Exam M Fall 2005
PRELIMINARY ANSWER KEY
Question #
Answer
Question #
Answer
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
C
C
C
D
C
B
A
D
B
A
A
A
D
C
A
D
D
D
B
B
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
E
B
E
E
C
E
E
D
A
D
A
A
B
C
A
A
C
C
E
B
**BEGINNING OF EXAMINATION**
1.
For a special whole life insurance on (x), you are given:
(i)
Z is the present value random variable for this insurance.
(ii)
Death benefits are paid at the moment of death.
(iii)
µ x ( t ) = 0.02,
(iv)
δ = 0.08
(v)
bt = e0.03t ,
t≥0
t≥0
Calculate Var ( Z ) .
(A)
0.075
(B)
0.080
(C)
0.085
(D)
0.090
(E)
0.095
Exam M: Fall 2005
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2.
For a whole life insurance of 1 on (x), you are given:
(i)
Benefits are payable at the moment of death.
(ii)
Level premiums are payable at the beginning of each year.
(iii)
Deaths are uniformly distributed over each year of age.
(iv)
i = 0.10
(v)
ax = 8
(vi)
ax +10 = 6
Calculate the 10th year terminal benefit reserve for this insurance.
(A)
0.18
(B)
0.25
(C)
0.26
(D)
0.27
(E)
0.30
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3.
A special whole life insurance of 100,000 payable at the moment of death of (x) includes a
double indemnity provision. This provision pays during the first ten years an additional
benefit of 100,000 at the moment of death for death by accidental means.
You are given:
(i)
(ii)
bg
µ b g bt g = 0.0002,
µ bxτ g t = 0.001, t ≥ 0
1
x
t ≥ 0 , where µ bx1g is the force of decrement due to death by
accidental means.
(iii)
δ = 0.06
Calculate the single benefit premium for this insurance.
(A)
1640
(B)
1710
(C)
1790
(D)
1870
(E)
1970
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4.
Kevin and Kira are modeling the future lifetime of (60).
(i)
Kevin uses a double decrement model:
x
τ
lx( )
d x( )
2
d x( )
60
1000
120
80
61
800
160
80
62
560
−
−
1
(ii)
Kira uses a non-homogeneous Markov model:
(a)
The states are 0 (alive), 1 (death due to cause 1), 2 (death due to cause 2).
Q60 is the transition matrix from age 60 to 61; Q61 is the transition matrix
(b)
from age 61 to 62.
(iii)
The two models produce equal probabilities of decrement.
Calculate Q61 .
(A)
⎛1.00 0.12 0.08 ⎞
⎜
⎟
0 ⎟
⎜ 0 1.00
⎜ 0
0 1.00 ⎟⎠
⎝
(B)
⎛ 0.80 0.12 0.08 ⎞
⎜
⎟
⎜ 0.56 0.16 0.08 ⎟
⎜ 0
0 1.00 ⎟⎠
⎝
(C)
⎛ 0.76 0.16 0.08 ⎞
⎜
⎟
1.00
0 ⎟
⎜ 0
⎜ 0
0 1.00 ⎟⎠
⎝
(D)
⎛ 0.70 0.20 0.10 ⎞
⎜
⎟
1.00
0 ⎟
⎜ 0
⎜ 0
0
1.00 ⎟⎠
⎝
(E)
⎛ 0.60 0.28 0.12 ⎞
⎜
⎟
0 ⎟
⎜ 0 1.00
⎜ 0
0 1.00 ⎟⎠
⎝
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5.
A certain species of flower has three states: sustainable, endangered and extinct. Transitions
between states are modeled as a non-homogeneous Markov chain with transition matrices Qi
as follows:
⎛
Sustainable ⎜
⎜
Q1 = Endangered⎜
⎜
Extinct
⎜
⎝
Sustainable Endangered Extinct
⎞
0.85
0.15
0 ⎟
⎟
0
0.7
0.3 ⎟
0
0
1 ⎟
⎟
⎠
⎛ 0.9 0.1 0 ⎞
⎜
⎟
Q2 = ⎜ 0.1 0.7 0.2 ⎟
⎜ 0
0
1 ⎟⎠
⎝
⎛ 0.95 0.05 0 ⎞
⎜
⎟
Q3 = ⎜ 0.2 0.7 0.1⎟
⎜ 0
0
1 ⎟⎠
⎝
⎛ 0.95 0.05 0 ⎞
⎜
⎟
Qi = ⎜ 0.5 0.5 0 ⎟ , i = 4,5,...
⎜ 0
0
1 ⎟⎠
⎝
Calculate the probability that a species endangered at the start of year 1 will ever become
extinct.
(A)
0.45
(B)
0.47
(C)
0.49
(D)
0.51
(E)
0.53
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6.
For a special 3-year term insurance:
(i)
Insureds may be in one of three states at the beginning of each year: active, disabled,
or dead. All insureds are initially active. The annual transition probabilities are as
follows:
Active
Disabled
Dead
Active
0.8
0.1
0.1
Disabled
0.1
0.7
0.2
Dead
0.0
0.0
1.0
(ii)
A 100,000 benefit is payable at the end of the year of death whether the insured was
active or disabled.
(iii)
Premiums are paid at the beginning of each year when active. Insureds do not pay
any annual premiums when they are disabled.
(iv)
d = 0.10
Calculate the level annual benefit premium for this insurance.
(A)
9,000
(B)
10,700
(C)
11,800
(D)
13,200
(E)
20,800
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7.
Customers arrive at a bank according to a Poisson process at the rate of 100 per hour. 20%
of them make only a deposit, 30% make only a withdrawal and the remaining 50% are there
only to complain. Deposit amounts are distributed with mean 8000 and standard deviation
1000. Withdrawal amounts have mean 5000 and standard deviation 2000.
The number of customers and their activities are mutually independent.
Using the normal approximation, calculate the probability that for an 8-hour day the total
withdrawals of the bank will exceed the total deposits.
(A)
0.27
(B)
0.30
(C)
0.33
(D)
0.36
(E)
0.39
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8.
A Mars probe has two batteries. Once a battery is activated, its future lifetime is exponential
with mean 1 year.
The first battery is activated when the probe lands on Mars. The second battery is activated
when the first fails.
Battery lifetimes after activation are independent.
The probe transmits data until both batteries have failed.
Calculate the probability that the probe is transmitting data three years after landing.
(A)
0.05
(B)
0.10
(C)
0.15
(D)
0.20
(E)
0.25
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9.
For a special fully discrete 30-payment whole life insurance on (45), you are given:
(i)
The death benefit of 1000 is payable at the end of the year of death.
(ii)
The benefit premium for this insurance is equal to 1000P45 for the first 15 years
followed by an increased level annual premium of π for the remaining 15 years.
(iii)
Mortality follows the Illustrative Life Table.
(iv)
i = 0.06
Calculate π .
(A)
16.8
(B)
17.3
(C)
17.8
(D)
18.3
(E)
18.8
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10.
