mc course3

**BEGINNING OF EXAMINATION**

COURSE/EXAM 3: MAY 2000 - 1 - GO ON TO NEXT PAGE

1. Given:

(i) eo0 25=

(ii) l x xx = - w w, 0

(iii) T xb g is the future lifetime random variable.

Calculate Var T 10b g .

(A) 65

(B) 93

(C) 133

(D) 178

(E) 333


2. Lucky Tom finds coins on his way to work at a Poisson rate of 0.5 coins/minute. Thedenominations are randomly distributed:

(i) 60% of the coins are worth 1;

(ii) 20% of the coins are worth 5; and

(iii) 20% of the coins are worth 10.

Calculate the conditional expected value of the coins Tom found during his one-hour walk today,given that among the coins he found exactly ten were worth 5 each.

(A) 108

(B) 115

(C) 128

(D) 165

(E) 180


3. For a fully discrete two-year term insurance of 400 on ( x):

(i) i = 01.

(ii) 400 74 3321Px: .=

(iii) 400 16581 21V x: .=

(iv) The contract premium equals the benefit premium.

Calculate the variance of the loss at issue.

(A) 21,615

(B) 23,125

(C) 27,450

(D) 31,175

(E) 34,150


4. You are given:

(i) The claim count N has a Poisson distribution with mean L .

(ii) L has a gamma distribution with mean 1 and variance 2.

Calculate the probability that N = 1.

(A) 0.19

(B) 0.24

(C) 0.31

(D) 0.34

(E) 0.37


5. An insurance company has agreed to make payments to a worker age x who was injured at work.

(i) The payments are 150,000 per year, paid annually, starting immediately and continuingfor the remainder of the workers life.

(ii) After the first 500,000 is paid by the insurance company, the remainder will be paid by areinsurance company.

(iii) t x

t

pt

t=

1

600 000600 000

03

Calculate the expected value of his bonus.

(A) 16,700

(B) 31,500

(C) 48,300

(D) 50,000

(E) 56,600


26. For a fully discrete 3-year endowment insurance of 1000 on xb g:(i) q qx x= =+1 020.

(ii) i = 0 06.

(iii) 1000 373633Px: .=

Calculate 1000 2 3 1 3V Vx x: : .-e j

(A) 320

(B) 325

(C) 330

(D) 335

(E) 340


27. For an insurer with initial surplus of 2:

(i) The annual aggregate claim amount distribution is:

Amount Probability038

0.60.30.1

(ii) Claims are paid at the end of the year.

(iii) A total premium of 2 is collected at the beginning of each year.

(iv) i = 0 08.

Calculate the probability that the insurer is surviving at the end of year 3.

(A) 0.74

(B) 0.77

(C) 0.80

(D) 0.85

(E) 0.86


28. For a mortality study on college students:

(i) Students entered the study on their birthdays in 1963.

(ii) You have no information about mortality before birthdays in 1963.

(iii) Dick, who turned 20 in 1963, died between his 32 nd and 33rd birthdays.

(iv) Jane, who turned 21 in 1963, was alive on her birthday in 1998, at which time she left thestudy.

(v) All lifetimes are independent.

(vi) Likelihoods are based upon the Illustrative Life Table.

Calculate the likelihood for these two students.

(A) 0.00138

(B) 0.00146

(C) 0.00149

(D) 0.00156

(E) 0.00169


29. For a whole life annuity-due of 1 on (x), payable annually:

(i) qx = 0 01.

(ii) qx+ =1 0 05.

(iii) i = 0 05.

(iv) && .ax+ =1 6 951

Calculate the change in the actuarial present value of this annuity-due if px+1 is increasedby 0.03.

(A) 0.16

(B) 0.17

(C) 0.18

(D) 0.19

(E) 0.20


30. X is a random variable for a loss.

Losses in the year 2000 have a distribution such that:

E X d d d d = - + - =0025 1475 2 25 10 11 12 262. . . , , , ,...,

Losses are uniformly 10% higher in 2001.

An insurance policy reimburses 100% of losses subject to a deductible of 11 up to a maximumreimbursement of 11.

Calculate the ratio of expected reimbursements in 2001 over expected reimbursements in theyear 2000.

(A) 110.0%

(B) 110.5%

(C) 111.0%

(D) 111.5%

(E) 112.0%


31. Company ABC issued a fully discrete three-year term insurance of 1000 on Pat whose stated ageat issue was 30. You are given:

(i)x qx30313233

0.010.020.030.04

(ii) i = 0 04.

(iii) Premiums are determined using the equivalence principle.

