BEGINNING OF EXAMINATIONFor a fully discrete whole life insurance of 100,000 on (35) you are given: (i) Percent of premium expenses are 10% per year. (ii) Per policy expenses are

Exam M: Spring 2005 - 1 - GO ON TO NEXT PAGE

**BEGINNING OF EXAMINATION**

1. For a fully discrete 3-year endowment insurance of 1000 on (x), you are given: (i) k L is the prospective loss random variable at time k.

(ii) i = 0.10

(iii) :3 2.70182xa =

(iv) Premiums are determined by the equivalence principle. Calculate 1L , given that (x) dies in the second year from issue. (A) 540

(B) 630

(C) 655

(D) 720

(E) 910


2. For a double-decrement model:

(i) ( )140' 1

60ttp = − , 0 60t≤ ≤

(ii) ( )240' 1

40ttp = − , 0 40t≤ ≤

Calculate ( ) ( )40 20τµ . (A) 0.025

(B) 0.038

(C) 0.050

(D) 0.063

(E) 0.075


3. For independent lives (35) and (45): (i) 5 35 0 90p = . (ii) 5 45 080p = . (iii) q40 0 03= . (iv) q50 0 05= . Calculate the probability that the last death of (35) and (45) occurs in the 6th year. (A) 0.0095 (B) 0.0105 (C) 0.0115 (D) 0.0125 (E) 0.0135


4. For a fully discrete whole life insurance of 100,000 on (35) you are given: (i) Percent of premium expenses are 10% per year.

(ii) Per policy expenses are 25 per year.

(iii) Per thousand expenses are 2.50 per year.

(iv) All expenses are paid at the beginning of the year.

(v) 351000 8.36P = Calculate the level annual expense-loaded premium using the equivalence principle. (A) 930

(B) 1041

(C) 1142

(D) 1234

(E) 1352


5. Kings of Fredonia drink glasses of wine at a Poisson rate of 2 glasses per day. Assassins attempt to poison the king’s wine glasses. There is a 0.01 probability that any given glass is poisoned. Drinking poisoned wine is always fatal instantly and is the only cause of death. The occurrences of poison in the glasses and the number of glasses drunk are independent events. Calculate the probability that the current king survives at least 30 days. (A) 0.40

(B) 0.45

(C) 0.50

(D) 0.55

(E) 0.60


6. Insurance losses are a compound Poisson process where: (i) The approvals of insurance applications arise in accordance with a Poisson process at

a rate of 1000 per day.

(ii) Each approved application has a 20% chance of being from a smoker and an 80% chance of being from a non-smoker.

(iii) The insurances are priced so that the expected loss on each approval is –100.

(iv) The variance of the loss amount is 5000 for a smoker and is 8000 for a non-smoker. Calculate the variance for the total losses on one day’s approvals. (A) 13,000,000 (B) 14,100,000 (C) 15,200,000 (D) 16,300,000 (E) 17,400,000


7. Z is the present-value random variable for a whole life insurance of b payable at the moment of death of (x). You are given: (i) δ = 0.04

(ii) ( ) 0.02x tµ = , 0t ≥

(iii) The single benefit premium for this insurance is equal to Var(Z). Calculate b. (A) 2.75

(B) 3.00

(C) 3.25

(D) 3.50

(E) 3.75


8. For a special 3-year term insurance on (30), you are given: (i) Premiums are payable semiannually.

(ii) Premiums are payable only in the first year.

(iii) Benefits, payable at the end of the year of death, are:

k 1kb + 0 1000 1 500 2 250

(iv) Mortality follows the Illustrative Life Table.

(v) Deaths are uniformly distributed within each year of age.

(vi) i = 0.06

Calculate the amount of each semiannual benefit premium for this insurance. (A) 1.3

(B) 1.4

(C) 1.5

(D) 1.6

(E) 1.7


9. A loss, X, follows a 2-parameter Pareto distribution with 2α = and unspecified parameter θ . You are given:

5E 100 100 E 50 503

X X X X⎡ − > ⎤ = ⎡ − > ⎤⎣ ⎦ ⎣ ⎦

Calculate E 150 150X X⎡ − > ⎤⎣ ⎦ . (A) 150

(B) 175

(C) 200

(D) 225

(E) 250


10. The scores on the final exam in Ms. B’s Latin class have a normal distribution with mean θ and standard deviation equal to 8. θ is a random variable with a normal distribution with mean equal to 75 and standard deviation equal to 6. Each year, Ms. B chooses a student at random and pays the student 1 times the student’s score. However, if the student fails the exam (score 65≤ ), then there is no payment. Calculate the conditional probability that the payment is less than 90, given that there is a payment. (A) 0.77

(B) 0.85

(C) 0.88

(D) 0.92

(E) 1.00


11. For a Markov model with three states, Healthy (0), Disabled (1), and Dead (2): (i) The annual transition matrix is given by

0 1 2

0 0.70 0.20 0.101 0.10 0.65 0.252 0 0 1

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

(ii) There are 100 lives at the start, all Healthy. Their future states are independent. Calculate the variance of the number of the original 100 lives who die within the first two years. (A) 11

(B) 14

(C) 17

(D) 20

(E) 23


12. An insurance company issues a special 3-year insurance to a high risk individual. You are given the following homogenous Markov chain model: (i) State 1: active

State 2: disabled State 3: withdrawn State 4: dead

Transition probability matrix:

1 2 3 41 0.4 0.2 0.3 0.12 0.2 0.5 0 0.33 0 0 1 04 0 0 0 1

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

(ii) Changes in state occur at the end of the year.

(iii) The death benefit is 1000, payable at the end of the year of death.

(iv) i = 0.05

(v) The insured is disabled at the end of year 1.

Calculate the actuarial present value of the prospective death benefits at the beginning of year 2. (A) 440

(B) 528

(C) 634

(D) 712

(E) 803


13. For a fully discrete whole life insurance of b on (x), you are given: (i) 9 0.02904xq + =

(ii) i = 0.03

(iii) The initial benefit reserve for policy year 10 is 343.

(iv) The net amount at risk for policy year 10 is 872.

