Math 630 Problem Set 2

Math 630 Problem Set 2

1. AN n-year term insurance payable at the moment of death has an actuarial present value (i.e. EPV) of 0.0572. Given ?x+t = 0.007 and = 0.05, find n.

(Answer: 11)

2. Given: A?1x:n| = 0.4275, = 0.055, ?x+t = 0.045. Calculate A?x:n|.

(Answer: 0.4775)

3. For a whole life insurance of 1000 on (x) with benefits payable at the moment of death,

given =

0.04, 0.05,

0 < t 10 t > 10

and ?x+t =

0.06, 0.07,

0 < t 10 t > 10

.

Calculate the single

benefit premium (i.e. EPV) for this insurance.

(Answer: 593.868)

4. Given: i = 5%, CFM, ex= 16.0. Find 20|A?x.

(Answer: 0.06064)

5. For a 5-year deferred whole life insurance of 1 payable at the moment of death, you are given that ? = 0.04 and = 0.1. Calculate the variance of the present value of the benefit.

(Answer: 0.030069)

6. A whole life insurance provides a death benefit of 1 at the moment of death plus a return of the net single premium with interest at = 0.08. The net single premium is calculated using ? = 0.04 and = 0.16. Calculate the net single premium.

(Answer: 3/10)

7. Given that mortality follows DML with = 100 and = 0.06. Calculate A?140:10|. (Answer: 0.12533)

8. Mortality for (40) follows DML with = 100. Z represents the present value of a whole life insurence payable at the moment of death. = 0.06. Calculate V(Z).

(Answer: 0.06578)

9. Given: A whole life insurance whose benefit is t for t 0; ? = 0.04; d = 0.0582355; benefits are payable at the moment of death. Calculate E(Z).

(Answer: 4)

10. CFM, = 0.06, 2A?x = 0.25. Calculate (I?A?)x.

(Answer: 4)

11. The purchase price of a washing machine is 600. Company ABC provides a 10-year warranty. In the event of failure for t 0, ABC will pay 600(1 - 0.1t)at time t of failure. There is a constant force of failure of 0.02, = 0.08. Calculate the APV (i.e. EPV) of the warranty given it is paid for at the time of purchase. (Answer: 44.14533)

12. Given: ? = 0.05, = 0.06, Z is the prsent value random variable of a 5-year deferred whole life insurance of 1000 on (x) payable at the moment of death. Calculate the 90th percentile of Z. (Answer: 628.19)

13. Z is the present value for a whole life insurance of 1 on (x) payable at the moment of death. Given that ?x+t = 0.01t and = 0.04, calculate the 80th percentile of Z. (Answer: 0.7655)

14. A maintenance contract on a hotel promises to replace burned out light bulbs at the end of each year for 3 years. The hotel has 10,000 light bulbs. The light bulbs are all new. If a replacement bulb burns out it will be replaced with a new bulb. Given that q0 = 0.1, q1 = 0.3, q2 = 0.5. Each light bulb costs 1. i = 0.05. Calculate the APV of this contract. (Answer: 6688.26)

15. Gary, age 30, is subject to ? = 0.12 and wants to buy a 3-year $1000 endowment insurance. = 0.09. Determine the net single premium for this insurance. (Answer: 787.61)

16. For a special 3-year term insurance on (x) you are given qx+k = 0.02(k + 1), k = 0, 1, 2. The following are death benefits payable at the end of year of death.

k

0

1

2

bk+1 300,000 350,000 400,000

i = 0.06. Calculate the EPV.

(Answer: 36829.06)

17. Given: DML with k|qx = 1/75 for all x, = 0.05, Z is the present value of a whole life insurance of $1 payable at the end of the year of death issued to (x). Determine V(Z).

(Answer: 0.06222)

18. For a 20-year pure endowment of 1 on (x), you are given 20px = 0.65 and V(Z) = 0.05E(Z) where Z is the present value. Calculate i.

(Answer: 0.102186)

19. Given: A40 = 0.3, A40:20| = 0.45, 20p40 = 0.9, i = 0.04. Calculate A60. (Answer: 0.634813)

20. Given: Ax+1 - Ax = 0.015, i = 0.06, qx = 0.05. Calculate Ax + Ax+1. (Answer: 1.18318)

21. Given: 1000(IA)50 = 4996.75, 1000A150:1| = 5.58, 1000A51 = 249.05, i = 6%. Calculate 1000(IA)51.

(Answer: 5073.07)

22. Deaths are UDD over each year of age. i = 0.10, qx = 0.05, qx+1 = 0.08. Calculate A?1x:2| (Answer: 0.113592)

23. Assuming UDD and A(x4) = 1.00248A(x2), find i.

(Answer: 0.02)

24. Problems from Chapter 5 of the text: Exercises 5.1?5.9.

Solutions or Hints to Selected Problems

2.

A?x:n| = A?1x:n| + nEx.

However, A?x = A?1x:n| + nExA?x+n

?

?

+ ? = 0.4275 + nEx + ?

0.45 = 0.4275 + nEx ? 0.45 nEx = 0.05. So, A?x:n| = 0.4275 + 0.05 = 0.4775.

3. A?x = A?1x:10| + v1010px ? A?x+10. We have v1010px = e-0.4e-0.6 = e-1 = 0.3678794412,

10

10

A?1x:10| =

vttpx?x+t dt =

e-0.04te-0.06t(0.06) dt = 0.3792723353

0

0

A?x+10

=

? +

?