During year 3, Company ABC discovers that Pat was really age 31 when the insurance wasissued. Using the equivalence principle, Company ABC adjusts the death benefit to the leveldeath benefit it should have been at issue, given the premium charged.

Calculate the adjusted death benefit.

(A) 646

(B) 664

(C) 712

(D) 750

(E) 963


32. Insurance for a citys snow removal costs covers four winter months.

(i) There is a deductible of 10,000 per month.

(ii) The insurer assumes that the citys monthly costs are independent and normallydistributed with mean 15,000 and standard deviation 2,000.

(iii) To simulate four months of claim costs, the insurer uses the Inverse Transform Method(where small random numbers correspond to low costs).

(iv) The four numbers drawn from the uniform distribution on [0,1] are:

0.5398 0.1151 0.0013 0.7881

Calculate the insurers simulated claim cost.

(A) 13,400

(B) 14,400

(C) 17,800

(D) 20,000

(E) 26,600


33. In the state of Elbonia all adults are drivers. It is illegal to drive drunk. If you are caught, yourdrivers license is suspended for the following year. Drivers licenses are suspended only fordrunk driving. If you are caught driving with a suspended license, your license is revoked andyou are imprisoned for one year. Licenses are reinstated upon release from prison.

Every year, 5% of adults with an active license have their license suspended for drunk driving.Every year, 40% of drivers with suspended licenses are caught driving.

Assume that all changes in driving status take place on January 1, all drivers act independently,and the adult population does not change.

Calculate the limiting probability of an Elbonian adult having a suspended license.

(A) 0.019

(B) 0.020

(C) 0.028

(D) 0.036

(E) 0.047


34. For a last-survivor whole life insurance of 1 on (x) and (y):

(i) The death benefit is payable at the moment of the second death.

(ii) The independent random variables T x T y Z* , * ,b g b g and are the components of acommon shock model.

(iii) T x*b g has an exponential distribution with mxT x t t* . , .b gb g = 003 0

(iv) T y*b g has an exponential distribution with myT y t t* . , .b g b g = 0 05 0(v) Z, the common shock random variable, has an exponential distribution with

mZ t tb g = 002 0. , .(vi) d = 006.

Calculate the actuarial present value of this insurance.

(A) 0.216

(B) 0.271

(C) 0.326

(D) 0.368

(E) 0.423


35. The distribution of Jacks future lifetime is a two-point mixture:

(i) With probability 0.60, Jacks future lifetime follows the Illustrative Life Table, withdeaths uniformly distributed over each year of age.

(ii) With probability 0.40, Jacks future lifetime follows a constant force of mortalitym= 002. .

A fully continuous whole life insurance of 1000 is issued on Jack at age 62.

Calculate the benefit premium for this insurance at i = 0 06. .

(A) 31

(B) 32

(C) 33

(D) 34

(E) 35


36. A new insurance salesperson has 10 friends, each of whom is considering buying a policy.

(i) Each policy is a whole life insurance of 1000, payable at the end of the year of death.

(ii) The friends are all age 22 and make their purchase decisions independently.

(iii) Each friend has a probability of 0.10 of buying a policy.

(iv) The 10 future lifetimes are independent.

(v) S is the random variable for the present value at issue of the total payments to those whopurchase the insurance.

(vi) Mortality follows the Illustrative Life Table.

(vii) i = 0 06.

Calculate the variance of S.

(A) 9,200

(B) 10,800

(C) 12,300

(D) 13,800

(E) 15,400


37. Given:

(i) pk denotes the probability that the number of claims equals k for k = 0,1,2,

(ii)pp

mn

n

m= !

! , m n 0 0,

Using the corresponding zero-modified claim count distribution with p pM M0 101= . , .calculate

(A) 0.1

(B) 0.3

(C) 0.5

(D) 0.7

(E) 0.9


38. For Shoestring Swim Club, with three possible financial states at the end of each year:

(i) State 0 means cash of 1500. If in state 0, aggregate member charges for the next year areset equal to operating expenses.

(ii) State 1 means cash of 500. If in state 1, aggregate member charges for the next year areset equal to operating expenses plus 1000, hoping to return the club to state 0.

(iii) State 2 means cash less than 0. If in state 2, the club is bankrupt and remains in state 2.

(iv) The club is subject to four risks each year. These risks are independent. Each of the fourrisks occurs at most once per year, but may recur in a subsequent year.

(v) Three of the four risks each have a cost of 1000 and a probability of occurrence 0.25 peryear.

(vi) The fourth risk has a cost of 2000 and a probability of occurrence 0.10 per year.

