Exercises in Life Insurance Mathematics
Transcription
Exercises in Life Insurance Mathematics
Exercises in Life Insurance Mathematics Edited by Bjarne Mess Jakob Christensen University of Copenhagen Laboratory of Actuarial Mathematics Introduction This collection of exercises in life insurance mathematics replaces the collection of Steen Pedersen and all other exercises and problems in any text or article in the FM0L curriculum. The following abbreviations are being used for the contributors of exercises: AM BM BS FW JC JH MSC MS RN SH SK SP SW Bowers et al. “Actuarial Mathematics”, Society of Actuaries, Itasca, Il 1986 Bjarne Mess Bo Søndergaard Flemming Windfeld Jakob Christensen Jan Hoem “Elementær rentelære”, Universitetsforlaget, Oslo, 1971. Michael Schou Christensen Mogens Steffensen Ragnar Norberg Svend Haastrup Stephen G. Kellison “The theory of Interest”, Richard D. Erwing, Inc., Homewood, Il 1970 Steen Pedersen “Opgaver i livsforsikringsmatematik” Schwartz “Numerical Analysis”, Wiley, 1989 Copenhagen, August 2, 1997 Bjarne Mess Jakob Christensen 4 Exercises in Life Insurance Mathematics Exercises 1. FM0 S91, 1 FM0 S93, 2 Interest FM0 S92, 1 FM0 S94, 1 FM0 S93, 1 Exercise 1.1 Show that an | < an| < än| when i > 0. (JH(1), 1971) Exercise 1.2 Show that an | /n decreases as n increases and i > 0. (JH(2), 1971) Exercise 1.3 Show that an | (i1 )/an| (i2 ) decreases as n increases if i1 > i2 . (JH(3), 1971) Exercise 1.4 A man needs approximately $2.500 and raises in that connection a loan in a bank. The principal, which is to be fully repaid after 6 months, is $2.600. From this amount the bank deducts the future interest $84,50 and other fees of $5,70, so that the bank pays out the man $2.509,80 cash. The interest rate of the bank is 6.5%. (a) What is the effective interest rate p. a. for the bank? (b) What is the effective interest rate p. a. for the borrower? (JH(4 rev.), 1971) Exercise 1.5 One day a company receives an american loan offer: Principal of $5.000.000, rate of course 99% and nominal interest rate 6.5%. The loan is free of installments for 5 years and is after that to be amortized over 15 years: Interests and installments are due annually. Assume that the dollar-rate of exchange decreases exponentially from DKK 6,75 at the initial time to DKK 4,75 at the end of the loan. How can one determine the effective interest rate of the loan? Interest 5 (SP(19)) Exercise 1.6 By calculation of the interest rate for a fraction of a year, a bank will usually calculate with linear payment of interest instead of exponential payment of interest. If the interest rate is i p. a. and we have to calculate interest for a period of time α (0 < α < 1), the bank will calculate interest as αi per kr. 1. – in capital, instead of calculating the interest as (1 + i)α − 1 per. kr. 1. – in capital. Is this for the benefit of the borrower? (JH(7), 1971) Exercise 1.7 A man is going to buy new furniture on an installment plan. In the hire-purchase agreement he finds the following account: Cash payment for the furniture − In advance to the salesman = Net balance + Installment fee for 18 mths. = Net balance for installment The monthly installments are $6.306,00 $2.217, 00 $4.089,00 $501, 00 $4.590, 00 $255, 00 What effective interest rate p. a. (JH(10), 1971) Exercise 1.8 A man has been promised some money. He can choose from two alternatives for the payment. Under alternative (i) A5 = $4.495 and A9 = $5.548 are paid out after 5 and 9 years respectively. Under alternative (ii) B7 = $10.000 is paid out after 10 years. Denote the market interest rate by i. For which value (values) of i is (i) just as good as (ii), and when is (i) more profitable for the man? What if A9 = $5.562, i. e. $14 more? (JH(11), 1971) 6 Exercises in Life Insurance Mathematics Exercise 1.9 A says to B: “I would like to borrow $208 in one year from today. In return for your kindness I will pay $100 cash now, and $108,15 in two years from today by the end of the loan.” What is the effective interest rate p. a. for B if he accepts this? (JH(12), 1971) Exercise 1.10 Consider a usual annuity loan with principal H, interest rate i and n installments. Show that the installment which falls due in period t is Ft = i(1 + i)t−1 H, (1 + i)n − 1 and find an expression for the remaining debt immediately after this period. (SP(6)) Exercise 1.11 Consider a linearly increasing annuity. At time t = 1, 2, . . . the amount t is being paid. The present value of this cash flow is denoted by Ian | . (a) Show that Ian| = n−1 X t| an−t| , t=0 and interpret this equation intuitively. (b) Give an explicit expression for Ian | . (c) What does the symbol Ia n| mean? Give expressions corresponding to the ones from (a) and (b). (SP(10)) Exercise 1.12 A person has a table of annual annutities with different interest rates and durations to his disposal. However, he needs some present values of half-year annuities. These annuities do all have the same interest rate and the corresponding whole-year annuities can be found in the table. What is the easiest way to find the desired annuities? (SP(11)) Interest 7 Exercise 1.13 A debtor is going to pay an amount of 1 some time in the future. He does not know this point of time in advance; he only knows that it is a stochastic variable T with a known distribution. Now consider the expected present value denoted by Aδ(T ) = E(e−δT ). (a) Show that the variance of the present value is given by δ 2 Var(e−δT ) = A2δ (T ) − (A(T ) ) . Now assume that creditor has to pay a continuous T -year annuity. The present value is aδT | . Let the expected present value be denoted by aδ(T ) = E(aδT | ). (b) Show that Aδ(T ) = 1 − δaδ(T ) and interpret the equation. (c) Give an expression for the variance Var(a δT | ) corresponding to the one from (a). (d) Show that Aδ(T ) > v ET , and find a similar inequality for aδ(T ) . (Hint: Use Jensen’s inequality.) (e) Find at least two situations where these considerations are relevant. (SP(16)) Exercise 1.14 Consider a loan, principal H, nominal interest rate i 1 , rate of course k and installment Ft in period t, where t = 1, . . . , N . Show that the effective interest rate ie satisfies 1−S i1 ie = k−S where N 1 X S= Ft vet . H t=1 (SP(20)) Exercise 1.15 A loan with principal H and fixed interest rate i 1 has to be amortized annually over a period of N years. The borrower can each year deduct half the interest 8 Exercises in Life Insurance Mathematics expenses on his tax declaration. Construct the installment plan in a way so that the amount of amortisation minus deductible (actual net payment) will be the same in all periods (assume tax is payable by the end of each year). Find the annual net installment. (RN “Opgaver til FM0 (rentelære)” 18.05.93) Exercise 1.16 Consider a general loan. Show that for t ≥ 1 we have At = (1 + i1 )Rt−1 − Rt Rt = n−t X At+j v1j (1.1) j=1 v1t Rt = H − t X Aj v1j (1.2) j=1 with the use of standard notation. Formula (1.1) is called the prospective formula for the remaining debt. Formula (1.2) is called the retrospective formula for the remaining debt. (JH(20), 1971) Exercise 1.17 A loan is being repaid by 15 annual payments. The first five installments are $400 each, the next five $300 each, and the final five are $200 each. Find expressions for the remaining debt immediately after the second $300 installment – (a) prospectively, (b) retrospectively. (SK(1) p. 122, 1970) Exercise 1.18 A loan of $1.000 is being repaid with annual installments for 20 years at effective interest of 5% . Show that the amount of interest in the 11th installment is 50 . 1 + v 10 (SK(10) p. 123, 1970) Exercise 1.19 A borrower has mortgage which calls for level annual payments of 1 at the end of each year for 20 years. At the time of the seventh regular payment he also Interest 9 makes an additional payment equal to the amount of principal that according to the original amortisation schedule would have been repaid by the eighth regular payment. If payments of 1 continue to be made at the end of the eighth and succeding years until the mortgage is fully repaid, show that the amount saved in interest payments over the full term of the mortgage is 1 − v 13 . (SK(16) p. 124, 1970) Exercise 1.20 A man has some money invested at an effective interest rate i. At the end of the first year he withdraws 162.5% of the interest earned, at the end of the second year he withdraws 325% of the interest earned, and so forth with the withdrawal factor increasing in arithmetic progression. At the end of 16 years the fund exhausted. Find i. (SK(40) p. 127, 1970) Exercise 1.21 A loan of a25 | is being repaid with continuous payments at the annual rate of 1 p. a. for 25 years. If the interest rate i is 0.05, find the total amount of interest paid during the 6th through the 10th years inclusive. (SK(42) p. 127, 1970) Exercise 1.22 After having made six payments of $100 each on a $1.000 loan at 4% effective, the borrower decides to repay the balance of the loan over the next five years by equal annual principal payments in addition to the annual interest due on the unpaid balance. If the lender insists on a yield rate of 5% over this five-year period, find the total payment, principal plus interest, for the ninth year. (SK(45) p. 127, 1970) Exercise 1.23 A student has heard of a bank that offers a study loan of L = 10.000 kr. The rate of interest is 3% p. a. and the student applicates for the loan on the following conditions: (i) The first m = 5 years he will only pay an interest of 300 kr. per year. (ii) After that period of time he will pay interests and installments of 900 kr. per year until the loan is fully amortized (the last installment may be reduced). (a) For how long N does he have to pay installments and how big is the last amount of amortisation? 10 Exercises in Life Insurance Mathematics (b) Which amount αt has to be paid at time t if the loan (with interest earned) is to be fully paid back at time t (t = 1, 2, . . . , N )? (Aktuarembetseksamen, Oslo 1960) Aggregate Mortality 11 2. Aggregate Mortality Exercise 2.1 Let T be a stochastic variable with distribution function F . Assume F is concentrated on the interval [a, b] and that F is continuous with continuous density f . Assume F (t) < 1 for t ∈ [a, b). Define µ(t) = f (t) , t ∈ [a, b). 1 − F (t) We say that µ is the intensity of F . (a) Show that Rb a µ(t)dt = ∞. (b) Can we conclude that µ(t) → ∞ for t → b − ? Let T be the life length of a newly born. Let a = 0 and b = ω where ω is the maximum life length. (c) Show that µ is the force of mortality. (HRH(1)) Exercise 2.2 Use the decrement tables of G82M to find the following probabilities: (a) The probability that a 1 year old person dies after his 50th year, but before his 60th year. (b) The probability that a 30 year old dies within the next 37 years. (c) The probability that two persons now 26 and 31 years old, and whose remaining life times are assumed to be stochastically independent, both are alive in 12 years. (SP(28)) Exercise 2.3 Explain why each of the following functions cannot serve in the role indicated by the symbol: µx = (1 + x)−3 , x ≥ 0 F (x) = 1 − 22x 11x2 7x3 + − , 0≤x≤3 12 8 24 12 Exercises in Life Insurance Mathematics f (x) = xn−1 e−x/2 , x ≥ 0, n ≥ 1. (AM(3.4) p. 77, 1986) Exercise 2.4 Consider a population, where the distribution functions for a man’s and a woman’s total life lengths are x q0M and x q0K respectively. Assume that these K probabilities are continuous so that the forces of mortality µ M x and µx are defined. Let s0 denote the probability that a newly born is a female. Assume moreover that s 0 and the forces of mortality are not being altered during the period of time considered in this exercise. (a) Find the distribution function x q0 for the total life time for a person of unknown sex and find t qx . Find the force of mortality µx for a person of unknown sex. (b) What is the probability sx that a person aged x is a woman? K Using decrement series `M x and `x for men and women respectively, work out a decrement serie `x for the total population: K (c) How should one appropriately choose ` M 0 and `0 ? K (d) Express ax in terms of aM x and ax . (SP(32)) Exercise 2.5 Consider a random survivorship group consisting of two subgroups: (1) The survivors of 1.600 births. (2) The survivors of 540 persons joining 10 years later at age 10. An excerpt from the appropriate mortality table for both subgroups follows: x 0 10 70 `x 40 39 26 If γ1 and γ2 are the numbers of survivors under the age of 70 out of subgroups (1) and (2) respectively, estimate a number c such that P (γ 1 + γ2 > c) = 0.05. Assume the lives are independent. Aggregate Mortality 13 (AM(3.13) p. 78, 1986) Exercise 2.6 When considering aggregate mortality the probability that an x-year old person is going to die between x + s and x + s + t, is denoted by the symbol s|t qx . (a) Express this probability by the distribution function of the person’s remaining life time. (b) Is there a connection between s|t qx and t qx ? (c) Show that s|t qx = Z s+t u px µx+u du, s and interpret this expression. When t = 1 we write s| qx = s|1 qx . (d) Show that for integer x and n we have n| qx (e) Show that s|t qx = dx+n . `x can be expressed similarly (use the function ` instead of d). (f) Prove the following identities: n|m qx = n px n| qx = n px n+m px = n px − n+m px · qx+n · m px+n . (SP(24)) Exercise 2.7 Let e◦x:n| denote the expected future lifetime of (x) between the ages of x and x + n. Show that e◦x:n| = = n Z tt px µx+t dt + nn px Z0 n 0 This is called the partial life expectancy. t px dt. 14 Exercises in Life Insurance Mathematics (AM(3.14) p. 78, 1986) Exercise 2.8 The force of mortality µ x is assumed to be µx = βcx . Three persons are x, y and z years old respectively. What is the probability of dying in the order