=

0.07 0.05 + 0.07

=

0.5833333333

Therefore, A?x = 0.3792723353 + (0.3678794412) ? (0.5833333333) = 0.593868676 The EPV of 1000 is 1000A?x = 593.87.

4. Under CFM, ex= 1/? ? = 1/16. i = 5% = ln(1.05).

20|A?x =

vttpx?x+t dt =

20

e-te-?t? dt = ? ? ? =

? e-20(+?) = 0.0606413904

20

+?

6. Let E be the single net premium. The present value of the benefit is then

Z = vTx (1 + Ee0.08Tx ) = vTx + Ee0.08Tx vTx

where v = e-0.16

E

=

E(Z )

=

E(vTx ) + EE(e0.08Tx vTx )

=

0.04 0.16+0.04

+

E

0.04 0.08+0.04

E=

4 20

+

4 12

E

E = 3/10.

Note that E(e0.08TxvTx) = E(e0.08Txe-0.16Tx) = E(e-0.08Tx) is the EPV of a continuous

whole

life

insurance

of

1

with

?

=

0.04

and

=

0.08,

which

is

equal

to

0.04 0.08+0.04

.

8.

First, recall

that for

DML,

tpx?x+t =

1 -x

.

So,

tp40?40+t

=

1 60

.

E(Z) =

60

vttp40?40+t dt =

0

60

e-0.06t

1

dt =

1

?

1 (1 - e-0.06?60) = 0.270188

0

60 60 0.06

E(Z2) =

60

v2ttp40?40+t dt =

0

60

e-0.12t

1

dt = 0.138785

0

60

V(Z) = 0.138785 - 0.2701882 = 0.06578.

9. First v = 1 - d = 1 - 0.0582355 = 0.9417645 = - ln v = 0.06

E(Z) = I?A? x =

tvttpx?x+t dt =

te-te-?t? dt = 0.04

te-0.1t dt

0

0

0

Use integration by parts to finish.

10.

First,

2A?x

=

? 2+?

0.25

=

? 0.12+?

? = 0.04.

I?A? = x

tvttpx?x+t dt =

te-0.06te-0.04t0.04 dt = 0, 04

te-0.1t dt

0

0

0

(Same as the previous problem.)

11. The is a 10-year term insurance with a variable warranty (i.e. benefit) payment of bt = 600(1 - 0.1t) at time t. The APV is

10

10

E(Z) =

btvttp0?0+t dt =

600(1 - 0.1t)e-0.08te-0.02t0.02 dt

0

0

13. First note that

tpx

= e-

t 0

?x+s

ds

= e-0.005t2

The present value is Z = vTx = e-0.04Tx. Let z0 = e-0.04t0 be the 80th percentile. Using a graph of z = e-0.04t, we have

0.8 = P (Z z0) = P (Tx t0) = t0px = e-0.005t20.

Solving the above equation for t0, we get t0 = 6.68047. So, the 80th percentile is z0 = e-0.04t0 = 0.7655.

14. Consider one bulb first. The possible replacement scenarios (outcomes) are listed in the following chart, where each "--" is the duration of a year and a dot means that a replacement occurs at that time.

Scenario ------ -- -- --? -- --? -- --? -- -- -- --? --? --? --? -- --? -- --? --? --? --?

PV of the Cost

0 v3 v2

v v2 + v3 v + v2 v + v3 v + v2 + v3

Probability

p0p1p2 = 0.315 p0p1q2 = 0.315 p0q1p0 = 0.243 q0p0p1 = 0.063 p0q1q0 = 0.027 q0q0p0 = 0.009 q0p0q1 = 0.027 q0q0q0 = 0.001

The average (mean) cost for one bulb is the sum of the products of the present value and its probability. Therefore, for 10000 bulbs the value is

10000 0.315v3 + 0.243v2 + ? ? ? + 0.027(v + v3) + 0.001(v + v2 + v3) = = 1000v + 2800v2 + 3700v3 = 6688.262607

Another solution is to use a chart to keep track of the replacements according to the given mortality rates (probabilities).

p1

? B ?

?

6300

$$$p2$$$ X q2 z

3150 3150

?

?

?

9000

p0

r r

rq1r

r r j

2700

$$$p0$$$ X q0 z

2430 270

10000

d

d

qd0

d

900 $$$p1$$$ X 630

p0

? B ?

?

?

q1 z 270

d

?

?

d 1000

r

r

rq0r

r r j

100

$$$p0$$$ X 90 q0 z 10

0

1

2

3

The numbers in red are the numbers of bulbs replaced at that times. The present value is then

1000v + (2700 + 100)v2 + (3150 + 270 + 270 + 10)v3 = 6688.262607

15. First, A130:3| = vq30 + v21|q30 + v32|q30, and 3E30 = v33p30. Then A30:3| = A130:3| + 3E30. The answer is 1000A30:3|.

17.

First

note

that

k|qx

=

k px

?

qx+k

=

1 -x

=

1 75

- x = 75, the maximal remaining

life time. Also note that v = e-0.05.

E(Z) =

74

vk+1k|qx

=

1 (v

75

+

v2

+

???

+ v75)

=

0.25394

k=0

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