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ACTL30002 Actuarial Modelling
Question One
[2+6=8 marks]
Consider a two-year mortality investigation over the period of 1 January 2004 to 31 December
2005. You are given the following complete data for three lives included in this investigation. The
age definition is “age last birthday at entry plus curtate duration at death”.
Date of Birth Date of Entry Date of Death Date Policy Effected
A 21 July 1975 1 July 2004 23 August 2005 12 May 2000
B 17 July 1972 1 May 2004 1 September 2006 1 August 1998
C 9 November 1958 30 June 2003 12 April 2005 2 September 1972
(a) Describe the rate interval relevant to this mortality investigation and specify the type of the rate
interval.
(b) For each life, calculate the contribution to the central exposed to risk () for ages (x) where
the contribution is greater than zero. Give your answer in days.
Solution:
(a) It is a policy year rate interval commencing policy anniversary by age last birthday.
(b)
For A:
From 1 July 2004 to 11 May 2005, (28) =31+31+30+31+30+31+31+28+31+30+11=315 days
From 12 May 2005 to 22 August 2005, (29) =31-11+30+31+22=103 days
For B:
From 1 May 2004 to 31 July 2004, (31) =31+30+31=92 days
From 1 August 2004 to 31 July 2005, (32) =365 days
From 1 August 2005 to 31 December 2005, (33) =31+30+31+30+31=153 days
For C:
From 1 January 2004 to 1 September 2004, (44) =366-29-31-30-31=245 days
From 2 September 2004 to 11 April 2005, (45) =29+31+30+31+31+28+31+11=222 days
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Question Two
[2+5+2=9 marks]
The State of Victoria carried out a mortality investigation over the period of 1 January 2017 to 31
December 2018. The data were recorded as follows.
() represents deaths aged (x) where (x) is the age last birthday at 1 April falling in the calendar
year of death.
()(0) represents the population at 1 January 2017 aged (x) last birthday.
()(1) represents the population at 1 January 2018 aged (x) next birthday.
()(2) represents the population at 31 December 2018 aged (x) nearest birthday.
(a) Describe the rate interval relevant to this mortality investigation and specify the type of the rate
interval.
(b) Develop a formula to calculate the - type mortality rate at age (x), ̂(). State any assumptions
needed. If no assumptions are necessary, then you must state this explicitly.
(c) State the age to which your estimated mortality in (b) relates. State any assumptions needed. If
no assumptions are necessary, then you must state this explicitly.
Solution:
(a) It is a calendar year rate interval commencing 1 January by age last birthday at the following 1
April.
(b) ̂() = ()()
() = 0.5(()∗ (0) + (+1)∗ (1)) + 0.5(()∗ (1) + ()∗ (2)) where we have assumed that birthdays,
new entrants, deaths and withdrawals occur uniformly over the calendar year.
At 1 January 2017, ()∗ (0) = 0.25(−1)(0) + 0.75()(0) assuming birthdays are uniformly
distributed over the calendar year.
At 1 January 2018, ()∗ (1) = 0.25()(1) + 0.75(+1)(1) assuming birthdays are uniformly
distributed over the calendar year.
At 31 December 2018, ()∗ (2) = 0.75(+1)(2) + 0.25(+2)(2) assuming birthdays are
uniformly distributed over the calendar year.
(c) ̂() relates to +0.75 assuming birthdays are uniformly distributed over the calendar year.
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Question Three
[1+2=3 Marks]
The Date of birth, Date of insurance policy effected, Date of death are given for two lives, Life A
and Life B in the table below.
The age definition is “age nearest birthday at the nearest 1 July at the nearest policy anniversary”.
Life Date of Birth Date policy effected Date of Death
A 15 August 1980 30 June 2004 31 August 2017
B 15 September 1960 15 October 1998 3 November 2010
(a) Write down the rate interval for the given age definition.
(b) Determine the age label according to the given age definition for two lives at the time of
death.
Solution:
(a) It is a policy year rate interval, commencing from half year before the PA at which the life
is aged x nearest birthday at the nearest 1 July.
(b) For Life A: 37; Life B: 50.
