Solutions Manual for Actuarial Mathematics for Life Contingent Risks (International Series on Actuar

SOLUTION MANUAL

Solutions Manual for Actuarial Mathematics for Life Contingent Risks This must-have manual provides detailed solutions to all of the 300 exercises in Dickson, Hardy and Waters’ Actuarial Mathematics for Life Contingent Risks, Third Edition.

This groundbreaking text on the modern mathematics of life insurance is required reading for the Society of Actuaries’ (SOA) LTAM Exam. The new edition treats a wide range of newer insurance contracts such as critical illness and long-term care insurance; pension valuation material has been expanded; and two new chapters have been added on developing models from mortality data and on changing mortality.

Beyond professional examinations, the textbook and solutions manual offer readers the opportunity to develop insight and understanding through guided hands-on work, and also offer practical advice for solving problems using straightforward, intuitive numerical methods. Companion Excel spreadsheets illustrating these techniques are available for free download. david c. m. dickson holds a PhD in Actuarial Science from Heriot-Watt University and is a Fellow of the Institute and Faculty of Actuaries and the Institute of Actuaries of Australia. David lectured for seven years at Heriot-Watt University before moving to the University of Melbourne in 1993. In 2000 he was appointed to the Chair in Actuarial Studies in Melbourne. He was Head of the Department of Economics from 2016 to 2018. He has twice been awarded the HM Jackson Prize of the Institute of Actuaries of Australia, most recently for his book Insurance Risk and Ruin. mary r. hardy is Professor of Actuarial Science at the University of Waterloo. She earned her PhD in Actuarial Science from Heriot-Watt University, where she lectured for 11 years before moving to the University of Waterloo in 1997. She is a Fellow of the Institute and Faculty of Actuaries, and of the Society of Actuaries. In 2007 she was awarded the Chartered Enterprise Risk Analyst designation of the Society of Actuaries, through their thought-leader recognition program. In 2013 she was awarded the Finlaison Medal of the Institute and Faculty of Actuaries for services to the actuarial profession, in research, teaching and governance. howard r. waters is Professor Emeritus at the Department of Actuarial Mathematics and Statistics at Heriot-Watt University. He holds a DPhil in mathematics from Oxford University, and worked as a consulting actuary for several years before joining HeriotWatt University. He is a Fellow of the Institute and Faculty of Actuaries. He was awarded the Finlaison Medal of the Institute of Actuaries in 2006 for services to actuarial research and education.

I N T E R NAT I O NA L S E R I E S O N AC T UA R I A L S C I E N C E Editorial Board Christopher Daykin (Independent Consultant and Actuary) Angus Macdonald (Heriot-Watt University) The International Series on Actuarial Science, published by Cambridge University Press in conjunction with the Institute and Faculty of Actuaries, contains textbooks for students taking courses in or related to actuarial science, as well as more advanced works designed for continuing professional development or for describing and synthesizing research. The series is a vehicle for publishing books that reflect changes and developments in the curriculum, that encourage the introduction of courses on actuarial science in universities, and that show how actuarial science can be used in all areas where there is long-term financial risk.

A complete list of books in the series can be found at www.cambridge.org/isas. Recent titles include the following: Actuarial Mathematics for Life Contingent Risks (3rd Edition) David C. M. Dickson, Mary R. Hardy & Howard R. Waters Modelling Mortality with Actuarial Applications Angus S. Macdonald, Stephen J. Richards & Iain D. Currie Claims Reserving in General Insurance David Hindley Financial Enterprise Risk Management (2nd Edition) Paul Sweeting Insurance Risk and Ruin (2nd Edition) David C.M. Dickson Predictive Modeling Applications in Actuarial Science, Volume 2: Case Studies in Insurance Edited by Edward W. Frees, Richard A. Derrig & Glenn Meyers Predictive Modeling Applications in Actuarial Science, Volume 1: Predictive Modeling Techniques Edited by Edward W. Frees, Richard A. Derrig & Glenn Meyers

SOLUTIONS MANUAL FOR ACTUARIAL MATHEMATICS FOR LIFE CONTINGENT RISKS third edition DAVID C. M. DICKSON University of Melbourne

MARY R. HARDY University of Waterloo, Ontario

HOWARD R. WATERS Heriot-Watt University, Edinburgh

University Printing House, Cambridge CB2 8BS, United Kingdom One Liberty Plaza, 20th Floor, New York, NY 10006, USA 477 Williamstown Road, Port Melbourne, VIC 3207, Australia 314–321, 3rd Floor, Plot 3, Splendor Forum, Jasola District Centre, New Delhi – 110025, India 79 Anson Road, #06–04/06, Singapore 079906 Cambridge University Press is part of the University of Cambridge.

It furthers the University’s mission by disseminating knowledge in the pursuit of education, learning, and research at the highest international levels of excellence. www.cambridge.org Information on this title: www.cambridge.org/9781108747615 DOI: 10.1017/9781108779432 © David C.M. Dickson, Mary R. Hardy and Howard R. Waters 2012, 2013, 2020 This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

First published 2012 Second Edition 2013 Third Edition 2020 Reprinted 2021 Printed in Great Britain by Ashford Colour Press Ltd. A catalogue record for this publication is available from the British Library.

ISBN 978-1-108-74761-5 Paperback Additional resources for this publication at www.cambridge.org/9781108747615.

Cambridge University Press has no responsibility for the persistence or accuracy of URLs for external or third-party internet websites referred to in this publication and does not guarantee that any content on such websites is, or will remain, accurate or appropriate.

Contents

page vii 1 11 24 38 62 77 98 126 155 168 183 199 210 239 248 260 268 278 289

Preface Solutions for Chapter 1 Solutions for Chapter 2 Solutions for Chapter 3 Solutions for Chapter 4 Solutions for Chapter 5 Solutions for Chapter 6 Solutions for Chapter 7 Solutions for Chapter 8 Solutions for Chapter 9 Solutions for Chapter 10 Solutions for Chapter 11 Solutions for Chapter 12 Solutions for Chapter 13 Solutions for Chapter 14 Solutions for Chapter 15 Solutions for Chapter 16 Solutions for Chapter 17 Solutions for Chapter 18 Solutions for Chapter 19

v

Preface

This manual presents solutions to all exercises from Actuarial Mathematics for Life Contingent Risks, third edition (AMLCR),

by David C. M. Dickson, Mary R. Hardy, Howard R. Waters, Cambridge University Press, 2020 ISBN 9781108478083 It should be read in conjunction with the spreadsheets posted at the website www.cambridge.org\9781108747615 which contain details of the calculations required. However, readers are encouraged to construct their own spreadsheets before looking at the authors’ approach. Spreadsheet solutions are provided for all the Excel-based exercises in AMLCR. In some cases the answers in the manual will differ from answers calculated using tables such as those provided in Appendix D of AMLCR. The differences arise from rounding errors. The numbers given in this manual are calculated directly, without using rounded table values, unless otherwise indicated.

From time to time, updates to www.cambridge.org\9781108747615.

vii

this

manual

may

appear

at

Solutions for Chapter 1

1.1 The principle of charging in advance for life insurance is to eliminate the potential for policyholders to benefit from short-term life insurance cover without paying for it. Suppose premiums were payable at the end of the policy year.

A life could sign up for the insurance, and lapse the contract at the end of the year. The life would have benefited from free insurance cover for that year.

In addition, life insurance involves significant acquisition expenses. The first premium is used to meet some or all of these expenses.

Background note: The fact that the insurance for a policyholder did not result in a claim does not make it free to the insurer. The insurer’s view is of a portfolio of contracts. Suppose 100 people buy term life insurance for one year, with a sum insured of $1000, at a premium of $11 each. The insurer expects a mortality rate of 1%, which means that, on average, one life out of the 100 dies. If all the policyholders pay their first year’s premiums in advance, and one life dies, then the insurer receives $1100 (plus some interest) and pays out $1000. On the other hand, if premiums were due at the year end, it is possible that many of the 99 expected to survive might decide not to pay. It would be difficult and expensive for the insurer to pursue payment. The policyholders have benefited collectively from the insurance and the insurer has not been appropriately compensated.

1.2 The purpose of whole life insurance is to provide tax efficient benefits on the death of the policyholder. However, the ability of the policyholder to maintain their policies, by paying the premiums on time, may decline with age, firstly because the policyholder’s income may decline, so that finding the funds to 1

2

Solutions for Chapter 1 pay premiums becomes harder, and secondly because at older ages individuals become less able to manage their financial affairs, due to age-related cognitive decline. So some insurers design products that obviate the need for policyholders to actively maintain their policies in old age by removing the need to pay premiums at older ages.

1.3 An insurance policy is lapse-supported if excess funds from lapsed contracts are used to subsidize the remaining policies. In term insurance, when a policyholder lapses there is no cash value payable, so there may be excess funds available from lapsing policies. However, for most common forms of term insurance, premiums are relatively low, and the funds released from lapsing policies are also small. Furthermore, because the premiums are low, and term insurance is typically bought to provide for the policyholder’s family in the event of the policyholder’s death, we do not expect a very significant number of these policies to lapse.

We also note that lapses in the early stages of the contract generate losses, not profits, as the insurer needs several years of premiums (typically three to seven) to recoup the initial expenses associated with writing the policy.

So pricing allowing for lapses may not significantly reduce the premium, and hence, although term insurance can be written and priced as a lapse-supported product, in practice the impact would be so small that it is typically not.

Note though that some niche term insurance policies sold for very long terms are lapse-supported. In Canada, some insurers sold term insurance policies terminating at age 100. These policies were essentially whole life policies without cash values, and were specifically designed to be lapse-supported.

1.4 Full life care provides for level fees (other than inflationary increases) throughout the resident’s future lifetime in a CCRC, for residents who enter into Independent Living Units. The CCRC’s costs associated with the care of each resident are expected to rise steeply as they progress through the different levels of care, so the full life care fee structure involves higher fees than costs in the early years, and lower fees than costs in later years.

Modified life care allows fees to increase when the resident moves between stages, but fees are level within each stage. The increase in fees applied when the resident moves from a lower level of care to a higher level is less than the increase in costs, so that the increase is predictable and manageable for the resident.

Solutions for Chapter 1

3

The fee-for-service plan sets fees approximately equal to the CCRC’s costs at each stage of care. In the early years, a life entering an Independent Living Unit would be expected to have the lowest level of fees, and as they move through to higher levels of care, the fees will rise substantially.

Hence, the initial fees for a life entering an Independent Living Unit will be highest under the full life care structure, and lowest under the fee-for-service structure, with modified life care falling in the middle.

1.5 (a) The insurer will calculate the premium for a term or whole life insurance policy assuming that the policyholder is in relatively good health; otherwise, if the insurer assumed that all purchasers were unhealthy, the cost of insurance would be prohibitive to those customers who are healthy. The assumption then is that claims will be relatively rare in the first few years of insurance, especially since most policies are sold to lives in their 30s and 40s.

This means that the price is too low for a life who is very unwell, for whom the risk of a claim shortly after purchase might be 10 or 100 times greater than for a healthy life. The insurer therefore needs evidence that the purchaser is in good health, to avoid the risk that insurance is bought too cheaply by lives who have a much higher probability of a claim.

The objective of underwriting is to produce a relatively homogeneous insured population when policies are issued. The risk that the policyholder purchases the insurance because they are aware that their individual risk is greater than that of the insured population used to calculate the premium is an example of adverse selection risk. Underwriting is a way of reducing the impact of adverse selection for life insurance.

Adverse selection for an annuity purchaser works in the other direction – a life might buy an annuity if they considered their mortality was lighter than the general population. But, since adverse selection is likely to affect all lives purchasing annuities, more or less, it does not generate heterogeneity, and the impact can be managed by assuming lower overall mortality rates for annuitants.

In addition, the difference in the net cost to the insurer arising from adverse selection will be smaller compared with the term insurance example. (b) The insurer will be more rigorous with underwriting for term insurance

4

Solutions for Chapter 1 than for whole life insurance because the potential financial consequence of adverse selection is greater. Note that the insurer expects few claims to arise from the term insurance portfolio. Premiums are small, relative to the death benefit, because the probability of payment of the death benefit is assumed to be small. For whole life insurance, premiums are substantially larger as payment of the death benefit is a certain event (ignoring surrenders). The only uncertainty is the timing of the benefit payment.

The main risk to the insurer is that a life with a very high mortality risk, much higher than the assumed insured population, purchases life insurance.

It is likely in this case that the life will pay very few premiums, and the policy will involve a large death benefit payout with very little premium income. Since term insurance has much lower premiums for a given sum insured than whole life insurance, it is likely that such a policyholder would choose term insurance. Hence, the risk of adverse selection is greater for term insurance than for whole life insurance, and underwriting is used to reduce the adverse selection risk.

1.6 (a) Without term insurance, the homeowner’s dependents may struggle to meet mortgage payments in the event of the homeowner’s death. The lending company wishes to reduce as far as possible the risk of having to foreclose on the loan. Foreclosure is expensive for the lender and creates hardship for the homeowner’s family at the worst possible time. Term insurance is used to pay off the mortgage balance in the event of the homeowner’s death, thus avoiding the foreclosure risk for both the lender and the homeowner’s family. (b) If the homeowner is paying regular instalments of capital and interest to pay off the mortgage, then the term insurance sum insured will decrease as the loan outstanding decreases. The reduction in loan outstanding is slow in the early years of, say, a 25-year mortgage, but speeds up later. The reduction in the term insurance sum insured is therefore not linear. Different loan provisions, including interest-only loan periods, cliff-edge repayment schedules (where the interest is very low for some period and then increases substantially), fixed or variable interest rates, and fixed or variable repayment instalments will all affect the sum insured. (c) In Section 1.5.2 it is noted that around 2%–3% of applicants for insurance are considered to be too high risk. If these lives are, in consequence, unable

Solutions for Chapter 1

5

to purchase property, then that is a social cost for these lives that may not be acceptable.

1.7 In with-profit whole life insurance, the insurer invests the premiums, and excess investment returns over the minimum required to fund the original benefits are shared between the policyholders and the insurer.

With a cash bonus, the policyholder’s share of profits can be paid out in cash, similar to a dividend on shares. In this case, the investments need to be realized (i.e. assets sold for cash). The payout is immediate.

With a reversionary bonus, the policyholder’s share of profits is used to increase the sum insured. The assets can remain in the capital markets until the sum insured is due.

Cash Bonus System – Insurer Perspective Advantages • Bonuses are transparent and easy to explain to policyholders.

• It does not involve maintenance of records of payouts and does not impact schedules for surrender values.

• The prospect of cash bonuses may persuade policyholders to continue with their policies rather than surrender.

Disadvantages • It creates a liquidity risk – that assets need to be sold to meet bonus expectations, possibly at unfavourable times.

• Investment proceeds are volatile; volatility in cash bonuses may be difficult to explain to policyholders. There may be a temptation to over-distribute in an attempt to smooth, that could cause long-term losses.

• There may be problems determining equitable payouts, resulting in possible policyholder grievances.

Cash Bonus System – Policyholder Perspective Advantages • Cash is immediate and it is easy to understand the distribution.

6

Solutions for Chapter 1 Disadvantages • May not be tax efficient.

• The risks to the insurer may lead to under-distribution to avoid risk.

• Possible volatility of bonuses.

Reversionary Bonus System – Insurer Perspective Advantages • Assets remain invested as long as a policy is in force, reducing liquidity risk.

• Bonuses appear larger as they are generally delayed many years.

• Bonuses may not be paid in full if a policy is surrendered subsequently, allowing higher rates of bonus to be declared for remaining policyholders.

• Over-distribution can be mitigated with lower bonuses between the declaration year and the claim event.

Disadvantages • More complex to value, to keep records.

• Policyholders may not understand the approach, and there may be resentment (e.g. on surrender).

• Difficult to determine an equitable distribution.

• Easy to over-declare, as profits are based on asset values which may subsequently decrease.

• It is difficult to reduce bonus rates, even when justified. This may lead to loss of new and existing business.

Reversionary Bonus System – Policyholder Perspective Advantages • It may be tax efficient to receive profit share with sum insured.

• The system allows more investment freedom for the insurer, with higher upside potential for the policyholder.

Disadvantages • Difficult to understand, especially ‘super-compound’ systems.

7

Solutions for Chapter 1 • Possible loss of profit share on surrender.

• Opaque system of distribution. It is difficult to compare how different companies perform.

1.8 Under a simple reversionary bonus of 5% per year, the sum insured in year t, for t = 1, 2, . . . , 5, is 50 000 (1 + 0.05(t − 1)).

Under a compound reversionary bonus of 5% per year, the sum insured in year t, for t = 1, 2, . . . , 5, is 50 000 (1.05t−1 ).

Consider now the super-compound reversionary bonus and let St denote the total sum insured in the tth year for t = 1, 2, . . . , 5, with S1 = 50 000. The sum insured in year t + 1 will consist of the sum insured in year t (St ), the bonus on the original sum insured (0.05S1 ) and the bonus on previously declared bonus (0.1(St − S1 )). Thus St+1 = St + 0.05S1 + 0.1(St − S1 ) = 1.1St − 0.05S1 , and so we can calculate the total sum insured recursively in this case. The table below shows rounded values of the different sums insured.

Year

Simple

Compound

Super-compound

1 2 3 4 5

50 000 52 500 55 000 57 500 60 000

50 000 52 500 55 125 57 881 60 775

50 000 52 500 55 250 58 275 61 603

Note that in each case the sum insured is 52 500 in year 2 as the bonus at the end of year 1 is 5% of the sum insured in year 1 under each bonus structure, but after that the different bonus structures yield different results, with the simple bonus giving the lowest sums insured, and the super-compound giving the highest.

1.9 The first period of sickness is from time 1.00 to time 1.25, and the second is from time 2.00 to time 3.50. These are separate periods of sickness because the off period is six months, and the time between these periods of sickness is nine months.

8

Solutions for Chapter 1 Because of the one-year waiting period, no sickness benefit is payable for the period of sickness from time 1.00 to time 1.25.

The third period of sickness commences 0.25 years (three months) after the end of the second period of sickness, so the third period of sickness is treated as a continuation of the second. So, allowing for the one-year waiting period which applies from time 2.00 to time 3.00, sickness benefit is payable from time 3.00 to time 3.50, and this benefit payment continues throughout the entire third period of sickness (for a total of 0.50 + 4.25 = 4.75 years, so the five-year term is not attained).

As the final period of sickness commences 0.75 years after the third period of sickness ended, this is treated as a new period of sickness (due to the six-month off period) to which the one-year waiting period applies, meaning that benefit is payable from time 9.75 to time 10.

The following table summarises the payments.

Time from inception

Benefit payments

3.00–3.50 3.75–8.00 9.75–10.00

6 months at $2000 per month 51 months at $2000 per month 3 months at $2000 per month

1.10 For a comprehensive answer, we need to understand Andrew’s age, health and family responsibilities and support. The answers for an average 65-year-old retiree in good health would be different from those for a 50-year-old retiree in poor health. Also, we should consider the impact of governmental benefits (old age pension, social security, health costs), and any potential support from family in the event that he faces financial ruin.

In the absence of more detailed information, we assume that Andrew is a person in average health at an average retirement age of, say, 65. We also assume that the $500 000 represents the capital on which he wishes to live reasonably comfortably for the remainder of his life. We also ignore tax issues, though these are likely to be very significant in this kind of decision in practice.

Consider the risks Andrew faces at retirement.

(1) Outliving his assets – this is the risk that at some point the funds are all spent and Andrew must live on whatever government benefit or family support that might be available.

Solutions for Chapter 1

9

(2) Inflation risk – that is, that his standard of living is gradually eroded by increases in the cost of living that are not matched by increases in his income.

(3) Catastrophe costs – this is the risk that a large liability arises and Andrew does not have the assets (or cannot access the assets) to meet the costs.

Examples might include the cost of health care for Andrew or a dependent (where health care is not freely available); catastrophic uninsured liability; cost of long-term care in older age.

Andrew may also have some ‘wants’ – for example (1) Bequest – Andrew may want to leave some assets to dependents if possible.

(2) Flexible spending – Andrew may want the freedom of full access to all his capital at all times.

We now consider the options listed in the question in light of the risks and potential ‘wants’ listed. (a) With a level life annuity, Andrew is assured of income for his whole life, and eliminates the risk of outliving his assets. However, he retains the inflation risk, and he may not have sufficient assets to meet catastrophe costs.

If he uses all his capital for an annuity, there will be no bequest funds available on his death, and no flexibility in spending during his lifetime. (b) As in (a), Andrew will not outlive his assets, and this option also covers inflation risk to some extent. There may be some residual inflation risk, as the cost of living increases that Andrew is exposed to may differ from the inflation adjustments applied to his annuity. In order to purchase the cost of living cover, Andrew will receive a significantly lower starting annual payment than under option (a). All other issues are similar to those under option (a). (c) A 20-year annuity-certain will offer a similar or slightly higher benefit to a life annuity for a 65-year-old man in average health. Andrew’s life expectancy might be around 18 years, so on average the annuity will be sufficient to give Andrew a life income and allow a small bequest. An annuitycertain can be reasonably easily converted to cash in the event of a catastrophe or a change in circumstances. However, there is a significant risk that Andrew will live more than 20 years, and it will be difficult to manage the dramatic change in income at such an advanced age.

10

Solutions for Chapter 1 (d) Investing the capital and living off the interest would involve much risk.

The interest income will be highly variable, and will be insufficient to live on in some years. If Andrew invests the capital in safe, stable long-term bonds, he might make only 2%–3% after expenses (or less, this figure has been highly variable over the last 20 years) which would be insufficient if it is his only income. There is also reinvestment risk, as he could live longer than the longest income he could lock-in in the market, and there will be counterparty risk (that is, the risk that the borrower will default on the interest and capital) if his investment is not in solid risk-free assets.

