introduction to actuarial mathematics

SOA Course 3Study program – Fall 2004 exams

Thank you for downloading the free copy of Part 1 of the BPP Course 3 Study Programfor the Fall 2004 exams. We are confident that you will find it helpful.

Contents

This file includes the following :

• BPP Course 3 Study Guide• Unit 1 – Actuarial models• Unit 2 – The life table• Unit 3 – Insurance functions• Unit 4 – Annuities• Question & Answer Bank Part 1 – Questions• Question & Answer Bank Part 1 – Solutions

The full BPP Course 3 Study Program includes:

• Course Notes on all syllabus topics (no other textbooks are required)

• An expanded Question & Answer Bank for each part that includes 50 questions with solutions

• Assignments following each part with 25 exam-style questions with full solutions

• Past exam pack includes full solutions to all SOA/CAS exams since May 2000

• Email Tutor Support, designed to help you if you get stuck.

You can order the full BPP Course 3 Study Program online at www.bpp.com

This page is intentionally left blank.

Course 3 Study Guide – Fall 2004

© BPP Professional Education: 2004 exams Page 1

Course 3Study Guide for the Fall 2004 exam

Overview

This Study Guide contains everything you need to know before starting to use the BPPCourse 3 study program, including:

• the structure of the course notes and practice questions• advice on how to study efficiently and prepare for the exam• a study plan for the session leading to the Fall 2004 exam• details of how to email a question to our tutors if you get stuck• a listing of additional study support available from BPP• the full Syllabus• the Tables for the Course 3 exam (produced by the profession)• binder labels.

How long will this Study Guide take to read?

It should take about 1 hour to read this Study Guide carefully. It will be time well spent,because you’ll be able to work through the course notes more efficiently if you understandhow the BPP study program is structured.

Study Guide – Fall 2004 Course 3

Page 2 © BPP Professional Education: 2004 exams

Contents

Section Page

1 Introduction 3

2 How to use the BPP study program 4

2.1 Structure of the course 42.2 Structure of each unit 62.3 Binder labels 7

3 Tutor support and optional products and services 9

3.1 Tutor support 93.2 Practice exams 103.3 Flashcards 103.4 Seminars 10

4 Study skills 11

4.1 Overall study plan 114.2 Study sessions 114.3 Order of study 124.4 Active study 13

5 Possible study plans 14

6 Syllabus 16

Appendix

Tables for the Course 3 examination



1 IntroductionWe’re confident that this BPP Course 3* study program will have a very positive impact onthe way you study for the actuarial exams. We believe that students should spend theirtime focusing on learning and preparing for the exam, not juggling their time betweentextbooks, study manuals to re-explain those textbooks and additional textbooks withsolutions. BPP’s courses contain everything you need to know. Just start at page 1 andenjoy a simpler approach to study.

We’ve designed the courses to cover the learning objectives in an order that’s right for you. Our notes include worked examples, self-test questions and review summaries to aid yourunderstanding. We also include a large number of brand new exam-style questions, whichare arranged clearly so that you can easily find the right questions to practice. And, as anadded feature, the BPP Past Exam Pack, with solutions to all of the Course 3 examsreleased by the SOA since May 2000 have been added to the study program. Ourexperienced tutors have worked hard on a course structure so you can focus on learning.

For BPP, providing a study program means providing a commitment to support students allthe way through to the exam. Our full-time tutors will answer any questions you may haveby email so that you always have the support you need. And we also offer seminars,which are co-ordinated with your study material to give you the very best chance ofsuccess in the exam.

Passing the actuarial exams is difficult enough. We understand that you’re competingagainst some very bright people and that it’s always hard to find time for studying betweenworking and socializing. The amount of material covered in Course 3 is daunting and thelast thing you need to worry about is how to approach the study. BPP’s study programsprovide the very best support to help you maximize your chances of success in the examand to make good progress towards qualification.

As you have more contact with BPP, you’ll find us to be a very friendly and proactivecompany. We are committed to providing high quality study support. We respond quicklyto students’ requests and will do all we can to meet your specific needs, eg offering aseminar near you. We look forward to helping you.

Good luck with your studying.

David CarrCEO, BPP Professional Education

* Note to CAS students: This study program is designed to help students prepare for the SOACourse 3 exam. Please check the BPP website for details of our CAS Exam 3 study program.



2 How to use the BPP study program

2.1 Structure of the course

Course notes

The course notes are divided into 5 parts to help you structure your progress through thesession leading to the Course 3 exam.

The following table shows how the parts and the units relate to each other and provides anestimate of the time you should plan for the first run-through of each unit:

Part Unit Title Time(hours)

1 Actuarial models 1.5

2 The life table 6

3 Insurance functions 41

4 Annuities 4

5 Premiums 2.5

6 Reserves 3

7 Joint life functions 82

8 Multiple decrement models 4

9 Introduction to loss distributions 1

10 Frequency distributions 3

11 Continuous distributions 4

12 Compound distributions 4

3

13 Reinsurance 3

14 Markov chains 4

15 The Poisson process 4416 Brownian motion 4

17 Ruin theory 5518 Simulation 3

Each unit of the course notes includes a unit summary – the summaries can serve as auseful review aid.



Throughout the course notes you will find short self-assessment questions, which aredesigned to test your understanding of the material and to ensure that you study actively.They are not intended to be representative of exam questions. Solutions are given at theend of each chapter.

Question & Answer Bank

The Question & Answer Bank is also divided into five parts. Each part of the Question &Answer Bank includes 50 new exam-style questions to test your understanding of theassociated part of the notes.

The Review Question & Answer Bank contains 75 new exam-style questions to test yourunderstanding of the whole course.

Assignments

There are five assignments to help you assess your level of understanding of theassociated part of the notes. Each assignment includes 25 new exam-style questions andshould take around 2 hours to complete.

Order of study

For each part of the course, we recommend that you:

• read the course notes carefully• review the course notes• attempt several questions from the Question & Answer Bank• attempt the assignment under exam conditions.

For more detail on how to structure your studying, see Section 4 on Study Skills.

Added Feature – Past exam packs

Our past exam packs contain all of the questions from the SOA released Course 3 exampapers from May 2000 to present and BPP’s full solutions, including comments to help youwith your exam technique, references to the course notes, and alternative approaches.



2.2 Structure of each unit

As you work through the course notes, you’ll see a variety of symbols in the margin. Thesymbols have been designed to help you find your way through the course notes quicklyand efficiently.

The symbols and their meanings are as follows:

An important point, often a definition or a new concept.

An example question with a fully worked solution.

A self-assessment question to test your understanding of the course. Fullsolutions are given at the end of the unit.

A multiple-choice question, which may be appropriate for the exam. Fullsolutions are given at the end of the unit.

You’ll also find the following symbols at the start of each unit:

An overview of the unit with advice on the main points to bear in mind.

A guide to the time required to study each unit on your first run-through.

The relevant learning objectives are also listed at the start of each unit.



2.3 Binder labels

Cut out the labels printed below and place them in the pockets on the spines of your BPPbinders.

Course 3Actuarial Models

Fall 2004

Volume 1

Course 3Actuarial Models

Fall 2004

Volume 2



This page has been left blank so thatyou can cut out and use the binder labels.



3 Tutor support and optional BPP products and services

3.1 Tutor support—included with every BPP study program

Our tutors are here to help you pass. If you get stuck, you can email technical questionsabout the course material to our team of Course 3 tutors. We’ll answer your questionsthoroughly and quickly so that you can get on with your studying with no worries.

If you have a question about the course material

From time to time you may come across something in the study material that is unclear toyou. The best way to solve such problems is often through discussion with friends andcolleagues – they will probably have had similar experiences while studying. What’s more,you’ll usually remember the answer more clearly if you worked it out for yourself, with orwithout help from your friends.

If you are still stuck, then you can email your questions to our team of Course 3 tutors. You should send an email to:

[email protected]

Please help us to help you by stating your name and your question clearly. Also, tell usyour fax number if you have one – our answers might include an actuarial formula that iseasier to fax than to type in an email!

We also put answers to frequently asked questions about the course material on ourwebsite at www.bpp.com. This information is updated regularly and is consolidated intothe rewriting of the course material each session.

If you have any non-technical questions about BPP’s study programs

If you have any non-technical questions or if you have any ideas, suggestions or feedbackon BPP’s study programs, you should send an email to:

[email protected]

Please don’t hesitate to contact us – we look forward to helping you. And, don’t forget toperiodically check our website at www.bpp.com for other useful study tips that will appearfrom time to time.



3.2 Practice exams

BPP has two Course 3 practice exams available to give you a realistic test of your exampreparation. Each practice exam contains brand new exam-style questions and is suppliedwith full solutions.

Each practice exam costs $25.

3.3 Flashcards

For Fall 2004, BPP is offering user-friendly flashcards to supplement the study program.Great for those on the go, they’ll help you remember the most important formulas,concepts, alternative solution methods and exam shortcuts.

3.4 Seminars

BPP offers a range of public seminars in major cities to help you prepare effectively for theexams. Our seminars have a maximum group size of 18 which means that every studentwill be able to ask questions and get personal help from the instructor.

Our seminars are designed to be interactive. They are challenging but enjoyable—apositive learning experience clearly focused on helping each student pass the exam.

We pride ourselves on the quality of our teaching. In feedback from past seminars, 96% ofstudents rated our instructors in the top two categories – Very Good (83%) or Good (13%).

The most up-to-date information on upcoming BPP seminars with dates, locations andseminar registration deadlines can be found on the BPP website at www.bpp.com.



4 Study skills

4.1 Overall study plan

Develop a realistic study plan, build in time for relaxation and allow some time forcontingencies. Be aware of busy times at work, when you may not be able to take asmuch study leave as you would like. Once you have set your plan, be determined to stickto it. (You don’t have to be too prescriptive at this stage about what precisely you do oneach study day. The main thing is to be clear that you will cover all the important activitiesin an appropriate manner.)

Manage your study to allow plenty of time for the concepts you meet in this course tobecome ingrained in your mind. Most successful students will complete the course nolater than the end of March, leaving at least a month for review. By finishing the course asquickly as possible, you will have a much clearer view of the big picture. It will also allowyou to structure your review so that you can concentrate on the important and difficultareas of the course. How often do you think “I’m just getting the hang of this, I wish theexam was two weeks later”?

Two possible study plans are included in Section 5 of this Study Guide.

4.2 Study sessions

Only do activities that will increase your chance of passing. Don’t include activities for thesake of it and don’t spend too much time reviewing material that you already understand. You will only improve your chances of passing the exam by getting on top of the materialthat you currently find difficult.

Each study session should have a specific purpose and be based on a specific task,eg “Finish reading Unit 4 and attempt Questions 21 through 25 from the Part 1 Questionand Answer Bank” not a specific amount of time, eg “Three hours studying the material inUnit 3”.

Study somewhere quiet and free from distractions (eg a library or a desk at homededicated to study). Find out when you operate at your peak, and try to study at thosetimes of the day. This might be between 8 am and 10 am or could be in the evening. Takeshort breaks during your study to remain focused – it’s definitely time for a short break ifyou find that your brain is tired and that your concentration has started to drift from theinformation in front of you.



4.3 Order of study

You should work through each of the units in turn. To get the maximum benefit from eachunit you should proceed in the following order:

1. Read the learning objectives. These are set out in the box at the beginning of eachunit.

2. Take a few minutes to look at the unit summary at the end of each unit. This will giveyou a useful overview of the material that you are about to study and help you toappreciate the context of the ideas that you meet.

3. Study the BPP course notes in detail, annotating the pages and possibly making yourown notes. Try the short self-assessment questions as you come to them. Oursuggested solutions are at the end of each unit.

4. Read the unit summary again carefully. If there are any ideas that you can’tremember covering in the notes, read the relevant section of the notes again torefresh your memory.

You may like to attempt some questions from the Question & Answer Bank when you havecompleted a part of the course. It’s a good idea to annotate the questions with details ofwhen you attempted each one and whether you got it right or wrong. This makes it easier toensure that you try all of the questions as part of your review without repeating any that yougot right first time.

Once you’ve read the relevant part of the notes and tried a selection of questions from theQuestion & Answer Bank, you should attempt the corresponding assignment. Try tocomplete the assignment under exam conditions – monitor the time you have taken and don’trefer to your notes. Although this seems difficult, it will really help you to prepare for theexam. You’ll learn much more from getting something wrong than from getting it right simplyby referring to the course notes.

It’s a fact that people are more likely to remember something if they review it periodically. So,do look over the units you have studied so far from time to time. It is useful to re-read theunit summaries or to try the self-assessment questions again a few days after reading theunit itself.

To be really prepared for the exam, you should not only know and understand the coursenotes but also be aware of what the examiners will expect. Your review program shouldinclude plenty of question practice so that you are aware of the typical format, content andlength of exam questions. You should attempt as many questions as you can from theQuestion & Answer Bank (including the review part) and the past exam pack.



4.4 Active study

Here are some techniques that will help you to study actively.

1. Don’t believe everything you read! Good students tend to question everything thatthey read. They will ask “why, when, how?” when confronted with a new concept,and they will apply their own judgement. The self-assessment questions will helpyou to achieve this.

2. As you read the course notes, think of possible questions that the examiners couldask. This will help you to understand the examiners’ point of view and should meanthat there are fewer nasty surprises in the exam room.

3. Annotate your notes with your own ideas and questions. This will make your studymore active and will help when you come to review the material. Do not simply copyout the notes without thinking about the issues.

4. Attempt the questions in the notes as you work through the course. Write downyour answer before you check against the solution.

5. Attempt other questions on a similar basis, ie write down your answer before lookingat the solution provided. Attempting the assignments under exam conditions forcesyou to think and act in a way that is similar to how you will behave in the exam.

6. Attend a BPP Course 3 review seminar. These focus on developing the skillsneeded to pass the exam. The registration deadlines for the Fall 2004 examseminars are shown on the BPP website at www.bpp.com.

7. Sit a practice exam a few weeks before the real exam to identify your weaknessesand work to improve them. Use the past exam pack and supplement that with oneor both of the practice exams written by BPP.



5 Possible study plansHere are possible study plans for students starting in May 2004 and July 2004.

For students starting to study in May 2004

Week#

Weekbeginning

Activity

1 May 17

2 May 24Read & review Part 1 of the course notes

3 May 31 Attempt questions from Q&A Bank Part 1 and complete Assignment 1

4 June 7

5 June 14Read & review Part 2 of the course notes

6 June 21 Attempt questions from Q&A Bank Part 2 and complete Assignment 2

7 June 28 Review Parts 1 & 2

8 July 5

9 July 12Read & review Part 3 of the course notes

10 July 19 Attempt questions from Q&A Bank Part 3 and complete Assignment 3

11 July 26

12 August 2*Read & review Part 4 of the course notes

13 August 9 Attempt questions from Q&A Bank Part 4 and complete Assignment 4

14 August 16 Review Parts 3 & 4

15 August 23

16 August 30Read & review Part 5 of the course notes

17 September 6 Attempt questions from Q&A Bank Part 5 and complete Assignment 5

18 September 13 Enjoy a well-deserved week off

19 September 20

20 September 27

21 October 4

22 October 11

23 October 18

24 October 25

Six-week review period:

Attempt all questions from Review Q&A Bank and all remainingquestion from Q&A Bank Parts 1 – 5

Attempt past exams or a BPP practice exam under exam conditions

Attend a BPP review seminar

Identify your weaknesses and work to improve them

25 November 1 Course 3 exam (November 2)

* Note: Check the BPP website for seminar registration deadlines.



For students starting to study in July 2004

Week#

Weekbeginning

Activity

1 July 5

2 July 12Read & review Part 1 of the course notesAttempt questions from Q&A Bank Part 1 and complete Assignment 1

3 July 19

4 July 26Read & review Part 2 of the course notesAttempt questions from Q&A Bank Part 2 and complete Assignment 2

5 August 2*

6 August 9Read & review Part 3 of the course notesAttempt questions from Q&A Bank Part 3 and complete Assignment 3

7 August 16

8 August 23Read & review Part 4 of the course notesAttempt questions from Q&A Bank Part 4 and complete Assignment 4

9 August 30

10 September 6Read & review Part 5 of the course notesAttempt questions from Q&A Bank Part 5 and complete Assignment 5

11 September 13 One spare week to allow for vacations & contingencies

12 September 20

13 September 27

14 October 4

15 October 11

16 October 18

17 October 25

Six-week review period:

Attempt all questions from Review Q&A Bank and all remainingquestion from Q&A Bank Parts 1 – 5

Attempt past exams or a BPP practice exam under exam conditions

Attend a BPP review seminar

Identify your weaknesses and work to improve them

18 November 1 Course 3 exam (November 2)

* Note: Check the BPP website for seminar registration deadlines.

Course 3 contains a large amount of material, including many new concepts that will taketime to understand. If you start to study in mid-July, you will be under enormouspressure to complete the reading quickly and leave an adequate amount of time forreview. You’ll need to work hard and work efficiently to complete each part thoroughly intwo weeks.



6 Syllabus

The Syllabus is an important part of your study. You should review this Syllabus now andwhen you have completed the course notes to make sure that you understand what youare required to know for the exam. The relevant learning objectives are listed at the startof each unit.

Understanding Actuarial Models

The candidate is expected to understand the models and techniques listed below and tobe able to apply them to solve problems set in a business context. The effects ofregulations, laws, accounting practices and competition on the results produced by thesemodels are not considered in this course. The candidate is expected to be able to:

1. Explain what a mathematical model is and, in particular, what an actuarial modelcan be.

2. Discuss the value of building models for such purposes as: forecasting, estimatingthe impact of making changes to the modeled situation, estimating the impact ofexternal changes on the modeled situation.

3. Identify the models and methods available, and understand the difference betweenthe models and the methods.

4. Explain the difference between a stochastic and a deterministic model and identifythe advantages/disadvantages of each.

5. Understand that all models presented (eg survival models, stochastic processes,aggregate loss models) are closely related.

6. Formulate a model for the present value, with respect to an assumed interest ratestructure, of a set of future contingent cash flows. The model may be stochastic ordeterministic.

7. Determine the characteristics of the components and the effects of changes to thecomponents of the model in 6. Components include:

• a deterministic interest rate structure;• a scheme for the amounts of the cash flows;• a probability distribution of the times of the cash flows; and• the probability distribution of the present value of the set of cash flows.



8. Apply a principle to a present value model to associate a cost or pattern of costs(possibly contingent) with a set of future contingent cash flows.

• Principles include: equivalence, exponential, standard deviation, variance,and percentile.

• Models include: present value models based on 9 –12 below.• Applications include: insurance, health care, credit risk, environmental risk,

consumer behavior (eg subscriptions), and warranties.

9.** Characterize discrete and continuous univariate probability distributions for failuretime random variables in terms of the life table functions, xl , xq , xp , n xq , n xp ,and |m n xq , the cumulative distribution function, the survival function, the probability

density function and the hazard function (force of mortality), as appropriate.

• Establish relations between the different functions.• Develop expressions, including recursion relations, in terms of the functions

for probabilities and moments associated with functions of failure timerandom variables, and calculate such quantities using simple failure timedistributions.

• Express the impact of explanatory variables on a failure time distribution interms of proportional hazards and accelerated failure time models.

10. Given the joint distribution of two failure times:

• Calculate probabilities and moments associated with functions of theserandom variables.

• Characterize the distribution of the smaller failure time (the joint life status)and the larger failure time (the last survivor status) in terms of functionsanalogous to those in 9, as appropriate.

• Develop expressions, including recursion relations, for probabilities andmoments of functions of the joint life status and the last survivor status, andexpress these in terms of the univariate functions in 9 in the case in whichthe two failure times are independent.

• Characterize the joint distribution of two failure times, the joint life status andthe last survivor status using the common shock model.



11. Characterize the joint distribution (pdf and cdf) of the time until failure and thecause of failure in the competing risk (multiple decrement) model, in terms of the

functions ( )xlt , ( )

t xq t , ( )t xp t , ( )

t xd t , ( )( )t x ttm .

• Establish relations between the functions.• Given the joint distribution of the time of failure and the cause of failure,

calculate probabilities and moments associated with functions of theserandom variables.

• Apply assumptions about the pattern of failures between integral ages toobtain the associated (discrete) single decrement models from a discretemultiple decrement model as well as the discrete multiple decrement modelthat results from two or more discrete single decrement models.

12.** Generalize the models of 9, 10, and 11 to multiple state models characterized interms of transition probability functions or transition intensity functions (forces oftransition).

13. Define a counting distribution (frequency distribution).

• Characterize the following distributions in terms of their parameters andmoments: Poisson, mixed Poisson, negative binomial and binomialdistributions.

• Identify the applications for which these distributions are used and thereasons why they are used.

• Given the parameters of a distribution, apply the distribution to anapplication.

14. Define a loss distribution.

• Characterize the following families of distributions in terms of theirparameters and moments: transformed beta, transformed gamma, inversetransformed gamma, lognormal and inverse Gaussian.

• Apply the following techniques for creating new families of distributions:multiplication by a constant, raising to a power, exponentiation, and mixing.

• Identify the applications in which these distributions are used and thereasons why they are used.

• Given the parameters of a distribution, apply the distribution to anapplication.

15. Define a compound distribution.

16.** Calculate probabilities associated with a compound distribution when thecompounding distribution is a member of the families in 13, and the compoundeddistribution is discrete or a discretization of a continuous distribution.



17. Adjust the calculation of 16 for the impact of policy modifications such asdeductibles, policy limits and coinsurance.

18. Define a stochastic process and distinguish between discrete-time and continuous-time processes.

19. Characterize a discrete-time Markov chain in terms of the transition probabilitymatrix.

• Use the Chapman-Kolmogorov equations to obtain probabilities associatedwith a discrete-time Markov chain.

• Classify the states of a discrete-time Markov chain.• Calculate the limiting probabilities of a discrete-time Markov chain.

20. Define a counting process.

21. Characterize a Poisson process in terms of

• the distribution of the waiting times between events,• the distribution of the process increments,• the behavior of the process over an infinitesimal time interval.

22. Define a nonhomogeneous Poisson process.

• Calculate probabilities associated with numbers of events and time periodsof interest.

23. Define a compound Poisson process.

• Calculate moments associated with the value of the process at a given time.• Characterize the value of the process at a given time as a compound

Poisson random variable.

24. Define a Brownian motion process.

• Determine the distribution of the value of the process at any time.• Determine the distribution of a hitting time.• Calculate the probability that one hitting time will be smaller than another.• Define a Brownian motion process with drift and a geometric Brownian

motion process.

25. For a discrete-time surplus process:

• Calculate the probability of ruin within a finite time by a recursion relation.• Analyze the probability of ultimate ruin via the adjustment coefficient and

establish bounds.



26. For a continuous-time Poisson surplus process:

• Derive an expression for the probability of ruin assuming that the claimamounts are combinations of exponential random variables.

• Calculate the probability that the surplus falls below its initial level,determine the deficit at the time this first occurs, and characterize themaximal aggregate loss as a compound geometric random variable.

• Approximate the probability of ruin using the compound geometricrecursion.

• Analyze the probability of ruin: analytically (eg adjustment coefficient);numerically; and by establishing bounds.

• Determine the characteristics of the distribution of the amount of surplus(deficit) at: first time below the initial level; and the lowest level (maximalaggregate loss).

27.** Analyze the impact of reinsurance on the probability of ruin and the expectedmaximum aggregate loss of a surplus process.

28. Generate discrete random variables using basic simulation methods.

29. Generate continuous random variables using basic simulation methods.

30. Construct an algorithm to appropriately simulate outcomes under a stochasticmodel.

Applications of Actuarial Models

The candidate is expected to be able to apply the models above to business applications.The candidate should be able to determine an appropriate model for a given businessproblem and be able to determine quantities that are important in making business decisions,given the values of the model parameters. Relevant business applications include, but arenot limited to:

• Premium (rate) for life insurance and annuity contracts,• Premium (rate) for accident and health insurance contracts,• Premium (rate) for casualty (liability) insurance contracts,• Premium (rate) for property insurance contracts,• Rates for coverages under group benefit plans,• Loss reserves for insurance contracts,• Benefit reserves for insurance contracts,• Resident fees for Continuing Care Retirement Communities (CCRCs),**• Cost of a warranty for manufactured goods,• Value of a financial instrument such as: a loan, a stock, an option, etc.,• Risk classification.• Solvency (ruin).

** Based upon the official SOA notice of 12/21/00, the learning objective/application stated here is notconsistent with the references listed in the SOA/CAS syllabus. The BPP Study Program is based on the mostcurrent SOA readings.

All study material produced byBPP Professional Education is copyright

and is sold for the exclusive use of the purchaser.

You may not hire out, lend, give out, sell,store or transmit electronically or photocopy

any part of the study material.

You must take care of your study material to ensurethat it is not used or copied by anybody else.

Legal action will be taken if these terms are infringed. In addition, we may seek to take disciplinary actionthrough the profession or through your employer.

Actuarial models Course 3

Fall 2004 exams © BPP Professional Education

Course 3Fall 2004 exams

Part 1Course Notes

Instructions

Read the Course Notes actively. Do not blindly accept what you read. You should thinkabout the ideas covered and decide for yourself whether you agree.

Attempt the self-assessment questions as you work through the Course Notesand write down your answer before you check against the solution.

When you have finished the Course Notes in this part of the course, attempt some of thequestions in the Question & Answer Bank.

Course 3 Professional models

© BPP Professional Education Fall 2004 exams






Legal action will be taken if these terms are infringed.In addition, we may seek to take disciplinary actionthrough the profession or through your employer.

Course 3, Unit 1 Actuarial models


Unit 1Actuarial models

Unit overview

This chapter introduces you to some of the actuarial models that are covered in Course 3.It also reminds you of some of the models that you should already have come across inCourses 1 and 2.

This chapter is an introduction to the whole of Course 3 and will give you a good overviewof the techniques you will be studying during the rest of the course.

Advice on study

This chapter is not as long as many of the other chapters in the course. It covers a widerange of ideas but is not very mathematical. You will have no formulae to learn, so youshould be able to read through the ideas relatively quickly.

How long will this unit take to study?

We suggest that you might need about 1½ hours to study the ideas outlined in thischapter.

Actuarial models Course 3, Unit 1


Learning objectives

1 Explain what a mathematical model is, and, in particular, what an actuarial modelcan be.

2 Discuss the value of building models for such purposes as: forecasting, estimatingthe impact of making changes to the modeled situation, estimating the impact ofexternal changes on the modeled situation.

4 Explain the difference between a stochastic and a deterministic model and identifythe advantages/disadvantages of each.



1 Introduction

A model is a theoretical representation of a practical situation.

Actuaries spend a considerable amount of time trying to predict what is likely to happen inthe future. In order to do so, they look at what has happened in the past, and try toquantify it. This process usually involves building a model of some kind.

The type of model used will depend very much on the practical situation being observed.It may also depend on the answer to some (or all) of the following questions.

• How accurate do I want my model to be?

• Is simplicity more important than accuracy?

• Who is going to interpret the results of the model?

• Are there any follow-up studies that need to be done based on the model I amusing?

In order to develop our ideas, let us look at a simple example that you should already befamiliar with.

Example 1.1

An actuary wishes to model the number of motor claims that are expected in the comingyears from a closed portfolio of policies. He has decided to use the Poisson distribution tomodel the number of claims. Outline how he might go about setting up the model.

Solution

We have deliberately chosen a relatively simple situation here. The stages he might gothrough are as follows:

• Collect some past data. The actuary will need to know how many claims havebeen received in the past.

• Decide on whether to stratify the data, or to look at the portfolio as a whole. Arethe claim numbers likely to be affected by factors such as type of car, address ofdriver and so on?



• If these factors are important, we must decide which to use and which to ignore.There are likely to be a number of factors whose importance is marginal.

• Subdivide the data into the different groups that are believed to be likely to havedifferent claim rates.

• Within each group, estimate the claim rate from the past data. This would involvecalculating a Poisson parameter for each group of data.

• Decide whether there are factors that are likely to change our experience in thefuture, and allow for these where possible.

• Predict what our claims experience is likely to be in future, allowing for any externalfactors.

• In due course, review the claims that come in subsequently, and update the modelin the light of this updated experience.

Notice that there are really three basic processes:

(i) Analyze the data that has been received in the past

(ii) Use the past data to predict what might be about to happen in the future.

(iii) Subsequently, update the experience to try to improve the accuracy of futurepredictions.

While not all actuarial models work in precisely this way, many do, and this procedure,which can be used in a wide variety of situations, is one example.

Question 1.1

How might the actuary test that the Poisson distribution was an appropriate model in theexample above?



2 Models you have seen before

2.1 The deterministic cash flow model

You have already used cashflows to model situations as part of your study of previouscourses. For example, you have used cashflow models to determine the rate of return onan investment.

Example 1.2

In return for an initial investment of $1,000, a company promises to pay an investor thesum of $200 at the end of a year, and to return the initial investment of $1,000 at the endof the second year. What is the return on this investment?

Solution

A simple equation of value will provide the answer here:

1000 200 1000 2= +v v ⇒ + − =5 5 02v v

Solving this equation of value, we find that:

1 1 4 5 5 0.9049910

v - ± - ¥ ¥ -= =

So the rate of return on the investment is:

1 1 10.5%0.90499

- =

2.2 Stochastic cashflow models

You may also have seen an example of a stochastic cashflow model. Here the amountand/or timing of the cashflows may be uncertain, and statistical analysis may be needed.Here is a very simple example.



Example 1.3

It is believed that the interest rate on a certain type of bond will be either 4% or 6% for thenext three years, but the actual value is unknown. Either possibility is believed to beequally likely. Assuming that the actual interest rate is constant throughout the timeperiod, find the expected accumulated value at time 3 of an initial investment of $1 at time0.

Solution

The accumulated amount of the investment will be either 1043. or 1063. . Since each ofthese is equally likely, the expected accumulated value will be:

3 3½ 1.04 ½ 1.06 1.15794¥ + ¥ =

So the expected accumulated value is $1.16.

Question 1.2

What is the variance of the accumulated amount at the end of three years, assuming thesame interest rate model as in the example above?

Of course, the distribution of the future interest rate is likely to be more complex than thatgiven here. For example we might want to use a continuous distribution taking values onthe range (4%, 8%) to get a more realistic model. It is also unlikely that the interest ratewould be fixed for the whole term of the investment.

Question 1.3

Suppose that the rate of return in each year had been equally likely to be 4% or 6%, withreturns in individual years being independent. What would the expected accumulationhave been at the end of three years in this case?



Models of future interest rate patterns can be much more complex than this. In the realworld there are likely to be interactions between interest rates, inflation, money supply andother economic variables. A more realistic interest rate model might have time relatedvariables for each of these parameters (ie a series of random variables ( )X t denotinginterest rates at time t , a series of random variables ( )I t denoting inflation at time t , andso on). We might then expect ( )X t to be dependent on inflation at earlier time periods, ie

( )X t will be dependent on ( 1)I t - , ( 2)I t - , as well as other random variables. You willsee more of time series models of this type in Course 4.

These two simple examples should remind you that you are already aware of a number ofdifferent possible methods for building a model.

We now outline briefly some of the models that you are likely to come across whilestudying Course 3.



3 Models used in Course 3

3.1 Survival models

Much of Course 3 is devoted to analysing cashflows contingent on human life. Forexample, we may agree to pay to a policyholder a fixed sum of $1,200 each month for therest of his life, in exchange for a single premium now of $200,000. This type of policy iscalled a life annuity.

How do we know the correct single premium to charge? We will need to ask the followingquestions (amongst others):

• How likely is the policyholder to live for a long time? The lower the mortality of thepolicyholder, the higher the premium we will need to charge.

