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Transcript of MLC Finan Solution
A Reading of the Theory of Life Contingency ModelsA Preparation for the Actuarial Exam MLC/3L
Marcel B. FinanArkansas Tech University
c©All Rights ReservedAnswers Key
2
3
The answer key manuscript is to help the reader to check his/her answers against mine. I am notin favor of providing complete and detailed solutions to every single problem in the book. Theworked out examples in the book are enough to provide the user with the skills needed to tacklethe practice problems.This manuscript should not be made public or shared with others. Best of wishes.
Marcel B. FinanRussellville, ArkansasJuly 2011
4
Section 18
18.1 No
18.2 III
18.3 A = 1, B = −1
18.4 0.033
18.5 0.9618
18.6 0.04757
18.7 s(0) = 1, s′(x) < 0, s(∞) = 0
18.8 0.149
18.9 1− e−0.34x, x ≥ 0
18.10 x2
100, x ≥ 0
18.11 I
18.12 (a 0.3 (b) 0.3
18.13 1− x108, x ≥ 0
18.14 (a)
18.15 (x+ 1)e−x
18.16 0.34e−0.34x
18.17 λe−λx
5
18.18
f(x) =
716, 0 < x < 1
3x8, 1 < x < 2
0, x > 2
18.19 Both functions represent the density of death at age x. The probability density function isunconditional (i.e., given only existence at age 0) whereas µ(x) is conditional on survival to age x
18.20 12(1− x)−1
18.21 f(x) = µ(x)SX(x) = µ(x)e−Λ(x)
18.22∫∞
0µ(x)dx = limR→∞
∫ R0µ(x)dx = − limR→∞ ln s(x) = −(−∞) =∞
18.23 0.34
18.24 s(x) = e−∫ x0 µ(s)ds = e−µx, F (x) = 1− s(x) = 1− e−µx, and f(x) = F ′(x) = µe−µx
18.25 1480
18.26 ln (x+ 1), x ≥ 0
18.27 2x4−x2 , 0 ≤ x < 2
18.28
s(x) =e−ΛX(x) = e−µx
F (x) =1− e−µx
f(x) =− S ′X(x) = −µe−µx
18.29 1.202553
18.30 0.2
18.31 (I) and (II)
18.32 2
18.33 24(2 + x)−4
6
18.34 2
18.35 4
18.36 kp
18.37 34k
18.38 45√
2
18.39 e0 = 60 and Var(X) = 450
18.40 (a) 7207
(b) 0.062
18.41 median = 0.51984 and mode = 0
7
Section 19
19.1 675
19.2 50
19.3 (a) µ(x) = − s′(x)s(x)
= 190−x . (b) F (x) = 1− s(x) = x
90. (c) f(x) = F ′(x) = 1
90. (d) Pr(20 < X <
50) = s(20)− s(50) = 5090− 20
90= 1
3
19.4 (a) F (x) = 1− s(x) = 1−(1− x
ω
)α(b) f(x) = F ′(x) = α
ω
(1− x
ω
)α−1
(c) µ(x) = f(x)s(x)
= αω
(1− x
ω
)−1
19.5 tω−x , 0 ≤ t ≤ ω − x.
19.6 1− tω−x , 0 ≤ t ≤ ω − x.
19.7 40
19.8 4
19.9 ln(
ωω−x
)19.10 0.449
19.11 0,1481
19.12 0.5
19.13 µ = 0.3054 and median = 2.27
19.14 1− e−µt
19.15 85.34
19.16 e0 = 60 and Var(X) = 3600
8
19.17 (III)
19.18 0.01837
19.19 s(x) = e−∫ x0 Bctdt = e
Bln c
(1−cx) and F (x) = 1− e Bln c
(1−cx)
19.20 f(x) = −s′(x) = BcxeBln c
(1−cx)
19.21 Λ(x) =∫ x
0Bctdt = Bct
ln c
∣∣∣x0
= Bln c
(cx − 1)
19.22 6.88
19.23 −3.008(1.05)x
19.24 1− e 0.0004ln 1.07
(1−1.07x)
19.25 f(x) = µ(x)s(x) = (A+Bcx)e−Ax−m(cx−1) where m = Bln c
19.26 1− e−Ax−m(cx−1)
19.27 f(x) = (0.31 + 0.45(2x))e−0.31x− 0.43ln 2
(2x−1) and F (x) = 1− e−0.31x− 0.43ln 2
(2x−1)
19.28 0.0005131
19.29 µ(x) = 0.31 + 0.43(2x)
19.30 0.111395
19.31 f(x) = kxne−kxn+1
n+1
19.32 k = 2 and n = 1
19.33 30
19.34 e−225
19.35 e−16
9
19.36 0.009831
19.37 n = 5.1285 and k = 1.5198× 10−11.
10
Section 20
20.1 sT (x)(t) = 1− tω−x , 0 ≤ t ≤ ω − x.
20.2 sT (x)(t) = 1− t75−x , 0 ≤ t ≤ 75− x and fT (t) = 1
75−x
20.3 m+npx = s(x+m+n)s(x)
= s(x+m+n)s(x+m)
· s(x+m)s(x)
= npx+m · mpx
20.4 0.9215
20.5 Induction on n and Problem 20.3
20.6 (a) 17p35 − 38p35 (b) 0.323
20.7 4t+4
20.8 0.9559
20.9 We have ∫ x+t
x
µ(y)dy =
∫ x+t
x
−[ln s(y)]′dy = ln
(s(x)
s(x+ t)
)so that
tpx =s(x+ t)
s(x)= e−
∫ x+tx µ(y)dy
20.10 0.59049
20.11 (a) 0.8795 (b) 0.9359
20.12 We have
∂
∂ttpx =
∂
∂t
(s(x+ t)
s(x)
)=s′(x+ t)
s(x)
=s′(x+ t)
s(x+ t)
s(x+ t)
s(x)= −tpxµ(x+ t)
20.13 µ(x) = −0.04 + 0.00189(1.056)x
11
20.14 We have
t|uqx =Pr(t < T (x) ≤ t+ u)
=FT (x)(t+ u)− FT (x)(t)
=t+uqx − tqx
=(1− t+upx)− (1− tpx)
=tpx − t+upx
20.15 We have
t|uqx =tpx − t+upx
=s(x+ t)
s(x)− s(x+ t+ u)
s(x)
=s(x+ t)− s(x+ t+ u)
s(x)
=
[s(x+ t)
s(x)
] [s(x+ t)− s(x+ t+ u)
s(x+ t)
]=tpxuqx+t
20.16 0.5714
20.17 0.9841
20.18 0.025
20.19 This follows from tqx = 1− tpx and Problem 20.12
20.20 1−(
αα+t
)β20.21 0.9913
20.22 2|q1 is the probability that a life currently age 1 will die between ages 3 and 4
20.23 0.23915
20.24 1−(
120120+t
)1.1
12
20.25 0.1694
20.26 0.6857
20.27 sT (x)(t) = 90−x−t90−x and fT (x)(t) = 1
90−x , 0 ≤ t ≤ 90− x
20.28 fT (x)(t) = 190−x , 0 ≤ t ≤ 90− x
20.29 0.633
20.30 fT (36)(t) = 0.0625
(64−t)12
20.31 fT (2)(t) = 2+t48
20.32 0.01433
20.33 ddt
(1− tpx) = ddt
(tqx) = tpxµ(x+ t)
20.34∫∞
0 tpxµ(x+ t)dx =∫∞
0fT (x)(x)dx = 1
20.35 µT (x)(t) =fT (x)(t)
tpx=
F ′T (x)
(t)
1−tqx =F ′T (x)
(t)
1−FT (x)(t)
20.36 µ(x+ t) = 1100−x−t , 0 ≤ t ≤ 100− x
20.37 µ(x+ t) = µ(x) = µ
20.38 0.015
20.39 5.25
20.40 0.3783
20.41 49.8
20.42 10510.341
20.43 300
13
20.44 50
20.45 pKx(k) = k−1px · qx+k−1 = k−1|qx
20.46 Pr(Kx ≥ k) = Pr(T (x) > k − 1) = sT (x)(k − 1) = k−1px
20.47 pK(x)(k) = kpx − k+1px =(
100−x−k100−x
)0.5 −(
100−x−k−1100−x
)0.5
20.48 ex = 99−x2
and◦ex=
100−x2
20.49 1
20.50 2
20.51 We have
ex =∞∑k=1
kpx = px +∞∑k=2
kpx
=px +∞∑k=2
pxk−1px+1
=px +∞∑k=1
pxkpx+1
=px(1 + ex+1)
20.52 T (x) = Kx − 1 + Sx = K(x) + Sx
20.53 1.07
20.54 9.5
14
Section 21
21.1 199.5−x
21.2 We have
nmx =
∫ x+n
xf(y)dy∫ x+n
xs(y)dy
=−∫ x+n
xs′(y)dy∫ x+n
xs(y)dy
=s(x)− s(x+ n)∫ x+n
xs(t)dt
21.3 nmx = 2n200−2nx−1
21.4 0.6039
21.5 175
21.6 m40 = 0.0096864 and 10m75 = 0.044548.