If Andrew needs a higher income, he will have to take more risk. For example, he might invest in corporate bonds with counterparty risk, or he might put some of his capital in stocks, which have upside potential but downside risk. Using riskier investments would increase the volatility of his income and threaten his capital. If he invests heavily in shares, he may see negative returns in some years. This strategy just might not be sustainable.

Income would also not be inflation hedged, in general.

On the other hand, the capital would be accessible in the event of a catastrophe or for flexible spending (although that would raise the risk of outliving assets). This system would allow for a significant bequest, assuming that Andrew managed to live on the investment, but at the expense of income level and stability for Andrew. Also, Andrew would have the added complication of managing a portfolio of assets, or paying someone to manage them for him. On the other hand, purchasing an annuity involves substantial hidden expenses that would not be incurred under this option. (e) $40 000 is 8% of the capital. If this rate is higher than the interest rate achievable on capital, then Andrew will be drawing down the capital and risks outliving his assets. The income is not inflation hedged, but the system does allow spending flexibility. Other issues are as for option (d).

Solutions for Chapter 2

2.1 (a) px+3 = 1 − qx+3 = 0.98. (b) 2 px = px px+1 = 0.99 × 0.985 = 0.97515. (c) As 3 px+1 = 2 px+1 px+3 , we have 2 px+1 = 0.95/0.98 = 0.96939. (d) 3 px = px 3 px+1 /px+3 = 0.95969. (e) 1 |2 qx = px (1 − 2 px+1 ) = 0.03031.

2.2 We have ex =

∞

t px = px +

t=1

∞

t px

t=2

= px + px ⎛ ⎜⎜ = px ⎜⎜⎜⎝1 +

∞

t=2 ∞

t−1 px+1

⎞ ⎟⎟⎟ ⎟⎠ t px+1 ⎟

t=1

= px (1 + ex+1 ).

So px = ex /(1 + ex+1 ) and hence 3 p60 = p60 p61 p62 =

e61 e62 e60 = 0.93834.

1 + e61 1 + e62 1 + e63

2.3 (a) From Example 2.3, 10 p40 = exp

−B 40 10 c c −1 log c 11

= 0.995078.

12

Solutions for Chapter 2 (b) Noting that t p40 = 1 − F40 (t), we have d t p40 = − f 40 (t) = −t p40 μ40+t . dt Evaluating this at t = 10 gives −0.000567.

2.4 The one-year survival probability is ⎧ ⎫ ⎪ ⎪ 1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ (0.002 + 0.001t) dt px = exp ⎪ − ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ 0

= exp{−0.002 − 0.0005} = 0.99750, giving qx = 0.00250.

2.5 We have ex:n = E[min(Kx , n)] n−1 ∞

k Pr[Kx = k] + n Pr[Kx = k] = k=0

=

n−1

k=n

k ( k |qx ) + n Pr[Kx ≥ n]

k=0

=

n−1

k ( k px − k+1 px ) + n n px

k=0

= ( 1 px − 2 px ) + 2( 2 px − 3 px ) + · · · + (n − 1)( n−1 px − n px ) + n n px = 1 px + 2 px + · · · + n px n

= k px . k=1

2.6 (a) G(x) can be written as G(x) =

(90 − x)(x + 200) .

18 000

At the limiting age, ω, we have G(ω) = 0, and since x ≥ 0, the limiting age is ω = 90.

13

Solutions for Chapter 2

A more complete statement of the resulting survival function S0 (x) then is ⎧ ⎪ 18 000 − 110x − x2 ⎪ ⎪ ⎪ ⎨ 18 000 S0 (x) = ⎪ ⎪ ⎪ ⎪ ⎩ 0

for 0 ≤ x < 90, for x ≥ 90.

(b) First, we have S0 (0) = 1. Next, we see that S0 (x) = 0 for all x ≥ 90. Third, the derivative of S0 (x) is ⎧ −110 − 2x ⎪ ⎪ ⎪ for 0 < x < 90, ⎪ ⎨ 18 000 ⎪ ⎪ ⎪ ⎪ ⎩ 0 for x > 90, which is negative for 0 < x < ω, indicating that the survival function decreases from age 0 to the limiting age of 90.

Hence all three conditions for a survival function are satisfied. (c) S0 (20)/S0 (0) = 0.8556. (d) The survival function for t < 70 is S20 (t) =

S0 (20 + t) S0 (20)

=

18 000 − 110 (20 + t) − (20 + t)2 18 000 − 110 (20) − 202

=

15 400 − 150 t − t 2 15 400

=1−

t2 3t − .

308 15 400

(e) (S0 (30) − S0 (40))/S0 (20) = 0.1169. (f) μx = −S0 (x)/S0 (x). Using part (b), for x < 90 we obtain

μx = = so that μ50 = 0.021.

110 + 2x 18 000

18 000 18 000 − 110x − x2

110 + 2x 18 000 − 110x − x2

14

Solutions for Chapter 2

2.7 (a) The survival function Sx is given by S0 (x + t) S0 (x) −λ (x+t) e = −λ x e −λt =e .

Sx (t) =

(b) μx = −

d d d log S0 (x) = − log e−λ x = λx = λ. dx dx dx

(c) As t px = e−λt , which is independent of x, we have ex =

∞

t px =

t=1

∞

e−λt

t=1

= e−λ + e−2λ + e−3λ + · · · e−λ 1 − eλ 1 = λ . e −1

=

(d) This lifetime distribution is unsuitable for human mortality as survival probabilities, and therefore expected future lifetimes, are independent of attained age. The force of mortality for this lifetime distribution is constant. The force of mortality for humans increases significantly with age.

2.8 (a) f0 (x) = −S0 (x) = 0.002 x e−0.001x .

2

(b) μx = f0 (x)/S0 (x) = 0.002 x. (c) 5 |15 q65 = 2.9 Write

S0 (70) − S0 (85) = 0.45937.

S0 (65) ⎧ t ⎫ ⎫ ⎧ x+t

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ ⎬ ⎨ μ ds μ ds − = exp . − ⎪ ⎪ ⎪ t px = exp ⎪ x+s s ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ ⎭ ⎩ x

0

In this case, we are treating t as fixed and x as variable. Let x+t μs ds .

h(x) = − x

15

Solutions for Chapter 2 Then −h(x) ⇒ t px = e

d

−h(x) = −h (x) t px . t px = −h (x) e dx

Now d h (x) = dx

x+t

⎛ x+t ⎞ x ⎜ ⎟⎟⎟ d ⎜⎜⎜⎜ ⎟ ⎜⎜⎜ μs ds − μs ds = μs ds⎟⎟⎟⎟ = μx+t − μx , ⎠ dx ⎝

x

0

0

giving the required result that

d t px = t px μx − μx+t . dx 2.10 Writing t px = st gc (c −1) , we have x

t

log 10 p50 = log 0.974054 = 10 log s + c50 (c10 − 1) log g , log 10 p60 = log 0.935938 = 10 log s + c60 (c10 − 1) log g , log 10 p70 = log 0.839838 = 10 log s + c70 (c10 − 1) log g .

Then log 10 p70 − log 10 p60 (c70 − c60 )(c10 − 1) log g = log 10 p60 − log 10 p50 (c60 − c50 )(c10 − 1) log g = c10 = 2.7141.

Hence c = 1.105.

2.11 (a) We have ◦

∞

ex =

1 t px dt =

0

∞ t px dt +

0

t px dt .

1

For the first term, we use the fact that t px ≤ 1 for all t ≥ 0, so that 1 t px dt ≤ 1.

0

16

Solutions for Chapter 2 For the second term, by changing the limits of integration we have ∞

∞

∞ t px dt = 1

r+1 px dr = 0

px r px+1 dr 0

◦

◦

= px ex+1 ≤ ex+1 . ◦

◦

So ex ≤ 1 + ex+1 as required. (b) We have ◦

∞

ex =

1 t px dt =

0

2 t px dt +

0

3 t px dt +

1

t px dt + · · · .

2

Since t px is a decreasing function of t, s t px dt ≥ s px , for s = 1, 2, . . . , s−1

so ◦

ex ≥ px + 2 px + 3 px + · · · = ex .

Alternatively, let denote the floor function. Then Kx = Tx ⇒ Kx ≤ Tx ⇒ E[Kx ] ≤ E[Tx ] ◦

⇒ ex ≤ ex . ◦

(c) We know that ex is the expected value of the future lifetime of (x), i.e. E[Tx ], and ex is the expected number of complete years of future life of (x), i.e.

E[Kx ]. If we assume that the difference between Tx and Kx is 0.5 years, on average (on the principle that it is roughly equally likely to be any number of years between 0 and 1), then we have ◦

ex ≈

1 + ex .

2

For a more mathematical explanation, we can use the repeated trapezium

17

Solutions for Chapter 2 rule for numerical integration as follows: ∞

◦

ex =

t px dt 0

1 ≈ (1 + 1 px + 1 px + 2 px + 2 px + · · · ) 2 1 ≈ + ex .

2 ◦

(d) It is almost always the case in practice that ex is a decreasing function of x, but, in principle, it need not be. Consider a hypothetical population where people die only at ages 1 or 50. Of all those born, precisely one half die at age 1 and the remainder all die at age 50. Then ◦

e0 =

1 (1 + 50) = 25.5 2

and

◦

e2 = 48.

2.12 (a) We have ∞

◦

ex =

r px dr 0

∞ =

S0 (x + r) dr S0 (x)

0

1 = S0 (x)

∞ S0 (t) dt. x

(b) Use the product rule to differentiate the answer to part (a): −S0 (x) d ◦ ex = dx S0 (x)2

∞

1 (−S0 (x)) S0 (x) x ⎞ ⎛ ⎜⎜ ∞ ⎟⎟⎟ −S0 (x) ⎜⎜⎜ 1 ⎟ ⎜⎜⎜ S0 (t) dt ⎟⎟⎟⎟ − 1 = ⎠ S0 (x) ⎝ S0 (x) S0 (t) dt +

x

◦

= μx ex − 1 since μx = −S0 (x)/S0 (x).

18

Solutions for Chapter 2 (c) We have d d ◦ ◦ ◦ x + ex = 1 + ex = μx ex > 0, dx dx ◦

so that x + ex increases with x. ◦

Note that x + ex is the expected age at death of a life, given that the life has survived to age x. This is clearly an increasing function of x.

2.13 (a) For fixed n > 0, note that T ≤ t if Tx ≤ n + t for t ≥ 0. Hence for t ≥ 0, Pr[T ≤ t] = Pr[Tx ≤ n + t] = Fx (n + t).

For t < 0, Pr[T ≤ t] = 0. Note that T has a mixed distribution with mass of probability Fx (n) at t = 0; if (x) does not survive to age x + n then (x) does not live for any period after age x + n. (b) The probability density function of T at t > 0 is fx (n + t). Thus, ∞ t fx (n + t) dt

E[T ] = 0

∞ =

(r − n) fx (r) dr n

∞ r r px μx+r dr − n Sx (n)

= n

∞ d =− r r px dr − n Sx (n) dr n ⎛ ⎞ ∞ ∞ ⎜⎜⎜ ⎟⎟⎟ ⎟ ⎜⎜⎜

= − ⎜⎜r r px − r px dr⎟⎟⎟⎟ − n Sx (n) ⎠ ⎝ n n

∞ =

r px dr

n ◦ ◦ = ex − ex:n

where we have used Assumption 2.2 from Chapter 2 of AMLCR and the fact that Sx (n) = n px .

19

Solutions for Chapter 2 2.14 (a) We start by finding Sx (t) for 0 < t ≤ ω − x as ⎧ x+t ⎧ x+t ⎫ ⎫

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ dr ⎪ ⎨ ⎨ ⎬ ⎬ − − = exp ⎪ μr dr⎪ Sx (t) = exp ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ω − r ⎩ ⎩ ⎭ ⎭ x

⎫ ⎧ x+t

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ d ⎬ ⎨ log(ω − r) dr⎪ = exp ⎪ ⎪ ⎪ ⎪ ⎪ dr ⎭ ⎩

x

x

= exp log(ω − x − t) − log(ω − x)

ω −x−t = exp log ω −x ω −x−t . = ω −x So Fx (t) = 1 − Sx (t) =

t , ω −x

which is the distribution function for the U(0, ω − x) distribution. (b) As Tx ∼ U(0, ω − x), we have e̊x = E[Tx ] = (ω − x)/2. We find ex as ex =

ω−x

t px =

t=1

ω−x

1−

t=1

=ω −x−

t ω −x

ω−x 1 t. ω − x t=1

As the sum of the first n integers is n (n + 1)/2, 1 (ω − x)(ω − x + 1) 1 = (ω − x) − .

2 (ω − x) 2 2

ex = ω − x −

Hence e̊x − ex = 12 . (Remark: we could have obtained the same result by setting the upper limit of summation in our expression for ex as ω − x − 1 since ω−x px = 0.)

Alternatively, note that Pr[Kx = k] =

1 ω −x

for k = 0, 1, 2, . . . , ω − x − 1

and so ex =

ω−x−1

k=0

1 k. ω − x k=0 ω−x−1

k Pr[Kx = k] =

20

Solutions for Chapter 2

2.15 (a) For 0 < t ≤ 10, the survival function is ⎫ ⎧ t ⎫ ⎧ t

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ dr ⎬ ⎨ ⎬ ⎨ μ dr S(t) = exp ⎪ = exp − − ⎪ ⎪ ⎪ r ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 4 (10 − r) ⎪ ⎭ ⎩ ⎭ ⎩ 0 0 ⎫ ⎧ t ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ d ⎬ ⎨1 log(10 − r) dr = exp ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ dr ⎭ ⎩4 0

1 log(10 − t) − log 10 = exp 4 1 10 − t log = exp 4 10

t 1/4 . = 1− 10 (b) First, 10 S(t) dt =

E[T ] = 0

10

1−

t 1/4 dt.

10

0

Making the substitution y = 1 − t/10 gives 1 y1/4 dy = 10 ×

E[T ] = 10

4 5/4 1 y = 8.

5 0

0

Next, following Example 2.6, 10 E[T ] = 2 2

10 t 1/4 t S(t) dt = 2 t 1 − dt, 10

0

0

and the same substitution gives 1 E[T ] = 20

10 (1 − y) y1/4 dy

2

0

1 = 200

(y1/4 − y5/4 ) dy 0

= 200

4 5/4 4 9/4 1

y − y 5 9 0

= 640/9.

Solutions for Chapter 2 Hence V[T ] =

64 640 − 82 = , 9 9

giving SD[T ] = 2 23 .

2.16 (a) We can check that S0 is a survival function as follows: S0 (0) = exp{0} = 1, lim S0 (x) = exp{−∞} = 0,

x→∞

and the derivative is

d 1 C C S0 (x) = −(A + Bx + CDx ) exp − Ax + Bx2 + Dx − dx 2 log D log D <0

for x > 0, as A, B, C, D > 0 .

Here we have used the result that d x d D = exp{x log D} = (log D) exp{x log D} = (log D) Dx . dx dx (b) The survival function Sx is given by Sx (t) =

S0 (x + t) S0 (x)

1 C C = exp −A(x + t) − B(x + t)2 − Dx+t + D 2 log D log D 1 C C + Ax + Bx2 + Dx − 2 log D log D C 1 Dx (Dt − 1) . = exp −At − B(2xt + t 2 ) − 2 log D (c) The force of mortality at age x is d 1 d S0 (x) = − log S0 (x) S0 (x) dx dx

d C C 1 Dx − = Ax + Bx2 + dx 2 log D log D x = A + Bx + CD .

μx = −

(d) The results below are obtained by numerical integration: (i) 20 |10 q30 = 0.1082. (ii) e70 = 13.046.

21

22

Solutions for Chapter 2 ◦

(iii) e70 = 13.544.

2.17 (a) We show an excerpt from the table below.

x

px

0 1 2 3 4 .. .

0.99954 0.99951 0.99948 0.99945 0.99942 .. .

(b) The probability that (70) dies at age 70 + k last birthday is Pr[K70 = k] where Kx is the curtate future lifetime. The most likely age at death is the value of k that maximizes Pr[K70 = k] = k |q70 .

The maximum value for k |q70 can be found by constructing a table of values and selecting the largest value; it is 3 |q70 = 0.05719, so the most likely age at death is 73. (c) The curtate expectation of life at age 70 is e70 =

∞

t p70 = 9.339.

t=1

(d) The complete expectation of life at age 70 is ∞

◦

e70 =

t p70 dt = 9.834, 0

using numerical integration.

2.18 (a) As μx∗ = 2μx for all x, we have ⎧ t ⎫ ⎫ ⎧

t ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ ⎬ ⎨ ∗ ∗ p = exp μ ds μ ds − = exp = ( t px )2 . −2 ⎪ ⎪ ⎪ ⎪ t x x+s x+s ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ ⎭ ⎩ 0

0

◦

◦∗

(b) Using numerical integration, e50 − e50 = 6.432.

Solutions for Chapter 2

23

(c) Using numerical integration, E[T502 ] = 575.40 and E[(T50∗ )2 ] = 298.25, giving V[T50 ] = 125.89 and V[T50∗ ] = 80.11.

Solutions for Chapter 3

3.1 Figures S3.1, S3.2 and S3.3 are graphs of μx , lx and dx , respectively, as functions of age x up to x = 100. Each graph has been drawn using the values from ELT 15, Males and Females. (a) The key feature of Figure S3.1 is that the value of μx is very low until around age 55, from where it increases steeply. Numerically, μx is very close to qx , provided qx is reasonably small, so that the features in Figure S3.1 are very similar to those shown in Figure 3.1 in AMLCR. The features at younger ages show up much better in Figure 3.1 in AMLCR because the y-axis there is on a logarithmic scale. Note that the near-linearity in Figure 3.1 in AMLCR for ages above 35 is equivalent to the near-exponential growth we observe in Figure S3.1. (b) The key feature of Figure S3.2 is that, apart from a barely perceptible drop in the first year due to mortality immediately following birth, the graph is more or less constant until around age 55, when it starts to fall at an increasing rate before converging towards zero at very high ages. This reflects the pattern seen in Figure S3.1.

24

25

Solutions for Chapter 3

0.50 0 45 0.45

Females Males

0.40 0.35 0.30

M[ 0.25 0.20 0.15 0.10 0.05 0.00 0

10

20

30

40

50 Age, [

60

70

80

90

100

Figure S3.1 A graph of μx as a function of x.

100000 90000 80000 70000 60000 O[ 50000

Females Males

40000 30000 20000 10000 0 0

10

20

30

40

50 Age, [

60

70

Figure S3.2 A graph of lx as a function of x.

80

90

100

26

Solutions for Chapter 3 4000 3500 3000 Females Males

2500 G[

2000

1500 1000 500 0 0

10

20

30

40

50 Age, [

60

70

80

90

100

Figure S3.3 A graph of dx as a function of x.

(c) The function dx is the expected number of deaths between exact ages x and x + 1 out of l0 lives aged 0. The relatively high mortality in the first year of life shows clearly in Figure S3.3, as does the increase in the expected number of deaths for males in the late teenage years – the so-called ‘accident hump’. See the comments on Figure 3.1 in AMLCR. The expected number of deaths increases gradually from around age 30 and sharply from around age 55. It reaches a peak at around ages 77 (males) and 87 (females) even though the force of mortality continues to increase beyond these ages, as can be seen in Figure S3.1. The peak occurs because the expected number of survivors to high ages, lx , is decreasing sharply (see Figure S3.2) and so the expected number of deaths from these survivors, dx , eventually has to fall as well.

3.2 The assumption of a uniform distribution of deaths between integer ages implies that lx+s = (1 − s) lx + s lx+1 for integer x and 0 ≤ s ≤ 1. This follows directly from formula (3.9) in AMLCR.

Solutions for Chapter 3

27

The assumption of a constant force of mortality between integer ages means that for integer x, and 0 ≤ s ≤ s + t ≤ 1, s s px+t = (px ) .

See Section 3.3.2 in AMLCR. Note that if s + t > 1, then we must calculate survival probabilities for fractions of a year at different (integer) ages. See part (c) for an example. (a) Using UDD, 0.2 q52.4 = 1 −

l52.6 0.4l52 + 0.6l53 =1− l52.4 0.6l52 + 0.4l53

= 0.001917.

Using the constant force assumption, 0.2 q52.4 = 1 − 0.2 p52.4

= 1 − (p52 )0.2

0.2 89 089 =1− = 0.001917.

89 948

(b) Using UDD, 5.7 p52.4 =

l58.1 0.9l58 + 0.1l59 = l52.4 0.6l52 + 0.4l53

= 0.935422.

Using the constant force assumption, 5.7 p52.4 = 0.6 p52.4 × 5 p53 × 0.1 p58

= (p52 )0.6 5 p53 (p58 )0.1 = 0.935423.

(c) Using UDD, l55.6 − l58.1 l52.4 0.4l55 + 0.6l56 − 0.9l58 − 0.1l59 = 0.6l52 + 0.4l53

3.2 |2.5 q52.4 =

= 0.030957.

28

Solutions for Chapter 3 Using the constant force assumption, 3.2 |2.5 q52.4 = 3.2 p52.4 (1 − 2.5 p55.6 )

= 0.6 p52.4 × 2 p53 × 0.6 p55 (1 − 0.4 p55.6 × 2 p56 × 0.1 p58 ) = (p52 )0.6 2 p53 (p55 )0.6 (1 − (p55 )0.4 2 p56 (p58 )0.1 ) = 0.030950.

3.3 We first note that under the constant force fractional age assumption, for 0 ≤ t < 1 and x = 0, 1, 2, . . .,

t S0 (x + 1) Sx (t) = Sx (1)t = .

S0 (x) This allows us to express S0 (x + t) in terms of S0 (x + 1) and S0 (x), as follows:

t S0 (x + 1) = (S0 (x))1−t (S0 (x + 1))t .