(vii) Aggregate member charges are received at the beginning of the year.

(viii) i = 0

Calculate the probability that the club is in state 2 at the end of three years, given that it is instate 0 at time 0.

(A) 0.24

(B) 0.27

(C) 0.30

(D) 0.37

(E) 0.56


39. For a continuous whole life annuity of 1 on (x):

(i) T xb g, the future lifetime of ( x), follows a constant force of mortality 0.06.(ii) The force of interest is 0.04.

Calculate Pr .a aT x xb ge j>

(A) 0.40

(B) 0.44

(C) 0.46

(D) 0.48

(E) 0.50

COURSE/EXAM 3: MAY 2000 - 40 - STOP

40. Rain is modeled as a Markov process with two states:

(i) If it rains today, the probability that it rains tomorrow is 0.50.

(ii) If it does not rain today, the probability that it rains tomorrow is 0.30.

Calculate the limiting probability that it rains on two consecutive days.

(A) 0.12

(B) 0.14

(C) 0.16

(D) 0.19

(E) 0.22

**END OF EXAMINATION**

Course 3 May 2000

Solutions to Course 3 Exam May 2000

Question # 1Key: C

et

dto

0

2

01

2 225 50= -FHG

IKJ = - = = =z w w ww w w

w

et

dto

10

2

0

401

4040

402 40

20= -FHGIKJ = - =z b gb g

Var T x tt

dt

t t

b g b g

b g

= -FHGIKJ -

= -

LNM

OQP

-

=

z2 1 40 202

2 3 4020

133

2

0

40

2 3

0

402

Question # 2Key: C

A priori, expect 30 coins: 18 worth one, 6 worth 5, 6 worth 10.

Given 10 worth 5, expect: 18 @ 1; 10 @ 5; 6 @ 10.

Total = 18 + 50 + 60 = 128

Course 3 May 2000

Question # 3Key: E

Need to determine q qx xand +1.

Formula 7.23: 1658400

1174 33 0251

11. .

. .= = - =+ +Vq

qx x

Formula 8.39: 0 0 400 1V vq v Vpx x= = - +p

qV

Vx= -

-=p 11

4000171

1

..

Loss (Example 8.51):

0 4001

1174 33 289 30

.. .FHG

IKJ - =

1 4001

117433 1

1

1118868

2

..

..FHG

IKJ - +

FHG

IKJ =

2 - +FHGIKJ = -7433 1

1

1114190.

..

Since E L = 0,

Var = + - + - - =289 30 017 18868 0 25 1 017 14190 1 017 1 0 25 341502 2 2. . . . . . . .b g b g b g b gb g b g b gb g

Course 3 May 2000

Question # 4Key: A

The distribution is negative binomial

E N = 1

Var Var Var

Var

N E N E N

E

= +

= += + =

L L

L L

L L

L L

c h c h

1 2 3

From the KPW appendix, we have:

rb = 1

rb b1 3+ =b g1

31

2+ = =b bb gr rb = =1 1

2

Pr .Nr

r= = += =+1

1

1

3019245

1 32

bbb g

Question # 5Key: B

If the worker survives for three years, the reinsurance will pay 100,000 at t=3, and everythingafter that. So the actuarial present value of the reinsurers portion of the claim

= FHGIKJ +

FHG

IKJ +

FHG

IKJ

FHG

IKJ100 000

7105

150 0007

1057

105

3 4 5

,..

,..

..

= 79 012,

Course 3 May 2000

Question # 6Key: D

Surplus = U t t S tb g b g= + -60 20 .

P P tS t

Ruinb g b g= < -FHGIKJ

60

20

Hence claims of 100, 200 causing ruin can only occur on the intervals

0100 60

202 0

200 60

207, , ,

- =LNMOQP

- =LNMOQP respectively.

P t P tt

Claim by no claimb g b g= - =+

11

Therefore,

P Ps

Ruin p s claim of size s causes ruinb g b g b g= =

++

+=60% 2

1 240%

71 7

75%

Question # 7Key: D

Ll l

qd

md

L

dd

md

L

dd

d d d d d

q

xx x

xi x

i

xd x

d

x

xd

xd

xw x

w

x

xw

xw

x xi

xd

xw

xi

xi

tt t

t

t

t

b gb g b g

b g b g

b g b gb g

b g b g

b g b gb g

b g b g

b g b g b g b g b g

b g

= + =

=

= = = =

= = = =

= = + + =

= =

+1

224 000

25 000

24 0000 02 480

24 0000 05 1200

2000 320

32025 000

00128

,

,

,.

,.

,.

