actu 357
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School
Concordia University *
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Course
357
Subject
Accounting
Date
Apr 3, 2024
Type
Pages
30
Uploaded by ElderWorld8925
Benefit Reserves
1. A fully continuous 20-payment years, 30-year term life insurance of 2000 is issued to
(35). You are given
n
A
1
35+
n
:
30
-
n
|
a
35+
n
:
20
-
n
|
0
0.068727
11.395336
10
0.097101
7.351745
25
0.079516
–
μ
(
x
)
=
0
.
00005
×
1
.
1
x
, x >
0
δ
=
6%
Hence
t
p
x
= exp
-
0
.
0005246
×
1
.
1
x
(
1
.
1
t
-
1
)
, t
≥
0
, x >
0
Questions:
(a) Find the premium rate. (Answer: 12.062328)
(b) Determine
j
L
for
j
= 10
,
25. Plot roughly
j
L
(function of
T
) for
j
= 10
,
25.
(c) Evaluate
10
L
if
T
= 12
.
3 and
25
L
if
T
= 27
.
4.
(Answers:
1716.283219 and
1731.775496)
(d) Find
E
[
j
L
|
T > j
] for
j
= 10
,
25. (Answers: 105.522838 and 159.031484)
(e) Give an expression for
V ar
[
j
L
|
T > j
],
j
= 10
,
25, in terms of actuarial symbols.
(f) Find Pr (
10
L
≤
1800
|
T >
10). (Answer: 0.993772)
2. A fully continuous 30-payment years, whole life insurance of 5000 is issued to (40).
You are given
n
A
40+
n
a
40+
n
:
30
-
n
|
0
0.168110
13.049173
10
0.267636
10.662510
25
0.480688
4.038099
μ
(
x
)
=
0
.
00005
×
1
.
1
x
, x >
0
δ
=
6%
Hence
t
p
x
= exp
-
0
.
0005246
×
1
.
1
x
(
1
.
1
t
-
1
)
, t
≥
0
, x >
0
Questions:
1
(a) Find the premium. (Answer: 64.413897)
(b) Determine
j
L
for
j
= 10
,
25. Plot roughly
j
L
(function of
T
) for
j
= 10
,
25.
(c) Find
10
L
if
T
= 12
.
3 and
25
L
if
T
= 27
.
4 . (Answers: 4217.109532 and 4185.460527)
(d) Find
E
[
j
L
|
T > j
] for
j
= 10
,
25. (Answers: 651.363944 and 2143.331917)
(e) Find Pr (
10
L
≤
4000
|
T >
10)). (Answer: 0.979835)
3. A fully continuous 15-payment years, 40-year endowment insurance of 3000 is issued
to (30). You are given
n
A
30+
n
:
40
-
n
|
a
30+
n
:
15
-
n
|
0
0.067292
9.800709
10
0.105196
4.292542
25
0.164050
–
μ
(
x
)
=
0
.
00005
×
1
.
1
x
, x >
0
δ
=
6%
Questions:
(a) Find the premium rate. (Answer: 20.597996)
(b) Determine
j
L
for
j
= 10, 25. Plot roughly
j
L
(function of
T
) for
j
= 10
,
25.
(c) Find
10
L
if
T
= 12
.
3 and
25
L
if
T
= 27
.
4 . (Answers: 2569.044265 and 2597.663244)
(d) Find
E
[
j
L
|
T > j
] for
j
= 10
,
25. (Answers: 227.170038 and 492.150521)
(e) Give an expression for
V ar
[
j
L
|
T > j
],
j
= 10
,
25, in terms of actuarial symbols.
(f) Find Pr (
10
L
≤
2000
|
T >
10). (Answer: 0.981720)
4. A fully continuous 20-payment years, 20-year deferred life insurance of 4000 is issued
to (45). You are given
n
20
-
n
|
A
45+
n
A
45+
n
a
45+
n
:
20
-
n
|
0
0.116304
0.213405
11.015769
10
0.225237
0.330918
25
–
0.562461
–
μ
(
x
)
=
0
.
00005
×
1
.
1
x
, x >
0
δ
=
6%
Questions:
(a) Find the premium rate. (Answer: 42.231885)
2
(b) Determine
j
L
for
j
= 10, 25. Plot roughly
j
L
(function of
T
) for
j
= 10
,
25.
(c) Find
10
L
if
T
= 12
.
3 and
25
L
if
T
= 27
.
4 . (Answers: -90.729087 and 3463.550992)
(d) Find
E
[
j
L
|
T > j
] for
j
= 10
,
25. (Answers: 601.276713 and 2249.84412)
(e) Find Pr (
10
L
≤
0
|
T >
10). (Answer: 0.150511)
5. A fully continuous whole life annuity of 12000 is issued to (45). The deferred period
is 20 years and the premiums are payable continuously during the first 15 years. You
are given
n
20
-
n
|
a
45+
n
a
45+
n
a
45+
n
:
15
-
n
|
0
2.094165
13.109934
9.524542
10
4.055593
11.151400
4.207874
25
–
7.292369
–
μ
(
x
)
=
0
.
00005
×
1
.
1
x
, x >
0
δ
=
6%
Questions:
(a) Find the premium rate. (Answer: 2638.44452)
(b) Determine
j
L
for
j
= 10, 25. Plot roughly
j
L
(function of
T
) for
j
= 10
,
25.
(c) Find
10
L
if
T
= 12
.
3 and
25
L
if
T
= 27
.
4 . (Answers: -5668.31584 and 26822.45)
(d) Find
E
[
j
L
|
T > j
] for
j
= 10
,
25. (Answers: 37564.86967 and 87508.42304)
(e) Give an expression for
V ar
[
j
L
|
T > j
],
j
= 10
,
25, in terms of actuarial symbols.
(f) Find Pr (
10
L
≤
0
|
T >
10). (Answer: 0.186985)
6. A fully continuous whole life annuity of 15000 is issued to (35). The deferred period
is 25 years and the premiums are payable continuously during the first 15 years. You
are given
(1)
μ
x
=
0
.
02
,
35
≤
x <
60
1
110
-
x
,
60
≤
x <
110
;
(2)
δ
= 5%.
Questions:
(a) Find the premium rate. (Answer: 3552.538621)
(b) Find the reserve at time 10. (Answer: 51448.50771)
(c) Find Pr (
10
L
≤ -
10000
|
T >
10).
3
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7. You are given
μ
(
x
)
=
0
.
02
40
≤
x <
60
1
110
-
x
60
≤
x <
110
δ
=
7%
Find
10
V
(
A
50:
20
|
). (Answer: 0.2920)
8. A fully discrete 20-payment years, 30-year term life insurance of 2000 is issued to (35).
You are given
n
A
1
35+
n
:
30
-
n
|
¨
a
35+
n
:
20
-
n
|
0
0.066714
11.760448
10
0.094256
7.595877
25
0.077183
–
μ
(
x
)
=
0
.
00005
×
1
.
1
x
, x >
0
δ
=
6%
Questions:
(a) Find the premium. (Answer: 11.345519)
(b) Determine
j
L
pour
j
= 10, 25.
(c) Find
10
L
if
T
= 12
.
3 and
25
L
if
T
= 27
.
4. (Answers: 1638.447524 and 1670.540423)
(d) Find
E
[
j
L
|
T > j
] for
j
= 10
,
25. (Answers: 102.3327081 and 154.365834)
9. A fully discrete 30-payment years, whole life insurance of 5000 is issued to (40). You
are given
n
A
40+
n
¨
a
40+
n
:
30
-
n
|
0
0.163154
13.497507
10
0.259735
11.060408
25
0.466430
4.223319
μ
(
x
)
=
0
.
00005
×
1
.
1
x
, x >
0
δ
=
6%
Questions:
(a) Find the premium. (Answer: 60.438497)
(b) Determine
j
L
for
j
= 10, 25.
4
(c) Find
10
L
if
T
= 12
.
3 and
25
L
if
T
= 27
.
4 . (Answers: 4005.589589 and 4005.589589)
(d) Find
E
[
j
L
|
T > j
] for
j
= 10
,
25. (Answers: 630.1993197 and 2076.900001)
10. A fully discrete 15-payment years, 40-year endowment insurance of 3000 is issued to
(30). You are given
n
A
30+
n
:
40
-
n
|
¨
a
30+
n
:
15
-
n
|
0
0.065320
10.106230
10
0.102112
4.428779
25
0.159236
–
μ
(
x
)
=
0
.
00005
×
1
.