S0 (x + t) = S0 (x) Sx (t) = S0 (x) S0 (x) Thus log S0 (x + t) = (1 − t) log S0 (x) + t log S0 (x + 1), meaning that h(s) = log s.

3.4 The required probabilities are calculated as follows. (a)

10 542 l85 = 0.66177. = l[75] 15 930

(b) 9 |2 q[75]+1 =

l85 − l87 10 542 − 9 064 = 0.09433. = l[75]+1 15 668

(c) 4 |2 q[77]+1 =

l82 − l84 12 576 − 11 250 = 0.08993. = l[77]+1 14 744

3.5 The required probabilities are calculated as follows. (a) 2 p[72] = (1 − q[72] )(1 − q[72]+1 ) = 0.994764 × 0.992544 = 0.987347. (b) 3 q[73]+2 = 1 − 3 p[73]+2 = 1 − (1 − q[73]+2 )(1 − q[73]+3 )(1 − q[73]+4 ) = 1 − 0.98863 × 0.985012 × 0.980684 = 0.044998.

Solutions for Chapter 3

29

(c) 1 |q[65]+4 = p[65]+4 × q70 = 0.992006 × 0.010599 = 0.010514. (d) 7 p[70] = (1 − q[70] )(1 − q[70]+1 ) · · · (1 − q[70]+4 )(1 − q75 )(1 − q76 ) = 0.995715 × 0.994033 × 0.991934 × 0.989371 × 0.986302 × 0.981226 × 0.978947 = 0.920271. (e) Probably the easiest way to calculate this probability is to express it first in terms of ls, as follows: l[70]+4 l[70]+0.2 l[70]+4 =1− using UDD 0.8 l[70] + 0.2 l[70]+1 4 p[70] =1− 0.8 + 0.2 p[70] (1 − q[70] )(1 − q[70]+1 )(1 − q[70]+2 )(1 − q[70]+3 ) =1− 0.8 + 0.2 (1 − q[70] ) = 0.027812.

3.8 q[70]+0.2 = 1 −

3.6 Working in terms of ls we need l56 /l53 . Setting l[50] = 100 000 we have l[50]+1 = l[50] (1 − q[50] ) = 98 399, l[50]+2 = l[50] 2 p[50] = 96 411, l53 = l[50]+2 − l[50] × 2 |q[50] = 94 001, l56 = l53 − l[50]+1 × 2 |3 q[50]+1 = 84 877.44.

Hence 3 p53 = 0.902942.

3.7 Our spline is Ca (x, t) = at 3 +bt 2 +ct +d, with derivative Ca (x, t) = 3at 2 +2bt +c.

The conditions we apply are as follows.

• The knot at time 0 (year 2020) is Ca (x, 0) = d = ϕ(x, 2020) = 0.02.

• The knot at time 20 is Ca (x, 20) = 8000a + 400b + 20c + d = ϕ(x, 2040) = 0.

• The gradient at time 0 is Ca (x, 0) = c = ϕ(x, 2020) − ϕ(x, 2019) = 0.

30

Solutions for Chapter 3 • The gradient at time 20 is Ca (x, 20) = 1200a + 40b + c = ϕ(x, 2041) − ϕ(x, 2040) = 0.

We immediately have c = 0 and d = 0.02, resulting in the following equations for a and b: 8000a + 400b + 0.02 = 0, 1200a + 40b = 0.

The solution is a = 5/106 and b = −1.5/104 , giving Ca (x, 4) = 0.01792.

3.8 Let S p denote the number of survivors from the preferred lives, and let Sn denote the number of survivors from the normal lives. Then S = S p + Sn is the total number of survivors from the group at the end of the year. (a) We know that S p ∼ B(20 000, 0.998), Sn ∼ B(80 000, 0.991), and S p and Sn are independent because the individual lives are independent. Then by independence V[S] = V[S p ] + V[Sn ] = 20 000 × 0.998 × 0.002 + 80 000 × 0.991 × 0.009 = 753.44, giving SD[S] = 27.45. (b) We have E[S p ] = 20 000 × 0.998 = 19 960 and similarly E[Sn ] = 79 280.

So the proportion of preferred lives expected in the survivor group is 19 960 = 0.201.

19 960 + 79 280 (c) The 90th percentile of the standard normal distribution is 1.2816, so our approximation to the 90th percentile of the number of survivors is E[S] + 1.2186 SD[S] = 99 275.

3.9 For integer x and 0 ≤ s < 1, we have s px = exp{−μx∗ s}, where px = exp{−μx∗ } =

lx+1 . lx

31

Solutions for Chapter 3 Then ∞ e̊x =

1 t px dt =

0

=

2 t px dt +

0

t px dt + · · · 1

j+1 ∞

t px dt

j=0 j ∞

j+1

=

j px t− j px+ j dt

j=0 j

=

=

∞

1 j px

s px+ j ds

j=0

0

∞

1

j=0

j px

∗ exp{−μx+ j s} ds

0

⎛ ⎞ ∗ ⎜⎜⎜ 1 − exp{−μx+ ⎟⎟⎟ j} ⎟ ⎜⎝ = ⎟⎠ j px ⎜ ∗ μx+ j j=0 ∞

=

∞

j px − j+1 px j=0

∗ μx+ j

.

∗ So a j = μx+ j = −log px+ j .

3.10 Use the superscript A to denote the mortality of country A. Other functions refer to US Life Tables, 2002, Females. The required probability is

A A A 10 p[30] = 5 p[30] 5 p35

where

A 5 p[30] = (1 − 6q30 )(1 − 5q31 )(1 − 4q32 )(1 − 3q33 )(1 − 2q34 )

= 0.98599

32

Solutions for Chapter 3 and

⎧ ⎫ ⎪ ⎪ 5 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ A A μ dt − ⎪ 5 p35 = exp ⎪ 35+t ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ 0

⎫ ⎧ ⎪ ⎪ 5 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎨ μ dt = exp ⎪ −1.5 × ⎪ 35+t ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎩ 0

= (5 p35 )

1.5

= 0.99139.

Hence 10 pA[30] = 0.977497.

3.11 (a) We start by showing that lx∗ = lx+1 for all x. We are given that this is true for x = 25. Consider first x = 26. We have ∗ ∗ l26 = p∗25 l25 = l27 ∗ = l26 . Similarly since p∗25 = p26 and l25 ∗ ∗ l27 = p∗26 l26 = l28

and the same argument gives lx∗ = lx+1

for x = 26, 27, 28, . . . .

Consider now lx∗ for x ≤ 24. We have ∗ ∗ l24 = l25 /p∗24

= l26 /p25 = l25 .

This argument gives lx∗ = lx+1 for x = 24, 23, 22, . . . , 0. ∗ The next step is to show that l[x]+2 = lx+2 for all x. As p∗[x]+2 = 2 px+2 , ∗ ∗ ∗ p∗[x]+2 l[x]+2 = l[x]+3 = lx+3 = lx+4 ∗ ⇒ 2 px+2 l[x]+2 = lx+4 ∗ ⇒ l[x]+2 = lx+4

lx+2 = lx+2 . lx+4

∗ Next we show that l[x]+1 = lx−1 for all x. As p∗[x]+1 = 3 px−1 , ∗ ∗ p∗[x]+1 l[x]+1 = l[x]+2 = lx+2 l x+2 ∗ ⇒ l[x]+1 = = lx−1 .

3 px−1

33

Solutions for Chapter 3 ∗ = lx−5 for all x. As p∗[x] = 4 px−5 , Finally we show that l[x] ∗ ∗ = l[x]+1 = lx−1 p∗[x] l[x] lx−1 ∗ ⇒ l[x] = = lx−5 .

4 px−5

In summary, ∗ = lx−5 , l[x]

∗ l[x]+1 = lx−1 ,

∗ l[x]+2 = lx+2 ,

lx∗ = lx+1 .

The required life table is as follows: x

∗ l[x]

∗ l[x]+1

∗ l[x]+2

∗ lx+3

20 21 22

99 180 99 135 99 079

98 942 98 866 98 785

98 700 98 615 98 529

98 529 98 444 98 363

(b) Using the results from part (a), ∗ ∗ ∗ − l62 )/l[21]+1 = 0.121265, (i) 2 |38 q∗[21]+1 = (l24 ∗ ∗ /l[22] = 0.872587, (ii) 40 p∗[22] = l62 ∗ ∗ /l[21]+1 = 0.874466, (iii) 40 p∗[21]+1 = l62 ∗ ∗ /l[20]+2 = 0.875937, (iv) 40 p∗[20]+2 = l62 ∗ ∗ /l22 = 0.876692. (v) 40 p∗22 = l62

Note that in this model there is ‘reverse selection’, i.e. the probabilities (ii) to (v) above are in ascending order of magnitude. ∗ denote the assumed constant force 3.12 (a) For integer x and k = 0, 1, . . ., let μx+k of mortality between ages x + k and x + k + 1. Then

⎧ ⎫ ⎪ ⎪ 1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ ∗ ∗ μ dt }. px+k = exp ⎪ − = exp{−μx+k ⎪ x+k ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ 0

34

Solutions for Chapter 3 The probability in formula (3.17) can be written as Pr[k < Tx ≤ k + s] Pr[k < Tx ≤ k + 1] k px s qx+k = k px (1 − px+k ) s qx+k = ∗ 1 − exp{−μx+k } ⎧ ⎫ s ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ ∗ 1 − exp ⎪ − μ dt ⎪ ⎪ ⎩ 0 x+k ⎪ ⎭ = ∗ 1 − exp{−μx+k }

Pr[Rx ≤ s|Kx = k] =

=

∗ 1 − exp{−μx+k s} , ∗ 1 − exp{−μx+k }

as required. (b) Suppose formula (3.17) holds for integer x and for k = 0, 1, 2, . . .. Then for 0 ≤ s ≤ 1, ∗ 1 − exp{−μx+k s} = Pr[Rx ≤ s|Kx = k] ∗ 1 − exp{−μx+k } Pr[k < Tx ≤ k + s] = Pr[k < Tx ≤ k + 1] k px s qx+k = k px (1 − px+k )

=

1 − s px+k , ∗ 1 − exp{−μx+k }

∗ which implies that s px+k = exp{−μx+k s}.

The force of mortality at age x+k+s can be written as μx+k+s = −

d ∗ log s px+k = μx+k ds

using the formula for s px+k derived above. Thus μx+k+s does not depend on s, as required.

35

Solutions for Chapter 3 3.13 First note that for any constant a, t

t a ds = s

0

exp{s log(a)} ds 0

1 (exp{t log(a)} − 1) log(a) at − 1 . = log(a)

=

Using this formula, we have for 0 ≤ t ≤ 2: ⎧ t ⎫

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ p = exp μ ds − ⎪ ⎪ t [x] [x]+s ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ 0 ⎫ ⎧ t

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎨ 2−s x+s 0.9 (A + Bc ) ds = exp ⎪ − ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎩ 0 ⎫ ⎧ t ⎪ ⎪ ⎪ ⎪ ⎪ ⎪

⎬ ⎨ 2 −s x s ) + Bc (c/0.9) A(0.9 ds = exp ⎪ −0.9 ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎩ 0

A(1 − 0.9−t ) Bcx ((c/0.9)t − 1) + log(0.9) log(c/0.9)

t Bcx (ct − 0.9t ) 2−t A(0.9 − 1) + = exp −0.9 log(0.9) − log(0.9/c)

A(1 − 0.9t ) Bcx (ct − 0.9t ) + , = exp 0.92−t log(0.9) log(0.9/c)

= exp −0.92

which is formula (3.15) in AMLCR. 3.14 Our age-based spline is Ca (x, t) = at 3 + bt 2 + ct + d, with derivative Ca (x, t) = 3at 2 + 2bt + c. The conditions we apply are as follows.

• The knot at time 0 (year 2015) is Ca (50, 0) = d = ϕ(50, 2015) = 0.0145.

• The knot at time 25 is Ca (50, 25) = 253 a + 252 b + 25c + d = ϕ(50, 2040) = 0.01.

36

Solutions for Chapter 3 • The gradient at time 0 is Ca (50, 0) = c = ϕ(50, 2015) − ϕ(50, 2014) = 0.0011.

• The gradient at time 25 is Ca (50, 25) = (252 × 3a) + 50b + c = ϕ(50, 2041) − ϕ(50, 2040) = 0.

We immediately have c = 0.0011 and d = 0.0145, resulting in the following equations for a and b: 253 a + 252 b + 0.032 = 0, (252 × 3a) + 50b + 0.0011 = 0.

The solution is a = 2.336/106 and b = −1.096/104 , giving Ca (50, 10) = 0.016876.

Our cohort-based spline is Cc (x, t) = a∗t 3 + b∗t 2 + c∗t + d ∗ , with derivative Cc (x, t) = 3a∗t 2 + 2b∗t + c∗ . The conditions we apply are as follows.

• The knot at time 0 (age 40, year 2015) is Cc (40, 0) = d ∗ = ϕ(40, 2015) = 0.0182.

• The knot at time 25 is Cc (65, 25) = 253 a∗ + 252 b∗ + 25c∗ + d ∗ = ϕ(65, 2040) = 0.01.

• The gradient at time 0 is Cc (40, 0) = c∗ = ϕ(40, 2015) − ϕ(39, 2014) = 0.0014.

• The gradient at time 25 is Cc (65, 25) = (252 × 3a∗ ) + 50b∗ + c∗ = ϕ(66, 2041) − ϕ(65, 2040) = 0.

We immediately have c∗ = 0.0014 and d ∗ = 0.0182, resulting in the following equations for a∗ and b∗ : 253 a∗ + 252 b∗ + 0.0432 = 0, (252 × 3a∗ ) + 50b∗ + 0.0014 = 0.

The solution is a∗ = 3.290/106 and b∗ = −1.514/104 , giving Cc (50, 10) = 0.020354.

Solutions for Chapter 3

37

The improvement factor for a life aged 50 in 2025 is the average of these two improvement factors, namely 0.018615.

Solutions for Chapter 4

4.1 (a) 5 E35 = v 5 l40 /l35 = 0.735942. (b) A 1

35: 5

= A35 − 5 E35 A40 = 0.012656.

(c) 5 |A35 = 5 E35 A40 = 0.138719. (d) Ā35: 5 = (i/δ ) A 1

35: 5

4.2 (a) A 1

30:20

+ 5 E35 = 0.748974.

= A30 − 20 E30 A50 = 0.07698 − 0.37254 × 0.18931 = 0.00645.

(b) Ā40:20 = Ā 1

40:20

+ 20 E40

=

i (A40 − 20 E40 A60 ) + 20 E40 δ

=

0.05 (0.12106 − 0.36663 × 0.29028) + 0.36663 log 1.05

= 0.38163. (c) 10 |A25 = 10 E25 A35 = 0.61198 × 0.09653 = 0.05907.

4.3 (a) The benefit is a death benefit, payable immediately on death, equal to t should (30) die at time t years from the issue of the policy if t < 25, and equal to 25 if death occurs more than 25 years after the issue of the policy. (b) (I¯Ā) 1

30:25

+ 25 25 E30 Ā55 .

38

39

Solutions for Chapter 4 4.4 The ordering is (4)

< A(12) < Āx:10 .

A 1x: 10 < Ax < Āx < A(4) x:10 x:10 The functions Āx:10 , A(12) and A(4) all value a 10-year endowment insurance x:10 x:10 with unit sum insured, the only difference being the timing of the payment of the death benefit. The payment of the death benefit is earliest under the continuous function and latest under the quarterly function, meaning that the EPV is greatest if the benefit is payable immediately on death, and least if it is payable at the end of the quarter year of death as vt is a decreasing function of t for i > 0.

Similarly, the relationship Ax < Āx follows as the payment is made sooner under Āx (on average half a year earlier) – indeed we know that under claims acceleration Āx ≈ (1 + i)1/2 Ax .

(4)

The smallest is A 1x: 10 as it is not certain that the benefit will be payable under 10-year term insurance.

As Ax and Āx value benefits under whole life insurance and Āx:10 , A(12) and A(4) x:10 x:10 value benefits under 10-year endowment insurance, the small probability of death within 10 years at ages at which insurance is sold means that the benefit will most likely be paid on survival for 10 years, leading to a higher EPV than under Ax or Āx .

4.5 The present value (PV) of the benefit is 100 000 vt if death occurs in the year from time t−1 to time t, for t = 1, 2, . . . , 5. So 100 000vt > 90 000 ⇔ vt > 0.9 ⇔ −t log 1.05 > log 0.9 log 0.9 = 2.16. ⇔t <− log 1.05 So the present value is greater than 90 000 if the benefit is payable at time 1 or time 2, with probability 2 q65 = 0.012494.

4.6 We have ⎧ ⎪ ⎪ ⎨ v Tx X =⎪ ⎪ ⎩ vn

if Tx ≤ n, if Tx > n,

and

⎧ ⎪ ⎪ ⎨ v Tx Y =⎪ ⎪ ⎩ 0

if Tx ≤ n, if Tx > n.

40

Solutions for Chapter 4 Let Z = X − Y be the present value random variable for a pure endowment: ⎧ ⎪ ⎪ if Tx ≤ n, ⎨ 0 Z=⎪ ⎪ ⎩ vn if Tx > n.

Since Y = X − Z, we know that E[X] = E[Y ] + E[Z] and V[Y ] = V[X] + V[Z] − 2 Cov[X, Z].

Now E[Z] = n Ex = v n n px = 0.3 × 0.8 = 0.24 and E[Z 2 ] = v 2n n px = 0.072, so that V[Z] = 0.0144.

The covariance term is E[XZ] − E[X] E[Z]. We have E[X] = E[Y ] + E[Z] = 0.28 and as

⎧ ⎪ ⎪ ⎨ 0 XZ = ⎪ ⎪ ⎩ v 2n

if Tx ≤ n, if Tx > n,

we have E[XZ] = E[Z 2 ] = 0.072, so that the covariance term is 0.072−0.28×0.24 = 0.0048, giving V[Y ] = 0.01.

Alternatively, V[Y ] = 2 Āx:1 n − Āx:1 n

2

= 2 Āx:1 n − E[Y ]2 = 2 Āx:1 n − 0.042 ,

and we can calculate 2 Āx:1 n by first noting that V[X] = 0.0052 = 2 Āx: n − Āx: n

2

,

where Āx: n = E[X] and E[X] = E[Y ] + v n n px = 0.04 + 0.3 × 0.8 = 0.28.

Thus 2 Āx: n = V[X] + 0.282 = 0.0836. We can then find 2 Āx:1 n from 2

Āx: n = 2 Āx:1 n + v 2n n px

⇒ 2 Āx:1 n = 0.0836 − 0.32 × 0.8 = 0.0116.

Hence V[Y ] = 2 Āx:1 n − Āx:1 n

2

= 0.01.

41

Solutions for Chapter 4

4.7 The present value (PV) of the benefit is 50 000 vt if death occurs at time t, so PV < 20 000 ⇔ vt < 0.4 ⇔ −t log 1.04 < log 0.4 log 0.4 = 23.36. ⇔t >− log 1.04 So the present value is less than 50 000 if the benefit is payable if (50) survives for at least 23.36 years, with probability 23.36 p50 = 0.887277.

4.8 Consider the whole life insurance as the sum of deferred one-year term insurances so that A(m) x =

∞

(m)

1 vt t px A x+t: .

1

t=0

Now for any integer age y, (m)

A

1

1

y: 1

2

m

= v m m1 qy + v m m1 | m1 qy + · · · + v m m−1 | m1 qy , m

and for r = 0, 1, 2, . . . , m − 1, | qy = r+1 qy − mr qy m

r 1 m m

r+1 r q y − qy m m qy = . m =

using UDD

Hence, A

(m)

1

y: 1

2 m 1 1 vm + vm + ··· + vm m

= qy

.

You might recognize the term in the { } parentheses as a(m) = 1

1−v iv = (m) . (m) i i

Alternatively, sum the geometric series to give 2 m 1 1−v 1 1 vm + vm + ··· + vm = vm 1 m m(1 − v m ) iv 1−v = (m) . = 1 i m((1 + i) m − 1)

42

Solutions for Chapter 4 So A(m) x =

∞

iv vt t px qx+t (m) i t=0

i t+1 i = (m) v t |qx = (m) Ax . i t=0 i ∞

4.9 We know that A50 = A 1

50:20

+ 20 E50 A70

and 20 E50 = A50:2 0 − A

1

50:20

= 0.27251.

Thus A70 =

A50 − A 1

50:2 0

20 E50

= 0.59704.

4.10 (a) Let S = 100 000. The present value random variable is Z = S v K30 +1 .

The EPV is S A30 = 105 × 0.07698 = 7698 and E[Z 2 ] is S2 2A30 = 1010 × 0.01109 = 110 900 000 so the standard deviation of the present value is √ 51 640 796 = 7186. (b) The present value random variable is ⎧ ⎪ ⎪ ⎨ S v K30 +1 if K30 ≤ 19, Z=⎪ ⎪ ⎩ 0 if K30 ≥ 20.

The EPV is SA1

30:20

= S (A30 − 20 E30 A50 ) = 105 (0.07698 − 0.37254 × 0.18931) = 645.45

Solutions for Chapter 4 and E[Z 2 ] is S2

2

A1

30:20

= S2 ( 2 A30 − v 20 20 E30 2 A50 ) = 1010 (0.01109 − 0.37689 × 0.37254 × 0.05108) = 39 180 407

so the standard deviation of the present value is !

39 180 407 − 645.452 = 6226.

4.11 We start from Ax − 20 Ex Ax+20 = Ax1 : 20 .

Adding 20 Ex to each side gives Ax − 20 Ex (Ax+20 − 1) = Ax : 20 so that 20

Ex =

Ax : 20 − Ax 0.55 − 0.25 = 0.5 = 1 − Ax+20 1 − 0.4

and so A1x:20 = Ax:20 − 20 Ex = 0.05. (a) Under claims acceleration, 10 000 Āx:20 = 10 000 (1 + i)1/2 Ax:1 20 + 20 Ex = 5507.44. (b) Under UDD, 10 000 Āx : 20 = 10 000

i δ

A1x: 20 + 20 Ex = 5507.46.

