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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) xd
[3] 2 Define spurious selection, giving two distinct examples.
[3] 3 Calculate the standardised mortality ratio for the population
of Urbania using the
following 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 benefit
payable 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 0.75p50.5 using two different methods. [5] 7 A company
is about to establish a pension scheme that will provide an age
retirement
benefit 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 be
shared 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: Rate of interest 4% per
annum
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) x x
x
9 A life insurance company models the experience of its pension
scheme contracts using the following three-state model:
(i) Derive the dependent probability of a life currently Active
and aged x retiring
in the year of age x to (x + 1) in terms of the transition
intensities. [2] (ii) Derive a formula for the independent
probability of a life currently Active and
aged x retiring in the year of age x to (x + 1) using the
dependent probabilities. [4] [Total 6] 10 The decrement table
extract below is based on the historical experience of a very
large
multinational 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 age x is:
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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CT5 A20105 PLEASE TURN OVER
12 On 1 January 2005, a life insurance company issued 1,000
10-year term assurance policies 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: Mortality AM92 Select 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 a male 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 the bid 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 is payable 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
Select Initial 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 policy to 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 maturity Future
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 25th policy 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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Faculty of Actuaries Institute of Actuaries
EXAMINERS REPORT
April 2010 Examinations
Subject CT5 Contingencies 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 Actuaries Institute 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 age x + r out of lx
lives alive at age x subject to select mortality.
(ii) The probability that a life age x will die between age x +
n and x + n + m. (iii) The number of lives that die between x and
(x + 1) out of xl lives alive at x.
Question generally answered well. 2 Spurious selection occurs
when mortality differences ascribed to groups are formed
by 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 confounded with 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 divided
by 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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Subject CT5 (Financial Mathematics Core Technical) April 2010
Examiners Report
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 as the multiplier
before the temporary assurance function rather then 9,900.
5 exp( )x tt x sxp ds+=
0.0002exp( ( 1) )x t Sx
e ds+=
0.0002exp( )x t x tSx x
e ds ds+ += +
0.0002( ) 0.0002
exp0.0002
x t xe et
+ = +
(i) Probability =
0.0002 70 0.0002 20
exp 500.0002
x xe e = +
0.6362=
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Subject CT5 (Financial Mathematics Core Technical) April 2010
Examiners Report
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 = 0.978850 p51 = 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 pp 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 one of 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
555554 55
(20 )1 40,00060
z raz raM Rs D
+
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Subject CT5 (Financial Mathematics Core Technical) April 2010
Examiners Report
Page 5
1 (20*128,026 963,869)40,00060 9.745*1,389
+= 173,584= (ii) Contributions
5554 55
40,000.s NK
s D
88,615.40,0009.745*1,389
K x= 261,868K= Therefore K = 173,584 / 261,868 i.e. 66.3% Most
students answered reasonably well. Most common error was the wrong
sx function. 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*
ap
=
45 66 2121:45 21:45 21:451 1 8695.6199*(1 * / ) * 1 0.17120*2 2
9976.3909
a a v l l a = =
21:45 0.42539a=
44 65 2121:45 21:448821.2612* / 21.045 .17805*
21.2029976.3909
a a v l l= + = + = 21:45 20.777a =
( )4552*1.04 (20.777)therefore fund 7, 240
8695.61999976.3909
= =
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Subject CT5 (Financial Mathematics Core Technical) April 2010
Examiners Report
Page 6
(ii) Let annuity be P per week. Then EPV of annuity at 66 is 10
10 66 7610
252 ( * . )3P 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 x and 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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Subject CT5 (Financial Mathematics Core Technical) April 2010
Examiners Report
Page 7
So
( ) (( ) ( ) )
( log(1 (( ) ( ) )))r r dx x xr d
x xaq aq aq
aq aq= + +
this can be rearranged to show
( ) log(1 (( ) ( ) ))( ) ( )
