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Faculty of Actuaries Institute of Actuaries

EXAMINATION

26 April 2010 (am)

Subject CT5 Contingencies

Core Technical

Time allowed: Three hours

INSTRUCTIONS TO THE CANDIDATE

1. Enter all the candidate and examination details as requested on the front of your answer

booklet.

2. You must not start writing your answers in the booklet until instructed to do so by the

supervisor.

3. Mark allocations are shown in brackets.

4. Attempt all 14 questions, beginning your answer to each question on a separate sheet.

5. Candidates should show calculations where this is appropriate.

Graph paper is NOT required for this paper.

AT THE END OF THE EXAMINATION

Hand in BOTH your answer booklet, with any additional sheets firmly attached, and this

question paper.

In addition to this paper you should have available the 2002 edition of the Formulae

and Tables and your own electronic calculator from the approved list.

Faculty of Actuaries

CT5 A2010 Institute of Actuaries

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CT5 A20102

1 Explain what the following represent:

(a) [ ]x rl +

(b) |n m xq

(c)x

d

[3]

2 Define spurious selection, giving two distinct examples. [3]

3 Calculate the standardised mortality ratio for the population of Urbania using thefollowing data:

Standard Population Urbania

Age Population Deaths Population Deaths

60 2,500,000 26,170 10,000 130

61 2,400,000 29,531 12,000 145

62 2,200,000 32,542 11,000 173

[3]

4 A life insurance company offers an increasing term assurance that provides a benefitpayable at the end of the year of death of 10,000 in the first year, increasing by 100 on

each policy anniversary.

Calculate the single premium for a five year policy issued to a life aged 50 exact.

Basis:

Rate of interest 4% per annum

Mortality AM92 Select

Expenses Nil

[4]

5 A population is subject to the force of mortality x= e0.0002x1.

Calculate the probability that a life now aged 20 exact:

(i) survives to age 70 exact [2]

(ii) dies between ages 60 exact and 70 exact [3]

[Total 5]

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CT5 A20103 PLEASE TURN OVER

6 You are provided with the following extract from a life table:

x lx

50 99,813

51 97,702

52 95,046

Calculate 0.75p50.5using two different methods. [5]

7 A company is about to establish a pension scheme that will provide an age retirementbenefit of n/60ths of final pensionable salary where n is total number of years of

service. Final pensionable salary is the average salary in the three years before

retirement.

An employee who will become a member of the pension scheme is currently aged 55

exact has and will be granted exactly 20 years of past service. The employees salary

in the year before the valuation date was 40,000.

(i) Calculate the present value of benefits for this member (including future

service). [3]

(ii) Calculate the contribution required to fund this benefit as a percentage of

future salaries. [3]

Basis:

Pension Scheme from the Formulae and Tables for Actuarial Examinations

[Total 6]

8 100 graduates aged 21 exact decide to place the sum of 1 per week into a fund to beshared on their retirement at age 66 exact.

(i) Show that each surviving member can expect to receive on retirement a fund

of approximately 7,240. [4]

Basis:


Mortality AM92 Ultimate

One of the survivors uses the accumulated fund to buy a weekly annuity payable for

10 years certain. After 10 years the annuity is payable at two-thirds of the initial level

for the rest of life.

(ii) Calculate the weekly amount of the annuity on the basis used in part (i). [2]

[Total 6]

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CT5 A20104

Active(A) Retired (R)

Dead (D)

xx

x

9 A life insurance company models the experience of its pension scheme contractsusing the following three-state model:

(i) Derive the dependent probability of a life currently Active and agedxretiring

in the year of agexto (x + 1) in terms of the transition intensities. [2]

(ii) Derive a formula for the independent probability of a life currently Active and

agedxretiring in the year of agexto (x + 1) using the dependent probabilities.

[4]

[Total 6]

10 The decrement table extract below is based on the historical experience of a very largemultinational companys workforce.

Age (x) Number of employees

( )xal

Deaths

( )dxad

Withdrawals

( )wxad

40 10,000 25 120

41 9,855 27 144

42 9,684

Recent changes in working conditions have resulted in an estimate that the annual

independent rate of withdrawal is now 75% of that previously used.

Calculate a revised table assuming no changes to the independent death rates, stating

your results to one decimal place. [7]

11 Thieles differential equation for the policy value at duration t (t> 0), t xV , of an

immediate life annuity payable continuously at a rate of 1 per annum from agexis:

1t x t x t xx tV V Vt

+

= +

(i) Derive this result algebraically showing all the steps in your working. [5]

(ii) Explain this result by general reasoning. [3]

[Total 8]

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12 On 1 January 2005, a life insurance company issued 1,000 10-year term assurancepolicies to lives aged 55 exact. For each policy, the sum assured is 50,000 for the

first five years and 25,000 thereafter. The sum assured is payable immediately on

death and level annual premiums are payable in advance throughout the term of this

policy or until earlier death.

The company uses the following basis for calculating premiums and reserves:


Interest 4% per annum

Expenses Nil

(i) Calculate the net premium retrospective reserve per policy as at 31 December

2009. [6]

(ii) (a) Give an explanation of your numerical answer to (i) above.

(b) Describe the main disadvantage to the insurance company of issuing

this policy.

(c) Give examples of how the terms of the policy could be altered so as to

remove this disadvantage.

[3]

There were, in total, 20 deaths during the years 2005 to 2008 inclusive and a further 8

deaths in 2009.

(iii) Calculate the total mortality profit or loss to the company during 2009. [3]

[Total 12]

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CT5 A20106

13 A life insurance company issues a 3-year unit-linked endowment assurance policy to amale life aged 45 exact.

Level premiums of 4,000 per annum are payable yearly in advance throughout the

term of the policy or until earlier death. 95% of the premium is allocated to units in

the first policy year, 100% in the second and 105% in the third. A policy fee of 50 is

deducted from the bid value of units at the start of each year. The units are subject to

a bid-offer spread of 5% on purchase. An annual management charge of 1.75% of thebid value of units is deducted at the end of each policy year.

Management charges are deducted from the unit fund before death, surrender and

maturity benefits are paid.

If the policyholder dies during the term of the policy, the death benefit of 125% of the

bid value of the units is payable at the end of the policy year of death. On maturity,

100% of the bid value of the units is payable.

The policyholder may surrender the policy only at the end of the first and second

policy years. On surrender, the bid value of the units less a surrender penalty ispayable at the end of the policy year of exit. The surrender penalty is 1,000 at the

end of the first policy year and 500 at the end of the second policy year.

The company uses the following assumptions in carrying out profit tests of this

contract:

Rate of growth on assets in the unit fund 5.5% per annum in year 1

5.25% per annum in year 2

5.0% per annum in year 3

Rate of interest on non-unit fund cash flows 4.0% per annum

Mortality AM92 SelectInitial expenses 200

Renewal expenses 50 per annum on the second and third

premium dates

Initial commission 15% of first premium

Renewal commission 2.0% of the second and third years

premiums

Rate of expense inflation 2.0% per annum

Risk discount rate 7.0% per annum

For renewal expenses, the amount quoted is at outset and the increases due to inflation

start immediately. In addition, you should assume that at the end of the first and

second policy years, 12% and 6% respectively of all policies still in force then

surrender immediately.

(i) Calculate the profit margin for the policy. [13]

(ii) Calculate the expected present value of profit for the policy if the company

assumed that there were no surrenders at the end of each of the first and

second policy years. [3]

[Total 16]

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CT5 A20107

14 A life insurance company issues a 30-year with profits endowment assurance policyto a life aged 35 exact. The sum assured of 100,000 plus declared reversionary

bonuses are payable on survival to the end of the term or immediately on death if

earlier.

(i) Show that the quarterly premium payable in advance throughout the term of

the policy or until earlier death is approximately 616.

Pricing basis:

Mortality: AM92 Select

Interest: 6% per annum

Initial commission: 100% of the first quarterly premium

Initial expenses: 250 paid at policy commencement date

Renewal commission: 2.5% of each quarterly premium from the start of the

second policy year

Renewal expenses: 45 at the start of the second and subsequent policy

years

Claim expense: 500 on death; 250 on maturityFuture reversionary bonus: 1.92308% of the sum assured, compounded and vesting

at the end of each policy year (i.e. the death benefit does

not include any bonus relating to the policy year of

death)

[10]

At the end of the 25thpolicy year, the actual past bonus additions to the policy have

been 145,000.

(ii) Calculate the gross prospective policy reserve at the end of that policy year

immediately before the premium then due.

Policy reserving basis:

Mortality: AM92 Ultimate

Interest: 4% per annum

Bonus loading: 4% of the sum assured and attaching bonuses,

compounded and vesting at the end of each policy year

Renewal commission: 2.5% of each quarterly premium

Renewal expenses: 90 at the start of each policy year

Claim expense: 1,000 on death; 500 on maturity

[6][Total 16]

END OF PAPER

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EXAMINERS REPORT

April 2010 Examinations


Core Technical

Introduction

The attached subject report has been written by the Principal Examiner with the aim of

helping candidates. The questions and comments are based around Core Reading as the

interpretation of the syllabus to which the examiners are working. They have however given

credit for any alternative approach or interpretation which they consider to be reasonable.

