chapter 8jbeckley/q/wd/stat 490/s17/stat 4… · january 14, 2016 copyright jeffrey beckley...

January 14, 2016 Copyright Jeffrey Beckley 2012-2016

Chapter 8

1. Mayfawny purchases a whole life insurance policy.

There are three ways that Mayfawny’s policy can terminate:

a. Death (1)

b. Diagnosis of a critical illness (2); and

c. Lapse (3).

The policy pays a death benefit of 10,000 at the moment of death. The policy will also pay a

critical illness benefit of 20,000 if Mayfawny is diagnosed with a critical illness. Only one benefit

will be paid.

There is no benefit paid upon lapse.

You are also given:

i. (1) 0.01x

ii. (2) 0.015x

iii. (3) 0.06x

iv. 0.035

Mayfawny pays a net premium continuously for her lifetime as long as the policy is inforce. The

net premium is determined using the equivalence principle.

Calculate the net premium paid by Mayfawny.


Solution:

( )

0 0

00 01 02 03

(0.01 0.015 0.06)00 ( ) 0.035

0 0 0

0.120.035 0.085 0.12

0 0 0

01

10,000 20,000 0

1

0.12 0.12

t t

x s

x x x x

ds dst t t

x t x

tt t t

x

PVP PVB Pa A A A

a v p dt e e dt e e dt

ee e dt e dt

A

( ) (1) 0.035 0.085

0 0

02 ( ) (2) 0.035 0.085

0 0

01 02 0300 01 02 03

00

0.010.01

0.12

0.0150.015

0.12

10,000 20,000 010,000 20,000 0

10,0

t t t

t x x

t t t

x t x x

x x xx x x x

x

v p dt e e dt

A v p dt e e dt

A A APa A A A P

a

0.01 0.01500 20,000 0

0.12 0.12400

1

0.12


2. Jeff is receiving a salary paid continuously for as long as he is in employed at Purdue. Jeff can

leave employment through death (1), retirement (2), or disability (3). Once Jeff leaves

employment, the salary stops.

You are given:

i. The salary pays at an annual rate of 70,000 per year.

ii. 0.05

iii. Jeff is currently age 59.

iv. Jeff will retire at age 65 if he is still teaching. He will not retire prior to age 65.

v. (1)

59 0.01 0.001t t

vi. (3)

59 0.03 0.001t t

Calculate the present value of Jeff’s future earnings while employed at Purdue.

Solution:

( ) (1) ( 2)59 59 59

0 0 0

6

( )

59

0

( ) (0.01 0.001 0.03 0.001 )0.05 ( )

59

70,000 The limits on the integral are determined by Jeff's retirement date.

;

t t t

s s s

t

t

ds ds t t dst t t

t

PV v p dt

v e e p e e e

0

0.040.04

6 6 6

( ) 0.05 0.04 0.09

59

0 0 0

60.09 0.09(6)

0

PV=70,000 70,000 70,000

170,000 70,000 324,529.14

0.09 0.09

t

dst

t t t t

t

t

e e

v p dt e e dt e dt

e e


3. You are given the following table where decrement (1) is death, decrement (2) is lapse, and

decrement (3) is diagnosis of critical illness:

x (1)

xq (2)

xq (3)

xq ( )

xq ( )

xp ( )

xl (1)

xd (2)

xd (3)

xd

55 0.02 0.15 0.010

56 0.03 0.06 0.015

57 0.04 0.04 0.020

58 0.05 0.03 0.025

59 0.06 0.02 0.030

a. Complete the table using a radix of 10,000.

See answers at end.

b. Calculate:

i. ( )

3 55p

Solution: ( )

( ) 583 55 ( )

55

6605.10.66051

10,000

lp

l

ii. (2)

2 56q

Solution: (2) (2)

(2) 56 572 56 ( )

56

492 293.560.0958

8200

d dq

l

iii. (3)

1|2 55q

Solution: (3) (3)

(3) 56 571|2 55 ( )

55

123 146.780.026978

10,000

d dq

l


iv. The probability that a person age 55 will decrement from death or critical illness

before age 60.

Solution: (1) (1) (1) (3) (3) (3)

55 56 59 55 56 59

( )

55

... ... 2136.7633050.21368

10,000

d d d d d d

l

c. Assuming uniform distribution of each decrement between integer ages, calculate:

i. (2)

0.25 55q

Solution:

(2) (2)

0.25 55 55

1500(0.25)( ) (0.25) 0.0375

10,000q q

ii. ( )

0.5 56p

Solution: ( ) ( ) ( )

0.5 56 0.5 56 561 1 (0.5)( ) 1 (0.5)(0.105) 0.9475p q q

iii. ( )

0.5 56.8p

Solution: ( ) ( ) ( )

( ) 57.3 57 580.5 56.8 ( ) ( ) ( )