rr dxx xr d
x x
aq aq aqaq aq
= ++ Given that: 1 ,rxq e
=
then
( )
(( ) ( ) )1 1 (( ) ( ) )rx
r dx x
aqr r d aq aqx 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 10 0 0
( ) ( ) expt
rx 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 qal 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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Subject CT5 (Financial Mathematics Core Technical) April 2010
Examiners Report
Page 8
This leads to two simultaneous equations:
( )1
dd xx r
x
aqqq
= and ( )
1
rr xx d
x
aqqq
= which results in a quadratic equation in rxq . (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 dx x x x x xq aq aq q aq aq= =
Calculate and d wx xq q 40
dq =.00250/(1.006) = .00252 and 41dq = .00274/(1.00731) =
.00276
40wq =.01200/(1.00125) = .01201 and 41
wq = .01461/(1.00137) = .01463 Adjusting for the 75% multiplier
of independent withdrawal decrements:
40
41
40
41
1 3( ) .00252* 1 * *.01201 .002512 41 3( ) .00276* 1 * *.01463
.002742 4
3 1( ) .01201* * 1 *.00252 .009004 23 1( ) .01463* * 1 *.00276
.010964 2
d
d
w
w
aq
aq
aq
aq
= = = =
= = = =
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Subject CT5 (Financial Mathematics Core Technical) April 2010
Examiners Report
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=108.342 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 a
rate of 1 per annum and secured by a single premium at age x is
given by:
0
st x x t s x tV a e p ds
+ += =
0 0
s st x x t s x t s x tV a e p ds e p dst t t t
+ + +
= = =
1 ln( ) (ln ln )s x t s x t x t s x t x t s x ts x t
p p l lp t t t+ + + + + + + ++
= = = +
( )s x t s x t x t s x tp pt + + + + + = +
0
( )st x s x t x t x t sV e p dst
+ + + +
=
0
sx t x t s x t x t sa e p ds
+ + + + +=
0
0
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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Subject CT5 (Financial Mathematics Core Technical) April 2010
Examiners Report
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 P for 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
2118.39 257.468.228
P = =
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Subject CT5 (Financial Mathematics Core Technical) April 2010
Examiners Report
Page 11
Net Premium Retrospective Reserves at the end of the fifth
policy year is given by:
[55]5 1[55]:5 [55]:560
(1 ) 50,000l
i Pa Al
+
[ ]9545.99291.21665 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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Subject CT5 (Financial Mathematics Core Technical) April 2010
Examiners Report
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:
x dxq
sxq
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 4200.000 B/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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Subject CT5 (Financial Mathematics Core Technical) April 2010
Examiners Report
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
132.020 Interest 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 P be 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=
-
Subject CT5 (Financial Mathematics Core Technical) April 2010
Examiners Report
Page 14
3 8821.261214.352 1 0.174118 9892.9151 =
14.0352=
EPV of benefits: 0.5 1.5 29 29.5[35] [35] [35]1 29100,000( (1 )
... (1 ) )q v q b v q b v+ + + + + 30 30 30 [35]100,000 (1 )b v p+
+ where b = 0.0192308
( )0.5 2 2 30 30[35] [35] [35]0.5 0.5 1 29100,000 (1.06) (1 ) (1
) ... (1 )(1 ) (1 ) q b v q b v q b vb b= + + + + + ++ +
30 3030 [35]100,000(1 )b v p+ +
0.5 1 30 30 [35][35]:30100,000 (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 0.308329892.9151
+ 4,742.594 27,492.112 32,234.706= + = where
1.06 1 0.041
ib
= =+ 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= + + +
-
Subject CT5 (Financial Mathematics Core Technical) April 2010
Examiners Report
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]:1 3 18a a p v=
3 9887.20691 1 0.9434 0.978578 9892.9151 = =
Equation of value gives: 56.1408 32,234.706 2.30566 906.322P P=
+ +
33,141.028 615.6053.8351
P = = (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 Pab
= + + + + + 1 5 5 6060:51000 500A v p+ +
(4)1 5 65
60:5 60:50.5 60:560
245,000 @ 245,000 @ 90 0.975 4(1.04)
lA i v i a Pal
= + +
0.5 5 565 6560:5
60 601000 1.04 500l lA v v
l l + +
(4) 5 65
60:560:560
3 3 8821.2612where 1 4.55 1 0.82193 4.46788 8 9287.2164
la a vl
= = =
4
601 060:5
60
1.04 465.9551and 1 0 @ 0.050171.04 9287.2164
tdi A i
l
+ = = = = =
0.5245,000 0.05017 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+ +
-
Subject CT5 (Financial Mathematics Core Technical) April 2010
Examiners Report
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
-
Faculty of Actuaries Institute of Actuaries
EXAMINATION
6 October 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 S2010 Institute of Actuaries
-
CT5 S20102
1 Calculate: (a) 20|10 [45]q (b) 30 [45]:[50]p Basis: AM92
Select [3] 2 Calculate 0.5p45.75 using the Uniform Distribution of
Deaths assumption. Basis: AM92 Ultimate [3] 3 Calculate the single
premium payable for a temporary reversionary annuity of
12,000 per annum payable monthly in arrear to a female life
currently aged 55 exact on the death of a male life currently aged
50 exact. No payment is made after 20 years from the date of
purchase.
