MLC Sample Solution
MLC Sample Solution
MLC Sample Solution
Some of the questions in this study note are taken from past SOA examinations.
MLC-09-08
PRINTED IN U.S.A.
Question #1
Key: E
2
Alternatively,
2 q30:34 = 2 q30 + 2 q34 2 q30:34
MLC-09-08
p30 ,
p34 ,
p30 ,
p34 )
-1-
Question #2
Key: E
1000 Ax = 1000 Ax1:10 + 10 Ax
10
= 1000 e 0.04t e 0.06t (0.06)dt + e0.4e0.6 e0.05t e0.07t (0.07)dt
0
10
= 1000 0.06 e 0.1t dt + e 1 (0.07) e0.12t dt
0
0
0.10 t 10
0.12 t
1
1.2
1 e 1 + 0.07
= 1000 0.06
0.12 e 1 e
0.10
Because this is a timed exam, many candidates will know common results for constant
force
and constant interest without integration.
Ax =
=e
(1 10 Ex )
10 ( + )
0.06
0.07
( 0.06+ 0.04 )10
0.06+ 0.04 )10
= 1000
+e (
1 e
0.07 + 0.05
0.06 + 0.04
= 593.86
Question #3
Key: D
1
dt
20
1 100
5
0.07 t
e
=
0
20 7
7
MLC-09-08
-2-
E Z 2 =
(b v )
t
1
1
dt = e0.09t dt
20
20 0
1 100 0.09t 5
=
e
0 = 9
20 9
5 5
Var [ Z ] = = 0.04535
9 7
Question #4
Key: C
Let ns = nonsmoker and s = smoker
k=
q xb + kg
ns
pxb + kg
q xb +gk
s
px( +)k
.05
0.95
0.10
0.90
.10
0.90
0.20
0.80
.15
0.85
0.30
0.70
1 ns
A x:2( ) =
ns
v q x( ) +
1
( 0.05)
1.02
A x:2( )
1 s
s
qx( )
1
1.02
ns
px( )
v2
1
1.022
ns
ns
+ v2
( 0.10 ) +
qx( +1)
(1.02 )2
s
px( )
q x( +)1
s
Question #5
Key: B
x( ) = x(1) + x( 2 ) + x( 3) = 0.0001045
t
px( ) = e 0.0001045t
MLC-09-08
-3-
1
APV Benefits = e t 1,000,000 t px ( ) x( ) dt
0
+ e
0
2
500,000 t px ( ) x( ) dt
3
+ e 200,000 t px ( ) x( ) dt
0
1,000,000 0.0601045t
500,000 0.0601045t
250,000 0.0601045t
e
dt +
e
dt +
e
dt
2,000,000 0
250,000 0
10,000 0
Question #6
Key: B
1
APV Benefits = 1000 A40:20
+
k E401000vq40+k
k = 20
k E401000vq40+k
k = 20
k = 20
20
1
= 1000 A40:
20
/ a&&40:20
= 5.11
1
and was structured to take
While this solution above recognized that = 1000 P40:20
advantage of that, it wasnt necessary, nor would it save much time. Instead, you could
do:
MLC-09-08
-4-
Question #7
Key: C
ln (1.06 )
( 0.53) = 0.5147
0.06
1 A70 1 0.5147
=
= 8.5736
a&&70 =
d
0.06 /1.06
0.97
a&&69 = 1 + vp69a&&70 = 1 +
( 8.5736 ) = 8.8457
1.06
A70 = A70 =
i
( 2)
= ( 2 ) a&&69 ( 2 ) = (1.00021)( 8.8457 ) 0.25739
a&&69
= 8.5902
This is the form of (a), (b) and (c) on page 119 of Bowers with x = k 1 . Thus, the
recursion could be:
MLC-09-08
-5-
Ax = vqx + vpx Ax +1
or
A1x: y x = vq x + vp x Ax1+1: y x 1
or
1
A69:1
= vq69 + vp69 (1)
u ( k 1) = Ax: y x is OK at x = 69 since
MLC-09-08
-6-
Question #9
Key: A
You arrive first if both (A) the first train to arrive is a local and (B) no express arrives in
the 12 minutes after the local arrives.
P ( A) = 0.75
Expresses arrive at Poisson rate of ( 0.25 )( 20 ) = 5 per hour, hence 1 per 12 minutes.
e 110
= 0.368
0!
A and B are independent, so
P ( A and B ) = ( 0.75 )( 0.368 ) = 0.276
f ( 0) =
Question #10
Key: E
d = 0.05 v = 0.095
At issue
49
10 L
K ( 40 ) 10 = 1000 A50
Revised
Revised
P40a&&50
where
Revised
A50
and
24
= v k +1 k q50
k =0
Revised
a&&50
=
MLC-09-08
Revised
(1 A
Revised
50
-7-
Question #11
Key: E
Let NS denote non-smokers and S denote smokers.
The shortest solution is based on the conditional variance formula
1 0.444
= 5.56
0.1
1 0.286
Similarly E aT Y = 0 =
= 7.14
0.1
=
( (
E E aT Y
( (
E E aT Y
))
( (
2
2
= 7.14 ( 0.70 ) + 5.56 ( 0.30 )
= 44.96
Var E aT Y
= 8.60
( )
( )
( )
(( )
E aT
NS = Var aT NS + E aT NS
( )
Var aT = E aT
E aT
MLC-09-08
-8-
= 0.30axS + 0.70axNS
=
0.30 1 AxS
) + 0.70 (1 A )
NS
x
0.1
0.1
0.30 (1 0.444 ) + 0.70 (1 0.286 )
=
= ( 0.30 )( 5.56 ) + ( 0.70 )( 7.14 )
0.1
= 1.67 + 5.00 = 6.67
( )
E aT
( (
) (
2
= 0.30 Var aT S + E aT S
+0.70 Var aT NS + E aT NS
2
2
= 0.30 8.818 + ( 5.56 ) + 0.70 8.503 + ( 7.14 )
1 vT
1 v
Var aT = Var
T
( )
vT
= Var
since Var ( X + constant ) = Var ( X )
=
Var vT
2
2
Ax ( Ax )
Var ( X )
( )
This could be transformed into 2Ax = 2 Var aT + Ax2 , which we will use to get
2
Ax NS and 2AxS .
MLC-09-08
-9-
Ax = E v 2T
2
+ 2Var aT S + AxS Prob ( S)
) (
) ( )
Ax = E vT
= E vT NS Prob ( NS) + E vT S Prob ( S)
= ( 0.286 )( 0.70 ) + ( 0.444 )( 0.30 )
= 0.3334
Ax ( Ax )
( )
Var aT =
=
2
0.20238 0.33342
= 9.12
0.01
Question #12
Key: A
To be a density function, the integral of f must be 1 (i.e., everyone dies eventually). The
solution is written for the general case, with upper limit . Given the distribution of
f 2 ( t ) , we could have used upper limit 100 here.
Preliminary calculations from the Illustrative Life Table:
l50
= 0.8951
l0
l40
= 0.9313
l0
MLC-09-08
- 10 -
50
50
1 = fT ( t ) dt = k f1 ( t ) dt + 1.2 f 2 ( t )dt
= k
50
f1 ( t ) dt + 1.2 f 2 ( t )dt
50
= k F1 ( 50 ) + 1.2 ( F2 ( ) F2 ( 50 ) )
= k (1 50 p0 ) + 1.2 (1 0.5 )
= k (1 0.8951) + 0.6
k=
1 0.6
= 3.813
1 0.8951
l
FT ( 40 ) = 3.813 1 40 = 3.813 (1 0.9313) = 0.262
l0
l
FT ( 50 ) = 3.813 1 50 = 3.813 (1 0.8951) = 0.400
l0
10
p40 =
1 FT ( 50 ) 1 0.400
=
= 0.813
1 FT ( 40 ) 1 0.262
Question #13
Key: D
Let NS denote non-smokers, S denote smokers.
+ 0.7e0.1t
MLC-09-08
- 11 -
This is quadratic, so x =
x = 0.3147
e 0.1t = 0.3147
so t = 11.56
Question #14
Key: A
P ( Ax ) = = 0.03
2
0.03
2 + 2 + 0.03
= 0.06
Ax = 0.20 =
Var ( 0 L ) =
where A =
Ax ( Ax )
( a )2
0.20 ( 13 )
( 0.06
0.09 )
0.03 1
=
0.09 3
a=
= 0.20
1
1
=
+ 0.09
Question #15
Key: C
Let N = number of sales on that day
S = aggregate prospective loss at issue on those sales
K = curtate future lifetime
N~Poisson (0.250)
0L
= 10,000v
K +1
500a&&K +1
500 K +1 500
0 L = 10,000 +
v
d
d
E [ N ] = Var [ N ] = 10
500 2
2
Var [ 0 L ] = 10,000 +
A65 ( A65 )
d
S = 0 L1 + 0 L2 + ... + 0 LN
E [ S ] = E [ N ] E [ 0 L]
MLC-09-08
- 12 -
0 E [S ]
Pr ( S < 0 ) = Pr Z <
Var
S
[
]
Substituting d = 0.06/(1+0.06), 2 A65 = 0.23603, A65 = 0.43980 and a&&65 = 9.8969 yields
E [ 0 L ] = 550.45
Var [ 0 L ] = 15,112,000
E [ S ] = 5504.5
Var [ S ] = 154,150,000
MLC-09-08
- 13 -
Question #16
Key: A
1000 P40 =
A40 161.32
=
= 10.89
a&&40 14.8166
a&&
11.1454
1000 20V40 = 1000 1 60 = 1000 1
= 247.78
14.8166
a&&40
( 20V + 5000 P40 ) (1 + i) 5000q60
21V =
P60
[Note: For this insurance, 20V = 1000 20V40 because retrospectively, this is identical to
whole life]
Though it would have taken much longer, you can do this as a prospective reserve.
The prospective solution is included for educational purposes, not to suggest it
would be suitable under exam time constraints.
20 E40 5 E60
a&&65
1
where A60:5
= A60 5 E60 A65 = 0.06674
1
= 4000 A60:5
+ 1000 A60 5000 P40 a&&60:5 5 E60 a&&65
20 V
MLC-09-08
- 14 -
Or we can continue to
21V
1
5000 A61:4
where
4 E61
21V
prospectively
l65 4 7,533,964
v =
( 0.79209 ) = 0.73898
l61
8,075, 403
1
A61:4
= A61 4 E61 A65 = 0.38279 0.73898 0.43980
a&&61:4
= 0.05779
= a&&61 4 E61 a&&65 = 10.9041 0.73898 9.8969
= 3.5905
21V
Finally. A moral victory. Under exam conditions since prospective benefit reserves
must equal retrospective benefit reserves, calculate whichever is simpler.
Question #17
Key: C
Var ( Z ) = 2 A41 ( A41 )
A41 = 0.21650
2
A41 = 0.07193
MLC-09-08
- 15 -
A41 )
Question #18
Key: D
This solution looks imposing because there is no standard notation. Try to focus on the
big picture ideas rather than starting with the details of the formulas.
Big picture ideas:
1.
We can express the present values of the perpetuity recursively.
2.
Because the interest rates follow a Markov process, the present value (at time t )
of the future payments at time t depends only on the state you are in at time t ,
not how you got there.
3.
Because the interest rates follow a Markov process, the present value of the
future payments at times t1 and t2 are equal if you are in the same state at times
t1 and t2 .
Method 1: Attack without considering the special characteristics of this transition matrix.
Let sk = state you are in at time k ( thus sk = 0, 1 or 2 )
Let Yk = present value, at time k , of the future payments.
Yk is a random variable because its value depends on the pattern of discount factors,
which are random. The expected value of Yk is not constant; it depends on what state
we are in at time k.
Recursively we can write
Yk = v (1 + Yk +1 ) , where it would be better to have notation that indicates the vs are not
constant, but are realizations of a random variable, where the random variable itself has
different distributions depending on what state were in. However, that would make the
notation so complex as to mask the simplicity of the relationship.
Every time we are in state 0 we have
E Yk sk = 0 = 0.95 1 + E Yk +1 sk = 0
( {(
))
(
= 1 ) Pr ob ( s
= 2 ) Pr ob ( s
)
= 1 s = 0])
= 2 s = 0])}
= 0.95 1 + E Yk +1 sk +1 = 0 Pr ob s k+1 = 0 sk = 0]
(
= ( E Y
+ E Yk +1 sk +1
k +1
sk +1
= 0.95 1 + E Yk +1 sk +1 = 1
MLC-09-08
k +1
k +1
)
- 16 -
That last step follows because from the transition matrix if we are in state 0, we always
move to state 1 one period later.
Similarly, every time we are in state 2 we have
E Yk sk = 2 = 0.93 1 + E Yk +1 sk = 2
= 0.93 1 + E Yk +1 sk +1 = 1
That last step follows because from the transition matrix if we are in state 2, we always
move to state 1 one period later.
Finally, every time we are in state 1 we have
E Yk sk = 1 = 0.94 1 + E Yk +1 sk = 1
( {
})
= 0.94 1 + E Yk +1 sk +1 = 0 Pr sk +1 = 0 sk = 1 + E Yk +1 sk +1 = 2 Pr sk +1 = 2 sk = 1
( {
})
Now lets write those last three paragraphs using this shorter notation:
xn = E Yk sk = n . We can do this because (big picture idea #3), the conditional
expected value is only a function of the state we are in, not when we are in it or how we
got there.
x0 = 0.95 (1 + x1 )
x1 = 0.94 (1 + 0.9 x0 + 0.1x2 )
x2 = 0.93 (1 + x1 )
Thats three equations in three unknowns. Solve (by substituting the first and third into
the second) to get x1 = 16.82 .