• How much interest can we earn on the single premium? If we can earn higherreturns on the money before we have to pay most of it back as income to thepolicyholder, then we can charge correspondingly less for the policy in the firstplace.

• What running costs are we likely to experience over the life of the policy? We willneed to charge an additional amount in the premium to cover our expenses.

• What are the other likely costs of the company, eg the tax position?

• How many policies of this type do we expect to sell? If we have some fixedexpenses (for example the costs of paying rental for our office premises), then thelarger the number of policies we sell, the smaller amount we need to charge thispolicy for its share of the fixed expenses.

• What other types of policies do we sell, and how profitable are they? Do we haveany particular profit target to be made on policies of this type?

• What premium rates are being charged by other companies in the market place?Can we match these rates? If not, are we realistically going to be able to sell manypolicies of this type?

We will not consider all of these problems in Course 3. However they are examples of thetypes of problems likely to face actuaries whose job it is to set premiums for life annuitypolicies.



However, the first of these problems (the policyholder’s likely future mortality) is central tothe ideas in Course 3. The first nine units of the course explain how we can calculatesimple premiums for policies like this, and also for policies that pay out benefits on death.For example, we will consider a term insurance policy that pays out a fixed sum if thepolicyholder dies within a certain term. This means that we need to make assumptions inour model about the likely future mortality of policyholders.

In order to do this, we need to make assumptions about the likely length of time for whicha person or group of people may live. Such a model is called a survival model.

For example, we might decide to use a random variable to represent the length of thefuture lifetime of a male life currently aged 40 exact.

Question 1.4

What range of values would this random variable take?

In fact random variables for the future lifetime form the basis of much of the theory in thefirst part of Course 3.

We use two random variables for future lifetime. We will use the random variable ( )T x tostand for the complete future lifetime of some individual. The complete future lifetime isthe actual length of time from now until death. The random variable in Question 1.4 aboveis an example of a complete future lifetime.

The other type of random variable we use for future lifetimes is the curtate future lifetime.This is a counting random variable that gives us the number of complete years from nowuntil death. So if a person aged 40 exact now dies at age 62½, say, then the value of thecurtate future lifetime random variable is 22.

We use the function ( )K x for the curtate future lifetime. In any particular case, K willtake the values 0, 1, 2, and so on, up to the maximum value that it is believed possiblethat K can take.

If we know the distribution of the future lifetime random variable, we can use it to find theprobability that the future lifetime of an individual exceeds any number x . This is calledthe survival function of the individual.

We can also calculate the expected value of the future lifetime and the variance of thefuture lifetime. We will come back to this in Units 2 through 8 of this course.



Question 1.5

The future lifetime of a certain type of electrical component is believed to have anexponential distribution with a mean of 200 hours. What proportion of components of thistype will we expect to last longer than 500 hours? If I have a box of ten of thesecomponents, what is the probability that exactly three of them last longer than 500 hours?

3.2 The life table

Consider a large number of people all born at the same moment. If we know their survivalfunction, then we can calculate the expected number of survivors, out of the initial group,who are alive at any point in the future. A table of the expected number of survivors ateach integer age in the future is called a life table.

For example, we could consider a sample of 10,000 policyholders who all take out acertain type of policy on a fixed date. We could then calculate, given information abouttheir future lifetimes, how many we would expect to be alive one year later, two yearslater, three years later, and so on.

Life tables are used extensively to calculate functions associated with survival. Forexample we could use a life table to find:

• the probability that a policyholder aged 50 dies in the next year

• the expected future length of life for a policyholder aged 60

• the number of policyholders in the group who would be expected to survive to age100.

A life table is a model of the likely future mortality of a group of individuals. As such, wecan use it to predict what may happen in the future to some other group of people, egthose who are applying to us for some particular type of insurance policy.

There is an example life table in the Tables for Course 3, which you should have a copy ofalready. (If not, you can obtain a copy by downloading it from the internet atwww.soa.org/eande/c3_tables.pdf) If you look at the first two columns of the IllustrativeLife Table in the Course 3 Tables, you will find that the table follows the mortalityexperience of a theoretical group of 10,000,000 lives, from birth (at age 0x = ) through tothe final age given (110, at which point there are only 11 out of the 10,000,000 lives leftalive). We can use the life table to predict the likely future mortality experience of anygroup of lives. However, we wouldn’t necessarily expect our group to behave in exactlythe same way, and there may be reasons why the group under consideration would havedifferent mortality characteristics. So care needs to be taken when considering whether amortality table is appropriate in any set of circumstances.



There are a number of other functions given in the Illustrative Life Table. You should befamiliar with all of these by the time you have finished the Part 1 of Course 3.

Question 1.6

A mortality study is observing the future lifetimes of a group of 10,000 people all currentlyaged exactly 30. If the mortality of this group follows that of the Illustrative Life Table, howmany of these lives do you expect to survive to age 60?

3.3 Annuities and Insurances

In order to decide how much to charge in premiums for a policy, we need to calculate thevalue of the benefits paid out. The premiums will depend also on mortality and interestrate assumptions.

For example, consider a term insurance policy. We may want to charge a monthlypremium in exchange for a promise to pay out $100,000 if the policyholder dies in the next20 years. Before we can charge a premium, we need to value the benefits, taking accountof the fact that we may pay out $100,000, or we may pay out nothing. The premium to becharged needs to take into account the likelihood that the benefit is paid.

Policies that pay out in this way are called insurance policies. Examples of insurancepolicies include:

• Whole life policies. These pay out on the death of the policyholder, whenever thistakes place.

• Term insurance policies. These pay out only if the policyholder dies within acertain period or term. If the policyholder survives to the end of the period, nobenefit is paid.

• Endowment policies. These will pay out on death within the term of the policy oron survival at the end of the term. They are really a combination of a terminsurance and a savings policy.

• Pure endowments. These pay out only on survival to the end of the term. If thepolicyholder dies sooner, no benefit payment is made. These are fairlyuncommon.

We shall see later on how to value the benefits arising under each of these types of policy.Once the benefits have been valued, we can then decide on an appropriate premium tocharge.



We will also be able to place a value on a stream of annuity payments. You have alreadycome across the idea of an annuity certain in Course 2. We now look at how to valueannuity payments that are dependent on survival (eg the policy mentioned at the start ofSection 3.1).

There are various different types of annuities that are dependent on survival:

• The amounts may be paid at different intervals (annually, monthly, weekly), andmay be paid in advance or in arrears.

• Amounts might increase over time. For example the monthly payment might startat $1,000 per month and increase in line with inflation, or at a fixed rate of 2% ayear.

• Amounts may be guaranteed for a fixed period. For example, an annuity might bepaid to a policyholder for life, but with a five year guarantee. This means that, ifthe policyholder were to die in the first five years, payments would continue to bemade up until the end of five years (probably to the husband or wife of the originalpolicyholder).

• A joint life annuity might be paid provided that the two lives who purchased itinitially were still alive. Alternatively it might be paid until the second death of thetwo.

• A reversionary annuity does not start to be paid until a death has occurred. Forexample, a policyholder might purchase an annuity that pays $1,000 each monthto his wife after his death, assuming that she is still alive.

We will see how to calculate the value of payments made under most of these types ofarrangement during our study of actuarial functions.

3.4 Premiums and reserves

We have outlined very briefly how we might calculate a premium for one of the policiesgiven above. But what does an insurer do with the premiums? Can an insurancecompany go out and spend the premiums it receives without worrying about the future?

The answer is no. It must retain the premiums (or most of them), since it expects to payout benefits in the future. In the meantime, it has a responsibility to invest the sum ofmoney that is associated with the particular group of policies. This sum is called areserve.

Companies must monitor the level of reserves that they need to hold. They have to beable to calculate the sum needed at any time in the future so that, under a certain set ofassumptions, there will be enough money available to pay out benefits. When all policieshave finished running, there might be some left over, in which case an additional profitmay emerge from this group of policies.



In Unit 6 we will describe how reserves are calculated for different types of policy. We willalso see how the size of a reserve is likely to vary over the term of a policy.

3.5 Joint life functions

Although we express the life table as the experience of a group of lives, we use it initiallyto find probabilities associated with a single life. We can extend the idea to estimateprobabilities associated with joint lives. For example, we could ask:

• What is the probability that two lives currently aged 60 both survive to age 80?

• Given two lives, one now aged 40 and one aged 50, what is the expected time ofthe second death?

• Given three lives all aged 40, what is the present value of a payment of $10,000made on the date of the second of the three deaths?

We see that, in this last example, we are combining the ideas of life expectancy andcashflows, ie we need a model that incorporates present values of future cashflows, aswell as survival probabilities. We will come across these ideas throughout the course.

We will also need the concept of independence. If we wish to use the individual survivalprobabilities to find the probability that two lives both survive, we will need to assume thatthe mortality of both lives is independent. Alternatively, if mortality is not independent, weneed to know exactly how the mortality of one life affects that of the other. This is usuallyvery difficult to quantify, and we will often make the assumption of independent mortalityrates, even though this is perhaps not very realistic.

Question 1.7

Why might you expect mortality rates not to be independent between two marriedpartners?

Even so, the assumption of independence of mortality rates is often made in practice.

Question 1.8

If the probability that A survives to age 60 is 0.8, and the probability that B survives to age60 is 0.9, what is the probability that (assuming independence):

(i) they both survive to age 60?

(ii) the both die before age 60?



3.6 Loss distributions

You have already met the idea of a continuous random variable. Later in the course wewill look at examples of a number of these distributions, and see how they are used tomodel the likely claims arising from different kinds of insurance portfolios. You will need tobe able to calculate means and variances of different distributions, as well as to usevarious techniques to derive further types of probability distributions. For example, if arandom variable X has normal distribution, then we need to be able to find thedistribution of the random variable e X .

Question 1.9

You have already come across the idea of a moment generating function. Explain what amoment generating function is, and how it can be used.

We will also meet the idea of a compound distribution. A compound random variable is arandom variable that can be expressed as the sum of a random number of randomvariables. For example, if N has a Poisson distribution with a mean of 50, and X haslognormal distribution with parameters µ = 6 and σ 2 2= , then we say that the randomvariable:

S X X XN= + + +1 2

has a compound Poisson distribution. We shall see how to analyze compounddistributions later in the course.

Compound distributions can provide a more realistic model of the likely future claimsexperience. When we look at the claims likely to arise under a group of policies, we donot know either the number of claims we are likely to receive, or the amounts of theindividual claims. Using a compound distribution, we model the uncertainty in the claimnumbers using, in the example above, a Poisson distribution, and the uncertainty in theclaim amounts by using a lognormal distribution.

We can also model the likely experience from a portfolio of policies by looking at eachpolicy, rather than by looking at each claim. In this case, if we had a portfolio of 10,000motor policies, then we could model the total aggregate claim amount by using therandom variable:

1 2 10,000S X X X= + + +



where iX is the amount paid during the course of the year on the i th policy. Of course,we would hope and expect that a large proportion of these iX ’s would be equal to zero (ifno claim was made on a particular policy during the year).

Question 1.10

What other distributions might be suitable for modeling the claim number distribution N ?

3.7 Stochastic processes

A large part of the final section of the course looks at the idea of a stochastic process.The random variables in the section above model the amount of claims received in a fixedunit of time (typically one year). A stochastic process might look at the likely future claimamount over a variable time period.

For example, consider the Poisson model mentioned earlier in this unit for the number ofclaims received in a year from a particular policy or group of policies. If we look at a fixedtime period of one year, we might decide that for a certain group, the total number ofclaims received might have a (50)Poisson distribution. However, if we want to look atclaims received over different time periods, the number of claims received will be different.For example, over a two year period the number of claims would be (100)Poisson , over afive year period it would be (250)Poisson , and over a three month period it would be

(12.5)Poisson .

A stochastic process assigns a random variable to each interval along a time line. Forevery interval of time, we can write down a distribution that represents the number ofclaims likely to arise in that time interval. So, in the example just given, we say that claimshave a (50 )Poisson t distribution, where t is the length of the time interval (in years).

We will look in Course 3 at various different types of stochastic processes, including theideas of a compound Poisson process, a Markov chain, and a Brownian motion process.All of these can be used to generalize the idea of the aggregate claim amount (or numberof claims) to a variable time period. A very brief summary of each type of process isincluded here.

3.8 Compound Poisson Processes

We saw earlier how we might model the aggregate claims from a portfolio by using acompound Poisson distribution.



Question 1.11

What is the defining equation for a compound Poisson random variable?

As you might have guessed, a compound Poisson process is a generalization of thissituation over time. So if we have a time period of length t , ( )S t represents theaggregate claims incurred over the whole t year period. ( )N t , the total number of claimsincurred over the period, will have a Poisson distribution with parameter tl , where l isthe rate at which claims arrive in a unit of time.

3.9 Markov Chains

A student is analyzing weather patterns. He classifies each day’s weather as sunny,cloudy or rainy. These three classifications can be modeled as the three states of aMarkov model.

The student may well believe that the type of weather on any particular day is a function ofwhat it was on the previous day. For example, he may believe that if the weather is cloudytoday, the probability that it is cloudy tomorrow is 0.6. But if it is sunny today, theprobability that it is cloudy tomorrow is only 0.4.

If probabilities are assigned to each possible transition from one state to another, then wecan use Markov chain analysis to answer various questions, eg:

• If it is sunny today, what is the probability that it will be sunny in three days’ time?

• What proportion of the time is the weather sunny?

• How long on average will it be until the next rainy day?

This type of analysis can be applied to any situation where something or someone ismoving between different states. For example, you might have a situation where yourmotor insurance premium is reduced in the following year if you make no claims this year.The policy might have various possible states (no discount, 25% discount, 50% discount),and policyholders would move between the states from year to year, depending onwhether they make claims in any year or not. This is another example of a Markov chain.



3.10 Brownian Motion

Brownian motion is an example of a statistical process that can take an infinite number ofstates, ie the state space is continuous.

In the Markov chain analysis above, we decided that there were only three possible typesof weather that we wished to model. However, consider trying to model the price of ashare on the NYSE. The number of prices possible is very large, and while in practice theprice may have to be a multiple of cents, because the number of possibilities is so great, itmay be more realistic to model the progress of a share price using the assumption that itcan take any value between, say zero and $100. The share price may then be modelledusing Brownian motion. We have:

• the price takes any value in a continuous range of values

• the monitoring of the price may also take place continuously, so that we are notjust looking at the price once per day, but every hour, or every minute, and so on.

This means that if we draw a graph of the share price, it will take the form of a continuouscurve over time. The unit on Brownian motion explains how we could model share pricesin this way.

3.11 Simulation

The final part of Course 3 looks at methods of simulating values from random variables.

In order to assess whether our models are realistic or not, we have to look and see whatsort of values are generated by them. These may be values from some particulardistribution, for example, or from a stochastic process. We need to be able to generatesequences of numbers that represent claim amounts, or claim numbers arising in a givenperiod, and so on.

There are various methods that can be used to generate variables of this kind. You mayhave a very simple random number generator on your calculator. This might, for example,generate numbers that are uniformly distributed in the range from zero to one. We shallsee later on in the course how we can use this random number generator to generatevalues from other types of distributions, for example a normal distribution or a Poissondistribution. We will also see how to generate values for a Markov chain, and a Brownianmotion process.

Question 1.12

Does the random number generator on your calculator generate numbers from acontinuous random variable?



4 ConclusionThis chapter gives you a very brief synopsis of the key ideas that you will come acrossduring your study of Course 3. You may wish to refer back to it as you reach the variousparts of the course that are mentioned here. At this stage, we can only give you a verybasic understanding of the key elements of the course. As you go through the course,watch out for links between the various different parts of the course, and try to see howthe ideas are connected. You should also try to look out for links between what you learnhere and what you can remember from your study of previous actuarial subjects. Againthere are connections which can help to deepen your understanding of the new material.

You will find further advice on how to study the course, and the order in which to attemptthings, in the Study Guide. If you have not already done so, please read the Study Guidebefore you proceed further with your reading of the chapters in Course 3. You willhopefully find that it should give you some help in deciding how to tackle the substantialtask of covering all the relevant material for the Course 3 exam.

You know this already, but the amount of material you need to learn for Course 3 isenormous. Get started as soon as you can, study for as long as you can, and as often asyou can. You will find that the exam is upon you before you are ready for it, unless youare ruthlessly efficient in organizing your time between now and the exam.

Good luck.



Unit 1 Summary

A model is an attempt to formalize mathematically the experiences that we may see goingon around us.

Actuarial models may have some or all of the following components:

• a cashflow component

• a survival function component

• a life table component

• a model for the size or number of future claim amounts

• a time related element.

When devising a model, an actuary might want to go through the following stages:

• decide on the purpose of the model

• collect an appropriate amount of relevant past data

• set up the model equations, using parameter values estimated from past data,adjusted if necessary for changes in likely future experience

• test the model to see if it produces realistic results

• watch the future experience as it materializes, and compare the actual futureexperience with the output from the model

• update the parameter values in the model (and the structure of the model itself ifnecessary) in the light of new past experience.

A survival model is a model of the future lifetime of a group of people. We will model boththe complete future lifetime (the exact time until the moment of death) and the curtatefuture lifetime (the time until death rounded to the next lower whole number of years).

We will be interested both in the probability density function of the future lifetime, and inthe distribution function.

A life table is a theoretical model of the likely future survival pattern of a group ofindividuals.

We can use a life table to calculate probabilities of survival or of death within a particularperiod.



By combining survival models with cashflow models, we can calculate expected presentvalues of benefits arising under insurance policies, for example the benefits paid underpure endowments, whole life policies, term insurance policies and endowment insurancepolicies.

By summing the values of individual payments, we can calculate the expected presentvalues of streams of payments (annuities).

These annuity payments may be made at various time intervals, may be paid in advanceor in arrears, may be contingent on the survival of one or more lives, or may bereversionary.

By looking at the expected present values of benefits, it is possible to calculate a premiumto be charged for a policy.

Once a policy is in force, an insurer needs to monitor the experience of policyholders. Iftoo many (or too few) policyholders die, there may be financial implications for thecompany.

The process of calculating a reserve for a group of policies is part of this monitoringprocess.

The amount to be paid out on a policy is often modeled using a continuous randomvariable. There are a variety of probability distributions that are useful for modeling theshape of a claim amount distribution.

An insurer will be interested in the mean and variance of the claim amount. It will also beinterested in the probability that individual claim amounts exceed certain large values (forexample, the probability that an individual claim amount exceeds $1,000,000).

The number of claims arising from a group of policies is also important. A frequencydistribution will be used to model this.

We can combine both of these ideas using compound distributions. A compounddistribution is the sum of a random number of random variables.

An insurer can also purchase reinsurance in order to reduce the mean amount that needsto be paid out on a group of policies. Reinsurance will also (probably) reduce the varianceof the claim outgo.

We can model the likely claims experience over time by using different types of stochasticprocesses.

A compound Poisson process is an extension in time of a compound Poisson distribution.We can use a compound Poisson process to model the aggregate claim payment over aperiod of time.



Markov Chains may be used to analyze situations where a variable moves between asmall fixed number of classes. Markov chains may be used to analyze how much time onaverage will be spent in each class.

Brownian motion is an example of a stochastic process with a continuous range of statespaces. It can be used to model share prices and other variables that can be deemed tovary continuously.

We need to be able to generate outcomes from our models, so that we can see whetherthey give us a realistic picture of what is actually happening in the real world. We do thisusing simulation techniques.

Simulation enables us to generate random samples from particular types of probabilitydistributions.

We can also develop methods that simulate stochastic processes, for example Markovchain and Brownian motion data.



This page has been left blank so that you can keep thesummary pages together to help you with your review.



Unit 1 Solutions

Solution 1.1

He might compare the numbers of claims received subsequently with those that werepredicted by the Poisson model. Some form of goodness of fit test, for example using thechi-square distribution, might be appropriate.

Note that, in real life, the Poisson distribution might well be too simple a model to fit theactual data well. For example, it is possible that individuals do make claims that have aPoisson distribution, but that the underlying Poisson parameter is different for differentclasses of individuals. In particular, we might find that policyholders in the age range 20-25 made claims that had a Poisson distribution with a mean of 0.3 claims per year,whereas policyholders in the 25-30 age group made claims that had a Poisson distributionwith a mean of 0.1 per year, and so on. In this case the overall claim number distributionwould be a mixture of different types of Poisson distributions. We cover mixturedistributions later in Course 3.

Solution 1.2

The variance of the amount will be:

6 6 2½ 1.04 ½ 1.06 1.15794 0.0010945¥ + ¥ - =

Solution 1.3

We now have independent interest rates from year to year. So the probability that theinterest rate is 4% in each of the three years is now 1/8. The probability that the rate is6% every year is also 1/8. We also have the possibility of two years at 4% and one yearat 6% (probability 3/8), and one year at 4% and two years at 6% (probability 3/8).

So the expected accumulated amount is now:

3 2 2 31 3 3 11.04 1.04 1.06 1.04 1.06 1.06 1.1576258 8 8 8

¥ + ¥ ¥ + ¥ ¥ + ¥ =

So the expected accumulated growth is now 15.76%.

Solution 1.4

The future lifetime of this individual could take any value within a particular range. So wemight model the future lifetime using a continuous random variable defined on the intervalfrom zero to 70 years, say. If we choose an upper limit of 70, we are assuming that it isnot possible to live to an age greater than 110, given that the life is currently aged 40.



Solution 1.5

The density function of the future lifetime is:

/ 2001( ) , 0200

tf t e t-= ≥

So the probability that a component lasts for more than 500 hours is:

/ 200 / 200 2.5500

500

1 0.08208200

t te dt e e• •- - -= - = =Ú

The number of components lasting more than 500 hours in a box of 10 has a binomialdistribution (assuming that the lifetimes of individual components are independent). Sothe probability that exactly 3 of the 10 last for more than 500 hours is:

3 7100.08208 (1 0.08208) 0.03644

3Ê ˆ

- =Á ˜Ë ¯

You should already have met both the exponential distribution and the binomial distributionin your study for previous exams.

Solution 1.6

In the illustrative life table, out of the 9,501,381 lives alive at age 30, 8,188,074 survive toage 60. So if this group of lives follows the same pattern, we would expect to have anumber of survivors equal to:

8,188,07410,000 8,6189,501,381

¥ =

So we would expect 8,618 of the lives to be still alive at age 60.

We shall develop these ideas further in Units 2 and 3 of the course.

Solution 1.7

For a number of reasons. Both partners might be exposed to the risk of accidents whichkill them both, for example being killed in the same car accident. Also, any health relatedmortality factors are likely to affect them both if they have similar diets, lifestyles and soon. Finally, bereavement is itself a very traumatic situation. If one partner dies, this mayaffect the health and consequently the survival chances of the remaining partner.



Solution 1.8

If mortality rates are independent, the probability that they both survive to age 60 is0.8 0.9 0.72¥ = . The probability that neither of them survives (ie they both die) is0.2 0.1 0.02¥ = .

Solution 1.9

The moment generating function of a random variable X is defined to be the expectationof tXe :

( ) ( )tXXM t E e=

We can expand this expression as a series to find the moments of the random variable( ( )E X , 2( )E X , 3( )E X and so on).

We will see more of this later on in the course.

Solution 1.10

Almost any frequency distribution. For example, the binomial distribution, the negativebinomial distribution, the geometric distribution. You will see many more of thesedistributions later on, if you have not met them already.

Solution 1.11

1 2 NS X X X= + + + , where N has a Poisson distribution. X can have any type ofdistribution.

Solution 1.12

In theory yes, but in practice no. The generator on my calculator gives me a numberrounded to three decimal places, ie I get a value from the frequency distribution defined onthe numbers 0.000, 0.001, 0.002, 0.003, …, 0.999, with the probability that each numberis generated is 0.001.

Course 3, Unit 2 The life table


Unit 2The life table

Unit overview

This unit looks at survival and death probabilities for integer and fractional ages. Itdevelops the ideas of the future lifetime random variable and expected future lifetimes.Some mortality laws are then considered, and we derive expressions for survivalprobabilities based on these laws. The unit finishes with an introduction to select andultimate mortality.

Advice on study

You should ensure that you are completely happy with the work and notation covered inthis unit before proceeding to later units.


We recommend that you should allow 6 hours to study this unit of the course.

The life table Course 3, Unit 2


Learning objectives

9 Characterize discrete and continuous univariate probability distributions for failuretime random variables in terms of the life table functions, , , , ,x x x n x n xl q p q p and

|m n xq , the cumulative distribution function, the survival function, the probability

density function and the hazard function (force of mortality), as appropriate

– Establish relations between the different functions.

– Develop expressions, including recursion relations, in terms of the functionsfor probabilities and moments associated with functions of failure timerandom variables, and calculate such quantities using simple failure timedistributions.



1 IntroductionIn order to plan finances in an insurance or pensions company you will need to be able toanswer questions such as: “How much will the pension payments for our customers be inthe coming year?” or “When will we have to pay out for John Doe’s life insurance policy?”.

We are not able to obtain exact answers to such questions, since time of death for mostpeople is unknown, but having a model for human mortality will enable us to find a goodapproximation.

Actuaries have studied human mortality for many years and have built up life tables fortypical populations showing how many people out of a particular cohort are expected tosurvive to each age.

In this unit we will cover the notation associated with mortality and develop the groundworkneeded for the subsequent units.

This unit introduces much new notation and many definitions. You may need to read itseveral times in order to be confident in the use of the symbols defined.



2 Future lifetimes

Definition

The age-at-death of a newborn baby is X.

The distribution function of X is ( )XF x .

The survival function ( )s x is defined to be >Pr( )X x .

In words, the survival function is the probability of a newborn baby surviving to age x.

Question 2.1

What is the connection between ( )s x and ( )XF x ?

Definition

The future lifetime of a life aged x is ( )T x , where = −( )T x X x , given that >X x .

t xq is the probability that a life aged x dies in the next t years.

t xp is the probability that a life aged x survives for the next t years.

By convention, if we are talking about a time period of 1 year then the probability notation isxq or xp .

Question 2.2

Define in words what the symbols xq and xp mean.

We can now link our definitions as follows:



t xp is the probability that a life aged x survives for the next t years. This is the same asthe probability that a new born baby survives to age +x t given that it survives to age x,ie:

= > = > + >Pr( ( ) ) Pr( | )t xp T x t X x t X x

But by the definition of conditional probabilities this can be written as:

+ − +> + += = =

> −0

0

1 ( )Pr( ) ( )Pr( ) ( ) 1 ( )

x t X

x X

p F x tX x t s x tX x p s x F x

Similarly:

< ≤ += ≤ = ≤ + > =

>+ −− +

= =−

Pr( )Pr( ( ) ) Pr( | )Pr( )

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

t x

X X

X

x X x tq T x t X x t X xX x

F x t F xs x s x ts x F x

Question 2.3

Define in words what the following notation means:

(i) 25 30q

(ii) 15 27p

(iii) <Pr( 60)X

(iv) < <Pr(32 79)X

(v) (15)XF

(vi) (60)T

(vii) (24)s

Question 2.4

Find two alternative expressions for +s t xp by splitting up the probability in two differentways.



3 Curtate future lifetime and curtate expectation of life

3.1 Curtate future lifetime

Definition

The curtate future lifetime is the complete number of years the life lives after age x. It isdenoted by ( )K x .

Question 2.5

Is ( )K x a discrete or continuous random variable?

We can express =Pr( ( ) )K x k in terms of the notation that we have already used:

+ +

= = ≤ < + = < ≤ += − =1

Pr( ( ) ) Pr( ( ) 1) Pr( ( ) 1)

k x k x k x x k

K x k k T x k k T x kp p p q

The second equality relies on the assumption that ( )T x is a continuous random variable(which is reasonable).

Notice that the last expression, +k x x kp q , is a useful way of thinking of =Pr( ( ) )K x k . Theprobability is equivalent to the probability that a life aged x survives to age +x k and thendies in the following year.



3.2 Curtate expectation of life

Definition

The curtate expectation of life is defined to be ( )( )E K x and it is denoted as xe .

We can find a simple expression for xe by using the definition of expectation:

( )∞ ∞

+= =

+ +

= = = =

= + +

∑ ∑0 0

1 1 2 2

( ) Pr( ( ) )

1 2

x k x x kk k

x x x x

e E K x k K x k k p q

p q p q

We can write this sum as:

+

+ +

+ + +

+ +

+ + +

+

1 1

2 2 2 2

3 3 3 3 3 3

x x

x x x x

x x x x x x

p qp q p qp q p q p q

If we then sum the columns we get:

∞ ∞

+= =

= ∑ ∑1

x j x x jk j k

e p q

Now ∞

+=∑ j x x jj k

p q represents the probability of dying at any age beyond +x k , which is

k xp .

Finally, therefore, our expression becomes:

∞

== ∑

1x k x

ke p



Curtate expectation of life

The curtate expectation of life, xe , is given by:∞

== ∑

1x k x

ke p

Question 2.6

Show algebraically that + = −1 1xx

x

eep

.

We can also derive an expression for the variance of the curtate expectation of life. Consider

( )2( )E K x :

( )∞

+=

= ∑2 2

0( ) k x x k

kE K x k p q

Using a similar argument to the one used in the expectation, summing in columns we get:

( )∞ ∞ ∞

+ + += = =

∞ ∞

+= =

∞

=

= + + +

= −

= −

∑ ∑ ∑

∑ ∑

∑

2

1 2 3

1

1

( ) 3 5

(2 1)

(2 1)

j x x j j x x j j x x jj j j

j x x jk j k

k xk

E K x p q p q p q

k p q

k p

Question 2.7

Show that ( )∞ ∞ ∞

+ + += = =

= + + +∑ ∑ ∑2

1 2 3( ) 3 5j x x j j x x j j x x j

j j jE K x p q p q p q .



The variance of ( )K x therefore is given by:

∞

=− −∑ 2

1(2 1)k x x

kk p e

Variance of curtate future lifetime

The variance of the curtate future lifetime is given by:

( )∞

== − −∑ 2

1var ( ) (2 1)k x x

kK x k p e



4 Force of mortalityThe force of mortality, µ( )x , is an instantaneous measure of mortality at age x. It is definedto be:

µ→

= < < + > 0

1( ) lim Pr( | )h

x x X x h X xh

ie ( )µ x h is the probability that an individual who survives to age x then dies in the next h

years, where h is a short period of time. The force of mortality is also known as the failurerate or the hazard rate.

Returning to the notation we used for the survival function:

µ→

− +=

0

1 ( ) ( )( ) lim( )h

s x s x hxh s x

But, →

+ − ′=0

( ) ( )lim ( )h

s x h s x s xh

, assuming that ( )s x is differentiable. So the force of

mortality becomes:

µ′

= −( )( )( )

s xxs x

We are now going to derive a series of results connecting the distribution function of X,survival probabilities and the force of mortality. You should be comfortable with the stages ofthe derivations, but we have also summarized the findings at the end so that you are awareof the key results.

µ′

= −( )( )( )

s xxs x

can be rewritten as µ = −( ) ln ( )dx s xdx

.

If we integrate both sides of this expression between +x t and x, we get:

[ ]µ+

+ +− = =∫

( )( ) ln ( ) ln( )

x tx tx

x

s x ty dy s xs x

which, by rearrangement, gives us:

µ+ + = −

∫

( ) exp ( )( )

x t

x

s x t y dys x



But +=

( )( ) t x

s x t ps x

, so we have the result that µ+ = −

∫exp ( )

x t

t xx

p y dy .