R D Muckart

Chairman of the Board of Examiners

July 2010

Comments

These are given in italics at the end of each question.

Faculty of ActuariesInstitute of Actuaries

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Subject CT5 (Financial Mathematics Core Technical) April 2010 Examiners Report

Page 2

1 (i) The number of lives still alive at agex + rout of lxlives alive at agexsubjectto select mortality.

(ii) The probability that a life agex will die between agex + nandx + n + m.

(iii) The number of lives that die betweenxand (x + 1) out of xl lives alive atx.

Question generally answered well.

2 Spurious selection occurs when mortality differences ascribed to groups are formedby factors which are not the true causes of these differences.

For example mortality differences by region may be put down to the actual class

structure of the region itself whereas a differing varying mix of occupations region by

region could be having a major effect. So Region is spurious and being confoundedwith occupation.

Another example might be in a company pension scheme which might be showing a

significant change in mortality experience which could be viewed as change over

time. However withdrawers from the scheme may be having an effect as their

mortality could be different. To that degree Time Selection may be spurious.

Question generally answered well. Credit was given for a wide range of valid examples.

3 The Standardised mortality ratio is the ratio of actual deaths in the population dividedby the expected number of deaths in the population if the population experienced

standard mortality.

Actual number of deaths for Urbania = 130+145+173 = 448

Mortality rates in standard population are:

Age 60: 26,170 / 2,500,000 = 0.0104680

Age 61: 29,531 / 2,400,000 = 0.0123046

Age 62: 32,542 / 2,200,000 = 0.0147918

Expected number of deaths for Urbania

= 0.010468 10,000 + 0.0123046 12,000 + 0.0147918 11,000 = 415

SMR = 448/415 = 107.95%

Question generally answered well.

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Page 3

4 1 1[50]:5 [50]:5

(10,000 100) 100( )EPV A IA= +

5 5[50] 5 [50] 55 [50] 5 [50] 55 559,900( ) 100(( ) (5 ( ) ))A v p A IA v p A IA= + +

5 9557.81799,900(0.32868 *0.38950)9706.0977

v=

5 9557.8179100* 8.5639 (5*0.38950 8.57976)9706.0977

v + +

132.96 4.34= +

137.30=

Many students answered the question well. The most common error was the use of 10,000 asthe multiplier before the temporary assurance function rather then 9,900.

5 exp( )x t

t x sx

p ds+

=

0.0002exp( ( 1) )x t S

xe ds

+=

0.0002exp( )x t x tSx x

e ds ds+ += +

0.0002( ) 0.0002

exp0.0002

x t xe e

t

+ = +

(i) Probability =

0.0002 70 0.0002 20

exp 500.0002

x xe e

= +

0.6362=

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Page 4

(ii) This is the probability that the life survives to 60 and then dies between 60 and

70

40 20 10 60Probability (1 )p p=

0.0002 60 0.0002 20 0.0002 70 0.0002 60

exp (60 20) . 1 exp (70 60)0.0002 0.0002

x x x xe e e e = + +

0.725 (1 0.8773)x=

0.0889=

This question was answered poorly overall. It was an unusual representation of the x

function but other than that was a straight forward probability and integration question.

6 p50= 97,702 / 99,813 = 0.978850p51= 95,046 / 97,702 = 0.972815

Uniform distribution of deaths

50 0.25 51 50 51

0.5 50 50

(1 0.25(1 )) 0.978850* (1 0.25*(1 0.972815))0.982588

(1 0.5(1 )) (1 0.5*(1 0.978850))

p p p p

p p

= = =

Constant force of mortality

t= ln(pt)

50 = ln(0.978850) = 0.021377

51= ln(0.972815) = 0.027561

0.5*0.021377 0.25*0.0275610.5 50 0.25 51* * 0.989368*0.993133 0.982574p p e e

= = =

Generally answered well. A limited number of students used the Balducci Assumption as oneof their answers. This is not in the CT5 Course whilst the above 2 methods clearly are. This

method was however credited solution not published as not in CT5

7 (i) Age retirement benefit

5555

54 55

(20 )140,000

60

z raz raM R

s D

+

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Page 5

1 (20*128,026 963,869)40,000

60 9.745*1,389

+=

173,584=

(ii) Contributions

55

54 55

40,000.

sN

Ks D

88,615.40,000

9.745*1,389K x=

261,868K=

Therefore K= 173,584 / 261,868 i.e. 66.3%

Most students answered reasonably well. Most common error was the wrong sxfunction.

Also some students included early retirement calculations which were not asked for.

Also students often did not include the past service benefits in the final contribution rate

believing the final result would have been too high (the question however was quite specific

on providing past benefits).

8 (i)(66 21)

21:45

45 21

1.04Fund 52*

a

p

=

4566 2121:45 21:45 21:45

1 1 8695.6199*(1 * / ) * 1 0.17120*

2 2 9976.3909a a v l l a

= =

21:45 0.42539a=

44 65 2121:45 21:44

8821.2612* / 21.045 .17805* 21.2029976.3909a a v l l= + = + =

21:4520.777a =

( )

4552*1.04 (20.777)therefore fund 7,240

8695.61999976.3909

= =

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Page 6

(ii) Let annuity be P per week. Then EPV of annuity at 66 is

1010 66 7610

252 ( * . )3

P a v p a+

10(1 ) 6589.9258252 *0.675564* (8.169 0.5)3ln(1.04) 8695.6199

vP = +

[ ]52 8.272 2.618]P= +

566.26P=

Therefore pension is given by

7,240 566.26P=

12.79P=

Many students struggled with this question and indeed a large number did not attempt it. As

will be seen from the solution above the actuarial mathematics involved are relatively

straightforward.

Note that 52.18 (i.e. 365.25/7) would have been an acceptable alternative to 52 as the

multiplier which will of course have adjusted the answer slightly.

9 (i) We are looking to derive ( )rxaq in terms of xand x

Use the Kolmogorov equations (assuming the transition intensities are

constant across a year age):

( )

( )

( )

( ) (1 )( )

r tt x

rx

aq et

aq e

+

+

=

= +

(ii) Similarly

( )( ) (1 )( )

dxaq e

+= +

Note that:

( )1 (( ) ( ) )r dx xaq aq e + + =

log(1 (( ) ( ) ))r dx xaq aq + = +

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Page 7

So

( ) (( ) ( ) )( log(1 (( ) ( ) )))

r r dx x xr d

x x

aq aq aqaq aq

= +

+

this can be rearranged to show

( )log(1 (( ) ( ) ))

( ) ( )

rr dxx xr d

x x

aqaq aq

aq aq = +

+

Given that:

1 ,rxq e=

then

( )

(( ) ( ) )1 1 (( ) ( ) )

rx

r dx x

aqr r d aq aq

x x xq aq aq + = +

In general this was poorly answered with most students making a limited inroad to the

question.

However, the question did not specify that constant forces must be assumed. So, a valid

alternative to part (i) is:

( )1 1

0 0 0

( ) ( ) exp

trx t x x t x r x r x taq ap dt dr dt + + + +

= = +

This makes no assumptions and provides an answer in the form asked for in the question, and

so would merit full marks. If constant forces are assumed, the above expression will turn into

the answer in the above solution.

For part (ii) a solution is only possible if some assumption is made. The following

alternatives could be valid:

(1) Assume dependent decrements are uniformly distributed over the year of age

With this assumption, deaths occur on average at age x + , so:

12 1

2 12

( ) ( ) ( )( ) ( )

( ) 1 ( )

r d r rx x xr r d r r x

x x x x x dx x

ad ad q aqq aq aq q q

al aq

+ = = + =

(This is covered by the Core Reading in Unit 8 Section 10.1.3.)

(2) Assume independent decrements are uniformly distributed over the year of age

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Page 8

This leads to two simultaneous equations:

( )

1

dd x

x rx

aqq

q=

and

( )

1

rr x

x dx

aqq

q=

which results in a quadratic equation inr

xq . (This is covered by the Core Reading Unit 8

Section 10.1.6.)

Whilst a full description has been given above to assist students, in reality those who

successfully attempted this question did assume constant forces.

10 First calculate ( ) and ( )d wx xaq aq

Age (x) Number of

employees( )xal

( )dxaq ( )wxaq

40 10,000 .00250 .01200

41 9,855 .00274 .01461

42 9,684

From this table and relationship

1 1( ) / (1 *( ) ) and ( ) / (1 *( ) )

2 2

d d w w w d x x x x x xq aq aq q aq aq= =

Calculate andd wx xq q

40d

q =.00250/(1.006) = .00252 and 41d

q = .00274/(1.00731) = .00276

40w

q =.01200/(1.00125) = .01201 and 41w

q = .01461/(1.00137) = .01463

Adjusting for the 75% multiplier of independent withdrawal decrements:

40

41

40

41

1 3( ) .00252* 1 * *.01201 .00251

2 4

1 3( ) .00276* 1 * *.01463 .00274

2 4

3 1( ) .01201* * 1 *.00252 .00900

4 2

3 1( ) .01463* * 1 *.00276 .01096

4 2

d

d

w

w

aq

aq

aq

aq

= = = =

= =

= =

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Page 9

Using the above data the Table can now be reconstructed

Age (x) Number of

employees

( )xal

Deaths

( )dxad

Withdrawals

( )wxad

40 10,000 (10000*.00251)=25.1 10000*.00900=90.0

41 9,884.9 (9,884.9*.00274)=27.1 9,884.9*.01096=108.3

42 9749.5

It should be noted that if more decimal places are used in the aq factors then the deaths at 40

become 25.0 so full credit was given for this answer also.