4.12 As vt = (1 + i)−t , d t v = −t (1 + i)−t−1 = −t vt+1 . di

43

44

Solutions for Chapter 4 Thus d d Āx = di di

∞ vt t px μx+t dt 0

∞ =

d (1 + i)−t t px μx+t dt di

0

∞ = − t vt+1 t px μx+t dt 0

= −v (I¯Ā)x , which is negative since v and (I¯Ā)x are positive. Thus Āx is a decreasing function of i. The result is intuitive if we think of Āx as the single premium required to provide a benefit of 1 immediately on the death of (x). The higher the rate of interest we can earn, the smaller the premium required.

4.13 The EPV of the benefit in the first year is 10 000 Ā 1

80:1

calculated under the heavier mortality. Using UDD, the EPV is i 1 i A = v (2 × q80 ) = 637.49 δ 80:1 δ since the mortality rate is doubled in the first year.

The EPV of the benefit after the first year, again using UDD, is v (1 − 2q80 ) Ā81 = v (1 − 2q80 )

i A81 = 55 632.51, δ

giving a total EPV of 56 270.00.

4.14 Starting from t (I¯Ā)x: t =

e−δ r r r px μx+r dr + t e−δt t px ,

0

we have d ¯ (I Ā)x: t = e−δt t t px μx+t + e−δt t px + t −δ e−δt t px − e−δt t px μx+t dt = e−δt t px (1 − δt).

45

Solutions for Chapter 4 We find 10 p40 = 0.99233 and hence with δ = log 1.05,

d ¯ (I Ā)40:t = 0.3120. dt t=10 4.15 The EPV is 100 000

∞

vt+1 1.03t t |q30 =

t=0

100 000 A30; j , 1.03

where A30; j is A30 calculated at rate of interest j = (0.05 − 0.03)/1.03. The EPV is $33 569.47.

4.16 We have 1 = (IA)x:n

n−1

v k+1 (k + 1) k |qx

k=0

=

n−1

v k+1 (n + 1) k |qx −

n−1

k=0

−

n−3

v k+1 k |qx −

k=0

v k+1 k |qx − · · · −

k=0

0

n−2

v k+1 k |qx

k=0

v k+1 k |qx

k=0

= (n + 1) Ax:1 n −

n

k=1

A1 . x: k

The benefit being valued is a payment at the end of the year of death if this occurs within n years. The amount is 1 in the first year, 2 in the second year, and so on. We can regard this as an amount n + 1 on death in any year minus n if death occurs in the first year, n − 1 if death occurs in the second year, and so on.

4.17 If failure occurs at time t ≥ 2, the present value of the benefit is 100 000 e−δt , and if failure occurs before time 2, the present value of the benefit is 0. As the present value (after time 2) is a decreasing function of t, we see that the highest 10% of present values must correspond to the set of times [2, x] in which there is a 0.1 probability of failure. We find x from x

x fT (t) dt =

0.1 = 2

0.4 −0.4 x 0.4 = 0.2 − , dt = t 2 x t2

2

yielding x = 4. Hence the 90th percentile of the present value is 100 000 e−0.05×4 = 81 873.

46

Solutions for Chapter 4

4.18 (a) We have A[40]+1: 4 =

3

vt+1 t |q[40]+1 + v 4 4 p[40]+1

t=0

d[40]+1 v + d[40]+2 v 2 + d[40]+3 v 3 + d44 v 4 + l45 v 4 = l[40]+1 = 0.79267. (b) Let Z denote the present value of the benefit. Then E[Z] = 100 000

4

vt+1 t |q[40] = 701.35

t=1

and E[Z 2 ] = 100 0002

4

vt+1

2

t |q[40] = 57 037 868 ,

t=1

giving a standard deviation of $7519.71. (c) The present value of the benefit is less than or equal to $85 000 if the life dies in the deferred period or survives five years (since in either case no benefit is payable), or if the benefit is payable at time t years and 100 000 vt < 85 000, which gives t as 3, 4 or 5. Thus the present value is strictly greater than $85 000 only if the payment is at time 2, the probability of which is 1 |q[40] .

Thus, the required probability is 1 − 1 |q[40] = 0.99825.

4.19 (a) Treating four months as one-third of a year, the EPV of the benefit is ∞ 10

1

1

1

vt+ 3 t p[50] μ[50]+t dt = 106 v 3 Ā[50] = 106 v 3

6 0

i A[50] , δ

where the last step follows by UDD. Using A[50] = 0.18913, the EPV is 190 693. Similarly, the second moment is ∞

2

2

v 2t+ 3 t p[50] μ[50]+t dt = 1012 v 3

12

10

0

2

2

Ā[50] = 1012 v 3

j δj

2

A[50] ,

47

Solutions for Chapter 4

where 1+ j = (1 + i)2 and δ j = log(1 + j). Using 2 A[50] = 0.05087 we find that the standard deviation is 123 938. (b) Let I be an indicator random variable that takes the value 0 if a claim is straightforward, and 1 if it is complex, and let PV denote the present value of the benefit. Then $ " " # #% E PV = E E PV|I , with

1 i " # E PV|I = 0 = 106 v 6 A[50] = 192 250.11, δ

(treating two months as one-sixth of a year) and " # i E PV|I = 1 = 106 v A[50] = 184 590.29. δ Hence " # E PV = 0.8 × 192 250.11 + 0.2 × 184 590.29 = 190 718.

Next,

where

and

$ " $ " " # #% #% V PV = E V PV|I + V E PV|I , 2 " # j E PV2 |I = 0 = 1012 v 6 δj

" # j E PV2 |I = 1 = 1012 v 2 δj

leading to " # V PV|I = 0 = 124 950.342

and

2

2

A[50]

A[50] ,

" # V PV|I = 1 = 119 971.952 .

Hence $ " #% E V PV|I = 0.8 × 124 950.342 + 0.2 × 119 971.952 = 123 970.662 .

Finally, $ " #% V E PV|I = 0.8 × 192 250.112 + 0.2 × 184 590.292 − 190 718.152 = 9 387 659.37, " # giving V PV = 124 0092 .

48

Solutions for Chapter 4

4.20 (a) The present value (PV) is 100 e−δ T50 if T50 < 20, and as 20 p50 = 0.9240 we see that there exists some τ < 20 such that τ p50 = 0.95. We start by finding τ. As l50+τ l50+τ = 0.95, = τ p50 = l50 98 576.37 it follows that l50+τ = 93 647.55 and by inspection of the Standard Ultimate Life Table, l66 = 94 020.33 and l67 = 93 398.05, so that 16 < τ < 17.

Applying UDD, l50+τ = 93 647.55 = l66 − t d66 , giving t = 0.599 and hence τ = 16.599. So we know that Pr[T50 > 16.599] = 0.95.

Let the 95th percentile of the present value be x. Then & x ' 1 −δ T50 Pr[100 e ≤ x] = 0.95 ⇒ Pr T50 ≥ − log = 0.95. δ 100 Thus,

x 1 log , δ 100 giving the 95th percentile of the present value as 44.492.

16.599 = −

(b) For each policy, the present value (PV) of the benefit is 100 v T50 if T50 < 20, and 0 otherwise. So, under UDD, E[PV] = 100 Ā 1

50:20

= 100

i 1 A δ 50:20

i (A50 − 20 E50 A70 ) δ i = 100 (0.18931 − 0.34824 × 0.42818) δ = 4.1197, = 100

and similarly, with 1+ j = (1+i)2 = eδ j , E[PV2 ] = 10 000 2 Ā 1

50:20

= 10 000

j 1 Ā δ j 50:20 j

j (A50 − 20 E50 A70 ) j δj j = 10 000 0.05108 − 0.34824 e−20δ × 0.21467 δj = 10 000

= 15.51122 .

49

Solutions for Chapter 4

Hence the variance of the present value for each policy is 223.63 = 14.952 .

As the 95th percentile of the standard normal distribution is 1.644854, the 95th percentile of the present value for the 10 000 policies is √ 10 000 × 4.1197 + 1.644854 × 100 × 14.95 = 43 657. (c) For a single policy, the distribution of the present value is a mixed distribution that does not resemble a normal distribution; note that with probability 20 p50 the present value is 0. From the central limit theorem, we know that the sum of a large number of independent and identically distributed random variables is approximately normal, even if the distribution of each random variable is not close to a normal distribution. As there are 10 000 policies in part (b), it is reasonable to apply this approximation.

4.21 (a) (i) Z1 is the present value of a benefit payable immediately on the death of (x), the amount of the benefit being 20 if death occurs before age x+15 and 10 if death occurs after age x+15. (ii) Z2 is the present value of a benefit of 10 payable immediately on the death of (x) should death occur between ages x+5 and x+15 or payable at age x+15 should (x) survive to that age. (b) (i) There are different ways in which the answer can be written. Viewing the benefit as being 10 immediately on death at any age plus an extra 10 on death before age x+15, we get ⎛ ∞ ⎞ 15 ⎜⎜⎜ ⎟⎟⎟ ⎜⎜⎜ ⎟ E[Z1 ] = 10 ⎜⎜ vt t px μx+t dt + vt t px μx+t dt ⎟⎟⎟⎟ . ⎝ ⎠ 0

0

Alternatively we can view the benefit as a term insurance plus a deferred whole life insurance giving 15 E[Z1 ] = 20

∞ v t px μx+t dt + 10

vt t px μx+t dt.

t

0

15

(ii) We have 15 E[Z2 ] = 10

vt t px μx+t dt + 10v 15 15 px .

5

50

Solutions for Chapter 4 (c) (i) Corresponding to the first integral expression, E[Z1 ] = 10 Āx + Āx:1 15 , and corresponding to the second we get E[Z1 ] = 20 Āx:1 15 + 10 15 Ex Ax+15 . (ii) Using formula (4.23) in AMLCR to value the death benefit, we get E[Z2 ] = 10 Āx:1 15 − Āx:1 5 + 15 Ex . (d) We have 15 E[Z12 ] = 400

−2δt

e

∞ t px μx+t dt + 100

0

e−2δt t px μx+t dt

15

= 100 2 Āx + 300 2 Āx:1 15 , and hence V[Z1 ] = 100 2 Āx + 300 2 Āx:1 15 − 100 Āx + Ā1x: 15

2

.

Similarly, ⎛ 15 ⎞ ⎜⎜⎜ ⎟⎟⎟ ⎟⎟ −2δt 30 e p μ dt + v p ⎟⎟⎠ t x x+t 15 x ⎟

⎜ E[Z22 ] = 100 ⎜⎜⎜⎜ ⎝

5

= 100 Āx:1 15 − 2 Āx:1 5 + v 30 15 px

j 2 1 2 1 30 = 100 Ax:15 − Ax:5 + v 15 px , δj 2

and hence

j V[Z2 ] = 100 δj

2

Ax:15 − Ax:5 + v 1

2

1

30

15 px

2

− 100 Āx:1 15 − Āx:1 5 + 15 Ex . (e) Since Cov[Z1 , Z2 ] = E[Z1 Z2 ] − E[Z1 ]E[Z2 ] and since, from part (c), we already have expressions in terms of standard actuarial functions for E[Z1 ] and E[Z2 ], it remains to find an expression for E[Z1 Z2 ].

51

Solutions for Chapter 4 From part (a) we have ⎧ ⎪ 0 ⎪ ⎪ ⎪ ⎨ Z1 Z2 = ⎪ 200 v 2Tx ⎪ ⎪ ⎪ ⎩ 100 v 15 v Tx

if Tx ≤ 5, if 5 < Tx ≤ 15, if 15 < Tx .

Hence E[Z1 Z2 ] = 200 2Āx:1 15 − 2Āx:1 5 + 100 v 15 Āx − Āx:1 15 and so Cov[Z1 , Z2 ] = 200 2Āx:1 15 − 2Āx:1 5 + 100 v 15 Āx − Āx:1 15 − 100 Āx + Āx:1 15

Āx:1 15 − Āx:1 5 + 15 Ex .

Note that there are several ways to express the covariance using standard actuarial functions. (f) With x = 40 and 1+ j = 1.052 = eδ j , we have (using UDD)

j 2 E[Z12 ] = 100 A40 + 3 2 A 1 40:15 δj

j 2 l55 2 = 100 A40 + 3 2 A40 − e−15δ j A55 δj l40 = 4.4872.

Similarly, with i = 0.05 and UDD, E[Z1 ] = 10 Ā40 + 10 Ā 1

40:15

= 1.3391,

giving SD[Z1 ] = 1.6414.

Again with x = 40 and 1+ j = 1.052 = eδ j , and assuming UDD, we have

j −5δ j l45 l55 l55 e A45 − e−15δ j A55 + v 30 E[Z22 ] = 100 δj l40 l40 l40 = 23.2252, and

E[Z2 ] = 10 Ā 1

40: 15

giving SD[Z2 ] = 0.3085.

− Ā 1

40: 5

+ 15 E40 = 4.8094,

52

Solutions for Chapter 4

4.22 (a) The present value is

⎧ ⎪ 0 ⎪ ⎪ ⎪ ⎨ Z=⎪ 100 000 v T40 ⎪ ⎪ ⎪ ⎩ 50 000 v 25

if T40 < 5, if 5 ≤ T40 < 25, if T40 ≥ 25.

(b) Figure S4.1 shows a plot of the present value of the benefit against time of death, t. If the life dies in the first five years, no benefit is payable, so the present value is 0. The maximum death benefit present value arises if the life dies at time 5; it is 100 000 v 5 = 78 353. The minimum present value of the death benefit, given that the life dies between time 5 and time 25, is 100 000 v 25 = 29 530. The survival benefit has present value 50 000 v 25 = 14 765.

Figure S4.1 Present value of the benefit, by time of death, for Exercise 4.22.

Solutions for Chapter 4

53

(c) The EPV of the benefit is

+ 0.5 20 E45 105 5 E40 Ā 1 45:20 i

A1 = 105 5 E40 + 0.5 20 E45 using UDD δ 45:20 i

(A45 − 20 E45 A65 ) + 0.5 20 E45 = 105 5 E40 δ

= 105 × 0.78113 1.02480 (0.15161 − 0.35994 × 0.35477)

+ 0.5 × 0.35994 = 15 972. (d) From the calculations for part (b) we see that the present value of the benefit is more than 16 000 only if the life dies between time 5 and time 25, with probability 5 |20 q40 = 5 p40 − 25 p40 = 0.0448.

(e) Let Z denote the present value of the benefit. The 60% quantile of Z is the value Q such that Pr[Z ≤ Q] = 0.6.

We can build up the distribution function for Z using the information in part (b). There are two probability masses for Z, one at Z = 0 and one at Z = 14 765. The rest of the distribution of Z is continuous, with 29 530 ≤ Z ≤ 78 353. So we have Pr[Z = 0] = 5 q40 = 0.00306, Pr[Z = 14 765] = 25 p40 = 0.95210, ⇒ Pr[Z ≤ 14 765] = 0.95516 and we see that the 60% quantile of the present value of the benefit is 14 765 (as are all quantiles greater than 0.306% and less than 95.516%).

4.23 (a) We have d d 1 Ā = dt 50: t dt

t v r r p50 μ50+r dr = vt t p50 μ50+t .

0

Evaluating this at t = 10 gives 0.004896.

54

Solutions for Chapter 4 (b) As t |Ā50 = Ā50 − Ā 1

50: t

, it follows that

d 1 d Ā = −vt t p50 μ50+t . t |Ā50 = − dt dt 50: t So the required answer is −0.004896.

4.24 Again we use the technique of considering the whole life insurance as the sum of deferred one-year term insurances so that Āx =

∞

1 v t px Āx+t: 1 t

1

and

Āy: 1 =

vt t py μy+t dt.

1

t=0

0

Under the assumption of a constant force of mortality, such that μy+t = νy for 0 ≤ t < 1, we have t py = exp{−νy t} for 0 ≤ t ≤ 1, so that 1 Ā y: 1 =

1 t py μy+t v dt = t

1

0

e 0

−νy t

νy = 1 − exp{−(δ + νy )} δ + νy νy = 1 − v py . δ + νy

1 νy v dt = νy t

e−(δ +νy ) t dt

0

Substituting this back into the original sum, replacing y with x + t, gives Āx =

∞

t=0

v t t px

νx+t (1 − v px+t ) . δ + νx+t

4.25 First note that Z3 = 10Z1 + 2Z2 , and that Z1 and Z2 are dependent variables.

Then E[Z3 ] = 10 E[Z1 ] + 2 E[Z2 ] = 66 + 215 = 281, V[Z3 ] = 100 V[Z1 ] + 4 V[Z2 ] + 40 Cov[Z1 , Z2 ] , where Cov[Z1 , Z2 ] = E[Z1 Z2 ] − E[Z1 ] E[Z2 ]. Note that Z1 Z2 = 0, as Z1 = 0 when Kx ≥ n and Z2 = 0 when Kx < n. So Cov[Z1 , Z2 ] = −E[Z1 ] E[Z2 ] = −709.5.

Inserting the numbers into the expression for V[Z3 ] results in SD[Z3 ] = 258.33.

55

Solutions for Chapter 4 4.26 (a) Starting from formula (4.19) in AMLCR, we have Ax: n = Ax:1 n + n Ex =

n−1

vt+1 t |qx + v n n px

t=0

=

n−2

vt+1 t |qx + v n n−1 |qx + v n n px

t=0

=

n−2

vt+1 t |qx + v n n−1 px

t=0

since n−1 |qx + n px = n−1 px . (b) Formula (4.19) in AMLCR splits the EPV into a death benefit, covering the n-year term, and a benefit on survival to age x + n. The formula in part (a) splits the benefit according to the possible payment times. Payments may be made at any of the times t = 1, 2, . . . , n − 1, if (x) dies in the year from t − 1 to t, and a payment is made at time n if (x) dies in the year from n − 1 to n, or if (x) survives to n.

That is, if (x) survives to the start of the final year (with probability n−1 px ), then the benefit will be paid at the year end, whether (x) dies or survives the year.

4.27 (a) The EPV is 2000 A

(4)

1

50:15

+ 1000 15 |A(4) 50 = 2000 A

(4)

= 1000 A

(4)

1

50:15

(4) 1 + 1000 A(4) 50 − A

50:15

+ 1000 A(4) 50 50:15

i 1 = 1000 (4) A + A50 using UDD 50:15 i = 218.88.

1

(4) (b) For the variance, letting Z denote the present value of the benefit and K50 denote the quarterly curtate future lifetime of (50), we have

⎧ 2 K (4) + 1 ⎪ ⎪ ⎨ 20002 v 50 4 Z =⎪ ⎪ ⎩ 10002 v 2 K50(4) + 14 2

(4) if K50 + 14 ≤ 15, (4) if K50 + 14 > 15 .

56

Solutions for Chapter 4 The expected value is

(4) E[Z 2 ] = 20002 2 A 1

50:15

2 (4) + 10002 2 A(4) 50 − A

50: 15

= 105 451 so that the standard deviation required is $239.88. (c) The cost of benefits will exceed the accumulated premium on death at time T , say, if the present value of the benefits up to T is greater than the premium.

During the first 15 years

( ) (4) 1 Pr[Z > 500] = Pr 2000 v K50 + 4 > 500 ( (4) 1 ) = Pr v K50 + 4 > 0.25 & ' 1 (4) = Pr K50 + < 28.4 4

so that the EPV of the benefit will exceed the single premium for any payment during the first 15 years.

If the benefit is paid after the first 15 years, the amount drops to $1000, and the maximum present value is 1000 v 15.25 = 475.2, which is less than the premium. That is, if (50) survives the first 15 years, the accumulated premium will exceed the value of the benefit whenever it is paid.

So, the probability that the EPV of benefit exceeds the premium is the probability that (50) dies in the first 15 years, which is 15 q50 = 0.04054.

4.28 The covariance is Cov[Z1 , Z2 ] = E[Z1 Z2 ] − E[Z1 ] E[Z2 ]. Assuming that the benefit is payable immediately on death, we have ⎧ ⎪ ⎪ if Tx ≤ n, ⎨ v Tx and Z2 = v Tx , Z1 = ⎪ ⎪ ⎩ 0 if Tx > n, with expected values E[Z1 ] = Āx:1 n and E[Z2 ] = Āx .

We see from the definitions that

⎧ ⎪ ⎪ ⎨ v 2Tx Z1 Z2 = ⎪ ⎪ ⎩ 0

if Tx ≤ n, if Tx > n,

so that E[Z1 Z2 ] = 2Āx:1 n

57

Solutions for Chapter 4 and so Cov[Z1 , Z2 ] = 2Āx:1 n − Āx:1 n Āx .

4.29 We can consider a whole life insurance as a sum of deferred one-year term insurances, and hence write A(m) x =

∞

(m)

1 vt t px A x+t: , 1

t=0

and similarly

IA

(m)

x =

∞

(m)

1 (t + 1) vt t px A x+t: .

1

t=0

Writing this equation in the following array form gives a clue as to how we can reorganize the expression: ⎧ (m) 1 ⎪ A x: 1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ (m) ⎨ IA x = ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

(m)

+ v px A x+1 1: 1 +

(m)

+

···

(m)

+

···

(m)

+

···

+

···

+ v 2 2 px A x+1 2: 1

(m)

+ v 2 2 px A x+1 2: 1

v px A x+1 1: 1

+ v 2 2 px A x+1 2: 1

Now, considering each row separately, (IA(m) )x = A(m) x + v px A x+1 1: 1 + v px+1 A x+1 2: 1 + v 2 2 px+1 A x+1 3: 1 + · · · (m)

(m)

(m)

+ v 2 2 px A x+1 2: 1 + v px+2 A x+1 3: 1 + v 2 2 px+2 A x+1 4: 1 + · · · (m)

(m)

(m)

+··· (m) (m) 2 = A(m) x + v px Ax+1 + v 2 px Ax+2 + · · · .

