SOCIETY OF ACTUARIES. EXAM MLC Models for Life Contingencies. EXAM
MLC SAMPLE SOLUTIONS. The following questions or solutions have been ...
SOCIETY OF ACTUARIES
EXAM MLC Models for Life Contingencies
EXAM MLC SAMPLE SOLUTIONS
The following questions or solutions have been modified since this document was prepared to use with the syllabus effective spring 2012, Prior to March 1, 2012: Questions: 151, 181, 289, 300 Solutions: 2, 284, 289, 290, 295, 300 Changed on March 19, 2012: Questions: 20, 158, 199 (all are minor edits) Changed on April 24, 2012: Solution: 292
Copyright 2011 by the Society of Actuaries
Some of the questions in this study note are taken from past SOA examinations.
MLC-09-11
PRINTED IN U.S.A.
Question #1 Answer: E 2
q30:34 2 p30:34 3 p30:34 2
p30 0.9 0.8 0.72
2
p34 0.5 0.4 0.20
2
p30:34 0.72 0.20 0.144 2
p30:34 0.72 0.20 0.144 0.776
3
p30 0.72 0.7 0.504
3
p34 0.20 0.3 0.06
3
p30:34 0.504 0.06 0.03024
3
p30:34 0.504 0.06 0.03024 0.53376
2
q30:34 0.776 0.53376 0.24224
Alternatively, 2 q30:34 2 q30 2 q34 2 q30:34
b
g
2 p30q32 2 p34q36 2 p30:34 1 p32:36 = (0.9)(0.8)(0.3) + (0.5)(0.4)(0.7) – (0.9)(0.8)(0.5)(0.4) [1-(0.7)(0.3)] = 0.216 + 0.140 – 0.144(0.79) = 0.24224 Alternatively, 2
q30:34 3 q30 3 q34 2 q30 2 q34 1 3 p30 1 3 p34 1 2 p30 1 2 p34 1 0.504 1 0.06 1 0.72 1 0.20 0.24224
(see first solution for
MLC‐09‐11
2
p30 ,
2
p34 ,
3
p30 ,
3
p34 )
1
Question #2 Answer: E
1000 Ax 1000 Ax1:10 10 Ax 10 1000 e0.04t e 0.06t (0.06)dt e 0.4 e 0.6 e 0.05t e 0.07t (0.07)dt 0 0 10 1000 0.06 e 0.1t dt e 1 (0.07) e 0.12t dt 0 0 0.10 t 10 0.12 t 1000 0.06 e0.10 e 1 (0.07) e0.12 0 0
1 1 e 1 0.07 1000 0.06 0.12 e 0.10 1000 0.37927 0.21460 593.87
Because this is a timed exam, many candidates will know common results for constant force and constant interest without integration.
For example Ax1:10 10 E x
Ax
e
1 10 Ex
10
With those relationships, the solution becomes 1000 Ax 1000 Ax1:10 10 Ex Ax 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
1000 0.60 1 e 1 0.5833 e 1 593.86
MLC‐09‐11
2
Question #3 Answer: D
1 dt 20 1 100 5 0.07 t e 0 20 7 7 2 1 1 bt vt t px x t dt e0.12t e 0.16t e0.05t dt e0.09t dt 0 20 20 0 1 100 0.09t 5 e 0 9 20 9
E Z bt v t t px x t dt e0.06 t e 0.08 t e 0.05 t 0
0
E Z 2
0
2
5 5 Var Z 0.04535 9 7
Question #4 Answer: C Let ns = nonsmoker and s = smoker k=
q xb kg ns
pxb kg
q xb gk s
p x k
0
.05
0.95
0.10
0.90
1
.10
0.90
0.20
0.80
2
.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
v
1 1.02
ns px
v2
1 1.022 s qx
ns
ns
0.95 0.10 0.1403
v2
0.10
qx 1
1
1.02 2
s px
q x 1 s
0.90 0.20 0.2710
A1x:2 weighted average = (0.75)(0.1403) + (0.25)(0.2710) = 0.1730
MLC‐09‐11
3
s
Question #5 Answer: B
x x1 x 2 x 3 0.0001045 t
px e 0.0001045t
1 APV Benefits e t 1,000,000 t px x dt 0
t
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
27.5 16.6377 457.54
Question #6 Answer: B EPV Benefits
1 1000 A40:20
EPV Premiums a40:20
k E401000vq40k
k 20
k E401000vq40k
k 20
Benefit premiums Equivalence principle 1 1000 A40:20
k E401000vq40k
k 20
a40:20 k E401000vq40 k 20
1 1000 A40: 20
/ a40:20
161.32 0.27414 369.13
14.8166 0.27414 11.1454
5.11 1 and was structured to take While this solution above recognized that 1000 P40:20
advantage of that, it wasn’t necessary, nor would it save much time. Instead, you could do:
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4
EPV Benefits 1000 A40 161.32 EPV Premiums = a40:20 a40:20
20 E40 k E60 1000vq60 k k 0
20 E40 1000 A60
14.8166 0.27414 11.1454 0.27414 369.13 11.7612 101.19 11.7612 101.19 161.32
161.32 101.19 5.11 11.7612
Question #7 Answer: C ln 1.06 0.53 0.5147 0.06 1 A70 1 0.5147 8.5736 a70 d 0.06 /1.06 0.97 a69 1 vp69a70 1 8.5736 8.8457 1.06
A70 A70 i
2 2 a69 2 1.00021 8.8457 0.25739 a69
8.5902
m 1 works well (is closest to the exact m Note that the approximation ax ax 2m answer, only off by less than 0.01). Since m = 2, this estimate becomes 1 8.8457 8.5957 4 Question #8 - Removed
Question #9 - Removed
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5
Question #10 Answer: E d = 0.05 v = 0.95 At issue
49
A40 v k 1 k q40 0.02 v1 ... v50 0.02v 1 v50 / d 0.35076 k 0
and a40 1 A40 / d 1 0.35076 / 0.05 12.9848 so P40
1000 A40 350.76 27.013 a40 12.9848
E 10 L K 40 10 1000 A50
Revised
P40 a50
549.18 27.013 9.0164 305.62
Revised
where Revised
A50
and
24
v k 1 k q50
Revised
k 0 Revised a50
1 A
Revised 50
0.04 v1 ... v 25 0.04v 1 v 25 / d 0.54918
/ d 1 0.54918 / 0.05 9.0164
Question #11 Answer: E Let NS denote non-smokers and S denote smokers. The shortest solution is based on the conditional variance formula
Var X E Var X Y Var E X Y
Let Y = 1 if smoker; Y = 0 if non-smoker 1 AxS E aT Y 1 axS
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 0 Prob Y=0 E E aT 1 Prob Y=1 7.14 0.70 5.56 0.30 6.67
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6
E E aT Y
2
2 2 7.14 0.70 5.56 0.30 44.96
44.96 6.672 0.47 E Var aT Y 8.503 0.70 8.818 0.30
Var E aT Y
8.60
Var aT 8.60 0.47 9.07 Alternatively, here is a solution based on
Var(Y ) E Y 2 E Y , a formula for the variance of any random variable. This can be 2
transformed into E Y 2 Var Y E Y which we will use in its conditional form
E aT
2
2
NS Var aT NS E aT NS
Var aT E aT
2
E aT
2
2
E aT E aT S Prob S E aT NS Prob NS 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
E aT 2 S Prob S E aT 2 NS Prob NS
2 0.30 Var aT S E aT S
0.70 Var aT NS E aT NS
2
2 2 0.30 8.818 5.56 0.70 8.503 7.14 11.919 + 41.638 = 53.557
Var aT 53.557 6.67 9.1 2
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7
Alternatively, here is a solution based on aT
1 vT
1 v Var aT Var T
vT Var since Var X constant Var X
since Var constant X constant
Var vT
2
2
Ax Ax
2
2
Var X
2
which is Bowers formula 5.2.9
This could be transformed into 2Ax 2 Var aT Ax2 , which we will use to get 2
Ax NS and 2AxS .
Ax E v 2T
2
E v 2T NS Prob NS E v 2T S Prob S 2 2 Var aT NS Ax NS Prob NS 2 2Var aT S AxS Prob S 0.01 8.503 0.2862 0.70 0.01 8.818 0.4442 0.30 0.16683 0.70 0.28532 0.30
0.20238 Ax E vT E vT NS Prob NS E vT S Prob S 0.286 0.70 0.444 0.30 0.3334
MLC‐09‐11
8
Ax Ax
2
Var aT
2
2
0.20238 0.33342 9.12 0.01 Question #12 - Removed
Question #13 Answer: D Let NS denote non-smokers, S denote smokers.
Prob T t Prob T t NS Prob NS P rob T t S P rob S
1 e0.1t 0.7 1 e0.2t 0.3 1 0.7e0.1t 0.3 e0.2t S0 (t ) 0.3e0.2 t 0.7e0.1t
Want tˆ such that 0.75 1 S0 tˆ or 0.25 S0 tˆ
ˆ
ˆ
ˆ 2
0.25 0.3e2t 0.7e 0.1t 0.3 e 0.1t
0.7e 0.1t
ˆ
Substitute: let x e0.1t 0.3 x 2 0.7 x 0.25 0
ˆ
This is quadratic, so x
0.7 0.49 0.3 0.25 4 2 0.3
x 0.3147
e 0.1t 0.3147 ˆ
MLC‐09‐11
so tˆ 11.56
9
Question #14 Answer: A
At a constant force of mortality, the benefit premium equals the force of mortality and so 0.03 . 0.03 2 Ax 0.20 2 2 0.03 0.06
Var 0 L where A
2
Ax Ax
a 2
2
0.20 13
0.06 0.09
0.03 1 0.09 3
2
2
a
0.20 1 1 0.09
Question #15 - Removed
Question #16 Answer: A 1000 P40
A40 161.32 10.89 a40 14.8166
a 11.1454 1000 20V40 1000 1 60 1000 1 247.78 14.8166 a40 20V 5000 P40 (1 i) 5000q60 21V P60
247.78 (5)(10.89) 1.06 5000 0.01376 255 1 0.01376
[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.
1000 P40 10.89 as above 1 1000 A40 4000 20 E40 A60:5 1000 P40 5000 P40 20 E40 a60:5
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10
20 E40 5 E60
a65
1 where A60:5 A60 5 E60 A65 0.06674
a40:20 a40 20 E40 a60 11.7612 a60:5 a60 5 E60 a65 4.3407
1000 0.16132 4000 0.27414 0.06674 10.89 11.7612 5 10.89 0.27414 4.3407 0.27414 0.68756 9.8969 161.32 73.18 128.08 64.79 1.86544 22.32
Having struggled to solve for , you could calculate above) calculate 21V recursively. 20V
20 V
prospectively then (as
1 4000 A60:5 1000 A60 5000 P40 a60:5 5 E60 a65
4000 0.06674 369.13 5000 0.01089 4.3407 22.32 0.68756 9.8969 247.86 (minor rounding difference from 1000 20V40 ) Or we can continue to 21V
21V
prospectively
1 5000 A61:4 1000 4 E61 A65 5000 P40 a61:4 4 E61 a65
where
4 E61
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
a61:4
0.05779 a61 4 E61 a65 10.9041 0.73898 9.8969 3.5905
21V
5000 0.05779 1000 0.73898 0.43980 5 10.89 3.5905 22.32 0.73898 9.8969 255
Finally. A moral victory. Under exam conditions since prospective benefit reserves must equal retrospective benefit reserves, calculate whichever is simpler.
MLC‐09‐11
11
Question #17 Answer: C Var Z 2 A41 A41
2
A41 A40 0.00822 A41 (vq40 vp40 A41 )
A41 (0.0028 /1.05 0.9972 /1.05 A41 )
A41 0.21650 2
A41 2 A40 0.00433 2 A41 v 2 q40 v 2 p40 2 A41
2 A41 (0.0028 /1.052 0.9972 /1.052 2
A41 0.07193
Var Z 0.07193 0.216502 0.02544
Question #18 - Removed
Question #19 - Removed
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12
2
A41 )
Question #20 Answer: D
x x1t x 2t 0.2 xt x 2t x 2t 0.8 xt '1 qx k
3
1
'1 px
1
1 e
0.2 k t 2 dt 0
1 e
0.2
k 3
0.04
ln 1 0.04 / 0.2 0.2041
k 0.6123 2 2 qx
2 2 px x dt 0.8
2
0 t
0
0.8 2 qx 0.8 1 2 px
px x t dt
x t d t 2
2
t
px e
0
2
e
k t2 d t 0
8 k e 3 8 0.6123 3 e
0.19538 2 2 qx
0.8 1 0.19538 0.644
MLC‐09‐11
13
Question #21 Answer: 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
0 0.18 0.864 4.536
2.124
5.580
E min( K ,3) 2.124
E min K ,3
2
5.580
Var min( K ,3) 5.580 2.1242 1.07
Note that E min( K ,3) is the temporary curtate life expectancy, ex:3 if the life is age x.
