where missing entries are denotedby ��—Calculate the expected number of deaths between ages 48 and 50.(A) 1280(B) 1290(C) 1300(D) 1310(E) 1320

191.18. You are given the following life table, where missing entries aredenoted by 720.0186———ex———14.8914.22Calculate the expected number of deaths between ages 67 and 69.(A) 2800(B) 2900(C) 3000(D) 3100(E) 32001.19. You are given:(i) lx 1000(ω - x) for 0 x ω(ii) µ30 0.0125Calculate: e̊40:20 .(A) 17.1(B) 17.6(C) 18.1(D) 18.6(E) 19.11.20. You are given the following life table, where missing values are indicated by .0——182.5—px0.875——0.680——Calculate 2 q0 .(A) 0.16(B) 0.17(C) 0.18(D) 0.19(E) 0.20

201.21. Woolhouse is currently age 40. Woolhouse’s mortality follows 110%of the Illustrative Life Table; that is, the probability that Woolhouse diesbetween ages x and x 1 is 110% of the probability of death between agesx and x 1 in the Illustrative Life Table for x 40, 41, ., 110.Calculate Woolhouse’s 4-year temporary curtate life expectancy.(A) 3.94(B) 3.95(C) 3.96(D) 3.97(E) 3.981.22. Suppose mortality follows the Illustrative Life Table, and deaths areuniformly distributed within each year of age.Calculate:4.5 q40.3 .(A) 0.0141(B) 0.0142(C) 0.0143(D) 0.0144(E) 0.01451.23. Suppose mortality follows the Illustrative Life Table, where deathsare assumed to be uniformly distributed between integer ages.Calculate the median future lifetime for (32).(A) 41(B) 43(C) 45(D) 47(E) 491.24. Suppose mortality follows the Illustrative Life Table with the assumption that deaths are uniformly distributed between integer ages.Calculate:(A) 0.01300.9 q60.6 .(B) 0.0131(C) 0.0132(D) 0.0133(E) 0.01341.25. You are given the mortality rates:q30 0.020, q31 0.019, q32 0.018.Assume deaths are uniformly distributed over each year of age.Calculate the 1.4-year temporary complete life expectancy for (30).(A) 1.36(B) 1.37(C) 1.38(D) 1.39(E) 1.40

211.26. (Exam MLC Only:) Suppose:(i) q70 0.04 and q71 0.05.(ii) Let UDD denote a uniform distribution of deaths assumption withineach year of age, and let CF denote a constant force of mortality within eachyear of age.Calculate the probability that (70.6) will die within the next 0.5 yearsunder UDD minus the probability that (70.6) will die within the next 0.5years under CF.(A) 0.00008(B) 0.00010(C) 0.00012(D) 0.000141.27. (Exam MLC Only:) You are given:(i) The force of mortality is constant between integer ages.(ii)0.3 qx 0.7 0.10Calculate: qx .(A) 0.24(B) 0.26(C) 0.28(D) 0.30(E) 0.32(E) 0.00016

22Answers to Exercises1.1. E1.26. A1.2. C1.27. D1.3. A1.4. B1.5. C1.6. B1.7. A1.8. C1.9. D1.10. B1.11. A1.12. C1.13. E1.14. D1.15. A1.16. B1.17. B1.18. E1.19. A1.20. D1.21. D1.22. E1.23. C1.24. A1.25. C

231.3Past Exam Questions Exam MLC, Spring 2012: #2 (MLC Only) Exam 3L, Spring 2012: #1, 2, 3 Exam MLC, Sample Questions: #13, 21, 22, 28, 32, 59, 65, 98, 106, 116,120, 131, 145, 155, 161, 171, 188, 189, 200, 201, 207, 219, 223, 267 (MLCOnly), 276 Exam 3L, Fall 2011: #1, 2 Exam 3L, Spring 2011: #1, 2, 3 Exam 3L, Fall 2010: #1, 2, 3 Exam 3L, Spring 2010: #1, 2, 3, 4 Exam 3L, Fall 2009: #1, 2, 3 Exam 3L, Spring 2009: #1, 3 Exam 3L, Fall 2008: #12, 13, 14 Exam 3L, Spring 2008: #13, 14, 15, 16 Exam MLC, Spring 2007: #1, 21

will have worked as many of the past Exam MLC and Exam 3L questions as possible. Exam 3L candidates should attempt problems from the prior 3L exams flrst before attempting any problems from prior MLC exams. Also, some sections are marked \Exam MLC Only;" as you may have guessed, candidates preparing for Exam 3L can skip those sections.