For a special fully discrete 2-year endowment insurance on (x):
(i)
The pure endowment is 2000.
(ii)
The death benefit for year k is (1000k ) plus the benefit reserve at the end of year k,
k = 1, 2 .
(iii)
π is the level annual benefit premium.
(iv)
i = 0.08
(v)
px + k −1 = 0.9, k = 1, 2
Calculate π .
(A)
1027
(B)
1047
(C)
1067
(D)
1087
(E)
1107
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11.
For a group of 250 individuals age x, you are given:
(i)
The future lifetimes are independent.
(ii)
Each individual is paid 500 at the beginning of each year, if living.
(iii)
Ax = 0.369131
(iv)
2
(v)
i = 0.06
Ax = 0.1774113
Using the normal approximation, calculate the size of the fund needed at inception in order to
be 90% certain of having enough money to pay the life annuities.
(A)
1.43 million
(B)
1.53 million
(C)
1.63 million
(D)
1.73 million
(E)
1.83 million
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12.
For a double decrement table, you are given:
Age
40
41
42
lx( )
1000
τ
d x( )
60
d x( )
55
−
750
−
70
−
−
1
2
Each decrement is uniformly distributed over each year of age in the double decrement table.
′(1) .
Calculate q41
(A)
0.077
(B)
0.078
(C)
0.079
(D)
0.080
(E)
0.081
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13.
The actuarial department for the SharpPoint Corporation models the lifetime of pencil
α
sharpeners from purchase using a generalized DeMoivre model with s ( x ) = (1 − x / ω ) , for
α > 0 and 0 ≤ x ≤ ω .
A senior actuary examining mortality tables for pencil sharpeners has determined that the
original value of α must change. You are given:
(i)
The new complete expectation of life at purchase is half what it was previously.
(ii)
The new force of mortality for pencil sharpeners is 2.25 times the previous force of
mortality for all durations.
(iii)
ω remains the same.
Calculate the original value of α .
(A)
1
(B)
2
(C)
3
(D)
4
(E)
5
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14.
You are given:
(i)
T is the future lifetime random variable.
(ii)
µ (t ) = µ ,
(iii)
Var [T ] = 100 .
t≥0
Calculate E [T ∧ 10] .
(A)
2.6
(B)
5.4
(C)
6.3
(D)
9.5
(E)
10.0
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15.
For a fully discrete 15-payment whole life insurance of 100,000 on (x), you are given:
(i)
The expense-loaded level annual premium using the equivalence principle is 4669.95.
(ii)
100,000 Ax = 51, 481.97
(iii)
ax:15 = 11.35
(iv)
d = 0.02913
(v)
Expenses are incurred at the beginning of the year.
(vi)
Percent of premium expenses are 10% in the first year and 2% thereafter.
(vii)
Per policy expenses are K in the first year and 5 in each year thereafter until death.
Calculate K.
(A)
10.0
(B)
16.5
(C)
23.0
(D)
29.5
(E)
36.5
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16.
For the future lifetimes of (x) and (y):
(i)
With probability 0.4, T ( x ) = T ( y ) (i.e., deaths occur simultaneously).
(ii)
With probability 0.6, the joint density function is
fT ( x ), T ( y ) (t , s ) = 0.0005 ,
0 < t < 40 ,
0 < s < 50
Calculate Prob ⎡⎣T ( x ) < T ( y ) ⎤⎦ .
(A)
0.30
(B)
0.32
(C)
0.34
(D)
0.36
(E)
0.38
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17.
The length of time, in years, that a person will remember an actuarial statistic is modeled by
an exponential distribution with mean 1Y . In a certain population, Y has a gamma
distribution with α = θ = 2 .
Calculate the probability that a person drawn at random from this population will remember
an actuarial statistic less than 1 2 year.
(A)
0.125
(B)
0.250
(C)
0.500
(D)
0.750
(E)
0.875
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18.
In a CCRC, residents start each month in one of the following three states: Independent
Living (State #1), Temporarily in a Health Center (State #2) or Permanently in a Health
Center (State #3). Transitions between states occur at the end of the month.
If a resident receives physical therapy, the number of sessions that the resident receives in a
month has a geometric distribution with a mean which depends on the state in which the
resident begins the month. The numbers of sessions received are independent. The number
in each state at the beginning of a given month, the probability of needing physical therapy in
the month, and the mean number of sessions received for residents receiving therapy are
displayed in the following table:
State #
1
Number in
state
400
Probability of
needing therapy
0.2
Mean number
of visits
2
2
300
0.5
15
3
200
0.3
9
Using the normal approximation for the aggregate distribution, calculate the probability that
more than 3000 physical therapy sessions will be required for the given month.
(A)
0.21
(B)
0.27
(C)
0.34
(D)
0.42
(E)
0.50
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19.
In a given week, the number of projects that require you to work overtime has a geometric
distribution with β = 2 . For each project, the distribution of the number of overtime hours in
the week is the following:
x
f ( x)
5
0.2
10
20
0.3
0.5
The number of projects and number of overtime hours are independent. You will get paid for
overtime hours in excess of 15 hours in the week.
Calculate the expected number of overtime hours for which you will get paid in the week.
(A)
18.5
(B)
18.8
(C)
22.1
(D)
26.2
(E)
28.0
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20.
For a group of lives age x, you are given:
(i)
Each member of the group has a constant force of mortality that is drawn from the
uniform distribution on [ 0.01, 0.02] .
(ii)
δ = 0.01
For a member selected at random from this group, calculate the actuarial present value of a
continuous lifetime annuity of 1 per year.
(A)
40.0
(B)
40.5
(C)
41.1
(D)
41.7
(E)
42.3
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21.
For a population whose mortality follows DeMoivre’s law, you are given:
D
D
D
D
(i)
e40:40 = 3 e60:60
(ii)
e20:20 = k e60:60
Calculate k.
(A)
3.0
(B)
3.5
(C)
4.0
(D)
4.5
(E)
5.0
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22.
For an insurance on (x) and (y):
(i)
Upon the first death, the survivor receives the single benefit premium for a whole life
insurance of 10,000 payable at the moment of death of the survivor.
(ii)
µ x ( t ) = µ y ( t ) = 0.06 while both are alive.
(iii)
µ x y ( t ) = 0.12
(iv)
After the first death, µ ( t ) = 0.10 for the survivor.
(v)
δ = 0.04
Calculate the actuarial present value of this insurance on (x) and (y).
(A)
4500
(B)
5400
(C)
6000
(D)
7100
(E)
7500
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23.
Kevin and Kira are in a history competition:
(i)
In each round, every child still in the contest faces one question. A child is out as
soon as he or she misses one question. The contest will last at least 5 rounds.
(ii)
For each question, Kevin’s probability and Kira’s probability of answering that
question correctly are each 0.8; their answers are independent.
Calculate the conditional probability that both Kevin and Kira are out by the start of round
five, given that at least one of them participates in round 3.