x, y, z? (Tentamen i försikringsmatematik, Stockholms Högskola 1954) Exercise 2.9 If F (x) = 1 − x/100, 0 ≤ x ≤ 100, find µ x , F (x), f (x) and P (10 < X < 40). (AM(3.5) p. 77, 1986) Exercise 2.10 If µx = 0.0001 for 20 ≤ x ≤ 25, evaluate 2|2 q20 . (AM(3.7) p. 77, 1986) Exercise 2.11 Assume that the force of mortality µ x is Gompertz-Makeham, i. e. µx = α+βcx . For at certain cause of death, the force of mortality is given by α 1 +β1 cx . Show that the probability of dying from the above disease for an x-year old is β1 α1 β − αβ1 ex . + β β (Tentamen i försikringsmatematik, Stockholms Högskola 1954) Exercise 2.12 Show that constants a and b can be determined so that µx = a log(1 − qx ) + b log(1 − qx+1 ). when µx can be put as a linear function for x < t < x + 2. (Tentamen i försikringsmatematik, Stockholms Högskola 1954) Exercise 2.13 Assuming the force of mortality to be Gompertz-Makeham, i. e. µx = α + βcx , show that for each age x we have − log c log(1 − qx ) < µx < − log(1 − qx ). c−1 Aggregate Mortality 15 (Tentamen i försikringsmatematik, Stockholms Högskola, 1954) Exercise 2.14 Given that `x+t is strictly decreasing for t ∈ [0, 1] show that (a) if `x+t is concave down, then qx > µx , (b) if `x+t is concave up, then qx < µx . (AM, 1986) Exercise 2.15 Prove the following expressions: d d `x µx < 0, when µx < µ2x dx dx d d `x µx = 0, when µx = µ2x dx dx d d `x µx > 0, when µx > µ2x . dx dx (AM(3.12) p. 77, 1986) Exercise 2.16 Show the following identities: ∂ t px ∂t ∂ t px ∂x = −µx+t · t px = (µx − µx+t ) · t px 1 = `x = Z Z ω−x t px µx+t dt 0 ω−x `x+t µx+t dt. 0 (SP(26)) Exercise 2.17 If the force of mortality µ x+t , 0 ≤ t ≤ 1, changes to µx+t − c where c is a positive constant, find the value of c for which the probability of (x) dying within a year will be halved. Express the answer in terms of q x . (AM(3.34) p. 80, 1986) Exercise 2.18 From a standard mortality table, a second table is prepared by doubling the force of mortality of the standard table. Is the rate of mortality, q x , at 16 Exercises in Life Insurance Mathematics any given age under the new table, more than double, exactly double or less than double the mortality rate, qx , of the standard table? (AM(3.35) p. 80, 1986) Exercise 2.19 If µx = Bcx , show that the function `x µx has a maximum at age x0 , where µx0 = log c. (Hint: Exercise 2.15). (AM(3.36) p. 80, 1986) Exercise 2.20 Assume µx = Acx 1 + Bcx for x > 0. (a) Find the survival function F (x). (b) Verify that the mode of the distribution of X, the age of death, is given by x0 = log(log c) − log A . log c (AM(3.37) p. 80, 1986) Exercise 2.21. (Interpolation in Life Annuity Tariffs) Consider a table with the present value ax:u−x| for an integer expiration age u with age at issue x = 0, 1, . . . , u and futhermore a table of the one year survival probabilities p x for the same ages. Let t be a real number, 0 ≤ t < 1. We are trying to find a way to determine a x+t:u−x−t| from the data of the table; this method will, of course, depend on how the mortality varies with the age. Assume that the force of mortality is constant on one year age intervals, i. e. µ x+t = µx for all t with 0 ≤ t < 1. (a) Find t−s px+s in terms of px for 0 ≤ s < t ≤ 1. (b) Find an expression for ax+t:1−t| . (c) Show that for every t there exists a λ so that ax+t:u−x−t| = λax:u−x| + (1 − λ)ax+1:u−x−1| , and express λ in terms of the discount rate v, t and p x . (d) How should one interpolate in a corresponding table for A x:u−x| . Aggregate Mortality 17 (FM1 exam, summer 1983) Exercise 2.22 For insurances where the policies are issued on aggravated circumstances, one operates with excess mortality. Let µ x be the force of mortality corresponding to the normal mortality. A person is said to have an excess mortality if his force of mortality is given by µ0x = (1 + k)µx . (a) Show that for all positive k, x and t we have 0 t qx < (1 + k)t qx . (b) Show that if there exists a constant ∆ so that µ 0x = µx+∆ is valid for all x then a0x = ax+∆ for all x; in this case the insurance is issued with an increase of age ∆. (c) Show that the condition in (b) is fulfilled if the mortality satisfies Gompertz’s law, i. e. there exist constants β and γ so that µ x = β exp(γx) for all x. (d) Show, oppositely, that if µx is strictly increasing in x and if there for any k ≥ 0 exists a constant ∆k so that µx+∆k = (1 + k)µx , then the mortality satisfies Gompertz’s law. (SP(43)) 18 Exercises in Life Insurance Mathematics 3. Insurance of a Single Life FM0 S92, 3 Exercise 3.1 Prove the identities: äx = 1 + vpx äx+1 1 − n Ex = äx:n| − ax:n| dax:n| = (µx + δ)ax:n| + n Ex − 1. dx (SP(29)) Exercise 3.2 Let µx be a weakly increasing function of x and assume that µ x → ∞ as x → ∞. (a) Show that ax → 0 as x → ∞. (b) Examine if Ax has a finite limit as x tends to infinity. (SP(31)) Exercise 3.3 Consider an insurance contract issued to an x-year old. At death within the first n years, the level continuous premium is paid back with interest and compound interest earned. Rewrite the integral expression for the present value after t years (t < h) of the future return of premium per unit of the premium in order to show that this value is Dx+h (st| + ax+t:n−t| ) − s , Dx+t h| where st| is defined by JH. Interpret the expression. (Eksamen i Forsikringsvidenskab og Statistik, KU, winter 1943-44) Exercise 3.4 Show that n Ex = 1 − iax:n| − (1 + i)A1x:n| , and interpret this formula (i is the interest rate). (Eksamen i Forsikringsvidenskab og Statistik (rev.), winter 1946-47) Insurance of a Single Life 19 Exercise 3.5 Assume there exist positive constants k and α, so that x `x = k(1 − )α ω for all x ∈ [0, ω]. (a) Find an expression for µx . (b) Find an expression for e◦x (see exercise 2.7). Now assume that α = 1. (c) Show that n| qx is independent of n. (d) Show that for n = ω − x we have ax = n − an| . nδ (SP(30)) Exercise 3.6 Ax:n| denotes the expected present value of a life insurance contract, where the amount of 1 is to be paid out by the end of the year in which the insured dies, not later than n years after the time of issue, or if he survives until the age of x + n. x is the age at entry. (a) Give an expression for Ax:n| and show that Ax:n| = 1 − däx:n| , where d is the discount rate. A1x:n| denotes the expected present value of a life insurance where the amount of 1 is paid out by the end of the year, during which he dies if he dies before the age of x + n. x is the age at entry. (b) Give an expression for A1x:n| and show that A1x:n| = 1 − n Ex − däx:n| . (c) Try to interpret the formulas in (a) and (b). (SP(33)) Exercise 3.7 Assume that active persons have force of mortality µ ax as a function of age and force of disability νx as a function of age. Assume moreover that disabled 20 Exercises in Life Insurance Mathematics persons have force of mortality µix . There is no recovery. The force of interest is denoted by δ. The four quantities defined below are the single net premiums an insured with age at entry x has to pay for a level continuous annuity with sum 1 p. a. The insurance cancels n years after issue. aix:n | The single net premium for an insured who is disabled at the time of issue. The annuity is payable from issue until the time of death of the insured. aax:n | Single net premium for active persons. The annuity is payable from the time of issue until death of the insured. aaa x:n | As above except that the premium is due for a contract that cancels by death or by disability of the insured. aai x:n | Single net premium for an active. The annuity is payable if the insured is being disabled within n years from time of issue. Expires if he dies. a i (a) Express aai x:n| in terms of µx , νx , µx and δ. aa i (b) Assume that µax = µix for all x. Express aai x:n| in terms of ax:n| and ax:n| . (c) Assume that µax = µix + ε and νx = ν where ε and ν are independent of x and aa i ε 6= ν. Express aai x:n| by ax:n| and ax:n| (and ε and ν). (Aktuarembetseksamen i Oslo (rev.), fall 1953) Exercise 3.8 Assume the force of mortality is a strictly increasing function of the age, when this is greater than or equal to a certain x 0 . Show that for x ≥ x0 and 0 < n ≤ ∞ the following inequalities hold: ax:n| ∂ax:n| ∂x < 1 − v n n px , µx + δ < 0, ax < 1 . δ (SP(37)) Insurance of a Single Life 21 Exercise 3.9 The quantity e◦x:n| = Z n t px dt 0 is the expected period of insurance for a term insurance or an n-year temporary annuity, age of entry x and age of expiration x + n. Define = ex:n| ëx:n| = n X t=1 n−1 X t px t px . t=0 Give a similar interpretation of these identities. Define at| = ät| = 1 − vt i 1 − vt d and show that for integer values of t the following inequalities are valid: ax:n| < aex:n || äx:n| < äëx:n |. | (SP(39)) Exercise 3.10 Show that ax = Z ∞ t px Ax+t dt. 0 (SP(41)) Exercise 3.11 Assume that µx is a weakly increasing function of x and consider for given x two insurance contracts with initial age x: First consider a whole-life life insurance with sum insured 1 and secondly a whole-life continuous annuity with level payment intensity determined so that the expected present value of the two insurance contracts are equal. The one with the biggest variance of the present value is naturally the one with the biggest risk for the company. Show that there exists an x0 ≥ 0 so the annuity is more risky than the life insurance iff x > x0 . 22 Exercises in Life Insurance Mathematics (SP(42)) Exercise 3.12. (Multiplicative Hazard Model) The mortality in a population varies from person to person; some has greater or lesser mortality than the average. This can be modelled as follows: There exists a underlying force of mortality µ x and for each person a constant θ independent of age exists so that the force of mortality for a person aged x is θµ x . The value of θ for a randomly chosen person is assumed to be a realisation of a stochastic variable Θ, and we assume moreover that EΘ = 1. Show that in this model the expected present value of a continuous temporary n-year annuity with payment intensity 1 is greater than or equal to ax:n| = Z n e−δt t px dt, 0 and examine under which conditions the two present values are equal. (SP(44)) Exercise 3.13 Consider an n-year endowment insurance, sum insured S, age at entry x and premium paid continuously during the entire period with level intensity p. Expenses are disregarded. (a) What is the surplus of this contract for the company in terms of the remaining life time of the insured? (b) Find the mean and variance of the surplus. (c) Explain how p should be determined so that the probalility of getting a negative surplus is lesser than a certain ε. (Hint: Tchebychev’s inequality.) (d) Show (by applying the central limit theorem) how it is possible to obtain a probability of a negative surplus for the entire portfolio lesser than ε by using a smaller p than the one found in (c). (SP(50)) We have so far worked with continuous insurance benefits - annuities that are due continuously and life insurances that are due upon death. The pure endowment seems to be of another origin, because the time of possible single payment is determined in advance. In the next exercise we will consider more general kinds of non-continuous or discrete benefits. For at start consider an x-year old whose remaining life time T Insurance of a Single Life 23 is determined by the survival function F (t | x) = e− R∞ 0 µx+τ dτ , where µx+t is the force of mortality at the age of x + t, t > 0. As usual let v denote the annual discount rate. The results of the following exercise will show that continuous benefits can be concidered as limits for discrete benefits. We will also see that both continuous and discrete annuities and life insurances are closely related to pure endowment benefits. Exercise 3.14 The present value of a t-year pure endowment with sum 1 is Cte = v t 1{T >t} . (a) Find the expectation, t Ex , of Cne and find Cov(Cse , Cte ) for s 6= t. A brute-forth generalisation of the pure endowment is produced by summing more benefits like this. A simple variant is the n-year temporary deferred annuity payable annually with fixed amounts as long as the insured is alive. This is the sum of n pure endowments with deferment times 1, . . . , n. The present value at time t = 0 is Cna(1) = n X Cte . t=1 If the annuity is payable h times a year with fixed amounts be Cna(h) = hn X 1 , h the present value will e Ct/h . t=1 a(h) (b) Find the expectation, a(h) . x:n| , and the variance of the present value C n An n-year temporary life insurance with sum insured 1, payable at the end of the year of death, has present value Cnti(1) = n X v t 1{t−1<T ≤t} , t=1 and the corresponding insurance payable at the end of the occurs, has present value Cnti(h) = hn X t=1 t v n 1{ t−1 t . h <T ≤ h } 1 th h year, in which death 24 Exercises in Life Insurance Mathematics (c) Express the present value Cnti(h) in terms of present values of annuities given by Cna(h) . Compare with similar expressions for continuous benefits. (h) (d) Find the expectation, Ax:n| , and the variance of the present value C nti(h) . (e) Use the results from (b) and (d) to prove the well known formulas ax:n| 1 Ax:n| = = n Z Z0 n 0 t v F (t | x)dt = Z n t Ex dt 0 v t F (t | x)µx+t dt = 1 − δax:n| − n Ex , for the expectations of Cna and Cnti and also to find their variances. (Hint: By monom tone convergence we have Cnα(2 ) % Cnα as m → ∞, α ∈ {a, ti}. Then use the monotone convergence for the expectation). (f) Use the technique in (e) to find formulas for the expectations and variances of continuous benefits in the usual Markov model. Consider an annuity, payable with level intensity of 1 by staying in state j, and an insurance where a sum of 1 is paid upon every transition j → k. (RN “Opgave E7” 29.01.90) Exercise 3.15 The functions that occur in insurance mathematics often depend on