56.8 56 57

(0.7) (0.3) (0.7)(7339) (0.3)(6605.1)0.94776

(0.2) (0.8) (0.2)(8200) (0.8)(7339)

l l lp

l l l

iv. (1)

0.5 55.6q

Solution: (1) (1)

(1) 0.4 55.6 0.1 560.5 55.6 ( )

55.6

(0.4)(200) (0.1)(246)0.01173

(0.4)(10,000) (0.6)(8200)

d dq

l


d. Assuming a constant force of decrement for each decrement between integer ages,

calculate:

i. (2)

0.25 55q

Solution:

(2)

(2) ( ) 0.25 0.25550.25 55 55( )

55

0.151 ( ) 1 (0.82) 0.04034

0.18

qq p

q

ii. ( )

0.5 56p

Solution: ( ) ( ) 0.5 0.5

0.5 56 56( ) (0.895) 0.94604p p

iii. ( )

0.5 56.8p

Solution: ( ) ( ) ( ) 0.3

( ) 57.3 57 0.3 570.5 56.8 ( ) ( ) ( ) 0.8

56.8 56 0.8 56

( ) (7339)(0.9)0.94763

( ) (8200)(0.895)

l l pp

l l p

iv. (1)

0.5 55.6q

Solution:

(1) (1) (1) (1) (1)(1) 0.4 55.6 0.1 56 55 0.6 55 0.1 56

0.5 55.6 ( ) ( ) ( ) 0.6

55.6 55 55

0.6 0.1

0.6

( )( )

0.02 0.03200 (10,000) 1 (0.82) (8200) 1 (0.895)

0.18 0.105

(10,000)(0.82)

101.11845370

8877.45156

d d d d dq

l l p

.011390


4. A fully discrete 3 year term pays a benefit of 1000 upon any death. It pays an additional 1000

(for a total of 2000) upon death from accident. You are given:

x 𝑞𝑥(1)

𝑞𝑥(2)

( )

xl (1)

xd (2)

xd

20 0.030 0.010 1000 30 10

21 0.025 0.020 960 24 19.2

22 0.020 0.030 916.8 18.336 27.504

Decrement (1) is death from accidental causes while decrement (2) is death from non-accidental

causes.

The annual effective interest rate is 10%.

a. Calculate the level annual net premium for this insurance.

Solution:

Columns above in yellow were added to facilitate the calculation.

2 2 3 2 3

2 3 2 3

2

(1000 960 916.8 ) 2000(30 24 18.336 ) 1000(10 19.2 27.504 )

2000(30 24 18.336 ) 1000(10 19.2 27.504 )63.64

(1000 960 916.8 )

PVP PVB

P v v v v v v v v

v v v v v vP

v v


b. Calculate the net premium reserve at the end of year 0, 1, 2, and 3.

Solution:

0 3

(1) (1) (2) (2)

0 1 11 ( )

12

By Definition ==> 0 and 0

We will use the recursive formula to get the other two reserves.

( )(1 )

(0 63.64)(1.1) (2000)(0.03) (1000)(0.01)0.00

1 0.03 0.01

(

x x

x

V V

V P i b q b qV

p

VV

(1) (1) (2) (2)

2 1 1 1

( )

1

)(1 )

(0 63.64)(1.1) (2000)(0.025) (1000)(0.02)0.00

1 0.025 0.02

Do not erroneously draw the conclusion that all reserves are zero.

It just so happens here. A poor

x x

x

P i b q b q

p

ly developed question. :)

5. You are given:

a. 𝑞𝑥′(1)

= 0.200

b. 𝑞𝑥′(2)

= 0.080

c. 𝑞𝑥′(3)

= 0.125

Assuming that each decrement is uniformly distributed over each year of age in the associated

single decrement table, calculate 𝑞𝑥(1)

.

Solution:

(1) (1) (2) (3) (2) (3)1 1/ 2 1/ 3

(0.2) 1 1/ 2 0.08 0.125 1/ 3 0.08 0.125 0.180167

x x x x x xq q q q q q


6. You are given:

a. 𝑞𝑥′(1)

= 0.200

b. 𝑞𝑥′(2)

= 0.080

c. 𝑞𝑥′(3)

= 0.125

Assuming that each decrement in the multiple decrement table is uniformly distributed over

each year of age, calculate 𝑞𝑥(1)

.

Solution:

(1)(1)

( )

( ) (1) (2) (3) (1) (2) (3)

(1) ( ) (1)1 0.644

(1 )(1 )(1 ) (0.8)(0.92)(0.875) 0.644

ln(0.8)0.8 (0.644) (1 0.644) 0.180520

ln(0.644)

xx

x

x x x x x x x

qq

qx x x

p p p p q q q

p p q

7. You are given the following for a double decrement table:

a. 𝑞𝑥′(1)

= 0.200

b. 𝑞𝑥′(2)

= 0.080

Assuming that each decrement in the multiple decrement table is uniformly distributed over

each year of age, calculate 0.4𝑞𝑥+0.4(1)

.