(v) 14.65976xa = Calculate the terminal benefit reserve for policy year 9. (A) 280

(B) 288

(C) 296

(D) 304

(E) 312


14. For a special fully discrete 2-year endowment insurance of 1000 on (x), you are given: (i) The first year benefit premium is 668.

(ii) The second year benefit premium is 258.

(iii) d = 0.06 Calculate the level annual premium using the equivalence principle. (A) 469

(B) 479

(C) 489

(D) 499

(E) 509


15. For an increasing 10-year term insurance, you are given: (i) ( )1 100,000 1kb k+ = + , k = 0, 1,…,9

(ii) Benefits are payable at the end of the year of death.

(iii) Mortality follows the Illustrative Life Table.

(iv) i = 0.06

(v) The single benefit premium for this insurance on (41) is 16,736. Calculate the single benefit premium for this insurance on (40). (A) 12,700

(B) 13,600

(C) 14,500

(D) 15,500

(E) 16,300


16. For a fully discrete whole life insurance of 1000 on (x): (i) Death is the only decrement.

(ii) The annual benefit premium is 80.

(iii) The annual contract premium is 100.

(iv) Expenses in year 1, payable at the start of the year, are 40% of contract premiums.

(v) i = 0.10

(vi) 11000 40xV = Calculate the asset share at the end of the first year. (A) 17

(B) 18

(C) 19

(D) 20

(E) 21


17. For a collective risk model the number of losses, X, has a Poisson distribution with 20λ = . The common distribution of the individual losses has the following characteristics: (i) [ ]E 70X =

(ii) [ ]E 30 25X ∧ =

(iii) ( )Pr 30 0.75X > =

(iv) 2E 30 9000X X⎡ ⎤> =⎣ ⎦

An insurance covers aggregate losses subject to an ordinary deductible of 30 per loss. Calculate the variance of the aggregate payments of the insurance. (A) 54,000

(B) 67,500

(C) 81,000

(D) 94,500

(E) 108,000


18. For a collective risk model: (i) The number of losses has a Poisson distribution with 2λ = .

(ii) The common distribution of the individual losses is:

x ( )xf x

1 0.6 2 0.4

An insurance covers aggregate losses subject to a deductible of 3. Calculate the expected aggregate payments of the insurance. (A) 0.74

(B) 0.79

(C) 0.84

(D) 0.89

(E) 0.94


19. A discrete probability distribution has the following properties:

(i) 111k kp c pk −

⎛ ⎞= +⎜ ⎟⎝ ⎠

for k = 1, 2,…

(ii) 0 0.5p = Calculate c. (A) 0.06

(B) 0.13

(C) 0.29

(D) 0.35

(E) 0.40


20. A fully discrete 3-year term insurance of 10,000 on (40) is based on a double-decrement model, death and withdrawal: (i) Decrement 1 is death.

(ii) ( ) ( )1

40 0.02tµ = , 0t ≥

(iii) Decrement 2 is withdrawal, which occurs at the end of the year.

(iv) ( )240' 0.04kq + = , k = 0, 1, 2

(v) 0.95v = Calculate the actuarial present value of the death benefits for this insurance. (A) 487

(B) 497

(C) 507

(D) 517

(E) 527


21. You are given:

(i) 30:40 27.692e =

(ii) ( ) 1 xs xω

= − , 0 x ω≤ ≤

(iii) ( )T x is the future lifetime random variable for (x). Calculate ( )( )Var 30T . (A) 332

(B) 352

(C) 372

(D) 392

(E) 412


22. For a fully discrete 5-payment 10-year decreasing term insurance on (60), you are given: (i) 1 1000kb + = ( )10 k− , k = 0, 1, 2,…, 9

(ii) Level benefit premiums are payable for five years and equal 218.15 each.

(iii) 60 0.02 0.001kq k+ = + , k = 0, 1, 2,…, 9

(iv) i = 0.06 Calculate 2 ,V the benefit reserve at the end of year 2. (A) 70

(B) 72

(C) 74

(D) 76

(E) 78


23. You are given: (i) ( )T x and ( )T y are not independent.

(ii) 0.05x k y kq q+ += = , k = 0, 1, 2,…

(iii) 1.02k xy k x k yp p p= , k = 1, 2, 3… Into which of the following ranges does :x ye , the curtate expectation of life of the last

survivor status, fall? (A) : 25.7x ye ≤

(B) :25.7 26.7x ye< ≤

(C) :26.7 27.7x ye< ≤

(D) :27.7 28.7x ye< ≤

(E) :28.7 x ye<


24. Subway trains arrive at your station at a Poisson rate of 20 per hour. 25% of the trains are express and 75% are local. The types and number of trains arriving are independent. An express gets you to work in 16 minutes and a local gets you there in 28 minutes. You always take the first train to arrive. Your co-worker always takes the first express. You are both waiting at the same station. Calculate the conditional probability that you arrive at work before your co-worker, given that a local arrives first. (A) 37%

(B) 40%

(C) 43%

(D) 46%

(E) 49%


25. Beginning with the first full moon in October deer are hit by cars at a Poisson rate of 20 per day. The time between when a deer is hit and when it is discovered by highway maintenance has an exponential distribution with a mean of 7 days. The number hit and the times until they are discovered are independent. Calculate the expected number of deer that will be discovered in the first 10 days following the first full moon in October. (A) 78

(B) 82

(C) 86

(D) 90

(E) 94


26. You are given: (i) ( ) 0.03x tµ = , 0t ≥

(ii) 0.05δ =

(iii) T(x) is the future lifetime random variable.

(iv) g is the standard deviation of ( )T x

a .

Calculate ( )( )Pr xT xa a g> − .

(A) 0.53

(B) 0.56

(C) 0.63

(D) 0.68

(E) 0.79


27. (50) is an employee of XYZ Corporation. Future employment with XYZ follows a double decrement model: (i) Decrement 1 is retirement.

(ii) ( ) ( )150

0.00 0 50.02 5

tt

tµ

≤ <⎧= ⎨ ≤⎩

(iii) Decrement 2 is leaving employment with XYZ for all other causes.