The explanation for this formula is that we can view an increasing whole life insurance as the sum of deferred whole life insurance policies with deferred periods 0, 1, 2, . . . years and sum insured $1, so that the death benefit in the tth policy year is t + 1. This argument applies whether the sum insured is payable at the moment of death, the end of the (1/m)th year of death, or at the end of the year of death.

58

Solutions for Chapter 4

4.30 (a) We have 1 = (IA)x:n

n−1

(k + 1) v k+1 k |qx

k=0

= v qx + 2v 2 1 |qx + 3v 3 2 |qx + · · · + nv n n−1 |qx = v qx + 2v 2 px qx+1 + 3v 3 2 px qx+2 + · · · + nv n n−1 px qx+n−1 = v qx + v px (2v qx+1 + 3v 2 px+1 qx+2 + · · · + nv n−1 n−2 px+1 qx+n−1 ) = v qx + v px (v qx+1 + 2v 2 px+1 qx+2 + · · ·+ (n − 1)v n−1 n−2 px+1 qx+n−1 ) + v px (v qx+1 + v 2 px+1 qx+2 + · · · + v n−1 n−2 px+1 qx+n−1 )

= v qx + v px (IA) 1 +A1 . x+1: n−1

x+1: n−1

(b) This formula splits the EPV according to whether (x) survives for one year or not. The contribution to the EPV resulting from death in the first policy year is v qx . If (x) survives the first policy year, the insurance benefit is an increasing benefit to a life now aged x + 1 with term n−1 years and benefit levels 2, 3, . . . , n. This part of the benefit can be expressed as an increasing benefit of 1, 2, . . . , n − 1, with EPV at age x + 1 of (IA) 1 , plus a level x+1: n−1 term benefit of 1 for the remaining n − 1 years of the contract, which has EPV at age x+1 of A 1 . To get the EPV at age x of the benefits payable x+1: n−1 from age x + 1, discount for interest and survival for one year using v px . (c) If we let n → ∞ in part (a) we get (IA)50 = v q50 + v p50 (IA)51 + A51 .

From the information given, we have 0.00558 = A 1

50: 1

= v q50 ⇒ p50 = 0.99409,

4.99675 = (IA)50 = 0.00558 + v 0.99409 (IA)51 + 0.24905 ⇒ (IA)51 = 5.07307.

4.31 We have ∞ (I¯Ā)x =

t vt t px μx+t dt 0 ∞

j+1

=

j=0 j

t vt t px μx+t dt

59

Solutions for Chapter 4 ∞

=

1 v j px

j=0 ∞

=

(r + j) v r r px+ j μx+ j+r dr

j

0

1 v j px qx+ j

j=0 ∞

=

(r + j) v r dr

j

under UDD

0

1 v

j+1

j px qx+ j

j=0

(r + j) eδ (1−r) dr.

0

Now 1 e

δ (1−r)

0

i dr = δ

1 and

r eδ (1−r) dr =

0

eδ − 1 − δ i−δ = 2 , 2 δ δ

giving ∞

1 v

j+1

j px qx+ j

j=0

(r + j) eδ (1−r) dr =

0

=

∞ i−δ i j+1 A + j v j px qx+ j x δ j=0 δ2

i−δ i Ax + ((IA)x − Ax ) .

2 δ δ

4.32 (a) The EPV of the benefit is 2

2 1.05 v t p70 μ70+t dt = 100 000 t

100 000

t p70 μ70+t dt

t

0

0

= 100 000 2 q70 = 2 196.19, where the first identity holds because 1.05v = 1 as the interest rate is 5%. (b) The EPV of the benefit is in the same form as in part (a), but now v is calculated at 4%, and with the constant force of mortality assumption, −0.012t for 0 ≤ t ≤ 2. So the EPV is t p70 = e 2

t 1.05 v4% e−0.012t 0.012 dt = 1 200

2

t

100 000 0

as v4% = e−0.039221 .

0

e−0.00243t dt = 2 394.18,

60

Solutions for Chapter 4

4.33 (a)

(i) We use the formula Ax = v qx + v px Ax+1 and start the recursion by setting Ax = v for a suitably large value of x such as 119 (i.e. assume ω = 120). (ii) We use the formula 1

1

(4) 4 4 A(4) x = v 41 qx + v 41 px Ax+ 1 4

1 4

and start the recursion by setting A(4) x = v for a suitably large value of x such as 119 34 . (For consistency with part (i), we have set the age to be 14 less than the value of ω assumed there.) (iii) We get the following values: A50 = 0.33587, (4)

A50 = 0.34330,

A100 = 0.87508, (4)

A100 = 0.89647.

(4) (b) Under UDD we have Ax = i/i(4) Ax and, when i = 0.06, we have i/i(4) = 1.02223, giving the approximate values as (4)

A50 = 0.34333

and

(4)

A100 = 0.89453.

(c) The approximation is very good at age 50 (with an error less than 0.01%) but not as good at age 100 (although still fairly accurate, with an error of around 0.2%). The reason for this is that UDD is fairly accurate when the mortality probabilities are low for adult ages, but UDD becomes rather less (4) accurate at older ages. In the approximation of A50 the older ages have less impact on the overall value, partly due to the dampening effect of discount(4) ing, but in the approximation of A100 the mortality rates are very high, the discount factors are low, and the impact of the UDD approximation is more significant.

4.34 (a) We use ⎧ ⎫ ⎧ t ⎫ ⎪ ⎪

h ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ ⎨ ⎬ − μ ds = p × exp μ ds − . ⎪ ⎪ ⎪ t p60 = exp ⎪ 60+s t−h 60 60+t−h+s ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ ⎪ ⎪ ⎩ ⎭ 0

0

We calculate the table of values for t p60 recursively, with 0 p60 = 1, and

61

Solutions for Chapter 4 using the trapezium rule for h μ60+t−h+s ds, 0

for each period of h = 1/40 years. An excerpt from the table is shown below. t

μ60+t

t p60

0 0.025 0.050 0.075 .. .

0.003850 0.003855 0.003860 0.003866 .. .

1.000000 0.999904 0.999807 0.999711 .. .

(b) We use 2 Ā

1

60 : 2

=

vt t p60 μ60+t dt 0

and evaluate the integral numerically using the repeated Simpson’s method, with a step of h = 1/40, giving

Ā 1 ≈ 1 × 0 p60 × μ60 × v 0 + 4 × h p60 × μ60+h × v h 60 : 2

+ 2 × 2h p60 × μ60+2h × v 2h + 4 × 3h p60 × μ60+3h × v 3h

+ · · · + 4 × 79h p60 × μ60+79h × v 79h + 1 × 2 p60 × μ62 × v 2 h/3 = 0.007725.

Solutions for Chapter 5

5.1 (a) The benefit is an annuity of 1 per year payable continuously to (x) for at most 15 years, with payments being made only when (x) is alive. The EPV is āx: 15 . (b) The benefit is an annuity of 1 per year payable annually in arrear to (x), with payments being guaranteed for the first 15 years and being payable thereafter only if (x) is alive. The EPV is ax:15 = a15 + 15 Ex ax+15 .

5.2 (a) The random variable Y is the present value of an annuity payable continuously at rate 1 per year following the death of (x) before time n (years).

The annuity payments are guaranteed to be paid from the death of (x) until time n, and cease at time n. If (x) survives to time n, there are no payments. (b) To determine the EPV of the benefit, we can sum the product of the amount payable, probability of payment and discount factor, over all possible payment dates. Because the benefit is payable continuously, the sum here is an integral.

Consider the interval (t, t + dt), for t < n. The probability that the annuity is paid is the probability that (x) has died at that time, which is t qx . The discount factor is vt and the amount of benefit is dt. The EPV therefore is n n n n t t t E[Y ] = t qx v dt = (1 − t px ) v dt = v dt − t px vt dt 0

0

0

= ā n − āx: n .

62

0

Solutions for Chapter 5

63

(c) The term ā n values a benefit payable continuously for n years, and the term āx: n values a benefit payable continuously for n years provided that (x) is alive. The difference between these two terms must therefore value a benefit payable continuously for n years provided that (x) is not alive.

5.3 We know that ä50: 10 = 1 + v p50 + v 2 2 p50 + · · · + v 9 9 p50 and a50: 10 = v p50 + v 2 2 p50 + · · · + v 9 9 p50 + v 10 10 p50 , giving ä50:10 − a50:10 = 1 − v 10 10 p50 .

The values in the question give v 10 = 0.675476, so i = 4.0014%.

5.4 We use Ax:1 10 = Ax: 10 − 10 Ex and Ax: 10 = 1 − d äx: 10 .

From the given information we calculate äx: 10 = äx − 10 |äx = 6, giving Ax: 10 = 1 − (1 − 0.94) 6 = 0.64 and hence Ax:1 10 = 0.64 − 0.375 = 0.265.

5.5 Using the recursive relationship ax = v px (1 + ax+1 ) we have ax , px = (1 + i) 1 + ax+1 giving p60 = 0.99147 and p61 = 0.99065, so that 2 p60 = 0.98220.

5.6 The present value random variable is Z = 50 000

1 − v K[60] +1 , d

64

Solutions for Chapter 5 which gives 2 50 000 V[v K[60] +1 ] V[Z] = d

2 50 000 2 = A[60] − A2[60] d

2 50 000 (0.10781 − 0.289842 ). = d

Hence, the standard deviation of Z is 161 996.

5.7 We use 1 − d (12) ä(12)

50:20

= A(12) , where 50:20

d (12) = 12 (1 − 1.06−1/12 ) = 0.05813, and A(12)

50:20

giving ä(12)

50:20

(12) = A(12) 50 − 20 E50 A70 + 20 E50 = 0.33749,

= 11.3974.

5.8 First, note that as i = 5%, we have 1.05v = 1. The EPV of the deferred annuity is ∞ ∞

vt 1.05t−5 t p70 = 10 000 v r+5 1.05r r+5 p70 10 000 r=0

t=5

= 10 000 v 5 5 p70

∞

(1.05v)r r p75

r=0

= 10 000 5 E70 (1 + e75 ) = 10 000 × 0.73295 × 15.102 = 110 690.

5.9 The EPV of the annuity is ä50 = 17.0245, and the total payment made under the annuity exceeds this if (at least) 18 payments are made. As payments are made at the start of each year, 18 payments will be made if the annuitant survives at least 17 years, the probability of which is 17 p50 = 0.94747.

5.10 (a) We have ä40:20 = ä40 − 20 E40 ä60 = 18.4578 − 0.36663 × 14.9041 = 12.994.

65

Solutions for Chapter 5 (b) Using Woolhouse’s formula with two terms ä(4)

40:20

3 = ä40:20 − (1 − 20 E40 ) 8 3 = 12.994 − (1 − 0.36663) 8 = 12.756.

(c) Under the UDD assumption ā25:10 = α(∞) ä25:10 − β (∞)(1 − 10 E25 ), where α(∞) =

id = 1.000198, δ2

β (∞) =

i−δ = 0.508232, δ2

which gives ā25:10 = 1.000198 (19.7090 − 0.61198 × 18.9728) − 0.508232 (1 − 0.61198) = 7.902. (d) Using Woolhouse’s formula with two terms a(12)

50:20

1 (1 − 20 E50 ) 12 11 1 = ä50:20 − (1 − 20 E50 ) − (1 − 20 E50 ) 24 12 13 = ä50 − 20 E50 ä70 − (1 − 20 E50 ) 24 13 = 17.0245 − 0.34824 × 12.0083 − (1 − 0.34824) 24 = 12.490. = ä(12)

50:20

−

(e) Under the UDD assumption, id i − i(12) α(12) = (12) (12) and β (12) = (12) (12) i d i d and 20 |ä45 = 20 E45 ä65 (12)

(12)

= 20 E45 α(12) ä65 − β (12)

= 0.35994 (1.000197 × 13.5498 − 0.466508) = 4.7102.

66

Solutions for Chapter 5 (f) We now have 20 |ä45 = 20 E45 ä65 (12)

(12)

143 11 − = 20 E45 ä65 − μ65 + δ 24 1728

11 143 = 0.35994 13.5498 − − 0.00560 + δ 24 1728 = 4.7105.

5.11 We have ä[x−1]+1:n = 17.5 =

n−1

vt t p[x−1]+1

t=0

=

n−1

vt

t=0

=1+

l[x−1]+1+t l[x−1]+1

n−1

vt

t=1

lx+t l[x−1]+1

since a two-year select period implies that l[x−1]+1+t = lx+t for t = 1, 2, 3, . . ..

Thus, n−1

16.5 l[x−1]+1 = vt lx+t . t=1

Similarly, äx:n = 17 = 1 +

n−1

vt

t=1

lx+t , lx

and 16 lx =

n−1

vt lx+t .

t=1

Hence p[x−1]+1 16.5 lx lx+1 lx = = = = 1.03125.

16 l[x−1]+1 l[x−1]+1 lx+1 px 5.12 Using to denote Scott’s mortality experience, the probability that he survives to age 41 is p 40 = e−0.05 p45

67

Solutions for Chapter 5 and the probability that he survives to age 40 + t for t = 2, 3, 4, . . . is

−0.05 p45 × t−1 p46 = e−0.05 t p45 . t p40 = p40 × t−1 p41 = e

Hence the EPV of the annuity is ⎛ ⎞ ⎛ ⎞ ∞ ∞

⎜⎜⎜ ⎟⎟ ⎜⎜⎜ ⎟⎟ t ⎟ −0.05 t ⎜ ⎟ ⎜ 10 000 ⎜⎝1 + v t p40 ⎟⎠ = 10 000 ⎜⎝1 + e v t p45 ⎟⎟⎟⎠ t=1

t=1

⎞ ⎛ ∞

⎟⎟ ⎜⎜⎜ −0.05 t ⎜ = 10 000 ⎜⎝1 + e v p45 v t p46 ⎟⎟⎟⎠ t=0

= 10 000 1 + e−0.05 v p45 ä46 = 10 000 1 + e−0.05 v p45 17.6706 = 169 960.

5.13 We first evaluate the mean and variance of Y , then assume that Y is normally distributed to find the deposit. Let Yi be the present value random variable for the ith life, for i = 1, 2, . . . , 200. Then Y =

200

Yi ,

i=1

with

⎡ 200 ⎤ 200 ⎢⎢⎢ ⎥⎥⎥ E[Y ] = E ⎢⎢⎣⎢ Yi ⎥⎥⎦⎥ = E[Yi ], i=1

i=1

and, as E[Yi ] = 100 äx = 1301.704, for i = 1, 2, . . . , 200, E[Y ] = 260 340.8.

By independence of the lives, V[Y ] =

200

V[Yi ] ,

i=1

and for i = 1, 2, . . . , 200 Ax − A2x , d2 with Ax = 1 − d äx = 0.49934 (calculated with i = 4%). The same calculation with i = 8.16% gives

0.0816 2 Ax = 1 − 9.59176 = 0.27636.

1.0816 V[Yi ] = 104

2

68

Solutions for Chapter 5 Hence V[Yi ] = 182 629.9 and V[Y ] = 6043.672 .

Now let D be the amount of the deposit in the trust fund. We require that Pr[Y < D] = 0.9, which, under our assumption of a normal distribution for Y , is the same as requiring & ' D − E[Y ] Pr Z < √ = 0.9, V[Y ] where Z ∼ N(0, 1). As Pr[Z < 1.2816] = 0.9, D = 1.2816 × 6043.67 + 260 340.8 = 268 086.

5.14 (a) ä[40]:4 =

l[40] + l[40]+1 v + l42 v 2 + l43 v 3 = 3.6664. l[40]

(b) a[40]+1:4 =

l42 v + l43 v 2 + l44 v 3 + l45 v 4 = 3.4506. l[40]+1

(c) (Ia)[40]:4 =

l[40]+1 v + 2l42 v 2 + 3l43 v 3 + 4l44 v 4 = 8.3750. l[40]

(d) (IA)[40]:4 =

d[40] v + 2d[40]+1 v 2 + 3d42 v 3 + 4d43 v 4 + 4l44 v 4 = 3.1630. l[40]

(e) Let Y denote the present value of the annuity. Then ⎧ ⎪ 1000 ä1 with probability q[41] , ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ 1000 ä2 with probability 1 |q[41] , Y =⎪ ⎪ ⎪ 1000 ä3 with probability 2 |q[41] , ⎪ ⎪ ⎪ ⎪ ⎩ 1000 ä with probability 3 p[41] .

4 So E[Y ] = 3665.58 and E[Y 2 ] = 13 450 684 which gives a standard deviation of 119.14. (f) At 6% per year, än < 3 if and only if 1 − vn < 3, 1−v which gives n < 3.19. As the term n must be an integer, the present value is less than 3 if at most three annuity payments are made (i.e. if the life does not survive to age 43 since payments are in advance), and the required

Solutions for Chapter 5 probability is 3 q[40] = 1 −

l43 = 0.00421. l[40]

5.15 (a) With H = min(Kx , n), we have aH = äH+1 − 1 and so V[aH ] = V[äH+1 − 1] = V[äH+1 ] & ' 1 − v min(Kx +1,n+1) =V d =

1 d2

2

Ax:n+1 − (Ax:n+1 )2 .

(b) We have (similarly to Exercise 4.26) Ax: n+1 = A 1

x: n+1

+ n+1 px v n+1

= Ax:1 n + n px qx+n v n+1 + n px px+n v n+1 = Ax:1 n + n px v n+1 .

Then

$ % 2 1 V aH = 2 2 Ax:1 n + v 2(n+1) n px − Ax:1 n + v n+1 n px d

2 1 = 2 2 Ax:1 n − Ax:1 n + v 2(n+1) n px − v 2(n+1) (n px )2 d 2 − 2 Ax:1 n v n+1 n px d 2

v 2n p (1 − p ) 2 1 + i 2 1 n x n x = Ax: n − Ax:1 n + i i2 2(1 + i) v n n px 1 − Ax: n . i2

69

70

Solutions for Chapter 5

5.16 If μx = 0.5 (μxA + μxB ), then ⎧ t ⎫

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ μ dr − ⎪ t px = exp ⎪ x+r ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ 0 ⎫ ⎧ t ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎨ A B + μ μ dr = exp ⎪ −0.5 ⎪ x+r x+r ⎪ ⎪ ⎪ ⎪ ⎭ ⎩ 0

A 1/2 t px

=

B 1/2 . t px

For positive α and β , αβ 1/2 ≤ (α + β )/2 since a geometric mean is less than an arithmetic mean. Alternatively, note that for positive α and β , α + β − 2(α β )1/2 = (α 1/2 − β 1/2 )2 ≥ 0.

Thus A 1/2 t px

B 1/2 ≤ 0.5 t px

∞

∞

A B t px + t px

and so ax =

vt t px =

t=1

vt

A 1/2 t px

B 1/2 t px

t=1

≤ 0.5 aAx + aBx .

Hence, the approximation overstates the true value.

5.17 The present value of the increasing annuity is (I ä)K +1 and the present value of x the death benefit is (Kx + 1) v Kx +1 . Also, $ % (I ä)x = E (I ä)K +1 x

and

We then have

$ % (IA)x = E (Kx + 1) v Kx +1 . ⎤ ⎡ ⎢⎢⎢ äK +1 − (Kx + 1) v Kx +1 ⎥⎥⎥ x ⎥⎥⎥ ⎢ (I ä)x = E ⎢⎢⎣ ⎦ d =

giving (IA)x = äx − d(I ä)x .

1 (äx − (IA)x ) , d

71

Solutions for Chapter 5

5.18 (a) For two random variables X and Y , Cov[X, Y ] = E[XY ] − E[X] E[Y ]. Thus, % $ % $ % $ % $ Cov āTx , v Tx = E āTx v Tx − E āTx E v Tx & ' 1 − v Tx Tx v − āx Āx =E δ 1 1 − Āx Āx − 2 Āx − Āx = δ δ = (b) As V[v Tx ] = 2 Āx − Āx

2

Āx

2

− 2 Āx δ

.

$ % > 0, we see that Cov āTx , v Tx < 0.

(c) As future lifetime increases, the annuity present value (Y ) increases (there are more payments under the annuity) whereas the insurance present value (Z) decreases (there is more time to earn interest to offset the cost of the death benefit), so Y and Z are negatively correlated and hence have a negative covariance.

5.19 Write everything in terms of δ : 1 1 i = δ + δ2 + δ3 + · · · , 2 6 1 2 1 3 d = δ − δ + δ − ··· , 2 6 1 2 1 3 (m) i =δ+ δ + δ + ··· , 2m 6m2 1 2 1 3 δ + δ − ··· . d (m) = δ − 2m 6m2 Then, ignoring terms in δ 4 and higher powers, id ≈ δ 2 and i(m) d (m) ≈ δ 2 so that id α(m) = (m) (m) ≈ 1. i d Similarly, ignoring terms in δ 3 and higher powers,

1 1 i − i(m) ≈ δ 2 1 − 2 m and so

1 1 m−1 i − i(m) .

1− = β (m) = (m) (m) ≈ 2 m 2m i d

72

Solutions for Chapter 5

5.20 (a) The EPV is 10 000 (40 ä60 + 30 ä70 + 10 ä80 ) = 10 418 970. (b) For a life aged x, the variance of the present value is 108 σx2 , where σx2 =

2

Ax − A2x . d2

By the independence of the lives, the variance of the present value is 2 2 2 + 30 σ70 + 10 σ80 108 40 σ60

and so the standard deviation is 0 2 2 2 + 30 σ70 + 10 σ80 = 311 537.

104 40 σ60 (c) The 95th percentile of the standard normal distribution is 1.644854, so the 95th percentile of the distribution of the present value is 10 418 961 + 1.644854 × 311 534 = 10 931 402.