1
x
, x >
0
δ
=
6%
Questions:
(a) Find the premium. (Answer: 19.389925)
(b) Determine
j
L
for
j
= 10, 25.
(c) Find
10
L
if
T
= 12
.
3 and
25
L
if
T
= 27
.
4 . (Answers: 2450.962645 and 2505.810634)
(d) Find
E
[
j
L
|
T > j
] for
j
= 10
,
25. (Answers: 220.462660 and 477.707182)
11. A fully discrete 20-payment years, 20-year deferred life insurance of 4000 is issued to
(45). You are given
n
20
-
n
|
A
45+
n
A
45+
n
¨
a
45+
n
:
20
-
n
|
0
0.112854
0.207110
11.398392
10
0.218556
0.321138
7.364010
25
–
0.545725
–
μ
(
x
)
=
0
.
00005
×
1
.
1
x
, x >
0
δ
=
6%
Questions:
(a) Find the premium (Answer: 39.603614)
(b) Determine
j
L
for
j
= 10, 25.
(c) Find
10
L
if
T
= 12
.
3 and
25
L
if
T
= 27
.
4 .
(Answers:
-112.0261477 and
3341.080846)
5
(d) Find
E
[
j
L
|
T > j
] for
j
= 10
,
25. (Answers: 582.580926 and 2182.899968)
12. A fully discrete whole life annuity of 12000 is issued to (45). The deferred period is 20
years and the premiums are payable during the first 15 years. You are given
n
20
-
n
|
¨
a
45+
n
¨
a
45+
n
¨
a
45+
n
:
15
-
n
|
0
2.216845
13.615237
9.847560
10
4.293178
11.657188
4.360644
25
–
7.800659
–
μ
(
x
)
=
0
.
00005
×
1
.
1
x
, x >
0
δ
=
6%
Questions:
(a) Find the premium (Answer: 2701.394551)
(b) Determine
j
L
for
j
= 10, 25.
(c) Find
10
L
if
T
= 12
.
3 and
25
L
if
T
= 27
.
4 .
(Answers:
-7641.394165 and
33944.21964)
(d) Find
E
[
j
L
|
T > j
] for
j
= 10
,
25. (Answers: 39738.31659 and 93607.90655)
13. A 4-year fully discrete contract is issued to (
x
) with a term insurance benefit of 1000
and a pure endowment benefit of 2000. You are given
i
= 0
.
2,
q
x
=
q
x
+1
= 0
.
25, and
q
x
+2
=
q
x
+3
= 0
.
5.
(a) Formulate the first, second and third year terminal prospective loss random vari-
ables
(b) Find the first, second and third year terminal benefit reserves as the expected
value of the loss
(c) Write the prospective form of first, second and third year terminal benefit reserve
and calculate it.
(d) Write the retrospective form of first, second and third year terminal benefit reserve
and calculate it.
14. You are given
10
V
(
A
x
:
25
|
) = 0
.
405,
10
V
x
:
25
|
= 0
.
4,
i
= 0
.
1 and UDD is assumed.
A 25-year fully discrete contract is issued to (
x
) with a term insurance benefit of 1
and a pure endowment benefit of 2. Find 10th year terminal reserve for this contract.
(Answer: 0.6984)
6
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15. You are given
P
1
40:
25
|
= 0
.
0051,
P
40:
25
|
= 0
.
02042,
P
65
= 0
.
02312, and
i
= 0
.
06. Find
¨
a
40
. (Answer: 15.50)
16. You are given
10
V
40
= 0
.
105,
10
V
40:
20
|
= 0
.
356, and
A
40
= 0
.
161. Find
20
10
V
40
. (Answer:
0.1454)
17. Show that the
t
-th year terminal prospective formula is equal to
t
-th year terminal
retrospective formula for
t
V
(
n
|
¨
a
x
).
18. You are given
(1)
n
A
45+
n
2
A
45+
n
A
45+
n
:
20
-
n
|
2
A
45+
n
:
20
-
n
|
A
1
45+
n
:
20
-
n
|
2
A
1
45+
n
:
20
-
n
|
0
0.187328
0.059900
0.321379
0.112660
0.085814
0.043715
10
0.297308
0.126462
0.559757
0.318244
0.094468
0.067632
20
0.435944
0.234743
–
–
–
–
(2)
n
20
-
n
E
45+
n
2
20
-
n
E
45+
n
¨
a
20
-
n
|
¨
a
45+
n
:
20
-
n
|
¨
a
45+
n
0
0.235565
0.068945
11.831563
11.310358
13.544539
10
0.465289
0.250612
7.689748
7.337388
11.711539
20
1.000000
1.000000
–
–
9.400936
(3)
d
= 6%
.
(a) A fully discrete whole life insurance of 200 is issued to (45). Find
E
[
10
L
|
K
≥
10]
and
V ar
(
10
L
|
K
≥
10). (Answers: 27.0664 and 2305.7471)
(b) A fully discrete 20-payment years, whole life insurance of 200 is issued to (45).
Find
E
[
10
L
|
K
≥
10],
V ar
(
10
L
|
K
≥
10),
E
[
20
L
|
K
≥
20] and
V ar
(
20
L
|
K
≥
20).
k
E
[
k
L
|
K
≥
10]
V ar
(
k
L
|
K
≥
k
)
10
35.1565
1768.8725
20
87.1888
1787.8332
(c) A fully discrete 20-year term life insurance of 200 is issued to (45). Find
E
[
10
L
|
K
≥
10]
and
V ar
(
10
L
|
K
≥
10). (Answers: 7.7595 and 2500.7004).
(d) A fully discrete 20-year endowment insurance of 200 is issued to (45).
Find
E
[
10
L
|
K
≥
10] and
V ar
(
10
L
|
K
≥
10).(Answers: 70.25365 and 426.9982).
(e) A fully discrete whole life annuity of 100 is issued to (45).
The deferred pe-
riod is 20 years. Find
E
[
10
L
|
K
≥
10],
V ar
(
10
L
|
K
≥
10),
E
[
20
L
|
K
≥
20], and
V ar
(
20
L
|
K
≥
20).
k
E
[
k
L
|
K
≥
k
]
V ar
(
k
L
|
K
≥
k
)
10
293.7515
55 755.28905
20
940.0936
124 155.0802
7
(f) A fully discrete 20-payment years, 20-year deferred life insurance of 200 is is-
sued to (45).
Find
E
[
10
L
|
K
≥
10],
V ar
(
10
L
|
K
≥
10),
E
[
20
L
|
K
≥
20], and
V ar
(
20
L
|
K
≥
20).
k
E
[
k
L
|
K
≥
k
]
V ar
(
k
L
|
K
≥
k
)
10
27.2439
660.0018
20
87.1888
1787.8332
19. A semicontinuous 20-payment years, 30-year term life insurance of 2000 with a true
quarterly premiums is issued to (35). You are given
n
A
1
35+
n
:
30
-
n
|
¨
a
(4)
35+
n
:
20
-
n
|
0
0.068727
11.485955
10
0.097101
7.412344
25
0.079516
–
μ
(
x
)
=
0
.
00005
×
1
.
1
x
, x >
0
δ
=
6%
Questions:
(a) Find
P
(4)
. (Answer: 11.967162)
(b) Find
j
L
pour
j
= 10, 25.
(c) Find
10
L
if
T
= 12
.
3 and
25
L
if
T
= 27
.
4 . (Answers: 1714.206326 and 1731.775496)
(d) Find
E
[
j
L
|
T > j
] for
j
= 10
,
25. (Answers: 105.497276 and 159.031484)
(e) Find
10
Δ such that Pr (
10
L >
10
Δ
|
T >
10) = 0
.
05. (Answer: 1086.82437)
20. A semicontinuous 30-payment years, whole life insurance of 5000 is issued to (40). You
are given
n
A
40+
n
¨
a
40+
n
:
30
-
n
|
0
0.168110
13.497507
10
0.267636
11.060408
25
0.480688
4.223319
μ
(
x
)
=
0
.
00005
×
1
.
1
x
, x >
0
δ
=
6%
Questions:
8
(a) Find the premium (Answer: 62.274322)
(b) Determine
j
L
for
j
= 10, 25.
(c) Find
10
L
if
T
= 12
.
3 and
25
L
if
T
= 27
.