several variables, e. g. m| ax:n| , and are often hard to tabulate. In order to solve this problem, we introduce the so-called commutation functions. In connection with life insurances of one life we consider the following: Cx Dx Mx Nx Rx Sx = = = = = = v x dx v x `x Cx Dx Mx Nx Rx Sx Pω v ξ dξ Pξ=x ω ξ ξ=x v `ξ Pω (ξ − x)v ξ dξ Pξ=x ω ξ ξ=x (ξ − x)v `ξ = = = = = = R x+1 ξ v ` µ dξ Rxx+1 ξ ξ ξ v ` dξ Rxω ξ ξ v ` µ dξ Rxω ξ ξ ξ v `ξ dξ Rxω (ξ − x)v ξ `ξ µξ dξ Rxω ξ x (ξ − x)v `ξ dξ. (a) Show that m| ax:n| = N x+m − N x+m+n . Dx 1 (b) Find corresponding expressions for a x:n| , Ax:n| , Ax , Ax:n| , n Ex , ax:n| and äx:n| . (c) What can we possibly use R x for? Insurance of a Single Life 25 (SP(35)) 26 Exercises in Life Insurance Mathematics 4. The Net Premium Reserve and Thiele’s Differential Equation FM0 S91, 1 FM0 S92, 2 FM0 S93, 2 FM0 S94, 2 FM0 S95, 1 Exercise 4.1 Consider an n-year pure endowment, sum insured S, premium payable continuously during the insurance period with level intensity π. Upon death two thirds of the premium reserve is being paid out. (a) Put up Thiele’s differential equation for V t . What are the boundary conditions? (b) Find an expression for the premium reserve at time t, t ∈ [0, n) (c) Determine the premium intensity π by adopting the equivalence principle. (SH and MSC, 1995) Exercise 4.2 We have the choice of two different premium payment schemes. • For an insurance of a single life a level continuous premium is due with force p as long as the insured is alive at the most for n years from the issue of the contract. • Every year an annual premium of the size p(1) = p · a1| is paid in advance. If the insured dies during the insurance period the return of premium is a | R = p(1) · θ = p · aθ| , a1| where θ denotes the remaining part of the year at time of death. (a) Show that these two premium payment schemes are equivalent in the manner that regardless of when the insured is going to die, the present values of the premium payments under the two schemes are equal. (b) Find the expected present value of the return of premium at the time of issue. Let the prospective reserves at time t from the time of issue of the two premium schemes be denoted by V t and Vt . The Net Premium Reserve and Thiele’s Differential Equation (c) Show that for t ≤ n 27 Vt = V t + p · a[t]−t | where [t] is the integer part of t. (SP(53)) Exercise 4.3 Consider a linearly increasing n-year term insurance. If the insured dies at time t after the time of issue where t < n, the amount tS is paid out. If the insured is alive at time n, the amount nS is paid out at this time. The age of the insured at entry is x. The net premium determined by the equivalence principle, is due continuously with level intensity p. (a) Find an expression for p. (b) Find a prospective and a retrospective expression for the reserve V t at any time t, 0 < t < n. (c) Show that the two expressions found in (b) are equal for all t, 0 ≤ t < n. (d) Derive Thieles differential equation. (e) Find the savings premium and the risk premium. (SP(54)) Exercise 4.4 An n-year insurance contract has been issued to a person (x). The premium is composed of a single premium π 0 at the beginning of the contract and by a continuous intensity (πt )t∈(0,n) as long as (x) is alive, at the most for n years. The benefits are a pure endowment, sum insured S n at time n, a term insurance, sum insured St at time t ∈ (0, n), and a continuous flow with intensity (s t )t∈(0,n) as long as (x) is alive during the insurance period. (a) Put up Thiele’s differential equation. (b) Find a boundary condition without assuming the equivalence principle. (c) Find a prospective expression for the premium reserve by solving the differential equation. (d) Adopt the equivalence principle and find an alternative boundary condition. (e) Find, by applying the new boundary condition a retrospective expression for the premium reserve. 28 Exercises in Life Insurance Mathematics Now assume that the benefits moreover consist of a pure endowment, sum insured S at time m (0 < m < n). (f) Has this altered Thiele’s differential equation? (g) Which extra boundary condition are now to be added in order to solve the differential equation? Assume that π0 = 0, πt = π for t ∈ (0, m) and that π is determined by the equivalence principle. (h) Find the net premium. (i) Find the premium reserve at any time. (SH “Opgave til 14/10-94” (rev.)) Exercise 4.5. (Prospective Widow Pension in Discrete Time) Consider a policy with widow pension, insurance period n years, issued to a man aged x and his wife aged y. During the insurance period the premium Π falls due annually in advance as long as both are alive. If the man dies, the benefit is a widow pension of sum 1 paid out on every following anniversary of the policy during the insurance period if the widow is still alive. All expenses are disregarded. (a) Put up an expression for the premium reserve for this policy at its tth anniversary. (b) Show how the premium reserve at any time t can be expressed in terms of the reserve at time t + 1 for t = 0, 1, . . . , n − 1 so that the premium reserves can be calculated recursively. (c) Define the savings premium and the risk premium and find an interpretable expression for the latter. (FM1 exam, summer 1977) Exercise 4.6 Consider an n-year term insurance, sum insured S, age at entry x. Continuous premium during the entire period with level intensity p determined by the equivalence principle. (a) Find Thieles differential equation for the premium reserve V t . (b) Which initial conditions would be natural to use for t = 0 and t = n − respectively? (c) Solve the differential equation with each of the initial conditions and compare the solutions. The Net Premium Reserve and Thiele’s Differential Equation 29 (SP(55)) Exercise 4.7 If the insured dies before time n (from the time of issue) the benefit is a continuous annuity with force 1, duration m from the time of death. If he is alive at time n, the benefit is a similar annuity from this time and if he is still alive at time n + m he receives a whole-life annuity with intensity 1. There is a single net premium at the time of issue and the equivalence principle is adopted. (a) Find the single net premium. (b) Find the prospective premium reserve at any time. (c) Derive Thieles differential equation. Let Rc denote the risk sum at time t. (d) Prove that Rt = m (1 − v )ax+t:n−t| − n+m−t| ax+t −n+m−t| ax+t −a x+t (t < n) (n ≤ t < n + m) (n + m ≤ t). In particular we have and since R0 = (1 − v m )ax:n| − n+m| ax , lim R0 = (1 − v m ax ) > 0, n→∞ we have R0 > 0 if only n is big enough. Assume that R 0 > 0. Because Rt is a strictly increasing continuous function on [0, n] and R n < 0, there exists a unique τ ∈ (0, n) so that Rτ = 0. (e) Show that this τ is determined by N x+τ = N x+n + N x+n+m . 1 − vm (SP(56)) Exercise 4.8 Consider a pension insurance contract, where the benefit is an n-year annuity of 1 deferred m years (expected present value m|n ax ). Premium is paid with level intensity c during the deferment period (expected present value ca x:m| ). (a) What is the equivalence premium c and the development of the reserve when x = 30, m = 30, n = 20 and the technical basis is G82M, i. e. i = 0.045 and µ x = 0.0005 + 10−4.12+0.038x . 30 Exercises in Life Insurance Mathematics (b) Do similar calculations as in (a) for an extended contract where k times the premium reserve is being paid out by possible death during the deferment period, k = 0.5, k = 1. (RN “Opgave til FM0” 15.10.93) Exercise 4.9 Consider a single-life status (x) with force of mortality µ x . Define the premium reserve by Vt = E(U[t,∞) | T > t) as usual. (a) Show that the premium reserve always is right continuous. (b) Discuss under which conditions it is left continuous. Consider the following benefits at time t • st dt1{T >t} , annuities • St 1{T ∈dt} , life insurances • Bt 1{T >t} , pure endowment, and show that if the premium is being paid with level intensity π and there isno lump sum at time t then Vt is left continuous. (c) If there is a lump sum at time t, what does V t − Vt− look like? NB: Assume that µx+t , st , St are continuous, and assume that there only exist a finite number of t’s where Bt 6= 0. (SH “Opgave til 13/10-95”) Exercise 4.10 Consider an insurance of a single life, age at entry x. At time t (t = 0, 1, 2, . . .) the premium Pt is being paid if the insured is still alive, and if he dies during [t − 1, t) then St is the benefit. The equivalence principle is adopted for the insurance and all expenses are disregarded. Let the premium reserve at time t be V t and let the stochastic variable Gt be given by Gt = 0 V + P − vS t t t+1 V + P − vV t t t+1 (the insured is dead at time t− ) (the insured dies during the interval [t, t + 1)) (the insured is alive at time t + 1). Let the stochastic variable Y be the present value at time 0 of the company’s surplus of the insurance. The Net Premium Reserve and Thiele’s Differential Equation (a) Interpret Gt and show that Y = ∞ X 31 v t Gt . t=0 (b) Show that VarY = ∞ X Var(v t Gt ). t=0 G1 , G2 , . . . are not necessarily stochastically independent, but note that (b) is valid regardless of whether G0 , G1 , . . . are stochastically independent or not. (c) Show that Hattendorf’s Formula VarY = ∞ X t=0 t px v 2t+2 px+t qx+t (St+1 − Vt+1 )2 is valid. (SP(58)) Exercise 4.11 In this exercise we are to study an endowment insurance with return of premium paid at death before expiration. Consider a person of age x who wishes to buy an insurance with age of expiration x + n, and where the premium is paid continuously with level intensity p as long as he is alive during the insurance period. The lump sum S = 1 is paid if he is alive at age x + n and if he dies before that the premium paid so far will be returned with interest (basic interest i) earned. We disregard expenses. (a) Show that the variable payment at death is given by Bt = pat| (1 + i)t . (b) Determine the continuous premium intensity p. (c) Find the net premium reserve Vt at time t, t ∈ [0, n). (d) Put up Thiele’s differential equation and determine the risk sum. (e) Comment on the results and evaluate whether or not you will recommend the insurance company to issue this kind of insurance. (FM1 exam, summer 1985) 32 Exercises in Life Insurance Mathematics Exercise 4.12 An n-year deferred whole-life annuity on the longest lasting life has been issued to two persons (x) and (y). Annual amount of 1 and level continuous premium on the longest lasting life with intensity π during the deferment period. The forces of mortality are denoted by µ x and νy Put up a retrospective expression for the net reserve at time t assuming that both are alive. (Eksamen i Forsikringsvidenskab og Statistik (rev.), winter 1944-45) Exercise 4.13 A family annuity is an insurance contract of one life that assures payment of a continuous annual annuity from the possible death of the insured during the insurance period until the expiration of the contract after n years. Force of mortality µx , interest rate i, age at entry x. (a) Put up the formulas for the net premium reserves, prospectively and retrospectively, with level continuous premium payment π. Show that if the insured does not die during the insurance period the reserve will at least once become negative. (b) Discuss how the total premium reserve of this contract for a portfolio of identical contract issued at the same time will develop during the insurance period. Show that this reserve never becomes negative. (The students of 1946 had 10 hours to complete this exercise!) (Aktuarembetseksamen (rev.), Oslo fall 1946) Exercise 4.14 During construction of a technical basis with mortality of death and mortality of survival it is a problem that the premium for an insurance can depend on whether the insurance stands alone or it is combined with other insurances. This exercise describes the attempts made under construction of G82 in order to solve this problem of additivity. Consider an insurance Ax:n| + s · n| ax issued against a single premium. Thiele’s differential equation for the net premium reserve is ∂Vx (t) = ∂t ( δVx (t) − µ̃x+t (1 − Vx (t)) (0 < t < n) δVx (t) − s + µ̃x+t Vx (t) (n < t) (4.1) The Net Premium Reserve and Thiele’s Differential Equation 33 where µ̃x+t is the actual expected force of mortality. Introduce the first order forces of mortality µ and µ̂ which satisfy µx+t < µ̃x+t < µ̂x+t , and replace (4.1) by ∂Vx (t) = ∂t ( (δ + µx+t )Vx (t) − µ̂x+t (0 < t < n) (δ + µx+t )Vx (t) − s (n < t). (4.2) Thus we get a smaller increase of the reserve and we get a technical basis “on the safe side”. (a) Determine the single premium on the first order technical basis by solving (4.2). Consider the special case µ̂x = (µx + g2 )(1 + g1 ), and let µ∗x = µx + g2 and δ ∗ = δ − g2 . (b) What will the single premium be in this case? Comment on the result. Now consider an educational endowment a x|n| . (c) Put up the differential equations (4.1) and (4.2) and solve (4.2) for the special case above. What is the problem in this case? Finally consider a survival annuity a x|y . (d) Answer the same question as in (c). (SP(99)) 34 Exercises in Life Insurance Mathematics 5. Expenses Exercise 5.1 Work out the details in RN in the case where α 00 = β 00 = γ 00 = 0. (RN(1) “Expenses, gross premiums and reserves” 12.10.90 rev. 13.03.93) Exercise 5.2 Express Vtg and cg in terms of Vtn and cn and α0 in the case where α00 = β 00 = γ 00 = γ 000 = 0. (RN(2) “Expenses, gross premiums and reserves” 12.10.90 rev. 13.03.93) Exercise 5.3 Treat the case of a level annuity payable upon death in (m, n) against level premiums in (0, m). (RN(3) “Expenses, gross premiums and reserves” 12.10.90 rev. 13.03.93) Exercise 5.4 Consider an n-year deferred whole-life annuity, age at entry x, payable