15
Section 22
22.1 (a) `x = 10− x (b) p2 = 78, q3 = 1
7, 3p7 = 0, 2q7 = 2
3
22.2 F (x) = `0−`x`0
22.3 t|uqx = `x+t−`x+t+u`x
22.4 We have
Age `x dx px qx0 100,000 501 0.99499 0.005011 99,499 504 0.99493 0.005062 98,995 506 0.99489 0.005113 98,489 509 0.99483 0.005174 97,980 512 0.99477 0.005235 97,468 514 NA NA
22.5 `t+x
22.6 (a) 9734 (b) 50 (c) 200 (d) 0.0211 (e) 0.0055
16
Section 23
23.1 µ(x) = 1ω−x , 0 ≤ x < ω
23.2 0.05
23.3 Let u = x+ t. Then
µ(x+ t) =µ(u) = −d`udu
`u
=−d`udt· dtdu
`u
=−d`x+tdt
`x+t
23.4 `x = 100− x
23.5 µ(x) = 13(90− x)−1
23.6 `x − `x+n =∫ x+n
x
[− ddy`y
]dy =
∫ x+n
x`yµ(y)dy
23.7 ddx`xµ(x) = d
dx
[− ddx`x]
= − d2
dx2`x
23.8 f(x) = 0.95(100− x)−1
23.9 5000
23.10 Using integration by parts we find∫ ∞0
xf(x)dx =− 1
`0
∫ ∞0
x`′xdx
=− [1
`0
[x`x|∞0 −
∫ ∞0
`xdx
]=
1
`0
∫ ∞0
`xdx
where we used the fact that `∞ = `0s(∞) = 0
17
23.11 Using integration by parts we find∫ ∞0
x2f(x)dx =− 1
`0
∫ ∞0
x2`′xdx
=− [1
`0
[x2`x
∣∣∞0− 2
∫ ∞0
x`xdx
]=
2
`0
∫ ∞0
x`xdx
where we used the fact that `∞ = `0s(∞) = 0
23.12 fT (x)(t) = ddt tqx = d
dt
[1− `x+t
`x
]= − 1
`xddt`x+t
23.13 Tx =∫ 10
x100(10− y)0.85dy = −100
[(10−y)1.85
1.85
]10
x= 100(10−x)1.85
1.85
23.14 5.405
23.15 48.881
23.16 5000(1 + x)−2
23.17 0.1
23.18 0.123
23.19 npx is the probability of surviving to age x + n + m. If we remove n+mpx, which is theprobability of surviving to x+ n+m years, then we have the probability of surviving to age x+ nbut dying by the age of x+ n+m which is n|mqx
23.20 6|10q64
23.21 mω−x
23.22 e−nµ − e−(n+m)µ
23.23 0.3064
23.24 20
18
23.25 800
23.26 400
23.27 1+x3
23.28 (1+x)2
3
23.29 29(1 + x)2
23.30 1373
23.31 352.083
23.32 The expected number of years (60) is expected to live in the next 25 years is 17.763
23.33 We have
ex:m+n =
∫ m+n
0tpxdt =
∫ m
0tpxdt+
∫ m+n
mtpxdt
=
∫ m
0tpxdt+
∫ n
0m+ypxdy =
∫ m
0tpxdt+
∫ n
0mpx · ypx+mdy
=ex:m + mpx · ex+m:n
23.34 We have
ex =
∫ ∞0
tpxdt =
∫ n
0tpxdt+
∫ ∞n
tpxdt
=
∫ n
0tpxdt+
∫ ∞0
y+npxdt =
∫ n
0tpxdt+
∫ ∞0
npx · ypx+ndt
=ex:n + npx · ◦ex+n
23.35 6.968
23.36 15.6
23.37 E[K(x)2] =∑∞
k=1(2k − 1)kpx = 1`x
∑∞k=1(2k − 1)`x+k
19
23.38 7.684
23.39 2.394
23.40 pK(20)(k) = kp20 − k+1p20 = e−0.05k − e−0.05(k+1) = e−0.05k(1− e−0.05)
23.41 0.905
23.42 We have Tx =∫∞x`ydy =
∑∞k=x
∫ k+1
k`ydy =
∑∞k=x
∫ 1
0`k+tdt =
∑∞k=x Lk
23.43 nLx = Tx − Tx+n = 200e−0.05x(1− e−0.05n)
23.44 nLx = Tx − Tx+n = 1000(x+ 1)−3 − 1000(x+ n+ 1)−3
23.45 577.190
23.46 Recall that
Lx = −(x− ω)− 1
2
so thatn∑k=1
Lk = L1 − Ln+1 = n
23.47 We have
Lx =
∫ x+1
x
`ydy = −∫ x
0
`ydy +
∫ x+1
0
`ydy
and therefored
dxLx = `x+1 − `x = −dx
23.48 We haved
dtLt = −dt = −mtLt.
Separating the variables and integrating both sides from x to x+ 1 we obtain
ln
(Lx+1
Lx
)= −
∫ x+1
x
mydy.
Solving for Lx+1 we find
Lx+1 = Lxe−
∫ x+1x mydy
20
23.49 (a) We have
dx =`x − `x+1 = 1
Lx =
∫ 1
0
`x+tdt =
∫ 1
0
(ω − x− t)dt = ω − x− 1
2
mx =dxLx
=1
ω − x− 0.5
(b) For DeMoivre’s Law RV we have µ(x) = 1ω−x .
(c)We havemx
1 + 0.5mx
=1
ω − x− 0.5· ω − x− 0.5
ω − x=
1
ω − x= µ(x)
23.50 (a) 2502.357 (b) 0.0159
23.51 (a) 10L20 = 750, 10d20 = 10 (b) 175
23.52 100
21
Section 24
24.1
`t =
100, 000− 501t 0 ≤ t ≤ 199, 499− 504(t− 1) 1 ≤ t ≤ 298, 995− 506(t− 2) 2 ≤ t ≤ 398, 489− 509(t− 3) 3 ≤ t ≤ 497, 980− 512(t− 4) 4 ≤ t ≤ 597, 468− 514(t− 5) 5 ≤ t ≤ 6
24.2
tp0 =
100,000−501t100,000
0 ≤ t ≤ 199,499−504(t−1)
100,0001 ≤ t ≤ 2
98,995−506(t−2)100,000
2 ≤ t ≤ 398,489−509(t−3)
100,0003 ≤ t ≤ 4
97,980−512(t−4)100,000
4 ≤ t ≤ 597,468−514(t−5)
100,0005 ≤ t ≤ 6
24.3 e0 = 4.92431 and e0 = 5.42431
24.4 We have
tqx+s =tqx+s =dx+s
`x+s
= t`x+s − `x+s+1
`x+1
=tdx
`x − sdx= t
dx`x
1− sdx`x
=tqx
1− sqx
24.5 0.95
24.6 (I) and (II)
24.7 1/9
24.8 r|hqx = `x+r−`x+r+h`x
= (`x−rdx)−(`x−(r+h)dx)`x
= hdx`x
= hqx
24.9 0.813
24.10 0.2942
22
24.11 0.5447
24.12 We have
t−sqx+s =1− t−spx+s
=1− e−∫ x+tx+s µ(y)dy
=1− e−∫ ts µ(x+r)dr
=1− e−(t−s)µx
24.13 112q90 = 0.02369 and 1
12q90+ 11
12= 0.02369
24.14 0.5qx = 0.0513 and 0.5qx+0.5 = 0.0513
24.15 L95 = 690.437 and m95 = `95−`96L95
= 200690.437
= 0.28967
24.16 We have
s−tqx+t =s(x+ t)− s(x+ s)
s(x+ t)= 1− s(x+ s)
s(x+ t)
=1− spx
