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Stat 475Life ContingenciesChapter 4: Life insurance

Review of (actuarial) interest theory — notationWe use i to denote an annual effective rate of interest.The one year present value (discount) factor is denoted byv 1/(1 i).i (m) is an annual nominal rate of interest, convertible m timesper year.The annual discount rate (a.k.a., interest rate in advance) isdenoted by d.d (m) is an annual nominal rate of discount, convertible mtimes per year.The force of interest is denoted by δ (or δt if it varies withtime).2

Review of (actuarial) interest theory — relationshipsTo accumulate for n periods, we can multiply by any of thequantities below; to discount for n periods, we would divide by anyof them.n Period Accumulation Factors!mni (m)n(1 i) 1 (1 d) n md (r )1 r! nr e δnIf the force of interest varies with time, we can discount from timen back to time 0 by multiplying bye Rn03δt dt

Valuation of life insurance benefitsThe timing of life insurance benefits generally depends on thesurvival status of the insured individual. Since the future lifetimeof the insured individual is a random variable, the present value oflife insurance benefits will also be a random variable.We’ll commonly denote the random variable representing the PVof a life insurance benefit by Z .Unless otherwise specified, assume a benefit amount of 1.We’re often interested in various properties (e.g., mean, variance)of Z .The mean of Z is referred to as the expected value of thepresent value, expected present value (EPV), actuarialpresent value (APV), or simply actuarial value.4

Whole life insurance — Benefits paid at moment of deathThe first type of life insurance we’ll consider is whole lifeinsurance.Consider the case where the benefit is paid at the moment ofdeath (this is sometimes referred to as the continuous case).For this case, the present value of the benefit is Z v Tx e δTx .The corresponding EPV is denoted by AxEPV for Whole Life Insurance — Continuous Casei Z hAx E [Z ] E e δTx e δt t px µx t dt05

Whole life insurance — Benefits paid at moment of deathWe can calculate the second moment for Z similarly: 2 Z 2 δTxE Z E e e (2δ)t t px µx t dt0We can find this second moment by computing the expectation attwice the force of interest, 2δ. When we calculate the expectationat twice the force of interest, we denote it with the symbol 2 Ax . Then for this case, we have E Z 2 2 Ax .Then we can calculate the variance of Z : 2 V [Z ] E Z 2 E [Z ]2 2 Ax AxWe may also be interested in various percentiles or functions of Z— all the usual rules of random variables apply.6

ExampleAssume that a particular individual, currently age x, has a futurelifetime described by a random variable with densityfx (t) 160for0 t 60This person wants to purchase insurance that will provide a benefitof 1 at the moment of death. Assuming a force of interest ofδ 0.06:1Find the EPV and variance for this death benefit.2Find the minimum value H such that P(Z H) 0.9.3Find the minimum single premium H that an insurancecompany must charge in order to be at least 90% certain thatthis premium will be adequate to fund the death claim, shouldtheir assets accumulate at δ 0.06.7

Whole life insurance —Benefits paid at end of the year of deathNext we consider a whole life insurance in which the death benefitis paid at the end of the year in which the insured dies (this issometimes called the annual case).Kx is the time corresponding to the beginning of the year of death;Kx 1 is the end of the year of death.Since the benefit is paid at the end of the year of death, thepresent value of the benefit is Z v Kx 1 .Then this is a discrete random variable.What does its pmf look like?8

Whole life insurance —Benefits paid at end of the year of deathWe can find the mean and variance of this random variable:EPV and Variance for Whole Life Insurance — Annual Case hi XE [Z ] E v Kx 1 v k 1 k qx Axk 02E [Z ] E vKx 1 2 Xv2 (k 1)k 0 V [Z ] E Z 2 E [Z ]2 2 Ax (Ax )29k qx 2 Ax

Whole life insurance exampleConsider a 50, 000 whole life insurance policy issued to (x), withdeath benefit paid at the end of the year of death. Let Z be thepresent value of the death benefit RV.We’re given:qx 0.01 qx 1 0.02 qx 2 0.03 qx 3 0.04 i 10%Find Pr [36, 000 Z 42, 000]10

Whole life insurance — Benefits paid mth lyNow we consider the case where the whole life death benefit is paidat the end of the 1/m period of a year in which the insured dies.(m)For this case, the present value of the benefit is Z v Kx1 mThen for this discrete random variable we have:EPV and Variance for Whole Life Insurance — mth ly Case h (m) 1 i Xk 1(m)v m k 1 qx AxE [Z ] E v Kx m mk 02E [Z ] E (m)vKx1 m 2 Xmv2k 0 ( k 1m )(m)km (m)(m) 2V [Z ] E Z 2 E [Z ]2 2 Ax Ax11 1 qx 2 Axm.