with level continuous intensity S. Level gross continuous premium intensity p payable during the deferment period. The premium is determined by the equivalence principle. For now assume that the expenses are initial expenses αS, loading for collection fees due continuously with level intensity βp and administration costs also due continuously with level intensity γS. (a) Find p and the prospective gross premium reserve V tg . Because of inflation, loading for collection fees and administration expenses are paid with intensities βf (t)p and γf (t)S at time t. (b) Put up exspressions for p and Vtg . (b) Find p and Vtg when f (t) = 1 + kt and where f (t) = exp(ct). (SP(62)) Exercise 5.5 Consider a whole-life life insurance, sum insured 1, age at entry x, single net premium B. (a) Show that the expected effective interest rate for the insured is Z ∞ 0 1 B − t t px µx+t dt − 1. Expenses 35 Assume that the company has some initial expenses α, but no other expenses. (b) What is the expected effective interest rate? (SP(60)) Exercise 5.6 A simple capital insurance, sum insured S, duration n, pays out S at time n from the time of issue no matter if the insured is alive or not. (a) Put up an expression for the net payment for this insurance and explain why it is independent of the age at entry. A simple capital insurance only makes sense if it is not paid by a single payment (when dealing with insurance). Assume that the premium is paid continuously during the entire insurance period with level intensity p, but only if the insured is alive. The premium is determined by the equivalence principle. (b) What will the net premium be? In the gross premium p, initial expenses αS are included as well as loading for collection fees βp and administration expenses γS paid continuously. (c) Determine p. (d) Put up an expression for the prospective gross premium reserve, both when the insured is alive as well as when he is dead. (SP(61)) The following exercise examines what happens to the insurance technical quantities when we bring surrender into consideration. Exercise 5.7 Consider an n-year endowment insurance, age of entry x, benefits are S1 if one dies during the insurance period and S 2 if one obtains the age of x + n. Life conditioned equivalence premium is paid continuously until time n (from the age of entry). Moreover assume that surrender can take place at any time during the premium payment period, and that the present value of the conventionally calculated gross premium reserve, liquidated by surrender, is positive. By surrender at time t the company pays out G(t). (a) Now disregard all expenses and assume that G(t) is lesser than or equal to the conventionally calculated (net) premium reserve at time t. Instead of using a conventional technique, the company could itself bring surrender into consideration and 36 Exercises in Life Insurance Mathematics into its own technical basis. Show that the equivalence principle then would lead to a premium P 0 ≤ P . Discuss conditions for P 0 = P and give a lower limit for how small P 0 can get when G(·) varies. Here and in the following it might be useful to study Thiele’s differential equation. Now assume that some administration costs are not neglectible. The expenses consist of the amount α in initial expenses, of γ in administration costs per time unit and of a fraction β of the actual annual gross premium P in loading for collection fees, P calculated conventionally. Upon surrender 100θ% of the gross premium reserve is paid out if the reserve is positive, 0 ≤ θ ≤ 1. By surrender where the gross premium reserve is positive, a fixed percentage of the reserve is deducted to cover the loss experienced by surrender where the gross premium reserve is negative. For now disregard expenses that fall upon surrender. This gross premium reserve is calculated without respect to surrender. (b) Show that you will get a lesser gross premium reserve if you bring surrender into consideration. Assume that θ is chosen so that the equivalence principle can be applied anyway. (c) Now assume that in the above situation a constant expense ξ is associated with the actual payment of G(t). If G(t) calculated in (b) is smaller than ξ, nothing is paid out by surrender. When the value upon surrender mentioned exceeds ξ the difference is paid out. Show that the actual gross premium reserve still will be lesser than the conventional when surrender is brought into consideration. Can θ still be determined so that the equivalence premium still can be applied? (FM1 exam (rev.), summer 1979) Exercise 5.8 It has been proposed that the administration expenses should be calculated as being proportional to the gross premium reserve instead of being proportional to the sum insured. Now consider an n-year endowment insurance, level continuous gross premium intensity p, sum insured S, age at entry x. Acquisition expenses αS. Loading for collection fees βp and γV t at time t, respectively. Vt denotes the gross premium reserve. Administration costs fall due continuously with level intensity γV t at time t. (a) Put up Thiele’s differential equation for V t . (b) Solve the differential equation with initial conditions for t = 0 and t = n − and show that the solutions can be expressed by expected present values for annuities with another interest rate than the interest rate of the technical basis. (c) Determine the equivalence premium. Expenses 37 In G82 the interest rate is i = 5% p. a. but gross premiums and gross reserves are calculated with an interest rate of 4.5% p. a. (d) What is the corresponding value of γ? (SP(64)) Exercise 5.9. (Equipment Insurance) By an equipment insurance, sum insured S, insurance period n, the sum S is paid out at time n if the insured is still alive; if he dies during the insurance period, the company returns the up till now paid premiums without interest earned. The level gross premium intensity p is payable during the entire insurance period. The expenses are initial expenses αS, loading for collection fees due with continuous level intensity βp and continuous administration costs due with level intensity γS. Put up an expression for p applying the equivalence principle. (SP(66)) Exercise 5.10. (Child’s Insurance) A person aged x has been issued a child’s insurance: If the insured dies during [x, y) the gross premium is paid back with earned interest according to the technical basis. If he dies during [y, u) the sum S is immediately paid out and if he is alive at age u, then S is paid out. The level gross premium p, the administration costs γS and loading for collection fees βp fall due continuously during the insurance period. Acquisition expenses are αS. (a) Put up Thiele’s differential equation for this insurance. (b) Find the prospective gross premium reserve at any time during the insurance period. (c) What is the gross (equivalence) premium intensity, and what is the risk sum at any time with this premium. (SP(68)) Exercise 5.11 Consider an n-year endowment insurance, sum insured S, age at issue x, premium payable until time m, initial expenses αS, administration costs and loading for collection fees due during the entire insurance period continuously with level intensities γS and βpg respsctively, where pg is the level gross premium intensity. Assume that m ≤ n and γ < δ. (a) Give an expression for pg applying the equivalence principle and prove that it can 38 Exercises in Life Insurance Mathematics be cast as pg = Ax:n| + γ̃ax:n| (1 − β̃)ax:m| S. (b) Show that β̃ > β and γ̃ > γ. The numerator of the expression is the so-called passive with added sum, because it is produced from the net passive Ax:n| increased by the present value of γ̃ during the S entire insurance period. This passive is denoted by A x:n| . Let Vt be the net premium reserve at time t (calculated from the time of issue) and let Vt1 be Vt increased by the reserve of the future administration costs. (c) Give expressions for Vt and Vt1 . The company ought to set aside the reserve V t1 but normally the reserve S Vt2 = SAx+t:n−1| − (1 − β̃)pg ax+t:m−t| is set aside. (d) Compare Vt , Vt1 and Vt2 and try to explain why one prefers to set aside V t2 instead of Vt1 . If the insured wishes to surrender his contract at time t, the company pays him the surrender value of the contract Gt , which is the reserve Vt1 less the part of the initial expenses that have not yet been amortized. (e) Show that the surrender value can be cast as Gt = S(Ax+t:n−t| + γax+t:n−t| ) − (1 − β)pg ax+t:m−t| . g The coefficient for S is the surrender value passive and is denoted by A x:n| . (f) Find an expression for the difference between the net premium reserve and the surrender value and prove that for m = n it is α(S − V t ). If the insured wishes to cancel the payment of premiums without entirely to surrender the contract, it is called a premium free policy. The size of this policy is determined by letting the surrender value of the new policy equal the surrender value of the original policy at the time of change. (g) Give an expression for the sum of the premium free policy and show that its reserve at the time of change is lesser than the reserve of the original policy. Expenses 39 Assume that the annual gross premium is given by p̈g = εpg , where ε is called the continuity factor. (h) Explain why the surrender value and the reserve can be cast as g SAx+t:n−t| − ζ p̈g ax+t:m−t| and S SAx+t:n−t| − η p̈g ax+t:m−t| , respectively. (SP(70 rev.)) Exercise 5.12 A married man considers a life insurance on the following conditions: (i) If he dies before time r from time of issue of the contract, the company has to pay a continuous pension with level intensity s in a period of time m. He furthermore considers a supplementary pension, also with level intensity s which is due to initiate right after the expiration of the pension (i). He considers two options: (ii) The supplementary pension is due as long as his wife lives. (iii) The supplementary pension is due as long as his wife lives, at the most until time n from the time of issue, n > m + r. As a second alternative he considers a survival annuity, also with payment intensity s. Here he considers two options: (iv) The annuity initiates if the man dies before time r from the time of issue and is due as long as his wife lives. (v) The annuity initiates if the man dies before time r from the time of issue and is due as long as his wife lives, at the most until time n from the time of issue, n > r. (a) Put up an expression for the single net premium for these five contracts. Assume that the above contracts are issued against an annual premium payment paid in advance as long as the man and his wife are alive, at the most until time r from the time of issue. If one of the two dies during the insurance period, the amount (aθ| /a1| )P is being returned (in Danish: Ristorno), where θ is the remaining part of the last premium payment period and P is the term premium. 40 Exercises in Life Insurance Mathematics (b) How would you calculate the annual net premiums? (c) Put up an expression for the net premium reserve at any time during the insurance period for the last insurance (v) applying the recently described premium payment principles. When calculating the gross premiums, the company uses the following expense rates: Initial expenses αS, loading for collection fees β times the gross premium, administration costs due continuously with intensityand γ times the gross premium reserve at any time, and finally payment costs of ε times the amount paid out. (d) Find the continuous gross premium intensity, applying the equivalence principle. (e) Find the gross premium when the premium payment takes place as described before (b). (SP(76) rev.) Select Mortality 41 6. Select Mortality Exercise 6.1 (a) What could be the meaning of the symbol (b) Put up an expression for period of selection is 5 years. 