(iv) ( ) ( )250

0.05 0 50.03 5

tt

tµ

≤ <⎧= ⎨ ≤⎩

(v) If (50) leaves employment with XYZ, he will never rejoin XYZ. Calculate the probability that (50) will retire from XYZ before age 60. (A) 0.069

(B) 0.074

(C) 0.079

(D) 0.084

(E) 0.089


28. For a life table with a one-year select period, you are given: (i) x [ ]xl [ ]xd 1xl +

[ ]xe

80 1000 90 − 8.5 81 920 90 − − (ii) Deaths are uniformly distributed over each year of age.

Calculate [ ]81e .

(A) 8.0

(B) 8.1

(C) 8.2

(D) 8.3

(E) 8.4


29. For a fully discrete 3-year endowment insurance of 1000 on (x): (i) 0.05i =

(ii) 1 0.7x xp p += = Calculate the second year terminal benefit reserve. (A) 526

(B) 632

(C) 739

(D) 845

(E) 952


30. For a fully discrete whole life insurance of 1000 on (50), you are given: (i) The annual per policy expense is 1.

(ii) There is an additional first year expense of 15.

(iii) The claim settlement expense of 50 is payable when the claim is paid.

(iv) All expenses, except the claim settlement expense, are paid at the beginning of the

year.

(v) Mortality follows De Moivre’s law with 100ω = .

(vi) i = 0.05 Calculate the level expense-loaded premium using the equivalence principle. (A) 27

(B) 28

(C) 29

(D) 30

(E) 31


31. The repair costs for boats in a marina have the following characteristics:

Boat type Number of

boats Probability that repair is needed

Mean of repair cost given a repair

Variance of repair cost given a repair

Power boats 100 0.3 300 10,000 Sailboats 300 0.1 1000 400,000 Luxury yachts 50 0.6 5000 2,000,000 At most one repair is required per boat each year. The marina budgets an amount, Y, equal to the aggregate mean repair costs plus the standard deviation of the aggregate repair costs. Calculate Y. (A) 200,000

(B) 210,000

(C) 220,000

(D) 230,000

(E) 240,000


32. For an insurance: (i) Losses can be 100, 200 or 300 with respective probabilities 0.2, 0.2, and 0.6.

(ii) The insurance has an ordinary deductible of 150 per loss.

(iii) PY is the claim payment per payment random variable. Calculate ( )Var PY .

(A) 1500

(B) 1875

(C) 2250

(D) 2625

(E) 3000


33. You are given:

( )0.05 50 600.04 60 70

xx

xµ

≤ <⎧= ⎨ ≤ <⎩

Calculate 504 14 q .

(A) 0.38

(B) 0.39

(C) 0.41

(D) 0.43

(E) 0.44


34. The distribution of a loss, X , is a two-point mixture: (i) With probability 0.8, X has a two-parameter Pareto distribution with

α θ= =2 100 and . (ii) With probability 0.2, X has a two-parameter Pareto distribution with

4 and 3000α θ= = . Calculate ( )Pr 200X ≤ . (A) 0.76 (B) 0.79 (C) 0.82 (D) 0.85 (E) 0.88


35. For a special fully discrete 5-year deferred whole life insurance of 100,000 on (40), you are given: (i) The death benefit during the 5-year deferral period is return of benefit premiums paid

without interest.

(ii) Annual benefit premiums are payable only during the deferral period.

(iii) Mortality follows the Illustrative Life Table.

(iv) i = 0.06

(v) ( )140:5 0.04042IA = Calculate the annual benefit premiums. (A) 3300

(B) 3320

(C) 3340

(D) 3360

(E) 3380


36. You are pricing a special 3-year annuity-due on two independent lives, both age 80. The annuity pays 30,000 if both persons are alive and 20,000 if only one person is alive. You are given: (i)

k 80k p 1 0.91 2 0.82 3 0.72

(ii) i = 0.05 Calculate the actuarial present value of this annuity. (A) 78,300

(B) 80,400

(C) 82,500

(D) 84,700

(E) 86,800


37. Company ABC sets the contract premium for a continuous life annuity of 1 per year on (x) equal to the single benefit premium calculated using: (i) 0.03δ =

(ii) ( ) 0.02x tµ = , 0t ≥ However, a revised mortality assumption reflects future mortality improvement and is given by

( )0.02 for 100.01 for 10x

tt

tµ

≤⎧= ⎨ >⎩

Calculate the expected loss at issue for ABC (using the revised mortality assumption) as a percentage of the contract premium. (A) 2%

(B) 8%

(C) 15%

(D) 20%

(E) 23%


38. A group of 1000 lives each age 30 sets up a fund to pay 1000 at the end of the first year for each member who dies in the first year, and 500 at the end of the second year for each member who dies in the second year. Each member pays into the fund an amount equal to the single benefit premium for a special 2-year term insurance, with: (i) Benefits:

k 1kb + 0 1000 1 500

(ii) Mortality follows the Illustrative Life Table.

(iii) i = 0.06 The actual experience of the fund is as follows:

k Interest Rate Earned Number of Deaths 0 0.070 1 1 0.069 1

Calculate the difference, at the end of the second year, between the expected size of the fund as projected at time 0 and the actual fund. (A) 840

(B) 870

(C) 900

(D) 930

(E) 960


39. In a certain town the number of common colds an individual will get in a year follows a Poisson distribution that depends on the individual’s age and smoking status. The distribution of the population and the mean number of colds are as follows:

Proportion of population Mean number of colds Children 0.30 3 Adult Non-Smokers 0.60 1 Adult Smokers 0.10 4

Calculate the conditional probability that a person with exactly 3 common colds in a year is an adult smoker. (A) 0.12

(B) 0.16

(C) 0.20

(D) 0.24

(E) 0.28

Exam M: Spring 2005 - 40 - STOP

40. For aggregate losses, S: (i) The number of losses has a negative binomial distribution with mean 3 and

variance 3.6. (ii) The common distribution of the independent individual loss amounts is uniform from

0 to 20. Calculate the 95th percentile of the distribution of S as approximated by the normal distribution. (A) 61