Basis: Rate of interest 4% per annum Mortality of male life
PMA92C20
Mortality of female life PFA92C20 Expenses Nil [4] 4 A gymnasium
offers membership for a three-year period at a fixed fee of 240
per
annum payable monthly in advance. The contract may only be
cancelled at a renewal anniversary. Monthly premiums cease
immediately on the death of the member.
Calculate the expected present value of membership fees if the
gymnasium sells 120
memberships: Basis: Rate of interest 6% per annum
Rate of mortality 1% per annum Probability of renewal 80% at
each anniversary Expenses Nil [5]
-
CT5 S20103 PLEASE TURN OVER
5 A pension scheme provides an age retirement benefit of n/80ths
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. Normal retirement age is 65 and age
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 and
their salary in the year before the valuation date was 25,000.
Give a formula for the expected cashflows between the 66th and 67th
birthdays as a
result of entitlement from this past service. [5] 6 Calculate:
(a) 30:40A
(b) 30:40:20a 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 the formula:
50,000 (1 0.1 ) 0,1,2........,9t t + =
where t denotes the curtate duration in years since the
inception of the policy. Level premiums are payable monthly in
advance throughout the term of the policy or
until earlier death. Calculate the monthly premium for this
policy using the following basis: Mortality AM92 Select Interest 4%
per annum Expenses Nil [6] 8 Describe the causal factors that
explain observed differences in mortality and
morbidity. [6]
-
CT5 S20104
9 The actuary advising a pension scheme has decided that the
independent mortality in the standard table for pension schemes
(PEN) from page 142 of the Formulae and Tables 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 revised
independent mortality rate at that age is 80% of the previous
independent mortality rate.
[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. The
following non-unit cash flows, tNUCF (t = 1,2,3,4), are obtained
at the end of each year t per 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
rate
of interest earned on non-unit reserves is 2.5% per annum. (ii)
Calculate the net present value of the profits after zeroisation
using a risk
discount rate of 6% per annum. [3] (iii) Comment on the results
obtained in (i) and (ii) above. [1] [Total 7]
-
CT5 S20105 PLEASE TURN OVER
12 A life insurance company issued a with profits whole life
policy to a life aged 40 exact on 1 January 2000. Under the policy,
the basic sum assured of 50,000 and attaching bonuses are payable
immediately on death. Level premiums are payable annually in
advance under the policy until age 65 or earlier death.
The company declares simple reversionary bonuses at the start of
each year including
the 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 policy
at the outset, defining symbols where necessary. [4] (ii)
Calculate the annual premium using the following assumptions:
Mortality AM92 Select Interest 6% per annum Bonus loading 2.5% per
annum simple Initial expenses 300 Renewal expenses 25 at the start
of the second and subsequent policy
years while the policy is in force Claim expenses 250
[4]
On 31 December 2009, the policy is still in force. Bonuses
declared to date total 13,750.