Question #22 Answer: B 0.1 60 0.08 60 e e S0 (60) 2 0.005354 0.1 61 0.08 61 e e S0 (60) 2 0.00492 0.00492 q60 1 0.081 0.005354
MLC‐09‐11
14
2
Question #23 Answer: D Let q64 for Michel equal the standard q64 plus c. We need to solve for c. Recursion formula for a standard insurance: 20V45
19V45 P45 1.03 q64 1 20V45
Recursion formula for Michel’s insurance 20V45
19V45 P45 0.01 1.03 q64 c (1 20V45 )
The values of
19 V45
and
20V45
are the same in the two equations because we are
told Michel’s benefit reserves are the same as for a standard insurance. Subtract the second equation from the first to get:
0 1.03 (0.01) c(1 20V45 ) c
(1.03) 0.01 1 20V45
0.0103 1 0.427 0.018
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15
Question #24 Answer: 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 aK 1 1 v K 1 d d 1 Ax E L Ax ax Ax d 0.75095 0.24905 0.025 0.082618 0.056604 Var L 1 d
2
2
Ax
Ax2
2
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
minimum number needed = 27
MLC‐09‐11
0.025 2 1 0.09476 0.24905 0.068034 0.056604
16
Question #25 Answer: D
1 v K 1 for K 0,1, 2,... d Z 2 Bv K 1 for K 0,1, 2,...
Z1 12,000
Annuity benefit: Death benefit:
1 v K 1 Bv K 1 d 12,000 12,000 K 1 B v d d
Z Z1 Z 2 12,000
New benefit:
2
12,000 K 1 Var( Z ) B Var v d 12,000 150,000 . Var Z 0 if B 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 Answer: A
x t: y t x t y t 0.08 0.04 0.12 Ax x t / x t 0.5714
Ay y t / y t 0.4
Axy x t: y t / x t: y t 0.6667
axy 1/ x t: y t 5.556 Axy Ax Ay Axy 0.5714 0.4 0.6667 0.3047 Premium = 0.304762/5.556 = 0.0549
MLC‐09‐11
17
Question #27 Answer: B
P40 A40 / a40 0.16132 /14.8166 0.0108878 P42 A42 / a42 0.17636 /14.5510 0.0121201 a45 a45 1 13.1121 E 3 L K 42 3 1000 A45 1000 P40 1000 P42 a45 201.20 10.89 12.12 13.1121 31.39 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 1000P40 instead of 1000 P42 .
Question #28 Answer: E
E min T ,40 40 0.005 40 32 2
32 t f t dt 40 f t dt w
40
0
40
t f t dt t f t dt 40 .6 w
w
0
40
86 tf t dt w
40
tf t dt 54 t 40 f t dt 54 40 .6 50 e w
40
w
40
40
MLC‐09‐11
s 40
.6
18
Question #29 Answer: B d 0.05 v 0.95
Step 1 Determine px from Kevin’s work:
608 350vpx 1000vqx 1000v 2 px px 1 qx 1
608 350 0.95 px 1000 0.95 1 px 1000 0.9025 px 1 608 332.5 px 950 1 px 902.5 px px 342 / 380 0.9
Step 2 Calculate 1000 Px:2 , as Kira did:
608 350 0.95 0.9 1000 Px:2 1 0.95 0.9 299.25 608 489.08 1000 Px:2 1.855
The first line of Kira’s solution is that the expected present value of Kevin’s benefit premiums is equal to the expected present value of Kira’s, since each must equal the expected present value of benefits. The expected present value of benefits would also have been easy to calculate as 1000 0.95 0.1 1000 0.952 0.9 907.25
Question #30 Answer: E Because no premiums are paid after year 10 for (x), 11Vx Ax 11 One of the recursive reserve formulas is 10V
h 1V
hV h 1 i bh1qx h
32,535 2,078 1.05 100,000 0.011 35,635.642
px h
0.989 35,635.642 0 1.05 100,000 0.012 36,657.31 A 11V x 11 0.988
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Question #31 Answer: B
t The survival function is S0 (t ) 1 : Then, x t ex and t px 1 2 x 105 45 30 e45 2 105 65 20 e65 2
e 45:65
40 0
t
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 .
e 45:65 e 45 e 65 e 45:65 30 20 1556 . 34
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. Question #32 Answer: E
4 S0 (4) / S0 (4)
d
e4 / 100
i
1 e4 / 100
e4 / 100 1 e4 / 100
e4 1.202553 100 e4
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20
Question # 33 Answer: A
L ln px big OP qb g LM ln e OP q xbi g q xb g M M ln p b g P x M ln e-b g P bi g
N
q xb g
Q
x
N
Q
bi g 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
q xb 2 g
b0.7769gb g b0.5gb0.7769g 2
b g
15 .
0.2590 Question # 34 Answer: D 22
A 60 v 3
B
B
q 60 2
B
pay at end of year 3
live then die 2 years in year 3
v4
3p 60
pay at end of year 4
2 p 60
1
1.03
3
live 3 years
q60 3
then die in year 4
1 0.09 1 0.11 0.13
1
1.03
4
1 0.09 1 0.11 1 0.13 0.15
0.19 MLC‐09‐11
21
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
b
gb g
a x 4.219 0.705 12.5 13.03 Question #36 Answer: D 1 2 px p ' x p ' x 0.8 0.7 0.56
1
qx
ln p ' 1 x q since UDD in double decrement table ln p x x ln 0.8 0.44 ln 0.56 0.1693
0.3 q x 0.1 1
0.3qx 1
0.053 1 0.1qx
To elaborate on the last step:
Number dying from cause 1 between x 0.1 and x 0.4 1 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‐09‐11
22
Question #37 Answer: E The benefit premium is oL
1 1 0.04 0.04333 ax 12
v T (0.04333 0.0066)aT 0.02 0.003aT 1 vT v 0.04693 0.02 0.04693 0.04693 v T 1 0.02 T
2
0.04693 Var o L Var v T 1 0.1(4.7230) 0.4723
Question #38 - Removed
Question # 39 - Removed
Question # 40 Answer: D Use Mod to designate values unique to this insured.
b
g
b
g b gb g 1000b0.36933 / 111418 . . g 3315
a60 1 A60 / d 1 0.36933 / 0.06 / 106 . 111418 .
1000 P60 1000 A60 / a60
d
i
Mod Mod p60 A60Mod v q60 A61
d
i
b
gb
g
1 . 01376 0.8624 0.383 0.44141 . 106
b
g
a Mod 1 A60Mod / d 1 0.44141 / 0.06 / 106 . 9.8684
d
Mod E 0 LMod 1000 A60Mod P60a60
i b
1000 0.44141 0.03315 9.8684
g
114.27
MLC‐09‐11
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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 a40:10
A A60 40 a40
10 E40 10 E50
P50Mod
a50:10 10 E50
1 A40 :20 20 E40
016132 . 7.70 7.57 0.06 P50Mod 14.8166 0.53667 0.51081 0.51081 0.27414
0.36913
b
gb
g
0.36913 0.30582 14.8196 P50Mod 0.21887 1000 P50Mod 19.04 Alternatively, you could equate the retrospective and prospective reserves at age 50. Your equation would be: A50
P50Mod
a50:10
1 A40 a40:10 A40:10 a40 10 E40 10 E40
1 where A40 A40 10 E40 A50 :10
b
gb
g
016132 . 0.53667 0.24905 0.02766
d
. 7.70 0.02766 ib g 14016132 .8166 0.53667 0.53667 . 1000 014437 b gb g 19.07
0.24905 P50Mod 7.57 1000 P50Mod
MLC‐09‐11
7.57
24
Alternatively, you could set the expected present value of benefits at age 40 to the expected present value of benefit premiums. The change at age 50 did not change the benefits, only the pattern of paying for them. A40 P40 a40:10 P50Mod
016132 .
10 E40
a50:10
. 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 xb1g 0.45 400 180 q x b 2 g
d xb 2 g
400
0.488 l b g d b1g 1000 180 x
x
pxb 2 g 1 0.488 0.512
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‐09‐11
25
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
1
q b 3g 1
5,
q1b 2 g
3,
q b3g 1
10 ,
0
1
1
b
9,
gb
q2b 2 g
1
8
q b 3g 1 2
gb
4
g
This gives p0b g 1 1 / 4 1 1 / 5 1 1 / 10 0.54
p1b g b1 1 / 5gb1 1 / 3gb1 1 / 9g 0.474 p2b g b1 1 / 3gb1 1 / 8gb1 1 / 4g 0.438 So 1b0 g 200, 11b g 200 b0.54g 108 , and 1b2 g 108 b0.474g 512 . q b1g log pb1g / log pb g q b g 2
2
2
2
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
g
Question #44 - Removed
MLC‐09‐11
26
Question #45 Answer: E
ex
For the given life table function:
x 2
1 k qx x x 1
Ax
k b
1 x 1 k 1 v x k b
a x
Ax ax
v k 1 k qx
x 1 Ax d
e50 25 100 for typical annuitants
e y 15 y Assumed age = 70 A70
a30
0.45883 30 a70 9.5607 500000 b a70 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 which should give 0.52604 a30:40:10 a30:40 10 E30:40 a3010:4010
b g b gb g b13.2068g b0.52604gb114784 . g a30:40 1 0.52604 a40:50 1 7.1687
MLC‐09‐11
27
10
p30
10
p40 v10 , all of
Question #47 Answer: A Equivalence Principle, where is annual benefit premium, gives
b g
d
1000 A35 IA
d
35
i
ax
1000 A35 1000 0.42898 (1199143 . . ) 616761 a35 IA 35
b gi
428.98 582382 . 73.66
We obtained a35 from
a35
1 A35 1 0.42898 1199143 . d 0.047619
Question #48 - Removed
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 . 0.05 019 . . 0.05 xy 014 xy 014 P
Axy a xy
Ax Ay Axy
MLC‐09‐11
a xy
b
g
2 0.5833 0.7368 0.0817 5.2632
28
Question #50 Answer: E
20V P20 1 i q40 1 21V
V
21
b0.49 0.01gb1 ig 0.022b1 0.545g 0.545 b1 ig b0.545gb1 0.022g 0.022 0.50
111 .
21V P20 1 i q41 1 22V 22V . g q b1 0.605g 0.605 b0.545.01gb111 41
q41
0.61605 0.605 0.395
0.028 Question #51 Answer: E
1000 P60 1000 A60 / a60
b
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 a61 60
60
61
60
61
Question #52 - Removed
MLC‐09‐11
29
Question #53 Answer: E g ln(0.96) 0.04082
x02t: y t 0.04082 0.01 0.03082 h ln(0.97) 0.03046
x01t: y t 0.03046 0.01 0.02046 5
5
pxy 5 pxy00 exp x01t: y t x02t: y t x03t: y t dt e 5(0.06128) 0.736 0
Question #54 - Removed
Question #55 Answer: B
lx x 105 x t p45 l45t / 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,
c
h
c
Prob K 19 13.66 Prob K 32.66
h
b g ProbbT 33g
Prob K 33 since K is an integer
33p45
l78 27 l45 60
0.450
MLC‐09‐11
30
Question #56 Answer: C
Ax
2
Ax
d IAi
z
E 0s x
zd
0
0.4
x
0.25 0.04
2
z
Ax ds
0 s
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
x
z z
0
0
bg b0.04g e
t t px x t e t dt t e 0.04 t
0.04
z
0
0.06 t
dt
t e 0.1t dt
(integration by parts, not shown) t 1 0.04 e 0.1 t 0 01 . 0.01 0.04 4 0.01
FG H
MLC‐09‐11
IJ K
31
Question #57 Answer: E Subscripts A and B here 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 10
0
IJ K
t t2 1 dt t 7 14
7
0
7
e AB
2 10
1
t 7
7
0
49
0
49 35 . 14
IJ FG 1 t IJ dt z FG 1 t t t IJ dt K H 10K H 10 7 70K 2
7
0
t2 t2 t3 t 20 14 210
7
0
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 . Question #58 Answer: A
xt 0.100 0.004 0.104 t
pxb g e 0.104 t
Expected present value (EPV) = EPV for cause 1 + EPV for cause 2.
z
5
b
z
g
b
5
g
2000 e 0.04 t e 0.104 t 0100 . dt 500,000 e 0.04 t e 0.104 t 0.400 dt 0
0
2000 0.10 500, 000 0.004 e 0.144t dt 5
0
MLC‐09‐11
32
2200 1 e 0.144 5 7841 0.144
Question #59 Answer: A
R 1 px q x S 1 px e
k
1
1
1
1
e 0
xt dt k dt xt k dt 0 since e 0 e 0
e 0
1
xt dt k dt
So S 0.75R 1 px e k 0.75q x
1 0.75q x px px 1 qx ek 1 0.75q x 1 0.75q x e k
k ln
LM 1 q OP N1 0.75q Q x
x
MLC‐09‐11
33
Question #60 Answer: C A60 0.36913 2
d 0.05660
A60 0.17741
and
2
2 A60 A60 0.202862
Expected Loss on one policy is E L 100,000 A60 d d 2 Variance on one policy is Var L 100,000 2 A60 A60 d On the 10000 lives, E S 10,000 E L and Var S 10,000 Var L 2
The is such that 0 E S / Var S 2.326 since 2.326 0.99 10,000 100,000 A60 d d 2.326 2 100 100,000 2 A60 A60 d 100 100,000 0.36913 d d 2.326 100,000 0.202862 d
0.63087
d
36913
100,000 0.63087
0.004719
d
36913 471.9 0.004719
d 36913 471.9 d 0.63087 0.004719 59706 59706 d 3379
MLC‐09‐11
d
34
Question #61 Answer: 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 *
1000 1000 1.052 q75 1.05q76 q77
1.05 1.05 3
2
1.05
1000 x 1 1.052 0.05169 1.05 0.05647 0.06168
3.310125 1000 1.17796 355.87 3.310125
* This equation is an algebraic manipulation of the three equations in three unknowns 1V , 2V , . One method – usually effective in problems where benefit = stated amount plus reserve, is to multiply the 1V equation by 1.052 , the 2V equation by 1.05, and add those two to the 3V equation: in the result, you can cancel out the 1V , and 2V terms. Or you can substitute the 1V equation into the 2 V equation, giving 2 V in terms of , and then substitute that into the 3V equation.
