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The Actuary’s Free Study Guide for Exam 3L – Second Edition – G. Stolyarov IISection 2Life-Table Probability Functions: PracticeProblems and Solutions – Part 2Here, we introduce another life-table probability function.t u q x is the probability that life (x) will survive for the next t years and die within thesubsequent u years. That is, t u q x is the probability that life (x) will die between ages x tand x t u.When u 1, this function can simply be written ast q x .Other expressions for t u q x are as follows:t u q x Pr[t T(x) t u]t u q x t u q xt u q x t p x- tqx- t upxWe can calculate t u q x as follows:t u q x (s(x t) - s(x t u))/s(x) t p x * u q x tSource: Bowers, Gerber, et. al. Actuarial Mathematics. 1986. First Edition. Society of Actuaries:Itasca, Illinois. pp. 47-48.Original Problems and Solutions from The Actuary's Free Study GuideProblem S3L2-1. The life of a triceratops has the following survival function associated with it:s(x) e-0.34x. Find 6 4 q 2 .Solution S3L2-1. We use the following formula:t u q x (s(x t) - s(x t u))/s(x).Thus,q 2 (s(2 6) - s(2 6 4))/s(2) (s(8) - s(12))/s(2) (e-0.34*8 - e-0.34*12)/(e-0.34*2) 6 4 q 2 about 0.0966554096 4Problem S3L2-2. The life of a giant pin-striped cockroach has the following survival functionassociated with it: s(x) 1 - x/94, for 0 x 94 and 0 otherwise. Find 5 5 q 22 .Solution S3L2-2. We use the following formula:6t u q x (s(x t) - s(x t u))/s(x).Thus,

The Actuary’s Free Study Guide for Exam 3L – Second Edition – G. Stolyarov II5 5 q 22 (s(22 5) - s(22 5 5))/s(22) (s(27) - s(32))/s(22) ((1 - 27/94) - (1 - 32/94))/(1 22/94) (5/94)/(72/94) 5 5 q 22 5/72 about 0.0694444444444Problem S3L2-3. Three-headed donkeys always survive until age 1. Thereafter, the survivalfunction for the life of a three-headed donkey is s(x) 1/x for all x 1. What is the probabilitythat a three-headed donkey that has survived to age 23 will senselessly perish (i.e., die) betweenthe ages of 45 and 69?Solution S3L2-3. We wish to find 22 24 q 23 (s(45) - s(69))/s(23) (1/45 - 1/69)/(1/23) 22 24 q 23 8/45 about 0.17777777778Problem S3L1-4. You know the following about the lives of jumping slugs. For a jumping slugthat has survived to age 39, the probability of surviving to age 45 is 0.835. For a jumping slugthat has survived to age 45, the probability of dying within the subsequent 8 years is 0.23. Whatis the probability that a jumping slug that has survived to age 39 will senselessly perish (i.e., die)between the ages of 45 and 53?Solution S3L1-4. We use the following formula: t u q x t p x * u q x t . We wish to find 6 8 q 39 6 p 39 * 8 q 45 , where we are given that 6 p 39 0.835 and 8 q 45 0.23. Thus, 0.835*0.23 6 8 q 39 0.19205Problem S3L1-5. There is a 0.05 probability that an ancient Greek god who has survived to age1245 will die between the ages of 1298 and 1403. Zeus is currently 1245 years old, and hisprobability of surviving to age 1298 is 0.95. What is the probability that, if Zeus survives to age1298, he will also survive past age 1403?Solution S3L1-5. We use the following formula: t u q x t p x * u q x t . We wish to findWe are given that 53 105 q 1245 0.05 and 1298 p 1245 0.95. We thus know that 0.05 0.95* 105 q 1298 , and so 105 q 1298 1/19. 1- 105 q 1298 105 p 1298 18/19 0.05263157897105 p 1298 .