In particular:

µ = − ∫00

exp ( )t

t p y dy

which gives us:

µ = − ∫0

( ) exp ( )t

s t y dy or µ = − ∫0

( ) exp ( )x

s x y dy

Differentiating this last expression with respect to x, we get:

µ µ µ ′ = − − = − ∫ 00

( ) exp ( ) ( ) ( )x

xs x y dy x p x

Since = −( ) 1 ( )s x F x , ′ = −( ) ( )s x f x , so:

µ= 0( ) ( )xf x p x

So we have found an expression for the probability density function of the future lifetimerandom variable for a new-born baby.

We can also find the probability density function of the future lifetime of a life aged x. Let usdefine ( )( )T xF t to be the cumulative distribution function of the future lifetime of a life aged

x, then since = ≤ =( )( ) Pr( ( ) )T x t xF t T x t q :

′ + += = = − = −

( ) ( )

( ) ( )( ) ( ) 1( ) ( )T x T x t x

d d d s x t s x tf t F t qdt dt dt s x s x

We can multiply the numerator and denominator of this fraction by +( )s x t , in order to get aform we can simplify:

µ′ ′ + + +

= − = − = + + ( )

( ) ( ) ( )( ) ( )( ) ( ) ( )T x t x

s x t s x t s x tf t p x ts x s x s x t



From this we have:

µ− = = +(1 ) ( )t x t x t xd dp q p x tdt dt

ie µ= − +( )t x t xd p p x tdt

Finally we can integrate both sides of the expression µ= +( )t x t xd q p x tdt

between the limits

0 and t to obtain:

µ= +∫0

( )t

t x s xq p x s ds

We now give a summary of the important formulas:

Summary of important formulas

µ+ = −

∫exp ( )

x t

t xx

p y dy

µ = − ∫0

( ) exp ( )x

s x y dy

µ= 0( ) ( )xf x p x

µ= +( )( ) ( )T x t xf t p x t

µ= − +( )t x t xd p p x tdt

µ= +∫0

( )t

t x s xq p x s ds



Question 2.8

If µ =( ) 0.002x for all ages from 20 to 60, calculate the probability that a life aged exactly28

(i) dies before he is 38

(ii) survives to his 60th birthday.

Question 2.9

Find an expression for ( )s x for a population where µ = +( ) 0.0001 0.05x x .



5 Deferred mortality

Deferred mortality

The probability that an individual aged x survives to age +x n , but then dies before age+ +x n m is written as |n m xq .

The period of n years, written before the line, is called the deferred period. We want theprobability of death in an m year period but starting in n years’ time.

From the definition it can be seen that:

+ + += = − = −|n m x n x m x n m n x n x n x m n xq p q q q p p

Question 2.10

What does |n xq represent? What other expressions are there for |n xq ?



6 Interpretation of life tables

Very often as actuaries we have to calculate probabilities such as t xp and t xq for aparticular individual. Rather than having to construct each value from first principles, lifetables have been created to help us. These contain tabulated values of several variablesagainst age. The problem with using the data from a mortality table is that the characteristicsof the group of lives represented in the table may not truly match those of the individual youare working with.

The lowest age in the table can vary depending on the purpose of the table; for example atable for retirements is very unlikely to start at age 0! There is one example of a life table, theIllustrative Life Table, which we will refer to from now on as the Tables. At the moment youneed to use the first three columns – you will meet the other notation in later units. Beforewe can look at an example of a life table we need some more definitions.

Consider a life table that starts at age 0 for a closed group of individuals:

Definition

0l is the number of newborns being considered (the radix of the table)

xl is the number of survivors at age x

xd is the number of deaths between ages x and +1x

n xd is the number of deaths between ages x and +x n

You may want to look at the figures in your Illustrative Life Table as you consider thesedefinitions.

We can now connect these new definitions to each other and to the other notation in thisunit.

It should be clear that:

+= − 1x x xd l l

+= −n x x x nd l l



Further:

= xx

x

dql

+−= − = = 11 x x x x

xx x x

d l d lpl l l

+−= =n x x x n

n xx x

d l lql l

The last equation then gives rise to:

+ +−= − = − =1 1 x x n x n

n x n xx x

l l lp ql l

Question 2.11

Use the Illustrative Life Table to calculate the probability that a life:

(i) aged 35 survives to 65

(ii) aged 27 dies before 47

(iii) aged 47 dies when they are 50

(iv) aged 49 lives to at least 100

(v) aged 40 dies between 75 and 80

Here the ages are exact, for example aged 35 means exact age 35.

Question 2.12

Sketch graphs of xl , xd and xq for ages 0 to 100.

These definitions and interpretation of the life table are known as the deterministicsurvivorship interpretation. An alternative interpretation is the random survivorshipinterpretation. This is where we consider 0l newborns each with probability of surviving toage x of ( )s x . The values in the life table now become expected numbers, for example xlis the expected number of survivors at age x.



The mathematical properties of the functions xq , xl and xd are the same under bothinterpretations but the random survivorship interpretation takes into account the inherentrandomness of the number of survivors.

The maximum age of people in the cohort is ω , which is called the limiting age. Typically,ω will be between 100 and 120 for a human population.

We can now link the life table functions to the probability functions and force of mortality.

Since += x nn x

x

lpl

, it follows that:

=00

xx

lpl

But =0 ( )x p s x , so:

= 0 ( )xl l s x

Consider the expression − 1 x

x

dll dx

. We can use the previous result as follows:

− = − = − = −0 0

0 0

[ ( )]1 1 ( ) 1 ( )( ) ( ) ( )

x

x

d l s x ldl ds x ds xl dx l s x dx l s x dx s x dx

But, from Section 4, we know that µ− =1 ( ) ( )( )

ds x xs x dx

, so:

µ− =1 ( )x

x

dl xl dx

If we integrate both sides of this expression between the limits of x and ∞ , we get:

µ∞

= ∫ ( )x yx

l l y dy

If we now write down the corresponding result for +x nl and take the difference, we get:

µ+

+− = ∫ ( )x n

x x n yx

l l l y dy



7 Complete expectation of lifeIn Section 3 we considered the curtate expectation of life. We now extend that idea to lookat the complete future lifetime and the complete expectation of life.

As a reminder, the future lifetime of a life is ( )T x , ie the length of life beyond age x. Alsoremember that ( )T x is a continuous random variable.

Definition

The complete expectation of life is defined to be ( )( )E T x and it is denoted as °xe .

We can find a simple expression for °xe by using the definition of expectation:

( ) µ∞ ∞ ∞

° = = = + = − ∫ ∫ ∫( )

0 0 0

( ) ( ) ( )x T x t x t xde E T x t f t dt t p x t dt t p dtdt

Using integration by parts, = −∫ ∫dv duu dx uv v dxdx dx

, where =u t , we get:

∞ ∞∞° = − + = ∫ ∫0

0 0x t x t x t xe t p p dt p dt

Note that this is analogous to the discrete result where we summed t xp instead ofintegrating it.

Question 2.13

Is it true that ° °= +0 4040e e ?

We can also derive an expression for the variance of the complete expectation of life.

Consider ( )2( )E T x :

( ) µ∞ ∞ ∞

= = + = − ∫ ∫ ∫2 2 2 2

( )0 0 0

( ) ( ) ( )T x t x t xdE T x t f t dt t p x t dt t p dtdt



since ∞ = 0xp .

Using integration by parts with = 2u t , we get:

( )∞ ∞∞

= − + = ∫ ∫2 20

0 0

( ) 2 2t x t x t xE T x t p t p dt t p dt

This means that the variance is given by:

( )∞

° = − ∫2

0

var ( ) 2 t x xT x t p dt e

Question 2.14

What is the standard deviation of future lifetime if µ =( ) 0.02x for all ages?



8 Recursive formulas for expectations of lifeIn this section we are going to develop recursive formulas for the curtate and completeexpectation of life.

Consider the curtate expectation of life, ∞

== ∑

1x k x

ke p . This sum can be expanded and

simplified:

+ + +

∞

+=

+

= + + +

= + + + +

= +

= +

∑

2 3

1 1 2 1 3 1

11

1

x x x x

x x x x x

x x k xk

x x x

e p p pp p p p p

p p p

p p e

Recursive formula

The recursive formula for the curtate expectation of life is:

+= + 1x x x xe p p e

This formula is used backwards ie you input a value of, say, 100e and get 99e . The starting

point is the limiting age of the table, since ω = 0e .

Question 2.15

Using the Illustrative Life Table, calculate 107e recursively.



Consider also the complete expectation of life, ∞

° = ∫0

x t xe p dt . This integral can be

manipulated as follows:

∞ ∞°

∞

− +

= = +

= +

∫ ∫ ∫

∫ ∫

1

0 0 11

1 10 1

x t x t x t x

t x x t x

e p dt p dt p dt

p dt p p dt

Changing the variable on the second integral, so that = −1s t , we get:

∞° °

+ += + = +∫ ∫ ∫1 1

1 10 0 0

x t x x s x t x x xe p dt p p ds p dt p e

Recursive formula

The recursive formula for the complete expectation of life is:

° °+= +∫

1

10

x t x x xe p dt p e

Question 2.16

For a certain population, = −exp( 0.02 )t xp t and ° =20 50e . Using the recursive formula

given, calculate °23e .



9 Median future lifetime

Definition

The median future lifetime, ( )m x , can be found by solving:

( )> =Pr ( ) ( ) 0.5T x m x

Since ( ) +> =

( )Pr ( )( )

s x tT x ts x

, we can also find the median future lifetime by solving:

+=

( ( )) 0.5( )

s x m xs x

Question 2.17

How would you find the quartiles of the future lifetime? How would you find the mode offuture lifetime?

Question 2.18

In a very poor country, the survival function is given by = −( ) exp( 0.03 )s x x . What is themedian and interquartile range of future lifetime of a life aged x?



10 Central death rate

Definition

The central death rate (over the interval from x to +1x ) is:

µ+

+

+

=∫

∫

1

01

0

( )x t

x

x t

l x t dt

m

l dt

The central death rate xm is a measure of the rate of mortality over the year from exact agex to exact age +1x . The formula can be interpreted as a weighted average of the force ofmortality over the next year of age.

We are going to introduce some more definitions in order to derive another expression forthe central death rate.

Definition

n xL is the total expected time spent alive between ages x and +x n by survivors to age xof the original group:

µ+ += + +∫0

( )n

n x x t x nL t l x t dt nl

Question 2.19

Explain the structure of the above equation.

Question 2.20

Use integration by parts to show that += ∫0

n

n x x tL l dt .



Question 2.21

Show that = ∫0

nn x

t xx

L p dtl

.

We now introduce another way of considering the last integral.

Definition

°=∫ :0

n

t x x np dt e is the n-year temporary complete life expectancy of a life aged x.

This is the expected number of years lived by a life aged x in the next n years.

Finally we return to the central death rate.

Question 2.22

Show that +−= 1x x

xx

l lmL

.

We can expand the idea of total time spent alive:

Definition

xT is the total number of years lived beyond age x by survivors to age x in the originalgroup.

µ∞

+= +∫0

( )x x tT t l x t dt

You should be able to see that →∞

= limx n xnT L .



Question 2.23

Show that °=xx

x

T el

?

The definition of xm can be extended to time periods longer than one year:

Definition

The average death rate (over the interval from x to +x n ) is:

µ++

+

+−

= =∫

∫

0

0

( )n

x tx x n

n x n n xx t

l x t dtl lm

Ll dt

Finally let us consider another life table function:

Definition

( )a x is the average number of years lived between x and +1x by those in the group whodie between those ages:

( )µ µ

µ µ

+

+

+ +

= = = <

+ +

∫ ∫

∫ ∫

1 1

0 01 1

0 0

( ) ( )

( ) ( ) | ( ) 1

( ) ( )

x t t x

x t t x

t l x t dt t p x t dt

a x E T x T x

l x t dt p x t dt

Question 2.24

For a particular population = −( ) 1500

xs x . Find µ(40) , µ(50) , 10 40q and 10 40m .

Comment on the values you obtain.



11 Fractional agesSo far we have assumed that the ages we are working with are integer ages, but we maywish to work with fractional ages to calculate such quantities as 3.5 46.5p .

Consider the expression given above. We can split it up as follows:

= ×3.5 46.5 0.5 46.5 3 47p p p

We can arrive at an approximation for the first factor if we make certain assumptions aboutthe mortality with the year from age 46 to age 47. The second factor can be calculateddirectly from the life table. In other words, we can always reduce an expression involvingfractional ages to one of the form t xp where ≤ ≤0 1t .

There are three different assumptions that we have to consider about the mortality during ayear. Under each assumption we can arrive at approximations for the life table functions.

11.1 Uniform distribution of deaths

We are assuming here that the probability density function of future lifetime, µ +( )t xp x t , isa constant for an integer value of x and ≤ ≤0 1t .

With this assumption the general relation ( )0

tt x s xq p x s dsµ= +∫ results in the following:

( )0 0

t tt x s xq p x s ds c ds ctµ= + = =∫ ∫

If we substitute = 1t into the above we see that xc q= and hence t x xq t q= . Since t xq is

the distribution function, by differentiating the relation t x xq t q= we see that the density

function is given by ( )t x xp x t qµ + = .

In summary, we have the following relations as a result of the assumption of a uniformdistribution of deaths:

( )

and 1

( )1 1

t x x t x x

t x x

x x x

t x t x x

q t q p t q

p x t q

q q qx tp q tq

µ

µ

= = −

+ =

+ = = =− −



Now consider the expression − +s t x tq where 0 1t s≤ < ≤ . We know that

− += ×s x t x s t x tp p p , so:

− + − += − = −1 1 s xs t x t s t x t

t x

pq p

p

Now we can use the approximation that we have already found:

− +− − −

= − = − =− − −

1 1 ( )1 11 1 1

s x x xs t x t

t x x x

q sq s t qqq tq tq

Note that − + +≠ −( )s t x t x tq s t q if ≠ 0t since +x t is not an integer.

To summarize the results for uniform distribution of deaths:

Results for uniform distribution of deaths

x is an integer and ≤ ≤0 1t

µ= +( )x t xq p x t

= −1t x xp tq

=t x xq tq

µ + =−

( )1

x

x

qx ttq

− +−

=−

( )1

xs t x t

x

s t qqtq



11.2 Constant force of mortality

We are assuming here that µ µ+ =( )x t , a constant for an integer value of x and ≤ ≤0 1t .

We know that exp ( )x t

t xx

p y dyµ+

= − ∫ , so under our new assumption:

[ ]µ µ µ+

+= − = − = −∫exp ( ) exp exp( )x t

x tt x x

x

p y dy y t

Substituting = 1t into this expression, we get µ= −exp( )xp , so we can deduce furtherresults:

µ= − −1 exp( )xq = ( )tt x xp p = −1 ( )tt x xq p µ = − ln xp

Now consider − +s t x tp where ≤ ≤ ≤0 1t s :

[ ]µ µ+

−− +

+

= − = − − = ∫exp exp ( ) ( )

x ss t

s t x t xx t

p dy s t p

Giving:

−− + = −1 ( )s t

s t x t xq p

Finally:

µ µ µ µ+ = = − = −( ) ( ) ln ( ) exp( )t tt x x x xp x t p p p t

To summarize the results for constant force of mortality:

Results for constant force of mortality

µ= − −1 exp( )xq

µ µ µ+ = − = −( ) ln ( ) exp( )tt x x xp x t p p t

= ( )tt x xp p = −1 ( )tt x xq p

µ µ+ =( )x t −− + = −1 ( )s t

s t x t xq p



11.3 Hyperbolic form for xt p

We are assuming here that = − +1 1

t x x

ttp p

for an integer value of x and ≤ ≤0 1t .

We can manipulate the expression for the assumption as follows:

− −= = = =

− + + − − + +1 1

(1 ) (1 ) 1x x x x

t xx x x x x x x

p q q ppt p t p t p q tq p tq

Therefore:

− + − += =

− + − −1 1

1 1 (1 )x x x x

t xx x x

q tq q tqqq tq t q

Further, we can find an expression for the force of mortality:

( )

µ

−

+ = −

= − − − − +

−=

− +

− + −=

− − +

= = =− + − − +

2

2

2

1( )

1 (1 )(1 )

(1 )1(1 )

1 (1 )1 (1 )

1 1 (1 )

t xt x

x x x xt x

x x

t x x x

x x x x

x x x

x x x

x x x x x

dx t pp dt

q q tq qp

q qp q tq

q tq q qq q tq

q q qq tq t q p tq

Finally an expression for − +s t x tq where ≤ ≤ ≤0 1t s can be derived as follows:

− +− − +

= − = − ×− + −

− += −

− +

− + − + −=

− +

−=

− −

1 11 1

1 1111

1 11

( )1 (1 )

s x x x xs t x t

t x x x x

x x

x x

x x x x

x x

x

x

p q q tqqp q sq q

q tqq sq

q sq q tqq sq

s t qs q



To summarize the results for the hyperbolic form for t xp :

Results for hyperbolic form

=− −1 (1 )

xt x

x

tqqt q

≤ ≤0 1t

−=

− +1

1x

t xx x

qpq tq

≤ ≤0 1t

µ + =− −

( )1 (1 )

x

x

qx tt q

≤ ≤0 1t

− +−

=− −( )

1 (1 )x

s t x tx

s t qqs q

≤ ≤ ≤0 1t s



Example 2.1

Calculate an approximate value for 4 73.5p using the Tables under the assumption that wehave:

(i) a uniform distribution of deaths between integer ages

(ii) a hyperbolic form for t xp between integer ages

Solution

(i) We have to split up the probability into factors that we can evaluate using theexpressions we have derived:

=4 73.5 0.5 73.5 3 74 0.5 77p p p p

This can be simplified as follows:

0.5 73.5 3 74 0.5 77 0.5 73.5 3 74 0.5 77

733 74 77

73

(1 ) (1 )

(1 0.5)1 (1 0.5 )

1 0.5

p p p q p q

qp q

q

= − −

−= − −

−

Using the Tables:

= − = − =7473

73

5,664,0511 1 0.04329835,920,394

lql

= = =773 74

74

4,828,182 0.85242565,664,051

lpl

= − = − =7877

77

4,530,3601 1 0.0616844,828,182

lq

l

Therefore =4 73.5 0.8079p .

(ii) We can use part of the above solution:

=

= − −

−= − − − − − −

4 73.5 0.5 73.5 3 74 0.5 77

0.5 73.5 3 74 0.5 77

73 773 74

73 77

(1 ) (1 )

(1 0.5) 0.51 11 (1 1) 1 (1 0.5)

p p p pq p q

q qpq q

Using the probabilities calculated above =4 73.5 0.8074p .



Question 2.25

The approximate value of 0.5 61.5q assuming a uniform distribution of deaths betweeninteger ages and the mortality of the Tables is:

A 0.00693B 0.00745C 0.00756D 0.00812E 0.01362

Question 2.26

The approximate value of 10 50.5p assuming a constant force of mortality between integerages and the mortality of the Tables is:

A 0.91112B 0.91114C 0.91116D 0.91118E 0.91120

Question 2.27

The approximate value of 10 50.5p assuming a hyperbolic form for t xp between integerages and the mortality of the Tables is:

A 0.91112B 0.91114C 0.91116D 0.91118E 0.91120



12 Mortality lawsIt can be useful to have the force of mortality expressed as a simple mathematical functionsince we can then calculate life table functions without using a life table. There is still debateabout whether human mortality follows such simple formulas but we are going to considersome of the laws here and look at how the laws can be applied.

De Moivre’s law

De Moivre’s mortality law states that µ ω −= − 1( ) ( )x x , where ω≤ <0 x .

Gompertz’s law

Gompertz’s mortality law states that µ =( ) xx Bc , where > > ≥0, 1, 0B c x .

Makeham’s law

Makeham’s mortality law states that µ = +( ) xx A Bc , where > 0B , ≥ − > ≥, 1, 0A B c x .

Weibull’s law

Weibull’s mortality law states that µ =( ) nx kx , where > > ≥0, 0, 0k n x .

The rationale behind Gompertz’ and Makeham’s laws is that it was observed that when µ( )xwas plotted on a logarithmic scale against age, the graph appears to follow a straight line formuch of the age range.

Question 2.28

Derive expressions for ( )s x and t xp for each of the mortality laws stated above.



Question 2.29 Over a period of time it has been found that, for a very large population whose mortalityconforms to Gompertz’s law, =10 15 0.957076p and =10 20 0.896579p . Find 17 18p forthis population.



13 Select mortalitySome life tables are select mortality tables. Functions in such a table depend upon age andduration since an event. Depending on the select period, at some point in the future mortalitywill be assumed to return to that of the general population. This is referred to as the ultimatepart of the table.

Let us think about particular ‘events’ to help clarify this issue.

The first event is taking out a life insurance policy. In order to be accepted for the policy, theperson will have to go through some sort of medical screening process, either by filling out aform or visiting a doctor for an examination. The mortality of this person straight afterpassing the medical screening will be lower than the mortality of the general population sincethey have been declared ‘fit’. Over a length of time, say 3 years, the person’s mortality willcome back into line with that of the general population. The select period in this case is 3years and beyond that time the person will experience ultimate mortality.

Another event could be giving up smoking. The mortality of the person straight after givingup will be higher than that of the general population. After 10 years, the harmful effects ofsmoking will have reduced to such an extent that the person’s mortality again returns to thatof the general population. The select period here is 10 years.

Notation for select mortality

Select mortality is indicated using square parentheses, for example:

[20]l , +[35] 2q , etc

If the table had a select period of 3 years, then the column headings for the number ofpeople alive would be:

[ ]xl , +[ ] 1xl , +[ ] 2xl , and +3xl ,

For example a table for life insurance with a three-year select period could be:

[ ]x [ ]xl +[ ] 1xl +[ ] 2xl +3xl + 3x

22 35,628.36 35,619.24 35,607.78 35,596.15 2523 35,615.35 35,604.89 35,590.01 35,581.18 2624 35,602.78 35,589.99 35,575.03 35,562.59 27



Notice that for durations beyond the select period, the square parenthesis notation isdropped, for example with a three-year select period, + += =[24] 3 27 [24] 7 31,l l l l . The table is

followed horizontally for a life until the end of the select period, when it is then followedvertically.

Example 2.2

Calculate +[23] 2q using the figures in the table above and explain what +[23] 2q means.

Solution

++

= − = − =26[23] 2

[23] 2

35,581.181 1 0.00024835,590.01

lql

+[23] 2q means the probability that a life currently aged 25, who was selected at age 23,

dies in the next year.

Question 2.30

The value of 2 [22]p using the above select mortality table is:

A 0.99815B 0.99928C 0.99934D 0.99942E 0.99981



Unit 2 Summary

Future lifetimes

The age-at-death of a newborn baby is X.

The survival function ( )s x is defined to be >Pr( )X x .

The future lifetime of a life aged x is ( )T x , where = − >( ) |T x X x X x .

t xq is the probability that a life aged x dies in the next t years.

t xp is the probability that a life aged x survives for the next t years.

Curtate future lifetime and curtate expectation of life

The curtate future lifetime is the complete number of years the life lives after age x. It isdenoted by ( )K x .

The curtate expectation of life is defined to be ( )( )E K x and it is denoted as xe , where:

∞

== ∑

1x k x

ke p

The variance of curtate future lifetime is given by:

( )∞

== − −∑ 2

1var ( ) (2 1)k x x

kK x k p e

Force of mortality

The force of mortality at age x is µ( )x .

µ+ = −

∫exp ( )

x t

t xx

p y dy µ = − ∫0

( ) exp ( )x

s x y dy

µ= 0( ) ( )xf x p x µ= +( )( ) ( )T x t xf t p x t

µ= − +( )t x t xd p p x tdt

µ= +∫0

( )t

t x s xq p x s ds



Deferred mortality

+=|n m x n x m x nq p q

Life tables

0l is the number of newborns being considered (the radix).

xl is the number of survivors at age x.

xd is the number of deaths between ages x and +1x .

n xd is the number of deaths between ages x and +x n .

Complete expectation of life

The complete future lifetime is the length of life life beyond age x. It is denoted by ( )T x .

The complete expectation of life is defined to be ( )( )E T x and it is denoted as °xe , where:

∞° = ∫

0x t xe p dt

The variance of complete future lifetime is given by:

( )∞

° = − ∫2

0

var ( ) 2 t x xT x t p dt e

Recursive formulas for expectations of life

The recursive formula for the curtate expectation of life is:

+= + 1x x x xe p p e

The recursive formula for the complete expectation of life is:

° °+= +∫

1

10

x t x x xe p dt p e



Median future lifetime

The median future lifetime of a life aged x, ( )m x , can be found by solving:

( )> =Pr ( ) ( ) 0.5T x m x

Central death rate

The central death rate (over the interval from x to +1x ) is:

µ+

+

+

=∫

∫

1

01

0

( )x t

x

x t

l x t dt

m

l dt

n xL is the total expected time spent alive between ages x and +x n by survivors to age xof the original group:

µ+ + += + + =∫ ∫0 0

( )n n

n x x t x n x tL t l x t dt nl l dt

The n-year temporary complete life expectancy of a life aged x is °=∫ :0

n

t x x np dt e .

xT is the total number of years lived beyond age x by survivors to age x in the originalgroup.

µ∞

+= +∫0

( )x x tT t l x t dt

The central death rate (over the interval from x to +x n ) is:

µ++

+

+−

= =∫

∫

0

0

( )n

x tx x n

n x n n xx t

l x t dtl lm

Ll dt



( )a x is the average number of years lived between x and +1x by those in the group whodie between those ages:

( )µ µ

µ µ

+

+

+ +

= = = <

+ +

∫ ∫

∫ ∫

1 1

0 01 1

0 0

( ) ( )

( ) ( ) | ( ) 1

( ) ( )

x t t x

x t t x

t l x t dt t p x t dt

a x E T x T x

l x t dt p x t dt

( )° = +1x x xe p q a x

Fractional ages

Assuming a uniform distribution of deaths between integer ages:

µ= +( )x t xq p x t = −1t x xp tq =t x xq tq

µ + =−

( )1

x

x

qx ttq − +

−=

−( )1

xs t x t

x

s t qqtq

Assuming a constant force of mortality between integer ages:

µ= − −1 exp( )xq = ( )tt x xp p = −1 ( )tt x xq p

µµ µ

+ = −

= −

( ) ln ( )exp( )

tt x x xp x t p p

tµ µ+ =( )x t −

− + = −1 ( )s ts t x t xq p

Assuming a hyperbolic form for t xp between integer ages:

=− −1 (1 )

xt x

x

tqqt q

−=

− +1

1x

t xx x

qpq tq

µ + =− −

( )1 (1 )

x

x

qx tt q − +

−=

− −( )

1 (1 )x

s t x tx

s t qqs q

ωω−

=( ) xs x ωω− +

=−( )

t xx tp

x



Mortality laws

De Moivre’s mortality law states that µ ω −= − 1( ) ( )x x , where ω≤ <0 x .

Under this law:

( ) ( )ωωω ω

− +−= =

−t xx txs x p

x

Gompertz’s mortality law states that µ =( ) xx Bc , where > > ≥0, 1, 0B c x .

Under this law:

( )= − −( ) exp ( 1)xs x m c ( )= − −exp ( 1)x tt xp mc c where =

lnBmc

Makeham’s mortality law states that µ = +( ) xx A Bc , where > 0B , ≥ −A B , > ≥1, 0c x .

Under this law:

( )= − − −( ) exp ( 1)xs x Ax m c ( )= − − −exp ( 1)x tt xp At mc c where =

lnBmc

Weibull’s mortality law states that µ =( ) nx kx , where > > ≥0, 0, 0k n x .

Under this law:

( )+= − 1( ) exp ns x ux ( )+ + = − + − 1 1exp ( )n n

t xp u x t x where =+1ku

n

Some life tables are select mortality tables. Functions in such a table depend upon ageand duration since an event. Select mortality is indicated using square parentheses, forexample [20]l , +[35] 2q .



This page has been left blank so that you can keep thesummary pages together to help you with your review.



Unit 2 Solutions

Solution 2.1

We know that = ≤( ) Pr( )XF x X x , so:

= −( ) 1 ( )XF x s x

Solution 2.2

xq is the probability that a life aged x dies in the next year.

xp is the probability that a life aged x survives to age +1x .

Solution 2.3

(i) The probability that a life aged 30 dies before his 55th birthday.

(ii) The probability that a life aged 27 survives to his 42nd birthday.

(iii) The probability that a newborn baby dies before he is 60.

(iv) The probability that a newborn baby dies between age 32 and age 79.

(v) = ≤(15) Pr( 15)XF X which is the probability that a newborn baby dies before he is15.

(vi) The future lifetime of a life aged 60.

(vii) = ≥(24) Pr( 24)s X , which is the probability that a newborn baby survives to at leastage 24.

Solution 2.4

This can be split up in two ways. Survival to age + +x s t is equivalent to survival to +x tfollowed by survival for another s years, or it is equivalent to survival to +x s followed bysurvival for another t years. In symbols:

+ + += =s t x t x s x t s x t x sp p p p p



Solution 2.5

( )K x is a discrete random variable.

Solution 2.6

If we substitute = 1s into the expression in Solution 2.4, we obtain:

+ +=1 1x t x t xp p p

Or rearranging this:

++ = 11

t xt x

x

pp

p

Since = + + +1 2 3x x x xe p p p , it follows that:

+ + + += + + +1 1 1 2 1 3 1x x x xe p p p

Using the previous expression:

+ + + += + + +

= + + +

+ + +=

−=

1 1 1 2 1 3 1

32 4

2 3 4

x x x x

xx x

x x x

x x x

x

x x

x

e p p ppp p

p p pp p p

pe p

p

From this it is clear that:

+ = −1 1xx

x

eep



Solution 2.7

This can be shown by expanding the expression for ( )2( )E K x :

( )∞

+=

+ + + +

+ + + +

+ + +

+ +

+

=

= + + + +

= + + + +

+ + + +

+ + +

+ +

∑2 2

0

1 1 2 2 3 3 4 4

1 1 2 2 3 3 4 4

2 2 3 3 4 4

3 3 4 4

4 4

( )

4 9 16

3 3 35 5

7

k x x kk

x x x x x x x x

x x x x x x x x

x x x x x x

x x x x

x x

E K x k p q

p q p q p q p qp q p q p q p q

p q p q p qp q p q

p q

Solution 2.8

(i) We can calculate this as follows:

[ ]

= −

= − −

= − −

= − −=

∫

10 28 10 2838

283828

1

1 exp 0.002

1 exp 0.002

1 exp( 0.02)0.0198

q p

dy

y

(ii) Similarly:

[ ]

= −

= −

= −=

∫60

32 2828

6028

exp 0.002

exp 0.002

exp( 0.064)0.9380

p dy

y



Solution 2.9

Using the definition of ( )s x :

( )

µ= −

= − +

= − − ×

= − −

∫

∫

0

0

1.5

01.5

( ) exp ( )

exp (0.0001 0.05 )

2exp 0.0001 0.053

exp 0.0001 0.033

x

x

x

s x y dy

y dy

y y

x x

Solution 2.10

|n xq is the probability that an individual survives to age +x n but dies before he is

+ +1x n .

It can be simplified as follows:

+ + += = − = −| 1 1n x n x x n n x n x n x n xq p q q q p p

Solution 2.11

(i) =65

350.799728

ll

(ii) − =47

271 0.047489l

l

(iii) − =

50 51

47 501 0.005831

l ll l

(iv) =100

490.004450

ll

(v) − =

75 80

40 751 0.159099

l ll l



Solution 2.12

For xl we are assuming that we have started with 100,000 lives. Notice that the numberof people alive decreases slowly at first but then decreases very quickly after the age ofabout 60. This graph is valid for a developed country, but for a third world country thesteep decline would take place at a much earlier age.