Question Four
[5 marks]
Insurance company ABC recently conducted a mortality investigation aiming to verify whether a
standard mortality table is suitable for modelling its own policyholders. The data collected cover
twenty ages for a large group of insurance policyholders, consisting of initial exposed to risk values
and numbers of deaths for ages 11 to 30. As an appointed actuary, you are asked to prepare a report
that addresses the above issue using the data provided.
Firstly, you decide to adopt the binomial model for the number of deaths, i.e. ∼ (,)
where denote the number of deaths, denote the initial exposed to risk of age x, and denote
the one year death probability at age x. You have calculated the following table of the individual
standardised deviations (ISD).
Age 11 12 13 14 15 16 17 18 19 20
ISD 0.49 -0.24 -0.11 -0.19 1.14 -1.26 0.54 2.47 -0.61 -1.06
Age 21 22 23 24 25 26 27 28 29 30
ISD -1.31 -0.30 1.96 -1.53 -2.13 1.08 -0.04 -0.89 0.80 1.29
Test the null hypothesis that the standard table mortality rates describe the mortality experience of
Insurer ABC. Use Runs test. Quote the null and the alternative hypotheses, the test statistic, the p-
4
value and the conclusion of your test. Conduct the test at the 5% significance level. Do NOT use
the normal approximation.
Solution:
H0: The standard mortality table describes the mortality experience of Insurer ABC.
H1: The standard mortality table doesn’t describe the mortality experience of Insurer ABC.
For Runs test, 0 = 12,1 = 8, = 20, = 6. Let T be the number of positive runs of ISDs. Then
we have
( = ) = �1 − 1 − 1 � �0 + 1 �
�
1
�
= � 7 − 1� �13 �
�208 �
Pr( 6)
7 13 7 13
6 7 7 8
1 Pr( 7) Pr( 8) 1
20 20
8 8
7 1716 1 12871 0.89443 0.05
125970 125970
p value T
T T
− = ≤
= − = − = = − −
× ×
= − − = >
We don’t reject H0.
Question Five
[5 marks]
A study was carried out to investigate the impact of a certain medicine. 30 patients joined the
investigation. The data are shown below, where for each patient, “+” means that the health
condition was improved and “-” means that the health condition got worse.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
+ + + - - - + + + + + - - - -
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
+ + - + + + + + + + - + + + +
Suppose your null hypothesis is that the medicine has no effect on changing health conditions.
Your alternative hypothesis is that the medicine improves health conditions. Use a sign test to test
the above null hypothesis. According to the purpose of this test, is it a one-sided or two-sided test?
Perform your test at the 5% significance level. Use the normal approximation and continuity
correction. State the value of the test statistic, the p-value and the conclusion of your test.
5
Solution:
There are 21 “+” out of 30. It is a one-sided test.
If ∼ Bio(30,0.5), then we have () = 15 and () = 7.5
p-value=( ≥ 21) = ( ≥ 20.5−15
√7.5 ) = ( ≥ 2.01) = 1 −Ф(2.01) = 1 − 0.9778 = 0.02 <0.05 so reject H0.
Question Six
[6+3=9 marks]
A Natural Cubic Spline Graduation with knots at ages 16, 18 and 19 is performed based on the
mortality data for seven ages shown below.
Age (x) Initial ETR Actual deaths
14 2823 0
15 3399.5 1
16 3888 0
17 4408.5 2
18 4928.5 0
19 5608.5 3
20 6354.5 5
Recall that the form of the natural cubic spline function with n knots 1,2..., is () = 0 +
1 + ∑ −2=1 () , where () = () − −−−1 −1() + −1−−−1 () and () =( − )3 for x greater than or equal to and zero otherwise.
The graduated values for all ages EXCEPT ages 14, 17, 19 and 20 are given in the table below.
Age 15 16 18
0.00011 0.00014 0.00030
(a) Calculate the graduated mortality rates at ages 14, 17, 19 and 20. Keep as many decimal places
as you can during your working for this part (a).
Note: Since you are not provided with the weights used in the parameter estimation, so you cannot
use the existing Excel Spreadsheet from Macquarie University to obtain the graduated rates. You
need to work out the expression for the natural cubic spline used in this question, then find out the
graduated rates.
(b) Test the graduation for adherence to data. Use the Chi-squared test. Conduct the test at the 5%
significance level. Quote the null and alternative hypotheses, the test statistic (to five decimal
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places), the critical value and the conclusion of your test.