8.5 Numerical Evaluation of Probabilities

Transcription

1 8.5 Numerical Evaluation of Probabilities 1 Density of event individual became disabled at time t is so probability is tp 7µ 1 7+t 16 tp 11 7+t 16.3e.4t e.16 t dt.3e.3 16 Density of event individual became disabled at time t is e.t dt.3. e.3 1 e We have tp 3µ 1 3+t 1 tp 11 3+t tp x tp 11 x e t.4+.3x+sds e.4+.3xt+.15t e t.+.x+sds e.+.xt+.1t so probability is te.496t+.15t e.641 t+.11 t dt 1 1 e e te.496t+.15t e.64+.t1 t+.11 t dt te t+.5t dt te t e e te t+8 dt 14 + t + 8e t+8 1 π Φ dt dt 11.6 e t+8 e dt Φ [ ] 1 1e t+8 + e 11.6 e 8 e 4 1

2 We have tp x tp 11 x e.4t e.3t Probability this happens from 1 transition: tp 7µ 1 7+t 16 tp 11 7+t dt.3e.4t e.316 t dt 16.3e.368 e.17t dt.3.17 e e.7 Probability this happens from 3 transitions: s t s t sp 7µ 1 7+s t sp 11 7+sµ 1 7+t u tp 7+tµ 1 7+u 16 up 11 7+u du dt ds.3.3e.4s+u t e.316 u+t s du dt ds.3.3e s t e.17s+u t du dt ds Making the substitution w s + u t, this becomes noting that J 1:.3.3e w 16+s w s 16 w.3.3e.368 e.17w s e.17w dt ds dw 16+s w e.368 w16 we.17w dw 1 dt ds dw In general, the probability that this happens from n + 1 transitions is n e.368 w n 16 w n n! e.17w dw

3 The total probability of this is therefore n e.368 w n 16 w n n! e.17w dw n Premiums 4 The rate of exit of state is.4+.3t t, so t p 4 e t.56+.3t dt e.56.15t. Premiums are payable while healthy. If the rate of premium is P, then the expected present value of the premium paid is P 5 P P e.3t e.56t.15t dt 5 5 e.356t.15t dt e.15t dt P e π P Φ Φ On the other hand, the expected death benefits for lives that are not critically ill first are: e.3t e.56t.15t.1 +.1t + 4 dt 5 5.1e e e.356t.15t t dt e.356t.15t.1t dt.3 5 t t+ e.3 dt e.15t+.3 [.3 ] e.356t.15t dt 3

4 If a life becomes critically ill at age x, the probability that it survives for t years is e t. dt e.t. The expected value of the benefits to such a life is therefore given by x 47 x e.3x+t 4 e.t dt 9e.3x 4 e.5t dt 18e.3x 4 1 e.547 x The expected value of death benefits to such an individual is 47 x 47 x.e.3x+t 4 e.t dt e.3x 4 e.5t dt 4e.3x 4 1 e.547 x So the total expected benefits paid to individuals who become disabled are e.56s.15s.3 +.s + 4e.35 s 1 e.55 s ds se s.15s 1 e.55 s ds se s.15s e s.15s ds s +.15s.15s s s s +.15s.15s s s So the expected benefits are se e.3s e e.3s ds e s e.3s ds 1.84e s e.3s 1.84e e π.3 π.3 Φ Φ [e.3s Φ Φ [e.3s ]5 ]5 4

5 The total expected benefit is therefore The premium is therefore x + 38x + x $ x + x 1469x + 37x + x3 3 67x +.5x 341x + 1.5x 74x + x We calculate the probability that the life is in each state at the end of each year: t tp 37 tp 1 37 tp EPV of death benefits is EPV of disability benefits is EPV of all benefits is EPV of unit premiums is So annual premium is $

