i 0.06 ªº ªº «»¬ ¼ ¬ ¼ AE A E A E 1000 i G P a E a a E a

March 23, 2015 Copyright Jeff Beckley 2015

1. A 20 year endowment insurance of 1000 on (60) has level annual benefit premiums payable at the

beginning of each year for 10 years. The death benefit is payable at the moment of death.

You are given that mortality follows the Illustrative Life Table with 0.06i and that deaths are

uniformly distributed between integral ages.

Calculate the annual net premium.

Solution:

60:2060:10

1

60:20 20 60 60 20 60 80 20 60

60 10 60 70 60 10 60 70

1000

1000 1000

1000 1.02971 0.36913 (0.14906)(0.66575) 0.1490658.66

11.1454 (0.45120)(8.5693)

Pa A

iA E A E A EP

a E a a E a


2. A fully discrete whole life policy on (25) provides a death benefit of 1 million until age 65 and a

death benefit of 0.5 million after age 65.

You are given that mortality follows the Illustrative Life table with 6%.i

Level annual net premiums are payable until age 65 at which time premiums stop.

Calculate the level annual net premiums.

Solution:

1

2525:40 25:40

25 40 25 65 25 25 40 25 65

500,000( )

500,000( )

500,000((2)(0.08165 (0.29873)(0.25634)(0.43980))

165.2242 (0.29873)(0.25634)(9.8969)

64,810.844190.45

15.46633

PVP PVB

Pa A A

P a E a A A E A

P

P


3. A special fully discrete whole life insurance is issued to (80). The special whole life has a death

benefit of 1000.

The annual net premiums for this whole life are level for the first 10 years. The annual net

premiums after 10 years are also level but the annual net premium after the tenth year is twice the

annual net premium during the first ten years.

You are given that mortality follows the Illustrative Life Table and 0.06i .

Calculate the annual benefit premium after 10 years.

Solution:

80 10 80 90 80

Let be the premium during the first ten years and 2 be the premium thereafter.

[ ] 1000

1000(0.66575)124.35

5.9050 (0.15100)(3.6488)

2 248.70

P P

PVP PVB

P a E a A

P

P


4. A fully discrete 20 year term insurance on (70) provides a death benefit of 100,000 for the first 10

years and a death benefit of 50,000 during the last 10 years.

Annual net premiums are payable for 20 years. The annual net premium for the first five years is

5000. The annual net premium for the last 15 years is equal to .

You are given that mortality follows the Illustrative Life Table with 0.06i .

Calculate .

Solution

1 1

70:20 70:10

70 20 70 90 70 10 70 80

70:20 70:5

70 20 70

50,000 50,000

50,000[ ] 50,000[ ]

50,000[0.51495 (0.04988)(0.79346)] 50,000[0.51495 (0.33037)(0.6655)]

38,518.92

(5000 )

(

PVB A A

A E A A E A

PVP a a

a E

90 70 5 70 75 70 5 70 75

5 70 75 20 70 90 70 5 70 75

) 5000( )

( ) 5000( )

((0.60946)(7.2170) (0.04988)(3.6488)) 5000((8.5693) (0.60946)(7.2170))

20,854.13 4.216470676

20,854.13 4.

a a E a a E a

E a E a a E a

216470676 38,518.92 4189.47


5. A special fully discrete 3 year endowment insurance on (76) has an annual net premium of that

is payable for two years. The death benefit is 40,000 plus the sum of the premiums paid at the

end of the year of death. At the end of three years, the policy endows for 40,000 plus the sum of

the premiums paid. (Note that this means the death benefit in the first year is 40,000 + and

that the death benefit in the second and third year as well as the endowment benefit is 40,000 +

2 .)

You are given that mortality follows the table immediately below and 0.90v .

x xl

75 2000

76 1600

77 960

78 480

79 120

80 0

Calculate .

Solution:

2 3

2 30 2 3

1600 960 640 (40,000 ) 480 (40,000 2 ) 480 (40,000 2 )

[1600 960 640 (2)(480) (2)(480) ] (40,000)[640 480 480 ]

52,5800,800128,090.41

410.56

PVP PVB

v v v v

v v v v v v v


6. Emily (25) purchases a special whole life insurance policy with increasing death benefits. The

death benefit is 50,000 for the first 10 years. It doubles at the end of 10 years to a death benefit

of 100,000 if Emily dies between the end of the 10th year and the end of the 20th year. Finally, it

doubles again at the end of the 20th year so the death benefit is 200,000 if Emily dies after the

end of the 20th year.