Because of the limited effect on the answer from the original table students were asked to

show the result to 1 decimal place. Many failed to do so and were penalised accordingly.

11 (i) Policy value at duration t of an immediate annuity payable continuously at arate of 1 per annum and secured by a single premium at age x is given by:

0

st x x t s x t V a e p ds

+ += =

0 0

s s

t x x t s x t s x t

V a e p ds e p dst t t t

+ + +

= = =

1ln( ) (ln ln )s x t s x t x t s x t x t s x t

s x t

p p l lp t t t

+ + + + + + + ++

= = = +

( )s x t s x t x t s x t p pt

+ + + + +

= +

0

( )s

t x s x t x t x t s

V e p dst

+ + + +

=

0

sx t x t s x t x t sa e p ds

+ + + + +=

00

s sx t x t s x t s x ta e p e p ds

+ + + +

=

1x t x t x ta a+ + += +

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Page 10

1x t t x t xV V+= +

(ii) Consider a short time interval (t, t + dt) then equation implies:

1 ( )t dt t x t t x t xV V V dt dt V dt o dt + + = + +

where

x t t xV dt+ = reserve released as a result of deaths in time interval

(t, t + dt)

1 dt = annuity payments made in time interval (t, t + dt)

t xV dt = interest earned on reserve over time interval (t, t + dt)

In general very poorly answered on what was a standard bookwork question.

12 (i) Annual premium Pfor the term assurance policy is given by:

1 1

[55]:10 [55]:5

[55]:10

25,000 25,000A AP

a

+=

where

1 1

[55]:10 [55]:525,000 25,000A A+

( )1/2 10 5[55] 10 [55] 65 [55] 5 [55] 6025,000 (1 ) ( ) ( )i A v p A A v p A= + +

8821.2612(0.38879 0.67556 0.52786)

9545.992925,000 1.019804

9287.2164(0.38879 0.82193 0.4564)9545.9929

=

+

( )25,495.10 (0.38879 0.32953) (0.38879 0.36496) 2118.39= + =

Therefore

2118.39257.46

8.228P = =

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Page 11

Net Premium Retrospective Reserves at the end of the fifth policy year is

given by:

[55]5 1

[55]:5 [55]:5

60

(1 ) 50,000l

i Pa Al

+

[ ]9545.9929

1.21665 257.46 4.59 50,000 1.019804 (0.38879 0.36496)9287.2164

=

41.71=

(ii) Explanation more cover provided in the first 5 years than is paid for by the

premiums in those years. Hence policyholder in debt at time 5, with size of

debt equal to negative reserve.

Disadvantage if policy lapsed during the first 5 years (and possibly longer),the company will suffer a loss which is not possible to recover from the

policyholder.

Possible alterations to policy structure

Collect premiums more quickly by shortening premium payment term or make

premiums larger in earlier years, smaller in later years

Change the pattern of benefits to reduce benefits in first 5 years and increase

them in last 5 years.

(iii) Mortality Profit = EDS ADS

Death strain at risk = 50,000 (42) = 50,042

59(1000 20) 50,042EDS q= 980 0.00714 50,042 350,154= =

8 50, 042 400, 336ADS= =

Total Mortality Profit = 350,154 400,336 = -50,182 (i.e. a mortality loss)

Quite reasonably answered by the well prepared student.

In (i) it should be noted that in this case the retrospective and prospective reserves are equal.

If the student recognised this, explicitly stated so and then did the easier prospective

calculation full marks were given. No credit was given for a prospective calculation without

explanation.

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Page 12

13

Annual premium 4000.00 Allocation % (1st yr) 95.0%

Risk discount rate 7.0% Allocation % (2nd yr) 100.0%

Interest on investments (1st yr) 5.5% Allocation % (3rd yr) 105.0%Interest on investments (2nd yr) 5.25% B/O spread 5.0%

Interest on investments (3rd yr) 5.0% Management charge 1.75%

Interest on non-unit funds 4.0% Surrender penalty (1st yr) 1000

Death benefit (% of bid value of units) 125% Surrender penalty (2nd yr) 500

Policy Fee 50

% prem

Initial expense 200 15.0%Renewal expense 50 2.0%

Expense inflation 2.0%

(i) Multiple decrement table:

xd

xq s

xq

45 0.001201 0.12

46 0.001557 0.06

47 0.001802 0.00

x ( )dxaq ( )

sxaq ( )ap 1( )t ap

45 0.001201 0.11986 0.878943 1.000000

46 0.001557 0.05991 0.938536 0.878943

47 0.001802 0.00000 0.998198 0.824920

Unit fund (per policy at start of year)

yr 1 yr 2 yr 3

value of units at start of year 0.000 3690.074 7693.641

Alloc 3800.000 4000.000 4200.000B/O 190.000 200.000 210.000

policy fee 50.000 50.000 50.000

Interest 195.800 390.604 581.682

management charge 65.727 137.037 213.768

value of units at year end 3690.074 7693.641 12001.554

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Page 13

Cash flows (per policy at start of year)

yr 1 yr 2 yr 3

unallocated premium + pol fee 250.000 50.000 150.000

B/O spread 190.000 200.000 210.000

expenses 800.000 131.000 132.020Interest 14.400 4.760 2.881

man charge 65.727 137.037 213.768

extra death benefit 1.108 2.995 5.407

surrender penalty 119.856 29.953 0.000

end of year cashflow 189.926 287.755 133.461

probability in force 1 0.878943 0.824920

discount factor 0.934579 0.873439 0.816298

expected p.v. of profit 133.280

premium signature 4000.000 3285.769 2882.069

expected p.v. of premiums 10167.837

profit

Margin 1.31%

(ii) Revised profit vector (309.781, 257.802, 133.461)

Revised profit signature (309.781, 257.492, 133.093)

Revised PVFNP = 289.515 + 224.904+ 108.643 = 44.032

Again most well prepared students made a good attempt at this question. The most common

error was to ignore dependent decrements.

Substantial credit was given to students who showed how they would tackle this question even

if they did not complete all the arithmetical calculations involved.

14

(i) Let Pbe the quarterly premium. Then:

EPV of premiums:

(4)

[35]:304 @6% 56.1408Pa P=

where

( )(4) 3030 [35][35]:30 [35]:30 3 18a a p v=

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Page 14

3 8821.261214.352 1 0.17411

8 9892.9151

=

14.0352=

EPV of benefits:

0.5 1.5 29 29.5[35] [35] [35]1 29

100,000( (1 ) ... (1 ) )q v q b v q b v+ + + + +

30 3030 [35]100,000 (1 )b v p+ +

where b= 0.0192308

( )

0.52 2 30 30

[35] [35] [35]0.5 0.5 1 29

100,000 (1.06)

(1 ) (1 ) ... (1 )(1 ) (1 ) q b v q b v q b vb b= + + + + + ++ +

30 30

30 [35]100, 000(1 )b v p+ +

0.5 1 30

30 [35][35]:30

100,000(1.06) @ 100,000 @

(1 )A i v p i

b = +

+

0.5100,000 (1.06) 8821.26120.32187 0.30832

(1 ) 9892.9151b

= +

8821.2612100,000 0.30832

9892.9151+

4,742.594 27,492.112 32,234.706= + =

where

1.061 0.04

1i

b= =

+

EPV of expenses (at 6%)

(4) (4)

[35]:30 [35]:1 [35]:30250 0.025 4 0.025 4 45 1P Pa Pa a = + + +

1 30

30 [35][35]:30500 250A v p+ +

250 0.025 56.1408 0.025 4 0.97857 45 13.352P P P= + + +

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Page 15

0.5 8821.2612 8821.2612500 1.06 0.18763 0.17411 250 0.174119892.9151 9892.9151

+ +

2.30566 906.322P= +

where

( )(4) [35][35]:1[35]:13

18

a a p v=

3 9887.20691 1 0.9434 0.97857

8 9892.9151

= =

Equation of value gives:

56.1408 32, 234.706 2.30566 906.322P P= + +

33,141.028615.60

53.8351P = =

(ii) Gross prospective policy value (calculated at 4%) is given by:

1prospective 1/2 5 (4) (4)

5 60 60:560:5 60:560:5

245,000(1 ) @ 245,000 @ 0.025 4 90 4

(1 )V i A i v p i Pa a Pa

b = + + + +

+

1 5

5 6060:51000 500A v p+ +

(4)1 5 6560:5 60:50.5 60:5

60

245,000@ 245,000 @ 90 0.975 4

(1.04)

lA i v i a Pa

l = + +

0.5 5 565 6560:5

60 60

1000 1.04 500l l

A v vl l

+ +

(4) 5 65

60:560:5 60

3 3 8821.2612where 1 4.55 1 0.82193 4.4678

8 8 9287.2164

la a v

l

= = =

4

601 060:5

60

1.04 465.9551and 1 0 @ 0.05017

1.04 9287.2164

td

i A il

+

= = = = =

0.5

245,0000.05017 245, 000 0.94983 90 4.55 0.975 4 615.60 4.4678

(1.04)= + +

( )

0.51000 1.04 0.82499 0.78069 500 0.78069+ +

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Page 16

12,052.954 232,708.35 409.5 10,726.473 45.177 390.345 234,880= + + + + =

Part (i) answered reasonably well. Students had more problems with (ii)

END OF EXAMINERS REPORT

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EXAMINATION

6 October 2010 (am)


Core Technical




booklet.


supervisor.







question paper.



