Chapter 9 - Purdue Universityjbeckley/WD/STAT 475/S18...The net annual premium rate for a fully continuous joint whole life on (60) and (60) with a death benefit of 1000. Solution:

January 22, 2018 Copyright Jeffrey Beckley 2012, 2015, 2016, 2018

Chapter 9

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

are independent and that deaths are uniformly distributed between integral ages, calculate:

a. 10 50:60q

Solution:

60 70 7010 50:60 10 50:60 10 50 10 60

50 60 50

6,616,1551 1 1 1 1 0.26084

8,950,901

l l lq p p p

l l l

b. 10| 50:60q

Solution:

8,075,403 6,396,60961 71(1 ) (1 0.26084)(1 * ) (1 0.26084)(1 * ) 0.0343610 | 50 : 60 10 50 : 60 60 : 70 8,188,074 6,616,165

60 70

l lq p p

l l

c. 60:60A

Solution:

60:60 0.47975

d. The net annual premium for a fully discrete joint whole life on (60) and (60) with a death benefit of

1000.

Solution:

e. 60:60A

Solution:

60:60 60:60 (1.02971)(0.47975) 0.49400i


f. The net annual premium rate for a fully continuous joint whole life on (60) and (60) with a death

benefit of 1000.

Solution:

60:60

60:6060:6060:60 1

60:60

(1.02971)(0.47975)( ) 0.056887

8.68381

0.056887*1000 56.89

i

i

Pa

g. 10 50:60p

Solution:

60 70 60 7010 10 50 10 60 10 50:6050:60

50 60 50 60

8,188,074 6,616,155 6,616,1550.98364

8,950,901 8,188,074 8,950,901

l l l lp p p p

l l l l

h. Calculate the probability that the survivor of (50) and (60) dies in year 11.

Solution:

10 11 11 50 11 60 11 50 11 6050:60 50:600.98364 ( )

8,075,403 6,396,609 8,075,403 6,396,6090.98364 [ * ]

8,950,901 8,188,074 8,950,901 8,188,074

0.98364 0.97860 0.00504

p p p p p p

i. Calculate the probability that exactly one life of (50) and (60) is alive after 10 years.

Solution:

10 10 50:6050:600.98364 [1 0.26084] 0.24448p p

j. 60:70

A

Solution:

60 70 60:7060:700.36913 0.51495 0.57228 0.31180A A A A


k. The net annual premium rate for a fully discrete survivor whole life on (60) and (70) with a death

benefit of 1000. Assume that the premium is paid until the second death.

Solution:

0.0660:70 1.06

60:70

60:70

( ) 0.31180( )0.02565

1 1 0.31180

0.02565*1000 25.65

dP

l.

Solution:

m.

Solution:

60 70 60:7060:70

60:70

0.061.06

11.1454 8.5693 7.5563 12.1584

1 1 0.3118012.1582

a a a a

or

d

n.

Solution:


2. A joint annuity on (50) and (60) pays a benefit of 1 at the beginning of each year if both annuitants are alive. The

annuity pays a benefit of 2/3 at the beginning of each year if one annuitant is alive.

You are given:

i. Mortality follows the Illustrative Life Table.

ii. (50) and (60) are independent lives.

iii. 0.06i

Calculate the actuarial present value of this annuity.

Solution:

a = 23

23

b 113

c a b

3. A joint annuity on (50) and (60) pays a benefit of 1 at the beginning of each year if both annuitants are alive. The

annuity pays a benefit of 2/3 at the beginning of each year if only (50) is alive. The annuity pays a benefit of ½ at

the beginning of each year if only (60) is alive.

You are given:

i. Mortality follows the Illustrative Life Table.

ii. (50) and (60) are independent lives.

iii. 0.06i


Solution:

a = 23

b = 12

2 1 113 2 6

c


4. You are given the following mortality table:

x xl xq xp

90 1000 0.10 0.90

91 900 0.20 0.80

92 720 0.40 0.60

93 432 0.50 0.50

94 216 1.00 0.00

95 0

Assume that deaths are uniformly distributed between integral ages and the lives are independent.