(iii) Calculate the gross premium prospective reserve for the
policy as at
31 December 2009 using the following assumptions: Mortality AM92
Ultimate Interest 4% per annum Bonus loading 3% per annum simple
Renewal expenses 35 at the start of each policy year while the
policy is in
force Claim expenses 250
[4] [Total 12]
-
CT5 S20106
13 On 1 January 2009, a life insurance company issued 10,000
joint life whole life assurance policies to couples. Each couple
comprised one male life aged 60 exact and one female life aged 55
exact when the policy commenced. Under each policy, a sum assured
of 100,000 is payable immediately on the death of the second of the
lives to die.
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 net
premium reserves: Mortality PMA92C20 for the male PFA92C20 for
the female Interest 4% per annum Expenses 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
a
policy where both the male and the female life died during the
year. In addition, there were 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 the
calendar year 2009. [10] [Total 14]
-
CT5 S20107
14 A life insurance company issues four-year without profits
endowment assurance policies to male lives aged 56 exact. The sum
assured is 21,500 payable on maturity or at the end of the year of
death if earlier. Premiums of 5,000 are payable annually in advance
throughout the term of the policy.
The company holds net premium reserves for these policies,
calculated using AM92
Ultimate mortality and interest of 4% per annum. Surrenders
occur only at the end of a year immediately before a premium is
paid. The
surrender value is 70% of the net premium reserve calculated at
the time the surrender value is payable.
The company uses the following assumptions in carrying out
profit tests of this
contract:
Rate of interest on cash flows 4% per annum Mortality AM92
Select Surrenders 10% of all policies still in force at the end of
each of
the first, second and third policy years Initial expenses 600
Renewal expenses 45 per annum on the second and subsequent
premium dates Risk discount rate 6% per annum
Calculate the expected profit margin for this contract. [15]
END OF PAPER
-
INSTITUTE AND FACULTY OF ACTUARIES
EXAMINERS REPORT
September 2010 examinations
Subject CT5 Contingencies 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. T J Birse
Chairman of the Board of Examiners December 2010
Institute and Faculty of Actuaries
-
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 5,266.4604 0.3809519,798.0837 9,706.0977
l lpl 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) .999227 p = =
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.
-
Subject CT5 (Contingencies Core Technical) September 2010
Examiners Report
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 1x =
Therefore
(12):1
1 11/ 24(1 0.99 /1.06) 0.96973x
= =
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 2500080 ( ) 80 ( )
t t t
tt
z r rl z r rls 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.
-
Subject CT5 (Contingencies Core Technical) September 2010
Examiners Report
Page 4
6 (a) ______
.04 .01 .02 .02 .01030:40
.05 .07 .06 .070
.05 .06 .070
.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
01.4
* *
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 P be the monthly premium. Then equating expected present
value of premiums
and 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 8.05624 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) 0.372845000
0.06044 5000 0.372812 568.99
8.05647.42
IA v p A
P
P
= =
+ = = =
In general question done well by well prepared students.
-
Subject CT5 (Contingencies Core Technical) September 2010
Examiners Report
Page 5
8 Occupation either because of environmental or lifestyle
factors mortality may be directly affected. Occupations may also
have health barriers to entry, e.g. airline pilots
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. by
reducing smoking Genetics there is genetic evidence of a
predisposition to contracting certain
illnesses, 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(( ) ))
xx
x
aqqaq
=
to derive the independent probabilities:
( ) (50 / 6548) 0.00809(1 0.5*((219 516) / 6548))(1 0.5(( )
))
dd xx d
x
aqqaq
= = = +
( ) (219 / 6548) 0.03496(1 0.5*((50 516) / 6548))(1 0.5(( )
))
ii xx i
x
aqqaq
= = = +
( ) (516 / 6548) 0.080455(1 0.5*((50 219) / 6548))(1 0.5(( )
))
rr xx r
x
aqqaq
= = = + Then the revised 80%*0.00809 0.006472dxq = = then use
the formula
1 1( ) (1 ( ...) ( . ...) ...)2 3x x x x x
aq q q q q = + + +
-
Subject CT5 (Contingencies Core Technical) September 2010