Question #62 Answer: D
A281:2
z
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
a28:2
2 t e 1 72 dt 0
500,000 A281:2 6643 a28:2
= 287 MLC‐09‐11
35
Question #63 Answer: D Let Ax and a x be calculated with x t and 0.06 Let Ax* and a x * be the corresponding values with x t increased by 0.03 and decreased by 0.03
ax
1 Ax
0.4 6.667 0.06
ax ax *
xs 0.03 ds 0.03t * 0 e dt Proof: ax 0 e t
0
e
t
xs ds 0.03t 0.03t 0
e
e
t
e
0
x s ds
0
e 0.06 t dt
ax
Ax* 1 0.03 a x * 1 0.03 a x
b gb
1 0.03 6.667
g
0.8
MLC‐09‐11
36
dt
Question #64 Answer: A bulb ages Year
0
1
0 1 2 3
10000 0 1000 9000 100+2700 900 280+270+3150
2
3
0 0 6300
0 0 0
The diagonals represent bulbs that don’t burn out. E.g., of the initial 10,000, (10,000) (1-0.1) = 9000 reach year 1. (9000) (1-0.3) = 6300 of those reach year 2. Replacement bulbs are new, so they start at age 0. At the end of year 1, that’s (10,000) (0.1) = 1000 At the end of 2, it’s (9000) (0.3) + (1000) (0.1) = 2700 + 100 At the end of 3, it’s (2800) (0.1) + (900) (0.3) + (6300) (0.5) = 3700
1000 2800 3700 105 . 105 . 2 105 . 3 6688
Expected present value
Question #65 Answer: E
e25:25
z z
15
0 t
p25dt 15 p25
15 .04 t
0
e
d
z
10 0t
FG z IJ e H Kz i e LMN.051 d1 e
dt e
15
0
.04 ds
1 .60 1 e.60 .04 . 112797 4.3187 15.60
MLC‐09‐11
p40 dt
10 .05t
0
dt
.50
iOPQ
37
# replaced 1000 2800 3700
Question #66 Answer: C 5
p 60 1
e1 q je1 q jb1 q gb1 q gb1 q g b0.89gb0.87gb0.85gb0.84gb0.83g 60 1
60 2
63
64
65
0.4589 Question # 67 Answer: E
1 0.08 0.04
12.50 a x Ax
0.5
1 2 Ax 2 3
e j
Var aT
2
Ax Ax2
2
1
1
3 4 52.083 0.0016 S.D. 52.083 7.217
MLC‐09‐11
38
Question # 68 Answer: D v 0.90 d 010 . Ax 1 dax 1 010 . 5 0.5
b gb g
Benefit premium
5000 Ax 5000vqx ax
b5000gb0.5g 5000b0.90gb0.05g 455 5
10th benefit reserve for fully discrete whole life 1 0.2 1
ax 10 ax
ax 10 ax 10 4 5
b gb g a b5000gb0.6g b455gb4g 1180
Ax 10 1 dax 10 1 010 . 4 0.6 10V
5000 Ax 10
x 10
Question #69 Answer: 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
d i
b g
0.3729
b g
b g
2
4
E Z2 v 2qx v 4 px qx 1 0.95 0.25 0.95 0.75 0.2 0.3478
b g d i c b gh
Var Z E Z2 E Z
MLC‐09‐11
2
b
g
2
0.3478 0.3729 0.21
39
Question #70 Answer: D Expected present value (EPV) 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
EPV of future premiums 1 1 0.005 0.04 /1.06 50 1.9009 50
95.05 E 1 L K55 1 111.77 95.05 16.72 Question #71 - Removed
Question #72 Answer: 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
z
t t
e e dt
5
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
2
2
2
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‐09‐11
40
Question #73 Answer: D Prob{only 1 survives} = 1-Prob{both survive}-Prob{neither survives}
b
ge
1 3p50 3p 50 13p50 1 3p 50
b
gb
gb
gb
j
gb
gb
g b
gb
1 0.9713 0.9698 0.9682 0.9849 0.9819 0.9682 1 0.912012 1 0.93632 0.912012
g
0.936320
0.140461 Question # 74 - Removed
Question #75 - Removed Question # 76 Answer: 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. EPV Prems P EPV Benefits 10q70 v 10 p70 q71v 2 Pp70 p71v 2
P
b10gb0.03318g b10gb1 0.03318gb0.03626g Pb1 0.03318gb1 0.03626g
. 108 . 2 108 0.3072 0.3006 0.7988 P 0.6078 P 3.02 0.2012
(EPV above means Expected Present Value).
MLC‐09‐11
41
. 2 108
Question #77 Answer: 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 nV And plug in to get
b
gb
g
0.090 Px1:n 0.00864 0.563 Px1:n 0.0851
Another approach is to think in terms of retrospective reserves. Here is one such solution:
V Px Px1:n sx:n
n
Px Px1:n
aE
x:n
n
P P P P Px Px1:n
x
ax:n 1
x:n
ax:n
1
x
x:n
1
x:n
e
j
0.563 0.090 Px1:n / 0.00864
b
gb
g
Px1:n 0.090 0.00864 0.563 0.0851
MLC‐09‐11
42
Question #78 Answer: A
b g
ln 1.05 0.04879
Ax
x t
0
x
0
px x t e t dt
1 t e dt for the given mortality function x
1 a x x
From here, many formulas for the reserve could be used. One approach is: Since
A50
a50 50
1 A50 18.71 0.3742 so a50 12.83 50
1 A40 19.40 0.3233 so a40 13.87 60 60 0.3233 P A40 0.02331 13.87 reserve A50 P A40 a50 0.3742 0.0233112.83 0.0751. A40
a60
Question #79 Answer: D
Ax E vTx E vTx NS Prob NS E vTx S Prob S 0.03 0.6 0.70 0.30 0.03 0.08 0.06 0.08 0.3195
Similarly, 2 Ax
F H
Var aT
I b gK x
MLC‐09‐11
2
FG 0.03 IJ 0.70 FG 0.06 IJ 0.30 01923 . . H 0.03 0.16K H 0.06 0.16K
Ax Ax2
2
0.31952 01923 . 141 .. 0.082
43
Question #80 Answer: B 2
q80:84 2 q80 2 q84 2 q80:84
0.5 0.4 1 0.6 0.2 0.15 1 0.1
= 0.10136 Using new p82 value of 0.3 0.5 0.4 1 0.3 0.2 0.15 1 0.1
= 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
0.20 0.03 0.20 0.03 0.224 3 p80 3 p84 3 p80 3 p84
0.12 0.003 0.12 0.003 0.12264 2 q 80:84 2 p80:84 3 p80:84 0.224 0.12264 0.10136
Revised 3 p80 0.20 0.30 0.06 3
p 80:84 0.06 0.003 0.06 0.003 0.06282 2 q 80:84 0.224 0.06282 0.16118
change 0.16118 0.10136 0.06
Question #81 - Removed
MLC‐09‐11
44
Question #82 Answer: 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 Answer: C Only decrement 1 operates before t = 0.7
b g b0.7g qb1g b0.7gb010 . g 0.07
1 0.7 q40
40
Probability of reaching t = 0.7
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
g
b2g 0.93 0125 q40 . 011625 .
MLC‐09‐11
45
Question #84 Answer: C
e
j
1 2 p80v 2 1000 A80
FG H
1
0.83910 1.062
vq80 v 3 2 p80q82 2 2
IJ 665.75 FG 0.08030 0.83910 0.09561IJ K . g 2b1.06g H 2b106 K 3
b g b b167524 . g 665.75
g
174680 . 665.75 0.07156 397.41
Where 2 p80
3,284 ,542 0.83910 3,914 ,365
b
gb
g
Or 2 p80 1 0.08030 1 0.08764 0.83910
Question #85 Answer: E
At issue, expected present value (EPV) of benefits
bt v t t p65 0
65 t
dt
1000 e0.04t e 0.04t t p65 65 t dt
0
1000
0
t
p65 65 t dt 1000 q65 1000
EPV of premiums a65
FG 1 IJ 16.667 H 0.04 0.02K
Benefit premium 1000 / 16.667 60 2V
z z
0
0
b g
b2u v u u p67 65 2 u du a67
b g b gb
g
1000 e0.04b 2uge 0.04u u p67 65 2 u du 60 16.667
1000e0.08
z
0 u
b g
p67 65 2 u du 1000
1083.29 q67 1000 1083.29 1000 83.29
MLC‐09‐11
46
Question #86 Answer: B (1)
a x:20 ax:20 1 20 Ex
(2)
ax:20
(3)
Ax:20
(4)
Ax A1x:20 20 Ex Ax 20
1 Ax:20 d A1x:20
Ax:201
b gb g
0.28 A1x:20 0.25 0.40 A1x:20 018 .
Now plug into (3):
Ax:20 018 . 0.25 0.43 ax:20
Now plug into (2):
Now plug into (1):
b
1 0.43 . 1197 0.05 / 105 .
g
a x:20 1197 . 1 0.25 1122 .
Question #87 - Removed
Question #88 Answer: B
ex px pxex 1 px
ex 8.83 0.95048 1 ex 1 9.29
ax 1 vpx v 2 2 px .... a
x:2
1 v v 2 2 p x ...
ax:2 ax vqx 5.6459 5.60 0.0459 v 1 0.95048 0.0459 v 0.9269 1 i 1 0.0789 v
MLC‐09‐11
47
Question #89 - Removed
Question #90 – Removed
Question #91 Answer: E
1 1 75 60 15 1 1 3 1 60F 85 60 15 5 25
60M
t 10 t 1 25
t
M p65 1
t
F p60
Let x denote the male and y denote the female. e x 5 mean for uniform distribution over 0,10 e y 12.5 mean for uniform distribution over 0,25 10 t t e xy 1 1 dt 0 10 25 10 7 t2 1 t dt 0 50 250
t
7 2 t3 t 100 750
10 0
10
7 1000 100 100 750
10 7
exy ex e y exy 5
MLC‐09‐11
4 13 3 3
25 13 30 75 26 1317 . 2 3 6
48
Question #92 Answer: 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
x
2
2 x
x
2
2
2
1 5
Question #93 Answer: A Let be the benefit premium Let kV denote the benefit reserve at the end of year k. For any n, nV 1 i q25 n n 1V p25 n n 1V
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
By induction (proof omitted) sn n V For n 35, n V a60 (expected present value of future benefits; there are no future premiums) a60 s35
a60 s35
For n 20,
MLC‐09‐11
20V
a s20 60 s35
s 20 49
Alternatively, as above nV 1 i n1V Write those equations, for n 0 to n 34 0 : 0V 1 i 1V 1: 1V 1 i 2V
2 : 2V 1 i 3V
34 : 34V 1 i 35V Multiply equation k by 1 i
34 k
and sum the results:
0V 1 i 35 1V 1 i 34 2V 1 i 33 34V 1 i 34 33 32 1V 1 i 2V 1 i 3V 1 i 34 V 1 i 35V For k 1, 2, , 34, the k V 1 i 0V
35 k
terms in both sides cancel, leaving
1 i 35 1 i 35 1 i 34 1 i 35V
Since 0V 0 s35 35V a60 (see above for remainder of solution)
MLC‐09‐11
50
Question #94 Answer: B
x t: y t
t qy t t qx
px x t t q x t p y y t
t p y t px t q y t px t p y
For (x) = (y) = (50)
50:50 10.5
where 10.5
p50
10.5 q50 10
2
l60 l61 12 8,188,074 8,075, 403 0.90848 l50
8,950,901
1 10.5 p50 0.09152
p50
10.5
1
10.5 q50 10 p50 q60 2 0.09152 0.91478 0.01376 2 0.0023 2 10.5 q50 10.5 p50 2 10.5 p50 0.09152 0.90848 2 0.908482
8,188,074 0.91478 8,950,901
p50 50 10.5 10 p50 q60
since UDD
Alternatively, 10t p50 10 p50 t p60 10t p50:50 10 p50
2
t p60
2
10t p 50:50 2 10 p50 t p60 10 p50
2
t p60
2
2 10 p50 1 tq60 10 p50 1 tq60 since UDD 2
2
Derivative 2 10 p50 q60 2 10 p50 1 tq60 q60 Derivative at 10 t 10.5 is 2
2 0.91478 0.01376 0.91478 1 0.5 0.01376 0.01376 0.0023 2
10.5
p 50:50 2 10.5 p50 10.5 p50
2
2 0.90848 0.90848
2
0.99162
(for any sort of lifetime)
MLC‐09‐11
dp dt 0.0023 0.0023 p 0.99162
51
Question #95 Answer: D
xt x1t x 2t 0.01 2.29 2.30
P P vt t px x 2t dt 50, 000 vt t p x x1t dt 50, 000 vt t px xt dt 2
2
0 2 0.1t 2.3t
0
P P e
e
0
2
2 0.1t 2.3t
2.29dt 50, 000 e 0
e
2
2 2.4 2 2.4 2 2.4 1 e 1 e e P 1 2.29 2.3 50000 0.01 2.4 2.4 2.4 P = 11,194
Question #96 Answer: B ex px 2 px 3 px ... 1105 .
b g
Annuity v 3 3 p x 1000 v 4 4 p x 1000 104 . ...