Thats the answer to the question, the expected present value of the future payments
given in state 1.
The solution above is almost exactly what we would have to do with any 3 3 transition
matrix. As we worked through, we put only the non-zero entries into our formulas. But
if for example the top row of the transition matrix had been ( 0.4 0.5 0.1) , then the first
of our three equations would have become x0 = 0.95 (1 + 0.4 x0 + 0.5 x1 + 0.1x2 ) , similar in
structure to our actual equation for x1 . We would still have ended up with three linear
equations in three unknowns, just more tedious ones to solve.
Method 2: Recognize the patterns of changes for this particular transition matrix.
MLC-09-08
- 17 -
This particular transition matrix has a recurring pattern that leads to a much quicker
solution. We are starting in state 1 and are guaranteed to be back in state 1 two steps
later, with the same prospective value then as we have now.
Thus,
E [Y ] = E Y first move is to 0 Pr [ first move is to 0] + E Y first move is to 2 Pr [first move is to 2 ]
(Note that the equation above is exactly what you get when you substitute x0 and x2
into the formula for x1 in Method 1.)
= 1.6497 + 0.8037 E [Y ] + 0.1814 + 0.0874 E [Y ]
E [Y ] =
1.6497 + 0.1814
(1 0.8037 0.0874 )
= 16.82
Question #19
Key: E
The number of problems solved in 10 minutes is Poisson with mean 2.
If she solves exactly one, there is 1/3 probability that it is #3.
If she solves exactly two, there is a 2/3 probability that she solved #3.
If she solves #3 or more, she got #3.
f(0) = 0.1353
f(1) = 0.2707
f(2) = 0.2707
1
2
P = ( 0.2707 ) + ( 0.2707 ) + (1 0.1353 0.2707 0.2707 ) = 0.594
3
3
MLC-09-08
- 18 -
Question #20
Key: D
x( ) = x(1) ( t ) + x( 2 ) ( t )
2
= 0.2 x( ) ( t ) + x( ) ( t )
2
x( ) ( t ) = 0.8 x( ) ( t )
1
0.2 k t
'1
'1
qx( ) = 1 px( ) = 1 e 0
dt
= 1 e
0.2
k
3
= 0.04
k = 0.6123
( 2)
2 qx
0 t
2
2
px( ) x( ) dt = 0.8
0
px( ) x( ) ( t ) dt
x (t ) d t
( )
0
2 px = e
2
kt
= e 0
dt
8 k
e 3
( 8 ) ( 0.6123)
3
e
=
= 0.19538
( 2)
2 qx
Question #21
Key: A
k
min(k ,3)
f(k)
f ( k ) ( min(k ,3) )
f ( k ) min ( k ,3)
0
1
2
3+
0
1
2
3
0.1
(0.9)(0.2) = 0.18
(0.72)(0.3) = 0.216
1-0.1-0.18-0.216 = 0.504
0
0.18
0.432
1.512
2.124
0
0.18
0.864
4.536
5.580
MLC-09-08
- 19 -
E min ( K ,3)
} = 5.580
Note that E [ min( K ,3) ] is the temporary curtate life expectancy, ex:3 if the life is age x.
Problem 3.17 in Bowers, pages 86 and 87, gives an alternative formula for the variance,
basing the calculation on k px rather than k q x .
Question #22
Key: B
0.1 60
0.08 60
e ( )( ) + e ( )( )
2
= 0.005354
s ( 60 ) =
0.1 61
0.08 61
e ( )( ) + e ( )( )
2
= 0.00492
0.00492
q60 = 1
= 0.081
0.005354
s ( 61) =
Question #23
Key: D
Let q64 for Michel equal the standard q64 plus c. We need to solve for c.
Recursion formula for a standard insurance:
20V45
The values of
19 V45
and
20V45
MLC-09-08
- 20 -
c=
(1.03) ( 0.01)
(1 20V45 )
0.0103
1 0.427
= 0.018
=
Question #24
Key: B
K is the curtate future lifetime for one insured.
L is the loss random variable for one insurance.
LAGG is the aggregate loss random variables for the individual insurances.
AGG is the standard deviation of LAGG .
M is the number of policies.
L = v K +1 a&&K +1 = 1 + v K +1
d
d
(1 Ax )
E [ L ] = ( Ax a&&x ) = Ax
d
0.75095
= 0.24905 0.025
= 0.082618
0.056604
Var [ L ] = 1 +
d
Ax
Ax2
E [ LAGG ] = M E [ L ] = 0.082618M
Var [ LAGG ] = M Var [ L ] = M (0.068034) AGG = 0.260833 M
L
E [ LAGG ] E ( LAGG )
>
Pr [ LAGG > 0] = AGG
AGG
AGG
0.082618M
Pr N (0,1) >
M ( 0.260833)
0.082618 M
0.260833
M = 26.97
1.645 =
MLC-09-08
0.025
2
= 1 +
0.09476 ( 0.24905 ) = 0.068034
0.056604
- 21 -
Question #25
Key: D
Death benefit:
1 v K +1
for K = 0,1, 2,...
Z1 = 12,000
d
Z 2 = Bv K +1
for K = 0,1, 2,...
New benefit:
Z = Z1 + Z 2 = 12,000
Annuity benefit:
1 v K +1
+ Bv K +1
d
12,000
12,000 K +1
=
+B
v
d
d
12,000
K +1
Var( Z ) = B
Var v
d
12,000
Var ( Z ) = 0 if B =
= 150,000 .
0.08
In the first formula for Var ( Z ) , we used the formula, valid for any constants a and b and
random variable X,
Var ( a + bX ) = b 2Var ( X )
Question #26
Key: A
Ay = y ( t ) / y ( t ) + = 0.4
Axy = xy ( t ) / xy ( t ) + = 0.6667
axy = 1/ xy ( t ) + = 5.556
Axy = Ax + Ay Axy = 0.5714 + 0.4 0.6667 = 0.3047
Premium = 0.304762/5.556 = 0.0549
MLC-09-08
- 22 -
Question #27
Key: B
Many similar formulas would work equally well. One possibility would be
1000 3V42 + (1000 P42 1000 P40 ) , because prospectively after duration 3, this differs from
the normal benefit reserve in that in the next year you collect 1000 P40 instead of
1000 P42 .
MLC-09-08
- 23 -
Question #28
Key: E
32 = t f ( t ) dt + 40 f ( t ) dt
40
40
= t f ( t ) dt t f ( t ) dt + 40 (.6 )
w
40
= 86 tf ( t ) dt
w
40
tf ( t )dt = 54
w
40
( t 40 ) f ( t ) dt = 54 40 (.6 ) = 50
w
e 40 =
40
s ( 40 )
.6
Question #29
Key: B
d = 0.05 v = 0.95
The first line of Kiras solution is that the actuarial present value of Kevins benefit
premiums is equal to the actuarial present value of Kiras, since each must equal the
actuarial present value of benefits. The actuarial present value of benefits would also
have been easy to calculate as
(1000 )( 0.95)( 0.1) + (1000 ) 0.952 ( 0.9 ) = 907.25
MLC-09-08
- 24 -
Question #30
Key: E
Because no premiums are paid after year 10 for (x), 11Vx = Ax +11
Rearranging 8.3.10 from Bowers, we get
10V
h +1V
( hV + h ) (1 + i ) bh+1qx+ h
px + h
0.989
( 35,635.642 + 0 ) (1.05) 100,000 0.012 = 36,657.31 = A
11V =
x +11
0.988
Question #31
Key: B
and t px = 1
2
x
105 45
= 30
e45 =
2
105 65
= 20
e65 =
2
ex =
e 45:65 =
=
40
0
p45:65dt =
40 60 t
0
FG
H
60
40 t
dt
40
60 + 40 2 1 3
1
60 40 t
t + t
2
3
60 40
IJ
K
40
0
= 1556
.
o
e 45:65 = e 45 + e 65 e 45:65
= 30 + 20 1556
. = 34
o
In the integral for e45:65 , the upper limit is 40 since 65 (and thus the joint status also)
can survive a maximum of 40 years.
MLC-09-08
- 25 -
Question #32
Answer: E
bg bg bg
d e / 100i
=
4 = s' 4 / s 4
4
1 e4 / 100
e4 / 100
1 e4 / 100
e4
100 e4
= 1202553
.
Question # 33
Answer: A
bi g
g L ln px O
ig
b
b
b g LM ln e OP
qx = qx M
=
q
P
x
M ln p b g P M ln e-b g P
bi g
b g
g
b
= q x b g
x b g = x b1g + x b 2 g + x b 3g = 15
.
q xb g = 1 e b g = 1 e 1.5
=0.7769
b0.7769g
q xb 2 g =
b g
b2g
b0.5gb0.7769g
15
.
= 0.2590
MLC-09-08
- 26 -
Question # 34
Answer: D
22
A 60 = v 3
q 60 + 2
live
then die
2 years in year 3
+ v4
3p 60
live
3 years
pay at end
of year 3
pay at end
of year 4
2 p 60
q60+ 3
then die
in year 4
1
. gb013
. g+
. gb1 013
. gb015
. g
1 0.09gb1 011
b
b1 0.09gb1 011
. g
. g
b103
b103
1
= 019
.
Question # 35
Answer: B
a x = a x:5 +5 E x a x +5
a x:5
5 Ex
1 e 0.07b5g
=
= 4.219 , where 0.07 = + for t < 5
0.07
= e 0.07b5g = 0.705
a x +5 =
1
= 12.5 , where 0.08 = + for t 5
0.08
gb g
MLC-09-08
- 27 -
Question #36
Key: D
1
2
px( ) = p ' x( ) p ' x( ) = 0.8 ( 0.7 ) = 0.56
( )
( )
ln p ' (1)
x
1
q( ) since UDD in double decrement table
qx( ) =
ln p( ) x
x
ln ( 0.8 )
=
0.44
ln ( 0.56 )
= 0.1693
(1)
0.3 q x + 0.1
0.3qx( )
1
= 0.053
1 0.1qx( )
0.3 q x + 0.1 =
Number alive at x + 0.1
Since UDD in double decrement,
1
l x( ) ( 0.3) qx( )
lx( ) 1 0.1qx( )
MLC-09-08
- 28 -
Question #37
Key: E
P ( Ax ) =
1
1
= 0.04 = 0.04333
12
ax
o Le = o L + E
= v T P ( Ax ) aT + co + ( g e ) aT
1 vT
1 vT
= v P ( Ax )
+ co + ( g e )
P ( Ax ) ( g e ) P ( Ax )
( g e)
= v T 1 +
+ co +
( )
Var ( o Le ) = Var v
P ( Ax ) ( g e )
1 +
Above step is because for any random variable X and constants a and b,
Var ( a X + b ) = a 2 Var ( X ) .
Apply that formula with X = v T .
Plugging in,
0.04333 ( 0.0030 0.0066 )
Var ( o Le ) = ( 0.10 ) 1 +
0.04
0.04
= ( 0.10 )( 2.17325 )
= 0.472
Question #38
Key: D
0.7 0.1 0.2
MLC-09-08
- 29 -
Actuarial present value (A.P.V.) prem = 800(1 + (0.7 + 0.1) + (0.52 + 0.13)) = 1,960
A.P.V. claim = 500(1 + 0.7 + 0.52) + 3000(0 + 0.1 + 0.13) = 1800
Difference = 160
MLC-09-08
- 30 -
Question # 39
Answer: D
b gb gb g
Per 10 minutes, find coins worth exactly 10 at Poisson rate 0.5 0.2 10 = 1
Per 10 minutes,
f
f
f
f
b0g = 0.3679
b1g = 0.3679
.
b2g = 01839
b3g = 0.0613
bg
bg
bg
bg
F 0 = 0.3679
F 1 = 0.7358
F 2 = 0.9197
F 3 = 0.9810
b gh b gc
b gh b
g b
gb
= 01965
.
bg
bg bg
b gb
Pr ob = F 1 = 0.7358
Pr ob = f 2 f 0
= 01839
.
0.3679 = 0.0677
Pr ob = 1 0.7358 0.0677
= 01965
.
Question # 40
Answer: D
g b gb g
= 1000b0.36933 / 111418
.
.
g = 3315
Mod
Mod
A60Mod = v q60
A61 =
+ p60
MLC-09-08
gb
1
.
01376
+ 0.8624 0.383 = 0.44141
.
106
- 31 -
Mod
E 0 LMod = 1000 A60Mod P60a&&60
i
b
= 114.27
Question # 41
Answer: D
The prospective reserve at age 60 per 1 of insurance is A60 , since there will be no
future premiums. Equating that to the retrospective reserve per 1 of coverage, we have:
&&s40:10
A60 = P40
+ P50Mod &&s50:10 20 k40
10 E50
A
A60 = 40
a&&40
0.36913 =
a&&40:10
10 E40 10 E50
P50Mod
a&&50:10
10 E50
1
A40
:20
20 E40
016132
.
7.70
7.57
0.06
+ P50Mod
gb
P50Mod
a&&50:10
1
A40 a&&40:10 A40:10
=
1
where A40
= A40 10 E40 A50
:10
gb
= 016132
.
0.53667 0.24905
= 0.02766
MLC-09-08
- 32 -
.
