Fundamentals of Actuarial Mathematics – Long Term
Solutions to Sample Multiple Choice Questions
June 7, 2023
Versions:
June 7, 2023
Fixed solution to 5.12.
March 6, 2023
Fixed answer key typo in sample solution to 7.34 to match the exam
answer key.
December 16, 2022
Added solutions for Fall 2022 exam questions
August 8, 2022
Revised solution to questions 18.1 and 18.4
July 29, 2022
Updated term “expense reserve” to “expense policy value”.
July 1, 2022
Original version for FAM-L.
These solutions are written to provide evidence that the stated answer is correct and to offer at
least one approach to solving the problem. There may be additional approaches that can be
employed.
Copyright © Society of Actuaries
Page 1
Question 1.1
Answer C
Answer C is false. If the purchaser of a single premium immediate annuity has higher mortality
than expected, this reduces the number of payments that will be paid. Therefore, the Actuarial
Present Value will be less and the insurance company will benefit. Therefore, single premium
life annuities do not need to be underwritten.
The other items are true.
A: Life insurance is typically underwritten to prevent adverse selection as higher mortality than
expected will result in the Actuarial Present Value of the benefits being higher than expected.
B: In some cases, such as direct marketed products for low face amounts, there may be very
limited underwriting. The actuary would assume that mortality will be higher than normal, but
the expenses related to selling the business will be low and partially offset the extra mortality.
D: If the insured’s occupation or hobby is hazardous, then the insured life may be rated.
E: If the purchaser of the pure endowment has higher mortality than expected, this reduces the
number of endowments that will be paid. Therefore, the Actuarial Present Value will be less and
the insurance company will benefit. Therefore, pure endowments do not need to be underwritten.
Question 1.2
Answer E
Insurers have an increased interest in combining savings and insurance products so Item E is
false.
The other items are all true.
Question 1.3
Answer C
Key is that the benefit is linked to the performance of an investment fund. That is true only for
C.
Question 1.4
Answer B
Key is that benefit is contingent upon being sick, but not related to long-term care. Replacing
salary and benefits tied to being unable to work also fit best with B.
Page 2
Question 2.1
Answer: B
1
t 4
1 ω − t 

Since S0 (t ) =
1 − F0 (t ) =
ln
.
1 −  , we have ln[S0 (t )] =
4  ω 
 ω
Then µt =
−
e106
1 1
1
1
1
d
log S0 (t ) =
, and µ65 = =
110.
⇒ω =
4 ω −t
180 4 ω − 65
dt
3
0
∑=
t p106 , since 4 p106
t =1
1/4
 106 + t 
1−
1/4

S0 (106 + t ) 
110 
 4−t 
=
=
=
t p106


1/4
S0 (106)
 4 
 106 
1 −

 110 
e106
=
i =4
∑
i =1
Page 3
t
p106
=
1
4
0.25
(1
0.25
+ 2 0.25 + 3 0.25=
) 2.4786
Question 2.2
Answer: D
This is a mixed distribution for the population, since the vaccine will apply to all once available.
Available?
(A)
Yes
No
S = # of survivors
Pr( A)
0.2
0.8
p| A
E ( S A)
Var ( S A)
E ( S 2 A)
0.9702
0.9604
97,020
96,040
E (S )
96,236
Var ( S )
SD( S )
2,891
3,803
9,412,883,291
9,223,685,403
E (S 2 )
9,261,524,981
2
157,285
397
As an example, the formulas for the “No” row are
Pr(No) = 1 – 0.2 = 0.8
2
p given No = (0.98 during year 1)(0.98 during year 2) = 0.9604.
E ( S | No), Var ( S | No) and E ( S 2 | No) are just binomial, n = 100, 000; p(success) = 0.9604
E ( S ), E ( S 2 ) are weighted averages,
Var
=
(S ) E (S 2 ) − E (S )2
Or, by the conditional variance formula:
=
Var ( S ) Var[ E ( S | A)] + E[Var ( S | A)]
= 0.2(0.8)(97, 020 − 96, 040) 2 + 0.2(2,891) + 0.8(3,803)
= 153, 664 + 3, 621= 157, 285
StdDev( S ) = 397
Page 4
Question 2.3
Answer: A
d
d  − B ( c x )( ct −1) 
f x (t ) =
− Sx (t ) =
−  e ln c

dt
dt 

=−e
= e
−
−
( )(
) 
B x t
c c −1
ln c
( )(
)
B x t
c c −1
ln c
B x t
⋅ −
⋅ c  ⋅ c ⋅ ln c
 ln c

⋅ Bc x +t
= 0.00027 ×1.1
x +t
⋅e
−
0.00027
1.1x 1.1t −1
ln (1.1)
( )(
50 +10
f50 (10) = 0.00027 ×1.1
⋅e
−
)
0.00027
1.150 1.110 −1
ln (1.1)
(
)(
)
= 0.04839
Alternative Solution:
f x=
(t )
t
px ⋅ µ x +t
=
=
A 0,=
B 0.00027 and
c 1.1
Then we can use the formulas given for Makeham with
 − 0.00027
(1.150 )(1.110 −1) 
ln (1.1)
=
f x (t )  e
0.00027 ×1.150+10 ) 0.04839
=

(


Page 5
Question 2.4
Answer: E

=
e75:10
=∫
10
0
∫
t =10
t =0
t
p75 dt
where t=
px
p0
=
x p0
t+x
(t + x )
1−
2
10000 − ( t + x )
10000
=
2
x
10000 − x 2
1−
10000
10000 − 752 − 150t − t 2
dt
10000 − 752
t =10
1 
t3 
=
⋅  4375t − 75t 2 −  = 8.21
4375 
3  t =0
Question 2.5
Answer: B
e40 = e40:20 + 20 p40 ⋅ e60
= 18 + (1 − 0.2 ) (25)
= 38
e40 = e40:1 + p40 ⋅ e41
⇒=
e41
e40 − e40:1 e40 − p40 38 − 0.997
=
=
= 37.11434
p40
p40
0.997
Question 2.6
Answer: C
d
x 
1 d 
ln 1 − 
− lnS0 ( x ) =
−
µx =
dx
3 d x  60 
−1
x 
1 
1
=
1 −  =
180  60 
3 ( 60 − x )
1
1000
= 13.3.
3 ( 25 )
75
µ35 (1000) =
Therefore, 1000=
Page 6
2
for 0 < t < 100 − x
Question 2.7
Answer: B
20
q30 =
S0 ( 30 ) − S0 ( 50 )
S0 ( 30 )
2
30    50   220 3

1
−
−
1
−


 
 250   100   250 − 4
=
=
220
30
1−
250
250
440 − 375 65 13
=
= = 0.1477
440
440 88
=
Question 2.8
Answer: C
The 20-year female survival probability = e −20 µ
The 20-year male survival probability = e −30 µ
We want 1-year female survival = e − µ
Suppose that there were M males and 3M females initially. After 20 years, there are expected to
be Me −30 µ and 3Me −20 µ survivors, respectively. At that time we have:
1
3Me −20 µ 85
85 17
 9  10
−µ
10 µ
e
⇒
=
e
=
⇒
=
=
  = 0.938
Me −30 µ 15
45 9
 17 
Question 2.9
Answer: E
The distribution is binomial with 10,000 trials.
Var[ L=
npq
= 10, 000(15 p50 )(15 q50 )
15 ]
[ − A (15) −
B
*c50 ( c15 −1)]
ln C
0.837445
e=
=
15 p50
15
1 −15 p50 =
0.162555
q50 =
000(0.837445)(0.162555) 1361.3
Var[ L15 ] 10,
=
Page 7
Question 2.10
Answer: A
10
10
− ∫ µx +t ⋅dt
− ∫ β t ⋅dt
3
10
lx +10
400
0
0
=
=
=
=
>
=
=
=
> 0.4 =
p
e
e
e− β t /3 |0
10 x
lx
1000
2
3
 1000 
e− β ⋅100 /3 =
=
=
> 0.4 =
=
> ln(0.4) =
−β 
=
>β =
− ln(0.4)(.003) =
0.0027489
=
 3 
Question 2.11
Answer: C
Let l80 = 1000 ⇒ l81 = 960 ⇒ l82 = 902.4
l81.1 − l81.6 (960)0.9 (902.4)0.1 − (960)0.4 (902.4)0.6
=
= 0.02978
l80.6
(0.4)(1000) + (0.6)(960)
=
Answer
The value for l80 is arbitrary. Any other starting value gives the same result.
Another form for the survivors between 81 and 82 would be
µ=
− ln(902 / 960) =
0.0095833;
l81.1 =
960 × e −0.1×0.0095833 =
959.08;
l81.6 =
960 × e −0.6×0.0095833 =
954.496.
Question 2.12
Answer: A
1
t2 + t
t3 t2 8
)dt → [72t − − ]0 = 5.1852
E[T0 ]= ∫ t p0 dt = ∫ (1 −
0
0
72
72
3 2
8
2 8
2
t4 t3 8
3
2
2
(72
)
[36
]0 30.815
E[=
T0 2 ] 2 ∫ ( t p0 ×=
t )dt
t
−
t
−
t
=
dt
t
−
−=
0
72 ∫0
72
4 3
3.9287
Var[T0 ] =
E[T0 2 ] − ( E[T0 ]) 2 =
8
Page 8
8
Question 3.1
Answer: B
Under constant force
over each year of age, lx + k (lx )1− k (lx +1 ) k for x an integer and 0 ≤ k ≤ 1.
=
23
q [60] + 0.75 =
l [60] + 2.75 − l [60] + 5.75
l [60] + 0.75
l [60] + 0.75 (80,
=
=
000)0.25 (79, 000)0.75 79, 249
l [60] + 2.75 = (77, 000)0.25 (74, 000)0.75 = 74, 739
l [60] + 5.75 (67,
000)0.25 (65, 000)0.75 65, 494
=
=
q
=
2 3 [60] + 0.75
l [60] + 2.75 − l [60] + 5.75 74, 739 − 65, 494
= = 0.11679
l [60] + 0.75
79, 249
1000 2 3 q [60] + 0.75 = 116.8
Question 3.2
Answer: D
l 65+1= 1000 − 40= 960
l 66+1 = 955 − 45 = 910

1
e[65] =
∫ t p [65]dt + p [65]
0
∫
1
0 t

p 66 dt + p [65] p 66 e 67
  1   40   960   1   50    960  910  
15.0 =
1 −  2   1000   + 1000 1 −  2  960   +  1000  960  e 67
 



  
  
15(1000) − (980 + 935)

e 67 = 14.37912
910
1
  1   45    910  


=
e[66] ∫ t p [66]dt + p [66] e 67 =
1 −  2   955   +  955  e 67
0
 

  
  1   45    910 

14.678
e[66] =
1 −  2   955   +  955  (14.37912 ) =
 

  
Note that because deaths are uniformly distributed over each year of age,
Page 9
∫
1
0 t
px dt = 1 − 0.5qx .
Question 3.3
Answer: 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
Question 3.4
Answer: B
Let S denote the number of survivors.
This is a binomial random variable with n = 4000 and success probability
21,178.3
= 0.21206
99,871.1
=
E ( S ) 4,=
000(0.21206) 848.24
The variance is Var ( S ) =
(0.21206)(1 − 0.21206)(4, 000) =
668.36
StdDev
=
(S ) =
668.36 25.853
The 90% percentile of the standard normal is 1.282
Let S* denote the normal distribution with mean 848.24 and standard deviation 25.853. Since S
is discrete and integer-valued, for any integer s,
For this probability to be at least 90%, we must have
⇒ s < 815.6
So s = 815 is the largest integer that works.
Page 10
s − 0.5 − 848.24
< −1.282
25.853
Question 3.5
Answer: 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
l 63.4 − l 65.9 62, 221.6 − 34, 444.1
=
= 0.277778
l 60
99,999
=
q
3.4 2.5 60
(a)
Using constant force
0.4
 l 64 
=
l 63.4 l =
l 630.6 l 640.4
63 
 l 63 
 