(B) 63

(C) 65

(D) 67

(E) 69

**END OF EXAMINATION**

Exam M: Spring 2005 - 41 -

Exam M Spring 2005

PRELIMINARY ANSWER KEY

Question # Answer Question # Answer

1 B 21 B 2 E 22 E 3 B 23 D 4 D 24 A 5 D 25 E 6 E 26 E 7 E 27 A 8 A 28 C 9 B 29 A

10 D 30 E 11 C 31 B 12 A 32 B 13 C 33 A 14 B 34 A 15 D 35 D 16 A 36 B 17 B 37 C 18 A 38 C 19 C 39 B 20 C 40 C

SPRING 2005 EXAM M SOLUTIONS

Question #1 Key: B Let K be the curtate future lifetime of (x + k)

1:3 11000 1000K

k x KL v P a++= − ×

When (as given in the problem), (x) dies in the second year from issue, the curtate future lifetime of ( )1x + is 0, so

1 :3 11000 1000 xL v P a= − 1000 279.211.1

629.88 630

= −

= ≈

The premium came from

:3:3

:3

xx

x

AP

a=

:3 :31x xA d a= −

:3:3

:3 :3

1 1279.21 xx

x x

d aP d

a a−

= = = −

Question #2 Key: E Note that above 40, decrement 1 is DeMoivre with omega = 100; decrement 2 is DeMoivre with omega = 80. That means ( ) ( ) ( ) ( )1 2

40 4020 1/ 40 0.025; 20 1/ 20 0.05µ µ= = = = ( ) ( )40 20 0.025 0.05 0.075τµ = + =

Or from basic definition of µ ,

( )2

4060 40 2400 100

60 40 2400tt t t tp τ − − − +

= × =

( )( ) ( )40 / 100 2 / 2400td p dt tτ = − +

at 20t = gives 60 / 2400 0.025− = ( ) ( ) ( )

( ) ( ) ( )( ) ( ) ( )20 40

40 40 20 40

2 / 3 * 1/ 2 1/ 3

20 / / 0.025/ 1/ 3 0.075t

p

d p dt p

τ

τ τ τµ

= =

⎡ ⎤= − = =⎢ ⎥⎣ ⎦

Question #3 Key: B 5 35 45 5 35 5 45 5 35 45

5 35 40 5 45 50 5 35 45 40 50

5 35 40 5 45 50 5 35 5 45 40 50

5 35 40 5 45 50 5 35 5 45 40 50

1

1

0 9 03 0 8 0 05 0 9 0 8 1 0 97 0 95

0 01048

q q q q

p q p q p qp q p q p p p

p q p q p p p p

: :

: :

:

. . . . . . . .

.

= + −

= + −

= + − × −

= + − × −

= + − −

=

b gb g

b gb g b gb g b gb g b gb g

Alternatively,

( )( )( )( )

6 35 5 35 40

6 45 5 45 50

0.90 1 0.03 0.873

0.80 1 0.05 0.76

p p p

p p p

= × = − =

= × = − =

( ) ( )5 65 35:45 35:45 35:45

5 35 5 45 5 35:45 6 35 6 45 6 35:45

q p p

p p p p p p

= −

= + − − + −

( ) ( )5 35 5 45 5 35 5 45 6 35 6 45 6 35 6 45p p p p p p p p= + + × − + − ×

( ) ( )0.90 0.80 0.90 0.80 0.873 0.76 0.873 0.76= + − × − + − ×

0.98 0.969520.01048

= −=

Question #4 Key: D Let G be the expense-loaded premium. Actuarial present value (APV) of benefits = 35100,000A APV of premiums = 35Ga

APV of expenses = ( )( ) 350.1 25 2.50 100G a+ +⎡ ⎤⎣ ⎦ Equivalence principle:

( )35 35 35

35

35

100,000 0.1 25 250

100,000 0.1 275

Ga A G aAG Ga

= + + +

= + +

( )( )350.9 100,000 275

100 8.36 2750.9

1234

G P

G

= +

+=

=

Question #5 Key: D Poisoned wine glasses are drunk at a Poisson rate of 2×0.01 = 0.02 per day. Number of glasses in 30 days is Poisson with 0.02 30 0.60λ = × = ( ) 0.600 0.55f e−= =

Question #6 Key: E View the compound Poisson process as two compound Poisson processes, one for smokers and one for non-smokers. These processes are independent, so the total variance is the sum of their variances. For smokers, ( )( )0.2 1000 200λ = =

Var(losses) ( ) ( )( )2Var X E Xλ ⎡ ⎤= +⎢ ⎥⎣ ⎦

( )2200 5000 100

3,000,000

⎡ ⎤= + −⎣ ⎦=

For non-smokers, ( )( )0.8 1000 800λ = =

Var(losses) ( ) ( )( )2Var X E Xλ ⎡ ⎤= +⎢ ⎥⎣ ⎦

( )2800 8000 100

14,400,000

⎡ ⎤= + −⎣ ⎦=

Total variance =3,000,000 + 14,400,000 =17,400,000

Question #7 Key: E [ ] xE Z b A=

since constant force /xA µ µ δ= +

E(Z) ( )( )

0.02/ 3

0.06bb bµ

µ δ= = =

+

[ ] ( )T 2 2 2 2

22

2 2

Var Var Var

2

2 1 410 9 45

x xZ bv b v b A A

b

b b

µ µµ δ µ δ

Τ⎡ ⎤ ⎡ ⎤= = = −⎣ ⎦ ⎣ ⎦⎛ ⎞⎛ ⎞⎜ ⎟= − ⎜ ⎟⎜ ⎟+ +⎝ ⎠⎝ ⎠⎡ ⎤ ⎛ ⎞= − = ⎜ ⎟⎢ ⎥⎣ ⎦ ⎝ ⎠

( ) ( )Var Z E Z=

2 4

45 34 1 3.7545 3

bb

b b

⎡ ⎤ =⎢ ⎥⎣ ⎦⎡ ⎤ = ⇒ =⎢ ⎥⎣ ⎦

Question #8 Key: A

( ) ( )( )

1 2 330 30 301 230:3

2 3

1000 500 250

1 1.53 1 1.61 1 1.701000 500 0.99847 250 0.99847 0.998391.06 1000 1.06 1000 1.06 1000

A vq v q v q= + +

⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

1.4434 0.71535 0.35572 2.51447= + + =

( ) ( ) ( )

( )( )