The death benefit is payable at the end of the year of death.

The policy has annual net benefit premiums that are non-level and payable for life. The net

benefit premiums is P for the first 20 years and then is 0.5P thereafter.

You are given :

a. Mortality follows the Illustrative Life Table.

b. 6%i .

Calculate P using the equivalence principle.

Solution:

25 10| 25 20| 25

25 10 25 35 20 25 45

25

50,000 50,000 100,000

50,000 50,000 100,000

(50,000)(0.08165) (50,000)(0.54997)(0.12872) (100,000)(0.29873)(0.20120) 13,632.55452

0.5 0.5

PVP PVB

PVB A A A

A E A E A

PVP Pa P

25 25 20 25 4525:200.5 ( )

0.5 [16.2242 16.2242 (0.29873)(14.1121)] 14.11634618

13,632.55452965.73

14.11634618

a P a a E a

P P

P


7. (Written Answer) Peterson Pet Insurance Company has developed the following life insurance

table for dogs:

Age xl Age

xl

0 2000 5 1200

1 1950 6 1000

2 1850 7 700

3 1600 8 300

4 1400 9 0

Peterson sells a pet insurance policy on a dog who is age 5. The policy pays a death benefit at

the end of the year of death of 1000. Level annual premiums are payable for the life of the dog.

The expenses associated with the policy are 10% of premium at the beginning of each year.

The interest rate used to calculate all values is 4%.

A. (7 points) Calculate the gross premium for this insurance using the Equivalence Principle.

Solution:

2 3 2 3 4 2 3

2 3 4

2 3

(1200 1000 700 300 ) 1000(200 300 400 300 ) 0.1 (1200 1000 700 300 )

1000(200 300 400 300 )390.81

0.9(1200 1000 700 300 )

PVP PVB PVE

P v v v v v v v P v v v

v v v vP

v v v


B. (5 points) Calculate the loss that Petereson will incur if the dog dies in the second year if the

gross premium is calculated using the Equivalence Principle.

Solution:

1 1

0 1 1 1

1 1

0 1 1

1000 0.1(390.81)a (390.81)a 1000 0.9(390.81) a

For 1 which means the dog dies in the second year

1000 0.9(390.81)a 234.63

x x

x x x

K Kg

K K K

x

g

L v v

K

L v

C. (5 points) Peterson decides to charge a gross premium so that the loss will be zero if the dog

dies in the second year. Determine the gross premium that Peterson decides to charge.

Solution:

0

1 1

0 1 1

2

2

We want 0 for K 1

1000 0.9( )a 0

1000523.71

0.9(a )

g

x

g

L

L v P

vP


D. (7 points) Estimate the monthly premium equivalent to the annual premium in Part C. by

using the two term Woolhouse formula.

Solution:

(12)

5 5

5 5

2 3

5

( ) ( )(12)( )

12 1(523.71)( ) ( )(12)

(2)(12)

(1200 1000 700 300 )2.562855599

1200

(523.71)(2.562855599)

12 1(12) 2.562855599

(2)(12)

13

Annual Monthly

Monthly

Monthly

P a P a

a P a

v v va

P

42.1953.15

25.25426719


8. Hassan, age 35, purchases a whole life policy with a death benefit of 1 million payable at the

moment of death. He will pay level annual premiums during his lifetime.

You are given:

a. Mortality follows the Illustrative Life Table

b. 6%i

c. Death are uniformly distributed between integer ages.

d. Expenses occurring at the beginning of the year are:

i. First year expenses of 800 per policy and 2 per 1000 of death benefit.

ii. Renewal expenses (occurring at the start of the second year and every year

thereafter) are 60 per policy.

iii. Commissions of 70% of premiums in the first year and 8% of premiums

thereafter.

iv. Premium Tax of 3% of premium.

e. The company will incur a termination expense at the moment of death of 1000.

Calculate the gross premium for this policy using the Equivalence Principle.

Solution:

35 35 35 35 35 35

35 35 35

35 35 35

35 35

35

1,000,000 740 (1000)(2) 60 0.62 0.08 0.03 1000

1,000,000 740 (1000)(2) 60 1000

0.62 0.08 0.03

1,001,000 2,740 60

(0.89)

PVP PVB PVE

Pa A a P Pa Pa A

A a AP

a a a

A a

a

1,001,000(1.02971)(0.12872) 2,740 60(15.3926)

0.62 (0.89)(15.3926) 0.62

10,424.04


9. Rafidah purchases a fully continuous whole life insurance policy with a death benefit of 50,000

payable at the moment of death. Level net premiums are paid continuously for as long as

Rafidah is alive.