4 . (Answers: 4179.33902 and 4153.284302)
(d) Find
E
[
j
L
|
T > j
] for
j
= 10
,
25. (Answers: 649.398353 and 2140.43726)
(e) Given the following information
n
5000
v
n
-
P
¨
a
n
|
7
2918.4959
8
2686.2615
9
2467.5515
10
2261.5782
Find Pr (
10
L
≤
2500
|
T >
10)). (Answer: 0.922106)
21. A semicontinuous 15-payment years, 40-year endowment insurance of 3000 with a true
sixthly premiums is issued to (30). You are given
n
A
30+
n
:
40
-
n
|
¨
a
(6)
30+
n
:
15
-
n
|
0
0.065320
9.851216
10
0.102112
4.315066
25
0.159236
–
μ
(
x
)
=
0
.
00005
×
1
.
1
x
, x >
0
δ
=
6%
Questions:
(a) Find the premium
P
(6)
. (Answer: 19.89186334)
(b) Determine
j
L
for
j
= 10, 25.
(c) Find
10
L
if
T
= 12
.
3 and
25
L
if
T
= 27
.
4 . (Answers: 2569.767353 and 2597.663244)
(d) Find
E
[
j
L
|
T > j
] for
j
= 10
,
25. (Answers: 220.501668 and 477.707182)
(e) Find
10
Δ such that Pr (
10
L >
10
Δ
|
T >
10) = 0
.
05. (Answer: 1367.513815)
22. A semicontinuous 20-payment years, 20-year deferred life insurance of 4000 with true
semiannual premium is issued to (45). You are given
n
20
-
n
|
A
45+
n
A
45+
n
¨
a
(2)
45+
n
:
20
-
n
|
0
0.116304
0.213405
11.206181
10
0.225237
0.330918
7.229287
25
–
0.562461
–
9
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μ
(
x
)
=
0
.
00005
×
1
.
1
x
, x >
0
δ
=
6%
Questions:
(a) Find the premium
P
(2)
(Answer: 41.514295)
(b) Determine
j
L
for
j
= 10, 25.
(c) Find
10
L
if
T
= 12
.
3 and
25
L
if
T
= 27
.
4 . (Answers: -97.829717 and 3463.550992)
(d) Find
E
[
j
L
|
T > j
] for
j
= 10
,
25. (Answers: 600.827298 and 2249.844124)
23. A fully discrete whole life annuity of 12000 with true monthly benefits (1000 per month)
is issued to (45). The deferred period is 20 years and the premiums are payable four
times per year during the first 15 years. You are given
n
20
-
n
|
¨
a
(12)
45+
n
¨
a
(12)
45+
n
¨
a
(4)
45+
n
:
15
-
n
|
0
2.104258
13.151637
9.604725
10
4.075139
11.193107
4.923740
25
–
7.334093
–
μ
(
x
)
=
0
.
00005
×
1
.
1
x
, x >
0
δ
=
6%
Questions:
(a) Find the premium
P
(4)
. (Answer: 2629.028267)
(b) Determine
j
L
for
j
= 10, 25.
(c) Find
10
L
if
T
= 12
.
3 and
25
L
if
T
= 27
.
4 . (Answers: -6149.26756 and 27063.08647)
(d) Find
E
[
j
L
|
T > j
] for
j
= 10
,
25. (Answers: 35957.01937 and 88009.11398)
(e) Find Pr (
j
L <
-
10000
|
T > j
) for
j
= 10
24. A semicontinuous 25-payment years, whole life insurance of 10000 is issued to (45).
We assume that the mortality follows De Moivre’s law with
ω
= 100 and
δ
= 7
.
5%.
The premium is 239.7868.
(a) Find the retrospective and prospective reserves at
t
= 15. (Answer : 1586.551474)
(b) Find
V ar
(
15
L
|
T >
15). (Answer: 8417596.441)
(c) Find the Pr(
15
L <
5000
|
T >
15))
10
Analysis of Benefit Reserves
1. A 10-year fully discrete term insurance contract provides a death benefits of 10 if death
occurs in the first year, with the benefit decreasing by 1 per year. The contract is paid
by annual premiums of
P
each, payable for 5 years. If
k
|
q
x
=0.02 (
k
= 0
,
1
,
2
, ...
) and
i
= 0 what is the benefit reserve at the end of 3 years? (Answer: )
2. A fully discrete 30-payment years whole life insurance is issued to (40) with the fol-
lowing characteristics:
(1)
b
K
+1
=
1000,
K
= 0, 1, ..., 24
500,
K
= 25, 26, ...
(2)
π
k
=
3
π, k
=0, 1, ..., 14
2
π
,
k
= 15, 16, ...,29
0,
k
= 30, 31, ...
.
You are given:
k
A
40+
k
A
1
40+
k
:
25
-
k
|
¨
a
40+
k
:
15
-
k
|
¨
a
40+
k
:
30
-
k
|
0
0.202
0.114
10.005
13.414
10
0.316
0.149
4.386
10.806
20
0.458
0.112
–
6.929
40
0.744
–
–
–
(a) Define
L
and
j
L
.
(b) Find
π
,
10
V
,
20
V
and
40
V
. (Answer: 4.29, 120.97, 225.25, 372)
3. A semicontinuous whole life insurance is issued to (45) with the following characteris-
tics:
b
T
=
800
0
< T <
20
600
T >
20
π
k
=
4
π
k
= 0
,
1
, ...,
19
π
k
= 20
,
21
, ...
11
You are given:
(1)
n
A
45+
n
A
1
45+
n
:
20
-
n
|
0
0.229
0.121
10
0.356
0.137
20
0.505
–
(2)
n
¨
a
45+
n
:
20
-
n
|
¨
a
45+
n
0
10.067
12.842
10
7.171
10.739
20
–
8.251
Find
π,
10
V
and
20
V
. (Answers: 3.7544, 119.9, 272.0)
4. A semicontinuous 20-payment years whole life insurance with true semiannual benefit
premiums is issued to (40). The contract also has the following characteristics:
(1)
b
T
=
500, 0
< T <
20
200,
T >
20
(2)
π
(2)
k
=
3
π
(2)
,
k
= 0, 1, ..., 9
π
(2)
,
k
= 10, 11, ..., 19
0,
k
= 20, 21, ...
.
You are given:
(1)
n
n
E
40
¨
a
(2)
40:
n
|
A
1
40:
n
|
10
0.531
7.676
0.035
15
0.376
10.005
0.063
20
0.257
11.634
0.085
25
0.166
12.727
0.113
∞
0.000
14.094
0.202
Find
π
(2)
,
15
V
(2)
and
25
V
(2)
. (Answers: 2.4420, 80.9, 107.2)
5. A fully continuous 30-payment years whole life insurance is issued to (40) with the
following characteristics:
(1)
b
t
=
600, 0
< t <
25
800,
t >
25.
(2)
π
t
=
2
π
, 0
< t <
15
π
, 15
< t <
30
0,
t >
30.
.
12
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You are given:
k
A
40+
k
A
1
40+
k
:
25
-
k
|
a
40+
k
:
15
-
k
|
a
40+
k
:
30
-
k
|
0
0.202
0.114
10.005
13.414
10
0.316
0.149
4.386
10.806
20
0.458
0.112
–
6.929
40
0.744
–
–
–
(a) Define
L
and
t
L
.
(b) Find
π
,
10
V
,
20
V
and
40
V
. (Answers: 5.9268, 133.0, 302.9, 595.2)
6. A special fully discrete 25-premium years whole life insurance is issued to (35). The
death benefit is give by
b
k
+1
= 50(
k
+ 1),
k
= 0
,
1
, ...
You are given:
(1)
10
V
35
= 0.111;
(2) (
IA
)
35
= 3.352;
(3)
A
35
= 0.142;
(4) (
IA
)
45
= 4.191;
(5)
10
V
35:
25
|
= 0.227.
Find
10
V
(Answer: 198.6)
7. A special fully discrete 15-premium years, 30-year endowment insurance is issued to
(40). The death benefit is 3000 for the first 20 years and 2000 afterward. The survival
benefit is 4000.
You are given:
(1)
15
P
1
40:
30
|
= 0.0080
(2)
15
P
1
40:
20
|
= 0.0050
(3)
15
P
40:
30
|
= 0.0166
(4)
15
P
40:
20
|
= 0.0273
(5)
5
P
1
50:
20
|
= 0.0271
(6)
5
P
1
50:
10
|
= 0.0130
(7)
5
P
50:
20
|
= 0.0667
(8)
5
P
50:
10
|
= 0.1161
(9) ¨
a
50:
5
|
= 4.3016
Find
P
and
10
V
. (Answers: 55.4 and 732.1)
8. A special fully discrete whole life insurance is issued to (30) with
b
K
+1
=
e
0
.
06(
K
+1)
,
K
= 0
,
1
,
2
, ...
