**BEGINNING OF EXAMINATION**
1.
For a fully discrete 3-year endowment insurance of 1000 on (x), you are given:
(i)
kL
is the prospective loss random variable at time k.
(ii)
i = 0.10
(iii)
ax:3 = 2.70182
(iv)
Premiums are determined by the equivalence principle.
Calculate 1 L , given that (x) dies in the second year from issue.
(A)
540
(B)
630
(C)
655
(D)
720
(E)
910
Exam M: Spring 2005
-1-
GO ON TO NEXT PAGE
2.
For a double-decrement model:
(i)
t
p' (40) = 1 −
t
,
60
0 ≤ t ≤ 60
(ii)
t
p' (40) = 1 −
t
,
40
0 ≤ t ≤ 40
1
2
( )
Calculate µ 40
( 20 ) .
τ
(A)
0.025
(B)
0.038
(C)
0.050
(D)
0.063
(E)
0.075
Exam M: Spring 2005
-2-
GO ON TO NEXT PAGE
3.
For independent lives (35) and (45):
(i)
5
p35 = 0.90
(ii)
5
p45 = 0.80
(iii)
q40 = 0.03
(iv)
q50 = 0.05
Calculate the probability that the last death of (35) and (45) occurs in the 6th year.
(A)
0.0095
(B)
0.0105
(C)
0.0115
(D)
0.0125
(E)
0.0135
Exam M: Spring 2005
-3-
GO ON TO NEXT PAGE
4.
For a fully discrete whole life insurance of 100,000 on (35) you are given:
(i)
Percent of premium expenses are 10% per year.
(ii)
Per policy expenses are 25 per year.
(iii)
Per thousand expenses are 2.50 per year.
(iv)
All expenses are paid at the beginning of the year.
(v)
1000 P35 = 8.36
Calculate the level annual expense-loaded premium using the equivalence principle.
(A)
930
(B)
1041
(C)
1142
(D)
1234
(E)
1352
Exam M: Spring 2005
-4-
GO ON TO NEXT PAGE
5.
Kings of Fredonia drink glasses of wine at a Poisson rate of 2 glasses per day.
Assassins attempt to poison the king’s wine glasses. There is a 0.01 probability that any
given glass is poisoned. Drinking poisoned wine is always fatal instantly and is the only
cause of death.
The occurrences of poison in the glasses and the number of glasses drunk are independent
events.
Calculate the probability that the current king survives at least 30 days.
(A)
0.40
(B)
0.45
(C)
0.50
(D)
0.55
(E)
0.60
Exam M: Spring 2005
-5-
GO ON TO NEXT PAGE
6.
Insurance losses are a compound Poisson process where:
(i)
The approvals of insurance applications arise in accordance with a Poisson process at
a rate of 1000 per day.
(ii)
Each approved application has a 20% chance of being from a smoker and an 80%
chance of being from a non-smoker.
(iii)
The insurances are priced so that the expected loss on each approval is –100.
(iv)
The variance of the loss amount is 5000 for a smoker and is 8000 for a non-smoker.
Calculate the variance for the total losses on one day’s approvals.
(A)
13,000,000
(B)
14,100,000
(C)
15,200,000
(D)
16,300,000
(E)
17,400,000
Exam M: Spring 2005
-6-
GO ON TO NEXT PAGE
7.
Z is the present-value random variable for a whole life insurance of b payable at the moment
of death of (x).
You are given:
(i)
δ = 0.04
(ii)
µ x ( t ) = 0.02 ,
(iii)
The single benefit premium for this insurance is equal to Var(Z).
t≥0
Calculate b.
(A)
2.75
(B)
3.00
(C)
3.25
(D)
3.50
(E)
3.75
Exam M: Spring 2005
-7-
GO ON TO NEXT PAGE
8.
For a special 3-year term insurance on (30), you are given:
(i)
Premiums are payable semiannually.
(ii)
Premiums are payable only in the first year.
(iii)
Benefits, payable at the end of the year of death, are:
k
bk +1
0
1
2
1000
500
250
(iv)
Mortality follows the Illustrative Life Table.
(v)
Deaths are uniformly distributed within each year of age.
(vi)
i = 0.06
Calculate the amount of each semiannual benefit premium for this insurance.
(A)
1.3
(B)
1.4
(C)
1.5
(D)
1.6
(E)
1.7
Exam M: Spring 2005
-8-
GO ON TO NEXT PAGE
9.
A loss, X, follows a 2-parameter Pareto distribution with α = 2 and unspecified
parameter θ . You are given:
5
E ⎡⎣ X − 100 X > 100 ⎤⎦ = E ⎡⎣ X − 50 X > 50 ⎤⎦
3
Calculate E ⎡⎣ X − 150 X > 150 ⎤⎦ .
(A)
150
(B)
175
(C)
200
(D)
225
(E)
250
Exam M: Spring 2005
-9-
GO ON TO NEXT PAGE
10.
The scores on the final exam in Ms. B’s Latin class have a normal distribution with mean θ
and standard deviation equal to 8. θ is a random variable with a normal distribution with
mean equal to 75 and standard deviation equal to 6.
Each year, Ms. B chooses a student at random and pays the student 1 times the student’s
score. However, if the student fails the exam (score ≤ 65 ), then there is no payment.
Calculate the conditional probability that the payment is less than 90, given that there is a
payment.
(A)
0.77
(B)
0.85
(C)
0.88
(D)
0.92
(E)
1.00
Exam M: Spring 2005
- 10 -
GO ON TO NEXT PAGE
11.
For a Markov model with three states, Healthy (0), Disabled (1), and Dead (2):
(i)
The annual transition matrix is given by
0
0
1
2
(ii)
1
2
⎡ 0.70 0.20 0.10 ⎤
⎢ 0.10 0.65 0.25 ⎥
⎢
⎥
⎢⎣ 0
0
1 ⎥⎦
There are 100 lives at the start, all Healthy. Their future states are independent.
Calculate the variance of the number of the original 100 lives who die within the first two
years.
(A)
11
(B)
14
(C)
17
(D)
20
(E)
23
Exam M: Spring 2005
- 11 -
GO ON TO NEXT PAGE
12.
An insurance company issues a special 3-year insurance to a high risk individual. You are
given the following homogenous Markov chain model:
(i)
State 1: active
State 2: disabled
State 3: withdrawn
State 4: dead
Transition probability matrix:
1
2
3
4
1 ⎡ 0.4 0.2 0.3 0.1⎤
2 ⎢ 0.2 0.5 0 0.3⎥
⎢
⎥
3 ⎢0
0
1
0⎥
⎢
⎥
4 ⎣0
0
0
1⎦
(ii)
Changes in state occur at the end of the year.
(iii)
The death benefit is 1000, payable at the end of the year of death.
(iv)
i = 0.05
(v)
The insured is disabled at the end of year 1.
Calculate the actuarial present value of the prospective death benefits at the beginning of
year 2.
(A)
440
(B)
528
(C)
634
(D)
712
(E)
803
Exam M: Spring 2005
- 12 -
GO ON TO NEXT PAGE
13.