2|6 q[30]+2 s|t q[x]+u ? in terms of ` under the assumption that the (c) Express the following three quantities with one symbol: • The probability that a person now 50 years old who got insured 3 years ago dies between the ages of 58 and 59, presuming the period of selection is 5 years, • the probability that a new born dies between 67 and 72 years of age, • the number of deaths between the ages of 29 and 30 in the third year of an insurance portfolio, presuming the period af selection now is 3 years. (SP(25)) Exercise 6.2 In this exercise we will try to explain the presence of select mortality for a portfolio of insured and study its properties. For two functions f and g we shall use the obvious notation f g(t) = f (t)g(t), (f + g)(t) = f (t) + g(t). The portfolio is assumed to be divided between the two states active and disabled according to the figure below where the course of events is modelled by a Markov process {Xt }t≥0 , and t is the age of the insured. σ(t) 1. Active K KKK o KK / ρ(t) µ(t)K KKK KK% 2. Disabled sss sss ν(t) ss s s ys s 3. Dead The transition probabilities of the model are denoted by pij (s, t) = P (Xt = j | Xs = i), s ≤ t, i, j = 1, 2, 3 42 Exercises in Life Insurance Mathematics and the intensities µij (t) are assumed to exist and are given by µij (t) = lim h&0 pij (t, t + h) , i 6= j. h We assume that the intensities are continuous functions. Define µ(t) = µ13 (t), ν(t) = µ23 (t), σ(t) = µ12 (t), ρ(t) = µ21 (t) and µ1 (t) = µ(t) + σ(t), µ2 (t) = ρ(t) + ν(t). We take it that µ(t) < ν(t), ∀t ≥ 0, i. e. the mortality for a disabled is always greater than for an active person. When we cannot observe whether an insured is active or disabled at any time after entry (as an active), one gets a filtration (of the above Markov model) which is determined by the force of mortality for a random insured. Let µ̃(τ, x) denote this intensity for an insured of age x with age of entry τ, τ ≤ x. (a) Explain that µ̃ is given by p12 p11 (τ, x) + ν(x) (τ, x) p11 + p12 p11 + p12 p12 = µ(x) + {ν(x) − µ(x)} (τ, x) p11 + p12 µ̃(τ, x) = µ(x) and thus give the grounds for the presence of select mortality. We obviously want τ → µ̃(τ, x) to be decreasing for fixed x which will be shown by differentiation in the following. (b) Explain that τ → µ̃(τ, x) is decreasing iff the fraction τ → (p 12 /p11 )(τ, x) is decreasing and give an interpretation of this. (c) Show that d dτ σ(τ )(p12 p21 − p11 p22 )(τ, x) p12 (τ, x) = p11 p211 (τ, x) and d (p12 p21 − p11 p22 )(τ, x) = {µ1 (τ ) + µ2 (τ )}(p12 p21 − p11 p22 )(τ, x), dτ and explain why τ → µ̃(τ, x) is decreasing. Let p̂ij (s, t) denote the transition probabilities corresponding to the model without recovery, i. e. ρ(t) = 0, ∀t ≥ 0. Select Mortality 43 (d) Find expressions for p̂11 (τ, x) and p̂12 (τ, x) as a function of the intensities and show that p12 p̂12 (τ, x) < (τ, x). p11 p̂11 Give an interpretation of this and explain how µ̃ is affected by changing to the model without recovery. It is a common opinion that the selection the insured goes through at entry disappears after a period of time, called the period of selection. We will try to explain this phenomenon mathematically. Lad τ0 be the age of entry for an insured. (e) Show that x → exp Z x τ0 µ1 (s)ds p11 (τ0 , x), x ≥ τ0 is an increasing function. Assume that µ 2 (t) ≥ µ1 (t), ∀t ≥ 0 and that there exists a w > τ0 so that Z w τ0 Show that (µ2 (t) − µ1 (t))dt = ∞. d lim x%w dτ monotonically with p12 p11 (τ0 , x) = 0 d p12 (τ0 , x) = 0, ∀x ≥ w, dτ p11 and explain why this verifies the presence of a period of selection of w. (FM1 exam, 1989-ordning, opgave 1, summer 1995) Exercise 6.3 Mortality in a portfolio of insured lives will usually be different from the mortality of the general population because the insured lives are a selected part of the population. We will in this exercise study one relationship that is assumed to contribute a great deal to the effect of selection, i. e. the fact that people with illnesses that cause severe excess mortality are not allowed to underwrite life insurances (under the usual terms). In the following such persons will be called “ill”. Thus the population can be divided according to the figure below. ρx / 1. Insured, not ill QQQ QQQ κ nnn κx nnnn QQQx n QQQ n QQQ nnn ( vnnn σx 4. DeadhP PPP mm6 m P m PPPλx λx mmm PPP mmm m PPP m m P m m 0. Not insured, not ill σx 3. Not insured, ill 2. Insured, ill 44 Exercises in Life Insurance Mathematics Now assume that every person enters state “0” at birth and that the transitions between the states afterwards go on as a time continuous Markov chain with transition intensities only dependent on the age x as indicated in the figure. Excess mortality for the ill persons means that λx ≥ κx , x > 0, (6.1) with “>” for some values of x. With the usual notation for the transition probabilities the following are satisfied p11 (x − t, x) = e p12 (x − t, x) = − Z e p02 (0, x) = p03 (0, x) = Z Rx 0 Rx 0 x e 0 0 x (σu +κu )du − x−t − Z x−t x p00 (0, x) = e− p01 (0, x) = e Rx Rz x−t , (σu +κu )du (σu +κu +ρu )du (σu +κu )du − e− Rz 0 Rz 0 (6.2) σz e− Rx z λu du dz, 0 < t < x; , (1 − e (σu +κu )du (6.3) (6.4) − Rx 0 (1 − e (σu +κu +ρu )du ρu du − Rz σz e− 0 ρu du Rx z ), (6.5) )σz e λu du − Rx z λu du dz, 0 < x. dz, (6.6) (6.7) (a) Prove the formulas (6.2) and (6.3) by putting up and solving differential equations. (b) Assume that (6.4) is given. Give direct, informal grounds for the expressions (6.5) – (6.7). The insured lives are either in state “1” or in state “2” (those in state “2” received the insurance contract before they were struck by illness). The insurance company does not observe in which of the two states the insured is. All the company knows is the time of entry and age. Let µ[x−t]+t denote the force of mortality for an insured of age x who received the insurance t years ago. (c) Derive an expression for µ[x−t]+t . Show that under the condition (6.1), µ [x−t]+t is a non-decreasing function of t for constant x (it might be desirable to express µ [x−t]+t as a weighted average of κx and λx ). How will you explain this result to a person who has no knowledge of actuarial science? (d) Discuss the formula for µ[x−t]+t to find theoretical explanations as to why insurance companies operate with a period of selection s so that the mortality is considered to be aggregate for t > s. Let µx denote the force of mortality for a randomly chosen person of age x in the population (that is, we do not observe in which of the states “0” – “3” the person is). Select Mortality 45 (e) Find an expression for µx . Show that under the condition (6.1) the inequality µx ≥ µ[x−t]+t , 0 < t < x is satisfied. The result clarifies the preliminary remarks of this exercise. Try to give an explanation that is comprehensible for a person without any knowledge of actuarial mathematics. (FM1 exam (1), winter 1985/86) 46 Exercises in Life Insurance Mathematics 7. Markov Chains in Life Insurance FM0 S94, 1 Exercise 7.1 Consider a model for competing risks with k + 1 states, 0: “alive” and 1, . . . , k denoting death from k different reasons; denote the partial probabilities of death by Z (j) t qx t = 1 − exp − 0 µ0j , x+τ dτ and define (j) t px = 1 − t qx(j) . (a) Prove that 00 t px = k Y (j) t px . j=1 (b) Prove that k X 0j t px = j=1 k X (j) t qx j=1 + − X X (i) (j) t qx t qx 1≤i<j≤k (h) (i) (j) t qx t qx t qx 1≤h<i<j≤k − · · · + (−1)k+1 t qx(1) · · · t qx(k) . Now assume that for given j there exist constants c and t 0 > 0 so that (j) 0j t px µx+t = c, for all t ∈ [0, t0 ]. (c) Prove that for all t ∈ [0, t0 ] (j) t qx = 0j t px = t · t q (j) t0 0 x Z t 1 πh6=j τ px (h) dτ. · t qx(j) t 0 (SP(83)) Exercise 7.2. (Life Insurance with Exemption from Payment of Premium by Disability) In this exercise we will consider a policy that can attend one of N states Markov Chains in Life Insurance 47 enumerated 1, . . . , N . The development of the policy is being described by a Markov process with transition intensities µ jk t for transition from state j to state k at time t from issue. In the contract it is stated that the amount B tjk is payable upon transition from state j to state k at time t. As long as the policy stays in state j, a continuous payment with intensity Btj is due, i. e. in the time interval [t, t + ∆t) the amount B tj ∆t + o(∆t) is paid out. Assume that all amounts B tjk are non-negative and that the force of interest δ is independent of time. For now we disregard administration expenses. (a) There are no assumptions regarding the sign of B tj . How should negative values of Btj be interpreted? Let Vtj denote the premium reserve in state j at time t. It is defined as the expected present value of the out payments in the time interval [t, ∞) discounted back until time t, given that the policy is in state j at time t. (b) Show that the premium reserve satisfies the differential equation system −Btj = X jk jk d j µt (Bt + Vtk − Vtj ), j = 1, . . . , N. Vt − δVtj + dt k6=j (c) Interpret this differential equation system intuitively in terms of savings premium and risk premium. A special case of the this general Markov model is the disability model. This model has three states, 1, 2 and 3, corresponding to active, disabled and dead. The insured’s age at entry is x and the intensities are denoted by µ12 = σx+t (from active to disabled), t µ21 = ρx+t (recovery), y µ13 = µx+t (dead as active), y µ23 = νx+t (dead as disabled). t All other transition intensities are 0. Apply this model for an investigation of an endowment insurance with exemption from payment of premiums by disability. Assume the insured is active at entry. The insurance period is n, so the insurance cancels when the insured has reached the age of x + n or if he dies before that. In question (d) it is assumed that the equivalence premium is paid continuously with level intensity π as long as the insured is active and that a constant sum insured S 48 Exercises in Life Insurance Mathematics is payable upon the expiration of the policy; hence, with the notation from above we have Bt1 = −π, Bt2 = 0, Bt13 = Bt23 = S. (d) Come up with formulas for π, Vt1 and Vt2 for 0 ≤ t < n in terms of the transition probabilities and transition intensities in the model and the discounting rate v, by direct prospective reasoning. If the premium and the benefits depend on the reserves, the premium and the premium reserve cannot be determined directly as in (d); instead the differential equation system from (b) must be solved with appropriate boundary conditions. Now it is assumed that the premium and the payment of benefit at age x + n are due as above, but upon death of the insured before time x + n, the premium reserve of the policy is paid out as a supplement to the sum insured; B tj3 = S +Vtj for j = 1, 2 and 0 < t < n. (e) Show that the differential equation system from (b) gives the grounds for a differential equation of first order in V t1 − Vt2 . What is the initial condition? Solve the differential equation. (f) What are π, Vt1 and Vt2 for 0 ≤ t < n. (g) Show that if νx+t ≥ µx+t for all t < n then Vt2 > Vt1 for all t < n. Interpret this result. Assume that we have the following expenses: Initial expenses due at time 0 with an amount of αS, loading for collection fees βp, where p is the level gross premium intensity and administration costs due with an intensity at time t equal to γV tj if the policy is in state j. (h) Explain, without performing any detailed calculations, what changes would follow from these assumptions in the theory discussed in (e) – (g). (FM1 exam , summer 1984) Exercise 7.3 An active person aged x considers a disability annuity, which falls due continuously with level intensity b upon disability before the age of x + n. Premium is payable at rate π as long as the person is active during the contract period. Assume that the state of the policy is S(t) at time t after the time of issue where {S(t)} t≥0 is a time continuous Markov model with state space and transitions as follows Markov Chains in Life Insurance 0. Active K 49 σ(x+t) KKK KK µ(x+t) K KKK KK% / 1. Disabled ttt ttt ν(x+t) tt t t yt t 2. Dead (a) Give, without any proof, expressions for the transition probabilities pjk (s, t) = P (S(t) = k | S(s) = j), 0 ≤ s ≤ t, j, k ∈ {0, 1, 2}. (7.1) The present value at time s of benefits less premiums in [s, n] can be cast as C(s) = Z n s v t−s (b1{S(t)=1} − π1{S(t)=0} )dt, where 1A is the indicator function for the event A. (b) Find, for 0 ≤ s ≤ n and j = 0, 1, 2, the conditional expection Vj (s) = E(C(s) | S(s) = j) (7.2) Zj = Var(C(s) | S(s) = j) (7.3) and the conditional variance as expressions of integrals of functions of the transition probabilities from (7.1). (c) Find EC(s) and VarC(s) in terms of the transition probabilities (7.1) above and the functions (7.2) and (7.3). (d) Now assume that the equivalence principle is being adopted, i. e. V 0 (0) = 0, where π is the net premium intensity and Vj (s) in (7.2) above is the net premium reserve in state j at time s. Does the net reserve ever become negative? (FM1 exam opgave 1, summer 1989) Exercise 7.4 The figure below illustrates an expansion of the model discussed in exercise 7.3. There are two states of disability i 1 and i2 representing two degrees of disability. Assume that ν2 (x+t) ≥ ν1 (x+t). Let {S̃}t≥0 be the corresponding Markov chain and define the stochastic process {S(t)} t≥0 by S(t) = S̃(t) for S̃(t) ∈ {a, d} and S(t) = i for S̃(t) ∈ {i1 , i2 }. If one does not know the degree of disability, {S(t)} is the observable process. 50 Exercises in Life Insurance Mathematics i1 . rr8 r r rr σ1 (x+t) r rrr r r r / σ2 (x+t) a. LL i2 . LLL rr r ν1 (x+t) LL r rr µ(x+t) ν2 (x+t) LLL r r LLL r & xrrr d. The transition intensities for the S(t)-process are µjk (t, {S(s)}s<t ) = lim dt&0 P (S(t + dt) = k | S(t) = j, {S(s)}s<t ) dt for j 6= k and any specification of {S(s)} s<t . (a) Prove that µai (t, {S(s)}s<t ) = σ1 (x + t) + σ2 (x + t) and µad (t, {S(s)}s<t ) = µ(x + t) (both independent of {S(s)}s<t ) and µid (t; S(s) = a for 0 ≤ s < τ, S(s) = i for τ ≤ s < t) = P2 j=1 σj (x + τ ) exp(− P2 j=1 R t−τ 0 σj (x + τ ) exp(− νj (x + τ + u)du)νj (x + t) R t−τ 0 νj (x + τ + u)du) (7.4) (a function of x + τ and t − τ ). Give a sufficient condition for {S(t)} t≥0 being Markov. (b) Interpret the expression in (7.4). Give sufficient conditions for finding an approximating function to (7.4) depensing only on x + t for sufficiently great t − τ , say t − τ ≥ s0 (the time of selection). (FM1 exam opgave 2, summer 1989) Exercise 7.5. (Markov Chains in Life Insurance) Give a detailed proof of the Kolmogorov backward differential equations (2.15) and the corresponding integral equations (2.21). (RN “Problems in Life Insurance” 14.10.93, problem 7) Exercise 7.6. (Markov Chains in Life Insurance) consider the model for disabilities, recoveries, and deaths in Paragraph H in RN “Markov Chains in Life Insurance” 04.03.94. Markov Chains in Life Insurance 51 (a) Find the second order differential equation for p 00 (s, ·) as outlined in the text, making appropriate assumptions about differentiability of the intensities. From now on consider the special case with constant intensities: (b) Find explicit solutions for p00 (s, t) and p01 (s, t). Note that the solution depends on s and t only through t − s, hence put s = 0. Discuss how the probabilities depend on t and the intensities. (c) Calculate for t = 0, 10, 20, . . . , 100 and draw graphs of the probabilities for some different choices of the intensities, including as key references 1. σ = 0.01, ρ = 0.01, µ = 0.005, ν = 0.005, 2. σ = 0.01, ρ = 0.01, µ = 0.005, ν = 0.01, 3. σ = 0.01, ρ = 0.01, µ = 0.005, ν = 0.1. (You can program the formulas and compute directly or you can employ the program ’retres’.) (d) Explain how the result 3. in item (c) above can be used to find the probabilites for the case σ = 0.1, ρ = 0.1, µ = 0.05, ν = 1. (The ratios between the intensities are the essential feature.) (e) The probability p02 (s, t) gives the mortality law for a person who is known to be active at age s. Discuss how it depends on the intensities, with special attention to the case where µ = ν. (f) Consider the special case with no mortality (µ = ν = 0), whereby the number of states essentially becomes 2. Find p 01 (0, t), and discuss how the expression depends on t and the intensities. Find the limit as t → ∞ and discuss the expression. (RN “Problems in Life Insurance” 14.10.93, problem 8) Exercise 7.7. (Markov Chains in Life Insurance) Formulate suitable continuous time Markov chain models for solution of the following problems, where N is assumed to be a Poisson variate with parameter 1: (a) Find the probability that N ∈ {0, 2, 4, . . .