= ( 66, 666 0.6 )( 55,555 0.4 )
= 61,977.2
0.1 0.9
=
l 65.9 l =
65 l 66
( 44, 444 )( 33,333 )
0.1
0.9
= 34,305.9
3.4 2.5
q 60 =
61,977.2 − 34,305.9
99.999
= 0.276716
100,
=
000(a − b) 100, 000(0.277778
=
− 0.276716) 106
Question 3.6
Answer: D
=
e[61] e[61]:3 + 3 p [61] (e 64 )
p [61] = 0.90,
=
0.9(0.88)
= 0.792,
2 p [61]
=
0.792(0.86)
= 0.68112
3 p [61]
3
e[61]:3 =
0.9 + 0.792 + 0.68112 =
2.37312
∑ k p[61] =
k =1
e[61] =
2.37312 + 0.68112e64 =
2.37312 + 0.68112(5.10) =
5.847
Page 11
(b)
Question 3.7
Answer: B
2.5
q [50]+ 0.4 =
1 − 2.5 p [50]+ 0.4 =
1 − 2.9 p [50] / ( p [50] )
{
1 {(1 − q
=−
0.4
}
1 − p [50] p [50]+1 ( p 52 )
/ (1 − q [50] )
=
0.9
[50]
{
)(1 − q
[50]+1
)(1 − q )
52
0.9
0.4
} / (1 − q
=−
1 (1 − 0.0050 )(1 − 0.0063)(1 − 0.0080 )
0.9
[50]
)
0.4
} / (1 − 0.0050)
0.4
= 0.01642
1000 2.5 q [50] + 0.4 = 16.42
Question 3.8
Answer: B
 85, 203.5 61,184.9 
E (=
N ) 1000 ( 40 p35 + 40 p=
1000 
+
=
45 )
 1473.65
 99,556.7 99, 033.9 
Var=
( N ) 1000 40 p35 (1 − 40 p35 ) + 1000 40 p45 (1 − 40 =
p45 ) 359.50
Since 1473.65 + 1.645 359.50 =
1504.84
N = 1505
Page 12
Question 3.9
Answer: E
From the SULT, we have:
25
p=
20
l45 99, 033.9
=
= 0.99034
l20 100, 000.0
25
p=
45
l70 91, 082.4
=
= 0.91971
l45 99, 033.9
The expected number of survivors from the sons is 1980.68 with variance 19.133.
The expected number of survivors from fathers is 1839.42 with variance 147.687.
The total expected number of survivors is therefore 3820.10.
The standard deviation of the total expected number of survivors is therefore
19.133 + 147.687 =
166.82 = 12.916
3850
The 99th percentile equals 3820.10 + (2.326)(12.916) =
Question 3.10
Answer: C
The number of left-handed members at the end of each year k is:
=
L0 75
=
and L1 (75)(0.75)
Thereafter, Lk = Lk −1 × 0.75 + 35 × 0.75 = 75 × 0.75k + 35 × ( 0.75 + 0.752 + ...0.75k −1 )
Similarly, the number of right-handed members after each year k is:
=
=
R0 25
and R1 (25)(0.5)
Thereafter, Rk = Rk −1 × 0.50 + 15 × 0.50 = 25 × 0.50k + 15 × ( 0.50 + 0.502 + ...0.50k −1 )
At the end of year 5, the number of left-handed members is expected to be 89.5752, and the
number of right-handed members is expected to be 14.8435.
The proportion of left-handed members at the end of year 5 is therefore
89.5752
= 0.8578
89.5752 + 14.8438
Page 13
Question 3.11
Answer: B
=
q
2.5 50
2
q50 + 2 p50
0.5 
q= 0.02 + ( 0.98 ) 
 ( 0.04 ) = 0.0298
 2 
0.5 52
Question 3.12
Answer: C
l 
=  65 
 l64 
0.5
 l64
l64 
−


 l[61] l[60]+1 
0.5
 4016   5737 5737 
−

 

 5737   8654 9600 
Question 3.13
Answer: B
e=
e[58]+2 + 0.5
58 + 2

=

l61
e60 2210
1
3904
1.100338
×=
×
=
=
l[58]+2 p60 3548 (2210 / 3904) 3549
e58
=
1.100338 +=
0.5 1.6
+2


Page 14
Question 3.14
Answer: C
We need to determine
q
=
3 2.5 90
3 2.5
q90 .
l 90+3 − l 90+3+ 2.5 l 93 − l 95.5 l 93 − (l 95 − 0.5d 95 ) 825 − [600 − 0.5(240)]
= =
=
= 0.3450
l 90
l 90
l 90
1, 000
where
=
l 90 1,=
000, l 93 825 , l=
97
=
l 95
d 97 72
l 97
72
= = 72 , =
l 96
= = 360 ,
q 97
1
p 96 0.2
l 96
360
=
= 600 , and d 95 = l 95 − l 96 = 600 − 360 = 240 .
p 95 1 − 0.4
Question 3.15
Answer: A
− s′(25) 0.5(0.01)(1 − 0.01× 25) −0.5
µ25 =
=
= 0.00667
s (25)
(1 − 0.01× 25)0.5
Question 3.16
Answer: C
[ x]
q[ x ]
q[ x ]+1
q[ x ]+ 2
qx +3
x+3
52
0.030
0.043
0.057
0.072
55
1000(0.6 | 1.5 q[52]+1.7 ) = 1000
l[52]+ 2.3 − l[52]+3.8
l[52]+1.7
l[52] = 1000 (arbitrary, for convenience; any other value would work and give the same answer)
l[52]+1 = 1000(1 − 0.03) = 970
l[52]+ 2 =970(1 − 0.043) =928.29
l55 = 928.29(1 − 0.057)= 875.3775
=
= 812.3503
l56 875.3775(1 − 0.072)
l[52]+ 2.3= (0.7)l[52]+ 2 + (0.3)l55= 912.42
(0.2)l55 + (0.8)l56 =
824.96
l[52]+3.8 =
l55.8 =
l[52]+1.7 = (0.3)l[52]+1 + (0.7)l[52]+ 2 = 940.80
912.42 − 824.96
=
= 92.96
1000(0.6 | 1.5 q[52]+1.7 ) 1000
940.80
Page 15
Question 3.17
Answer: C
15
p
Alive at time 0
Alive at time 15
A
e −15(0.01) = 0.8607
8,000
6,885.66
B
e −15(0.02) = 0.7409
2,000
1,481.64
10,000
8,367.30
Total
Fraction A =
Page 16
6,885.66
= 0.823
8,367.30
Question 4.1
Answer: A
E[ Z ] =2 ⋅ A40 − 20 E40 A60 =
(2)(0.36987) − (0.51276)(0.62567) =
0.41892
E[ Z 2 ] = 0.24954 which is given in the problem.
Var ( Z ) =E[ Z 2 ] − ( E[ Z ]) 2 =
0.24954 − 0.418922 =
0.07405
=
SD( Z )
=
0.07405 0.27212
1
An alternative way to obtain the mean is=
E[ Z ] 2 A40:20
+ 20| A40 . Had the problem asked
for the evaluation of the second moment, a formula is
1
=
E[ Z 2 ] (22 ) ( 2 A40:20
) + (v 2 )20 ( 20 p40 ) ( 2 A60 )
Question 4.2
Answer: D
Half-year
=
300, 000v 0.5 (300,
=
000)(1.09) −1 275, 229
1
2
PV of Benefit
=
=
330, 000v1 (330,
000)(1.09) −2 277, 754
3
360, 000v1.5 (360,
000)(1.09) −3 277,986
=
=
4
=
390, 000v 2 (390,
=
000)(1.09) −4 276, 286
Under CF assumption,=
0.5 p x
=
0.5 p x +1
(
0.5
PV > 277, 000
if and only if
(x) dies in the
2nd or 3rd half
years.
0.5
p x + 0.5 (0.84)
0.9165 and
=
=
0.5
0.5
=
p x +1.5 (0.77)
=
0.8775 Then the probability of dying in the 2nd or 3rd half-years is
0.5
p x )(1 − 0.5 p x + 0.5 ) + ( p x )(1 −=
(0.9165)(0.0835) + (0.84)(0.1225)
= 0.1794
0.5 p x +1 )
Page 17
Question 4.3
Answer: D
A60:3= q60 v + (1 − q60 )q60+1v 2 + (1 − q60 )(1 − q60+1 )v 3= 0.86545
q 60+1
0.01 0.99
A60:3 − q60 v − (1 − q60 )v 3 0.86545 − 1.05 − 1.05 3
= = 0.017 when v = 1/1.05.
0.99
0.99
(1 − q60 )v 2 − (1 − q60 )v 3
−
2
1.05
1.05 3
The primes indicate calculations at 4.5% interest.
′ = q60 v′ + (1 − q60 )q60+1v′ 2 + (1 − q60 ) (1 − q60+1 )v′ 3
A60:3
0.01 0.99(0.017) 0.99(0.983)
=+
+
1.045
1.045 2
1.0453
= 0.87777
Question 4.4
Answer: A
=
Var
(Z ) E (Z 2 ) − E (Z )2
E (Z ) =
E (1 + 0.2T )(1 + 0.2T ) −2  =
E (1 + 0.2T ) −1 
40
1
1 40 1
=
f
t
dt
dt
(
)
T
∫0 (1 + 0.2t )
40 ∫0 1 + 0.2t
1 1
1
40
=
ln (1 + 0.2t ) 0 = ln(9) = 0.27465
40 0.2
8
2
2
−2 2
E (Z ) =
E{(1 + 0.2T ) [(1 + 0.2T ) ] } =
E[(1 + 0.2T ) −2 ]
=
40
1
1 1  −1 
=∫
fT (t ) =
2
0 (1 + 0.2t )
40 0.2  (1 + 0.2t )  0
1 1 1
= 1 −  = =0.11111
8 9 9
40
Var ( Z ) =
0.11111 − (0.27465) 2 =
0.03568
Page 18
Question 4.5
Answer: C
The earlier the death (before year 30), the larger the loss. Since we are looking for the 95th
percentile of the present value of benefits random variable, we must find the time at which 5% of
the insureds have died. The present value of the death benefit for that insured is what is being
asked for.
l45= 99, 033.9 ⇒ 0.95l45= 94, 082.2
l65 = 94,579.7
l66 = 94, 020.3
So, the time is between ages 65 and 66, i.e., time 20 and time 21.
l65 =
− l66 94,579.7 − 94, 020.3
= 559.4
l65+t =
− l66 94,579.7 − 94, 082.2
= 497.5
497.5 / 559.4 = 0.8893
The time just before the last 5% of deaths is expected to occur is: 20 + 0.8893 = 20.8893
The present value of death benefits at this time is:
100, 000e −20.8893(0.05) = 35,188
Question 4.6
Answer: B
Time
Age
q xSULT
Improvement
factor
qx
0
70
0.010413
100.00%
0.010413
1
71
0.011670
95.00%
0.011087
2
72
0.013081
90.25%
0.011806
=
v 1/1.05
= 0.952381
EPV =
1, 000[0.010413v + 0.989587(0.011087)v 2 + 0.989587(0.988913)(0.011806)v 3 ]
= 29.85
Page 19
Question 4.7
Answer: B
Var ( Z ) =0.10 E[ Z ] ⇒ v 50 25 px (1 − 25 px ) =0.10 ⋅ v 25 25 px
⇒
(1 − 0.57)
1
0.10
=×
50
(1 + i )
(1 + i ) 25
⇒ (1 + i )25=
0.43
= 4.3 ⇒ i= 0.06
0.10
Question 4.8
Answer: C
Let A51SULT designate A51 using the Standard Ultimate Life Table at 5%.
 1 
SULT
APV
(insurance) 1000 
=
 ( q 50 + p 50 A51 )
 1.04 
 1 
= 1000 
 0.001209 + (1 − 0.001209 )( 0.19780 ) 
 1.04 
= 191.12
Question 4.9
Answer: D
1
=
A35 A35:15
+ A35:151 A50
0.32
= 0.25 + 0.14 A50
=
A50
Page 20
0.07
= 0.50
0.14
Question 4.10
Answer: D
Drawing the benefit payment pattern:
x
x + 10
E[ Z ] = 10 Ex × A x +10 +
20
x + 20
Ex × A x + 20 − 2
30
x + 30
Ex × A x +30
Question 4.11
Answer: A
Var (=
Z2 )
(1000)
2
=
(1000)
2
= (1000 )
( )
+ A ) − (1000 )  A

2
2A − A =
 15,000
x:n
 x : n

(
2
2 2
A1x : n
2
2
1
x:n
A1x : n + (1000 )
(
(
− 1000 A x1: n
(
)
2
)  + (1000
2
(
2
A x :1n − (1000 ) A1x : n
2
22
( )
− (1000 ) A x 1n
2
1
xn
1
x:n
) (
)
A x :1n − 1000 A x :1n
2
)(1000 A )
) − (1000 A ) − (1000 A )
−(2) (1000 A )(1000 A )
− (2) 1000 A x1: n
=
V ( Z 1 ) + (1000) 1000 2 A x : n1
2
( )
− 2(1000) ( A )( A )
2 2
2
= (1000) 2  2 Ax1: n − Ax1: n

+ A x 1n 

1
x:n
1
xn
1
xn
2
1
x: n
1
x:n
2
1
xn
15,000= Var ( Z 1 ) + (1000)(136) − (209) 2 − 2(528)(209)
Therefore, Var ( Z 1 ) = 15,000 − 136,000 + 43,681 + 220,704 = 143,385.
Page 21
2
Question 4.12
Answer: C
(
=
Z 3 2 Z 1 + Z 2 so that Var ( Z 3 ) = 4Var ( Z1 ) + Var ( Z 2 ) + 4Cov Z 1 , Z 2
(
)
)
−(1.65)(10.75)
E  Z 1 Z 2  − E  Z 1  E  Z 2  =
where Cov Z 1 , Z 2 =


=0
Var ( Z 3=
) 4(46.75) + 50.78 − 4(1.65)(10.75)
= 166.83
Question 4.13
Answer: C
v3
A
=
2 2 65
payment year 3
+
v4
× q [65]+ 2
2 p [ 65]


Lives 2 years
payment year 4
Die year 3
× q 65+3
3 p[ 65]


Lives 3 years
Die year 4
3
 1 
=
 (0.92)(0.9)(0.12)
 1.04 
4
 1 
+
 (0.92)(0.9)(0.88)(0.14)
 1.04 
= 0.088 + 0.087 = 0.176
352 .
The actuarial present value of this insurance is therefore 2000 × 0.176 =
Page 22
Question 4.14
Answer: E
l85
 61,184.9 
=
=
400
400 
Out of 400 lives initially, we expect 400=
25 p60
 253.26 survivors
l60
 96, 634.1 
The standard deviation of the number of survivors is
400 25 p60 (1 − 25 p60 ) =
9.639
To ensure 86% funding, using the normal distribution table, we plan for
253.26 + 1.08 ( 9.639 ) =
263.67
25
 1 
The initial fund must
therefore be F (264)(5000)
=
=
389,800.