(A)
0.13
(B)
0.16
(C)
0.19
(D)
0.22
(E)
0.25
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24.
For a special increasing whole life annuity-due on (40), you are given:
(i)
Y is the present-value random variable.
(ii)
Payments are made once every 30 years, beginning immediately.
(iii)
The payment in year 1 is 10, and payments increase by 10 every 30 years.
(iv)
Mortality follows DeMoivre’s law, with ω = 110 .
(v)
i = 0.04
Calculate Var (Y ) .
(A)
10.5
(B)
11.0
(C)
11.5
(D)
12.0
(E)
12.5
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25.
For a special 3-year term insurance on ( x ) , you are given:
(i)
Z is the present-value random variable for this insurance.
(ii)
qx + k = 0.02( k + 1) ,
(iii)
The following benefits are payable at the end of the year of death:
(iv)
k = 0, 1, 2
k
bk+1
0
300
1
350
2
400
i = 0.06
bg
Calculate Var Z .
(A)
9,600
(B)
10,000
(C)
10,400
(D)
10,800
(E)
11,200
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26.
For an insurance:
(i)
Losses have density function
⎧0.02 x 0 < x < 10
fX ( x) = ⎨
elsewhere
⎩0
(ii)
The insurance has an ordinary deductible of 4 per loss.
(iii)
Y P is the claim payment per payment random variable.
Calculate E ⎡⎣Y P ⎤⎦ .
(A)
2.9
(B)
3.0
(C)
3.2
(D)
3.3
(E)
3.4
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27.
An actuary has created a compound claims frequency model with the following properties:
(i)
The primary distribution is the negative binomial with probability generating function
−2
P ( z ) = ⎡⎣1 − 3 ( z − 1) ⎤⎦ .
(ii)
The secondary distribution is the Poisson with probability generating function
λ z −1
P( z) = e ( ) .
(iii)
The probability of no claims equals 0.067.
Calculate λ .
(A)
0.1
(B)
0.4
(C)
1.6
(D)
2.7
(E)
3.1
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28.
In 2005 a risk has a two-parameter Pareto distribution with α = 2 and θ = 3000 . In 2006
losses inflate by 20%.
An insurance on the risk has a deductible of 600 in each year. Pi , the premium in year i,
equals 1.2 times the expected claims.
The risk is reinsured with a deductible that stays the same in each year. Ri , the reinsurance
premium in year i, equals 1.1 times the expected reinsured claims.
R 2005
P 2005
Calculate
= 0.55
R 2006
(A)
0.46
(B)
0.52
(C)
0.55
(D)
0.58
(E)
0.66
P2006
.
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29.
For a fully discrete whole life insurance of 1000 on (60), you are given:
(i)
The expenses, payable at the beginning of the year, are:
Expense Type
% of Premium
Per Policy
First Year
20%
8
(ii)
The level expense-loaded premium is 41.20.
(iii)
i = 0.05
Renewal Years
6%
2
Calculate the value of the expense augmented loss variable, 0 Le , if the insured dies in the
third policy year.
(A)
770
(B)
790
(C)
810
(D)
830
(E)
850
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30.
For a fully discrete whole life insurance of 1000 on (45), you are given:
t
22
23
24
1000t V45
235
255
272
q45+t
0.015
0.020
0.025
Calculate 100025V45 .
(A)
279
(B)
282
(C)
284
(D)
286
(E)
288
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31.
The graph of a piecewise linear survival function, s ( x ) , consists of 3 line segments with
endpoints (0, 1), (25, 0.50), (75, 0.40), (100, 0).
Calculate
20 55 q15
55 q35
(A)
0.69
(B)
0.71
(C)
0.73
(D)
0.75
(E)
0.77
.
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32.
For a group of lives aged 30, containing an equal number of smokers and non-smokers, you
are given:
(i)
For non-smokers, µ n ( x ) = 0.08 ,
x ≥ 30
(ii)
For smokers, µ s ( x ) = 0.16,
x ≥ 30
Calculate q80 for a life randomly selected from those surviving to age 80.
(A)
0.078
(B)
0.086
(C)
0.095
(D)
0.104
(E)
0.112
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33.
For a 3-year fully discrete term insurance of 1000 on (40), subject to a double decrement
model:
(i)
x
τ
lx( )
d x( )
1
2
d x( )
40
2000
20
60
41
−
30
50
42
−
40
−
(ii)
Decrement 1 is death. Decrement 2 is withdrawal.
(iii)
There are no withdrawal benefits.
(iv)
i = 0.05
Calculate the level annual benefit premium for this insurance.
(A)
14.3
(B)
14.7
(C)
15.1
(D)
15.5
(E)
15.7
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34.
Each life within a group medical expense policy has loss amounts which follow a compound
Poisson process with λ = 0.16 . Given a loss, the probability that it is for Disease 1 is 116 .
Loss amount distributions have the following parameters:
Disease 1
Other diseases
Mean per loss
5
Standard
Deviation per loss
50
10
20
Premiums for a group of 100 independent lives are set at a level such that the probability
(using the normal approximation to the distribution for aggregate losses) that aggregate
losses for the group will exceed aggregate premiums for the group is 0.24.
A vaccine which will eliminate Disease 1 and costs 0.15 per person has been discovered.
Define:
A = the aggregate premium assuming that no one obtains the vaccine, and
B = the aggregate premium assuming that everyone obtains the vaccine and the cost of the
vaccine is a covered loss.
Calculate A/B.
(A)
0.94
(B)
0.97
(C)
1.00
(D)
1.03
(E)
1.06
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35.
An actuary for a medical device manufacturer initially models the failure time for a particular
device with an exponential distribution with mean 4 years.
This distribution is replaced with a spliced model whose density function:
(i)
is uniform over [0, 3]
(ii)
is proportional to the initial modeled density function after 3 years
(iii)
is continuous
Calculate the probability of failure in the first 3 years under the revised distribution.
(A)
0.43
(B)
0.45
(C)
0.47
(D)
0.49
(E)
0.51
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36.
For a fully continuous whole life insurance of 1 on (30), you are given:
(i)
The force of mortality is 0.05 in the first 10 years and 0.08 thereafter.
(ii)
δ = 0.08
Calculate the benefit reserve at time 10 for this insurance.
(A)
0.144
(B)
0.155
(C)
0.166
(D)
0.177
(E)
0.188
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37.
For a 10-payment, 20-year term insurance of 100,000 on Pat:
(i)
Death benefits are payable at the moment of death.
(ii)
Contract premiums of 1600 are payable annually at the beginning of each year for 10
years.
(iii)
i = 0.05
(iv)
L is the loss random variable at the time of issue.
Calculate the minimum value of L as a function of the time of death of Pat.
(A)
− 21,000
(B)
− 17,000
(C)
− 13,000
(D)
− 12,400
(E)
− 12,000
Exam M: Fall 2005
-37-
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38.
For an insurance:
(i)
The number of losses per year has a Poisson distribution with λ = 10 .