tpx= 1−
pxs+(1−s)px
pxt+(1−t)px
=1− t+ (1− t)pxs+ (1− s)px
=(s− t)(1− px)s+ (1− s)px
=(s− t)qx
1− (1− s)qx
24.17 0.75p80 = 0.95857 and 2.25p80 = 0.87372
24.18 13440
24.19 0.00057
24.20 (i) 1.475801 (ii) 1.475741
23
Section 25
25.1 84
25.2 8.2
25.3 8056
25.4 0.4589
25.5 0.0103
24
Section 26
26.1 0.03125
26.2 A20 = 0.4988, 2A20 = 0.2998, Var(Z20) = 0.0501
26.3 3.75
26.4 0.04
26.5 116.09
26.6 14.10573
26.7 A 125:10
= 0.0885, 2A 125:10
= 0.0685, Var(Z 125:10
) = 0.0607
26.8 A 130:20
= 0.3167, 2A 130:20
= 0.1987, Var(Z 130:20
) = 0.0984
26.9 0.2378
26.10 0.4305
26.11 11+e−(µ+δ)
26.12 0.05
26.13 A 130:20
= 0.2628, 2A 130:20
= 0.0967, Var(Z 130:20
) = 0.0276
26.14 A 130:20
= 0.2231, 2A 130:20
= 0.0821, Var(Z 130:20
) = 0.0323
26.15 0.02497
26.16 0.7409
26.17 mean = 1051.43 and the standard deviation is 197.94
26.18 (a) 590.41 (b) 376.89
25
26.19 0.2793
26.20 0.4775
26.21 0.73418
26.22 2Ax:n = 2A1x:n + 2A 1
x:n =∫ n
0ν2t
tpxµ(x+ t)dt+ ν2nnpx
26.23 2m|Ax = ν2m
mpx2Ax+m
26.24 0.1647
26.25 0.0253
26.26 0.0873
26.27 0.0154
26
Section 27
27.1 A30 = 0.3168, 2A30 = 0.1805, Var(Z30) = 0.0801
27.2 A 130:10
= 0.2461, 2A 130:10
= 0.1657, Var(Z 130:10
) = 0.1051
27.3 10|A30 = 0.1544, 210|A30 = 0.02981, Var(10|Z30) = 0.00597
27.4 A30:10 = 0.4692, 2A30:10 = 0.2478, Var(Z30:10 ) = 0.0277
27.5 1730.10
27.6 0.19026
27.7 0.9396
27
Section 28
28.1 0.671
28.2 We have
Ax =∞∑k=0
νk+1kpxqx+k = A1
x:n +∞∑k=n
νk+1kpxqx+k
=A1x:n +
∞∑k=0
νk+1+nk+npxqx+k+n
=A1x:n + νnnpx
∞∑k=0
νk+1kpx+nqx+k+n
=A1x:n + νnnpxAx+n
28.3 0.0081
28.4 0.00242
28.5 Using (i) and (ii), we can rewrite the given relation as
u(k − 1) = u(k)νpk−1 + νqk−1.
Now, we have
u(70) =1
u(69) =νp69 + νq69 = A69:1
u(68) =[νp69 + νq69]νp68 + νq68 = ν2p68p69 + ν2p68q69 + νq68 = A68:2
u(67) =[ν2p68p69 + ν2p68q69 + νq68]νp67 + νq67
=ν3p67p68p69 + ν3p67p68q69 + ν2p67q68 + νq67 = A67:3
... =...
u(40) =A40:30
28
28.6 We have
2Ax − νnnEx2Ax+n + νnnEx =∞∑k=0
ν2(k+1)kpxqx+k −
∞∑k=0
ν2(k+1+n)npxkpx+nqx+n+k + ν2n
npx
=∞∑k=0
ν2(k+1)kpxqx+k −
∞∑k=0
ν2(k+1+n)n+kpxqx+n+k + ν2n
npx
=∞∑k=0
ν2(k+1)kpxqx+k −
∞∑k=n
ν2(k+1)kpxqx+k + ν2n
npx
=n−1∑k=0
ν2(k+1)kpxqx+k + ν2n
npx = 2Ax:n
28.7 0.02544
28.8 2.981%
29
Section 29
29.1 11772.61
29.2 10416.22
29.3 E(Z) = 20.3201, E(Z2) = 2683.7471, Var(Z) = 2270.8406
29.4 87.35
29.5 12.14
29.6 4
29.7 0.3403
29.8 1
29.9 (IA)30 = −∑29
k=0(k + 1)e−0.02(k+1)
29.10 (IA)x =∫∞
0e−δte−µt(µ)dt = µ
(µ+δ)2= (µ+ δ)−1Ax
29.11 1.9541
29.12 We have
(IA)x =
∫ ∞0
tνttpxµ(x+ t)dt
=
∫ ∞0
(∫ t
0
ds
)νttpxµ(x+ t)dt
=
∫ ∞0
∫ ∞s
νttpxµ(x+ t)dtds
=
∫ ∞0
s|Axds
30
29.13 We have
(IA)1x:n + (DA)1
x:n =
∫ n
0
bt+ 1cνttpxµ(x+ t)dt+
∫ n
0
(n− btc)νttpxµ(x+ t)dt
=(n+ 1)
∫ n
0
νttpxµ(x+ t)dt
=(n+ 1)A1x:n
where we used the fact that bt+ 1c − btc = 1 for k − 1 ≤ t ≤ k.
29.14 We have
(IA)1x:n =
n−1∑k=0
(k + 1)νk+1k|qx =
n−1∑k=0
(k + 1)νk+1kpxqx+k
=n−1∑k=0
kνk+1kpxqx+k +
n−1∑k=0
νk+1kpxqx+k
=n−1∑k=0
νk+1kpxqx+k + ν
n−1∑k=1
kνkpxk−1px+1qx+k
=n−1∑k=0
νk+1kpxqx+k + νpx
n−2∑k=0
(k + 1)νk+1kpx+1qx+k+1
=∞∑k=0
νk+1kpxqx+k + νpx
n−2∑k=0
(k + 1)νk+1k|qx+1
=A1x:n + νpx(IA) 1
x+1:n−1 .
29.15 5.0623
29.16 This follows from the two recursion relations
A1x:n = νqx + νpxA
1x+1:n−1
and
(IA)1x:n = A1
x:n + νpx(IA) 1x+1:n−1 .
31
29.17 We have
(IA)1x:n + (DA)1
x:n =n−1∑k=0
(k + 1)νk+1k|qx +
n−1∑k=0
(n− k)νk+1k|qx
=n−1∑k=0
[k + 1 + n− k]νk+1k|qx
=(n+ 1)n−1∑k=0
νk+1k|qx
=(n+ 1)n−1∑k=0
νk+1kpxqx+k
=(n+ 1)A1x:n .
29.18 12.2665.
32
Section 30
30.1 115.10
30.2 543.33
30.3 2758.99
30.4 We have
(IA)x =E[bT + 1cνT ] = E[(K + 1)νK+1νS−1]
=E[(K + 1)νK+1]E[νS−1]
=i
δ(IA)x.
30.5 We have
E[(S − 1)νS−1] =
∫ 1
0
(s− 1)(1 + i)1−sds
=−∫ 1
0
s(1 + i)sds
=− 1
δesδ∣∣∣∣10
+1
δ
∫ 1
0
esδds
=−(
1 + i
δ− i
δ2
).
30.6 We have
(IA)x =E(TνT ) = E[(K + 1 + S − 1)νT ]
=E[(K + 1)νT ] + E[(S − 1)νT ]
=E[bT + 1cνT ] + E[(S − 1)νT ]
=(IA)x + E[(S − 1)νK+1νS−1]
=i
δ(IA)x + E[νK+1]E[(S − 1)νS−1]
=i
δ(IA)x + AxE[(S − 1)νS−1]
=i
δ(IA)x −
(1 + i
δ− i
δ2
)Ax.
33
Section 31
31.1 A(2)69 = 0.5020 is the actuarial present value of a whole life insurance of $1 issued to (69)
with death benefit paid at the end of the semiannual in the year of death.