 1.05 
Question 4.15
Answer: E
∞
∞
0
0
E [ Z ] = ∫ bt ⋅ vt ⋅t px ⋅ µ x +t dt = ∫ e0.02t ⋅ e −0.06t ⋅ e −0.04t ⋅ 0.04 dt
0.04 1
=
0.08 2
∞
t
= 0.04 ∫ e −0.08=
dt
0
E  Z 2  =
∫ (b ⋅ v )
∞
0
t
t 2
2
⋅t px ⋅ µ x +t dt =
1 1
1
Var [ Z ] =−   ==
0.0833
3  2  12
Page 23
∫ ( e )( e
∞
0
0.04 t
−0.12 t
)( 0.04e ) dt =
−0.04
0.04 1
=
0.12 3
Question 4.16
Answer: D
A[501 ]:3 =+
vq[50] v 2 p[50]q[50]+1 + v3 p[50] p[50]+1q52
where: v =
1
1.04
=
=
q[50] 0.7
( 0.045) 0.0315
1 − q[50] =
0.9685
p[50] =
=
=
q[50]+1 0.8
( 0.050 ) 0.040
1 − q[50]+1 =
0.960
p[50]+1 =
q52 = 0.055
So: A[501 ]:3 = 0.1116
Question 4.17
Answer: A
The median of 𝐾𝐾48 is the integer 𝑚𝑚 for which
P ( K 48 < m ) ≤ 0.5 and P ( K 48 > m ) ≤ 0.5.
This is equivalent to finding 𝑚𝑚 for which
l48+ m
l48
≥ 0.5 and
l48+ m +1
l48
≤ 0.5.
Based on the SULT
and l48 (0.5) (98,
=
=
783.9)(0.5) 49,391.95 , we have m = 40 since
l88 ≥ 49, 391.95 and l89 ≤ 49, 391.95.
So: APV= 5000 A48 + 500040 E48 A88
= 5000 A48 + 5000 ⋅20 E48 ⋅20 E68 ⋅ A88
=
5000 ( 0.17330 ) + 5000 ( 0.35370 ) (0.20343) ( 0.72349 ) =
1126.79
Page 24
Question 4.18
Answer: A
The present value random
variable PV 1, 000, 000e −0.05T , 2 ≤ T ≤ 10 is a decreasing function of T
=
so that its 90th percentile is
1, 000, 000e −0.05 p where
∫
is the solution to
∫
p
2
0.4t −2 dt = 0.10 .
p
 t −1 
1 1 
−0.4   =
0.4t dt =
0.4  −  =
0.10
2 p
 −1  2
p
−2
2
p=4
1, 000, 000e −0.05×4 = 81,873.08
Question 4.19
Answer: B
Superscript SULT refers to values from the SULT. Values without superscripts refer to this
select life.
=
q80 0.8q80SULT
= 0.0261264 ⇒ =
p80 0.9738736
=
A80 vq80 + vp80 A81SULT
(1.05) −1 (0.0261264) + (1.05) −1 (0.9738736)(0.60984) =
0.59051
=
100, 000 A80 = 59, 051
Question 4.20
Answer: D
∞
∫
∫
E (=
Z)
0 t
=
∞
0
px × µ × e0.05×t e −δ ×t dt
e −0.03×t × 0.03 × e0.05×t e −0.06×t dt
0.03
= × e −0.04t
0.04
∫
∫
2
)
E ( Z=
∞
0 t
=
∞
0
∞
0.75
=
0
px × µ × e0.05×2×t e −δ ×2×t dt
e −0.03×t × 0.03 × e0.05×2×t e −0.06×2×t dt
0.03
= × e −0.05t
0.04
∞
0.6
=
0
0.6 − 0.752 =
0.375
( Z ) E ( Z 2 ) − [ E ( Z )]2 =
Var
=
Page 25
Question 4.21
Answer: E
0.04  1
1
i

(0.04)
(0.96)(0.06)  =
E ( Z ) = vqx + v 2 px qx +1  =
+
0.09353831
2

ln(1.04) 1.04
1.04
δ

(1 + i ) 2 − 1 2
(1.042 − 1)  1
1

v qx + v 4 px q=

E (=
Z2)
+
=
(0.04)
(0.96)(0.06)
0.08969072
x +1 
2
4

2δ
2 ln(1.04) 1.04
1.04

Var
=
0.08969072 − (0.09353831) 2 =
0.0809413
( Z ) E ( Z 2 ) − [ E ( Z )]2 =
Question 4.22
Answer: A
Probability (50) survives one year under Standard Ultimate Life Table = 1 – 0.001209 =
0.998791
Probability (50) survives one year following surgery = 0.55 × 0.998791 =
0.5493 =
p50
q50 =
1 − p50 =
0.4507
 1 
 1 
 1 
A50:3 =
q50 
+ p50 p51q52 
 + p50 q51 
2 
3 
 1.05 
 1.05 
 1.05 
 1 
 1 
 1 
(0.4507) 
=
+ (0.5493)(0.99867)(0.001469) 
 + (0.5493)(0.001331) 
2 
3 
 1.05 
 1.05 
 1.05 
= 0.4306
Therefore, answer = 100, 000 × 0.4306 ≈ 43, 000
Question 5.1
Answer: A
E (Y=
) a10 + e −δ (10) e − µ (10) a x +10
(1 − e ) + e
− 0.6
=
0.06
= 14.6139
− 0.7
1
0.07
 1 − e −0.06T 
Y > E (Y ) ⇒ 
 > 14.6139
 0.06 
⇒ T > 34.90
Pr[Y > E (Y )] =Pr(T > 34.90) =e −34.90(0.01) =0.705
Page 26
(A)
Question 5.2
Answer: B
A x:1n = n E x
A x A 1x:n + n E x A x + n
=
0.3 = A1x:n + (0.35)(0.4) ⇒ A1x:n = 0.16
A x:n = A x1:n + n E x = 0.16 + 0.35 = 0.51
=
ax:n
1 − A x:n
1 − 0.51
=
= 10.29
(0.05 /1.05)
d
a x:=
ax:n − 1 + n E=
10.29 − 0.65
= 9.64
x
n
Question 5.3
Answer: C
( Ia )
t
40 : t
= ∫ s s p40 v s ds ⇒
d ( Ia )
0
40 : t = t p vt
t 40
dt
At t = 10.5,
10.5 10.5 E40 = 10.5 10 p40 0.5 p50 v10.5
= 10.5 10 E40 0.5 p50 v 0.5
= 10.5 × 0.60920 × (1 − 0.5 × 0.001209)(0.975900073)
= 6.239
Page 27
Question 5.4
Answer: A
1
= 50 So, receive K for 50 years guaranteed and for life thereafter.

e 40=
µ
10,
=
000 K  a50 + 50| a40 
50
e −δ t
∫=
a 50
=
0
=
50| a40
=
K
50
1 − e −50δ 1 − e −50(0.01)
=
= 39.35
0.01
δ
1
1
−1.5
E
e − (δ + µ )50
7.44
=
= e=
40 a40 + 50
0.03
µ +δ
10, 000
= 213.7
39.35 + 7.44
Question 5.5
Answer: A
S
S SULT
a45
= 1 + vp45
a46
1
S
p45
= e
∫
S
− µ45
+ t dt
0
1
= e
∫
SULT
− ( µ 45
+ t + 0.05) dt
0
1
= e
∫
1
∫
SULT
− ( µ 45
+ t ) dt − (0.05) dt
0
e
0
 98,957.6  −0.05
SULT
= 0.9504966
= p45
⋅ e −0.05 = 
e
 99,033.9 
S
S SULT
=
1 + vp45
1 + (1.05) −1 (0.9504966)(17.6706) = 17.00
a45
a46 =
S
100a45
= 1700
Page 28
Question 5.6
Answer: D
Let Yi be the present value random variable of the payment to life i.
2
1 − Ax
E[Y=
a=
= 11.55
i]
x
d
Var[Y=
i]
Ax − ( Ax )
=
d2
2
0.22 − 0.45 2
7.7175
=
2
( 0.05 /1.05)
100
Then Y = ∑ Yi is the present value of the aggregate payments.
i =1
=
E[Y ] 100
=
E[Yi ] 1155 and=
Var[Y ] 100=
Var[Yi ] 771.75
F − 1155 
F − 1155

= 1.645
Pr[Y ≤ F ] = Pr  Z ≤
 = 0.95 ⇒
771.75 
771.75

⇒ F= 1155 + 1.645 771.75= 1200.699
Question 5.7
Answer: C
(2)
a35:30
≈ a35:30 −
( m − 1)
(1 − v 30 30 p35 )
2m
1
1 − A35:30
1 − A35:30
− 30 E35
=
d
d
1 − ( A35 − 30 E35 × A65 ) − 30 E35
=
d
=
a35:30
30
=
v=
Since
0.2722, then
30 E 35
30 p 35
a35:30 =
1 − ( A35 − v 30
30
p 35 × A65 ) − v 30
30
p 35
d
1 − (0.188 − (0.2722)(0.498)) − 0.2722
=
(0.04 /1.04)
= 17.5592
(2)
a35:30
≈ 17.5592 −
1
(1 − 0.2722) =
17.38
4
(2)
1000 a35:30
≈ 1000 × 17.38 =
17,380
Page 29
Question 5.8
Answer: C
The expected present value is:
a5 + 5 E55 a60 = 4.54595 + 0.77382 ×14.9041 = 16.07904
The probability that the sum of the undiscounted payments will exceed the expected present
value is the probability that at least 17 payments will be made. This will occur if (55) survives to
age 71. The probability is therefore:
p=
55
16
l71 90,134.0
=
= 0.92118
l55 97,846.2
Question 5.9
Answer: C
(
)
(
)
a[ x]:n =1 + vp[ x] ax +1:n −1 =1 + (1 + k ) vpx ax +1:n −1 =1 + (1 + k ) ax:n − 1
Therefore, we have
=
k
a[ x ]:n − 1
21.167
=
−1
=
− 1 0.015
20.854
ax:n − 1
Question 5.10
Answer: C
a65:10 = a10 + 10 E65 × a75 = 8.10782 + 0.55305 ×10.3178 = 13.8141
Assuming payments of 1 (any other payment amount would just cancel, giving the same number
of years), 14 payments are needed for the sum of payments to exceed 13.8141. That requires
surviving to the start of year 14, thus surviving 13 years.
13
p=
65
l78 80, 006.2
=
= 0.846
l65 94,579.7
Page 30
Question 5.11
Answer: B
Discounted Payment
Probability
10
(1-0.95) = 0.05
10+10v = 19.3
(0.95)(1-0.9) = 0.095
10+10v+10v2=27.949
(0.95)(0.9) = 0.855
Var
=
( X ) E ( X 2 ) − [ E ( X )]2
E ( X ) = 10(1 − 0.95) + [10 + 10v](0.95)(1 − 0.9) + [10 + 10v + 10v 2 ](0.95)(0.9) = 26.23
E ( X 2 )= 102 (1 − 0.95) + [10 + 10v]2 (0.95)(1 − 0.9) + [10 + 10v + 10v 2 ]2 (0.95)(0.9)= 708.27
Var ( X ) =708.27 − (26.23) 2 =20.26
1/2
Standard De
=
viation (20.26)
=
4.5
It’s not significantly less work, but since everyone receives the first 10, you could use the
formula Var(X-constant) = Var(X), and work with the three payment amounts as 0, 10v,
10v+10v2, and you would get the same answer.
Question 5.12
Answer: E
Any of these ways of viewing the benefit structure would be fine; all give the same answer:
(i) A temporary life annuity-due of X plus a deferred life annuity-due of 0.75X.
(ii) A whole life annuity-due of 0.75X plus a temporary deferred life annuity-due of 0.25X
(iii) A whole life annuity-due of X minus a deferred life annuity-due of 0.25X
This solution views it the first way.
Xa55:10 + 0.75 Xa65  10 E55 =
8.0192 X + (0.75 X )(0.59342)(13.5498) =
14.05 X =
250, 000
17, 794
⇒X=
Page 31
Question 6.1
Answer: D
The equation of value is given by
Actuarial Present Value of Premiums = Actuarial Present Value of Death Benefits.
The death benefit in the first year is 1000 + P . The death benefit in the second year is 1000 + 2P .
1 .
1
+ P( IA) 80:2
The formula =
is Pa80:2 1000 A80:2
Solving for P we obtain P =
a80:2 =
1 + p 80 v =
1+
1
1000 A80:2
1
a80:2 − ( IA) 80:2
.
0.967342
1.93917
=
1.03
 0.032658 (0.967342)(0.036607) 
1
1000 A80:2
= 1000 ( vq 80 + v 2 p 80 q 81 ) = 1000 
+
 = 65.08552
1.03 2
 1.03