7.70
0.02766
ib g 14016132
.8166 0.53667 0.53667
1000
014437
.
b gb
g = 19.07
=
7.57
Alternatively, you could set the actuarial present value of benefits at age 40 to the
actuarial present value of benefit premiums. The change at age 50 did not change the
benefits, only the pattern of paying for them.
A40 = P40 a&&40:10 + P50Mod
016132
.
=
10 E40
a&&50:10
.
FG 016132
IJ b7.70g + d P ib0.53667gb7.57g
H 14.8166K
b1000gb0.07748g = 19.07
=
Mod
50
1000 P50Mod
4.0626
Question # 42
Answer: A
d xb 2 g = q xb 2 g lxb g = 400
b g
d xb 2 g
lxb g d xb1g
400
= 0.488
1000 180
Note: The UDD assumption was not critical except to have all deaths during the year so
that 1000 - 180 lives are subject to decrement 2.
MLC-09-08
- 33 -
Question #43
Answer: D
Use age subscripts for years completed in program. E.g., p0 applies to a person newly
hired (age 0).
Let decrement 1 = fail, 2 = resign, 3 = other.
Then q0b1g = 1 4 , q1 b1g = 15 , q2b1g = 1 3
q0b 2 g =
q b 3g = 1
5,
q1b 2 g =
3,
q b3g = 1
10 ,
9,
gb
q2b 2 g =
q b 3g = 1
2
gb
b gb gb g
pb g = b1 1 / 3gb1 1 / 8gb1 1 / 4g = 0.438
So 1b g = 200, 1b g = 200 b0.54g = 108 , and 1b g = 108 b0.474g = 512
.
p1b g = 1 1 / 5 1 1 / 3 1 1 / 9 = 0.474
c h / logb0.438g 1 0.438
= b0.405 / 0.826gb0.562g
q2b1g = log
2
3
= 0.276
d2b1g = l2b g q2b1g
= 512
. 0.276 = 14
b gb
Question #44
Answer: C
Let:
G =
N = number
X = profit
S = aggregate profit
subscripts G = good, B = bad, AB = accepted bad
c hb60g = 40
2
3
MLC-09-08
- 34 -
AB =
c hc hb60g = 10
1
2
1
3
b g b g b g b g b g
= b40gb10,000g + b40gd300 i = 4,000,000
Var b S g = E b N g Var b X g + Var b N g E b X g
= b10gb90,000g + b10gb100g = 1,000,000
S and S are independent, so
Var b S g = Var b S g + Var b S g = 4,000,000 + 1,000,000
Var SG = E N G Var X G + Var N G E X G
AB
AB
AB
AB
AB
AB
AB
= 5,000,000
If you dont treat it as three streams (goods, accepted bads, rejected bads), you
can compute the mean and variance of the profit per bad received.
B = 13 60 = 20
c hb g
b g b g
= 90,000 + b100g = 100,000
d i
b g
b
g c
h
= b0.5gb 100g + b0.5gb0g = 50
E d X i = b0.5gb100,000g + b0.5gb0g = 50,000
g c
2
B
Likewise,
Now Var S B = E N B Var X B + Var N B E X B
b g b g b g b g b g
= b20gb47,500g + b20gd50 i = 1,000,000
2
bg
b g
b g
Question #45
Key: E
MLC-09-08
- 35 -
ex =
x
2
qx =
x 1
Ax =
k +1
k =b
1 x 1 k +1
v
k qx =
x k =b
a x
Ax =
a&&x =
1
x
x
1 Ax
d
e y = 15 y = Assumed age = 70
A70 =
a30
= 0.45883
30
a&&70 = 9.5607
500000 = b a&&20 b = 52, 297
Question #46
Answer: B
10 E30:40 = 10 p30 10
d p v id p v ib1 + i g
= b E gb E gb1 + i g
= b0.54733gb0.53667 gb179085
.
g
p40 v10 =
10
10
30
10
10
10
40
10
10
30
10
40
= 0.52604
The above is only one of many possible ways to evaluate
should give 0.52604
a30:40:10 = a30:40 10 E30:40 a30+10:40+10
b
g b
gb
g
= b13.2068g b0.52604gb114784
.
g
= a&&30:40 1 0.52604 a&&40:50 1
= 7.1687
MLC-09-08
- 36 -
10
p30
10
Question #47
Answer: A
Equivalence Principle, where is annual benefit premium, gives
b g
1000 A35 + IA
35
= a&&x
1000 A35
1000 0.42898
=
(1199143
.
.
)
616761
a&&35 IA 35
b gi
428.98
.
582382
= 73.66
=
a&&35 =
1 A35 1 0.42898
=
= 1199143
.
0.047619
d
MLC-09-08
- 37 -
Question #48
Answer: C
b gb g b gb g
Co-worker:
Question #49
Answer: C
xy = x + y = 014
.
0.07
Ax = Ay =
=
= 0.5833
+ 0.07 + 0.05
xy
014
014
1
1
.
.
Axy =
=
=
= 0.7368 and a xy =
=
= 5.2632
xy + 014
xy + 014
. + 0.05 019
.
. + 0.05
P=
Axy
a xy
Ax + Ay Axy
MLC-09-08
a xy
2 0.5833 0.7368
= 0.0817
5.2632
- 38 -
Question #50
Answer: E
b V + P gb1 + ig q b1 V g= V
b0.49 + 0.01gb1 + ig 0.022b1 0.545g = 0.545
b1 + ig = b0.545gb1 0.022g + 0.022
20 20
20
40
21 20
21 20
0.50
= 111
.
b V + P gb1 + ig q b1 V g= V
. g q b1 0.605g = 0.605
b0.545+.01gb111
21 20
20
41
22 20
22 20
41
q41 =
0.61605 0.605
0.395
= 0.028
Question #51
Answer: E
gb
g
= 1000bq + p A g / b106
. + p a&& g
= c15 + b0.985gb382.79gh / c106
. + b0.985gb10.9041gh = 33.22
= 1000 v q60 + p60 A61 / 1 + p60 v a&&61
60
MLC-09-08
60
61
60
61
- 39 -
Question #52
Key: D
Since the rate of depletion is constant there are only 2 ways the reservoir can be empty
sometime within the next 10 days.
Way #1:
There is no rainfall within the next 5 days
Way #2
There is one rainfall in the next 5 days
And it is a normal rainfall
And there are no further rainfalls for the next five days
Prob (Way #1) = Prob(0 in 5 days) = exp(-0.2*5) = 0.3679
Prob (Way #2) = Prob(1 in 5 days) 0.8 Prob(0 in 5 days)
= 5*0.2 exp(-0.2* 5)* 0.8 * exp(-0.2* 5)
= 1 exp(-1) * 0.8 * exp(-1) = 0.1083
Hence Prob empty at some time = 0.3679 + 0.1083 = 0.476
Question #53
Key: E
+
0.96 = e ( 1 )
1 + = ln ( 0.96 ) = 0.04082
1 = 0.04082 = 0.04082 0.01 = 0.03082
Similarly
5 0.06128 )
pxy = e ( ) (
= e 0.3064 = 0.736
Question #54
Answer: B
Transform these scenarios into a four-state Markov chain, where the final disposition of
rates in any scenario is that they decrease, rather than if rates increase, as what is
given.
MLC-09-08
- 40 -
from year t 3
to year t 2
Decrease
from year t 2
to year t 1
Decrease
Increase
Decrease
0.6
Decrease
Increase
0.75
Increase
Increase
0.9
State
LM0.80
0.60
Transition matrix is M
MM0.00
N0.00
0.00
0.00
0.75
0.90
0.20
0.40
0.00
0.00
OP
PP
PQ
0.00
0.00
0.25
.
010
MLC-09-08
- 41 -
Question #55
Answer: B
lx = x = 105 x
t P45 = l45+ t / l45 = 60 t / 60
Let K be the curtate future lifetime of (45). Then the sum of the payments is 0 if K 19
and is K 19 if K 20 .
F 60 K IJ 1
60 K
b40 + 39+...+1g = b40gb41g = 13.66
=
60
2b60g
&&
20 a45 =
1 GH
60
K = 20
Hence,
b g
= ProbbT 33g
= 33p45 =
l78 27
=
l45 60
= 0.450
MLC-09-08
- 42 -
Question #56
Answer: C
Ax =
Ax =
d IAi
E
0s x
zd
= 0.4
= 0.25 = 0.04
+ 2
0 s
Ax ds
Ax ds
ib g
e 0.1s 0.4 ds
F e I
= b0.4 gG
H 01. JK
0.1s
=
0
0.4
=4
01
.
Alternatively, using a more fundamental formula but requiring more difficult integration.
c IA h
=
=
z
z
bg
b0.04g e
t t px x t e t dt
t e 0.04 t
= 0.04
0.06 t
dt
t e 0.1t dt
t
1
= 0.04
e 0.1 t
0
01
. 0.01
0.04
=
=4
0.01
FG
H
MLC-09-08
IJ
K
- 43 -
Question #57
Answer: E
Subscripts A and B here just distinguish between the tools and do not represent ages.
We have to find e AB
eA =
eB =
FG 1 t IJ dt = t t
H 10K
20
z
z FGH
z FGH
e AB =
10
2 10
IJ
K
t
t2
dt = t
7
14
t
7
= 49
49
= 35
.
14
IJ FG 1 t IJ dt = z FG 1 t t + t IJ dt
K H 10K
H 10 7 70K
7
t2 t2
t3
=t +
20 14 210
= 7
49 49 343
+
= 2.683
20 14 210
= 10 5 = 5
e AB = e A + e B e AB
= 5 + 35
. 2.683 = 5817
.
MLC-09-08
- 44 -
Question #58
Answer: A
bg
bx g t = 0100
.
+ 0.004 = 0104
.
t
pxb g = e 0.104 t
Actuarial present value (APV) = APV for cause 1 + APV for cause 2.
c b g
= 2000 010
. + 500,000 0.004
=
ghz e
5 0.144 t
dt
2200
1 e 0.144b5g = 7841
.
0144
Question #59
Answer: A
R = 1 px = q x
S = 1 px
eb k g since
d
e z
bg i
x t + k dt
=e z
=e z
bg
bg
e z
x t dt k dt
0
x t dt k dt
0
So S = 0.75R 1 px e k = 0.75q x
1 0.75q x
px
1 qx
px
ek =
=
1 0.75q x 1 0.75q x
e k =
k = ln
LM 1 q OP
N1 0.75q Q
x
MLC-09-08
- 45 -
Question #60
Key: C
A60 = 0.36913
2
d = 0.05660
A60 = 0.17741
and
2
A60 A60
= 0.202862
2
Variance on one policy is Var L ( ) = 100,000 + 2 A60 A60
d
2
100 100,000 + 2 A60 A60
d
100 100,000 + ( 0.36913)
d
d
= 2.326
100,000 + ( 0.202862 )
d
0.63087
36913
100,000 +
0.63087
= 0.004719
d
36913 + 471.9
=
d 0.63087 0.004719
= 59706
= 59706 d = 3379
MLC-09-08
- 46 -
Question #61
Key: C
= ( 0V + ) (1 + i ) (1000 + 1V 1V ) q75
1V
= 1.05 1000q75
Similarly,
2 V = ( 1V + ) 1.05 1000q76
3V
= ( 2V + ) 1.05 1000q77
1000 =3V = 1.053 + 1.052 + 1.05 1000 q75 1.052 1000 1.05 q76 1000 q77 *
=
=
=
(1.05) + (1.05)
3
+ 1.05
3.310125
1000 1.17796
= 355.87
3.310125
Question #62
Answer: D
A281:2 =
3V
d
i
FG IJ
H K
b g
1
.
= 0.05827
1 e 2 = 0.02622 since = ln 106
72
71
= 1+ v
= 19303
.
72
=
a&&28:2
2 t
e 1 72 dt
0
= 287
MLC-09-08
- 47 -
Question #63
Answer: D
bg
bg
ax =
1 Ax
ax = ax
0.4
= 6.667
0.06
LM Proof: a
N
*
x
z
z
z
=
=
e z
z
e
d
e z
t
bg
x s + 0.03 ds 0.03t
bg
x s ds 0.03t 0.03t
0
t
bg
0 x s ds 0.06 t
dt
dt
dt
= ax
b gb
= 1 0.03 6.667
= 0.8
Question #64
Answer: A
bulb ages
Year
0
1
2
3
10000
0
1000
9000
100+2700
900
280+270+3150
0
0
6300
0
0
0
MLC-09-08
- 48 -
#
replaced
1000
2800
3700
Question #65
Key: E
o
e25:25 =
z
z
15
0 t
10
0t
p40 dt
F
IJ e
dt + G e z
= e
H
Kz
1
L1
1 e i + e M d1 e
=
d
.04
N.05
15
15 .04 t
10 .05t
.04 ds
.60
.60
dt
.50
= 112797
+ 4.3187
.
= 15.60
iOPQ
Question #66
Key: C
p 60 +1 =
60 + 2
63
64
65
= 0.4589
MLC-09-08
- 49 -
Question # 67
Key: E
1
+ = 0.08 = = 0.04
+
12.50 = a x =
Ax =
= 0.5
+
1
2
Ax =
=
+ 2 3
e j
Var aT =
Ax Ax2
= 3 4 = 52.083
0.0016
S.D. = 52.083 = 7.217
Question # 68
Key: D
v = 0.90 d = 010
.