Examiners Report
Page 6
to derive dependent probabilities:
1 1( ) (1 ( ) ( . )) 0.00610462 3
d d i r i rx x x x x xaq q q q q q= + + =
1 1( ) (1 ( ) ( . )) 0.03344652 3
i i d r d rx x x x x xaq q q q q q= + + =
1 1( ) (1 ( ) ( . )) 0.07879482 3
r 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
xcx t
x
E m
E=
where ,
cx tE is central exposed to risk in population between age x and
x+t
mx,t is central rate of mortality in population between age x
and x+t
(b) Indirectly standardised mortality rate
, ,
,
, ,
, ,
s c sx t x t
xs c
x tx
c sx t x t
xcx t x t
x
E m
E
E m
E m
=
-
Subject CT5 (Contingencies Core Technical) September 2010
Examiners Report
Page 7
,s c
x tE is central exposed to risk in standard population between
age x and x+t ,
sx tm is central rate of mortality in standard population
between age x and x+t
This question generally done well. Other symbol notation was
accepted provided it was consistent and properly defined. 11
Year t
xq xp 1t xp tNUCF Profit Signature
1 0.01 0.99 1 50.2 50.2 2 0.01 0.99 0.99 43.1 42.7 3 0.01 0.99
0.9801 32.1 31.5 4 0.01 0.99 0.9703 145.5 141.2
(i) PV of profit @ 6%
2 3 450.2 42.7 31.5 141.247.4 38.0 26.4 111.8
0.0 6%
v v v v
IRR
= += += =
(ii) 232.1 31.3
1.025V = =
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= = = and NPV of profit = 121.8/1.06 +
111.8 = -3.1 (iii) As expected, the NPV after zeroisation is
smaller because the emergence of
the 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).
-
Subject CT5 (Contingencies Core Technical) September 2010
Examiners Report
Page 8
12 (i) The gross future loss random variable is ( ) 40 40
40 4040 1 min( 1,25)50,000 1 1 ( )T TK Kb K v I e ea fv Pa+ + +
+ + + +
Note: select functions also acceptable where b is the annual
rate of bonus I is the initial expense e is the annual renewal
expense payable in the 2nd and
subsequent years f is the claim expense P is the gross annual
premium K40(T40) is the curtate (complete) random future lifetime
of a life
currently aged 40 (ii) The annual premium P is 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.
-
Subject CT5 (Contingencies Core Technical) September 2010
Examiners Report
Page 9
13 (i) Let P be the annual premium. Then equating expected
present value of premiums and benefits gives:
60 :55 60 :55100000m f m fPa A=
where 60 55 60 :5560 :55 15.632 18.210 14.756 19.086m f m fm fa
a a a= + = + = 0.5 0.5
60 :55 60 :55 60 :551.04 1.04 (1 )m f m f m fA 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 20475.941.04
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 6247.121.04
f fA Pa =
-
Subject CT5 (Contingencies Core Technical) September 2010
Examiners Report
Page 10
Mortality Profit = Expected Death Strain Actual Death Strain (a)
Both lives die during 2009 = 1 actual claim. Mortality Profit
( ) ( )( ) ( )
0.560 5510,000 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 need
to 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 we
need 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 if students used only joint life situations.
-
Subject CT5 (Contingencies Core Technical) September 2010
Examiners Report
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 0.233643.745
1.9551 1 0.477973.745
1.01 1 0.732983.745
aV
a
aV
a
aV
a
aV
a
= =
= = =
= = =
= = =
Multiple decrement table:
T [56] 1tdq + [56] 1t
sq + [56] 1( ) tdaq + [56] 1( ) t
saq + [56] 1( ) tap + 1 [56]( )t ap
1 0.003742 0.1 0.003742 0.09963 0.896632 1.000000 2 0.005507 0.1
0.005507 0.09945 0.895044 0.896632 3 0.006352 0.1 0.006352 0.09936
0.894283 0.802525 4 0.007140 0.0 0.007140 0.0 0.992860 0.717685
Probability in force [56] 1 [56] 1 [56] 1( ) (1 ) (1 )
d st t tap q q+ + + =
The calculations of the profit vector, profit signature and NPV
are set out in the table
below: 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 4145.23 2 5000 45.00
198.20 118.40 0.00 715.38 4319.42 3 5000 45.00 198.20 136.57 0.00
1096.1613 3920.475 4 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 338.47 2 4174.53 200.93
345.82 0.89663 0.890000 275.96 3 3816.72 411.05 514.84 0.80253
0.839619 346.91 4 15759.07 630.36 42.63 0.71768 0.792094 24.24
Total NPV = 308.63
-
Subject CT5 (Contingencies Core Technical) September 2010
Examiners Report
Page 12
The calculations of the premium signature and profit margin are
set out in the table below:
Policy year 1 2 3 4 Premium 5000.00 5000.00 5000.00 5000.00
probability in force 1.00000 0.89663 0.80253 0.71768 discount
factor 1.00000 0.943396 0.890000 0.839619 p.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.