0.032658
(0.967342)(0.036607)
1
( IA) 80:2
vq 80 + 2v 2 p 80 q 81 =
0.09846
=
+ (2)
=
1.03
1.03 2
P=
65.08552
= 35.36 → D
1.93917 − 0.09846
Question 6.2
Answer: E
x:10 100, 000 Ax1:10 + G ( IA)1x:10 + 0.45G + 0.05Gax:10 + 200ax:10
Ga
=
G
(100, 000)(0.17094) + 200(6.8865)
= 3604.23
(1 − 0.05)(6.8865) − 0.96728 − 0.45
Page 32
Question 6.3
Answer: C
Let C be the annual contribution, then C =
20
E 45 a65
a45:20
Let K 65 be the curtate future lifetime of (65). The required probability is
 20 E 45 a65 a45:20

 Ca45:20

> aK +1  = Pr 
> aK +1  = Pr a65 > aK +1 = Pr 13.5498 > aK +1
Pr 
 E

65
65
65
65
 a45:20 20 E45

 20 45



(
)
(
=
=
Thus, since a21 13.4622
and a22 13.8212 we have
(
Pr aK
65 +1
)
< 13.5498
= Pr ( K 65 + 1 ≤ 21) =−
1 21 p65 =−
1
l86
57, 656.7
=−
1
=
0.390
l65
94,579.7
Question 6.4
Answer: E
Let X i be the present value of a life annuity of 1/12 per month on life i for i = 1, 2,..., 200.
200
Let S = ∑ X i be the present value of all the annuity payments.
i =1
(12)
E[=
X i ] a=
62
(
)
2
( )
( )
A62
− A62
0.2105 − (0.4075) 2
13.15255
=
=
2
(12 ) 2
(0.05813)
d
2
Var ( X i )
=
(12)
1 − A62
1 − 0.4075
=
= 10.19267
(12)
0.05813
d
12
(
12
)
E[ S ] (200)(180)(10.19267)
= 366,936.12
2
(13.15255) 85, 228,524
Var ( S ) (200)(180)
=
=
With the normal approximation, for Pr( S ≤ M ) =
0.90
366,936.12 + 1.282 85, 228,524 =
378, 771.45
M = E[ S ] + 1.282 Var ( S ) =
So π
=
378, 771.45
= 1893.86
200
Page 33
)
Question 6.5
Answer: D
Let k be the policy year, so that the mortality rate during that year is q30+ k −1 . The objective is to
determine the smallest value of k such that
v k −1 ( k −1 p30 ) (1000 P30 ) < v k ( k −1 p30 ) q30+ k −1 (1000)
P30 < vq30+ k −1
0.07698 q29+ k
<
19.3834 1.05
q29+ k > 0.00417
29 + k > 61 ⇒ k > 32
Therefore, the smallest value that meets the condition is 33.
Question 6.6
Answer: B
=
Net Premium 10,
=
000 A62 / a62 10, 000(0.31495)
=
/14.3861 218.93
=
G 218.93(1.03)
= 225.50
Let 0 L* be the present value of future loss at issue for one policy.
*
10, 000v K +1 − (G − 5)aK +1 + 0.05G
=
0L
1 − v K +1
+ 0.05(225.50)
d
= (10, 000 + 4630.50 ) v K +1 − 4630.50 + 11.28
= 10, 000v K +1 − (225.50 − 5)
= 14, 630.50v K +1 − 4619.22
E ( 0 L* ) =
14, 630.50 A62 − 4619.22 =
14, 630.50(0.31495) − 4619.22 =
−11.34
=
Var ( 0 L* ) (14, 630.50) 2 =
− 0.31495 2 ) 5,536, 763
( 2 A62 − A622 ) (14, 630.50)2 ( 0.12506=
Let 0 L be the aggregate loss for 600 such policies.
600 E ( 0 L* ) =
600(−11.34) =
E ( 0 L) =
−6804
763) 3,322, 057,800
Var ( 0 L) 600
Var ( 0 L* ) 600(5,536,
=
=
=
057,8000.5 57, 637
StdDev( 0 L) 3,322,
=
=
 40, 000 + 6804 
Pr ( 0 L < 40, 000 ) =
Φ
Φ ( 0.81) =
0.7910
=
57, 637


Page 34
Question 6.7
Answer: C
There are four ways to approach this problem. In all cases, let π denote the net premium.
The first approach is an intuitive result. The key is that in addition to the pure endowment, there
is a benefit equal in value to a temporary interest only annuity due with annual payment π.
However, if the insured survives the 20 years, the value of the annuity is not received.
π a=
100, 000
40:20
20
40:20 − 20 p40 a
20 5%π
E40 + π a
Based on this equation,
=
π
100, 000 20 E40 100, 000v 20 100, 000 100, 000
=
= =
= 2880


a20
s20
34.71925
20 p40 a20
The second approach is also intuitive. If you set an equation of value at the end of 20 years, the
present value of benefits is 100,000 for all the people who are alive at that time. The people who
have died have had their premiums returned with interest. Therefore, the premiums plus interest
that the company has are only the premiums for those alive at the end of 20 years. The people
who are alive have paid 20 premiums. Therefore π 
s20 = 100, 000 .
The third approach uses random variables to derive the expected present value of the return of
premium benefit. Let K be the curtate future lifetime of (40). The present value random variable
is then
π 
s v K +1 , K < 20
Y =  K +1
K ≥ 20
0,
π a , K < 20
=  K +1
K ≥ 20
0,
K < 20
π a − 0,
=  K +1
.


−
≥
π
a
π
a
K
,
20
20
 20
The first term is the random variable that corresponds to a 20-year temporary annuity. The
second term is the random variable that corresponds to a payment with a present value of π a20
contingent on surviving 20 years. The expected present value is then π a40:20 − 20 p40 a20 π .
Page 35
The fourth approach takes the most steps.
π a
19
E +π ∑v
100, 000
=
k +1

s
19
100, 000 E + π ∑ v
q =
20 40
20 40
k +1 k 40
40:20
k 0=
k 0
=
= 100, 000 20 E40 +
= 100, 000
= 100, 000
= 100, 000
20
π
E40 +
19
∑
d 
k =0
π
(
d
20
k

q40 − v k +1 k q=
100, 000
40 

q40 − 1 + da40:20 + v 20 20 p40
20
E40 +
k +1
(1 + i ) k +1 − 1
q
k 40
d
π
(
d
20
1
q40 − A40:20
)
)
1 − v 20
d
− 20 p40 a20 6%π .

20 E40 + π a 40:20 − π 20 p40
20
E40 + π a40:20
Question 6.8
Answer: B
a60:10 = 7.9555
a60:20 = 12.3816
Annual level amount
=
40 + 5a60:10 + 5a60:20 141.686
= = 9.51
a60
14.9041
Question 6.9
Answer: D
a50:10 = 8.0550
1
A50:20
= A50:20 − 20 E 50 = 0.38844 − 0.34824 = 0.04020
a50:20 = 12.8428
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
8.0550G =
4020 + 1.2666G + 371.07
6.7883G = 4391.07
⇒G =
646.86
Page 36
Question 6.10
Answer: D
=
a x:3
Actuarial PV of the benefit 152.85
= = 2.727
Level Annual Premium
56.05
0.975 0.975 ( p x +1 )
+
=
1+
2.727
ax:3 =
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
Question 6.11
Answer: C
For calculating P
A50 = vq 50 + vp 50 A51 = v(0.0048) + v(1 − 0.0048)(0.39788) = 0.38536
a50 =
15.981
(1 − A50 ) / d =

=
P A=
0.02411
50 / a 50
For this particular life,
A′50= vq′50 + vp′50 A51= v(0.048) + (1 − 0.048)(0.39788)= 0.41037
a′50 =
15.330
(1 − A′50 ) / d =
Expected PV of loss = A′50 − Pa′50= 0.41037 − 0.02411(15.330)
= 0.0408
Page 37
Question 6.12
Answer: E
1,020 in the solution is the 1,000 death benefit plus the 20 death benefit claim expense.
1 − dax =
1 − d (12.0) =
0.320755
Ax =
=
Gax 1, 020 A x + 0.65G + 0.10Gax + 8 + 2ax
G
1, 020 A x + 8 + 2ax 1, 020(0.320755) + 8 + 2(12.0)
=
= 35.38622
12.0 − 0.65 − 0.10(12.0)
ax − 0.65 − 0.10ax
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
x +1
denote the present value random variable for a life annuity-due of 1 on (x).
L 1, 020 Z + 0.65G + 0.10GY + 8 + 2Y − GY
=
= 1, 020 Z + (2 − 0.9G )Y + 0.65G + 8
1 − v K x +1
+ 0.65G + 8
d
0.9G − 2  K x +1 2 − 0.9G

= 1, 020 +
+
+ 0.65G + 8
v
d
d


= 1, 020v K x +1 + (2 − 0.9G )
0.9G − 2 

 Ax − ( Ax )  1, 020 +
Var ( L) =

d


2
2
2
0.9(35.38622) − 2 

(0.14 − 0.320755 ) 1, 020 +
=

d


= 0.037116(2,394,161)
= 88,861
2
Page 38
2
Question 6.13
Answer: D
If T45 = 10.5 , then K 45 = 10 and K 45 + 1 =
11.
=
10, 000v
0L
K 45 +1
− G (1 − 0.10)aK
45 +1
+ G (0.80 − 0.10)
= 10, 000v 11 − 0.9Ga11 + 0.7G
4953= 10, 000(0.58468) − 0.9G (8.72173) + 0.7G
G=
(5846.8 − 4953) / (7.14956) =
125.01
=
E ( 0 L) 10, 000 A45 − (1 − 0.1)Ga45 + (0.8 − 0.1)G
=
(10, 000)(0.15161) − (0.9)(125.01)(17.8162) + (0.7)(125.01)
E ( 0 L) = −400.87
Question 6.14
Answer: D
=
A40 P[a40:10 + 0.5 10 a40:10 ]
100, 000
=
P
100, 000 A40
100, 000(0.12106)
12,106
=
= = 1148.59
a40:10 + 0.5 10 a40:10 8.0863 + 0.5(4.9071) 10.53985
where
=
a40:10
10
=
E 40  a50:10  0.60920
=
[8.0550] 4.9071