(ii)
Loss amounts are uniformly distributed on (0, 10).
(iii)
Loss amounts and the number of losses are mutually independent.
(iv)
There is an ordinary deductible of 4 per loss.
Calculate the variance of aggregate payments in a year.
(A)
36
(B)
48
(C)
72
(D)
96
(E)
120
Exam M: Fall 2005
-38-
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39.
For an insurance portfolio:
(i)
The number of claims has the probability distribution
n
pn
0
1
2
3
0.1
0.4
0.3
0.2
(ii)
Each claim amount has a Poisson distribution with mean 3; and
(iii)
The number of claims and claim amounts are mutually independent.
Calculate the variance of aggregate claims.
(A)
4.8
(B)
6.4
(C)
8.0
(D)
10.2
(E)
12.4
Exam M: Fall 2005
-39-
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40.
Lucky Tom deposits the coins he finds on the way to work according to a Poisson process
with a mean of 22 deposits per month.
5% of the time, Tom deposits coins worth a total of 10.
15% of the time, Tom deposits coins worth a total of 5.
80% of the time, Tom deposits coins worth a total of 1.
The amounts deposited are independent, and are independent of the number of deposits.
Calculate the variance in the total of the monthly deposits.
(A)
180
(B)
210
(C)
240
(D)
270
(E)
300
**END OF EXAMINATION**
Exam M: Fall 2005
-40-
STOP
Fall 2005
Exam M Solutions
Question #1
Key: C
Var [ Z ] = E ⎡⎣ Z 2 ⎤⎦ − E [ Z ]
2
E [Z ] = ∫
∞
0
=∫
∞
0
(v b )
∞
t
t t
p x µ x ( t ) dt = ∫ e −0.08t e0.03t e −0.02t ( 0.02 ) dt
0
( 0.02 ) e−0.07t dt =
E ⎡⎣ Z 2 ⎤⎦ = ∫
∞
0
∞
( vt bt )
2
t
0.02 2
=
7
0.07
px µ x ( t ) dt = ∫
0
= ∫ 0.02 e −0.12t µ x ( t ) dt = 2
0
( )
1
Var [ Z ] = − 2
7
6
2
∞
(e
12
−0.05t
=1
)
2
e −0.02t ( 0.02 ) dt
6
1 4
= −
= 0.08503
6 49
Question #2
Key: C
0.1
(8) = 311
1.1
0.1
= 1−
( 6 ) = 511
1.1
we have Ax = 1 −
From Ax = 1 − d ax
Ax +10
Ax = Ax × i
Ax =
3
0.1
×
= 0.2861
11 ln (1.1)
Ax +10 =
10Vx
δ
5
0.1
×
= 0.4769
11 ln (1.1)
= Ax +10 − P ( Ax ) × ax +10
⎛ 0.2861 ⎞
= 0.4769 − ⎜
⎟6
⎝ 8 ⎠
= 0.2623
There are many other equivalent formulas that could be used.
Question #3
Key: C
∞
Regular death benefit = ∫ 100,000 × e −0.06t × e −0.001t 0.001dt
0
0.001 ⎞
⎛
= 100,000 ⎜
⎟
⎝ 0.06 + 0.001 ⎠
= 1639.34
Accidental death
= ∫ 100,000 e −0.06t e −0.001t ( 0.0002 ) dt
10
0
10
= 20 ∫ e −0.061t dt
0
⎡1 − e −0.61 ⎤
= 20 ⎢
⎥ = 149.72
0.061
⎣
⎦
Actuarial Present Value = 1639.34 + 149.72 = 1789.06
Question #4
Key: D
Once you are dead, you are dead. Thus, you never leave state 2 or 3, and rows 2 and 3 of the
matrix must be (0 1 0) and (0 0 1).
Probability of dying from cause 1 within the year, given alive at age 61, is 160/800 = 0.20.
Probability of dying from cause 2 within the year, given alive at age 61, is 80/800 = 0.10
Probability of surviving to 62, given alive at 61, is 560/800 = 0.70
(alternatively, 1 – 0.20 – 0.10), so correct answer is D.
Question #5
Key: C
This first solution uses the method on the top of page 9 of the study note.
Note that if the species is it is not extinct after Q3 it will never be extinct.
This solution parallels the example at the top of page 9 of the Daniel study note. We want the
second entry of the product ( Q1 × Q2 × Q3 ) e3 which is equal to Q1 × ( Q2 × ( Q3 × e3 ) ) .
0
0
Q3 0 = 0.1
1
1
0
0.01
Q2 0.1 = 0.27
1
1
0.01
0.049
Q1 0.27 = 0.489
1
1
The second entry is 0.489; that’s our answer.
Alternatively, start with the row matrix (0 1 0) and project it forward 3 years.
(0
(0
(0.07
1
0.70
0.49
0
)
0.30)
0.44)
Q1 = (0.00
Q2 = (0.07
Q3 = (0.16
0.70
0.49
0.35
0.30)
0.44)
0.49)
Thus, the probability that it is in state 3 after three transitions is 0.49.
Yet another approach would be to multiply Q1 × Q2 × Q3 , and take the entry in row 2, column 3.
That would work but it requires more effort.
Question #6
Key: B
Probabilities of being in each state at time t:
t
0
1
2
3
Active
1.0
0.8
0.65
not needed
Disabled
0.0
0.1
0.15
not needed
Dead
0.0
0.1
0.2
0.295
Deaths
0.1
0.1
0.095
We built the Active Disabled Dead columns of that table by multiplying each row times the
transition matrix. E.g., to move from t = 1 to t = 2, (0.8 0.1 0.1) Q = (0.65 0.15 0.2)
The deaths column is just the increase in Dead. E.g., for t = 2, 0.2 – 0.1 = 0.1.
v = 0.9
APV of death benefits = 100,000* 0.1v + 0.1v 2 + 0.095v3 = 24,025.5
(
APV of $1 of premium = 1 + 0.8v + 0.65v 2 = 2.2465
Benefit premium =
24,025.5
= 10,695
2.2465
)
Question #7
Key: A
Split into three independent processes:
Deposits, with λ * = ( 0.2 )(100 )( 8 ) = 160 per day
Withdrawals, with λ * = ( 0.3)(100 )( 8 ) = 240 per day
Complaints. Ignore, no cash impact.
For aggregate deposits,
E ( D ) = (160 )( 8000 ) = 1, 280,000
Var ( D ) = (160 )(1000 ) + (160 )( 8000 )
2
2
= 1.04 × 1010
For aggregate withdrawals
E (W ) = ( 240 )( 5000 ) = 1, 200,000
Var (W ) = ( 240 )( 2000 ) + ( 240 )( 5000 )
2
2
= 0.696 ×1010
E (W − D ) = 1, 200,000 − 1, 280,000 = −80,000
Var (W − D ) = 0.696 ×1010 + 1.04 × 1010 = 1.736 × 1010
SD (W − D ) = 131,757
⎛ W − D + 80,000 80,000 ⎞
Pr (W > D ) = Pr (W − D > 0 ) = Pr ⎜
>
131,757
131,757 ⎟⎠
⎝
= 1 − Φ ( 0.607 )
= 0.27
Question #8
Key: D
Exponential inter-event times and independent implies Poisson process (imagine additional
batteries being activated as necessary; we don’t care what happens after two have failed).