The Actuary’s Free Study Guide for Exam 3L – Second Edition – G. Stolyarov IISection 3Curate-Future-LifetimeThe curate-future-lifetime of life (x) is the number of future years that life (x) completes beforedeath. The curate-future-lifetime of (x) is represented by the random variable K(x).The probability density function of K(x) can be expressed in the following ways:Pr[K(x) k] Pr[k T(x) k 1]Pr[K(x) k] k p x -k 1 p xPr[K(x) k] k p x *q x kPr[K(x) k] k q xfor all nonnegative integer values of k.The cumulative distribution function of K(x) can be expressed as follows:Pr[K(x) k] k 1 q x for all nonnegative integer values of k.Source: Bowers, Gerber, et. al. Actuarial Mathematics. 1986. First Edition. Society of Actuaries:Itasca, Illinois. p. 48.Original Problems and Solutions from The Actuary's Free Study GuideProblem S3L3-1. The life of a triceratops has the following survival function associated with it:s(x) e-0.34x. Find the probability that the curate-future-lifetime of a triceratops life (34) is 7years.Solution S3L3-1. We want to find Pr[K(34) 7].We use the formula Pr[K(x) k] k q x . In this case,7 q 34 (s(34 7) - s(34 7 1))/s(34) (s(41) - s(42))/s(34) (e-0.34*41 - e-0.34*42)/(e-0.34*34) Pr[K(34) 7] 0.0266758231Problem S3L3-2. Every year, 0.26 of the lemming population senselessly perishes (i.e., dies),and every lemming has the same likelihood of senselessly perishing. Find the probability that thecurate-future-lifetime of a lemming currently aged 3 is 2 years.Solution S3L3-2. We use the formula Pr[K(x) k] k p x - k 1 p x . We want to find Pr[K(3) 2].Thus, we need to find 2 p 3 and 3 p 3 . The probability of a lemming of any age surviving two moreyears is8

The Actuary’s Free Study Guide for Exam 3L – Second Edition – G. Stolyarov II(1 - 0.26)2 0.5476 so 2 p 3 0.5476. The probability of a lemming of any age surviving threemore years is (1 - 0.26)3 0.405224 so 3 p 3 0.405224. Thus, Pr[K(3) 2] 2 p 3 - 3 p 3 0.5476 0.405224 0.142376Problem S3L3-3. The life of a giant pin-striped cockroach has the following survival functionassociated with it: s(x) 1 - x/94, for 0 x 94 and 0 otherwise. Find the cumulativedistribution function of the curate-future-lifetime of life (36). Use k as your only variable.Solution S3L3-3. The cumulative distribution function of K(36) can be expressed as follows:Pr[K(36) k] k 1 q 36 . We recall from Section 1 that t q x 1 - s(x t)/s(x) and sok 1 q 36 1 - s(37 k)/s(36) 1 - (1 - (37 k)/94)/(1 - 36/94) 1 - ((94 - 37 - k)/94)/(58/94) 1 - (57 - k)/58 (58 - 57 k)/58 Pr[K(36) k] (1 k)/58Problem S3L3-4. For white elephants, the probability that the curate-future-lifetime of life (123)is 4 years is 0.536. The probability that a white elephant that has survived for 127 years willsurvive for another year is 0.356. Find the probability that a white elephant that has survived for123 years will survive to age 127.Solution S3L3-4. We use the formula Pr[K(x) k] k p x *q x k . We are given that Pr[K(123) 4]is 0.536 and that p 127 is 0.356. We can quickly find q 127 1 - 0.356 0.644. Our goal is to find4 p 123 . By the formula above, 4 p 123 Pr[K(123) 4]/q 127 0.536/0.644 4 p 123 0.8322981366.Problem S3L3-5. Three-headed donkeys always survive until age 1. Thereafter, the survivalfunction for the life of a three-headed donkey is s(x) 1/x for all x 1. Eight-tailed oxen alsoalways survive until age 1. Thereafter, the survival function for the life of an eight-tailed ox is1/x2 for all x 1. Find the probability that the curate-future-lifetime of a three-headed donkeylife (3) is 5 years and that the curate-future-lifetime of an eight-tailed ox life (3) is 5 years.Solution S3L3-5.For three-headed donkeys, Pr[K(3) 5] 5 q 3 (s(8) - s(9))/s(3) (1/8 - 1/9)/(1/3) 1/24For eight-tailed oxen, Pr[K(3) 5] 5 q 3 (s(8) - s(9))/s(3) (1/82 - 1/92)/(1/32) 17/576To get the probability that both of these curate-future-lifetimes are 5 years, we multiply theindividual probabilities and obtain our desired answer: (1/24)(17/576) 17/13824 about0.0012297454.9