Solution:

(1)(1)

( )

( ) (1) (2) (1) (2)

(1) ( ) (1)1 0.736

(1)(1) 0.4 0.4

0.4 0.4 ( )

0.4

(1 )(1 ) (0.8)(0.92) 0.736

ln(0.8)0.8 (0.736) (1 0.736) 0.192186173

ln(0.736)

(0.4)(0.19218

xx

x

x x x x x

qq

qx x x

xx

x

p p p q q

p p q

dq

l

( )6173)0.085951 <==This assumes 1

1 (0.4)(1 0.736)xl


8. You are given:

a. 𝑞𝑥′(1)

= 0.200

b. 𝑞𝑥′(2)

= 0.080

Decrement (1) is uniformly distributed over the year. Decrement (2) occurs at time 0.6.

Calculate 𝑞𝑥(2)

.

Solution:

Worked in class.

9. For a double decrement table with 𝑙40𝜏 = 2000:

x 𝑞𝑥(1)

𝑞𝑥(2)

𝑞𝑥′(1)

𝑞𝑥′(2)

40 0.24 0.10 0.25 y

41 -- -- 0.20 2y

Calculate 𝑙42𝜏 .

Solution: ( ) (1) (2) (1) (2)

40 40 40 40 40

( ) ( ) ( ) ( ) ( ) (1) (2) (1) (2)

42 40 40 41 40 40 40 41 41

1 1 0.24 0.10 0.66 (1 0.25)(1 )

0.661 0.12 2 0.24

0.75

2000(1 0.25)(1 0.12)(1 0.2)(1 0.24)

p q q p p y

y y

l l p p l p p p p

802.56


10. For iPhones, the phone may cease service for mechanical failure or for other reasons (lost,

stolen, dropped in a pitcher of beer, etc). You are given the following double decrement table:

Year of Service

For an iPhone at the beginning of the year of service, probability of

Mechanical Failure Failure for Other

Reasons Survival through the

year of service

1 0.2 0.30 --

2 -- 0.40 --

3 -- -- 0.20

You are also given:

a. The number of iPhones that terminate for other reasons in year 3 is 40% of the number

of iPhones that survive to the end of year 2.

b. The number of iPhones that terminate for other reasons in year 2 is 80% of the number

of iPhones that survive to the end of year 2.

Calculate the probability that an iPhone will cease to function due to mechanical failure during

the three year period following its entry into service.


Solution:

( ) ( )

Let (m) be mechanical failure and (o) be failure for other reasons.

Now we will build a table assuming that we start with 1000 phones.

Surviving Phones

1 200 300 500

2 50 200 250

3 100 100 50

m o

t t

a b c

f d e

i g h

Year d d

( )

1

(0)

1

(0)

1

1000( ) (1000)(0.2) 200

1000( ) (1000)(0.3) 300

1000 200 300 500

500( ) (500)(0.4) 200

From Given (b)==>200 (0.8)( ) 200 / 0.8 250

500 200 250 50

From Given (a)==> (

ma q

b q

c

d q

e e

f

g g

( ) ( ) ( )( ) 1 2 3

3 ( )

0

0.4)(250) 100

250(0.2) 50

250 100 50 100

200 50 1000.35

1000

m m mm

x

h

i

d d dq

l


11. *Your actuarial student has constructed a multiple decrement table using independent mortality

and lapse tables. The multiple decrement table values, where decrement d is death and

decrement w is lapse , are as follows:

( )

60l ( )

60

dd ( )

60

wd ( )

61l

950,000 2,580 94,742 852,678

You discover that an incorrect value of ( )

60

wq was taken from the independent lapse table.

The correct value is 0.05.

Decrements are uniformly distributed over each year of age in the multiple decrement

table.

You correct the multiple decrement table, keeping ( )

60 950,000l .

Calculate the correct values of ( )

60

wd .


Solution:

( )60

( )60

( )60

( )60

( ) ( )

60 60

( )( ) ( ) ( ) 6160 60 60 ( )

60

( )( ) 6160 ( )

60

(2580/950,000) (1 852,678/950,000)

( ) ( )

60 60

1

852,6780.997138907

950,000

Note that we can use the inc

d

d

q

d q

d w

q

d q

p p

lq q q

l

lp

l

p p

( ) ( )60 60( )60

( )

60

( ) ( )(1 (0.997138907)(1 0.05))

60 60

( )

60

orrect values to derive since this value

was correct in the calculations.

0.95 (0.997138907)(1 0.05)

ln(0.95)

ln (0.997138907)(1

w w

d

q qw q

w

p

p p

q

( )

60

1 (0.997138907)(1 0.05) 0.0499290.05)

(950,000)(0.049929) 47,433wd

chapter 8jbeckley/q/wd/stat 490/s17/stat 4… · january 14, 2016 copyright jeffrey beckley...

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