10
There are several other ways to write the right-hand side of the first equation.
Page 39
Question 6.15
Answer: B
Woolhouse:
W
3
(4)
3.4611 −= 3.0861
a=
x
8
=
ax(4) α (4)ax − β (4)
UDD
UDD: = 1.00019(3.4611) − 0.38272
= 3.0790
and
=
P (W )
Ax =
1 − d ax =
1 − (0.04762)(3.4611) =
0.83518
1000(0.83518)
= 270.63
3.0861
1000(0.83518)
= 271.25
3.0790
=
P (UDD )
P (UDD ) 271.25
=
= 1.0023
P (W )
270.63
Question 6.16
Answer: A
P30:20
=
1
a30:20
−d ⇒
2,143
1
+ 0.05
=
⇒ a30:20
= 14
100, 000
a30:20
A30:20 =
1 − d a30:20 =
1 − 0.05(14) =
0.3
=
Ga30:20 100, 000 A30:20 + (200 + 50a30:20 ) + (0.33G + 0.06G a30:20 )
=
14G 100, 000(0.3) + [200 + 50(14)] + (0.33G + 0.84G )
12.83 G = 30,900
G = 2408
Page 40
Question 6.17
Answer: A
1 − e −0.1 =
0.095
q xNS =
q xNS
+1 =
Then the annual premium for the non-smoker policies is P NS , where
NS
100, 000vq xNS + 100, 000v 2 p xNS q xNS+1 + 30, 000v 2 p xNS p xNS+1
P NS (1 + vp
=
x )
P
NS
100, 000(0.926) (0.095) + 100, 000(0.857) (0.905) (0.095) + 30, 000(0.857) (0.905) 2
=
1 + (0.926) (0.905)
P NS = 20, 251
And then P S = 40,502.
qxS =qxS+1 =1.5(1 − e −0.1 ) =0.143
=
EPV
( LS ) 100, 000vq xS + 100, 000v 2 p xS q xS+1 + 30, 000v 2 p xS p xS+1 − P S − P Svp xS
= 100, 000(0.926)(0.143) + 100, 000(0.857)(0.857)(0.143)
+ 30, 000(0.857)(0.857) 2 − 40,502 − 40,502(0.926)(0.857)
= −30, 017
Question 6.18
Answer: D
=
P 30, 000
20
1
a40 + PA40:20
⇒
=
P 30, 000
20
1
a40 / 1 − A40:20
(
)
= 30, 000(5.46429) / (1 − 0.0146346)
= 166,363
Page 41
Question 6.19
Answer: C
Let π be the annual premium, so that π a50 =
50 + 0.19
A50 + 0.01a
⇒
=
π
A50 + 0.19
a50
0.01
+=
0.18931 + 0.19
0.01 0.03228
+=
17.0245
(
)
Loss at issue: L 0 = v k +1 − (π − 0.01) ak +1 1 − v k +1 / d + 0.19
2
 (π − 0.01)  2
2
⇒ Var  L 0  =
1 +
 ( A50 − A50 )
d


= (2.15467)(0.05108 − 0.18931 2 )
= (2.15467)(0.015242)
= 0.033
Question 6.20
Answer: B
EPV(premiums) = EPV(benefits)
P (1 + vp x + v 2 2 p x =
) P ( vq x + 2v 2 p x q x+1 ) + 10000 ( v 3 2 p x q x+2 )
0.9 0.9 × 0.88 

 0.1 2 × 0.9 × 0.12 
 0.9 × 0.88 × 0.15 
P 1 +
+
=
+
 P
 + 10000 

2
2
1.04
1.04
1.04 3


 1.04

 1.04

=
2.5976 P 0.29588 P + 1056.13
P = 459
Question 6.21
Answer: C
(
)
1
1
P=
× a75:15 1000 A75:15
+ 15 × P × A75:15
=
→P
1
A75:15
=A75:15 − A75:151 =0.7 − 0.11 =0.59
1 − A75:15
a75:15 = =
(1 − 0.7) / 0.04 =
7.5
d
=
So P
590
= 100.85
7.5 − 15(0.11)
Page 42
1
1000 A75:15
a75:15 − 15 × A75:151
Question 6.22
Answer: C
Let the monthly net premium = π
12π =
α (12) = 1.00020
β (12) = 0.46651
100, 000 A45
(12)
a45:20
i
δ
= 1.02480
i
=
100, 000 A45 100,
=
000 A45 (1.02480)(15,161)
= 15,536.99
δ
(12)
a45:20
= α (12)a45:20 − β (12) (1 − 20 E 45 )
= 1.00020 [12.9391] − 0.46651(1 − 0.35994)
= 12.6431
15,536.99
12.6431
12π = 1228.891
π = 102.41
12π =
Question 6.23
Answer: D
=
Gax:30 APV
=
[gross premium] APV [ Benefits + expenses]
(
= FA x + ( 30 + 30ax ) + G 0.6 + 0.10ax:30 + 0.10ax:15
G=
=
=
FA x + 30 + 30ax
ax:30 − 0.6 − 0.1ax:30 − 0.1ax:15
FA x + 30 + 30(15.3926)
14.0145 − 0.6 − 0.1(14.0145) − 0.1(10.1329)
FA x + 491.78
10.9998
FA x
FA x
491.78
=
+
=
+ 44.71
10.9998 10.9998 10.9998
⇒h=
44.71
Page 43
)
Question 6.24
Answer: E
In general, the loss at issue random variable can be expressed as:
 1− Zx 
 P P
L = Z x − P ×Yx = Z x − P × 
 = Z x × 1 +  −
 δ δ
 δ 
Using actuarial equivalence to determine the premium rate:
=
P
Ax
=
ax
0.3
= 0.03
(1 − 0.3) / 0.07
2
2
P

 0.03 
Var ( L) =
0.18
1 +  × Var ( Z x ) =
1 +
 × Var ( Z x ) =
 δ
 0.07 
0.18
0.088
=
Var ( Z x ) =
2
 0.03 
1 +

 0.07 
2
 P*
 0.06 
Var ( L*) =
1 +
 × Var ( Z x ) =
1 +

δ 

 0.07 
2
0.304
( 0.088) =
Question 6.25
Answer: C
Need EPV(Ben + Exp) − EPV(Prem) =
−800
55:10 8.0192G
EPV(Prem)
= Ga
=
(12)
+ 300a55
EPV(Ben + Exp)=12,000 10 a55
(12)
= 12, 000 10 E55 a65
+ 300a55
m −1 

= 12, 000 10 E55  a65 −
 + 300a55
2m 

= 12, 000(0.59342) 13.5498 − 11
+ 300(16.0599)
24
= 98, 042.83
(
Therefore,
98, 042.83 − 8.0192G =
−800
G = 12,326
Page 44
)
Question 6.26
Answer: D
EPV (Premiums) = Pa=
90 −=
P(a
1) (4.1835) P
90
EPV(Benefits)
= 1000
=
A90 1000(0.75317)
= 753.17
Therefore,
=
P
753.17
= 180.03
4.1835
Question 6.27
Answer: D
EPV(Premiums) = EPV(Benefits)
EPV(Premiums)
= 3P a x − 2 P 20 E x a x + 20
(
= 3P 1
(
)
(
−20( µ +δ )
) 1µ + δ
µ + δ − 2P (e
= 3P 1
0.09
= 29.66 P
) − 2Pe
−1.8
−1
)
0.09
EPV(Benefits)
= 1, 000, 000 Ax − 500, 000 20 Ex Ax + 20
(
= 1, 000, 000 µ
(
)
−20( µ +δ ) µ
µ + δ − 500, 000e
µ +δ
= 1, 000, 000 0.03
0.09
) − 500, 000e
−1.8
0.03
0.09
= 305, 783.5
29.66 P = 305,783.5
305,783.5
29.66
P = 10,309.62
P=
Question 6.28
Answer: B
40:5 1000 A40 + 0.15G + 0.05Ga40:5 + 5 + 5a40:5
Ga
=
a40:5 =
a40 − 5 E 40  a45 =
18.4578 − (0.78113)(17.8162) =
4.5410
G
121.06 + 5 + 5(4.5410)
= 35.73
−0.15 + 0.95(4.5410)
Page 45
Page 46
Question 6.29
Answer: B
Per equivalence Principle:
=
Ga35 100,000 A35 + 0.4G + 150 + 0.1Ga35 + 50a35
1770
=
a35 100,000 (1 − da35 ) + 0.4(1770) + 150 + 0.1(1770)a35 + 50a35

 0.035  
1770=
a35 100,000 + 708 + 150 + a35 177 + 50 − 100,000 

 1.035  

35 , we have
Solving for a
100,858
100,858
=
a
= = 20.48
1770 + 3154.64 4924.64
Question 6.30
Answer: A
The loss at issue is given by:
L0 = 100v K +1 + 0.05G + 0.05G aK +1 − G aK +1
= 100v
K +1
 1 − v K +1 
+ 0.05G − 0.95G 

 d 
0.95G  K +1
G

=100 +
 v + 0.05G − 0.95
d 
d

Thus, the variance is
0.95 ( 2.338 ) 

Var ( L0 ) =
100 +

0.04 /1.04 

2
( A −(A ) )
2
2
x
0.95 ( 2.338 ) 

= 100 +

0.04 /1.04 

= 908.1414
Page 47
x
2
2

 0.04
 
0.17
−
1
−
16.50
(
)  


 1.04
 

Question 6.31
Answer: D
(
A35 =
1 − e 35(
a 35
=
P
=
35
−
µ +δ )
) ×  µ µ+ δ  + e
−35( µ +δ )
A70 =
0.063421 + 0.146257 =
0.209679
1 − A35 1 − 0.209679
=
= 15.80642
0.05
δ
A35 0.209679
=
= 0.0132654
a 35 15.80642
1,326.54
The annual net premium for this policy is therefore 100, 000 × 0.0132654 =
Question 6.32
Answer: C
Assuming UDD
Let P = monthly net premium
12 Pax(
EPV(premiums) =
12 )
≅ 12 P α (12 ) a x − β (12 ) 
12 P 1.00020 ( 9.19 ) − 0.46651
= 104.7039P
EPV(benefits) = 100, 000 Ax
i
i
Ax 100,000 (1 − dax )
= 100,000
=
δ
= 100, 000
0.05  0.05
( 9.19 ) 
1 −
ln (1.05 )  1.05

= 57, 632.62
=
P
57, 632.62
= 550.43
104.7039
Page 48
δ
Question 6.33
Answer: B
The probability that the endowment payment will be made for a given contract is:
(
15
=
p x exp − ∫ 0.02t dt
15
(
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,
2
Var ( 0 L ) 500 (10, 000v 15 ) ( 15 p x )(1 − 15 p x ) 


= 1,942,329, 000
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 

Question 6.34
Answer: A
Let B be the amount of death benefit.
= 500
=
a61 500(14.6491)
= 7324.55
EPV(Premiums)
EPV(Benefits) =B ⋅ A61 =(0.30243)B
EPV(Expenses) =
(0.12)(500) + (0.03)(500)a61 =
(0.12)(500) + (0.03)(7324.55) =
279.74
= EPV(Benefits) + EPV(Expenses)
EPV(Premiums)
7324.55 = (0.30243)B + 279.74
7044.81 = (0.30243)B
B = 23, 294
Page 49
Question 6.35
Answer: D
Let G be the annual gross premium. By the equivalence principle, we have
35 100,000 A35 + 0.15G + 0.04Ga35
Ga
=
so that
=
G
100, 000 ( 0.09653)
100, 000 A35
=
= 534.38
0.96a35 − 0.15 0.96 (18.9728 ) − 0.15
Question 6.36
Answer: B
By the equivalence principle,
=
4500a x:20 100,000 A x1:20 + Ra x:20
where
µ
(1 − e
µ +δ
A x1:20 =
− 20( µ +δ )
) =0.04
(1 − e
0.12
− 20(0.12)
) =0.3031
1 − e − 20( µ +δ ) 1 − e − 20(0.12)
a x:20
=
= = 7.5774
0.12
µ +δ
Solving for R, we have
 0.3031 
R=
4500 − 100,000 
500
=
 7.5774 
Question 6.37
Answer: D
By the equivalence principle, we have
Ga35:10
= 50,000 A35 + 100a35 + 100 A35
so
G
50,100 A35 + 100 ( a35 − 1) 50,100 ( 0.09653) + 100 (17.9728 )
=
= 819.69
a35:10
8.0926
Page 50
Question 6.38
Answer: B
Let P be the annual net premium
=
P
1000 A x:n 1000(0.192)
=
ax:n
ax:n
where
ax=
:n
A x:n
=
1 − A x:n (1.05)
=
1 − A1x:n − Ax:n1
d
(0.05)
(
i
δ
(A )+
0.192
⇒=
1
x:n
n
)
Ex
(
)
0.05
A 1 + 0.172
0.04879 x:n
⇒ A 1x:n =
0.019516
⇒ ax:n =
1.05
(1 − 0.019516 − 0.172)= 16.978
0.05
Therefore, we have
=
P
1000 ( 0.192 )
= 11.31
16.978
Page 51
Question 6.39
Answer: A
1000 A40 121.06
Premium at issue for (40): =
= 6.5587
a40
18.4578
Premium at issue for (80):
1000 A80 592.93
= = 69.3615
a80
8.5484
Lives in force after ten years:
10,000 ×
Issued at age 40: 10,000 10 p40 =
Issued at age 80:
98,576.4
=
9923.30
99,338.3
10, 000 10 p80 =
10, 000 ×
41,841.1
=
5530.35
75, 657.2
The total number of lives after ten years is therefore: 9923.30 + 5530.35 =
15, 453.65
The average premium after ten years is therefore:
(6.5587 × 9923.30) + (69.3615 × 5530.35)
= 29.03
15, 453.65
Page 52
Question 6.40
Answer: C
Let P be the annual net premium at x+1. Also, let A*y be the expected present value for the
special insurance described in the problem issued to ( y ) .
Pax +1 1000
=
∑ k =0 (1.03) v k +1 k| qx+1 1000 Ax*+1
∞
k +1
We are given
110ax 1000
=
∑ k =0 (1.03) v k +1 k| qx 1000 Ax*
∞
k +1
Which implies that
*
110 (1 + vpx a=
x +1 ) 1000 (1.03vq x + 1.03vp x Ax +1 )
Solving for Ax*+1 , we get
110
1 + v ( 0.95 )( 7 )  − 1.03v ( 0.05 )
1000
= 0.8141032
1.03v ( 0.95 )
Ax*+1
Thus, we have
=
P
1000 ( 0.8141032 )
= 116.3005
7
Question 6.41
Answer: B
Let P be the net premium for year 1.
Then:
P + 1.01Pvpx= 100, 000vqx + (1.01)(100, 000)v 2 px qx +1
 0.01 (1.01)(0.99)(0.02) 
 1.01