For a fully discrete whole life insurance of b on (x), you are given:
(i)
qx +9 = 0.02904
(ii)
i = 0.03
(iii)
The initial benefit reserve for policy year 10 is 343.
(iv)
The net amount at risk for policy year 10 is 872.
(v)
ax = 14.65976
Calculate the terminal benefit reserve for policy year 9.
(A)
280
(B)
288
(C)
296
(D)
304
(E)
312
Exam M: Spring 2005
- 13 -
GO ON TO NEXT PAGE
14.
For a special fully discrete 2-year endowment insurance of 1000 on (x), you are given:
(i)
The first year benefit premium is 668.
(ii)
The second year benefit premium is 258.
(iii)
d = 0.06
Calculate the level annual premium using the equivalence principle.
(A)
469
(B)
479
(C)
489
(D)
499
(E)
509
Exam M: Spring 2005
- 14 -
GO ON TO NEXT PAGE
15.
For an increasing 10-year term insurance, you are given:
(i)
bk +1 = 100,000 (1 + k ) ,
(ii)
Benefits are payable at the end of the year of death.
(iii)
Mortality follows the Illustrative Life Table.
(iv)
i = 0.06
(v)
The single benefit premium for this insurance on (41) is 16,736.
k = 0, 1,…,9
Calculate the single benefit premium for this insurance on (40).
(A)
12,700
(B)
13,600
(C)
14,500
(D)
15,500
(E)
16,300
Exam M: Spring 2005
- 15 -
GO ON TO NEXT PAGE
16.
For a fully discrete whole life insurance of 1000 on (x):
(i)
Death is the only decrement.
(ii)
The annual benefit premium is 80.
(iii)
The annual contract premium is 100.
(iv)
Expenses in year 1, payable at the start of the year, are 40% of contract premiums.
(v)
i = 0.10
(vi)
10001Vx = 40
Calculate the asset share at the end of the first year.
(A)
17
(B)
18
(C)
19
(D)
20
(E)
21
Exam M: Spring 2005
- 16 -
GO ON TO NEXT PAGE
17.
For a collective risk model the number of losses, X, has a Poisson distribution with λ = 20 .
The common distribution of the individual losses has the following characteristics:
(i)
E [ X ] = 70
(ii)
E [ X ∧ 30] = 25
(iii)
Pr ( X > 30 ) = 0.75
(iv)
E ⎡⎣ X 2 X > 30 ⎤⎦ = 9000
An insurance covers aggregate losses subject to an ordinary deductible of 30 per loss.
Calculate the variance of the aggregate payments of the insurance.
(A)
54,000
(B)
67,500
(C)
81,000
(D)
94,500
(E)
108,000
Exam M: Spring 2005
- 17 -
GO ON TO NEXT PAGE
18.
For a collective risk model:
(i)
The number of losses has a Poisson distribution with λ = 2 .
(ii)
The common distribution of the individual losses is:
x
fx ( x)
1
2
0.6
0.4
An insurance covers aggregate losses subject to a deductible of 3.
Calculate the expected aggregate payments of the insurance.
(A)
0.74
(B)
0.79
(C)
0.84
(D)
0.89
(E)
0.94
Exam M: Spring 2005
- 18 -
GO ON TO NEXT PAGE
19.
A discrete probability distribution has the following properties:
(i)
⎛ 1⎞
pk = c ⎜1 + ⎟ pk −1
⎝ k⎠
(ii)
p0 = 0.5
for
k = 1, 2,…
Calculate c.
(A)
0.06
(B)
0.13
(C)
0.29
(D)
0.35
(E)
0.40
Exam M: Spring 2005
- 19 -
GO ON TO NEXT PAGE
20.
A fully discrete 3-year term insurance of 10,000 on (40) is based on a double-decrement
model, death and withdrawal:
(i)
Decrement 1 is death.
(ii)
(1)
µ 40
( t ) = 0.02 ,
(iii)
Decrement 2 is withdrawal, which occurs at the end of the year.
(iv)
q '(40)+ k = 0.04 ,
(v)
v = 0.95
2
t≥0
k = 0, 1, 2
Calculate the actuarial present value of the death benefits for this insurance.
(A)
487
(B)
497
(C)
507
(D)
517
(E)
527
Exam M: Spring 2005
- 20 -
GO ON TO NEXT PAGE
21.
You are given:
D
(i)
e30:40 = 27.692
(ii)
s ( x) = 1−
(iii)
T ( x ) is the future lifetime random variable for (x).
x
ω
,
0 ≤ x ≤ω
Calculate Var (T ( 30 ) ) .
(A)
332
(B)
352
(C)
372
(D)
392
(E)
412
Exam M: Spring 2005
- 21 -
GO ON TO NEXT PAGE
22.
For a fully discrete 5-payment 10-year decreasing term insurance on (60), you are given:
(i)
bk +1 = 1000 (10 − k ) ,
(ii)
Level benefit premiums are payable for five years and equal 218.15 each.
(iii)
q60+ k = 0.02 + 0.001 k ,
(iv)
i = 0.06
k = 0, 1, 2,…, 9
k = 0, 1, 2,…, 9
Calculate 2V , the benefit reserve at the end of year 2.
(A)
70
(B)
72
(C)
74
(D)
76
(E)
78
Exam M: Spring 2005
- 22 -
GO ON TO NEXT PAGE
23.
You are given:
(i)
T ( x ) and T ( y ) are not independent.
(ii)
qx + k = q y + k = 0.05 ,
(iii)
k
pxy = 1.02 k px k p y ,
k = 0, 1, 2,…
k = 1, 2, 3…
Into which of the following ranges does ex: y , the curtate expectation of life of the last
survivor status, fall?
(A)
ex: y ≤ 25.7
(B)
25.7 < ex: y ≤ 26.7
(C)
26.7 < ex: y ≤ 27.7
(D)
27.7 < ex: y ≤ 28.7
(E)
28.7 < ex: y
Exam M: Spring 2005
- 23 -
GO ON TO NEXT PAGE
24.
Subway trains arrive at your station at a Poisson rate of 20 per hour. 25% of the trains are
express and 75% are local. The types and number of trains arriving are independent. An
express gets you to work in 16 minutes and a local gets you there in 28 minutes. You always
take the first train to arrive. Your co-worker always takes the first express. You are both
waiting at the same station.
Calculate the conditional probability that you arrive at work before your co-worker, given
that a local arrives first.
(A)
37%
(B)
40%
(C)
43%
(D)
46%
(E)
49%
Exam M: Spring 2005
- 24 -
GO ON TO NEXT PAGE
25.
Beginning with the first full moon in October deer are hit by cars at a Poisson rate of 20 per
day. The time between when a deer is hit and when it is discovered by highway maintenance
has an exponential distribution with a mean of 7 days. The number hit and the times until
they are discovered are independent.
Calculate the expected number of deer that will be discovered in the first 10 days following
the first full moon in October.
(A)
78
(B)
82
(C)
86
(D)
90
(E)
94
Exam M: Spring 2005
- 25 -
GO ON TO NEXT PAGE
26.
