Solutions to the May 2013 Course MLC Examination
by Krzysztof Ostaszewski, http://www.krzysio.net, krzysio@krzysio.net
Copyright © 2013 by Krzysztof Ostaszewski
All rights reserved. No reproduction in any form is permitted without explicit
permission of the copyright owner.
Dr. Ostaszewski’s manual for Course MLC is available from Actex Publications
(http://www.actexmadriver.com) and the Actuarial Bookstore
(http://www.actuarialbookstore.com)
Exam MLC seminar at Illinois State University:
http://math.illinoisstate.edu/actuary/exams/prep_courses.shtml
May 2013 Course MLC Examination, Problem No. 1
For a fully discrete whole life insurance of 1000 on (30), you are given:
(i) Mortality follows the Illustrative Life Table.
(ii) i = 0.06.
(iii) The premium is the benefit premium.
Calculate the first year for which the expected present value at issue of that year’s
premium is less than the expected present value at issue of that year’s benefit.
A. 11
B. 15
C. 19
D. 23
E. 27
Solution.
Let us write s for the policy year. Then the mortality rate during year s is q30+s−1 . We are
looking for the smallest value of s such that the expected present value at issue of that
year’s premium is less than the expected present value at issue of that year’s benefit, or
1.06 −( s−1) ⋅ s−1 p30 ⋅1000P30 < 1.06 − s ⋅ s−1 p30 ⋅ q30+s−1 ⋅1000 .



expected present value at issue of year s premium
expected present value at issue of year s benefit
Since the interest rate is 6%, and mortality follows the Illustrative Life Table, we can get
the value of 1000P30 from values given in the table
1000 A30 102.48
=
.
a30
15.8561
Hence we have this inequality
102.48
1.06 −( s−1) ⋅ s−1 p30 ⋅
< 1.06 − s ⋅ s−1 p30 ⋅ q30+s−1 ⋅1000,
15.8561
or
102.48
1000q30+s−1 > 1.06 ⋅
≈ 6.85091542.
15.8561
The first age in the table where 1000 times mortality rate exceeds 6.85091542 is age 52.
We set 30 + s – 1 = 52, and conclude that s = 23.
Answer D.
1000P30 =
Copyright © 2013 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without
explicit permission of the copyright owner.
May 2013 Course MLC Examination, Problem No. 2
P&C Insurance Company is pricing a special fully discrete 3-year term insurance policy
on (70). The policy will pay a benefit if and only if the insured dies as a result of an
automobile accident. You are given:
(i)
x
Benefit
l x(τ )
d x(1)
d x( 2 )
d x( 3)
70
1000
80
10
40
5.000
71
870
94
15
60
7,500
72
701
108
18
82
10,000
(1)
(2)
where d x represents deaths from cancer, d x represents deaths from automobile
accidents, and d x( 3) represents deaths from all other causes.
(ii) i = 0.06.
(iii) Level premiums are determined using the equivalence principle.
Calculate the annual premium.
A. 122
B. 133
C. 144
D. 155
E. 166
Solution.
The expected present value of benefits is
5000 10
7500 15 10000 18
⋅
+
⋅
+
⋅
≈ 298.42.
1.06 1000 1.06 2 1000 1.06 3 1000
Let P be the annual premium sought. The expected present value of premiums is
P 870
P
701
P+
⋅
+
⋅
≈ 2.4446P.
2
1.06 1000 1.06 1000
Since the premium is set by the equivalence principle, those two quantities must be equal,
and therefore
298.42
P≈
≈ 122.
2.4446
Answer A.
May 2013 Course MLC Examination, Problem No. 3
For a special fully discrete 20-year endowment insurance on (40), you are given:
(i) The only death benefit is the return of annual benefit premiums accumulated with
interest at 6% to the end of the year of death.
(ii) The endowment benefit is 100,000.
(iii) Mortality follows the Illustrative Life Table.
(iv) i = 0.06.
Calculate the annual benefit premium.
A. 2365
B. 2465
C. 2565
D. 2665
E. 2765
Solution.
Copyright © 2013 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without
explicit permission of the copyright owner.
Let us write P for the annual benefit premium sought. Note that in case of death, the
death benefit is the accumulated value of all premiums paid up to that point, with interest,
at the end of the year of death, so all dying policyholders simply get their own premiums
with interest back, and any remaining premiums are those for surviving policyholders. On
the other hand, reserve is only calculated for surviving policyholders. This is the central
idea of the most efficient solution. At policy duration 20, the benefit reserve equals,
prospectively, the actuarial present value of future benefits minus the actuarial present
value of future premiums, and since there are no future premiums, it simply equals
20V = 100,000. On the other hand, retrospectively, the reserve equals the actuarial
accumulated value of past premiums minus the accumulated actuarial value of past
benefits. But the past benefits were paid to those policyholders who died by a return of
their own premiums with interest, which means that any past premiums left are the ones
paid by policyholders surviving till policy duration 20, and since the reserve is calculated
s20 6% . Equating the two
only for surviving policyholders, the reserve equals 20V = P
formulas for the same reserve at policy duration 20, we obtain
100,000 = Ps20 6% .
Based on this,
100,000
P=
≈ 2,564.58.