31.2 0.0695
31.3 0.9137
31.4 0.5217
31.5 0.8494
34
Section 32
32.1 280.65
32.2 248.67
32.3 0.6614
32.4 FT (20)(t) = 1 + ln t4.
32.5 0.8187
32.6 1,430,000
35
Section 33
33.1 12
33.2 E(Y 2x ) = 1
δ2[1− 2Ax + 2Ax]
33.3 2.8
33.4 7.217
33.5 Pr(Yx > ax) =(
µµ+δ
)µδ
33.6 13.027
33.7 112µ2
33.8 13.96966
33.9 0.7901
33.10 0.8
33.11 65098.637
33.12 19.0042586
36
Section 34
34.1 ax:n = 1−e−(µ+δ)n
µ+δ
34.2 2.16166
34.3 We have
ax =
∫ ∞0
tExdt
=
∫ n
0tExdt+
∫ ∞n
tExdt
=ax:n +
∫ ∞n
νtnpxt−npx+ndt
=ax:n + νnpx
∫ ∞0
νttpx+ndt
=ax:n + νnnpxax+n
34.4 7.8202
34.5 We have
ax:m+n =
∫ m+n
0
νttpxdt
=
∫ m
0
νttpxdt+
∫ m+n
m
νttpxdt
=ax:m +
∫ m+n
m
νtt−mpx+mmpxdt
=ax:m + νmmpx
∫ m+n
m
νt−mt−mpx+mdt
=ax:m + νmmpx
∫ n
0
νttpx+mdt
=ax:m + mExax+m:n
37
Section 35
35.1 We have
n|ax =
∫ ∞n
e−δte−µtdt =
∫ ∞n
e−t(δ+µdt
= −e−(µ+δ)t
µ+ δ
∣∣∣∣∞n
=e−n(µ+δ)
µ+ δ
35.2 0.3319
35.3 For T (x) ≤ n, we have
n|Yx = Z 1x:n = n|Zx = 0.
For T (x) > n we have Z 1x:n = νn and n|Zx = νT . Thus,
Z 1x:n − n|Zx
δ=νn − νT
δ= νn
1− νT−n
δ= n|Yx.
35.4 We have
E[(n|Yx)2] =
∫ ∞n
ν2n(at−n )2tpxµ(x+ t)dt
=ν2n
[− (at−n )2
tpx∣∣∞n
+ 2
∫ ∞n
νt−ntpxdt
]=2ν2n
npx
∫ ∞0
νtat tpx+ndt
35.5 20|ax = 1.2235 and Var(20|Yx) = 0.9753
38
Section 36
36.1 20|a50 = 0.3319 and a50:20
= 8.9785
36.2 We have
ax:n =E(Yx:n )
=
∫ n
0
an tpxµ(x+ t)dt+
∫ ∞n
at tpxµ(x+ t)dt
=an [−tpx]n0 +
∫ ∞n
at tpxµ(x+ t)dt
=an nqx +
∫ ∞n
at tpxµ(x+ t)dt.
36.3 We have
ax:n =an nqx +
∫ ∞n
at tpxµ(x+ t)dt
=an nqx − at tpx|∞n +
∫ ∞n
νttpxdt
=an + +
∫ ∞n
νttpxdt.
36.4 This follows fromVar(Yx:n ) = Var(an + n|Yx) = Var(n|Yx)
36.5 From Section 35, we have that
n|ax = nExax+n.
Therefore,ax:n = an + nExax+n.
36.6 From Problem 34.3, we have that
ax = ax:n + nExax+n.
Therefore,ax:n = an + (ax − ax:n ).
39
Section 37
37.1 0.5235
37.2 13.78
37.3 We have
ax =∞∑k=0
νkkpx = 1 +∞∑k=1
νkkpx = 1 + νpx
∞∑k=1
νk−1k−1px+1
=1 + νpx
∞∑k=0
νkkpx+1 = 1 + νpxax+1.
37.4 0.364
37.5 7%
37.6 150,000
37.7 52,297.43
37.8 1296.375
37.9 We have
ax:n =n−1∑k=0
νkkpx
=1 +n−1∑k=1
νkkpx
=1 + νpx
n−1∑k=1
νk−1k−1px+1
=1 + νpx
n−2∑k=0
νkkpx+1
=1 + νpxax+1:n−1 .
40
37.10 264.2196
37.11 114.1785
37.12 0.2991
37.13 280.41
37.14 49.483
37.15 10.3723
37.16 Recall the followingAx = A1
x:n + nExAx+n
andAx:n = A1
x:n + nEx.
Thus,
ax =1− Axd
=1− A1
x:n − nExAx+n
d
=1− Ax:n + nEx − nExAx+n
d
=1− Ax:n
d+ nEx
(1− Ax+n
d
)=ax:n + nExax+n.
37.17 Recall that
Z 1x:n =
{0 T ≤ nνn T > n
and
n|Zx =
{0 K ≤ n− 1
νK+1 K ≥ n
Thus, if K ≤ n− 1 then T ≤ n and therefore Z 1x:n = n|Zx = n|Zx = 0. If K ≥ n then T > n so that
Z 1x:n = νn and n|Zx = νK+1. Thus,
Z 1x:n −n|Zx
d= νn−νK+1
d= νaK+1−n = n|Zx.
37.18 This follows from the previous problem by taking expectation of both sides.
41
37.19 n|ax = nExax+n = νn(px)n
1−e−(δ+µ) = e−n(δ+µ)
1−e−(δ+µ) .
37.20 0.4151
37.21 0.45
37.22 16.6087
37.23 15.2736
37.24 58.36
37.25 3.30467
37.26 We have
ex =∞∑k=1
kpx =∞∑k=1
pxk−1px+1 = px + px
∞∑k=2
k−1px+1 = p(x)(1 + ex+1).
(b) 0.0789.
37.27 ax = e−(µ+δ)
1−e−(µ+δ)
37.28 We have
ax =∞∑k=1
νkkPx =∞∑k=1
νkpxk−1px+1 = νpx
∞∑k=1
νk−1k−1px+1
=νpx(1 +∞∑k=2
νk−1k−1px+1) = νpx(1 +
∞∑k=1
νkkpx+1) = νpx(1 + ax+1)
37.29 7.6
37.30 0.1782
37.31 ax:n = e−(µ+δ)(
1−e−n(µ+δ)1−e−(µ+δ)
)37.32 11.22
42
Section 38
38.1 12.885
38.2 13.135
38.3 A80 = 0.8162 and a80 = 2.5018
38.4 15.5
38.5 8.59
38.6 We have
a(m)x:n =
1
m
mn−1∑k=0
νkm kmpx
=1
m+
1
m
mn∑k=1
νkm kmpx −
1
mνnnpx
=a(m)x:n +
1
m(1− nEx).
38.7 We have
a(m)x:n =a(m)
x − n|a(m)x
≈ i
i(m)
d
d(m)ax −
i− i(m)
i(m)d(m)− i
i(m)
d
d(m) n|ax +i− i(m)
i(m)d(m) nEx
=i
i(m)
d
d(m)(ax − n|ax)−
i− i(m)
i(m)d(m)(1− nEx)
=i
i(m)
d
d(m)axn −
i− i(m)
i(m)d(m)(1− nEx).
43
38.8 (a) We have
a(m)x =a(m)
x − 1
m
≈ i
i(m)
d
d(m)ax −
i− i(m)
i(m)d(m)− 1
m
=i
i(m)
d
d(m)(ax + 1)− i− i(m)
i(m)d(m)− 1
m
=i
i(m)
d
d(m)ax +
i
i(m)
d
d(m)− i− i(m)
i(m)d(m)− (1− ν 1
m )i(m)
i(m)d(m)
=i
i(m)
d
d(m)ax +
d(m) − di(m)d(m)
.
(b) We have
a(m)x:n =a
(m)x:n −
1
m(1− nEx)
≈ i
i(m)
d
d(m)axn −
i− i(m)
i(m)d(m)(1− nEx)−
1
m(1− nEx)
=i
i(m)
d
d(m)(1− nEx + ax:n )− i− i(m)
i(m)d(m)(1− nEx)−
1
m(1− nEx)
=i
i(m)
d
d(m)ax:n
d(m) − di(m)d(m)
(1− nEx).
38.9 (a) We have
a(m)x:n =a(m)
x − n|a(m)x
≈ax −m− 1
2m− n|ax +
m− 1
2mnEx
=ax:n −m− 1
2m(1− nEx).
(b) We have
a(m)x =a(m)
x − 1
m
≈ax −m− 1
2m− 1
m
=ax + 1− m− 1
2m− 1
m
=ax +m− 1
2m.
44
(c) We have
n|a(m)x =nExa
(m)x+n
≈nEx(ax+n +
m− 1
2m
)=nExax+n +
m− 1
2mnEx
=n|ax +m− 1
2mnEx.
(d) We have
a(m)x:n =a(m)
x − n|ax
≈ax +m− 1
2m− n|ax −
m− 1
2mnEx
=ax:n +m− 1
2m(1− nEx).
38.10 (a) We have
a(m)x:n =a(m)
x − nExa(m)x+n
≈ax −m− 1
2m− m2 − 1
12m(µ(x) + δ)
−nEx(ax+n −m− 1
2m− m2 − 1
12m(µ(x+ n) + δ)
=ax:n −(m− 1
m
)(1− nEx)−
m2 − 1
12m(δ + µ(x)− nEx(δ + µ(x+ n))).