You are given that
δ
= 8% and
10
V
30
A
30
A
40
@
δ
= 2%
0.094
?
0.688
@
δ
= 6%
0.045
0.359
?
@
δ
= 8%
0.033
?
?
Define
10
L
and find
10
V
. (Answer: 0.6196)
13
9. A special fully continuous 15-payment years whole life insurance contract is issued to
(40). The death benefit is
$
1,000 during the first 10 years and it increases by
$
1,000
every 10 years. You are also given that
π
t
=
Pe
0
.
1
t
, 0
< t <
15 ,
μ
(
x
) = 0
.
03, and
δ
= 0
.
07.
(a) Find
P
.
(b) Define the prospective loss random variable at time 7.3.
(c) Find the reserve (retrospective and prospective) at time 7.3.
(d) Given that reserve at time 26 is
V
, give the reserve at time 7.3 in terms of actuarial
symbols.
(e) Find
P
[
7
.
3
L
≥
1000
|
T
(40)
≥
7
.
3] using numerical methods.
10. You are given
i
= 10%,
q
x
= 0
.
1,
q
x
+1
= 0
.
2, and
q
x
+2
= 0
.
25. A 3-year fully discrete
endowment insurance has a death benefit of 1000 and an endowment amount of 2000.
The annual benefit premium is 491.59. Find the expected value of the unconditional
third year net cash loss
C
2
(Answer: 791.51)
11. You are given
10
V
= 2000,
b
11
= 10
,
000,
π
10
= 15,
q
x
+10
= 0
.
012 and
i
= 0
.
06. The
policyholder is given the option of cancelling the premium
π
10
and reducing the death
benefit for the coming year to 10000-C. The arrangement is set up so that
11
V
will be
the same either way. Find C. (Answer:1325)
12. A fully discrete 3-year endowment insurance is issued to (
x
). The benefit is equal to
1000+terminal reserve in the year of death. The endowment benefit is 1000. You are
given
i
= 10%,
q
x
= 0
.
25,
q
x
+1
= 0
.
5, and
q
x
+2
= 0
.
8.
Find the first year terminal
benefit reserve. (Answer: 551.36)
13. A fully discrete 20-payment years, 20-year deferred insurance of 100 is issued to (40).
The policyholders have the choice between two types of death benefit during the de-
ferred period:
•
death benefit I = sum of the premiums accumulated at the rate
j
= 80%
i
;
•
death benefit II = 80%
×
[sum of the premiums accumulated at the rate
j
=
i
].
You are given:
(1)
A
x
= 0.4,
x
= 0
,
1
,
2
,
· · ·
(2)
i
= 6%.
Find the premium and
5
V
for both options.(Answers:
5
V
I
= 5
.
8312 and
5
V
II
= 5
.
5241)
14
14. A fully discrete whole life annuity of 1000 is issued to (35). The deferred period is 25
years. The death benefit during the deferred period is
100%
×
[terminal reserve in the year of death]
.
You are given:
(1) ¨
a
60
= 12;
(2)
d
= 6%.
Find the premium and
10
V
(determine the prospective and retrospective formulas for
the reserve
j
V
(
j
= 0, 1, ..., 24)). (Answers: 194.7621 and 2781)
15. A fully discrete whole life annuity of
$
g is issued to (35). The deferred period is 30
years.
The death benefit during the deferred period (30 years) is equal to 60% of terminal
reserve in the year of death plus 200
$
.
You are given:
(1)
A
x
= 0.4,
x
= 0
,
1
, ...
(2)
i
= 6%.
Find g if we assume that the premium is
$
20. (Answer: 136.2931)
16. A fully discrete 30-year endowment insurance of 25000 is issued to (35). You are given:
(1)
i
= 7.5%;
(2)
29
V
= 22971.71.
Find
P
. (Answer: 284.1)
17. A semicontinuous whole life insurance of 10000 is issued to (30).
You are given:
(1)
10
V
(
A
30
) = 0.0801
(2)
11
V
(
A
30
) = 0.0901
(2)
P
(
A
30
) = 0.0074
(3)
δ
= 5%
(4)
q
40
= 0.0020
(5)
q
41
= 0.0022
(a) Find
10
1
4
V
based on the classical approximation. (Answer: 881.5)
(b) Find
10
1
4
V
and
11
2
3
V
based on the UDD assumption.(Answers: 881.265 and 994.593)
(c) Find
10
1
4
V
and
11
2
3
V
based on the constant force of mortality assumption.(Answers:
881.262 and 994.455)
15
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18. A semicontinuous 20-year endowment insurance of 10000 with true semiannual benefit
premiums is issued to (45).
You are given:
(1)
10
V
(2)
(
A
45:
20
|
) = 0.3568
(2)
11
V
(2)
(
A
45:
20
|
) = 0.4030
(2)
P
(2)
(
A
45:
20
|
) = 0.0328
(3)
δ
= 6%
(4)
q
55
= 0.0181
(5)
q
56
= 0.0200
(6)
q
64
= 0.0423
(a) Find
10
1
4
V
(2)
and
10
2
3
V
(2)
based on the classical approximation.(Answers: 3765.50
and 3985.33)
(b) Find
11
1
4
V
(2)
,
11
2
3
V
(2)
,
19
V
(2)
,
19
1
3
V
(2)
and
19
3
4
V
(2)
based on UDD assumption.
(c) Find
11
1
4
V
(2)
,
11
2
3
V
(2)
,
19
V
(2)
,
19
1
3
V
(2)
and
19
3
4
V
(2)
based on constant force of
mortality assumption.
t
UDD
Constant force
11
1
4
4228
.
15
4227.92
11
2
3
4452.48
4452.21
19
9110.05
9110.18
19
1
3
9452.26
9452.25
19
3
4
9851.93
9851.92
Multiple Life Functions
1. You are given:
(1)
20
p
x
= 0.9
(2)
20
p
y
= 0.8
T
(
x
) and
T
(
y
) are independent.
Find
20
p
xy
and
20
p
xy
(Answers: 0.72 and 0.98).
2. You are given:
•
f
T
(
x
)
T
(
y
)
(
t, s
) =
1
4
α
2
e
-
α
(
t
+
s
)
+
3
4
β
2
e
-
β
(
t
+
s
)
,
t >
0 and
s >
0;
•
α
= 0.01 and
β
= 0.02
Find
P
(20
< T
(
xy
)
<
50) and
P
(20
< T
(
xy
)
<
50). (Answers : 0.311 and 0.249).
3. You are given:
16
•
10
p
x
= 0
.
9 and
30
p
x
= 0
.
5
•
10
q
y
= 0
.
15 and
30
q
y
= 0
.
45
T
(
x
) and
T
(
y
) are independent.
Find ...
•
Probability that the first death between (
x
) and (
y
) happens in the time interval
(10,30) (Answer : 0.49);
•
Probability that no more than one death between (
x
) and (
y
) happens in the time
interval (10,30) (Answer : 0.88);
•
Probability that at least one of the lives (
x
) and (
y
) will die in the time interval
(10,30)(Answer : 0.58);
•
Probability that the second death between (
x
) and (
y
) happens in the time interval
(10,30)(Answer : 0.21);
•
Probability that only one death between (
x
) and (
y
) happens in the time interval
(10,30)(Answer : 0.46).
4. You are given:
t
p
x
= exp(
-
0
.
02
t
)
, t >
0
t
p
y
= exp(
-
0
.
04
t
)
, t >
0.
δ
= 6%.
T
(
x
) and
T
(
y
) are independent.
Find
a
xy
and
a
xy
.
(Answers : 14.17 and 8.33)
5. You are given:
•
20
q
40:40
= 0
.
0121
•
40
q
40:40
= 0
.
3538
•
q
60:60
= 0
.
0002
•
q
80:80
= 0
.
0077
T
(
x
) and
T
(
y
) are independent.
Find...
(a)
P
((40) or (60) die in the next 20 years)
(b)
P
((40) or (60) die in the 21th year)(Answers :0.595 and 0.05203).
6. You are given:
•
f
T
(
x
)
T
(
y
)
(
t, s
) =
1
4
α
2
e
-
α
(
t
+
s
)
+
3
4
β
2
e
-
β
(
t
+
s
)
,
t >
0 and
s >
0;
17
•
α
= 0.01 and
β
= 0.02;
•
δ
= 5%;
•
x
= 30 and
y
= 40.
Find
A
x
,
A
xy
,
A
xy
,
a
xy
, and
a
xy
(Answers : 0.2559, 0.4048, 0.1071, 17.857, 11.9048).