You are given:
(i)
µ x ( t ) = 0.03 ,
(ii)
δ = 0.05
(iii)
T(x) is the future lifetime random variable.
(iv)
g is the standard deviation of aT ( x ) .
(
t≥0
)
Calculate Pr aT ( x ) > ax − g .
(A)
0.53
(B)
0.56
(C)
0.63
(D)
0.68
(E)
0.79
Exam M: Spring 2005
- 26 -
GO ON TO NEXT PAGE
27.
(50) is an employee of XYZ Corporation. Future employment with XYZ follows a double
decrement model:
(i)
Decrement 1 is retirement.
(ii)
(1)
µ50
(t ) = ⎨
(iii)
Decrement 2 is leaving employment with XYZ for all other causes.
(iv)
( 2)
µ50
(t ) = ⎨
(v)
If (50) leaves employment with XYZ, he will never rejoin XYZ.
⎧0.00
⎩0.02
⎧0.05
⎩0.03
0≤t <5
5≤t
0≤t <5
5≤t
Calculate the probability that (50) will retire from XYZ before age 60.
(A)
0.069
(B)
0.074
(C)
0.079
(D)
0.084
(E)
0.089
Exam M: Spring 2005
- 27 -
GO ON TO NEXT PAGE
28.
For a life table with a one-year select period, you are given:
(i)
(ii)
D
x
l[ x]
d[ x]
l x +1
e[ x]
80
81
1000
920
90
90
−
8.5
−
−
Deaths are uniformly distributed over each year of age.
D
Calculate e[81] .
(A)
8.0
(B)
8.1
(C)
8.2
(D)
8.3
(E)
8.4
Exam M: Spring 2005
- 28 -
GO ON TO NEXT PAGE
29.
For a fully discrete 3-year endowment insurance of 1000 on (x):
(i)
i = 0.05
(ii)
px = px +1 = 0.7
Calculate the second year terminal benefit reserve.
(A)
526
(B)
632
(C)
739
(D)
845
(E)
952
Exam M: Spring 2005
- 29 -
GO ON TO NEXT PAGE
30.
For a fully discrete whole life insurance of 1000 on (50), you are given:
(i)
The annual per policy expense is 1.
(ii)
There is an additional first year expense of 15.
(iii)
The claim settlement expense of 50 is payable when the claim is paid.
(iv)
All expenses, except the claim settlement expense, are paid at the beginning of the
year.
(v)
Mortality follows De Moivre’s law with ω = 100 .
(vi)
i = 0.05
Calculate the level expense-loaded premium using the equivalence principle.
(A)
27
(B)
28
(C)
29
(D)
30
(E)
31
Exam M: Spring 2005
- 30 -
GO ON TO NEXT PAGE
31.
The repair costs for boats in a marina have the following characteristics:
Boat type
Power boats
Sailboats
Luxury yachts
Number of
boats
100
Probability that
repair is needed
0.3
Mean of repair cost
given a repair
300
Variance of repair
cost given a repair
10,000
300
0.1
1000
400,000
50
0.6
5000
2,000,000
At most one repair is required per boat each year.
The marina budgets an amount, Y, equal to the aggregate mean repair costs plus the standard
deviation of the aggregate repair costs.
Calculate Y.
(A)
200,000
(B)
210,000
(C)
220,000
(D)
230,000
(E)
240,000
Exam M: Spring 2005
- 31 -
GO ON TO NEXT PAGE
32.
For an insurance:
(i)
Losses can be 100, 200 or 300 with respective probabilities 0.2, 0.2, and 0.6.
(ii)
The insurance has an ordinary deductible of 150 per loss.
(iii)
Y P is the claim payment per payment random variable.
( )
Calculate Var Y P .
(A)
1500
(B)
1875
(C)
2250
(D)
2625
(E)
3000
Exam M: Spring 2005
- 32 -
GO ON TO NEXT PAGE
33.
You are given:
⎧0.05
⎩0.04
µ ( x) = ⎨
Calculate
50 ≤ x < 60
60 ≤ x < 70
4 14 q50 .
(A)
0.38
(B)
0.39
(C)
0.41
(D)
0.43
(E)
0.44
Exam M: Spring 2005
- 33 -
GO ON TO NEXT PAGE
34.
The distribution of a loss, X , is a two-point mixture:
(i)
With probability 0.8, X has a two-parameter Pareto distribution with
α = 2 and θ = 100 .
(ii)
With probability 0.2, X has a two-parameter Pareto distribution with
α = 4 and θ = 3000 .
Calculate Pr ( X ≤ 200 ) .
(A)
0.76
(B)
0.79
(C)
0.82
(D)
0.85
(E)
0.88
Exam M: Spring 2005
- 34 -
GO ON TO NEXT PAGE
35.
For a special fully discrete 5-year deferred whole life insurance of 100,000 on (40), you are
given:
(i)
The death benefit during the 5-year deferral period is return of benefit premiums paid
without interest.
(ii)
Annual benefit premiums are payable only during the deferral period.
(iii)
Mortality follows the Illustrative Life Table.
(iv)
i = 0.06
(v)
( IA)140:5
= 0.04042
Calculate the annual benefit premiums.
(A)
3300
(B)
3320
(C)
3340
(D)
3360
(E)
3380
Exam M: Spring 2005
- 35 -
GO ON TO NEXT PAGE
36.
You are pricing a special 3-year annuity-due on two independent lives, both age 80. The
annuity pays 30,000 if both persons are alive and 20,000 if only one person is alive.
You are given:
(i)
(ii)
p80
k
k
1
0.91
2
0.82
3
0.72
i = 0.05
Calculate the actuarial present value of this annuity.
(A)
78,300
(B)
80,400
(C)
82,500
(D)
84,700
(E)
86,800
Exam M: Spring 2005
- 36 -
GO ON TO NEXT PAGE
37.
Company ABC sets the contract premium for a continuous life annuity of 1 per year on (x)
equal to the single benefit premium calculated using:
(i)
δ = 0.03
(ii)
µ x ( t ) = 0.02 ,
t≥0
However, a revised mortality assumption reflects future mortality improvement and is given
by
⎧0.02 for t ≤ 10
⎩0.01 for t > 10
µ x (t ) = ⎨
Calculate the expected loss at issue for ABC (using the revised mortality assumption) as a
percentage of the contract premium.
(A)
2%
(B)
8%
(C)
15%
(D)
20%
(E)
23%
Exam M: Spring 2005
- 37 -
GO ON TO NEXT PAGE
38.
A group of 1000 lives each age 30 sets up a fund to pay 1000 at the end of the first year for
each member who dies in the first year, and 500 at the end of the second year for each
member who dies in the second year. Each member pays into the fund an amount equal to
the single benefit premium for a special 2-year term insurance, with:
(i)
Benefits:
k
bk +1
0
1
1000
500
(ii)
Mortality follows the Illustrative Life Table.
(iii)
i = 0.06
The actual experience of the fund is as follows:
k
Interest Rate Earned
Number of Deaths
0
1
0.070
0.069
1
1
Calculate the difference, at the end of the second year, between the expected size of the fund
as projected at time 0 and the actual fund.
