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 We have that 4 p 1 37 p 37 p p 1 37 p For the numbers in the table, this gives So the second actuary s calculations cannot be right. Furthermore, since µ x < µ 1 x for all x, we should have 4p 37 < 4 p 1 37, which rules out the first actuary s calculations. This means that only Actuary III s calculations might be correct. [Indeed these are the correct values.] 8.6 Premiums 5 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 P P e.3t e.56t.15t dt 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: 3

4 1 1 1 e.3t e.56t.15t.1 +.1t + 4 dt.1e e e.356t.15t t dt e.356t.15t.1t dt.3 t t+ e.3 dt e.15t+.3 [.3 ] e.356t.15t dt 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

5 So the expected benefits are se e.3s e e.3s ds 1.84e 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 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 5

6 EPV of disability benefits is EPV of all benefits is EPV of unit premiums is So annual premium is $ If the rate of premium is P, the EPV of total premiums received is P a 34:1 P a 34 1 p 34e 1δ a 44 1 p 34e 1δ a e e.3.11P P The total EPV of benefits are 8a 1 34:1 + 8A 34:1 8 a p 34a p 1 34a A 34 1 p 34e.3 A 44 1 p 1 34e.3 A e.3.43.e e.3.19.e The premium is therefore $ We calculate a x:5 e.3t.11349e.67351t e t dt.11349e t dt e t dt e e

7 The EPV of the benefits to lives who die accidentally from State are given by e.67351t e t e.3t dt.11349e.67351t e t e.3t dt The EPV of the benefits to lives who die otherwise from State are given by e.67351t e t e.3t dt.11349e.67351t e t e.3t dt The EPV of the benefits to lives who die accidentally from State 1 are given by e t.88675e.67351t e.3t dt.88675e t.88675e t dt e e The EPV of the benefits to lives who die otherwise from State 1 are given by e t.88675e.67351t e.3t dt.88675e t.88675e t dt The total EPV of benefits is therefore $,

8 The annual rate of premium is $ a In this case, additional premiums are due for the first three months of each period of sickness. We have tp 11 x a 11 e.6t s x:s e.6t e.3t dt 1 e.36s.36 The EPV of premiums at rate 1 due during the first 3 months of sickness is given by e.9.36 e δt tp x µ 1 1 e.36.5 x+t dt e δt tp x+tµ 1 1 e.365 t x+t.36 e.3t.11349e.67351t e t.1 dt + e.3t.11349e.67351t e t.1 1 e.365 t dt e e e e t e t dt e e t e t e.365 t dt.11349e e e.7351t e.7351t dt e e e e e e e The new annuity value is therefore , so the annual rate of premium is $ b This situation is more complicated. One natural approach would be to add an additional state to represent the off-period i.e. a new state representing healthy lives who had just recovered from disability, and would therefore have no waiting time, or a reduced waiting period if they become disabled. The trouble with this approach is that transitions from this off-period state are not Markovian lives transition to healthy once they have been in this state for 6 months. As a simplifying approximation, we will assume that all periods of disability last at least 3 months and so use up all the waiting time. Since the time spent in the disabled state is exponentially distributed with 8

9 λ , the probability that a period of disability lasts for less than 3 months is 1 e , so this approximation should not have a big effect on our estimates. Suppose the life is healthy at time t. We want to calculate the distribution of the time since they were last disabled if ever. The density of the time they last stopped being disabled is given by fs s p 1 x µ 1 x+st sp x+s.e.4t s.88675e s.88675e.67351s tp.11349e.67351t e t x The probability p d t that the individual was disabled at some point during the preceding 6 months is therefore for t >.5 p d t t t.5 fs ds.e.4t.11349e.67351t e t e.4t.11349e.67351t e t e.7351t e.7351t e.7351t e.7351t e.7351t e.7351t t t.5 t t.5 t t e.7351t.53418e.7351t.11349e.7351t e.7351t e.7351t.53418e.7351t.11349e.7351t e.7351t e t.53418e.67351t tp x e.4s.88675e s.88675e.67351s ds e.7351s e.7351s ds e.7351s e.7351s ds e.7351t e.7351t.5 e.7351t.5 e.7351t e e.7351t.3666 e e.7351t