7. The insured
x
and his partner
y
can choose between options A and B:
•
Option A: a fully continuous annuity of
$
80,000 is payable during the first 5 years
if only one of (
x
) and (
y
) is alive, after 5 years, the annuity is payable if (
x
) or
(
y
) are alive.
•
Option B: a fully continuous annuity of
K
$
is payable if (
x
) and (
y
) are alive. The
annuity is reducing to
$
90%K on the death of (
y
) and to
$
60%K on the death of
(
x
).
You are given for
x
= 60 and
y
= 65:
¯
a
x
= 8
.
544
¯
a
y
= 8
.
701
¯
a
xy
= 7
.
356
5
p
x
= 0
.
923
¯
a
x
+5
= 7
.
666
¯
a
y
+5
= 7
.
854
¯
a
x
+5:
y
+5
= 6
.
326
5
p
y
= 0
.
929
v
5
= 0
.
650
T
(
x
) and
T
(
y
) are independent
If options A and B have the same APV, find
K
. (Answer:52501)
8. You are given:
•
t
p
xy
=
1
20
t
p
x
+
1
20
t
p
y
+
9
10
t
p
x t
p
y
, t
≥
0
•
t
p
x
=
e
-
0
.
006
t
, t
≥
0
•
t
p
y
=
e
-
0
.
008
t
, t
≥
0
•
δ
= 5%
Find
t
p
xy
μ
xy
(
t
)
,
A
xy
and
A
xy
(Answers :
A
xy
=0.2091 and
A
xy
=0.03595 ).
9. Consider a fully continuous annuity of 1000 per annum payable while at least one of
two lives
x
and
y
is living. Let
Z
be the present value random variable of the contract
where
E
[
Z
] = 1000
a
xy
.
You are given:
(1)
μ
x
+
t
=
B
1
c
x
+
t
1
,
t >
0;
(2)
μ
y
+
t
=
B
2
c
y
+
t
2
,
t >
0;
(3)
x
= 60,
y
= 55;
(4)
B
1
= 0
.
00005
, c
1
= 10
0
.
05
;
(5)
B
2
= 0
.
00004
, c
2
= 10
0
.
04
;
(6)
δ
= 8%.
18
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T
(
x
) and
T
(
y
) are independent.
Find
P
(
Z
≤
5000) (Answer: 0.020157).
10. You are given:
•
F
T
(
x
)
T
(
y
)
(
t, s
) =
F
T
(
x
)
(
t
)
F
T
(
y
)
(
s
) +
1
4
F
T
(
x
)
(
t
)
F
T
(
y
)
(
s
)
×
(1
-
F
T
(
x
)
(
t
))(1
-
F
T
(
y
)
(
s
)),
t >
0 and
s >
0;
•
F
T
(
x
)
(
t
) = 1
-
exp
(
-
0
.
02
t
)
, t >
0;
•
F
T
(
y
)
(
t
) = 1
-
exp
(
-
0
.
04
t
)
, t >
0;
•
δ
= 5%;
•
x
= 30 and
y
= 40.
Find
a
(
xy
):
30
|
(Answers : 14.02689).
11. Let
a
y
|
x
be the actuarial present value of a fully continuous reversionary annuity (see
section 9.7.3 from the Bowers), which represents an annuity of 1 per year payable
continuously during the existence of the status (
x
), but only after the failure of the
second status (
y
).
•
Example 1:
a
y
|
x
=
E
[
Z
] where
Z
=
a
T
(
x
)
|
-
a
T
(
xy
)
|
. We have
a
y
|
x
=
E
[
a
T
(
x
)
|
-
a
T
(
xy
)
|
]
=
E
[
a
T
(
x
)
|
]
-
E
[
a
T
(
xy
)
|
]
=
a
x
-
a
xy
.
•
Example 2: Consider the
APV
of fully continuous annuity of
$
R payable to (
x
)
as long as he lives jointly with (
y
).
The annuity is reducing to
$
90%R on the
death of (
y
) and to
$
60%R on the death of (
x
).
APV
=
R
a
xy
+ 0
.
9
R
a
y
|
x
+ 0
.
6
R
a
x
|
y
=
R
a
xy
+ 0
.
9
R
(
a
x
-
a
xy
) + 0
.
6
R
(
a
y
-
a
xy
)
•
Example 3: We also have
a
y
:
n
||
x
=
a
x
-
a
xy
:
n
|
a
y
|
x
:
n
|
=
a
x
:
n
|
-
a
xy
:
n
|
and
a
y
:
n
||
x
=
a
x
-
a
x
:(
y
:
n
|
)
=
a
x
-
(
a
xy
+
a
x
:
n
|
-
a
xy
:
n
|
)
.
19
You are given:
t
p
x
= exp(
-
0
.
02
t
)
, t >
0.
t
p
y
= exp(
-
0
.
04
t
)
, t >
0.
δ
= 6%.
T
(
x
) and
T
(
y
) are independent.
(a) Find
a
y
|
x
and the
APV
in examples 2 and 3 when
n
= 15 (Answers:
Ex1=
4.16667, Ex2= 13.08333, Ex3= 2.38744).
12. You are given:
(1)
T
(
x
) = min(
T
*(
x
),
Z
);
(2)
T
(
y
) = min(
T
*(
y
),
Z
);
(3)
T
*(
x
),
T
*(
y
),
Z
are independent random variables;
(4)
P
(
Z
≤
t
) =
exp
(
-
0
.
02
t
) ,
t >
0;
(5)
P
(
T
*(
x
)
> t
) =
S
T
*
(
x
)
(
t
) =
t
p
*
x
=
100
-
x
-
t
100
-
x
, 0
< t <
100
-
x
;
(6)
P
(
T
*(
y
)
> t
) =
S
T
*
(
y
)
(
t
) =
t
p
*
y
=
120
-
y
-
t
120
-
y
,0
< t <
120
-
y
;
Define
S
T
(
x
)
T
(
y
)
(
t, s
),
f
T
(
x
)
T
(
y
)
(
t, s
),
S
T
(
xy
)
(
t
), and
S
T
(
xy
)
(
t
).
13. You are given:
(1)
T
(
x
) = min(
T
*(
x
),
Z
1
);
(2)
T
(
y
) = min(
T
*(
y
),
Z
1
);
(3)
T
*(
x
),
T
*(
y
),
Z
1
are independent random variables;
(4)
P
(
Z
1
> t
) =
exp
(
-
λt
) ,
t >
0;
(5)
P
(
T
*(
x
)
> t
) =
S
T
*
(
x
)
(
t
) =
t
p
*
x
= e
-
0
.
020
t
,
t >
0;
(6)
P
(
T
*(
y
)
> t
) =
S
T
*
(
y
)
(
t
) =
t
p
*
y
= e
-
0
.
015
t
,
t >
0;
(7)
λ
= 0.005;
(8)
δ
= 6%
.
(a) Find
20
q
1
xy
,
20
q
2
xy
,
o
e
xy
,
E
(
Z
) =
a
xy
,
V ar
(
Z
) and
P
(
Z
≤
5).
(Answers : 0.2753, 0.049299, 25, 10, 25, 0.2116)
(b) Find
E
[
Z
] =
A
(
xy
):
10
|
and
V ar
[
Z
] (Answers : 0.6207 and 0.01612).
20
14. You are given:
(1)
T
(
x
) = min(
T
*(
x
),
Z
1
);
(2)
T
(
y
) = min(
T
*(
y
),
Z
1
);
(3)
T
*(
x
),
T
*(
y
),
Z
1
are independent random variables;
(4)
P
(
Z
1
> t
) = exp(-
λt
) ,
t >
0;
(5)
δ
4.8269%
5.8269%
A
*
x
0.185
0.161
A
*
y
0.271
0.249
A
*
xy
0.322
0.294
A
*
1
xy
0.085
0.074
A
*
x
1
y
0.237
0.220
(6)
20
q
*
x
= 0.10 and
20
q
*
y
= 0.15
(7)
λ
= 0.01
(8)
δ
= 4
.
8269%
Find ...
(a)
A
xy
and
A
xy
(Answers = 0.415 and 0.268) ;
(b)
20
q
x
,
20
q
y
,
20
q
xy
and
20
q
xy
(Answers = 0.263, 0.304, 0.373, 0.194).
(c)
A
2
xy
(Answer = 0.1098).
15. You are given:
(1)
T
(
x
) = min(
T
*(
x
),
Z
);
(2)
T
(
y
) = min(
T
*(
y
),
Z
);
(3)
T
*(
x
),
T
*(
y
),
Z
are independent random variables;
(4)
P
(
Z > t
) =
exp
(
-
0
.