Poisson rate of 1 per year implies failures in 3 years is Poisson with λ = 3 .
f ( x)
0.050
0.149
x
0
1
F(x)
0.050
0.199
Probe works provided that there have been fewer than two failures, so we want F(1) = 0.199.
Alternatively, the sum of two independent exponential θ = 1 random variables is Gamma with
α = 2, θ = 1
1 3 −t
F ( 3) = Γ ( 2;3) =
t e dt
Γ ( 2 ) ∫0
= ( −t − 1) e −t
= 1− 4e
3
0
−3
= 0.80 is probability 2 have occurred
1 – 0.80 = 0.20
Question #9
Key: B
1000 P45a45:15 + π a60:15 × 15 E45 = 1000 A45
1000
A45
( a45 − 15 E45 a60 ) + π ( a60 − 15 E60 a75 )( 15 E45 ) = 1000 A45
a45
201.20
(14.1121 − ( 0.72988)( 0.51081)(11.1454 )
14.1121
+π (11.1454 − ( 0.68756 )( 0.39994 )( 7.2170 ) ) × ( 0.72988 )( 0.51081) = 201.20
where
15 E x
was evaluated as 5 Ex × 10 Ex +5
14.2573 ( 9.9568 ) + (π )( 3.4154 ) = 201.20
π = 17.346
Question #10
Key: A
1V
= ( 0 V + π ) (1 + i ) − (1000 + 1V − 1V ) qx
2V
= ( 1V + π )(1 + i ) − ( 2000 + 2V − 2V ) qx +1 = 2000
( (π (1 + i ) − 1000q ) +π )(1 + i ) − 2000q = 2000
( (π (1.08) − 1000 × 0.1) + π ) (1.08) − 2000 × 0.1 = 2000
x
π = 1027.42
x +1
Question #11
Key: A
Let Y be the present value of payments to 1 person.
Let S be the present value of the aggregate payments.
E [Y ] = 500 ax = 500
σ Y = Var [Y ] =
(1 − Ax ) = 5572.68
d
( 500 )2
1
d2
(
2
)
Ax − Ax2 = 1791.96
S = Y1 + Y2 + ... + Y250
E ( S ) = 250 E [Y ] = 1,393,170
σ S = 250 × σ Y = 15.811388σ Y = 28,333
⎡ S − 1,393,170 F − 1,393,170 ⎤
0.90 = Pr ( S ≤ F ) = Pr ⎢
≤
28,333 ⎥⎦
⎣ 28,333
F − 1,393,170 ⎤
⎡
≈ Pr ⎢ N ( 0,1) ≤
28,333 ⎥⎦
⎣
0.90 = Pr ( N ( 0,1) ≤ 1.28 )
F = 1,393,170 + 1.28 ( 28,333)
=1.43 million
Question #12
Key: A
q ′ b1g = 1 − p′ b1g = 1 − e p bτ g j
41
41
bg
q41 1
41
bg
q41 τ
τ
τ
1
2
l41( ) = l40( ) − d 40( ) − d 40( ) = 1000 − 60 − 55 = 885
1
τ
2
τ
d 41( ) = l41( ) − d 41( ) − l42( ) = 885 − 70 − 750 = 65
q41( )
1
750
τ
p41( ) =
885
(τ )
q41
750 ⎞
′ (1) = 1 − ⎛⎜
q41
⎟
⎝ 885 ⎠
65
135
=
65
135
= 0.0766
Question #13
Key: D
α
x⎞
⎛
s ( x ) = ⎜1 − ⎟
⎝ ω⎠
d
α
µ ( x) =
log ( s ( x ) ) =
dx
ω−x
D
ex = ∫
0
α
t ⎞
ω−x
⎜1 −
⎟ dt =
α +1
⎝ ω −x⎠
ω −x ⎛
1
ω
ω
D
= new
⇒ α new = 2α old + 1
e0new = × old
2 α +1 α
+1
2α old + 1 9 α old
µ (0new ) =
= ×
⇒ α old = 4
ω
4 ω
Question #14
Key: C
Constant force implies exponential lifetime
where θ = 1/ µ using the Loss Models parameterization
2
2
2 ⎛1⎞
1
Var [T ] = E ⎡⎣T 2 ⎤⎦ − ( E [T ]) = 2 − ⎜ ⎟ = 2 = 100
µ ⎝µ⎠ µ
µ = 0.1
10
E [T ∧ 10] = ∫ e− µ t dt =
0
1 − e −10 µ
µ
= 6.3
Alternatively, the formula for E ( X ∧ x ) is given in the tables handout.
Note that since T is the future lifetime random variable, E (T ∧ 10 ) can also be written as
D
ex:10 , which for the exponential distribution (constant force of mortality) is independent of x.
Question #15
Key: A
Gax:15
% premium amount for 15 years
= 100,000 Ax + 0.08G + 0.02Gax:15 + ( ( x − 5 ) + 5ax )
(
)
Per policy for life
4669.95 (11.35 ) = 51, 481.97 + ( 0.08 )( 4669.95 ) + ( 0.02 )(11.35 )( 4669.95 ) + ( ( x − 5 ) + 5ax )
ax =
1 − Ax 1 − 0.5148197
=
= 16.66
d
0.02913
53,003.93 = 51, 481.97 + 1433.67 + ( x − 5 ) + 83.30
4.99 = ( x − 5 )
x = 9.99
The % of premium expenses could equally well have been expressed as 0.10G + 0.02G ax:14 .
The per policy expenses could also be expressed in terms of an annuity-immediate.
Question #16
Key: D
For the density where T ( x ) ≠ T ( y ) ,
Pr (T ( x ) < T ( y ) ) = ∫
40
y
0.0005dxdy
y =0 ∫x =0
=∫
40
y =0
y
0.0005 x dy
0
40
= ∫ 0.0005ydy
0
=
0.0005 y 2
2
40
0
= 0.40
For the overall density,
Pr (T ( x ) < T ( y ) ) = 0.4 × 0 + 0.6 × 0.4 = 0.24
where the first 0.4 is the probability that T ( x ) = T ( y ) and 0.6 is the probability that
T ( x) ≠ T ( y) .
Question #17
Key: D
The following derives the general formula for the statistic to be forgotten by time x. It would
1
work fine, and the equations would look simpler, if you immediately plugged in x = , the only
2
1
value you want. Then the x + becomes 1.