}. (b) Find the probability that N ∈ {0, 3, 6, . . .}. Think of other variations of the problem. 52 Exercises in Life Insurance Mathematics (RN “Problems in Life Insurance” 14.10.93, problem 9) Exercise 7.8. (Markov Chains in Life Insurance) Prove that the four definitions (2.1), (2.2), (2.3), and (2.5) of the Markov property are equivalent (assuming that the sample paths of the process are as stated in Paragraph 1A). (RN “Problems in Life Insurance” 14.10.93, problem 10) Exercise 7.9. (Reserves) At time 0 a person (x) aged x buys a standard pension insurance policy specifying that, conditional on survival, premiums are payable with level intensity c from time 0 to time m and pensions are payable continuously with level intensity b from time m to time n, m < n. There are two states, 0: “alive” and 1: “dead”. Let µx+t be the force of mortality at age x + t, and denote by Rt p = exp(− µ t x 0 x+s ds) the probability that (x) survives to age x + t. Assume that interest is earned at a constant rate δ so that v(t) = v t , with v = e−δ the annual discount rate. Throughout b is taken as fixed and c is to be determined by the equivalence principle. (a) Put up prospective and retrospective expressions for the reserves in both states at any time t ∈ [0, n) after issue of the policy. As an exercise, find the reserves also by solving the appropriate differential equations. Determine c. (b) Find the conditional variances of the individual reserves (the present values of future and past payments) at time t, given the state of the policy at time t. Henceforth the standard policy is referred to as policy ’S’. Consider a modified policy ’P’, by which the prospective reserve in state 0 is to be repaid upon death of (x) during the period [0, n). (c) Find the statewise reserves for ’P’. (Differential equations must now be used.) Determine c. (d) Find the conditional variances of the individual reserves for ’P’, corresponding to those in (b). Consider another modified policy ’R’, by which the retrospective reserve in state 0 is to be paid upon the death of (x) during [0, n). (e) Find the statewise reserves for ’R’ and determine c. (f) Find the conditional variances of the individual reserves by policy ’R’, corresponding to those in (b). (g) Compare the results obtained for ’S’, ’P’, and ’R’. Markov Chains in Life Insurance 53 Finally, consider a policy ’M’ with a mixed rule for repayment of the reserve, by which the retrospective reserve in state 0 is to be repaid upon the death of (x) during the premium period [0, m], whilst the prospective reserve in state 0 is to be repaid upon death during the pension period [m, n]. (h) Find the statewise reserves for the policy ’M’, and determine c. (RN “Problems in Life Insurance” 14.10.93, problem 15) Exercise 7.10. (Reserves) At time 0 a person buys a life insurance policy specifying that an amount b (the sum insured) is provided immediately upon death before time n and premiums are payable with level intensity c as long as the person is alive and active up to time n (premium waiver by disability). Assuming that recovery is impossible, the relevant Markov model can be sketced below. Assume that the discount function is v(t) = v t = e−δt . The premium c is to be determined by the equivalence principle. σ(x) 0. Active J JJJ JJ µ(x)J JJJ JJ % / 1. Disabled ttt ttt ν(x) tt t t yt t 2. Dead (a) Put up integral expressions for the transition probabilities. (b) Put up expressions for the prospective and retrosprective reserves in all states at any time t ∈ [0, n). Find the reserves also by solving the appropriate differential equations. Determine the premium intensity c. Suppose that instead of full premium waiver, the premium during disability is made dependent on the past savings on the contract. More specifically, assume that premiums during disability fall due with intensity c − c 0 V1− (t) at time t if the policy then is in state 1. (c) Which relation must c and c0 satisfy? (RN “Problems in Life Insurance” 14.10.93, problem 16) 54 Exercises in Life Insurance Mathematics Exercise 7.11 Consider the usual disability model without recovery: σ(x) 0. Active J JJJ JJ µ(x)J / 1. Disabled ttt ttt ν(x) ttt t t yt JJJ JJ% 2. Dead An insurance is issued to a person (x). Premium is payed continuously with level intensity π as long as (x) is alive, at the most for n years. As long as (x) is disabled, he receives a payment with level intensity c until his death. If (x) dies before the age of x + n the sum D is paid out. If (x) is active at the time m ∈ (0, n) the sum S is paid out. (a) Put up the statewise reserves. (b) Put up Thiele’s differential equation and determine the risk sums. (BS “Opgave til FM0” 1995) Exercise 7.12 A person considers a life insurance policy specifying that an amount S is paid immediately upon death before time of expiration n. If the insured is alive at time n he is also provided the sum insured S. Premiums are payable with level intensity p as long as the insured is active up to time n. That is, the insured has the right to exemption from payment of premium if he is disabled. Interest is earned at a constant rate δ. The Markov model used is illustrated below. σ(x) 0. Active J JJJ o JJ / ρ(x) µ(x)J JJJ JJ% 1. Disabled ttt ttt ν(x) ttt t t yt 2. Dead (a) Put up the following: 1. Thiele’s differential equation for the statewise reserves V 0 (t) and V1 (t). 2. The statewise reserve V0 (t) as an integral function of p00 (t, u) and V1 (t). 3. The statewise reserve V0 (t) as an integral function of p00 (t, u) and p01 (t, u). 4. The statewise reserve V1 (t) as an integral function of p00 (t, u) and V0 (t). 5. The statewise reserve V1 (t) as an integral function of p11 (t, u) and p10 (t, u). Markov Chains in Life Insurance 55 (b) Differentiate both expressions for V 0 (t) in order to verify Thiele’s differential equation. (c) How will you in practice find the equivalence premium for this kind of insurance? (d) Let µt = νt . Put up a differential equation for V 1 (t) − V0 (t). Solve it and use the result to find out how σt and ρt ought to be chosen in relation to σt0 and ρ0t so that the effect will be the desired increase of the premium. Give an interpretation of all results. (e) Find expressions for the safety loadings for the two states and consider the difference. Which sign does it have with a reasonable choice of parameters? (f) Expand the model with the state “surrender”. What would you consider a reasonable payment in connection with surrender from the state of active and disabled respectively. How should the first order intensity of surrender be put in relation to the intensity on the second order basis with your choice of surrender value? (MS “Opgave til FM1” 19.09.95) Exercise 7.13 Consider the usual disability model with four level transition intensities µ, ν, σ and ρ. Let α = µ + σ, κ = ν + ρ and assume that α 6= κ. (a) Show that the probability for an active person being active in t years, after having been disabled once and only once is σρ e−αt − e−κt te−αt + κ−α α−κ . (b) Find the probability for an active person being disabled for the second time in t years. (c) Explain how one relying, on information about death, disability and recovery in some population, can estimate the probabilities in (a) and (b) in two different ways. (HRH “Opgaver til FM1”) 56 Exercises in Life Insurance Mathematics 8. Bonus Schemes FM0 S95, 1 Exercise 8.1 An insurance company uses the following technical basis: Force of mortality µx = 0.0005+105.88+0.038x−10 (G82M), interest rate i = 5% p. a., acquisition costs 3% of the sum insured, loading for collecting fees 5% of the premium and administration expenses due continuously with intensity γV tg where Vtg is the gross premium reserve and 1, 05 γ = log . 1, 045 The technical basis of second order has interest rate ĩ = 5.5% p. a. and force of mortality µ̃x = µx − (δ̃ − δ), where δ and δ̃ are the forces of interest corresponding to i and ĩ respectively. Same expenses as above. A 30-year old person considers an endowment insurance, insurance period n years. The premium, calculated according to the equivalence principle, is due continuously with level intensity during the entire period of insurance. (a) Calculate the gross premium intensity and the gross premium reserve after 20 and 40 years, respectively. (b) Put up a differential equation for the safety loading S t Assume for the time being that the bonus fund is entirely paid out after 40 years. (c) Confirm that this gives the greatest possible bonus fund during the entire insurance period. (d) Calculate this bonus fund after 20 and 40 years. Now assume that the bonus scheme consists of two discrete payments: One after 20 years and one again after 40 years. (e) Calculate these two payments. Finally assume that the bonus scheme is as previously, but that the payment after 20 years is used as a deposit for a life conditioned capital insurance, duration n years, calculated on the technical basis of second order without expense contributions. (f) Show that the amount, payable at the 40th year, is the same as the payment according to the first bonus scheme and explain the difference between the two schemes. Bonus Schemes 57 (g) Critisize the considered bonus schemes especially related to the insured that passes away during the insurance period and try to suggest a more reasonable bonus scheme. (SP(74)) Exercise 8.2 Consider two lives (x) and (y) ages x and y respectively with remaining life times Tx and Ty . Assume Tx and Ty to be stochastically independent. For premium calculation the company has established a technical basis of first order; (x) force of interest δ, force of mortality µ x+t for (x) at age x + t and force of mortality (y) µy+t for (y) at age y + t. (x) and (y) consider a widow insurance (whole-life annuity). The possible whole-life annuity is due continuously with intensity 1 as long as (y) is alive and (x) is dead. (a) What is the present value of the possible whole-life annuity? (b) Derive an expression for the expected present value of the possible whole-life annuity in terms of one-life and both-life annuity expressions. Give a direct interpretation of this expression. For the possible annuity a level continuous premium is due as long as they both are alive. (c) What is the equivalence premium intensity? (d) Give an expression for the premium reserve by prospective reasoning. (e) Derive Thiele’s differential equation. Introducing the technical basis of second order, the interest rate is δ̃ and the forces (y) (x) of mortality are µ̃x+t and µ̃y+t . (f) Which principles in general should be the basis for choosing the technical basis of second order? How would you determine the elements of the technical basis of second order compared to the ones of the first order for the possible whole-life annuity? (g) Derive an expression for the safety loading. We now introduce the following bonus scheme: As long as (x) and (y) both are alive, no return of safety margin is due to payment. If (y) dies before (x) no return is due either. If (x) dies before (y), a continuous amount B is added to the annuity payment of one. The amount B is paid out as a continuous bonus. (h) Put up an expression for determination of B. 58 Exercises in Life Insurance Mathematics Instead of the bonus scheme above, we would like a return of an amount equal to the present value of the added amount B at the time of (x)’s death if (x) dies before (y). The payment of bonus consists of a continuous payment of an amount B as long as the whole-life annuity is due. (i) What is the difference between the two bonus schemes? Which Bonus scheme would be preferable from a safety margin view? (FM100, Oslo 1987 05.12.87 (ex. 1)) Exercise 8.3 By an endowment insurance an amount of S is paid either upon death before year n after the time of issue or at the latest n years after the time of issue. The premium is due with level intensity B during the insurance period, the force of interest is δ, the force of mortality µ x , expenses αS by issue and continuous expenses with intensity γS + βB per year during the insurance period. (a) Determine the premium B by applying the equivalence principle. Determine the net and gross premium reserves at any time during the insurance period. Find a connection between the two reserves and explain the reason for this difference. The expenses α, β, γ do not contain any safety loadings. On the contrary there are assumed to be safety loadings κ and λ included in the force of interest and in the force of mortality respectively. Hence, a realistic force of interest of second order is δ̃ = δ + κ and a realistic force of mortality is µ̃ x = µx − λ. (b) Put up the expression for the safety loading, the policy contributes at any time during the insurance period. The return of premium scheme is as follows: (i) An amount (κ − λ)V t is paid out during (t, t + dt) during the insurance