005
t
) ,
t >
0;
(5)
t
p
*
x
= e
-
0
.
020
t
,
t >
0;
(6)
t
p
*
y
= e
-
0
.
015
t
,
t >
0;
(7)
δ
= 6%
.
(a) Find
a
y
|
x
(Answer : 1.7647).
(b) Find
a
y
:
15
||
x
and
a
y
:
15
||
x
(Answers : 3.996 and 1.05602).
16. You are given that:
•
The force of mortality for (
x
) and (
y
) follows a De Moivre law
•
T
(
x
) and
T
(
y
) are independent random variables;
•
∞
q
1
xy
=
2
3
where
x
= 80 and
y
= 60
21
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•
i
= 5%
Find
A
xy
and
A
2
xy
where
x
= 50 and
y
= 70 (Answers : 0.47339 and 0.0999341).
17. You are given that:
•
μ
x
= 0
.
03
,
x >
20
•
μ
y
= 0
.
05
,
y >
20
•
A
=
APV
of a continuous 30-year term life insurance issued to (
x
) and (
y
) that
pays
$
30,000
$
indexed by a force of interest of 5% until the first death and payable
at the second death.
•
B
=
APV
of a continuous 30-year term life insurance issued to (
x
) and (
y
) that
pays
M
indexed by a force of interest of 5% beginning at the first death and
payable at the second death.
•
x
= 35,
y
= 40,
A
=
B
and
δ
= 8%.
•
T
(
x
) and
T
(
y
) are independent random variables.
Find
M
(Answer : 27325).
18. A fully discrete special whole life insurance of 5,000 is issued to (
x
) and (
y
). The death
benefit (5,000) is payable if and only if (
x
) and (
y
) die during the same year.
You are given:
•
The force of mortality for (
x
) follows a De Moivre law with
ω
= 115
•
The force of mortality for (
y
) follows a De Moivre law with
ω
= 105
•
d
= 6
.
5%
•
Let
Z
be the present value random variable of this special insurance contract.
•
x
= 42 and
y
= 39
.
•
T
(
x
) and
T
(
y
) are independent.
Define
Z
and find
E
(
Z
) and
V ar
(
Z
) (Answers : 14.7511 and 35843.68).
19. You are given:
•
x
= 61 and
y
= 60
•
i
= 6%;
•
μ
(
x
) = 0
.
02 and
μ
(
y
) = 0
.
04
•
T
(
x
) and
T
(
y
) are independent.
Find ¨a
(
xy
):
5
|
(Answer: 15.02102).
22
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20. An annuity-due of
$
R is payable to (
x
) as long as he lives jointly with (
y
). The annuity
is reducing to
$
85%R on the death of (
y
) and to
$
55%R on the death of (
x
). You are
given:
•
x
= 64 and
y
= 60
•
i
= 6%
•
μ
(
x
) = 0
.
02 and
μ
(
y
) = 0
.
04
•
T
(
x
) and
T
(
y
) are independent.
Find the
APV
if
R
= 10000 (Answer : 135808.7).
21. An insurance contract is issued to (
x
) and (
y
). The death benefit is given by:
(1) 10000
$
if (
x
) dies first
(2) 8000
$
if (
y
) dies first
(3) 5000
$
at the moment of the second death.
You are given:
(1)
A
x
= 0.102;
(2)
A
y
= 0.161;
(3)
A
1
xy
= 0.056
(4)
A
x
1
y
= 0.140.
Find
APV
of this contract (Answer : 2015).
Multiple Decrement Models
1. For a double-decrement model, you are given:
•
μ
(
j
)
x
(
t
) =
j
+1
60+
t
,
t >
0 and
j
= 1
,
2
Calculate
f
T,J
(20
,
2),
f
T
(20),
f
J
(2), and
f
J
|
T
(1
|
10) .
(Answers: 0.0089, 0.0148, 0.6,
and 0.4)
2. You are given three separate two-decrement models:
(a)
μ
(1)
x
(
t
) =
c
and
μ
(2)
x
(
t
) =
d
(b)
μ
(1)
x
(
t
) =
1
50
-
t
and
μ
(2)
x
(
t
) =
1
50
-
t
for 0
< t <
50
(c)
μ
(1)
x
(
t
) =
a
and
μ
(2)
x
(
t
) =
1
50
-
t
for 0
< t <
50
For which of the models are the marginal distributions of
T
(
x
) and
J
independent?
(Answer: (a) and (b) )
23
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3. The joint density function of two decrements
T
1
(
x
) and
T
2
(
x
) is
f
T
1
,T
2
(
s, t
) = 4
st
for
0
≤
s
≤
1 and 0
≤
t
≤
1.
(a) Find
f
J
(1) the probability that decrement 1 occurs before 2. (Answer: 0.5)
(b) Find
f
T
(
t
) the marginal density of
T
=
min
(
T
1
, T
2
) (Answer:
f
T
(
t
) = 4
t
(1
-
t
2
)
0
≤
t
≤
1)
(c) Find
f
T,J
and show that
T
and
J
are independent.
(d) Find
E
[
T
](Answer: 8/15)
4. For a double-decrement model, you are given:
•
μ
(1)
10
(
t
) =
1
30
-
t
, 0
≤
t <
30
•
μ
(
τ
)
10
(
t
) =
50
-
2
t
600
-
50
t
+
t
2
, 0
≤
t <
20
Calculate the probability that (10) will terminate from cause 2 during the 6th year.
(Answer: 49/1200)
5. A multiple decrement model has 2 decrements. You are given
q
0
(1)
x
= 0
.
2 and
q
0
(2)
x
= 0
.
4.
Assuming constant force of decrement in each year of age, find
0
.
5
|
0
.
5
q
(2)
x
.
(Answer:
0.148)
6. A multi-decrement model with two causes of decrements has forces of decrement given
by
•
μ
(1)
x
(
t
) =
1
100
-
(
x
+
t
)
•
μ
(2)
x
(
t
) =
2
100
-
x
(select and ultimate table)
Calculate
3
p
(
τ
)
65
,
3
|
q
(1)
65
, and
3
q
(2)
65
. (Answers: 0.77024, 0.23395, and 0.15098)
7. You are given the following absolute rates
x
q
0
(1)
x
q
0
(2)
x
q
0
(3)
x
62
0.020
0.030
0.20
63
0.022
0.034
0.10
64
0.028
0.040
0.12
(a) Assuming the constant force for multiple decrement, find
0
.
5
p
(
τ
)
63
,
0
.
5
q
(2)
63
, and
0
.
5
q
(2)
63
.
5
(Answers: 0.9221, 0.01661, 0.0166129)
(b) Assuming the uniform distribution of decrement for the associated single decre-
ments, find
0
.
5
p
(
τ
)
63
,
0
.
5
q
(2)
63
, and
0
.
5
q
(2)
63
.
5
(Answers: 0.9235776, 0.01648462, 0.01674609)
24
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8. In the single decrement tables associated with double decrement table we are given
model
q
0
(1)
x
= 0
.
2 and
q
0
(2)
x
= 0
.
3.
Actuary A calculates
0
.
5
q
(
τ
)
A
x
assuming that each
of the decrements is uniformly distributed in the multiple decrement table. Actuary
B calculates
0
.
5
q
(
τ
)
B
x
assuming that each the associated single tables are uniformly
distributed. Find
0
.
5
q
(
τ
)
B
x
-
0
.
5
q
(
τ
)
A
x
(Answer: 0.015)
9. In considering a 3-decrement model, Actuaries A and B both model decrement 3 as
a discrete decrement, and decrements 1 and 2 as continuous decrements uniformly
distributed in their associated single decrement tables. Actuary A assumes decrement
3 occurs at the start of the year and calculate
q
(3)
A
x
. Actuary B assumes that decrement
3 occurs at the end of the year and also
q
(3)
B
x
. Suppose that
q
0
(1)
x
=
a
,
q
0
(2)
x
=
b
, and
q
0
(3)
x
=
c
. Find
q
(3)
A
x
-
q
(3)
B
x
(Answer: c(a+b-ab))
10. In a two decrement table, decrement 1 has constant force of decrement in each year of
age and decrement 2 is uniformly distributed in its associated single decrement table.
Given that
q
0
(1)
x
=
q
0
(2)
x
= 0
.