s20 6%
Answer C.
May 2013 Course MLC Examination, Problem No. 4
Employment for Joe is modeled according to a two-state homogeneous Markov model
with states: Actuary (Ac) and Professional Hockey Player (H). You are given:
(i) Transitions occur December 31 of each year. The one-year transition probabilities are:
Ac
H
Ac ⎡ 0.4 0.6 ⎤
⎢
⎥
H ⎣ 0.8 0.2 ⎦
Ac
= 0.10 + 0.05k for k = 0, 1, 2,
(ii) Mortality for Joe depends on his employment: q35+k
H
q35+k
= 0.25 + 0.05k for k = 0, 1, 2.
(iii) i = 0.08.
On January 1, 2013, Joe turned 35 years old and was employed as an actuary. On that
date, he purchased a 3-year pure endowment of 100,000. Calculate the expected present
value at issue of the pure endowment.
A. 32,510
B. 36,430
C. 40,350
D. 44,470
E. 48,580
Solution.
Ac
Ac
= 0.10, q36
= 0.15,
The specific mortalities given by the formulas in (ii) are: q35
Ac
H
H
H
q37
= 0.2, q35
= 0.25, q36
= 0.3, q37
= 0.35. Over the next three years, Joe has the
following paths to collection of the endowment benefit at the end of that period (with
Copyright © 2013 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without
explicit permission of the copyright owner.
probabilities for each step indicated below):
1/1/13  12/31/13  1/1/14  12/31/14  1/1/15  12/31/15
Actuary
Alive
Actuary
Alive
Actuary
Alive
0.9
0.4
0.85
0.4
0.8
1/1/13  12/31/13  1/1/14  12/31/14  1/1/15  12/31/15
Actuary
Alive
Actuary
Alive
Hockey
Alive
0.9
0.4
0.85
0.6
0.65
1/1/13  12/31/13  1/1/14  12/31/14  1/1/15  12/31/15
Actuary
Alive
Hockey
Alive
Hockey
Alive
0.9
0.6
0.7
0.2
0.65
1/1/13  12/31/13  1/1/14  12/31/14  1/1/15  12/31/15
Actuary
Alive
Hockey
Alive
Actuary
Alive
0.9
0.6
0.7
0.8
0.8
Therefore, the probability of collecting the 100,000 endowment benefit is
0.9 ⋅ 0.4 ⋅ 0.85 ⋅ 0.4 ⋅ 0.8 + 0.9 ⋅ 0.4 ⋅ 0.85 ⋅ 0.6 ⋅ 0.65 +
+ 0.9 ⋅ 0.6 ⋅ 0.7 ⋅ 0.2 ⋅ 0.65 + 0.9 ⋅ 0.6 ⋅ 0.7 ⋅ 0.8 ⋅ 0.8 = 0.50832.
The expected present value at issue of the pure endowment is
100,000
0.50832 ⋅
≈ 40, 352.08.
1.08 3
Answer C.
5.
May 2013 Course MLC Examination, Problem No. 5
lifetime
of Kevin,
Kira,
is modeled
TheThe
jointjoint
lifetime
of Kevin,
ageage
65, 65,
andand
Kira,
ageage
60, 60,
is modeled
as: as:
State 0
µ 01
State 1
Kevin alive
Kira alive
µ
02
Kevin alive
Kira dead
µ
State 2
Kevin dead
Kira alive
03
µ 13
State 3
µ
23
Kevin dead
Kira dead
You are given the following constant transition intensities:
You are given the following constant transition intensities:
(i) µ 01 = 0.004.
(ii) µ 02 = 0.005.
(i)
(ii)
µ 01 = 0.004
µ 02 = 0.005
Copyright © 2013 by Krzysztof
03 Ostaszewski. All rights reserved. No reproduction in any form is permitted without
= 0.001
(iii) of theµcopyright
explicit permission
owner.
(iv)
µ 13 = 0.010
(v)
µ 23 = 0.008
(iii) µ 03 = 0.001.
(iv) µ13 = 0.010.
(v) µ 23 = 0.008.
Calculate
10
A. 0.046
02
p65:60
.
B. 0.048
C. 0.050
D. 0.052
E. 0.054
Solution.
02
The probability 10 p65:60
is the probability of transition from state 0 to state 2 in such a
way that the joint life status (65:60) is in state 0 at time 0 and is in state 2 at time 10. The
subscript 65:60 indicates that the probability refers to the joint life status (65:60). The
probability sought equals (remember that under constant forces of transition, the
probability of remaining in state j over period of time t equals the expression of the form
exp(–t times the sum of forces of transition out of state j) )
10
02
10 p65:60 =
∫
µ 02 dt