(A)
840
(B)
870
(C)
900
(D)
930
(E)
960
Exam M: Spring 2005
- 38 -
GO ON TO NEXT PAGE
39.
In a certain town the number of common colds an individual will get in a year follows a
Poisson distribution that depends on the individual’s age and smoking status. The
distribution of the population and the mean number of colds are as follows:
Proportion of population
Mean number of colds
Children
0.30
3
Adult Non-Smokers
0.60
1
Adult Smokers
0.10
4
Calculate the conditional probability that a person with exactly 3 common colds in a year is
an adult smoker.
(A)
0.12
(B)
0.16
(C)
0.20
(D)
0.24
(E)
0.28
Exam M: Spring 2005
- 39 -
GO ON TO NEXT PAGE
40.
For aggregate losses, S:
(i)
The number of losses has a negative binomial distribution with mean 3 and
variance 3.6.
(ii)
The common distribution of the independent individual loss amounts is uniform from
0 to 20.
Calculate the 95th percentile of the distribution of S as approximated by the normal
distribution.
(A)
61
(B)
63
(C)
65
(D)
67
(E)
69
**END OF EXAMINATION**
Exam M: Spring 2005
- 40 -
STOP
Exam M
Spring 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
B
E
B
D
D
E
E
A
B
D
C
A
C
B
D
A
B
A
C
C
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
B
E
D
A
E
E
A
C
A
E
B
B
A
A
D
B
C
C
B
C
Exam M: Spring 2005
- 41 -
SPRING 2005
EXAM M SOLUTIONS
Question #1
Key: B
Let K be the curtate future lifetime of (x + k)
kL
= 1000v K +1 − 1000 Px:3 × aK +1
When (as given in the problem), (x) dies in the second year from issue, the curtate future lifetime
of ( x + 1) is 0, so
1L
= 1000v − 1000 Px:3 a1
1000
− 279.21
1.1
= 629.88 ≈ 630
=
The premium came from
A
Px:3 = x:3
a x:3
Ax:3 = 1 − d ax:3
Px:3 = 279.21 =
1 − d ax:3
ax:3
=
1
−d
ax:3
Question #2
Key: E
Note that above 40, decrement 1 is DeMoivre with omega = 100; decrement 2 is DeMoivre with
omega = 80.
(1)
That means µ 40
( 20 ) = 1/ 40 = 0.025; µ 40( 2) ( 20 ) = 1/ 20 = 0.05
(τ )
µ 40
( 20 ) = 0.025 + 0.05 = 0.075
Or from basic definition of µ ,
( )
t p40 =
τ
d
(
)
( )
p40
/ dt = ( −100 + 2t ) / 2400
τ
t
60 − t 40 − t 2400 − 100t + t 2
×
=
60
40
2400
at t = 20 gives −60 / 2400 = 0.025
( )
p40
= ( 2 / 3) * (1/ 2 ) = 1/ 3
τ
20
(τ )
µ 40
( 20 ) = ⎡⎢ −d
⎣
(
t
)
(τ )
(τ )
p40
/ dt ⎤⎥ / 20 p40
= 0.025 / (1/ 3) = 0.075
⎦
Question #3
Key: B
5
q35:45 = 5 q35 + 5 q45 − 5 q35:45
= 5 p35q40 + 5 p45q50 − 5 p35:45q40:50
b
g
= p q + p q − p × p b1 − p p g
= b0.9gb.03g + b0.8gb0.05g − b0.9gb0.8g 1 − b0.97gb0.95g
= 5 p35q40 + 5 p45q50 − 5 p35 × 5 p45 1 − p40:50
5 35 40
5 45 50
5 35
5 45
40
50
= 0.01048
Alternatively,
6
p35 = 5 p35 × p40 = ( 0.90 )(1 − 0.03) = 0.873
6
p45 = 5 p45 × p50 = ( 0.80 )(1 − 0.05 ) = 0.76
5 q35:45
= 5 p35:45 − 6 p35:45
= ( 5 p35 + 5 p45 − 5 p35:45 ) − ( 6 p35 + 6 p45 − 6 p35:45 )
= ( 5 p35 + 5 p45 + 5 p35 × 5 p45 ) − ( 6 p35 + 6 p45 − 6 p35 × 6 p45 )
= ( 0.90 + 0.80 − 0.90 × 0.80 ) − ( 0.873 + 0.76 − 0.873 × 0.76 )
= 0.98 − 0.96952
= 0.01048
Question #4
Key: D
Let G be the expense-loaded premium.
Actuarial present value (APV) of benefits = 100,000A35
APV of premiums = Ga35
APV of expenses = ⎡⎣ 0.1G + 25 + ( 2.50 )(100 ) ⎤⎦ a35
Equivalence principle:
Ga35 = 100,000 A35 + ( 0.1G + 25 + 250 ) a35
A35
+ 0.1G + 275
a35
0.9G = 100,000 P35 + 275
G = 100,000
G=
(100 )(8.36 ) + 275
0.9
= 1234
Question #5
Key: D
Poisoned wine glasses are drunk at a Poisson rate of 2 × 0.01 = 0.02 per day.
Number of glasses in 30 days is Poisson with λ = 0.02 × 30 = 0.60
f ( 0 ) = e −0.60 = 0.55
Question #6
Key: E
View the compound Poisson process as two compound Poisson processes, one for smokers and
one for non-smokers. These processes are independent, so the total variance is the sum of their
variances.