6 8.7 Policy Values and Thiele s Differential Equation 6 Thiele s differential equation gives d dt t v 1 δ t v 1 + P 1 B 1 µ 1j x+ts 1j + t v j t v 1 j 1.3 t v 1 9 µ 1 x+t1 t v 1.3 t v t v 1.5 t v 1 9 tv e.5t 5 d dt t v δ t v + P B µ j x+ts j + t v j t v j.3 t v µ 1 x+t t v 1 t v µ x+t1 t v.3 t v t t v 1 t v t1 t v.3 t v t1841 e.5t 5 t v t1 t v t1 e.5t 5.1t t v +.3t t v e.5t t te.5t t v +.3t t v e.5t Multiple Decrement Tables 7 We have t p x e.3t.5x+t x, so we have a 1 e.4t e.3t.536+t 36 dt 36:1 1 e 466 e.466t.5t dt 1 1 πe 466 e t+466 dt Φ Φ

7 1 36:1.1 +.te.4t e.3t.536+t 36 dt A 1 e e 466.e t + 51e t+466 dt 1 So the annual rate of premium is t + 466e t+466 dt 415 [ 1e t+466 ] π $ e t+466 dt Φ Φ 1 1 b For policies which do not lapse, we only need to consider the mortality rate. This gives 1 t e.4s e.3s.536+t+s 36+t ds 36+t:1 t a 1 t e 41 e.41+.1ts.5s ds 1 t 1 πe t+41 e s+t+41 ds 41.1 t + 41 Φ Φ and So the policy value is A 36+t:1 t 1.4a 36+t:1 t tv 31.4a 36+t:1 t 1 t p 36+t e.41 t P a 36+t:1 t P a 3 36+t:1 t 1 tp 36+t e.41 t P 1 πe t Φ Φ 1 at + btp t t p 36+t e.41 t 7

8 for some functions at and bt. From a, the expected death benefit of the policy is $ The expected surrender benefit for the policy is 1 1 e.4t tp 36 µ 1 36+ttV dt e.4t e.3t+.5t +7t t t V dt e.5t +853t.1164 t t V dt e.5t +853t.1164 tat + btp dt e.5t +853t.1164 tat dt P Using numerical integration. We note that this requires the policy value to always be positive. We therefore have 1 e.5t +853t.1164 tbtp 8.743P P P P $ We have In particular, tp x We therefore calculate e t.3+.1x+s ds e.3+.1xt+.5t a On the other hand we have tp 9 e.59t+.5t 1 e.5t e.59t+.5t dt 9:1 1 e 559 e.559t+.5t dt 1 e t+559 dt e π Φ Φ

9 A 1 1 9:1.3e.5t e.59t+.5t dt.3 1.3e 559 e.559t+.5t dt 1 e t+559 dt.3e π Φ Φ 1 1 and.3367 A 1 9:1.9 + te.5t e.59t+.5t dt te.559t+.5t dt.e 559.e t e t The EPV of benefits is therefore dt e 559 e π Φ Φ 1 1 and the premium is 9 1 We calculate $ A :5 1 A :5 1 ä 4:

10 So the net premium is $ a Using UDD in the multiple decrement table, the probability that the policy is still in force at age 46 years 3 months is The probability that the life then dies in an accident before age 47 is The probability that the life dies in an accident between ages 46 and 3 months and 47 is therefore The probability that the life survives to age 47 is The probability that a life aged 47 dies in an accident before age 47 and 5 months is The probability that the life dies in an accident between ages 47 and 47 and 5 months is therefore The total probability is therefore b Using constant transition intensities, the rate of decrement for a life aged 4 43 is log The probability that a policy starting at age 4 is still in force at age 4 and 4 months is e The intensity of decrement aged is log This gives that l e The probability that the policy is still in force at age 46 and 3 months is therefore The intensity of accidental deaths between ages is e x dx and the probability of accidental death between ages 46 and 3 months and 47 is therefore e x dx.75 The probability of surviving to age 47 is The intensity of decrements between ages is log The intensity of accidental deaths between ages is e x dx.5 1 e e e.64x dx and the probability of accidental death between ages 47 and 47 and 5 months is therefore e x dx 5 1 e.64x dx.4 1 e e The total probability of dying in an accident between ages 46 years 3 months and 47 years 5 months is therefore