(b) The result follows by letting m→∞ in the 3-term Woolhouse formula.(c) The result follows by letting m→∞ in (a)
45
Section 39
39.1 218.79
39.2 8.56
39.3 5.1029
39.4 5.7341
39.5 5.3465
39.6 204.08
39.7 4.4561
39.8 (Ia)x =∫∞
0dteνtipxdt
39.9 (Ia)x =∫ n
0dteνtipxdt
39.10 (Da)x:n =∫ n
0dn− teνtipxdt
46
Section 40
40.1 0.2
40.2 P (A75) = 0.02901 and Var(Lx) = 0.15940
40.3 0.2
40.4 0.7125
40.5 0.1
40.6 0.05137
40.7 P (A1x:n ) = 0.02 and Var(L 1
75:20) = 0.1553
40.8 P (A1x:n ) =
1δ(ω−x) (1−e−nδ)
1δ (1− 1
δ(ω−x) (1−e−nδ)−e−nδ(1− nω−x))
and
Var(L1x:n ) =
1 +
1δ(ω−x)
(1− e−nδ)(1− 1
δ(ω−x)(1− e−nδ)− e−nδ
(1− n
ω−x
))2
×[
1
2δ(ω − x)(1− e−nδ)− 1
δ2(ω − x)2(1− e−nδ)2
]40.9 P (A 1
75:20) = 0.02402 and Var(L 1
75:20) = 0.13694
40.10 0.25285
40.11 0.47355
40.12 0.04291
40.13 0.09998
40.14 We have
P (Ax:n ) =
[1
δ(ω−x)(1− e−nδ) + e−nδ
(1− n
ω−x
)]δ
1− 1δ(ω−x)
(1− e−nδ)− e−nδ(1− n
ω−x
)
47
Var(Lx:n ) =
12δ(ω−x)
(1− e−2nδ) + e−2nδ(1− n
ω−x
)−(
1δ(ω−x)
(1− e−nδ) + e−nδ(1− n
ω−x
))2
(1− 1
δ(ω−x)(1− e−nδ)− e−nδ
(1− n
ω−x
))2
40.15 0.04498
40.16 0.10775
40.17 0.06626
40.18 0.4661
40.19 0.0229
40.20 0.42341
40.21 We have
P (A1x:n ) + P (A 1
x:n )Ax+n =A1x:n
ax:n
+A 1x:n
ax:n
Ax+n
=A1x:n + n|Axax:n
Ax+n
=Axax:n
=nP (Ax)
40.22 We have
P (Ax:n ) + P (A1x:n ) =
Ax:n
ax:n
+A1x:n
ax:n
=Ax:n + A1
x:n
ax:n
=A 1x:n
ax:n
= P (A 1x:n )
40.23 0.01657
40.24 0.03363
48
40.25 0.0498
40.26 1.778
40.27 0.7696
40.28 −5.43
40.29 −14.09
40.30 0.005
49
Section 41
41.1 12381.06
41.2 124.33
41.3 P (Ax) = Axax
=qxqx+i1+iqx+i
= νqx
41.4 16076.12
41.5 33.15
41.6 4105
41.7 From the definition of P (Ax) and the relation Ax + dax = 1 we can write
P (Ax) =Axax
=1− daxax
P (Ax)ax =1− daxax(P (Ax) + d) =1
ax =1
P (Ax) + d
41.8 33.22
41.9 We have
L1x:n =Z1
x:n − PYx:n
=Z1x:n − P
(1− Zx:n
d
)=Z1
x:n − P(
1− Z1x:n − Z 1
x:n
d
)=
(1 +
P
d
)Z1x:n +
P
dZ 1x:n −
P
d
41.10 0.317
41.11 2410.53
50
41.12 0.0368
41.13 281.88
41.14 −10877.55
41.15 261.14
41.16 0.2005
41.17 0.087
41.18 This follows easily by dividing
Ax:n = A1x:n ) + A 1
x:n
by ax:n
41.19 We have
P (A1x:n ) + P (A 1
x:n )Ax+n =A1x:n
ax:n
+A 1x:n
ax:n
Ax+n
=A1x:n + A 1
x:nAx+n
ax:n
=Axax:n
= nP (Ax)
41.20 0.00435
41.21 0.03196
41.22 0.03524
41.23 0.51711
51
41.24 We have
n|Lx =n|Zx − P(
1− Zxd
)= n|Zx −
P
d+P
d
(Z1x:n + n|Zx
)=
(1 +
P
d
)n|Zx +
P
dZ1x:n −
P
d
41.25 Note first that
Z1x:n n|Zx = νK+1I(K ≥ n)νK+1I(K ≤ n− 1) = 0.
Thus,
E
[(n|Lx +
P
d
)2]
=E
[(P
d
)2
(Z1x:n )2 +
(1 +
P
d
)2
(n|Zx)2
]
=
(P
d
)2
(2A1x:n +
(1 +
P
d
)22n|Ax
41.26 The loss random variable is
νK+1I(K ≥ n)− P amin (K+1,t) = n|Zx − PYx:t .
The actuarial present value is
n|Ax − P ax:t
41.27 The benefit premium which satisfies the equivalence principle is
tP (n|Ax) =n|Axax:t
41.28 0.01567
41.29 13092.43
41.30 0.024969
52
Section 42
42.1 0.0193
42.2 0.0256
42.3 0.0347
42.4 This is the benefit premium for a 20-payment, semi-continuous whole life insurance issuedto (40) with face value of 1000
42.5 0.04575
42.6 0.0193
42.7 0.0289
42.8 0.829
42.9 0.0069
42.10 11.183
42.11 −12972.51
42.12 0.0414
42.13 0.0620
42.14 0.0860
53
42.15 We have
P (Ax:n − nP (Ax)
P (A 1x:n )
=Ax:n − AxA 1x:n
=A1x:n + A 1
x:n − AxA 1x:n
=A 1x:n − nAxA 1x:n
=A 1x:n − A 1
x:n Ax+n
A 1x:n
= 1− Ax+n.
42.16 This follows from the formula Ax:n = A1x:n + A 1
x:n
42.17 0.0096
42.18 0.0092
42.19 We have
P (n|Ax) =A 1x:n Ax+n
ax
=A 1x:n Ax+n
ax:n + nExax+n
.
42.20 77079
54
Section 43
43.1 231.64
43.2 122.14
43.3 331.83
43.4 493.58
43.5 94.83
43.6 224.45
43.7 117.52
43.8 325.19
43.9 484.32
55
Section 44
44.1 7.747π
44.2 102
44.3 0.078π
44.4 0.88π
44.5 15.02
44.6 5.1
44.7 19.07
44.8 73.66
44.9 397.41
44.10 1.276
44.11 478.98
44.12 3362.51
44.13 900.20
44.14 17.346
44.15 3.007986
44.16 15513.82
56
Section 45
45.1 0.0363
45.2 0.0259
45.3 0.049
45.4 0.07707
45.5 0.02174
45.6 (a) E(Lx) = bAx − πax (b) Var(Lx) =(b+ π
d
)2[2Ax − (Ax)
2]
45.7 33023.89
45.8 27
45.9 0.208765
45.10 36.77
57
Section 46
46.1 We have
tV (Ax) =Ax+t − P (Ax)ax+t
=(1− δax+t)−(
1− δaxax
)ax+t
=1− δax+t −ax+t
ax+ δax+t
=1− ax+t
ax.
46.2 8.333
46.3 0.04
46.4 0.0654
46.5 1.6667
46.6 0.1667
46.7 0.47213
46.8 0.20
46.9 0.14375
46.10 0.3
46.11 0.1184
46.12 0.1667
46.13 0.1183
46.14 0.1183
58
46.15 we have
tV (Ax) =Ax+t − P (Ax)ax+t
=ax+t
(Ax+t
ax+t
− P (Ax)
)=ax+t
(P (Ax+t)− P (Ax)
)46.16 0.1183
46.17 0.0654
46.18 The prospective formula is
10V (A50) = A60 − P (A50)a60.