. g 1000b104
k 3 k
v
k px
k 3
1000v 3
k px
k 3
F 1 IJ 1000v be 0.99 0.98g 1000G H 104 . K 3
x
3
9.08 8072
Let = benefit premium.
e
j
1 0.99v 0.98v 2 8072 2.8580 8072 2824
MLC‐09‐11
0.01dt 50, 000 e 0.1t e 2.3t 2.3dt
52
Question #97 Answer: B
b g
b ge
1 a30:10 1000 A30 P IA 30 10 :10
10
A30
j
1000 A30
b g
a30:10 IA
1 3010 :
1010 A30
b
g b
1000 0102 . 7.747 0.078 10 0.088
g
102 6.789 15.024
Question #98 Answer: E
For the general survival function S0 (t ) 1
FG1 t IJ dt H 30K L t OP Mt N 2b 30g Q
e 30
z
t
,0t ,
30
0
2
30
0
30 2
100 30 35 2
Prior to medical breakthrough
100 e30
After medical breakthrough
e 30 e 30 4 39
so
e30 39
30 2
108
Question #99 Answer: A
L 100, 000v 2.5 4000a3
@5%
= 77,079 MLC‐09‐11
53
Question #100 Answer: D
accid 0.001 total 0.01 other 0.01 0.001 0.009
Expected present value 500,000 e 0.05t e0.01t 0.009 dt 0
10 50,000 e0.04t e0.05t e 0.01t 0.001 dt 0
0.009 0.001 500, 000 100, 000 0.06 0.02 Question #101 Removed
Question #102 Answer: D
1000 20V 1000 Ax 20 ax 20
1000 19V 20 Px 1.06 qx 19 1000 px 19
342.03 13.72 1.06 0.01254 1000 369.18 0.98746
1 0.36918 11.1445 0.06 /1.06
so 1000 Px 20 1000
MLC‐09‐11
Ax 20 369.18 33.1 ax 20 11.1445
54
Question #103 Answer: B k
1
k
x t dt 2 xt dt e 0 e 0
k px
x1t dt e 0 k
2
k px where k px is from Illustrative Life Table, since follows I.L.T. 2
10
6,616,155 0.80802 8,188,074 6,396,609 0.78121 8,188,074
p60
11 p60
10
1
b g p b g p b g q60 10 60 11 60
b
10
p60
g b 2
11 p60
g
2
from I.L.T.
0.80802 2 0.781212 0.0426
Question #104 Answer: C
Ps
1 d , where s can stand for any of the statuses under consideration. a s
as
1 Ps d
1 6.25 . 0.06 01 1 axy 8.333 0.06 0.06 ax ay
axy axy ax ay
axy 6.25 6.25 8.333 4.167 Pxy
1 . 0.06 018 4.167
MLC‐09‐11
55
Question #105 Answer: A
z
b g g j 48
d 0b g 1000 e b 0.04 gt 0.04 dt 1
0
1000 1 e b 0.04
e
e b 0.04 g 0.952
b
0.04 ln 0.952
g
0.049 0.009
z
b
g
d 3b1g 1000 e 0.049 t 0.009 dt 4
3
1000
0.009 b 0.049 gb 3g b 0.049 gb 4 g e e 7.6 0.049
e
j
Question #106 Answer: B This is a graph of lx x .
b
g
x would be increasing in the interval 80,100 . The graphs of lx px , lx and lx2 would be decreasing everywhere.
Question #107 Answer: B Expected value v15 15 px
Variance v30 15 px 15 qx
v30 15 px 15 qx 0.065 v15 15 px v15 15qx 0.065 15 qx 0.3157 Since is constant 15 q x
px
15
1 px
15
0.6843
px 0.975 qx 0.025 MLC‐09‐11
56
Question #108 Answer: E
1 2
11V
A
11V
B
1 2
10V
11V
A
10V
A
B
0
p
1 i
B
qx 10 1000 px 10
x 10
p
1 i
11V B
x 10
10V
qx 10 1000 px 10
10V B B
A
101.35 8.36
p
1 i x 10
1.06 1 0.004
98.97
Question #109 Answer: A 2
EPV (x’s benefits) v k 1bk 1 k px q x k k 0
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
Question #110 Answer: E
denotes benefit premium V EPV future benefits - EPV future premiums 1 0.6 0.326 108 . . q65 10 10V 108 11V p65
19
b
gb g b gb g
. g b010 . gb10g b5.0 0.326gb108 1 010 .
=5.28 MLC‐09‐11
57
Question #111 Answer: A Expected present value Benefits
0.8 0.110,000 0.8 0.9 0.097 9,000 1.062
1, 239.75
1.063
0.8 0.8 0.9 1, 239.75 P 1 1.062 1.06 P 2.3955 P 517.53 518
Question #112 Answer: A
1180 70a30 50a40 20a30:40
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
Question #113 Answer: B
z
bg
a a f t dt t
z
1 e 0.05t
1 0.05
0.05
zb
te
t
1 te t dt 2
bg
i
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 MN H 105 Q
t 1 1 t 1 et e 1.05t 2 . 0.05 105 . 105 2
20,000 185941 . 37,188
MLC‐09‐11
58
OP Q
0
Question #114 Answer: C
Event x0
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
E X 0.05 15 0.095 33.87 0.855 56.12 51.95 E X 2 0.05 15 0.095 33.87 0.855 56.12 2813.01 2
2
2
Var X E X 2 E X 2813.01 51.95 114.2 2
2
Question #115 Answer: B Let K be the curtate future lifetime of (x + k) kL
1000v K 1 1000 Px:3 aK 1
When (as given in the problem), (x) dies in the second year from issue, the curtate future lifetime of x 1 is 0, so 1L
1000v 1000 Px:3 a1
1000 279.21 1.1 629.88 630
The premium came from A Px:3 x:3 ax:3
Ax:3 1 d ax:3 Px:3 279.21
MLC‐09‐11
1 d ax:3 ax:3
1 d ax:3
59
Question #116 Answer: D Let M = the force of mortality of an individual drawn at random; and T future lifetime of the individual.
Pr T 1
n
E Pr T 1 M
z zz zd
0
0 0
0
bg
Pr T 1 M f M d
2 1
2
s
1 2
e t dt d
1 e
i 21 du 21 d2 e
2
0.56767 Question #117 Answer: E For this model: 1/ 60 1 (1) 40(1)t ; 40 20 1/ 40 0.025 1 t / 60 60 t 1/ 40 1 (1) 40(2)t ; 40 20 1/ 20 0.05 1 t / 40 40 t 40 20 0.025 0.05 0.075
MLC‐09‐11
i 21 d1 e i
1
60
2
Question #118 Answer: D Let benefit premium Expected present value of benefits = 0.03 200,000 v 0.97 0.06 150,000 v 2 0.97 0.94 0.09 100,000 v 3
b gb
g b gb gb
g b gb gb gb
5660.38 7769.67 6890.08 20,32013 .
Expected present value of benefit premiums ax:3
b gb g
1 0.97v 0.97 0.94 v 2 2.7266 20,32013 . 7452.55 2.7266 1V
. g b200,000gb0.03g b7452.55gb106 1 0.03
1958.46
Initial reserve, year 2 = 1V = 1958.56 + 7452.55 = 9411.01
Question #119 Answer: A Let denote the premium.
b g
L bT v T aT 1 i
T
v T aT
1 aT E L 1 ax 0
L 1 aT 1
b
g
aT ax
1
ax
d
ax 1 v T
i
ax
v T 1 ax v T Ax ax 1 Ax MLC‐09‐11
61
g
Question #120 Answer: D
(0, 1)
(1, 0.9) (1.5, 0.8775) (2, 0.885)
tp1
1
2
t
1
p1 (1 01 . ) 0.9
2
p1 0.9 1 0.05 0.855
b gb
g
b
g
since uniform, 1.5 p1 0.9 0.855 / 2 0.8775
e1:1.5 Area between t 0 and t 15 .
FG 1 0.9 IJ b1g FG 0.9 0.8775IJ b0.5g H 2 K H 2 K
0.95 0.444 . 1394
MLC‐09‐11
62
Alternatively,
e11: .5
z z zb 1.5
0 t
p1dt
1
z
0.5
p dt 1p1 x 0 t 1 0 1
0
g
p2 dx
1 01 . t dt 0.9
t 0.12t
2
1 0
z
0.5
0
b1 0.05xgdx
0.9 x 0.052 x
0.5
2
0
. 0.95 0.444 1394
Question #121 Answer: A
b g
10,000 A63 112 . 5233 A63 0.4672 Ax 1 A64
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 gross premium at 65 = (1.12) (10,000) (0.4955) = 5550
b1 ig
2
5550 5233
i
5550 1 0.02984 5233
Question #122A Answer: C Because your original survival function for (x) was correct, you must have x t 0.06 x02t: y t x03t: y t x02t: y t 0.02
x02t: y t 0.04 Similarly, for (y) y t 0.06 x01t: y t x03t: y t x01t: y t 0.02
x01t: y t 0.04 MLC‐09‐11
63
The first-to-die insurance pays as soon as State 0 is left, regardless of which state is next. The force of transition from State 0 is x01t: y t x02t: y t x03t: y t 0.04 0.04 0.02 0.10 . With a constant force of transition, the expected present value is 0.10 t 0.05t 0.10 t 00 01 02 03 e p dt ( ) t xy x t : y t x t : y t x t : y t 0 0 e e (0.10)dt 0.15
Question #122B Answer: E Because (x) is to have a constant force of 0.06 regardless of (y)’s status (and vice23 versa) it must be that 13 x t: y t x t : y t 0.06 . There are three mutually exclusive ways in which both will be dead by the end of year 3: 1: Transition from State 0 directly to State 3 within 3 years. The probability of this is 3
0.02 0.10t 0.3 0 t p dt 0 e 0.02dt 0.10 e 0 0.2(1 e ) 0.0518 2: Transition from State 0 to State 1 and then to State 3 all within 3 years. The probability of this is 3
00 xy
3
0
3
03 x t: y t
0.10 t
3
t
0.10 t pxy00 x01t: y t 3t p13 0.04(1 e 0.06(3t ) )dt x t : y t dt e 0
3
0.04 e 0 0.4(1 e
0.10 t
0.3
e
)e
0.18 0.04 t
0.18
e
(1 e
0.04 0.10t 0.04e 0.18 0.04t e e 0.10 0.04
0.12
3
0
) 0.00922
3: Transition from State 0 to State 2 and then to State 3 all within 3 years. By symmetry, this probability is 0.00922. The answer is then 0.0518 + 2(0.00922) = 0.0702.
MLC‐09‐11
64
Question #122C Answer: D Because the original survival function continues to hold for the individual lives, with a constant force of mortality of 0.06 and a constant force of interest of 0.05, the expected present values of the individual insurances are
Ax Ay
0.06 0.54545 , 0.06 0.05
Then, Axy Ax Ay Axy 0.54545 0.54545 0.66667 0.42423 Alternatively, the answer can be obtained be using the three mutually exclusive outcomes used in the solution to Question 122B. 1:
0
e 0.05t t pxy00 x03t: y t dt e0.05t e0.10t 0.02dt
2 and 3:
0
0
0.02 0.13333 0.15
13 e 0.05t t pxy00 x01t: y t e 0.05 r r p11 x t : y t x t r : y t r drdt 0
0
0
e 0.05t e 0.10t 0.04 e 0.05 r e 0.06 r 0.06drdt
0.04 0.06 0.14545 0.15 0.11
The solution is 0.13333 + 2(0.14545) = 0.42423. The fact that the double integral factors into two components is due to the memoryless property of the exponential transition distributions.
MLC‐09‐11
65
Question #123 Answer: B 5
q35:45 5 q35 5 q45 5 q35:45 5 p35q40 5 p45q50 5 p35:45q40:50
b g p q p q p p b1 p p g b0.9gb.03g b0.8gb0.05g b0.9gb0.8g 1 b0.97gb0.95g 5 p35q40 5 p45q50 5 p35 5 p45 1 p40:50 5 35 40
5 45 50
5 35
5 45
40
50
0.01048
Alternatively, 6
p35 5 p35 p40 0.90 1 0.03 0.873
6
p45 5 p45 p50 0.80 1 0.05 0.76
5 q35:45
5 p35:45 6 p35:45 5 p35 5 p45 5 p35:45 6 p35 6 p45 6 p35:45 5 p35 5 p45 5 p35 5 p45 6 p35 6 p45 6 p35 6 p45 0.90 0.80 0.90 0.80 0.873 0.76 0.873 0.76 0.98 0.96952 0.01048
Question #124 – Removed
Question #125 - Removed
MLC‐09‐11
66
Question #126 Answer: E Let
Y = present value random variable for payments on one life S Y present value random variable for all payments
E Y 10a40 148166 . Var Y 10
2
d
2
2 A40 A40
d
d
i
2
ib
g
2 . . / 0.06 100 0.04863 016132 106
2
. 70555 E S 100 E Y 14,816.6 Var S 100 Var Y 70,555
Standard deviation S 70,555 265.62 By normal approximation, need E [S] + 1.645 Standard deviations = 14,816.6 + (1.645) (265.62) = 15,254
Question #127 Answer: B
e
5 A30 4 A301 :20
Initial Benefit Prem
Where
e
j
5a30:35 4a30:20
b
g b g b
g
4 0.02933 5 010248 . 5 14.835 4 11959 .
0.5124 011732 0.39508 . 0.015 74.175 47.836 26.339
b
g
j
1 A30 A30:20 A30:201 0.32307 0.29374 0.02933 :20
and
a30:20
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 MLC‐09‐11
67
Question #128 Answer: B 0.75
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
x
0.75
y
01097 .