10 For t <.5 we have p d t t fs ds t.e.4t tp x e.4t tp x e.4s.88675e s.88675e.67351s ds t t e.7351s e.7351s ds e.4t e.7351s tp e.7351s ds x e.4t e.7351t 1 tp 1 e.7351t x e t e.4t tp x.7351 e.4t e.67351t e t e.67351t e.4t tp x 1 e.4t tp x Now we will use the same method as in a, except that whenever the life becomes sick, the waiting period is with probability p d t and.5 with probability 1 p d t. We already calculated in a, that given W, the expected premiums during this waiting period are given by a 11 x:.5 1 e e As in part a we calculate the EPV of premiums during the first 3 months of disability that are a continuation of an earlier period of disability so this is the difference between the answer and the answer to part a: 1

11 .5 p d te δt tp x µ 1 1 e.36.5 x+t dt e.9.5 e.3t tp x.1 dt.36 1 e p d te δt tp x µ 1 1 e.36.5 x+t dt + p d te δt tp x µ 1 1 e.365 t x+t e.3t e.4t.1 dt e.3t.49999e t.53418e.67351t.1 dt + e.3t.49999e t.53418e.67351t.1 1 e.365 t e e e e e e e t.53418e t dt e t.53418e t e.365 t dt e e e e e dt e e e e Therefore, under this model, we have a x: The premium is therefore $ Policy Values and Thiele s Differential Equation 1 Thiele s differential equation gives e e

12 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 11 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

13 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 13

14 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 e.5t +853t.1164 tbtp dt Using numerical integration. We note that this requires the policy value to always be positive. We therefore have 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 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

15 On the other hand we have and 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 :1.9 + te.5t e.59t+.5t dt A 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 We calculate $

16 A :5 1 A :5 1 ä 4: 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 16

17 and the probability of accidental death between ages 47 and 47 and 5 months is therefore e x dx 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 Constructing a Multiple Decrement Table 16 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 17

18 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 y x y+z. The table is therefore the same as in part a. 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 y+z 18

19 x l x d x Joint Life and Last Survivor Benefits 18 Advantages Annuity value does not depend on time of death Disadvantages Value of benefit varies with time of death 9.4 Independent Future Lifetimes 19 ä 63,6:

20 ä 63,6: So the premium is $.9. 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

21 For the premiums, we have ä 53,64: So the net annual premium is $ 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 1 1 1

22 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 3 a Year PHusband Dies PWife Survives PWife Survives to 49 PHusband Dies and Wife Survives to b Suppose the probability of the wife surviving while the husband is alive is p 1 and while the husband is dead is p. If the husband dies at time t during the year, the wife s probability of surviving the year is 1 tq 1 p 1 tq

23 The probability of the wife s surviving the year if the husband dies at a uniformly distributed time during the year is therefore 1 1 tq 1 p 1 tq q1 dt p + q q 1 log1 q q Year PHusband Dies PWife Survives PWife survives year PWife Survives to 49 PHusband Dies and to start of year given Husband dies 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 1 tq tq tq 1 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 3

24 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 4

25 This gives Year Both Alive Husband Dead, Wife Alive Wife Alive a 39: A 39: So the annual premium is $ The Common Shock Model 5 We have tp 5:56 e t s+.5+s+.15+s +.56+s ds e t s+.3s ds e.34316t+.8715t +.1t 3 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 6 If the husband dies first after time t, then the present value at time of husband s death of the life annuity 5 e.5s e.4s ds

26 The total transition intensity out of state is t t t t +.4.3t +.13t The expected value of the life annuity is therefore $84, t te.3t +.13t e.4t dt.5 + t te.3t +.413t dt Numerically integrated The expected present value of the premiums is P P P e.3t +.13t e.4t dt e.3t +.413t dt e.3t dt P e e.3t+67.5 dt P e π P 1 Φ 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 6

27 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

28 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 SN 4. Mortality Improvement Scales $63, We first interpolate the age effect. That is, we set φ45, + t f 45 t at 3 + bt + ct + d. We have f f f5.1 f 5 Substituting the above formula gives the equations subtracting twice 3 from 5 times 4 gives d c a + 65b + 5c + d a + 5b + c a 5c + d a b This gives f For the cohort effect, we set φ 3 + t, + t gt a t 3 + b t + c t + d. Setting up the same equations gives d c a + 65b + 5c + d a + 5b + c 8 8