The retrospective formula is
10V (A50) =P (A50)a50:10 − A 1
50:10
10E50
46.19 We have
tV (Ax) =P (Ax)ax:n − A1
x:n
nEx
=P (Ax)−
A1x:nax:n
nExax:n
=P (Ax)− P (A1
x:n )
P (A 1x:n )
46.20 True
46.21 0.0851
59
46.22 We have
tV (A1x:n ) =A 1
x+t:n−t − P (A1x:n )ax+t:n−t
=µ
µ+ δ(1− e(n−t)(µ+δ))− µ
[1− Ax:n
δ
]=
µ
µ+ δ(1− e(n−t)(µ+δ))− µ
[1− µ
µ+δ(1− e(n−t)(µ+δ))− e(n−t)(µ+δ)
δ
]
=
(µ
µ+ δ− µ
δ+
µ2
δ(µ+ δ)
)(1− e(n−t)(µ+δ))
=0× (1− e(n−t)(µ+δ)) = 0
46.23 Follows from the previous problem.
46.24 0.0294
46.25 tV (A1x:n ) = ax+t,n−t [P (A 1
x+t:n−t )− P (A1x:n )]
46.26 tV (A1x:n ) = A 1
x+t:n−t
[1− P (A1
x:n )
P (A 1x+t:n−t )
]46.27 0.4207
46.28 0.3317
46.29 Recall that
ax:n =1− Ax:n
δ.
Thus,
tV (Ax:n ) =Ax+t:n−t − P (Ax:n )ax+t:n−t
=Ax+t:n−t −Ax:n
ax:n
ax+t:n−t
=Ax+t:n−t −Ax:n
1−Ax:nδ
· 1− Ax+t:n−t
δ
=Ax+t:n−t − Ax:n
1− Ax:n
60
46.30 0.3431
46.31
tV (A 1x:n ) =
{A 1x+t:n−t − P (A 1
x:n )ax+t:n−t , t < n
1, t = n.
46.32 0.7939
46.33 1
46.34 0.3088
46.35 0.2307
46.36 This is the 10th year benefit reserve for a fully continuous 20-year pure endowment of unitbenefit issued to (75).
46.37 We have
tV (A 1x:n ) + 2A 1
x+t:n−t =A 1x+t:n−t − P (A 1
x:n )ax+t:n−t + 2A 1x+t:n−t
=Ax+t:n−t − A 1x+t:n−t − (P (Ax:n ) + P (A1
x:n )ax+t:n−t + 2A 1x+t:n−t
=[Ax+t:n−t − P (Ax:n )ax+t:n−t ] + [A 1x+t:n−t − P (A1
x:n )]ax+t:n−t
=tV (Ax:n ) + tV (A1x:n ).
46.38 This follows from the previous problem and Problem 46.20.
46.39 24
46.40 4.6362
46.41 5.9055
46.42 14.2857
46.43 14.2857
61
Section 47
47.1 (a) 0.0533 (b) 0.1251
47.2 We have
kV (Ax) =Ax+k − P (Ax)ax+k
=1− dax+k −(1− dax)
axax+k
=1− dax+k −ax+k
ax+ dax+k
=1− ax+k
ax
47.3 0.053
47.4 We have
kV (Ax) =Ax+k − P (Ax)ax+k
=P (Ax+k)ax+k − P (Ax)ax+k
=(P (Ax+k)− P (Ax))ax+k
47.5 0.0534
47.6 We have
kV (Ax) =Ax+k − P (Ax)ax+k
=Ax+k
(1− P (Ax)
ax+k
Ax+k
)=Ax+k
(1− P (Ax)
P (Ax+k)
)
47.7 0.0534
62
47.8 We have
kV (Ax) =1− ax+k
ax
=1−1−Ax+k
d1−Axd
=1− 1− Ax+k
1− Ax=Ax+k − Ax
1− Ax47.9 0.053
47.10 We have
kV (Ax) =Ax+k − P (Ax)ax+k
=Ax+k − P (Ax)ax+k +P (Ax)ax − Ax
kEx
=P (Ax)
(ax − kExax+k
kEx
)−(Ax − kExAx+k
kEx
)=P (Ax)
(ax:k
kEx
)−
(A1x:k
kEx
)
=P (Ax)sx:k −A1x:k
kEx
47.11 0.053
47.12 We have
P (Ax+k) =Ax+k
d−1(1− Ax+k)=⇒ Ax+k
P (Ax+k)=
1
P (Ax+k) + d.
Thus,
kV (Ax) =Ax+k
(1− P (Ax)
P (Ax+k)
)=
[P (Ax+k)− P (Ax)]Ax+k
P (Ax+k)
=P (Ax+k)− P (Ax)
P (Ax+k) + d
63
47.13 305.651
47.14 114.2984
47.15 0.0851
47.16 171.985
47.17 4420.403
47.18 0.0042
47.19 −0.0826
47.20 0.1587
47.21 0.2757
47.22 0.0138
47.23 629.89
47.24 528.48
47.25 (a) For a fully discrete n−year pure endowment, the insurer’s prospective loss at time k(or at age x+ k) is:
kL(A 1x:n ) = νn−kI(K ≥ n)− P (A 1
x:n )amin{(K−k+1,n−k)} , k < n
and nL(A 1x:n ) = 1.
(b) The prospective benefit reserve is
kV (A 1x:n ) =
{A 1x+k:n−k − P (A 1
x:n )ax+k:n−k k < n
1 k = n.
47.26 0.23426
47.27 8119.54
64
47.28 7.2170
47.29 (a) The prospective formula is
3V (15|a65) = 12E68a80 − P (15|a65)a68:12 .
(b) The retrospective formula is
3V (15|a65) =P (15|a65)a65:3
3E65
47.30 kV (n|ax) =P (n|ax)ax:n
kEx− na
x:k−nkEx
47.31 3.3086
65
Section 48
48.1 0.0828
48.2 (a) The kth terminal prospective loss random variable for an n−year term insurance con-tract
kL(A1x:n ) = Z 1
x+k:n−k − P (A1x:n )Yx+k:n−k .
(b) The kth terminal prospective reserve is given by
kV (A1x:n ) = A 1
x+k:n−k − P (A1x:n )ax+k:n−k
48.3 (a) The prospective loss random variable is
hkL(A1
x:n ) =
{Z 1x+k:n−k − hP (A1
x:n )Yx+k:h−k k < h < n
Z 1x+k:n−k h < k < n.
(b) The kth terminal prospective reserve for this contract
hkV (A1
x:n ) =
{A 1x+k:n−k − hP (A1
x:n )ax+k:h−k k < h < n
A 1x+k:n−k h < k < n
48.4 (a) The prospective loss random variable is
kL(Ax:n ) = Zx+k:n−k − P (Ax:n )Yx+k:n−k .
(b) The kth terminal prospective reserve for this contract
kV (Ax:n ) = Ax+k:n−k − P (Ax:n )ax+k:n−k .
48.5 (a) The prospective loss random variable is
hkL(Ax:n ) =
Zx+k:n−k − hP (Ax:n )Yx+k:h−k k < h < nZx+k:n−k h ≤ k < n1 k = n.
(b) The kth terminal prospective reserve for this contract is
hkV (Ax:n ) =
Ax+k:n−k − hP (Ax:n )ax+k:h−k k < h < nAx+k:n−k h ≤ k < n1 k = n
66
48.6 Recall that under UDD, we have
Ax+k:n−k =i
δAx+k:n−k + n−kEx+k
hP (Ax:n ) =Ax:n
ax:h
=A1x:n + nExax:h
=iδA1x:n + nEx
ax:h
=i
δhP (A1
x:n ) + hP (A 1x:n ).
Thus,
Ax+k:n−k − hP (Ax:n )ax+k:h−k =i
δAx+k:n−k + n−kEx+k
−(i
δhP (A1
x:n ) + hP (A 1x:n ))ax+k:h−k
=i
δhkV (A1
x:n ) + hkV (A 1
x:n ).
48.7 Recall the following expressions:
Ax =A1x:k
+ kExAx+k
ax =ax:k + kExax+k.
Thus,
kV (Ax) =Ax+k − P (Ax)ax+k
=Ax+k − P (Ax)ax+k +P (Ax)ax − Ax
kEx
=Ax+k − P (Ax)ax+k +P (Ax)[ax:k + kExax+k]
kEx−
[A1x:k
+ kExAx+k]
kEx
=P (Ax)ax:k
kEx−A1x:k
kEx
=P (Ax)sx:k −A1x:k
kEx.
67
Section 49
49.1 0.342035
49.2 0.0840
49.3 We will prove (b) and leave (a) to the reader. We have
kV(m)(Ax)− kV (Ax) =P (Ax)ax+k − P (m)(Ax)[α(m)ax+k − β(m)]
=P (m)(Ax)a
(m)x
axax+k − P (m)(Ax)[α(m)ax+k − β(m)]
=P (m)
[a
(m)x
axax+k − α(m)ax+k + β(m)
]
=P (m)
[α(m)ax − β(m)
axax+k − α(m)ax+k + β(m)
]=β(m)P (m)
[1− ax+k
ax
]=β(m)P (m)(Ax)kV (Ax).