12

2 1 1 12 2 2 3030:1

1 1 1 0.001531 0.97129 11.06 2 2 2

1 1 0.97129 0.9992352 20.985273

a q⎛ ⎞ ⎛ ⎞= + − = + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= +

=

Annualized premium 2.514470.985273

=

2.552=

Each semiannual premium 2.5522

=

1.28=

Question #9 Key: B E x d x d⎡ − > ⎤⎣ ⎦ is the expected payment per payment with an ordinary deductible of d It can be evaluated (for Pareto) as

( ) ( )( )

1

11 1

11 1

dE x E x dF d

d

α

α

θ θ θα α θ

θθ

−⎡ ⎤⎛ ⎞− −⎢ ⎥⎜ ⎟− − +⎝ ⎠− ∧ ⎢ ⎥⎣ ⎦=− ⎡ ⎤⎛ ⎞− −⎢ ⎥⎜ ⎟+⎝ ⎠⎢ ⎥⎣ ⎦

1

1 d

d

α

α

θ θα θ

θθ

−⎛ ⎞⎜ ⎟− +⎝ ⎠=

⎛ ⎞⎜ ⎟+⎝ ⎠

1

d θα+

=−

d θ= + in this problem, since 2α =

( )

53

53

100 100 50 50

100 50

E x x E x x

θ θ

⎡ − > ⎤ = ⎡ − > ⎤⎣ ⎦ ⎣ ⎦+ = +

300 3 250 5

25θ θ

θ+ = +

= =

150 150 150

150 25175

E x x θ⎡ − > ⎤ = +⎣ ⎦= +=

Question #10 Key: D Let S = score

( ) ( )( ) ( )

( ) ( ) ( )75E S E E S E

Var S E Var S Var E S

θ θ

θ θ

= = =

⎡ ⎤ ⎡ ⎤= +⎣ ⎦ ⎣ ⎦

( ) ( )28E Var θ= + 264 6

100= +=

S is normally distributed (a normal mixture of normal distributions with constant variance is normal; see Example 4.30 in Loss Models for the specific case, as we have here, with a normally distributed mean and constant variance)

( ) ( )( )

90 65Pr ob 90 65

1 65F F

S SF−

⎡ < > ⎤ =⎣ ⎦ −

90 75 65 7510 10

65 75110

− −⎛ ⎞ ⎛ ⎞Φ −Φ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠=

−⎛ ⎞−Φ⎜ ⎟⎝ ⎠

( ) ( )

( )( )

( )1.5 1.0 0.9332 1 0.8413 0.7745 0.92061 1.0 1 1 0.8413 0.8413

Φ −Φ − − −= = =

−Φ − − −

Note that (though this insight is unnecessary here), this is equivalent to per payment model with a franchise deductible of 65.

Question #11 Key: C Ways to go 0→2 in 2 years

( )( )( )( )( )( )

0 0 2; 0.7 0.1 0.07

0 1 2; 0.2 0.25 0.05

0 2 2; 0.1 1 0.1

p

p

p

− − = =

− − = =

− − = =

Total = 0.22 Binomial m = 100 q = 0.22 Var = (100) (0.22) (0.78) = 17 Question #12 Key: A For death occurring in year 2

0.3 1000 285.711.05

APV ×= =

For death occurring in year 3, two cases: (1) State 2 State 1 State 4: (0.2 0.1)→ → × = 0.02 (2) State 2 State 2 State 4: (0.5 0.3)→ → × = 0.15 Total 0.17

APV = 20.17 1000 154.20

1.05×

=

Total. APV = 285.71 + 154.20 = 439.91

Question #13 Key: C ( )( ) ( )

( )9 9 9 10

9 10 10

1.03 1x x

x

V P q b q V

q b V V+ +

+

+ = + −

= − +

( )( ) ( ) 10343 1.03 0.02904 872 V= + 10 327.97V⇒ =

( )10 10 872 327.97 1199.97

1 1 0.03120014.65976 1.03x

b b V V

P b da

= − + = + =

⎛ ⎞ ⎛ ⎞= − = −⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

= 46.92 9V = initial reserve – P = 343 – 46.92 = 296.08 Question #14 Key: B d = 0.06 ⇒ V = 0.94 Step 1 Determine xp

( )

( ) ( )( ) ( ) ( )( )

21 1668 258 1000 1000

668 258 0.94 1000 0.94 1 1000 0.8836 1

668 242.52 940 1 883.6

x x x x x

x x x

x x x

vp vq v p p q

p p p

p p p

+ ++ = + +

+ = − +

+ = − +

272 / 298.92 0.91xp = = Step 2 Determine :21000 xP

( )( ) ( )( )[ ]

:2

:2

668 258 0.94 0.91 1000 1 0.94 0.91

220.69 6681000 479

1.8554

x

x

P

P

+ = +⎡ ⎤⎣ ⎦+

= =

Question #15 Key: D

( ) ( ) ( ) [ ]

( )

( )

1 1 10 140 9 41 50 40:1040:10 41:10

10

100,000 100,000 10 100,000 see comment

8,950,901100.99722 9,287,264100,000 0.16736 0.00592

1.06 1.06

0.02766 100,000

IA v p IA v p q A⎡ ⎤= − +⎣ ⎦⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟

⎝ ⎠⎢ ⎥= − ×⎢ ⎥⎢ ⎥⎣ ⎦

+ ×

=15,513 Where 1

40 10 40 5040:10A A E A= −

( )( )0.16132 0.53667 0.249050.02766

= −

=

Comment: the first line comes from comparing the benefits of the two insurances. At each of age 40, 41, 42,…,49 ( )140:10IA provides a death benefit 1 greater than ( )141:10IA . Hence the 1

40:10A

term. But ( )141:10IA provides a death benefit at 50 of 10, while ( )140:10IA provides 0. Hence a

term involving 41 9 41 509 q p q= . The various v’s and p’s just get all actuarial present values at

age 40.