Course 3 May 2000

Question # 8Key: C

Zv k

kE Z E Z S E Z N

vq v q vq v q

k

xs

x

s

x xN

x

N

===

RST= +

= + + +

+10

0

0 1

2 31 3 2 3

10 1 3 2 3

5 1

5 21

21

,

, ,...

e j e j{ }

q px x= -1 p ex =-FHG

IKJ

1

q

t

1 2q p px x x= - 2

2

p ex =-FHG

IKJq

t

Smokers Nonsmokersp x 0.64118 0.77880

2 p x 0.16901 0.36788

Now plug in

= + FHGIKJ

LNMM

OQPP + +

FHG

IKJ

LNMM

OQPP

RS|T|

UV|W|

= + =

101

105035882

1

1050 47217 1 3

1

10502212

1

105041092 2 3

58 338 2 3 77 000 1 3 64 559

52 2

..

..

..

..

, , ,

b g b g b g b g

b g b g

Course 3 May 2000

Question # 9Key: A

510

10 35 10 35 35 5V a P a se j e j= :

=P aa

s10 3545

35 10e j

:

a v p dt p dt i

et

tt t45 45 45

0

40

0

40

45

2

0

40

0

140

402

20

= = =

= -FHG

IKJ

zz= =

b go

s a E a v p

t dtll

35 10 35 10 10 35 35 1010

10 35

35

450

10150

50

45040

: : :/ /= =

= -

=

z b g

= =P a10 35

20450

40 1778e j .

510

10 35 35 535

400

5

2

0

5

1778

5050

1

501778

50

4550

2938

V P a s t dtll

tt

= = -

= FHGIKJ -FHG

IKJ =

ze j b g b gb g

:

.

. .

Course 3 May 2000

Question # 10Key: D

Let X(t) be the number leaving by cab in a t hour interval and let Yi denote the numberof people in the ith group. Then:

E Y

E Y

i

i

= + + =

= + + =

1 0 6 2 0 3 3 01 15

1 06 2 0 3 3 01 2 72 2 2 2. . . .

. . . .

b g b g b gc h c h c h

E X

Var X

72 10 72 15 1 080

72 10 72 2 7 1944

b gb g

= =

= =

. ,

. ,

P X 72 1050b g , = P X 72 10495b g .=

-

-LNM

OQPP

X 72 1080

1944

10495 1080

1944

b g .

= - -1 0 691754F .b g= F 0 691754.b g= 0 7553.

The answer is D with or without using the 0.5 continuity correction.

Course 3 May 2000

Question # 11Key: E

Calculate convolutions of f xx b g:x f f *2

1020

0.30.3

0.000.09

n 0 1 2 3 4Pr N n=b g 0.37 0.37 0.18 0.06 0.02

f

f

f

s

s

s

0 037

10 037 03 011

20 018 009 03 037 013

b gb gb g

=

= =

= + =

.

. . .

. . . . .

F

F

F

s

s

s

0 037

10 037 011 0 48

20 0 48 013 0 61

b gb gb g

=

= + =

= + =

.

. . .

. . .

E S = + + =1 03 10 03 20 0 4 50 29. . .b gb g b gb g b gb g

E S E S F F FS S S- = - - - - - -

= - - - - = - =+30 10 1 0 10 1 10 10 1 20

29 10 3 037 048 061 29 154 136

b g b gc h b gc h b gc hb g. . . . .

Question # 12Key: A

0 0408 8151 1 2

0040081

8181. . .= = -

=mb g b gq

qq

Similarly, q and q80 820 0200 0 0600= =. .

2 80 5 1 2 80 5 1 2 80 5 81 81 1 2 82

0010 99

098099

004 096 0 03 0 0782

q q p q p q. / . / . /..

.

.. . . .

= + +

= + + =b g

Course 3 May 2000

Question # 13Key: A

E Z e

e

b g b g=+

= FHGIKJ

=

- +

-

1000

1000512

1255

10

1 2

mm d

m d

.

.

Var Z e eb g b g=+

- FHGIKJ

FHG

IKJ

=

- + -10002

5

12

23 61016

2 10 22

2 4mm d

m d .

, .