General strategy for calculating EPVWe can always find the EPV of any life-contingent payment (notjust life insurance benefits) by summing over all possible paymenttimes the product of:1The amount of the payment2An appropriate present value (discount) factor3The probability that the payment will be madeAll of the EPV formulas for life insurance benefits we’ve seen arespecific cases of this principle.Note that this formula only works for calculating EPVs.12

Recursion formulasUsing the summations given above, we can derive variousrecursion formulas for the expected present value of life insurancebenefits, i.e., formulas relating successive values of the EPV.We’ll encounter these types of formulas in many contexts.These formulas are useful for a number of reasons.For the annual case, we haveEPV of Whole Life Insurance — Annual CaseAx v qx v px Ax 1For the mth ly case, we haveEPV of Whole Life Insurance — mth ly Case(m)Ax(m) v 1/m 1 qx v 1/m 1 px Ax 1mmm13

Relating the whole life EPV valuesNote that in order to calculate Ax , we only need the information ina life table.However, in order to calculate Ax , we need the full survival model.If we’re not given this information (i.e., if we only have a lifetable), then we’ll have to make some sort of fractional ageassumption as before.Making the UDD assumption gives the relationships:Relationships Between Whole Life EPV Values under UDDUDD i(m) UDD iAx AxAx (m) AxδiThese relationships are often used as approximations, but are onlyexact under UDD.14

Numerical EPV values for whole life insurance(12)xAxAx204, 9225, 0335, 0434012, 10612, 37912, 4046029, 02829, 68329, 7438059, 29360, 64160, 76410087, 06889, 15889, 341Note the patternbetween the values forthe three cases, at eachgiven age.AxUsing the UDDapproximations, we cancalculate approximate(12)values for Ax and Ax .Then we can comparethem to the actual valuesshown in the table.Table 4.3 from Dickson et al.:EPV values for a whole life insurance with a death benefitof 100, 000, using Makeham’smortality model and i 5%.15

SULT Whole Life ExampleLet Z be the PV of a 100,000 whole life insurance (annual case)issued to (45). Let i 5% and mortality be given by the StandardUltimate Life Table (SULT).(a) Calculate E [Z ](b) Calculate the standard deviation of Z .(c) Recalculate E [Z ] for the monthly case, i.e., if the deathbenefit was payable at the end of the month of death. Use theUDD fractional age assumption.16

Term life insurance — Benefits paid at moment of deathThe next type of life insurance we’ll consider is (n-year) term lifeinsurance.First consider the continuous case, where the benefit is paidat the moment of death.For this case, the present value of the benefit is Tv x if Tx nZ 0if Tx n1The corresponding EPV is denoted by Āx:nEPV for n-year Term Life Insurance — Continuous CaseZ n1Āx:n E [Z ] e δt t px µx t dt017

Term life insurance — Benefits paid at moment of deathWe can also calculate the second moment for Z :Z n 2 1E Z e 2δt t px µx t dt 2 Āx:n0Some term life insurance example problems:1How would you expect the EPV for a term insurance to varyas n increases?2Redo the whole life example, using the same survival model asbefore, but this time assuming that the person wishes topurchase a 20-year term insurance (with benefit payable atthe moment of death) rather than a whole life insurancepolicy. Compare the answers for the two insurances.18

Term life insurance —Benefits paid at end of the year of deathNext we consider the annual case for an n-year term life insurance.For this case, the present value of the benefit is K 1v xif Kx n 1Z 0if Kx n1Then the EPV is denoted by Ax:nEPV for Term Life Insurance — Annual Case1Ax:n E [Z ] n 1Xv k 1 k qxk 019

Term life insurance — Benefits paid mth lyNow we consider the case where the term life death benefit is paidat the end of the 1/m period of a year in which the insured dies.For this case, the present value of the benefit is((m)1(m)v Kx m if Kx n Z (m)0if Kx nThen the EPV is denoted by A(m)1x:nEPV for Term Life Insurance — mth ly CaseA(m)1x:n E [Z ] nm 1Xk 0v (k 1)/m k 1 qxm20m1m

Term life insurance example and relationshipsA person age x wishes to purchase a 3-year term life insurancepolicy with benefit amount 400, 000 payable at the end of theyear of death.Find the EPV of this benefit, assuming thatpx 0.97px 1 0.96px 2 0.94i 0.10It turns out that the relationships between EPV values derived forwhole life insurance also work for term life insurance.They’re exact under UDD and approximations otherwise.Relationships Between Term Life EPV Values under UDDUDD iUDD i111Āx:n Ax:nA(m)1x:n (m) Ax:nδi21