P 1 + =
+
=
0.99  100, 000 
 ⇒ P 1416.93
2
1.05
(1.05)
 1.05



Page 53
Question 6.42
Answer: D
The policy is fully discrete, so all cash flows occur at the start or end of a year.
Die Year 1 ==> L0 =1000v − 315.80 =625.96
Die Year 2 ==>=
L0 1000v 2 − 315.80(1 +=
v) 273.71
Survive Year 2 ==> L0 =1000v3 − 315.80(1 + v + v 2 ) =−58.03
There is a loss if death occurs in year 1 or year 2, otherwise the policy was profitable.
Pr ( death in year 1 or 2 ) =
1 − e −2 µ =
0.113
Question 6.43
Answer: C
APV ( expenses
=
) 0.35G + 8 + 0.15Ga30:4 + 4a30:9
= 0.20G + 4 + 0.15G a30:5 + 4a30:10
1
G a=
0.20G + 4 + 0.15G a30:5 + 4a30:10 + 200,000 A30:10
30:5
G=
1
200,000 A30:10
+ 4 + 4a30:10
0.85a30:5 − 0.20
1
=
200, 000 A30:10
200, 000  A30:10 − 10 E30 
= 200,000(0.61447 − 0.61152)
= 590
G
590 + 4 + 4 ( 8.0961)
= 171.07
0.85 ( 4.5431) − 0.20
Page 54
Question# 6.44
Answer: D
Let P be the premium per 1 of insurance.
1
Pa50:10 P ( I A )50:10
=
+ 10 E50 A60
a50:10 =a50 − 10 E50 a60 =17.0 − 0.60 ×15.0 =8
 0.05 
A60 =
1 − da60 =
1− 
0.285714
15 =
 1.05 
(
)
1
P a50:10 − ( I A )50:10
=
10 E50 A60
P=
0.6 × 0.285714
10 E50 A60
=
= 0.021838
8 − 0.15
a50:10 − ( I A )501 :10
100 P = 2.18
Page 55
Question 6.45
Answer: E
560  −δ T 560

L0= 100, 000vT − 560aT= 100, 000 +
e −
δ 
δ

Since L0 is a decreasing function of
that Pr[T35 > t ] =
0.75 .
, the 75th percentile of L0 is L0 (t ) where t is such
l35+t
= 0.75
l35
l35+t =
0.75l35 =
0.75 × 99,556.7 =
74, 667.5
l81 < 74, 667.5 < l80
t = (80 − 35) + s
l80+ s = sl81 + (1 − s )l80
74, 667.5 =73,186.3s + 75, 657.2(1 − s )
s = 0.40054
t = 45.40054

560  −45.40054ln(1.05)
560
= 100, 000 +
L0 (45.40054)
−
=
689.25
e
ln(1.05) 
ln(1.05)

Question 6.46
Answer: E
Let P be the premium per 1 of insurance.
=
Pa55:10 0.51213P + v10 10 p55 a65
a55 =
a55:10 + v10 10 p55 a65 ⇒ v10 10 p55 a65 =
12.2758 − 7.4575 =
4.8183
7.4575
=
P 0.51213P + 4.8183=
⇒ P 0.693742738
300 P = 208.12
Page 56
Question 6.47
Answer: D
Ga70:10
= 100, 00010 E70 a80 + 0.05Ga70:10 + 0.7G
=
7.6491G (100, 000)(0.50994)(8.5484) + 0.05G (7.6491) + 0.7G
⇒G =
66,383.54
Question 6.48
Answer: A
Actuarial present value of insured benefits:
 0.95 × 0.02 0.95 × 0.98 × 0.03 0.95 × 0.98 × 0.97 × 0.04 
+
+
100, 000 
5, 463.32
6
 =
1.067
1.068
 1.06
 0.95 
⇒ P 1 + =
⇒ P 3,195.12
5, 463.32=
5 
 1.06 
Question 6.49
Answer: C
i
(12)
Ga40:20
=
100, 000 
δ

(12)
 A40 + 200 + 0.04Ga40:20

(12)
a40:20
=
α (12)a40:20 − (1 − 20 E40 ) β (12)
= 1.00020 ⋅12.9935 − (1 − 0.36663) ⋅ 0.46651
= 12.700625
G
(100, 000)(1.02480)(0.12106)+200
= 1033.92
0.96 ×12.700625
⇒ G/12 = 86.16
Page 57
Question 6.50
Answer: A
A35
96.53
=
=
= 5.0878
1, 000 P 1, 000
a35 18.9728
Benefits paid during July 2018:
= 3910
10,000 ×1,000 × q=
10, 000 × 0.391
35
Premiums payable during July 2018:
= 9,996.09 × 5.0878
= 50,858.10
10,000 × (1 − q35 ) × 5.0878
Cash flow during July 2018:
3910 − 50,858 =
−46,948
Question 6.51
Answer: D
Under the Equivalence Principle
(
) (
1
Pa62:10
= 50, 000 a62 − a62:10 + P ( IA) 62:10
1
1
= 11A62:10
−
where ( IA) 62:10
So
=
P
Page 58
(
10
∑A
k =1
)
1
62: k
)
= 11(0.091) − 0.4891 = 0.5119
50,000 a62 − a62:10
50,000(12.2758 − 7.4574)
=
= 34,687
1
7.4574 − 0.5119
a62:10 − ( IA)62:10
Question 6.52
Answer: E
Ga45:10= HA45 + G + 0.05Ga45:10 + 80 + 10a45 + 10a45:10
G=
G=
=
G
HA45 + 80 + 10 ( a45 + a45:10 )
0.95a45:10 − 1
HA45 + 80 + 10 (17.8162 + 8.0751)
(0.95 × 8.0751) − 1
80 + 10 (17.8162 + 8.0751)
A45
H+
(0.95 × 8.0751) − 1
(0.95 × 8.0751) − 1
g=
A45
(0.95 × 8.0751) − 1
f
80 + 10 (17.8162 + 8.0751)
= 50.80
(0.95 × 8.0751) − 1
Question 6.53
Answer: D
 0.1 0.9 × 0.1 0.9 × 0.9 × 0.1 
G=
+
+
0.35G + 2000 

1.082
1.083
 1.08

0.65G = 468.107
G = 720.16
Page 59
Question 6.54
Answer: A
2
2

A45  2
 P 2
2
2
On a unit basis, Var ( L0 ) =
1 +
  A45 − ( A45 ) 
1 +   A45 − ( A45 )  =
 d
 da45 
2
 da + 1 − da45  2
=  45
A45 − ( A45 ) 2 
 =

da45


2
A45 − ( A45 ) 2
(da) 2
0.03463 − 0.151612
=
0.016178038
2
 0.05

×17.8162 

 1.05

The standard deviation of L0 = 0.127193
(200, 000)(The standard deviation of L0 ) = 25, 439
Question 6.55
Answer: D
1
EPV(benefits) = 1000 × A45:20
+ 2500 × 20 a45
= 1000(0.15161 − 0.35994 × 0.35477) + 2500 × 0.35994 ×13.5498 =
12, 216.7
P ×12.9391
EPV(premiums) = P × (17.8161 − 0.35994 ×13.5498) =
=
P
12216.7
= 944.17
12.9391
Page 60
Question 6.56
Answer: D
A35:20 = 0.37981
a35:20 = 13.0240
APV
of Premium APV of Benefits + APV of Expenses
=
APV of Benefits + APV of $ Expenses = 1, 000, 000 × A35:20 + 50 × a35:20 + 100 =
380,561.20
APV of Premium − APV of % Expenses = (0.95) × P × a35:20 − (0.5) × P = 11.8728 × P
=
P
380,561.20
= 32, 053.20
11.8728
Page 61
Question 7.1
Answer: C
10
V= 50, 000 ( A50 + 10 E50 A60 ) − (875)  a50:10 


50, 000[0.18931 + (0.60182)(0.29028)] − 875[8.0550]
=
= 11,152
Question 7.2
Answer: C
V =0
0
V = 2000
2
Year 1:
(
V + P ) (1
=
+ i ) qx ( 2000 + 1V ) + px 1V
0
P=
(1.1) 0.15(2000 + 1V ) + 0.85(1V )
1.1P − 300 =
1V
Year 2:
(1 + i )
( V + P )=
1
qx +1 (2000 + 2V ) + px +1 (2000)
(1.1P − 300 + P=
)(1.1) 0.165(2000 + 2000) + 0.835(2000)
2.31P − 330 =
2330
P
=
Page 62
2330 + 330
= 1152
2.31
Question 7.3
Answer: E
means an interest rate of j 0.02 per quarter. This problem can be done with
i (4) 0.08
=
two quarterly recursions or a single calculation.
Using two recursions:
10.75
V=
[10.5V + 60(1 − 0.1)](1.02) −
753.72
=
10.5
V
10.25
706
800
800 − 706
(1000)
800
[10.5V + 54](1.02) − 117.50
=
⇒10.5 V 713.31
0.8825
898 − 800
(1000)
[ V ](1.02) − 109.13
898
=
⇒ 713.31 10.25
800
0.8909
898
[10.25V ](1.02) −
V = 730.02
Using a single step, 10.25V is the EPV of cash flows through time 10.75 plus
0.5
E80.25 times
the EPV of cash flows thereafter (that is, 10.75V ).
 800   706 
 898 − 800 800 − 706 
=
+
− (60)(1 − 0.1) 
(1000) 
(753.72) = 730
10.25 V
+
2
2
 898(1.02) 898(1.02) 
 898(1.02)   898(1.02) 
Page 63
Question 7.4
Answer: B
EPV of benefits at=
issue 1000 A40 + 4 11 E40 (1000 A51 )
=
121.06 + (4)(0.57949)(197.80) =
579.55
EPV of expenses at issue = 100 + 10(a40 − 1) = 100 + 10(17.4578) = 274.58
π=
(579.55 + 274.58) / a40 =
854.13 /18.4578 =
46.27
=
G 1.02
=
π 47.20
EPV of benefits at time 1 = 1000 A41 + 4 10 E 41 ×1000 A51
=
126.65 + (4)(0.60879)(197.80) =
608.32
EPV of expenses at time
=
1 10(
=
a41 ) 10(18.3403)
= 183.40
Gross Prem Policy Value = 608.32 + 183.40 − Ga41= 791.72 − 47.20(18.3403)= −73.94
Question 7.5
Answer: E
V=
v 0.5 0.5 p x + 4.5 5V + v 0.5
4.5
=
0.5 q x + 4.5
0.5
0.5
q x + 4.5 b, where b =
10, 000 is the death benefit during year 5
0.5(0.04561)
0.5 q x + 4
=
= 0.02334
1 − 0.5 q x + 4 1 − 0.5(0.04561)
p x + 4.5 = 0.97666
V=
5
V=
5
V
4.5
( 4V + P) (1.03) − q x + 4b
p x+4
(1, 405.08 + 647.46) (1.03) − 0.04561(10, 000)
= 1, 737.25
0.95439
(1.03) −0.5 (0.97666) (1, 737.25) + (1.03) −0.5 (0.02334) (10, 000)
= 1, 671.81 + 229.98 = 1,902
V can also be calculated recursively:
4.5
=
0.5(0.04561)
= 0.02281
0.5 q x + 4
4.5V
(1, 405.08 + 647.16)(1.03)0.5 − 0.02281(10, 000) / (1.03)0.5
= 1,902
1 − 0.02281
The interest adjustment on the death benefit term is needed because the death benefit will not be
paid for another one-half year.
Page 64
Question 7.6
Answer: E
Gross premium = G
  2000 
   2000 

