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