00
t p65:60 ⋅

22
⋅ 10−t p65+t:60+t
=

0 Remain in state Transition from
Remain in state 2 til
state 0 to state 2
0 for t years
the end of 10 years
in period ( t ,t+dt )
10
= ∫e
(
) ⋅ µ 02 ⋅ e− µ
− µ 01 + µ 02 + µ 03 t
23
(10−t )
0
10
dt = ∫ e−0.01t ⋅ 0.005 ⋅ e−0.008(10−t ) dt =
0
10
= 0.005 ⋅ ∫ e
10
−0.01t
⋅e
−0.08
⋅e
0
0.008t
0.005
dt = 0.08 ⋅ ∫ e−0.01t ⋅ e0.008t dt =
e
0
0.005
0.005
0.005 1− e−0.02
−0.002t
⋅
e
dt
=
⋅
a
=
⋅
≈ 0.04569732.
e0.08 ∫0
e0.08 10 0.2%
e0.08
0.002
10
=
Answer A.
May 2013 Course MLC Examination, Problem No. 6
For a wife and husband ages 50 and 55, with independent future lifetimes, you are given:
1
, for 0 ≤ t < 50.
(i) The force of mortality on (50) is µ50+t =
50 − t
(ii) The force of mortality on (55) is µ55+t = 0.04, for t > 0.
(iii) For a single premium of 60, an insurer issues a policy that pays 100 at the moment of
the first death of (50) and (55).
(iv) δ = 0.05.
Calculate the probability that the insurer sustains a positive loss on the policy.
A. 0.45
B. 0.47
C. 0.49
D. 0.51
E. 0.53
Solution.
The force of mortality on (50) represents De Moivre’s Law, so that t p50 =
50 − t
t
= 1−
50
50
Copyright © 2013 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without
explicit permission of the copyright owner.
for 0 ≤ t ≤ 50. The force of mortality on (55) is constant, so that t p55 = e−0.04t for t ≥ 0.
t ⎞
⎛
Since the future lifetimes are independent, t p50:55 = ⎜ 1− ⎟ e−0.04t for 0 ≤ t ≤ 50, but for
⎝ 50 ⎠
t > 50, t p50:55 = 0 because (50) will be dead with certainty for t > 50. Let us write (using
two possible notations for the future lifespan)
T50:55 = T ( 50 : 55 ) = min (T50 ,T55 ) = min (T ( 50 ) ,T ( 55 ))
for the future lifespan of the joint life status (50:55). The loss of the insurer on this policy
is given by the formula
L = 100e−0.05T50:55 − 60.
We want to know the probability that L > 0. We have
Pr ( L > 0 ) = Pr (100e−0.05T50:55 − 60 > 0 ) = Pr (100e−0.05T50:55 > 60 ) =
= Pr ( e−0.05T50:55 > 0.6 ) = Pr ( ln e−0.05T50:55 > ln 0.6 ) =
ln 0.6 ⎞
⎛
= Pr ( −0.05T50:55 > ln 0.6 ) = Pr ⎜ T50:55 <
⎟=
⎝
−0.05 ⎠
ln 0.6 ⎞
⎛
⎛ ln 0.6 ⎞
ln 0.6 ⎞
⎛
⎜
⎟ −0.04⋅⎜⎝ −0.05 ⎟⎠
−0.05
= 1− Pr ⎜ T50:55 ≥
e
=
⎟ = 1− ⎜ 1−
⎝
−0.05 ⎠
50 ⎟
⎜⎝
⎟⎠
⎛ ln 0.6 ⎞ ln 0.6 0.05
⎛ ln 0.6 ⎞
0.8
= 1− ⎜ 1+
e ) = 1− ⎜ 1+
(
⎟
⎟ 0.6 ≈ 0.47124578.
⎝
⎝
2.5 ⎠
2.5 ⎠
0.04
Answer B.
May 2013 Course MLC Examination, Problem No. 7
You are given:
(i) q60 = 0.01.
(ii) Using i = 0.05, A60:3 = 0.86545.
Using i = 0.045 calculate A60:3 .
A. 0.866
B. 0.870
C. 0.874
D. 0.878
E. 0.882
Solution.
Using i = 0.05,
q60 (1− q60 ) q61 (1− q60 ) (1− q61 )
+
+
.
1.05
1.05 2
1.05 3
But we are given that q60 = 0.01, and using that, we obtain
0.01 0.99q61 0.99 (1− q61 )
0.86545 =
+
+
,
1.05 1.05 2
1.05 3
so that
0.86545 ⋅1.05 3 = 0.01⋅1.05 2 + 0.99q61 ⋅1.05 + 0.99 (1− q61 ) ,
0.86545 = A60:3 =
Copyright © 2013 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without
explicit permission of the copyright owner.
0.86545 ⋅1.05 3 − 0.01⋅1.05 2 − 0.99
≈ 0.01700114.
0.99 ⋅1.05 − 0.99
Now we calculate using i = 0.045
(1− q60 ) q61 + (1− q60 )(1− q61 ) ≈
q
A60:3 = 60 +
1.045
1.045 2
1.045 3
0.01 0.99 ⋅ 0.01700114 0.99 ⋅ 0.98299886
≈
+
+
≈ 0.87776672.
1.045
1.045 2
1.045 3
Answer D.
q61 =
May 2013 Course MLC Examination, Problem No. 8
For a special increasing whole life insurance on (40), payable at the moment of death,
you are given:
(i) The death benefit at time t is bt = 1+ 0.2t, t ≥ 0.
(ii) The interest discount factor at time t is v ( t ) = (1+ 0.2t ) , t ≥ 0.
−2
⎧⎪ 0.025, 0 ≤ t < 40
(iii) t p40 ⋅ µ 40+t = ⎨
otherwise.
⎩⎪ 0,
(iv) Z is the present value random variable for this insurance.
Calculate Var(Z).
A. 0.036
B. 0.038
C. 0.040
Solution.
We have
D. 0.042
(
E ( Z ) = E ( bT ⋅ v (T )) = E (1+ 0.2T ) (1+ 0.2T )
−2
E. 0.044
) = E ((1+ 0.2T ) ) =
−1
1
t=40
40
fT ( t )
1
40
=∫
dt = ∫
dt =
ln (1+ 0.2t )
=
1+ 0.2t
1+ 0.2t
40 ⋅ 0.2
t=0
0
0
40
= 0.125 ( ln 9 − ln1) ≈ 0.27465307.
Also,
(
) (
E ( Z 2 ) = E ( bT ⋅ v (T )) = E
40
=
2
fT ( t )
∫ (1+ 0.2t )
0
40
2
dt =
∫
0
((1+ 0.2T )(1+ 0.2T )−2 )
2
)
(
= E (1+ 0.2T )
−2
)=
t=40
−1
1
1 ⎛ (1+ 0.2t ) ⎞
−2
=
(1+ 0.2t ) dt = ⎜
40
40 ⎝ −1⋅ 0.2 ⎟⎠
t=0
⎛ 9 −1 1−1 ⎞
⎛ 1 ⎞
= 0.125 ⎜
− ⎟ = 0.125 ⎜ − + 1⎟ ≈ 0.11111111.
⎝ 9 ⎠
⎝ −1 −1 ⎠
Therefore,
(
)
Var ( Z ) = E ( Z 2 ) − E ( Z 2 ) ≈ 0.11111111− 0.27465307 2 ≈ 0.0356768.
Answer A.
Copyright © 2013 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without
explicit permission of the copyright owner.
May 2013 Course MLC Examination, Problem No. 9
For a fully discrete whole life insurance of 10,000 on (x), you are given:
(i) Deaths are uniformly distributed over each year of age.
(ii) The benefit premium is 647.46.
(iii) The benefit reserve at the end of year 4 is 1405.08.
(iv) qx+4 = 0.04561.
(v) i = 0.03.
Calculate the benefit reserve at the end of 4.5 years.
A. 1570
B. 1680
C. 1750
D. 1830
E. 1900
Solution.
Let us write tV for the benefit reserve at policy duration t, and P for the bene. We use a
slightly modified standard recursive reserve formula
0.5
+ P ) ⋅ (1+ i )
(4V

Reserve from policy duration 4
with premium accumulated with
interest pays for the items on the
right-hand side
=
p ⋅ 4.5V
x+4