11 8.1 Constructing a Multiple Decrement Table 1 a Under UDD, suppose there are x policies at the start of the year, and during the year, y surrender and z z die. Assuming UDD, the number of policies still in force at time t is x y +zt, so the rate of dying is x y+zt. Assume that the deaths in the new table are uniformly distributed over the year. This is inconsistent with our uniform distribution assumption for policies with surrender, but what the hell. If the updated mortality v table has w deaths from v lives, then the rate of death at time t is w vt. With this update, let x t be the number of policies still in force at time t. We have dx t dt y x y + zt + v x t w vt This gives y x y+zt + x 1 xe 1 x y + z x y v w vt dt e [y logx y+zt+v logw vt]1 y+z w v w e y y+z x y+z log x +log w v w In the new table, we get x y + z x y y+z x y + z x y y + z y y + z y y+z w v w x y+z x log log x y+z x y y+z log x y+z x y y+z log x y+z x y y y +z y+z w v w + log w v w Using this, we can calculate the probabilities on the following slide. b Suppose there are x policies at the start of the year, and during the year, y surrender and z die. Under constant transition probabilities, we have the number of policies still in force at time t is xe x x y z t. x If the constant rate of surrender is µ 1, then y 1 µ 1 xe x y z log x t dt µ 1 1 x y z x x logx logx y z y + z µ1 logx logx y z x y z log x t This gives µ 1 ylogx logx y z y+z, so without deaths, the probability of surrender is e logx logx y z y x y z x y y+z. The table is therefore the same as in part a. 11 y+z

12 c If each independent decrement satisfies UDD, then suppose the probabilities for surrender and death as p q only decrements are p and q respectively, then the rate of surrender is 1 pt and the rate of death is 1 qt. p Now in a model with two decrements, the total rate of decrement is 1 pt + q 1 qt, so the total probability of no decrement in the year is e 1 p 1 pt + q 1 qt dt e [log1 pt+log1 qt]1 1 p1 q Similarly, the probability of no decrement before time t is 1 pt1 qt. The probability of surrender is 1 p 1 pt1 qt 1 pt dt 1 p1 qt dt p 1 q Similarly, the probability of death is q 1 p. If we are given the probabilities of surrender and death in the multiple decrement model are a and b respectively, then we have to solve p 1 q a q 1 p b p q a b p p a p a b p + b a p + a p a + b ± a + b 8a a + b ± a + b + 4 ab 4a 4b q b + a ± a + b + 4 ab 4a 4b x l x d x Joint Life and Last Survivor Benefits 14 Advantages Annuity value does not depend on time of death Disadvantages Value of benefit varies with time of death 1

13 9.4 Independent Future Lifetimes 15 ä 63,6: Ä 63,6: So the premium is $ If the Husband is dead and the wife is alive at the end of the policy with probability , then the wife receives a reversionary annuity with value The expected present value of payments after the end of the policy is therefore Payments during the term of the policy are received if and only if the husband is dead and the wife is alive. The expected present value of payments received during the term of the policy is therefore So the total expected benefit is For the premiums, we have ä 53,64: So the net annual premium is $

14 17 The probabilities of each being dead after n years are: Years PHusband Dead PWife Dead PSurvivor EPV benefit total The EPV of benefits after years is , so total EPV is Net premium is $77, Year PHusband Dies PWife Dies PBoth die POne dies Annual EPV from one dying: Annual EPV from both dying.8636 x 1 + i

15 y x x x x x x x x x x x y xy x + x + + x x x y 144x 1 1 1i 1 i y x x x1 3x 1 1 x11 1 x 1 3i + x11 1 x 1 A i ,76:7 i 1 A 1 45,76: d This gives ä 1 45,76: So the monthly premiums are $ A Model with Dependent Future Lifetimes 19 a Year PHusband Dies PWife Survives PWife Survives to 49 PHusband Dies and Wife Survives to b ap1 +1 ap aq 1 +1 aq ar 1 +1 ar a 3 p 1 q 1 r 1 + a 1 ap 1 q 1 r a 3 p q r 15