Ax = 1 da&&x = 1 010
. 5 = 0.5
b gb g
Benefit premium =
5000 Ax 5000vqx
a&&x
a&&x +10
a&&x
a&&
0.2 = 1 x +10 a&&x +10 = 4
5
10Vx
= 1
b gb g
a&&
= b5000gb0.6g b455gb4g = 1180
= 5000 Ax +10
MLC-09-08
x +10
- 50 -
Question #69
Key: D
v is the lowest premium to ensure a zero % chance of loss in year 1 (The present value
of the payment upon death is v, so you must collect at least v to avoid a loss should
death occur).
Thus v = 0.95.
2
E Z = vqx + v 2 px qx +1 = 0.95 0.25 + 0.95 0.75 0.2
bg
b g
= 0.3729
b g
d i
b g
b g d i c b gh
Var Z = E Z2 E Z
Question #70
Key: D
Actuarial present value (APV) of future benefits =
= ( 0.005 2000 + 0.04 1000 ) /1.06 + (1 0.005 0.04 )( 0.008 2000 + 0.06 1000 ) /1.062
= 47.17 + 64.60
= 111.77
= 95.05
E 1 L K ( 55 ) 1 = 111.77 95.05 = 16.72
Question #71
Key: A
(t)
MLC-09-08
- 51 -
1 ( t ) dt =
Average =
2
1 ( 3 + 3t )dt
2
3t 2
= 3t +
= 7.5
e 7.5 7.52
= 0.0156
f ( 2) =
2!
Question #72
Key: A
Let Z be the present value random variable for one life.
Let S be the present value random variable for the 100 lives.
bg
E Z = 10
= 10
t t
e e dt
e b + g 5
= 2.426
FG IJ e b g
H 2 + K
F 0.04IJ de i = 11233
= 10 G
.
H 0.16K
Var b Z g = E d Z i c E b Z gh
d i
2 + 5
E Z 2 = 102
0.8
.
= 11233
2.4262
= 5.348
bg
bg
VarbSg = 100 Varb Zg = 534.8
E S = 100 E Z = 242.6
F 242.6
.
= 1645
F = 281
534.8
MLC-09-08
- 52 -
Question #73
Key: D
Prob{only 1 survives} = 1-Prob{both survive}-Prob{neither survives}
ge
= 1 3p50 3p 50 1 3p50 1 3p 50
gb
gb
gb
gb
g b
gb
gb
0.936320
= 0.140461
Question # 74
Key: C
The tyrannosaur dies at the end of the first day if it eats no scientists that day. It dies at
the end of the second day if it eats exactly one the first day and none the second day. If
it does not die by the end of the second day, it will have at least 10,000 calories then,
and will survive beyond 2.5.
b g bg b g
b gb g
Prob (dies) = f 0 + f 1 f 0
= 0.368 + 0.368 0.368
= 0.503
1 0
b g e 01! = 0.368
e 1
f b1g =
= 0.368
1!
since f 0 =
1 1
Question #75
Key: B
Let X = expected scientists eaten.
For each period, E X = E X dead Prob already dead + E X alive Prob alive
b g
b g
Day 1, E X1 = 1
1 1
g b g e 0!0 = 0.368
= 0 0.368 + 1 b1 0.368g = 0.632
Day 2, E X 2
Prob (dead at end of day 2) = 0.503
[per problem 10]
MLC-09-08
- 53 -
b g
day in period.
Question # 76
Key: C
This solution applies the equivalence principle to each life. Applying the equivalence
principle to the 100 life group just multiplies both sides of the first equation by 100,
producing the same result for P.
b
g
g
b10gb0.03318g + b10gb1 0.03318gb0.03626g + Pb1 0.03318gb1 0.03626g
P=
108
.
108
. 2
= 0.3072 + 0.3006 + 0.7988 P
0.6078
= 3.02
P=
0.2012
108
. 2
Question #77
Key: E
Level benefit premiums can be split into two pieces: one piece to provide term
insurance
for n years; one to fund the reserve for those who survive.
Then,
Px = Px1:n + Px:n1 nVx
gb
MLC-09-08
- 54 -
e
j
a&&
= eP P j
E
= Px Px1:n &&sx:n
x:n
x:n
= Px Px1:n
jP
a&&x:n
1
x:n
a&&x:n
eP P j
=
eP j
1
x:n
x:n
gb
Question #78
Key: A
b g
= ln 1.05 = 0.04879
Ax =
=
=
z
z
bg
px x t e t dt
1
e t dt for DeMoivre
x
1
a
x x
10 V
c A h could be used.
40
Since
MLC-09-08
- 55 -
A50 =
A40 =
a50
50
a60
60
FG
H
F1 A
=G
H
IJ
K
IJ = 1387
K .
18.71
1 A50
= 0.3742 so a50 =
= 12.83
50
19.40
= 0.3233 so a40
60
40
.3233
= 0.02331
c h 01387
.
V c A h = A P c A ha = 0.3742 b0.02331gb12.83g = 0.0751.
so P A40 =
10
40
50
40
50
Question #79
Key: D
b g
bg
F 0.03 IJ 0.70 + FG 0.6 IJ 0.30
=G
H 0.03 + 0.08 K
H 0.06 + 0.08 K
Ax = E v T b x g = E v T b x g NS Prob NS + E v T b x g S Prob S
= 0.3195
Similarly, 2 Ax =
F I=
Var a
H b gK
T x
Ax Ax2
0.31952
01923
.
=
= 141
..
0.082
Question #80
Key: B
= 0.10136
Using new p82 value of 0.3
0.5 0.4 (1 0.3) + 0.2 0.15 (1 0.1)
MLC-09-08
- 56 -
= 0.16118
Change = 0.16118 0.10136 = 0.06
Alternatively,
2 p80 = 0.5 0.4 = 0.20
3 p80 = 2 p80 0.6 = 0.12
2 p84 = 0.20 0.15 = 0.03
3 p84 = 2 p84 0.10 = 0.003
2 p80:84 = 2 p80 + 2 p84 2 p80 2 p84 since independent
3
p80:84
Revised
3 p80 = 0.20 0.30 = 0.06
3
Question #81
Key: D
Poisson processes are separable. The aggregate claims process is therefore
equivalent to two independent processes, one for Type I claims with expected
frequency
MLC-09-08
- 57 -
b g
b g
b g b g b g b g b g
= b1000gb0g + b1000gb10g
= 100,000
bg
b g
b g
2,100,000 = 100,000 + Var b S g
Var b S g = 2 ,000,000
II
Question #82
Key: A
5
b g = pb1g pb2g
p50
5 50 5 50
FG 100 55IJ e b gb g
H 100 50K
= b0.9gb0.7788g = 0.7009
0.05 5
Similarly
10
FG 100 60IJ e b g b g
H 100 50 K
= b0.8gb0.6065g = 0.4852
b g =
p50
0.05 10
b g = pb g pb g = 0.7009 0.4852
5 5 q50
5 50
10 50
= 0.2157
Question #83
Key: C
Only decrement 1 operates before t = 0.7
1
0.7 q40
40
since UDD
is
1-0.07 = 0.93
Decrement 2 operates only at t = 0.7, eliminating 0.125 of those who reached 0.7
b gb
- 58 -
Question #84
Key: C
FG
H
1+
0.83910
1.062
vq80 v 3 2 p80q82
+
2
2
b
g
b
b167524
.
g = 665.75
174680
.
= 665.75 + 0.07156
= 397.41
3,284 ,542
= 0.83910
3,914 ,365
Where 2 p80 =
gb
z
z
bt v t t p65
65
bt gdt
d id
bg
= 1000
p
0 t 65
bg
FG 1 IJ = 16.667
H 0.04 + 0.02K
=
=
z
z
b g
b g b gb
= 1000e0.08
0 u
b g
p67 65 2 + u du 1000
MLC-09-08
- 59 -
Question #86
Key: B
(1)
a x:20 = a&&x:20 1+ 20 Ex
(2)
a&&x:20 =
(3)
Ax:20 =
(4)
Ax = A1x:20 + 20 Ex Ax + 20
1 Ax:20
d
A1x:20
+ Ax:201
b gb g
Ax:20 = 018
. + 0.25 = 0.43
a&&x:20 =
1 0.43
= 1197
.
0.05 / 105
.
a x:20 = 1197
. 1 + 0.25 = 1122
.
Question #87
Key: A
p1 = p (1 ) f ( ) d =
( 2 )
e 1 ( / 2 ) e
1!
(1)
1 32
= e d
2 0
[Integrate by parts; not shown]
1 2 32 4 23
= e e
2 3
9
MLC-09-08
2
= 0.22
9
- 60 -
Question #88
Key: B
ex = px + pxex +1 px =
ex
8.83
=
= 0.95048
1 + ex +1 9.29
x:2
= 1 + v + v 2 2 p x + ...
Question #89
Key: E
M = Initial state matrix = 1 0 0 0
LM0.20
0.50
T = One year transition matrix = M
MM0.75
N1.00
0.80
0.50
OP
0 P
0.25P
P
0 Q
0
M T = 0.20 0.80 0 0
.
0.40 0
b M T g T = 0.44 016
.
cb M T g T h T = 0.468 0.352 0.08 010
ib g
MLC-09-08
- 61 -
Question #90
Key: B
Let Yi be the number of claims in the ith envelope.
b g
g b
g b g b g
E Y = b1 0.2g + b4 0.25g + b9 0.4g + b16 015
. g = 7.2
E X b13g = 50 13 2.5 = 1625
Var X b13g = 50 13 7.2 = 4680
. g
ProbmXb13g Zr = 0.90 = b1282
R X b13g 1625 1282
U
Prob S
. V
T 4680
W
X b13g 1712.7
E Yi = 1 0.2 + 2 0.25 + 3 0.4 + 4 015
. = 2.5
2
b g
Note: The formula for Var X 13 took advantage of the frequencys being
Poisson.
The more general formula for the variance of a compound distribution,
2
Var S = E N Var X + Var N E X , would give the same result.
bg b g b g
b gb g
Question #91
Key: E
b g 1 60 = 75 1 60 = 151
b60g = 1 60 = 151 53 = 251 = 85
M 60 =
F
t
10
t
= 1
25
M
p65
= 1
F
p60
MLC-09-08
- 62 -
eo x
eo y
eo xy
z
z
10
10
10
0
= 10 7 +
exy = ex + e y exy = 5 +
4 13
=
3 3
25 13 30 + 75 26
=
= 1317
.
2
3
6
Question #92
Key: B
1
=
+ 3
1
2
Ax =
=
+ 2 5
Ax =
c h
P Ax = = 0.04
F Pc A hI A A
Var b Lg = G 1 +
j
H JK e
F 0.04 IJ FG 1 FG 1IJ IJ
= G1 +
H 0.08 K H 5 H 3K K
F 3I F 4 I
=G J G J
H 2 K H 45K
2
2
x
1
5
Question #93
Key: A
MLC-09-08
- 63 -
=
Thus 1V = ( 0 V + ) (1 + i )
2V
3V
n +1V
= ( 1V + )(1 + i ) = ( (1 + i ) + ) (1 + i ) = &&
s2
= ( 2V + )(1 + i ) = &&
s2 + (1 + i ) = &&
s3
20 V
= &&
s20
a&&
= 60 &&
s
&&
20
s
35
Alternatively, as above
( nV + ) (1 + i ) = n+1V
Write those equations, for n = 0 to n = 34
0 : ( 0V + ) (1 + i ) = 1V
1: ( 1V + )(1 + i ) = 2V
2 : ( 2V + )(1 + i ) = 3V
M
34 : ( 34V + ) (1 + i ) = 35V
Multiply equation k by (1 + i )
34 k
MLC-09-08
- 64 -
35 k
Since 0V = 0
&&s35 = 35V
= a&&60
(see above for remainder of solution)
Question #94
Key: B
xy ( t ) =
t qy t
t qx
px ( x + t ) + t qx t p y ( y + t )
t p y + t px t q y + t px t p y
50:50 (10.5 ) =
where
10.5
p50 =
10.5 q50
10
8,950,901
p50 =
10.5
8,188,074
= 0.91478
8,950,901
since UDD
( t p60 )
( t p60 )
MLC-09-08
- 65 -
10.5
= 2 ( 0.90848 ) ( 0.90848 )
= 0.99162
dp
dt = ( 0.0023) = 0.0023
p
0.99162
Question #95
Key: D
2
1
P = P vt t px( ) x( ) ( t ) dt + 50, 000 vt t px( ) x( ) ( t ) dt + 50, 000 vt t px( ) x( ) ( t ) dt
0
2 0.1t 2.3t
P = P e
2 0.1t 2.3t
2( 2.4 )
1 e
P 1 2.29
2.4
P = 11,194
2( 2.4 )
1 e
= 50000 0.01
2.4
+ 2.3
2( 2.4 )
Question #96
Key: B
ex = px + 2 px + 3 px + ... = 1105
.
b g
. g
1000b104
k 3 k
k px
k =3
= 1000v 3
k px
k =3
FG 1 IJ
H 104
. K
9.08 = 8072
MLC-09-08
- 66 -
2.4
2.8580 = 8072
= 2824
Question #97
Key B
b g
b ge
1
a&&30:10 = 1000 A30 + P IA 30
+ 10
:10
10
A30
1000 A30
b g
a&&30:10 IA
1
30:10
1010 A30
g
b
1000 0102
.