29 subtracting twice 7 from 5 times 8 gives 1565a 5c + d a b.3339 Substituting these gives φ 45, 13 g The overall scale function is the average of these two values φ45, The Lee Carter Model 9 16 a We have logm4, 34 α 4 +β 4 K 34. We also have that K 34 K c+σ k i Z 17+i. A sum of normal distributions is normal, so K 34 follows a normal distribution with mean K c and variance 17σ k. Substituting this into the above formula gives us logm4, 34 N 5.767, The mean of a log-normal distribution with parameters µ and σ σ.9988 µ+ is e e , while the expected square of the log-normal distribution is e µ+σ e The variance of m4, 34 is therefore , so the standard deviation is The 5th percentile of logm4, 34 is Φ Therefore the 5th percentile of m4, 34 is e b The life will be 4 in 7 years, so we need to calculate m4, 4. Using the same methods as in part a, we get that logm4, 4 N 5.7, Under the UDD assumption, suppose we are given q x, we have t p x 1 tq x, so 1 t p x dt 1 1 tq x dt 1 qx. This gives q x m x 1 qx q x q x 1 m x 1 q x q x m x + m x m x q x m x + m x We have qx mx log log N 5.7 log, qx so qx has a logit-normal distribution. There is no analytic solution for the mean, but numerically, we can get Eq x.55. Using this, the premium is $1,

30 c The EPVFL of the policy is 3q4, , so the EPVFL exceeds $5 if 3q4, q4, q4, m4, logm4, logm4, so the probability is 1 Φ Cairns-Blake-Dowd Models 3 a We have that q33, 48 log 1 q33, 48 K K 48 K K c1 19c σ k1 Z 1 17+i 19σ k Z i1 31 i1 17+i σ k1 Z 1 17+i 19σ k Z 17+i Now we have that so so Varσ k1 Z 1 17+i 19σ k Z 17+i σ k σ k 38σ k1 σ k ρ Var σ k1 Z 1 17+i 19σ k Z 17+i i1 The median is therefore given by solving q33, 48 log N 7.14, q33, 48 3

31 q33, 48 log q33, 48 q33, 48 1 q33, q33, q33, q33, The 95th percentile of log q33,48 1 q33,48 is 7.14+Φ , so the 95th percentile of q33, 48 is given by solving q33, 48 log q33, 48 q33, 48 1 q33, q33, q33, q33, b From the simulated values we get: t 1t K 1 t K t

32 This gives us q6, 17 log 1 q6, 17 q7, 18 log 1 q7, 18 q8, 19 log 1 q8, 19 q9, log 1 q9, q3, 1 log 1 q3, and thus q6, q7, q8, q9, q3, For these values we get ä 6: and A 6: , and 5p , so A 6: The premium is therefore LM 1 Empirical Estimation LM 1.1 The Empirical Distribution 31 The probability mass function is The cumulative Hazard rate is n P X n $

33 x Hx Hx The Nelson-Åalen estimate is H , so this gives S5 e LM 1. The Empirical Distributions for Grouped Data 33 The total number of policies is are less than $1,, and 558 are less than $5,, so the empirical estimates are F and F The ogive then gives F So the probability that a random policy would be affected by this tax is See slides. 35 over the observations, the total of all values of X 6 is 181 3, 6,. There are 3 observations for which X 6 6. The sum of these is therefore 3 6 1, 8,. The total of the 17 observations where X is less than 6, is therefore 3, 6, 1, 8, 1, 8,. The total of the 3 observations between 6, and 7, is,, so the total of the 1,73 observations below 7, is,,. The total of X 7 for the 7 observations above 7, is 7 7 1, 89, so the total of all observations of X 7 is , so the average EX , The total number of observations is +x+y. The number of observations less than 5 is 36. The number of observations less than 15 is 36 + x. The number of observations less than 5 is 36 + x + y. Therefore 36 F n 5 + x + y F n x + x + y F n x + y + x + y x F n 9 + x + y.1 F n x +.6y + x + y.51 We therefore need to solve the equations 33