END OF EXAMINERS REPORT
-
INSTITUTE AND FACULTY OF ACTUARIES
EXAMINATION
26 April 2011 (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 13 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.
CT5 A2011 Institute and Faculty of Actuaries
-
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) 65:10s Basis: Mortality PMA92C20 Rate of interest 4% per
annum [4] 3 Calculate ( )xIa Basis: x = 0.02 for all x = 4% per
annum [4] 4 Outline the benefits that are usually provided by a
pension scheme on retirement due
to ill health. [5] 5 A pension scheme uses the following model
to calculate probabilities, where the
transition 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
-
CT5 A20113 PLEASE TURN OVER
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 life
aged 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: Mortality AM92
Select 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]
-
CT5 A20114
9 A male life aged 52 exact and a female life aged 50 exact take
out a whole life assurance 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 an
endowment 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: Mortality AM92 Select Rate of interest 4% per annum
Expenses Nil [8] 11 A life insurance company issues a 4-year
unit-linked endowment policy to a life aged
61 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.
-
CT5 A20115 PLEASE TURN OVER
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 and
subsequent 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]
-
CT5 A20116
12 On 1 April 1988, a life insurance company issued a 25-year
term assurance policy to a life 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:
Mortality AM92 Select
Rate of interest 4% per annum 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]
-
CT5 A20117
13 (i) Explain, including formulae, the following expressions
assuming that the sum assured 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 Rate of
interest 4% per annum Expenses Nil [11] [Total 17]
END OF PAPER
-
INSTITUTE AND FACULTY OF ACTUARIES
EXAMINERS REPORT
April 2011 examinations
Subject CT5 Contingencies 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. T J Birse
Chairman of the Board of Examiners July 2011
Institute and Faculty of Actuaries
-
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
3534.054 0.3663079647.797
lpl
= = =
(b) 70 7510|5 6060
( ) (9238.134 8405.160) 0.0847719826.131
l lql = = =
(c)
( )( )
10 10 1065:10 65 10 65 7565:10
10 65 10 65
10 10
(1 ) (1 ) ( )
8,405.160(1.04) (13.666 (1.04 ) 9.456)9,647.7978,405.160
9,647.7971.48024 (13.666 0.67556 0.87120 9.456) /
0.8712013.764
i a i a v p asp p
+ + = =
=
= =
This question was generally done well for parts (a) and (b) but
students struggled more with part (c). 3
1 2 3
0 1 2.04 .02 .06
0.06 .06 2 .06 3 1
.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 0.970591286.19
=
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
compute
0( ) tx t xIa tv p dt
= . Only minimal credit was given for this.
-
Subject CT5 (Contingencies Core Technical) Examiners Report,
April 2011
Page 3
4 Schemes usually allow members to retire on grounds of
ill-health and receive a pension benefit after a minimum length of
scheme service.
Benefits are usually related to salary at the date of ill-health
retirement in similar ways
to age retirement benefits. However, pensionable service is
usually more generous than under age retirement
with years beyond those served in the scheme being credited to
the member e.g. actual pensionable service subject to a minimum of
20 years, or pensionable service that would have been completed by
normal retirement age. A lump sum may be payable on retirement and
a spouse pension on death after retirement.
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 0.050.07502
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.
-
Subject CT5 (Contingencies Core Technical) Examiners Report,
April 2011
Page 4
6 (i) The definition of the uniform distribution of deaths (UDD)
is .s x xq s q= (alternatively t x x tp + 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
66 1 1 0.75 1 0.75 0.02711 p 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
pq 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 awareness
campaigns. 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.
