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
................
................
In order to avoid copyright disputes, this page is only a partial summary.
To fulfill the demand for quickly locating and searching documents.
It is intelligent file search solution for home and business.
Related download
- impedance matching transformers allied elec
- compatibility matrix
- standard 701901 08
- army command policy
- aluminum conductor mc cable priority wire
- ieee standard ratings current and voltage transformers cts and vts abb
- leaves and passes united states army
- personnel readiness processing united states army
- personnel processing in out soldier readiness and deployment cycle
- section 4 sheet 1 c l x type mc hl xhhw 2 okonite
Related searches
- math word problem worksheets pdf
- problem set 7
- math word problem worksheets
- math word problem solver
- free math word problem solver
- chemistry stoichiometry problem sheet 2 key
- college math word problem solver
- solve my math word problem free
- 8th grade math word problem worksheets
- math word problem vocabulary terms
- math word problem vocabulary list
- math word problem practice