4, find
q
(1)
x
. (Answer: 0.3269)
11. You are given the following probabilities of decrement
x
q
(1)
x
q
(2)
x
62
0.020
0.08
63
0.022
0.34
64
0.028
0.40
65
0.000
1.00
(a) Find
2
p
(
τ
)
62
,
d
(1)
63
,
l
(2)
62
if
l
(
τ
)
62
= 100
,
000 (Answers: 0.5742, 1980, 94412)
(b) Find the joint probability function of
K
(62) and
J
.
(c) Find the probability function of
K
(62).
(d) Find the probability function of
J
. (Answer:Pr[
J
= 1]=0.055878)
(e) Find the conditional probability functions of
K
(62)
|
J
= 1 and
K
(62)
|
J
= 2.
(f) Find the conditional probability functions of
J
|
K
(62) = 0,
J
|
K
(62) = 1,
J
|
K
(62) =
2 and
J
|
K
(62) = 3.
(g) Find
E
[
K
(62)],
E
[
K
(62)
|
J
= 1] and
E
[
K
(62)
|
J
= 2].
(Answers:
1.802642,
0.929804, and 1.85430)
(h) Find APV at age 62 for a 3-year discrete term benefit that pays 1000 at the end
of the year of decrement if due to decrement 1 if
i
= 0
.
05. (Answer: 50.8952)
12. For a triple decrement model, you are given:
•
q
0
(1)
65
= 0
.
035 and
q
0
(2)
65
= 0
.
02
•
Decrements 1 and 2 are assumed to be uniformly distributed over each year age
in the associated single decrement tables.
25
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•
t
p
0
(3)
65
=
1
-
0
.
1
t
0
≤
t <
1
0
t
= 1
Calculate
q
(3)
65
(Answer: 0.948)
13. In a 3-decrement model we have
(a) decrement 1 is discrete, with all occurrences of decrement taking place at the start
of the year age
(b) decrement 2 is discrete, with all occurrences of decrement 2 taking place at time
α
(0
< α <
1)
(c) decrement 3 is UDD in its associated single decrement table.
Find
q
(3)
x
in terms of absolute rates of decrement (Answer:
q
0
(3)
x
(1
-
q
0
(1)
x
)[
α
+(1
-
α
)(1
-
q
0
(2)
x
)])
14. For a triple-decrement table, you are given:
•
Decrement 1 is UDD over each year of age;
•
Decrement 2 occurs at the end of the year;
•
Decrement 3 occurs only at the beginning of the year;
•
x
l
(
τ
)
x
q
0
(1)
x
q
0
(2)
x
q
0
(3)
x
60
100000
0.0625
0.20
0.20
61
-
-
0.20
0.15
62
37536
-
-
-
Calculate
q
(1)
61
,
q
(1)
60
,
q
(2)
60
, and
q
(3)
60
.(Answers: 0.068, 0.05, 0.15 and 0.2)
15. A multiple decrement model has 2 decrements
•
q
(1)
x
= 0
.
2
q
(
τ
)
x
,
x
= 20
,
21
, ...,
41
•
q
(1)
x
= 0
.
1
q
(
τ
)
x
,
x
= 42
,
43
, ...
•
x
l
x
40
9313
41
9287
42
9260
43
9230
59
8293
60
8188
26
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(a) Assuming constant force of decrement in each year of age, find
20
q
0
(2)
40
,
20
q
(2)
40
, and
a
0
(1)
40:
1
|
if
δ
= 7% (Answers: 0.1088983, 0.108149897, and 0.9655)
(b) Assuming each decrement is uniformly distributed in the multiple decrement, find
20
q
0
(1)
40
and
20
q
(1)
40
. (Answers: 0.0133549 and 0.01264899)
16. You are given
•
t
q
0
(1)
x
= 0
.
1
t
, 0
≤
t
≤
1
•
t
q
0
(2)
x
=
0
.
2
t
0
≤
t <
1
1
t
= 1
•
μ
(3)
x
+
t
= 0
.
05, 0
≤
t
≤
1
•
δ
= 7%
Find
A
(2)
1
x
:
1
|
,
A
(2)
1
x
:
1
|
, and
A
(1)
1
x
:
1
|
. (Answers: 0.8178138, 0.8114574 , and 0.08499809)
17. A special whole life insurance is issued to (
x
).
The basic death benefit is 1 during
policy year one and 2 thereafter.
An additional benefit of 2 is provided if death is
accidental. You are given:
•
Benefit are payable at the moment of death
•
the force of mortality due to accidental death is
μ
(
ad
)
x
(
t
) = 0
.
005
•
μ
(
τ
)
x
(
t
) = 0
.
04
•
δ
= 0
.
06
Find the single benefit premium for this insurance.(Answer: 0.862)
18. A fully continuous whole life insurance with double indemnity provision is issued to
(40). There is only one premium payable at the beginning of the contract. The benefit
amount is
•
B
x
+
t
=
e
0
.
1
t
upon death due to cause
J
= 1 before age 65
•
B
x
+
t
= 2 upon death due to cause
J
= 2 before age 65
•
B
x
+
t
= 1 upon death after age 65
We also suppose that
μ
(1)
x
(
t
) = 0
.
02,
μ
(2)
x
(
t
) = 0
.
03 and
δ
= 5%. Find the
A
(the APV)
and the variance of
L
(the prospective loss). (Answers: 1.091792 and 0.5931536)
19. A fully discrete 4-year term life insurance with a double indemnity provision is issued
to (
x
). The benefit amount is
B
(1)
x
+
k
+1
=
k
+ 1 for decrement
J
= 1 and
B
(2)
x
+
k
+1
= 2 for
decrement
J
= 2. We also suppose
q
(1)
x
= 0
.
02,
q
(2)
x
= 0
.
03 and
i
= 5%. Find the level
premiums (payable for the first 4 years) and the 2nd year terminal reserve. (Answers:
0.1023857 and 0.0399002)
27
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Multiple State Models
1. You are given the 4-state Markov model to model multiple lives. State 1 refers to both
x
and
y
alive, State 2 refers to
x
alive and
y
failed, State 3 refers to
y
alive and
x
failed, and State 4 refers to both
x
and
y
failed. The force of force of transitions are:
λ
1
i
(
t
) = 0
.
01
i
, for
i
= 2
,
3,
λ
24
(
t
) = 0
.
02 + 0
.
02
t
, and
λ
34
(
t
) = 0
.
02
t
.
(a) Find
q
xy
.
(b) Find the probability that
x
dies first.
(c) Find
A
xy
:
3
|
if
δ
= 5%.
2. You are given the 4-state Markov model to model sickness-death events.
State 0
refers to healthy, State 1 refers to light disability, State 2 refers to heavy disability,
and State 3 refers to dead.
The force of transitions are:
λ
0
i
(
t
) = 0
.
01, for
i
= 1
,
3,
λ
1
i
(
t
) = 0
.
01 +
i/
100 for
i
= 0
,
2
,
3, and
λ
2
i
(
t
) =
i
2
/
100 for
i
= 1
,
3.
(a) Write the Kolmogorov forward differential equations with the initial condition for
t
p
(0)
00
,
t
p
(0)
01
,
t
p
(0)
02
, and
t
p
(0)
03
.
(b) A discrete 2-year sickness and term insurance contract is issued to a healthy (
x
)
at time 0.
The benefit is 100
i
if (
x
) is in state
i
at the end of the year.
The
premiums are paid at the beginning each year if (
x
) is healthy. Using the Euler’s
approximation (step size of 0.5) and
d
= 0
.
95, find the premium and the reserve
at time 1.
3. You are given the 3-state Markov model to model sickness-death events
Healthy
0
Sick
1
→
Dead
2
The force of transitions are:
λ
01
(
t
) = 0
.
1 + 0
.
1
t
,
λ
10
(
t
) = 0
.
1 + 0
.
05
t
, and
λ
12
(
t
) = 0
.
1
t
.
(a) Using the Euler’s approximation (step size of 0.5), find the probability that a life
who is healthy at 0 is healthy at time 1. (Ans: 0.8819)
(b) Using the Euler approximation (step size of 0.5), find the probability that a life
who is sick at 0 is healthy at time 1.(Ans: 0.1056)
(c) Using the Euler approximation (step size of 0.5), find the probability that a life
who is sick at 0.5 is dead at time 1.5.(Ans: 0.070625)
(d) A discrete 2-year sickness and term insurance contract is issued to a sick (
x
) at
time 0. The benefit is 100 is payable at the end of the year if (
x
) is sick at the
end of the year. A death benefit of 200 is payable at the end of the year if (
x
)
dies during the year. The premium is paid at thew beginning and also at time 1
if (
x
) is healthy. Using the Euler’s approximation (step size of 0.5) and
d
= 0
.