Faculty of ActuariesCT5 S2010 Institute of Actuaries

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CT5 S20102

1 Calculate:

(a) 20|10 [45]q

(b) 30 [45]:[50]p

Basis: AM92 Select[3]

2 Calculate 0.5p45.75using the Uniform Distribution of Deaths assumption.

Basis: AM92 Ultimate[3]

3 Calculate the single premium payable for a temporary reversionary annuity of12,000 per annum payable monthly in arrear to a female life currently aged 55 exacton the death of a male life currently aged 50 exact. No payment is made after 20years from the date of purchase.

Basis:

Rate of interest 4% per annumMortality of male life PMA92C20Mortality of female life PFA92C20Expenses Nil

[4]

4 A gymnasium offers membership for a three-year period at a fixed fee of 240 perannum payable monthly in advance. The contract may only be cancelled at a renewalanniversary. Monthly premiums cease immediately on the death of the member.

Calculate the expected present value of membership fees if the gymnasium sells 120memberships:

Basis:

Rate of interest 6% per annumRate of mortality 1% per annumProbability of renewal 80% at each anniversaryExpenses Nil

[5]

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CT5 S20103 PLEASE TURN OVER

5 A pension scheme provides an age retirement benefit of n/80ths of final pensionablesalary where n is total number of years of service. Final pensionable salary is theaverage salary in the three years before retirement. Normal retirement age is 65 andage retirement is only permitted between ages 60 and 65 exact.

A member of the pension scheme currently aged 45 exact has 12 years of service andtheir salary in the year before the valuation date was 25,000.

Give a formula for the expected cashflows between the 66thand 67thbirthdays as aresult of entitlement from this past service. [5]

6 Calculate:

(a)30:40

A

(b)30:40:20

a

Basis:

= 0.01 throughout for the life aged 30 now

= 0.02 throughout for the life aged 40 now

= 4% per annum[6]

7 A life insurance company issues a 10-year term assurance policy to a life aged 55

exact. The sum assured which is payable immediately on death is given by theformula:

50,000 (1 0.1 ) 0,1,2........,9t t + =

where tdenotes the curtate duration in years since the inception of the policy.

Level premiums are payable monthly in advance throughout the term of the policy oruntil earlier death.

Calculate the monthly premium for this policy using the following basis:

Mortality AM92 SelectInterest 4% per annumExpenses Nil

[6]

8 Describe the causal factors that explain observed differences in mortality andmorbidity. [6]

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CT5 S20104

9 The actuary advising a pension scheme has decided that the independent mortality inthe standard table for pension schemes (PEN) from page 142 of the Formulae andTables for Actuarial Examinations is no longer appropriate for that pension scheme.

Calculate the revised row of the service table for age 61, assuming that the revisedindependent mortality rate at that age is 80% of the previous independent mortalityrate.

[7]

10 Define the following terms, giving formulae and defining all notation used:

(a) Crude mortality rate(b) Indirectly standardised mortality rate

[7]

11 A life insurance company issues a four-year unit-linked policy to a male life. Thefollowing non-unit cash flows, tNUCF(t= 1,2,3,4), are obtained at the end of each

year tper policy in force at the start of the year t:

Year t 1 2 3 4

tNUCF 50.2 43.1 32.1 145.5

Assume that the annual mortality rate for the male life is constant at 1% at all ages.

(i) Show that the annual internal rate of return is 6%. [3]

The company sets up reserves in order to zeroise future negative cash flows. The rateof interest earned on non-unit reserves is 2.5% per annum.

(ii) Calculate the net present value of the profits after zeroisation using a riskdiscount rate of 6% per annum. [3]

(iii) Comment on the results obtained in (i) and (ii) above. [1][Total 7]

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CT5 S20105 PLEASE TURN OVER

12 A life insurance company issued a with profits whole life policy to a life aged 40exact on 1 January 2000. Under the policy, the basic sum assured of 50,000 andattaching bonuses are payable immediately on death. Level premiums are payableannually in advance under the policy until age 65 or earlier death.

The company declares simple reversionary bonuses at the start of each year includingthe first year and the bonus entitlement on the policy is earned immediately the bonus

is declared.

(i) Give an expression for the gross future loss random variable under the policyat the outset, defining symbols where necessary. [4]

(ii) Calculate the annual premium using the following assumptions:

Mortality AM92 SelectInterest 6% per annumBonus loading 2.5% per annum simpleInitial expenses 300

Renewal expenses 25 at the start of the second and subsequent policyyears while the policy is in force

Claim expenses 250[4]

On 31 December 2009, the policy is still in force. Bonuses declared to date total13,750.

(iii) Calculate the gross premium prospective reserve for the policy as at31 December 2009 using the following assumptions:

Mortality AM92 UltimateInterest 4% per annumBonus loading 3% per annum simpleRenewal expenses 35 at the start of each policy year while the policy is in

forceClaim expenses 250

[4][Total 12]

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CT5 S20106

13 On 1 January 2009, a life insurance company issued 10,000 joint life whole lifeassurance policies to couples. Each couple comprised one male life aged 60 exact andone female life aged 55 exact when the policy commenced. Under each policy, a sumassured of 100,000 is payable immediately on the death of the second of the lives todie.

Premiums under each policy are payable annually in advance while at least one of the

lives is alive.

The life insurance company uses the following basis for calculating premiums and netpremium reserves:

Mortality PMA92C20 for the malePFA92C20 for the female

Interest 4% per annumExpenses Nil

(i) Calculate the annual premium payable under each policy. [4]

During the calendar year 2009, there was one claim for death benefit, in respect of apolicy where both the male and the female life died during the year. In addition, therewere 20 males and 10 females who died during the year.

(ii) Calculate the mortality profit or loss for the group of 10,000 policies for thecalendar year 2009. [10]

[Total 14]

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CT5 S20107

14 A life insurance company issues four-year without profits endowment assurancepolicies to male lives aged 56 exact. The sum assured is 21,500 payable on maturityor at the end of the year of death if earlier. Premiums of 5,000 are payable annuallyin advance throughout the term of the policy.

The company holds net premium reserves for these policies, calculated using AM92Ultimate mortality and interest of 4% per annum.

Surrenders occur only at the end of a year immediately before a premium is paid. Thesurrender value is 70% of the net premium reserve calculated at the time the surrendervalue is payable.

The company uses the following assumptions in carrying out profit tests of thiscontract:

Rate of interest on cash flows 4% per annumMortality AM92 SelectSurrenders 10% of all policies still in force at the end of each of

the first, second and third policy yearsInitial expenses 600Renewal expenses 45 per annum on the second and subsequent

premium datesRisk discount rate 6% per annum

Calculate the expected profit margin for this contract. [15]

END OF PAPER

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INSTITUTE AND FACULTY OF ACTUARIES

EXAMINERS REPORT

September 2010 examinations


Core Technical

Introduction

The attached subject report has been written by the Principal Examiner with the aim of

helping candidates. The questions and comments are based around Core Reading as theinterpretation of the syllabus to which the examiners are working. They have however givencredit for any alternative approach or interpretation which they consider to be reasonable.

T J BirseChairman of the Board of Examiners

December 2010

Institute and Faculty of Actuaries

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Subject CT5 (Contingencies Core Technical) September 2010 Examiners Report

Page 2

1 (a) 20|10 [45] 65 75 [45]( ) /

(8,821.2612 6,879.1673) / 9,798.0837 0.198212

q l l l=

= =

(b) 75 8030 [45]:[50][45] [50]

6,879.1673 5,266.46040.380951

9,798.0837 9,706.0977

l lp

l l

= = =

Question generally done well.

2 .5 45.75 .25 45.75 .25 46*p p p=

.25 45.75 45 45.25* / (1 .75* ) .25*.001465 / (1 .75*.001465)q q q= =

.000367 by UDD=

.25 46 46.25* .25*.001622 .000406q q= = =

.5 45.75Hence (1 .000367)*(1 .000406) .999227p = =

In general question done well. However many students did not appreciate the split in line 1

above and attempted to apply formula directly.

3 Value of Single Premium is:

( )( ) ( ) ( ) ( )( )

( ) ( )

( )

(12) (12)

55:20 50:55:20

20 2055 20 55 75 50:55 20 50:55 70:75

20

20

12 1,000

13 13 13 1312,000 24 24 24 24

8784.95513 1312,000 18.210 10.93324 249917.623

8784.955 921316.909 24 9917.623

a a

v p v p

v

v

=

=

( )38.134 138.792 249941.92312,000((17.668 4.201) (16.367 3.099))

2,388

=

=Many students struggled with how to break down the monthly annuity functions into those

which could then utilise the Tables. However question generally done well by well prepared

students.