Question #129 Answer: D Let G be the expense-loaded premium. Expected present value (EPV) of benefits = 100,000A35 EPV of premiums = Ga35 EPV of expenses = 0.1G 25 2.50 100 a35 Equivalence principle: Ga35 100,000 A35 0.1G 25 250 a35
A35 0.1G 275 a35 0.9G 100,000 P35 275 G 100,000
G
100 8.36 275 0.9
1234
MLC‐09‐11
68
Question #130 Answer: A The person receives K per year guaranteed for 10 years Ka10 8.4353K The person receives K per years alive starting 10 years from now 10 a40 K
b
g
*Hence we have 10000 8.435310 E40a50 K Derive
10 E40: 1 A40 A4010 :
10 E40
Derive a50
b
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
c
b gb
gh
18.5753K K 538.35
MLC‐09‐11
69
Question #131 Answer: D
z
11
STANDARD: e25:11
MODIFIED:
p25
0
e z
FG1 t IJ dt t t H 75K 2 75 2
1
0.1ds 0
e2511 :
11
101933 .
0
e .1 0.90484
z
1
p dt 0 t 25
p25
z FGH z FGH 10
0
1
IJ K
t dt 74
IJ K F t I 1 e e Gt 01 . H 2 74 JK 0.95163 0.90484b9.32432g 9.3886
z
1 0.1t
0
e
dt e0.1
0.1
0.1
10
0
1
t dt 74
2
10 0
Difference
=0.8047
Question #132 Answer: 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, C’s reserve is lower than C*’s which is lower than B’s. Comparing B to E: Reserves on E are constant at 0.
MLC‐09‐11
70
Question #133 Answer: C 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
b
gb
g
Probability of remaining through the year is p60 . 0.84645 b1g p60 b 2g p60 b3g 1 0.01 1 0.05 1 010
b
gb
gb
g
Probability of exiting at the end of the year is b3g 0.9405 0.84645 0.09405 q60
Question #134 - Removed
Question #135 Answer: D EPV of regular death benefit
zb
0
gd
ib
gd
zb
0
gd ib
gd i
100000 e- t 0.008 e- t dt
i
100000 e-0.06t 0.008 e-0.008t dt
b
g
100000 0.008 / 0.06 0.008 11,764.71
EPV of accidental death benefit
zb 30
0
gd
ib
gd
i
zb 30
0
gd ib
100000 e-0.06t 0.001 e-0.008t dt
100 1 e-2.04 / 0.068 1,279.37 Total EPV 11765 1279 13044
MLC‐09‐11
gd i
100000 e- t 0.001 e- t dt
71
Question #136 Answer: B
b gb
g b gb
g
g b gb
g
l 60 .6 .6 79,954 .4 80,625 80,222.4
b gb
l 60 1.5 .5 79,954 .5 78,839 79,396.5 0.9 q 60 .6
80222.4 79,396.5 80,222.4 0.0103
Question #137 - Removed
Question #138 Answer: A
b g qb1g qb2g 0.34 q40 40 40 1 2 1 p40 b g p40 b g
0.34 1 0.75 p40 b 2g p40 b 2g 0.88
q40 . y 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
MLC‐09‐11
72
Question #139 Answer: C
b g
Pr L ' 0 0.5
Pr 10,000v K 1 ' aK 1 0 0.5 From Illustrative Life Table,
47
p30 0.50816 and
48
p30 .47681
Since L is a decreasing function of K, to have Pr L 0 0.5 means we must have L 0 for K 47.
b g
b g
b g
Highest value of L for K 47 is at K = 47.
b g
L at K 47 10,000 v 471 a471
b g
609.98 16.589
b
g
L 0 609.98 16.589 0
609.98 36.77 16.589
Question #140 Answer: 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
2
2
2
2
2
MLC‐09‐11
73
Question #141 Answer: E E Z b Ax
since constant force Ax /( ) E(Z)
b 0.02 b b/3 0.06
Var Z Var b vT b 2 Var v b 2
2
Ax Ax2
2 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
Question #142 Answer: B
b g b g c AA h 2 2
In general Var L 1 P
c h
Here P Ax
x
2 x
1 1 .08 .12 ax 5
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
2 x
x
2
5 4
2
x
2 x
15 2 8 12 2 8
x
MLC‐09‐11
x
x
74
Question #143 - Removed
Question #144 Answer: B Let l0b g number of students entering year 1 superscript (f) denote academic failure superscript (w) denote withdrawal subscript is “age” at start of year; equals year - 1 p0b g 1 0.40 0.20 0.40 l2b g 10 l2b g q2b f g q2b f g 01 .
b
g
q2b wg q2b g q2b f g 10 . 0.6 01 . 0.3 l1b gq1b f g 0.4 l1b g 1 q1b f g q1b wg
e
j
q1b f g 0.4 1 q1b f g 0.3
e
q1b f g
j
0.28 0.2 14 .
p1b g 1 q1b f g q1b wg 1 0.2 0.3 0.5
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
0
0
1
0
1
2
b gb g b gb gb g
0.2 0.4 0.3 0.4 0.5 0.3 0.38
MLC‐09‐11
75
Question #145 Answer: D
e25 p25 (1 e26 ) M e26N e26 due to having the same 1 M 0.05 p25N exp 25Mt 0.1(1 t )dt p25 e 0 M M e25N p25N (1 e26 ) e 0.05 p25 (1 e26 ) 0.951e25 9.51
Question #146 Answer: D
b g F c1 A hI 10,000,000 100b10,000gG H JK
E YAGG 100E Y 100 10,000 a x
x
b10,000g 1 c A A h b10,000g b0.25g b016 . g 50,000
Y Var Y
2
2
x
2
b
2 x
g
AGG 100 Y 10 50,000 500,000 0.90 Pr
LM F E Y N
AGG
0
AGG
1282 .
OP Q
F E YAGG
AGG
F 1282 . AGG E YAGG
b
g
F 1282 . 500,000 10,000,000 10,641,000
MLC‐09‐11
76
Question #147 Answer: A 1 A30:3 1000vq30 500v 2 1 q30 250v3 2 q30 2
3
1 1.53 1 1.61 1 1.70 1000 500 0.99847 250 0.99847 0.99839 1.06 1000 1.06 1000 1.06 1000
1.4434 0.71535 0.35572 2.51447 1
1 1 1 2 0.00153 1 a30:1 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
Each semiannual premium
Question: #148 Answer: E
b DAg
1 80:20
q80 .2
q80 .1
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 13
. 106
Question #149 - Removed
MLC‐09‐11
77
Question #150 Answer: A t
LM N
px exp
z
OP Q
b
ds exp ln 100 x s 0 100 x s t
g
t 0
100 x t 100 x
e50:60 e50 e60 e50:60
e50
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 t 1 F 2000t 45t G 2000 H 3 JK 50
50 50 t
0
t2 1 50t dt 50 50 2
0
2 40
40
0
0
e50:60
40
40
0
0
2
3
2
40 0
e50:60 25 20 14.67 30.33
Question #151 Answer: C Ways to go 0 2 in 2 years 0 0 2; p 0.7 0.1 0.07
0 1 2; p 0.2 0.25 0.05 0 2 2; p 0.11 0.1 Total = 0.22 Binomial m = 100 q = 0.22 Var = (100) (0.22) (0.78) = 17
MLC‐09‐11
78
Question #152 Answer: A For death occurring in year 2 0.3 1000 EPV 285.71 1.05 For death occurring in year 3, two cases:
(1) State 2 State 1 State 4: (0.2 0.1) = 0.02 (2) State 2 State 2 State 4: (0.5 0.3) = 0.15 Total 0.17 EPV =
0.17 1000 154.20 1.052
Total. EPV = 285.71 + 154.20 = 439.91
Question #153 - Removed
Question #154 Answer: C Let denote the single benefit premium. 30 a35 A351 :30
30
a35
1 1 A35 :30
eA
35:30
j
1 a65 A35 :30
1 1 A35 :30
b.21.07g9.9 b1.07g
1.386 .93 1.49
MLC‐09‐11
79
Question #155 Answer: E 0.4 p0
.5 e z
0. 4
0
e F e2 x jdx
LM e22x OP.4 N Q0 e 0.8 .4 F F e 2 1I H K e .4 F
.5 e.4 F .6128
bg
ln .5 .4 F .6128 .6931 .4F .6128 F 0.20
Question #156 Answer: C
9V P 1.03 qx9b 1 qx9 10V qx 9 b 10V 10V 3431.03 0.02904 872 10V 10V 327.97
b b 10V 10V 872 327.97 1199.97 1 1 0.03 P b d 1200 14.65976 1.03 ax = 46.92 9V benefit reserve at the start of year ten – P = 343 – 46.92 = 296.08
Question #157 Answer: B d = 0.06 V = 0.94 Step 1 Determine px
668 258vpx 1000vqx 1000v 2 px p x 1 qx 1 668 258 0.94 px 1000 0.94 1 px 1000 0.8836 px 1 668 242.52 px 940 1 px 883.6 px
px 272 / 298.92 0.91 MLC‐09‐11
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Step 2 Determine 1000 Px:2
668 258 0.94 0.91 1000 Px:2 1 0.94 0.91 220.69 668 479 1000 Px:2 1.8554
Question #158 Answer: D 1 1 1 100,000 IA 40:10 100,000 v p40 IA 41:10 10 v10 9 p41 q50 A40:10 100,000 8,950,901 10 0.99722 9, 287, 264 0.16736 0.00592 100,000 1.06 1.0610 0.02766 100,000
see comment
=15,513 1 A40 10 E40 A50 Where A40:10
0.16132 0.53667 0.24905 0.02766
Comment: the first line comes from comparing the benefits of the two insurances. At each of age 40, 41, 42,…,49 IA 40:10 provides a death benefit 1 greater than 1
IA141:10 .
1 Hence the A40:10 term. But IA 41:10 provides a death benefit at 50 of 10, 1
while IA 40:10 provides 0. Hence a term involving 9 q41 9 p41 q50 . The various v’s 1
and p’s just get all expected present values at age 40.
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Question #159 Answer: A 1000 1V 1 i qx 1000 1000 1V
40 80 1.1 qx 1000 40 qx
88 40 0.05 960
1 AS
G
expenses 1 i 1000qx px
100 0.4 100 1.1 1000 0.05 1 0.05
60 1.1 50 16.8 0.95
Question #160 Answer: 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.
Expected present value (EPV) at the start of each year for that year’s death benefits = 10,000*0.0198 v = 188.1 px 0.9802*0.96 0.9410 Ex px v 0.941 v 0.941*0.95 0.8940
EPV of death benefit for 3 years 188.1 E40 *188.1 E40 * E41 *188.1 506.60
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Question #161 Answer: 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 a linear survival function, it may be simpler to draw a picture:
0
p30 1
40
30
70
e30:40 area =27.692 40 40
p30
p30 0.3846
1 40 p30 2
70 0.3846 30 95 t
p30
65 t 65
Var E T E T 2
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83
Using a common formula for the second moment:
Var T 2 t t px dt ex2
0
65 t t 2 t 1 dt 1 dt 2 65 0 0 65 65
2* 2112.5 1408.333 65 65 / 2
2
1408.333 1056.25 352.08
Another way, easy to calculate for a linear survival function is
Var T t 2 t px x t dt 0
65
t2 0
t3 3 65
0
t t px x t dt
1 1 65 dt t dt 0 65 65
65 0
t2 2 65
65 0
2
2
2
1408.33 32.5 352.08 2
With a linear survival function and a maximum future lifetime of 65 years, you probably didn’t need to integrate to get E T30 e30 32.5
Likewise, if you realize (after getting 95 ) that T30 is uniformly distributed on (0, 65), its variance is just the variance of a continuous uniform random variable: 2 65 0 Var
12
352.08
Question #162 Answer: E
218.15 1.06 10,000 0.02 31.88 1 0.02 31.88 218.151.06 9,000 0.021 77.66 2V 1 0.021
1V
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Question #163 Answer: D
ex e y t px 0.95 0.952 ... k 1
0.95 19 1 0.95 exy pxy 2 pxy ...
1.02 0.95 0.95 1.02 0.95 0.95 ... 2
1.02 0.95
1.02 0.95 0.95 ... 1 0.952 exy ex e y exy 28.56 2
4
2
2
9.44152
Question #164 - Removed
Question #165 - Removed
Question #166 Answer: E
ax e 0.08t dt 12.5 0
3 0.375 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 Ax e 0.08t 0.03 dt 0
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
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Question #167 Answer: A 5
0.05 5 p50 e e 0.25 0.7788
1 5 q55
5 1 0.03 0.02 t 55 dt 0.02 / 0.05 e 0.05t t e 0
0.4 1 e
0.25
= 0.0885 Probability of retiring before 60 5 p50 5 q55 = 0.7788*0.0885 = 0.0689
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5 0
Question #168 Answer: C Complete the table: l81 l80 d80 910
l82 l81 d81 830 (not really needed) 1 1 e x ex since UDD 2 2
e x e x 1 2 l l l 1 e x 81 82 83 l80 2 e 1 l l l Call this equation (A) 80 2 80 81 82 e 1 l l Formula like (A), one age later. Call this (B) 81 2 81 82
Subtract equation (B) from equation (A) to get 1 1 l81 e80 l80 e81 l81 2 2 910 8.5 0.51000 e81 0.5 920
e81
8000 460 910 8.21 920
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Alternatively, and more straightforward,
910 0.91 1000 830 p81 0.902 920 830 p81 0.912 910 p80
e80 1 q 80 p 80 1 e 81 2
1 where q80 contributes since UDD 2 1 8.5 1 0.91 0.91 1 e81 2
e81 8.291
1 e81 q81 p81 1 e82 2 1 8.291 1 0.912 0.912 1 e82 2
e82 8.043 1 e81 q81 p81 1 e82 2 1 1 0.902 0.902 1 8.043 2 8.206
Or, do all the recursions in terms of e, not e , starting with e80 8.5 0.5 8.0 , then
final step e81 e81 0.5
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Question #169 Answer: A
T
p x t
0 1 2 3
px
vt
vt t px
0.7
1
1
0.7
0.7 0.49
0.95238 0.90703 –
1 0.6667 0.4444
t
2
From above ax:3 vt t px 2.1111 t 0
a 1000 2Vx:3 1000 1 x 2:1 ax:3
1 1000 1 526 2.1111
Alternatively,
Px:3
1 d 0.4261 ax:3
1000 2Vx:3 1000 v Px:3
1000 0.95238 0.4261 526
You could also calculate Ax:3 and use it to calculate Px:3 .