Ga=
2000 A45 + 1
45
 + 20  +  0.5 
 + 10  a45 + 0.20G + 0.05Ga45
1000 
 1000
  





22
( 0.95a
G
45
11
− 0.20=
) G 2000 A45 + 22 + 11a45
45 2000(0.15161) + 22 + 11(17.8162)
2000 A45 + 22 + 11a
=
= 31.16
45 − 0.20
0.95a
0.95(17.8162) − 0.20
There are two ways to proceed. The first is to calculate the gross premium policy value (with
equivalence principle gross premium and original assumptions, both of which do apply here) and
the net premium policy value and take the difference.
2000 A45
a45
=
The net premium
is
2000(0.15161)
= 17.02
17.8162
The net premium policy value is 2000 A55 − 17.02
=
a55 2000(0.23524) − 17.02(16.0599)
= 197.14
The gross premium policy value is
2000 A55 + [0.05(31.16) + 0.5(2000 /1000) + 10]a55 − 31.16a55
= 2000(0.23524) + (12.56 − 31.16)(16.0599)
= 171.77
Expense policy value is 171.77 – 197.14 = –25
The second is to calculate the expense policy value directly based on the pattern of expenses.
The first step is to determine the expense premium.
The present value of expenses is
[0.05G + 0.5(2000 /1000) + 10]a45 + 0.20G + 1.0(2000 /1000) + 20
28.232 251.97
= 12.558(17.8162) +=
The expense premium is 251.97/17.8162=14.14
The expense policy value is the expected present value of future expenses less future expense
premiums, that is,
[0.05G + 0.5(2000 /1000) + 10]a55 − 14.14a55 =
−1.582(16.0599) =
−25
Page 65
There is a shortcut with the second approach based on recognizing that expenses that are level
throughout create no expense policy value (the level expense premium equals the actual
expenses). Therefore, the expense policy value in this case is created entirely from the extra first
year expenses. They occur only at issue, so the expected present value is
0.20(31.16)+1.0(2000/1000)+20 = 28.232. The expense premium for those expenses is then
28.232/17.8162 = 1.585 and the expense policy value is the present value of future non-level
expenses (0) less the present value of those future expense premiums, which is 1.585(16.0599) =
25 for a reserve of –25.
Question 7.7
Answer: D
ax +10 =
(1 − Ax +10 ) / d =
(1 − 0.4) / (0.05 /1.05) =
12.6
ax(12)
12.142
+10 ≈ 12.6 − 11/ 24 =
V
= 10, 000 A x +10 + 100ax +10 − 12ax(12)
+10 (30)(1 − 0.05)
10
V= 10, 000(0.4) + 100(12.6) − 12(12.142)(28.50)
10
V = 1107
10
Question 7.8
Answer: C
The simplest solution is recursive:
0
V = 0 since the policy values are net premium policy values.
=
q [70] (0.7)(0.010413)
= 0.007289
V
1
(0 + 35.168)(1.05) − (1000)(0.007289)
= 29.86
1 − 0.007289
=
q [70]+1 (0.8)(0.011670)
= 0.009336;
=
q [70]+ 2 (0.9)(0.013081)
= 0.011773
Prospectively,
=
A[70]+1 (0.009336)v + (1 − 0.009336)(0.011773)v 2
+ (1 − 0.009336)(1 − 0.011773)(0.47580)v 2 =
0.44197
(
)
a[70]+1 =
1 − A[70]+1 / d =
(1 − 0.44197) / (0.05 /1.05) =
11.7186
V =(1000)(0.44197) − (11.7186)(35.168) =29.85
1
Page 66
Question 7.9
Answer: A
Let P = 0.00253 be the monthly net premium per 1 of insurance.
i 1

=
+ A55:101 − 12 Pa(12)
100, 000  A55:10
10V
55:10 
δ

= 100, 000 [1.02480(0.02471) + 0.59342 − (12)(0.00253)(7.8311) ]
≈ 38,100
Where
1
A55:10
= A55:10 − 10 E 55 = 0.61813 − 0.59342 = 0.02471
1
A=
55:10
E 55 0.59342
=
10
a55:10 = 8.0192
= α (12)a55:10 − β (12) 1 − 10 E 55 
a(12)
55:10
= 1.00020(8.0192) − 0.46651(1 − 0.59342)
= 7.8311
Page 67
Question 7.10
Answer: C
Use superscript g for gross premiums and gross premium policy values.
Use superscript n (representing “net”) for net premiums and net premium policy values.
Use superscript e for expense premiums and expense policy values.
P g = 977.60 (given)
0.58 P g + 450 + ( 0.02 P g + 50 ) a45
e
P =
a45
0.58(977.60) + 450 + [0.02(977.60) + 50]17.8162
= 126.64
17.8162
Alternatively,
Pn =
100, 000 A45
45
a
= 850.97
P e = P g − P n = 126.63
Ve=
−972
[0.02(977.60) + 50](17.0245) − 126.64(17.0245) =
( 0.02 P g + 50 ) a50 − P e a50 =
5
Alternatively,
n
=
100, 000 A50 − P n a50
5V
= 100, 000(0.18931) − 850.97(17.0245) = 4443.66
g
=
100, 000 A50 + ( 50 + 0.02 P g − P g ) a50
5V
= 100, 000(0.18931) + [50 + 0.02(977.60) − 977.60](17.0245)
= 3471.93
g
− 5V n =
−972
Ve=
5V
5
Question 7.11
Answer: B
L= 10, 000v
K 45 +1
K
− Pa
45 +1
11
= 10, 000v 11 − Pa
=
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
55 (10, 000)(0.36093) − (160.15)(13.4205)
=
10, 000 A55 − Pa
=
= 1, 460
10V
Page 68
Question 7.12
Answer: E
In the final year: ( 24V + P)(1=
+ i) b 25 (q 68 ) + 1( p 68 )
Since b 25 = 1, this reduces to ( 24V + P)(1 + i ) =1 ⇒ (0.6 + P)(1.04) =1 ⇒ P =0.36154
Looking back to the 12th year: ( 11V + P)(1=
+ i ) b12 (q 55 ) + 12V ( p 55 )
⇒ (5.36154)(1.04)
= 14(0.15) + 12V (0.85) ⇒ 12=
V 4.089
Question 7.13
Answer: A
This first solution recognizes that the full preliminary term reserve at the end of year 10 for a 30
year endowment insurance on (40) is the same as the net premium policy value at the end of year
9 for a 29 year endowment insurance on (41). Then, using superscripts of FPT for full
preliminary term reserve and NLP for net premium policy value to distinguish the symbols, we
have
FPT
NLP
=
=
1000
1000
1000( A50:20 − P41:29 a50:20 )
10V
9V
=
1000[0.38844 − 0.01622(12.8428)] =
180
 a
or = 1000 1 − 50:20
 a41:29


 12.8428 
 = 1000 1 −
 = 180

15.6640 


where
= a41 − 29 E 41 a70
a41:29
= 18.3403 − (0.2228726)(12.0083)
= 15.6640
A41:29 =
1 − d (15.6640) =
0.254095
 l 70 
 91, 082.4 
29
=
=
E
v
  (0.242946)  =
29
41
 0.2228726
 99, 285.9 
 l 41 
0.254095
=
= 0.01622
P41:29
15.6640
Page 69
=0
Alternatively, working from the definition of full preliminary term reserves as having 1V
and the discussion of modified net premium reserves in the Notation and Terminology Study
Note, let α be the valuation premium in year 1 and β be the valuation premium thereafter.
Then (with some of the values taken from above),
FPT
vq40 0.5019
=
=
α 1000
APV (valuation premiums) = APV (benefits)
1000 A40:30
α + 1 E40 (a41:29 ) β =
0.5019 + 0.95188(15.6640) β =
242.37
242.37 − 0.5019
=
β = 16.22
14.9102
Where
1
E40 =
(1 − 0.000527)v =
0.95188
A40:30 =
A40 + 20 E40 ( 10 E60 )(1 − A70 )
=0.12106 + 0.36663(0.57864)(1 − 0.42818) =0.24237
= 180
V = 1000 A50:20 − β a50:20= 1000(0.38844) − 16.22(12.8427)
FPT
10
Question 7.14
Answer: A
=
( V + 0.96G − 50 ) (1.05)
5
G − 50 ) (1.05)
( 5500 + 0.96=
q 50 (100, 200) + p 50 6V
(0.009)(100, 200) + (1 − 0.009)(7100)
(1.05)(0.96)G + 5722.5 =
7937.9
(1.05)(0.96)G = 2215.4
G = 2197.8
Question 7.15
Answer: E
V (1=
+ i)
0.4
15.6
0.4
px +15.6
V + 0.4 qx +15.6 100
16
0.4
=
0.957447(49.78) + 0.042553(100)
15.6V (1.05)
V = 50.91
15.6
Page 70
Question 7.16
Answer: D
APV future benefits
= 1000 0.04v + 0.05 × 0.96v 2 + 0.96 × 0.95 × (0.06 + 0.94 × 0.683)
=
v 3  630.25
APV future premiums =130(1 + 0.96v) =248.56
E[ 3 L] = 630.25 − 248.56 = 381.69
Question 7.17
Answer: D
2
V  10 L 
V [ 11 L ]
 p 2
2
1 +  ( A x +10 − A x +10 )
d

=
2
 p 2
2
1 +  ( A x +11 − A x +11 )
d


=
A
vqx +10 + vp x +10 A x +11
x +10
1
1
= (0.90703) 2 (0.02067) + (0.90703) 2 (1 − 0.02067)(0.52536)
= 0.50969
2
A x +10 v 2 q x +10 + v 2 p x +10 2 A x +11
=
= (0.90703)(0.02067) + (0.90703)(1 − 0.02067)(0.30783)
= 0.29219
Var ( k L )
(0.29219) − (0.50969) 2 0.03241
⇒
=
=
=
1.018
2
0.03183
Var ( k +1 L ) (0.30783) − (0.52536)
Page 71
Question 7.18
Answer: A
x +1 =
x +1 − Px a
x +1
Vx =
Ax +1 − Px a
1 − da
1

=
1 − ( Px + d ) ax +1 =
1 − a x +1 a



(
x
)
⇒ ax 1 − 1Vx =
ax +1
Since ax = 1 + vp x ax +1 substituting we get
a − 1
ax (1 − 1Vx ) = x
⇒ ax (1 − 1Vx ) vpx =ax − 1
vpx
x , we get ax =
Solving for a
1
1 − (1 − 1Vx ) vp x
1
=
 1 
 (1 − 0.009 )
 1.04 
1 − (1 − 0.012) 
= 17.07942
Question 7.19
Answer: D
Let G be the annual gross premium.
Using the equivalence principle, 0.90Ga40 − 0.40
=
G 100, 000 A40 + 300
So G
100, 000(0.12106) + 300
= 765.2347
0.90(18.4578) − 0.40
The gross premium policy value after the first year and immediately after the second premium
and associated expenses are paid is
41 − 1)
100, 000 A41 − 0.90G ( a
= 12, 665 − 0.90(765.2347)(17.3403)
= 723
Page 72
Question 7.20
Answer: A
If G denotes the gross premium, then
G
35 + 270 1000(0.09653) + 30(18.9728) + 270
1000 A35 + 30a
=
= 52.12
35 − 0.26
0.96a
0.96(18.9728) − 0.26
So that,
=
R 1000 A36 + (30 − 0.96G )a36
=
1000(0.10101) + (30 − 0.96 × 52.12)(18.8788) =
−277.23
Note that S = 0 as per definition of FPT reserve.
Question 7.21
Answer: D
π
=
=
9V
1000 10 a55
1000 ( 0.59342 )(13.5498 )
=
= 1021.46
1
8.0192 − 0.14743
a55:10 − ( IA )55:10
1000
= 1000
1
1
a64 + 10π A64:1
− π a64:1
1  94,579.7 
1
13.5498 + 10 (1021.46 )
( 0.005288) − 1021.46


1.05  95, 082.5 
1.05
= 11,866
Question 7.22
Answer: C
P1 

V  L 0 #1 =+
 B1

d 

2
A x − Ax2
=
2
(
2
Ax − A
20.55
)=
 1.25(1.06) 
=
=
> 8 +

0.06 

2
(
2
A x − Ax2 ) =
20.55
20.55
0.0227
=
2
 1.25(1.06) 
8 +

0.06 

1.875(1.06) 

