November 2013 MLC Solutions
1.
Key: B
A x:1n = n E x
A x = Ax1:n + n E x A x + n
0.3 = A1x:n + (0.35)(0.4)  A1x:n = 0.16
A x :n = A 1x :n + n E x = 0.16 + 0.35 = 0.51
a x:n
2.
1 − A x:n
1 − 0.51
= 10.29
d
(0.05 /1.05)
= ax:n − 1 + n E x = 10.29 − 0.65 = 9.64
ax:n =
=
Key: C
a xy = a x + a y − a xy = 10.06 + 11.95 − 12.59 = 9.42
a xy =
1 − A xy
δ
1 − A xy
 A xy = 0.34
0.07
A xy = A 1xy + A xy1
9.42 =
0.34 = A1xy + 0.09  A1xy = 0.25
3.
Key: E
2.2
q[51]+ 0.5 =
l[51]+ 0.5 − l53.7
l[51]+ 0.5
l[51]+ 0.5 = 0.5l[51] + 0.5l[51]+1 = 0.5(97, 000) + 0.5(93, 000) = 95, 000
l53.7 = 0.3l53 + 0.7l54 = 0.3(89, 000) + 0.7(83, 000) = 84,800
95, 000 − 84,800
= 0.1074
95, 000
10, 0002.2 q[51]+ 0.5 = 1, 074
2.2
q[51]+ 0.5 =
Page 1 of 13
4.
Key: B
 0.05 


Prob( H → D in 2 months) = (0.75 0.2 0.05)  0.20  = 0.1275
1 


You could do more extensive matrix multiplication and also obtain the probability that it
is H after 2 or it is S after 2, but those aren’t needed.
Let D be the number of deaths within 2 years out of 10 lives
Then D~binomial with n = 10,
p = 0.1275
 10 
P ( D = 4) =   (0.1275) 4 (1 − 0.1275)6 = 0.0245
4 
5.
Key: A

e 40 =
1
= 50 So receive K for 50 years guaranteed and for life thereafter.
μ
10, 000 = K  a50 + 50| a40 
50
a 50 =  e −δ t =
0
50|
a40 =
K=
50
1 − e −50δ
δ
=
1 − e−50(0.01)
= 39.35
0.01
E40 a40+50 = e − (δ + μ )50
1
1
= e−1.5
= 7.44
μ +δ
0.03
10, 000
= 213.7
39.35 + 7.44
Page 2 of 13
6.
Key: C
Simplest solution is retrospective:
q [55] = (0.7)(0.00896) = 0.006272
V=
1
(24.453)(1.06) − (1000)(0.006272)
= 19.77
1 − 0.006272
Prospectively, q [55]+1 = (0.8)(0.00975) = 0.0078; q [55]+ 2 = (0.9)(0.01062) = 0.009558
A[55]+1 = (1000)(0.0078)v + (1000)(1 − 0.0078)(0.009558)v 2
(
+ (1 − 0.0078)(1 − 0.009558)(342.65)v 2 = 315.49
)
a[55]+1 = 1 − A[55]+1 / d = (1 − 0.31549) / (0.06 /1.06) = 12.093
V = 315.49 − (12.093)(24.453) = 19.78
1
7.
Key: A
Let P = 0.00258 be the monthly benefit premium per 1 of insurance.
i 1