05,
find the premium and the reserve at time 1. (Ans: State0 =-130.54 and State 1=
97.34)
28
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Expenses
1. A semicontinuous 10-year term life insurance of 1000 is issued to (40). The expenses
are given in the following table:
First Year
Renewal Years
% of
G
Constant
% of
G
Constant
20%
5
5%
2
We also suppose
μ
x
(
t
) = 0
.
02 and
i
= 5%. Find
G
the level expense-loaded premiums
and the 5th year terminal reserve. (Answers: 23.365285 and -3.80627888)
2. A fully discrete 20-year term life insurance of
$
1,000,000 is issued to (40). The expenses
are 4% of
G
per year plus
$
500 per year. There is also an expense of
$
10,000 for future
investigation and lawyers, which is assumed to be payable at the same time as the
death benefit. We also suppose
q
x
= 0
.
02 and
i
= 5%. Find
G
the level expense-loaded
premiums and the 10th year terminal reserve. (Answers: 20560.515873 and 0)
3. A special semi-continuous 15-premium years, 5-year deferred 20-year term life insurance
contract is issued to (40). The annual expenses are 2% of the premium paid G for the
first three years and 1% of the premium paid G afterward. In addition, the annual per
contract expense is
$
2 per year and the claim settlement expense is
$
50. You are given
that
δ
is 7% and the death benefit amount is
b
t
=
e
0
.
03
t
for 5
≤
t
≤
25.
(a) Give the expression for the gross premium G in terms of actuarial symbols.
(b) Give the expression for
2
V
in terms of
17
V
and actuarial symbols.
Universal Life
1. For a universal life policy with death benefit of 150,000 plus the account value on (40),
you are given that the interest rate is 4% and
Year
Annual
Premium
Cost of Insurance
Annual
End of Year
Premium
Charge
Rate per 1000
Expense
Account Value
1
3000
0.3
1.22
75
-
2
3500
0.1
1.27
R
-
3
2500
0.1
1.32
R
7400
Find R. (Ans: 14.54)
2. For a universal life policy with death benefit of 50,000 on (60), you are given that the
credited interest rate is 5%, the corridor factor is 2.2, and
Annual
Premium
Cost of Insurance
Annual
Surrender
Interest for
Premium
Charge
Rate per 1000
Expense
Charge
COI
10000
0.1
3
100
0.4
0.06
29
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(a) Find the cost of insurance for first and fourth years (Ans: 115.40 and 135.12).
(b) Find the cash surrender value at time 4. (Ans: 23870.99)
30
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1/2 Yr
Full-Yr
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0.9578
0.9174
0.9578
0.9174
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0.9174
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1.8753
1.7591
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0.7722
2.7540
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0.6499
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0.9174
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1.7591
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0.7722
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0.7084
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1.0000 1.0000 1.0000 1.0000
1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000
2.0300 2.0350 2.0400 2.0450 2.0500 2.0550 2.0600 2.0650 2.0700 2.0750 2.0800 2.0850 2.0900
3.0150 3.0301 3.0452 3.0604 3.0756
3.0909 3.1062 3.1216 3.1370 3.1525 3.1680 3.1836 3.1992 3.2149 3.2306 3.2464
4.0301 4.0604
3.2622
4.0909 4.1216 4.1525 4.1836 4.2149 4.2465 4.2782 4.3101
4.3423
4.3746 4.4072 4.4399 4.4729 4.5061 4.5395
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Needvexcel solution...
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1
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Annuity X and Y provide the following payments:
End of Year
Y
1-10
1
K
11-20
21-30
1
K
Annuity X and Y have equal present values at an effective annual interest rate of 7.1773%. What is
present value of X? What is K?
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Present value of an ordinary
annuity.
Fill in the missing present values in the following table for an ordinary annuity.
Number of
Payments or
Years
Annual
Interest Rate
Future Value
Annuity
Present Value
5
8%
0
$213.22
?
16
15%
0
$3,317.78
?
29
4.5%
0
$674.57
?
300
1%
0
$2,538.86
?
Number of
Payments or
Years
Annual
Interest Rate
Future Value
Annuity
Present Value
5
8%
0
$213.22
$nothing
(Round to the nearest cent.)
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Future value of an ordinary annuity. Fill in the missing future values in the following table for an ordinary annuity.
Number of
Payments or
Years
Annual
Interest Rate
Present Value
Annuity
Future Value
10
9%
0
$286.87
?
18
16%
0
$1,397.76
?
30
2.5%
0
$721.92
?
280
1%
0
$553.71
?
Number of
Payments or
Years
Annual
Interest Rate
Present Value
Annuity
Future Value
10
9%
0
$286.87
$nothing
(Round to the nearest cent.)
18
16%
0
$1,397.76
$nothing
(Round to the nearest cent.)
30
2.5%
0
$721.92
$nothing
(Round to the nearest cent.)
280
1%
0
$553.71
$nothing
(Round to the nearest cent.)
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Use the following Annuity Table for questions 1 through 6.
Future Value of Ordinary Annuity of 1
Period 5% 6% 8% 10% 12%
11.000001.000001.000001.000001.00000
22.050002.060002.080002.100002.12000
33.152503.183603.246403.310003.37440
44.310134.374624.506114.641004.77933
55.525635.637095.866606.105106.35285
66.801916.975327.335927.715618.11519
78.142018.393848.922809.4871710.08901
89.549119.8974710.6366311.4358912.29969
911.0265611.4913212.4875613.5794814.77566
1012.5778913.1807914.4865615.9374317.54874
Present Value of an Ordinary Annuity of 1
Period 5% 6% 8% 10% 12%
1.95238.94340.92593.90909.89286
21.859411.833391.783261.735541.69005
32.723252.673012.577102.486852.40183
43.545953.465113.312133.169863.03735
54.329484.212363.992713.790793.60478
65.075694.917324.622884.355264.11141
75.786375.582385.206374.868424.56376
86.463216.209795.746645.334934.96764
97.107826.801696.246895.759025.32825
107.721737.360096.710086.144575.65022
Use the following…
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Use the following Annuity Table for questions 1 through 6.
Future Value of Ordinary Annuity of 1
Period 5% 6% 8% 10% 12%
11.000001.000001.000001.000001.00000
22.050002.060002.080002.100002.12000
33.152503.183603.246403.310003.37440
44.310134.374624.506114.641004.77933
55.525635.637095.866606.105106.35285
66.801916.975327.335927.715618.11519
78.142018.393848.922809.4871710.08901
89.549119.8974710.6366311.4358912.29969
911.0265611.4913212.4875613.5794814.77566
1012.5778913.1807914.4865615.9374317.54874
Present Value of an Ordinary Annuity of 1
Period 5% 6% 8% 10% 12%
1.95238.94340.92593.90909.89286
21.859411.833391.783261.735541.69005
32.723252.673012.577102.486852.40183
43.545953.465113.312133.169863.03735
54.329484.212363.992713.790793.60478
65.075694.917324.622884.355264.11141
75.786375.582385.206374.868424.56376
86.463216.209795.746645.334934.96764
97.107826.801696.246895.759025.32825
107.721737.360096.710086.144575.65022
Use the following…
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Use the following Annuity Table for questions 1 through 6.
Future Value of Ordinary Annuity of 1
Period 5% 6% 8% 10% 12%
11.000001.000001.000001.000001.00000
22.050002.060002.080002.100002.12000
33.152503.183603.246403.310003.37440
44.310134.374624.506114.641004.77933
55.525635.637095.866606.105106.35285
66.801916.975327.335927.715618.11519
78.142018.393848.922809.4871710.08901
89.549119.8974710.6366311.4358912.29969
911.0265611.4913212.4875613.5794814.77566
1012.5778913.1807914.4865615.9374317.54874
Present Value of an Ordinary Annuity of 1
Period 5% 6% 8% 10% 12%
1.95238.94340.92593.90909.89286
21.859411.833391.783261.735541.69005
32.723252.673012.577102.486852.40183
43.545953.465113.312133.169863.03735
54.329484.212363.992713.790793.60478
65.075694.917324.622884.355264.11141
75.786375.582385.206374.868424.56376
86.463216.209795.746645.334934.96764
97.107826.801696.246895.759025.32825
107.721737.360096.710086.144575.65022
Use the following…
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Complete the ordinary annuity. (Please use the following provided Table.) (Do not round intermediate calculations. Round
your answer to the nearest cent.)
Amount of
Payment payable
Years
Interest rate
Value of annuity
payment
12,800
Semiannually
8 %
acer
Σ
%24
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