0.5
Reserve at policy duration
4.5, of course only held for
surviving policyholders
+ 0.5 qx+4 ⋅10,000 ⋅ (1+ i ) .


−0.5
Death benefits for those who die by
policy duration 4.5, note that we have
to discount for half a year, as the death
benefit is paid at the end of the year
We substitute known values and obtain
(1405.08 + 647.46 ) ⋅1.030.5 = 0.5 px+4 ⋅ 4.5V + 0.5 qx+4 ⋅10,000 ⋅1.03−0.5.
We are not given 0.5 qx+4 but we know qx+4 = 0.04561 and we know that the UDD
assumption applies. Hence
0.5 q x+4 = 0.5q x+4 = 0.022805
UDD
and
px+4 = 1− 0.5 qx+4 = 0.977195.
Substituting this, we obtain
(1405.08 + 647.46 ) ⋅1.030.5 = 0.977195 ⋅ 4.5V + 0.022805 ⋅10,000 ⋅1.03−0.5 ,
so that
(1405.08 + 647.46 ) ⋅1.030.5 − 0.022805 ⋅10,000 ⋅1.03−0.5 ≈ 1,901.77.
V
=
4.5
0.977195
Answer E.
0.5
10.
May 2013 Course MLC Examination, Problem No. 10
A multi-state
modelmodel
is being
used to
value
sickness
benefit
insurance:
A multi-state
is being
used
to value
sickness
benefit
insurance:
healthy (h)
sick (s)
dead (d)
For a policy on (x) you are given:
For a (i)
policy
on (x) are
youpayable
are given:
Premiums
continuously at the rate of P per year while the policyholder is
healthy.
(i)
Premiums are payable continuously at the rate of P per year while the
policyholder is healthy.
Copyright © 2013 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without
explicit
of thebenefits
copyright owner.
(ii) permission
Sickness
are payable continuously at the rate of B per year while
the policyholder is sick.
(iii)
There are no death benefits.
(ii) Sickness benefits are payable continuously at the rate of B per year while the
policyholder is sick.
(iii) There are no death benefits.
ij
(iv) µ x+t
denotes the intensity rate for transition from i to j, where i, j = s, h or d.
(v) δ is the force of interest.
(vi) tV (i ) is the reserve at time t for an insured in state i where i = s, h or d.
Which of the following gives Thiele’s differential equation for the reserve that the
insurance company needs to hold while the policyholder is sick?
A.
B.
C.
D.
E.
d (s)
sh
= δ tV ( s ) + B − µ x+t
tV
dt
d (s)
sh
= δ tV ( s ) − B − µ x+t
tV
dt
d (s)
sh
= δ tV ( s ) + B − µ x+t
tV
dt
d (s)
sh
= δ tV ( s ) − B − µ x+t
tV
dt
d (s)
sh
= δ tV ( s ) − B − µ x+t
tV
dt
( V( ) − V( ))
( V( ) − V( ))
( V( ) − V( ))− µ
( V( ) − V( ))− µ
( V( ) − V( ))+ µ
h
t
s
t
h
t
s
t
h
t
h
t
sd
x+t t
s
sd
x+t t
s
sd
x+t t
t
h
t
V (s)
s
t
t
V (s)
V (s)
Solution.
The rate of change of reserve held while the policyholder is sick,
d ( s)
V , has the
dt t
following components:
• Instantaneous interest on the current reserve: δ tV ( ) ,
• Rate of premiums received while in state s: 0 (premium is paid only while the
policyholder is healthy),
• Rate of benefits paid (outgo, so a negative component) while in state s: –B,
• Transition intensity for transition to state h, times the change in reserve upon transition
(as a result of such transition, the insurer must hold reserve for h and release reserve for
s
(
)
sh
⋅ tV ( h ) − tV ( s ) ,
s): − µ x+t
• Transition intensity for transition to state d, times the change in reserve upon transition
(as a result of such transition, the insurer releases reserve for s, and there is no reserve for
(
)
sd
⋅ 0 − tV ( s ) .
state d, because the policy ends): − µ x+t
By adding all of those terms, we obtain
d (s)
(h)
(s)
sh
sd
= δ tV ( s ) − B − µ x+t
− tV ( s ) + µ x+t
tV
tV
tV .
dt
Answer E.
(
)
May 2013 Course MLC Examination, Problem No. 11
For a one-year term insurance on (45), whose mortality follows a double decrement
Copyright © 2013 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without
explicit permission of the copyright owner.
model, you are given:
(i) The death benefit for cause (1) is 1000 and for cause (2) is F.
(ii) Death benefits are payable at the end of the year of death.
(1)
(2)
(iii) q45
= 0.04 and q45
= 0.20.
(iv) i = 0.06
(v) Z is the present value random variable for this insurance.
Calculate the value of F that minimizes Var(Z).
A. 0
B. 50
C. 167
D. 200
E. 500
Solution.
Z can have the following values:
1000 ⋅1.06 −1 with probability 0.04,
F ⋅1.06 −1
with probability 0.20,
0
with probability 1 – 0.04 – 0.20 = 0.76.
Therefore, the mean of Z is
0.04 ⋅1000 ⋅1.06 −1 + 0.20 ⋅ F ⋅1.06 −1 ,
while the second moment of Z is
0.04 ⋅ (1000 ⋅1.06 −1 ) + 0.20 ⋅ ( F ⋅1.06 −1 ) .
2
2
The variance of Z is
Var ( Z ) =
= 0.04 ⋅ (1000 ⋅1.06 −1 ) + 0.20 ⋅ ( F ⋅1.06 −1 ) − ( 0.04 ⋅1000 ⋅1.06 −1 + 0.20 ⋅ F ⋅1.06 −1 ) =
2
2
2
0.04 ⋅ (1000 ) + 0.20 ⋅ F 2 − 0.04 2 ⋅1000 2 − 0.20 2 ⋅ F 2 − 2 ⋅ 0.04 ⋅ 0.20 ⋅1000F
=
1.06 2
0.16F 2 − 16F + 38400
=
.
1.06 2
This numerator of this expression is a quadratic function of F, which is minimized for
−16
F=−
= 50.
2 ⋅ 0.16
Answer B.
2
=
May 2013 Course MLC Examination, Problem No. 12
Russell entered a defined benefit pension plan on January 1, 2000, with a starting salary
of 50,000. You are given:
(i) The annual retirement benefit is 1.7% of the final three-year average salary for each
year of service.
(ii) His normal retirement date is December 31, 2029.
(iii) The reduction in the benefit for early retirement is 5% for each year prior to his
normal retirement date.
(iv) Every January 1, each employee receives a 4% increase in salary.
(v) Russell retires on December 31, 2026.
Copyright © 2013 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without
explicit permission of the copyright owner.