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Question #170 Answer: E Let G be the gross premium. Expected present value (EPV) of benefits = 1000A50 . EPV of expenses, except claim expense = 15 1 a50 EPV of claim expense = 50A50 (50 is paid when the claim is paid) EPV of premiums = G a50 Equivalence principle: Ga50 1000 A50 15 1 a50 50 A50 1050 A50 15 a50 G a50 For the given survival function, a50 1 1.0550 1 50 k 1 50 k ( ) (2) 0.36512 A50 v l l v 50k 1 50k 100 50 0.05(50) l50 k 1 k 1 1 A50 a50 13.33248 d Solving for G,
G = 30.88
Question #171 Answer: A 4
0.05 4 p50 e 0.8187
10 8
0.05 10 p50 e 0.6065
0.04 8 p60 e 0.7261
18
p50 10 p50 8 p60 0.6065 0.7261
0.4404 414 q50 4 p50 18 p50 0.8187 0.4404 0.3783
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Question #172 Answer: D
a40:5 a40 5 E40 a45
14.8166 0.73529 14.1121 4.4401
a40:5 100,000 A45 v5 5 p40 IA 40:5 1
1 100,000 A45 5 E40 / a40:5 IA 40:5
100,000 0.20120 0.73529 / 4.4401 0.04042 3363 Question #173 Answer: B Calculate the probability that both are alive or both are dead. P(both alive) = k pxy k px k p y P(both dead) = k qxy k q x k q y P(exactly one alive) = 1 k pxy k qxy Only have to do two year’s worth so have table Pr(both alive)
Pr(both dead)
Pr(only one alive)
1
0
0
(0.91)(0.91) = 0.8281
(0.09)(0.09) = 0.0081
0.1638
(0.82)(0.82) = 0.6724
(0.18)(0.18) = 0.0324
0.2952
0.8281 0.6724 0.1638 0.2952 1 0 EPV Annuity 30, 000 20, 000 80, 431 0 1 2 0 1.05 1.05 1.051 1.052 1.05 1.05 Alternatively, 0.8281 0.6724 2.3986 1.05 1.052 0.91 0.82 ax:3 ay:3 1 2.6104 1.05 1.052 EPV 20, 000 ax:3 20, 000 ay:3 10, 000 axy:3 axy:3 1
(it pays 20,000 if x alive and 20,000 if y alive, but 10,000 less than that if both are alive) 20,000 2.6104 20,000 2.6104 10,000 2.3986 80, 430 MLC‐09‐11
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Question #174 Answer: C Let P denote the gross premium.
P ax e E L
0 axIMP 10
axIMP
t t
e
e
dt e 0.05t dt 20 0
P
0.03t 0.02t
e
dt e
0.0310 0.0210
e
0
e
0.03t 0.01t
e
dt
0
l e 0.5 e0.5 23 0.05 0.04 E L 23 20 3
E L 3 15% P 20
Question #175 Answer: C 1 A30:2 1000vq30 500v 2 1 q 30 2
1 1 1000 0.00153 500 0.99847 0.00161 1.06 1.06 2.15875 Initial fund 2.15875 1000 participants = 2158.75
Let Fn denote the size of Fund 1 at the end of year n. F1 2158.75 1.07 1000 1309.86 F2 1309.86 1.065 500 895.00
Expected size of Fund 2 at end of year 2 = 0 (since the amount paid was the single benefit premium). Difference is 895.
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Question #176 Answer: C 2 Var Z E Z 2 E Z
E Z
t
t t
0
v b
0
E Z 2
px x t dt e 0.08t e0.03t e 0.02t 0.02 dt 0
0.02 2 0.02 e0.07t dt 7 0.07
0
vt bt
2 t
px x t dt
0
e
0.02 e 0.12t x t dt 2 0
Var Z
1 2 7 6
2
12
0.05t
1
2
e0.02t 0.02 dt
6
1 4 0.08503 6 49
Question #177 Answer: 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
10V
5 0.1 0.4769 11 ln 1.1
Ax 10 P Ax ax 10 0.2861 0.4769 6 8 = 0.2623
There are many other equivalent formulas that could be used.
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Question #178 Answer: C
100,000 e 0.06t e 0.001t 0.001dt
Regular death benefit
0
0.001 100,000 0.06 0.001 1639.34 Accidental death
100,000 e 0.06t e 0.001t 0.0002 dt 10
0
10
20 e 0.061t dt 0
1 e 0.61 20 149.72 0.061
Expected Present Value 1639.34 149.72 1789.06
Question #179 Answer: C 00 p61 560 / 800 0.70 01 p61 160 / 800 0.20 10 p61 0, once dead, stays dead 11 p61 1, once dead by cause 1, stays dead by cause 1
00 01 10 11 p61 p61 p61 p61 0.70 0.20 0 1 1.90
Question #180 - Removed
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Question #181 Answer: B
Pr(dies in year 1) p 02 0.1 Pr(dies in year 2) p 00 p 02 p 01 p12 0.8(0.1) 0.1(0.2) 0.10 Pr(dies in year 3) p 00 p 00 p 02 p 00 p 01 p12 p 01 p11 p12 p 01 p10 p 02 0.095 EPV (benefits) 100, 000[0.9(0.1) 0.92 (0.10) 0.93 (0.095)] 24, 025.5 Pr(in State 0 at time 0) 1 Pr(in State 0 at time 1) p 00 0.8 Pr(in State 0 at time 2) p 00 p 00 p 01 p10 0.8(0.8) 0.1(0.1) 0.65 EPV ($1 of premium) 1 0.9(0.8) 0.92 (0.65) 2.2465
Benefit premium = 24,025.5/2.2465 = 10,695.
Question #182 - Removed Answer: A
Question #183 - Removed
Question #184 Answer: B
1000 P45a45:15 a60:15 15 E45 1000 A45 1000
A45 a45 15 E45 a60 a60 15 E60 a75 15 E45 1000 A45 a45
201.20 14.1121 0.72988 0.5108111.1454 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
14.2573 9.9568 3.4154 201.20
17.346
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Question #185 Answer: A 1V
0 V 1 i 1000 1V 1V qx
2V
1V 1 i 2000 2V 2V qx 1 2000
1 i 1000q 1 i 2000q 2000 1.08 1000 0.1 1.08 2000 0.1 2000 x 1
x
1027.42 Question #186 Answer: A Let Y be the present value of payments to 1 person. Let S be the present value of the aggregate payments.
E Y 500 ax 500
Y Var Y
1 Ax 5572.68 d
500 2
1 d2
2
Ax Ax2 1791.96
S Y1 Y2 ... Y250
E S 250 E Y 1,393,170
S 250 Y 15.811388 Y 28,333 S 1,393,170 F 1,393,170 0.90 Pr S F Pr 28,333 28,333 F 1,393,170 Pr N 0,1 28,333 0.90 Pr N 0,1 1.28
F 1,393,170 1.28 28,333
=1.43 million
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Question #187 Answer: 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 2 d 41 l41 d 41 l42 885 70 750 65
q41 1
750 p41 885
q41
750 1 1 q41 885
65
135
65 135
0.0766
Question #188 Answer: D
t S0 (t ) 1 d t log S0 (t ) t dt
ex
0
t x 1 dt 1 x
x
1 e0new old new new 2 old 1 2 1 1 old old 2 1 9 old 4 0new 4
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Question #189 Answer: C Constant force implies exponential lifetime 2
2 2 1 1 Var T E T 2 E T 2 2 100 0.1
E min T ,10 t 0.1e .1t dt 10 0.1e.1t dt 10
0
10
te .1t 10e 1
.1t 10 0
10e.1t
1
10e 10e 10 10e
10 1
10 1 e 1 6.3 Question #190 Answer: A
Gax:15
% premium amount for 15 years 100,000 Ax 0.08G 0.02Gax:15 x 5 5ax
Per policy for life 4669.95 11.35 51, 481.97 0.08 4669.95 0.02 11.35 4669.95 x 5 5ax
ax
1 Ax 1 0.5148197 16.66 d 0.02913
53,003.93 51, 481.97 1433.67 x 5 83.30
4.99 x 5 x 9.99 The % of premium expenses could equally well have been expressed as 0.10G 0.02G ax:14 . The per policy expenses could also be expressed in terms of an annuityimmediate.
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Question #191 Answer: D For the density where Tx Ty ,
Pr Tx Ty
40
y
50
40
0.0005dxdy 0.0005dxdy y 0 x 0 y 40 x 0
40 y 0
y
0.0005 x dy 0
y 40
40
50
0
y 40
0.0005 ydy
50
0.0005 y 2 2
40 0
0.02 y
0.0005 x
40
dy
0
0.02 dy 50 40
0.40 + 0.20 = 0.60
For the overall density, Pr Tx Ty 0.4 0 0.6 0.6 0.36
where the first 0.4 is the probability that Tx Ty and the first 0.6 is the probability that Tx Ty .
Question #192 Answer: B The conditional expected value of the annuity, given , is
1 . 0.01
The unconditional expected value is 0.02 1 0.01 0.02 ax 100 d 100 ln 40.5 0.01 0.01 0.01 0.01 100 is the constant density of on the interval 0.01,0.02 . If the density were not constant, it would have to go inside the integral.
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Question #193 Answer: E
Recall ex
x 2
ex:x ex ex ex:x
ex:x
t t dt 1 1 x y
x
0
Performing the integration we obtain x ex:x 3 2 x ex:x 3
2 2a 2 3a 3 2 7a 3 3 2 3a 2 a k 3 3 3.5a a k 3.5a 3a k 5
(i) (ii)
The solution assumes that all lifetimes are independent.
Question #194 Answer: B Although simultaneous death is impossible, the times of death are dependent as the force of mortality increases after the first death. There are two ways for the benefit to be paid. The first is to have (x) die prior to (y). That is, the transitions are State 0 to State 2 to State 3. The EPV can be written with a double integral
0
e 0.04t t pxy00 x02t: y t e 0.04 r r px22t: y t x23t r: y t r drdt 0
0
0
e 0.04t e 0.12t 0.06 e 0.04 r e 0.10 r 0.10drdt
0.06 0.10 0.26786 0.16 0.14
By symmetry, the second case (State 0 to State 1 to State 3) has the same EPV. Thus the total EPV is 10,000(0.26786+0.26786) = 5,357.
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Question #195 Answer: E Let
k
p0 Probability someone answers the first k problems correctly.
2
p0 0.8 0.64
4
p0 0.8 0.41
2
p0:0 2 p0 0.642 0.41
4
p0:0 0.41 0.168
2
p0:0 2 p0 2 p0 2 p0:0 0.87
4
p0:0 0.41 0.41 0.168 0.652
2
4
2
2
Prob(second child loses in round 3 or 4) = 2 p0:0 4 p0:0 = 0.87-0.652 = 0.218 Prob(second loses in round 3 or 4 second loses after round 2) =
2
p0:0 4 p0:0 2
=
p0:0
0.218 0.25 0.87
Question #196 Answer: E If (40) dies before 70, he receives one payment of 10, and Y = 10. 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. If (40) survives to 100, he receives 3 payments. Y 10 20v30 30v 60 19.01819 The probability of this is 1 – 3/7 – 3/7 = 1/7 E Y 3/ 7 10 3/ 7 16.16637 1/ 7 19.01819 13.93104
E Y 2 3/ 7 102 3/ 7 16.16637 2 1/ 7 19.018192 206.53515
Var Y E Y 2 E Y 12.46 Since everyone receives the first payment of 10, you could have ignored it in the calculation.