Question #16 Key: A

( ) ( )1 11000 1 1000 1000x x xV i q Vπ= + − −

( ) ( )40 80 1.1 1000 4088 40 0.05

960

x

x

q

q

= − −

−= =

( )( )

1G expenses 1 1000 x

x

i qAS

p− + −

=

( )( )( )( ) ( )( )

( )

100 0.4 100 1.1 1000 0.051 0.05

60 1.1 5016.8

0.95

− −=

−−

= =

Question #17 Key: B (Referring to the number of losses, X, was a mistake. X is the random variable for the loss amount, the severity distribution). Losses in excess of the deductible occur at a Poisson rate of ( )( )* 1 30 0.75 20 15Fλ λ= − = × =

( ) 70 25 4530 30 600.75 0.75

E X X −− > = = =

( ) ( )( )2* 30 30Var S E X Xλ= × − >

( ) ( )( )2 215 60 900 30 15 60 30 900 30E X X X E X X X= − + > = − − − >

( )15 9,000 60 60 90067,500

= − × −

=

Question #18 Key: A

S ( )3S+

− ( ) [ ] ( ) ( ) ( )3 3 3 0 2 1 1 2S S SE S E S f f f+⎡ ⎤− = − + + +⎣ ⎦

0 0 [ ] [ ]2 0.6 2 0.4 2.8E S = × + × = 1 0 ( ) 20Sf e−= 2 0 ( ) ( )2 21 2 0.6 1.2Sf e e− −= × × = 3 0

( ) ( ) ( )2 2

22 222 2 0.4 0.6 1.522S

ef e e−

− −= × + × =

4 1 5 2 6 3

( ) 2 2 2

2

3 2.8 3 3 2 1.2 1 1.52

0.2 6.920.7365

E S e e e

e

− − −+

−

⎡ ⎤− = − + × + × + ×⎣ ⎦

= − +=

Question #19 Key: C

Write (i) as 1

k

k

p ccp k+

= +

This is an (a, b, 0) distribution with a = b = c. Which? 1. If Poisson, a = 0, so c = 0 and b = 0

1 2 ... 0p p= = =

0 0.5p = 'kp s do not sum to 1. Impossible. Thus not Poisson

2. If Geometric, b = 0, so c = 0 and a = 0 By same reasoning as #1, impossible, so not Geometric.

3. If binomial, a and b have opposite signs. But here a = b, so not binomial. 4. Thus negative binomial.

( )( ) ( )

/ 1 111 / 1 1

ab r r

β ββ β+

= = =− − −

so r = 2 ( ) ( ) 2

0 0.5 1 1rp β β− −= = + = +

1 2 1.414

2 1 0.414

β

β

+ = =

= − =

( )/ 1 0.29c a β β= = + =

Question #20 Key: C At any age, ( )1 0.02 0.9802xp e−′ = =

( )1 1 0.9802 0.0198xq′ = − = , which is also ( )1xq , since decrement 2 occurs only at the end of the

year. Actuarial present value (APV) at the start of each year for that year’s death benefits = 10,000*0.0198 v = 188.1

( ) 0.9802*0.96 0.9410xp τ = = ( ) 0.941 0.941*0.95 0.8940x xE p v vτ= = = =

APV of death benefit for 3 years 40 40 41188.1 *188.1 * *188.1 506.60E E E+ + =

Question #21 Key: B

( )

40

3030:40040

02 40

0

3030

2 3080040

3027.692

te p dt

t dt

tt

ωω

ω

ω

=

− −=

−

= −−

= −−

=

∫

∫

95ω =

Or, with De Moivre’s law, it may be simpler to draw a picture: 0 30p = 1 40 30p 30 70

( )40 3030:40

1area =27.692 40

2p

e+

= =

40 30 0.3846p =

70 0.384630

95

ωω

ω

−=

−=

3065

65ttp −

=

( ) ( )( )22Var E T E T= −

One way to evaluate this expression is based on Equation 3.5.4 in Actuarial Mathematics

( ) 20

2 t x xVar T t p dt e∞

= −∫

65 65

2

0 0

2 1 165 65t tt dt dt

⎛⎛ ⎞ ⎛ ⎞= − − −⎜⎜ ⎟ ⎜ ⎟⎜⎝ ⎠ ⎝ ⎠⎝∫ ∫i

( ) ( )22* 2112.5 1408.333 65 65/ 21408.333 1056.25 352.08

= − − −

= − =

Another way, easy to calculate for De Moivre’s law is

( ) ( ) ( )( )22

0 0

265 6520 0

1 165 65

t x x t x xVar T t p t dt t p t dt

t dt t dt

µ µ∞ ∞

= −

⎛ ⎞= × − ×⎜ ⎟⎝ ⎠

∫ ∫

∫ ∫

23 265 65

0 03 65 2 65t t⎛ ⎞

= − ⎜ ⎟× ×⎝ ⎠

( )21408.33 32.5 352.08= − = With De Moivre’s law and a maximum future lifetime of 65 years, you probably didn’t need to

integrate to get ( )( ) 3030 32.5E T e= = Likewise, if you realize (after getting 95ω = ) that ( )30T is uniformly on (0, 65), its variance is just the variance of a continuous uniform random variable:

( )265 0352.08

12Var

−= =

Question #22 Key: E

( ) ( )1

218.15 1.06 10,000 0.0231.88

1 0.02V

−= =

−

( )( ) ( )( )2

31.88 218.15 1.06 9,000 0.02177.66

1 0.021V

+ −= =

−

Question #23 Key: D

2

10.95 0.95 ...x y t x

ke e p

∞

=

= = = + +∑

0.95 191 0.95

= =−

2 ...xy xy xye p p= + +

( )( ) ( ) ( )2 21.02 0.95 0.95 1.02 0.95 0.95 ...= + +

( )22 4

21.02 0.95

1.02 0.95 0.95 ... 9.441521 0.95

⎡ ⎤= + + = =⎣ ⎦ −

28.56x y xyxye e e e= + − =

Question #24 Key: A Local comes first. I board So I get there first if he waits more than 28 – 16 = 12 minutes after the local arrived. His wait time is exponential with mean 12 The wait before the local arrived is irrelevant; the exponential distribution is memoryless Prob(exp with mean 12>12) = = =

− −e e1212 1 36 8%.