2
Let X be the random variable for when the statistic is forgotten. Then FX ( x y ) = 1 − e − xy
For the unconditional distribution of X, integrate with respect to y
∞
(
FX ( x ) = ∫ 1 − e
− xy
0
)
2
1 ⎛ y⎞ − y2
⎜ ⎟ e dy
Γ ( 2) y ⎝ 2 ⎠
∞
1
− y x+ 1
= 1 − ∫ y e ( 2 )d y
40
= 1−
F ( 12 ) = 1 −
1
4 ( x + 12 )
2
1
4 ( 12 + 12 )
2
= 0.75
There are various ways to evaluate the integral in the second line:
1.
Calculus, integration by parts
2.
Recognize that
∞
∫0
⎛
1⎞
1 ⎞ − y⎜ x + ⎟
⎛
y ⎜ x + ⎟ e ⎝ 2 ⎠ dy
2⎠
⎝
is the expected value of an exponential random variable with θ =
1
x+
2
⎛
1⎞
1
2
− y⎜ x + ⎟
1⎞
⎛
3.
Recognize that Γ ( 2 ) ⎜ x + ⎟ y e ⎝ 2 ⎠ is the density function for a Gamma random
2⎠
⎝
1
, so it would integrate to 1.
variable with α = 2 and θ =
1
x+
2
1
(Approaches 2 and 3 would also work if you had plugged in x = at the start. The resulting θ
2
becomes 1).
Question #18
Key: D
State#
1
2
3
Number Probability Mean
of needing Number
Therapy
of visits
E(X)
400
0.2
2
300
0.5
15
200
0.3
9
E(N)
Var(N)
Var(X)
E(S)
Var(S)
80
150
60
64
75
42
6
240
90
160
2,250
540
2,950
736
52,875
8,802
62,413
Std Dev ( S ) = 62413 = 250
⎛ S − 2950 50 ⎞
Pr ( S > 3000 ) = Pr ⎜
>
⎟ = 1 − Φ ( 0.2 ) = 0.42
250 ⎠
⎝ 250
The Var ( X ) column came from the formulas for mean and variance of a geometric
distribution.
Using the continuity correction, solving for Pr ( S > 3000.5 ) , is theoretically better but
does not affect the rounded answer.
Question #19
Key: B
Frequency is geometric with β = 2 , so
p0 = 1/ 3, p1 = 2 / 9, p2 = 4 / 27
Convolutions of f X ( x) needed are
x
f
5
10
0.2
0.3
so f S ( 0 ) = 1/ 3,
f *2
0
0.04
f S ( 5 ) = 2 / 9 ( 0.2 ) = 0.044 , f S (10 ) = 2 / 9 ( 0.3) + 4 / 27 ( 0.04 ) = 0.073
E ( X ) = ( 0.2 )( 5 ) + ( 0.3)(10 ) + ( 0.5 )( 20 ) = 14
E [ S ] = 2 E ( X ) = 28
E [ S − 15]+ = E [ S ] − 5 (1 − F ( 0 ) ) − 5 (1 − F ( 5 ) ) − 5 (1 − F (10 ) )
= 28 − 5 (1 − 1/ 3) − 5 (1 − 1/ 3 − 0.044 ) − 5 (1 − 1/ 3 − 0.044 − 0.073)
= 28 − 3.33 − 3.11 − 2.75 = 18.81
Alternatively,
E [ S − 15]+ = E [ S ] − 15 + 15 f S ( 0 ) + 10 f S ( 5 ) + 5 f S (10 )
⎛1⎞
= 28 − 15 + (15 ) ⎜ ⎟ + 10 ( 0.044 ) + 5 ( 0.073)
⎝ 3⎠
= 18.81
Question #20
Key: B
The conditional expected value of the annuity, given µ , is
1
.
0.01 + µ
The unconditional expected value is
0.02
1
⎛ 0.01 + 0.02 ⎞
ax = 100∫
d µ = 100 ln ⎜
⎟ = 40.5
0.01 0.01 + µ
⎝ 0.01 + 0.01 ⎠
100 is the constant density of µ on the internal [ 0.01,0.02] . If the density were not constant, it
would have to go inside the integral.
Question #21
Key: E
D
Recall ex =
ω−x
D
2
D
D
D
ex:x = ex + ex − ex:x
D
ex:x = ∫
t ⎞⎛
t ⎞
⎟ dt
⎜1 −
⎟ ⎜1 −
⎝ ω − x ⎠⎝ ω − y ⎠
ω−x ⎛
0
Performing the integration we obtain
ω−x
D
ex:x =
3
2
(ω − x )
D
ex:x =
3
(i)
(ii)
2 (ω − 2a )
2 (ω − 3a )
= 3×
⇒ 2ω = 7a
3
3
2 (ω − 3a )
2
(ω − a ) = k ×
3
3
3.5a − a = k ( 3.5a − 3a )
k =5
The solution assumes that all lifetimes are independent.
Question #22
Key: B
µ
0.10 ⎞
⎛
= 10,000 ⎜
⎟ = 7143
µ +δ
⎝ 0.10 + 0.04 ⎠
The actuarial present value of the insurance of 7143 is
µ xy
0.12 ⎞
⎛
= ( 7,143) ⎜
7,143
⎟ = 5357
µ xy + δ
⎝ 0.12 + 0.04 ⎠
Upon the first death, the survivor receives 10,000
If the force of mortality were not constant during each insurance period, integrals would be
required to express the actuarial present value.
Question #23
Key: E
Let k p0 = Probability someone answers the first k problems correctly.
2
p0 = ( 0.8 ) = 0.64
4
p0 = ( 0.8 ) = 0.41
2
p0:0 = ( 2 p0 ) = 0.642 = 0.41
4
p0:0 = ( 0.41) = 0.168
2
p0:0 = 2 p0 + 2 p0 − 2 p0:0 = 0.87
4
p0:0 = 0.41 + 0.41 − 0.168 = 0.652
2
4
2
2
Prob(second child loses in round 3 or 4) = 2 p0:0 − 4 p0:0
= 0.87-0.652
= 0.218
Prob(second loses in round 3 or 4 second loses after round 2) =
2
p0:0 − 4 p0:0
2
=
p0:0
0.218
= 0.25
0.87
Question #24
Key: E
If (40) dies before 70, he receives one payment of 10, and Y = 10. Under DeMoivre, the
probability of this is (70 – 40)/(110 – 40) = 3/7
If (40) reaches 70 but dies before 100, he receives 2 payments.
Y = 10 + 20v30 = 16.16637
The probability of this is also 3/7. (Under DeMoivre, all intervals of the same length, here 30
years, have the same probability).
If (40) survives to 100, he receives 3 payments.
Y = 10 + 20v30 + 30v 60 = 19.01819
The probability of this is 1 – 3/7 – 3/7 = 1/7
E (Y ) = ( 3/ 7 ) ×10 + ( 3/ 7 ) × 16.16637 + (1/ 7 ) × 19.01819 = 13.93104
( )
E Y 2 = ( 3/ 7 ) × 102 + ( 3/ 7 ) × 16.16637 2 + (1/ 7 ) × 19.018192 = 206.53515
( )
Var (Y ) = E Y 2 − ⎡⎣ E (Y ) ⎤⎦ = 12.46
Since everyone receives the first payment of 10, you could have ignored it in the calculation.