-
Subject CT5 (Contingencies Core Technical) Examiners Report,
April 2011
Page 5
8 Let b be the simple bonus rate (expressed as a percentage of
the sum assured). Then the equation of value at 4% p.a. interest is
(where P = 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 0.549,833.6179
23,579.5423 8,115.9564 1,906
24,348.
P a A b IA P
P
b Pb
b
+ = + + + + + =
+ + + + = + + =
0756 3.008,115.9564
=
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 of P
(12) (12) (12)5 57:5557:5552:5052:50:5 52:50
(12)52:5052:50
(12)57:5557:55
5 57:5552:50
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)
52:50:512 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 1st death 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.
-
Subject CT5 (Contingencies Core Technical) Examiners Report,
April 2011
Page 6
None the less many students struggled with this question 10
Expected present value is [40]:15A where
1 1[40]:[40]:15 [40]:15 15A 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 0.555269,854.3036
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 15A 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 0.30832
9,854.3036 9,854.3036 = +
0.06775 0.05318 0.299040.31361
= +=
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.
-
Subject CT5 (Contingencies Core Technical) Examiners Report,
April 2011
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
Total Initial expense 325 10.0% 575 Renewal expense 74 2.5%
136.5
(i) Multiple decrement table:
x dxq
sxq
61 0.006433 0.06 62 0.009696 0.06 63 0.011344 0.06 64 0.012716
0.06
x ( )dxaq ( )
sxaq ( )ap 1( )t ap
61 0.006240 0.05981 0.933953 1.00000062 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
137.50 137.50 137.50interest 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
-
Subject CT5 (Contingencies Core Technical) Examiners Report,
April 2011
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
250.00B/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 profit
margin 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) Let P be the monthly premium. Then: EPV of premiums: (12)
[40]:2512 @ 4% 186.996Pa P=
where
( )(12) 2525 [40][40]:25 [40]:25 11 124a a p v=
11 8821.261215.887 1 0.3751224 9854.3036
= 15.583=
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Subject CT5 (Contingencies Core Technical) Examiners Report,
April 2011
Page 9
EPV of benefits: 0.5 1.5 24 24.5[40] [40] [40]1 2475,000( (1 )
... (1 ) )q v q b v q b v+ + + + + where b = 0.04
0.5 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 ib b
+ + = = + +
0.575,000 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 0.054226
400 4.6749 0.2946 1745.6558 16.26
P Pa Pa a
A
P P P
P P P
= + + + +
= + + + + = + + + +
610.3803 2161.9218P= +
where
( )(12) [40][40]:1[40]:1 11 124
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 23.275419854.3036
la l l e el l+
= + + =
= =
-
Subject CT5 (Contingencies Core Technical) Examiners Report,
April 2011
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 0.054229854.3036
lA A A vl
= = = =
Equation of value gives:
186.996 7709.6880 10.3803 2161.9218
9871.6098 55.89176.6157
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 6563:263:2
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 8821.2612where 1 1.951 1 0.9245624 24 9037.3973
la a vl
+ + + +
= = 1.90629
4,335.0628 6.8932 31.9628 367.6104 1, 278.51063,463.02
=
= + + + =
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 of
the sum assured (i.e. the present value at time t + 1 of all
benefits payable on death 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 the
life insurance company expects to pay extra to the end of year
provision for the policy.
i.e. EDS for year t + 1 1( )tq S V+=
-
Subject CT5 (Contingencies Core Technical) Examiners Report,
April 2011
Page 11
The actual death strain for year t + 1 (t = 0, 1, 2) is the
observed value at t+1 of the death strain random variable
i.e. ADS for year t + 1 1( )tS V+= if the life died in the year
t to t+1 = 0 if the life survived to t + 1 Note: Full credit given
if definition of death strain is given for a block of
policies rather than for a single policy as per above. (ii) (a)
Annual premium for endowment assurance with 100,000 sum assured
given by:
35:2535:25
100,000 100,000 0.38359 2,393.4016.027
EAP Aa
= = = Annual premium for term assurance with 200,000 sum assured
given
by:
140:25
40:25
1 2525 4040:25 40:25
200,000
where
8,821.26120.38907 0.37512 0.38907 0.33573 0.053349,856.2863
200,000 0.05334 671.6215.884
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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Subject CT5 (Contingencies Core Technical) Examiners Report,
April 2011
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 = 528,538.9 Hence, total mortality profit =
528,538.9 475,183.5 = 53,355.4 (c) Although there is an overall
mortality profit in 2010, the actual number
of deaths for the endowment assurances is approximately 25%
higher than expected, which is a concern. Further investigation
would be required to determine reasons for poor mortality
experience for the endowment assurances, e.g. there may have been
limited underwriting requirements 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.