period, where V t is the premium reserve for covering the outstanding claims and return of premiums (i. e. the insured earns an interest (κ − λ)dt of his part of the total fund); (ii) He receives an additional amount K to the sum insured upon death during the insurance period. (c) Determine K so the total safety loadings are being allotted to the insured. (FM1, winter 1985-86 (ex. 2)) Exercise 8.4 Consider an endowment insurance, sum insured S, duration n, age of entry x. The company adopts the following technical basis for calculation of premiums and premium reserves: Force of interest δ, force of mortality µ x and expenses βπ paid continuously where π is the level net premium, paid continuously during the entire period. In the technical basis of first order we assume the force of surrender to be zero. Bonus Schemes 59 (a) Put up Thiele’s differential equation and an expression for the equivalence premium intensity. The actual development with respect to the technical basis of second order for the company is as follows: The interest rate has dropped to δ 0 and the administration expenses have dropped to β 0 . Now assume that the force of surrender exists and is denoted by νx+t t years after the time of issue where ν x is independent of the time the person has been insured. (b) Discuss if this assumption of independence i reasonable. Until now the company has used the safety loadings to bring down the future premiums every year, or bonus has been paid out in connection with death or surrender, where the total reserve is allotted. A new bonus scheme is allotting bonus cash every whole year or by death or surrender. (c) Put up Thiele’s differential equation for the premium reserve on the technical basis of second order. What are the boundary conditions? Intuitively, why does ν x vanish? (d) Put up the differential equations for the bonus funds. What are the boundary conditions for the two bonus schemes? (e) Try to figure out the variance of the bonus funds of the 2 bonus schemes. (FW (rev.), 1994) 60 Exercises in Life Insurance Mathematics 9. Moments of Present Values Exercise 9.1 Consider the standard setup, where the development of the life insurance is described by a continuous Markov Chain on a state space J = {0, . . . , J}, and the contract specifies that a0g (t)dt is payable if the policy stays in state g in the time interval (t, t + dt) and a0gh (t) is payable upon transition from state g to state h at time t. (More general annuity payments can easily be dealt with, but the expression becomes more messy.) Assume for the time being that the discount function v is deterministic. To calculate the variance of V = Z ∞ v(τ )A(dτ ), 0 the present value at time 0 of future benefits less premiums, one needs to find EV 2 = E = E ∞ Z v(τ )A(dτ ) Z0 ∞ 2 2 2 v (τ )(A(dτ )) + 2 0 Z ∞ v(τ )A(dτ ) 0 Z v(ϑ)A(dϑ) . ϑ>τ (9.1) (a) Prove that (9.1) can be recast as EV 2 = Z ∞ v 2 (τ ) 0 + X p0g (0, τ ){2a0g Vg+ (τ ) g X µgh (τ )a0gh (τ )(a0gh (τ ) + 2Vh+ (τ ))}dτ, (9.2) h6=g where Vg+ denotes the prospective reserve in state g at time t. The variance is obtained by subtracting the square of the mean present value from the mean of the square. Formula (9.2) appears to offer an escape from the double integration that has to be performed in (9.1). It requires that the prospective reserve in different states be computed (they are essentially the inner integral, of course) and stored in memory beforehand. However, we have standard programs for that. (b) Use (9.2) to calculate the variance for a simple term insurance and for a simple life annuity, for which the results are well-known. Another simple formula for the variance, which shall not be proved here, is VarV = Z ∞ 0 v 2 (τ ) X g6=h p0g (0, τ )µgh (τ )(a0gh (τ ) + Vh+ (τ ) − Vg+ (τ ))2 dτ (9.3) (c) Prove that (9.2) essentially remains valid by stochastic discount function v(t) = exp(−∆(t)) if the process {∆(t)}t≥0 has independent increments. Moments of Present Values 61 (RN “Problems in Life Insurance Mathematics” 14.10.93 (Problem 1)) Exercise 9.2 Consider a temporary life insurance issued to a person at age x. The policy specifies that the sum insured S is payable immedieately upon (possible) death of the insured before time n and that premium is due with fixed amount c at times 0, 1, . . . , n − 1 as long as the insured is alive. Assume that administration costs incur continuously with constant intensity γS throughout the duration of the policy and that the force of interest δ is fixed. (a) Determine the premium c as a function of S and γ by the equivalence principle. Find an expression for the prospective reserve of the policy at time t ∈ [0, n). (b) Find the variances and the covariances of the present values at time 0 of the insurance payment, the premiums and the administration costs. Find the variance of the present value at time 0 of the total cash flow of payments generated by the policy. To be continued as exercise 11.3, page 71. (RN “Problems in Life Insurance Mathematics” 14.10.93 (Problem 2)) Exercise 9.3 Consider an n-year term insurance with equivalence premium payable continuously with level intensity π throughout the duration of the policy. (a) Put up the prospective reserve and the variance of the present value at issue at time 0 of benefits less premiums. (b) Suppose that in case of surrender the insured immedieately gets the current net value of the policy defined as the prospective reserve at the time of surrender. Assume that surrender takes place with intensity γ(t) at time t < n. (Thus we consider an extended model with three states “insured”, “withdrawn” and “dead”.) Find the prospective reserve in state “insured” and the variance of the present value at issue of benefits less premiums. Compare with the results in (a). (RN “Problems in Life Insurance Mathematics” 14.10.93 (Problem 4)) It is possible to find differential equations which determines any n-order moment (n) Vk (t) recursively in any state k for a generalized Markov-model. In the next exercise we consider the 2nd order moment for an endowment insurance. Exercise 9.4 Assume that a person at age x buys an n-year endownment insurance with level premium intensity π payable as long as the insured is alive. Upon death the amount S is paid out immediately. The force of mortality is given by µ x at age x. Let as usual Ut denote the present value at time t of benefits less premiums in the 62 Exercises in Life Insurance Mathematics future. (a) Find an expression for V (2) (t) = E(Ut2 | T > t). (b) Derive the expression from (a) with respect to t to find that you will get a differential equation which determines V (2) (t) recursively from V (t). (BM and JC, 1995) Inference in the Markov Model 10. 63 Inference in the Markov Model Exercise 10.1 The time continuous Markov Model shown below 1 kk5 kkk 0 E E µ (x) kk 1 kkk µ2 (x) EE EE EE EE / 2 .. EE . EE EE EE E" µh (x) E h is called the model for competing risks with h reasons for resignation. The function P µ = hi=1 µi is the total intensity of resignation. Define t px = p00 (x, x + t) and t qx(k) = p0k (x, x + t). (a) Prove that t px = p00 (x, x + t) = e− and (k) t qx = p0k (x, x + t) = Z Rt o µ(x+s)ds x+t p00 (x, s)µk (s)ds, for k = 1, 2, . . . , h. x Assume that L independent persons of the same age are being observed during one year. Let the state “0” in the model above represent the state “alive” and let the h reasons of resignation be reasons of deaths. It is possible to assume the forces of mortality, µ1 , . . . , µh , to be constant and the same for all L persons. The number of deaths from the h reasons are denoted D 1 , . . . , Dh and the sum of life time is T . (b) Determine the maximum likelihood estimators µ̂ 1 , . . . , µ̂h and the asymptotic distributions of the estimators. (HRH “Opgaver til FM3” (ex. 2)) Exercise 10.2 Consider the time continuous Markov model as follows 64 Exercises in Life Insurance Mathematics σ(x) / 0. Working KKK o KKK µ(x)K KKK KK % 1. Not working rrr rrr ρ(x) µ(x) rrr r r yr 2. Dead (a) Find explicit expressions for the transition probabilities. (b) Assume that all members in a portfolio of N independent lives are being observed during the period of time [0, 1] and assume that all transition intensities are constant during the time of observation. How would you estimate σ, ρ and µ? Determine the asymptotic distributions of the estimators. (HRH “Opgaver til FM3” (ex. 12)) Exercise 10.3 An insurance company is to perform a mortality study based on complete records for n life insurance policies with unlimited term period. Policy number i was issued zi years ago to a person who was then aged x i . The actuary sets out to maximize the likelihood n Y µ(xi + Ti , θ) Di exp i=1 Z xi +Ti ! µ(s, θ)ds , xi where the notation is obvious. One employee in the department objects that the method represents a neglect of information; it is known that the insured have survived, not only the period they were insured, but also the period from birth until entry into the scheme. Thus, he claims, the appropriate likelihood is rather n Y i=1 µ(xi + Ti , θ) Di exp Z xi +Ti ! µ(s, θ)ds . 0 Settle this apparent paradox. (A suitible framework for discussing the problem is an enriched model with three states, “uninsured”, “insured” and “dead”.) (RN “Problems in Life Insurance Mathematics” 14.10.93 (Problem 5)) Exercise 10.4 Reference is to Paragraph 2C in the paper RN “Inference in the Markov Model”, 09.02.93, the Gompertz-Makeham mortality study. Inference in the Markov Model 65 (a) Modify the formulas to the situation where person number i entered the study z i years ago at age xi . (b) Find explicit expressions for the entries of the asymptotic covariance matrix of the MLE. (RN “Problems in Life Insurance Mathematics” 14.10.93 (Problem 6)) Exercise 10.5. (Transformation of Data and Analytical Smoothening) Consider the mortality model where n independent persons are being observed during the interval of age [x, x + z), where x and z are integers. Exact times of death are observed and we disregard censoring during the period except by age z. The force of mortality µ(t) at the age of t is assumed to be piecewise constant over one-year intervals of age so that µ(t) = µk , for t ∈ [k, k + 1), k = x, . . . , x + z + 1. We introduce µ = (µx , . . . , µx+z−1 )t . Let further more Dk and Tk denote the number of deaths occured and the observed time of risk in [k, k + 1) respectively and let µ̂ = (µˆx , . . . , µ̂x+z−1 )t , where µ̂k = Dk , k = x, . . . , x + z − 1. Tk (a) Assume that µk can be cast as µk = gk (θ), k = x, . . . , x + z − 1, where gk (θ) = g(ξk ; θ) and ξk ∈ [k, k + 1). Here g(t; θ) denotes a function of the age t and of an unknown parameter θ = (θ1 , . . . , θp )t , p < z. Assume moreover that the exists and has full rank p for g(θ) = (g x (θ), . . . , gx+z−1 (θ))t . Let Jacobian Dg = Dg(θ) dθ Rz+ = {η = (ηx , . . . , ηx+z−1 ) ∈ Rz | ηk > 0, k = x, . . . , x + z − 1}, and let L : Rz+ → Rz be a differentiable mapping, so that the z × z Jacobian DL = DL(η) has full rank z. Define at last α̂ = (α̂ x , . . . , α̂x+z−1 )t and g̃(θ) by dη α̂ = L(µ̂) and g̃(θ) = L ◦ g(θ) = L(g(θ)). The parameter θ can be determined in two ways by analytical smoothening. You can either by modified χ2 -minimizing determine the value of θ that brings g(θ) “closest” to µ̂, or you can by modified χ2 -minimizing determine the value of θ that brings g̃(θ) “closest” to α̂. Denote these two estimators by θ̂ and θ̈ respectively and show that θ̂ and θ̈ has the same asymptotic variance. Is θ̂ = θ̈? 66 Exercises in Life Insurance Mathematics (b) Now assume that µk = βcξk , k = x, . . . , x + z − 1. Show how it is possible to use the theory in (a) to construct an estimator for (β, c) that is easy to calculate and give the asymptotic variance of the estimators. (FM3 exam, winter 1985-86) Exercise 10.6 Verify (1.13) – (1.16) in the paper RN “Inference in the Markov Model”, 09.02.93. (RN “Inference in the Markov Model” 09.02.93 (Problem 1)) Exercise 10.7 In the situation of paragraph 1E in the paper RN “Inference in the Markov Model”, 09.02.93, consider the problem of estimating µ from the D i alone, the interpretation being that it is only observed whether survival to z takes place or not. Show that the likelihood based on D i , i = 1, . . . , n, is q N (1 − q)n−N , with q = 1 − e−µz , the probability of death before z. (Trivial: It is the binomial situation.) Note that N is now sufficient, and that the class of distributions is a regular exponential class. The MLE of q is N q∗ = n with the first two moments Eq ∗ = q, Varq ∗ = q(1 − q) . n The MLE of µ = − log(1 − q)/z is µ∗ = − log(1 − q ∗ )/z. Apply (6.6) in the Appendix of the paper to show that µ∗ ∼as N µ, q . 