You are given:

a. Rafidah is now age 50.

b. Mortality follows the Illustrative Life Table

c. 6%i

d. Deaths are uniformly distributed between integer ages.

Calculate the Variance for the loss random variable, 0

nL , for this policy.

Solution:

50 50 50 505050

50 5050 50

50,000( / )50,000 50,000 50,00050,000

1 ( / )1 1

50,000(0.06)(0.24905)1004.84

1 (1.02971)(0.24905)

PVP PVB

i AA A APa A P

a i AA A

2 2 2

2 2 2 2

50 50

1004.84 (1.06) 1( ) 50,000 0.09476 (1.02971) (0.24905)

0.05827 2(0.05827)

157,062,748.5

PVar S A A


10. You are given the following mortality table:

x xl

90 1000

91 900

92 700

93 400

94 100

95 0

For a whole life to (92) with a death benefit of 50,000 payable at the end of the year of death

and level annual premiums, the expenses are 400 per policy at issue and 50 per policy at the

beginning of each year including the first year.

You are given that 5%i .

a. Calculate the level gross premium using the equivalence principle.

Solution:

2 2 3 2

1 2 3 2

1 2

(700 400 100 ) 50,000(300 300 100 ) 400(700) 50(700 400 100 )

50,000(300(1.05) 300(1.05) 100(1.05) ) 400(700) 50(700 400 100 )

700 400(1.05) 100(1.05)

27,780.29

P v v v v v v v

v vP

b. Complete the following table:

xK 0

gL Probability

0 50,000/1.05+450-27,780.29

=20,288.76

300/700=3/7

1 50,000/1.052+400

-(27,780.29-50)(1+1.05-1)

=-8,388.62

300/700=3/7

2 50,000/1.053+400

-(27,780.29-50)(1+1.05-1+1.05-2)

=-35,700.40

100/700=1/7

c. Calculate the variance of the loss at issue random variable.


2 2

0

2[ ] (20,288.76) (3 / 7) ( 8388.62) ( 35,700.40) (1/ 73 / )7) (

388,646,614

Var L

d. Calculate the expected value and the variance of the loss at issue random variable if

the gross premium was 30,000.

Solution:

xK 0

gL Probability

0 50,000/1.05+450-30,000

=18,069.05

300/700=3/7

1 50,000/1.052+400

-(30,000-50)(1+1.05-1)

=-12,722.34

300/700=3/7

2 50,000/1.053-

30,000(1+1.05-1+1.05-2)

=-42,047.46

100/700=1/7

2

2

0

0

2 2 2

0

[ ] (18,069.05)(3 / 7) ( 12,722.34)(3 / 7) (

[ ] (17,619.05) (3 / 7) (

42,047.46)(1/ 7)

3715.33

42,590.43) (1/ 7)

461,862,

13,219.95) (3

057

[ ] 461,862,057 ( 3715.33) 448,058,3

7 (

8

)

0

/

Var

E L

L

E L


11. Chao (30) buys a 40-year endowment insurance with a death benefit of 40,000 from Li Life

Insurance Company. The death benefit is payable at the end of the year of death.

You are given :

a. Mortality follows the Illustrative Life Table.

b. 6%i .

c. Deaths are uniformly distributed between integral ages.

d. The net annual benefit premium for this policy is 354.33.

A. Calculate the variance of the loss random variable 0

nL for this policy with annual premiums.

B. If the death benefit on the above policy was payable at the moment of death and the

premiums were paid monthly, calculate what the monthly net benefit premium would be

under the equivalence principle.

Solution:

Part A

2

2 2

0 30:40 30:40

30 40 30 70 40 3030:40

2 2 40 2 40

30 40 30 70 40 3030:40

40

[ ] ( )

0.10248 (0.29374)(0.23047)(0.51495) (0.29374)(0.23047) 0.13531704

0.02531 (1.06) (0.29374)(0.23

n PVar L S A A

d

A A E A E

A A v E A v E

40

2

2

0

047)(0.30642) (1.06) (0.29374)(0.23047) 0.029874986

354.33[ ] 40,000 0.029874986 (0.13531704) 24,747,243.94

0.06 /1.06

nVar L


Part B:

(12) (12) (12)30:40 30 40 30 70 30 40 30 70 40 3030:40

30 40 30 70 30 40 30 70 40 30

30 4

12 40,000 12 [ ] 40,000[ ]

12 { (12) (12) [ (12) (12)]} 40,000[( / ) ( / ) ]

40,000[( / )

PVP PVB

Pa A P a E a A E A E

P a E a i A E i A E

i AP

0 30 70 40 30

30 40 30 70

( / ) ]

12{ (12) (12) [ (12) (12)]}

40,000[(1.02971)(0.10248) (0.29374)(0.23047)(1.02971)(0.51495) (0.29374)(0.23047)]

12{(1.00028)(15.8561) 0.46812 (0.29374)(0.23047)[(1

E i A E

a E a

30.84

.00028)(8.5693) 0.46812]}


12. A fully discrete whole life insurance policy on (70) pays a death benefit of 100,000. The policy

has annual net premiums determined by the equivalence principal.

You are given:

a. Mortality follows a 2 year select and ultimate mortality table such that

i. [ ] 0.5 where is based on the Illustrative Life Table.x x xq q q

ii. [ ] 1 1 10.8 where is based on the Illustrative Life Table.x x xq q q

iii. Ultimate mortality follows the Illustrative Life Table

iv. 0.06i


Solution:

2 2

[70] [70] [70] [70] 1 [70] [70] 1 72

2 2

70 70 70 1 70 70 1 72

We need to find values in terms of the ILT.

(0.5)( ) [1 (0.5)( )](0.8)( ) [1 (0.5)( )][1 (0.8)( )]

(0.5)(0.03318) [1 (0.5)

1.06

PVP PVB

A vq v p q v p p A

v q v q q v q q A

2

2

[70]

[70]

[70]

[70

(0.03318)](0.8)(0.03626)

(1.06)

[1 (0.5)(0.03318)][1 (0.8)(0.03626)](0.54560)

(1.06)

0.504714

1 1 0.5047148.750056

0.06 1.06

100,000

Aa

d

AP

a

]

50,471.405768.12

8.750056


13. You are given:

i. 40 14a

ii. 60 9a

iii. 40:20

11.75a

iv. 6%i

v. Deaths is uniformly distributed between integral ages.

For a fully discrete 20 year term insurance on (40) with a death benefit of 56,000, the net

premium is paid quarterly during the life of the policy.

Calculate quarterly net premium.

Solution:

1

(4) 1 40:20 40 20 40 60

(4)40:20 40:2020 4040:20 40:20

40 20 40 60 20 40 20 4040:20

40 40

56,000 14,000[ ]4 56,000

4 (4) (4)(1 )

14 11.7511.75 14 (9) 0.25

9

0.01 1

PVP PVB

A A E APa A P

a a E

a a E a E E

A da

60 60

6(14) 0.20754717

1.06

0.061 1 (9) 0.490566038

1.06

14,000[0.20754717 (0.25)(0.20754717)]103.40

(1.00027)(11.75) (0.38424)(1 0.25)

A da

P


14. Surin purchases a fully discrete 2 year term insurance with a death benefit of 10,000. Surin is

(80).

You are given that:

i. 80 0.08q

ii. 81 0.12q

iii. 0.9v

0

nL is the loss-at-issue random variable based on the net premium.

Calculate the 0[ ]nVar L .

Solution:

2

1

0

0

(1 ) 10,000( )

10,000[(0.9)(0.08) (0.81)(0.92)(0.12)]883.06

1 (0.9)(0.92)

There are three possible events:

Die Year 1==> (10,000)(0.9) 883.06 8116.94 Prob=0.08

Die Year 2==> (10,

x x x xP vp vq v p q

P

L

L

0

0

2 2 2

0 0

000)(0.81) 883.06(1 0.9) 6422.19 Prob=(0.92)(0.12)=0.1104

Live 2 years==> 0 883.06(1 0.9) 1677.81 Prob=(0.92)(0.88)=0.8096

[ ] 0

[ ] [ ] (0.08)(8116.94) (0.1104)(6422.19) (0.8096)(

L

E L

Var L E L

21677.81)

12,103,234


15. Matt who is (45) purchases a deferred annuity which will pay benefits beginning at age 65. The

annuity will pay 10,000 at the beginning of each year if Matt is alive. If Matt dies during the

deferral period, no benefits will be paid.

Matt will pay annual net premiums for during the 20 year deferral period.

You are given:

i. Mortality follows the Illustrative Life Table

ii. 0.06i

Calculate the annual net premium for this annuity.