The Actuary’s Free Study Guide for Exam 3L – Second Edition – G. Stolyarov IISection 4Force of MortalityThis section will explore the force of mortality, alternatively known as the failure rate or thehazard rate function. It is abbreviated as μ x . The force of mortality describes an "instantaneousrate of death" and can be found via the following formulas:μ x f(x)/(1 - F(x))μ x -s'(x)/s(x)The force of mortality is always greater than or equal to zero, since s(x) is always nonnegativeand its derivative, s'(x), is always non-positive, so -s'(x) is always nonnegative.Meaning of Terms:X refers to the age of death of a particular life.F(x) is the cumulative distribution function of X.s(x) is the survival function of X.f(x) is the probability density function of X.Source: Bowers, Gerber, et. al. Actuarial Mathematics. 1986. First Edition. Society of Actuaries:Itasca, Illinois. p. 48-49.Original Problems and Solutions from The Actuary's Free Study GuideProblem S3L4-1. The life of a triceratops has the following survival function associated with it:s(x) e-0.34x. Find μ x .Solution S3L4-1. We use the formula μ x -s'(x)/s(x). Since s(x) e-0.34x, it follows thats'(x) -0.34e-0.34x and so μ x (0.34e-0.34x)/(e-0.34x) μ x 0.34Problem S3L4-2. The life of a giant pin-striped cockroach has the following survival functionassociated with it: s(x) 1 - x/94, for 0 x 94 and 0 otherwise. Find μ x .Solution S3L4-2. We use the formula μ x -s'(x)/s(x). Since s(x) 1 - x/94, it follows thats'(x) -1/94 and so μ x (1/94)/(1 - x/94) (1/94)/((94 - x)/94) μ x 1/(94 - x)10

The Actuary’s Free Study Guide for Exam 3L – Second Edition – G. Stolyarov IIProblem S3L4-3. Three-headed donkeys always survive until age 1. Thereafter, the survivalfunction for the life of a three-headed donkey is s(x) 1/x for all x 1. What is the force ofmortality for three-headed donkeys? Warning: Make sure you examine all possible ages!Solution S3L4-3. Prior to age 1, no three-headed donkeys die, so the force of mortality is 0. (Wecan confirm this by recognizing that s(x) is 1 for x 1 and so s'(x) 0 and thus μ x 0).After age 1, s(x) is 1/x, so s'(x) -1/x2. We now apply the formula μ x -s'(x)/s(x) (1/x2)/(1/x) x/x2 μ x 1/x.Thus, μ x 0, for 0 x 1 and μ x 1/x for x 1.Problem S3L4-4. Black swans always survive until age 16. After age 16, the lifetime of a blackswan can be modeled by the cumulative distribution function F(x) 1 - 4x-1/2, x 16. What is theforce of mortality for black swans? Warning: Make sure you examine all possible ages!Solution S3L4-4. Prior to age 16, no black swans die, so the force of mortality for x 16 is 0.After age 16, F(x) 1 - 4x-1/2, and f(x) F'(x) (-1/2)(-4)x-3/2 f(x) 2x-3/2. So we can use theformula μ x f(x)/(1 - F(x)) (2x-3/2)/(4x-1/2) 1/(2x).Thus, μ x 0, for 0 x 16 and μ x 1/(2x) for x 16.Problem S3L4-5. The lives of Unicorn-Pegasi can be modeled by the following cumulativedistribution function: F(x) x/36 - e-0.45x. If any Unicorn-Pegasus reaches the age of 35, it willget to live forever. What is the force of mortality for Unicorn-Pegasi? Warning: Make sure youexamine all possible ages!Solution S3L4-5. For all x 35, Unicorn-Pegasi are still mortal. It is given that F(x) x/36 - e, and so f(x) F'(x) 1/36 0.45e-0.45x. So we can use the formula0.45xμ x f(x)/(1 - F(x)) (1/36 0.45e-0.45x)/(1 - x/36 e-0.45x) (1 16.2e-0.45x)/(36 - x 36e-0.45x).After reaching age 35, Unicorn-Pegasi no longer die, so their force of mortality becomes 0.Thus, μ x (1 16.2e-0.45x)/(36 - x 36e-0.45x), for 0 x 35 and μ x 0 for x 35.11