V [ L0 # 2] = 12 +

0.06


Page 73
2
x
2
(
2
2
1.875(1.06) 

A x − A ) = 12 +
 ( 0.0227 ) = 46.24
0.06


2
x
Question 7.23
Answer: D
We have Present Value of Modified Premiums = Present Value of level net premiums
vqx + β ( a25:20 − 1) + P ⋅ 20 E25 ⋅ a45:20 =
Pa25:40
⇒β
P ( a25:40 ) − P ⋅ 20 E25 ⋅ a45:20 − vqx P a25:20 − vqx
=
a25:20 − 1
a25:20 − 1
We are given that P = 0.0216
⇒β
0.0216(11.087) − (1.04) −1 ( 0.005 )
= 0.023265
11.087 − 1
For insurance of 10,000, β = 233.
Question 7.24
Answer: C
g
P=
Pn + Pe
where P e is the expense loading
A
 0.18931 
000, 000 50 1, 000, 000
=
P n 1,=
=

 11,119.86
a50
 17.0245 
P e = P g − P n = 11,800 − 11,120 = 680
Question 7.25
Answer: B
FPT
=
100,000 A[55]+3 − 100,000 P[55]+1 a[55]+3
3V
= 100,000 A58 − 100,000
A[55]+1
a[55]+1


0.24
= 100,000  0.27 −
1 − 0.24

d

= 3947.37
Page 74
a58


1 − 0.27 

d


Question 7.26
Answer: D
V=
( 0V + P )(1 + i ) − ( 25, 000 + 1V − 1V ) qx =
V=
( 1V + P )(1 + i ) − ( 50,000 + 2V − 2V ) qx+1 =
1
2
P(1 + i ) − (25, 000)qx
50,000
(( P (1 + i ) − 25,000q ) + P ) (1 + i ) − 50,000q =50,000
50,000
(( P (1.05) − 25,000 ( 0.15) ) + P ) (1.05) − 50,000 ( 0.15) =
x +1
x
Solving for P, we get
=
P
61,437.50
= 28,542.39
2.1525
Question 7.27
Answer: B
Since G is determined using the equivalence principle, 0V = 0
P






 0 + G − 187 − 0.25G − 10  (1.03)



= −38.7
Then, 1V e = 
0.992
e
=
⇒ 0.75G
−38.7 ( 0.992 )
+=
187 + 10 159.72
1.03
⇒G =
212.97
Page 75
Question 7.28
Answer: D
20
V=
0=
=
> 1000 A65 =+
( P W ) × a65
At issue, present value of benefits must equal present value of premium, so:
1000 A45 =
Pa45 + W 20 E45 × a65
354.77 = ( P + W )(13.5498) ⇒ P + W = 26.182674 ⇒ P = 26.182674 − W
151.61 17.8162 P + W (0.35994)(13.5498)
=
151.61 17.8162(26.182674 − W ) + W (0.35994)(13.5498)
=
⇒W =
24.33447
Page 76
Question 7.29
Answer: E
 a 
 11.4 
V10= 2, 290= B 1 − x +10 = B 1 −
 ⇒ B= 9,968.24
ax 
 14.8 

Gax = 25 + 5ax + B × Ax
 0.04

1 − dax =
1− 
0.430769231
Ax =
× 14.8  =
 1.04

G × 14.8 = 25 + 5 × 14.8 + 9,968.24 × 0.430769231
296.82
⇒G=
g
=
9,968.24 Ax +10 + 5ax +10 − 296.82ax +10
10V
 0.04

1 − dax +10 =
1− 
×11.4  =
0.561538462
Ax +10 =
 1.04

g
=
9,968.24 × 0.561538462 + 5 ×11.4 − 296.82 ×11.4
10V
⇒ 10V g =
2, 270.80
Alternatively, the expense net premium is based on the extra expenses in year 1, so
P e =−
(30 5) / 14.8 =
1.68919
10
Ve =
0 − 1.68919(11.4) =
−19.26
10
V g =10 V n +10 V e = 2290 − 19.26 = 2270.74
Question 7.30
Answer: E
000 A35 965.30
L10 10,
=
=
L*10 = 10, 000
10, 000 − 965.30
L*10 − L=
= 9034.70
10
Page 77
Question 7.31
Answer: E
=
2 0.08G + 5
Future expenses at x +
Expense load at x + 2 =
Pe
−23.64
= (0.08G + 5) − P e
58.08
⇒ Pe =
1000 Px:3 = 368.05 − 58.08 = 309.97
Question 7.32
Answer: B
 10  T 10
LA =
v T − 0.10a T =
1 +  v −
6
 6
2
 10 
T
0.455 ⇒ Var (vT ) =
0.06398
Var ( LA ) =
1 +  Var (v ) =
6


16  T 16

2v T − 0.16a T =
LB =
2 + v −
6
6

2
2
 16 
 16 
T
1.39
Var ( LB ) =
 2 +  Var (v ) =
 2 +  (0.06398) =
6
6


Question 7.33
Answer: C
1
A80:10
= A80:10 − 10 E80 = 0.67674 − 0.33952 = 0.33722
1
=
A80:10
0.05
=
0.33722 0.34559
ln(1.05)
EPV future benefits = 1000 (0.34559) + 500 (0.33952) = 515.35
EPV future net premiums
= a=
P 6.7885 ( 35.26
=
) 239.36
80:10
Policy value = 515.36 – 239.36 = 276
Page 78
Question 7.34
Answer: D
(10.5V + P)(1 + i=
)0.25 1, 000, 000 × 0.25 q50.5 + 0.25 p50.5 × 50.75V
 (1 − 0.75 × 0.01)   (1 − 0.75 × 0.01) 
0.25
=
(10, 000 + 750)(1 + 0.05)
1, 000, 000 × 1 −
+
 × 50.75V
(1 − 0.5 × 0.01)   (1 − 0.5 × 0.01) 

50.75V = 8,390
Question 7.35
Answer: D
The modified net premium reserve at duration 2=
is: 2V mod 100000 A62 − P mod a62
A62
 0.31495 
=
P mod 100000
=
100000 =
 2189
a62
 14.3861 
Question 7.36
Answer: B
There is no change to 9V because there were no errors after time 9.
An equation for 8Vorig is 8Vorig + P – e = 1000 (0.003736) / 1.05 + 100 (0.996264) / 1.05
An equation for 8Vnow is 8Vnow + P – e = 1000 (0.002736) / 1.05 + 100 (0.997264) / 1.05
By subtraction, 8Vnow – 8Vorig = 1000 (–0.001) / 1.05 + 100 (0.001) / 1.05 = –0.86
now
8V
= 86.74 – 0.86 = 85.88 (B)
In those equations, P – e, which cancels, represents the gross premium less all expenses. You
could think of P as the gross premium less commissions, and e as the flat expenses. However
you think of representing them, they cancel.
Question 7.37
Answer: D
a65:5 = a65 − 5 E65 × a70 = 13.5498 − 0.75455(12.0083) = 4.4889
EPV future net premiums = 4.4889 (5,808) = 26,072
130,580 = EPV future benefits – EPV future net premiums = EPV future benefits – 26,072
EPV future benefits = 156,652
Page 79
Question 7.38
Answer: A
There is no change in the EPV of future benefits or of future gross premiums. Each year that a
premium is due, the expense changes by (3% - 2%) of 16, or 0.16. Since at the end of year 10,
the insured is 60 with 20 remaining premiums, the EPV of future expense rises by
=
0.16a60:20 0.16(12.3816)
= 1.981
The new reserve is 110 + 1.981 = 112
Question 7.39
Answer: C
(18V + P )(1 +=
i ) p[45]+18 × 19V
V = 0.976956 × 575, 000 /1.05 − 20, 000= 515, 000
18
Question 7.40
Answer: A
1000 A62 − ( P mod )(a62 ) =
0
The policy value at duration 2 is 2V mod =
=
P mod 1000
=
A62 / a62 1000(0.31495)
=
/14.3861 21.89
V mod =
1000 A65 − ( P mod )(a65 ) =
1000(0.35477) − 21.89(13.5498) =
58
5
Question 18.1
Answer: A
 50 − 10 − 1 
=
Sˆ (10.8) Sˆ=
(10.0) 
 0.897
 50 − 10 
Question 18.2
Answer: D
Sˆ (1) = 0.8
S (1)(1 − S (1) (0.8)(0.2)
=
≈
0.012652
n
1000
⇒ 95% CI is approximately (0.8 ± 1.96(0.01265)) =
(0.775,0.825)
=
Var[ S (1)]
Page 80
Question 18.3
Answer: D
1 3
Hˆ (1.5) =
+ = 0.048148
90 81
− Hˆ (1.5)
ˆ (1.5) e=
0.9530
S=
Question 18.4
Answer: D
59 59 − 8 + 1 − 1 51 − 6 + 7 − 2 50 − 7 + 7 − 1 49 − 6 + 5 − 1
Sˆ (21.0) =
×
×
×
×
=
0.8899
60 59 − 8 + 1
51 − 6 + 7
50 − 7 + 7
49 − 6 + 5
1
2
1
1 
 1
Var[ Sˆ (21.0)] ≈ 0.88992 
+
+
+
+
0.00181
=
 60 × 59 52 × 51 52 × 50 50 × 49 48 × 47 
The upper limit of the 80% linear confidence interval is
0.8899 + 1.282(0.00181)0.5 =
0.944.
Question # 18.5
Answer: D
100 − 10 − 14 − 16
100 − 10 − 14 − 16 − 20
=
= 0.60,
=
= 0.40
Sˆ (30)
Sˆ (40)
100
100
Use linear interpolation to find Sˆ (32)
 40 − 32  ˆ
 32 − 30  ˆ
=
Sˆ (32) 
S (40)+ 0.8(0.60) + 0.2(0.40) = 0.56
 S (30) + 
=
 40 − 30 
 40 − 30 
Page 81
Question 18.6
Answer: A
The contribution from Life 1 is
contribution is
20
p70 . With Gompertz and the selected parameters, the
20
 B 70 20

 0.000003 70

1.1 (1.120 − 1)  = 0.86730.
p70 = exp  −
c (c − 1)  = exp  −
ln1.1
 ln c



The contribution from Life 2 is
19
p70 − 20 p70 . The contribution is
 B 70 19

 0.000003 70

p70 − 20 p70 = exp  −
c (c − 1)  − 0.86730= exp  −
1.1 (1.119 − 1)  − 0.86730
ln1.1
 ln c



= 0.88058 − 0.86730 = 0.01328.
19
The contribution to the likelihood is 0.86730(0.01328) = 0.01152.
Question 18.7
Answer: D
The contribution from Life 1 is
contribution is
14.5
p60 . With Gompertz and the selected parameters, the
 B 60 14.5

 0.000004

p60 = exp  −
c (c − 1)  = exp  −
1.1260 (1.1214.5 − 1)  = 0.87619.
 ln c

 ln1.12

The contribution from Life 2 is
14.5
14.5
14.5
p60 × µ74.5 . The contribution is
= 0.87619 × 0.000004 ×1.1274.5
= 0.01627.
p60 × µ74.5
The contribution to the likelihood is 0.87619(0.01627) = 0.01426.
The contribution to the log-likelihood is ln(0.01426) = –4.25.
Question 18.8
Answer: A
1 1 1 1
+ + + = 0.50397
Hˆ (50) =
12 9 7 6
−0.50397
=
Sˆ (50) e= 0.60413
11 8 6 5
+ 3 + 3 + 3 = 0.05798
V [ Hˆ (50)] =
3
12 9 7 6
2
2
V [ Sˆ (50)] Sˆ=
(50) × V [ Hˆ (50)] = 0.02116 0.1455
Page 82
Question 18.9
Answer: B
r1 = 80 − 20 + 4 = 64; r2 = 63 − 2 + 3 = 64; r3 = 63 − 6 = 57
63 63 56
Sˆ (2) = Sˆ (t(3) ) =
× ×
= 0.951994
64 64 57
Question 18.10
Answer: E
r1 = 50 − 4 = 46; r2 = 45 − 2 + 3 = 46; r3 = 45 − 5 = 40
1
1
1
Hˆ (2) =
+ +
= 0.06848
46 46 40

− H (2)
=
Sˆ (2) e=
0.9338
Question 18.11
Answer: C
sd  Sˆ (25)  ≈ Sˆ (25)
dj
t( j ) ≤ 25 rj ( rj − d j )
∑
Sˆ (25)= pˆ1 × pˆ 2 = 0.8940
d1 r1 (1=
d 2 r2 (1 =
=
− pˆ1 ) 1;=
− pˆ 2 ) 2
2 
 1
⇒ sd  Sˆ (25)  ≈ 0.8940 
+
0.0579
=
 29 × 28 27 × 25 
Page 83