Fundamentals of Actuarial Mathematics- Solutions to Chapter 2
2.1 (a) 10v(1, 3)v(3, 6)v(6, 8) = 10(0.9)(0.8)(1.2)−1 = 6
(b) 100v(8, 6)v(6, 3) = 100(1.2)(0.8)−1 = 150
2.2 Equating values at time 1 gives (2 + 2 + 3/2) = (4 + K + 1/2), so K = 1
2.3 If c = (2, 0, −3, 5, 10), We want
Val1 (c) = 2(1.25) − 3(1.25)−1 + 5(1.25)−1 (1/2) + 10(1.25)−1 (1/4) = 4.1
2.4 (a) v = (1, 2/3, 4/9, 8/27, 2/9, 1/6)
(b) 6 + (2/3)6 + (4/9)6 + (8/27)9 + (2/9)(9) + (1/6)12 = 20 23
(c) 20 32 (3/2)2 = 46.5
2.5 v = (1, 0.8, 0.64, 0.48, 0.36, 0.18)
(a) 3 V = 3 − 8v(3, 4) + 12v(3, 4)v(4, 5) = 3 − 8(3/4) + 12(3/4)(1/2) = 1.5
1
(1 + 2(0.8) + 4(0.64)) = 10.75
b) B3 = 0.48
2.6 (a) 3 V = −3 + 3v(4)/v(3) + 5v(5)/v(3) = −3 + 3(1/2) + 5(1/12) = −19/12
(b)
1
1 − 2v(1) + 4v(2) = 16(1 − 5/16 + 2/3) = 40/3
B3 =
v(3)
2.7 (a) 9 V (c) = (100 + 60)/0.8 = 200, 10 V (c) = (200 + 70)/0.75 = 360
(b) B10 (c) = Val10 (c) + −10 V (c) = 40/0.5) +10 V (c) = 440
2.8 v ◦ 3(0) = 1, v ◦ 3(1) = v(4)/v(3) = 6/7, v ◦ 3(2) = v(5)/v(3) = 5/7,
c ◦ 3 = (7, 7, 7),
3 c = (1, 1, 2),
ä(c) = (1 + 2(.9) + 2(.8) + 7(.7) + 7(.6) + 7(.5) = 17 ä(3 c) = 1 + 2(.9) + 2(.8) = 4.4
ä(c ◦ 3 : v ◦ 3) = 7(1 + 6/7 + 5/7) = 18
To verify (2.27) note that 17 = 4.4 + 0.7(18)
2.9 v(2, 1)v(1, 0) = 1/4 6= v(2, 0) = 1/3
2.10 For all s, t, u
v(s, t)v(t, u) =
g(t) g(u)
g(u)
=
= v(s, u)
g(s) g(t)
g(s)
1
each year. The interest payable at time j will be (n − j)d. So the total repayment
vector is dk n + (0, 1n ). Equating the value at time 0 of this vector with n gives the
desired formula.
2.14 Replace the payments at time 7,8,9 by a single payment at time 7 of 4 + 8v(7, 8) +
3v(7, 9) = 4 + 8v(2, 8)/v(2, 7) + 3v(2, 9/v(2, 7) = 4 + 6.4 + 1.8 = 12.2
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2.15 Let s = v(6) + v(7) + . . . + v(n + 4) . Then
2300 = 230v(5) + 230s = 115v(5) + 240s.
Solving, v(5) = 0.8
2.16 Yearly payment is 20, 000/¨(0, 1n )at6% = 2717.16
(a) 2717.16 ä(14 ) at 6% = 9980.89
(b) 2717.16ä(14 )at 5% = 10,117.40
2.17 The 5 year bond will mature for 1000(1.05)5 = 1338.23. The 10 year bond will have a
maturity value of 1000(1.05)1 0 = 1790.85
(a) Value at time 5 of the 10 year bond at 4% = 1790.85(1.04)−5 = 1471.95. Investor
is ahead by 133.72.
(b) Value at time 5 of the 10 year bond at 7% = 1790.85(1.04)−5 = 1276.85. Investor
is worse off by 61.38.
2.18
15 = 1 + 5v(0, 1) + v(0, 1)v(1, 2)ä(c ◦ 2)
(1)
12.6 = 1 + +5v(0, 1) + v(0, 1)[v(1, 2) − 0.1]ä(c ◦ 2)
(2)
13 = 1 + +5[v(0, 1) − 0.1] + v(0, 1)v(1, 2)ä(c ◦ 2)
(3)
Substracting (2) from (1) gives
v(0, 1)ä(c ◦ 2) = 24
(4)
Substracting (3) from (1) and rearranging gives
v(1, 2)ä(c ◦ 2) = 15
(5)
Substituting (4) into (1) and solving, v(0, 1) = 0.7 and then dividing (3) by (4) gives
v(1, 2) = (5/8)v(0, 1) = 0.4375
2.19 Let c be the the constant entry in the vector c. Since d is nondecreasing, there must
be an integer m such that di ≤ c for i < m and di > c for i ≥ m. Then for k < m,
k V ((c − d)) = Bk (c − d) is the value at time k of a cash flow vector, all of whose
entries are nonegative, and which therefore is nonnegative For k ≥ m, k V (c − d) is the
negative of the value at time k of a cash flow vector all of whose entries are negative
, so it is again positive.
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