The Actuary’s Free Study Guide for Exam 3L – Second Edition – G. Stolyarov IISection 5Identities Involving Life Table Functions andForce of MortalityHere, we will explore several useful relationships between the force of mortality, discussed inSection 4, and various life table functions:npx exp(- x x n μ y dy)npx exp(- 0 n μ x s ds), where s y - x.xp0 s(x) exp(- 0 x μ s ds)F'(x) f(x) x p 0 *μ xd(1 - t p x )/dt -d( t p x )/dt t p x μ x tMeaning of Terms:X refers to the age of death of a particular life.F(x) is the cumulative distribution function of X.s(x) is the survival function of X.f(x) is the probability density function of X.tpxis the probability that life (x) will survive during the next t years and reach the age x t.μ x is the force of mortality for life (x).Source: Bowers, Gerber, et. al. Actuarial Mathematics. 1986. First Edition. Society of Actuaries:Itasca, Illinois. p. 49-50.Original Problems and Solutions from The Actuary's Free Study GuideProblem S3L5-1. Tyrannosaurs have a force of mortality μ y 1/y2. Find12 p 34for tyrannosaurs.Solution S3L5-1. We can let n 12 and x 34 and use the formulanpx exp(- x x n μ y dy) exp(- 34 46 (1/y2)dy) exp(-0.0076726343) 12 p 34 0.9923567253.12

The Actuary’s Free Study Guide for Exam 3L – Second Edition – G. Stolyarov IIProblem S3L5-2. The force of mortality for octahedral armadillos is μ s 0 for s 1 and 1/(95s)for s 1. What is the survival function for octahedral armadillos?Solution S3L5-2. Since the force of mortality is 0 for s 1, it follows from this that s(x) 1 forx 1.For x 1, we use the formula s(x) exp(- 1 x μ s ds) exp(- 1 x 1/(95s)ds) exp(-ln(s)/95 1 x) exp(-ln(x)/95 ln(1)/95) exp(-ln(x)/95 0/95) exp(-ln(x)/95) exp(ln(x-1/95)) s(x) x-1/95.Thus, s(x) 1 for x 1 and s(x) x-1/95 for x 1.Problem S3L5-3. The force of mortality for octahedral armadillos is μ s 0 for s 1 and 1/(95s)for s 1. What is the probability density function for the lives of octahedral armadillos?Solution S3L5-3. We use the formula f(x) x p 0 *μ x . We are given that μ s 0 for s 1, so f(x) isalso 0 for s 1. From Solution S3L5-2, we know that x p 1 , the survival function for x 1, is x1/95.Also, we know that for x 1, μ x 1/(95x). Thus, we have f(x) x-1/95*1/(95x) (1/95)x-1/96.Therefore, f(x) 0 for x 1 and f(x) (1/95)x-1/96 for x 1.Problem S3L5-4. Tyrannosaurs have a force of mortality μ y 1/y2. Find -d( 32 p 56 )/dt fortyrannosaurs.Solution S3L5-4. We use the formula -d( t p x )/dt t p x μ x t for x 56 and t 32. Thus, we need tofind μ 88 1/882 and 32 p 56 exp(- 56 88 (1/y2)dy) exp(-0.0064935065) 0.9935275308. So d( t p x )/dt 0.9935275308/882 about 0.000128296427Problem S3L5-5. For orange rabbits, 18 p 12 0.836, and μ y has the form 1/(k - y). Find k.Solution S3L5-5. We use the formula n p x exp(- x x n μ y dy), which in this particular casetranslates to3018 p 12 0.836 exp(- 12 (1/(k - y)dy). Thus,0.836 exp(ln(k - y) 12 30)0.836 exp(ln(k - 30) - ln(k - 12))0.836 exp(ln[(k-30)/(k-12)])0.836 (k-30)/(k-12)0.836(k-12) (k-30)0.836k - 10.032 k - 3019.968 0.164kk 121.756097613

The Actuary’s Free Study Guide for Exam 3L – Second Edition – G. Stolyarov IISection 6Life-Table Probability Functions: PracticeProblems and Solutions – Part 3If we have a particular group of lives, to which no new members are added, we can let l 0 be thenumber of lives in the group at time 0. We can also let ℓ(x) be the number of survivors in thisgroup at age x. Then the expected number of survivors E(ℓ(x)) can be expressed as follows:E(ℓ(x)) l x l 0 *s(x)If we assume that the indicators for the survival of each life are independent, then ℓ(x) has abinomial distribution with parameters n l 0 and p s(x). Recall that a binomial probabilitymass function is of the following form: p(r) C(n, r)pr(1-p)n-rWe can also let n D x be the number of deaths between ages x and x n and we can define theexpected number of such deaths as E( n D x ) n d x . When n 1, n d x is simply written as d x .The following formulas hold with respect to l x and n d x :n d x l 0 [s(x) - s(x n)]n d x l x - l x n(-1/l x )(dl x /dx) (-1/[s(x)])ds(x)/dx μ x-dl x l x *μ x dxMeanings of Terms:X refers to the age of death of a particular life.s(x) is the survival function of X.μ x is the force of mortality for life (x).Source: Bowers, Gerber, et. al. Actuarial Mathematics. 1986. First Edition. Society of Actuaries:Itasca, Illinois. p. 52-53.Original Problems and Solutions from The Actuary's Free Study GuideProblem S3L6-1. The life of a triceratops has the following survival function associated with it:s(x) e-0.34x. Out of a group of 9035 newborn triceratopses, how many would you expect toremain alive at age 8? Round down to the nearest whole triceratops.Solution S3L6-1. We want to find l 8 , given l 0 9035 and s(x) e-0.34x. We use the formula l x l 0 *s(x). Thus, l 8 l 0 *s(8) 9035* e-0.34*8 595.1784062 about 595 triceratopses.14