Calculate at 4%i :

a. 91:92A

Solution:

l90:91 = (1000)(900) = 900,000

l91:92 = (900)(720) = 648,000

l92:93 = (720)(432) = 311,040

l93:94 = (432)(216) = 93,312

l94:95 = (216)(0) = 0

2 31 1 1

1.04 1.04 1.0491:92

(648,000 311,040)( ) (311,040 93,312)( ) (93,312)( )0.93867

648,000

b.

Solution:

1

2 31 1 11.04 1.04 1.04

90:91:3

(900,000 648,000)( ) (648,000 311,040)( ) (311,040 93,312)( )

900,000

0.83045

A

c.

Solution:


5. * You are given:

i. 3 40 0.990p

ii. 6 40 0.980p

iii. 9 40 0.965p

iv. 12 40 0.945p

v. 15 40 0.920p

vi. 18 40 0.890p

For two independent lives aged 40, calculate the probability that the first death occurs after 6 years, but

before 12 years.

Solution:

2 2

6 40:40 12 40:40 6 40 6 40 12 40 12 40 (0.98) (0.945) 0.067375p p p p p p

6. * For a last survivor insurance of 10,000 on independent lives (70) and (80), you are given:

i. The benefit, payable at the end of the year of death, is paid only if the second death occurs

during year 5.

ii. Mortality follows the Illustrative Life Table

iii. 0.03i

Calculate the actuarial present value of this insurance.

Solution:

5 5 5

4 5 4 5 5 470:80 70:80 70:80 70:80 70:80 70:80

5 75 85 74 845 70 5 80 4 70 80 5

70 80 70 80

( )(10,000) [1 (1 )](10,000) [ ](10,000)

10,000[ 4 ](10,000) [(1 )(1 ) (1 )(1 )] 234.82

(1.03)

PV v p p v q q v q q

l l l lv q q q q

l l l l


7. * For a temporary life annuity-immediate on independent lives (30) and (40):

i. Mortality follows the Illustrative Life Table

ii. 0.06i

Calculate a

30:40:10

Solution:

50

30

10

30:40 10 30:40 40:50 30:40 10 30 10 40 40:5030:40:10

10 10 8,950,9011 11.06 1.06 9,501,381

( 1) ( 1)

(14.2068 1) ( ) ( )(12.4784 1) 13.2068 ( ) (11.4784) 7.1687l

l

a a a a v p p a

8. Two lives, (x) and (y), are subject to a common shock model. You are given:

i. ( ) 0.04x t

ii. ( ) 0.06y t

iii. ( )x t and ( )y t do not reflect the mortality from the common shock.

iv. The mortality from the common shock is a constant force of 0.01.

v. 0.03

Calculate:

a. t xp

Solution:

0

0

0.050.05

0.04 0.01 0.05

t

x s

t

ds

t x

x swithout commoncommon shockshock

dst

t x

p e

p e e



b. t yp

Solution:

0

0.070.070.06 0.01 0.07

t

dst

y s t yp e e

c. t xyp

Solution:

0

0.110.11

: 0.04 0.06 0.01 0.11

t

dst

x t y t x t x t cs t xyp e e

d. xe

Solution:

0.05 0.051 100.05 0.05

0 0

( | ) (0 1) 20t tx t xe p dt e dt e

e. ye

Solution:

0.07

0

114.286

0.07

tye e dt

f. xye

Solution:

0.11

0 0

19.091

0.11

txy t xye p dt e dt


g. xye

Solution:

We cannot calculate directly, but can calculate using the box formula

20 14.286 9.091 25.195xy x y xye e e e

h. xyA

Solution:

0.11 (0.03 0.11) 0.11: 0.14

0 0 0

(0.11) (0.11) 0.78571t t t t

xy t xy x t y tv p dt e e dt e dt

i. xyA

Solution:

0.03 0.05

0

0.03 0.07

0

0.05 0.07 0.110.08 0.10 0.14

0.05(0.05) 0.625

0.08

0.07(0.07) 0.700

0.10

0.53929

x y xyxy

t t

x

t t

y

xy

e e dt

e e dt


9. You are given:

i. ( ) 0.02x t t

ii. 1( ) (40 )y t t

iii. The lives are subject to an common shock model with 0.015

iv. ( )x t and ( )y t incorporate deaths from the common shock.

