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ACTL30002 Actuarial Modelling II
End of semester Examination 2021


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-


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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.


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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.

Solution:

(a) 1 = 16, 2 = 18, 3 = 19 are the knots. The expressions used in this solution are given
in page 24 in week 7 lecture slides.

For (−∞, 16], the expression is 0 + 1. Then 1 = 0.00014−0.0001116−15 = 0.00003.
0 + 0.00003 ⋅ 15 = 0.00011 ⇒ 0 = −0.00034 ⇒ 14 = −0.00034 + 0.00003 ⋅ 14 =0.00008.

For [16, 18], the expression is
−0.00034 + 0.00003 ⋅ + 1( − 16)3.
By using

18 = 0.0003 , we have (18 − 16)31 = 0.0003 + 0.00034 − 0.00003 ⋅ 18 =0.0001 ⇒ 1 = 0.0000125 ⇒ 17 = −0.00034 + 0.00003 ⋅ 17 + 0.0000125(17 − 16)3 =0.0001825.

For [18,19], the expression is
−0.00034 + 0.00003 ⋅ + 0.0000125( − 16)3 − 1−3
2−3
⋅ 0.0000125( − 18)3.
Then we have

19 = −0.00034 + 0.00003 ⋅ 19 + 0.0000125(19 − 16)3 − 3 ⋅0.0000125(19 − 18)3 = 0.00053

For [19, 20], the expression is
−0.00034 + 0.00003 ⋅ + 0.0000125( − 16)3 − 1−3
2−3
⋅ 1( − 18)3 − 1−23−2 ⋅ 1( − 19)3.
Then we plug in x=20 then we have

20 = 0.000785.

(b) The 2 for 14, 17, 19 and 20 is 0.2259, 1.7766, 0.000254 and 0 respectively. Then the total
2=5.0744. The degree of freedom is 7-3=4. 42 = 9.486>5.074, so we don’t reject.

















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Question Seven

[3+2+1=6 marks]

You decide to fit a graduation by a mathematical formula of the form = + . Assume that
the Poisson model of deaths applies. You have available observed death data and observed
central exposed to risk data for all ages from 1 to inclusive. Your crude estimate ̂()
relates to exact age x.

(a) Derive all equations that you would need to solve for the parameters a and b in order to fit this
model using the method of maximum likelihood estimation.

After you have obtained the above equations, you suddenly realised that you were given more
details about the data. You are told that at each age , i =1, 2,…, n, the proportion of male
policyholders is . That is, of the population aged are male, and 1- of the population aged
are female. So the total number of deaths is composed of number of male deaths denoted by ,
and the number of female deaths.

(b) Write down the distribution of under the assumption that the central exposed to risk and the
force of mortality for a life aged are
and . Using the result, find expressions for ()
and ().

(c) You are also given that for age , i =1, 2… n, the observed number of male deaths is , so
the observed number of female deaths is -. You decide to revise your answer to (a). Derive
all equations that you would need to solve in order to fit this model using the method of maximum
likelihood estimation and compare your result with (a). Do you think the more detailed information
is useful in the parameter estimation?

Solution:

(a) ∼ ()
= ∏1 ()
! −
= 1{ [( + )] − ( + ) − !}


= 1(
+ − )


= 1(
+ − )
(b) ∼ ( ) ⇒ = =
(c) They would be the same. The detailed result is not useful.



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Question Eight

[2+4=6 marks]

The following Lee-Carter model has been fitted to an Australian mortality data for age 30 and a 5-
year time period from 2013 to 2017:
log, = + + ,.

(a) State the constraints that are imposed on parameters and , and explain why we need
the constraints.

(b) Suppose 30 = −6.1 , 30 = 0.5431 and 2017 = −0.4821 . Provide a forecast of
30,2020 and 95% confidence interval of �30,2020 based on a random walk model of :

+1 = + +

where ̂ = −0.1947 and �2 = 0.023 , for t=1, 2, …is a sequence of independent and
identically distributed random variables having zero mean and unit variance. Give your result
to 5 decimal places.

Solution:

The solution of this question is included in the excel file. Please note that you are not allowed
to use excel in 2022 final exam. You need to use a calculator to produce the results.

Question Nine

[2+1+3=6 marks]

You are expected to use the Reduction Factor Approach to do mortality forecasting. You are
given the following information:

• q54,1990=0.00891;
• 54=0.13;
• n=20;
• fn,54=0.65;

(a) Compute the forecasted rate for q54,2020.

(b) Find the long-term mortality rate at age 54, based on the 1990 mortality rate.

(c) You are given the updated information:
• q54,2005=0.00336;
• q54,2020=0.00319;



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Rather than relying on expert opinion, derive an updated estimation which fit the recent data,
using the updated information provided.

Solution:

The solution of this question is included in the excel file. Please note that you are not allowed
to use excel in 2022 final exam. You need to use a calculator to produce the results.


Question Ten

[6 marks]

A machine learning model is trained to predict tumor in patients. The test dataset consists of 200
people. In the dataset, you are provided with the following information:

• 14 people who have tumors are predicted positively by the model.
• 6 people who have tumors are predicted negatively by the model.
• 145 people who don’t have tumors are predicted negatively by the model.

(a) Find out the number of people who don’t have tumors that are predicted positively by the
model.

(b) Write out the Confusion Matrix for the model.

(c) Calculate the accuracy, the true positive rate (TPR), the F1 score and the false positive
score (FPR). Give your result in 2 decimal places.

Solution:

The solution of this question is included in the excel file. Please note that you are not allowed
to use excel in 2022 final exam. You need to use a calculator to produce the results.



Question Eleven [4+2+1=7 marks]
A questionnaire has been designed to find out people’s hobby to like playing video games or not.
The factors to consider are age, gender (male or female) and marriage status (married or not). The
dataset from 15 lives is given in below.

Life Age Gender Marriage status Like playing
video games?
The outcomes.
1 12 Male Not married Yes
2 10 Female Not married Yes
3 20 Male Not married No


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4 25 Female Married Yes
5 40 Male Married No
6 18 Male Not married Yes
7 32 Female Married Yes
8 60 Female Married No
9 55 Female Not married No
10 8 Female Not married Yes
11 23 Male Married No
12 37 Female Married No
13 42 Male Not married Yes
14 46 Female Not married No
15 35 Male Not married Yes

A decision tree is trained to predict people’s likes on playing video games. The splitting rules are
shown below.


Yes No




Male Female




Not married Married





(a) Fill out the Leaf Nodes A, B, C and D in the tree with the outcomes from the 15 lives.
(b) Compute the Gini index of the decision tree.
(c) Based on your result in Part (b), comment on the splitting rules.

Solution:

(a) & (b) & (c)
Leaf A:1Y, 2Y, 6Y, 10Y Index=0
Leaf B: 3N, 5N, 11N, 13Y, 15Y Index=2*2/5*3/5=12/25
Leaf C: 9N, 14N Index=0
Leaf D: 4Y, 7Y, 8N, 12N Index=2*2/4*2/4=1/2

Total index=5/15 * 12/25 + 4/15 * 1/2=0.16+0.13333=0.29333 which is close to 0 and be
Age ≤ 18?
Leaf Node A
Gender?
Leaf Node B Marriage status?
Leaf Node C Leaf Node D


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better than binary classifier. This number is closer to 0.5 than to 0, so not perform well.



[Total 70 Marks]
[End of Examination]