2
Question #25
Key: C
2
(
E ( Z ) = ∑ v k +1bk +1
k =0
)
px qx+ k
k
= ⎡⎣ v ( 300 ) × 0.02 + v 2 ( 350 )( 0.98 )( 0.04 ) + v3 400 ( 0.98 )( 0.96 )( 0.06 ) ⎤⎦
= 36.8
( )
2
(
E Z 2 = ∑ v k +1bk +1
k =0
)
2
k
px qx + k
= v 2 ( 300 ) × 0.02 + v 4 ( 350 ) ( 0.98 )( 0.04 ) + v 6 4002 ( 0.98 )( 0.96 ) 0.06
2
2
= 11,773
( )
Var [ Z ] = E Z 2 − E ( Z )
2
= 11,773 − 36.82
= 10, 419
Question #26
Key: E
S X ( 4 ) = 1 − ∫ f X ( x ) dx = 1 − ∫ 0.02 x dx
4
4
0
0
= 1 − 0.01 x 2
4
0
= 0.84
f ( y + 4 ) 0.02 ( y + 4 )
2
fY p ( y ) = X
=
= 0.0238 ( y + 4 )
S X ( 4)
0.84
⎛ y3 4 y 2 ⎞
6
E Y p = ∫ y ( 0.0238 ( y + 4 ) ) dy = 0.0238 ⎜ +
⎟
0
2 ⎠
⎝ 3
= 3.4272
( )
6
0
Question #27
Key: E
By Theorem 4.51 (on page 93 of the second edition of Loss Models), probability of zero
claims = pgf of negative binomial applied to the probability that Poisson equals 0.
For the Poisson, f ( 0 ) = e − λ
(
)
So 0.067 = ⎡1 − β e − λ − 1 ⎤
⎣
⎦
Solving gives λ = 3
−r
(
)
= ⎡1 − 3 e − λ − 1 ⎤
⎣
⎦
−2
Question #28
Key: D
For any deductible d and the given severity distribution
E ( X − d )+ = E ( X ) − E ( X ∧ d )
3000 ⎞
⎛
= 3000 − 3000 ⎜1 −
⎟
⎝ 3000 + d ⎠
⎛ 3000 ⎞
= ( 3000 ) ⎜
⎟
⎝ 3000 + d ⎠
⎛ 3000 ⎞
So P2005 = (1.2 )( 3000 ) ⎜
⎟ = 3000
⎝ 3600 ⎠
The following paragraph just clarifies the notation in the rest of the solution:
Let r denote the reinsurer’s deductible relative to losses (not relative to reinsured claims). Thus
if r = 1000 (we are about to solve for r), then on a loss of 4000, the insured collects
4000 – 600 = 3400, the reinsurer pays 4000 – 1000 = 3000, leaving the primary insurer
paying 400.
Another way, exactly equivalent, to express that reinsurance is that the primary company pays
the insured 3400. The reinsurer reimburses the primary company for its claims less a deductible
of 400 applied to claims. So the reinsurer pays 3400 – 400 = 3000, the same as before.
Expected reinsured claims in 2005
⎛ 3000 ⎞ 9,000,000
= ( 3000 ) ⎜
⎟=
⎝ 3000 + r ⎠ 3000 + r
⎛ 9,000,000 ⎞
R2005 = (1.1) ⎜
⎟ = ( 0.55 ) P2005
⎝ 3000 + r ⎠
9,900,000
= ( 0.55 )( 3000 ) = 1650
3000 + r
r = 3000
In 2006, after 20% inflation, losses will have a two-parameter Pareto distribution with α = 2 and
θ = (1.2 )( 3000 ) = 3600 .
The general formula for claims will be
⎛ 3600 ⎞ 12,960,000
E ( X − d )+ = ( 3600 ) ⎜
⎟=
3600 + d
⎝ 3600 + d ⎠
⎛ 12,960,000 ⎞
P2006 = 1.2 ⎜
⎟ = 3703
⎝ 3000 + 600 ⎠
⎛ 12,960,000 ⎞
R2006 = 1.1⎜
⎟ = 2160
⎝ 3600 + 3000 ⎠
R2006 / P2006 = 0.5833
[If you applied the reinsurer’s deductible to the primary insurer’s claims, you would solve that
the deductible is 2400, and the answer to the problem is the same].
Question #29
Key: A
Benefits +
3
0 Le = 1000v +
at G = 41.20 and i = 0.05,
0 Le
( for K = 2 ) = 770.59
Expenses
( 0.20G + 8 ) + ( 0.06G + 2 ) v + ( 0.06G + 2 ) v 2
– Premiums
− G a3
Question #30
Key: D
P = 1000 P40
( 235 + P )(1 + i ) − 0.015 (1000 − 255) = 255
[A]
( 255 + P )(1 + i ) − 0.020 (1000 − 272 ) = 272
[B]
Subtract [A] from [B]
20 (1 + i ) − 3.385 = 17
1+ i =
Plug into [A]
20.385
= 1.01925
20
( 235 + P )(1.01925 ) − 0.015 (1000 − 255) = 255
235 + P =
255 + 11.175
1.01925
P = 261.15 − 235 = 26.15
1000 25V40 =
( 272 + 26.15 )(1.01925 ) − ( 0.025)(1000 )
1 − 0.025
= 286
Question #31
Key: A
1.2
1
1
0.8
0.6
0.5
0.4
0.4
0.2
0
0
10
20
30
Given
40
50
60
Given
70
80
90
0
100
Given
Given
x
0
15
25
35
75
90
100
s ( x)
1
0.70
Linear
Interpolation
0.50
0.48
Linear
Interpolation
0.4
0.16
Linear
Interpolation
0
55 q35
20 55 q15
20 55 q15
55 q35
= 1−
s ( 90 )
0.16 32
= 1−
=
= 0.6667
s ( 35 )
0.48 48
=
s ( 35 ) − s ( 90 ) 0.48 − 0.16 32
=
=
= 0.4571
s (15 )
0.70
70
=
0.4571
= 0.6856
0.6667
Alternatively,
20 55 q15
55 q35
=
20 p15 × 55 q35
55 q35
0.48
0.70
= 0.6856
=
=
20 p15
=
s ( 35 )
s (15 )
Question #32
Key: A
s ( 80 ) = 1 * ( e ^ ( −0.16*50 ) + e ^ ( −0.08*50 ) ) = 0.00932555
2
s ( 81) = 1 * ( e ^ ( −0.16*51) + e ^ ( −0.08*51) ) = 0.008596664
2
p80 = s ( 81) / s ( 80 ) = 0.008596664 / 0.00932555 = 0.9218
q80 = 1 − 0.9218 = 0.078
Alternatively (and equivalent to the above)
For non-smokers, px = e −0.08 = 0.923116
50
px =
−0.16
0.018316
For smokers, px = e
= 0.852144
50 p x = 0.000335
So the probability of dying at 80, weighted by the probability of surviving to 80, is
0.018316 × (1 − 0.923116 ) + 0.000335 × (1 − 0.852144 )
= 0.078
0.018316 + 0.000335
Question #33
Key: B
x
40
41
42
lx( )
2000
1920
1840
τ
because 2000 − 20 − 60 = 1920 ;
d x( )
20
30
40
1
d x( )
60
50
2
1920 − 30 − 50 = 1840
Let premium = P
1840 2 ⎞
⎛ 2000 1920
APV premiums = ⎜
+
v+
v ⎟ P = 2.749 P
2000 ⎠
⎝ 2000 2000
30 2
40 3 ⎞
⎛ 20
v+
v +
v ⎟ = 40.41
APV benefits = 1000 ⎜
2000
2000 ⎠
⎝ 2000
40.41
P=
= 14.7
2.749
Question #34
Key: C
Consider Disease 1 and other Diseases as independent Poisson processes with respective
⎛1⎞
⎛ 15 ⎞
λ ' s = ( 0.16 ) ⎜ ⎟ = 0.01 and ( 0.16 ) ⎜ ⎟ = 0.15 respectively. Let S1 = aggregate losses from
⎝ 16 ⎠
⎝ 16 ⎠
Disease 1; S2 = aggregate losses from other diseases.