Pure endowmentA pure endowment is a type of contract that pays a benefit atthe end of a fixed time period (e.g., n years) if the policyholder isstill alive at that time.This type of policy is not typically sold by itself, but is nonethelessimportant:It can be combined with other types of insurance.It can be used to find the EPV of life contingent payments.For this case, the present value of the benefit is 0if Tx nZ nif Tx nv22

Pure endowmentThe EPV of an n-year pure endowment is denoted by Ax:n1Note that there are not separate continuous and mth ly casesfor a pure endowment.The alternate (more convenient) notation n Ex is also used todenote the EPV of an n-year pure endowment.We will commonly use n Ex as a sort of general “life-contingentdiscount factor”.EPV for Pure EndowmentAx:n1 n Ex E [Z ] v n n px23

SULT Pure Endowment ExampleLet Z by the PV of a 100,000 10-year pure endowment issued to(45). Let i 5% and mortality be given by the Standard UltimateLife Table (SULT).(a) Calculate E [Z ](b) Redo part (a) using i 9%24

Endowment insuranceEndowment insurance combines term insurance with a pureendowment.An n-year endowment insurance pays a benefit if the insureddies within n years.It also pays a benefit (of the same amount) at the end of nyears if the person is alive at that point.In the case where the death benefit portion is paid at the momentof death, the present value of the entire benefit (assuming as usuala benefit amount of 1) is Tv x if Tx nZ vnif Tx n v min(Tx ,n)25

Endowment insuranceFor the continuous case, the EPV is denoted Āx:n , which we canwrite in terms of other EPV symbols:EPV of Endowment Insurance — Continuous Case1Āx:n E [Z ] Āx:n Ax:n1For the annual case, the PV of the benefit is K 1v xif Kx n 1Z nvif Kx n v min(Kx 1,n)Then the EPV for this case is denoted by Ax:nEPV of Endowment Insurance — Annual Case1Ax:n E [Z ] Ax:n Ax:n126

Endowment insuranceFor the mth ly case, the PV of the benefit is((m)1(m)v Kx m if Kx n Z (m)vnif Kx n v1m (m)1,nmin Kx m(m)Then the EPV for the mth ly case is denoted by Ax:nEPV of Endowment Insurance — mth ly Case(m)Ax:n E [Z ] A(m)1x:n Ax:n127

Relationships for endowment insuranceIt’s important to note that the UDD relationships we developed forthe whole life and term insurance work only for death benefits, notfor endowment benefits.Therefore, in order to apply the UDD approximation, we must firstsplit the term insurance benefit from the endowment portion:Relationships Between Term Life EPV Values under UDDUDD i(m) UDD i1Āx:n A1 Ax:n1Ax:n (m) Ax:n Ax:n1δ x:niAs before, these relationships are exact under the UDDassumption, and approximate otherwise.28

SULT Term ExampleLet Z by the PV of a 100,000 20-year (annual case) terminsurance issued to (45). Let i 5% and mortality be given by theStandard Ultimate Life Table (SULT).(a) Calculate E [Z ](b) Calculate P(Z 0)(c) Recalculate E [Z ] for the continuous case, i.e., if the deathbenefit was payable at the moment of death. Use the UDDfractional age assumption.29

Deferred insurance benefitsIn all of the types of insurances we’ve discussed thus far, the deathbenefit period starts immediately at the time of purchase.It’s also possible to defer the coverage until some future time.For example, consider the continuous case of a whole lifeinsurance. Suppose we wanted to defer this insurance for u years.The PV of the benefit would be 0if Tx uZ v Tx if Tx uWe denote this deferment of benefits in much the same way as wedid for a deferred mortality probability.EPV for Deferred Whole Life Insurancesu Āx u Ex Āx uu Ax u Ex Ax u30(m)u Ax(m) u Ex Ax u

Deferred insurance benefits — term insuranceSimilarly, we can also consider deferred term life insurances. If wewere to defer an n year continuous term insurance by u years, thePV of the benefit would be 0if Tx u or Tx u nZ v Tx if u Tx u n1The corresponding EPV is denoted by u Āx:nEPV for Deferred Term Life Insurance — Continuous Case1u Āx:n1 u Ex Āx u:nThe annual and mth ly cases work similarly.31