7.747 0.078 10 0.088
102
6.789
= 15.024
=
Test Question: 98
Key: E
FG1 t IJ dt
H 30K
L t OP
= Mt
N 2b 30g Q
e 30 =
30
30
30
2
100 30
= 35
2
= 100 eo 30 =
e 30 = e 30 + 4 = 39
so
MLC-09-08
e30
= 39 =
30
2
= 108
- 67 -
Test Question: 99
0L
Key: A
@5%
= 77,079
Question #100
Key: D
( accid ) = 0.001
( total ) = 0.01
( other ) = 0.01 0.001 = 0.009
0.009 0.001
= 500, 000
+
= 100, 000
0.06 0.02
MLC-09-08
- 68 -
Key: E
b gb g
b gb g b gb g b gb g
= b0.6gb1g + b0.2gb25g + b0.2gb100g = 25.6
E N = Var N = 60 0.5 = 30
E X = 0.6 1 + 0.2 5 + 0.2 10 = 3.6
E X2
d i
b g
bg b g
Var = b60gb0.5gb0.6g1 + b60gb0.5gb0.2g5 + b60gb0.5gb0.2gb10g
2
= 768
Test Question: 102
1000
20
20Vx
= 1000 Ax +20 =
=
a&&x +20 =
Key: D
1000
20
19Vx + 20 Px
. g q b1000g
ib106
x +19
px +19
. g 0.01254b1000g
b342.03 + 13.72gb106
.
= 36918
0.98746
1 0.36918
.
= 111445
.
0.06 / 106
MLC-09-08
.
Ax +20 36918
.
=
= 331
.
a&&x +20 111445
- 69 -
pxb g = e
bx g t dt
bg
Key: B
z b gb g
F b g b g IJ
= Ge z
H
K
= b p g where
=e
2 x1 t dt
x1 t dt
10
6,616,155
= 0.80802
8,188,074
6,396,609
=
= 0.78121
8,188,074
p60 =
11 p60
10
k px
b g = p b g p b g
q60
10 60 11 60
=
10
p60
g b
2
11 p60
from I.L.T.
Ps =
Key: C
1
d , where s can stand for any of the statuses under consideration.
a&& s
a&&s =
1
Ps + d
1
= 6.25
. + 0.06
01
1
= 8.333
a&&xy =
0.06 + 0.06
a&&x = a&&y =
1
0.06 = 018
.
4.167
MLC-09-08
- 70 -
Key: A
b
g
g j = 48
= 1000 1 e b + 0.04
e b + 0.04 g = 0.952
+ 0.04 = ln 0.952
= 0.049
= 0.009
= 1000
Question #106
Key: B
bg
This is a graph of lx x .
bg
lx2
Question #107
Key: B
Expected value = v15 15 px
Variance = v30 15 px 15 qx
( px )
15
= 1 ( px )
15
= 0.6843
px = 0.975
qx = 0.025
MLC-09-08
- 71 -
Question #108
Key: E
(1)
( 2)
11V
11V
(1) ( 2 )
10V
11V
10V
+0
) (p
1+ i)
+ B
x +10
) (p
11V B =
qx +10
1000
px +10
1+ i)
x +10
10V
qx +10
1000
px +10
10V B B
= (101.35 8.36 )
) (p
1+ i)
x +10
(1.06 )
1 0.004
= 98.97
Key: A
b g
b gb g
b gb gb g
= 1000 300v 0.02 + 350v 2 0.98 0.04 + 400v 3 0.98 0.96 0.06
= 36,829
MLC-09-08
- 72 -
Key: E
gb g b gb g
gb g b gb g
Question #111
Key: A
= 1, 239.75
1.062
1.062
1.06
= P ( 2.3955 )
P = 517.53 518
Key: A
b gb g b gb g
1180 = 70 12 + 50 10 20a30:40
a30:40 = 8
a30:40 = a30 + a40 a30:40 = 12 + 10 8 = 14
100a30:40 = 1400
MLC-09-08
Key: B
- 73 -
1.063
bg
a = a f t dt =
o t
1 e 0.05t
1
0.05
0.05
zb
te
1
te t dt
2
bg
te 1.05t dt
LM b g FG
IJ
H
K
N
1 L F 1 I O
.
=
M1 G . JK PP = 185941
0.05 NM H 105
Q
1
1
t
=
t + 1 et +
+
e 1.05t
.
0.05
105
105
. 2
OP
Q
20,000 185941
.
= 37,188
Question #114
Key: C
Event
x=0
Prob
( 0.05)
( 0.95)( 0.10 ) = 0.095
( 0.95)( 0.90 ) = 0.855
x =1
x2
Present Value
15
15 + 20 /1.06 = 33.87
15 + 20 /1.06 + 25 /1.062 = 56.12
( )
Question #115
Key: B
Let K be the curtate future lifetime of (x + k)
kL
When (as given in the problem), (x) dies in the second year from issue, the curtate
future lifetime of ( x + 1) is 0, so
MLC-09-08
- 74 -
1L
1000
279.21
1.1
= 629.88 630
=
Ax:3 = 1 d a&&x:3
Px:3 = 279.21 =
1 d a&&x:3
a&&x:3
1
d
a&&x:3
Key: D
Let M = the force of mortality of an individual drawn at random; and T = future lifetime
of the individual.
Pr T 1
= E Pr T 1 M
=
=
=
z
zz
zd
0 0
bg
Pr T 1 M = f M d
2 1
1
2
e t dt d
1 e
i 21 du = 21 d2 + e
i 21 d1 + e i
1 =
= 0.56767
Question #117
Key: E
Note that above 40, decrement 1 is DeMoivre with omega = 100; decrement 2 is
DeMoivre with omega = 80.
(1)
That means 40
( 20 ) = 1/ 40 = 0.025; 40( 2) ( 20 ) = 1/ 20 = 0.05
( )
40
( 20 ) = 0.025 + 0.05 = 0.075
60 t 40 t 2400 100t + t 2
=
t p40 =
60
40
2400
MLC-09-08
( )
- 75 -
( )
p40
/ dt = ( 100 + 2t ) / 2400
20
( )
40
( 20 ) = d
( )
( )
p40
/ dt / 20 p40
= 0.025 / (1/ 3) = 0.075
Key: D
b gb
g b gb gb
g b gb gb gb
b gb g
. g b200,000gb0.03g
b7452.55gb106
1 0.03
= 1958.46
Key: A
b g
L = bT v T aT = 1+ i
v T aT
= 1 aT
E L = 1 ax = 0
MLC-09-08
ax
- 76 -
L = 1 aT = 1
=
aT
ax
ax 1 v T
ax
v T 1 ax
v T Ax
=
1 Ax
ax
Key: D
(0, 1)
(1, 0.9)
(1.5, 0.8775)
(2, 0.885)
tp1
p1 = (1 01
. ) = 0.9
b gb
= 0.95 + 0.444
= 1394
.
Alternatively,
MLC-09-08
- 77 -
e11: .5 =
=
=
z
z
zb
1.5
0 t
p1dt
0.5
p1dt +1p1 x
t
0
0
1
0
p2 dx
1 01
. t dt + 0.9
= t 0.12t
1
0
0.5
b1 0.05xgdx
+ 0.9 x 0.052 x
0.5
Key: A
b g
b g
Ax 1 + i qx
px
. g 0.01788
b0.4672gb105
1 0.01788
= 0.4813
A65 =
. g 0.01952
b0.4813gb105
1 0.01952
= 0.4955
Single contract premium at 65 = (1.12) (10,000) (0.4955)
= 5550
b1 + ig
MLC-09-08
5550
5233
i=
5550
1 = 0.02984
5233
- 78 -
Key: B
x = 0.06
y = 0.06
xy = 0.06 + 0.06 = 012
.
x
0.06
Ax =
=
= 0.54545
x + 0.06 + 0.05
y
0.06
Ay =
=
= 0.54545
y + 0.06 + 0.05
xy
012
.
Axy =
=
= 0.70588
xy + 012
. + 0.05
Axy = Ax + Ay Axy = 0.54545 + 0.54545 0.70588 = 0.38502
Revised Calculation (common shock model):
x = 0.06, Tx *b x g = 0.04
MLC-09-08
- 79 -
Question #123
Key: B
q35:45 = 5 q35 + 5 q45 5 q35:45
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
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
bg
Key: C
bg
t dt = 6 so N 3 is Poisson with = 6.
g g
P is Poisson with mean 3 (with mean 3 since Prob yi < 500 = 0.5
P and Q are independent, so the mean of P is 3, no matter what the value of Q is.
MLC-09-08
- 80 -
Key: A
At age x:
Actuarial Present value (APV) of future benefits =
APV of future premiums =
1000
5
1000
4
10V
FG 4 a&& IJ
H5 K
FG 1 A IJ 1000
H5 K
x
4
A25 = a&&25 by equivalence principle
5
A25
1
8165
.
= =
= 1258
.
a&&25
4 16.2242
1000
4
A35 a&&35
5
5
4
1
15.3926
.
= 128.72 1258
5
5
= 10.25
g b gb
Let
Key: E
E Y = 10a&&40 = 148166
.
Var Y = 10
2
A40 A40
ib
2
= 100 0.04863 016132
106
.
. / 0.06
= 70555
.
E S = 100 E Y = 14,816.6
Var S = 100 Var Y = 70,555
MLC-09-08
- 81 -
Key: B
Where
5a&&30:35 4a&&30:20
g b
g b
4 0.02933
5 010248
.
5 14.835 4 11959
.
0.5124 011732
0.39508
.
=
= 0.015
74.175 47.836
26.339
1
A30
= A30:20 A30:201 = 0.32307 0.29374 = 0.02933
:20
and
a&&30:20 =
1 A30:20
d
1 0.32307
= 11959
.
0.06
106
.
FG IJ
H K
Comment: the numerator could equally well have been calculated as A30 + 4 20 E30 A50
= 0.10248 + (4) (0.29374) (0.24905)
= 0.39510
Key: B
b gb g
px = 1 0.75 0.05
= 0.9625
0.75
b gb g
p y = 1 0.75 .10
= 0.925
0.75 qxy = 1 0.75 pxy
b p gd p i since independent
= 1- b0.9625gb0.925g
= 1
0.75
0.75
= 01097
.
Question #129
Key: D
Let G be the expense-loaded premium.
Actuarial present value (APV) of benefits = 100,000A35
APV of premiums = Ga&&35
APV of expenses = 0.1G + 25 + ( 2.50 )(100 ) a&&35
Equivalence principle:
MCL-09-08
- 82 -
G=
= 1234
Test Question:
130 Key:
The person receives K per year guaranteed for 10 years Ka&&10 = 8.4353K
The person receives K per years alive starting 10 years from now 10 a&&40 K
10 E40:
1
A40 = A4010
+
:
10 E40
Derive a&&50 =
10 E40
1
A40 A40
:10
A50
gA
50
0.30 0.09
= 0.60
0.35
1 A50 1 0.35
=
= 16.90
.04
d
104
.
Plug in values:
10,000 = 8.4353 + 0.60 16.90 K
b gb
gh
= 18.5753K
K = 538.35
Test Question:
131 Key:
11
STANDARD: eo 25:11 =
MODIFIED:
p25
o
=e z
e2511
: =
MCL-09-08
FG1 t IJ dt = t t
H 75K
2 75
0.1ds
0
D
2
= e .1 = 0.90484
p dt
0 t 25
+ p25
z FGH
10
IJ
K
t
dt
74
- 83 -
11
0
= 101933
.
IJ
K
F t I
1 e
=
+ e Gt
01
.
H 2 74 JK
= 0.95163 + 0.90484b9.32432g = 9.3886
=
1 0.1t
dt + e0.1
0.1
z FGH
10
0.1
t
dt
74
10
0
Difference
=0.8047
Test Question:
132
Key: B
Comparing B & D: Prospectively at time 2, they have the same future benefits. At issue,
B has the lower benefit premium. Thus, by formula 7.2.2, B has the higher reserve.
Comparing A to B: use formula 7.3.5. At issue, B has the higher benefit premium. Until
time 2, they have had the same benefits, so B has the higher reserve.
Comparing B to C: Visualize a graph C* that matches graph B on one side of t=2 and
matches graph C on the other side. By using the logic of the two preceding paragraphs,
Cs reserve is lower than C*s which is lower than Bs.
Comparing B to E: Reserves on E are constant at 0.
Test Question:
133 Key:
Since only decrements (1) and (2) occur during the year, probability of reaching the end
of the year is
p60
b1g p60
b 2g = 1 0.01 1 0.05 = 0.9405
gb
gb
gb
MCL-09-08
- 84 -
Question #134
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 = 0.02 30 = 0.60
f ( 0 ) = e 0.60 = 0.55
Test Question:
135 Key:
zb
zb
gd
ib
gd
gd ib
gd i
100000 e- t 0.008 e- t dt
zb
30
gd
ib
gd
zb
30
gd ib
Test Question:
b gb
136 Key:
g b gb
g b gb
b gb
80222.4 79,396.5
80,222.4
= 0.0103
=
P0 = 111 = 9.0909%
MCL-09-08
gd i
100000 e- t 0.001 e- t dt
- 85 -
Question #137
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, = ( 0.2 )(1000 ) = 200
2
Var(losses) = Var ( X ) + ( E ( X ) )
2
= 200 5000 + ( 100 )
= 3,000,000
2
= 800 8000 + ( 100 )
= 14, 400,000
Test Question:
138 Key:
. =y
q40
b 2g = 012
q41
b 2g = 2 y = 0.24
b gb g
l b g = 2000b1 0.34gb1 0.392g = 803
b g = 1 0.8 1 0.24 = 0.392
q41
42
MCL-09-08
- 86 -
Test Question:
139 Key:
b g
47
48
p30 =.47681
b g
b g
b g
b g
b g
= 609.98 16.589
L 0 609.98 16.589 0
>
609.98
= 36.77
16.589
Test Question:
140
Key: B
b g
Prb K = 1g = p p = 0.9 0.81 = 0.09
Prb K > 1g = p = 0.81
E bY g = .1 1+.09 187
. +.81 2.72 = 2.4715
E dY i = .1 1 +.09 187
. +.81 2.72 = 6.407
VARbY g = 6.407 2.4715 = 0.299
Pr K = 0 = 1 px = 01
.