5.21 (a) We know that d t px = t px (μx − μx+t ). dx Then ∞ ∞

d t d äx = v t px = vt t px (μx − μx+t ) dx dx t=0 t=0

= μx äx −

∞

vt t px μx+t .

t=0

(b) Similarly, n−1

d äx:n = μx äx:n − vt t px μx+t . dx t=0

5.22 (a) The benefit is a deferred annuity of 1000 per year payable continuously to (x), with a deferred period of five years. (b) The simplest expression is 1000 5 |āx .

73

Solutions for Chapter 5 (c)

(i) We have (Y | I=1) = 1000 v 5 āTx+5 = 1000 v 5

1 − v Tx+5 , δ

and so V[Y |I=1] = 106 v 10

2 Āx+5 − Ā2x+5 V[v Tx+5 ] 6 10 = 10 v . δ2 δ2

(ii) We use the result (given in Appendix A.4) $ " #% $ % V[Y ] = E V Y | I + V E[Y | I] .

Now (Y | I=0) = 0, so E[Y | I=0] = V[Y | I=0] = 0. Then $ % E V[Y | I] = 5 px V[Y | I=1] + 5 qx × 0.

As E[Y | I=1] = 1000 v 5 āx+5 , we can write E[Y | I] = I 1000 v 5 āx+5 .

Note that I ∼ B(1, 5 px ), so V[I] = 5 px 5 qx , and $ % 2 2 V E[Y |I] = 1000 v 5 āx+5 V[I] = 106 v 5 āx+5 5 px 5 qx ⎛ ⎞ 2 Āx+5 − Ā2x+5 ⎟⎟⎟ ⎜⎜ 5 2 6 ⎜ 10 ⇒ V[Y ] = 10 ⎝⎜ v āx+5 5 px 5 qx + 5 px v ⎠⎟ . (5.1) δ2 (d)

(i) First, E[Y ] = 1000 5 |ā65 = 1000 5 E65 ā70 , and under UDD, ā70 =

id i−δ ä70 − 2 = 1.000198 × 12.0083 − 0.508232 = 11.50245, δ2 δ

giving E[Y ] = 1000 × 0.75455 × 11.50245 = 8679.

To reduce use of tables, rewrite the first term in parentheses in (5.1) as v 5 ā70

2

v 5 ā70

2

5 p65 (1 − 5 p65 )

= (ā70 ) v

10

5 p65 (1 − 5 p65 )

5 p65 5 q65 =

2

2 = (ā70 )2 v 5 5 E65 − 5 E65

= 2.8927.

74

Solutions for Chapter 5 The second term in parentheses in (5.1) is 2 10 5 p65 v

2 Ā70 − Ā270 Ā70 − Ā270 5 = v E 5 65 δ2 δ2 = 8.1833,

where 1.052 − 1 2 0.05 A70 .

A70 and Ā70 = 2δ δ Hence the standard deviation of Y is √ 1 000 2.8927 + 8.1833 = 3328.

2

Ā70 =

(ii) If the life dies at time t > 5, the present value of the annuity payments is 1000 v 5 āt −5 . So the present value is greater than 8679 if death occurs after time τ, where 1 000 v 5

1 − v τ−5 = 8679. δ

Rearranging, we obtain v τ−5 = 0.45995 giving τ =5+

log 0.45995 = 20.936. log v

So the present value exceeds 8 679 if the life survives for at least 20.936 years, with probability 20.936 p65 . Using life table functions, 20 p65 = 0.64691 and 21 p65 = 0.60961, giving 20.936 p65 = 0.064 20 p65 + 0.936 21 p65 = 0.6120.

(iii) Let Ỹ denote the present value of the payments from the 10 000 policies. As the lives are independent, the distribution % of Ỹ is approxi$ mately normal, and so our estimate of Pr Ỹ > E[Ỹ ] is 0.5.

5.23 We have

& ' $ % # " 1 − e−δ Tx ax = E āTx = E = E f (Tx ) , δ

where f (x) = (1 − e−δ x )/δ .

Hence f (x) > 0 and f (x) < 0, and (as given in the question) Jensen’s inequality for concave functions gives " # E f (Tx ) ≤ f (E[Tx ])

75

Solutions for Chapter 5 so ax ≤ f (E[Tx ]) = āE[T ] . x

5.24 (a) The formula for the EPV is EPV =

10

(I ä)t t−1 |q50 + 10 p50 (I ä)10 , t=1

and the formula for the variance is 10

2

(I ä)t

2 2 t−1 |q[50] + 10 p[50] ((I ä)10 ) − (EPV) .

t=1

For a recursive calculation we can use (I ä)t = (I ä)t−1 + t vt−1 with (I ä)1 = 1.

The EPV is 13.550 and the variance is 12.497, giving a standard deviation of 3.535. (b) The present value of the annuity if it were payable for t years certain would be PVt = 1 + 1.03v + 1.032 v 2 + · · · + 1.03t−1 vt−1 =

1 − 1.03t vt = ät j , 1 − 1.03v

where ät j is the present value of an annuity-certain evaluated at effective interest rate j, where 1 + j = (1 + i)/1.03. Hence the formulae for the first two moments of the present value are respectively 10

ät j

t−1 |q50 + 10 p50 ä10 j

t=1

and 10

ät j

2

t−1 |q50 + 10 p50

2

ä10 j .

t=1

For a recursive calculation, ä1 j = 1 and for t = 2, 3, . . . , ät j = ät−1 j + 1.03t−1 vt−1 .

The first two moments of the present value are 13.814 and 199.209, leading to a standard deviation of 2.895.

76

Solutions for Chapter 5

5.25 (a) The EPV is calculated as ∞

ät t−1 |q65 = 13.550

t=1

and the second moment is calculated as ∞

2 ät t−1 |q65 = 196.094, t=1

which gives a standard deviation of 3.535. (b) The EPV is calculated as 10

t−1 |q65 ä10 +

t=1

∞

t−1 |q65 ät

t=11 ∞

= 10 q65 ä10 +

t−1 |q65 ät

= 13.814

t=11

and the second moment is calculated as 10

t−1 |q65

ä10

2

+

t=1

= 10 q65 ä10

2

∞

t−1 |q65

t=11 ∞

+

t−1 |q65

ät ät

2

2

t=11

= 199.209, which gives a standard deviation of 2.895.

The EPV is greater in part (b) because of the guarantee. The value of the benefit is greater for those who die in the first 10 years, and the same for those who survive 10 years, compared with the annuity in part (a), so, overall, the expected value must be larger with the guarantee than without.

However, the guarantee reduces the standard deviation of the present value as there is less variability in the payment terms. In part (a) the possible payment terms are 1, 2, 3, . . . years, whereas in part (b) they are 10, 11, 12, . . . years.

Solutions for Chapter 6

6.1 (a) Let P be the annual premium. Then L0 = 200 000 v K[30] +1 − P ämin(K

[30]+1 ,20)

.

(b) The equation of value is P ä[30]:20 = 200 000 A[30] , giving P = $1179.74 since ä[30]:20 = ä[30] − 20 E[30] ä[50] = 13.0419 and A[30] = 0.07693. (c) We want the smallest integer, n, such that the accumulation of premiums to time n exceeds the sum insured. Thus, assuming n > 20, we want n such that P s̈20 1.05n−20 > 200 000, which gives n = 53. The probability of a profit is thus 52 p[30] = 0.70704, since the death benefit will be paid at time 53 or later if the life survives for 52 years from issue of the contract.

6.2 (a) Let P be the single premium. Then ⎧ ⎪ ⎪ ⎨ 106 v T[40] − P L0 = ⎪ ⎪ ⎩ −P 77

if T[40] ≤ 5, if T[40] > 5.

78

Solutions for Chapter 6 (b) The equation of value is P = 106 Ā 1

[40]:5

and, under UDD, Ā 1

[40]:5

=

i 1 i A = A[40] − 5 E[40] A45 = 0.002594, δ [40]:5 δ

leading to P = $2594. (c) Note that P (1.055 ) < 106 , meaning that there is a loss if the death benefit is payable at time five years (or earlier). Hence the probability of a profit is 5 p[40] = 0.99704.

6.3 (a) Let S be the sum insured. Then 350 ä[41]: 3 = S A 1

[41]: 3

gives S = $216 326.38 since ä[41]: 3 =

2

vt t p[41] = 2.82965

t=0

and A1

[41]: 3

=

2

vt+1 t |q[41] = 0.00458.

t=0

(b) The possible values of L0 and the associated probabilities are as follows: S v − 350 S v 2 − 350 ä2 S v 3 − 350 ä3 −350 ä3

with probability q[41] , with probability 1 |q[41] , with probability 2 |q[41] , with probability 3 p[41] .

Now E[L0 ] = 0 since the calculation in part (a) uses the equivalence principle and so V[L0 ] = E[L02 ] 3

2 2 = S vt − 350 ät t−1 |q[41] + 350 ä3 3 p[41] t=1

= 188 541 300, giving the standard deviation as $13 731.03.

79

Solutions for Chapter 6

(c) We can see from part (b) that L0 will be positive if and only if the life dies before age 44, so Pr[L0 > 0] = 1 − 3 p[41] = 0.0052.

6.4 (a) Let P be the total premium each year.

EPV of premiums less maintenance expenses:

15 3 − 0.95P ä(4) = 0.95P ä − δ + μ [65] [65] [65] 8 192 = 0.95P × 13.1852 = 12.5260P. EPV of death benefit: 5 (4) (4) 105 A(4) ä[65] = 36 060.

[65] = 10 1 − d

As the initial expenses are 0.01 × 105 = 103 , the equation of value is 12.5260P = 36 060 + 1 000, giving P = $2 958.63. (b) Let P̃ be the revised premium. The policy now provides 1.1P̃ at issue, so the revised equation of value is 12.5260P̃ = 36 060 + 1 000 + 1.1P̃, giving P̃ = $3 243.47.

6.5 (a) Let P be the annual premium. Then ⎧ K[50] +1 ⎪ − P ämin(K +1,10) ⎪ ⎨ 100 000 v [50] L0 = ⎪ ⎪ ⎩ −P ä min(K[50] +1,10)

if K[50] < 10, if K[50] ≥ 10,

or, using an indicator random variable, L0 = 100 000 v K[50] +1 I(K[50] < 10) − P ämin(K

[50] +1,10)

(b) The equation of value is P ä[50]:10 = 100 000 A 1

[50]:10

and as ä[50]:10 = ä[50] − 10 E[50] ä60 = 8.05643

.

80

Solutions for Chapter 6 and A1

[50]:10

= A[50]:10 − 10 E[50] = 1 − d ä[50]:10 − 10 E[50] = 0.01438,

we find P = $178.50.

6.6 Let P be the annual premium. We equate the EPV of premiums with the EPV of the sum insured plus expenses. When the cash flows are complicated, it is often convenient to value each element separately.

EPV of premiums: P ä[45]:20 = 12.9409 P. EPV of endowment insurance benefit: 100 000 A[45]:20 = 38 376.55.

EPV of initial and renewal expenses: 0.08P + 42 + (0.02P + 8) ä[45]:20 = 0.3388 P + 145.53, where we have split the initial expenses as 0.1P = 0.08P + 0.02P and 50 = 42 + 8. The equation of value is 12.9409 P = 38 376.55 + 0.3388 P + 145.53, giving P = $3056.80.

6.7 Let P be the annual premium.

EPV of premiums: P ä[40]: 10 = 8.08705 P. EPV of death benefit: 100 000 A 1

[40]: 20

= 1453.58.

EPV of taxes and commission: 0.09 P ä[40]: 10 + 0.2P = 0.92783 P. EPV of policy maintenance costs: 5 ä[40]: 20 + 5 = 69.97.

Equate the EPVs of income and outgo to give P = $212.81.

Solutions for Chapter 6

81

6.8 We know that

L0 = v Kx +1 − P äK +1 = v Kx +1 1 + P/d − P/d x

and

L0∗ = v Kx +1 − P∗ äK +1 = v Kx +1 1 + P∗ /d − P∗ /d. x

Hence

%

$ V[L0 ] = 1 + P/d 2 V v Kx +1

so that

and

%

$ V[L0∗ ] = 1 + P∗ /d 2 V v Kx +1

2

1 + P∗ /d 2 d + P∗ V[L ] = V[L0 ].

V[L0∗ ] = 0

d+P 1 + P/d 2

Also E[L0 ] = 0 = Ax − P äx = 1 − (P + d) äx and E[L0∗ ] = −0.5 = Ax − P∗ äx = 1 − (P∗ + d) äx so that P+d 1 (P + d) äx = ∗ . = ∗ (P + d) äx 1.5 P + d Thus V[L0∗ ] = 1.52 V[L0 ] = 1.6875.

6.9 Let P denote the total annual premium.

EPV of premiums less premium expenses:

P 0.95 ä(12) − 0.15 ä(12) = 7.3321 P. [50]:10

[50]: 1

Note that the premium related expenses are 20% of all the premiums in the first year, not just the first premium.

Calculations are under UDD, so, for example,

ä(12) = α(12) ä[50] − 10 E[50] ä60 − β (12)(1 − 10 E[50] ) = 7.8723.

[50]:10

EPV of death benefit + claim expenses + other expenses: (100 000 + 250) Ā 1

[50]:10

1

v 12 + 100.

82

Solutions for Chapter 6 The v

1 12

term allows for the 1-month delay in paying claims.

Using UDD, we have i 1 A = 0.01474 δ [50]:10 which gives the EPV of the benefits and non-premium expenses of 1572.09, leading to P = $214.41. The monthly premium is thus $17.87.

Ā 1

[50]: 10

=

6.10 Let P denote the annual premium. Then P äx = 150 000 Ax .

Now L0n = 150 000 v Kx +1 − P äK +1 x

P P Kx +1 =v 150 000 + − d d so that

$ % P 2 $ Kx +1 % V L0n = 150 000 + V v d

P 2 2 = 150 000 + Ax − (Ax )2 d

P 2 0.0143 − 0.06532 . = 150 000 + d To find P/d we recall that 1 − Ax äx = d so we may write the equation of value as P Ax = 150 000 = 10 479.30. d 1 − Ax Hence

$ % V L0n = 258 460 863

and the standard deviation of L0n is $16 076.72.

6.11 Let P denote the initial annual premium.

EPV of premiums less premium expenses: P ä[55]: 10 + 0.5P 10 |ä[55] 0.97 − 0.22P = 0.5P ä[55]: 10 + 0.5P ä[55] 0.97 − 0.22P = 11.4625 P.

83

Solutions for Chapter 6 EPV of death benefit: 50 000 A[55] + 50 000 A 1

[55]:10

= 12 965.36.

Hence the initial annual premium is P = $1131.11.

6.12 (a) Let P be the annual premium. Treating the premium related expenses as 3% of each premium plus an additional 17% of the first premium, we have L0 = 100 000 v min(K[35] +1, 20) + 3000 + 0.17P − 0.97P ämin(K

[35] +1, 20)

.

(b) The equation of value is 100 000 A[35]: 20 + 3000 = P 0.97 ä[35]: 20 − 0.17 , giving P = $3287.55, since ä[35]:20 = 13.0250 and A[35]:20 = 0.37977. (c) We can write

0.97P 0.97P L0 = v min(K[35] +1, 20) 100 000 + − 3000 − 0.17P − d d

so that

2 $ % 0.97P V[L0 ] = 100 000 + V v min(K[35] +1, 20) d

2

2 0.97P 2 A[35]:20 − A[35]:20 . = 100 000 + d

We determine that 2 A[35]:20 = 0.14511, giving a standard deviation of $4964. (d) If the sum insured is payable at time k, there is a profit if the present value of premiums is greater than the present value of the sum insured and expenses, i.e. if 0.97 P äk > 100 000 v k + 3000 + 0.17 P, which gives 1.05k >

100 000 + 0.97 (P/d) = 2.6332 ⇒ k > 19.8.

0.97 (P/d) − 0.17 P − 3000

The payment date must be an integer, so there is a profit only if the benefit is payable (on death or survival) at time k = 20, i.e. if the life survives to time 19. The required probability is 19 p[35] = 0.98466.

84

Solutions for Chapter 6

6.13 (a) Let P be the single premium. Then ⎧ K[45] +1 ⎪ ⎪ ⎪ ⎨ Pv −P L0 = ⎪ ⎪ ⎪ ⎩ 40 000 äK +1 − ämin(K

[45] +1,20)

[45]

if K[45] < 20, −P

if K[45] ≥ 20.

(b) EPV of the death benefit: PA 1

[45]:20

= 0.02378 P.

EPV of the deferred annuity: 40 000 20 |ä[45] = 195 098.74.

Hence P = 0.02378 P + 195 098.74, giving P = $199 850. (c) Let P̂ be the new single premium. The EPV of the deferred annuity is now 40 000 20 E[45] (ä5 + 5 E65 ä70 ) = 195 933.28 so that the equation of value is P̂ = 0.02378 P̂ + 195 933.28, giving P̂ = $200 705.

6.14 Let P denote the annual premium.

EPV of premiums less initial expenses and renewal commission: Treating the expenses/commission as 2.5% of all premiums and 57.5% of the first premium, the EPV is 0.975P ä[40]:25 − 0.575P = 13.7083P, since ä[40]:25 = ä[40] − v 25

l65 ä65 = 14.6496. l[40]

EPV of the death benefit: 200 000

∞

vt+1 t |q[40] (1.022t ) =

t=0

since j = 1.05/1.022 − 1 = 0.0274.

200 000 A[40] j = 57 978, 1.022

85

Solutions for Chapter 6 EPV of other maintenance expenses: 5

∞

vt t p[40] (1.02t ) = 5 ä[40] j∗

t=0

where j∗ = 1.05/1.02 − 1 = 0.0294, with d j∗ = 0.0294/1.0294. Then the EPV is

1 − A40] j∗ 5 = 127.30. d j∗ Equating the EPVs of income and outgo gives P = $4 238.72.

6.15 (a) Let P denote the single premium. Then P = 20 000 30 |ä[35] + P A 1

[35]:30

.

We find that 30 |ä[35] = 30 E[35] ä65 = 2.97842

and A 1

[35]: 30

= 0.01848, giving P=

59 568 = $60 691.

1 − 0.01848

(b) Because there is an extra benefit, the revised premium will be greater than $60 691. However, the extra benefit, a deferred decreasing term insurance, is unlikely to have much effect on the EPV of the benefits and hence is likely to increase the premium by only a small amount. Assuming that the revised single premium remains less than $80 000, then the additional death benefit would be payable on death during the first three years of the annuity payout phase of the contract.

So, we let P̃ denote the revised premium and assume, first, that the additional death benefit will apply for three years from age 65 to age 68. Then we have P̃ = 20 000 30 |ä[35] + P̃ A 1

[35]:30

+ (P̃ − 20 000)v 31 30 |q[35] + (P̃ − 40 000)v 32 31 |q[35] + (P̃ − 60 000)v 33 32 |q[35] .

86

Solutions for Chapter 6 That is P̃ = 20 000 30 |ä[35] + P̃ A 1

[35]:30

= 20 000 30 |ä[35] + P̃ A

1

[35]:33

where A 1

[35]:33

30 |(IA)

+ P̃ 30 |A 1

[35]: 3

− 20 000 30 |(IA) 1

[35]: 3

− 20 000 30 |(IA)

1

[35]: 3

,

= 0.02442, and a first-principles calculation gives 1

[35]: 3

= v 31

d65 d66 d67 + 2 v 32 + 3 v 33 = 0.00803. l[35] l[35] l[35]

P̃ =

59 568 − 160.60 = $60 770.

1 − 0.02242

Then

We can see from the answer that we were correct in our assumption that the premium would not increase by a large amount and that the extra death benefit would apply only for three years. Note that, if the premium calculated had been greater than $80 000, then we would need to change the assumption about the term of the death benefit after the start of the annuity payment period, and repeat the calculations, until the answer is consistent with the assumption used.

6.16 (a) Let P be the single premium. Then P = 30 000 a[60] = $417 402. (b) The present value of profit if death occurs at age 60 + t for t = 1, 2, 3, . . . is P − 30 000 at and this is positive if at < 13.9134, which holds for t = 1, 2, 3, . . . , 24.

Thus, the probability that the present value of profit is positive is 1 − 25 p[60] = 0.36641. (c) The present value of the profit is P − 30 000 aK[60] = P − 30 000 (äK

[60] +1

− 1)

and so the variance of the present value of the profit is & ' ( ) 1 − v K[60]+1 2 2 30 000 V äK +1 = 30 000 V [60] d 30 0002 V[v K[60]+1 ] = d2 30 0002 2 A[60] − (A[60] )2 . = d2

Solutions for Chapter 6

87

Hence the standard deviation of the present value of the profit is 0 30 000 !

30 000 2 A[60] − (A[60] )2 = 0.10781 − 0.289842 d d = 97 197. (d) Let X denote the present value of profit on the 1000 policies. By the central limit theorem we have (approximately) X ∼ N(1 000(P − 417 402), 1 000 × 97 1972 ) and so

&

' −1 000 (P − 417 402) Pr[X > 0] = Pr Z > , √ 97 197 1 000

where Z ∼ N(0, 1). As Pr[Z > −1.6449] = 0.95 we find P = $412 346.

6.17 (a) The EPV of the benefits and expenses is 50 000 ä65 + 3 000 + 0.1G and the premium is 110% of this EPV, so G = 1.1 (50 000 ä65 + 3 000 + 0.1G) = 680 490 + 0.11G, giving the single premium as G = $841 055. (b) If the annuity is payable for t years (t = 1, 2, 3, . . .), then the present value of the profit is 0.9G − 3 000 − 50 000 ät .

This present value is positive if 1 − vt . d Equivalently, vt > 0.2820 or t < 25.95. So the present value is positive if the annuity is payable (in advance) for at most 25 years, meaning that it is positive if (65) does not survive 25 years, the probability of which is 25 q65 = 0.55761.