Calculate Russell’s annual retirement benefit.
A. 49,000
B. 52,000
C. 55,000
D. 58,000
E. 61,000
Solution.
The final three-year average salary is
1.04 26 + 1.04 25 + 1.04 24
50,000 ⋅
≈ 133, 360.20.
3
Based on this, the annual retirement benefit is approximately
3
0.017
 ⋅ 27
 ⋅133,
 ≈ 52,030.
360.20
 ⋅ 0.95
1.7% for each Number of
Final average salary 5% reduction for
year of service years of service
each of three years
of early retirement
Answer B.
May 2013 Course MLC Examination, Problem No. 13
An automobile insurance company classifies its insured drivers into three risk categories.
The risk categories and expected annual claim costs are as follows:
Risk Category
Expected Annual Claim Cost
Low
100
Medium
300
High
600
The pricing model assumes:
• At the end of each year, 75% of insured drivers in each risk category will renew their
insurance.
• i = 0.06.
• All claim costs are incurred mid-year.
For those renewing, 70% of Low Risk drivers remain Low Risk, and 30% become
Medium Risk. 40% of Medium Risk drivers remain Medium Risk, 20% become Low
Risk, and 40% become High Risk. All High Risk drivers remain High Risk. Today the
Company requires that all new insured drivers be Low Risk. The present value of
expected claim costs for the first three years for a Low Risk driver is 317. Next year the
company will allow 10% of new insured drivers to be Medium Risk. Calculate the
percentage increase in the present value of expected claim costs for the first three years
per new insured driver due to the change.
A. 14%
B. 16%
C. 19%
D. 21%
E. 23%
Solution.
There are three states: Low Risk, Medium Risk and High Risk. The transition matrix is
⎡ 0.7 0.3 0 ⎤
Q = ⎢ 0.2 0.4 0.4 ⎥
⎢
⎥
0
1 ⎦⎥
⎣⎢ 0
and the two-step transition matrix is
Copyright © 2013 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without
explicit permission of the copyright owner.
⎡ 0.55 0.33 0.12 ⎤
⎢
⎥
Q = ⎢ 0.22 0.22 0.56 ⎥ .
⎢⎣ 0
0
1 ⎥⎦
We know that the present value of expected claim costs for the first three years for a Low
Risk driver is 317. The present value of expected claim costs for the first three years for a
Medium Risk driver is
−1.5
300 ⋅1.06 −0.5 + (100 ⋅ 0.2 + 300 ⋅ 0.4 + 600 ⋅ 0.4 ) ⋅ 0.75
 ⋅1.06 +
2
Probability
of renewing
2
−2.5
+ (100 ⋅ 0.22 + 300 ⋅ 0.22 + 600 ⋅ 0.56 ) ⋅ 0.75
 ⋅1.06 ≈ 758.7025.
Probability
of renewing
twice
Therefore, the present value of costs for the new portfolio is
0.9 ⋅ 317 + 0.1⋅ 758.7025 ≈ 361.170254.
The rate of increase is approximately
361.170254
− 1 ≈ 13.933834.
317
Answer A.
May 2013 Course MLC Examination, Problem No. 14
For a universal life insurance policy with a death benefit of 150,000, you are given:
(i)
Policy
Monthly
Percent of
Monthly Cost
Monthly
Year
Premium
Premium
of Insurance
Expense
Charge
Rate per 1000
Charge
1
2000
3.5%
1.00
50
(12 )
(ii) i = 0.06.
(iii) The account value at the end of month 11 is 25,000.
Calculate the account value at the end of month 12.
A. 26,830
B. 26,850
C. 26,870
D. 26,890
E. 26,910
Solution.
We have the standard recursive formula (we have only one interest rate here, no
difference between credited rate and rate used for discounting)
AVEnd = ( AVStart + P (1− f ) − e − COI ) (1+ i ) ,
where
DBEnd − AVEnd
COI =
⋅ ( COI rate ) .
1+ i
Let us write AVt for the account value at the end of month t. Then
AV12 = ( AV11 + P (1− f ) − e − COI ) (1+ i ) =
= ( 25,000 + 2000 ⋅ (1− 0.035 ) − 50 − COI ) ⋅1.005 = 27014.40 − 1.005COI.
Copyright © 2013 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without
explicit permission of the copyright owner.
Also,
COI =
150,000 − ( 27014.40 − 1.005COI )
⋅ 0.001 ≈ 122.373731− 0.001COI,
1.005
so that
122.373731
≈ 122.25148.
1.001
We conclude that
AV12 = 27014.40 − 1.005COI ≈ 27014.40 − 1.005 ⋅122.25148 ≈ 26891.5373.
Answer D.
COI ≈
May 2013 Course MLC Examination, Problem No. 15
For fully discrete whole life insurance policies of 10,000 issued on 600 lives with
independent future lifetimes, each age 62, you are given:
(i) Mortality follows the Illustrative Life Table.
(ii) i = 0.06.
(iii) Expenses of 5% of the first year gross premium are incurred at issue.
(iv) Expenses of 5 per policy are incurred at the beginning of each policy year.
(v) The gross premium is 102% of the benefit premium.
(vi) 0 L is the aggregate present value of future loss at issue random variable.
Calculate Pr( 0 L < 60,000), using the normal approximation.
A. 0.74
B. 0.78
C. 0.82
D. 0.86
E. 0.90
Solution.
The aggregate present value of future loss random variable is the sum of 600 individual
policies present value of future loss random variables, which are independent and
identically distributed, thus by the Central Limit Theorem the aggregate present value of
future loss random variable can be approximated by a normal random variable with the
same mean and variance. Let us begin by finding the mean and variance of 0 L. We will
need to know the gross premium for this calculation. The benefit premium is
10,000A62 10 ⋅ 396.70
=
≈ 372.1947.
ILT 10.6584
a62
Therefore, the gross premium is
G ≈ 1.02 ⋅ 372.1947 ≈ 379.638595.
Let us denote by 0 LSingle the present value of future loss at issue random variable for a
single policy. Then
Copyright © 2013 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without
explicit permission of the copyright owner.
0
G − 5)
(
LSingle = 10000 ⋅1.06 −( K +1) −
⋅ aK +1 +
Gross premium after
recurrent expense of 5
≈ 10000 ⋅1.06 −( K +1) − 374.638595 ⋅
0.05G