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Page 3

4 The value of 1 per annum payable monthly for 1 year is

(12) (12)(12)1 :1:1

. 11/ 24(1 . )x x xx xx v p v p+= =

:1Where 1

x =

Therefore

(12)

:11 11 /24(1 0.99 /1.06) 0.96973

x = =

The probability of reaching the beginning of each year is :

Year 1=1

Year 2 = 0.99*0.8 = 0.792 Year 3 = 0.792 * 0.792 = 0.6273

The value is therefore

2120 240 0.96973 (1 0.792 /1.06 0.6273 / (1.06) ) + + 64,388=

This question was overall done very poorly with few students realising that the key

element to the calculation involved a one year annuity due payable monthly.

5 The formula is:

1945 0.5 45 66 65 65 66

44 45 45 0.5 44 45 6515

12 ( ) 12 ( )25000 25000

80 ( ) 80 ( )t t t

tt

z r rl z r rl

s l rl s l rl

+ + + +

+ +=

+

Question done very poorly. Many students attempted to use annuity functions

whereas the question sought was a pure cash flow one.

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Page 4

6 (a)

______.04 .01 .02 .02 .01

030:40

.05 .07 .06 .07

0

.05 .06 .07

0

.05

{ (1 )*.01 (1 )*.02}

{.01*( ) .02*( )}

(.01* .02* .03* )

.01 .0 *

.05

t t t t t

t t t t

t t t

t

A e e e e e dt

e e e e dt

e e e dt

e

= +

= +

= +

=

.06 .07

0

2 .03* *

.06 .07

(1/ 5 1/ 3 3 / 7) .10476

t te e

+

= + =

(b)20 .04 .01 .02

30:40:20 0

20 .070

20.07

0

1.4

* *

1

.07

(1/ .07) / .07) 10.763

t t t

t

t

a e e e dt

e dt

e

e

=

=

=

= =

Question generally done well.

7 Let Pbe the monthly premium. Then equating expected present value of premiumsand benefits gives:

(12) 1 1[55]:10 [55]:10[55]:10

12 45000 5000( )Pa A IA= +

where

( )

( ) ( )( )

(12) 1010 [55][55]:10[55]:10

1 0.5 10 0.510 [55][55]:10 [55]:10

1 0.5 1010 [55][55]:10 [55]

11 8821.26121 8.228 0.458 1 .67556 8.056

24 9545.9929

1.04 1.04 0.68354 0.62427 0.06044

( ) 1.04

a a v p

A A v p

IA IA v p

= = =

= = =

=

( )( )10 10 [55] 65650.5

10

1.04 (8.58908 0.62427 7.89442 10 0.62427 0.52786) 0.3728

45000 0.06044 5000 0.372812 568.99

8.056

47.42

IA v p A

P

P

= =

+ = =

=

In general question done well by well prepared students.

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Page 5

8 Occupation either because of environmental or lifestyle factors mortality may bedirectly affected. Occupations may also have health barriers to entry, e.g. airlinepilots

Nutrition poor quality nutrition increases morbidity and hence mortality

Housing standard of housing (reflecting poverty) increases morbidity

Climate climate can influence morbidity and may also be linked to natural disaster

Education linked to occupation but better education can reduce morbidity, e.g. byreducing smoking

Genetics there is genetic evidence of a predisposition to contracting certainillnesses, even if this has no predictive capability

A straightforward bookwork question generally done well although not all students captured

the full range. All valid examples not shown above were credited.

Students who misunderstood the question and tried to answer using Class, Time, Temporary

Initial Selection were given no credit.

9 Use the formula

( )

(1 0.5(( ) ))

x

xx

aq

q aq

=

to derive the independent probabilities:

( ) (50 / 6548)0.00809

(1 0.5*((219 516) / 6548))(1 0.5(( ) ))

dd x

x dx

aqq

aq = = =

+

( ) (219 / 6548)0.03496

(1 0.5*((50 516) / 6548))(1 0.5(( ) ))

ii xx i

x

aqq

aq = = =

+

( ) (516 / 6548)0.080455

(1 0.5*((50 219) / 6548))(1 0.5(( ) ))

rr x

x rx

aqq

aq = = =

+

Then the revised 80%*0.00809 0.006472dxq = = then use the formula

1 1( ) (1 ( ...) ( . ...) ...)

2 3x x x x xaq q q q q = + + +

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Page 6

to derive dependent probabilities:

1 1( ) (1 ( ) ( . )) 0.0061046

2 3d d i r i r x x x x x xaq q q q q q= + + =

1 1( ) (1 ( ) ( . )) 0.0334465

2 3i i d r d r x x x x x xaq q q q q q= + + =

1 1( ) (1 ( ) ( . )) 0.0787948

2 3r r d i d ix x x x x xaq q q q q q= + + =

The resulting service table is:

lx dx ix rx

6,548 40 219 516

This question was done poorly. Many students appeared not to remember the derivation

process for multiple decrements etc. Some students wrote down the final table without

showing intermediate working. This gained only a proportion of the marks.

10 (a) Crude mortality rate = actual deaths / total exposed to risk

, ,

,

cx t x t

xc

x t

x

E m

E=

where

,c

x tE is central exposed to risk in population between agexandx+t

mx,tis central rate of mortality in population between agexandx+t

(b) Indirectly standardised mortality rate

, ,

,

, ,

, ,

s c sx t x t

xs c

x t

xc s

x t x t

xc

x t x t

x

E m

E

E m

E m

=

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Page 7

,s c

x tE is central exposed to risk in standard population between agexandx+t

,s

x tm is central rate of mortality in standard population between agexandx+t

This question generally done well. Other symbol notation was accepted provided it wasconsistent and properly defined.

11

Year t xq xp 1t xp tNUCF Profit Signature

1 0.01 0.99 1 50.2 50.22 0.01 0.99 0.99 43.1 42.7

3 0.01 0.99 0.9801 32.1 31.54 0.01 0.99 0.9703 145.5 141.2

(i) PV of profit @ 6%

2 3 450.2 42.7 31.5 141.2

47.4 38.0 26.4 111.8

0.0 6%

v v v v

IRR

= +

= +

= =

(ii) 2

32.1

31.31.025V= =

1 2 11.025 43.1 72.3xV p V V = =

revised cash flow in year 1 150.2 50.2 71.6 121.8xp V= = =

andNPVof profit = 121.8/1.06 + 111.8 = -3.1

(iii) As expected, theNPVafter zeroisation is smaller because the emergence ofthe non- unit cash flow losses have been accelerated and the risk discount rate

is greater than the accumulation rate.

Parts (i) and (iii) done well generally. In Part (ii) many students failed to develop the

formulae properly although they realised the effect in (iii).

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Page 8

12 (i) The gross future loss random variable is

( ) 40 4040 40

40 1 min( 1,25)50,000 1 1 ( )T T

K Kb K v I e ea fv Pa+ + + + + + +

Note: select functions also acceptable

where b is the annual rate of bonusI is the initial expensee is the annual renewal expense payable in the 2ndand

subsequent yearsf is the claim expenseP is the gross annual premiumK40(T40) is the curtate (complete) random future lifetime of a life

currently aged 40

(ii) The annual premium Pis given by

[ ] ( )[ ] [40]40[40]:25 4050,250 1,250 300 25( 1)Pa A IA a= + + +

0.5 0.513.29 50, 250 1.06 0.12296 1, 250 1.06 3.85489

300 25(15.494 1)

P = +

+ +

13.29 6361.402 4961.065 300 362.35P = + + +

901.79P =

(iii) The required reserve is

( )50 50 50:155064,000 1,500 35 901.79A IA a a+ +

0.5 0.564,000 1.04 0.32907 1,500 1.04 8.55929

35 17.444 901.79 11.253

= +

+

21, 477.560 13,093.196 610.54 10,147.84= + +

25,033.32=

In general question done well by well prepared students. In (i) credit also given if the

formulae included a limited term on the expense element although in reality this is unlikely.

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Page 9

13 (i) Let Pbe the annual premium. Then equating expected present value ofpremiums and benefits gives:

60 :55 60 :55100000m f m f Pa A=

where60 55 60 :5560 :55

15.632 18.210 14.756 19.086m f m f m fa a a a= + = + =

0.5 0.5

60 :55 60 :55 60 :551.04 1.04 (1 )

m f m f m f A A d a= =

0.51.04 (1 0.038462 19.086) 0.2711804= =

19.086 100000*0.2711804P =

1,420.83P = .

(ii) Reserves at the end of the first policy year:

Where both lives are alive:

0.5 61 :56

60 :55

100000 1.04 1m f

m f

a

a

0.5 15.254 17.917 14.356100000 1.04 1 1448.0115.632 18.210 14.756

+ = = +

Where the male life is alive only:

61 61

0.5

100000

0.04100000 1.04 1 15.254 1420.83 15.254 20475.94

1.04

m mA Pa

=

Where the female life is alive only:

56 56

0.5

100000

0.04100000 1.04 1 17.917 1420.83 17.917 6247.12

1.04

f fA Pa

=

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Page 10

Mortality Profit = Expected Death Strain Actual Death Strain

(a) Both lives die during 2009 = 1 actual claim.