2 nz (1 − q) The asymptotic efficiency of µ̂ relative to µ ∗ is asVarµ∗ = asVarµ̂ e µz 2 − e− µz µz 2 !2 = sinh(µz/2) µz/2 2 (sinh is the hyperbolic sine function defined by sinh(x) = (e x − e−x )/2). This function measures the loss of information suffered by observing only death/survival by age z Inference in the Markov Model 67 as compared to inference based on complete obervation throughout the time interval (0, z). It is ≥ 1 and increases from 1 to ∞ as µz increses from 0 to ∞. Thus, for small µz, the number of deaths is all that matters, whereas for large µz, the life lengths are all that matters. Reflect over these findings. (RN “Inference in the Markov Model” 09.02.93 (Problem 2)) Exercise 10.8 Use the general theory of Section 2 of the paper RN “Inference in the Markov Model”, 09.02.93, to prove the special results in Section 1. (RN “Inference in the Markov Model” 09.02.93 (Problem 3)) Exercise 10.9 Work out the details leading to (2.9) – (2.11) in the paper RN “Inference in the Markov Model”, 09.02.93. (RN “Inference in the Markov Model” 09.02.93 (Problem 5)) 68 Exercises in Life Insurance Mathematics 11. Numerical Methods In this chapter we will use some numerical methods on the theory of life inurance that we have already seen. One does not always have a program that can calculate the premiums, the development of the reserves etc. and it can be necessary to develop your own programs. Inspired by SW let us at first deal with numerical integration. We wish to calculate the definite integral I= Z b f (x)dx. (11.1) a A commonly used numerical method for evaluating (11.1) is the summed Simpson’s rule. In all it’s simplicity is states that I= Z b a N −1 X h f (x)dx ' f (a) + 4f (x1 ) + f (b) + 2 (f (x2k ) + 2f (x2k+1 )) 3 k=1 " # with h = (b − a)/2N, xj = a + jh, j = 1, 2, . . . , 2N − 1. The accuracy is good for small values of h and the implementation of this method is straightforward. Exercise 11.1 A man aged 25 years considers a pure endowment of 1.000.000, age of expiration 60. The technical basis of the company is G82M, i. e. the interest rate is i = 4.5% and the force of mortality is µx = 0.0005 + 105.88+0.038x−10 . The premium is a net continuous premium with level intensity π. Disregard expenses. (a) Put up Thiele’s differential equation and the expression for the equivalence premium π. (b) Apply the summed Simpson’s rule in order to calculate π numerically. (c) Solve Thiele’s differential equation and apply the same algorithm as in (b) to evaluate the premium reserve at times t = 10, 20, 30. Some expenses are, however, not neglectible and the insured has to pay some expenses during the insurance period in order to cover the administration expenses β, some fraction of the net premium intensity π. Administration costs are due with a continuous intensity γVt . Assume that γ is lesser than the force of interest δ. (d) How is Thiele’s differential equation and the equivalence premium (which is now the gross premium) modified? (e) Use your program in (c) to calculate the gross premium and the gross premium reserve at times t = 10, 20, 30. Numerical Methods 69 (BM og JC, 1995) The following is inspired by SW. Now consider the function f . When evaluating the net premium reserve in the above we solved the differential equation theoretically and then applied Simpson for evaluation. This is not always possible. Another way is to solve Thiele’s differential equation numerically; methods for this are plentyful and the d Runge-Kutta method of fourth degree is highly recommended. Let dx y = f (x, y(x)), xk = x0 + kh, k integer and define k1 = hf (xk , yk ), k2 = hf (xk + 0.5h, yk + 0.5k1 ), k3 = hf (xk + 0.5h, yk + 0.5k2 ), k4 = hf (xk + h, yk + k3 ) then (k1 + k2 + 2k3 + k4 ) . 6 The initial condition is y0 = x0 . It turns out that a surprisingly big h can be chosen when the slope is not too big. A sixth order Runge-Kutta can also be applied, but the difference from the fourth order R-K is really not that great. It is possible to use the difference between the fourth and the sixth order Runge-Kutta in order to find appropriate and varying h’s. The sixth order Runge-Kutta looks like this: y(xk+1 ) ' y(xk ) + k1 = hf (xk , yk ), k2 = hf (xk + 0.5h, yk + 0.5k1 ), k3 = hf (xk + 0.5h, yk + 0.5k2 ), k4 = hf (xk + h, yk + k3 ), 7 10 1 2 k5 = hf (xk + h, yk + k1 + k2 + k4 ), 3 27 27 27 1 28 1 546 54 378 k6 = hf (x + h, yk + k1 − k2 + k3 + k4 − k5 ), 5 625 5 625 625 625 and hence 5 27 125 1 k1 + k4 + k5 + k6 . 24 48 56 336 In both the fourth and the sixth order R-K, we have initial condition y 0 = x0 . y(xk+1 ) ' y(xk ) + It is easy to generalisize the R-K to a system of differential equations. For such a system of differential equations y(x) = y1 (x) y2 (x) .. . yn (x) f(x, y) = f1 (x, y1 (x), . . . , yn (x)) f2 (x, y1 (x), . . . , yn (x)) .. . fn (x, y1 (x), . . . , yn (x)) y = 0 (1) y0 (2) y0 .. . (n) y0 70 Exercises in Life Insurance Mathematics we have, similar to the fourth order R-K, k 1 = hf (xk , y k ), k 2 = hf (xk + 0.5h, y k + 0.5k1 ), k 3 = hf (xk + 0.5h, y k + 0.5k2 ), k 4 = hf (xk + h, y k + k3 ), y(xk+1 ) ' y(xk ) + (k 1 + k 2 + 2k 3 + k 4 ) , 6 d y(x) = f(x, y(x)) and y(x0 ) = y 0 . Remember this method when evaluating where dx a system of simultaneous differential equations. It is obvious that this method for solving differential equation systems simultaneously has a great applicability when studying the simultaneous development of the statewise reserves in a general Markov model for a policy. The functions are allowed to depend on each other just as the statewise reserves depend on each other, compare with the expression for the statewise reserves. Exercise 11.2 Consider an endowment insurance, duration 20 years, age at entry x = 30, interest rate i = 4.5%, force of mortality µx = 0.0005 + 105.88+0.038x−10 . The sum insured is 500.000. The only premium is a single net premium Π upon issue of the contract. The equivalence principle is adopted. (a) Use the fourth order Runge-Kutta to find this single net premium Π. Use the same program to study the development of the net premium reserve. Instead the man does not want to compose any initial capital, but a level continuous premium with intensity π during the insurance period. (b) Use Runge-Kutta to calculate this premium. (c) Our man is having a hard time deciding, but he chooses to compose an initial capital of 5.000 and then a level continuous premium with intensity π during the insurance period. What will π be, again applying Runge-Kutta? (BM and JC, 1995) Exercise 11.3. (Continued from exercise 9.2) Compute quantities in items (a) and (b) numerically in the case with the G82M mortality, δ = log(1.045), x = 30, n = 10 and b = 1. (A numerical integration must be performed to find the second order moments. Recall formulas (9.1)-(9.2) from exercise 9.1, page 61. Formula (9.1) requires Numerical Methods 71 integration in two dimensions. Formulas (9.2) and (9.3) require integration in one dimension when a table of reserves has been generated.) (RN “Problems in Life Insurance Mathematics” 14.10.93 (Problem 2)) Exercise 11.4. (Moments of Present Values) Consider the Markov model for disabilities, recoveries and deaths with intensities as in G82M technical basis; no recoveries, non-differential mortality intensity µ(x) = 0.0005 + 10 −4.12+0.038x at age x and disability intensity σ(x) = 0.0004 + 10 −5.46+0.06x at age x. As annual interest rate use 4.5%. (a) Compute the expected value and standard deviation of the present value at time 0 of benefits less premiums for a disability pension insurance issued to an active person at age x = 30, with insurance period n = 20 years and specifying that pensions are payable continuously with intensity 1 during disability and premiums determined by the equivalence principle. Perform the calculations also for x = 30, n = 40 and for x = 50, n = 20. (b) Now consider a portfolio of I insurance contracts and let V denote the present value √ of future benefits less premiums for the entire insurance portfolio. Suppose V + 2 V is to be provided as a reserve. Assume all I contracts are identical pension insurance policies as described above, with x = 30 and n = 20 and that the individual life histories √ are stochastically independent. Study e. g. the “fluctuation loading per policy”, 2 V /I, as function of I. (c) Perform calculations parallel to those in (a) for a modified contract where the benefit, instead of pensions during disability, consists of a lump sum payment of Z n v u−t e− t Ru t µ(x+s)ds du upon onset of disability at time t < n. (That is, the sum paid is the value of the pension described in (a), capitalised upon onset of disability.) Take x = 30 and n = 20. Compare with the corresponding result in (a) and comment. (RN “Problems in Life Insurance Mathematics” 14.10.93 (Problem 3)) Exercise 11.5 Consider a 30-year term insurance, issued on G82M, sum insured DKK 100.000, age at entry 40. The equivalence principle is adopted. Assume at first that we have a single net premium. (a) What is the single net premium? (b) Put up an expression for the premium reserve V t at time t and calculate it for t = 0, 5, 10, 15, 20, 25 and 30. 72 Exercises in Life Insurance Mathematics (c) Now assume that the premium is paid continuously during the entire insurance period with level intensity π. What is π? (d) Put up an expression for the premium reserve and evaluate it using the same values of t as in (b). (SP(57)) Exercise 11.6 A man aged 40 years has been issued a 20-year term insurance with sum insured 200.000 and level continuous premium intensity during the insurance period. Find the premium intensity and the net premium reserve 10 years after the time of issue of the contract using the technical basis G82, 4.5% net, assuming the equivalence princple is adopted. (SP(52)) Exercise 11.7 Consider the disability model outline below with recovery and excemption from payment of premium by disability. The insurance contract is a 40-year term insurance, sum insured S = 800.000, age of entry x = 25, level premium intensity π as long as the person is active at the most for 40 years. The interest rate is 4.5%. σ(x) 0. Active J JJJ o JJ / ρ(x) µ(x)J JJJ JJ% 1. Disabled ttt ttt ν(x) tt t t yt t 2. Dead Assume that the intensities are µx = 0.0005 + 105.88+0.038x−10 , σx = 0.0004 + 104.54+0.06x−10 , ρx = 0.15, νx = 10 · µx , where µx and σx correspond to the technical basis G82M (including GA82M). (a) Put up differential equations for the statwise reserves and for the transition probabilities. What are the boundary conditions? Numerical Methods 73 (b) Find the equivalence premium intensity π applying Runge-Kutta to finde the transition probabilities and some numerical integration method for evaluating the integrals. (c) Study the development of the reserves simultaneously, again applying RungeKutta. What are the statewise reserves at the times t = 10, 20, 30? (BM and JC, 1995) Exercise 11.8 (a) Construct graphs for the reserves for the following four life insurances 1. Pure endowment, sum 1 against single net premium. 2. Pure endowment, sum 1 against level continuous premium during the entire period. 3. Term insurance, sum 1 upon death before time n against level continuous premium. 4. Endowment insurance, sum 1 upon death or at time x + n, if the insured is still alive, against level continuous premiums. All insurances are issued on G82, i. e. µ x = 0.0005 + 10−4.12+0.038x and i = 0.045. (b) Find the premiums above. (c) What is the expected insurance period for product 3? (d) Consider product 1. Assume that 50% of the reserve is paid out upon death before the age of x + n. Construct graphs of the reserve and calculate the single net premium. (MSC “Tillæg til opgave E3” FM0 12.10.94) Exercise 11.9 We consider a force of mortality which is Gompertz-Makeham, i. e. µx = α + βcx . As usual the survival function is denoted by F . Consider an x-year old person with remaining life time T . The survival function for the person is F (t | x) = F (x + t) . F (x) The person can sign different kinds of insurance contracts: 74 Exercises in Life Insurance Mathematics • A pure endowment, where an amount of 1 is payable at time x + n if the person then is alive. Expected present value for this benefit is n Ex = v n F (n | x), where v = 1/(1 + i) is the one-year discount rate at interest rate i. • An n-year temporary annuity, payable continuously with force 1 per year until death, at the most for n years. Expected present value for this contract is ax:n| = Z n 0 t v F (t | x)dt = Z n t Ex dt. 0 • An n-year term insurance, where an amount of 1 is payable upon death before age x + n. The expected present value of this contract is 1 Ax:n| = Z n 0 v t F (t | x)µx+t dt = 1 − δax:n| − n Ex , where δ = log(1 + i) is the force of interest. • An endowment insurance which is the sum of a pure endowment and a term insurance. The expected present value of this benefit is Ax:n| = 1 − δax:n| . (a) Work out tables for µx , F (x) and the density f (x) = F (x)µx for x = 0, 1, . . . , 100. Use the values in the Danish technical basis G82, i. e. α = 0.0005, β = 10 −4.12 , c = 100.038 and interest rate i = 0.045. (b) Calculate the expected present values for the contracts above for x = 30, i = 0.045. (c) Redo the calculations in (b) using other values of x and n. In particular, let n vary for fixed x = 30 and let x vary for fixed n = 30. (d) Construct tables that show how the expected present values above depend on i, α, β and c. (RN FM1 89/90 opg. E3 05.10.89) Numerical Methods 75 76 Exercises in Life Insurance Mathematics Insurance Terms Aggravated circumstance Allot Amount allotted Annuity Bonus scheme Capital insurance Child’s insurance Collection costs Deduct Discount rate Down payment Duration Endowment insurance Entry Exemption from payment of premium Excess mortality Force of interest Force of mortality Hire-purchase agreement Installment Insurance period Interest rate Issue Life annuity Loading for collection costs Portfolio Premium free policy Principal Pure endowment Rate of course Return of premium Second order technical basis Single net premium Sum insured Surrender Surrender value Technical basis Term insurance Time of expiry Underwriting Skærpede vilkår Tilbageføre Tilbageføringsbeløb Annuitet, rente Bonusplan Kapitalforsikring Børneforsikring Inkassoomkostninger At fratrække (skat) Diskonteringsfaktor Kontant udbetaling (på lån) Varighed Livsforsikring med udbetaling evt. sammensat livsforsikring Indtrædelse Præmiefritagelse Overdødelighed Renteintensitet Dødelighedsintensitet Købskontrakt Afdrag Forsikringstid Rentefod Udstedelse Livrente Inkassotillæg Portefølje, bestand Fripolice Hovedstol Ren oplevelsesforsikring Kurs Ristorno Teknisk grundlag af anden orden Nettoengangspræmie Forsikringssum Tilbagekøb, genkøb Tilbagekøbsværdi, genkøbsværdi Teknisk rundlag, beregningsgrundlag Ophørende livsforsikring Udløbsdato Processen at tegne en forsikring