16 We have Year PHusband Dies PWife Survives PWife Survives to 49 PHusband Dies and Wife Survives to By summing over times the husband dies, we calculate the following lifetable for the wife. Suppose the mortality for the wife while the husband is alive is q 1, the mortality while the husband is dead is q, and the mortality of the husband is q 3. Conditional on the husband dying at time t in the year, the probability that the wife dies during the year is tq tq q tq tq 1 1 tq 1 tq 1 tq The total probability that the wife dies during the year if the husband is alive at the start of the year is therefore 16

17 1 q 3 q q 3 q 1 + q 1q 3 + q q 3 1 q 3 q 1 + q 1q 3 + q q 3 1 q 3 q 1 + q 1q 3 + q q 3 1 q 3 q 1 + q 1q 3 + q q 3 q 3 tq tq 1 1 tq 1 tq q 3 q 1 + q q 1 + q 1 dt 1 t 1 tq 1 dt 1 tq q 1 t + q 1 t 1 tq dt 1 + q1 + q1 q q q 1 q 1 q 3 q 1 + q q 1 + q 1 q 1 + q 3 q1 + q q q 1+q 1+ q 1 q q 1 + q 1 + q1 q q 1 t + 1 dt q q 1 tq q [ 1 + q1 q log1 tq q q q + q q 1 + q1 q q q 3 q q q 1 q q 1 + q 3 q + q 1 q + q 1 q q q log1 q log1 log1 q log1 q ] 1 If the probability that the husband and wife are both alive at the start of the year is p 1 and the probability that the wife is alive, but the husband is dead is p, then the probability that both are alive at the end of the year is p 1 1 q 1 1 q 3. The probability that the wife is alive and the husband is dead at the end of the year is therefore: q1 + q p 1 q + p 1 1 q 1 + q 3 q q1 + q p 1 q + p 1 q 3 1 q 1 q 3 q p 1 q p 1 q 3 q 1 + q 1 + q + q 3 q + q 1 q + q 1 q q log1 q 1 q 1 1 q 3 log1 q q log1 q + q 3 q + q 1 q + q 1 q q + q 1 q + q 1 q q 17

18 This gives Year Both Alive Husband Dead, Wife Alive Wife Alive a 39: A 39: So the annual premium is $ The Common Shock Model 1 We have tp 5:56 e t s+.5+s+.15+s +.56+s ds e t s+.3s ds e So the probability is given by 1 1 e.34316t+.8715t +.1t t te t 3 3 dt e t.49985t.1t t +.t dt is If the husband dies first after time t, then the present value at time of husband s death of the life annuity 5 The total transition intensity out of state is e.5s e.4s ds

19 t t t t +.4.3t +.13t The expected value of the life annuity is therefore t te.3t +.13t e.4t dt Numerically integrated The expected present value of the premiums is P e.3t +.13t e.4t dt P e.3t +.413t dt P e e.3t+67.5 dt P e π P 1 Φ P e.3t+6 We therefore get that P P $3, We have tp xy e t.1y+s+.1x+s+.1 ds e.1y+.1x+.1s+.1515s Numerically integrating gives the following lifetable 19

20 Year Both H alive H dead Both alive W dead W alive dead Year Both H alive H dead Both alive W dead W alive dead

21 a Summing up we get that A 75, This gives a 75, so the premium is $16, b After 1 years, if the husband is dead, but the wife is alive, then the mortality is µ 3 y.y. The probability that she survives for s years is therefore e s.39+s ds e.78s.1s This gives A 39 i1 1.6 i e.78i 1.1i 1 e.78i.1i.5969, so ä This means that the policy value is The Salary Scale Function $63, a Average Salary from age 6 65 is given by $111, b Average Salary from age 6 65 is given by $139, 9 c for scale in a we have $11, for scale in b, we can use linear interpolation to estimate s This gives a final average salary of $138, Setting the DC Contribution 5 If current salary is 1, final average salary is The replacement ratio means the original annuity is worth , and the reversionary annuity is worth tp 65 e [ ] t.elog t dt e e log1.93s t log1.93 e elog1.93t The value of the life annuity is given by a 65 The value of the reversionary annuity is given by a 65 6 e.7837elog1.93t 1 e log1.4t dt e.7837elog1.93t 1 e elog1.93t 1 e log1.4t dt So the EPV of the benefits at the time of retirement is If first monthly salary is x, then we have