The Actuary’s Free Study Guide for Exam 3L – Second Edition – G. Stolyarov IIProblem S3L6-2. The lives of Unicorn-Pegasi can be modeled by the following cumulativedistribution function: F(x) x/36 - e-0.45x. If any Unicorn-Pegasus reaches the age of 35, it willget to live forever. For a group of newborn 3612 Unicorn-Pegasi, find 2 d 7 . Round down to thenearest whole Unicorn-Pegasus.Solution S3L6-2. We first find s(x) 1 - F(x) 1 - x/36 e-0.45x. We are also given l 0 3612.So we use the formula n d x l 0 [s(x) - s(x n)]. 2 d 7 3612[s(7) - s(9)] 3612[(1 - 7/36 e-0.45*7) - (1 - 9/36 e-0.45*9)] 292.5189317 about 292 Unicorn-Pegasi.Problem S3L6-3. Black swans always survive until age 16. After age 16, the lifetime of a blackswan can be modeled by the cumulative distribution function F(x) 1 - 4x-1/2, x 16. There is acohort of 25 newborn black swans. Assuming that the survival of every black swan isindependent of the survival of every other black swan, what is the probability that exactly 9black swans of this group are alive at age 23? (Hint: The answer is extremely small.)Solution S3L6-3. We want to find Pr(ℓ(23) 9). We know that ℓ(23) follows a binomialdistribution with n l 0 25 and p s(23) 1 - F(23) 4*23-1/2 0.8340576562.We use the binomial probability mass function p(r) C(n, r)pr(1-p)n-r, for r 9.p(9) C(25, 9)*0.83405765629*(1 - 0.8340576562)16p(9) about 1.319298986*10-7Problem S3L6-4. For a particular group of polka-dotted zebras, you are given l x 95/x4, for allx 1. Find μ x for all x 1.Solution S3L6-4. We use the formula (-1/l x )(dl x /dx) μ x .(dl x /dx) -4*95x-5 -380x-5Thus, μ x (-x4/95)(-380x-5) μ x 4/x, for x 1.Problem S3L6-5. The lives of Unicorn-Pegasi can be modeled by the following cumulativedistribution function: F(x) x/36 - e-0.45x. If any Unicorn-Pegasus reaches the age of 35, it willget to live forever. For a group of 69 newborn Unicorn-Pegasi, find the probability that at leastone of them will get to live forever.Solution S3L6-5. We want to find 1 - Pr(ℓ(35) 0). We know that ℓ(35) follows a binomialdistribution with n l 0 69 and p s(35) 1 - 35/36 e-0.45*35 p 0.0277779223.We use the binomial probability mass function p(r) C(n, r)pr(1-p)n-r, for r 0.p(0) C(69, 0)*p0(1-p)69p(0) (1-p)69p(0) (1-0.0277779223)69p(0) 0.1431588013Our desired answer is 1 - p(0) 0.856841198715

The Actuary’s Free Study Guide for Exam 3L – Second Edition – G. Stolyarov IISection 7Life-Table Probability Functions: PracticeProblems and Solutions – Part 4A requirement for the force-of-mortality function is useful to remember. For every va

The Actuary's Free Study Guide for Exam 3L - Second Edition - G. Stolyarov II 3 Section 42: Discrete Temporary Life Annuities 119 . Section 70: Pooled Sample Variances and Covariances for Multiple-Life Statuses 222 Section 71: The Coefficient of Determination in Linear Regression, Fully Continuous Benefit