15.079 > ät =

(c) Let L0,i denote the loss at issue random variable for the ith policy when the premium is G∗ . Then L0,i = 50 000 äK +1 + 3 000 − 0.9 G∗ 65

1 − v K65 +1 + 3 000 − 0.9 G∗ . = 50 000 d

88

Solutions for Chapter 6 Then E[L0,i ] = 50 000 ä65 + 3 000 − 0.9G∗ = 680 490 − 0.9G∗ and

2 2 50 000 50 000 K65 +1 2 V[L0,i ] = V[v ]= A65 − (A65 )2 , d d giving SD[L0,i ] = 176 757.

1 ∗ Let L = 8000 i=1 L0,i . Then we set G such that

−E[L] Pr[L < 0] = Φ √ = 0.9, V[L] where Φ is the standard normal cumulative distribution function, and Φ(1.2816) = 0.9. So 1.2816 =

8000(0.9G∗ − 680490) √ 176 757 8000

(1.2816)(176 757) + 680 490 √ 8000 ⇒ G∗ = 758 914.

⇒ 0.9G∗ =

6.18 We note first that the survival function for the life subject to extra risk is ⎧ t ⎫ ⎫ ⎧ t

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪

⎨ ⎬ ⎬ ⎨

μx+r + φ dr⎪ − μx+r dr⎪ = exp ⎪ = t px e−φt − t px = exp ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ ⎭ ⎩ 0

0

so that Ā x =

∞

vt t p x μx+t dt

0

∞ =

vt t px e−φt (μx+t + φ ) dt

0

∞ =

−(δ +φ )t

e 0

∞ t px μx+t dt + φ

e−(δ +φ )t t px dt

0

= Āxj + φ āxj , where 1+ j = eδ +φ .

6.19 (a) First, we note that if the policyholder’s curtate future lifetime, K[30] , is k years, where k = 0, 1, 2, . . . , 24, then the number of bonus additions is k, the death benefit is payable k + 1 years from issue, and hence the present

89

Solutions for Chapter 6

value of the death benefit is 250 000 (1.025K[30] ) v K[30] +1 . However, if the policyholder survives for 25 years, then 25 bonuses are added. Thus the present value of the endowment insurance benefit is 250 000Z1 , where ⎧ ⎪ ⎪ ⎨ (1.025K[30] ) v K[30] +1 if K[30] ≤ 24, Z1 = ⎪ ⎪ ⎩ (1.02525 ) v 25 if K[30] ≥ 25.

As P denotes the annual premium, we have L0 = 250 000 Z1 + 1200 + 0.39P − 0.99P ämin(K

[30] +1,25)

= 250 000 Z1 + 1200 + 0.39P − 0.99P

1−v

min(K[30] +1,25)

d 0.99P 0.99P Z2 + 1200 + 0.39P − , = 250 000 Z1 + d d

where Z2 = v min(K[30] +1,25) . (b) We calculate P by taking the expected value of the equation for L0 in part (a), and setting it equal to 0.

Now E[Z1 ] =

1 A1 + v 25 j 25 p[30] , 1.025 [30]: 25 j

where j denotes calculation at interest rate j = (1+i)/1.025−1 = 0.02439. = 0.012707 and so E[Z1 ] = 0.54958. Also, We find A 1 [30]: 25 j

E[Z2 ] = A[30]: 25 = 0.29852 (at i = 5%).

So E[L0 ] = 0 ⇒ P =

250 000 × 0.549579 + 1200 = $9764.44.

(0.99/d) × (1 − 0.298517) − 0.39

(c) We have 0.992 P2 0.99P Cov[Z1 , Z2 ].

V[Z2 ] + 500 000 d d2 Let us calculate each term in turn. First, ⎧ ⎪ ⎪ ⎨ ((1.025 v)2 )K[30] +1 /1.0252 if K[30] ≤ 24, 2 Z1 = ⎪ ⎪ ⎩ ((1.025 v)2 )25 if K[30] ≥ 25.

V[L0 ] = 250 0002 V[Z1 ] +

So E[Z12 ] =

1 A1 + v 25 j∗ 25 p[30] = 0.30251, 1.0252 [30]: 25 j∗

90

Solutions for Chapter 6 where j∗ denotes calculation at interest rate j∗ = (1 + j)2 − 1 = 0.04938.

This gives us V[Z1 ] = 0.00047. Secondly, E[Z22 ] = 2 A[30]: 25 = 0.09020, giving V[Z2 ] = 0.090198 − 0.2985172 = 0.00109.

Thirdly,

⎧ ⎪ ⎪ ⎨ (1.025 v 2 )K[30] +1 /1.025 Z1 Z2 = ⎪ ⎪ ⎩ (1.025 v 2 )25

if K[30] ≤ 24, if K[30] ≥ 25.

So E[Z1 Z2 ] =

1 A1 + v 25 p = 0.16477, j† 25 [30] 1.025 [30]: 25 j†

where j† denotes calculation at interest rate j† =

1.052 = 0.07561.

1.025 − 1

Subtracting E[Z1 ] E[Z2 ] gives Cov[Z1 , Z2 ] = 0.00071. Inserting these values into the formula for V[L0 ] gives V[L0 ] = 146 786 651 = 12 115.552 . (d) Suppose that the benefit is payable at time t years (t = 1, 2, . . . , 25) on account of the policyholder’s death. Then the accumulation of premiums less expenses is 0.99P s̈t − (1 200 + 0.39P)(1 + i)t and the death benefit is 250 000(1.025t−1 ). Thus, there is a profit at time t years if 0.99P s̈t − (1 200 + 0.39P)(1 + i)t − 250 000(1.025t−1 ) > 0.

A straightforward calculation on a spreadsheet shows that when t = 24, 0.99P s̈t − (1 200 + 0.39P)(1 + i)t − 250 000(1.025t−1 ) = −5603.08 and when t = 25, 0.99P s̈t − (1 200 + 0.39P)(1 + i)t − 250 000(1.025t−1 ) = 15 295.72.

Hence there is a profit if the death benefit is payable at time 25. Similarly,

Solutions for Chapter 6

91

there is a profit if the policyholder survives 25 years, the amount of this profit being 0.99P s̈25 − (1 200 + 0.39P)(1 + i)25 − 250 000(1.02525 ) = 3991.18.

Thus, there is a profit if the policyholder survives 24 years and pays the premium at the start of the 25th policy year. Hence the probability of a profit is 24 p[30] = 0.98297.

6.20 (a) Let P be the annual premium. Then P ä[40]: 20 = 250 000 A[40]: 20 gives P = $7 333.81 since ä[40]:20 = 12.9947 and A[40]:20 = 0.38120. (b) We have L0n = 250 000 v min(Kx +1,20) − P ämin(K +1,20) x

P min(Kx +1,20) P = 250 000 + − . v d d Since the premium has been calculated by the equivalence principle, E[L0n ] = 0, and we have

P 2 $ min(Kx +1,20) % V[L0n ] = 250 000 + V v d

2 P 2 2 = 250 000 + A[40]:20 − A[40]:20 d

P 2 = 250 000 + 0.14660 − 0.381202 d = 209 966 499.

Thus, the standard deviation of L0n is $14 490. (c) The sum of 10 000 independent random variables each having the same distribution as L0n has approximately a normal distribution with mean 0 and standard deviation 1 449 022. As the 99th percentile of the standard normal distribution is 2.326, the 99th percentile of the net future loss from the 10 000 policies is 2.326 × 1 449 022 = 3 370 929.

92

Solutions for Chapter 6 (d) Let P̃ be the premium calculated by the portfolio percentile principle. Then E[L] = 10 000 (250 000 A[40]: 20 − P̃ ä[40]: 20 ) = 10 000 (95 300.76 − 12.9947P̃) and

P 2 (0.001286).

V[L] = 10 000 250 000 + d

We set P̃ such that

−E[L] Pr[L < 0] = Φ √ = 0.99, V[L]

where Φ is the standard normal cumulative distribution function, and Φ(2.326) = 0.99. So we find that P̃ = $7 351 by solving −10 000 (95 300.76 − 12.9947P̃) 2.326 = 0 .

2 10 000 250 000 + P̃/d (0.001286) 6.21 (a) If death occurs in year 1, K35 = 0 and the death benefit is X ä20 . Similarly, if death occurs in year 2, K35 = 1 and the death benefit is X ä19 . Extending this, we see that if death occurs in year k (for k = 3, 4, . . . , 20), then K35 = k−1 and the death benefit is X ä21−k . So, the death benefit at time k is X ä21−k = X

1 − v 21−k . d

(b) The present value random variable is ⎧

⎪ 1 − v (20−K35 ) K35 +1 ⎪ ⎪ ⎪ if K35 < 20, v ⎨ X Z=⎪ d ⎪ ⎪ ⎪ 0 ⎩ if K35 ≥ 20, ⎧ ⎪ X ⎪ K35 +1 ⎪ − v 21 if K35 < 20, ⎪ ⎨ d v =⎪ ⎪ ⎪ ⎪ ⎩ 0 if K35 ≥ 20. (c) From the second definition of Z in part (b), X k+1 (v − v 21 ) k |q35 d k=0

X 1 = − v 21 20 q35 .

A d 35:20 19

E[Z] =

93

Solutions for Chapter 6

(d) We find X by setting the value of the repayments equal to the amount of the loan, so 250 000 = X a20 at 5% gives X = $20 060.65. (e) Let P be the net annual premium. Then P ä35:20 = E[Z] =

X 1 − v 21 20 q35 .

A d 35:20

We calculate 20 q35 = 0.01718, ä35:20 = ä35 − 20 E35 ä55 = 13.0241 and A1

35:20

= A35 − 20 E35 A55 = 0.00939,

giving P = $104.40. (f) The policy provides a decreasing death benefit, paid for by level premiums. So there will be a point in time after which the premium exceeds the expected death benefit in a year, in which case it would be cheaper for a policyholder to lapse and to buy different insurance cover, such as one-year term insurance.

6.22 (a) Let P denote the annual premium. Then the equation of value for P, at 6% interest, is P ä[80]:2 = 1 000 v q[80] + 600 v 2 1 |q[80] + 200 v 3 2 |q[80] , giving P = 51.72/2.7767 = $18.63. (b) The table below shows the possible values of L0 and the associated probabilities: L0

Probability

1 000 v − P = 933.75 600 v 2 − P ä[80]:1 = 507.85 200 v 3 − P ä[80]:2 = 119.50 −P ä[80]:2 = −53.27

q[80] = 0.02799 1 |q[80] = 0.03384 2 |q[80] = 0.03849 3 p[80] = 0.89968

Hence Pr[L0 > 0] = 3 q[80] = 0.10032.

94

Solutions for Chapter 6 (c) As E[L0 ] = 0, we have V[L0 ] = E[L02 ], and E[L02 ] = 0.02799(933.752 ) + 0.03384(507.852 ) + 0.03849(119.502 ) + 0.89968(−53.272 ) = 36 235.82, giving SD[L0 ] = 190.357. (d) As the policies are independent, E[L] = 0 and V[L] = 6 000 × 190.3572 .

The 95th percentile of the standard normal distribution is 1.6449, and so the 95th percentile of L is !

1.6449 6 000 × 190.3572 = 24 253.36. (e) For a life with four policies, the benefit amount is four times that for a life with a single policy, so for these lives E[L0 ] = 0 and V[L0 ] = 16 × 190.3572 .

So E[L] = 0, and now (again by independence) V[L] = 4 800 (190.3572 ) + 300 (16 × 190.3572 ) = 9 600 (190.3572 ).

The 95th percentile of L is therefore !

1.6449 9 600 × 190.3572 = 30 678.35. (f) For each of the lives with four policies, there is much greater variability in the possible claim amount compared with a life with a single policy, which leads to greater variability in L0 for these policyholders. The greater variability in claim amounts dominates the variability in claim numbers, leading to an increased variance for L.

6.23 Let P be the monthly premium. Then EPV of premiums less premium expenses: 0.95 × 12P ä(12)

[40]:20

as ä(12)

[40]:20

− 0.15P = 144.65P

= 12.7019.

EPV of expenses on death: 20 q[40] v (1.03) + 1 |q[40] v 2 (1.032 ) + · · · = 20A[40] i∗ = 8.33, where i∗ = 1.05/1.03 − 1, and A[40] i∗ = 0.416425.

Solutions for Chapter 6

95

EPV of other expenses: 0.025 × 50 000 = 1250.

EPV of annuity benefit: 50 000 20 p[40] v 20 + (1.02) v 21 21 p[40] + (1.022 ) v 22 22 p[40] + · · · = 50 000 20 E[40] ä60 ( j) = 342 689, where ä60 ( j) is calculated at rate j = 1.05/1.02 − 1, giving ä60 ( j) = 18.6920.

The pure endowment factor 20 E[40] = 0.366669 is evaluated at the original rate of 5%.

Equating the EPVs gives a monthly premium of P = $2377.75.

6.24 Let Pn denote the net annual premium, and Pg denote the gross annual premium.

We calculate the select insurance and annuity functions by constructing the life table for [40]. We calculate the ultimate one-year survival probabilities for ages 40 and over using Makeham’s formula, then calculate the select rates using p[40] = 1 − q[40] = 1 − 0.75 (1 − p40 ) = 0.25 + 0.75p40 and p[40]+1 = 1 − q[40]+1 = 1 − 0.9 (1 − p41 ) = 0.1 + 0.9p41 .

Using this life table in a spreadsheet format, we can calculate all the required functions. We have assumed uniform distribution of deaths; other fractional age assumptions would give very similar results.

EPV of death benefit: 20 000 Ā[40] + 80 000 Ā 1

[40]: 20

= 15 152.74.

EPV of net premiums: Pn ä[40]:20 = 11.2962 Pn .

EPV of gross premiums less premium expenses: 0.97 Pg ä[40]:20 − 0.27Pg = 10.6874 Pg .

96

Solutions for Chapter 6 EPV of other expenses: 10 ä[40]: 20 j = 141.74,

where

j=

1.06 − 1.

1.03

Then Pn =

15 152.74 = $1341.40 11.2962

and Pg =

15 152.74 + 141.74 = $1431.08.

10.6874

6.25 (a) For non-smokers aged 50, the probability of death during the 20-year term is 20 q[50] = 0.0758. For smokers, the one-year survival probability at age 50 + t is ps50+t = 1 − 2.5 q[50]+t for t = 0, 1, 2, . . . , 19, where the superscript s denotes smoker mortality. Then s s 20 q50 = 1 − 20 p50 = 1 −

19 2

ps50+t = 0.1795.

t=0

(b) Let P denote the annual net premium for a non-smoker. Then P ä[50]:20 = 1 000 000 A 1

,

[50]:20

giving P = $3 181.41 since ä[50]:20 = 13.3378 and A 1

[50]:20

= 0.04243 at

4.5%.

Similarly if Ps denotes the annual net premium for a smoker, then Ps ä50:20 = 1 000 000 A 1

50:20

,

where calculations using smoker mortality give ä50:20 = 12.9707 and A1 = 0.10125, leading to Ps = $7 806.11.

50:20

The difference in premiums is therefore $4 624.70. (c) The expected value of the loss at issue is 0 for the policies that are correctly classified. The expected value of the loss at issue for an incorrectly classified policy is 1 000 000 A 1

50:20

− P ä50:20 = 59 985.41,

where we use smoker mortality and the non-smoker premium from part (b) of P = $3 181.41.

Solutions for Chapter 6

97

Per policy issued, the expected value of the loss at issue is 0 × 0.95 + 59 985.41 × 0.05 = 2 999.27. (d) If the death benefit is payable at time t, then the deduction from the sum insured is (Ps − P) s̈t at 4.5%. Hence the EPV of the deductions is (Ps − P)

19

vt+1 t |qs50 s̈t+1 = 8 406.33.

t=0

The expected value of the loss at issue for an incorrectly classified policy then becomes 59 985.41 − 8 406.33 = 51 579.08 and the expected value of the loss at issue, per policy issued, becomes 0.05 × 51 579.08 = 2 578.95. (e) An adjustment needs to be made for all misclassified policyholders, not just those who die. If the interest assumption is borne out in practice, then the misclassified policyholders who die are effectively paying the correct premium after the deduction has been made from their death benefits, but the misclassified policyholders who survive are still being subsidized.

Solutions for Chapter 7

7.1 The net annual premium, P, is calculated as P ä40:20 = 100 000 A40 − 50 000 A 1

40:20

where A40 = 0.12106, A 1

40:20

⇒ P = $875.38,

= 0.01463 and ä40:20 = 12.99348.

The policy value at duration 10 years, 10V , is calculated from − P ä50:10 , 10V = 100 000 A50 − 50 000 A 50:10 1

giving 10V = $11 149.02.

7.2 Let P = $134 denote the net premium. As 0V n = 0, we have 1.045 P = 10 000 q[x] + 1V n p[x] , which gives 1V n = $129.57.

7.3 (a) Let P denote the total premium per year, payable in quarterly instalments.

Then (4) P ä(4) 70 = 500 000 A70 ,

where, under UDD, i A(4) 70 = (4) A70 = 0.43613, i and ä(4) 70 = α(4) ä70 − β (4) = 11.6278, giving P = 18 754. The policy value at time 10 is (4) (4) 10V = 500 000 A80 − P ä80 = $148 799,

98

Solutions for Chapter 7

99

(4) since A(4) 80 = 0.60393 and ä80 = 8.1673.

(b) Let P̃ denote the new annual premium. Then 1

4 P̃ ä(4) 70 = 500 000 v Ā70 ,

giving P̃ = 18 640. The policy value at time 10 is 1 (4) 10V = 500 000 v 4 Ā80 − P̃ ä80 = $147 895.

7.4 Let P denote the premium payable in the second and subsequent years under the Full Preliminary Term (FPT) approach. Then the equation of value for P is P ä71:14 = 500 000 A71 , where ä71:14 = ä71 − v 14

l85 ä85 = 9.3492. l71

Hence P = $23 734.24. Then the FPT reserve at the end of the fifth year is 5V

FPT

= 500 000 A75 − P ä75:10 = $80 598.09

since ä75:10 = ä75 − v 10

l85 ä85 = 7.3203. l75

7.5 (a) (i) The random variable representing the present value of the deferred annuity benefit can be written as the difference between an immediate whole life annuity and an immediate term annuity, as follows: ⎞ ⎛ ⎟⎟⎟ ⎜⎜ ⎟⎟⎠ .

10 000 ⎜⎜⎜⎝ä(12) − ä(12) (12) 1 K[30] + 12

(12) 1 min K[30] + 12 , 30

The random variable representing the present value of the return of premiums on death before age 60 is as follows: 0 T[30] P v T[30] 10P v T[30]

if if if

T[30] ≥ 30, T[30] < 10, 10 ≤ T[30] < 30.

This can be written in a single expression using indicator random variables as T[30] P v T[30] I(T[30] < 10) + 10P v T[30] I(10 ≤ T[30] < 30).

100

Solutions for Chapter 7 (ii) The loss at issue random variable, L0 , is the present value of the benefits, as in part (i), minus the present value of the premiums. Hence ⎞ ⎛ ⎟⎟⎟ ⎜⎜⎜ (12) (12) ⎟⎟⎠ ⎜ L0 = 10 000 ⎜⎝ä − ä (12) 1 K[30] + 12

(12) 1 min K[30] + 12 ,30

+ T[30] P v T[30] I T[30] < 10

+ 10P v T[30] I 10 ≤ T[30] < 30 − P ā

min(T[30] ,10)

.

(b) By the equivalence principle, E[L0 ] = 0. Hence

(12) − ä + 10P 10 E[30] Ā 1 + P (I¯Ā) 1 0 = 10 000 ä(12) [30] [30]:10

[30]:30

so that P=

40:20

(12) 10 000 ä(12) [30] − ä

[30]:30

ā[30]:10 − (I¯Ā) 1

[30]:30

− P ā[30]:10

.

− 10P 10 E[30] Ā 1

40:20

(c) When writing down the expression for L5 we need to be careful to value the return on death before age 60 of the premiums already paid, 5P, as well as those yet to be paid. The required expression is ⎞ ⎛ ⎟⎟⎟ ⎜⎜⎜ (12) (12) − ä L5 = 10 000 ⎜⎜⎝ä ⎟⎟⎠ + 5P v T35 I (T35 < 25) (12) 1 K35 + 12

(12) 1 min K35 + 12 ,25

+ T35 P v T35 I (T35 < 5) + 5P v T35 I (5 ≤ T35 < 25) − P āmin(T ,5) .

35

(d) Recalling that 5V = E[L5 ], we have

(12) (12) 5V = 10 000 ä35 − ä 35:25

+ 5P Ā

1

35:25

+ P (I¯Ā) 1

35:5

+ 5P 5 E35 Ā 1

40:20

− P ā35:5 .

7.6 (a) The policy value at time 10 is 10V = 100 000 Ā80 − 0.95 × 4 000 ā80 ,

where ā80 = ä80 −

1 1 − (μ80 + δ ) = 8.0417, 2 12

101

Solutions for Chapter 7 and hence Ā80 = 1 − δ ā80 = 0.60764.

Thus, the policy value is 10V = $30 205.74. (b) Thiele’s differential equation gives d t V = δ t V + 0.95 × 4 000 − μ80 (100 000 − 10V ). dt Evaluating the right-hand side at t = 10 gives 3 088.25.

7.7 The contribution to the surplus from mortality is (100 q65 − 1)(100 000 + 200 − 6V ) = −$26 504.04.

Note that this uses the assumed, rather than the actual, expenses.

The contribution from interest is 100 ( 5V + 0.95 × 5 200)(0.065 − 0.05) = $51 011.26.

The contribution from expenses is 100 (−0.01) 5 200 × 1.065 − 250 + 200 = −$5 588.00.