22 x x x The accumulated value of all salary paid monthly in arrear at the end of 35 years is i i i1 So the percentage of salary needed each month is i % i For questions a, b, e, f and g, the employee s accumulated total is still , and the final average salary is For a, the replacement ratio is % 6.% For b, the value of the reversionary annuity is a e.7837elog1.93t 1 e elog1.93t 1 e log1.4t dt so the replacement ratio is % 6.66% For e the replacement ratio is % 6.79% For f we have a 65 The value of the reversionary annuity is given by a 65 6 e.7837elog1.93t 1 e log1.3t dt e.7837elog1.93t 1 e elog1.93t 1 e log1.4t dt so the replacement ratio is For g we have and a 65 6 a % 5.8% e elog1.143t 1 e log1.4t dt e elog1.143t 1 e elog1.93t 1 e log1.4t dt 1.311

23 so the replacement ratio is For c, the accumulated value of the investments is % 79.73% So the replacement ratio is % 7.94%. For e, the accumulated value of the investments is but the final average salary is So the new replacement ratio is % 4.9%. 1.5 The Service Table 7 a See slide b Valuation of Benefits 8 a ä n 1 e 1 n t dt n 1 e n n n The member s final average salary is $148, The EPV of the accrued benefit conditional on the individual retiring at age 65 is therefore: b $113, ä n 1 e.31.1 n n log The member s final average salary is $11, The EPV of the accrued benefit conditional on the individual retiring at age 6 is therefore: log c $19,

24 age probability of retirement EPV of benefits Probability times EPV total So the EPV of accrued benefits is $11, We calculate t p 65 e t t dt e t 1 log1.1 e t 1 ä n 1 n e n a If he withdraws today, he receives an annual pension of $34, The EPV of this is p We have p 43 e t dt e log1.1 So the EPV is $86, b If the employee withdraws after t years,then his annual salary is t, so his accrued withdrawl benefits have present value t t t p 43+t 1.4. Conditional on the employee withdrawing before age 6, the EPV of the accrued withdrawl benefits is tp 43 t 1.5 e.743+t e e.7t e.7times17.7 dt $15, We calculate t p 65 e t t dt e t 1 log1.1 ä n e n n If the member withdraws at age x, then the salary is x 46 e t , so with the COLA, the accrued pension has an annual value of x x x 46, 1. so the value at age 65 is x Discounting at 5% to the age of withdrawl and at 6% to the present day, gives a conditional present value of x x The EPV is therefore x xp 46+x We have 65 x p 46+x e x 65 log1.1 e x 65 so the EPV is x 46 e.31 The probability that the member is still enrolled in the plan at age x is e x 46 e.7y y dy e e 3. e.7x x the EPV of accrued pension benefits paid to early withdrawls is therefore 4

25 x 46 e x 65 e.7x e e 3. e.7x x The probability he is still employed at age 6 is e 6 46 e.7x x dx e e 3. e log If this happens, then his final average salary is He has probability.3 of retiring at age 6, in which case the expected value of the accrued pension is ä The EPV of pension benefits from retirements at age 6 is therefore To simplify, we assume the remaining retirements, except at age 65 happen in the middle of their year. We get the following: age Pretire ä 1 x S Fin EPVPension Benefits total So the total EPV of accrued pension benefits is $1, Funding the Benefits a Under the projected unit method, the final average salary is expected to be We have ä Therefore the EPV of accrued benefits for an individual who reaches retirement age is The probability that this individual reaches retirement age is p 45 e log So the EPV of benefits is After another year, the projected final average salary will still be , so the EPV conditional on surviving to retirement age will be The probability of surviving to retirement age is e log The EPV of benefits at the end of the year is therefore The accumulated value of the reserves at the begining of the year is , so the annual contribution is $15, b Under the traditional unit method, the final average salary is The value in the current year is therefore p If the member survives the year, the final average salary in one year s time is so EPV at the end of next year is p