For smokers, λ = ( 0.2 )(1000 ) = 200
2
Var(losses) = λ ⎡⎢ Var ( X ) + ( E ( X ) ) ⎤⎥
⎣
⎦
2
= 200 ⎡5000 + ( −100 ) ⎤
⎣
⎦
= 3,000,000
For non-smokers, λ = ( 0.8 )(1000 ) = 800
2
Var(losses) = λ ⎡⎢ Var ( X ) + ( E ( X ) ) ⎤⎥
⎣
⎦
2
= 800 ⎡8000 + ( −100 ) ⎤
⎣
⎦
= 14, 400,000
Total variance = 3,000,000 + 14,400,000
= 17,400,000
Question #7
Key: E
E [ Z ] = b Ax
since constant force Ax = µ / µ + δ
E(Z) =
b ( 0.02 )
bµ
=
= b/3
µ +δ
( 0.06 )
Var [ Z ] = Var ⎡⎣b v T ⎤⎦ = b 2 Var ⎡⎣ v Τ ⎤⎦ = b 2
2
⎛ µ
⎛ µ ⎞ ⎞
⎟
=b ⎜
−
⎜ µ + 2δ ⎝⎜ µ + δ ⎠⎟ ⎟
⎝
⎠
⎡ 2 1⎤
⎛ 4⎞
= b2 ⎢ − ⎥ = b2 ⎜ ⎟
⎣10 9 ⎦
⎝ 45 ⎠
2
Var ( Z ) = E ( Z )
⎡4⎤ b
b2 ⎢ ⎥ =
⎣ 45 ⎦ 3
⎡4⎤ 1
b ⎢ ⎥ = ⇒ b = 3.75
⎣ 45 ⎦ 3
(
2
Ax − Ax2
)
Question #8
Key: A
1
A30:3
= 1000vq30 + 500v 2 1 q30 + 250v3 2 q30
2
3
⎛ 1 ⎞⎛ 1.53 ⎞
⎛ 1 ⎞
⎛ 1.61 ⎞
⎛ 1 ⎞
⎛ 1.70 ⎞
= 1000 ⎜
⎟⎜
⎟ + 500 ⎜
⎟ ( 0.99847 ) ⎜
⎟ + 250 ⎜
⎟ ( 0.99847 )( 0.99839 ) ⎜
⎟
⎝ 1.06 ⎠⎝ 1000 ⎠
⎝ 1.06 ⎠
⎝ 1000 ⎠
⎝ 1.06 ⎠
⎝ 1000 ⎠
= 1.4434 + 0.71535 + 0.35572 = 2.51447
1
1 1
⎛ 1 ⎞2
⎛ 0.00153 ⎞
1
a30:1 = 2 + 2 ⎜
⎟ (1 − 2 q30 ) = + ( 0.97129 ) ⎜1 −
⎟
2 2
2
⎝ 1.06 ⎠
⎝
⎠
1 1
= + ( 0.97129 )( 0.999235 )
2 2
= 0.985273
2.51447
Annualized premium =
0.985273
= 2.552
( 2)
1
1
2.552
2
= 1.28
Each semiannual premium =
Question #9
Key: B
E ⎡⎣ x − d x > d ⎤⎦ is the expected payment per payment with an ordinary deductible of d
It can be evaluated (for Pareto) as
α −1
θ
θ ⎡ ⎛ θ ⎞ ⎤
−
⎢1 − ⎜
⎟ ⎥
E ( x ) − E ( x ∧ d ) α − 1 α − 1 ⎣⎢ ⎝ d + θ ⎠ ⎦⎥
=
1− F (d )
⎡ ⎛ θ ⎞α ⎤
1 − ⎢1 − ⎜
⎟ ⎥
⎣⎢ ⎝ d + θ ⎠ ⎦⎥
α −1
θ ⎛ θ ⎞
α − 1 ⎜⎝ d + θ ⎟⎠
=
α
⎛ θ ⎞
⎜
⎟
⎝ d +θ ⎠
d +θ
=
α −1
= d + θ in this problem, since α = 2
E ⎡⎣ x − 100 x > 100 ⎤⎦ = 5 3 E ⎡⎣ x − 50 x > 50⎤⎦
100 + θ = 5 3 ( 50 + θ )
300 + 3θ = 250 + 5θ
= θ = 25
E ⎡⎣ x − 150 x > 150 ⎤⎦ = 150 + θ
= 150 + 25
= 175
Question #10
Key: D
Let S = score
E ( S ) = E E ( S θ ) = E (θ ) = 75
(
)
Var ( S ) = E ⎡⎣Var ( S θ ) ⎤⎦ + Var ⎡⎣ E ( S θ ) ⎤⎦
( )
= E 82 + Var (θ )
= 64 + 62
= 100
S is normally distributed (a normal mixture of normal distributions with constant variance is
normal; see Example 4.30 in Loss Models for the specific case, as we have here, with a normally
distributed mean and constant variance)
Pr ob ⎡⎣ S < 90 S > 65⎤⎦ =
F ( 90 ) − F ( 65 )
1 − F ( 65 )
⎛ 90 − 75 ⎞
⎛ 65 − 75 ⎞
Φ⎜
⎟ − Φ⎜
⎟
10 ⎠
⎝ 10 ⎠
= ⎝
⎛ 65 − 75 ⎞
1− Φ ⎜
⎟
⎝ 10 ⎠
Φ (1.5 ) − Φ ( −1.0 ) 0.9332 − (1 − 0.8413) 0.7745
=
=
= 0.9206
1 − Φ ( −1.0 )
1 − (1 − 0.8413)
0.8413
Note that (though this insight is unnecessary here), this is equivalent to per payment model with
a franchise deductible of 65.
Question #11
Key: C
Ways to go 0 → 2 in 2 years
0 − 0 − 2; p = ( 0.7 )( 0.1) = 0.07
0 − 1 − 2; p = ( 0.2 )( 0.25 ) = 0.05
0 − 2 − 2; p = ( 0.1)(1) = 0.1
Total = 0.22
Binomial m = 100
q = 0.22
Var = (100) (0.22) (0.78) = 17
Question #12
Key: A
For death occurring in year 2
0.3 × 1000
APV =
= 285.71
1.05
For death occurring in year 3, two cases:
(1) State 2 → State 1 → State 4: (0.2 × 0.1) = 0.02
(2) State 2 → State 2 → State 4: (0.5 × 0.3) = 0.15
0.17
Total
APV =
0.17 ×1000
= 154.20
1.052
Total. APV = 285.71 + 154.20 = 439.91
Question #13
Key: C
( 9V + P ) (1.03) = qx+9b + (1 − qx+9 ) 10V
= qx +9 ( b − 10V ) + 10V
( 343)(1.03) = 0.02904 ( 872 ) + 10V
⇒ 10V = 327.97
b = ( b − 10V ) + 10V = 872 + 327.97 = 1199.97
⎛ 1
⎞
1
0.03 ⎞
⎛
−
P = b ⎜ − d ⎟ = 1200 ⎜
⎟
⎝ 14.65976 1.03 ⎠
⎝ ax
⎠
= 46.92
V
=
initial
reserve
– P = 343 – 46.92 = 296.08
9
Question #14
Key: B
d = 0.06 ⇒ V = 0.94
Step 1 Determine px
668 + 258vpx = 1000vqx + 1000v 2 px ( p x +1 + qx +1 )
668 + 258 ( 0.94 ) px = 1000 ( 0.94 ) (1 − px ) + 1000 ( 0.8836 ) px (1)
668 + 242.52 px = 940 (1 − px ) + 883.6 px
px = 272 / 298.92 = 0.91
Step 2 Determine 1000 Px:2
668 + 258 ( 0.94 )( 0.91) = 1000 Px:2 ⎡⎣1 + ( 0.94 )( 0.91) ⎤⎦
[ 220.69 + 668] = 479
1000 Px:2 =
1.8554
Question #15
Key: D
1
1
1
100,000 ( IA )40:10 = 100,000 v p40 ⎡( IA )41:10 − 10 v10 9 p41 q50 ⎤ + A40:10
(100,000 )
⎣
⎦
⎡
⎤
⎛ 8,950,901 ⎞
10 ⎜
⎢
⎥
⎟
0.99722 ⎢
9, 287, 264 ⎠
0.16736 − ⎝
0.00592
= 100,000
×
(
)⎥
1.06 ⎢
1.0610
⎥
⎢
⎥
⎣
⎦
+ ( 0.02766 × 100,000 )
=15,513
[see comment ]
1
= A40 − 10 E40 A50
Where A40:10
= 0.16132 − ( 0.53667 )( 0.24905 )
= 0.02766
Comment: the first line comes from comparing the benefits of the two insurances. At each of
1
age 40, 41, 42,…,49 ( IA )40:10 provides a death benefit 1 greater than ( IA )41:10 . Hence the A40:10
1
1
term. But ( IA )41:10 provides a death benefit at 50 of 10, while ( IA )40:10 provides 0. Hence a
1
1
term involving 9 q41 = 9 p41 q50 . The various v’s and p’s just get all actuarial present values at
age 40.