Calculate 3| xyq

Solution:

1

0 0 0 0 0 0 0

2 2

( ) ( ) ( ) 0.015 0.02 (40 ) 0.015

0.01 ln(40 ) ln(40) 0.015 0.01 0.0154040

0.01(9) 373| 3 4

( ) ( ) ( ) 0.015

t t t t t t t

xy x y

xy x y

s ds s ds s ds ds sds s ds ds

t xy

t t t t tt

xy xy xy

t t t

p e e e e e e e

e e e e e

q p p e

0.045 0.01(16) 0.063640 40

0.06994e e e


10. A male age 65 and a female age 60 are subject a common shock model.

You are given:

i. Male mortality follows the Illustrative Life Table.

ii. Female mortality follows the Illustrative Life Table but for a live five years

younger. In other words, the mortality for a female age 60 is equal to the

mortality for a person 55 in the Illustrative Life Table.

iii. The above mortality rates do not reflect the common shock risk.

iv. Both lives are subject to a common shock based on a constant force of mortality

of 0.01.

v. 8%i - NOTE that it is not 6%.

Calculate:

a. 20 65E where (65) is a male.

Solution: 20 20 (0.01)(20) 20 0.22,358,2461 1

20 65 20 65 20 651.08 1.08 7,533,964( ) ( )( ) ( ) ( )( ) 0.05498m m

commontablestock

v p p e e

b. 20 60E where (60) is a female.

Solution: 20 20 (0.01)(20) 20 0.25,396,0811 1

20 60 20 60 20 551.08 1.08 8,640,861( ) ( )( ) ( ) ( )( ) 0.10970f f

commontablestock

v p p e e

c. 20 65:60E where (65) is a male and (60) is a female.

Solution:

20E65:

m

60

f = v20

20 p65:

m

60

f = v20 ( 20 p65 )maletable

( 20 p55 )femaletable

( e-0.2

commonstock

) = ( 11.08 )20 ( 2,358,246

7,533,964 )( 5,396,0818,640,861)(e-0.2 ) = 0.03434

d. 20

E65:60

where (65) is a male and (60) is a female.

Solution:

Use Box formula

20 20 65 20 65 20 65:6065:600.05498 0.10970 0.03434 0.13034m f m f m f


11. * 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 that 0.05i , 1 80 0.91p , 2 80 0.82p and 3 80 0.72p .


Solution:

80:3 80:3 80:80:3

2 1 2

80 2 8080:3

2 1 2 2 2

80:80 2 80:8080:80:3

20,000 20,000 10,000

1 1 (1.05) (0.91) (1.05) (0.82) 2.61043

1 1 (0.91) (0.82) 2.39855

40,000(2.61043) 10,000(2.3985

PV a a a

a vp v p

a vp v p v v

PV

5) 80,431.70

12. *For a special continuous joint life annuity on ( x ) and ( y ), you are given:

i. The annuity payments are 25,000 per year while both are alive and 15,000 per

year when only one is alive.

ii. The annuity also pays a death benefit of 30,000 upon the first death.

iii. 0.06i

iv. 8xya

v. 10xy

a

Calculate the actuarial present value of this special annuity.

Solution:

15,000 15,000 5,000 30,000

15,000( ) 5,000 30,000(1 )

15,000(8 10) 5,000(8) 30,000(1 ln(1.06)(8)) 246,015,46

x y xy xy

xy xy xyxy

PV a a a

a a a a


13. Lindsay and Rehan purchase a two year joint term policy. The policy pays a death benefit of

1,000,000 upon the first death. Annual premiums are paid for two years for this policy.

The probability that Lindsay will die during the next two years is 0.05 and 0.08. The probability

that Rehan will die during the next two years is 0.07 and 0.10.

Lindsay and Rehan are independent lives.

You are given that 0.10d .