0

10000

20000

30000

40000

50000

60000

70000

80000

90000

100000

0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100

Age x

lx

The number of deaths will be (relatively) high at first due to deaths during birth and earlyinfant months. The number then decreases during childhood and there is another rise atthe late teenage years. This increase is due to deaths by drugs, suicide and vehicleaccidents. The number then falls again, starting to rise at about age 30, and continues torise with age until the seventies. The number of deaths will reach its peak among theelderly (perhaps between ages 75 and 85) and then drop to zero as age increasesbecause the number of lives still alive is much reduced. Again this graph is only valid for adeveloped country.

0

500

1000

1500

2000

2500

3000

3500

4000

0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100

Age x

dx



The graph for xq has many of the same features as the graph for the number of deaths,such as the teenage ‘accident hump’ and high infant mortality; as age increases, theprobability of death in the coming year increases steadily. We have used a logarithmicscale for the xq axis so that the main features are visible.

-4.0

-3.5

-3.0

-2.5

-2.0

-1.5

-1.0

-0.5

0.0

0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100

Age x

log qx

Solution 2.13

This relationship is not true. It would only be true if the probability of dying before the ageof 40 is 0.

Solution 2.14

We know that:

µ+ = −

∫exp ( )

x t

t xx

p y dy

Which here is:

[ ] += − = −exp 0.02 exp( 0.02 )x tt x xp y t

We need to find the complete expectation of life:

∞ ∞° − = = = − ∫

00

exp( 0.02 ) 500.02x t x

te p dt



Then we can use the formula for variance using integration by parts with = 2u t :

( )∞

°

∞∞

∞

= − −

− = × + − − −

− = − −

= − =

∫

∫

2

0

2

0 0

2

0

2

var ( ) 2 exp( 0.02 )

exp( 0.02 ) 22 exp( 0.02 ) 500.02 0.02

exp( 0.02 )100 500.02

1100 500.02

2500

xT x t t dt e

tt t dt

t

Thus the standard deviation is 50.

Solution 2.15

The limiting age of the table is 110 since there are no survivors after this age, therefore=110 0e .

We can calculate that =1091136

p , =10836

108p and =107

108292

p , so:

= + =109 11011 11 0.305636 36

e e

= + =108 10936 36 0.4352

108 108e e

= + =107 108108 108 0.5308292 292

e e



Solution 2.16

Since = −exp( 0.02 )t xp t , = − =exp( 0.02) 0.98012xp .

Using the recursive formula:

° °+

°+

°+

°+

= − +

− = + − −

= − +

= +

∫1

10

1

10

1

1

exp( 0.02 ) 0.980199

exp( 0.02 ) 0.9801990.02

1 exp( 0.02) 0.9801990.02 0.020.990066 0.980199

x x

x

x

x

e t dt e

t e

e

e

We can start with ° =20 50e to give:

° =21 50e

° =22 50e

° =23 50e

Solution 2.17

To find the lower and upper quartiles you would solve the following equations:

+=

( ( )) 0.75( )

s x l xs x

+=

( ( )) 0.25( )

s x u xs x

where ( )l x and ( )u x are the lower and upper quartiles respectively.

To find the mode, set the derivative of ( )( )T xf t with respect to t equal to 0, and solve the

resulting equation.



Solution 2.18

If = −( ) exp( 0.03 )s x x , then the equation for the median is:

+ − += =

−( ( )) exp( 0.03[ ( )]) 0.5

( ) exp( 0.03 )s x m x x m x

s x x

This can be solved as follows:

− += − =

−

⇒ = =−

exp( 0.03[ ( )]) exp( 0.03 ( )) 0.5exp( 0.03 )

ln0.5( ) 23.10.03

x m x m xx

m x

ie the median future lifetime is 23.1 years.

Similarly, the lower quartile and upper quartile are:

= =−ln0.75( ) 9.59

0.03l x = =

−ln0.25( ) 46.2

0.03u x

so the interquartile range is 36.6 years.

Solution 2.19

We need to work out the total expected time spent alive between x and +x n , so we mustconsider two groups of lives separately, namely those that survive to +x n and those thatdon’t.

Those that survive to +x n have a time spent alive of n and there are expected to be +x nlof them. Altogether they contribute +x nnl to the total expected time spent alive.

The length of time spent alive by those who die before +x n is expected to be

∫ ( )0

( )n

T xt f t dt .

There were xl people alive at age x, so the total contribution from this group is:

µ µ+= + = +∫ ∫ ∫( )0 0 0

( ) ( ) ( )n n n

x T x x t x x tl t f t dt l t p x t dt t l x t dt

Summing gives the required result.



Solution 2.20

We know that µ= − +( )t x t xd p p x tdt

, therefore µ+ += − +( )x t x td l l x tdt

.

If we perform integration by parts using =u t , we get:

+ + + += − + + = ∫ ∫00 0

n nn

n x x t x t x n x tL t l l dt nl l dt

Solution 2.21

++= = =

∫∫ ∫0

0 0

n

x t n nx tn x

t xx x x

l dtlL dt p dt

l l l

Solution 2.22

µ µ++

+

+ +−

= = = =∫ ∫

∫

1 1

0 0 11

0

( ) ( )x t x t xx x x x

xx x x

x t

l x t dt l p x t dtl q l lm

L L Ll dt



Solution 2.23

µ

µ

∞

+ ∞+

= = +∫

∫0

0

( )

( )x t

xt x

x x

t l x t dtT t p x t dtl l

Using integration by parts with = −u t , we get:

∞ ∞∞= − + = ∫ ∫0

0 0

xt x t x t x

x

T t p p dt p dtl

But ∞

°=∫0

t x xp dt e , so °=xx

x

T el

.

Solution 2.24

We can find the values of µ as follows:

µ′−

= =−

( ) 1 500( )( ) 1 500

s xxs x x

Giving:

µ

µ

= =−

= =−

1 500(40) 0.002171 40 500

1 500(50) 0.002221 50 500

Now for n xq :

+= − = −

( )1 1( )n x n x

s x nq ps x

Giving:

−= − = − =

−10 40(50) 1 50 5001 1 0.02174(40) 1 40 500

sqs



Finally for n xm we divide the expression for n xm by xl giving:

( )− − −

= = =− + + × − − − −

∫ ∫2

0 0 0

1 1 1

1 500 1 500 11 500 1 500 2 500

n x n x n xn x n n n

t x

p p pmx t x tp dt dt

x x

Putting in the values in the question:

( )− ×

= = =− + × − − −

10 4010 40 10 2 22

0

1 0.92 0.02174 0.002198250 0.92 0.901 500 401

1 40 500 2 500

pm

t

Solution 2.25

Answer C.

Using the assumption of uniform distribution of deaths, we have:

−=

−61

0.5 61.561

(1 0.5)1 0.5

qq

q

where = − = − =6261

61

7,954,1791 1 0.01501158,075,403

lq

l.

This gives =0.5 61.5 0.00756q .



Solution 2.26

Answer C.

We simplify the probability as follows (whatever our assumption):

=10 50.5 0.5 50.5 9 51 0.5 60p p p p

Under the assumption of constant force of mortality this simplifies to:

=

=

=

0.5 0.510 50.5 50 9 51 60

0.5 0.551 61

9 5150 60

( ) ( )

0.91116

p p p p

l lp

l l

Solution 2.27

Answer B.

The probability simplifies in the same way as Question 2.26, ie

=10 50.5 0.5 50.5 9 51 0.5 60p p p p

Under the assumption of a hyperbolic form for t xp , this simplifies to:

( )

= − −

= − −

− =

10 50.5 0.5 50.5 9 51 0.5 60

60 6050

51 60

(1 ) (1 )

0.51 0.5 1

1 0.5

0.91114

p q p q

l qq

l q



Solution 2.28

De Moivre’s law:

[ ] ωµ ω ωω

− − = − = − − = − = ∫ ∫ 1

00 0

( ) exp ( ) exp ( ) exp ln( )x x

x xs x y dy y dy y

ωω ωω ω ω

+ − +− + −= = =

−( ) ( )( )

( )t xs x t x tx t xp

s x x

Gompertz’s law:

µ = −

− = − = = − + = − −

∫

∫

0

0 0

( ) exp ( )

exp exp exp exp ( 1)ln ln ln ln

x

xx y xy x

s x y dy

Bc Bc B BBc dy cc c c c

This is often written as ( )− −exp ( 1)xm c where =lnBmc

.

( ) ( )( ) ( )

+

+

+= = − − − −

= − + + − = − −

( ) exp ( 1) exp ( 1)( )

exp exp ( 1)

x t xt x

x t x x t

s x tp m c m cs x

mc m mc m mc c

Makeham’s law:

( ) ( )

µ = −

= − − = − −

= − − + = − − −

∫

∫

0

0 0

( ) exp ( )

exp expln

exp exp ( 1)

x

xx yy

x x

s x y dy

BcA Bc dy Ayc

Ax mc m Ax m c

where =lnBmc

.



( ) ( )( )( )

+

+

+= = − + − − − − −

= − + − + + + −

= − − −

( ) exp ( ) ( 1) exp ( 1)( )

exp ( )

exp ( 1)

x t xt x

x t x

x t

s x tp A x t m c Ax m cs x

A x t mc m Ax mc m

At mc c

Weibull’s law:

µ

+ +

= −

= − = − = − + +

∫

∫

0

1 1

0 0

( ) exp ( )

exp exp exp1 1

x

xx n nn

s x y dy

ky kxky dyn n

This is sometimes written as ( )+− 1exp nux where =+1ku

n.

( ) ( )( ) ( )

+ +

+ + + +

+= = − + −

= − + + = − + −

1 1

1 1 1 1

( ) exp ( ) exp( )

exp ( ) exp ( )

n nt x

n n n n

s x tp u x t uxs x

u x t ux u x t x



Solution 2.29

We first need to find the parameters for Gompertz’s law. From the previous question:

( )= − −exp ( 1)x tt xp mc c

So we have simultaneous equations:

( )= − − =15 1010 15 exp ( 1) 0.957076p mc c

( )= − − =20 1010 20 exp ( 1) 0.896579p mc c

Taking logs and dividing the two equations:

− −= =− −

15 10

20 10 5ln0.957076 ( 1) 1ln0.896579 ( 1)

mc cmc c c

From this we get that =1.2c . By substituting this value back into the equation we can getthat = 0.00054848m , ie = 0.0001B .

We require 17 18p , which is:

= − − =

18 1717 18

0.0001exp 1.2 (1.2 1) 0.733911ln1.2

p

Solution 2.30

Answer D.

+= = =[22] 22 [22]

[22]

35,607.78 0.9994235,628.36

lp

l

Course 3, Unit 3 Insurance functions


Unit 3Insurance functions

Unit overview

In this unit we look at the various life insurance benefits. These are benefits paid on orafter the death of the policyholder. In some cases benefits are paid immediately on death. In other cases benefits are paid at the end of year of death or at some other convenientpoint in the year. The benefit can also be paid on survival to a certain date.

First we describe each benefit in practical terms and then we introduce the new notationthat is required for this unit. Next we derive the actuarial present value of each benefit byconsidering random variables.

The unit concludes with sections looking at benefits that change over time, recursiverelationships to find term insurance benefits and the relationships between benefitspayable at different times.

Advice on study

This is another unit that introduces new notation. Spend plenty of time familiarizingyourself with the different insurance benefits and how they are calculated.



Insurance functions Course 3, Unit 3


Learning objectives

6 Formulate a model for the present value, with respect to an assumed interest ratestructure, of a set of future contingent cash flows. The model may be stochastic ordeterministic.

8 Apply a principle to a present value model to associate a cost or pattern of costs(possibly contingent) with a set of future contingent cashflows.

– Principles include: equivalence, exponential, standard deviation, varianceand percentile.

– Models include: present value models based on learning objectives9-12.

– Applications include: insurance, health care, credit risk, environmental risk,consumer behavior (eg subscriptions) and warranties.



1 IntroductionIn this unit we are going to look at life insurance payments, ie payments made on thedeath of the policyholder. Policyholders pay premiums in order to receive a benefitpayment on death. Here we have two unknowns; the length of time before death (whichwill affect the number of premiums that will be received) and the rate of interest that isearned on the premiums that are invested.



2 Types of benefitThere are several types of death benefit. You need to be familiar with each one, noting ifand when payments are made. In this section we describe how claims occur under eachtype of policy. The next section covers the notation used for each policy.

2.1 Whole life insurance

A whole life insurance policy provides a payment on the death of the policyholder at anypoint in the future. The payment can be made immediately on death (although in practicethere will be a short delay while the insurance company processes the claim and posts thecheck!) at the end of the year of death or at other times.

2.2 Pure endowment

A pure endowment policy provides a payment if the policyholder survives to a certain date.No payment is made otherwise.

2.3 Term insurance

A term insurance policy provides a payment on the death of the policyholder as long as hedies within a specified time period. If the policyholder dies after this time period haselapsed, no payment is made. The payment can be made immediately on death or at theend of the year of death or at other times.

2.4 Endowment insurance

An endowment insurance policy provides a payment on the death of the policyholder if hedies within a specified time period or if he survives to the end of that period. It is acombination of a pure endowment and a term insurance. The death payment can bemade immediately on death or at the end of the year of death.

2.5 Deferred insurance

A deferred insurance policy provides a payment on the death of the policyholder as longas the death occurs after a specified time period (which is called the deferred period).



3 NotationInternational Actuarial Notation is a system of communication used by actuaries globally inorder to ensure that actuaries in different countries can understand each other’s workings.You have already met the notation for annuities in Course 2, namely na , na , na etc, as

well as in Unit 2 eg t xp , t xq .

Question 3.1

What does the bar on a mean?

Question 3.2

What does the symbol |m na mean?

In order to understand the symbols associated with life insurance, it is necessary for us tointroduce some further notation. Consider the subscript that follows the ‘a’ symbol in theannuity functions above. It describes a status that at any stage is either active or inactive.When the status is active, payments are made. In this context n means that the conditionis active for n years, ie payments are made for n years.

n is usually used for a definite number of years, whereas x is used for a life. When wetalk about x being active it means that the person is alive.

We will now detail some alternative subscripts that we will be using in connection withinsurance:

Subscript Meaninguv Payment is made whilst both u and v are activeuv Payment is made whilst at least one of u and v is active|u v Payment is made whilst u is inactive but v is active

1:u v Payment is made only if u becomes inactive first

1:u v Payment is made only if v becomes inactive first



The symbol for the actuarial present value of death benefits paid at the end of year ofdeath is A. If benefits are paid immediately on death the symbol is A . We will now look atthis new notation in the context of the policies covered in Section 2. The notation assumesthat payment is made immediately on death and in each case refers to the actuarialpresent value of the benefits concerned.


The symbol used for whole life insurance is xA . The subscript x means that the paymentis made on the death of the policyholder at any point in the future.

3.2 Pure endowment

The symbol for a pure endowment with a term of n years is 1:x n

A . The 1 above the n

means that the payment is made on the expiry of n years but only if the policyholder is stillalive at that point. In other words, the status n has to be the first condition to becomeinactive in order for the payment to be made.

3.3 Term insurance

The symbol for a term insurance (with a term of n years) is 1:x n

A . The 1 above the x

means that the policyholder must come to an end (ie die!) before the n years have beencompleted in order for payment to be made.


The symbol for endowment assurance (with a term of n years) is :x nA . The subscript

:x n means that payment is made on death if death occurs in the first n years or onsurvival to n years. The payment is made on the first of the two statuses to terminate.


The symbol for deferred whole life insurance (with a deferred period of m years) is |m xA .

Note that this refers to a policyholder who is currently aged x, so he will have to survive atleast m years (ie until age +x m ) in order to receive the payment on his eventual death.

Deferred term insurance and endowments also exist.



4 Evaluating benefitsWe will study each type of benefit in detail and express the present value of the benefit asa random variable. We can then develop expressions for the mean and variance of theserandom variables. In each section we consider benefits that are paid at the end of theyear of death and benefits that are paid immediately. We will assume throughout that thepolicyholder is aged x when he takes out his policy and will use the notation (x) torepresent a life aged x.


Benefit paid at the end of the year of death

Deterministic model

Consider a whole life insurance with a benefit of 1 paid at the end of the year of death.

The benefit will be paid at the end of the first year if the policyholder dies in the first year. The benefit will be paid at the end of the second year if the policyholder survives the firstyear and then dies in the second year, and so on. If we know survival and deathprobabilities, then an expression for the actuarial present value of the benefit is:

∞+

+ +=

= + + + = ∑2 3 11 2 2 |

0

kx x x x x x k x

kA v q v p q v p q v q

Stochastic model

We will now write ( )K x , the curtate future lifetime of (x) as K for convenience. Therandom variable for the present value of this benefit is then given by:

+= 1KZ v

The actuarial present value is the expected value of this random variable:

∞+ +

== = = =∑1 1

0( ) ( ) Pr( )K k

xk

A E Z E v v K k

The stochastic and deterministic formulas are the same since, from Unit 2,= = |Pr( ) k xK k q .



The variance is given by:

[ ]

( ) ( ) ( ) ( )

( )

+ +

= −

= − = −

= −

22

2 21 2 2 1

22

var[ ] ( ) ( )

( ) ( )K Kx x

x x

Z E Z E Z

E v A E v A

A A

The first term in this expression is in the same form as the present value of a whole lifeinsurance, but this time v has been replaced by 2v . Effectively this means that the firstterm is the present value of a whole life insurance calculated at a different rate of interest,namely + −2(1 ) 1i .

Question 3.3

Explain why the new interest rate is + −2(1 ) 1i .

Question 3.4

What is the equivalent new force of interest?

Benefit paid immediately on death

Deterministic model

Consider a whole life insurance policy with a benefit of 1 payable immediately on death. Consider the short time interval +( , )t t dt . The probability that (x) survives to age +x tand then dies during the interval +( , )t t dt is:

+t x dt x tp q

So the actuarial present value of the benefit would be:

++

t dtt x dt x tv p q

But µ+ ≈ +( )dt x tq x t dt for small time intervals, so the probability is approximately:

µ +( )t xp x t dt



The actuarial present value is approximately:

µ +( )tt xv p x t dt

Considering all possible times of death gives:

µ∞

= +∫0

( )tx t xA v p x t dt

Stochastic model

The random variable Z, which represents the present value of the above benefit, is ( )T xvwhere ( )T x (≥ 0 ) is the future lifetime of the policyholder. We will drop the ( )x notationfor convenience.

The actuarial present value of this benefit is ( )E Z :

∞

= = = ∫0

( ) ( ) ( )T tx TA E Z E v v f t dt

Question 3.5

Do the stochastic and deterministic models give the same value for the actuarial presentvalue of this benefit?

Now consider the variance:

[ ] ( ) ( )

( ) ( )

= −

= −

22

22

var

( )

T T

T T

Z E v E v

E v E v

The actuarial notation for 2( )TE v is 2xA . Therefore:

[ ] ( )= −22var x xZ A A



4.2 Pure endowment

Deterministic model

For a pure endowment policy on (x) with a benefit of 1 and a term of n years, the benefit ispaid if the policyholder survives to at least age +x n . The actuarial present value of thisbenefit is:

= + =1:

0 n nn x n x n xx n

A q v p v p

Stochastic model

The random variable representing the present value of this benefit is:

≤= >

0n

T nZ

v T n

or equivalently:

<= ≥

0n

K nZ

v K n

The actuarial present value of the benefit is ( )E Z :

= = + ≥1:

[ ] 0 Pr( )nx n

A E Z v K n

The versions obtained using the stochastic and deterministic approaches are equivalentsince, from Unit 2, ≥ =Pr( ) n xK n p .

Notice that 1:x n

A is meaningless since the benefit is only ever paid at time n.


( ) [ ]

( ) ( )( ) ( )

= −

= −

= −

22

22 1:

22 1:

var[ ] ( )

nn x x n

nn x x n

Z E Z E Z

v p A

v p A



The first term is the actuarial present value of a pure endowment calculated at a rate ofinterest + −2(1 ) 1i , so the variance is:

( )= −22 1 1

: :var[ ]

x n x nZ A A

Question 3.6

Suppose that Z is the present value of a 10-year pure endowment policy issued to aperson aged 52. The sum insured is $100,000. Find the expected value and the standarddeviation of Z, assuming that interest is 4% a year effective and that mortality follows theTables.

4.3 Term life insurance


Deterministic model

Consider a term life insurance with a benefit of 1 and a term of n years. Suppose that thebenefit is payable at the end of the year of death.

The benefit will be paid at the end of the first year if the policyholder dies in the first year. The benefit will be paid at the end of the second year if the policyholder survives the firstyear and then dies in the second year and so on up to the nth year. If we know survivaland death probabilities, then an expression for the actuarial present value of the benefit is:

−+

+ + − + −=

= + + + + = ∑1

1 2 3 11 2 2 1 1 |:

0

nn k

x x x x x n x x n k xx nk

A vq v p q v p q v p q v q

Stochastic model

For this policy the random variable which represents the present value of the benefit is:

+ <= ≥

1

0

K K nvZK n

The actuarial present value is the expected value of this random variable:

− −+ +

= == = = + × ≥ = =∑ ∑

1 11 1 1

:0 0

( ) Pr( ) 0 Pr( ) Pr( )n n

k kx n

k kA E Z v K k K n v K k



The stochastic and deterministic formulas are the same since, from Unit 2,= = |Pr( ) k xK k q .

The variance is given by:

( ) [ ]

( ) ( )

( )

−+

=− +

=

= −

= −

= −

∑

∑

1 2 21|

01 212 1

| :0

22 1 1: :

var[ ] ( )n

kk x

kn k

k x x nk

x n x n

Z v q E Z

v q A

A A

Benefit paid immediately

Deterministic model

Consider a term life insurance with a term of n years and a benefit of 1 payableimmediately on death. Using a similar method as for the whole life insurance:

µ= +∫1:

0

( )n

tt xx n

A v p x t dt

This is the same integral as for the whole life insurance, but this one has an upper limit ofn since no benefit is paid if death occurs after n years.

Stochastic model

The random variable Z, which represents the present value of the above benefit is:

≤= > 0

T T nvZT n


µ∞

= = + = +∫ ∫ ∫1:

0 0

( ) ( ) 0 ( ) ( )n n

t tT T t xx n

n

A E Z v f t dt f t dt v p x t dt




( ) [ ] ( ) ( )µ µ= + − = + −∫ ∫22 2 2 1

:0 0

var[ ] ( ) ( ) ( )n n tt

t x t x x nZ v p x t dt E Z v p x t dt A

The first term is the actuarial present value of a term life insurance with v replaced by 2v .

Therefore:

( )= −22 1 1

: :var[ ]

x n x nZ A A

Note that there is an alternative way of setting up the random variable that you may see inother books or the exams.

Define tb to be the benefit function, tv to be the discount function and Z to be the randomvariable that represents the present value of the benefit. Hence:

= t tZ b v

For the term life insurance we have been considering:

≤= >

10t

t nb

t n= t

tv v , where ≥ 0t ≤=

> 0

T T nvZT n



Deterministic model

Consider an endowment insurance policy with a benefit of 1 and a term of n years wherethe benefit is payable at the end of the year of death.

The benefit will be paid at the end of the first year if the policyholder dies in the first year. The benefit will be paid at the end of the second year if the policyholder survives the firstyear and then dies in the second year, and so on up to the nth year. The benefit will alsobe paid if the policyholder survives for n years. If we know survival and death probabilities,then an expression for the actuarial present value of the benefit is:

−+

== + ∑

11

|:0

nn k

n x k xx nk

A v p v q



Stochastic model

Again consider the situation where the benefit is paid at the end of year of death.

Question 3.7

What is Z, the random variable representing the present value of this benefit? What are( )E Z and var[ ]Z ?

Benefit paid immediately

Deterministic model

Consider an endowment insurance policy with a term of n years and a benefit of 1 wherethe benefit is payable immediately on death. Endowment insurance is the combination of aterm life insurance and a pure endowment, ie:

= +1 1: : :x n x n x n

A A A

An expression for the actuarial present value of the endowment insurance is then:

µ= + +∫:0

( )n

t nt x n xx n

A v p x t dt v p

Stochastic model


≤= >

T

nv T n

ZT nv


= + >∫0

( ) ( ) Pr( )n

t nTE Z v f t dt v T n




( ) ( ) [ ]

( ) ( ) ( )

= + > −

= + > −

∫

∫

2 2 2

0

22 2:

0

var[ ] ( ) Pr( ) ( )

( ) Pr( )

nt n

T

n t nT x n

Z v f t dt v T n E Z

v f t dt v T n A

The first two terms are the present value of an endowment insurance policy with vreplaced by 2v . Therefore:

( )= −22

: :var[ ] x n x nZ A A


Benefit paid immediately on death

Deterministic model

Consider a deferred whole life insurance with a benefit of 1 payable immediately on deathwith a deferred period of m years. The actuarial present value is similar to the whole lifeinsurance covered earlier. However, the earliest time any benefit can be paid is m, so therequired expression is:

µ∞

= +∫| ( )tm x t x

m

A v p x t dt

Stochastic model


≤= >

0T

T mZ

T mv


|( ) ( )tm x T

m

E Z A v f t dt∞

= = ∫




( ) [ ]

( ) ( )

∞

∞

= −

= −

∫

∫

2 2

22|

var[ ] ( ) ( )

( )

tT

m

tT m x

m

Z v f t dt E Z

v f t dt A

The first term is the present value of a deferred insurance policy with v replaced by 2v .Therefore:

( )= −22

| |var[ ] m x m xZ A A

Question 3.8

Suppose that Z is the random variable representing a deferred insurance with a benefit of1 payable at the end of year of death and a deferred term of m years. Write down anexpression for Z in terms of future lifetime random variables. What is ( )E Z ? What isvar[ ]Z ?

Question 3.9

For the same deferred insurance as defined in Question 3.8, find an expression for theactuarial present value using a deterministic approach.

We can also have deferred term life insurance and deferred endowment insurance.



5 Valuing insurance where benefits change over time

5.1 Increasing insurances

Consider a whole life insurance where a benefit of t is paid immediately on death at time t. Using T to denote the complete future lifetime, the random variable representing thepresent value is:

= TZ Tv where ≥ 0T

The actuarial present value of this benefit is written as ( )xI A .

This is a continuously increasing benefit and is referred to as a continuously increasingwhole life insurance.

Question 3.10

Write down integral expressions for ( )E Z and var[ ]Z .

In the discussion above, the increases took place continuously. Alternatively the increasescould take place at the beginning of each year. This time a benefit of + 1t is paid

immediately on death at time t. Recall that x represents the integer part of x. This

policy pays 1 immediately on death in the first year, 2 immediately on death in the secondyear etc. The random variable representing the present value is:

= + 1TZ T v where ≥ 0T

The notation for the actuarial present value of this policy is ( )xIA .

Question 3.11

Write down an integral expression for ( )E Z .



In fact the increases could take place more frequently than once a year. If increases of

amount 1m

take place m times per year at the start of each period, with a first benefit of

1m

, then the random variable representing the present value of the benefit is:

+ =1 TTm

Z vm

where ≥ 0T

The actuarial notation for the actuarial present value of this benefit is ( )( )mxI A .

Question 3.12

What happens when →∞m in the above expression?

Question 3.13

What do you think 1:

( )x n

IA represents? Write down an expression for Z in terms of T.

Question 3.14

What do you think :

( )x n

IA represents? Write down an expression for Z in terms of T.

If you are confident working with double integrals, you can also show that∞

= ∫ |0

( )x s xI A A ds .

Question 3.15

Explain what the actuarial symbols ( )xIA , 1:

( )x n

IA and :

( )x n

IA mean. In each case, write

down an expression for Z in terms of T. what is an expression for ( )E Z ?



5.2 Decreasing insurance

A policy that pays n immediately on death in the first year, −1n immediately on death inthe second year etc with no benefit after n years is called an annually decreasing n-yearterm life insurance. The random variable representing the present value of this benefit is:

≤ − = >

( )0

T T nv n TZT n

The notation for the actuarial present value of this benefit is 1:

( )x n

DA .

Question 3.16

Write down an integral expression for ( )E Z .

Question 3.17

Suppose that Z represents a continuously decreasing n-year term life insurance. Writedown an expression for Z in terms of T.

Question 3.18

What is +1 1: :

( ) ( )x n x n

DA IA equal to?

Question 3.19

Explain what the actuarial symbol 1:

( )x n

DA means. Write down an expression for Z in

terms of K. What is an expression for [ ]E Z ?



6 Recursive relationshipsRecursive relationships were derived for life expectancies in Unit 2. We can also deriverecursive relationships for the value of term insurance benefits.

Recall that:

− −+ +

+= =

= =∑ ∑1 1

1 1 1|:

0 0

n nk k

k x k x x kx nk k

A v q v p q

This can be re-expressed:

−+

+=

−+

+ + +=

+ −

= +

= +

= +

∑

∑

11 1

:1

21

1 1011: 1

nk

x k x x kx nk

ni

x x i x x ii

x x x n

A vq v p q

vq vp v p q

vq vp A

where ω= −…0,1, , 1x .

To use this formula we define +

=1:0

0x n

A .

Question 3.20

Calculate the value of 150:5

A recursively using an interest rate of 8% per annum and

assuming mortality follows the Tables.

Similarly:

+ −= +

: 1: 1x xx n x nA vq vp A where =

:00

xA

+= + 1x x x xA vq vp A where ω= −…0,1, , 1x , ω = 0A

Question 3.21

Find a recursive formula for ( )xIA where ω= −…0,1, , 1x .



7 Relationship between benefits payable at different timesIn this section we consider a whole life insurance and make an assumption of a uniformdistribution of deaths to arrive at an approximate relationship between xA and xA .

Consider a whole life insurance as the sum of deferred term life insurances:

+ +

= + + + +

= + + +

1 1 1 10| 1| 2| 3|:1 :1 :1 :1

1 1 2 12:1 1:1 2:1

x x x x x

x xx x x

A A A A A

A vp A v p A

Now we know from Unit 2 that, under a uniform distribution of deaths between integerages:

µ+ ++= + + =∫ ∫

1 11

:10 0

( )t tt x k x kx k

A v p x k t dt v q dt

Further:

δ−

= =∫1

10

1t vv dt a

This means that we can simplify +1

:1x kA :

δ+ ++−

= =∫1

1:1

0

1tx k x kx k

vA q v dt q

Which in turn means that we can simplify xA :

( )( )( )

δ

δ

δ

δ

+ +

+ +

+ +

−≈ + + +

= + + +

= + + +

=

21 2 2

21 2 2

2 31 2 2

1x x x x x x

x x x x x

x x x x x

x

vA q vp q v p q

iv q vp q v p q

i vq v p q v p q

i A



Question 3.22

Show that δ

≈1 1: :x n x n

iA A .

Similarly it can be shown that the following relationships hold:

δ≈1 1

: :( ) ( )

x n x niIA IA

δ δ ≈ − −

1 1( ) ( )x x xiI A IA A

d

7.1 Insurances payable mthly

Suppose we have a whole life insurance for a life aged exactly x where the benefit of 1 ispayable at the end of the quarter when death occurs. The actuarial present value of thisbenefit is written as (4)

xA .

In general we have ( )mxA for a whole life insurance for a life aged exactly x where the

benefit of 1 is payable at the end of the mthly period during which death occurs.

Clearly we should be able to approximate ( )mxA in terms of xA . It can be shown that:

=( )( )

mx xm

iA Ai

Question 3.23

A person aged exactly 45 takes out a whole life insurance policy with a benefit of $60,000payable at the end of the month of death. Assuming interest rates of 6% a year effectiveand mortality as given in the Tables find an approximate value for the actuarial presentvalue of this benefit.