A55:10 + A55:101 − 12 Pa(12)
10V = 100, 000 
55:10 
δ

= 100, 000 [1.02971(0.09102) + 0.48686 − (12 × 0.00258)(7.21928) ]
≈ 35, 700
Where
1
= A55 − 10 E 55 A65 = 0.30514 − (0.48686)(0.43980) = 0.09102
A55:10
A55:101 = 10 E 55 = 0.48686
a55:10 = a55 − 10 E 55 a65 = 12.2758 − (0.48686)(9.8969) = 7.45740
a(12)
= α (12)a55:10 − β (12) 1 − 10 E 55 
55:10
= 1.00028(7.45740) − 0.46812(1 − 0.48686) = 7.21928
Page 3 of 13
8.
Key: C
Use superscript g for gross premiums and gross premium reserves.
Use superscript n (representing “net”) for benefit premiums and benefit reserves.
Use superscript e for expense premiums and expense reserves.
P g = 1, 605.72 (given)
e
P =
0.58P g + 450 + ( 0.02 P g + 50 ) a45
a45
0.58(1, 605.72) + 450 + [0.02(1, 605.72) + 50]14.1121
= 180.00
14.1121
Alternatively,
100, 000 A45
Pn =
= 1425.73
P e = P g − P n = 180
a45
=
V e = ( 0.02 P g + 50 ) a50 − P e a50 = [0.02(1, 605.72) + 50](13.2668) − 180(13.2668) = −1, 298.63
5
Alternatively,
n
= 100, 000 A50 − P n a50
5V
= 100, 000(0.24905) − 1, 425.73(13.2668) = 5,990.13
V g = 100, 000 A50 + ( 50 + 0.02 P g − P g ) a50
5
= 100, 000(0.24905) + [50 + 0.02(1, 605.72) − 1, 605.72](13.2668) = 4, 691.63
V e = 5V g − 5V n = −1298.51
5
9.
Key: D
a50:10 = a50 − 10 E 50 a60 = 13.2668 − 0.51081(11.1454) = 7.5736
1
A50:20
= A50 − 20 E 50 A70 = 0.24905 − 0.23047(0.51495) = 0.13037
a50:20 = a50 − 20 E 50 a70 = 13.2668 − 0.23047(8.5693) = 11.2918
APV of Premiums = APV Death Benefit + APV Commission and Taxes + APV
Maintenance
1
Ga50:10 = 100, 000 A50:20
+ 0.12Ga50:10 + 0.3G + 25a50:20 + 50
7.5736G = 13, 037 + 1.2088G + 332.30
6.3648G = 13,369
 G = 2,100
Page 4 of 13
10.
Key: E
Let A denote Alive, which is equivalent to not Dead. It is also equivalent to Healthy or
Disabled. Let H denote Healthy. The conditional probability is:
P( H and A)
P( H )
P( H A) =
=
,
P ( A)
P( H ) + P (Disabled)
Where
10
10
−  ( μ 01 + μ 02 ) ds
−  (0.05) ds
00
0
0
P ( H ) = 10 p = e
=e
= e − 0.5 = 0.607
And
u
10
01
02
−
μ12 ds
10 −  ( μ + μ ) ds
01
01  u
0
P (Disabled) = 10 p =  e
μ e
du
0
10
=  e −0.05u (0.02)e 0.05u −0.5 du
0
10
=  (0.02)e −0.5 du
0
= (0.02)e −0.5 = 0.121
Then
P( H A) =
11.
P( H )
0.607
=
= 0.83
P( H ) + P(Disabled) 0.607 + 0.121
Key: A
We need to adjust the cash flows at time 4 and time 5.
a80:5 = 4.3868 +
870  1
1 
×
−
= 4.3589
4
1, 000 1.05 1.044 
1
= 0.1655 +
A80:5
50  1
1  60  1
1 
×
−
+
×
−
= 0.1594
4
4
5
1000 1.05 1.04  1000 1.06 1.04 5 
Benefit Premium = 100, 000 ×
0.1594
= 3657
4.3589
Page 5 of 13
12.
Key: B
The probability that the endowment payment will be made for a given contract is:
15
(
15
p x = exp −  0.02t dt
(
0
= exp −0.01t 2
15
0
)
)
= exp ( −0.01(15) 2 )
= 0.1054
Because the premium is set by the equivalence principle, we have E [ 0 L ] = 0. Further,
Var ( 0 L ) = 500 (10, 000v 15 )

= 1,942,329, 000
2
(
15
p x )(1 − 15 p x ) 

Then, using the normal approximation, the approximate probability that the aggregate
losses exceed 50,000 is