1 x
2 x
2 x
Question #141
Key: E
E [ Z ] = b Ax
b ( 0.02 )
b
=
= b/3
+
( 0.06 )
MCL-09-08
- 87 -
Ax Ax2
=b
+ 2 +
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
Test Question:
142 Key:
b g b g c AA h
2 2
In general Var L = 1 + P
c h
Here P Ax =
2
x
1
1
= .08 =.12
5
ax
F .12 I
So Var b Lg = G 1 + J c A A h =.5625
H .08K
F b.12gIJ c A A h
and Var b L *g = G 1 +
H .08 K
b1 + g b0.5625g =.744
So Var b L *g =
b1 + g
E L * = A .15a = 1 a b +.15g = 1 5b.23g = .15
E L * + Var b L *g =.7125
2
2
x
5
4
2
x
15 2
8
12 2
8
Test Question:
143 Key:
Serious claims are reported according to a Poisson process at an average rate of 2 per
month. The chance of seeing at least 3 claims is (1 the chance of seeing 0, 1, or 2
claims).
b g
b g
b g bg b g
- 88 -
01
. e + e + 2 / 2 e
The expected value is 2 per month, so we would expect it to be at least 2 months
=4 .
Plug in and try
e4 + 4e4 + 42 / 2 e4 =.238, too high, so try 3 months = 6
e6 + 6e6
d i
+ d6 / 2ie
2
[While 2 is a reasonable first guess, it was not critical to the solution. Wherever you
start, you should conclude 2 is too few, and 3 is enough].
Test Question:
144 Key:
q1b f g =
0.28
= 0.2
14
.
b g = q b w g + p b g q b w g + p b g p b g q b w g
w
3 q0
b gb g b gb gb g
MCL-09-08
- 89 -
Test Question:
145 Key:
N
M
since same
e26
= e26
z
=e z
N
p25
=e
bg c h
M
t + 0.1 1 t dt
25
bg
bg
e z
c h
M
25
t dt 0.1 1 t dt
0
=e z
c h
M
25
t dt 0.1 1 t dt
0
L F
M MN H
= p25 e
0.1 t t2
I OP1
K Q0
M
= e0.05 p25
N
N
e25
= p25
1 + e26
M
1 + e26
= e0.05 p25
g
gb g
M
= 0.951 10.0 = 9.5
= 0.951 e25
MCL-09-08
- 90 -
Test Question:
146 Key:
b g
F c1 A hI = 10,000,000
= 100b10,000gG
H JK
b10,000g 1 c A A h
b10,000g b0.25g b016
=
. g = 50,000
Y = Var Y =
2
x
LM F E Y
N
AGG
>0
AGG
1282
=
.
OP
Q
F E YAGG
AGG
F = 1282
. AGG + E YAGG
F = 1282
.
500,000 + 10,000,000 = 10,641,000
Question #147
Key: A
1
A30:3
= 1000vq30 + 500v 2 1 q30 + 250v3 2 q30
2
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 1
1 2
0.00153
1
a30:1 = 1 2 + 1 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 )
2.552
2
= 1.28
MCL-09-08
- 91 -
Test Question:
148
b DAg
1
80:20
Key: E
eb g j
= 20vq80 + vp80 DA
bg
1
8119
:
b g
b g
b g
b g b gb g
2+.9b12.225g
=
= 12.267
20 .2
.8
1
+
DA 81
:19
106
.
106
.
13 106
. 4
1
DA 8119
=
= 12.225
:
.8
DA801 :20 = 20 v .1 + v .9 12.225
q80 =.2
13 =
q80 =.1
106
.
Test Question:
149
Key: B
Let T denote the random variable of time until the college graduate finds a job
Let N t , t 0 denote the job offer process
m bg
R|
S|Type II T
m b gr
- reject with probability b1 pg mN bt gr
2
= 01587
.
1 = 01587
.
2 = 0.3174
T has an exponential distribution with =
bg
1
= 315
.
.3174
Pr T > 3 = 1 F 3
3
= e 3.15 = 0.386
MCL-09-08
- 92 -
Test Question:
LM
N
px = exp
150 Key:
OP
Q
ds
= exp ln 100 x s
0 100 x s
t
t
0
100 x t
100 x
e50
o
e60
LM
OP = 25
=z
N
Q
t O
40 t
1 L
40t P = 20
dt =
=z
M
40
40 N
2Q
F 50 t IJ FG 40 t IJ dt = z 1 d2000 90t + t idt
=z G
H 50 K H 40 K
2000
I = 14.67
1 F
t
2000
45
t
t
=
+
2000 GH
3 JK
50
50 50 t
t2
1
50t
dt =
50
50
2
2 40
40
e50:60
40
40
40
0
Question #151
Key: C
Ways to go 0 2 in 2 years
0 0 2; p = ( 0.7 )( 0.1) = 0.07
Question #152
Key: A
For death occurring in year 2
0.3 1000
APV =
= 285.71
1.05
For death occurring in year 3, two cases:
MCL-09-08
- 93 -
0.17 1000
= 154.20
1.052
Test Question:
b g
Var b g
Var 0 L
0
153
Key: E
b g
b g since Varb g = 0
= v bb V g p q
b10,000 3,209g b0.00832gb0.99168g
=
= Var 0 + v 2 Var 1
1 1
50 50
2
103
. 2
= 358664.09
b g
Var 1
= v b2 2V
p50q51 p51
= 101075.09
b g
Var 0 L
= 358664.09 +
103
. 2
101075.09
= 453937.06
103
. 2
Alternative solution:
MCL-09-08
- 94 -
50
50 51
Test Question:
154 Key:
30
a&&35
eA
=
35:30
1
1 A35
:30
1
A35
a&&65
:30
1
=
1
A35
:30
b.21.07g9.9
b1.07g
1.386
.93
= 1.49
=
Test Question:
0.4 p0
=.5 = e z
0. 4
155 Key:
e F +e2 x jdx
LM e22x OP.4
N Q0
=e
0.8
.4 F F e 2 1 I
H
K
=e
.4 F
.5 = e.4 F .6128
MCL-09-08
- 95 -
bg
ln .5 = .4 F .6128
.6931 = .4F .6128
F = 0.20
Question #156
Key: C
1
0.03
P = b d = 1200
14.65976 1.03
a&&x
= 46.92
V
=
initial
reserve
P = 343 46.92 = 296.08
9
Question #157
Key: B
d = 0.06 V = 0.94
Step 1 Determine px
MCL-09-08
- 96 -
Question #158
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
= 100,000
0.16736
0.00592
(
)
1.06
1.0610
+ ( 0.02766 100,000 )
=15,513
[see comment ]
1
Where A40:10
= A40 10 E40 A50
Comment: the first line comes from comparing the benefits of the two insurances. At
each of age 40, 41, 42,,49 ( IA )40:10 provides a death benefit 1 greater than ( IA )41:10 .
1
1
Hence the A40:10
term. But ( IA )41:10 provides a death benefit at 50 of 10, while ( IA )40:10
1
provides 0. Hence a term involving 9 q41 = 9 p41 q50 . The various vs and ps just get all
actuarial present values at age 40.
Question #159
Key: A
10001Vx = (1 + i ) qx (1000 10001Vx )
40 = 80 (1.1) qx (1000 40 )
qx =
88 40
= 0.05
960
1 AS
=
=
=
(G
60 (1.1) 50
= 16.8
0.95
MCL-09-08
- 97 -
Question #160
Key: C
1
At any age, px ( ) = e 0.02 = 0.9802
1
1
qx ( ) = 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 years death benefits
= 10,000*0.0198 v = 188.1
APV of death benefit for 3 years 188.1 + E40 *188.1 + E40 * E41 *188.1 = 506.60
Question #161
Key: B
40
e30:40 =
t p30dt
0
30 t
dt
30
0
40
=t
t2
2 ( 30 )
40
0
800
30
= 27.692
= 40
= 95
Or, with De Moivres law, it may be simpler to draw a picture:
p30 = 1
40
30
MCL-09-08
p30
70
- 98 -
(1 + 40 p30 )
2
p30 = 0.3846
70
= 0.3846
30
= 95
65 t
t p30 =
65
40
Var = E (T ) ( E (T ) )
2
Var (T ) = 2 t t px dt ex2
o
65
t
t
= 2 t 1 dt 1 dt 2
65
0
0 65
65
= 2* ( 2112.5 1408.333) ( 65 65 / 2 )
Var (T ) = t 2 t px x ( t ) dt
0
t t px x ( t ) dt
1
1
65
t dt t dt
0
65
65
65 2
t3
=
3 65
65
0
t2
2 65
65
0
With De Moivres law and a maximum future lifetime of 65 years, you probably didnt
need to integrate to get E (T ( 30 ) ) = e30 = 32.5
o
Likewise, if you realize (after getting = 95 ) that T ( 30 ) is uniformly on (0, 65), its
variance is just the variance of a continuous uniform random variable:
MCL-09-08
- 99 -
2
65 0 )
(
Var =
= 352.08
12
Question #162
Key: E
1V
Question #163
Key: D
0.95
= 19
1 0.95
exy = pxy + 2 pxy + ...
=
1.02 ( 0.95 )
= 9.44152
Question #164
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%
MCL-09-08
- 100 -
Question #165
Key: E
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 1 e
(10 s )
7
ds
s 10
= 20 10 7e 7
= 20 10 7 + 7e
10
0
10 7
) = 94
Question #166
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
aT = Var aT =
Ax ( Ax ) = 400 0.23077 ( 0.375 ) = 6.0048
2
Pr aT > ax aT = Pr aT > 12.5 6.0048
1 vT
= Pr
> 6.4952 = Pr 0.67524 > e0.05T
0.05
ln 0.67524
= Pr T >
= Pr [T > 7.85374]
0.05
= e 0.037.85374 = 0.79
Question #167
Key: A
Ax = e 0.08t ( 0.03) dt =
( )
( )
MCL-09-08
- 101 -
0.05 5
( )
p50
= e ( )( ) = e 0.25 = 0.7788
(1)
5 q55
5 (1)
= 55
( t ) e( 0.03+0.02)t dt = ( 0.02 / 0.05 ) e0.05t
0
= 0.4 1 e
0.25
5
0
= 0.0885
( )
()
Probability of retiring before 60 = 5 p50
5 q55
= 0.7788*0.0885
= 0.0689
Question #168
Key: C
Complete the table:
l81 = l[80] d[80] = 910
e x = ex +
since UDD
2
2
e[ x] = e[ x] + 1 2
l + l + l +K 1
o
+
e[ x] = 81 82 83
l[80]
1
1
o
o
2
2
MCL-09-08
- 102 -
e[81] =
910
= 0.91
1000
830
p[81] =
= 0.902
920
830
p81 =
= 0.912
910
p[80] =
o
e[80] = 1 q 80 + p 80 1 + eo 81
[ ]
2 [ ]
1
where q[80] contributes
since UDD
2
1
o
8.5 = (1 0.91) + ( 0.91) 1 + e81
2
e81 = 8.291
1
o
o
e81 = q81 + p81 1 + e82
2
1
o
8.291 = (1 0.912 ) + 0.912 1 + e82
2
e82 = 8.043
1
o
o
e[81] = q[81] + p[81] 1 + e82
2
1
= (1 0.902 ) + ( 0.902 )(1 + 8.043)
2
= 8.206
o
Or, do all the recursions in terms of e, not e , starting with e[80] = 8.5 0.5 = 8.0 , then final
o
MCL-09-08
- 103 -
Question #169
Key: A
px +t
0
1
2
3
px
vt
vt t px
0.7
0.7
0.7
0.49
0.95238
0.90703
1
0.6667
0.4444
a&&
1000 2Vx:3 = 1000 1 x + 2:1
a&&x:3
= 1000 1
= 526
2.1111
Alternatively,
Px:3 =
1
d = 0.4261
a&&x:3
MCL-09-08
- 104 -
Question #170
Key: E
Let G be the expense-loaded premium.
Actuarial present value (APV) of benefits = 1000A50 .
APV of expenses, except claim expense = 15 + 1 a&&50
APV of claim expense = 50A50 (50 is paid when the claim is paid)
APV of premiums = G a&&50
Equivalence principle: Ga&&50 = 1000 A50 + 15 + 1 a&&50 + 50 A50
1050 A50 + 15 + a&&50
G=
a&&50
a
For De Moivres with = 100, x = 50 A50 = 50 = 0.36512
50
1 A50
= 13.33248
a&&50 =
d
Solving for G,
G = 30.88
Question #171
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
414 q50
MCL-09-08
- 105 -
Question #172
Key: D
1
= 100,000 A45 5 E40 / a&&40:5 ( IA )40:5
Question #173
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 years worth so have table
Pr(both alive)
Pr(both dead)
(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
a&&x = a&&y = 1 +
+
= 2.6104
1.05 1.052
APV = 20,000 a&&x + 20,000 a&&y 10,000 a&&xy
a&&xy = 1 +
(it pays 20,000 if x alive and 20,000 if y alive, but 10,000 less than that if both are
alive)
MCL-09-08
- 106 -
= 80, 430
Other alternatives also work.