Question #16
Key: A
1000 1Vx = π (1 + i ) − qx (1000 − 10001Vx )
40 = 80 (1.1) − qx (1000 − 40 )
qx =
88 − 40
= 0.05
960
1 AS
=
=
=
(G −
expenses )(1 + i ) − 1000qx
px
(100 − ( 0.4 )(100 ) ) (1.1) − (1000 )( 0.05)
1 − 0.05
60 (1.1) − 50
= 16.8
0.95
Question #17
Key: B
(Referring to the number of losses, X, was a mistake. X is the random variable for the loss
amount, the severity distribution).
Losses in excess of the deductible occur at a Poisson rate
of λ * = (1 − F ( 30 ) ) λ = 0.75 × 20 = 15
E ( X − 30 X > 30 ) =
Var ( S ) = λ * × E
(
70 − 25
45
=
= 60
0.75
0.75
(( X − 30)
2
X > 30
)
)
(
= 15E X 2 − 60 X + 900 X > 30 = 15 E X 2 − 60 ( X − 30 ) − 900 X > 30
= 15 ( 9,000 − 60 × 60 − 900 )
= 67,500
)
Question #18
Key: A
S
( S − 3) +
0
0
E [ S ] = 2 × [ 0.6 + 2 × 0.4] = 2.8
1
0
f S ( 0 ) = e −2
2
0
f S (1) = e−2 × 2 × ( 0.6 ) = 1.2e −2
3
0
f S ( 2 ) = e −2 × 2 ( 0.4 ) +
4
5
6
1
2
3
E ⎡⎣( S − 3)+ ⎤⎦ = E [ S ] − 3 + 3 f S ( 0 ) + 2 f S (1) + 1 f S ( 2 )
E ⎡⎣( S − 3)+ ⎤⎦ = 2.8 − 3 + 3 × e −2 + 2 × 1.2 e −2 + 1× 1.52 e−2
= −0.2 + 6.92 e −2
= 0.7365
e −2 22
2
× ( 0.6 ) = 1.52 e −2
2
Question #19
Key: C
pk
c
=c+
pk +1
k
This is an (a, b, 0) distribution with a = b = c.
Which?
1.
If Poisson, a = 0, so c = 0 and b = 0
p1 = p2 = ... = 0
p0 = 0.5
pk ' s do not sum to 1. Impossible. Thus not Poisson
2.
If Geometric, b = 0, so c = 0 and a = 0
By same reasoning as #1, impossible, so not Geometric.
3.
If binomial, a and b have opposite signs. But here a = b, so not binomial.
4.
Thus negative binomial.
β / (1 + β )
a
1
1= =
=
b ( r − 1) β / (1 − β ) r − 1
so r = 2
Write (i) as
p0 = 0.5 = (1 + β )
−r
= (1 + β )
1 + β = 2 = 1.414
β = 2 − 1 = 0.414
c = a = β / (1 + β ) = 0.29
−2
Question #20
Key: C
1
At any age, p′x ( ) = e −0.02 = 0.9802
1
1
q′x ( ) = 1 − 0.9802 = 0.0198 , which is also qx( ) , since decrement 2 occurs only at the end of the
year.
Actuarial present value (APV) at the start of each year for that year’s death benefits
= 10,000*0.0198 v = 188.1
τ
px( ) = 0.9802*0.96 = 0.9410
τ
Ex = px( )v = 0.941 v = 0.941*0.95 = 0.8940
APV of death benefit for 3 years 188.1 + E40 *188.1 + E40 * E41 *188.1 = 506.60
Question #21
Key: B
40
e30:40 =
∫ t p30dt
0
ω − 30 − t
dt
ω − 30
0
40
=
∫
=t−
t2
2 (ω − 30 )
40
0
800
ω − 30
= 27.692
= 40 −
ω = 95
Or, with De Moivre’s law, it may be simpler to draw a picture:
0
p30 = 1
40
30
p30
70
e30:40 = area =27.692 = 40
p30 = 0.3846
ω − 70
= 0.3846
ω − 30
ω = 95
65 − t
t p30 =
65
(1 + 40 p30 )
2
40
Var = E (T ) − ( E (T ) )
2
2
One way to evaluate this expression is based on Equation 3.5.4 in Actuarial Mathematics
∞
Var (T ) = ∫ 2 t t px dt − ex2
0
⎛ 65 ⎛
t ⎞
t ⎞
⎛
= 2 ∫ t i⎜1 − ⎟dt − ⎜ ∫ ⎜1 − ⎟ dt 2
⎜
65 ⎠
0 ⎝
⎝ 0 ⎝ 65 ⎠
65
= 2* ( 2112.5 − 1408.333) − ( 65 − 65 / 2 )
2
= 1408.333 − 1056.25 = 352.08
Another way, easy to calculate for De Moivre’s law is
∞
Var (T ) = ∫ t 2 t px µ x ( t ) dt −
0
=∫
(∫
∞
0
t t px µ x ( t ) dt
1
1 ⎞
⎛ 65
t × dt − ⎜ ∫ t × dt ⎟
0
65
65 ⎠
⎝
65 2
0
t3
=
3 × 65
65
0
⎛ t2
−⎜
⎝ 2 × 65
65 ⎞
0
)
2
2
2
⎟
⎠
= 1408.33 − ( 32.5 ) = 352.08
2
With De Moivre’s law and a maximum future lifetime of 65 years, you probably didn’t need to
integrate to get E (T ( 30 ) ) = e30 = 32.5
Likewise, if you realize (after getting ω = 95 ) that T ( 30 ) is uniformly on (0, 65), its variance is
just the variance of a continuous uniform random variable:
Var =
( 65 − 0 )2 = 352.08
12
Question #22
Key: E
218.15 (1.06 ) − 10,000 ( 0.02 )
= 31.88
1 − 0.02
( 31.88 + 218.15)(1.06 ) − ( 9,000 )( 0.021) = 77.66
2V =
1 − 0.021
1V
=
Question #23
Key: D
∞
ex = e y = ∑ t px = 0.95 + 0.952 + ...
k =1
0.95
= 19
1 − 0.95
exy = pxy + 2 pxy + ...
=
= 1.02 ( 0.95 )( 0.95 ) + 1.02 ( 0.95 ) ( 0.95 ) + ...
2
2
1.02 ( 0.95 )
= 1.02 ⎡⎣0.952 + 0.954 + ...⎤⎦ =
= 9.44152
1 − 0.952
exy = ex + e y − exy = 28.56
2
Question #24
Key: A
Local comes first. I board
So I get there first if he waits more than 28 – 16 = 12 minutes after the local arrived.