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101
Question #197 Answer: C 2
E Z v k 1bk 1 k 0
k
px qxk
v 300 0.02 v 2 350 0.98 0.04 v3 400 0.98 0.96 0.06 36.8
2
E Z 2 v k 1bk 1 k 0
2 k
px qx k
v 2 300 0.02 v 4 350 0.98 0.04 v 6 4002 0.98 0.96 0.06 2
2
11,773
Var Z E Z 2 E Z
2
11,773 36.82 10, 419
Question #198 Answer: A Benefits + 3 0 Le 1000v
Expenses 0.20G 8 0.06G 2 v 0.06G 2 v 2
at G 41.20 and i 0.05, 0L
for K 2 770.59
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– Premiums G a3
Question #199 Answer: D
P 1000 P40
235 P 1 i 0.015 1000 255 255
[A]
255 P 1 i 0.020 1000 272 272
[B]
Subtract [A] from [B] 20 1 i 3.385 17
1 i Plug into [A]
20.385 1.01925 20
235 P 1.01925 0.015 1000 255 255 235 P
255 11.175 1.01925
P 261.15 235 26.15
1000 25V
272 26.151.01925 0.0251000 1 0.025 286
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Question #200 Answer: A
1.2 1
1 0.8 0.6
0.5 0.4
0.4 0.2 0 0
10
x
Give n 0
s x
1
55 q35
20 55 q15
20 55 q15 55 q35
1
20
30
15 0.70 Linear Interpolation
40
50
Give n 25
60
35
0.50
0.48 Linear Interpolation
s 90 0.16 32 1 0.6667 0.48 48 s 35
s 35 s 90 0.48 0.16 32 0.4571 s 15 0.70 70
0.4571 0.6856 0.6667
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70
Give n 75 0.4
80
90
0 100
Given 90
100
0.16 Linear Interpolation
0
Alternatively, 20 55 q15 55 q35
20 p15 55 q35 55 q35
20 p15
s 35 s 15
0.48 0.70 0.6856
Question #201 Answer: A
S0 (80) 1 * e ^ 0.16*50 e ^ 0.08*50 0.00932555 2 S0 (81) 1 * e ^ 0.16*51 e ^ 0.08*51 0.008596664 2 p80 s 81 / s 80 0.008596664 / 0.00932555 0.9218
q80 1 0.9218 0.078 Alternatively (and equivalent to the above) For non-smokers, px e 0.08 0.923116 50
px 0.018316
For smokers, px e 0.16 0.852144 50 p x 0.000335 So the probability of dying at 80, weighted by the probability of surviving to 80, is
0.018316 1 0.923116 0.000335 1 0.852144 = 0.078 0.018316 0.000335
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Question #202 Answer: B x 40 41 42
lx 2000 1920 1840
because 2000 20 60 1920 ;
d x 20 30 40
d x 60 50
1
2
1920 30 50 1840
Let premium = P
1840 2 2000 1920 v v P 2.749 P EPV premiums = 2000 2000 2000 30 2 40 3 20 EPV benefits = 1000 v v v 40.41 2000 2000 2000 40.41 P 14.7 2.749
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Question #203 Answer: A
10
a30 e0.08t e 0.05dt 10 Ex e 0.08t e 0.08t dt 0 10 0.13t
e 0
dt e
0 1.3 0.16
0 e
e 0.13t 10 e 0.16t e 1.3 0.13 0 0.16 1.3 1.3 e 1 e 0.13 0.13 0.16 = 7.2992
dt
0
A30 e 0.08t e 0.05t 0.05 dt e 1.3 e 0.16t 0.08 dt 10
0
0
1 e e 0.05 0.08 0.16 0.13 0.13 0.41606 1.3
1.3
= P A30
A30 0.41606 0.057 a30 7.29923 1 1 a40 0.08 0.08 0.16 A40 1 a40 1 0.08 / 0.16 0.5
10V
A30 A40 P A30 a40
0.5
0.057 0.14375 0.16
Question #204 Answer: C Let T be the future lifetime of Pat, and [T] denote the greatest integer in T. ([T] is the same as K, the curtate future lifetime). L 100,000 vT 1600 a T
1
100,000vT 1600 a10 1600 a10 Minimum is 1600 a10 12,973 MLC‐09‐11
0 T 10 10 t 20
201
1L
v
Px1:2
0 Px1:2
Prob qx 1
1 qx 1
So Var 1 L v 2 qx 1 1 qx 1 0.1296 MLC‐09‐11
116
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, evaluate Px1:2 0.1303
0.9 0.1303 0.7697 if K x 1 1 L 0 0.1303 0.1303 if K x 1 1L
E 1 L 0.2 0.7697 0.8 0.1303 0.0497
E 1 L2 0.2 0.7697 0.8 0.1303 0.1320 2
2
Var 1 L 0.1320 0.0497 0.1295 2
Question #228 Answer: 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
2
2
2
2
2
Ax Ax2
Ax Ax2
Ax Ax2 0.08888
Var L 1
2
2
Ax Ax2
16 1 0.08888 45 0.1 2
1 4 0.1 0.1 2
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Question #229 Answer: E
Seek g such that Pr aT 40 g 0.25
aT is a strictly increasing function of T. Pr T 40 60 0.25 since
Pr aT
40
60
p40
100 40 0.25 120 40
a60 0.25
g a60 19.00
Question 230 Answer: B
0.05 A51:9 1 da51:9 1 7.1 0.6619 1.05 11V
2000 0.6619 100 7.1 613.80
10V P 1.05 11V q50 2000 11V
10V 100 1.05 613.80 0.011 2000 613.80 10V
499.09
where q50 0.00110 0.001 0.011
Alternatively, you could have used recursion to calculate A50:10 from A51:9 , then
a50:10 from A50:10 , and used the prospective reserve formula for 10V .
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Question #231 Answer: 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
q80 0.5q80 0.05 1000 A80 1000vq80 vp80 1000 A81 1000
0.05 0.95 689.52 665.14 1.06 1.06
Question #232 Answer: D
lx 776 752
d x 8 8
42 43
d x 16 16
1
2
1 2 l42 and l43 came from lx 1 lx d x d x
2000 8v 8v 2 1000 16v 16v 2 EPV Benefits =
776
76.40
776 752v EPV Premiums 34 34 1.92 = 65.28 776 2V
76.40 65.28 11.12
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Question #233 Answer: B
pxx 1 qxx 0.96 px 0.96 0.9798
px 1:x 1 1 qx 1:x 1 0.99 px 1 0.99 0.995
ax 1 vpx v 2 2 px 1
0.9798 0.9798 0.995 1.05 1.052
2.8174 0.96 0.96 0.99 2.7763 1.05 1.052 EPV = 2000 ax 2000 ax 6000 axx axx 1 vpxx v 2 2 pxx 1
4000 2.8174 6000 2.7763
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.
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Question #234 Answer: B t
1 1 1 px x t qx 0.20
t
2 2 px 1 tqx 1 0.08t
t
3 3 px 1 tqx 1 0.125t
qx 1
1 0t
1 1 2 3 1 1 px x t dt t px t px t px xt dt 0
1 0.08t 1 0.125t 0.20 dt 1
0
1
0.2 1 0.205t 0.01t 2 dt 0
1
0.205t 2 0.01t 3 0.2 t 2 3 0 0.01 0.2 1 0.1025 0.1802 3
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Question #235 Answer: B
1
d w G 0.1G 1.50 1 1.06 1000q40 2.93 q40 AS 1 q40 q40 d
w
0.9G 1.50 1.06 1000 0.00278 2.93 0.2 1 0.00278 0.2
0.954G 1.59 2.78 0.59 0.79722
1.197G 6.22
2
d w 1 AS G 0.1G 1.50 1 1.06 1000q41 2 CV q41 AS 1 q41 q41 d
w
1.197G 6.22 G 0.1G 1.50 1.06 1000 0.00298 2 CV 0 1 0.00298 0
2.097G 7.72 1.06 2.98 0.99702
2.229G 11.20
2.229G 11.20 24 G 15.8
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Question #236 Answer: A
5
1 2 4 AS G 1 c4 e4 1 i 1000q x 4 5 CV q x 4 AS
1 qx4 qx 4 1
2
396.63 281.77 1 0.05 7 1 i 90 572.12 0.26 1 0.09 0.26
657.311 i 90 148.75 0.65
694.50
657.311 i 90 148.75 0.65 694.50 690.18 1.05 657.31 i 0.05
1 i
Question #237 - Removed
Question #238 – Removed
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Question #239 Answer: B
Let P denote the annual premium The EPV of benefits is 25, 000 Ax:20 25, 000(0.4058) 10,145 . The EPV of premiums is Pax:20 12.522 P The EPV of expenses is 25, 000 (15 3) 3ax:20 1, 000 0.20 P 0.6261P 194.025 12 37.566 0.8261P 243.591
0.25 0.05 P 0.05 P ax:20
2.00 0.50 0.50 ax:20
Equivalence principle: 12.522 P 10,145 0.8261 P 243.591 10,389.591 P 12.522 0.8261 888.31
Question #240 Answer: D Let G denote the premium. Expected present value (EPV) of benefits = 1000A40:20 EPV of premiums = G a40:10 EPV of expenses 0.04 0.25 G 10 0.04 0.05 G a40:9 5a40:19
0.29G 10 0.09G a40:9 5a40:19 0.2G 10 0.09G a40:10 5a40:19 (The above step is getting an a40:10 term since all the answer choices have one. It could equally well have been done later on).
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Equivalence principle: G a40:10 1000 A40:20 0.2G 10 0.09G a40:10 5 a40:19
G a40:10 0.2 0.09 a40:10 1000 A40:20 10 5 a40:19 G
1000 A40:20 10 5 a40:19 0.91a40:10 0.2
Question #241 - Removed
Question #242 Answer: C
11
d w 10 AS G c10 G e10 1 i 10,000q x 10 11 CV q x 10 AS
1 qx10 qx10 d
w
1600 200 0.04 200 70 1.05 10,000 0.02 1700 0.18
1 0.02 0.18 1302.1 0.8
1627.63
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Question #243 Answer: E The benefit reserve at the end of year 9 is the certain payment of the benefit one year from now, less the premium paid at time 9. Thus, it is 10,000v – 76.87. The gross premium reserve adds expenses paid at the beginning of the tenth year and subtracts the gross premium rather than the benefit premium. Thus it is 10,000v + 5 + 0.1G – G where G is the gross premium. Then, 10, 000v 76.87 (10, 000v 5 0.9G ) 1.67 0.9G 81.87 1.67 0.9G 83.54 G 92.82
Question #244 Answer: C
4
d w 3 AS G c4G e4 1 i 1000q x 3 4 CVq x 3 AS
1 qx 3 qx 3 d
w
Plugging in the given values: 4 AS
25.22 30 0.02 30 5 1.05 1000 0.013 75 0.05 1 0.013 0.05
35.351 0.937
= 37.73
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With higher expenses and withdrawals: 4 AS
revised
25.22 30 1.2 0.02 30 5 1.05 1000 0.013 75 1.2 0.05 1 0.013 1.2 0.05
48.51.05 13 4.5 0.927 33.425 0.927
= 36.06 4 AS
4 AS revised 37.73 36.06 = 1.67
Question #245 Answer: E
Let G denote the gross premium. EPV (expected present value) of benefits 100010 20 A30 . EPV of premiums Ga30:5 .
EPV of expenses 0.05 0.25 G 20 first year 0.05 0.10 G 10 a 30:4 years 2-5
10 5 a35:4 years 6-10 (there is no premium) 0.30G 0.15G a30:4 20 10 a30:4 10 5 a 30:5 0.15G 0.15Ga30:5 20 10 a30:9 (The step above is motivated by the form of the answer. You could equally well put it that form later). Equivalence principle: Ga30:5 100010 20 A30 0.15G 0.15G a30:5 20 10 a30:9 G
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10 20 A30
20 10 a 30:9
20 10 a30:9
1 0.15 a30:5 0.15
1000
10 20 A30
0.85 a30:5 0.15 127
Question #246 Answer: E
Let G denote the gross premium EPV (expected present value) of benefits 0.1 3000 v 0.9 0.2 2000 v 2 0.9 0.8 1000v 2
300 360 720 1286.98 2 1.04 1.04 1.042
EPV of premium = G EPV of expenses 0.02G 0.03G 15 0.9 2 v 0.05G 16.73 Equivalence principle: G 1286.98 0.05G 16.73 1303.71 G 1372.33 1 0.05
Question #247 Answer: C
EPV (expected present value) of benefits = 3499 (given) EPV of premiums G 0.9 G v G
0.9G 1.8571G 1.05
EPV of expenses, except settlement expenses, 25 4.5 10 0.2 G 0.9 10 1.5 10 0.1G v 0.9 0.85 10 1.5 10 v 2 0.9 25 0.1G 0.765 25 1.05 1.052 =108.78+0.2857G 70 0.2G
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Settlement expenses are 20 110 30 , payable at the same time the death benefit is paid. 30 So EPV of settlement expenses EPV of benefits 10,000 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 Answer: D
a50:20 a50 20 E50 a70
13.2668 0.23047 8.5693 11.2918
0.06 A50:20 1 d a50:20 1 11.2918 1.06 0.36084 Expected present value (EPV) of benefits 10,000A50:20 3608.40
EPV of premiums 495a50:20 5589.44 EPV of expenses 0.35 495 20 15 10 0.05 495 5 1.50 10 a50:19 343.25 44.75 11.2918 1 803.81
EPV of amounts available for profit and contingencies = EPV premium – EPV benefits – APV expenses = 5589.44 – 3608.40 – 803.81 = 1177.23
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Question #249 Answer: B 1
1
0
0
q1xy t pxy x t dt t px t p y x t dt 1
qx e 0.25t dt (under UDD, t px x t qx ) 0
1
0.125 qx (4e 0.25t ) qx (4)(1 e 0.25 ) 0.8848qx 0
qx 0.1413
Question #250 Answer: C 2
p[11x ]1 p[11x ]1 p[11x ] 2 p[12x ]1 p[21x ] 2 0.1 0.1 0.1 0.2 0.7 0.7 0.3 0.4 2 3 2 3 0.75(0.7333) 0.25(0.3333) 0.6333
Note that Anne might have changed states many times during each year, but the annual transition probabilities incorporate those possibilities.
Questions #251-260 – Removed
Question #261 Answer: A The insurance is payable on the death of (y) provided (x) is already dead.