26 The accumulated value of the assets funding the benefit at the start of the year is , so the contribution is $13, The rate of exit for ages below 6 is µ 1 x + µ x.e.4x x so the probability of the employee remaining employed at age x is e x 46.e.4t t dt.187 [ log1.13 e 1.13t..4 e.4t ] x 46 e x e.4x 46 If the individual has retired, is at age x and has passed the guaranteed time of the pension, the value of the pension at that time is R i ip x 1.5 i where R is the regular pension payment. We have that This gives that the value of the pension is tp x e x t es log1.13 ds e x 1.13 t 1 log1.13 R i e x 1.13 i 1 log i If the individual retires at age x, with pension value R then the present value of the pension at that time is R ä e x 1.13 i 1 log i and the present value is R i5 ä i5 e x 1.13 i+65 x 1 log i If the member exits at age x, then the final average salary is x x 46. If the member withdraws at age x, then the final average salary is also x 46, and with COLA, the eventual accrued pension benefit is x x x per 1 years of service. Since the employee has 1 years of service, the accrued pension beneifit if he withdraws at age x is x The overall expected pension benefit for an individual who withdraws is therefore e.4x e x e.4x x 1.5 ä e x 1.13 i+65 x 1 log i dx i5 6

27 After another year, the expected pension benefit to an individual who withdraws will be 6 47.e.4x e x e.4x x ä i+65 x 1 19 log i dx i5 e x In addition, annother year will have accrued, so the expected benefit is Valued at the present time, not at age 47. If the individual is still employed at age 6, and retires at age x, then the present value of the accrued pension is x ä 46 x e x 1.13 i 1 log i Denote this value by P x. We then have that the expected pension payments to individuals who retire are 14p 46.8P We have that 5.1e.1t t 1 log1.13 i5 P 6 + t dt + 19 p 46 P p 46 14p 46 e e So the expected payments for individuals who retire are $ The expected payments conditional on being employed at the end of the year are p Multiplying by to account for the additional year of service, we get Finally, conditional on the individual surviving for the year, the expected death benefits decrease by the expected death benefits for the year, which is t e.187 log t 1 5e e.4t t dt The increase in the present value of expected benefits is therefore $ The employee contribution is 4% of 87, which is $3,48. 7

28 11. The Yield Curve , so the annual forward rate is 5.3% Valuation of Insurances and Life Annuities 35 See slide See slide Diversifiable and Non-diversifiable Risks 37 a If the interest rate changes to.4 after 1 year, then the expected accumulated value of the premiums received at the end of the term is The expected accumulated value of the death benefits is So the expected loss is $11.15 The present value of this loss is b If the interest rate changes to.6 after 1 year, then the expected accumulated value of the premiums received at the end of the term is The expected accumulated value of the death benefits is So the expected profit is $1.1. The present value of this profit is If the interest rate in 1 years time is i.5, then the variance of the present value of aggregate loss on N policies is conditional on dying: EL

29 and EL So Conditional on surviving and VarL EL Overall, we get VarL VarL For the aggregate loss on N policies, we have EL N VarL N Similarly, if interest rates fall to 1.4 in one year, PV future loss has EL N If interest rates increase to 1.6 in one year: VarL N EL N VarL N The overall variance of the aggregate loss is therefore N N N N 39 The premium is $6.95. We have the following: 9

30 q 58 probability EL VarL So the variance of the aggregate loss for N policies is given by: So the aggregate variance is N N Monte Carlo Simulation 4 Probability of surviving t years is e 47+t x dx e log t 1 e t 1, so for the simulations e t 1 u t 1 logu 1.88 t 1 logu log 1 logu t log1.88 For the interest rate, we have that logi log.4 is normally distributed with mean and standard deviation.4. i.4e.4φ 1 v We therefore have the following: No. u t v i Premiums Benefits PVFL Standard deviation of simulation mean is Using a normal approximation, a 95% confidence interval is 1.96 standard deviations to either side of the mean. That is [ , ] [ , ] 3

