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 µx+t = 0.007 and δ = 0.05, find n.
(Answer: 11)
2. Given: Ā1x:n| = 0.4275, δ = 0.055, µx+t = 0.045. Calculate Āx:n| .
(Answer: 0.4775)
3. For a whole
life insurance of 1000 on (x) withbenefits payable at the moment of death,
0.04, 0 < t ≤ 10
0.06, 0 < t ≤ 10
given δ =
and µx+t =
. Calculate the single
0.05, t > 10
0.07, t > 10
benefit premium (i.e. EPV) for this insurance.
(Answer: 593.868)
◦
4. Given: i = 5%, CFM, ex = 16.0. Find
20| Āx .
(Answer: 0.06064)
5. For a 5-year deferred whole life insurance of 1 payable at the moment of death, you
are given that µ = 0.04 and δ = 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 δ = 0.08. The net single premium is
calculated using µ = 0.04 and δ = 0.16. Calculate the net single premium.
(Answer: 3/10)
7. Given that mortality follows DML with ω = 100 and δ = 0.06. Calculate Ā140:10| .
(Answer: 0.12533)
8. Mortality for (40) follows DML with ω = 100. Z represents the present value of a
whole life insurence payable at the moment of death. δ = 0.06. Calculate V(Z).
(Answer: 0.06578)
9. Given: A whole life insurance whose benefit is t for t ≥ 0; µ = 0.04; d = 0.0582355;
benefits are payable at the moment of death. Calculate E(Z).
(Answer: 4)
10. CFM, δ = 0.06, 2Āx = 0.25. Calculate (I¯Ā)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, δ = 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: µ = 0.05, δ = 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 µx+t = 0.01t and δ = 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 µ = 0.12 and wants to buy a 3-year $1000 endowment
insurance. δ = 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
bk+1
0
300,000
1
350,000
2
400,000
i = 0.06. Calculate the EPV.
(Answer: 36829.06)
17. Given: DML with k| qx = 1/75 for all x, δ = 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
0.05E(Z) where Z is the present value. Calculate i.
20 px
= 0.65 and V(Z) =
(Answer: 0.102186)
19. Given: A40 = 0.3, A40:20| = 0.45,
20 p40
= 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 Ā1x:2|
(Answer: 0.113592)
(4)
(2)
23. Assuming UDD and Ax = 1.00248Ax , find i.
(Answer: 0.02)
24. Problems from Chapter 5 of the text: Exercises 5.1–5.9.
Solutions or Hints to Selected Problems
µ
µ
= 0.4275 + n Ex
δ+µ
δ+µ
= 0.4275 + 0.05 = 0.4775.
2. Āx:n| = Ā1x:n| + n Ex . However, Āx = Ā1x:n| + n Ex Āx+n ⇒
⇒ 0.45 = 0.4275 + n Ex · 0.45 ⇒ n Ex = 0.05. So, Āx:n|
3.
Āx = Ā1x:10| + v 10 10 px · Āx+10 .
We have v 10 10 px = e−0.4 e−0.6 = e−1 = 0.3678794412,
Z 10
Z 10
t
1
e−0.04t e−0.06t (0.06) dt = 0.3792723353
v t px µx+t dt =
Āx:10| =
0
0
Āx+10 =
µ
0.07
=
= 0.5833333333
δ+µ
0.05 + 0.07
Therefore, Āx = 0.3792723353 + (0.3678794412) · (0.5833333333) = 0.593868676 ⇒
The EPV of 1000 is 1000Āx = 593.87.
◦
4. Under CFM, ex = 1/µ ⇒ µ = 1/16. i = 5% ⇒ δ = ln(1.05).
Z ∞
Z ∞
µ −20(δ+µ)
t
e−δt e−µt µ dt = · · · =
v t px µx+t dt =
e
= 0.0606413904
20| Āx =
δ+µ
20
20
6. Let E be the single net premium. The present value of the benefit is then
Z = v Tx (1 + Ee0.08Tx ) = v Tx + Ee0.08Tx v Tx
where v = e−0.16 ⇒ E = E(Z) = E(v Tx ) + EE(e0.08Tx v Tx ) =
4
4
E = 20
+ 12
E ⇒ E = 3/10.
0.04
0.16+0.04
0.04
+ E 0.08+0.04
⇒
Note that E(e0.08Tx v Tx ) = E(e0.08Tx e−0.16Tx ) = E(e−0.08Tx ) is the EPV of a continuous
0.04
whole life insurance of 1 with µ = 0.04 and δ = 0.08, which is equal to 0.08+0.04
.
1
1
8. First, recall that for DML, t px µx+t = ω−x
. So, t p40 µ40+t = 60
.