50, 000 − 0 
P ( 0 L > 50, 000 ) = P  Z >
 = P ( Z > 1.13) = 0.13
1,942,329, 000 

13.
Key: B
Time
Age
q xILT
0
1
2
70
71
72
0.03318
0.03626
0.03962
Improvement
factor
100.00%
95.00%
90.25%
qx
0.03318
0.03445
0.03576
v = 1/1.06 = 0.943396
EPV = 1, 000[0.03318v + 0.96682(0.03445)v 2 + 0.96682(0.96555)(0.03576)v 3 ]
= 88.97
Page 6 of 13
14.
Key: D
Under the Equivalence Principle
(
) (
1
Pa62:10 = 50, 000 a62 − a62:10 + P ( IA) 62:10
)
10
1
1
1
where ( IA) 62:10
= 11A62:10
−  A62:
= 11(0.091) − 0.4891 = 0.5119
k
So
15.
P=
(
k =1
50, 000 a62 − a62:10
1
a62:10 − ( IA)62:10
) = 50, 000(12.2758 − 7.4574) = 34, 687
7.4574 − 0.5119
Key: D
ax :3 =
Actuarial PV of the benefit 152.85
=
= 2.727
Level Annual Premium
56.05
ax:3 = 1 +
0.975 0.975 ( p x +1 )
+
= 2.727
1.06
(1.06) 2
 p x +1 = 0.93
Actuarial PV of the benefit =
 0.025 0.975(1 − 0.93) 0.975(0.93) ( q x + 2 ) 

+
+
152.85 = 1, 000 
(1.06) 2
(1.06)3
 1.06

 q x + 2 = 0.09  px + 2 = 0.91
16.
Key: C
For calculating P
A50 = vq 50 + vp 50 A51 = v (0.0048) + v (1 − 0.0048)(0.39788) = 0.38536
a50 = (1 − A50 ) / d = 15.981
P = A50 / a50 = 0.02411
For this particular life,
A′50 = vq′50 + vp′50 A51 = v (0.048) + (1 − 0.048)(0.39788) = 0.41037
a′50 = (1 − A′50 ) / d = 15.330
Expected PV of loss = A′50 − Pa′50 = 0.41037 − 0.02411(15.330) = 0.0408
Page 7 of 13
17.
Key: B
L = 10, 000v
K 45 +1
− PaK
= 10, 000v 11 − Pa11
45 +1
4450 = 10, 000(0.58468) − 8.7217P
P = (5,846.8 − 4, 450) / 8.7217 = 160.15
A55 = 1 − da55 = 1 − (0.05 / 1.05)(13.4205) = 0.36093
V = 10, 000 A55 − Pa55 = (10, 000)(0.36093) − (160.15)(13.4205) = 1, 460
10
18.
Key: E
1,020 in the solution is the 1,000 death benefit plus the 20 death benefit claim expense.
A x = 1 − da x = 1 − d (12.0) = 0.320755
Ga x = 1, 020 A x + 0.65G + 0.10Ga x + 8 + 2a x
G=
1, 020 A x + 8 + 2ax
ax − 0.65 − 0.10ax
=
1, 020(0.320755) + 8 + 2(12.0)
= 35.38622
12.0 − 0.65 − 0.10(12.0)
Let Z = v K x +1 denote the present value random variable for a whole life insurance of 1 on
(x).
Let Y = aK +1 denote the present value random variable for a life annuity-due of 1 on (x).
x
L = 1, 020Z + 0.65G + 0.10GY + 8 + 2Y − GY
= 1, 020 Z + (2 − 0.9G )Y + 0.65G + 8
1 − v K x +1
= 1, 020v
+ (2 − 0.9G )
+ 0.65G + 8
d
0.9G − 2  K x +1 2 − 0.9G

= 1, 020 +
+
+ 0.65G + 8
v
d
d


K x +1
0.9G − 2 

Var ( L) =  2 Ax − ( Ax ) 2  1, 020 +

d


2
0.9(35.38622) − 2 

= (0.14 − 0.320755 ) 1, 020 +

d


= 0.037116(2,394,161)
2
= 88,861
Page 8 of 13
2
19.
Key: D
If T45 = 10.5 , then K 45 = 10 and K 45 + 1 = 11.
0
L = 10, 000v
K 45 +1
− G (1 − 0.10)aK
45 +1
+ G (0.80 − 0.10) = 10, 000v 11 − 0.9Ga11 + 0.7G
3, 767 = 10, 000(0.52679) − 0.9G (8.3601) + 0.7G
G = (5, 267.9 − 3767) / (6.8241) = 219.94
E ( 0 L) = 10, 000 A45 − (1 − 0.1)Ga45 + (0.8 − 0.1)G
= (10, 000)(0.20120) − (0.9)(219.94)(14.1121) + (0.7)(219.94)
E ( 0 L) = −627.48
20.
Key: E
Defined Benefit:
0.015 × Final Average Earnings × Years of Service
(
)
= 0.015 × 50, 000 × (1.05 19 + 1.05 18 + 1.05 17 ) / 3 × 20 = 36,128 per year
APV at 65 of Defined Benefit = 36,128a65 = 36,1268(10.0) = 361, 280.
Defined Contribution accumulated value at65:
X % × 50, 000 × 1.05 20 + X % × (50, 000 × 1.05) × 1.05 19 +  + X % × ( 50, 000 × 1.05 19 ) × 1.05
= X % × 50, 000 ×1.05 20 × 20 = X %(2, 653, 298)
Therefore,
361, 280 = X %(2, 653, 298)
X % = 0.136
X = 13.6
Page 9 of 13
21.
Key: A
t
t
t
p x00 = exp  −  ( μ x01+ s + μ x02+ s ) ds  = exp  −  ( 0.20 + 0.10s + 0.05 + 0.05s ) ds 
 0