Question #25 Key: E This problem is a direct application of Example 5.18 in Probability Models (p. 308); it follows from proposition 5.3 (p. 303). Deer hit at time s are found by time t (here, t = 10) with probability F(t – s), where F is the exponential distribution with mean 7 days. We can split the Poisson process “deer being hit” into “deer hit, not found by day 10” and “deer hit, found by day 10”. By proposition 5.3, these processes are independent Poisson processes. Deer hit, found by day 10, at time s has Poisson rate 20×F(t – s). The expected number hit and found by day 10 is its integral from 0 to 10.

( )( ) ( )0

20t

E N t F t s ds= −∫

( )( )( )

( )107

10107

0

10 107

0

10 20 1

20 10 7

20 10 7 7 94

s

s

E N e ds

e

e

− −

−

−

= −

⎛ ⎞= −⎜ ⎟⎜ ⎟

⎝ ⎠

= − + =

∫

Question #26 Key: E

0.080

12.5txa e dt

∞ −= =∫

( )

( )

0.080

2 0.130

30.03 0.3758

30.03 0.2307713

tx

tx

A e dt

A e dt

∞ −

∞ −

= = =

= = =

∫

∫

( ) ( ) ( )2 222

1Var 400 0.23077 0.375 6.0048x xT Ta a A Aσδ

⎡ ⎤ ⎡ ⎤⎡ ⎤= = − = − =⎣ ⎦ ⎢ ⎥ ⎣ ⎦⎣ ⎦

( )Pr Pr 12.5 6.0048xT T Ta a a aσ⎡ ⎤ ⎡ ⎤> − = > −⎣ ⎦⎣ ⎦

0.051Pr 6.4952 Pr 0.675240.05

TTv e−

⎡ ⎤− ⎡ ⎤= > = >⎢ ⎥ ⎣ ⎦⎣ ⎦

[ ]ln 0.67524Pr Pr 7.853740.05

T T−⎡ ⎤= > = >⎢ ⎥⎣ ⎦

0.03 7.85374 0.79e− ×= = Question #27 Key: A

( ) ( )( )0.05 5 0.255 50 0.7788p e eτ − −= = =

( ) ( ) ( ) ( ) ( )551 1 0.03 0.02 0.05

5 55 550 00.02 / 0.05t tq t e dt eµ − + −= × = −∫

( )0.250.4 1 e−= −

= 0.0885 Probability of retiring before ( ) ( )1

5 50 5 5560 p qτ= × = 0.7788*0.0885 = 0.0689

Question #28 Key: C Complete the table:

[ ] [ ]81 80 80 910l l d= − =

[ ] [ ]82 81 81 830l l d= − = (not really needed)

12

x xe e= + 1 since UDD2

⎛ ⎞⎜ ⎟⎝ ⎠

[ ] [ ]1

2x xe e= +

[ ][ ]

81 82 83

80

12x

l l lel

⎡ ⎤+ + +⎢ ⎥= +⎢ ⎥⎣ ⎦

…

[ ] [ ] 81 8280 8012

e l l l⎡ ⎤⎢ ⎥− = + +⎢ ⎥⎣ ⎦

…Call this equation (A)

[ ] [ ] 8281 8112

e l l⎡ ⎤⎢ ⎥− = +⎢ ⎥⎣ ⎦

…Formula like (A), one age later. Call this (B)

Subtract equation (B) from equation (A) to get

[ ] [ ] [ ] [ ]81 80 80 81 811 12 2

l e l e l⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= − − −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

[ ] [ ]81910 8.5 0.5 1000 0.5 920e⎡ ⎤= − − −⎢ ⎥⎣ ⎦

[ ]818000 460 910 8.21

920e + −

= =

Alternatively, and more straightforward,

[ ]

[ ]

80

81

81

910 0.911000830 0.902920

830 0.912910

p

p

p

= =

= =

= =

[ ] [ ] [ ]80 8180 801 12

e q p e⎛ ⎞= + +⎜ ⎟⎝ ⎠

where [ ]80q contributes 12

since UDD

( ) ( ) 81

81

18.5 1 0.91 0.91 12

8.291

e

e

⎛ ⎞= − + +⎜ ⎟⎝ ⎠

=

( )

81 81 81 82

82

82

1 12

18.291 1 0.912 0.912 12

8.043

e q p e

e

e

⎛ ⎞= + +⎜ ⎟⎝ ⎠

⎛ ⎞= − + +⎜ ⎟⎝ ⎠

=

[ ] [ ] [ ]

( ) ( )( )

8281 81 811 121 1 0.902 0.902 1 8.04328.206

e q p e⎛ ⎞= + +⎜ ⎟⎝ ⎠

= − + +

=

Or, do all the recursions in terms of e, not e , starting with [ ]80 8.5 0.5 8.0e = − = , then final step

[ ] [ ]81 81 0.5e e= +

Question #29 Key: A

t x tp + t xp tv tt xv p

0 0.7 1 1 1 1 0.7 0.7 0.95238 0.6667

2 − 0.49 0.90703 0.4444

3 − − – −

From above 2

:30

2.1111tt xx

ta v p

=

= =∑

2:12 :3

:3

11000 1000 1 1000 1 5262.1111

xx

x

aV

a+

⎛ ⎞ ⎛ ⎞= − = − =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

Alternatively,

:3:3

1 0.4261xx

P da

= − =

( )( )

2 :3 :31000 1000

1000 0.95238 0.4261526

x xV v P= −

= −

=

You could also calculate :3xA and use it to calculate :3xP .