Mortality Profit

( ) ( )( ) ( )

0.560 55

10,000 1 100000 1.04 1448.01

10, 000 0.002451 0.001046 1 100532.38 97954.99

m fq q=

= =

(b) Males only die during 2009 = 20 actual deaths (and therefore we needto change reserve from joint life to female only surviving).

Mortality Profit

( ) ( )( ) ( )

55 6010,000 20 6247.12 1448.01

10, 000 0.998954 0.002451 20 4799.11 21520.95

f mp q= = =

(c) Females only die during 2009 = 10 actual deaths (and therefore weneed to change reserve from joint life to male only surviving).

Mortality Profit

( ) ( )

( ) ( )

60 5510,000 10 20475.94 1448.01

10, 000 0.997549 0.001046 10 19027.93 8265.02

m fp q=

= =

Hence overall total mortality profit

97954.99 21520.95 8265.02 68,169.01= + + =

i.e. a mortality loss

Part (i) generally done well. Part (ii) was challenging and few students realised the full

implications of reserve change on 1st

death. Only limited partial credit was given ifstudents used only joint life situations.

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Page 11

14 Reserves required on the policy per unit sum assured are:

56:40 56:4

56:4

57:31 56:4

56:4

58:22 56:4

56:4

59:13 56:4

56:4

1 0

2.8701 1 0.23364

3.745

1.9551 1 0.47797

3.745

1.01 1 0.73298

3.745

aV

a

aV

a

aV

a

aV

a

= =

= = =

= = =

= = =

Multiple decrement table:

T[56] 1t

dq

+

[56] 1t

sq

+

[56] 1( )

t

daq

+

[56] 1( )

t

saq

+ [56] 1( ) tap + 1 [56]( )t ap

1 0.003742 0.1 0.003742 0.09963 0.896632 1.0000002 0.005507 0.1 0.005507 0.09945 0.895044 0.8966323 0.006352 0.1 0.006352 0.09936 0.894283 0.8025254 0.007140 0.0 0.007140 0.0 0.992860 0.717685

Probability in force [56] 1 [56] 1 [56] 1( ) (1 ) (1 )d s

t t tap q q+ + + =

The calculations of the profit vector, profit signature and NPV are set out in the tablebelow:

Policy

year Premium Expenses Interest

Death

claim

Maturity

claim

Surrender

claim

In force

cash flow

1 5000 600.00 176.00 80.45 0.00 350.31 4145.232 5000 45.00 198.20 118.40 0.00 715.38 4319.423 5000 45.00 198.20 136.57 0.00 1096.1613 3920.4754 5000 45.00 198.20 153.51 21346.49 0.00 16346.80

Policy

year

Increase in

reserves

Interest on

reserves Profit vector

Cum probability

of survival

Discount

factor

NPV

profit

1 4504.02 0.00 358.78 1.00000 0.943396 338.472 4174.53 200.93 345.82 0.89663 0.890000 275.963 3816.72 411.05 514.84 0.80253 0.839619 346.914 15759.07 630.36 42.63 0.71768 0.792094 24.24

Total NPV= 308.63

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Page 12

The calculations of the premium signature and profit margin are set out in the tablebelow:

Policy year 1 2 3 4

Premium 5000.00 5000.00 5000.00 5000.00probability in force 1.00000 0.89663 0.80253 0.71768discount factor 1.00000 0.943396 0.890000 0.839619p.v. of premium signature 5000.000 4229.40 3571.22 3012.91=> expected p.v. of premiums 15813.53

profit margin = 2.0%

Many well prepared students were able to outline the process required without being totally

accurate on the calculation. Significant credit was awarded in such situation.

Many students failed to appreciate the multiple decrement element.


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EXAMINATION

26 April 2011 (am)


Core Technical




booklet.


supervisor.







question paper.



CT5 A2011 Institute and Faculty of Actuaries

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CT5 A20112

1 Give a different example of selection shown by each of the following mortality tables:

(a) ELT15

(b) PMA92

(c) AM92

[3]

2 Calculate:

(a) 23 65p

(b) 10|5 60q

(c)65:10

s

Basis:

Mortality PMA92C20

Rate of interest 4% per annum [4]

3 Calculate ( )xIa

Basis: x = 0.02 for allx

= 4% per annum

[4]

4 Outline the benefits that are usually provided by a pension scheme on retirement dueto ill health. [5]

5 A pension scheme uses the following model to calculate probabilities, where thetransition intensities are = 0.05 and = 0.08.

Calculate:

(a) the dependent probability of retirement

(b) the independent probability of death from active service

using the Kolmogorov equations. [5]

Active Retired

Dead

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6 (i) Define uniform distribution of deaths [2]

(ii) Using the method in (i) above calculate 1.25q65.5 [4]

Basis:

Mortality ELT15(Males) [Total 6]

7 Explain how education influences morbidity. [6]

8 A life insurance company issues a with profits whole life assurance policy to a lifeaged 40 exact. The sum assured of 100,000 plus declared reversionary bonuses are

payable immediately on death. Level premiums are payable annually in advance to

age 65 or until earlier death.

A simple bonus, expressed as a percentage of the sum assured, is added to the policy

at the start of each year (i.e. the death benefit includes the bonus relating to the policy

year of death).

The following basis is used to price this policy:


Rate of Interest 4% per annum

Initial expenses 300 plus 50% of the first annual premium, incurred at the

policy commencement date

Renewal commission 2.5% of each premium from the start of the second policy

year

Claim expense 350 at termination of the contract

Using the principle of equivalence, calculate the level simple bonus rate that can be

supported each year on this policy if the annual premium is 3,212. [6]

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CT5 A20114

9 A male life aged 52 exact and a female life aged 50 exact take out a whole lifeassurance policy. The policy pays a sum assured of 100,000 immediately on first

death. Premiums are payable for a period of five years, monthly in advance.

Calculate the monthly premium payable.

Basis:

Mortality PMA92C20 (male life), PFA92C20 (female life)Rate of interest 4% per annum

Expenses Nil [7]

10 Calculate the expected present value and variance of the present value of anendowment assurance of 1 payable at the end of the year of death for a life aged 40

exact, with a term of 15 years.

Basis:


Rate of interest 4% per annumExpenses Nil [8]

11 A life insurance company issues a 4-year unit-linked endowment policy to a life aged61 exact under which level premiums of 2,500 are payable yearly in advance

throughout the term of the policy or until earlier death. In the first policy year 40% of

the premium is allocated to units, while in the second and subsequent policy years

110% of the premium is allocated to units. The unit prices are subject to a bid-offer

spread of 5%.

If the policyholder dies during the term of the policy, the death benefit of 10,000 or

the bid value of the units, whichever is higher, is payable at the end of the policy year

of death.

The policyholder may surrender the policy, in which case a value equal to a fixed

percentage of the total premiums paid on the policy is payable at the end of the policy

year of surrender. The percentage is based on the policy year of surrender as follows:

Policy year % of total premiums payable

as a surrender value

1 0

2 25

3 50

4 75

On maturity, 105% of the bid value of units is payable.

An annual management charge of 0.5% of the bid value of units is deducted at the end

of each policy year before death, surrender and maturity benefits are paid.

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The company uses the following assumptions in carrying out profit tests of this

contract:

Rate of growth on assets in the unit fund 4.25% per annum

Rate of interest on non-unit fund cash-flows 3.5% per annum

Independent rate of mortality AM92 Select

Independent rate of surrender 6% per annum

Initial expenses 325

Renewal expenses 74 per annum on the second

and subsequent premium dates

Initial commission 10% of first premium

Renewal commission 2.5% of the second andsubsequent years premiums

Risk discount rate 5.5% per annum

(i) Construct a multiple decrement table for this policy assuming that there is a

uniform distribution of both decrements over each year of age in the single

decrement table. [3]

(ii) Construct tables showing the growth of the unit fund and the non-unit fund.

Include all commissions in the non-unit fund. [7]

(iii) Calculate the profit margin for this policy on the assumption that the company

does not zeroise future expected negative cashflows. [3]

[Total 13]

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CT5 A20116

12 On 1April 1988, a life insurance company issued a 25-year term assurance policy to alife then aged 40 exact. The initial sum assured was 75,000 which increased by 4%

per annum compound at the beginning of the second and each subsequent policy year.

The sum assured is payable immediately on death and level monthly premiums are

payable in advance throughout the term of the policy or until earlier death.

The company uses the following basis for calculating premiums and reserves:



Initial commission 50% of the total premium payable in the first policy year

Initial expenses 400 paid at the policy commencement date

Renewal commission 2.5% of each premium from the start of the second policy

year

Renewal expenses 75 per annum, inflating at 4% per annum compound, at the

start of the second and subsequent policy years (the renewal

expense quoted is as at the start of the policy and the

increases due to inflation start immediately)

Claim expense 300 on termination (the claim expense is fixed over the

duration of the policy)

(i) Show that the monthly premium for the policy is approximately 56. [10]

(ii) Calculate the gross premium prospective reserve as at 31 March 2011. [6][Total 16]

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CT5 A20117

13 (i) Explain, including formulae, the following expressions assuming that the sumassured is payable at the end of the year of death:

death strain at risk

expected death strain

actual death strain [6]

(ii) A life insurance company issues the following policies:

25-year term assurances with a sum assured of 200,000

25-year endowment assurances with a sum assured of 100,000

The death benefit under each type of policy is payable at the end of year of

death.