Unit 3 Summary

Whole life insurance

A whole life insurance policy provides a payment on the death of the policyholder at anypoint in the future.

If the benefit is payable at the end of year of death, we have:

∞+

== ∑ 1

|k

x k xk o

A v q or ∞

+

== =∑ 1

0Pr( )k

xk

A v K k

[ ] ( )= − 22var x xZ A A

If the benefit is payable immediately on death, we have:

µ∞

= +∫0

( )tx t xA v p x t dt or

∞

= ∫0

( )tx TA v f t dt

[ ] ( )= −22var x xZ A A

Pure endowment

A pure endowment policy provides a payment if the policyholder survives to a certain date. No payment is made otherwise.

=1:

nn xx n

A v p

( )= −22 1 1

: :var[ ]

x n x nZ A A

Notice that 1:x n

A is meaningless since the benefit is only ever paid at time n. The

alternative notation n xE is sometimes used for 1:x n

A . (See the Illustrative Life Table used

in conjunction with the Course 3 exam.)



Term insurance

A term life insurance policy provides a payment on the death of the policyholder as long ashe dies within a specified time period. If the policyholder dies after that time has elapsed,no payment is made.


−+

== ∑

11 1

|:0

nk

k xx nk

A v q or −

+

== =∑

11 1

:0

Pr( )n

kx n

kA v K k

[ ] ( )= −22 1 1

: :var

x n x nZ A A


µ= +∫1:

0

( )n

tt xx n

A v p x t dt or = ∫1:

0

( )n

tTx n

A v f t dt

[ ] ( )= −22 1 1

: :var

x n x nZ A A

Endowment insurance

An endowment insurance policy provides a payment on the death of the policyholder if hedies within a specified time period or if he survives to the end of that period. It is acombination of a pure endowment and a term life insurance.


−+

== + ∑

11

|:0

nn k

n x k xx nk

A v p v q or −

+

== > + =∑

11

:0

Pr( ) Pr( )n

n kx n

kA v K n v K k

[ ] ( )= −22

: :var

x n x nZ A A


µ= + +∫:0

( )n

n tn x t xx nA v p v p x t dt or = > + ∫:

0

Pr( ) ( )n

n tTx n

A v T n v f t dt

[ ] ( )= −22

: :var

x n x nZ A A



Deferred insuranceA deferred insurance policy provides a payment on the death of the policyholder as longas the death occurs after the deferred period.

If the benefit is payable at the end of year of death, for a deferred whole life insurance wehave:

∞+

== =∑ 1

| Pr( )km x

k mA v K k

[ ] ( )= −22

| |var m x m xZ A A

If the benefit is payable immediately on death, for a deferred whole life insurance we have:

∞

= ∫| ( )tm x T

m

A v f t dt

[ ] ( )= −22

| |var m x m xZ A A

We can also have deferred term life insurance and deferred endowment insurance.

Valuing insurances where benefits change over time

A continuously increasing whole life insurance is where a benefit of t is paid immediatelyon death at time t. We have:

µ∞

= +∫0

( ) ( )tx t xI A tv p x t dt

An annually increasing whole life insurance is where a benefit of + 1t is paid

immediately on death at time t. We have:

µ∞

= + + ∫0

( ) 1 ( )tx t xI A t v p x t dt



An annually decreasing n-year term life insurance is where a benefit of − ( )n t is paid

immediately on death at time t. We have:

µ= − + ∫1:

0

( ) ( ) ( )n

tt xx n

DA v n t p x t dt

Other formulas are constructed in a similar way.

Notice also that: + = +1 1 1: : :

( ) ( ) ( 1)x n x n x n

DA IA n A

Recursive relationships

+ −= +1 1

: 1: 1x xx n x nA vq vp A where =1

:00

xA

+ −= +

: 1: 1x xx n x nA vq vp A where =

:00

xA

+= + 1x x x xA vq vp A where ω= −…0,1, , 1x , ω = 0A

+ += + +1 1( ) ( )x x x x x xIA vq vp A vp IA

Relationships between benefits payable at different times

Approximate relationships hold between the present values of insurance benefits payableimmediately on death and at the end of year of death:

δ≈x x

iA A

δ≈1 1

: :x n x niA A

δ≈1 1

: :( ) ( )

x n x niIA IA

δ δ ≈ − −

1 1( ) ( )x x xiI A IA A

d

Insurances payable mthly

=( )( )

mx xm

iA Ai



Unit 3 Solutions

Solution 3.1

It means that the payments are made continuously.

Solution 3.2

This is the present value of an annuity of 1 payable at the end of each year starting in myears’ time and continuing for a further n years.

Solution 3.3

If we say that the new interest rate is I, associated with V, then:

= − = − = + −22

1 11 1 (1 ) 1I iV v

as required.

Solution 3.4

If we are using a force of interest, δ , then δ−=T Tv e . Thus:

( ) ( ) ( )δ δ− −= =2 2 2 TT Tv e e

The equivalent new force of interest is therefore δ2 .

Solution 3.5

Yes, the stochastic and deterministic models give the same result for the actuarial presentvalue. Recall from Unit 2 that:

µ= +( ) ( )T t xf t p x t



Solution 3.6

For a 10-year pure endowment policy on a life aged 52:

≤= >

100 10

100,000 10

TZ

v T

= >10[ ] Pr( 10)E Z v T and ( )= > −102 2 2var[ ] 100,000 Pr( 10) ( )Z v T E Z

If mortality follows the Tables we can use the fact that > = 10 52Pr( 10)T p to evaluatethese. We get:

−= × =10 62

52[ ] 100,000 1.04 $60,781.56

lE Z

l

( )−= − =22 20 62

52var[ ] 100,000 1.04 60,781.56 411,786,125

lZ

l

Therefore the standard deviation is $20,292.51.

Solution 3.7

The random variable representing the present value of an endowment insurance withbenefit paid at the end of year of death is:

+ <= ≥

1K

nv K n

ZK nv

The expectation and variance are as follows:

−+

== = + ≥∑

11

0( ) Pr( ) Pr( )

nk n

kE Z v K k v K n

[ ]−

+

== = + ≥ −∑

121 2 2

0var[ ] ( ) Pr( ) ( ) Pr( ) ( )

nk n

kZ v K k v K n E Z

The variance can be written as ( )−22

: :x n x nA A .



Solution 3.8

The random variable representing the present value of a deferred insurance paid at theend of year of death is:

+

<= ≥

10

KK m

Zv K m

The expectation is as follows:

∞+

== =∑ 1( ) Pr( )k

k mE Z v K k

The variance of the random variable is:

[ ] [ ]∞ ∞

+ +

= == = − = = −∑ ∑2 21 2 2 1var[ ] ( ) Pr( ) ( ) ( ) Pr( ) ( )k k

k m k mZ v K k E Z v K k E Z

The first term is just the actuarial present value of a deferred insurance evaluated at the

new rate of interest + −2(1 ) 1i , so we can write ( )= −22

| |var[ ] m x m xZ A A .

Solution 3.9

The benefit will be paid after +1m years if death occurs during year m to +1m and after+ 2m years if death occurs during year +1m to + 2m and so on. The actuarial present

value is then:

1 2 3| 1 1 2 2

m m mm x m x x m m x x m m x x mA v p q v p q v p q+ + +

+ + + + + + += + + +

Solution 3.10

If = TZ Tv , then:

∞

= ∫0

( ) ( )tTE Z tv f t dt



( ) [ ]∞

= −∫2 2

0

var[ ] ( ) ( )tTZ tv f t dt E Z

Solution 3.11

If = + 1TZ T v , then:

∞

= + ∫0

[ ] 1 ( )tTE Z t v f t dt

Solution 3.12

When →∞m , + →

1TmT

m, therefore the actuarial present value of the benefit becomes

( )xI A .

Rather than thinking about this mathematically, you can arrive at this conclusion fromgeneral reasoning. m is the number of times per year that the increases take place. Asthis value becomes very large, it is equivalent to the increases taking place continuously,ie a continuously increasing whole life insurance.

Solution 3.13

1:

( )x n

IA is the notation for the actuarial present value an increasing term life insurance

where the increases take place at the beginning of each year and the benefit is paidimmediately on death. Therefore on death in year 1 the benefit would be 1, on death inyear 2 the benefit would be 2, and so on.

The random variable that represents the present value of this benefit is:

≤ + = >

10

T T nT vZT n



Solution 3.14

:( )

x nIA is the notation for the actuarial present value of an increasing endowment

insurance where the increases take place at the beginning of each year and the benefit ispaid immediately on death. Therefore on death in year 1 the benefit would be 1, on deathin year 2 the benefit would be 2 and so on. If the policyholder survives to age +x n , thenhe receives a benefit of n.

The random variable that represents the present value of this benefit is:

+ ≤ = >

1 T

nT v T n

Znv T n

Solution 3.15

( )xIA represents the actuarial present value of an increasing whole life insurance wherethe increases take place at the beginning of each year and the benefit is paid at the end ofthe year of death.

The random variable representing the present value of the benefit is += + 1( 1) KZ K v where= …0,1,K .

The expected value of Z is:

∞+

== + =∑ 1

0[ ] ( 1) Pr( )k

kE Z k v K k

1:

( )x n

IA represents the actuarial present value of an increasing term life insurance where

the increases take place at the beginning of each year and the benefit is paid at the end ofthe year of death provided death occurs within n years.

The random variable representing the present value is:

+ < += ≥

1( 1)0

K K nK vZK n


−+

== + =∑

11

0[ ] ( 1) Pr( )

nk

kE Z k v K k



:( )

x nIA represents the actuarial present value of an increasing endowment insurance

where the increases take place at the beginning of each year and the benefit is paid at theend of year of death if death occurs within n years or on survival for n years.

The random variable representing this present value is:

+ + <= ≥

1( 1) K

nK v K n

Znv K n


−+

== + = + ≥∑

11

0( ) ( 1) Pr( ) Pr( )

nk n

kE Z k v K k nv K n

Solution 3.16

If Z is given by:

≤ − = >

( )0

T T nv n TZT n

Then ( )E Z is given by:

= − ∫0

[ ] ( ) ( )n

tTE Z v n t f t dt

Solution 3.17

The random variable representing the present value of a continuously decreasing n-yearterm life insurance is:

≤ −= >

( )0

T T nv n TZT n



Solution 3.18

If we calculate the sum of an increasing and a decreasing term insurance we get a levelterm insurance:

+ = +1 1 1: : :

( ) ( ) ( 1)x n x n x n

DA IA n A

Notice that the amount of the benefit in the level insurance is +1n and not n since thebenefit in, for example, year 1 is n from the decreasing insurance and 1 from theincreasing insurance.

Solution 3.19

1:

( )x n

DA represents the actuarial present value of a decreasing term insurance where the

decreases take place at the beginning of each year and the benefit is paid at the end ofthe year of death provided death occurs within n years. There is a benefit of n if deathoccurs in the first year.

The random variable representing the present value is:

+ < −= ≥

1( )0

K K nn K vZK n


−+

== − =∑

11

0[ ] ( ) Pr( )

nk

kE Z n k v K k



Solution 3.20

Using the recursive formula, we have:

= +1 150 5050:5 51:4

A vq vp A

= +1 151 5151:4 52:3

A vq vp A

= +1 152 5252:3 53:2

A vq vp A

= +1 153 5353:2 54:1

A vq vp A

= + = − =

1 154 5454:1 55:0

1 8,640,8611 0.0076261.08 8,712,621

A vq vp A

Working back through the recursive formulas:

= − + × × =

153:2

1 8,712,621 1 8,712,6211 0.007626 0.014021.08 8,779,128 1.08 8,779,128

A

= − + × × =

152:3

1 8,779,128 1 8,779,1281 0.01402 0.019351.08 8,840,770 1.08 8,840,770

A

= − + × × =

151:4

1 8,840,770 1 8,840,7701 0.01935 0.023751.08 8,897,913 1.08 8,897,913

A

= − + × × =

150:5

1 8,897,913 1 8,897,9131 0.02375 0.027341.08 8,950,901 1.08 8,950,901

A



Solution 3.21

( )xIA can be written as follows:

+ += + + +2 31 2 2( ) 2 3x x x x x xIA vq v p q v p q

This can be split up:

+ +

+ +

= + + +

+ + +

2 31 2 2

2 31 2 2

( )

2x x x x x x

x x x x

IA vq v p q v p q

v p q v p q

Common factors can then be taken:

+ + +

+ + +

= + + + + + +

21 1 2

21 1 2

( )

2

x x x x x x

x x x x

IA vq vp vq v p q

vp vq v p q

This can be written as:

+ += + +1 1( ) ( )x x x x x xIA vq vp A vp IA

So the recursive formula is + += + +1 1( ) ( )x x x x x xIA vq vp A vp IA .



Solution 3.22

Using a similar argument to the one used in the notes, we can proceed as follows.

Consider a term life insurance as the sum of deferred term life insurances:

−

−− + −+ +

= + + + +

= + + + +

1 1 1 1 10| 1| 2| 1|: :1 :1 :1 :1

1 1 2 1 1 12 1 1:1:1 1:1 2:1

nx n x x x xn

x x n x x nx x x

A A A A A

A vp A v p A v p A

Now we know that under a uniform distribution of deaths between integer ages:

µ+ ++= + + =∫ ∫

1 11

:10 0

( )t tt x k x kx k

A v p x k t dt v q dt

Further:

δ−

= =∫1

10

1t vv dt a

This means that we can simplify +1

:1x kA :

δ+ ++−

= =∫1

1:1

0

1tx k x kx k

vA q v dt q

Which in turn means that we can simplify xA :

( )( )δ

δ

δ

−+ + − + −

+ + − + −

−≈ + + + +

= + + + +

=

1 2 11 2 2 1 1:

2 31 2 2 1 1

1:

1 nx x x x x n x x nx n

nx x x x x n x x n

x n

vA q vp q v p q v p q

i vq v p q v p q v p q

i A



Solution 3.23

We require (12)4560,000A . This can be calculated as follows:

( ) ( )≈ =

− −

(12)45 4545 1 12 1 12

0.06 0.0660,000 60,000 60 1,00012 1.06 1 12 1.06 1

A A A

Using the values from the Tables, =451,000 201.20A , which makes the actuarial presentvalue $12,400.49.

Course 3, Unit 4 Annuities


Unit 4Annuities

Unit overview

In this unit we will study life annuities. We consider whole of life annuities, temporaryannuities, deferred annuities and guaranteed annuities. Deterministic and stochasticmodels are developed for each type of annuity. We also consider the situations where theannuity payments are made annually in advance, annually in arrears, m times a year (inadvance and in arrears), and continuously throughout the year.

Advice on study

In this unit we assume knowledge of annuities-certain and lifetime random variables. Youmay wish to revise these topics before continuing.



Annuities Course 3, Unit 4


Learning objectives

6 Formulate a model for the present value, with respect to an assumed interest ratestructure, of a set of future contingent cash flows. The model may be stochastic ordeterministic.

8 Apply a principle to a present value model to associate a cost or pattern of costs(possibly contingent) with a set of future contingent cashflows.

− Principles include: equivalence, exponential, standard deviation, varianceand percentile.

− Models include: present value models based on learning objectives 9−12.

− Applications include: insurance, health care, credit risk, environmental risk,consumer behavior (eg subscriptions) and warrants.



1 IntroductionIn the previous unit we studied life insurance contracts. Under the terms of such contracts,benefits are payable upon the death of the policyholder. In this unit we consider policiesthat provide (regular) benefit payments as long as the policyholder is alive. Such policiesare known as annuities or life annuities.

You should already be familiar with the concept of annuities-certain. Recall that anannuity-certain provides a series of payments made at regular time intervals or a paymentstream payable continuously over some period of time. These payments do not dependon the survival of the policyholder.

A life annuity is similar but, when calculating the present value of the series of payments,we must make an allowance for the probability of each payment being made, ie theprobability that the policyholder is still alive.

A pension plan is an example of an annuity. Once a person retires they receive a series ofregular payments until they die. (They may also receive other benefits, depending on theexact terms of the plan.)



2 Types of annuitiesThere are several types of annuities that you could be asked about in the Course 3examination. This section covers the details of the policies. The next section introducesthe notation that is used for annuities.

2.1 Life annuities

A life annuity is a series of payments made at regular time intervals, or a payment streampayable continuously, throughout the lifetime of the policyholder.

2.2 Temporary and whole life annuities

An annuity is said to be temporary if the payments are limited to a given number of years.Otherwise, it is said to be a whole life annuity.

2.3 Annuities-due

If a payment is made at the beginning of each time interval, then the payments are said tobe made in advance. An annuity payable in advance is often referred to as an annuity-due.

2.4 Immediate annuities

If a payment is made at the end of each time interval, then the payments are said to bemade in arrears. An annuity payable in arrears is often referred to as an immediateannuity.

2.5 Deferred annuities

A deferred life annuity is one that starts to be paid after a given period of time (thedeferment period) has elapsed.

2.6 Guaranteed annuities

If the annuity payments are certain to be paid for the first n years and are contingent onlife thereafter, then the annuity is said to be guaranteed for n years.

2.7 Annuities payable m times a year

If annuity payments are made m times every year, then the annuity is said to be payablem thly. For example, an annuity that provides payments at the end of every 3 months issaid to be payable quarterly in arrears.



2.8 Premiums for annuities and insurance policies

Life annuities are usually purchased by a single premium, ie a single lump sum payment.To determine the amount of the premium, we have to be able to calculate the expectedpresent value, or the actuarial present value (APV), of the annuity.

On the other hand, most life insurance products are purchased by a series of regular(usually monthly) premiums, rather than by a single lump sum. In the next unit we will seehow to calculate such premiums.



3 NotationIn this section we introduce the notation that is used to represent annuity functions. Youmay wish to reread Section 3 of Unit 3 to remind yourself of the conventions used inInternational Actuarial Notation.

3.1 Whole life annuities

The symbol used for a whole life annuity payable annually in arrears is xa . The subscriptx denotes the age of the policyholder when he purchases his annuity.

If the payments are made annually in advance, then we put a double dot above the a . Sothe symbol used for a whole life annuity-due is xa .

If the annuity is payable continuously, we put a bar above the a . So the symbol used for acontinuously payable whole life annuity is xa .

3.2 Temporary annuities

The symbol used for a temporary immediate annuity, payable for a maximum of n years is

:x na .

3.3 Deferred annuities

The symbol for a deferred whole life immediate annuity is |m xa . The subscript of “ |m ” in

front of the a tells us that the annuity is deferred for a period of m years. (Since thisannuity is payable in arrears, the first payment will be made when the policyholder reachesage 1x m+ + . If he dies before that age, no benefit will be payable.)

3.4 Guaranteed annuities

The symbol for an annuity that is guaranteed for n years is :x n

a .

3.5 Annuities payable m times a year

The symbol for a whole life annuity that is payable m times a year is ( )mxa .

3.6 Actuarial present values

More precisely, each of the symbols we have described above represents the actuarialpresent value of a given type of annuity.



In the sections that follow, we will develop a deterministic model and a stochastic modelfor each type of annuity, and show how the actuarial present values are calculated. Unless otherwise stated, we will assume a constant rate of interest throughout.



4 Whole of life annuities with annual paymentsIn this section we will consider whole of life annuities that are payable annually at the rateof one unit each year. We begin with the case when the payments are made annually inarrears.

4.1 Payments made annually in arrears

Suppose that a policyholder, who is currently age x , is due to receive payments of oneunit at the end of each year for the remainder of his life. Since we don’t know how long thepolicyholder will live, we don’t know how many annuity payments will be made. Thismeans that the present value of the annuity cannot be known with certainty at the outset(ie when the policy is purchased).

In the deterministic model described below, we assume that the probability of thepolicyholder surviving to each future age is known. These values are then used tocalculate the actuarial present value of the annuity.

Deterministic model

Since the annuity payments are made annually in arrears, the first payment is made in oneyear’s time. The present value of this first payment is v . However, this payment will onlybe made if the policyholder survives to age x +1. Since the probability of survival is px ,the actuarial present value of the first payment is:

v px

Now consider the second payment, which is due in two years’ time. The present value ofthis payment is v 2 and the probability that it is made is just the probability that thepolicyholder survives to age x + 2 , ie 2 px . The actuarial present value of the secondpayment is therefore:

v px2

2

Continuing in this way, we can find a formula for the actuarial present value of the entirepayment stream. The symbol that is used to denote this actuarial present value is ax . Sowe have:

a vp v p v p v px x x xk

k xk

= + + + ==

∞

∑22

33

1



Stochastic model

We now consider an alternative way of modeling the present value of the annuity. Thestochastic model expresses the present value of the life annuity in terms of a lifetimerandom variable. The random variable needed here is the curtate future lifetime, ( )K x . Inthe discussion that follows, the age x has been dropped in order to simplify the notation.

Since payments are made at the end of each year, as long as the policyholder is still alive,the last payment will be made (with certainty) in K years’ time. (Recall that, if K k= , thenthe policyholder is still alive at age x k+ but dies before age x k+ + 1. So the lastpayment would be made at age x k+ , ie in k years’ time.) The payments are illustratedon the timeline below.

The present value of this series of payments is therefore:

v v v aKK+ + + =2

which is clearly a function of the random variable K .

Actuarial present value

The actuarial present value of the annuity is then given by:

K k kk k

E a a K k a K k0 1

( ) Pr( ) Pr( )∞ ∞

= == = = =∑ ∑

since a0 0= .

It is not immediately obvious that the deterministic and stochastic models give the same

actuarial present value. We now explain why a K k v pkk

kk x

kPr( )= =

=

∞

=

∞

∑ ∑0 1

.

agex x + 1 x + 2 … x K+ x K+ + 1

1 1 … 1



By substituting a vkj

j

k=

=∑

1 into the expression for the actuarial present value, we obtain:

kj

Kk j

E a v K k1 1

( ) Pr( )∞

= == =∑ ∑

Next we have to change the order of summation. Note that, since j k≤ in the first doublesummation above, we must ensure that k j≥ in the rearranged expression. We thereforehave to sum over values of k from j to ∞ . This gives:

E a v K kKj

k jj( ) Pr( )= =

=

∞

=

∞

∑∑1

Finally, pulling the v j term out from the inner sum, we see that:

∞ ∞ ∞ ∞

= = = == = = ≥ = =∑ ∑ ∑ ∑j j j

j x xKj k j j j

E a v K k v K j v p a1 1 1

( ) Pr( ) Pr( )

Variance

Using the stochastic approach it is quite simple to derive an expression for the variance ofthe present value of the annuity.

Recall from your knowledge of annuities-certain that:

a viK

K=

−1

Hence:

( )2 21 1 1var( ) var var 1 var( )

KK K

Kva v vi i i

Ê ˆ-= = - =Á ˜Ë ¯

However, var( )aK is more commonly expressed in terms of life insurance functions.



In the last unit we saw that:

• the present value of a whole of life insurance with unit sum assured payable at theend of the year of death is v K +1

• E v AKx( )+ =1

• ( ) ( )2 21 2( 1) 1 2var( ) ( )K K K

x xv E v E v A A+ + +È ˘= - = -Î ˚

Remember that the superscript in front of the first A indicates that the insurance functionshould be evaluated at twice the standard force of interest, or equivalently, at rate ofinterest ( )1 12+ −i .

It now follows that:

2 211

2 2 2 2( )1 1var( ) var var( )

KK x x

KA Ava v

vi i v d

++Ê ˆ -

= = =Á ˜Ë ¯

(Recall that 2var( ) var( )cX c X= for any constant c and random variable X .)

4.2 Payments made annually in advance

Now suppose that payments of one unit are made at the beginning of every yearthroughout the life of a policyholder that is now aged x . The actuarial present value ofthis annuity-due is denoted by ax . The only difference between the annuity-due and theimmediate annuity described in the previous section is the payment of one unit that ismade at age x . Since the present value of this first payment is 1, and it is made withcertainty, the actuarial present value of the annuity-due satisfies the equation:

a ax x= +1

Deterministic model

Assuming that the survival probabilities are known, the actuarial present value of theannuity-due may be calculated using the equation:

a vp v p v p v px x x xk

kk x= + + + + =

=

∞

∑1 22

33

0



Question 4.1

Show that the actuarial present value of the annuity-due satisfies the recursive formula:

11x x xa v p a += +

Example 4.1

Calculate the value of 30a given that a29 17.200= , 29 0.999p = and 0.05i = .

Solution

From the recursive formula we have:

29 29 301a v p a= +

So:

2930

29

1 1.05 16.200 17.0270.999

aa

v p- ¥= = =

Stochastic model

Again we want to express the present value of the annuity in terms of a future lifetimerandom variable.

With an annuity-due, the policyholder receives a payment of one unit at the start of everyyear for the whole of his life. The payments made under this annuity are illustrated on thetimeline below.

agex x + 1 x + 2 x K+ x K+ + 1

1 1 1 … 1



As we can see from the picture, there are K +1 payments altogether. The present valueof the annuity-due can therefore be expressed as:

1 21+ + + + =

+v v v aK

K


The actuarial present value is given by:

E a a K kK kk

( ) Pr( )+ +

=

∞= =∑1 1

0

It can be shown (using the same argument that was used in the case of an immediateannuity) that the deterministic and stochastic models give the same result, ie that:

Pr( )a K k v pkk

kk x

k+

=

∞

=

∞= =∑ ∑1

0 0

Variance

Writing the annuity-certain as:

a vdK

K

+

+=

−1

11

we find that:

var( ) var( )( )

ad

vA A

dKK x x

++= =

−1 2

12 2

21

Notice that var( ) var( )a aK K+=1 . This result is to be expected since the only difference

between these annuities is the payment that is made (with certainty) at age x under theannuity-due.

Tables

Values of ax are given in the Tables for ages 0 to 110. Values of ax are not listed sincethey are easily calculated from the values of ax . All the functions in the Tables arecalculated using an interest rate of 6% a year.



Question 4.2

A life aged exactly 40 purchases a whole life annuity under which payments of $5,000 aremade annually in arrears. The expectation and standard deviation of the present value ofthis annuity, calculated assuming interest and mortality as given in the Tables are,respectively:

A $69,083 and $188B $69,083 and $13,281C $69,083 and $35,282D $74,083 and $13,281E $74,083 and $35,282

4.3 Extra risk

We have seen above that the actuarial present value of an annuity can be expressed as asum of terms of the form v pk

k x . Sometimes, however, we need to find the present valueof an annuity payable to a life that experiences non-standard mortality. In other words, wehave to make an adjustment to the survival probabilities before we can work out theactuarial present value of the annuity. A person whose survival probabilities are lowerthan the standard values for his age (according to some life table) is said to be an impairedlife.

One way to adjust the standard mortality to allow for impairment is to apply an age rating. To do this we treat the life under consideration, assumed to be age x , as if he wereactually aged x s+ . (Note that the value of s could be either positive or negative,depending on our beliefs about the life’s mortality.)

Alternatively, we could make an addition (or subtraction) to the standard force of mortality.Recall that:

exp ( )x k

k x xp y dym

+Ï ¸= -Ì ˝Ó ˛Ú

where ( )ym denotes the (standard) force of mortality at age y .

Now define:

* ( ) ( )x xm m m= + for all ages x



If we use the symbol k xp * to denote the probability of survival from age x to age x k+calculated using force of mortality * ( )xm , then:

* exp ( ( ) ) exp ( )x k x kk k

k x k xx xp y dt e y dt e pm mm m m

+ +- -Ï ¸ Ï ¸= - + = - =Ì ˝ Ì ˝Ó ˛ Ó ˛Ú Ú

Also, if δ denotes the standard force of interest, we have:

v e= −δ

From the discussion above, we know that the actuarial present value of an annuity is asum of terms of the form k

k xv p .

We can now write:

v p e p v pkk x

kk x

kk x

* ( ) ( *)= =− +δ µ

where v * is defined by the equation:

v e ve* ( )= =− + −δ µ µ

So, when we are calculating the actuarial present value of an annuity, we can treat aconstant addition of µ to the standard force of mortality as a constant addition of µ to thestandard force of interest.

Example 4.2

A life aged 65 purchases a whole life annuity-due with an annual payment of 8,000. Calculate the actuarial present value of this annuity assuming an interest rate of 5% and aconstant addition of 0.0094787 to the standard force of mortality used in the Tables.

Solution

The actuarial present value of the annuity is *a65 , evaluated using an interest rate of 5%and non-standard mortality. However:

@ @*a a65 655% 6%=

since:

v ve e i*.

. * ..= = = ⇒ =− −µ 1105

0 9433963 0 060 0094787



So the actuarial present value of the annuity is:

8 000 6% 8 000 9 8969 79175 265, @ , . , .a = × =

Note that this method of switching the addition from the force of mortality to the interestrate does not work for insurance functions since they cannot be expressed solely in termsof v pk

k x . However, insurance functions can be dealt with using a premium conversionformula.

4.4 Premium conversion formula for whole life policies

We have already remarked that:

a vdK

K

+

+=

−1

11 and E v AKx( )+ =1

If we take the expected value of each side of the first equation above, we obtain:

aA

dxx=

−1

This can be rearranged to give the premium conversion formula for whole life policies:

Premium conversion formula for whole life policies

A dax x= −1



Example 4.3

Verify this formula for the values of a40 and A40 given in the Tables.

Solution

From the Tables we have:

.a40 14 8166= A40 016132= . 0.06 0.05660381 1.06

idi

= = =+

So:

40 401 1 (0.0566038 14.8166) 0.161324da A- = - ¥ = =

Question 4.3

The value of A50 , calculated assuming an interest rate of 5% a year and a constantaddition of 0.0094787 to the standard force of mortality used in the Tables, is:

A 0.24905B 0.36825C 0.63175D 0.75095E none of the above



5 Temporary annuities with annual paymentsNow suppose that, instead of being payable for the whole of a policyholder’s life, paymentsmade under an annuity are limited to a maximum of n years.

5.1 Temporary annuities payable annually in arrears

Consider a policyholder that purchases an annuity at age x and receives annuitypayments annually in arrears. Even if the policyholder survives past the age of x n+ , hewould receive the final payment at age x n+ .

Once again, we can model the present value of the annuity in two ways. The notationused to denote the actuarial present value of this annuity is ax n: .

Deterministic model

If the survival probabilities are known, then the actuarial present value of the n -year lifeannuity is given by:

a v p v p v p v px n x xn

n xk

k

n

k x: = + + + ==∑2

21

Example 4.4

Calculate the actuarial present value of a 3-year annuity, payable to a life now aged 60 onsurvival to the end of each year, given that:

(i) the annual payment is $10,000

(ii) p p p60 61 62 0 99= = = .

(iii) the interest rate is 5% a year.

Solution

The actuarial present value is:

( )2 360 2 60 3 6060:3

2 3

2 3

10,000 10,000

0.99 0.99 0.9910,0001.05 1.05 1.05

26,700.17

a vp v p v p= + +

Ê ˆ= + +Á ˜

Ë ¯=



Question 4.4 Now suppose that the annuity payments in Example 4.4 are made annually in advance. The actuarial present value of the annuity is then:

A $26,969.87B $28,035.18C $28,062.48D $28,318.37E $36,700.17

Stochastic model

Since there is now a maximum of n payments and each payment is contingent onsurvival, the total number of annuity payments that will be made is min{ , }K n . The presentvalue of the annuity is therefore:

v v v aK nK n+ + + =2 min{ , }

min{ , }


The actuarial present value of the annuity may therefore be expressed as:

E a a K kK n k nk

( ) Pr( )min{ , } min{ , }= ==

∞

∑0

Again, it is not immediately obvious that the deterministic and stochastic models lead tothe same result. We will now show that this is indeed the case.