≈
One time expense of 5% of gross
premium at issue, it contributes to the loss
1− 1.06 −( K +1)
+ 0.05 ⋅ 379.638595 ≈
0.06
1.06




aK +1 =
1−v K +1
d
⎛
⎞
⎛
⎞
374.638595 ⎟
⎜
⎜ 374.638595
⎟
−( K +1)
≈ ⎜ 10000 +
−⎜
− 0.05 ⋅ 379.638595 ⎟ ≈
⎟ ⋅1.06
0.06
0.06
⎜⎝
⎟⎠
⎜⎝
⎟⎠
1.06
1.06
≈ 16618.6152 ⋅1.06 −( K +1) − 6599.63325.
The expected value of 0 LSingle is
E ( 0 LSingle ) ≈ 16618.6152 ⋅ A62 − 6599.63325 =
ILT
= 16.6186152 ⋅ 396.7 − 6599.63325 ≈ −7.0286073.
ILT
Single
The variance of 0 L
Var ( 0 L
Single
is
) ≈ 16618.6152 ⋅ ( A − A ) =
= 16618.6152 ⋅ ( 0.19941− 0.3967 ) ≈ 11,610,292.90,
2
2
62
2
62
ILT
2
2
ILT
so that the standard deviation is approximately 11,610,292.90 ≈ 3, 407.3880. The
aggregate loss present value of future loss at issue random variable 0 L is the sum of six
hundred independent identically distributed random variables with the mean and variance
we have just calculated. Therefore,
E ( 0 L ) ≈ 600 ⋅ ( −7.0286073) ≈ −4217.1644,
and
Var ( 0 L ) ≈ 600 ⋅ 3, 407.3880 ≈ 83, 463.6194.
Therefore, if we write Z for a standard normal random variable, and Φ for its cumulative
distribution function, we obtain
⎛ L − ( −4217.1644 ) 60,000 − ( −4217.1644 ) ⎞
Pr ( 0 L < 60,000 ) ≈ Pr ⎜ 0
<
⎟⎠ ≈
⎝ 83, 463.6194
83, 463.6194
60,000 − ( −4217.1644 ) ⎞
⎛
≈ Pr ⎜ Z <
⎟⎠ ≈ Pr ( Z < 0.76940306 ) ≈
⎝
83, 463.6194
≈ Φ ( 0.76940306 ) ≈ 0.7792.
Answer B.
May 2013 Course MLC Examination, Problem No. 16
For a fully discrete whole life insurance policy of 2000 on (45), you are given:
(i) The gross premium is calculated using the equivalence principle.
Copyright © 2013 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without
explicit permission of the copyright owner.
(ii) Expenses, payable at the beginning of the year, are:
% of Premium
Per 1000
First year
25%
1.5
Renewal years
5%
0.5
(iii) Mortality follows the Illustrative Life Table.
(iv) i = 0.06.
Calculate the expense reserve at the end of policy year 10.
A. −2
B. −10
C. −14
D. −19
Per Policy
30
10
E. −27
Solution.
While this is not stated clearly, it is implied by the structure of expenses that all
premiums are level annual premiums. The expense reserve equals the actuarial present
value of future expenses minus the actuarial present value of future expense premiums.
Or, equivalently, it equals the actuarial accumulated value of past expense premiums
minus the actuarial accumulated value of past expenses. In order to know the expense
premiums, we must know the gross and net (i.e., benefit) premium. The benefit premium
is simpler to calculate:
2000A45 2 ⋅ 201.20
=
≈ 28.5145.
a45 ILT 14.1121
Since the gross premium is calculated using the equivalence principle, it must satisfy the
equation
2000
⎛ 2000
⎞ ⎛
⎞
G
a45 = 2000A45 + ⎜ 1⋅
+ 20 ⎟ + ⎜ 0.5 ⋅
+ 10 ⎟ ⋅ a45 +
⎝ 1000
⎠ ⎝
⎠
1000

 

First year extra expenses,
per thousand and per policy
+
0.20G

First year percentage
of premium expenses
in excess of such expenses
in all renewal years
+
Recurring expenses,
per thousand and per policy
0.05G
a45 .