49.4 We have
kV(m)(Ax)− kV (Ax)
kV (m)(Ax)− kV (Ax)=P (m)(Ax)
P (m)(Ax)=
Ax
a(m)x
Ax
a(m)x
=AxAx
=AxiδAx
=δ
i
68
49.5 We have
kV (Ax) =Ax+k − P (m)(Ax)a(m)x+k
=Ax+k − P (m)(Ax)a(m)x+k +
P (m)(Ax)a(m)x − Ax
kEx
=Ax+k − P (m)(Ax)a(m)x+k +
P (m)(Ax)[a(m)
x:k+ kExa
(m)x+k]
kEx−
[A1x:k
+ kExAx+k]
kEx
=P (m)(Ax)a
(m)
x:k
kEx−A1x:k
kEx
=P (m)(Ax)s(m)
x:k−A1x:k
kEx
69
Section 50
50.1 The insurer’s prospective loss random variable is
hL =
{0 K(x) < h
bK(x)+1+hνK(x)+1−h −
∑K(x)j=h πjν
j−h K(x) ≥ h
50.2 564.46
50.3 1027.42
50.4 30.395
50.5 (a) 30.926 (b) 129.66 (c) 382.44
50.6 255.064
50.7 31.39
50.8 499.102
70
Section 51
51.1 The prospective loss of this contract at time t is
tL = PVFB− PVFP =
{0 T (x) ≤ t
bT (x)νT (x)−t −
∫ T (x)
tπuν
u−tdu T (x) > t.
51.2 We have
tV =E[tL|T (x) > t] = E
[b(T (x)−t)+tν
T (x)−t −∫ T (x)−t
0
πt+rνrdr|T (x) > t
]
=E
[bT (x+t)+tν
T (x+t) −∫ T (x+t)
0
πt+rνrdr
]
=
∫ ∞0
[bu+tν
u −∫ u
0
πt+rνrdr
]fT (x+t)(u)du
=
∫ ∞0
[bu+tν
u −∫ u
0
πt+rνrdr
]upx+tµ(x+ t+ u)du
=
∫ ∞0
bu+tνuupx+tµ(x+ t+ u)du−
∫ ∞0
∫ u
0
πt+rνrdrupx+tµ(x+ t+ u)du
=
∫ ∞0
bu+tνuupx+tµ(x+ t+ u)du+
∫ ∞0
∫ u
0
πt+rνrdr
d
du[upx+t]du
=
∫ ∞0
bu+tνuupx+tµ(x+ t+ u)du−
∫ ∞0
πt+rνrupx+tdr
=APV of future benefits− APV of future benefit premiums
51.3 95.96
51.4 1055.79
51.5 we have
0 = 0V =
∫ ∞0
buνuupxµ(x+ u)du−
∫ ∞0
πuνuupxdu.
Thus, ∫ t
0
(πuνuupx − buνuupxµ(x+ u))du =
∫ ∞t
(buνuupxµ(x+ u)− πuνuupx)du.
71
Letting u = t+ y on the right integral, we obtain∫ t
0
(πuνuupx − buνuupxµ(x+ u))du =
∫ ∞0
(bt+yνt+y
t+ypxµ(x+ t+ y)− πt+yνt+yt+ypx)dy
=νttpx
[∫ ∞0
(bt+yνyypx+tµ(x+ t+ y)− πt+yνyypx+y)dy
]=tEx
[∫ ∞0
(bt+yνyypx+tµ(x+ t+ y)− πt+yνyypx+y)dy
].
Now, the result follows by dividing both sides by tEx
72
Section 52
52.1 (a) 564.470 (b) 2000
52.2 324.70
52.3 77.66
52.4 −4.33
52.5 0.015
52.6 36657.31
52.7 0.017975
52.8 0.028
52.9 355.87
52.10 For any n, we have
(nV + π)(1 + i) = q25+nn+1V + p25+nn+1V = n+1V.
Thus,34∑n=0
(nV + π)(1 + i)35−n =34∑n=0
n+1V (1 + i)34−n
which implies
0V (1 + i)35 + πs35 = 35V.
But 0V = 0 and 35V = a60 (actuarial present value of future benefits; there are no future premiums).Thus,
π =a60
s35
.
Likewise,19∑n=0
(nV + π)(1 + i)20−n =19∑n=0
n+1V (1 + i)19−n
73
which implies
0V (1 + i)20 + πs20 = 20V.
Hence,
20V =
(a60
s35
)s20 .
52.11 5.28
52.12 9411.01
52.13 296.08
52.14 1027.42
52.15 286.04
52.16 (a) 0.091 (b) 101.05
Section 53
53.1 0
74
Section 54
54.1 0.25904
54.2 302.31
54.3 1799.037
54.4 697.27
54.5 495.80
54.6 (a) 0.0505 (b) 110.85
75
Section 55
55.1 The expected value is 0.37704 and the variance is 0.03987
55.2 0.458431
55.3 5.4
55.4 0.1296
76
Section 56
56.1 We have
tqxy =1− tpxy = 1− tpx tpy
=1− (1− tqx)(1− tqy)
=1− (1− tqx − tqy + tqx tqy)
=tqx + tqy − tqx tqy
56.2 We have
Pr[(T (x) > n) ∪ (T (y) > n)] =Pr[T (x) > n] + Pr[T (y) > n]− Pr[(T (x) > n) ∩ (T (y) > n)]
=npx + npy − npxnpy = npx + npy − npxy
56.3 0.2
56.4 13
56.5 0.067375
56.6 We have
tqxy =1− tpxy = 1− tpx tpy
=1− (1− tqx)(1− tqy)
=1− (1− tqx − tqy + tqx tqy)
=tqx + tqy − tqx tqy
56.7 0.10969
56.8 n|mqxy = n+mqxy − tqxy = tpxy − n+mpxy
56.9 0.03436
77
56.10 We have
13qxy =1− 1
3pxy = 1− (1− 1
3qx)(1− 1
3qy)
=1−(
1− 1
3qx
)(1− 1
3qy
)=
1
3qx +
1
3qy −
1
9qxqy
12qxy =
1
2qx +
1
2qy −
1
4qxqy.
Thus,18 1
3qxy − 12 1
2qxy = 6qx + 6qy − 2qxqy − 6qx − 6qy + 3qxqy = qxqy
56.11 0.08
56.12 23
56.13 4× 10−8
56.14 0.06
56.15 0.10
56.16 10.42
56.17 12.5
56.18 0.21337
56.19 0.36913
56.20 2.916667
56.21 160.11
78
Section 57
57.1 54.16667
57.2 0.9167
57.3 5.41667
57.4 0.05739
57.5 0.961742
57.6 0.24224
57.7 34
57.8 40.8333
57.9 0.05982
57.10 We have
µxy(t) =−ddt tpxy
tpxy=
ddt tqxy
1− tqxy
=ddt tqxtqy
1− tqxy=
tpxtqyµ(x+ t) + tpytqxµ(y + t)
1− tqxtqy
57.11 0.0023
57.12 13.17
57.13 30.33
57.14 5
57.15 28.5585
57.16 1/14
79
57.17 (a) 0.155 (b) 30
57.18 1.25
80
Section 58
58.1 We have
Cov(T (xy), T (xy)) =E[T (xy) · T (xy)]− E[T (xy)]E[T (xy)]
=E[T (x)T (y)]− E[T (xy)]E[T (x) + T (y)− T (xy)]
=E[T (x)T (y)]− E[T (x)E[T (y)]− E[T (xy)](E[T (x)] + E[T (y)]− E[T (xy)])
+E[T (x)E[T (y)]
=Cov(T (x), T (y))− exy (ex + ey − exy) + exey
=Cov(T (x), T (y)) + (ex − exy)(ey − exy)
58.2 3.7
58.3 4.3
58.4 400
58.5 We have
Cov(T (xy), T (xy)) =(ex − exy)(ey − exy)=exey − exy (ex + ey − exy)=exey − exyexy
81
Section 59
59.1 0.6
59.2 0.030873
59.3 We have
nq1xy + n
1qxy=
∫ n
0tpxyµ(x+ t)dt+
∫ 10
0tpxyµ(y + t)dt
=
∫ n
0tpxy[µ(x+ t) + µ(y + t)]dt =
∫ n
0tpxyµxy(t)dt
=
∫ n
0
∫ n
0
fT (xy)(t)dt = nqxy
59.4 0.0099
59.5 0.0001467
59.6 0.141
59.7 1−enµµ(95−x)
59.8 0.0134
82
Section 60
60.1 4.2739
60.2 0.9231
60.3 0.06
60.4 0.1345
60.5 11.27
60.6 0.0549
60.7 0.0817
60.8 0.18
60.9 27927.51
60.10 (a) 115,714.29 (b) 14.4P .