Question #30 Key: E Let G be the expense-loaded premium. Actuarial present value (APV) of benefits = 501000A . APV of expenses, except claim expense = 5015 1 a+ × APV of claim expense = 5050A (50 is paid when the claim is paid) APV of premiums = 50G a Equivalence principle: 50 50 50 501000 15 1 50Ga A a A= + + × +

50 50

50

1050 15A aGa+ +

=

For De Moivre’s with ω = 100, x = 50 5050 0.36512

50a

A = =

5050

1 13.33248Aad−

= =

Solving for G, G = 30.88 Question #31 Key: B The variance calculation assumes independence, which should have been explicitly stated. ( ) ( ) ( )( ) ( ) ( ) ( )2Var Var( ) Var

E S E N E X

S E N X E X N

=

= +

( )E N ( )Var N ( )E X ( )Var X E(S) ( )Var S P.B 30 21 300 10,000 9,000 62.19 10× S.B 30 27 1000 400,000 30,000 639 10× L.Y 30 12 5000 2,000,000 150,000 6360 10× 189,000 6400 10

(rounded)×

Standard deviation = 6400 10 20,000× = 189,000 + 20,000 = 209,000

Question #32 Key: B

( )150 1 0.2 0.8XS = − =

( ) ( )( )

150150p

X

XY

f yf y

S+

= So ( ) 0.250 0.250.8pYf = =

( ) 0.6150 0.750.8pYf = =

( ) ( )( ) ( )( )0.25 50 0.75 150 125pE Y = + =

( ) ( )( ) ( )( )2 220.25 50 0.75 150 17,500pE Y⎡ ⎤ = + =⎢ ⎥⎣ ⎦

( ) ( ) ( ) 22 2Var 17,500 125 1875p p pY E Y E Y⎡ ⎤ ⎡ ⎤= − = − =⎢ ⎥ ⎣ ⎦⎣ ⎦

Slight time saver, if you happened to recognize it:

( ) ( )50p pVar Y Var Y= − since subtracting a constant does not change variance,

regardless of the distribution But 50pY − takes on values only 0 and 100, so it can be expressed as 100 times a binomial random variable with n = 1, q = 0.75

( )( )( )( )2100 1 0.25 0.75 1875Var = =

Question #33 Key: A

( )( )0.05 44 50 0.8187p e−= =

( )( )0.05 1010 50 0.6065p e−= =

( )( )0.04 88 60 0.7261p e−= =

( )( )18 50 10 50 8 60 0.6065 0.72610.4404

p p p= = ×

=

50 4 50 18 50414 0.8187 0.4404 0.3783q p p= − = − =

Question #34 Key: A Model Solution: X denotes the loss variable.

1X denotes Pareto with α = 2 ; X 2 denotes Pareto with α = 4

1 2(200) 0.8 (200) 0.2 (200)X X XF F F= +

2 4100 30000.8 1 0.2 1

200 100 3000 200

⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞= − + −⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

2 41 151 0.8 0.2

3 16⎛ ⎞ ⎛ ⎞= − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

0.7566=

Question #35 Key: D

( )( )40 5 40 4540:5

14.8166 0.73529 14.11214.4401

a a E a= −

= −

=

( )1545 5 4040:5 40:5100,000a A v p IAπ π= +

( )( )( )( ) ( )

145 5 40 40:5 40:5100,000 /

100,000 0.20120 0.73529 / 4.4401 0.040423363

A E a IAπ = × −

= −

=

Question #36 Key: B Calculate the probability that both are alive or both are dead. P(both alive) = k xy k x k yp p p= ⋅ P(both dead) = k k x k yxyq q q=

P(exactly one alive) = 1 k xy k xyp q− −

Only have to do two year’s worth so have table

Pr(both alive) Pr(both dead) Pr(only one alive) 1 0 0 (0.91)(0.91) = 0.8281 (0.09)(0.09) = 0.0081 0.1638 (0.82)(0.82) = 0.6724 (0.18)(0.18) = 0.0324 0.2952

0 1 2 0 1 21 0.8281 0.6724 0 0.1638 0.295230,000 20,000 80,431

1.05 1.05 1.05 1.05 1.05 1.05APV Annuity ⎛ ⎞ ⎛ ⎞= + + + + + =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

Alternatively,

20.8281 0.67241 2.39861.05 1.05xya = + + =

20.91 0.821 2.61041.05 1.05x ya a= = + + =

20,000 20,000 10,000x y xyAPV a a a= + − (it pays 20,000 if x alive and 20,000 if y alive, but 10,000 less than that if both are alive) ( )( ) ( )( ) ( )20,000 2.6104 20,000 2.6104 10,000 2.3986= + − 80,430= Other alternatives also work.

Question #37 Key: C Let P denote the contract premium.

0.05

0 0

20t t txP a e e dt e dtδ µ

∞ ∞− − −= = = =∫ ∫

[ ] IMPxE L a P= −

( ) ( )10

0.03 10 0.02 100.03 0.02 0.03 0.01

0 0

IMP t t t txa e e dt e e e e dt

∞− −− − − −= +∫ ∫

0.5 0.5

230.05 0.04

l e e− −−= + =

[ ] 23 20 3E L = − =

[ ] 3 15%20

E LP

= =

Question #38 Key: C

( ) ( )( )

1 230 30130:2

2

1000 500

1 11000 0.00153 500 0.99847 0.001611.06 1.06

2.15875

A vq v q= +

⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

=

Initial fund 2.15875 1000= × participants = 2158.75 Let nF denote the size of Fund 1 at the end of year n.

( )1 2158.75 1.07 1000 1309.86F = − =

( )2 1309.86 1.065 500 895.00F = − = Expected size of Fund 2 at end of year 2 = 0 (since the amount paid was the single benefit premium). Difference is 895.

Question #39 Key: B Let c denote child; ANS denote Adult Non-Smoker; AS denote Adult Smoker.

( ) ( )

( ) ( )

( ) ( )

3 3

1

3 4

33 0.3 0.0673!

13 0.6 0.0373!

43 0.1 0.0203!

eP c P c

eP ANS P ANS

eP AS P AS

−

−

−

= × =

= × =

= × =

( ) ( )0.0203 0.16

0.067 0.037 0.020P AS N = = =

+ +

Question #40 Key: C [ ] [ ] [ ] 3 10 30E S E N E X= = × =

( ) [ ] ( ) [ ] ( )2

4003 100 3.6 100 360 46012

Var S E N Var X E X Var N= +

= × + × = + =

For 95th percentile, [ ] ( )1.645 30 460 1.645 65.28E S Var S+ = + × =

Documents

**BEGINNING OF EXAMINATION**For a fully discrete whole life insurance of 100,000 on (35) you are given: (i) Percent of premium expenses are 10% per year. (ii) Per policy expenses are

BEGINNING OF EXAMINATIONFor a fully discrete whole life insurance of 100,000 on (35) you are given: (i) Percent of premium expenses are 10% per year. (ii) Per policy expenses are