His wait time is exponential with mean 12
The wait before the local arrived is irrelevant; the exponential distribution is memoryless
−12
Prob(exp with mean 12>12) = e 12 = e −1 = 36.8%
Question #25
Key: E
This problem is a direct application of Example 5.18 in Probability Models (p. 308); it follows
from proposition 5.3 (p. 303).
Deer hit at time s are found by time t (here, t = 10) with probability F(t – s), where F is the
exponential distribution with mean 7 days.
We can split the Poisson process “deer being hit” into “deer hit, not found by day 10” and “deer
hit, found by day 10”. By proposition 5.3, these processes are independent Poisson processes.
Deer hit, found by day 10, at time s has Poisson rate 20× F(t – s). The expected number hit and
found by day 10 is its integral from 0 to 10.
t
E ( N ( t ) ) = 20∫ F ( t − s )ds
0
10
E ( N (10 ) ) = 20 ∫
−(10 − s )
1 − e 7 ds
0
s −10
⎛
= 20 ⎜ 10 − 7e 7
⎜
⎝
(
= 20 10 − 7 + 7e
10 ⎞
0
−10 7
⎟
⎟
⎠
) = 94
Question #26
Key: E
∞
ax = ∫ e −0.08t dt = 12.5
0
∞
3
= 0.375
0
8
∞
3
2
Ax = ∫ e −0.13t ( 0.03) dt = = 0.23077
0
13
2
1 ⎡2
2
σ aT = Var ⎡⎣ aT ⎤⎦ =
Ax − ( Ax ) ⎤ = 400 ⎡ 0.23077 − ( 0.375 ) ⎤ = 6.0048
2 ⎢
⎥
⎣
⎦
⎦
δ ⎣
Pr ⎡⎣ aT > ax − σ aT ⎤⎦ = Pr ⎡⎣ aT > 12.5 − 6.0048⎤⎦
⎡ 1 − vT
⎤
= Pr ⎢
> 6.4952 ⎥ = Pr ⎡⎣ 0.67524 > e−0.05T ⎤⎦
⎣ 0.05
⎦
− ln 0.67524 ⎤
⎡
= Pr ⎢T >
⎥⎦ = Pr [T > 7.85374]
0.05
⎣
= e −0.03×7.85374 = 0.79
Ax = ∫ e −0.08t ( 0.03) dt =
( )
( )
Question #27
Key: A
5
− 0.05 5
(τ )
p50
= e ( )( ) = e −0.25 = 0.7788
(1)
5 q55
5 (1)
= ∫ µ55
( t ) × e−( 0.03+0.02)t dt = − ( 0.02 / 0.05 ) e−0.05t
0
(
= 0.4 1 − e
−0.25
)
= 0.0885
( )
()
Probability of retiring before 60 = 5 p50
× 5 q55
= 0.7788*0.0885
= 0.0689
τ
1
5
0
Question #28
Key: C
Complete the table:
l81 = l[80] − d[80] = 910
l82 = l[81] − d[81] = 830 (not really needed)
1
⎛1
⎞
e x = ex +
⎜ since UDD ⎟
2
⎝2
⎠
e[ x] = e[ x] + 1 2
⎡ l + l + l +…⎤ 1
⎥+
e[ x] = ⎢ 81 82 83
l[80]
⎢⎣
⎥⎦ 2
⎡
⎤
⎢ e − 1 ⎥ l = l + l + … Call this equation (A)
⎢ [80] 2 ⎥ [80] 81 82
⎣
⎦
⎡
⎤
⎢ e − 1 ⎥ l = l + … Formula like (A), one age later. Call this (B)
⎢ [81] 2 ⎥ [81] 82
⎣
⎦
Subtract equation (B) from equation (A) to get
⎡
⎤
⎡
⎤
1⎥
1⎥
⎢
⎢
l81 = e[80] − l[80] − e[81] − l[81]
⎢
⎢
2⎥
2⎥
⎣
⎦
⎣
⎦
⎡
⎤
910 = [8.5 − 0.5]1000 − ⎢e[81] − 0.5⎥ 920
⎣
⎦
e[81] =
8000 + 460 − 910
= 8.21
920
Alternatively, and more straightforward,
910
= 0.91
1000
830
p[81] =
= 0.902
920
830
p81 =
= 0.912
910
p[80] =
e[80] = 1 q 80 + p 80 ⎛⎜1 + e81 ⎞⎟
[ ]
2 [ ]
⎝
⎠
1
where q[80] contributes since UDD
2
1
⎛
⎞
8.5 = (1 − 0.91) + ( 0.91) ⎜1 + e81 ⎟
2
⎝
⎠
e81 = 8.291
1
⎛
⎞
e81 = q81 + p81 ⎜ 1 + e82 ⎟
2
⎝
⎠
1
⎛
⎞
8.291 = (1 − 0.912 ) + 0.912 ⎜ 1 + e82 ⎟
2
⎝
⎠
e82 = 8.043
1
⎛
⎞
e[81] = q[81] + p[81] ⎜ 1 + e82 ⎟
2
⎝
⎠
1
= (1 − 0.902 ) + ( 0.902 )(1 + 8.043)
2
= 8.206
Or, do all the recursions in terms of e, not e , starting with e[80] = 8.5 − 0.5 = 8.0 , then final step
e[81] = e[81] + 0.5
Question #29
Key: A
px
vt
vt t px
0.7
1
1
0.7
−
0.7
0.49
−
−
0.95238
0.90703
–
1
0.6667
0.4444
t
px +t
0
1
2
3
t
2
From above ax:3 = ∑ vt t px = 2.1111
t =0
⎛ a
1000 2Vx:3 = 1000 ⎜ 1 − x + 2:1
⎜
ax:3
⎝
⎞
1 ⎞
⎛
⎟ = 1000 ⎜1 −
⎟ = 526
⎟
⎝ 2.1111 ⎠
⎠
Alternatively,
Px:3 =
1
− d = 0.4261
ax:3
(
1000 2Vx:3 = 1000 v − Px:3
)
= 1000 ( 0.95238 − 0.4261)
= 526
You could also calculate Ax:3 and use it to calculate Px:3 .
−
Question #30
Key: E
Let G be the expense-loaded premium.
Actuarial present value (APV) of benefits = 1000A50 .
APV of expenses, except claim expense = 15 + 1× a50
APV of claim expense = 50A50 (50 is paid when the claim is paid)
APV of premiums = G a50
Equivalence principle: Ga50 = 1000 A50 + 15 + 1× a50 + 50 A50
1050 A50 + 15 + a50
G=
a50
a
For De Moivre’s with ω = 100, x = 50 A50 = 50 = 0.36512
50
1 − A50
a50 =
= 13.33248
d
Solving for G, G = 30.88
Question #31
Key: B
The variance calculation assumes independence, which should have been explicitly stated.