E ( Z ) Axy2 e t t qx t p y y t dt 0
e
0.06 t
0
(1 e 0.07 t )e 0.09t 0.09dt
0.09 e 0.15t e 0.22t dt 0
1 1 0.09 0.191 0.15 0.22
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Question #262 Answer: C
t
px
95 x t 1 , x t , 95 x 95 x t
t
p y e t
Pr(x dies within n years and before y)
n
0
n
0
t
px t p y x t dt
n 95 x t t 1 1 1 e n t e dt e dt (95 x) 95 x 95 x t 95 x 0
Question #263 Answer: A 2 0.25 q30.5:40.5
0.25
0
t
0.25
0.4 0.6t dt 1 0.5(0.4) 1 0.5(0.6)
0
p30.5 30.5t t q40.5 dt
0.4(0.6) t 2 0.8(0.7) 2
0.25
0.0134 0
Question #264 – Removed
Question #265 Answer: D
t 5rdr e 2.5t 2 p exp t x 0 0.5t 2 t t p y exp 0 rdr e 1
1
0
0
1
qx1: y t p y t px x t dt e 0.5t e 2.5t 5tdt 5 e 3t tdt 2
2
0
1
5 2 5 e 3t (1 e 3 ) 0.7918 6 6 0
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2
Question #266 Answer: B 5
G
5
0
1 t 1 t2 dt 30 25 30(25)(2) 0 60
H 5 p80:85 10 p80:85 GH
25 20 20 15 200 4 30 25 30 25 750 15
1 16 17 0.2833 60 60 60
Question #267 Answer: D t t S0 (t ) exp (80 x) 0.5 dx exp 2(80 x)0.5 exp 2((80 t )0.5 800.5 ) 0 0 F S0 (10.5) exp 2(69.50.5 800.5 ) 0.29665
S0 (10) 0.31495 S0 (11) 0.27935 G S0 (10.5)exp [0.31495(0.27935)]0.5 0.29662 F G 0.00003 Question #268 Answer: A 4
4
E ( Z ) 500 0.2(1 0.25t )dt 1000 0.25(0.2t )dt 0
0
500(0.2) t 0.125t 2 1000(0.25) (0.1t 2 ) 4
4
0
0
100(4 2) 250(1.6) 600
Question #269 Answer: A 10
q30:50 10 q30 10 q50 (1 e 0.5 )(1 e 0.5 ) 0.1548
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Question #270 Answer: C
e30:50 e30 e50 e30:50
e30 e50 e 0.05t dt 20
0
e30:50 e 0.10t dt 10
0
e30:50 20 20 10 30
Question #271 Answer: B
0
0
1 A30:50 e t t p30 t p50 30t dt e 0.03t e 0.05t e 0.05t 0.05dt
0.05 0.3846 0.13
Question #272 Answer: B
T30:50 has the exponential distribution with parameter 0.05 + 0.05 = 0.10 and so its mean is 10 and its variance is 100.
Question #273 Answer: D
Cov[T30:50 , T30:50 ] e30 e30:50 e50 e30:50 (20 10)(20 10) 100 See solution to Question #270 for the individual values.
Question #274 Answer: E
V ( 2V 3 )(1 i3 ) qx 2 (b3 3V )
3
96 (84 18)(1.07) qx 2 (240 96) qx 2 13.14 /144 0.09125
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Question #275 Answer: A
( 3V 4 )(1 i4 ) qx 3b4 (96 24)(1.06) 0.101(360) 101.05 px 3 0.899
V
4
Question #276 Answer: D Under UDD: 0.5qx 3 0.5(0.101) 0.0532 0.5 q x 3.5 1 0.5qx 3 1 0.5(0.101)
Question #277 Answer: E
V v 0.5 0.5 p3.5 4V v 0.5 0.5 q3.5b4
3.5
1.060.5 (0.9468)(101.05) 1.060.5 (0.0532)(360) 111.53
Question #278 Answer: D
1 10 p30:40 1 10 p30 10 p40 1
60 50 2 70 60 7
Question #279 Answer: A
10
2 q30:40
10
0
t
p30 30t (1 t p40 )dt
10
0
1 t 50 0.0119 dt 70 60 70(60)
Question #280 Answer: A
20
10 t
p30 30t t q40 dt
20
10
20
10 t
p40 40t t q30 dt
20 1 t 1 t 1 400 100 1 400 100 dt dt 0.0714 10 60 70 70 60 70 2(60) 60 2(70)
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Question #281 Answer: C
140, 000
30 t
0
p30 30t t p40 dt 180, 000
30
0
t
p40 40t t p30 dt
30 1 70 t 1 60 t dt 180, 000 dt 0 70 60 0 60 70 1 602 302 1 702 402 140, 000 180, 000 115, 714 70 2(60) 60 2(70)
140, 000
30
Question #282 Answer: B
P
20
0
t
p30 t p40 dt P
20
0
70 t 60 t P 20 dt 4200 130t t 2 dt 0 70 60 4200
P [4200(20) 130(200) 8000 / 3] 14.444 P 4200
Question #283 Answer: A Note that this is the same as Question 33, but using multi-state notation rather than multiple-decrement notation. The only way to be in State 2 one year from now is to stay in State 0 and then make a single transition to State 2 during the year. 1
1
1
p t p dt e 02 x
0
00 x
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0
(0.3 0.5 0.7) t
e 1.5t 1 (1 e 1.5 ) 0.259 0.5dt 0.5 1.5 0 3
135
Question #284 Answer: C
m 1 m2 1 ( 80 ) . Then, 2m 12m 2 2 1 22 1 4 1 42 1 (4) a80(2) a80 a80 ( ) a ( 80 ) 80 80 2(2) 12(2) 2 2(4) 12(4) 2 8.29340 8.16715 (1/ 4) (3 / 8) [(3 / 48) (15 /192)]( 80 )
(m) Woolhouse’s formula to three terms is a80 a80
0.00125 0.015625( 80 )
80 0.08 (2) a80 a80 1/ 4 (3 / 48)(0.08) 8.5484.
The answer is: (12) a80 a80
12 1 122 1 11 143 ( 80 ) 8.54840 (0.08) 8.08345. 2 2(12) 12(12) 24 1728
Question #285 Answer: B 1
| a70:2 v 2 2 p70 v3 3 p70 B x t c ( c 1) At ln c
t
p70 e
2
p70 e 0.0002(2) (0.9754483)1.211 0.9943956
3
p70 e0.0002(3) (0.9754483)1.3311 0.9912108
1
e
e
0.000003 70 t 1.1 (1.1 1) 0.0002 t ln(1.1)
e
e0.0002t (0.9754483)1.1 1 t
| a70:2 0.9943956 /1.052 0.9912108 /1.053 1.7582.
Question #286 Answer: E 1 A50:2 vq50 v 2 p50 q51
px e
B x c ( c 1) ln c
e
0.000005 x 1.2 (0.2) ln(1.2)
e 0.0000054848(1.2)
50
p50 0.951311 p51 0.941861 1 A50:2 0.048689 /1.03 0.951311(0.058139) /1.032 0.09940.
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Question #287 Answer: E
The reserve at the end of year 2 is 10, 000vq52 3 where 3 is the benefit premium in year 3. From the equivalence principle at time zero: 10, 000(vq50 v 2 p50 q51 v 3 p50 p51q52 ) 0.5 3 0.5 3vp50 3v 2 p50 p51 0.95 0.95(0.94) 0.05 0.95(0.06) 0.95(0.94)(0.07) 10, 000 0.5 0.5 3 2 3 1.04 1.04 1.04 1.042 1.04 1563.4779 1.78236 3
3 877.1953 V 10, 000(0.7) /1.04 877.1953 204.12
2
Question #288 Answer: C
The benefit premium in year one is vq50 . By the equivalence principle: 10, 000(vq50 v 2 p50 q51 v 3 p50 p51q52 ) 10, 000vq50 (vp50 v 2 p50 p51 ) 0.95(0.06) 0.95(0.94)(0.07) 0.95 0.95(0.94) 10, 000 2 1.043 1.042 1.04 1.04 1082.708665 1.739090 622.5719 2V 10, 000(0.07) /1.04 622.5719 50.51.
Question #289 Answer: E t-th year 0* 1 2 3
V
P
E
I
EDB
E tV
0
0
1000
0
0
0
-1000 1.000
0 700 700
14500 14500 14500
100 100 100
864 906 906
14000 15000 16000
690.2 689.5 0
573.8 1.000 316.5 0.986 6 0.971
t 1
Pr
t 1
px
P 1000.0 573.8 312.1 5.8
NPV 1000 573.8v 312.1v 2 5.8v 3 216.08 using a 10% discount rate.
*The 1000 at time 0 is neither accumulated nor discounted. The value is treated as occurring at the end of time 0 and not as occurring at the beginning of year 1.
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Question #290 Answer: C NPV 140v 0.95(135) p67 v 2 (0.95) 2 (130) 2 p67 v 3 314.09 at a 10% discount rate.
Question #291 Answer: B Solution Pr (6, 000 700 10)(1.06)1/12 0.002(200, 000) 0.0015(50, 000 15, 000) 0.9965(6, 200) 46.7.
Question #292 Answer: C
245 p40 274 and 300 2 p40 395 Present value of premium: 1000[1 (245 / 274)(1/1.12) (300 / 395)(1/1.122 )] 2403.821. Present value of profit: 400 150 / 1.12 245 / 1.12 2 300 / 1.123 142.775. PV Profit / PV premium = 5.94%
Question #293 Answer: B
P[1 0.97(0.98624) 0.92(0.98624)(0.98499)] 100, 000[0.97(0.01376) 0.92(0.98264)(0.01501) 0.87(0.98264)(0.98499)(0.01638)] P 1431.74.
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Question #294 Answer: A Let IF equal 1 if the index drops below its current level in the next year and equals 0 otherwise. E[ X N | I F 1] N (1000)vqx 48.54 N E[ X N | I F 0] 0 E[ X N ] 0.1(48.54 N ) 0 4.854 N Var[ X N | I F 1] (1000v) 2 Nqx (1 qx ) 44, 773.31N Var[ X N | I F 0] 0 E[Var ( X N | I F )] 0.1(44, 773.31N ) 0 4, 477.33 N Var[ E ( X N | I F )] 0.1(48.54 N ) 2 0 (4.854 N ) 2 212.05 N 2 Var[ X N ] 4, 477.33 N 212.05 N 2 Var ( X 10 ) 10 lim
25.69
Var ( X N )
N
lim
N N 25.69 14.56 11.13.
4, 477.33 N 212.05 N 2 212.05 14.56 N
Question #295 Answer: E The actuarial present value of the death benefit is
S62 (4)d 62(2) v 0.5 S63 (4)d 63(2) v1.5 S64 (4)d 64(2) v 2.5 100, 000 S62l62 100, 000
3.589(4)(213)v 0.5 3.643(4)(214)v1.5 3.698(4)(215)v 2.5 3.589(52,860)
4,585.
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Question #296 Answer: A Policy 1: AV371 [ AV36 G E (100, 000 AV371 ) 1/12 q63 /1.004](1.004)
( AV36 G E )1.004 100, 000 1/12 q63 AV371 1/12 q63 [( AV36 G E )1.004 100, 000 1/12 q63 ] / (1 1/12 q63 ) Policy 2 AV372 [ AV36 G E 100, 000 1/12 q63 /1.004](1.004)
( AV36 G E )1.004 100, 000 1/12 q63 Because the starting account value, G and E are identical for both policies: AV361 / AV362 1/ (1 1/12 q63 ) 1 / [1 (1 / 12)(0.01788)] 1.0015.
Question #297 Answer: E The recursive formula for the account values is: AVk 1 ( AVk 0.95G 50)1.06 (100, 000 AVk 1 )q50 k . This is identical to the recursive formula for benefit reserves for a 20-year term insurance where the benefit premium is 0.95G – 50. Because the benefit reserve is zero after 20 years, using this premium will ensure that the account value is zero after 20 years. Therefore,
0.95G 50 100, 000
1 A50:20 A E A 0.24905 0.23047(0.51495) 50 20 50 70 100, 000 1154.55 13.2668 0.23047(8.5693) a50:20 a50 20 E50 a70
G = (1154.55+50)/0.95 = 1,267.95.
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Question #298 Answer: A 2
5, 000 E ( Z ) (5, 000 /1.030) (0.005) (1 0.005)(0.006) 1.030(1.032) 2
2
2
5, 000 (1 0.005)(1 0.006)(0.007) 1.030(1.032)(1.035) 392,917
Question #299 Answer: E 44.75 0.00004(1.144.75 ) 0.002847
44.5 0.00004(1.144.5 ) 0.002780 1000 E ( 4.75 L) 0.25{0.04 E ( 4.75 L) 150 150(0.05) 0.002847[10, 000 100 E ( 4.75 L)]} E ( 4.75 L) {1000 0.25[150 150(0.05) 0.002847(10, 000 100)]} / [1 0.25(0.04 0.002847)] 961.27 961.27 E ( 4.5 L) 0.25{0.04 E ( 4.5 L) 150 150(0.05) 0.002780[10, 000 100 E ( 4.5 L)]} E ( 4.5 L) {961.27 0.25[150 150(0.05) 0.002780(10, 000 100)]} / [1 0.25(0.04 0.002780)] 922.795
Question #300 Answer: B Assumed profit = 990*[(100 + 90 – 90*0.030)*(1.05) – 0.003*10,000 – (1 – 0.003)*125] = 41,619.60 Actual profit = 990*[(100 + 90 – 90*0.025)*(1.04) – 0.002*10000 – (1 – 0.002)*125] = 50,004.90 Total gain from mortality, interest and expense = 50,004.90 – 41,619.60 = 8,385.30.
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