31 1.3 Profit Testing a Term Insurance Policy 4 See slide. 43 See slide. 44 See slide. 1.5 Profit Measures 45 See slide. 1.6 Using the Profit Test to Calculate the Premium 46 At a risk discount rate of 1%, the NPV is 1.1 P The premium should be chosen to make this equal to zero. That is P P P $ Using the Profit Test to Calculate Reserves 47 Recalling Question 4, with a premium of $9, The net outflows without reserves are all positive except for the final year, where the net outflow is To correct for this, the reserve R must satisfy 1.4R This means R $4.43. This reserve is a negative cash flow at the end of year 9, so the net cash flow from year 9 is To prevent this negative cash flow, the reserve needs to satisfy 1.4R 3.4, so R 3.7. This is a negative cash flow at the end of year 8, so the net cash-flow in that year is , which is positive, so reserves of $3.7 in year 9 and $4.43 in year 1 are needed. 1.8 Profit Testing for Multiple-State Models 48 Recall from Question 48 that the probabilities of the life being in each state are: 31

32 t tp 37 tp 1 37 tp If the life is in the healthy state at the start of year i, and is then aged x, the probability of being sick at the end is e x + x x x + x x x 75 + x Probabilities of being sick or dead at the end of each year for a life alive at start of year are t p 37+t p 1 37+t p 37+t p 1 37+t p t p 1 37+t t Premium Exp Interest Expected Expected Net Cash Disability Death Flow Benefit Benefit

33 t Premium Exp Interest Expected Expected Net Cash Disability Death Flow Benefit Benefit We see that expected cash flows are all negative if the life starts the year in the sick state, and positive if the life starts the year in the healthy state. We need to calculate separate reserves for each state in the usual way by working backwards. For the 1th year, if the life is in the sick state, the expected net cash flow is $79, 55.34, so the reserve needed is For a life in the sick state at the start of year 9, the probability that the life is in the sick state at the end of year 9 is , so the expected reserves needed are $ For a life in the healthy state at the start of year 9, the probability of being in the sick state at the end is.394, so the additional expected cashflow is $ , making the total expected net cash flow for that year $9.8. We proceed up the table in this way. t Reserve Premium Exp Interest Expected Expected Expected Net Cash Disability Death Reserve Flow Benefit Benefit There is no need to calculate the reserves for the first year, since we know that the life is in the healthy state at the start of year 1. Knowing the reserves needed if the life is in the sick state, we can calculate the reserves as extra expenses in the healthy state: 33

34 t Reserve Premium Exp Interest Expected Expected Expected Expected Net Cash Disability Death Reserve Reserve Flow Benefit Benefit Sick Healthy The resulting profit signature is then Universal Life Insurance 49 See slide. 5 Profit signature is given by 34

35 Year Profit vector Probability in force Profit signature At the risk discount rate the NPV is The EPV of the premiums received is The profit margin is We have CoI t q t AV t CoI t q t CoI t q t AV t q t CoI t 4 AV t q t q t AV t AV t AV t q t 5 Profit signature is given by 35

36 Year Profit vector Probability in force Profit signature At the risk discount rate the NPV is The EPV of the premiums received is The profit margin is After the premium is payed, the account value is $13,389. The Expense charge is $1,33.89, so after this charge, the account value is $1, After applying the cost of insurance and interest, the account value is $13, CoI. The death benefit is the larger of.313, CoI and $3,. We first suppose that the death benefit is $3,, and calculate the cost of insurance charge. The Cost of insurance satisfies: CoI , CoI CoI CoI CoI With this cost of insurance, the account value at the end of the year is 13, CoI This means that using the corridor factor requirement, the minimum death benefit is $3, 519.5, so the corridor factor requirement means that the death benefit is increased. We need to calculate the Cost of Insurance based on the new death benefit. 36

37 CoI , CoI CoI CoI CoI Deterministic Profit Testing for Equity-Linked Insurance 54 See slide. 55 See slide Stochastic Profit Testing 56 See slide. 57 See slides Stochastic Pricing 58 See slide Stochastic Reserving 59 a b See slide. 37