On 1 January 2000, the company sold 10,000 term assurance policies to male

lives then aged 40 exact and 20,000 endowment assurance policies to male

lives then aged 35 exact. For each type of policy, premiums are payable

annually in advance.

During the first ten years, there were 145 actual deaths from the term

assurance policies written and 232 actual deaths from the endowment

assurance policies written.

(a) Calculate the death strain at risk for each type of policy during 2010.

During 2010, there were 22 actual deaths from the term assurance policies and

36 actual deaths from the endowment assurance policies.

Assume that there were no lapses/withdrawals on each type of policy during

the first eleven years.

(b) Calculate the total mortality profit or loss to the office in the year 2010.

(c) Comment on the results obtained in (b) above.

Basis:

Mortality AM92 Ultimate


Expenses Nil [11]

[Total 17]

END OF PAPER

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EXAMINERS REPORT

April 2011 examinations

Subject CT5 ContingenciesCore Technical

Introduction

The attached subject report has been written by the Principal Examiner with the aim ofhelping candidates. The questions and comments are based around Core Reading as theinterpretation of the syllabus to which the examiners are working. They have however givencredit for any alternative approach or interpretation which they consider to be reasonable.

T J BirseChairman of the Board of Examiners

July 2011

Institute and Faculty of Actuaries

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Subject CT5 (Contingencies Core Technical) Examiners Report, April 2011

Page 2

1 (a) Time selection because it is based on a period of three calendar years(b) Class selection applies only to male pensioners(c) Temporary initial selection as there are select rates

Other valid answers acceptable

This question was generally done well. However some students did not supply different

selection types for each part and this was penalised.

2 (a) 8823 6565

3534.0540.366307

9647.797

lp

l= = =

(b) 70 7510|5 6060

( ) (9238.134 8405.160)0.084771

9826.131

l lq

l

= = =

(c)

( )( )

10 10 1065:10 65 10 65 75

65:1010 65 10 65

10 10

(1 ) (1 ) ( )

8,405.160(1.04) (13.666 (1.04 ) 9.456)9,647.797

8,405.1609,647.797

1.48024 (13.666 0.67556 0.87120 9.456) / 0.87120

13.764

i a i a v p as

p p

+ + = =

=

=

=

This question was generally done well for parts (a) and (b) but students struggled more withpart (c).

31 2 3

0 1 2

.04 .02 .06

0.06 .06 2 .06 31

.06 2 .06

( ) 2 3 .......

Now * throughout.

Hence

( ) (1 2 3( ) 4( ) .........) at force of interest 6%

(1/(1 )) ((1 ) / .06)

294.

t t tx t x t x t x

x

x

Ia v p dt v p dt v p dt

vp e e e

Ia e e e a

e e

= + + +

= =

= + + + +

=

=

8662 0.970591

286.19

=

This question was not done well. The majority of students failed to realise that the increasing

function I was not continuous, although the payment is continuous. Instead most attempted

to compute0

( ) tx t xIa tv p dt

= . Only minimal credit was given for this.

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Page 3

4 Schemes usually allow members to retire on grounds of ill-health and receive apension benefit after a minimum length of scheme service.

Benefits are usually related to salary at the date of ill-health retirement in similar waysto age retirement benefits.

However, pensionable service is usually more generous than under age retirementwith years beyond those served in the scheme being credited to the member e.g. actualpensionable service subject to a minimum of 20 years, or pensionable service thatwould have been completed by normal retirement age.

A lump sum may be payable on retirement and a spouse pension on death afterretirement.

Other valid points were credited. Generally this bookwork question was done well.

5 The Kolmogorov equations in this case are:

( )

( )

( )

( )

r tt x

d tt x

aq et

aq et

+

+

=

=

For the case where t= 1 the solution for the dependent probability of retirement is:

( )( ) (1 )rxaq e +=

+

Hence the dependent probability of retirement is

(0.05 0.08)0.08( ) (1 )0.08 0.05

0.07502

rxaq e

+= +

=

The formula for the independent probability of death is

1dxq e=

Hence the independent probability of death is:

0.051 0.04877dxq e= =

Generally this question was completed satisfactorily by well prepared students.

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Page 4

6 (i) The definition of the uniform distribution of deaths (UDD) is .s x xq s q=

(alternatively t x x t p + is constant).

(ii) We have

1.25 65.5 0.5 65.5 0.75 66 p p p=

0.5 65.5 0.5 65.5 65 65(1 ) (1 (0.5 / (1 0.5 ))) by UDDp q q q= = = (1 ((0.5 0.02447) / (1 0.5 0.02447))) = 0.98761

0.75 66 0.75 66 661 1 0.75 1 0.75 0.02711p q q= = = = 0.97967

Hence

1.25 65.5

1.25 65.5 1.25 65.5

= 0.98761 0.97967 0.96753

1 1 0.96753

= 0.03247

p

q p

=

= =

A straightforward question that was generally done well.

7 Education influences the awareness of a healthy lifestyle, which reduces morbidity.

Education includes formal and informal processes, such as public health awarenesscampaigns.

Shows in:

Increased income Better diet Increased exercise Better health care Reduced alcohol and tobacco consumption

Lower levels of illicit drug use Safer sexual practices

Some effects are direct (e.g. drug use); some are indirect (e.g. exercise)

Students generally scored on a range of points but in most cases did not write enough of them

to gain all the marks.

Students who mentioned over indulgence risks for the better educated were given credit.

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Page 5

8 Let bbe the simple bonus rate (expressed as a percentage of the sum assured). Thenthe equation of value at 4% p.a. interest is (whereP = 3,212):

( ) ( )

[40] [40][40]:25

0.5 0.5

(.975 0.025) (100,000 350) 1,000 ( ) 300 0.5

(.975 15.887 0.025)

(100,000 350) 1.04 0.23041 1,000 1.04 7.95835 300 0.5

49,833.6179 23,579.5423 8,115.9564 1,906

24,348.

P a A b IA P

P

b P

b

b

+ = + + + +

+ =+ + + +

= + +

=

07563.00

8,115.9564=

i.e. a simple bonus rate of 3% per annum

Generally done well although some students treated b as not vesting in the first year.

9 Value of benefits using premium conversion

1/252:50 52:50

1/252:50

100,000 100,000 (1.04)

100,000 (1.04) (1 (0.04 /1.04) )

101,980.4 (1 0.0384615 17.295)

34,143.89

A A

a

=

=

=

=

Value of monthly premium ofP

(12) (12) (12)5 57:5557:5552:5052:50:5 52:50

(12)52:5052:50

(12)57:5557:55

5 57:55

52:50

12 12

11 / 24 17.295 0.458 16.837

11 / 24 15.558 0.458 15.100

(0.82193 9,880.196 9,917

lP P v

l

lv

l

=

= = =

= = =

=

.623) / (9,930.244 9,952.697)

0.81491

=

Hence (12)52:50:5

12 12 (16.837 0.81491 15.100) 54.3823P P P= =

Therefore:

34,143.89 / 54.3823 627.85P= =

There was an anomaly in this question in that it was not fully clear that the premium paying

period ceased on 1stdeath within the 5 year period. Even though the vast majority of students

who completed this question used the above solution a small minority used

(12)

512Pa

i.e.ignoring the joint life contingency. This was credited.

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Page 6

None the less many students struggled with this question

10 Expected present value is[40]:15

A where

1 1[40]:[40]:15 [40]:15 15

A A A= +

15 15[40] 15 [40] 55 15 [40]A v p A v p= +

9,557.8179 9,557.81790.23041 0.55526 0.38950 0.55526

9,854.3036 9,854.3036

= +

0.23041 0.20977 0.53855= +

0.55919=

Variance

2 2[40]:15 [40]:15

( )A A=

2 2 1 2 1[40]:[40]:15 [40]:15 15

A A A= +

2 2 15 2 2 15[40][40] 15 55 15 [40]( ) ( )A v p A v p= +

9,557.8179 9,557.81790.06775 0.30832 0.17785 0.308329,854.3036 9,854.3036

= +

0.06775 0.05318 0.29904

0.31361

= +

=

2So variance 0.31361 0.55919 0.000917= =

Note answers are sensitive to number of decimal places used.

Question done well by well prepared students. Many students failed to realise that the

endowment function needed to be split into the term and pure endowment portions.