Question #174
Key: C
Let P denote the contract premium.
0
axIMP
P = ax = e t e t dt = e0.05t dt = 20
E [ L] =
10
axIMP
= e
0.03t 0.02t
dt + e
0.03(10 ) 0.02(10 )
0.03t 0.01t
dt
l e 0.5 e0.5
+
= 23
0.05
0.04
E [ L ] = 23 20 = 3
=
E [ L] 3
=
= 15%
P
20
Question #175
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
Expected size of Fund 2 at end of year 2 = 0 (since the amount paid was the single
benefit premium). Difference is 895.
MCL-09-08
- 107 -
Question #176
Key: C
2
Var [ Z ] = E Z 2 E [ Z ]
E [Z ] =
t t
(v b )
0.02 2
=
( 0.02 ) e0.07t dt =
7
0.07
0
(e
= 0.02 e 0.12t x ( t ) dt = 2
12
E Z 2 =
( vt bt ) t px x ( t ) dt =
Var [ Z ] =
( )
1 2
7
6
0.05t
=1
e 0.02t ( 0.02 ) dt
1 4
= 0.08503
6 49
Question #177
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
5
0.1
= 0.4769
11 ln (1.1)
MCL-09-08
- 108 -
Question #178
Key: C
0.001
= 100,000
0.06 + 0.001
= 1639.34
Accidental death
10
= 20 e 0.061t dt
0
1 e 0.61
= 20
= 149.72
0.061
Question #179
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.
MCL-09-08
- 109 -
Question #180
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
Q3 0 = 0.1
1
1
0
0.01
Q2 0.1 = 0.27
1
0.01
1
0.049
Q1 0.27 = 0.489
1
1
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)
MCL-09-08
- 110 -
Question #181
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
Benefit premium =
24,025.5
= 10,695
2.2465
Question #182
Key: A
Split into three independent processes:
Deposits, with * = ( 0.2 )(100 )( 8 ) = 160 per day
Withdrawals, with * = ( 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
= 1.04 1010
= 0.696 1010
MCL-09-08
- 111 -
W D + 80,000 80,000
Pr (W > D ) = Pr (W D > 0 ) = Pr
>
131,757
131,757
= 1 ( 0.607 )
= 0.27
Question #183
Key: D
Exponential inter-event times and independent implies Poisson process (imagine
additional batteries being activated as necessary; we dont care what happens after two
have failed).
Poisson rate of 1 per year implies failures in 3 years is Poisson with = 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 = 1 random variables is Gamma
with = 2, = 1
1 3 t
F ( 3) = ( 2;3) =
t e dt
( 2 ) 0
= ( t 1) e t
= 1 4e
3
0
MCL-09-08
- 112 -
Question #184
Key: B
A45
( a&&45 15 E45 a&&60 ) + ( a&&60 15 E60 a&&75 )( 15 E45 ) = 1000 A45
a&&45
201.20
(14.1121 ( 0.72988)( 0.51081)(11.1454 )
14.1121
+ (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
= 17.346
Question #185
Key: A
1V
= ( 0 V + ) (1 + i ) (1000 + 1V 1V ) qx
2V
x +1
= 1027.42
MCL-09-08
- 113 -
Question #186
Key: A
Let Y be the present value of payments to 1 person.
Let S be the present value of the aggregate payments.
Y = Var [Y ] =
(1 Ax ) = 5572.68
d
( 500 )2
1
d2
Ax Ax2 = 1791.96
S = Y1 + Y2 + ... + Y250
E ( S ) = 250 E [Y ] = 1,393,170
28,333
28,333
F 1,393,170
Pr N ( 0,1)
28,333
=1.43 million
Question #187
Key: A
q b1g = 1 p b1g = 1 e p b g j
41
41
bg
q41 1
41
bg
q41
1
2
l41( ) = l40( ) d 40( ) d 40( ) = 1000 60 55 = 885
1
( )
p41
q41( )
1
750
=
885
750
(1) = 1
q41
885
MCL-09-08
( )
q41
65
135
65
135
= 0.0766
- 114 -
Question #188
Key: D
s ( x ) = 1
d
( x) =
log ( s ( x ) ) =
dx
x
ex =
o
t
x
1
dt =
+1
x
o
= new
new = 2 old + 1
e0new = old
2 +1
+1
old
old
2 + 1 9
(0new ) =
=
old = 4
4
Question #189
Key: C
Constant force implies exponential lifetime
Var [T ] = E T ( E [T ])
2
= 0.1
1
1
= 2 = 2 = 100
2
= te
10
.1t
10e
.1t 10
0
1
10e.1t
10
1
= 10 (1 e 1 ) = 6.3
Question #190
MCL-09-08
- 115 -
Key: A
Ga&&x:15
a&&x =
1 Ax 1 0.5148197
=
= 16.66
d
0.02913
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 #191
Key: D
For the density where T ( x ) T ( y ) ,
Pr (T ( x ) < T ( y ) ) =
40
50
40
y =0
0.0005 x dy +
0
50
y = 40
40
50
y = 40
= 0.0005 ydy +
=
40
0.0005dxdy +
0.0005dxdy
y = 0 x = 0
y = 40 x = 0
0.0005 y 2
2
40
0
+ 0.02 y
0.0005 x
0.02 dy
50
40
MCL-09-08
- 116 -
40
0
dy
where the first 0.4 is the probability that T ( x ) = T ( y ) and the first 0.6 is the probability
that T ( x ) T ( y ) .
Question #192
Key: B
The conditional expected value of the annuity, given , is
1
.
0.01 +
Question #193
Key: E
Recall ex =
o
x
2
o
ex:x = ex + ex ex:x
ex:x =
o
t
t
dt
1
1
x y
2 ( 2a )
2 ( 3a )
= 3
2 = 7a
3
3
2 ( 3a )
2
( a ) = k
3
3
3.5a a = k ( 3.5a 3a )
MCL-09-08
- 117 -
k =5
Question #194
Key: B
0.10
= 10,000
= 7143
+
0.10 + 0.04
The actuarial present value of the insurance of 7143 is
xy
0.12
7,143
= ( 7,143)
= 5357
xy +
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 #195
Key: E
Let
p0 = ( 0.8 ) = 0.64
p0 = ( 0.8 ) = 0.41
p0:0 4 p0:0
2
Question #196
Key: E
MCL-09-08
- 118 -
0.218
= 0.25
0.87
p0:0
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
( )
( )
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 #197
Key: C
2
E ( Z ) = v k +1bk +1
k =0
px qx+ k
( ) = (v
E Z
k +1
k =0
bk +1
2
k
px qx + k
= 11,773
( )
Var [ Z ] = E Z 2 E ( Z )
= 11,773 36.82
= 10, 419
MCL-09-08
- 119 -
Question #198
Key: A
Benefits +
3
0 Le = 1000v +
Expenses
( 0.20G + 8) + ( 0.06G + 2 ) v + ( 0.06G + 2 ) v 2
( for K = 2 ) = 770.59
Question #199
Key: D
P = 1000 P40
[A]
[B]
1+ i =
Plug into [A]
20.385
= 1.01925
20
255 + 11.175
1.01925
1000 25V40 =
MCL-09-08
- 120 -
Premiums
G a&&3
Question #200
Key: A
1.2
1
1
0.8
0.6
0.5
0.4
0.4
0.2
0
0
10
Give
n
0
s ( x)
55 q35
20 55 q15
20 55 q15
55 q35
= 1
20
30
40
50
Give
n
25
15
0.70
Linear
Interpolation
35
0.50
0.48
Linear
Interpolation
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
60
20 p15 55 q35
55 q35
20 p15
s ( 35 )
s (15 )
0.48
0.70
= 0.6856
=
MCL-09-08
- 121 -
70
Give
n
75
0.4
80
90
0
100
Given
90
100
0.16
Linear
Interpolation
Question #201
Key: A
px = 0.018316
Question #202
Key: B
x
40
41
42
lx( )
2000
1920
1840
d x( )
20
30
40
d x( )
60
50
1920 30 50 = 1840
Let premium = P
1840 2
2000 1920
+
v+
v P = 2.749 P
APV premiums =
2000
2000 2000
30 2
40 3
20
APV benefits = 1000
v+
v +
v = 40.41
2000
2000
2000
40.41
P=
= 14.7
2.749
MCL-09-08
- 122 -
Question #203
Key: A
10
= 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
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 a40
= 1 ( 0.08 / 0.16 ) = 0.5
10V
= 0.5
( 0.057 ) = 0.14375
0.16
Question #204
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
MCL-09-08
0 < T 10
10 < t 20
20<t
when evaluated at i = 0.05
- 123 -
Question #205
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
( )
= ( 22 )( 5.3475 ) + ( 22 ) 2.052
= 210.1
Question #206
Key: A
P
= 500000
2500
2500
P ( 2.77 ) = 2550.68
p = 921
MCL-09-08
- 124 -
Question #207
Key: D
30 s ( x ) dx =
80
e30:50 =
s ( 30 )
80
x2
30 1 10,000 dx
2
30
1
100
x3 80
x
30,000 30
0.91
33.833
=
0.91
= 37.18
Question #208
Key: B
= 10.147
100010V50 = 1000 A60 1000 P50 a&&60
= 425.66 10.147 25
= 172
MCL-09-08
- 125 -
Question #209
Key: E
Let Portfolio be the present value random variable for the aggregate payments.
Let Y65 = present value random variable for an annuity due of one on one life age 65.
Thus E (Y65 ) = a&&65
Let Y75 = present value random variable for an annuity due of one on one life age 75.
Thus E (Y75 ) = a&&75
Var (Portfolio) = 50 22Var [Y65 ] + 30 (1) Var [Y75 ] = 200 (13.2996 ) + 30 (11.5339 ) = 3005.94
2
1
d2
1
d2
2
A65 A65
=
2
A75 A75
=
( 0.05660 )
1
( 0.05660 )
X E ( Portfolio )
Question #210
Key: C
a = e t e t dt =
0
APV = 50,000
1
0.5
1
+
1
0.045 + 1
= 100,000 ln
0.045 + 0.5
= 65,099
MCL-09-08
- 126 -
Question #211
Key: E
The process described, where a key feature is the exponential time between events, is
a
Poisson process with = 1 5 per minute.
The number of claims in any interval of length n minutes has a Poisson distribution with
mean
n = n / 5 .
Here n = 10. So parameter = 10/5 = 2
Pr ( N 2 ) = 1 Pr ( N = 0 ) Pr ( N = 1)
= 1 e 2 e 2 2
= 1 0.135 0.271 = 0.594
Question #212
Key: D
The payouts in any time period of length t have a Poisson distribution with parameter
5t .
The payouts can be grouped by size. For each i, the number of payouts of size i is a
Poisson random variable with mean 5t / 2i , and these random variables are
independent.
Since they are independent Poisson random variables, the sum of the payouts of size 1,
5t 5t 5t 35t
2 or 3 is a Poisson random variable with mean + + =
2 4 8 8
For t = 1/ 3 hour, the mean is
35 1
= 1.4583
8 3
MCL-09-08
- 127 -
Question #213
Key: D
How long was the expected wait during first 45 minutes? In that interval, wait is
exponential with
= 30, so
1 30x
1 30x
E min ( X , 45 ) = x e dx + 45 e dx
0
45
30
30
45
= 30 1 e 30 = 23.31
45
e
45
= 1.5 , so f ( 0 trains ) =
Expected trains =
30
1.5
(1.5)0 = 0.223
0!
1/ 30, 0 t < 45
where x ( t ) =
1/15, t 45
o
= 10,000
A45:20
a&&45:20
MCL-09-08
- 128 -
A60:51 = 0.68756
A60:5 = 0.06674 + 0.68756 = 0.7543
a&&60:5 = 11.1454 0.68756 9.8969 = 4.3407
1
After the change, expected prospective loss = 10,000 A60:5
+ (Reduced Amount) A60:51
where
5( 0.04 + 0.02 )
= 0.7408
5 Ex = e
0.04
Ax1:5 =
(1 0.7408 ) = 0.1728
0.04 + 0.02
7 ( 0.05+ 0.02 )
= 0.6126
7 E x +5 = e
0.05
Ax +1 5 :7 =
(1 0.6126 ) = 0.2767
0.05 + 0.02
12 E x = 5 E x 7 E x + 5 = 0.7408 0.6126 = 0.4538
0.05
Ax +12 =
= 0.625
0.05 + 0.03
Ax = 0.1728 + ( 0.7408 )( 0.2767 ) + ( 0.4538 )( 0.625 )
= 0.6614
Question #216
Key: A
APV of Accidental death benefit and related settlement expense =
0.004
= 89.36
( 2000 1.05)
0.004 + 0.04 + 0.05
0.04
APV of other DB and related settlement expense = (1000 1.05 )
= 446.81
0.094
MCL-09-08
- 129 -
3
= 31.91
0.094
100
= 1063.83
0.094
APV of 0 Le = 89.36 + 446.81 + 50 + 31.91 1063.83
= 445.75
Question #217
Key: C
Compute the probabilities of moving from healthy to NH. There are three paths.