END OF EXAMINERS REPORT
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INSTITUTE AND FACULTY OF ACTUARIES
EXAMINATION
4 October 2011 (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.
CT5 S2011 Institute and Faculty of Actuaries
-
CT5 S20112
1 Calculate: (a) 10|1 [50]q (b) 10 [60] 1p + (c) (12)
[40]:20a
Basis: Mortality AM92 Rate 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 following
relationships are true for all ages:
0.5 [ ]
0.5 [ ] 0.5
0.25
0.45
x x
x x
q q
q q+
==
Express p[x] in terms of px . [3] 4 A term assurance contract
with a term of 20 years pays a sum assured of 1
immediately on death to a life now aged 30 exact. Calculate the
expected value and variance of this contract. Basis: Mortality AM92
Ultimate Rate 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: x = 0.02 for all x y = 0.03 for all y = 4%
per annum [5]
-
CT5 S20113 PLEASE TURN OVER
6 Explain why it is necessary to have different mortality tables
for different classes of lives. [6]
7 A special joint life last survivor annuity of 10,000 per annum
is payable
continuously in respect of a male and female life each aged 60
exact. Payments commence on the first death and continue for 5
years after the second death.
Calculate the expected present value of this annuity. Basis:
Mortality PMA92C20 (male life), PFA92C20 (female life) Rate of
interest 4% per annum Expenses Nil [6] 8 The following data is
extracted from a population census:
All Professions Profession A Age
Population Deaths Population Deaths
2029 120,000 256 12,500 30 3039 178,000 458 15,000 40 4049
156,000 502 16,000 50 5064 123,000 600 14,000 60
(a) Calculate the area comparability factor for Profession A
using the data for All
Professions as the standard population. (b) Hence or otherwise
derive the standardised mortality ratio and the indirectly
standardised mortality rate. [6] 9 Members of a pension scheme
are subject to three decrements: (a) Deaths - with independent
decrement rates that are assumed to follow
ELT15(Males) (b) Ill-health retirement - with an independent
decrement rate of 0.01 at age 50
exact increasing by 0.005 for each additional year of age (so
the ill-health independent decrement at age 53 exact is 0.025)
(c) Age retirement - with an independent decrement rate of 0.2
at each age from
60 to 64 all exact. Age retirements are assumed to take place on
the attainment of the exact age, whilst
other decrements act uniformly across the year of age. Calculate
the probability that a member currently aged 59 exact will retire
at age 62
exact. [6]
-
CT5 S20114
10 (i) Five years ago a with profits whole life assurance policy
was sold to a life then aged 30 exact.
The sum assured is 150,000 payable at the end of year of death
and
premiums are payable annually in advance throughout life. The
super compound method of adding bonuses to the policy is used as
follows:
each year there is a simple bonus of 2.5% on the sum assured
and an additional bonus of 5% on all existing bonuses (excluding
the
simple bonus relating to that policy year)
Assume that bonuses vest at the start of each policy year and
that the actual past bonus additions have followed the assumptions
stated above.
Calculate the net premium policy value just before payment of
the 6th
premium. Basis: Mortality AM92 Select Rate of interest 4% per
annum [5]
(b) Suggest two reasons why a life insurance company might use
the super
compound method of adding bonuses to with profits policies, as
opposed to the compound method. [2]
[Total 7] 11 A pension scheme provides a pension on retirement
of 1% of final pensionable salary
for each completed year of pensionable service. On retirement
due to ill-health, pensionabl