Z 60
Z 60
1
1
1
t
e−0.06t dt =
v t p40 µ40+t dt =
·
(1 − e−0.06·60 ) = 0.270188
E(Z) =
60
60
0.06
0
0
Z 60
Z 60
1
2
2t
E(Z ) =
v t p40 µ40+t dt =
e−0.12t dt = 0.138785
60
0
0
2
⇒ V(Z) = 0.138785 − 0.270188 = 0.06578.
9. First v = 1 − d = 1 − 0.0582355 = 0.9417645 ⇒ δ = − ln v = 0.06
Z ∞
Z
Z ∞
t
−δt −µt
¯
E(Z) = I Ā x =
tv t px µx+t dt =
te e µ dt = 0.04
0
0
∞
te−0.1t dt
0
Use integration by parts to finish.
10. First, 2Āx =
µ
2δ+µ
I¯Ā x =
⇒ 0.25 =
µ
0.12+µ
⇒ µ = 0.04.
Z
Z
∞
∞
t
−0.06t −0.04t
tv t px µx+t dt =
te
e
0.04 dt = 0, 04
Z
0
0
∞
te−0.1t dt
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
Z 10
Z 10
t
600(1 − 0.1t)e−0.08t e−0.02t 0.02 dt
bt v t p0 µ0+t dt =
E(Z) =
0
0
13. First note that
t px
= e−
Rt
0
µx+s ds
2
= e−0.005t
The present value is Z = v Tx = e−0.04Tx . Let z0 = e−0.04t0 be the 80th percentile. Using
a graph of z = e−0.04t , we have
2
0.8 = P (Z ≤ z0 ) = P (Tx ≥ t0 ) = t0 px = e−0.005t0 .
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
2
— —· —·
v + v3
—· —· —
v + v2
—· — —·
v + v3
—· —· —·
v + v2 + v3
Probability
p0 p1 p2 = 0.315
p0 p1 q2 = 0.315
p0 q1 p0 = 0.243
q0 p0 p1 = 0.063
p0 q1 q0 = 0.027
q0 q0 p0 = 0.009
q0 p0 q1 = 0.027
q0 q0 q0 = 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 0.315v 3 + 0.243v 2 + · · · + 0.027(v + v 3 ) + 0.001(v + v 2 + v 3 ) =
= 1000v + 2800v 2 + 3700v 3 = 6688.262607
Another solution is to use a chart to keep track of the replacements according to the
given mortality rates (probabilities).
:
p2
6300 XXXX
p1*
z
q2 X X
3150
:
p0
j 2700 XX
H
XXX
q
z
X
2430
9000 H
HH
q1 HH
p0
3150
0
270
:
p1
XXX
q XXX
z
630
p0 :
XXX
q XXX
z
90
10000
@
@
q@
0
p0
@
@
R
@
*
900
1
1000 H
HH
q0 HH
j
H
100
0
0
1
2
270
10
3
The numbers in red are the numbers of bulbs replaced at that times. The present value
is then
1000v + (2700 + 100)v 2 + (3150 + 270 + 270 + 10)v 3 = 6688.262607
15. First, A130:3| = vq30 + v 2 1| q30 + v 3 2| q30 , and 3 E30 = v 3 3 p30 . Then A30:3| = A130:3| + 3 E30 .
The answer is 1000A30:3| .
1
17. First note that k| qx = k px · qx+k = ω−x
=
−0.05
life time. Also note that v = e
.
E(Z) =
74
X
k=0
v k+1 k| qx =
1
75
⇒ ω − x = 75, the maximal remaining
1
(v + v 2 + · · · + v 75 ) = 0.25394
75
2
E(Z ) =
74
X
v 2(k+1) k| qx =
k=0
1 2
(v + v 4 + · · · + v 150 ) = 0.126708
75
⇒ V(Z) = 0.126708 − 0.253942 = 0.06222.
19. Use A40 = A140:20| + v 20 20 p40 A60 = (A40:20| − v 20 20 p40 ) + v 20 20 p40 A60
21. First, A150:1| = vq50 ⇒ q50 = A150:1| (1 + i) ⇒ p50 = 1 − q50 . Then use
(IA)50 = A50 + vp50 (IA)51 = (A150:1| + vp50 A51 ) + vp50 (IA)51
22. Ā1x:2| = δi A1x:2| where A1x:2| = vqx + v 2 1| qx = vqx + v 2 px qx+1 .
(4)
23. Ax =
i(2) (2)
A
i(4) x
⇒
i(2)
i(4)
= 1.00248 ⇒
√
2( 1+i−1)
√
4( 4 1+i−1
= 1.00248 ⇒
√
4
1+i+1
2
= 1.00248 ⇒ i