Percentage of premium
expenses that are the
same in all years
From this, we can solve for G and obtain
G
a45 = 2000A45 + 22 + 11
a45 + 0.20G + 0.05G
a45 ,
and
2000A45 + 22 + 11
a45 2 ⋅ 201.20 + 22 + 11⋅14.1121
G=
=
≈ 43.8900.
0.95
a45 − 0.20 ILT
0.95 ⋅14.1121− 0.20
Based on this, the expense premium is approximately 43.8900 – 28.5145 = 15.3755. At
policy duration 10, the expense reserve equals
⎛
⎞
2000
a55 − 15.3755
a55 ≈
⎜⎝ 0.5⋅ 1000 + 10⎟⎠ ⋅ a55 + 0.05G
≈ (11+ 0.05⋅ 43.8900 − 15.3755) ⋅ a55 =
ILT
= (11+ 0.05⋅ 43.8900 − 15.3755) ⋅12.2758 ≈ −26.77.
ILT
Of course, you should know that there is nothing strange about having a negative expense
reserve.
Answer E.
Copyright © 2013 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without
explicit permission of the copyright owner.
May 2013 Course MLC Examination, Problem No. 17
You are profit testing a fully discrete whole life insurance of 10,000 on (70). You are
given:
(i) Reserves are benefit reserves based on the Illustrative Life Table and 6% interest.
(ii) The gross premium is 800.
(iii) The only expenses are commissions, which are a percentage of gross premiums.
(iv) There are no withdrawal benefits.
(v)
(
)
( death )
q70+k−1
Policy Year k
Commission Rate
Interest Rate
q70+k−1
1
0.02
0.20
0.80
0.07
2
0.03
0.04
0.10
0.07
Calculate the expected profit in policy year 2 for a policy in force at the start of year 2.
withdrawal
A. 180
B. 190
C. 200
D. 210
E. 220
Solution.
The benefit premium (used for benefit reserves calculations) is
10,000A70 10 ⋅ 514.95
10000P70 =
=
≈ 600.92423.
ILT
a70
8.5693
The benefit reserve at policy duration 1 is
71 ) = 10 ⋅ 530.26 − 600.92423⋅ 8.2988 ≈ 315.65.
1V70 = 10000 ( A71 − P71a
ILT
The benefit reserve at policy duration 2 is
a71 = 10 ⋅ 545.60 − 600.92423⋅ 8.0278 ≈ 631.90.
2V70 = 10000A72 − P
ILT
The general recursive formula, which shows emergence of the expected profit in policy
year t is:
P + t−1V − Et 1+ i = Sqx+t−1 + tVpx+t−1 + Prt .
The version of it that applies to policy year 2 in this problem is:
death
withdrawal)
P + V − E 1+ i = Sq( ) + 0 ⋅ q(
+ Vp + Pr .
(
(
)( )
1
1
)( )
Therefore,
71
70+k−1
(
2
71
2
)
(death )
(death )
( withdrawal)
Pr2 = ( P + 1V − E1 ) (1+ i ) − Sq71
− 2V 1− q71
− q70+k−1
=
≈ (800 + 315.65 − 0.1⋅800 ) ⋅1.07 − 10,000 ⋅0.03− 631.90 ⋅ (1− 0.03− 0.04 ) ≈
≈ 220.48.
Answer E.
May 2013 Course MLC Examination, Problem No. 18
An insurance company sells special fully discrete two-year endowment insurance policies
to smokers (S) and non-smokers (NS) age x. You are given:
(i) The death benefit is 100,000. The maturity benefit is 30,000.
(ii) The level annual premium for non-smoker policies is determined by the equivalence
principle.
Copyright © 2013 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without
explicit permission of the copyright owner.
(iii) The annual premium for smoker policies is twice the non-smoker annual premium.
NS
(iv) µ x+t
= 0.1, t > 0.
S
NS
= 1.5qx+
(v) qx+k
j , for k = 0,1.
(vi) i = 0.08.
Calculate the expected present value of the loss at issue random variable on a smoker
policy.
A. −30,000
B. −29,000
C. −28,000
D. −27,000
E. −26,000
Solution.
Given that the force of mortality for non-smokers is constant,
NS
qxNS = qx+1
= 1− e−0.1 ≈ 0.09516258.
And, of course,
S
NS
qx+k
= 1.5qx+
j ≈ 1.5 ⋅ 0.09516258 ≈ 0.14274387.
Since we are not told that the premium for smokers is determined by the equivalence
principle, while it is the case for non-smokers, we must conclude that the premium for
smokers is not derived from the equivalence principle, so the only way to find it is by
finding the premium for non-smokers and doubling it. Let us find the premium for nonsmokers, P NS , which is determined by the equivalence principle:
P NSax:2 = P NS (1+ 1.08 −1 ⋅ pxNS ) =
NS
NS
= 100,000 ⋅1.08 −1 ⋅ qxNS + 100,000 ⋅1.08 −2 ⋅ pxNS ⋅ qx+1
+ 30,000 ⋅1.08 −2 ⋅ pxNS ⋅ px+1
,
so that
NS
NS
100,000 ⋅1.08−1 ⋅ qxNS + 100,000 ⋅1.08−2 ⋅ pxNS ⋅ qx+1
+ 30,000 ⋅1.08−2 ⋅ pxNS ⋅ qx+1
P =
.
1+ 1.08−1 ⋅ pxNS
We calculate the numerator of the above as
100,000
100,000
⋅0.09516258 +
⋅ 1− 0.09516258 ⋅0.09516258 +
1.08
1.082
2
40,000
+
⋅ 1− 0.09516258 ≈ 37.251.4986.
2
1.08
We calculate the denominator as
1− 0.09516258
1+
≈ 1.83781242.
1.08
Therefore,
37.251.4986
P NS ≈
≈ 20,269.48, 1.83781242
and
PS = 2P NS ≈ 2 ⋅ 20,269.48 ≈ 40,538.96. The expected present value of the loss at issue random variable on a smoker policy is
therefore, approximately,
NS
(
(
)
)
Copyright © 2013 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without
explicit permission of the copyright owner.
NS
NS
100,000 ⋅1.08−1 ⋅ qxS + 100,000 ⋅1.08−2 ⋅ pxNS ⋅ qx+1
+ 30,000 ⋅1.08−2 ⋅ pxNS ⋅ qx+1
−
100,000
⋅0.14274387 +
1.08
2
100,000
30,000
+
⋅0.14274387 ⋅ (1− 0.14274387 ) +
⋅ (1− 0.14274387 ) −
2
2
1.08
1.08
40,538.96
− 40,538.96 −
⋅ (1− 0.14274387 ) ≈ −30,107.43.
1.08
− PS − PS ⋅1.08−1 ⋅ pxNS ≈
Answer A.
May 2013 Course MLC Examination, Problem No. 19
You are given:
(i) The following extract from a mortality table with a one-year select period:
x
x+1
l[ x ]
d[ x ]
lx+1
65
1000
40
66
955
45
(ii) Deaths are uniformly distributed over each year of age.
---
66
67
o
(iii) e[ 65 ] = 15.0.
o
Calculate e[ 66 ] .
A. 14.1
B. 14.3
C. 14.5
D. 14.7
E. 14.9
Solution.
We note immediately that l66 = l[ 65 ]+1 = 1000 − 40 = 960, and l67 = l[ 66 ]+1 = 955 − 45 = 910,
so that d66 = l66 − l67 = 960 − 910 = 50. We also have
o
1
1
0
0
o
15 = e[ 65 ] = ∫ t p[ 65 ] dt + p[ 65 ] ⋅ ∫ t p66 dt + p[ 65 ] ⋅ p66 ⋅ e67 =
∫(
1
=
UDD
0
)
1
UDD
1− tq[ 65 ] dt + p[ 65 ] ⋅ ∫ (1− tq66 ) dt + p[ 65 ] ⋅ p66 ⋅ e67 =
o
0
o
⎛ 1
⎞
⎛ 1 ⎞
= ⎜ 1− q[ 65 ] ⎟ + p[ 65 ] ⋅ ⎜ 1− q66 ⎟ + p[ 65 ] ⋅ p66 ⋅ e67 =
⎝ 2
⎠
⎝ 2 ⎠
⎛ 1 40 ⎞ 960 ⎛ 1 50 ⎞ 960 910 o
= ⎜ 1− ⋅
+
⋅ 1− ⋅
+
⋅
⋅e .
⎝ 2 1000 ⎟⎠ 1000 ⎜⎝ 2 960 ⎟⎠ 1000 960 67
Therefore,
⎛ 1 40 ⎞ 960 ⎛ 1 50 ⎞
15 − ⎜ 1− ⋅
−
⋅ 1− ⋅
o
⎝ 2 1000 ⎟⎠ 1000 ⎜⎝ 2 960 ⎟⎠
e67 =
≈ 14.3791209.
960 910
⋅
1000 960
We also have
Copyright © 2013 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without
explicit permission of the copyright owner.
o
1
o
e[ 66 ] = ∫ t p[ 66 ] dt + p[ 66 ] ⋅ e67 =
UDD
0
∫(
1
)
o
1− tq[ 66 ] dt + p[ 66 ] ⋅ e67 =
0
o
⎛ 1
⎞
⎛ 1 45 ⎞ 910
= ⎜ 1− q[ 66 ] ⎟ + p[ 66 ] ⋅ e67 = ⎜ 1− ⋅
+
⋅14.3791209 ≈ 14.6780105.
⎝ 2
⎠
⎝ 2 955 ⎟⎠ 955
Answer D.
May 2013 Course MLC Examination, Problem No. 20
Scientists are searching for a vaccine for a disease. You are given:
(i) 100,000 lives age x are exposed to the disease.
(ii) Future lifetimes are independent, except that the vaccine, if available, will be given to
all at the end of year 1.
(iii) The probability that the vaccine will be available is 0.2.
(iv) For each life during year 1, qx = 0.02.
(v) For each life during year 2, qx+1 = 0.01, if the vaccine has been given, and qx+1 = 0.02,
if it has not been given.
Calculate the standard deviation of the number of survivors at the end of year 2.
A. 100
B. 200
C. 300
D. 400
E. 500
Solution.
The probability distribution of the random number of survivors is mixed, with probability
0.2 of having lower mortality due to the vaccine, and with probability 0.8 of having the
same mortality as the first year. Let A be the event that the vaccine becomes available.
Let N be the random number of survivors at the end of year 2. If event A happens, for
each of the original 100,000 lives, that person is still alive at the end of year 2 with
probability 0.98 ⋅ 0.99. This means that for each person, the number of survivors for that
person is 1 with probability 0.98 ⋅ 0.99, and 0 with probability 1− 0.98 ⋅ 0.99, so it is a
Bernoulli Trial with p = 0.98 ⋅ 0.99. The number of survivors for all 100,000 lives insured
is binomial with n = 100,000 and p = 0.98 ⋅ 0.99. This means that its first moment is
100,000 ⋅ 0.98 ⋅ 0.99 = 97020, and the second moment is
2
100,000 ⋅ 0.98 ⋅ 0.99 ⋅ (1− 0.98 ⋅ 0.99 ) +
97020
= 9412883291.196.
 