34 36 +.4x.1 + x + y 19x 1y x +.6y.51 + x + y 49x + 9y 66 1x 144 x 1 y 8 37 Suppose we are estimating the survival function at x which is in the interval c 1, c ]. The estimate is Sx c x c c 1 Sc 1 + x c 1 c c 1 Sc Let X be the number of observations from a sample of n observations that are less than c 1, and let Y be the number that are between c 1 and c. We then have Ŝx c x n X + x c 1 n X Y c c 1 n c c 1 n 1 Xc x + X + Y x c 1 c c 1 1 X n x c 1 c c 1 Y We therefore have that x c VarX + 1 VarŜx c c 1 VarY + x c 1 c c 1 CovX, Y We also have that X and Y are multinomially distributed with probabilities 1 Sc 1 and Sc 1 Sc respectively. This means n VarX nsc 1 1 Sc 1 VarY nsc 1 Sc 1 + Sc Sc 1 CovX, Y n1 Sc 1 Sc 1 Sc This gives that VarŜx c c 1 Sc 1 1 Sc 1 c c 1 x c 1 1 Sc 1 Sc 1 Sc + x c 1 Sc 1 Sc 1 + Sc Sc 1 nc c 1 c c 1 1 Sc 1 c c 1 Sc 1 x c 1 Sc 1 Sc +x c 1 Sc 1 Sc x c 1 1+Sc Sc 1 c c 1 1 Sc 1 c c 1 1 Sc 1 c xsc 1 +x c 1 Sc +x c 1 Sc 1 Sc x c 1 1 Sc 1 c x1 Sc

35 We compute Ŝ and Ŝ This gives Ŝ The variances are given by VarŜ VarŜ CovŜ1, Ŝ VarŜ The standard deviation is A 95% confidence interval is therefore ± [.4437, ]. LM 1.3&1.5 Empirical Estimation with Modified Data 39 The probability that a randomly chosen individual survives to more than 1.6 is expressed as the product < 1 So the median is y The cumulative hazard rate funtion is given by H The survival function is therefore S1.6 e The Kaplan-Meier estimator gives S n and S n

36 So the conditional probability is Let the dying times be t 1,..., t n, and let the corresponding risk sets be r 1,..., r n. Let the number of people surviving at each dying time be X i. Suppose that the true probability of surviving at time t i is p i. The Kaplan-Meier estimate of the survival probability is therefore n X i i1 r i. Since the X i are independent, we have E n i1 X i r i 3 11 n E i1 n i1 p i Xi r i so the Kaplan-Meier estimate is unbiassed. We also have: E n i1 X i r i n E i1 Xi r i n pi 1 p i + p i i1 r i so the variance is n pi 1 p i + p i i1 r i n n n p i p i i1 i1 i p i 1 p i r i If we let s i be the total survivial probability up to time i, so that s i i j1 p j, then this becomes n s n 1 + s i 1 s i 1 s i r i i1 44 Greenwoods formula gives that the variance is 36

37 j VarS n y j Ŝy j s i r i1 i r i s i The 95% confidence interval is therefore 45 The confidence interval is ± [ , ] where So the confidence interval is [S n 1 1 U, Sn 1 U ] U e log [ , ] [.11719,.7576] 46 The Nelson-Åalen estimator is H5 6 The variance of this estimator is then 6 We therefore have logĥ VarlogĤ So a 95% confidence interval for logh5 is ± [ , ] The corresponding interval for H5 is [1.4584, ] and the corresponding interval for S5 is [.46184, ] 37

38 LM 1.7 Approximations for Large Data Sets 47 a The exact exposure is years. There are two deaths in the interval. The estimate for the hazard rate is therefore , and the probability of dying in the year is 1 e b The actuarial exposure is , so the estimate for the probability of dying is Using insuring ages, the table looks like this: entry death exit entry death exit a Now the exact exposure is given by , so the estimated hazard rate is and the estimated probability of dying is 1 e The actuarial exposure is given by so the estimated probability of dying is b Using an anniversary-to-anniversary study, we ignore all partial units of exposure, so the exposure is 11, which makes q See next slide. 5 a The exact exposure is The hazard rate is therefore and the probability of dying during the year is therefore 1 e b The actuarial exposure is and the probability of dying during the year is therefore Now death is the censoring event and withdrawl is the event we are trying to estimate. The exposure is and the probability of withdrawing is therefore LM 1.9 Estimation of Transition Intensities 5 MLE estimates are based on exact exposure. We calculate the total exposure in each state for the data: 38

39 State Exposure Transition to No. observed Intensity Disabled Healthy Surrender Dead Disabled The Salary Scale Function Healthy Surrender Dead 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 54 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 x x x

40 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 so the replacement ratio is For c, the accumulated value of the investments is % 79.73% So the replacement ratio is % 7.94%. 4

41 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 56 a See slide b Valuation of Benefits 57 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 $18, 15.3 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: $1, c age probability of retirement EPV of benefits Probability times EPV total log