E S E Zb g b g= =400 50 200,

Var VarS Zb g b g= =400 9 444 064, ,

09550 200

9 444 0641645 9 444 064 50 200 55 255. Pr

,

, ,. , , , ,=

-

-FHG

IKJ

= + =S E S

S

kk

b gb gVar

Course 3 May 2000

Question # 14Key: B

According to the theorem on p. 67, the expected number of interations is c 178. , as follows:

f xg x

x x x

ddx

f xg x

x x

x x

x

b gb g d ib gb g d i

b gb g

= - +

FHG

IKJ = - + =

- - =

=

12 2

12 1 4 3 0

12 3 1 1 0

1

3

2 3

2

ddx

f xg x

2

21

3

b gb g

FHG

IKJ = - + = - + = -

Course 3 May 2000

Question # 15Key: B

p p p30 301

302 09 07 063tb g b g b g b gb g= = =. . .

l p l31 30 30 630t t tb g b g b g= =

d q l311

1 301

30 75b g b g b g= =t

d l l31 31 32 630 472 158t t tb g b g b g= - = - =

d d d312

31 311 158 75 83b g b g b g= - = - =t

qd

l312 31

2

31

83630

01317 013b gb gb g= = = t . .

Question # 16Key: C

Let S = the aggregate loss.

E Sb g = 8 10 000,

Var Sb g b g b g= + =8 3937 3 10 000 32 0002 2 2 2, ,

P S E S PS

P Z

> = - >FHGIKJ

= >

= -

= -

1580 000

32 000

40 000

32 000

125

125

1 125

.,

,

,

,

.

.

.

b gc hb gb g

b gF

F

Course 3 May 2000

Question # 17Key: E

s x X x E X x e d

e e

x

x

x x

b g b g c hc h

= > = > =

=-

-

- -

zPr Pr ..

q qq01

01

1

11

11

So: s 05 01205. .b g =

Question # 18Key: A

pm

md m

d m

md m

d m= = +

+

= =- +

- +zz

1000 1000

11000 1500

0

e dt

e dt

t

t

b g

b g

Expected loss = 1000 80 80A a- p .

1000 1000 1029708672 66575 6855380 80Ai

A= = =d

. . .b gb g

a801 068553

53969= - =. .d

Expected loss = 68553 150 53969 124. .- = -b g

Course 3 May 2000

Question # 19Key: D

E S E N X E X Nb g b g b g b g b g= +Var Var2

N E N nq N nq qis binomial Varb g b g b g= = -, 1 , where q is the probability of a claim.

X E X Xis uniform. Max Var Maxb g b g= =2 12

2

,

Hence,

Total variance

= + - FHGIKJ

RS|T|UV|W|

= + FHGIKJ

LNMM

OQPP

+ + FHGIKJ

LNMM

OQPP

=

#policies Max12Max

policies

q q q2 2

2 2 2 2

12

100 005400

12005 0 95

400

2200 0 06

300

12006 094

300

2

600 46667

b g

b gb g b gb g. . . . . ., .

Course 3 May 2000

Question # 20Key: A

APV v p p= -5 4 70 80 5 70 80: :c h

4 7074

70

4 8084

80

4 70 80 4 4 80 4 70 4 80

5 7075

70

5 8085

80

5 70 80 5 70 5 80 5 70 5 80

56 64051

66 161550856094

26 607343914365

0679736

0953912

53 960816616155

0815592

23 58246

39143650 602459

0926690

70

pll

pll

p p p p p

pll

pll

p p p p p

= = =

= = =

= + - =

= = =

= = =

= + - =

, .

, ..

, ., .

.

.

, ., .

.

, .

, ..

.

:

:

APV v= - = =0953912 0926690 0027222103

0 023485 5. ..

..b g

Course 3 May 2000

Question # 21Key: E

The expected values at the end of 20 years are 11 11620 20 20. . .b g b g= -e m

Actuarial present values at time zero are 1120 20.b g v and 116 20 20 20.b g e v- m

So 11 116. .= -e m

So m = FHGIKJ =log

.

..

116

110 0531

S x e xx0 5 0 50 5 0505 06931

005311305. .. .

log . .

..b g = = = - = =-m

m

Question # 22Key: B

Note: These values of l are 1/100 of the ones in the Exam booklet, which does not affect theanswer.

l

l

l l

K

30

30

81 82

95 014

1 0 63 35155

35155

30 51

=- =< = - >FHG

IKJ

= > -+

FHG

IKJ

LNM

OQP

= = -+

FHG

IKJ

=

=+

FHG

IKJ

= FHGIKJ =

-

e j 1

1

1

6

1004648

0

0

32

0

d

dm

d m

dm

d m

mm d

m

md

where

Course 3 May 2000

Question # 40Key: D

p = limiting probability of rain.

p p p

p

= + -

= =

05 0 3 1

0 3

080 375

. .

.

..

b g

Prob (rain on two consecutive days) = Prob (rain on first) x Prob (rain on second, given rain on first)

= (0.375)(0.5)= 0.1875

mc course3

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