 Square
of the first monent
Variance
If event A does not happen, the number of survivors for all 100,000 lives insured is
binomial with n = 100,000 and p = 0.98 ⋅ 0.98, with mean 100,000 ⋅ 0.98 ⋅ 0.98 = 96040,
and the second moment
2
100,000 ⋅ 0.98 ⋅ 0.98 ⋅ (1− 0.98 ⋅ 0.98 ) +
96040
= 9223685403.184.
 

 Square
of the first monent
Variance
This implies that
E ( N ) = 0.2 ⋅ 97020 + 0.8 ⋅ 96040 = 96236,
E ( N 2 ) = 0.2 ⋅ 9412883291.196 + 0.8 ⋅ 9223685403.184 = 9261524980.7864,
and therefore
Copyright © 2013 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without
explicit permission of the copyright owner.
Var ( N ) = E ( N 2 ) − ( E ( N )) = 9261524980.7864 − 96236 2 ,
2
and the standard deviation of N is
Var ( N ) ≈ 9261524980.7864 − 96236 2 ≈ 396.5914603.
Answer D.
May 2013 Course MLC Examination, Problem No. 21
You are given:
(i) δ t = 0.06, t ≥ 0.
(ii) µ x ( t ) = 0.01, t ≥ 0.
(iii) Y is the present value random variable for a continuous annuity of 1 per year, payable
for the lifetime of (x) with 10 years certain.
Calculate Pr (Y > E (Y )) .
A. 0.705
B. 0.710
C. 0.715
D. 0.720
E. 0.725
Solution.
We have
E (Y ) = a10 + e−0.06⋅10 ⋅ e−0.01⋅10 ⋅ ax+10 =
1− e−0.06⋅10
1
+ e−0.7 ⋅
≈ 14.61388183.
0.06
0.06 + 0.01
Therefore,
⎛ 1− e−0.06T
⎞
Pr (Y > E (Y )) ≈ Pr (Y > 14.61388183) = Pr ⎜
> 14.61388183⎟ =
⎝ 0.06
⎠
ln (1− 0.06 ⋅14.61388183) ⎞
⎛
= Pr ⎜ T >
⎟⎠ ≈ Pr (T > 34.9035565 ) =
⎝
−0.06
= e−0.01⋅34.9035565 ≈ 0.705368043.
Answer A.
May 2013 Course MLC Examination, Problem No. 22
For a whole life insurance of 10,000 on (x), you are given:
(i) Death benefits are payable at the end of the year of death.
(ii) A premium of 30 is payable at the start of each month.
(iii) Commissions are 5% of each premium.
(iv) Expenses of 100 are payable at the start of each year.
(v) i = 0.05.
(vi) 1000Ax+10 = 400.
(vii) 10V is the gross premium reserve at the end of year 10 for this insurance.
Calculate 10V using the two-term Woolhouse formula for annuities.
A. 950
B. 980
C. 1010
D. 1110
E. 1140
Copyright © 2013 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without
explicit permission of the copyright owner.
Solution.
The most common form of the Woolhouse formula is
m − 1 m2 − 1
m
ax( ) ≈ ax −
−
(δ + µx ).
2m 12m2
But this problem asks us to use the two-term Woolhouse formula, i.e.,
m −1
m
ax( ) ≈ ax −
.
2m
In this problem, we have
1− Ax+10 1− 0.4
ax+10 =
=
= 12.6,
0.05
d
1.05
and
12 − 1
11
(12) 
ax+10
≈ ax+10 −
= 12.6 −
≈ 12.14166667.
2 ⋅12
24
The gross premium reserve sought is
(12 )
100
ax+10 − (12 ⋅ 30 ) ⋅ (1− 0.05 ) ⋅ ax+10
≈
10V = 10,000Ax+10 +


 



Actuarial present value
of future benefits
Actuarial present value
of future expenses
Actuarial present value of future premiums
after commissions
≈ 10,000 ⋅ 0.4 + 100 ⋅12.6 − 360 ⋅ 0.95 ⋅12.14166667 ≈ 1107.55.
Answer D.
May 2013 Course MLC Examination, Problem No. 23
For an increasing two-year term insurance on (x), you are given:
(i) The death benefit during year k is 2000k, k = 1, 2.
(ii) Death benefits are payable at the end of the year of death.
(iii) qx+k−1 = 0.02k, k = 1,2.
(iv) The following information about zero coupon bonds of 100 at t = 0:
Maturity (in years)
Price
1
97.00
2
92.00
(v) Z is the present value random variable for this insurance.
Calculate Var(Z).
A. 569,600
B. 570,600
C. 571,600
D. 572,600
E. 573,600
Solution.
The possible values of the random variable Z are:
• In case of death in the first year,
97
with probability 0.02,
Z = 2000 ⋅
100
• In case of death in the second year,
92
with probability 0.98 ⋅ 0.04,
Z = 4000 ⋅
100
Copyright © 2013 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without
explicit permission of the copyright owner.
and otherwise Z = 0. Therefore,
97
92
E ( Z ) = 2000 ⋅
⋅ 0.02 + 4000 ⋅
⋅ 0.98 ⋅ 0.04 = 183.056,
100
100
2
2
97 ⎞
92 ⎞
⎛
⎛
E ( Z 2 ) = ⎜ 2000 ⋅
⋅
0.02
+
4000
⋅
⎟
⎜⎝
⎟ ⋅ 0.98 ⋅ 0.04 = 606134.08,
⎝
100 ⎠
100 ⎠
Var ( Z ) = E ( Z 2 ) − ( E ( Z )) = 606134.08 − 183.056 2 = 572624.580864.
2
Answer D.
May 2013 Course MLC Examination, Problem No. 24
For a fully discrete whole life insurance, you are given:
(i) First year expenses are 10% of the gross premium and 5 per policy.
(ii) Renewal expenses are 3% of the gross premium and 2 per policy.
(iii) Expenses are incurred at the start of each policy year.
(iv) There are no deaths or withdrawals in the first two policy years.
(v) i = 0.05.
(vi) The asset share at time 0 is 0. The asset share at the end of the second policy year is
64.11.
Calculate the gross premium.
A. 32.7
B. 34.2
C. 35.7
D. 37.2
E. 38.7
Solution.
The basic asset shares recursive formula is
(1)
(2)
(τ )
h−1 AS + G (1 − ch−1 ) − eh−1 ⋅ (1 + i ) = q x + h−1 + q x + h−1 ⋅ h CV + h AS ⋅ p x + h−1 .
First we use it for year 1, with h = 1, and obtain
( 0 AS + G (1− 0.10 ) − 5 ) ⋅1.05 = 0 + 0 ⋅ 1 CV + 1 AS ⋅ (1− 0 − 0 ),
resulting in
1 AS = ( 0.9G − 5 ) ⋅1.05 = 0.945G − 5.25.
Then we write the recursive formula for the second policy year and obtain
( 1 AS + G (1− 0.03) − 2 ) ⋅1.05 = 0 + 0 ⋅ 2 CV + 2 AS ⋅ (1− 0 − 0 ),
or
(1.915G − 7.25 ) ⋅1.05 = 2 AS = 64.11.
From this
64.11+ 7.623
G=
≈ 35.67474823.
2.01075
Answer C.
(
)
May 2013 Course MLC Examination, Problem No. 25
For a fully discrete whole life insurance on (60), you are given:
(i) Mortality follows the Illustrative Life Table
Copyright © 2013 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without
explicit permission of the copyright owner.
(ii) i = 0.06.
(iii) The expected company expenses, payable at the beginning of the year, are: 50 in the
first year, 10 in years 2 through 10, 5 in years 11 through 20, 0 after year 20.
Calculate the level annual amount that is actuarially equivalent to the expected company
expenses.
A. 8.5
B. 11.5
C. 12.0
D. 13.5
E. 15.0
Solution.
The expected company expenses are:
50 + 1 E60 ⋅10
a61:9 + 10 E60 ⋅ 5
a70:9 = 40 + 5
a60:10 + 5
a60:20 =
= 40 + 5 ( a60 − 10 E60 ⋅ a70 ) + 5 ( a60 − 20 E60 ⋅ a80 ) =
ILT
= 40 + 5 (11.1454 − 0.4512 ⋅ 8.5693) + 5 (11.1454 − 0.14906 ⋅ 5.905 ) ≈


ILT
≈7.27893184
≈10.2652007
≈ 127.720663.
If this were paid as a level amount instead, the annual payment would be approximately
127.720663 127.720663
=
≈ 11.4594956.
a60
11.1454
Answer B.
Copyright © 2013 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without
explicit permission of the copyright owner.