Solution:

20| 4545:20

20| 45 20 45 65

45 20 45 6545:20

10,000

10,000 (10,000)( )( ) (10,000)(0.25634)(9.8969)2191.74

( )( ) 14.1121 (0.25634)(9.8969)

PVP PVB

Pa a

a E aP

a a E a


16. Matt who is (45) purchases an annuity which will pay benefits beginning at age 65. The annuity

will pay 10,000 at the beginning of each year with the first ten payments guaranteed. These first

ten payments will be paid even if Matt dies before the payments begin. The payments after ten

years will continue for as long Matt lives.


You are given:


ii. 0.06i


Solutions:

10

30| 4545:20 10

10

30| 4510

45:20

The first ten payments are guaranteed to be made so they are only discounted

for interest and not for survival.

10,000 ( )

10,000 ( ) (10,000)(

PVP PVB

Pa v a a

v a a vP

a

10

30 45 7510

45 20 45 65

)( )

( )( )

(10,000)[4.356424 (0.25634)(0.39994)(7.2170)]4402.82

14.1121 (0.25634)(9.8969)

a E a

a E a


17. Matt who is (45) purchases a deferred annuity which will pay benefits beginning at age 65. The

annuity will pay 10,000 at the beginning of each year if Matt is alive. If Matt dies during the

deferral period, his premiums plus interest will be paid as a death benefit but no annuity payments

will be made


You are given:


ii. 0.06i


Solution:

The formula below veiws the equality as of age 65. The premiums that have been

accumulated are with interest only because all those that died got their premium and

interest earned returned. Therefore

6520

65

20

, the left side is the amount that has been accumulated

for those alive at age 65. The right hand side is the present value of future benefits at age 65.

10,000

10,000 (10,000)

Ps a

aP

s

(9.8969)2538.14

38.992727


18. You are given:

i. 1000 600xA

ii. 0.05i

0

nL is the loss-at-issue random variable for a fully discrete whole life of 10,000 on (x) based on

the net premium.

Calculate the value of 0

nL if death occurs during the 20th year of the policy.

Solution:

20

0 20

20

0 20

10,000

10,000 10,000 10,000(0.600)849.06

1 1 0.600

0.06 1.06

10,000 (849.06) 3118.05 10,322.97 7204.92

n

x x

xx

n

L v Pa

A AP

Aa

d

L v a


19. For a fully discrete whole life of 100 on (65), you are given:

i. Maintenance expenses are 2.50 per policy at the start of each year

ii. Issue expenses at the start of the first policy year of 1.

iii. Mortality follows the Illustrative Life Table.

iv. 6%i

The gross premium is determined using the equivalence principle.

Calculate the variance of the loss-at-issue random variable.

Solution:

1

0 1

65 65 65

1 1

0 1 1

1

0

100 1 (2.50 )

100(0.43980) 1 2.50(9.8969)100 1 2.50 7.044857

9.8969

100 1 (2.50 7.044857) 100 1 (4.544857)

[ ] [100 1 (4.544857)

x

x

x x

x x

x

K

K

K K

K K

K

L v P a

Pa A a P

L v a v a

Var L Var v

1

11 1

1

]

1 4.544857 4.544857100 1 (4.544857) 100 1

0.061.06

4.544857100 since the Variance of a constant is zero

0.061.06

= 100

x

x

x x

x

K

KK K

K

a

vVar v Var v

d d

Var v

2 2

21 2

65 65

2 2

4.544857 4.544857[ ] 100

0.06 0.061.06 1.06

(180.292474) [0.23603 (0.43980) ] 1384.92

xKVar v A A


20. For a special fully discrete 2 year endowment insurance on (80), you are given:

i. The death benefit 10,000.

ii. The endowment benefit is a return of the premiums paid without interest.

iii. Mortality follows the Illustrative Life Table.

iv. 0.05i


Solution:

2 2

80 80 80 81 80 81

2

80 80 81

2

80 80 81

1 2

1 2

(1 ) 10,000 10,000 2

10,000 10,000

1 2

10,000[(1.05) (0.0803) (1.05) (1 0.0803)(0.08764)]

1 (1.05) (1 0.0803) (2)(1.05) (1 0.0803)(1 0.0876

PVP PVB

P vp vq v p q Pv p p

vq v p qP

vp v p p

4)

1495.8504134228.76

0.353732441

Documents

i 0.06 ªº ªº «»¬ ¼ ¬ ¼ AE A E A E 1000 i G P a E a a E a