 0

t

= exp  − ( 0.25s + 0.075s 2 )  = exp ( −0.25t + 0.075t 2 )
0

3
p x00 = exp ( −0.25 × 3 + 0.075 × 9 ) = exp(−1.425) = 0.2405
−δ t
dt
EPV =  g (t )dt , where g (t ) = 10, 000 × ( t p x00 μ x02+ t + t p x01μ 12
x+t ) e
n
0
g (3) = 10, 000 × 0.2405 × ( 0.05 + 0.05 × 3) + 0.4174 × ( 0.15 + 0.01× 3 2 )  × e −3×0.02 = 1, 400
22.
Key: B
AV11 = 1500
10, 000(0.003)
= 29.88
1.004
AV12 = (1, 500 + 100(1 − 0.15) − 10 − 29.88)(1.004) = 1, 551.3
COI 12 =
SV12 = 1551.3 − 400 = 1151.3
Then
1,151.3 = X a10 + 10 E 61 a71 = X (7.8017 + 0.44231(8.2988)) = 11.4723 X
(
)
X = 100.35
Page 10 of 13
23.
Key: E
AV 0 = 0
150, 000(0.00122) 

AV1 = 3, 000(1 − 0.7) − 75 −
 (1.04) = 675
1.04

150, 000(0.00127) 

AV 2 = 675 + 3, 000(1 − 0.1) − R −
 (1.04)
1.04

= [ (3375 − R ) − 183.17 ] (1.04)
= 3319.50 − R(1.04)

150, 000(0.00133) 
AV 3 = 6, 028.95 = 3,319.50 − R(1.04) + 3, 000(1 − 0.1) − R −
 (1.04)
1.04

 [ 6, 019.50 − 2.04 R − 191.83] (1.04)
6060.78 − 2.12 R = 6028.95
 R = 15
Page 11 of 13
24.
Key: B
Let S denote the number of survivors.
This is a binomial random variable with n = 4000 and success probability
2,358, 246
= 0.24655
9,565, 017
E ( S ) = 4, 000(0.24655) = 986.2
The variance is Var ( S ) = (0.24655)(1 − 0.24655)(4, 000) = 743.05
StdDev ( S ) = 743.05 = 27.259
The 90% percentile of the standard normal is 1.282
Let S* denote the normal distribution with mean 986.2 and standard deviation 27.259.
Since S is discrete and integer-valued, for any integer s,
Pr( S ≥ s ) = Pr( S > s − 0.5) ≈ Pr( S * > 0.5)
 S * −986.2 s − 0.5 − 986.2 
=
>

27.259
 27.259

For 90%,
s − 0.5 − 986.2
< −1.282
27.259
 s < 951.754
So
s = 951 is the largest integer that works
s = 950 is the largest from the list
Page 12 of 13
25.
Key: E
Using UDD
l 63.4 = (0.6)66,666 + (0.4)(55,555) = 62, 221.6
l 65.9 = (0.1)(44, 444) + (0.9)(33,333) = 34, 444.1
q =
3.4 2.5 60
l 63.4 − l 65.9
l 60
=
62, 221.6 − 34, 444.1
= 0.277778
99,999
(a)
Using constant force
l 63.4
 l 64 
= l 63  
 l 63 
 
0.4
= l 630.6 l 640.4
= ( 66, 666 0.6 )( 55,555 0.4 )
= 61,977.2
l 65.9 = l 650.1 l 660.9 = ( 44, 444 0.1 )( 33,333 0.9 )
= 34,305.9
3.4 2.5
61,977.2 − 34,305.9
99.999
= 0.276716
q 60 =
(b)
100,000(a − b) = 100,000(0.277778 − 0.276716) = 106
Page 13 of 13