Let 0

nL be the loss at issue random variable for this insurance policy calculated on a net

premium basis.

a. Calculate the annual net benefit premium.

b. Calculate the standard deviation of 0

nL .


Solution:

2

2

Part a.

(1 0.9(1 0.05)(1 0.07))

1,000,000[(0.9)(1 (1 0.05)(1 0.07)) (0.9) (1 0.05)(1 0.07)(1 (1 0.08)(1 0.10))]

1,000,000[(0.9)(1 (1 0.05)(1 0.07)) (0.9) (1 0.05)(1 0.07)(1 (1 0.08)(1

PVP PVB

P

P

0.10))]

1 0.9(1 0.05)(1 0.07)

227,939.22126,975.03

1.79515

Part b.

We have to calculate this from first principals. There are three possible outcomes. These are

listed below with their probabilities:

Die

0

Year 1 Prob==>1 (1 0.05)(1 0.07) 0.1165

Die Year 2 Prob==>(1 0.05)(1 0.07)(1 (1 0.08)(1 0.10)) 0.151962

Live 2 Years Prob==>(1 0.05)(1 0.07)(1 0.08)(1 0.10) 0.731538

(1,000,000)(0.9) 126,975.0nL

2

0

0

2 2

0

3 773,024.97 if Die Year 1

(1,000,000)(0.9) 126,975.03(1 0.9) 568,747.44 if Die Year 2

126,975.03(1 0.9) 241,252.56 if Live 2 Years

[ ] (773,024.97) (0.1165) (568,747.44) (0.151962) (

n

n

n

L

L

Var L

2

2 2 2

241,252.56) (0.731538)

(773,024.97) (0.1165) (568,747.44) (0.151962) ( 241,252.56) (0.731538)

401,683.81

SD


14. Oscar is age 50 and Cora is age 60. They have independent future lifetimes. They purchase a

special 10 year deferred joint and survivor annuity policy. The policy make annual payments

beginning at age 60 for Oscar and age 70 for Cora.

You are given:

a. The annuity payments are paid at the beginning of each year after the ten year deferral

period as long as at least one annuitant is alive.

b. The benefit is 14,000 if both Oscar and Cora are alive. It is 10,000 if only Cora is alive

and it is 5000 if only Oscar is alive.

c. The premium of P is paid at the start of each year during the deferral provided both

Oscar and Cora are alive.

d. Mortality follows the Illustrative Life Table.

e. 0.06i

Calculate P .

Solution:

50:60 10 50:60 60:70 10 60 70 10 50 60 10 50:60 60:70

10 60 70 10 50 60 10 50 10 60 60:70

50:60 10 50 10 60 60:70

10,000 5000 1000

10,000 5000 1000

(10,000)(0.45120)(8.

PVP PVB

P a E a E a E a E a

E a E a E p aP

a E p a

5693) (5000)(0.51081)(11.1454) (1000)(0.51081)(6,616,155 / 8,188,074)(7.5563)

10.1944 (0.51081)(6,616,155 / 8,188,074)(7.5563)

64,011.754789046.88

7.075564314


Answers

1.

a. 0.26084

b. 0.03436

c. 0.47975

d. 52.20

e. 0.49400

f. 56.89

g. 0.98364

h. 0.00504

i. 0.24448

j. 0.31180

k. 25.65

l. 13.0997

m. 12.1584

n. 3.5891

2. 12.8768

3. 12.7182

4.

a. 0.93867

b. 0.83045

c. 1.28846

5. 0.067375

6. 234.82

7. 7.1687

8.

a. 0.05te

b. 0.07te

c. 0.11te

d. 20

e. 14.286

f. 9.091

g. 25.195

h. 0.78571

i. 0.53929

9. 0.06994


10.

a. 0.05498

b. 0.10970

c. 0.03434

d. 0.13034

11. 80,431.70

12. 246,015.46

13.

a. 126,975.03

b. 401,700

14. 9046.88

Documents

Chapter 9 - Purdue Universityjbeckley/WD/STAT 475/S18...The net annual premium rate for a fully continuous joint whole life on (60) and (60) with a death benefit of 1000. Solution: