COURSE 3 MULTIPLE-CHOICE ANSWER KEY - SOA · 14 b 39 e 15 a 40 b 16 b 17 a 18 e 19 c 20 c 21 e 22 d 23 a 24 b 25 c course 3 may 2001 multiple-choice answer key. solutions for may

COURSE 3: MAY 2001 - 41 – STOP

1 E 26 B2 A 27 D3 C 28 D4 E 29 E5 D 30 D6 B 31 D7 C 32 A8 D 33 D9 C 34 A10 A 35 B11 B 36 E12 A 37 C13 E 38 A14 B 39 E15 A 40 B16 B17 A18 E19 C20 C21 E22 D23 A24 B25 C

COURSE 3MAY 2001

MULTIPLE-CHOICE ANSWER KEY

SOLUTIONS FOR MAY 2001 COURSE 3 EXAM

Test Question: 1 Key: E

For de Moivre’s law,

et

dt

tt

o30

0

30

2

0

30

130

2 30

302

= −−

FHG

IKJ

= −−

LNM

OQP

= −

−

−

z ω

ω

ω

ω

ω

b g

Prior to medical breakthrough ω = ⇒ = − =100100 30

23530eo

After medical breakthrough ′ = + =e eo o30 30 4 39

so ′ = = ′ − ⇒ ′ =e30 3930

2108

o ω ω

Test Question: 2 Key: A

02 5

3100 000 4000L v a= −, &&. @5%

= 77,079

Test Question: 3 Key: C

Ε Ν Ε Ε Ν Λ Ε ΛΛ Λ= = = 2

Var Var Var E NΝ Ε Ν Λ ΛΛ Λ= +

=Ε Λ ΛΛ Λ+ = + =Var 2 2 4

Distribution is negative binomial (Loss Models, 3.3.2)

Per supplied tablesmean r

Var r

r

r

= == + =

+ ====

ββ β

βββ

2

1 4

1 2

1

2

2

b gb g

From tables

pr r r

r3

3

3

3

5

1 2

3 1

2 3 4 1

3 2432

0125=+ +

+= = =+

b gb gb g

b gb gb gβ

β! !.

1000 p3 125=


E N Var N= = =60 05 30b gb g.

E X = + + =0 6 1 0 2 5 02 10 36. . . .b gb g b gb g b gb gE X 2 06 1 02 25 02 100 256= + + =. . . .b gb g b gb g b gb gVar X = 256 36 12 642. . .− =

For any compound distribution, per Loss Models

Var S E N Var X Var N E X= + c h2= (30) (12.64) + (30) 362.d i= 768

For specifically Compound Poisson, per Probability Models

Var S t E X= =λ 2 (60) (0.5) (25.6) = 768

Alternatively, consider this as 3 Compound Poisson processes (coins worth 1; worth 5; worth

10), where for each Var Xb g = 0 , thus for each Var S Var N E Xb g b g= 2 .Processes are independent, so total Var is

Var = + +=

60 05 0 6 1 60 05 0 2 5 60 05 02 10

768

2 2 2b gb gb g b gb gb g b gb gb gb g. . . . . .

Test Question: 5 Key: D

1000 10001000 106 1000

34203 1372 106 0 01254 1000

09874636918

1 0 36918006 106

111445

1000 100036918111445

331

2020

201920

20 19

19

20

2020

20

V AV P q

p

a

so PAa

x xx x x

x

x

xx

x

= =+ −

=+ −

=

= − =

= = =

++

+

+

++

+

d ib g b g

b gb g b g

b g

.

. . . .

..

&&.

. / ..

&&

..

.

Test Question: 6 Key: B

k xt dt t dt

t dt

k x k x

p e e

e

p p

xk

xk

xk

τ µ µ

µ

τ

µ

b g b g b g

b g

b g

b g b g

b g

b g

= z = z

= zFHG

IKJ

=

− −

−

0

1

0

1

0

2

2

2 where is from Illustrative Life Table, since follows I.L.T.1

10 60

11 60

10 60 10 60 11 60

10 602

11 602

2 2

6 616 1558188 074

080802

6 396 6098188 074

0 78121

080802 078121 0 0426

p

p

q p p

p p

= =

= =

= −

= −

= − =

, ,, ,

.

, ,, ,

.

. . .

τ τ τb g b g b g

b g b g from I.L.T.


State 1: light TrainingState 2: heavy Training

P11 04 05 0 6 0 0 2= × + × =. . . .PP

P

12

21

22

04 05 0 6 1 0808 05 02 0 0 4

08 05 02 1 0 6

= × + × == × + × == × + × =

. . . .

. . . .

. . . .

P = LNM

OQP

0 2 08

0 4 06

. .

. .

π π ππ π ππ π

1 1 2

2 1 2

1 2

0 2 04

08 0 61

= += ++ =

. .

. .

⇒ − = − + × = + ⇒ = = =1 02 1 04 02 0 2 12 088

12232 2 2 2 2 2π π π π π π. . . . . . .b g

Note: the notation in Probability Models would label the states 0 and 1, and would label thetop row and left column of the matrix P with subscript 0. The underlying calculationsare the same. The matrix P would look different, but the result would be the same, ifyou chose to make “heavy” the lower-numbered state.


σ = =0 0004 0 02. .Y 1b g is normal 0 01 0 0004. , .b g0.1587 corresponds to -1 standard deviation ⇒

Y (1) = 0.01-(1)(0.02) = -0.01Y Y( ) ( )2 1− is normal (0.01, 0.0004)0.9332 corresponds to +1.5 standard deviation ⇒

Y Y( ) ( ) . ( . )( . )2 1 0 01 15 002= + +

= − + +=

0 01 001 0 03

0 03

. . .

.

F e e

G e e

G F

Y

Y

= = =

= = =− =

−100 100 99 00

100 100 10305

405

1 0 01

2 0 03

b g

b g

.

.

.

.

.


Pa

dss

= −1&&

, where s can stand for any of the statuses under consideration.

&&

&& &&. .

.

&&. .

.

aP d

a a

a

ss

x y

xy

=+

= =+

=

=+

=

1

101 0 06

625

10 06 006

8 333

&& && && &&a a a axy xy x y+ = +

&& . . . .axy = + − =625 625 8333 4167

Pxy = − =14167

0 06 018.

. .


d e dt

e

t0

0 04

0

1

0 04

1000 0 04

1000 1 48

τ µ

µ

µb g b g

b g

b ge j

= +

= − =

− +

− +

z .

.

.

e− + =+ = −

=

µ

µ

0 04 0 952

004 0 952

0049

. .

. ln .

.

b g

b g

µ = 0 009.

d e dt

e e

t31 0 049

3

4

0 049 3 0 049 4

1000 0 009

100000090049

7 6

b g

b gb g b gb g

b g

e j

=

= − =

−

− −

z .

. .

.

.

..


2 70

2 70

0 701

1 01 01 01 05 0 2

0 020

1000 943 0 0 30

p

F p

F v F F q

τ

τ

b g

b g

b gb g b g

= − − − − =

= =

= = + =

. . . . .

( ) .

( ) .

F v F F q

F v F F q

F v F F q F

1100 979 943 040

1100 1038 979 050

1200 1068 1038 100 1

21 70

1

0 702

21 70

2

d i b g b gb g b gd i b g b g b g

b g

b g

b g

= = + =

= = + =

= = + = =

.

( ) .

. good; must have maximum possible

F(943) < random number < F(979), so choose 979


Let Zi be random variable indicating death; Wi be random variable indicating lapse for policy.Let U denote the random number used.

policy # 1: q100 0 40812= . from Illustrative Life TableU = <03 0 40812. . Z1 1= W1 0=

policy # 2: q91 0 20493= . from Illustrative Life TableU = >05 0 20493. . Z2 0=next checking lapse U = <01 015. . (surrender rate) ⇒ =W2 1

policy # 3 q96 030445= .U = >04 0 30445. . Z3 0=next checking lapse U W= > ⇒ =08 015 03. .

⇒ total Death and Surrender Benefits = 10+20+0 = 30


2

3

1 01 0 2 07

0 7 03 04

p

px

x

= − − =

= − =

. . .

. . .

Use lx = 1 (arbitrary, doesn’t affect solution)

so l lx x+ += =2 307 04. .

By hyperbolic 1

51

51

2 5 2 3l l lx x x+ + += +

.. .

= + =..

..

.57

54

19643

l px x+ = =2 5 2 505091. ..

2 5 1 05091 04909. . .qx = − =

Prob (all 3 failed) = =04909 01183. .b g


This is a graph of l xxµb g.µ xb g would be increasing in the interval 80 100,b g .

The graphs of l px x , lx and lx2 would be decreasing everywhere.

The graph shown is comparable to Figure 3.3.2 on page 65 of Actuarial Mathematics


Using the conditional mean and variance formulas:

E N E N= Λ Λc h

Var N Var E N E Var N= +Λ ΛΛ Λc hd j c hd jSince N, given lambda, is just a Poisson distribution, this simplifies to:

E N E= Λ Λb gVar N Var E= +Λ ΛΛ Λb g b gWe are given that E N = 0 2. and Var N = 04. , subtraction gives Var Λb g = 0.2


N = number of salmonX = eggs from one salmonS = total eggs.E(N) = 100tVar(N) = 900t

E S E N E X tb g b g b g= = 500

Var S E N Var X E X Var N t t tb g b g b g b g b g= + = ⋅ + ⋅ =2 100 5 25 900 23 000,

P S PS t

tt

t> =

−>

−FHG

IKJ = ⇒10 000

50023 000

10 000 50023 000

95,,

,,

.b g10 000 500 1645 23000 250

40 2

2 40 0

1 1 3204

4 73

22 4

2

, .

.

.

− = − ⋅ = −

− = −

− − =

=± +

=

=

t t t

t t

t t

t

t

d i

round up to 23


A P V (x’s benefits) = ++

=+∑v b p qk

k k xk

x k1

10

2

= + +

=

1000 300 0 02 350 098 004 400 098 0 96 006

36 829

2 3v v v. . . . . .

,

b g b gb g b gb gb g


π denotes benefit premium19V APV= future benefits - APV future premiums

0 61

1080326.

..= − ⇒ =π π

1110 65

65

108 10V

V qp

=+ −πb gb g b gb g.

=+ −

−5 0 0326 108 010 10

1 010. . . .

.b gb g b gb g

=5.28


X = losses on one lifeE X = + +

=03 1 0 2 2 01 3

1

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

S = total lossesE S E X

E S E S F

E S f

s

s

= =

− = − −

= − −

= − −

= −=

+

3 3

1 1 1 0

1 1 0

3 1 1 04

3 0 936

2 064

3

b g b gc hb g b gc h

b gd i.

.

.


M r E e

e e e

Me e e

xrx

r r r

x

b g

b g

=

= + +

= + + =

2 3

0 5 1 53

053

2 95. .. .

p E X11 2 3

32= = + + =

λ M r crxb g − =1

Since λ = 2 and r = 0.5,2 05 1 05

2 2 95 1 05

39 05

7 8

M c

c

c

c

x . .

. .

. .

.

b gb g

− =

− === = premium rate per period


Simple’s surplus at the end of each year follows a Markov process with four states:State 0: out of businessState 1: ending surplus 1State 2: ending surplus 2State 3: ending surplus 3 (after dividend, if any)

State 0 is absorbing (recurrent). All the other states are transient states.Thus eventually Simple must reach state 0.


(See solution to problem 21 for definition of states).

t = 0

0 0 0 1

10 0 0 000 000

01 05 025 015

01 0 0 050 0 40

0 0 01 000 090

00 01 00 09 1

. . . .

. . . .

. . . .

. . . .

. . . .

L

N

MMMM

O

Q

PPPP= =at t

t = 1

00 01 0 0 09

10 00 0 00 000

01 05 0 25 015

01 00 050 0 40

00 01 0 00 090

0 01 014 0 025 0825 2. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

L

N

MMMM

O

Q

PPPP= =at t

Expected dividend at the end of the third year =

k=∑

0

3

(probability in state k at t = 2) × (expected dividend if in state k)

0.01*0 + 0.14*0 + 0.025(0*0.85 + 1*0.15) + 0.825*(0*0.6 +1*0.25 + 2*0.15) = 0.4575


1180 70 50 20

1180 70 12 50 10 20

8

12 10 8 14

100 1400

30 40 30 40

30 40

30 40

30 40 30 40 30 40

30 40

= + −= + −== + − = + − =

=

a a a

a

a

a a a a

a

:

:

:

: :

:

b gb g b gb g


a a f t dtt

= ¬∞zo b g =

− −∞ −z 10 05

12

0 05ete dt

tt

.

. Γb go

= −−∞ −z10 05

105

..te t tte dtb i

o

= − + + +FHG

IKJ

LNM

OQP

− −∞

10 05

1105

11052

1 05

0. . ..t e

tet tb g

= − FHG

IKJ

LNMM

OQPP

=10 05

11

105185941

2

. ..

20 000 185941 37 188, . ,× =


p kk

p k

kp k

b g b g

b g

= −

= +LNM

OQP −

21

02

1

Thus an (a, b, 0) distribution with a = 0, b = 2.

Thus Poisson with λ = 2 .

pe

42

40 09

2 4

b g =

=

−

!.


By the memoryless property, the distribution of amounts paid in excess of 100 is stillexponential with mean 200.

With the deductible, the probability that the amount paid is 0 is F e100 1 0 393100 200b g = − =− / . .

Thus the average amount paid per loss is (0.393) (0) + (0.607) (200) = 121.4

The expected number of losses is (20) (0.8) = 16.

The expected amount paid is (16) (121.4) = 1942.


From UDD ll l

96 596 97

2. = +

480600

236097

97= + ⇒ =ll

Likewise, from l l97 97 5360 288= =and . ,we get l98 216=

For constant force,

ell

p e

l l

−

−

= = =

= = =

= = =

µ

µ

98

97

0 5 975

97 5 97

216360

0 6

0 6 07746

07746 0 7746 360 27886

12

.

. .

. . ..

.

.

b gb g b gb g


Let M = the force of mortality of an individual drawn at random; and T = future lifetime of theindividual.

Pr T ≤ 1 = ≤E T MPr 1n s= ≤ =

∞z Pr T M f dM10

µ µ µb g= −zz µ µµe dt dt 1

20

1

0

2

= − = + − = +− − −z 112

12

2 112

12 20

2e du e eµd i d i d i

= 056767.


E N

E N

E X

= + =

= + =

= − =

= + =

08 1 02 2 12

08 1 0 2 4 16

16 12 016

70 100 170

2

2

. . .

. . .

. . .

b gb g b gb gb g b gb g

b gVar N

Var Xb g b g= − = + − =E X E X2 2 27000 100 000 170 78100, ,

E S E N E X= = =12 170 204. b gVar S Var X Var Nb g b g b g b g b g= + = + =E N E X 2 212 78100 170 016 98 344. , . ,

Std dev Sb g = =98 344 3136, .So B = 204 + 314 = 518


f

f

f

s

s

s

1000 08 01 0 2 2 02 01 0088

1100 02 2 07 01 0028

2000 02 01 0 0022

b g b gb g b gb gb gb gb g b gb gb gb gb g b gb g

= + =

= =

= =

. . . . . .

. . . .

. . .

E S − = + +

=+200 0 088 800 0028 900 0002 1800

99 2

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

.With 175% relative security loading, cost = (2.75) (99.2) = 272.8

Alternatively,

f F

f

F

s s

s

s

0 0 08 0 2 02 02 0168

100 08 07 02 2 02 07 0 616

100 0168 0616 0 784

2b g b g b gb g b gb gb g b gb g b gb gb gb gb g

= = + =

= + =

= + =

. . . . .

. . . . . .

. . .

E S = 204 [from problem 29]

E S E S F

E S F F

s

s s

− = − − −

= − − − −

= − − − −=

+ +200 100 100 1 100

100 1 0 100 1 100

204 100 1 0168 100 1 0 784

99 2

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

.cost = (2.75) (99.2) = 272.8


Let π = benefit premium

Actuarial present value of benefits == + += + +=

003 200 000 0 97 006 150 000 097 094 0 09 100 000

566038 7769 67 689008

20 32013

2 3. , . . , . . . ,

. . .

, .

b gb g b gb gb g b gb gb gb gv v v

Actuarial present value of benefit premiums=

= + +

=

= =

&&

. . .

.

, ..

.

:a

v v

x 3

21 0 97 097 0 94

2 7266

20 320132 7266

745255

π

π

π

π

b gb g

1745255 106 200 000 0 03

1 003195846

V =−

−=

. . , .

..

b gb g b gb g

Initial reserve, year 2 = 1V + π = 1958.56 + 7452.55 = 9411.01


Let π denote the premium.

L b v a i v a

a

TT

TT T

T

T

= − = + × −

= −

π π

π

1

1

b g

E L ax ax= − = ⇒ =1 0 1π π

⇒ = − = − =− −

=− −

= −−

L aa

a

a v

a

v a

av A

A

TT

x

xT

xT

x

x

Tx

x

1 11

1

1

πδ

δ

δδ

d i

b g


1 1

2 1

1 01 0 9

09 1 005 0855

p

p

= − == − =

( . ) .

. . .b gb g

since uniform, 1 5 1 09 0855 2

08775. . . /

.

p = +=b g

e t to1 1 5 0 15

1 0 92

10 9 08775

205

0 95 0 444

1394

: . .

. . ..

. .

.

= = =

= +FHG

IKJ + +F

HGIKJ

= +=

Area between and

b g b g

Alternatively,

e p dt

p dt p p dx

t dt x dx

t x

t

t x

t x

o11 5 10

1 5

10

11 1 20

0 5

0

1

0

0 5

0 12

0

10 05

20

0 5

1 01 09 1 0 05

09

095 0 444 1394

2 2

: ..

.

.

. ..

. . .

.

. . .

=

= +

= − + −

= − + −

= + =

zz zz zb g b g

tp1

t1 2

(1.5, 0.8775)

(2, 0.885)

(0, 1) (1, 0.9)


10 000 112 523363, .A b g =A

AA i q

p

A

A

xx x

x

63

1

64

65

0 4672

1

04672 105 0017881 0 01788

04813

04813 105 0 01952

1 00195204955

=

=+ −

=−

−=

=−

−=

+

.

. . ..

.

. . .

..

b g

b gb g

b gb g

Single contract premium at 65 = (1.12) (10,000) (0.4955) = 5550

155505233

55505233

1 0 029842+ = = − =i ib g .


Original Calculation (assuming independence):

µµ

µ

x

y

xy

==

= + =

006

006

0 06 006 012

.

.

. . .

A

A

xx

x

yy

y

=+

=+

=

=+

=+

=

µµ δ

µµ δ

0060 06 0 05

054545

0 06006 0 05

054545

.. .

.

.. .

.

A

A A A A

xyxy

xy

xy x y xy

=+

=+

=

= + − = + − =

µµ δ

012012 005

0 70588

054545 054545 070588 038502

.. .

.

. . . .

Revised Calculation (common shock model):

µ µ

µ µx x

T x

y yT y

= =

= =

006 0 04

006 004

. , .

. , .

*

*

b g

b g

µ µ µ µxy xT x

yT y Z= + + + + + =* * . . . .b g b g 004 0 04 0 02 010

A

A

xx

x

yy

y

=+

=+

=

= =+

=+

µµ δµ

µ δ

0060 06 005

0 54545

0 06006 005

054545

.. .

.

.. .

.

A

A A A A

xyxy

xy

xy x y xy

=+

=+

=

= + − = + − =

µµ δ

010010 005

0 66667

054545 054545 066667 042423

.. .

.

. . . .

Difference = − =0 42423 038502 003921. . .


Treat as three independent Poisson variables, corresponding to 1, 2 or 3 claimants.

rate

rate

rate2

3

1 612

12

4

2

= = ×LNM

OQP

==

Var

Var

Var

1

2

3

=

= = ×

=

6

16 4 2

18

2

total Var = 6 16 18 40+ + = , since independent.

Alternatively,

E X 22 2 212

23

36

103

d i = + + =

For compound Poisson, Var S E N E X= 2

= FHG

IKJ =12

103

40b g


λ λt dt Nb g b g= =z 6 3 60

3so is Poisson with .

P is Poisson with mean 3 (with mean 3 since Prob yi < =500 05b g g.

P and Q are independent, so the mean of P is 3, no matter what the value of Q is.


At age x :

Actuarial Present value (APV) of future benefits = 15

1000AxFHG

IKJ

APV of future premiums = 45

&&axFHG

IKJ π

10005

4525 25A a= π && by equivalence principle

10004

14

816516 2242

125825

25

Aa&&

..

.= ⇒ = × =π π

10V = APV (Future benefits) – APV (Future benefit premiums)

= −

= −

=

10005

45

15

128 7245

1258 153926

1025

35 35A aπ &&

. . .

.

b g b gb g


Let Y = present value random variable for payments on one lifeS Y= =∑ present value random variable for all payments

E Y a= =10 14816640&& .

Var YA A

d=

−

= −

=

10

100 004863 016132 106 006

70555

22

40 402

2

2 2

d i

d ib g. . . / .

.

E S E Y

S Y

= =

= =

100 14 816 6

100 70 555

, .

,Var Var

Standard deviation S = =70 555 26562, .

By normal approximation, needE [S] + 1.645 Standard deviations = 14,816.6 + (1.645) (265.62)

= 15,254


Initial Benefit Prem =−

−

5 4

5 430 30 20

1

30 35 30 20

A A

a a:

: :&& &&

e j

=

−−

= −−

= =

5 010248 4 002933

5 14 835 4 11959

05124 01173274175 47 836

03950826339

0015

. .

. .

. .. .

..

.

b g b gb g b g

Where

A A A30 201

30 20 30 201 032307 029374 002933: : : . . .= − = − =e j

and

&&...

.::a

A

d30 2030 201 1 0 32307

0 06106

11959=−

= −FHG

IKJ

=

Comment: the numerator could equally well have been calculated as A E A30 20 30 504+= 0.10248 + (4) (0.29374) (0.24905)= 0.39510

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COURSE 3 MULTIPLE-CHOICE ANSWER KEY - SOA · 14 b 39 e 15 a 40 b 16 b 17 a 18 e 19 c 20 c 21 e 22 d 23 a 24 b 25 c course 3 may 2001 multiple-choice answer key. solutions for may