First observe that:

E a a K k

a K k a K k

a K k a K n

K n k nk

kk

n

nk n

kk

n

n

( ) Pr( )

Pr( ) Pr( )

Pr( ) Pr( )

min{ , } min{ , }= =

= = + =

= = + ≥ +

=

∞

= = +

∞

=

∑

∑ ∑

∑

0

0 1

01



Now replace ak by v j

j

k

=∑

1 in the first term and change the order of summation. We then

have:

a K k v K k

v K k

kk

nj

j

k

k

n

j

k j

n

j

n

Pr( ) Pr( )

Pr( )

= = =

= =

= ==

==

∑ ∑∑

∑∑

0 10

1

Pulling the v j term outside the inner summation gives:

a K k v K k

v j K n

kk

nj

k j

n

j

n

j

j

n

Pr( ) Pr( )

Pr( )

= = =

= ≤ ≤

= ==

=

∑ ∑∑

∑

0 1

1

If we write Pr( ) Pr( ) Pr( )j K n K j K n≤ ≤ = ≥ − ≥ +1 , we then see that:

a K k v K j v K n

v K j a K n

kk

nj

j

nj

j

n

j

j

n

n

Pr( ) Pr( ) Pr( )

Pr( ) Pr( )

= = ≥ − ≥ +

= ≥ − ≥ +

= = =

=

∑ ∑ ∑

∑

0 1 1

1

1

1

Finally, substituting this last expression into the formula for the actuarial present value, weobtain:

E a v K j a K n a K n

v p

a

K nj

j

n

n n

jj x

j

n

x n

( ) Pr( ) Pr( ) Pr( )min{ , }

:

= ≥ − ≥ + + ≥ +

=

=

=

=

∑

∑

1

1

1 1

So the deterministic and stochastic models do give the same result.



Variance

As in previous cases, we can use our knowledge of annuities-certain to express thepresent value of the temporary annuity as:

a viK n

K n

min{ , }

min{ , }=

−1

Then:

min{ , }min{ , }

min{ , } 21 1var( ) var var( )

K nK n

K nva v

i i

Ê ˆ-= =Á ˜Ë ¯

From the last unit, we know that:

: ( )x nA E Z=

where:

1 if

if

K

nv K n

Zv K n

+Ï <Ô= Ì≥ÔÓ

Another way of expressing this result is as follows:

min{ 1, }: ( )K n

x nA E v +=

So the variance of the temporary annuity is:

var( ) var( )( )

min{ , }min{ , } : :

ai v

vA A

dK nK n x n x n

= =−

+ + + +12 2

1 12

1 12

2

Again, the superscript of 2 in front of the first insurance symbol indicates that it should becalculated using twice the standard force of interest.

5.2 Temporary annuities payable annually in advance

The notation used to denote the actuarial present value of an n -year life annuity payableto a life currently aged x is :ax n . As the maximum number of payments made under this

annuity is n , the latest possible payment date is at age x n+ − 1.



Deterministic model

If the survival probabilities are known, then the actuarial present value of the annuity is:

:a vp v p v p v px n x xn

n xk

k xk

n= + + + + =−

−=

−

∑1 22

11

0

1

Stochastic model

Since payments continue throughout the policyholder’s life, up to and including agex n+ − 1, the present value of the annuity may be expressed as:

1 2 11+ + + + =−

+v v v aK n

K nmin{ , }

min{ , }


From the discussion above it follows that:

E a a K kK n k nk

( ) Pr( )min{ , } min{ , }+ +=

∞= =∑1 1

0

This can be shown to be equivalent to :ax n .

Variance

The variance of the present value is:

var( )( )

min{ , }: :

aA A

dK nx n x n

+=

−

1

2 2

2

Question 4.5

Show that var( )( )

min{ , }: :

aA A

dK nx n x n

+=

−

1

2 2

2.



Tables

Values of :ax n and ax n: are not given in the Tables. However, we can easily work them

out from the values of ax since:

nx n x x nx na a v p a: += −

and

n nx n x x n x n x x nx na a v p a a v p a: ( 1) ( 1)+ += - = - - -

Question 4.6

Show that nx n x x nx na a v p a: += − .

The notation n xE is sometimes used instead of v pnn x . Values of xE51000 , xE101000

and xE201000 are given in the Tables for ages 0 to 110.

Example 4.5

Calculate the value of :a40 25 assuming mortality and interest as used in the Tables.

Solution

Using the formula given above, we have:

. ( . ) , ,, ,

.

.

:a a v p a40 25 4025

25 40 65

2514 8166 106 7 533 9649 313166

9 8969

12 9512

= −

= − × ×

=

−



5.3 Premium conversion formula

There is also a premium conversion formula for temporary annuities and endowmentinsurances.

Premium conversion formula for temporary annuities

A dax n x n: := −1

Question 4.7

Show that A dax n x n: := −1 .



6 Deferred life annuitiesPayments made under an annuity do not necessarily have to begin in the first year of thepolicy. In some cases, annuity payments are deferred for a number of years.

6.1 Deferred whole life annuities payable annually in arrears

Consider a life aged x , who is due to receive an annuity payable annually in arrears. Ifthis annuity is deferred for a period of m years, then the first payment will be made whenthe policyholder reaches age x m+ + 1 and annual payments will be made throughout lifethereafter. Note that, if the policyholder dies before age x m+ + 1, no payments will bemade.

As usual, we would like to develop a formula for the present value of this annuity, whichwe will denote by the symbol m xa| .

Deterministic model

Since the payments start at age x m+ + 1 and are contingent on survival, the actuarialpresent value is given by:

m xm

m xm

m xm

m xa v p v p v p| = + + +++

++

++

11

22

33

This can be rewritten as:

( )2 3| 2 3

mm x m x x m x m x m

mm x x m

m x x m

a v p v p v p v p

v p aE a

+ + +

+

+

= + + +

==

This last expression may also be deduced by general reasoning:

• The probability that the policyholder is still alive at the end of the m -year defermentperiod is m xp .

• Given that he does survive to age x m+ , the actuarial present value at age x m+of the whole of life annuity is ax m+ .

• However, we want the actuarial present value at age x , so we have to discount thevalue at age x m+ by multiplying by a factor of v m . Since v p Em

m x m x= , wehave the required result.



Another way we can express |m xa is as follows:

|1

1 1

:

km x k x

k m

mk k

k x k xk k

x x m

a v p

v p v p

a a

•

= +

•

= =

=

= -

= -

Â

Â Â

Stochastic model

If we now think in terms of future lifetime random variables, we know that the last paymentwill be made at age x K+ . The payment series is illustrated on the timeline below.

In the discussion above, we saw that a deferred annuity can be viewed as the differencebetween a whole life annuity and a temporary annuity. Hence, the present value of awhole life annuity, deferred for m years, is given by:

a aK K m− min{ , }



E a a E a E aK K m K K m( ) ( ) ( )min{ , } min{ , }− = −

We have already seen that:

E a v pKk

kk x( ) =

=

∞

∑1

and E a v pK mk

k

m

k x( )min{ , } ==∑

1

age

x x m+ 1x m+ + 2x m+ + x K+ x K+ + 1

1 1 … 1



So:

E a a v p v p v p aK K mk

kk x

k

k

m

k xk

k mk x m x( )min{ , } |− = − = =

=

∞

= = +

∞

∑ ∑ ∑1 1 1

Once again we have shown that the deterministic and stochastic models give us the sameresult.

Variance

Recall that for any two random variables X and Y , the variance of X Y- is given by:

var( ) var( ) var( ) 2cov( , )X Y X Y X Y- = + -

where cov( , )X Y denotes the covariance of X and Y . Covariance is defined by theequation:

cov( , ) ( ) ( ) ( )X Y E XY E X E Y= -

Recall also that, given random variables X and Y , and a constant k :

cov( , ) 0k X =

and

cov( , ) cov( , )kX Y k X Y=

We will use these results in the derivation of the variance of the deferred annuity.

First note that the variance is given by:

var( ) var( ) var( ) cov( , )min{ , } min{ , } min{ , }a a a a a aK K m K K m K K m− = + − 2

We have already seen that:

var( )( )

aA A

dKx x=−2 2

2

and:

var( )( )

min{ , }: :

aA A

dK mx m x m

=−

+ +2

1 12

2



In addition:

min{ , }

min{ , }

min{ , }2

1 min{ 1, 1}2

1 1cov( , ) cov ,

1 cov( , )

1 cov( , )

K K m

K K m

K K m

K K m

v va ai i

v vi

v vd

+ + +

Ê ˆ- -= Á ˜Ë ¯

=

=

and:

cov( , ) ( ) ( ) ( )

( )

min{ , } min{ , } min{ , }

min{ , }:

v v E v v E v E v

E v A A

K K m K K m K K m

K K mx x m

+ + + + + + + + +

+ + + ++

= −

= −

1 1 1 1 1 1 1 1 1

1 1 11

Now we can write:

2( 1)1 min{ 1, 1}

2if

if

KK K m

K mv K m

vv K m

W Z

++ + + +

+ +

Ï £Ô= Ì>ÔÓ

= +

where:

2( 1) if 0 if

Kv K mWK m

+ÏÔ £= Ì>ÔÓ

and 20 if

if K mK m

Zv K m+ +

£ÏÔ= Ì>ÔÓ

From Unit 3, we know that:

2 1|: 1

( )x m

E W A+

= and 11|( ) m

m xE Z v A++=

Hence:

1 min{ 1, 1} 2 1 11||: 1

( )K K m mm xx m

E v A v A+ + + + +++

= +

and:

{2 2 2 22min{ , } : 1 : 1

2 1 11|| : 1: 1

1var( ) ( ) ( )

2

x xK K m x m x m

mm x x x mx m

a a A A A Ad

A v A A A

+ +

++ ++

− = − + −

− + −



6.2 Deferred whole life annuities payable annually in advance

Now consider the case of a deferred annuity-due. If the deferment period is m years, thenthe first annuity payment will be made at age x m+ , provided that the policyholder is stillalive.

Deterministic model

The actuarial present value of this deferred annuity is:

m xm

m xm

m xm

m xm

m x x m

x x m

a v p v p v p

v p aa a

|

:

= + + +

== −

++

++

+

11

22

Stochastic model

The present value of this deferred annuity may be expressed in terms of the curtate futurelifetime as:

min{ , }a aK K m+ +−1 1


As in the case of a deferred annuity payable annually in arrears, we can show that thedeterministic and stochastic models give the same result, ie:

E a a a aK K m x x m( )min{ , } :+ +− = −1 1

Variance

The variance of this deferred annuity-due is given by:

{2 2 2 2min{ , } : :2

2 1|| ::

1var( ) ( ) ( )

2

x xK K m x m x m

m x x x mx m

a a A A A Ad

A A A A

- = - + -

¸Ê ˆ- + - ˝Á ˜Ë ¯˛

The proof of this result is similar to that for a deferred annuity payable annually in arrears.



Example 4.6

Calculate the values of 20 40|a and 25 35|a using the Tables.

Solution

We have:

20| 40 20 40 60 0.27414 11.1454 3.0554a E a= = ¥ =

and

25 256025| 35 25 35 60 60

35

8,188,074(1.06) 11.1454 2.25719,420,657

la E a v a

l-= = = ¥ ¥ =

Note that, since 25 35E is not given in the Tables, we had to work it out from firstprinciples.

Question 4.8

A life aged exactly 25 has purchased a deferred, whole life annuity-due, under whichannual payments of one unit will be made. You are given that:

(i) the deferment period is 30 years

(ii) A25 30 0 25: .=

(iii) A25 301 0 05

:.=

(iv) a55 10 85= . .

The actuarial present value of this annuity is:

A 2.07B 2.17C 2.27D 2.37E 2.47



6.3 Deferred temporary annuities with annual payments

Now consider an n -year temporary annuity that is deferred for a period of m years. If thepayments are made annually in advance, then the actuarial present value of this annuity isdenoted by m x na| : . If they are made annually in arrears, then the notation used is m x na| : .

The actuarial present value can be evaluated using the formula:

m x n m x x m na E a| : :=+

or m m xx n x m na E a| : :+=

To see why these formulas are correct, consider a life aged x who purchases an n -yearannuity-due with a deferment period of m years.

• The first annuity payment will be made at age x m+ if the policyholder lives to thatage. Otherwise no benefit will be paid. The probability of survival is m xp .

• Starting at age x m+ , and continuing for n years, an annual annuity payment ofone unit is made. These payments are again contingent on survival. The actuarial

present value of this annuity at age x m+ is :ax m n+ .

• Discounting back to age x and incorporating the probability of survival, we see thatthe actuarial present value at age x of this annuity is m

m x x m nv p a :+.

Question 4.9

Which of the following statements are correct?

I m x n x m x m na a a| : : :+ =+

II m x n m x nma a v| : | := +

III m x m x x ma E a| = + + +1 1

A none of themB I and II onlyC I and III onlyD II and III onlyE I, II and III



7 Guaranteed annuitiesNow suppose that a policyholder, currently aged x , buys a life annuity under whichpayments are guaranteed for the first m years of the policy. (Payments after the age ofx m+ are contingent on survival.) This annuity can be considered to be the sum of an m -year annuity-certain and a deferred life annuity.

7.1 Guaranteed annuities payable annually in arrears

The actuarial present value of an immediate annuity payable at the rate of one unit a yearto a life currently aged x , and guaranteed for a period of m years, is denoted by a

x m:.

Deterministic model

If the survival probabilities are known, then the deterministic model can be used tocalculate the actuarial present value of the guaranteed annuity.

The actuarial present value may be expressed in many ways:

m m mm x m xx m

mm x x mm

m x x mm

m xm

a v v v v p v p

a v p a

a E a

a a

2 1 21 2:

|

+ ++ +

+

+

= + + + + + +

= +

= +

= +

Stochastic model

Alternatively, the present value can be expressed in terms of the curtate future lifetime:

a a am K K m+ − min{ , }


Using the results obtained above, we see that the actuarial present value then satisfies theequation:

E a a a a E a E a

a a a

a a

m K K m m K K m

m x x m

m m x

( ) ( ) ( )min{ , } min{ , }

:

|

+ − = + −

= + −

= +



Variance

Since am is a constant, the variance of the present value satisfies the equation:

var( ) var( )min{ , } min{ , }a a a a am K K m K K m+ − = −

So the variance of a guaranteed annuity is the same as the variance of a deferred annuity,provided that the guarantee period and deferment period are the same.

7.2 Guaranteed annuities payable annually in advance

Similar logic can be applied when the payments are made annually in advance.

Question 4.10

Write down formulas for the present value and actuarial present value of an annuity-dueunder which the payments are guaranteed for the first m years.

Question 4.11

A life aged 40 purchases an annuity-due. Annual payments of $12,000 are guaranteed forthe first 10 years of the policy and are payable for life thereafter. Assuming mortality andinterest are as given in the Tables, the actuarial present value of this annuity is:

A $179,059B $183,741C $189,040D $266,358E $288,382



8 Annuities payable m times a yearIn this section we study annuities that are payable m times every year. We consider anannuity paid at the start (or end) of each 1/ m th of a year. Each annuity payment is nowof amount 1/ m , so that we end up with a total payment of one unit each year. In thiscase we would say that we have an annuity of one unit a year, payable m thly in advance(or arrears). For example, if we receive payments of $1,000 at the end of every month, wewould say that an annuity of $12,000 a year is payable monthly in arrears.

We shall now develop the models assuming that the payments are made in arrears.

8.1 Whole of life annuities payable m times a year

Suppose that we have a policyholder, who is currently aged x , and is due to receivepayments of 1/ m payable m thly in arrears throughout his life.

Deterministic model

The actuarial present value of this annuity is denoted by axm( ) . In terms of survival

probabilities it is given by:

am

v pm

v pm

v pxm m

m xm

m xk m

k m xk

( ) //

//

//= + + =

=

∞

∑1 1 111

22

1

Stochastic model

In order to develop a stochastic model we need more information than just an individual’scurtate future lifetime. In fact, what we need is a random variable to denote the age at thestart period of length 1/ m during which death occurred.

We now define the random variable J by the equation:

( )J T K m= -Í ˙Î ˚

ie J is the integer part of ( )T K m− , where T denotes the complete future lifetime and Kdenotes the curtate future lifetime. By this definition, J is the number of complete m ths ofa year lived in the year of death.

The following example should help you to understand what is going on here.



Example 4.7

Suppose that a policyholder receives a monthly immediate annuity. If this policyholder hasa complete future lifetime of 5.8 years, calculate the value of J and interpret the result.

Solution

We have m = 12 and T = 5 8. . Hence:

0.8 12 9.6 9J = ¥ = =Í ˙ Í ˙Î ˚ Î ˚

This is saying that the policyholder survives for 9 whole months in his year of death, butdies before the end of the 10th month. Altogether, he would receive 69 annuity payments.

We can now express the present value of an immediate annuity with m thly payments interms of the random variables K and J . The formula is:

1 1 11 2m

vm

vm

v am m K J mK J mm/ / /

/( )+ + + =++



1( ) ( )

/ /0 0

( ) Pr[( ) ( )]m

m mK J m k j m

k jE a a K k J j

• -

+ += =

= = « =Â Â

Now define:

H Km J= +

Note that H represents the number of complete periods of length m1/ years for which thepolicyholder survives.

Then:

m m m mh m h mK J m H m

h hE a E a a H h a H h( ) ( ) ( ) ( )

/ // /0 1

( ) ( ) Pr( ) Pr( )• •

+= =

= = = = =Â Â

since ma( )0

0= .



If we replace mh m

a( )/

by h

i m

iv

m/

1

1

=∑ , we then obtain:

hm i m

K J mh i

E a v H hm

( ) //

1 1

1( ) Pr( )•

+= =

= =Â Â

Next, we change the order of summation, noting that i h≤ . This gives:

m i mK J m

i h iE a v H h

m( ) /

/1

1( ) Pr( )• •

+= =

= =Â Â

Pulling the 1m

v i m/ term outside of the inner summation, we obtain:

m i mK J m i h i

i m

i

i m

i

i m

i

i mi m x

i

E a v H hm

v H im

v Km J im

v K J m i mm

v pm

( ) // 1

/

1

/

1

/

1

//

1

1( ) Pr( )

1 Pr( )

1 Pr( )

1 Pr( / / )

1

∞ ∞

+= =∞

=∞

=∞

=∞

=

= =

= ≥

= + ≥

= + ≥

=

∑ ∑

∑

∑

∑

∑

Again, this is the result that we obtained using the deterministic model.

Variance

The variance is obtained in the usual way:

/( ) ( 1) /

( ) ( ) 1/ 2/1 1var( ) var var( )

( )

K J mm K J m

m m mK J mva vi i v

++ +

+

Ê ˆ-= =Á ˜Ë ¯

It is easily checked that:

( ) 1/ ( )m m mi v d=



Also, since v K J m+ +( ) /1 is the present value of a whole life insurance with unit sumassured payable at the end of the period of length 1/ m in which death occurs, we have:

( )2( ) ( )2

( )( ) 2/

var( )( )

m mx xm

mK J m

A Aa

d+

-=

We can obtain similar results for annuities-due payable m times a year. If the annuitypayments are made at the start of each period, then the symbol used to denote the

actuarial present value is ( )axm .

Question 4.12

Write down an expression in terms of survival probabilities for the actuarial present valueof a life annuity of one unit a year, payable m thly in advance, to a life currently aged x .

Tables

Values of axm( ) and ( )ax

m are not given in the Tables. However, they may be calculatedusing the formulas below.

Evaluating mthly life annuities

Assuming that deaths occurs uniformly over each year of age, the actuarial present valueof a whole of life annuity, payable m thly in advance, satisfies the equation:

( ) ( ) ( )mx xa m a ma b= -

Here:

α( )( ) ( )

m idi dm m= and β( )

( )

( ) ( )m i i

i d

m

m m=−

Also, since the only difference between ( )axm and ax

m( ) is that ( )axm has an extra payment

of amount 1/ m at age x , values of axm( ) can be calculated using the identity:

a amx

mxm( ) ( )= −

1



Note that the values of ( )mi , ( )md , ( )ma and ( )mb are given in the Tables for1,2,4,12m = and • .

In order to prove the above result, we first note that there is a premium conversion formulathat holds for m thly annuities.

Premium conversion formula for mthly annuities

( ) ( )( )1m mmx xA d a= -

Question 4.13

Show that ( ) ( )( )1m mmx xA d a= - .

Subtracting ( ) ( )( )1m mmx xA d a= - from the equation:

x xA da1= −

we obtain:

m mmx xx xA A d a da( ) ( )( )− = −

This can be rearranged as:

d a da A Amxm

x x xm( ) ( ) ( )= + −

Dividing through by d m( ) then gives:

( )( ) ( )( ) ( )

1m mx x x xm m

da a A Ad d

= - -



Now, if we assume that deaths are uniformly distributed across each year of age, we have:

A ii

Axm

m x( )

( )=

Substituting this in the above, we obtain:

( )( )

( ) ( ) ( ) ( ) ( ) ( )1 m

mx x x x x xm m m m m m

d i d i ia a A A a Ad d i d i d

Ê ˆ -= - - = -Á ˜Ë ¯

Finally, replacing Ax by 1− dax gives:

( )( )

( ) ( ) ( )

( ) ( )

( ) ( ) ( ) ( ) ( )

( )

( ) ( ) ( ) ( )

(1 )

( )

( ) ( )

mm

x x xm m m

m mxm m m m m

mxm m m m

x

d i ia a dad i d

d d i i i iad i d i d

id i iai d i d

m a ma b

-= - -

Ê ˆ- -= + -Á ˜Ë ¯

-= -

= -

Example 4.8

Use the Tables to calculate the value of ( )a4012 . Assume that deaths are uniformly

distributed between integer ages.

Solution

From the Tables, we have:

α( ) .12 100028= and β( ) .12 0 46812=

Hence:

( ) ( ) ( . . ) . .( )a a4012

4012 12 100028 14 8166 0 46812 14 353= − = × − =α β



Question 4.14

Verify the values of α( )12 and β( )12 given in the Tables.

Question 4.15

A policyholder now aged exactly 60 has purchased a whole of life annuity under which shereceives payments of $5,000 at the end of every three months, provided she is still alive atthat time. The actuarial present value of this annuity, calculated using the values given inthe Tables, is:

A $195,283B $210,283C $215,283D $220,283E $235,283

8.2 Temporary annuities payable m times a year

The formula:

( ) ( ) ( )mx xa m a ma b= -

only works for whole life annuities. To value a temporary annuity that is payable m timesa year, we must first express the temporary annuity in terms of whole life annuities. Thiscan be done as follows:

( ) ( ) ( ):m m m

x n x x nx na a E a += -

Then each of the whole life terms can be evaluated using the α β, formula.



Question 4.16

A life office sells a 10-year temporary annuity to a life now aged 65. The policy pays out$2,000 at the start of every month. The actuarial present value of the annuity, calculatedusing the values given in the Tables is:

A $13,463B $64,337C $91,928D $135,886E $161,559



9 Continuous life annuitiesSuppose now that, instead of payments being made at the beginning or end of each year,a policyholder receives a continuous stream of payments until he dies. In practice thepayments may be made weekly rather than continuously. A continuous payment stream isa theoretical concept that is often used to model a series of payments that are made manytimes a year.

9.1 Whole of life annuities

Deterministic model

As we are now working in continuous time, we can no longer just work out the actuarialpresent value of each payment and add them all together. The continuous analog ofsummation is integration, and that’s what we must do here.

Consider the short time interval ( , )t t dt+ . If the rate of payment is one unit per annum,then over a time interval of length dt , the total amount paid will be dt . If dt is small thenthe value at time 0 (ie when the policyholder is aged x ) of this payment stream isapproximately:

v dtt

Also the probability that the policyholder is alive to receive this payment is approximately:

t xp

So the actuarial present value of the payment stream payable from time t to time t dt+ isapproximately:

v p dttt x

Considering all possible values of t and letting dt → 0 , we find that the actuarial presentvalue of the annuity is:

0t

x t xa v p dt•

= Ú

Stochastic model

Since we are now working in continuous time, we must write the present value of theannuity in terms of ( )T x , the complete future lifetime random variable. For notationalconvenience we will drop the x here.



Also, as the annuity is payable continuously at the rate of one unit a year throughout thelifetime of the policyholder, the present value of the annuity is aT .


The actuarial present value may be expressed as:

0( ) ( )TT tE a a f t dt

•= Ú

where f tT ( ) is the PDF of T .

Question 4.17

Show that 0

( ) tt xTE a v p dt

•= Ú .

Variance

Since:

a vT

T=

−1δ

it follows that:

21 1var( ) var var( )

TT

Tva vd d

Ê ˆ-= =Á ˜Ë ¯

In the previous unit we saw that:

var( ) ( )v A ATx x= −2 2

So we now have:

2 2

2( )var( ) x x

TA Aa

d-=



Tables

Values of ax are not given in the Tables. However, values of α( )∞ and β( )∞ are listed,so we can obtain values of ax by letting m → ∞ in the formula:

( ) ( ) ( )mx xa m a ma b= -

Hence:

( ) ( )x xa aa b= • - •

This formula makes the implicit assumption that deaths are uniformly distributed betweeninteger ages.

The following approximation is also sometimes used:

a ax x≈ −12

This follows since ( ) 1a • ª and 1( )2

b • ª .

Example 4.9

Calculate the value of a55 assuming interest and mortality as given in the Tables.

Solution

From the formula above we have:

a a55 55 100028 12 2758 0 50985 117694= ∞ − ∞ = × − =α β( ) ( ) ( . . ) . .



9.2 Temporary annuities

Deterministic model

We can apply similar logic to that used in previous sections to calculate the actuarialpresent values of temporary, guaranteed and deferred annuities that are payablecontinuously.

In particular, we can express the actuarial present value of a continuously payabletemporary annuity in terms of whole life annuities as follows:

a a E ax n x n x x n: = − +

The integral representation of ax n: is therefore:

: 0 0

0 0

0

0

t n tt x n x t x nx n

t n tt x n t x

t tt x t xn

n tt x

a v p dt v p v p dt

v p dt v p dt

v p dt v p dt

v p dt

• •+

• • ++

• •

= -

= -

= -

=

Ú Ú

Ú Ú

Ú Ú

Ú

Similarly, we can write:

|t

m x t xma v p dt

•= Ú

and

|: 0

n t tn x t xnx n n

a a a v dt v p dt•

= + = +Ú Ú



Question 4.18

The value of a30 25: , calculated using the interest and mortality assumed in the Tables is:

A 12.8568B 13.2250C 14.4541D 16.0986E 17.8446

As in the annual payments case, the actuarial present value of a whole of life annuity maybe calculated using a recursive formula. The formula is:

a a vp ax x x x= + +:1 1

We prove this as follows:

1

0 1

1:1 0

1:1

t tx t x t x

sx s xx

x xx

a v p dt v p dt

a v p v p ds

a v p a

•

•+

+

= +

= +

= +

Ú Ú

Ú

Example 4.10

Calculate the value of a30 given that a31 15.206= , 0.05i = and x( ) 0.01=µ forx30 31≤ < .

Solution

From the recursive formula we have:

a a v p a30 30 3130:1= +

Now:

p e 0.0130 0.99005−= =



Also:

tt

t t

t

a v p dt

e e dt

e

e

1

3030:101

0.01

0

1( 0.01)0

(ln1.05 0.01)

10.01

1 1ln1.05 0.01

0.97117

δ

δδ

− −

− +

− +

=

=

= − +

= − +

=

∫

∫

So:

a3010.97117 0.99005 15.206

1.0515.309

= + × ×

=

Stochastic model

The present value of a continuously payable, n -year temporary annuity is:

aa T n

a T nT nT

nmin{ , } =

≤

>

RS|T|

if

if



E a a f t dtT n t n T( ) ( )min{ , } min{ , }=∞z0

It is not obvious that this is the same as the result given for the deterministic model. Weprove this result below.



First observe that:

min{ , } 0

0

( ) ( ) ( )

( ) Pr( )

nT TT n t nn

nTt n

E a a f t dt a f t dt

a f t dt a T n

•= +

= + >

Ú Ú

Ú

Since a v dstst

= z0 , we may write the first term as:

0 0 0( ) ( )

n n t sT Tta f t dt v f t ds dt=Ú Ú Ú

Changing the order of integration (and noting that s t≤ ) gives:

0 0( ) ( )

n n n sT Tt s

a f t dt v f t dt ds=Ú Ú Ú

We can now pull the v s term outside the inner integral to obtain:

0 0

0

( ) ( )

Pr( )

n n nsT Tt s

n s

a f t dt v f t dt ds

v s T n ds

=

= £ £

Ú Ú Ú

Ú

Now, writing:

Pr( ) Pr( ) Pr( ) Pr( ) Pr( )s T n T s T n T s T n≤ ≤ = ≥ − > = > − >

gives:

0 0 0

0 0

0

( ) Pr( ) Pr( )n n ns s

Tt

n ns ss x n x

n ss x n x n

a f t dt v T s ds v T n ds

v p ds p v ds

v p ds p a

= > - >

= -

= -

Ú Ú Ú

Ú ÚÚ



Substituting this back into the original expression for the actuarial present value, we have:

E a v p ds p a p a

v p ds

T ns

s xn

n x n n x n

ss x

n

( )min{ , } = − +

=

zz0

0

which was the integral expression obtained using the deterministic model.

Variance

Since:

a vT n

T n

min{ , }

min{ , }=

−1δ

it follows that:

var( ) var( )min{ , }min{ , } : :

a vA A

T nT n x n x n

= =− FH

IK1

2

22

2δ δ

9.3 Premium conversion formulas for continuously payable annuities

Premium conversion formulas also exist for whole life and temporary annuities that arepayable continuously. The formulas are given below.

Premium conversion formulas for continuous annuities

A ax x= −1 δ

A ax n x n: := −1 δ

The proofs of these results are similar to those for annuities-due.



Question 4.19

You are given that i = 0 05. , a30 10 7 8: .= , 10 30 0 99p = . . The value of A30 101

: is:

A 0.00552B 0.01166C 0.01466D 0.02080E 0.61944



10 AccumulationsIn the final section of this unit we consider accumulated fund values.

Suppose that:

• N policyholders, all aged exactly x , contribute one unit to a fund at the start ofeach year for the next n years

• each contribution is contingent on survival

• At the end of the n -year period, the fund is divided equally amongst the survivingpolicyholders.

We want to know:

• the accumulated value of the fund at the end of the n -year period

• the share of the fund per surviving policyholder.

First of all, we can say that the actuarial present value of the contributions from eachindividual policyholder is :ax n . Since we start with N policyholders, the actuarial present

value of the fund is Nax n: .

If the rate of interest is i each year, then the accumulated value of the fund at the end ofn years will be Na ix n

n( ): 1+ .

At the outset, the number of surviving policyholders is unknown − it is a random variable. However, if N is large, we can approximate the actual number of survivors by theexpected number, ie N pn x . Then the expected share per surviving policyholder isapproximately:

Na i

N p

a

v p

a

Ex n

n

n x

x nn

n x

x n

n x

( ): : :1+= =

The notation :sx n is sometimes used instead of :a

Ex n

n x.



Question 4.20

A number of lives, all aged exactly 45 at the outset, contribute to a fund at the rate of 100 ayear payable continuously for the first 5 years and 150 a year payable continuously for thefollowing 5 years. At the end of 10 years the fund is divided equally amongst thesurvivors. Assuming interest and mortality as given in the Tables, the amount received byeach survivor is:

A 1,703B 1,707C 1,709D 1,803E 1,813



End of Part 1

Congratulations! You have completed the Course Notes in Part 1 of this study program.

