September 29, 2020 - Purdue University

STAT 472 Test 1

Fall 2020

September 29, 2020

1. You are given that mortality follows the following mortality table:

Age x

q x

100

0.20

101

0.30

102

0.50

103

0.75

104

1.00

You are also given that d 0.08 which means that v 0.92 . Further, you are given that

deaths are uniformly distributed between integral ages for ages 100 and 101 and between ages 101 and 102. For ages over 102, mortality follows a constant force of mortality between integral ages.

a. (6 points) Calculate 0.8 p101.6 .

Solution:

p ll 0.8 101.6

102.4 101.6

l100 1000;l101 (1000)(1 0.2) 800;l102 (800)(1 0.3) 560;l103 560(0.5) 280

l104 (280)(1 0.75) 70;l105 0

p 0.8 101.6

l102.4 l101.6

(560)0.6 (280)0.4 (800)(0.4) (560)(0.6)

0.64695

b. Let Z be the present value random variable for a whole life insurance policy to (100)

with a death benefit of 10,000 paid at the end of the year of death.

i. (4 points) Write an expression of Z .

Solution:

Z 10, 000vK1001 (10, 000)(0.92)K1001

ii. (8 points) The expected present value which is 10, 000A100 is 8000 to the nearest 100. Calculate 10, 000A100 to the nearest 1.

Solutions:

Using the ls from Part a:

1000A100 200v 240v2 280v3 210v4 70v2 801.7469 A100 0.8017469 10, 000A100 8017.47

iii. (8 points) Calculate Var[Z] .

Solution:

1000

A 2 100

200v2 240v4 280v6 210v8 70v10 2 A100 0.64918295

Var[Z ] (10, 000)2[0.64918295 (0.8017469)2] 638, 486

2. (8 points) You are given the following select and ultimate mortality table:

[x]

q[ x]

q[ x]1

qx2

x2

75

0.10

0.20

0.4

77

76

0.15

0.30

0.6

78

77

0.20

0.40

0.8

79

78

0.25

0.55

0.9

80

79

0.30

0.70

1.00

81

You are also given that d 0.1. Calculate 100, 000 A which is a four year endowment insurance to a newly underwritten

[ 77 ]:4

life age 77 with a death benefit of 100,000 paid at the end of the year of death. Solution: l[77] 100, 000;l[77]1 (100, 000)(0.8) 80, 000;l79 (80, 000)(0.6) 48, 000 l80 48, 000(0.2) 9600;l81 9600(0.1) 96;l82 0

100, 000A (100, 000 80, 000)(0.9) (80, 000 48, 000)(0.9)2 [ 77 ]:4 (48, 000 9600)(0.9)3 9600v4 78, 212.34

3. You are given that Sx (t) (1 0.001t3) for 0 t 10 .

a. (6 points) Calculate x5 .

Solution:

xt

d dt

Sx

(t)

Sx (t)

(0.003t2 ) (1 0.001t3)

x5

(0.003(5)2 ) (1 0.001(5)3)

0.08571

b. (6 points) Calculate px6 .

Solution:

px6

Sx (7) Sx (6)

1 0.001(7)3 1 0.001(6)3

0.83801

c. (8 points) Calculate the Var[Tx ] . Solution: Var[Tx ] E[Tx2 ] (E[Tx ])2

E[Tx ]

10 0

t

px dt

10

(1 0.001t3)dt

0

t

0.001t 4 4

10

0

7.5

10

E[Tx2 ] 2 t t

0

pxdt

10

2 t(1 0.001t3)dt

0

2

t2 2

0.001t 5

5

10 0

60

Var[Tx ] 60 (7.5)2 3.75

4. (8 points) Let Z be the present value for a whole life insurance to (50) with a death benefit of 1

paid at the end of the year of death. You are given:

1. A50 0.3 2. i 0.05 3. q50 0.0018 and q51 0.0020 and q52 0.0022 Determine 1000A52 Solution: A50 vq50 vp50 A51 0.3 (1.05)1(0.018) (1.05)1(1 0.018)( A51) A51 0.31376

A51 vq51 vp51 A52 0.31376 (1.05)1(0.002) (1.05)1(1 0.002)( A52) A52 0.32810

1000A52 328.10

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