E ( S1 ) = 100 × 0.01× 5 = 5
(
)
Var ( S1 ) = 100 × 0.01× 502 + 52 = 2525
E ( S 2 ) = 100 × 0.15 × 10 = 150
(
)
Var ( S2 ) = 100 × 0.15 × 202 + 102 = 7500
If no one gets the vaccine:
E ( S ) = 5 + 150 = 155
Var ( S ) = 2525 + 7500 = 10,025
Φ ( 0.7 ) = 1 − 0.24
A = 155 + 0.7 10,025 = 225.08
If all get the vaccine, vaccine cost = (100 )( 0.15 ) = 15
No cost or variance from Disease 1
B = 15 + 150 + 0.7 7500 = 225.62
A / B = 0.998
Question #35
Key: A
bg
For current model f x =
1
4e
− 4x
Let g(x) be the new density function, which has
0≤ x≤3
(i)
g(x) = c,
−x/4
g ( x ) = ke
, x > 3*
(ii)
(iii)
c = ke −3/ 4 , since continuous at x = 3
Since g is density function, it must integrate to 1.
∞
1 = 3c + ∫ ke
−x
4
d x = 3ke
−3
4
+ 4ke
−3
4
3
3
31
0
07
F ( 3) = ∫ cd x = ∫
= 3c + 4c ⇒ c = 1
7
d x = 3 7 = 0.43
⎛1
⎞
*This could equally well have been written g ( x ) = d × ⎜ e− x / 4 ⎟ , then let k = d/4, or even carry
⎝4
⎠
the d/4 throughout.
Question #36
Key: A
∞
10
a30 = ∫ e−0.08t e −0.05dt + 10 Ex ∫ e−0.08t e−0.08t dt
0
10 −0.13t
=∫ e
0
dt + e
0
∞
−1.3
−0.16
∫0 e
−e −0.13t 10
−e −0.16t
+ e −1.3
0.13 0
0.16
1.3
−1.3
−e
1
e
=
+
+
0.13 0.13 0.16
= 7.2992
(
)
dt
∞
0
∞
A30 = ∫ e −0.08t e −0.05t ( 0.05 ) dt + e−1.3 ∫ e−0.16t ( 0.08 ) dt
10
0
0
⎛ 1
e ⎞
e
= 0.05 ⎜
−
⎟ + ( 0.08 )
0.16
⎝ 0.13 0.13 ⎠
= 0.41606
−1.3
−1.3
= P ( A30 ) =
A30 0.41606
=
= 0.057
a30 7.29923
1
1
a40 =
=
0.08 + 0.08 0.16
A40 = 1 − δ a40
= 1 − ( 0.08 / 0.16 ) = 0.5
10V
( A40 ) = A40 − P ( A40 ) a40
= 0.5 −
( 0.057 ) = 0.14375
0.16
Question #37
Key: C
Let T be the future lifetime of Pat, and [T] denote the greatest integer in T. ([T] is the same as K,
the curtate future lifetime).
L = 100,000 vT − 1600 a T
[ ]+1
= 100,000vT − 1600 a10
−1600 a10
Minimum is −1600 a10
= −12,973
0 < T ≤10
10 < t ≤ 20
20<t
when evaluated at i = 0.05
Question #38
Key: C
Since loss amounts are uniform on (0, 10), 40% of losses are below the deductible (4), and 60%
are above. Thus, claims occur at a Poisson rate λ * = ( 0.6 )(10 ) = 6 .
Since loss amounts were uniform on (0, 10), claims are uniform on (0, 6).
Let N = number of claims; X = claim amount; S = aggregate claims.
E ( N ) = Var ( N ) = λ * = 6
E ( X ) = ( 6 − 0) / 2 = 3
Var ( X ) = ( 6 − 0 ) /12 = 3
2
Var ( S ) = E ( N )Var ( X ) + Var ( N ) ⎡⎣ E ( X ) ⎤⎦
2
= 6*3 + 6*32
= 72
Question #39
Key: E
n
pn
n × pn
n 2 × pn
0
1
2
3
0.1
0.4
0.3
0.2
0
0.4
0.6
0.6
0
0.4
1.2
1.8
E [ N ] = 1.6
E ⎡⎣ N 2 ⎤⎦ = 3.4
Var ( N ) = 3.4 − 1.62 = 0.84
E[X ] = λ = 3
Var ( X ) = λ = 3
Var ( S ) =
= E [ N ]Var ( X ) + E ( X ) × Var ( N )
2
= 1.6 ( 3) + ( 3) ( 0.84 )
2
= 12.36
Question #40
Key: B
Method 1: as three independent processes, based on the amount deposited. Within each
process, since the amount deposited is always the same, Var ( X ) = 0 .
Rate of depositing 10 = 0.05 * 22 = 1.1
Rate of depositing 5 = 0.15 * 22 = 3.3
Rate of depositing 1 = 0.80 * 22 = 17.6
Variance of depositing 10 = 1.1 * 10 * 10 = 110
Variance of depositing 5 = 3.3 * 5 * 5 = 82.5
Variance of depositing 1 = 17.6 * 1 *1 = 17.6
Total Variance = 110 + 82.5 + 17.6 = 210.1
Method 2: as a single compound Poisson process
E ( X ) = 0.8 × 1 + 0.15 × 5 + 0.05 × 10 = 2.05
( )
E X 2 = 0.8 ×12 + 0.15 × 52 + 0.05 × 102 = 9.55
Var ( S ) = E ( N )Var ( X ) + Var ( N ) ( E ( X ) )
(
= ( 22 )( 5.3475 ) + ( 22 ) 2.052
= 210.1
)
2