Chapter 8

[Pages:22]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) x

0.01

ii.

(2) x

0.015

iii.

(3) x

0.06

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.

February 17, 2015 Copyright Jeffrey Beckley 2012

Solution:

PVP PVB Pax00 10, 000 Ax01 20, 000 Ax02 0 Ax03

t

t

( x

) s

ds

(0.010.0150.06)ds

ax00 vt t px( ) dt et e 0

dt e0.035t e 0

dt

0

0

0

e0.035t

0

e0.085t

dt

e0.12t

0

dt

e0.12t

0.12

0

1 0.12

Ax01

vt t

0

px(

)

(1) x

dt

e0.035t e0.085t 0.01 dt

0

0.01 0.12

Ax02

vt t

0

px(

)

(2) x

dt

e0.035t e0.085t 0.015 dt

0

0.015 0.12

Pax00

10, 000 Ax01

20, 000 Ax02

0 Ax03

P

10, 000 Ax01

20, 000 Ax02 ax00

0 Ax03

10,

000

0.01 0.12

20,

000

0.015 0.12

0

400

1

0.12

February 17, 2015 Copyright Jeffrey Beckley 2012

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) 59t

0.01 0.001t

vi.

(3) 59t

0.03 0.001t

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

Solution:

6

PV 70, 000 vt t p5(9) dt The limits on the integral are determined by Jeff's retirement date. 0

t

t

t

t

5(9)s ds

( 5(19)s 5(92)s )ds

(0.010.001t 0.030.001t )ds

0.04ds

vt

et

e0.05t ;t

p( ) 59

e

0

e 0

e 0

e 0

e0.04t

6

6

6

PV=70,000

vt

t

p( ) 59

dt

70, 000

e0.05t e0.04t dt 70, 000

e0.09t dt

0

0

0

e0.09t 6

e0.09(6) 1

70,

000

0.09

0

70, 000

0.09

324,529.14

February 17, 2015 Copyright Jeffrey Beckley 2012

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

q(1) x

q(2) x

q(3) x

qx( )

px( )

lx( )

d (1) x

d(2) x

d (3) x

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.

p( )

3 55

Solution:

3

p( ) 55

l( )

58

l( )

55

6605.1 10, 000

0.66051

ii.

q(2)

2 56

Solution:

q(2)

2 56

d(2) 56

d(2) 57

l( )

56

492 293.56 8200

0.0958

iii.

q(3)

1|2 55

Solution:

q(3)

1|2 55

d(3) 56

d(3) 57

l( )

55

123 146.78 10, 000

0.026978

February 17, 2015 Copyright Jeffrey Beckley 2012

iv. The probability that a person age 55 will decrement from death or critical illness before age 60.

Solution:

d (1) 55

d (1) 56

...

d (1) 59

d (3) 55

d (3) 56

...

d (3) 59

l ( )

55

2136.763305 0.21368 10, 000

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

i.

q(2)

0.25 55

Solution:

0.25

q(2) 55

(0.25)(q5(52) )

(0.25)

1500 10, 000

0.0375

ii.

p( )

0.5 56

Solution:

0.5

p( ) 56

1 0.5

q( ) 56

1 (0.5)(q5(6) )

1 (0.5)(0.105)

0.9475

iii.

p( )

0.5 56.8

Solution:

p( )

0.5 56.8

l( )

57.3

l( )

56.8

(0.7)l5(7 ) (0.2)l5(6 )

(0.3)l5(8) (0.8)l5(7)

(0.7)(7339) (0.3)(6605.1) (0.2)(8200) (0.8)(7339)

0.94776

iv.

q(1)

0.5 55.6

Solution:

q(1)

0.5 55.6

d d (1)

(1)

0.4 55.6 0.1 56

l( )

55.6

(0.4)(200) (0.1)(246) (0.4)(10,000) (0.6)(8200)

0.01173

February 17, 2015 Copyright Jeffrey Beckley 2012

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

calculate:

i.

q(2)

0.25 55

Solution:

(2)

q (2)

55

q 0.25 55

( )

q55

1

(

p( ) 55

)0.25

0.15 1 (0.82)0.25 0.18

0.04034

ii.

p( )

0.5 56

Solution:

0.5

p( ) 56

(

p( ) 56

)0.5

(0.895)0.5

0.94604

iii.

p( )

0.5 56.8

Solution:

p( )

0.5 56.8

l( )

57.3

l( )

56.8

l( )

57

l( )

56

(0.3 (0.8

p( ) 57

p( ) 56

) )

(7339)(0.9)0.3 (8200)(0.895)0.8

0.94763

iv.

q(1)

0.5 55.6

Solution:

q(1)

0.5 55.6

d d (1)

(1)

0.4 55.6 0.1 56

l( )

55.6

d d d (1)

(1)

(1)

55 0.6 55 0.1 56

(l5(5

)

)(

p( ) 55

)0.6

200

(10,

000)

0.02 0.18

1 (0.82)0.6

(8200)

0.03 0.105

1 (0.895)0.1

(10, 000)(0.82)0.6

101.1184537 0.011390 8877.45156

February 17, 2015 Copyright Jeffrey Beckley 2012

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

(1)

(2)

lx( )

d (1) x

d (2) x

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.

PVP PVB

P(1000 960v 916.8v2 ) 2000(30v 24v2 18.336v3 ) 1000(10v 19.2v2 27.504v3)

2000(30v 24v2 18.336v3 ) 1000(10v 19.2v2 27.504v3 )

P

(1000 960v 916.8v2 )

63.64

February 17, 2015 Copyright Jeffrey Beckley 2012

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

By Definition ==>0V 0 and 3V 0

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

1V

(0V

P)(1

i)

b(1) 1

px( )

q(1) x

b1(2)

q(2) x

(0 63.64)(1.1) (2000)(0.03) (1000)(0.01) 0.00 1 0.03 0.01

2V

(1V

P)(1

i)

b(1) 2

q(1) x 1

p( ) x 1

b1(2)

q(2) x 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 poorly developed question. :)

5. You are given: a. (1) = 0.200 b. (2) = 0.080 c. (3) = 0.125

Assuming that each decrement is uniformly distributed over each year of age in the associated single decrement table, calculate (1).

Solution:

q(1) x

qx(1)

11 / 2 qx(2) qx(3) 1 / 3 qx(2) qx(3)

(0.2)11/ 20.08 0.125 1/ 30.08 0.125 0.180167

February 17, 2015 Copyright Jeffrey Beckley 2012

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