Time to practice questions

We recommend that you now work through a selection of questions from Part 1 of theQuestion & Answer Bank and then attempt the Part 1 Assignment. This will test yourunderstanding of the material you have covered so far.

We also recommend that you attempt some genuine past exam questions in order to gainan understanding of the style, format and difficulty of Course 3 exam questions. Aftercompleting Part 1 of this study program, you should be able to attempt the following examquestions from the Past Exam Pack:

Past exam paper Questions

May 2000 1, 5, 8, 12, 13, 21, 29, 36, 39

November 2000 3, 4, 10, 20, 25, 26, 28, 31, 36

May 2001 1, 13, 14, 17, 24, 27, 33, 34, 39

November 2001 1, 2, 3, 8, 17, 26, 37

November 2002 1, 3, 4, 31, 32, 35, 39

Planning the rest of your study

You should make steady progress through the remaining units of this study program toleave adequate time for review and to practice questions. Our Study Guide includesillustrative study schedules to help you plan your time between now and the exam.

In order to help you remember the material covered in Part 1, we recommend that youreview it regularly. You may find it useful to re-read the concise summaries or attempt theself-assessment questions. Our Course 3 flashcards are also an excellent review of thematerial. Visit our website at www.bpp.com for full details.



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Unit 4 Summary

Whole of life annuities

Payable annually in arrears Payable annually in advance

PV aK= PV aK=+1

kx k x

kAPV a v p

1

∞

== = ∑ k

x k xk

APV a v p0

∞

== = ∑

var( )( )

PVA A

dx x=−2 2

2var( )

( )PV

A Ad

x x=−2 2

2

Values of ax are given in the Tables. Values of ax can be calculated using the formula:

a ax x= − 1

Recursive formula

11x x xa v p a += +

Extra risk

Let * ( ) ( )y ym m m= + for all 0y > . In addition, let *xa denote the actuarial present value of

a whole life annuity-due payable to a life aged x , who is subject to force of mortality *m . Then:

* @ @ *x xa i a i=

where * 1i ed m+= - .

In other words, a constant addition to the standard force of mortality can be treated as aconstant addition to the standard force of interest.



Temporary annuities


PV a K n= min{ , } PV a K n=+min{ , }1

nk

k xx nk

APV a v p:1=

= = ∑n

kk xx n

kAPV a v p

1

:0

−

== = ∑

var( )( ): :

PVA A

d

x n x n=

−+ +

21 1

2

2var( )

( ): :PV

A A

d

x n x n=

−2 2

2

Values of :ax n are not given in the Tables, but can be calculated using the formula:

:a a E ax n x n x x n= − +

Similarly:


Deferred whole of life annuities (m-year deferment period)

Payable annually in arrears

PV a aK K m= − min{ , }

m x m x x m x x mAPV a E a a a| :+= = = -

{2 2 2 2: 1 : 12

2 11|| : 1: 1

1var( ) ( ) ( )

2

x x x m x m

m x x x mx m

PV A A A Ad

A A A A

+ +

+ ++

= - + -

¸Ê ˆ- + - ˝Á ˜Ë ¯˛



Payable annually in advance

PV a aK K m= −+ +min{ , }1 1

m x m x x m x x mAPV a E a a a| :+= = = -

{2 2 2 2: :2

2 1|| ::

1var( ) ( ) ( )

2

x x x m x m

m x x x mx m

PV A A A Ad

A A A A

= - + -

¸Ê ˆ- + - ˝Á ˜Ë ¯˛

Deferred temporary annuities (m-year deferment period, n-year term)


m x n

m x x m n

APV a

E a

| :

:+

=

=

m x n

m x x m n

APV a

E a

| :

:+

=

=

Guaranteed annuities (whole life, m-year guarantee)


PV a a am K K m= + − min{ , } PV a a am K K m= + −+ +min{ , }1 1

m xmx mAPV a a a|:

= = + m xmx mAPV a a a|:

= = +

The variances are the same as those for deferred annuities with a deferment period of myears.



Whole life annuities payable m times a year

Define J to be the whole number of periods of length 1/ m survived in the year of death,ie ( )J T K m= -Í ˙Î ˚ .

Payable m-thly in arrears Payable m-thly in advance

PV aK J mm=+ /

( ) PV aK J mm=+ +( ) /

( )1

m k mx k m x

kAPV a v p

m( ) /

/1

1•

== = Â m k m

x k m xk

mx

APV a v pm

am

( ) //

0

( )

1

1

•

== =

= +

Â

var( )( )

( )

( ) ( )

( )PV

A Ad

xm

xm

m=−2 2

2var( )

( )

( )

( ) ( )

( )PV

A Ad

xm

xm

m=−2 2

2

Values of ( )axm are not given in the Tables, but can be calculated using the formula:

( ) ( )( )a m a mxm

x= −α β

where:

α( )( ) ( )

m idi dm m= and β( )

( )

( ) ( )m i i

i d

m

m m=−

This formula assumes that deaths occur uniformly between integer ages. Values of α( )mand β( )m are listed in the Tables for m = 1 2 4 12, , , and ∞ .

Temporary annuities payable m times a year

( ) ( ) ( ):m m m

x n x x nx na a E a += - ( ) ( ) ( )

:m m m

x n x x nx na a E a += -



Whole life annuities payable continuously

PV aT=

tx t xAPV a v p dt

0

∞= = ∫

var( )( )

PVA Ax x=

−2 2

2δ

Values of ax can be calculated using the formula:

a ax x= ∞ − ∞α β( ) ( )

Temporary annuities payable continuously (n-year term)

PV a T n= min{ , }

n tt xx nAPV a v p dt: 0

= = ∫

var( )( ): :

PVA Ax n x n

=−2 2

2δ

Values of ax n: can be calculated using the formula:


Recursive formulas

a v p ax x x= + +1 1

1:1x x xxa a v p a += +



Premium conversion formulas

A d ax x= −1 A d ax n x n: := −1

A d axm m

xm( ) ( ) ( )= −1 A d a

x nm m

x nm

:( ) ( )

:( )= −1

A ax x= −1 δ A ax n x n: := −1 δ

Note that the formula A d ax nm m

x nm

:( ) ( )

:( )= −1 is not used in this unit. It is listed here for the

sake of completeness.

Accumulations

Suppose that a number of lives, all aged exactly x , contribute one unit to a fund at thestart of every year for a period of n years and the contributions are contingent on survival.If the fund is divided equally amongst the survivors at the end of the n -year period, then

the share per survivor is approximately :a

Ex n

n x.



Unit 4 Solutions

Solution 4.1

From the deterministic model, we have:

a v p v p v p

v p v p v p

v p a

x x x x

x x x

x x

= + + + +

= + + + +

= +

+ +

+

1

1 1

1

22

33

12

2 1

1

e j

Solution 4.2

Answer B.

The actuarial present value of the annuity is:

5 000 5 000 14 8166 1 69 08340, , ( . ) ,a = × − =

The variance is:

5 00025 000 1000 25 1000

25 000 48 63 25 161320 05660

176 388 483 3

22

40 402

2

240 40

2

2

2

2

,( ) ( , , ) ( , )

( , . ) ( . ).

, , .

A Ad

A Ad

−FHG

IKJ=

× × −

=× − ×

=

So the standard deviation is 176 388 483 3 13 28113, , . , .=



Solution 4.3

Answer B.

In order to calculate the value of A50 we have to use the premium conversion formula:

A da50 501= −

From Example 4.2, we know that:

@ @*a a50 505% 6%=

where * denotes non-standard mortality. Hence:

A da50 501 1 0 05105

13 2668 0 36825* * ..

. .= − = − ×FHG

IKJ =

Note that d is calculated using the standard rate of interest, ie 5% a year.

Solution 4.4

Answer D.


10 000 10 000 1

10 000 1 0 99105

0 99105

28 318 37

60 3 602

2 60

2

2

, ,

, ..

..

, .

:a v p v p= + +

= + +FHG

IKJ

=

e j



Solution 4.5

From our knowledge of annuities-certain we have:

min{ , }

min{ , }a v

dK n

K n

+

+=

−1

11

So:

var( ) var

var( )

( )

( )

min{ , }

min{ , }

min{ , }

min{ , } min{ , }

: :

a vd

dv

dE v E v

A A

d

K n

K n

K n

K n K n

x n x n

+

+

+

+ +

=−FHG

IKJ

=

= −RST

UVW

=−

1

1

21

22 1 1 2

2 2

2

1

1

1 e j e j

Solution 4.6

We have:

:

a v p a v p v p v p

v p v p

v p v p

v p

a

xn

n x x nk

k xk

nn x

kk x n

k

kk x

k

n kn k x

k

kk x

k

jj x

j n

kk x

k

n

x n

− = −

= −

= −

=

=

+=

∞

+=

∞

=

∞+

+=

∞

=

∞

=

∞

=

−

∑ ∑

∑ ∑

∑ ∑

∑

0 0

0 0

0

0

1



Solution 4.7

From Section 5.2, we know that:

( ): min{ , }a E ax n K n=+1

So:

min{ 1, } min{ 1, } ::

11 1 ( )K n K n x nx n

Av E va Ed d d

+ + -Ê ˆ- -= = =Á ˜Ë ¯

Rearranging gives:

A dax n x n: := −1

Solution 4.8

Answer D.


30 25 30 25 55 25 30 25 301

55 1 0 20 1185 2 37| : : |( ) ( ) . . .a E a A A a= = − × + = × =

Solution 4.9

Answer C.

Statement I

a a v p v p v p

v p v p v p

v p v p v p

a

x m n x m x xm n

m n x

x xm

m x

mm x

mm x

m nm n x

m x n

: :

| :

( )

++

+

++

++

++

− = + + +

− + + +

= + + +

=

22

22

11

22



Rearranging this gives:

m x n x m x m na a a| : : :+ =+

So I is correct.

Statement II

m x nm

m xm

m xm n

m n xa v p v p v p| : = + + +++

+ −+ −

11

11

But:

m x nm m

m xm

m xm n

m n xma v v p v p v p v| : + = + + + ++

++

++

+1

12

2

So II is incorrect.

Statement III

m x x mm

m x x m x m

mm x

mm x

mm x

m x

E a v p v p v p

v p v p v p

a

+ + ++

+ + + + +

++

++

++

= + + +

= + + +

=

1 11

1 12

2 1

11

22

33

1( )

|

So III is correct.

Solution 4.10

The present value is:

PV a a am K K m= + −+ +min{ , }1 1


: |a a a a E ax m m m x m m x x m= + = + +



Solution 4.11

Answer A.


10 40 501012,000 a E aÊ ˆ+Ë ¯

The factors are:

( . ). / .

.a vd10

10 101 1 1060 06 106

7 80169=−

=−

=−

10 40 0 53667E = . 4

.a50 13 2668=

So the actuarial present value is:

12 000 7 80169 0 53667 13 2668 179 059 03, . ( . . ) , .+ × =

ie $179,059 to the nearest dollar.

(Note that you get 179,060 if you use 1010 40v p instead of 10 40E . This is due to

rounding.)

Solution 4.12


( ) 1/ 2 /1/ 2 /

//

0

( )

1 1 1

1

1

m m mx m x m x

k mk m x

k

mx

a v p v pm m m

v pm

am

•

=

= + + +

=

= +

Â



Solution 4.13

We can write:

( 1) /( ) ( )( 1) /

( ) ( ) ( )1 1 11 ( ) 1

K J mm mK J m

x xm m mva E E v A

d d d

+ ++ +Ê ˆ- È ˘È ˘= = - = -Á ˜ Î ˚ Î ˚Ë ¯

and rearranging this gives:

( ) ( )( )1m mmx xA d a= -

Solution 4.14

The formulas for α( )12 and β( )12 give:

(12) (12)0.06 (0.06 /1.06)(12) 1.00028

0.058410607 0.058127667id

i da ¥= = =

¥

(12)

(12) (12)0.06 0.058410607(12) 0.46812

0.058410607 0.058127667i i

i db - -= = =

¥

NB These calculations are very sensitive to rounding. You have to use lots of decimalplaces in the intermediate steps to produce accurate answers for (12)a and (12)b . In theexamination, you should use the values of a and b given in the Tables.

Solution 4.15

Answer B.

Since the policyholder receives $5,000 at the end of every three months, the total paymentreceived in a year is 4 5,000 $20,000¥ = . The actuarial present value is given by:

( )

(4) (4)60 60

60

120,000 20,0004

120,000 (4) (4)4

20,000 (1.00027 11.1454) 0.38424 0.25210,283.39

a a

aa b

Ê ˆ= -Á ˜Ë ¯

Ê ˆ= - -Á ˜Ë ¯

= ¥ - -=



Solution 4.16

Answer E.


(12) (12) (12)10 65 756565:10

24,000 24,000a a E aÈ ˘= -Î ˚

The factors are:

( ) ( ) ( . . ) . .( )a a6512

6512 12 100028 9 8969 0 46812 9 4316= − = × − =α β

( ) ( ) ( . . ) . .( )a a7512

7512 12 100028 7 2170 0 46812 6 7509= − = × − =α β

10 65 0 39994E = .

So the actuarial present value is:

24 000 9 4316 0 39994 6 7509 161559 47, . ( . . ) , .− × =

ie $161,559 to the nearest dollar.

Solution 4.17

By definition:

0( ) ( )TT tE a a f t dt

•= Ú

Writing a v dstst

= z0 , we obtain:

0 0( ) ( )

t sTTE a v f t ds dt

•= Ú Ú

Changing the order of integration (noting that s t£ ) gives:

0( ) ( )s

TT sE a v f t dt ds

• •= Ú Ú



We can now pull the v s term outside the first integral and simplify. We obtain:

0 0 0( ) ( ) Pr( )s s s

T s xT sE a v f t dt ds v T s ds v p ds

• • • •Ê ˆ= = > =Á ˜Ë ¯Ú Ú Ú Ú

as required.

Solution 4.18

Answer A.

We can write:

a a E a30 25 30 25 30 55: = −

The individual factors are:

a a30 30 100028 15 8561 0 50985 15 3507= ∞ − ∞ = × − =α β( ) ( ) . . . .

a a55 55 100028 12 2758 0 50985 117694= ∞ − ∞ = × − =α β( ) ( ) . . . .

25 3025 55

30

25106 8 640 8619 501381

0 211896E vll

= = =−( . ) , ,, ,

.

So:

a30 25 15 3507 0 211896 117694 12 8568: . ( . . ) .= − × =

Solution 4.19

Answer B.

First note that:

1110 3030:10 30:1030:10 30:10

A A A A E= - = -

By premium conversion:

A a30 10 30 101 1 105 7 8 0 619437: : (ln . . ) .= − = − × =δ



Also:

10 3010

10 3010105 0 99 0 607774E v p= = × =−( . ) . .

So:

A30 101 0 619437 0 607774 0 011663

:. . .= − =

Solution 4.20

Answer A.

The actuarial present value of the fund per original member is:

100 5045 10 5 45 50 5a E a: :+ × ×

We can calculate a45 10: as follows:

a a E a

a E a45 10 45 10 45 55

45 10 45 55

100028 141121 0 50985 0 52652 100028 12 2758 0 509857 4094

:

( ) ( ) ( ) ( )

. . . . . . ..

= − ×

= ∞ − ∞ − ∞ − ∞

= × − − × −

=

α β α β

Similarly:

a50 5 4 2706: .=

So the actuarial present value per original member is:

( . ) ( . . ) .100 7 4094 50 0 72988 4 2706 896 79× + × × =

The share of the fund per surviving policyholder is then:

10 45

896.79 896.79 1,703.240.52652E

= =

Actuarial models Course 3

Fall 2004 exams © BPP Professional Education

Course 3Fall 2004 exams

Part 1Question & Answer Bank

Instructions

This Question & Answer Bank contains 50 exam-style questions.

We recommend that you attempt a selection of these questions after completingPart 1 of the Course Notes and the remaining questions as part of your review.

It’s a good idea to annotate the questions with details of when you attempted each oneand whether you got it right or wrong. This makes it easier to ensure that you complete all

of questions and identify the areas of the course that you are finding most difficult.

Course 3 Professional models

© BPP Professional Education Fall 2004 exams






Legal action will be taken if these terms are infringed.In addition, we may seek to take disciplinary actionthrough the profession or through your employer.

Course 3 Question and Answer Bank Part 1 – Questions


Part 1 – QuestionsQuestion 1

The annual rate of interest earned on a certain investment is equally likely to be 5%, 6%or 7% in each year. Tony invests $1,000 at the start of each of the next three years. Assuming that returns in individual years are independent, Tony’s expected return at theend of three years is:

A $3,183.60B $3,183.67C $3,374.62D $3,374.89E $3,395.11

Question 2

Suzy invests $1,000 into the investment described in Question 1. If she makes noadditions or withdrawals, the standard deviation of her return at the end of three years is:

A $15.89B $45.59C $53.00D $65.72E $166.54

Question 3

Which of the following expressions are equivalent to |n k x tq + ?

I [ ]Pr ( )n T x t k< + £

II x t n k x t n

x t

l ll

+ + + + +

+

-

III n x t n k x tp p+ + +-

A I onlyB II onlyC III onlyD I and II onlyE II and III only

Question and Answer Bank Part 1 – Questions Course 3


Question 4

In a certain population:

- < <Ï

Ô= - £ <ÌÔ ≥Ó

30

1 0.01 for 0 501.75 0.025 for 50 700 for 70

t

t tp t t

t

The curtate expectation of life of (30) is:

A 27 yearsB 35 yearsC 40 yearsD 42 yearsE 50 years

Question 5

Given mortality as in Question 4, the standard deviation of the curtate future lifetime is:

A 20.7 yearsB 26.1 yearsC 35.0 yearsD 43.9 yearsE 46.8 years

Question 6

Given mortality as in Question 4, calculate 20 70p .

A 0.333B 0.350C 0.417D 0.583E 0.650



Question 7

Given mortality as in Question 4, calculate 10 30m .

A 0.0100B 0.0101C 0.0105D 0.0111E 0.0115

Question 8

Calculate 0.75 55.5p using the information given in the Tables and assuming a uniformdistribution of deaths between integer ages.

A 0.99307B 0.99309C 0.99317D 0.99327E 0.99329

Question 9

Repeat Question 8 assuming a constant force of mortality between integer ages.

A 0.99307B 0.99309C 0.99317D 0.99327E 0.99329



Question 10

You are given the following information:

• 42 35.0e =

• 40 0.995p =

• 41 0.993p =

The value of 40e is:

A 36.6B 36.7C 36.8D 36.9E 37.0

Question 11

If 0.01( ) 0.001 xx x em = , then the value of (50)s is:

A 0.0162B 0.0824C 0.1727D 0.8273E 0.9704

Question 12

In a certain population, the survival function is given by:

ÏÊ ˆÔ - £ £Á ˜= Ë ¯ÌÔÓ

21 for 0 100( ) 100

0 otherwise

x xs x

The value of [ ](25)E T is:

A 24.5 yearsB 25.0 yearsC 25.5 yearsD 37.5 yearsE 50.0 years



Question 13

In a certain population:

• (60.75) 0.0105m =

• (61.75) 0.0318m =

• deaths are uniformly distributed between integer ages.

The probability that a person aged 60.25 will die within 18 months is:

A 0.0306B 0.0309C 0.0312D 0.0314E 0.0317

Question 14

You are given that:

• 0.05( ) xx A em = + , 0x ≥

• (40) 0.85s =

The value of A is:

A −4.20B −3.19C 0.00D 3.19E 4.20

Question 15

You are given that 70 0.953p = . Assuming a hyperbolic form for t xp , the value of

0.75 70p is:

A 0.948B 0.956C 0.964D 0.976E 0.988



Question 16

Given mortality as in Question 15, the value of 0.25 70.75q is:

A 0.0118B 0.0119C 0.0120D 0.0121E 0.0122

Question 17

You are given the following extract from a select-and-ultimate mortality table with a 2-yearselect period:

x [ ]xl [ ] 1xl + 2xl + 2x +

50 80,625 79,954 78,839 52

51 79,137 78,402 77,252 53

52 77,575 76,770 75,578 54

The probability that a life aged exactly 52, who became select one year ago, survives toage 53 is:

A 0.9792B 0.9799C 0.9844D 0.9853E 0.9896

Question 18

Use the life table in Question 17 to calculate the value of 0.5 [51] 0.8q + . Assume that the

force of mortality is constant between integer ages.

A 0.00462B 0.00464C 0.00520D 0.00625E 0.00628



Question 19

Calculate the median future lifetime of (25), given that the lifetime distribution isexponential with a mean of 70.

A 22.5B 24.3C 35.0D 45.0E 48.5

Question 20

You are given that:

• (90) 0.32m =

• the force of mortality is constant between integer ages.

Calculate the average number of years lived between the ages of 90 and 91 for lives thatdie between those two ages.

A 0.23B 0.32C 0.47D 0.50E 0.73

Question 21

Assuming that mortality conforms to de Moivre’s law and that (80) 0.15m = , calculate thevalue of 20 60p .

A 0.1125B 0.1534C 0.2308D 0.2500E 0.4724



Question 22

Consider the following three statements about :x nA :

I : 1: 1xx n x nA vq A + -= +

II |: x n x n xx nA A A E= - +

III2

11:

0

nk n

k x x k n xx nk

A v p q v p-

++ -

== +Â

Which of these statements are correct?


Question 23

Assuming mortality and interest as given in the Tables, the value of 110| 35:20A is:

A 0.0481B 0.0495C 0.0509D 0.0886E 0.0991

Question 24

A life aged 50 purchases a whole life insurance. A benefit of $75,000 is payable at theend of the year of death if death occurs before age 60. A benefit of $50,000 is payable atthe end of the year of death if death occurs after age 60. Assuming mortality and interestas given in the Tables, the actuarial present value of the benefit is:

A $9,451B $10,237C $13,526D $13,965E $17,166



Question 25

Assuming mortality and interest as given in the Tables, the standard deviation of thepresent value of a 10-year term life insurance on (50) that pays $75,000 at the end of theyear of death is:

A $1,639B $3,037C $4,510D $15,093E $15,876

Question 26

Assuming mortality and interest as given in the Tables, the standard deviation of thepresent value of a deferred whole life insurance on (50) with a deferment period of 10years, which pays $50,000 at the end of the year of death is:

A $1,181B $5,181C $6,134D $10,143E $11,733

Question 27

Assuming mortality and interest as given in the Tables, the standard deviation of thepresent value of the benefit described in Question 24 is:

A $8,146B $13,412C $14,920D $16,291E $18,734



Question 28

If mortality is the same as that given in the Tables and 0.05i = for the next 6 years and

0.06i = thereafter, then 132:10A is equal to:

A 0.01037B 0.01601C 0.57784D 0.57948E 0.87836

Question 29

Consider the following three statements concerning ( )xIA :

I ( ) ( )x xiIA IAd

ª

II0

( ) ( )tx t xIA t v p x t dtm

•= +Ú

III1

10( ) ( ) ( )t

x t x x xIA t v p x t dt vp IAm += + +Ú

Which of these statements are correct?


Question 30

A life aged 30 takes out a 30-year term life assurance under which a benefit of $10,000 ispayable two months after the date of death. Assuming mortality and interest as given inthe Tables, the expected present value of this benefit is:

A 480B 482C 483D 485E 490



Question 31

The value of 45:20A , calculated using mortality and interest as given in the Tables and

assuming a uniform distribution of deaths between integer ages, is:

A 0.3223B 0.3448C 0.3474D 0.3550E 0.4029

Question 32

Calculate the value of (4)5| 35A assuming interest and mortality as given in the Tables and a

uniform distribution of deaths between integer ages.

A 0.11658B 0.11917C 0.12077D 0.12094E 0.12182

Question 33

A life aged 60 has purchased an annuity that provides payments at the start of each yearfor the next 5 years. The amount of the first payment is $10,000. Each subsequentpayment is $2,000 higher than the last. Assuming that mortality follows that in the Tablesand 0.05i = , the expected present value of the annuity is:

A $56,852B $57,820C $59,694D $60,711E $75,129



Question 34

The standard deviation of the present value of the annuity described in Question 33 is:

A $624B $8,129C $8,729D $27,817E $29,486

Question 35

You are given that:

• 40 13.300a =

• 0.04i =

• 40 0.995p =

The value of 41a is then:

A 12.725B 12.856C 13.300D 13.681E 13.901

Question 36

A whole life annuity is payable annually in advance to (60) at the rate of 10 a year. Assuming that mortality and interest conform to that given in the Tables, the variance ofthe present value of this annuity is:

A 72.7B 128.4C 1,143.1D 1,242.1E 1,284.4



Question 37

A life aged 50 wishes to purchase a policy with the following terms:

• a death benefit of 25,000 is payable immediately if death occurs before age 60

• on survival to age 60, an annuity of 5,000 is payable annually in advance

• the annuity payments are guaranteed for the first five years

• payments are contingent on survival thereafter

Assuming that interest and mortality conform to that given in the Tables, the expectedpresent value of the benefit is:

A 30,296B 30,341C 31,707D 35,195E 38,239

Question 38

You are given the following information:

• 30 16.250a =

• 40 15.125a =

• 10 30 0.987p =

• 0.04i =

The value of 130:10A is:

A 0.0088B 0.0130C 0.0866D 0.0961E 0.1461



Question 39

One hundred lives, all aged exactly 50, pay $200 into a fund at the beginning of eachmonth for the next 15 years. The fund earns interest at the effective annual rate of 6%. At the end of 15 years the fund is divided equally amongst the survivors. Assumingmortality as in the Tables, the share per survivor is:

A $54,040B $59,214C $64,850D $70,351E $86,276

Question 40

A life aged 68 purchases a temporary life annuity under which payments of $10,000 aremade at the start of each of the next three years. Calculate the expected present value ofthis annuity assuming an effective annual rate of interest of 6% and a constant deductionof 0.00947843 from the standard force of mortality used in the Tables.

A $27,303B $27,332C $27,493D $27,562E $27,810

Question 41

A life aged 47 receives an annuity payable continuously at the rate of M a year. Theannuity is guaranteed for the next 13 years and continues for life thereafter. The expectedpresent value of the annuity is 100,000, assuming mortality and interest as given in theTables. The value of M (correct to the nearest $1) is:

A 7,090B 7,096C 7,122D 7,348E 7,374



Question 42

A life aged 60 has just purchased a whole of life annuity-due. You are given the followinginformation:

• 60 0.98p =

• 2 60 0.93p =

• 0.04i =

• 61 11.60a =

If 61p increases by 0.02, then the expected present value of the annuity changes by:

A 0.19B 0.20C 0.21D 0.22E 0.23

Question 43

You are given that 55 0.01t q t= for 1 10t≤ ≤ . Calculate 60:5a using an effective annual

rate of interest of 5%.

A 4.1972B 4.2038C 4.2054D 4.4032E 4.4549

Question 44

A life aged 70 purchases a whole life annuity, which is payable twice yearly in arrears. Ifthe total annual payment is 10,000 and mortality and interest follow those given in theTables, then the actuarial present value of the annuity is:

A 78,137B 83,137C 83,285D 88,137E 88,285



Question 45

Mark is aged 45. He buys a policy that provides:

• a benefit of $40,000 at the end of the year of death, if death occurs before age 65

• a benefit of $15,000 payable on survival to age 65

• a life annuity of $10,000 a year, payable annually in arrears from age 65 onwards.

Assuming that mortality and interest conform to those given in the Tables, the probabilitythat Mark will receive benefits with a present value of less than $40,000 is:

A 0.0563B 0.2341C 0.2573D 0.7097E 0.7427

Question 46

You are given that:

• 28:30 14.197a =

• 30 28 0.885p =

• 0.05i =

The value of 128:30A is:

A 0.0423B 0.1026C 0.1192D 0.2048E 0.3240



Question 47

You are given the following extract from a select-and-ultimate mortality table with a 2-yearselect period:

x [ ]xl [ ] 1xl + 2xl + 2x +

50 80,625 79,954 78,839 52

51 79,137 78,402 77,252 53

52 77,575 76,770 75,578 54

Given that 0.05i = , the value of [51]:3A is:

A 0.82499B 0.86158C 0.86529D 0.86541E 0.88586

Question 48

Assuming that 0.05i = and that mortality follows the life table in Question 47, calculatethe value of 52:3a .

A 2.6692B 2.7764C 2.8027D 2.8262E 2.8314



Question 49

A life aged 50 purchases a 10-year endowment assurance with a sum assured of$20,000. The benefit is paid on survival to age 60 or at the end of the year of earlierdeath. You are given the following information:

• 0.04i =

• “standard” mortality is that given in the Tables

• standard mortality applies between ages 50 and 53

• between ages 53 and 60, there is a constant addition of 0.019048194 to thestandard force of mortality.

Calculate the expected present value of the endowment assurance.

A $12,091B $13,165C $13,815D $13,919E $15,911

Question 50

Five hundred lives, all aged exactly 45, pay $1,000 into a fund at the beginning of eachyear for the next 10 years. The fund earns interest at the effective annual rate of 6%. Atthe end of 10 years the fund is divided up. Each surviving member receives an amountS . In addition, 1

2S is paid in respect of each member who died. Mortality is expected to

conform to that given in the Tables. Calculate the expected share per survivor.

A $14,100B $14,382C $14,527D $14,954E $15,407

Course 3 Question and Answer Bank Part 1 – Solutions


Part 1 – SolutionsSolution 1

C Tony’s return R is given by:

[ ]1 2 3 2 3 31,000 (1 )(1 )(1 ) (1 )(1 ) (1 )R i i i i i i= + + + + + + + +

where ki denotes the rate of interest earned in year k . The expected return is therefore:

( ) ( )1 2 3 2 3 3( ) 1,000 (1 )(1 )(1 ) (1 )(1 ) (1 )E R E i i i E i i E iÈ ˘= + + + + + + + +Î ˚

Since the ki are independent and identically distributed:

3 2( ) 1,000 (1 ) (1 ) (1 )E R j j jÈ ˘= + + + + +Î ˚

where ( )kj E i= .

Also, since the annual interest rate is equally likely to be 5%, 6% or 7%, it follows that0.06j = .

So Tony’s expected return is given by:

3 2( ) 1,000 (1.06) (1.06) (1.06) $3,374.62E R È ˘= + + =Î ˚

Solution 2

A Suzy’s return S is given by:

1 2 31,000(1 )(1 )(1 )S i i i= + + +

We can calculate the variance of Suzy’s return using the formula:

[ ]22var( ) ( ) ( )S E S E S= -

Since the ki are independent, we have:

2 2 2 2 21 2 3

2 2 2 21 2 3

( ) 1,000 (1 ) (1 ) (1 )

1,000 (1 ) (1 ) (1 )

E S E i i i

E i E i E i

È ˘= + + +Î ˚È ˘ È ˘ È ˘= + + +Î ˚ Î ˚ Î ˚

Question and Answer Bank Part 1 – Solutions Course 3


Also, since the ki are identically distributed:

( ) [ ]22

2

2 2

1 var(1 ) (1 )

var( ) (1 )

(1 )

k k k

k

E i i E i

i j

js

È ˘+ = + + +Í ˙Î ˚

= + +

= + +

where 2 var( )kis = .

We can calculate 2s as follows:

( ) ( )2 2 222 2 20.05 0.06 0.07 0.06 0.000066667

3kkE i E is + +È ˘= - = - =Î ˚

So:

( ) ( )

( ) ( )

3 62 2 2

3 62 2

var( ) 1,000 (1 ) 1

1,000 0.000066667 (1.06) 1.06

252.510

S j jsÈ ˘= + + - +Í ˙

Î ˚È ˘

= + -Í ˙Î ˚

=

Hence the standard deviation of Suzy’s return is 252.510 $15.89= .