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Page 7

11 Summary of assumptions:

Annual premium 2,500.00 Allocation % (1st yr) 40%Risk discount rate 5.5% Allocation % (2nd yr +) 110%Interest on investments 4.25% Man charge 0.5%Interest on sterling provisions 3.5% B/O spread 5.0%

Minimum death benefit 10,000.00 Maturity benefit 105%

% prm TotalInitial expense 325 10.0% 575Renewal expense 74 2.5% 136.5

(i) Multiple decrement table:

x dxq sxq

61 0.006433 0.0662 0.009696 0.0663 0.011344 0.0664 0.012716 0.06

x ( )dxaq ( )

sxaq ( )ap 1( )t ap

61 0.006240 0.05981 0.933953 1.000000

62 0.009405 0.05971 0.930886 0.93395363 0.011004 0.05966 0.929337 0.86940464 0.012335 0.05962 0.928047 0.807969

(ii) Unit fund (per policy at start of year)

yr 1 yr 2 yr 3 yr 4

value of units at start of year 0.00 985.42 3,732.08 6,581.15alloc 1,000.00 2,750.00 2,750.00 2,750.00B/O 50.00 137.50 137.50 137.50

interest 40.37 152.91 269.65 390.73management charge 4.95 18.75 33.07 47.92value of units at year end 985.42 3,732.08 6,581.15 9,536.46

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Page 8

Non-unit fund (per policy at start of year)

yr 1 yr 2 yr 3 yr 4

unallocated premium 1,500.00 250.00 250.00 250.00

B/O spread 50.00 137.50 137.50 137.50expenses/commission 575.00 136.50 136.50 136.50interest 34.12 8.72 8.72 8.72man charge 4.95 18.75 33.07 47.92extra death benefit 56.25 58.95 37.62 5.72extra surrender benefit 58.94 148.20 168.91 121.41extra maturity benefit 0.00 0.00 0.00 442.51end of year cashflow 1,016.76 149.71 93.36 536.62

(iii)

probability in force 1 0.933953 0.869404 0.807969discount factor 0.947867 0.898452 0.851614 0.807217

expected p.v. of profit 419.03

premium signature 2,500.00 2,213.16 1,952.79 1,720.19expected p.v. of

premiums 8,386.15 profitmargin 5.00%

Credit was given to students who showed good understanding of the processes involved even

if the calculations were not correct. Generally well prepared students did this question quite

well.

12 (i) LetPbe the monthly premium. Then:

EPV of premiums:

(12)[40]:25

12 @ 4% 186.996Pa P=

where

( )(12) 2525 [40][40]:25 [40]:2511

124

a a p v=

11 8821.261215.887 1 0.37512

24 9854.3036

=

15.583=

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Page 9

EPV of benefits:

0.5 1.5 24 24.5[40] [40] [40]1 24

75,000( (1 ) ... (1 ) )q v q b v q b v+ + + + +

where b= 0.040.5 0.5

1 / 25 /[40] 25 [40] 65[40]:25

75,000 (1 ) 75,000 (1 )@ @

(1 ) (1 )

i iA i A p v A i

b b

+ + = = + +

0.5

75,000 8821.26121 1 1

(1.04) 9854.3036

=

7709.6880=

where

/ /1.04 1 0.00 i.e. 0%1

i ib

= = =+

EPV of expenses (at 4% unless otherwise stated

(12) (12) @0%

[40]:25 [40]:1 [40]:25

1

[40]:25

0.5 12 400 0.025 12 0.025 12 75 1

300

6 400 0.025 12 15.583 0.025 12 0.982025 75 23.27542300 0.05422

6 400 4.6749 0.2946 1745.6558 16.26

P Pa Pa a

A

P P P

P P P

= + + +

+

= + + + +

= + + + +

6

10.3803 2161.9218P= +

where

( )(12) [40][40]:1[40]:111

124

11 9846.53841 1 0.96154 0.98202524 9854.3036

a a p v=

= =

( )@0% 64[40] 1 64 [40] 64[40]:25[40] [40]

11 ....

8934.877139.071 17.421 23.27541

9854.3036

la l l e e

l l+ = + + =

= =

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Page 10

1 0.5 1 0.5 25 65[40]:25 [40]:25 [40]:25

[40]

0.5

1.04 1.04

8821.26121.04 0.38896 0.37512 0.05422

9854.3036

lA A A v

l

= =

= =

Equation of value gives:

186.996 7709.6880 10.3803 2161.9218

9871.609855.89

176.6157

P P

P

= + +

= =

(ii) Gross prospective policy value at t= 23 (calculated at 4%) is given by:

[ ] [ ]

[ ]

[ ]

(12)prospective 23 0.5 0.563 63 64 63 63 64 63:2

(12)2363 63:2

75,000 (1.04) (1.04) 300 0.025 12

75 (1.04) 1 (1.04) 12

184,853.66 0.98058 0.011344 (1.04) 0.988656 0.012716 0.96154

300 0.9

V v q p q v v q p q v Pa

p v Pa

= + + + +

+ +

= +

+

[ ]

[ ]

(12) 2 65

63:263:2 63

8058 0.011344 0.988656 0.012716 0.96154 0.025 12 55.89 1.90629

184.854 1 (1.04) 0.988656 0.96154 12 55.89 1.90629

11 11 8821.2612

where 1 1.951 1 0.9245624 24 9037.3973

la a v

l

+ +

+ +

= =

1.90629

4,335.0628 6.8932 31.9628 367.6104 1,278.5106

3,463.02

=

= + + +

=

This question was generally not done well especially part (ii). In part (i) although it was

commonly recognised that a resultant rate of interest of 0% emerged students did not often

seem to know how to progress from there.

13 (i) The death strain at risk for a policy for year t + 1 (t = 0, 1, 2) is the excess ofthe sum assured (i.e. the present value at time t + 1 of all benefits payable ondeath during the year t + 1) over the end of year provision.

i.e. DSAR for year t + 1 1tS V+=

The expected death strain for year t + 1 (t = 0, 1, 2) is the amount that thelife insurance company expects to pay extra to the end of year provision forthe policy.

i.e. EDS for year t + 1 1( )tq S V+=

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Page 11

The actual death strain for year t+ 1 (t= 0, 1, 2) is the observed value att+1 of the death strain random variable

i.e. ADS for year t + 1 1( )tS V+= if the life died in the year tto t+1= 0 if the life survived to t + 1

Note: Full credit given if definition of death strain is given for a block ofpolicies rather than for a single policy as per above.

(ii) (a) Annual premium for endowment assurance with 100,000 sum assuredgiven by:

35:2535:25

100,000 100,0000.38359 2,393.40

16.027EAP A

a= = =

Annual premium for term assurance with 200,000 sum assured givenby:

140:25

40:25

1 2525 4040:25 40:25

200,000

where

8,821.26120.38907 0.37512 0.38907 0.33573 0.05334

9,856.2863

200,000 0.05334671.62

15.884

TA

TA

AP

a

A A v p

P

=

=

= = =

= =

Reserves at the end of the 11th year:

for endowment assurance with 100,000 sum assured given by:

11 46:14 46:14100,000

100,000 0.58393 2,393.40 10.818

58,393.0 25,891.8 32,501.2

EA EAV A P a=

=

= =

for term assurance with 200,000 sum assured given by:

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Page 12

111 51:14 51:14

1 1414 5151:14 51:14

11

200,000

where

8,821.26120.58884 0.57748 0.58884 0.52583 0.063019,687.7149

200,000 0.06301 671.62 10.69

12,602.0 7,179.6 5,422.4

TA TA

TA

V A P a

A A v p

V

=

=

= = =

=

= =

Therefore, sums at risk are:

Endowment assurance: DSAR = 100,000 32,501.2 = 67,498.8

Term assurance: DSAR = 200,000 5,422.4 = 194,577.6

(b) Mortality profit = EDS ADS

For endowment assurance

4519768 67, 498.8 19768 0.001465 67, 498.8 1, 954, 773.3

36 67,498.8 2,429,956.8

EDS q

ADS

= = =

= =

mortality profit = 475,183.5 (i.e. a loss)

For term assurance

509,855 194,577.6 9,855 .002508 194,577.6 4,809, 246.1

22 194,577.6 4,280,707.2

EDS q

ADS

= = =

= =

mortality profit = 528,538.9Hence, total mortality profit = 528,538.9 475,183.5 = 53,355.4

(c) Although there is an overall mortality profit in 2010, the actual numberof deaths for the endowment assurances is approximately 25% higherthan expected, which is a concern. Further investigation would berequired to determine reasons for poor mortality experience for theendowment assurances, e.g. there may have been limited underwritingrequirements applied to this type of contract when they were written.

Generally (a) was done well. The most common error in (b) was to assume reserves at 10

years rather than 11. On the whole well prepared students coped with (b) well. Many

students did not attempt (c) or at best gave a somewhat sketchy answer.


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EXAMINATION

4 October 2011 (am)


Core Technical




booklet.


supervisor.







question paper.



CT5 S2011 Institute and Faculty of Actuaries

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CT5 S20112

1 Calculate:

(a) 10|1 [50]q

(b) 10 [60] 1p +

(c) (12)[40]:20

a

Basis:Mortality AM92Rate of interest 6% per annum [3]

2 Calculate 0.5 75.25q using the assumption of a constant force of mortality.

Basis:Mortality AM92 [3]

3 In a special mortality table with a select period of one year, the followingrelationships are true for all ages:

0.5 [ ]

0.5 [ ] 0.5

0.25

0.45

x x

x x

q q

q q+

=

=

Expressp[x] in terms ofpx . [3]

4 A term assurance contract with a term of 20 years pays a sum assured of 1immediately on death to a life now aged 30 exact.

Calculate the expected value and variance of this contract.

Basis:Mortality AM92 UltimateRate of interest 4% per annum [4]

5 (a) Write down the random variable form of 1:x yA .

(b) Calculate 1:x yA on the following assumptions:

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