42 So the EPV of accrued benefits is $115, 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 75 $33, The EPV of this is p We have p 43 e t dt e log1.1 So the EPV is $7, 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 tp 43. The probability density of withdrawl after t years is e.743+t tp 43 The EPV of the accrued withdrawl benefits paid upon withdrawl before age 6 is tp 43 t 1.5 e.743+t tp 43 dt t 1.5 e.743+t dt e 3.1 e log1.5 log1.t e.7t dt e 3.1 e log1.5 log1..7t dt e 3.1 [ e log1.5 log1..7t log1.5 log e 3.1 e17log1.5 log log1.5 log1..7 $ 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 1 ] , 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 4

43 We have 65 x p 46+x e log1.1 x 65 e x 65 so the EPV is x 46 e x 65 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 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: by age Pretire ä 1 x S Fin EPVPension Benefits total So the total EPV of accrued pension benefits is $1, If the individual retires or withdraws at age x, then career average earnings time years of service is given 18 x y dy 18 log x 38 1 The life s current total pensionable earnings are therefore given by 18 log $1, 17, 658 so at an accrual rate of 4%, the annual pension is based on an annual pension rate of $48, We will apply COLA of % to this, so if the life starts receiving the pension in t years, then the EPV of the accrued pension at that time is 48, t a 44+t. Mortality is given by µ x.31.1 x, so tp x e x+t x.31.1 y dy e.31.1x 1.1 t log This means that a x e.31.1 x log t t dt. 43

44 Numerically, we calculate a , so the accrued pension benefits for a life who starts receiving pension benefits at age 65 are given by , $1, 44, 58. The probability density that the life withdraws at age 44 + t t < 16 is tp 44 e t e.744+s ds e.744+t t p 44 e e e.7t.7 e.744+t so the EPV of withdrawl benefits is given by tp 44 e e p 44 1 e.7t.7 e e e.7t.7 e.744+t 1 tp 44+t 1.6 t 1.5 t 1 dt e.744+t 1.6 t 1.5 t 1 dt We can evaluate this expression numerically to get 16 e e e.7t.7 This means the EPV of accrued withdrawl benefits is given by e.744+t 1.6 t 1.5 t 1 dt e log1.1 1 $13, 79. The probability that the life is still employed at age 6 is 16p 44 e e e e log1.1 1 e e e This means the probability of retiring at age 6 is We calculate numerically a , so the EPV of accrued benefits from retirement at age 6 is The probability of continuing to work past 6 is Given that an employee continues to work past 6, the probability that they retire at age 65 is 5 p 6 e.6 5 e log1.1 1 e , so the probability that the life retires at age 65 is , so the EPV of accrued benefits from retirement at 65 is $99, For retirements between ages 6 and 65, conditional on being alive, retirements happen following an exponential distribution, so for an individual alive at age 6 + t, who continued working past 6, the probability of not being retired is e.6t, while the density of having retired at age 6 + s is.6e.6s. We can evaluate the EPV of retirement benefits for individuals aged 6 65 by integrating over all payment times t, and all possible retirement ages s as follows: 44

45 t tp t.6e.6s 1.5 s t s ds dt 16+t 1. t s t 1.5 tp 6.6e.6s ds dt t t tp log e.6+log t dt t t t log 1.6 tp 6 1 e.6t dt t 1.5 t log 1.6 tp 6 e.6t dt So the EPV of payments made between ages 6 and 65 is $5, For payments made after age 6 to individuals who retire before age 65, we use a similar integral, but with the limits of the integral for s ranging from to 5 since we have already accounted for individuals who retire at age 65. We calculate 5 tp log t e.6+log tp s 1.5.6e.6s ds dt t 1. dt tp t 1. dt 1.5 The EPV of benefits paid after age 65 to individuals who retire between ages 6 and 65 is therefore The total EPV of all accrued benefits is therefore 1.7 Funding the Benefits $41, 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 So the EPV of benefits is log

46 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 The accumulated value of the assets funding the benefit at the start of the year is , so the contribution is $13, Current reserve Pension Benefits: If the individual retires after t years, their final average salary is t +1.6 t t t The pension benefits are therefore t.1ä 46+t:5 These need to be discounted by t years at i.5. We then take the expectation over possible retirement times to calculate the total EPV of retirement benefits as tp 46µ 46+t t.1ä 46+t:5 1.5 t dt Deferred Pension Benefits: Death Benefits: Next year reserve Pension Benefits: Deferred Pension Benefits: Death Benefits: Benefits for exits during year Deferred Pension Benefits: Death Benefits: 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 log