Solution 3

C The symbol |n k x tq + denotes the probability that ( x t+ ) survives for n years and then

dies some time during the subsequent k years, that is the probability that ( x t+ ) diesbetween the ages of x t n+ + and x t n k+ + + .

The expression in I is the probability that ( x t+ ) dies between the ages of x t n+ + andx t k+ + . So this is not equal to |n k x tq + .

The expression in II is negative since x t n k x t nl l+ + + + +< . So this is not equal to |n k x tq + .

The expression in III is the probability that ( x t+ ) survives to age x t n+ + , but not to agex t n k+ + + . This is equal to |n k x tq + .



Solution 4

D The curtate expectation of life of (30) is:

69 49 69

30 301 1 50

(1 0.01 ) (1.75 0.025 )kk k k

e p k k= = =

= = - + -Â Â Â

Using the well-known identity1

1 ( 1)2

n

kk n n

== +Â , this becomes:

301 1 149 0.01 49 50 35 0.025 69 70 49 502 2 2

49 12.25 35 29.7542

e È ˘ È ˘= - ¥ ¥ + - ¥ ¥ - ¥ ¥Í ˙ Í ˙Î ˚ Î ˚= - + -=

Solution 5

A The variance is given by:

( )69

230 30

149 69

2

1 5049 69

2 2 2

1 50

var (30) (2 1)

(2 1)(1 0.01 ) (2 1)(1.75 0.025 ) 42

(2.01 0.02 1) (3.525 0.05 1.75) 42

kk

k k

k k

K k p e

k k k k

k k k k

=

= =

= =

= - -

= - - + - - -

= - - + - - -

Â

Â Â

Â Â

Using the formulas 1

1 ( 1)2

n

kk n n

== +Â and 2

1

1 ( 1)(2 1)6

n

kk n n n

== + +Â , we find that:

492

1

1 1(2.01 0.02 1) 2.01 49 50 0.02 49 50 99 49 1,604.752 6k

k k=

Ê ˆ Ê ˆ- - = ¥ ¥ - ¥ ¥ ¥ - =Á ˜ Á ˜Ë ¯ Ë ¯Â



Similarly:

692

50

1 1(3.525 0.05 1.75) 3.525 69 70 49 502 2

1 10.05 69 70 139 49 50 996 6

(20 1.75)

4,194.75 3,573.5 35

586.25

kk k

=

Ê ˆ- - = ¥ ¥ - ¥ ¥Á ˜Ë ¯

Ê ˆ- ¥ ¥ ¥ - ¥ ¥ ¥Á ˜Ë ¯

- ¥

= - -

=

Â

So:

( ) 2var (30) 1,604.75 586.25 42 427K = + - =

So the standard deviation of the curtate future lifetime is 427 20.7= years.

Solution 6

C 60 3020 70

40 30

1.75 0.025 60 0.4171 0.01 40

pp

p- ¥= = =- ¥

Solution 7

C From the definition of t xm , we have:

30 40 30 4010 30 1010 30

300 t

l l l lm

L l dt+

- -= =

Ú

Dividing the numerator and denominator by 30l gives:

10 3010 30 10

300 t

qm

p dt=

Ú



Since 30 1 0.01t p t= - for 0 100t< < , it follows that:

10 30 0.01 10 0.1q = ¥ =

( )1010 10 2 2300 0 0

(1 0.01 ) 0.005 10 0.005 10 9.5t p dt t dt t tÈ ˘= - = - = - ¥ =Î ˚Ú Ú

Hence:

10 300.1 0.01059.5

m = =

Solution 8

A Since the uniform distribution of deaths assumption applies between integer ages, wehave to write the survival probability 0.75 55.5p as 0.5 55.5 0.25 56p p¥ and calculate each ofthese terms separately. We have:

( )

0.25 56 0.25 56

56

11 0.25 by UDD assumption1 0.25 0.009750.997563

p qq

= -= -

= - ¥=

Also:

550.5 55.5

0.5 55

1 0.00896 0.9955001 0.5 0.00896

pp

p-= = =

- ¥

Multiplying these together gives:

0.75 55.5 0.997563 0.995500 0.99307p = ¥ =



Solution 9

A Again we have to write

0.75 55.5 0.5 55.5 0.25 56p p p= ¥

Then:

( )1 14 40.25 56 56 (0.99025) 0.997554p p= = =

( )1 12 20.5 55.5 55 (0.99104) 0.995510p p= = =

So:

0.75 55.5 0.997554 0.995510 0.99307p = ¥ =

Solution 10

A From the definition of the curtate expectation of life, we have:

40 401

40 2 40 403

40 2 40 2 40 2 423

40 2 40 2 40 421

40 2 40 2 40 42

0.995 (0.995 0.993) (0.995 0.993 35.0)

36.6

kk

kk

kk

kk

e p

p p p

p p p p

p p p p

p p p e

•

=•

=•

-=

•

=

=

= + +

= + + ¥

= + +

= + +

= + ¥ + ¥ ¥

=

Â

Â

Â

Â



Solution 11

C By definition:

5050 0 0

(50) exp ( )s p x dxmÏ ¸= = -Ì ˝Ó ˛Ú

If 0.01( ) 0.001 xx x em = , then integrating by parts, we obtain:

50 50 0.010 0

50 500.01 0.0100

500.5 0.01

00.5 0.5

( ) 0.001

0.001 0.0010.01 0.01

0.1(0.1 50 ) 00.01

5 10 10

1.75639

x

x x

x

x dx x e dx

x e e dx

e e

e e

m =

È ˘= -Í ˙Î ˚

È ˘È ˘= ¥ ¥ - - Í ˙Î ˚ Î ˚È ˘= - -Î ˚

=

Ú Ú

Ú

Hence:

1.75639(50) 0.1727s e-= =

Solution 12

B The complete expectation of life at age 25 is given by:

[ ] 25 250(25) tE T e p dt

•∞= = Ú

From the survival function:

21 for 0 100( ) 100

0 otherwise

x xs xÏÊ ˆÔ - < <Á ˜= Ë ¯ÌÔÓ

we obtain:

2

25(25 ) 1 for 0 75

75(25)0 otherwise

t

ts t tps

ÏÊ ˆ+ Ô - < <Á ˜= = Ë ¯ÌÔÓ



Hence:

[ ]275

0

1 20

13

0

(25) 175

75 (substituting 1 )75

753

25

tE T dt

tu du u

u

Ê ˆ= -Á ˜Ë ¯

= = -

È ˘= Í ˙Î ˚=

Ú

Ú

Solution 13

B We want to calculate 1.5 60.25q . This can be expressed as:

( )1.5 60.25 1.5 60.25 0.75 60.25 0.75 611 1q p p p= - = - ¥

By the UDD assumption:

0.75 61 611 0.75p q= -

and:

161 61 610

(61 ) (61 )t tq p t dt p tm m= + = +Ú for all t , 0 1t< <

In particular:

61 0.75 61 61(61.75) (1 0.75 ) (61.75)q p qm m= = -

Rearranging gives:

61(61.75) 0.0318 0.03106

1 0.75 (61.75) 1 (0.75 0.0318)q m

m= = =

+ + ¥

So:

0.75 61 1 (0.75 0.03106) 0.97671p = - ¥ =

We can then calculate the value of 0.75 60.25p using the UDD assumption as follows:

60 60 600.75 60.25

0.25 60 0.25 60 60

1 11 1 0.25

p q qp

p q q- -

= = =- -



Using the same method as for 61q , we obtain:

60(60.75) 0.0105 0.01042

1 0.75 (60.75) 1 (0.75 0.0105)q m

m= = =

+ + ¥

So:

0.75 60.251 0.01042 0.99217

1 (0.25 0.01042)p -= =

- ¥

Finally, we have:

( )1.5 60.25 1 0.99217 0.97671 0.0309q = - ¥ =

Solution 14

B From the definition of ( )s x , we have:

( )

{ }

40

0

40 0.050

400.050

2

(40) exp ( )

exp

exp 20

exp 40 20 20

x

x

s x dx

A e dx

Ax e

A e

mÏ ¸= -Ì ˝Ó ˛Ï ¸= - +Ì ˝Ó ˛Ï ¸È ˘= - +Ì ˝Î ˚Ó ˛

= - - +

Ú

Ú

Now (40) 0.85s = implies that:

240 20 20 ln0.85A e- - + =

Rearranging gives:

2ln0.85 20 20 3.1940

eA - += - = -



Solution 15

C If t xp has a hyperbolic form, then:

11 (1 )

x xt x

x x x

q ppt q p tq

-= =

- - +

for integer values of x and 0 1t£ £ .

So:

700.75 70

70 70

0.953 0.9640.75 0.953 (0.75 0.047)p

pp q

= = =+ + ¥

Solution 16

A If t xp has a hyperbolic form, then:

( ) ( )1 (1 )

x xs t x t

x x x

s t q s t qqs q p sq- +

- -= =

- - +

for integer values of x and 0 1t£ £ .

So:

700.25 70.75

70 70

0.25 0.25 0.047 0.01181

qq

p q¥= = =

+

Solution 17

D The required survival probability is:

53[51] 1

[51] 1

77,252 0.985378,402

lp

l++

= = =



Solution 18

E The probability 0.5 [51] 0.8q + can be expressed as:

0.5 [51] 0.8 0.5 [51] 0.8 0.2 [51] 0.8 0.3 [51] 11 1q p p p+ + + += - = - ¥

Now:

{ }

( )

0.20.2 [51] 0.8 0

0.2[51]

0.2

exp ([51] 0.8 )

exp 0.2 ([51]) using the constant force of mortality assumption

= p

78,40279,137

0.998136

p t dtm

m

+Ï ¸= - + +Ì ˝Ó ˛

= -

Ê ˆ= Á ˜Ë ¯

=

Ú

Similarly:

( )0.30.3

0.3 [51] 1 [51] 177,252 0.99557778,402

p p+ +Ê ˆ= = =Á ˜Ë ¯

Hence:

0.5 [51] 0.8 1 0.998136 0.995577 0.00628q + = - ¥ =

Solution 19

E The median future lifetime is the value of (25)m that satisfies the equation:

Pr( (25) (25)) 0.5T m> =

Now:

(25 (25)) / 70

25 / 70

(25) / 70

(25 (25))Pr( (25) (25))(25)

m

m

s mT ms

ee

e

- +

-

-

+> =

=

=



So the median future lifetime is:

(25) 70ln0.5 48.5m = - =

Solution 20

C The average number of years lived between the ages of 90 and 91 by lives that diebetween those ages is:

1900

1900

(90 )(90)

(90 )

t

t

t p t dta

p t dt

m

m

+=

+

ÚÚ

Since (90) 0.32m = and the force of mortality is constant between 90 and 91:

1 0.3201 0.320

0.32(90)

0.32

t

t

t e dta

e dt

−

−=∫∫

Integrating by parts gives:

1 10.32 0.3200

10.320

10.32 0.320

0.32

0.32 0.32

0.32

(90)

10.32

11 1

0.321

0.47

t t

t

t

t e e dta

e

e e

e

e e

e

- -

-

- -

-

- -

-

È ˘- +Î ˚=È ˘-Î ˚

È ˘- - Î ˚=

-

È ˘- - -Î ˚=-

=

Ú



Solution 21

D According to de Moivre’s law:

1( )xx

mw

=-

for 0 x w£ <

So:

{ }

0

0

exp ( )

1exp

exp ln( ) ln( )

tt x

t

p x s ds

dsx s

x t x

x tx

m

w

w w

ww

Ï ¸= - +Ì ˝Ó ˛

Ï ¸= -Ì ˝- -Ó ˛

= - - - -

- -=-

Ú

Ú

Since (80) 0.15m = , it follows that:

1 80 86.66670.15

w = + =

Hence:

20 6086.6667 60 20 0.2500

86.6667 60p - -= =

-



Solution 22

E :x nA can be expressed as:

11

:0

11

12

21 1

02

11 1

0

1: 1

(setting 1)

nk n

k x x k n xx nk

nk n

x k x x k n xkn

j nx j x x j n x

jn

j nx x j x x j n x

j

x x x n

A v p q v p

vq v p q v p

vq v p q v p j k

vq vp v p q v p

vq v p A

-+

+=

-+

+=-

++ + +

=-

++ + +

=

+ -

= +

= + +

= + + = -

= + +

= +

Â

Â

Â

Â

Statement I is incorrect since the factor of xvp has been omitted.

Now consider the RHS of statement II. This can be expanded as follows:

1 1|

01

1

0

:

k k nx n x n x k x x k k x x k n x

k k nn

k nk x x k n x

k

x n

A A E v p q v p q v p

v p q v p

A

• •+ +

+ += =-

++

=

- + = - +

= +

=

Â Â

Â

So statement II is correct.

Finally, we can also write:

11

:02

11 1

02

11

0

nk n

k x x k n xx nkn

k n nk x x k n x x n n x

kn

k nk x x k n x

k

A v p q v p

v p q v p q v p

v p q v p

-+

+=-

++ - + -

=-

++ -

=

= +

= + +

= +

Â

Â

Â

So statement III is also correct.



Solution 23

B The value of 110| 35:20A can be calculated as follows:

( )

( )

1 110| 10 3535:20 45:20

110 35 45:20

10 35 45 20 45 65

0.060.54318 0.20120 0.25634 0.43980ln1.06

0.0495

A E A

iE A

iE A E A

d

d

= ¥

ª ¥ ¥

= ¥ ¥ - ¥

= ¥ ¥ - ¥

=

Solution 24

D The actuarial present value of the benefit is:

( ) ( )

110| 50 50 10 50 6050:10

75,000 50,000 75,000 25,000

75 249.05 25 510.81 0.3691313,964.87

A A A E A+ = -

= ¥ - ¥ ¥=

Solution 25

D The present value of the benefit can be expressed as:

175,000 if 90 if 10

Kv KYK

+ÏÔ £= Ì≥ÔÓ

We can calculate var( )Y as follows:

( )( )

( )

21 12 250:10 50:10

22 2 20 250 10 50 60 50 10 50 60

2 10

22

var( ) 75,000

75,000

75 1,000 94.76 [(1.06) 0.51081 177.41]

75 249.05 [0.51081 369.13]

248,381,441.1 20,585,302.3227,796,13

Y A A

A v p A A E A

-

È ˘= -Í ˙

Î ˚È ˘= - - -Í ˙Î ˚

È ˘= ¥ - ¥ ¥Î ˚È ˘- - ¥Í ˙Î ˚

= -= 8.8



So the standard deviation of the present value of the benefit is 227,796,138.8 15,093= .

Solution 26

C The present value of the benefit can be expressed as:

1

0 if 9

50,000 if 10KK

Zv K+

£ÏÔ= Ì≥ÔÓ

The variance of Z is given by:

( )( )

( )

22 250 10| 5010|

22 10 210 50 60 10 50 60

22 10 2

var( ) 50,000

50,000

50 (1.06) 510.81 177.41 50 0.51081 369.13

37,625,499.9

Z A A

v E A E A

-

È ˘= -Í ˙Î ˚È ˘= -Í ˙Î ˚

È ˘= ¥ ¥ - ¥Î ˚=

So the standard deviation of Z is 37,625,499.9 6,134= .

Solution 27

B The variance of the present value is given by:

var( ) var( ) var( ) 2cov( , )Y Z Y Z Y Z+ = + +

From Solution 25 and Solution 26, we know that var( ) 227,769,138.8Y = andvar( ) 37,625,499.9Z = .

The covariance term is:

110| 5050:10

cov( , ) ( ) ( ) ( )

0 75,000 50,000

Y Z E YZ E Y E Z

A A

= -

= - ¥



Since:

( )( )

150 10 50 6050:10

75,000 75,000

75 249.05 0.51081 369.134,537.1

A A E A= -

= - ¥=

and:

10| 50 10 50 6050,000 50,000

50 0.51081 369.139,427.8

A E A=

= ¥ ¥=

we have:

cov( , ) 42,774,738.4Y Z = -

Hence:

var( ) 227,796,138.8 37,625,499.9 (2 42,774,738.4)179,872,161.9

Y Z+ = + - ¥=

So the standard deviation of Y Z+ is 179,872,161.9 13,412=

Solution 28

C The symbol 132:10A denotes the expected present value of a benefit of 1 payable to (32)

on survival to age 42. So:

1 6 4 6 410 3232:10

9,259,571(1.05) (1.06) (1.05) (1.06) 0.577849,471,591

A p- - - -= = ¥ =

Solution 29

A ( )xIA can be expressed in integral form as follows:

0( ) 1 ( )t

x t xIA t v p x t dtm•

= + +Í ˙Î ˚Ú



This can be manipulated as follows:

1

0

1

00

1

00

10

0

1

0

( ) ( 1) ( )

( 1) ( ) letting

( 1) by UDD

( 1)

1 ( 1)

1 ( 1)

( )

k tx t xk

k

k sk x s x k

k

k sk x x k

k

kk x x k

k

kk x x k

k

kk x x k

k

IA k v p x t dt

k v p v p x k s ds s t k

k v p q v ds

k v p q a

v k v p q

v k v p qv

i IA

m

m

d

d

d

• +

=•

+=•

+=•

+=

•

+=•

++

=

= + +

= + + + = -

+

= +

-= +

-= +

=

Â Ú

Â Ú

Â Ú

Â

Â

Â

x

So statement I is correct.

Statement II is incorrect since in II the benefit amount increases continuously throughtime.

Also:

1

0 11 1

10 01

10 01

10 0

( ) ( ) 1 ( )

( ) 2 ( 1) letting 1

( ) 2 ( 1 )

( ) ( 1 )

t tx t x t x

t rt x r x

t rt x x r x

t tt x x t x

IA v p x t dt t v p x t dt

v p x t dt r v p x r dr r t

v p x t dt v p r v p x r dr

v p x t dt v p v p x t dt

v p

m m

m m

m m

m m

•

• ++

•+

•+

= + + + +Í ˙Î ˚

= + + + + + = -Í ˙Î ˚

= + + + + +Í ˙Î ˚

= + + + +

+

Ú ÚÚ ÚÚ ÚÚ Ú

( )10

11 10

1 ( 1 )

( ) ( )

rx r x

tt x x x x

r v p x r dr

v p x t dt v p A IA

m

m

•+

+ +

+ + +Í ˙Î ˚

= + + +

ÚÚ

So statement III is incorrect since the 1x xv p A + term is missing.



Solution 30

A If the benefit were payable immediately on death, then the expected present value would

be 130:30

10,000A . However, as there is a two-month delay in the payment, the expected

present value is:

( )12 /12 2 / 12 3030 30 30 6030:30

2 / 12

30

10,000 10,000

0.0610,000 (1.06) 0.10248ln1.06

8,188,074(1.06) 0.369139,501,381

480.25

iv A v A v p Ad-

-

¥ -

Ê= ¥ ¥ ÁË

ˆ- ¥ ¥ ˜̄

=

Solution 31

C Since only the death benefit can be accelerated and we are assuming a uniformdistribution of deaths between integer ages:

( )

( )

1 145:20 45:20 45:20

45 20 45 65 20 45

0.06 0.20120 0.25634 0.43980 0.25634ln1.060.3474

iA A A

i A E A E

d

d

= +

= - +

= - ¥ +

=

Solution 32

E We can express (4)5| 35A as follows:

(4) (4)5| 5 35 5 35 4035 40 (4)

0.73873 1.02223 0.16132 0.12182iA E A E Ai

= = ¥ ¥ = ¥ ¥ =



Solution 33

D Working in $1,000s, the expected present value of the annuity is:

2 3 460 2 60 3 60 4 60

2 3

4

10 12 14 16 18

12 8,075,403 14 7,954,179 16 7,823,879101.05 8,188,074 8,188,074 8,188,0741.05 1.05

18 7,683,9798,188,0741.05

10 11.2713 12.3357 13.2066 13

v p v p v p v p+ + + +

= + × + × + ×

+ ×

= + + + + .8970

60.7106

$60,711

=

=

Solution 34

C To calculate the variance of the PV, we require the PV of the benefit and Pr( )K k= foreach 0,1, 2,3k = and Pr( 4)K ≥ . These are shown in the table below:

k Pr( )K k= PV benefit ( )2PV benefit

0 60 0.01376q = 10 100

1 60 61 0.01480p q = 10 12 21.4286v+ = 459.1837

2 2 60 62 0.01591p q = 210 12 14 34.1270v v+ + = 1,164.6510

3 3 60 63 0.01709p q = 2 310 12 14 16 47.9484v v v+ + + = 2,299.0477

4≥ 4 60 0.93844p = 2 3 410 12 14 16 18 62.7570v v v v+ + + + = 3,938.4448

Now:

2( ) (100 0.01376) (459.1837 0.01480) (1,164.6510 0.01591)

(2,299.0477 0.01709) (3,938.4448 0.93844)

3,761.965

E PV = ¥ + ¥ + ¥

+ ¥ + ¥

=



So:

2var( ) 3,761.965 60.7106 76.1896PV = - =

Hence, the standard deviation of the present value is 76.1896 8.729 $8,729= = .

Solution 35

E From the recursive formula, we have:

40 40 411a v p a= +

Rearranging this gives:

40 4041

40 40

1 13.300 1.04 13.9010.995

a aa

v p v p− ×= = = =

Solution 36

E If the rate of payment were 1 a year, then the variance would be:

( )22 260 602 2

0.17741 (0.36913) 12.844(0.06 /1.06)

A A

d

− −= =

Since the rate of payment is 10 units a year, the variance is

( )2260 60

2100 1284.4

A A

d

−× =

Solution 37

B The EPV of the death benefit is:

( )

( )

150 10 50 6050:10

25,000 25,000

0.0625,000 0.24905 (0.51081 0.36913)ln1.06

1,557.3

iA A E Aδ

≈ −

= − ×

=



The EPV of the guaranteed part of the annuity is:

510 50 5

1 (1.06)5,000 5,000 0.51081(0.06 /1.06)

11,404.1

E a−−= × ×

=

The EPV of the life annuity is:

10 50 5 60 655,000 5,000 0.51081 0.68756 9.896917,379.6

E E a× × = × × ×=

So the EPV of the total benefit is:

1,557.3 11,404.1 17,379.6 30,341+ + =

Solution 38

D We can write:

110 3030:1030:10

A A E= −


( )30:10 30:10

30 10 30 40

10

1

1

0.04 0.9871 16.250 15.1251.04 1.04

0.76289

A da

d a E a

= −

= − −

= − − ×

=

So:

11030:10

0.9870.76289 0.09611.04

A = − =



Solution 39

C The EPV of the contributions per original member is (12)50:15

2,400a . So the share per

survivor is (12)50:15

15 50

2,400a

E.

We can calculate 15 50E as follows:

15 50 10 50 5 60 0.51081 0.68756 0.35121E E E= × = × =

The annuity factor can be expressed as:

(12) (12) (12)15 5050 6550:15

a a E a= −

Now:

(12)5050 (12) (12) 1.00028 13.2668 0.46812 12.8024a aα β= − = × − =

and:

(12)6565 (12) (12) 1.00028 9.8969 0.46812 9.4316a aα β= − = × − =

So:

( )(12)50:15

12.8024 0.35121 9.4316 9.4899a = − × =

It follows that the share per survivor is:

2,400 9.4899 64,8500.35121

× =

Solution 40

E The force of mortality at age x for this life is given by:

0.00947843x xµ µ∗ = −

So the probability of surviving from age x to age x n+ is:

0.009478430

expn n

n x x t n xp dt p eµ∗ ∗+

= − = ∫



It follows that:

0.00947843 ( )n n n nn x n x n xv p v p e v p∗ ∗= =

where:

0.00947843 10.952380653 0.051

v v e ii

∗ ∗∗

= = = ⇒ =+

So:

68:3 68:32

68 2 68

2

@6% @5%

11 6,823,367 1 6,616,1551

1.05 7,018,432 7,018,4321.052.78095

a a

v p v p

∗ =

= + +

= + +

=

The EPV of the annuity is therefore:

68:310,000 27,810a∗ =

Solution 41

D The equation of value is:

( )+ =13 47 6013 100,000M a E a

The factors in this expression are:

13 13

131 1 1.06 9.1157

ln1.06vaδ

−− −= = =

13 1313 47 13 47

8,188,0741.06 0.422419,088,049

E v p −= = × =

60 60( ) ( ) (1.00028 11.1454) 0.50985 10.6387a aα β= ∞ − ∞ = × − =

So:

( )9.1157 4.4939 100,000 7,348M M+ = ⇒ =



Solution 42

C Before the survival probability is increased, we have:

60 60 6111 1 0.98 11.60 11.931

1.04a v p a Ê ˆ= + = + ¥ ¥ =Á ˜Ë ¯

and:

2 6061

60

0.93 0.948980.98

pp

p= = =

Increasing the value of 61p by 0.02 and using * to denote the adjusted functions, gives:

61 0.96898p* =

and:

61 61 62620.968981 1

1.04a v p a a* *= + = +

However:

61 61 62 62 620.948981 1 11.60 11.617

1.04a v p a a a= + = + = fi =

So:

610.968981 11.617 11.823

1.04a* = + ¥ =

and:

60 60 610.981 1 11.823 12.1411.04

a v p a* *= + = + ¥ =

Hence the expected present value of the annuity is increased by 12.141 11.931 0.21- =



Solution 43

A We can calculate the EPV of the annuity using life table functions:

2 3 4 560 2 60 3 60 4 60 5 6060:5a v p v p v p v p v p= + + + +

To calculate the survival probabilities, we can construct a life table using the rule55 0.01t q t= for 1 10t≤ ≤ . Since we are given information relating to age 55, we will take

this to be the first age in the table.

According to the rule, the probability of dying between the ages of 55 and 56 is 0.01. So ifwe use a radix of 100, we will have 99 survivors at age 56.

If we then let 2t = , we find that the probability of dying between the ages of 55 and 57 is0.02. So out of the 100 starters at age 55, we have 98 survivors at age 57.

Continuing in this way, we obtain the following life table:

x xl xd

55 100 156 99 157 98 1

x 155 x− 1

65 90

From the table, we see that 6160

60

9495

lp

l= = , 62

2 6060

9395

lp

l= = , and so on.

Hence:

60:5 2 3 4 594 93 92 91 90

1.05 95 1.05 95 1.05 95 1.05 95 1.05 95

4.1972

a = + + + +× × × × ×

=



Solution 44

A The actuarial present value of the annuity is:

(2) (2) 170 70 2

170 2

12

10,000 10,000

10,000 (2) (2)

10,000 1.00021 8.5693 0.25739

78,137

a a

aa b

È ˘= -Í ˙Î ˚È ˘= - -Î ˚È ˘= ¥ - -Î ˚

=

Solution 45

E if Mark dies before age 60, then a benefit of $40,000 will be paid at the end of the year ofhis death. The present value of this benefit would be less than $40,000.

If Mark dies between the ages of 65 and 66, then he will receive a benefit of $15,000 atage 65. The present value of this benefit would be less than $40,000.

If Mark dies between the ages of 65 n+ and 65 1n+ + , then he will receive a benefit of$15,000 at age 65, followed by annuity payments of $10,000 at ages 66, 67, …, 65 n+ . So the present value of the benefit received would be:

20 2015,000 10,000 nv v a+

To calculate the probability that this PV is less than $40,000, we need to find the smallestvalue of n such that:

20 2015,000 10,000 40,000nv v a+ ≥

Rearranging this inequality gives:

2040,000(1.06) 15,000 11.3285410,000na

-≥ =

Since 1 n

nvai-= , it follows that:

1 (11.32854 0.06) 0.32029nv £ - ¥ =

Taking logs, we obtain:

ln1.06 ln0.32029n- £



So:

19.54n ≥

Hence the present value of the benefit will exceed $40,000 if and only if Mark survives toage 85. The probability that the present value of the benefit is less than $40,000 istherefore:

8540 45

45

2,358,2461 1 0.74279,164,051

lq

l= - = - =

Solution 46

C The present value of the term insurance can be written as:

1 128:30 28:30 28:30A A A= -


28:3028:300.051 1 14.197 0.323951.05

A da= - = - ¥ =

In addition:

1 30 3030 2828:30

(1.05) 0.885 0.20477A v p -= = ¥ =

So:

128:30

0.32395 0.20477 0.11918A = - =

Solution 47

C The present value of this endowment insurance is:

2 3[51] [51] [51] 1 2 [51][51]:3

[51] [51] 1 [51] 1 53 532 3[51] [51] [51]

2 3

1 1 11.05 1.05 1.05

1 79,137 78,402 1 78,402 77,252 1 77,2521.05 79,137 79,137 79,1371.05 1.05

A v q v p q v p

l l l l ll l l

+

+ +

= + +

Ê ˆ Ê ˆ Ê ˆ- -= + +Á ˜ Á ˜ Á ˜

Ë ¯ Ë ¯ Ë ¯

- -Ê ˆ Ê ˆ= + +Á ˜ Á ˜Ë ¯ Ë ¯0.86529

Ê ˆÁ ˜Ë ¯

=



Solution 48

C The present value of the annuity is:

252 2 5252:3

2

1

1 77,252 1 75,57811.05 78,839 78,8391.05

2.8027

a v p v p= + +

Ê ˆÊ ˆ= + ¥ + ¥Á ˜ Á ˜Ë ¯ Ë ¯=

Solution 49

D The expected present value of the endowment insurance is:

( )*= +1 33 5050:10 50:3 53:7

20,000 20,000A A v p A

where * denotes non-standard mortality.

We can calculate the first term as follows:

1 2 350 50 51 2 50 5250:3

2 31 1 10.00591985 0.00638405 0.00688681

1.04 1.04 1.040.01772

A vq v p q v p q= + +

Ê ˆ Ê ˆÊ ˆ= ¥ + ¥ + ¥Á ˜ Á ˜ Á ˜Ë ¯ Ë ¯ Ë ¯=

(To reduce rounding error, the probabilities in the above equation have been calculatedusing the tabulated values of xl .)

By premium conversion, we can write:

53:7 53:71A da* *= -

In the annuity term, a constant addition to the force of mortality can be treated as a

constant addition to the force of interest. We can therefore value 53:7a * using standard

mortality and force of interest *d , where:

* 0.019048194 ln1.04 0.019048194 0.058268907d d= + = + =

It then follows that:

** 1 0.06i ed= - =



So:

53:7 53:77

53 7 53 60

7

@ 4% @6%

1 8,188,07412.6896 11.14548,779,1281.06

5.7763

a a

a v p a

* =

= -

= - ¥ ¥

=

It follows that:

53:70.041 5.7763 0.777831.04

A * = - ¥ =

Substituting these values into the equation for the endowment insurance gives:

50:10 31 8,779,12820,000 20,000 0.01772 0.77783

8,950,9011.0413,919

AÊ ˆ

= + ¥ ¥Á ˜Ë ¯=

Solution 50

A The total fund at the end of 10 years is:

( ) [ ][ ]

10 1045 10 45 5545:10

10

500 1,000 1.06 500,000 1.06

500,000 1.06 14.1121 (0.52652 12.2758)6,848,779.84

a a E a¥ = ¥ -

= ¥ - ¥

=

The expected number of survivors will be:

10 458,640,861500 500 471.459,164,051

p = ¥ =

The expected share per survivor can then be calculated from the equation:

12471.45 (500 471.45) 6,848,779.84S S+ - ¥ =

This simplifies to give $14,100S = .

introduction to actuarial mathematics

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