60.11 0.38
60.12 600
60.13 0.191
83
Section 61
61.1 12.8767
61.2 12.7182
61.3 5.95238
61.4 4.7
61.5 We know that
ay =1− Ayδ
and
axy =1− Axy
δ.
Hence,
ax|y =1− Ayδ− 1− Axy
δ=Axy − Ay
δ
84
Section 62
62.1 0.069944
62.2 1.441188
62.3 0.082667
62.4 1691.92
62.5 0.736
85
Section 63
63.1 ty36
+ 112
63.2 2.75
63.3 We have
fT (x)(tx)fT (y)(ty) =
(tx36
+1
12
)(ty36
+1
12
)6= tx + ty
216= fT (x)T (y)(tx, ty)
63.4 4.4375
63.5 1.6136
63.6 1
63.7 (6−n)2(6+n)216
63.8 0.8102
86
Section 64
64.1 e−0.06t
64.2 (a) We have
fT,J(t, j) = tp(τ)50 µ
(j)(x+ t) =j
503(50− t)2, j = 1, 2.
(b) fT (t) =∑2
j=1 fT (x),J(x)(t, j) = 3503
(50− t)2.
(c) fJ(j) =∫ 50
0fs,J(x)(t, j)ds = j
503
∫ 50
0(50− t)2dt = j
3, j = 1, 2.
(d) fJ |T (j|t) = µ(j)(50+t)
µ(τ)(50+t)= 1
3j, j = 1, 2 64.3 0.12
64.4 11.11
64.5 (a) 0.00446 (b) 1/3
64.6 0.259
64.7 0.0689
87
Section 65
65.1 0.60
65.2 0.4082483
65.3 0.12531
65.4 0.0198
65.5 0.216
65.6 0.644
65.7 0.512195
88
Section 66
66.1 A decrease of 10 in the value of d(1)26 .
66.2 119
66.3 0.05
66.4 0.0555
66.5 0.2634
66.6 0.0426
66.7 7.6
66.8 803
66.9 0.38
89
Section 67
67.1 0.154103
67.2 0.04525
67.3 0.0205
67.4 0.02214
67.5 25.537
67.6 0.053
67.7 14.1255563
67.8 0.09405
67.9 0.0766
67.10 0.1802
90
Section 68
68.1 3000
68.2 1.90
68.3 1
68.4 53,045.10
68.5 40.41
68.6 457.54
68.7 7841.28
91
Section 69
69.1 18,837.04
69.2 2.5
69.3 120 is payable for the next 10 years and 100 is payable after 10 years
69.4 0
69.5 14.7
69.6 11,194.0199
69.7 922.014
92
Section 70
70.1 19.88
70.2 (a) 10.8915 (b) 17.6572 (c) 6.7657 (d) 104.297 (e) 104.5549 (f) 0.00027
70.3 8.8932
70.4 10.0094
70.5 888.225
70.6 0.472
70.7 −445.75
70.8 92.82
70.9 1371.72
70.10 2302.52
70.11 1177.23
93
Section 71
71.1 (a) 23.88 (b) 5.655
71.2 30.88
71.3 (a) 883.9871 (b) 903.9871
71.4 887.145
71.5 4.2379
71.6 G =100010|20A30+20+10a
30:90.85a
30:5−0.15
94
Section 72
72.1 1750.03
72.2 414.82
72.3 16.8421
72.4 Multiplying the equation
k+1AS`(τ)x+k+1 = (kAS +G− ckG− ek)(1 + i)`
(τ)x+k − bk+1d
(d)x+k − k+1CV d
(w)x+k.
by νk+1 we obtain
k+1ASνk+1`
(τ)x+k+1 − kASν
k`(τ)x+k = G(1− ck)νk`(τ)
x+k − ekνk`
(τ)x+k − (bk+1d
(d)x+k + k+1CV d
(w)x+k)ν
k+1.
Using the fact that 0AS = 0 and summing this telescoping series gives
nASνn`
(τ)x+n =
n−1∑k=0
[k+1ASνk+1`
(τ)x+k+1 − kASν
k`(τ)x+k]
=Gn−1∑k=0
(1− ck)νk`(τ)x+k −
n−1∑k=0
ekνk`
(τ)x+k
−n−1∑k=0
(bk+1d(d)x+k + k+1CV d
(w)x+k)ν
k+1
72.5 10AS1 − 10AS2 = (G1 −G2)∑9
k=0(1− ck)νk−9
(`(τ)x+k
`(τ)x+10
)72.6 1627.63
72.7 1.67
95
Section 73
73.1 We have ∑j∈E
Qn(i, j) =∑j∈E
Pr(Xn+1 = j|Xn = i) = Pr(Xn+1 ∈ E|Xn = i) = 1
73.2 The entries in the second row do not sum up to 1. Therefore, the given matrix can not be atransition matrix.
73.3
Q =
0 1 00 1
212
13
0 23
73.4 The transition diagram is
The transition matrix is
Q =
(0.8 0.20.6 0.4
)73.5
Qn =
(px+n qx+n
0 1
)73.6 The transition probabilities areQn(0, 0) = p
(τ)x+n, Qn(0, j) = q
(j)x+n for j = 1, 2, · · · ,m, Qn(j, j) =
1 for j = 1, 2, · · · ,m, Qn(i, j) = 0 for all other values of i and j
73.7
Q61 =
0.20 0.10 0.700 1 00 0 1
96
Section 74
74.1 At time t = 1 we have
Q =
0.92 0.05 0.030.00 0.76 0.240.00 0.00 1.00
At time t = 2 we have
2Q = Q2 =
0.8464 0.084 0.06960.00 0.5776 0.42240.00 0.00 1.00
At time t = 2 we have
3Q = Q3 =
0.778688 0.106160 0.1151520.00 0.438976 0.5610240.00 0.00 1.00
74.2 (a) 0.70 (b) 0.3125
74.3 0.056
74.4 0.892
74.5 0.0016
74.6 0.489
74.7 4.40
97
Section 75
75.1 16.82
75.2 170.586
75.3 1960
75.4 185.11
75.5 0.34
75.6 10,694.64
75.7 160
98
Section 76
76.1 Since N(s) and N(t)−N(s) are independent, we have Cov(N(s), N(t)−N(s)) = 0. Thus,
Cov(N(s), N(t)) =Cov(N(s), N(s) +N(t)−N(s)) = Cov(N(s), N(s)) + Cov(N(s), N(t)−N(s))
=Cov(N(s), N(s)) = Var(N(s)) = λs
76.2 8.338× 10−4
76.3 e−6 65
5!
76.4 E[2N(3)− 4N(5)] = −28 and Var[2N(3)− 4N(5)] = 88
76.5 We have
Pr(N(t) = k|N(s+ t) = n) =Pr(N(t) = k,N(s+ t) = n)
Pr(N(s+ t) = n)
=Pr(N(t) = k,N(s+ t)−N(t) = n− k)
Pr(N(s+ t) = n)
=e−λt (λt)k
k!e−λs (λs)n−k
(n−k)!
e−λ(s+t) [λ(s+t)]n
n!
=
(nk
)(t
t+ s
)k (s
t+ s
)n−k76.6 0.2963
76.7 0.593994
76.8 (a) 0.503 (b) 18805
99
Section 77
77.1 If Sn ≤ n then the nth event happens before time t. This means that there are n or moreevents in the interval [0, t] which implies that N(t) ≥ n
77.2 0.0000393
77.3 The expected value is 1 and the variance is 1/3
77.4 fT5(t) = 3e−3t, t ≥ 0
77.5 1/9
77.6 0.6321
77.7 Both expected arrival time are the same.
100
Section 78
78.1 0.0025
78.2 0.2048
78.3 (a) 48 (b)0.04262 (c) 100 minutes
78.4 768
78.5 2,000,000
78.6 0.5
78.7 0.276
78.8 0.1965
78.9 0.55
78.10 0.3859
78.11 0.3679
78.12 210.10
78.13 0.23
101
Section 79
79.1 0.1954
79.2 ∞
79.3 0.016
79.4 (a) 3 (b) 1.875
79.5 0.03642
79.6 93.55
102
Section 80
80.1 The mean is 225,000 and the variance is 45,000,000
80.2 29
80.3 The mean is 25 and the variance is 215/3
80.4 The mean is 40,000 and the variance is 160,000,000
80.5 (a) 100,000,000 (b) 0.0228
80.6 0.6712