Deferred insurance benefits — relationshipsUsing the principle of benefit deferment allows us to develop someuseful relationships among the various insurance EPV values:1u Ax:n 1Ax:u n 1Ax:u1Ax:n n 1X1r Ax:1r 0Ax X1r Ax:11Ax Ax:n n Axr 0There are analogous continuous and mth ly versions of theserelationships as well.32

Another SULT Term ExampleLet Z by the PV of a 100,000 17-year (annual case) terminsurance issued to (45). Let i 5% and mortality be given by theStandard Ultimate Life Table (SULT).(a) Calculate E [Z ](b) Calculate the standard deviation of Z(c) Redo part (a) for a 30-year term product, leaving everythingelse the same.33

Variable benefits — Arithmetically IncreasingWe can find the EPV for benefits with various patterns.One pattern that deserves special mention is the case of anarithmetically increasing benefit, one in which the benefitincreases by a constant amount each year:Z n 1 I Ā x:n t e δt t px µx t dt I Ā x 1(IA) x:n0 Zt e δt t px µx t dt0 n 1X(k 1) v k 1 k qxk 0(IA)x X(k 1) v k 1 k qxk 034

Variable benefits — Geometrically IncreasingAnother pattern that could be useful is that of a geometricallyincreasing benefit, one in which the benefit increases by a constantpercentage each year.Example: Consider a 20-year term insurance issued to (x) inwhich the death benefit is paid at the end of the year of death.The amount of the death benefit is 100, 000 if the insured dies inthe first year, and rises by 3% each subsequent year. Find anexpression for the expected present value of this benefit.We could use similar logic to find the EPV of a benefit that wasarithmetically / geometrically decreasing.35

Valuing Insurance Benefits under a Select-and-UltimateMortality ModelWe can calculate present values and EPVs for all of the insuranceswe’ve seen using a select-and-ultimate mortality model; we simplyhave to use the correct mortality values.For example, under a 2-year select-and-ultimate mortality model,the EPV of a whole life insurance issued to [x] would beA[x] v q[x] v 2 p[x] q[x] 1 v 3 p[x] p[x] 1 qx 2 v 4 p[x] p[x] 1 px 2 qx 3 · · ·and we can also write1 n E[x] Ax nA[x] A[x]:n(assuming n 2 here)36

Example (AMLCR Exercise 4.1)x xAx35100, 000.000.1513753699, 737.150.1582453799, 455.910.1653863899, 154.720.1728043998, 831.910.1805054098, 485.680.188492Assuming i 0.06, calculate3715 E3521A35:535 A354Ā35:5 assuming UDD

SOA Example Multiple Choice #3For a special whole life insurance on (x), payable at the moment ofdeath:µx t 0.05, t 0δ 0.08The death benefit at time t is bt e 0.06t , t 0.Z is the present value random variable for this insurance atissue.Calculate Var [Z ]. [0.04535]38

SOA Example Multiple Choice #4For a group of individuals all age x, 25% are smokers (s), 75% arenonsmokers (ns), i 0.02k012sqx k0.100.200.30nsqx k0.050.100.151Calculate 10, 000Ax:2for an individual chosen at random from thisgroup. [0.1730]39

SOA Example Multiple Choice #17For a whole life insurance of 1 on (41) with death benefit payableat the end of year of death, you are given:i 0.05p40 0.9972A41 A40 0.008222A41 2 A40 0.00433Z is the present value random variable for this insurance.Calculate Var [Z ]. [0.02544]40

SOA Example Multiple Choice #34Assuming i 0.03, you are given the following select and ultimatemortality table with a select period of three years.x6061626364q[x]0.090.100.110.120.13q[x] 10.110.120.130.140.15q[x] 20.130.140.150.160.171Calculate 2 A[60]:2. [0.19]41qx 30.150.160.170.180.19x 36364656667

SOA Example Written Answer #7For a special deferred term insurance on (40) with death benefitspayable at the end of the year of death, you are given:The death benefit is 0 in years 1-10; 1000 in years 11-20; 2000in years 21-30; 0 thereafter.Mortality follows the Illustrative Life Table.i 0.06.The random variable Z is the present value, at age 40, of thedeath benefits.E [Z ] 107.1Write an expression for Z in terms of K40 .2Calculate Pr(Z 0). [0.75]3Calculate Pr(Z 400). [0.1392]4Calculate Var [Z ]. [36,046]42

The timing of life insurance bene ts generally depends on the survival status of the insured individual. Since the future lifetime of the insured individual is a random variable, the present value of life insurance bene ts will also be a random variable. We'll commonly denote the random variable representing the PV of a life insurance bene t .