H to H to NH: (0.8)(0.05) = 0.04
H to HHC to NH: (0.15)(0.05) = 0.0075
H to NH to NH: (0.05)(1) = 0.05
Summing, we get 0.0975 as the probability for each member.
Variance for m members = mpq, here = 50*(0.0975)(0.9025) = 4.40
Question #218
Key: C
0.6 0.3 0.1
Q0 = 0
0
1
0
0
1
Q0 Q1 = 0
0
1
0
0
1
0 0.108 0.892
Q0 Q1 Q2 = 0
0
1
0
0
1
MCL-09-08
- 130 -
Alternatively, the same effort here but often shorter when everyone is in the same initial
state:
(1.00
( 0.60
( 0.36
This method just calculates the top row of the cumulative transition matrix. It gives the
same elements you use if you calculate the complete cumulative transition matrix, so
you finish the problem the same way as before.
Question #219
Key: E
0.25 1.5 q x
0.25
px 1.75 px
ln ( 0.1)
= 0.5756
=
4
0.25
1
px = e 4( 0.5756 ) = 0.8660
1.75
px = px 0.75 px +1 = e e
1.5 q x + 0.25 =
0.251.5 q x
0.25
px
0.25
3
( 3 )
4
=e
13
( 0.5756 )
4
= 0.1540
Question #220
Key: C
The form of lx for non-smokers matches DeMoivres law, so
1
xNS =
110 x
2
= 1 xS xS =
2
110 x
l xS = l0S (110 x ) [see note below]
2
MCL-09-08
- 131 -
S
t p20
Thus
S
l20
( 90 t )
= S+t =
902
l20
NS
t p25 =
e 20:25 =
o
85
0 t
NS
l25
(85 t )
+t
=
NS
85
l25
p20:25dt
85
pS
0 t 20 t
NS
p25
dt
2
90 t ) ( 85 t )
(
dt
=
0
( 90 )2 85
85
85
1
( 90 t )2 ( 90 t 5) dt
0
688,500
1
85 ( 90 t )3 dt 5 85 ( 90 t )2 dt
=
0
688,500 0
85
( 90 t )4 5 ( 90 t )3
1
=
+
688,500
4
3
0
1
=
[ 156.25 + 208.33 + 16, 402,500 1, 215,000]
688,500
= 22.1
[There are other ways to evaluate the integral, leading to the same result].
Note: the solution above assumes the candidate will recognize that the smoker mortality
is modified DeMoivre and can proceed directly to the lx or s ( x ) form. The s ( x ) form is
derived as s ( x ) = e
x 2
0 110 t
dt = e
2ln (110t )
110 x
=
110
Question #221
Key: B
MCL-09-08
- 132 -
1
1
= 1000 A40:10
+ 2000 10 E30 A50:10
MCL-09-08
- 133 -
Question #222
Key: A
s25:15
15V25 = P25 &&
1
P25:15
= P25:15
1
A25:15
15 E25
P25:151 = 0.05332 0.05107
= 0.00225
=
0.05107 = P25:151 =
1
A25:15
15 E25
a&&25:15
15 E25
a&&25:15
1
A25:15
/ a&&25:15
15 E25
1
A25:15
/ a&&25:15
1
&&
s25:15
0.00225
= 0.04406
0.05107
0.01128
= 0.22087
0.05107
25,00015V25 = 25,000 ( 0.22087 0.04406 ) = 25,000 ( 0.17681) = 4420
P25 &&
s25:15 =
There are other ways of getting to the answer, for example writing
A: the retrospective reserve formula for 15V25 .
1
, which = 0
B: the retrospective reserve formula for 15V25:15
Question #223
Key: C
ILT:
We have p70 = 6,396,609 / 6,616,155 = 0.96682
2 p70 = 6,164,663/ 6,616,155 = 0.93176
MCL-09-08
p70 for the DM model equal the ILT, therefore l72 for the DM
- 134 -
also equals the ILT. For DM we have l70 l71 = l71 l72 l71(
DM )
= 6,390, 409
order p70 .
Question #224
Key: D
( )
l60
= 1000
( )
l61
= 1000 ( 0.99 )( 0.97 )( 0.90 ) = 864.27
( )
d 60
= 1000 864.27 = 135.73
( )
d 60
= 135.73
3
ln ( 0.9 )
0.1054
=
= 98.05
ln ( 0.99 )( 0.97 )( 0.9 ) 0.1459
( )
l62
= 864.27 ( 0.987 )( 0.95 )( 0.80 ) = 648.31
( )
d 61
= 864.27 648.31 = 215.96
( )
d 61
= 215.96
3
ln ( 0.80 )
0.2231
=
= 167.58
ln ( 0.987 )( 0.95 )( 0.80 ) 0.2875
( )
( )
+ d 61
= 98.05 + 167.58 = 265.63
So d60
3
Question #225
Key: B
t
p40 = e 0.05t
p50 = ( 60 t ) / 60
50+t = 1/ ( 60 t )
MCL-09-08
- 135 -
10
p40:50 50+t dt =
10 e
10
0.05t
1 e0.05t
dt =
60
60 ( 0.05 ) 0
=
20
1 e 0.5 = 0.13115
60
Question #226
Key: A
Actual payment (in millions) =
q3 = 1
1 q3
3
5
+ 2 = 6.860
1.1 1.1
0.30
= 0.5
0.60
0.30 0.10
= 0.333
0.60
0.5 0.333
+
= 7.298
Expected payment = 10
2
1.1 1.1
Ratio
6.860
= 94%
7.298
Question #227
Key: E
At duration 1
K ( x)
1L
>1
0 Px1:2
Px1:2
Prob
qx +1
1 qx +1
So Var ( 1 L ) = v 2 qx +1 (1 qx +1 ) = 0.1296
That really short formula takes advantage of
Var ( a X + b ) = a 2Var ( X ) , if a and b are constants.
Here a = v; b = Px1:2 ; X is binomial with p ( X = 1) = qx +1 .
Alternatively, that same formula for Var arises from Hattendorf, since
MCL-09-08
- 136 -
= 0 and Var ( 2 L ) = 0
2V
1L
( )
Question #228
Key: C
P ( Ax ) =
Ax
Ax
Ax ( 0.1) ( 13 )
=
=
=
= 0.05
ax 1 Ax 1 Ax
(1 13 )
P ( Ax )
Var ( L ) = 1 +
1 0.05
= 1 +
5 0.10
Ax Ax2
Ax Ax2
Ax Ax2 = 0.08888
Var [ L] = 1 +
Ax Ax2
16
= 1 +
( 0.08888 )
45 0.1
2
1 +
=4
0.1
= 0.1
2
Question #229
Key: E
MCL-09-08
- 137 -
Pr aT
( 40 )
60
p40 =
100 40
= 0.25
120 40
g = a60 = 19.00
Question 230
Key: B
0.05
A51:9 = 1 da&&51:9 = 1
( 7.1) = 0.6619
1.05
11V
= 499.09
Alternatively, you could have used recursion to calculate A50:10 from A51:9 , then a&&50:10
from A50:10 , and used the prospective reserve formula for 10V .
Question #231
Key: C
1000 A81 = (1000 A80 ) (1 + i ) q80 (1000 A81 )
689.52 = ( 679.80 )(1.06 ) q80 (1000 689.52 )
q80 =
720.59 689.52
= 0.10
310.48
MCL-09-08
- 138 -
0.05
0.95
+ 689.52
= 665.14
1.06
1.06
Question #232
Key: D
lx( )
776
752
d x( )
8
8
42
43
d x( )
16
16
1
2
( )
( )
l42
and l43
came from lx( +1) = lx( ) d x( ) d x( )
776
) = 76.40
776 + 752v
APV Premiums = 34
= ( 34 )(1.92 ) = 65.28
776
2V
Question #233
Key: B
a&&x = 1 + vpx + v 2 2 px = 1 +
= 2.8174
MCL-09-08
- 139 -
= 27,927
Notes: The solution assumes that the future lifetimes are identically distributed. The
precise description of the benefit would be a special 3-year temporary life annuity-due.
Question #234
Key: B
t
1
1
1
px( ) x( ) ( t ) = qx( ) = 0.20
2
2
px( ) = 1 tqx( ) = 1 0.08t
3
3
px( ) = 1 tqx( ) = 1 0.125t
qx( ) =
1
1
0t
1
1
2
3
1
1
0.205t 2 0.01t 3
= 0.2 t
+
2
3 0
0.01
= ( 0.2 ) 1 0.1025 +
= 0.1802
3
Question #235
Key: B
1V40
= 1
1 CV40
a&&41
14.6864
= 1
= 0.00879
a&&40
14.8166
(1000 )(1)
MCL-09-08
( 0.00879 ) = 2.93
- 140 -
(d )
( w)
G 0.1G (1.50 )(1) ) (1.06 ) 1000q40
1CV40 q40
(
AS =
( )
( )
1 q40
q40
d
= 1.197G 6.22
(d )
( w)
(
1 AS + G 0.1G (1.50 )(1) ) (1.06 ) 1000q41 2 CV40 q41
AS =
( )
( )
1 q41
q41
d
= 2.229G 11.20
2.229G 11.20 = 24
G = 15.8
Question #236
Key: A
(1)
( 2)
(
4 AS + G (1 c4 ) e4 ) (1 + i ) 1000q x + 4 5 CV q x + 4
AS =
1 qx(+)4 qx(+ 4)
1
( 657.31)(1 + i ) 90 148.75
0.65
= 694.50
MCL-09-08
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1+ i =
Question #237
Key: C
Excluding per policy expenses, policy fee, and expenses associated with policy fee.
APV (actuarial present value) of benefits = 25,000 Ax:20 = ( 25,000 )( 0.4058 ) = 10,145
Let G denote the expense-loaded premium, excluding policy fee.
APV of expenses = ( 0.25 0.05 ) G + 0.05G a&&x:20 + ( 2.00 0.50 ) + 0.50 a&&x:20 ( 25,000 /1000 )
= 0.20 + ( 0.05 )(12.522 ) G + 1.50 + ( 0.50 )(12.522 ) 25
= 0.8261G + 194.025
APV of premiums = G a&&x:20 = 12.522 G
Equivalence principle:
APV premium = APV benefits + APV expenses
12.522 G = 10,145 + 0.8261G + 194.025
10,339.025
G=
= 883.99
(12.522 0.8261)
This G is the premium excluding policy fee.
Now consider only year 1 per policy expenses, the year one policy fee (call it F1 ), and
expenses associated with F1 .
APV benefits = 0
APV premium = F1
Equivalence principle
F1 = 15 + 0.25 F1
15
F1 =
= 20
0.75
MCL-09-08
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Question #238
Key: B
MCL-09-08
- 143 -
Question #239
Key: B
= 0.8261 P + 243.59
Equivalence principle:
12.522 P = 10,145 + 0.8261 P + 244
10,389
P=
12.522 0.8261
= 888
Question #240
Key: D
MCL-09-08
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Question #241
Key: C
0.04
= 100,000 1
(10.8 )
1.04
= 58, 462
Equivalence principle:
10.8 G = 58, 462 + 0.892 G + 690
59,152
G=
= 5970.13
9.908
Let F denote the policy fee.
APV of benefits = 0
APV of premiums = F a&&x = 10.8 F
APV of expenses = 150 + 25 ax + 0.5 F + 0.04 F ax
= 395 + 0.892 F
Equivalence principle:
MCL-09-08
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Question #242
Key: C
11
(d )
( w)
(
10 AS + G c10 G e10 ) (1 + i ) 10,000q x +10 11 CV q x +10
AS =
1 qx(+10) qx(+10)
d
MCL-09-08
1 0.02 0.18
- 146 -
1302.1
0.8
= 1627.63
Question #243
Key: E
G = 1000 P35:10 + e
= 76.87 + 15.95
= 92.82
MCL-09-08
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Question #244
Key: C
4 AS =
35.351
0.937
= 37.73
With higher expenses and withdrawals:
4 AS
revised
( 48.5)(1.05) 13 4.5
0.927
33.425
0.927
= 36.06
4 AS
MCL-09-08
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Question #245
Key: E
MCL-09-08
(1000
10 20 A30
+ 20 + 10 a 30:9
+ 20 + 10 a30:9
(1000
10 20 A30
- 149 -
Question #246
Key: E
300 360
720
+
+
= 1286.98
2
1.04 1.04 1.042
APV of premium = G
APV of expenses = 0.02G + 0.03G + 15 + ( 0.9 )( 2 ) v
16.8
1.04
= 0.05G + 16.15
= 0.05G +
Question #247
Key: C
0.9G
= 1.8571G
1.05
Settlement expenses are 20 + (1)(10 ) = 30 , payable at the same time the death benefit is
paid.
30
So APV of settlement expenses =
APV of benefits
10,000
MLC-09-08
- 150 -
= ( 0.003)( 3499 )
= 10.50
Equivalence principle:
1.8571G = 3499 + 108.78 + 0.2857G + 10.50
3618.28
G=
= 2302.59
1.8571 0.2857
Question #248
Key: D
0.06
A50:20 = 1 d a&&50:20 = 1
(11.2918 )
1.06
= 0.36084
Actuarial present value (APV) of benefits = 10,000 A50:20
= 3608.40
MLC-09-08
- 151 -