E (S ) = E (N ) E ( X )
Var ( S ) = E ( N ) Var( X ) + E 2 ( X ) Var ( N )
P.B
S.B
L.Y
E(N)
Var ( N )
E(X )
Var ( X )
E(S)
Var ( S )
30
30
30
21
27
12
300
1000
5000
10,000
400,000
2,000,000
9,000
30,000
150,000
189,000
2.19 × 106
39 × 106
360 × 106
400 × 106
(rounded)
Standard deviation =
400 × 106 = 20,000
189,000 + 20,000 = 209,000
Question #32
Key: B
S X (150 ) = 1 − 0.2 = 0.8
fY p ( y ) =
f X ( y + 150 )
0.2
= 0.25
So fY p ( 50 ) =
0.8
S X (150 )
0.6
= 0.75
0.8
= ( 0.25 )( 50 ) + ( 0.75 )(150 ) = 125
fY p (150 ) =
( )
E ⎡(Y ) ⎤ = ( 0.25 ) ( 50 ) + ( 0.75 )(150 )
⎢⎣
⎥⎦
E Yp
p
2
2
( )
( )
2
= 17,500
( )
2
2
Var Y p = E ⎡ Y p ⎤ − ⎡ E Y p ⎤ = 17,500 − 1252 = 1875
⎢⎣
⎥⎦ ⎣
⎦
Slight time saver, if you happened to recognize it:
Var Y p = Var Y p − 50
since subtracting a constant does not change variance,
( )
(
)
regardless of the distribution
But Y − 50 takes on values only 0 and 100, so it can be expressed as 100 times a
binomial random variable with n = 1, q = 0.75
Var = 1002 (1)( 0.25 )( 0.75 ) = 1875
p
(
)
Question #33
Key: A
4
− 0.05 4
p50 = e ( )( ) = 0.8187
10
8
− 0.05 10
p50 = e ( )( ) = 0.6065
− 0.04 8
p60 = e ( )( ) = 0.7261
18
p50 = ( 10 p50 )( 8 p60 ) = 0.6065 × 0.7261
= 0.4404
414 q50
= 4 p50 − 18 p50 = 0.8187 − 0.4404 = 0.3783
Question #34
Key: A
Model Solution:
X denotes the loss variable.
X1 denotes Pareto with α = 2 ; X 2 denotes Pareto with α = 4
FX (200) = 0.8 FX1 (200) + 0.2 FX 2 (200)
⎡ ⎛ 100 ⎞2 ⎤
⎡ ⎛ 3000 ⎞ 4 ⎤
= 0.8 ⎢1 − ⎜
⎟ ⎥ + 0.2 ⎢1 − ⎜
⎟ ⎥
⎣⎢ ⎝ 200 + 100 ⎠ ⎦⎥
⎣⎢ ⎝ 3000 + 200 ⎠ ⎦⎥
2
⎛1⎞
⎛ 15 ⎞
= 1 − 0.8 ⎜ ⎟ − 0.2 ⎜ ⎟
⎝ 3⎠
⎝ 16 ⎠
4
= 0.7566
Question #35
Key: D
a40:5 = a40 − 5 E40 a45
= 14.8166 − ( 0.73529 )(14.1121)
= 4.4401
π a40:5 = 100,000 A45 v5 5 p40 + π ( IA )40:5
1
(
π = 100,000 A45 × 5 E40 / a40:5 − ( IA )40:5
1
)
= 100,000 ( 0.20120 )( 0.73529 ) / ( 4.4401 − 0.04042 )
= 3363
Question #36
Key: B
Calculate the probability that both are alive or both are dead.
P(both alive) = k pxy = k px ⋅ k p y
P(both dead) = k qxy = k q x k q y
P(exactly one alive) = 1 − k pxy − k qxy
Only have to do two year’s worth so have table
Pr(both alive)
Pr(both dead)
Pr(only one alive)
1
0
0
(0.91)(0.91) = 0.8281
(0.09)(0.09) = 0.0081
0.1638
(0.82)(0.82) = 0.6724
(0.18)(0.18) = 0.0324
0.2952
0.8281 0.6724 ⎞
0.1638 0.2952 ⎞
⎛ 1
⎛ 0
APV Annuity = 30,000 ⎜
+
+
+ 20,000 ⎜
+
+
⎟ = 80, 431
0
1
2 ⎟
0
1.05
1.05 ⎠
1.051
1.052 ⎠
⎝ 1.05
⎝ 1.05
Alternatively,
0.8281 0.6724
+
= 2.3986
1.05
1.052
0.91 0.82
ax = a y = 1 +
+
= 2.6104
1.05 1.052
APV = 20,000 ax + 20,000 a y − 10,000 axy
axy = 1 +
(it pays 20,000 if x alive and 20,000 if y alive, but 10,000 less than that if both are alive)
= ( 20,000 )( 2.6104 ) + ( 20,000 )( 2.6104 ) − (10,000 ) 2.3986
= 80, 430
Other alternatives also work.
Question #37
Key: C
Let P denote the contract premium.
∞
P = ax = ∫ e
E [ L] =
0
axIMP
10
axIMP
−δ t − µ t
= ∫e
e
∞
dt = ∫ e−0.05t dt = 20
0
−P
−0.03t −0.02t
e
dt + e
−0.03(10 ) −0.02(10 )
e
0
∞
∫e
−0.03t −0.01t
e
dt
0
l − e −0.5 e−0.5
+
= 23
0.05
0.04
E [ L ] = 23 − 20 = 3
=
E [ L] 3
=
= 15%
P
20
Question #38
Key: C
1
A30:2
= 1000vq30 + 500v 2 1 q 30
2
⎛ 1 ⎞
⎛ 1 ⎞
= 1000 ⎜
⎟ ( 0.00153) + 500 ⎜
⎟ ( 0.99847 )( 0.00161)
⎝ 1.06 ⎠
⎝ 1.06 ⎠
= 2.15875
Initial fund = 2.15875 × 1000 participants = 2158.75
Let Fn denote the size of Fund 1 at the end of year n.
F1 = 2158.75 (1.07 ) − 1000 = 1309.86
F2 = 1309.86 (1.065 ) − 500 = 895.00
Expected size of Fund 2 at end of year 2 = 0 (since the amount paid was the single benefit
premium). Difference is 895.
Question #39
Key: B
Let c denote child; ANS denote Adult Non-Smoker; AS denote Adult Smoker.
33 e−3
× 0.3 = 0.067
3!
1e −1
× 0.6 = 0.037
P ( 3 ANS ) P ( ANS ) =
3!
43 e −4
× 0.1 = 0.020
P ( 3 AS ) P ( AS ) =
3!
P (3 c ) P (c ) =
P ( AS N = 3) =
0.020
= 0.16
( 0.067 + 0.037 + 0.020 )
Question #40
Key: C
E [ S ] = E [ N ] E [ X ] = 3 ×10 = 30
Var ( S ) = E [ N ]Var ( X ) + E [ X ] Var ( N )
2
400
+ 100 × 3.6 = 100 + 360 = 460
12
For 95th percentile, E [ S ] + 1.645 Var ( S ) = 30 + 460 × 1.645 = 65.28
= 3×