7.8 (a) The annual premium, P, is calculated using the equivalence principle from P ä[41]: 3 = 200 000 A 1

[41]: 3

,

giving P = $323.59. (b) The value of the random variable L1 depends on whether the policyholder dies in the coming year, dies in the following year or survives for two years.

The distribution of L1 is as follows: ⎧ ⎪ 200 000 v − P = $188 355.66 w.p. q[41]+1 , ⎪ ⎪ ⎪ ⎨ 2 L1 = ⎪ 200 000 v − P (1 + v) = $177 370.43 w.p.

1 |q[41]+1 , ⎪ ⎪ ⎪ ⎩ −P (1 + v) = −$628.85 w.p. 2 p[41]+1 .

Hence E[L1 ] = 188 355.66 × 0.001876 + 177 370.43 × 0.002197 − 628.85 × 0.995927 = $116.68,

102

Solutions for Chapter 7 and V[L1 ] = 188 355.662 × 0.001876 + 177 370.432 × 0.002197 + (−628.85)2 × 0.995927 − 116.682 = 11 663.782 so that the standard deviation of L1 is $11 663.78. (c) The sum insured, S, is calculated using the equivalence principle. As P ä[41]: 3 = S A[41]: 3 we have S = $1090.26. (d) For the endowment insurance, the distribution of L1 is: ⎧ ⎪ ⎪ w.p. q[41]+1 , ⎨ 1090.26 v − P = $704.96 L1 = ⎪ 2 ⎪ ⎩ 1090.26 v − P (1 + v) = $341.47 w.p. p[41]+1 .

Note that the loss is the same whether the policyholder dies between ages 42 and 43 or survives to age 43.

Hence, E[L1 ] = $342.15 and SD[L1 ] = $15.73. (e) The value of E[L1 ] is greater for the endowment insurance than for the term insurance; this is typical, as for endowment insurance a large portion of the premium is needed to fund the maturity benefit. The difference between the standard deviations is more substantial. The value for the term insurance, $11 663.78, is considerably larger than that for the endowment insurance, $15.73. The future cash flows for the term insurance are much more uncertain than those for the endowment insurance. For the term insurance L1 takes very different values, $177 370.43 and −$628.85, depending on whether the policyholder dies between ages 42 and 43 or survives to age 43, whereas for the endowment insurance the value is the same, $341.47.

Put more simply, for the endowment insurance the insurer knows that the sum insured will be paid at some time within the next two years, the only uncertainty is over the timing; for the term insurance it is not certain that the sum insured will ever be paid.

7.9 (a) The equation for the gross annual premium, Pg , is 0.95Pg ä[35] − 0.35Pg = 40 ä[35] + 85 + 100 000 A[35] , which gives Pg = $469.81.

Solutions for Chapter 7

103

(b) The net premium policy value is calculated using the net premium, Pn , which is calculated as Pn ä[35] = 100 000 A[35] , giving Pn = $391.22.

The net premium policy value at time t = 1, 1V n , is then given by 1V

n

= 100 000 A[35]+1 − Pn ä[35]+1 = $381.39.

(c) The gross premium policy value at time t = 1 is calculated using the gross premium, Pg = $469.81, and is given by 1V

g

= 100 000 A[35]+1 + 40 ä[35]+1 − 0.95Pg ä[35]+1 = $132.91.

(d) Both the net and gross premium policy values use the same future benefits.

The difference is that the gross premium policy value includes future expenses, and deducts the value of the future gross premiums, while the net premium policy value does not include future expenses, but only deducts the value of the future net premiums. The expense loading in the gross premium is Pe = Pg − Pn . The gross premium policy value is less than the net premium policy value when the expected present value of future expenses is less than the expected present value of future expense loadings.

We can write the gross premium policy value as 1V

g

= EPV Future Benefits + EPV Future Expenses − EPV Net Premiums − EPV Expense Loadings

and the net premium policy value as 1V

n

= EPV Future Benefits − EPV Net Premiums

so the difference, 1V n − 1V g , is EPV Expense Loadings − EPV Future Expenses.

This is generally greater than zero as the expense loadings include the amortized initial expenses. That is, the gross premium policy value is less than the net premium policy value because it allows for the recovery of the initial expenses from future premiums.

104

Solutions for Chapter 7 (e) In this case the asset share is given by AS1 =

1.1 (0.6 × 469.81 − 125 − 25) − 100 000 × 0.0012 = $25.10.

1 − 0.0012

(f) Per policy issued, the insurer’s surplus at the end of the first year will be 1.1 (0.6 × 469.81 − 125 − 25) − 100 000 × 0.0012 − 1V (1 − 0.0012) = −$107.67. (g) To calculate the contribution to the surplus from interest, we assume mortality and expenses are as in the premium basis. This gives the actual interest minus the expected interest as (0.1 − 0.06)(0.6P − 125) = $6.28.

The contribution to surplus from mortality is (q[35] − 0.0012)(100 000 − 1V ) = −$86.45.

The contribution from expenses, allowing now for the actual interest earned, is 1.1 (125 − 150) = −$27.50.

Note that the total is −$107.67 as required.

7.10 (a) We have Ltn = S v T[x]+t − P āT[x]+t

1 − v T[x]+t T[x]+t = Sv −P δ

P T[x]+t P , =− + S+ v δ δ so

P 2 V[v T[x]+t ] V[Ltn ] = S + δ

P 2 2 Ā[x]+t − (Ā[x]+t )2 . = S+ δ (b) The premium equation is 1200 ā[55] = S Ā[55] .

Solutions for Chapter 7

105

Thus, 1200 × 15.56159 = 0.240747 S, giving S = $77 566.44. (c) The standard deviation of Ltn is calculated by taking the square root of V[Ltn ], calculated from the formula in part (a). We need the following values, calculated using Woolhouse’s three-term formula for the associated annuities: Ā[55] = 0.240747,

2

Ā[55] = 0.078216,

Ā60 = 0.297434,

2

Ā60 = 0.113739,

Ā65 = 0.363520,

2

Ā65 = 0.161893.

Using these values, we have SD[L0n ] = $14 540.32, SD[L5n ] = $16 240.72, n SD[L10 ] = $17 619.98.

The values of SD[Ltn ] are increasing as t increases from 0 to 10. This is not surprising. As t increases from 0, the time until the sum insured is likely to be paid decreases and so the present value of the loss increases. This will increase the standard deviation of the present value of the loss provided there is still considerable uncertainty about when the policyholder is likely to die, as will be the case for the range of values of t being considered here.

7.11 (a) (i) Let P be the gross annual premium. Then the equation of value is P ä60:20 = 100 000 A60 + 0.3P + 0.1P ä60:20 + 450 + 50ä60 , where ä60:20 = ä60 − 20 E60 ä80 = 12.3816.

This gives P = $2787.23. (ii) The gross premium policy value at time 2 is 2V = 100 000 A62 + 0.1P ä62:18 + 50 ä62 − P ä62:18 ,

where ä62:18 = ä62 − v 18 This gives 2V = $3 153.24.

l80 ä80 = 11.5850. l62

106

Solutions for Chapter 7 (b) (i) Let Pm denote the modified premium in the second and subsequent years. Then the equation of value is Pm ä60:20 − 0.5 Pm = 100 000 A60 , which gives Pm = $2 443.10, and so the first-year modified net premium is $1 221.55. (ii) The modified net premium policy value at time 2 is 2V

m

= 100 000 A62 − Pm ä62:18 = $3 191.66.

(c) (i) The first-year modified net premium under the FPT method is 100 000 v q60 = $323.65. (ii) Let PFPT be the FPT premium in years 2 onwards. Then the FPT reserve at time 2 is 2V

FPT

= 100 000 A62 − PFPT ä62:18 .

This is smaller than the modified net premium policy value from part (b) since PFPT > Pm .

This follows because the EPV of modified net premiums equals the EPV of FPT premiums (and both equal the EPV of the death benefit).

As the FPT premium in year 1 is considerably less than the modified net premium in year 1, it follows that the FPT premiums in years 2 onwards must be greater than the modified net premiums in those years.

7.12 The premium equation is P ä(12) = S A(12) , [x]:n [x]:n giving ⎛ ⎞ ⎜⎜⎜ 1 − d (12) ä(12) ⎟⎟⎟ [x]:n ⎟⎟⎟ .

P = S ⎜⎜⎜⎜⎝ ⎟⎠ ä(12) [x]:n

107

Solutions for Chapter 7 The policy value just before time t, when a monthly premium is due, is (12) − P ä(12) tV = S A [x]+t:n−t [x]+t:n−t

= S 1 − d (12) ä(12)

[x]+t:n−t

⎛ ⎞ ⎜⎜⎜ ⎟⎟ ä(12) [x]+t:n−t ⎟ ⎜ ⎟⎟⎟ , = S ⎜⎜⎜⎝1 − ⎟⎠ (12) ä

⎛ ⎞ ⎜⎜⎜ 1 − d (12) ä(12) ⎟⎟⎟ [x]:n ⎟⎟⎟ ä(12) − S ⎜⎜⎜⎜⎝ ⎟⎠ [x]+t:n−t ä(12) [x]:n

[x]:n

as required.

7.13 (a) The equation for the annual premium, P, is 0.95 P ä[40]:10 = 10 000 A[40]:10 + 10 000 A 1

[40]:10

,

giving P = $807.71. (b) The fifth premium is paid at duration 4, so that − 0.95 P ä44:6 = $3429.68.

4V = 10 000 A44:6 + 10 000 A 44:6 1

(c) Let the revised death benefit be S. Then, equating policy values before and after the alteration, we have 3 429.68 = 0.5 S A44:6 + 0.5 S A 1

44:6

− 0.95 × 0.5 P ä44:6

giving S = $14 565.95.

7.14 (a) The equation for the annual premium, P, is P ä[40] = 50 000 A[40] − 49 000 A 1

[40]:3

giving P = $323.83. We calculated A 1

[40]:3

A1

[40]:3

=

2

,

from first principles as

vt+1 t |q[40] = 0.001441.

t=0

(b) The formula for the policy value at integer duration t ≥ 3 is t V = 50 000 A40+t − P ä40+t .

Note that, since the select period for the survival model is two years, the life is no longer select at age 40 + t for t ≥ 3.

108

Solutions for Chapter 7 (c) We insert the following values into the formula in part (b): P = 323.83, ä43 = 18.0895, A43 = 0.13859.

This gives 3V = $1071.41. (d) The recurrence relation for policy values in this case is 1.05 ( 2V + P) = 1 000 q42 + 3V p42 .

Inserting the values for P, 3V and p42 (= 1 − q42 = 0.999392), gives 2V = $696.52. (e) The interest profit is 985 (2V + P)(0.06 − 0.055) = $5025.21, and the mortality profit is (985 q42 − 4)(1000 − 3V ) = $242.85, giving a total profit of $5268.06.

7.15 (a) The recurrence relation for t = 1, 2, 3, . . . , 19 is (1 + i) t V = q[60]+t + p[60]+t t+1V. For time 0 we have (1 + i)( 0V + P) = q[60] + p[60] 1V, where P is the single premium and 0V = 0.

For t = 1, 2, 3, . . . , 19 the explanation is as follows: t V represents the expected value (at time t) of the present value (at time t) of the future net cash flows from the insurer, which in this case are benefits only. The left-hand side gives the accumulated proceeds at time t+1 of the cash flow at time t.

The right-hand side gives the expected value of the cash flow required at time t+1, given that the policy is in force at time t. We can calculate this expected value by considering what can happen in the year (t, t+1). Either the policyholder dies – with probability q[60]+t – in which case the only future net cash flow is the sum insured, $1, payable at time t+1, or the policyholder survives – with probability p[60]+t – in which case the expected present value at time t+1 of the future net cash flows is t+1V .

For time 0, the explanation is the same except that the amount we accumulate for one year is the premium rather than the policy value.

Solutions for Chapter 7

109

(b) In this case the recurrence relation for t = h, 2h, 3h, . . . , 20 − h is (1 + i)h t V = h q[60]+t + h p[60]+t t+hV ⇒ (1 + i)h t V = t+hV + h q[60]+t (1 − t+hV ) .

(7.1)

For time 0 we have (1 + i)h ( 0V + P) = hV + h q[60] (1 − hV ) . (c) First, we use equation (7.1) to construct an equation for (t+hV − t V )/h.

Then, we can take the limit as h → 0+ to get the differential equation.

Substituting 1+i = eδ in equation (7.1), and rearranging by first subtracting t V from each side then dividing throughout by h, we have t+hV − t V

h

=

eδ h − 1 h q[60]+t (1 − t+hV ) . tV − h h

(7.2)

Now, (1 + δ h + δ 2 h2 /2 + · · · ) − 1 eδ h − 1 = lim+ = δ. h→0 h→0 h h lim+

Also lim+

h→0

h qy

h

= lim+ h→0

1 Pr[Ty ≤ h] = μy h

(from the definition of μy in equation (2.6) of AMLCR).

So, taking limh→0+ of equation (7.2), we get d t V = δ t V − μ[60]+t (1 − t V ) dt

= μ[60]+t + δ t V − μ[60]+t .

The boundary conditions are limt→0+ t V = P and limt→20− t V = 0. (d) To check that t V = Ā

1

[60]+t: 20−t

is the solution to the differential equation in part (c), we need to check that it satisfies the boundary conditions and that d Ā 1 = (μ[60]+t + δ ) Ā 1 − μ[60]+t .

[60]+t: 20−t dt [60]+t: 20−t

(7.3)

110

Solutions for Chapter 7 To check that the proposed solution satisfies (7.3), we use the arguments from Section 7.4.1 of AMLCR. We have 20−t

Ā

1

[60]+t: 20−t

=

e−δ r r p[60]+t μ[60]+t+r dr.

0

Now let s = r + t, so that t = s − r and ds = dr. Then 20 Ā

=

1

[60]+t: 20−t

e−δ (s−t) s−t p[60]+t μ[60]+s ds

t

and as s−t p[60]+t =

s p[60] t p[60]

,

we have −δt

e

20 t p[60] Ā

1

[60]+t:20−t

=

e−δ s s p[60] μ[60]+s ds .

t

Differentiating the right-hand side with respect to t we get −e−δt t p[60] μ[60]+t , and differentiating the left-hand side we get Ā

1

[60]+t:20−t

d −δt d . e t p[60] + e−δt t p[60] Ā 1 dt dt [60]+t:20−t

Now d −δt d e t p[60] = −δ e−δt t p[60] + e−δt t p[60] dt dt = −δ e−δt t p[60] − e−δt t p[60] μ[60]+t , so we have, for the derivative of the left-hand side, − δ e−δt t p[60] + e−δt t p[60] μ[60]+t Ā 1

[60]+t:20−t

+ e−δt t p[60]

d Ā 1 dt [60]+t:20−t

= −e−δt t p[60] μ[60]+t .

Dividing throughout by e−δt t p[60] , then rearranging, we obtain d Ā 1 = (δ + μ[60]+t ) Ā 1 − μ[60]+t , [60]+t:20−t dt [60]+t:20−t as required.

111

Solutions for Chapter 7 The boundary conditions are clearly satisfied – that is, lim Ā

t→20−

1

[60]+t:20−t

= 0 and P = Ā 1

[60]:20

.

7.16 (a) The annual premium, P, is calculated from P ä60:10 = 50 000 A 1

60:10

+ 10 000 10 |ä60 ,

giving P = $9001. (b) The recurrence relations for policy values are 1.05 ( t V + P) = 50 000 q60+t + t+1V p60+t 1.05 ( t V − 10 000) = t+1V p60+t

for t = 0, 1, . . . , 9, for t = 10, 11, . . . .

(c) The death strain at risk in the second year of the contract, for a policy in force at the start of the second year, is (50 000 − 2V ). We can calculate the policy value at the end of the second year from the recurrence relations in part (b), or using + 10 000 8 |ä62 = $19 113.

2V = 50 000 A 62:8 1

So, the death strain at risk is 50 000 − 19 113 = $30 887. (d) In the 13th year there is no benefit payable on death. Hence, the death strain at risk is (− 13V ), which is calculated as − 13V = −10 000 ä73 = −$110 081. (e) The mortality profit in the second year is 98 ( 1V + P) × 1.05 − 2 × 50 000 − 96 2V = −$50 297. (f) The mortality profit in the 13th year is 80 ( 12V − 10 000) × 1.05 − 76 13V = $325 130.

7.17 (a) Let P denote the net single premium. We calculate P as the EPV of the deferred annuity, 50 000 20 |āx , plus the EPV of the return of the premium.

Suppose the premium is returned, with interest, at time s years. The amount refunded will be P (1+i)s and its present value will be P (1+i)s (1+i)−s = P. The probability that the premium will be refunded is 20 qx so that the EPV of the refund is P 20 qx . Hence P = 50 000 20 |āx + P 20 qx ,

112

Solutions for Chapter 7 giving P=

50 000 20 |āx .

20 px

(i) Using the same argument as above, the EPV at duration t of the return of the premium immediately on death at time s years, measured from the start of the policy, and where t < s < 20, is P (1 + i)t 20−t qx+t .

Hence, the prospective policy value is t t V = 50 000 20−t |āx+t + P (1 + i) 20−t qx+t .

(ii) For t ≥ 20 the prospective policy value at duration t is t V = 50 000 āx+t .

(b) (i) For t < 20, the retrospective policy value is given by P − P t qx . t Ex (ii) For t ≥ 20, the retrospective policy value is given by P − P 20 qx − 50 000 20 |āx: t−20 . t Ex (c) The prospective and retrospective net premium policy values are equal without any further conditions since the (net) premium is calculated using the equivalence principle with the same basis as for the policy values.

To show that the prospective and retrospective policy value formulae are equal, we start with the equation for the net premium and split the EPVs into values before and after time t. (i) For t < 20 we have P = 50 000 20 |āx + P 20 qx = 50 000 t Ex 20−t |āx+t + P ( t qx + t px 20−t px+t )

= t Ex 50 000 20−t |āx+t + P (1 + i)t 20−t qx + P t qx .

Rearranging this last formula, we get P (1 − t qx ) = 50 000 20−t |āx+t + P (1 + i)t 20−t qx E t x as required.

113

Solutions for Chapter 7 (ii) For t ≥ 20 we have P = 50 000 20 |āx + P 20 qx = 50 000

t Ex āx+t + 20 |āx:t−20

+ P 20 qx .

Rearranging this last formula, we get P (1 − 20 qx ) − 50 000 20 |āx: t−20 t Ex

= 50 000 āx+t

as required.

7.18 (a) Let Pg denote the gross premium payable in the first year. Then

+A1 0.9 × 0.5 Pg ä[40]:5 + ä[40]:20 − 0.4 Pg = 200 + 300 000 A 1 [40]:5

[40]:20

giving Pg = $710.30 since ä[40]:5 = 4.541, ä[40]:20 = 12.995, A 1

[40]:5

= 0.00253, A 1

[40]:20

= 0.01454.

(b) Using a recursive approach, the gross premium policy values are

g 1 +A1 + 0.4 Pg 0V = 200 + 300 000 A [40]:5

− 0.9 × 0.5 P

g

[40]:20

ä[40]:5 + ä[40]:20

= $96.93, ( 0V g − 200 + 0.5 Pg ) × 1.045 − 600 000 q[40] p[40] = − $6.95,

1V

g

=

2V

g

=

( 1V g + 0.9 Pg ) × 1.045 − 600 000 q[40]+1 p[40]+1 = $338.85,

and for the final year, 1.045 (19V g + 0.5 × 0.9 Pg ) = 300 000 q59 gives 19V g = $555.40. (c) Let Pn denote the net premium payable in the first five years for the policy.

We assume that it reduces by 50%, from the sixth premium, in proportion

,

114

Solutions for Chapter 7 to the gross premium. Then

300 000 A 1

Pn =

[40]:5

+A 1

[40]:20

0.5 ä[40]:5 + ä[40]:20

= $596.47.

Note that the net premium is calculated using an interest rate of 4.5% per year.

Again using a recursive approach, the net premium policy values are

n 1 1 V = 300 000 A + A − 0.5 Pn ä[40]:5 + ä[40]:20 0 [40]:5

[40]:20

= $0, ( 0V n + Pn ) × 1.045 − 600 000 q[40] p[40] = $353.08,

n 1V =

2V

n

( 1V n + Pn ) × 1.045 − 600 000 q[40]+1 p[40]+1 = $670.50, =

and for the final year, 1.045 (19V n + 0.5 × Pn ) = 300 000 q59 gives 19V n = $576.81. (d)

(i) Using ä[40]:n = 1 + v p[40] ä[40]+1:n−1 we obtain (from the values in the question) ä[40]+1:4 = 3.7458

and

ä[40]+1:19 = 13.0650.

(ii) Let PFPT denote the net premium payable from duration 1 (which we assume halves from duration 5) under the FPT approach, so that

+A 1 300 000 A 1 [40]+1:4 [40]+1:19 PFPT = 0.5 ä[40]+1:4 + ä[40]+1:19 = $638.47. (iii) First, let 1 P denote the cost of insurance in the first year, calculated as 1 P = 600 000 v q[40] = $258.74.

Solutions for Chapter 7

115

Again using a recursive approach for policy values,

FPT = 300 000 A 1 +A 1 − 1P 0V [40]:5

− v p[40] 0.5 P

FPT

[40]:20

ä[40]+1:4 + ä[40]+1:19

= $0, ( 0V FPT + 1 P) × 1.045 − 600 000 q[40] p[40] = $0,

1V

FPT

=

2V

FPT

=

( 1V FPT + PFPT ) × 1.045 − 600 000 q[40]+1 p[40]+1 = $345.25.

For duration 19 years, we use 1.045 19V FPT + 0.5 × PFPT = 300 000 q59 , which gives 19V FPT = $555.81. (e) The rationale for the FPT approach is explained in Section 7.8 of AMLCR. In brief, it is intended to combine the computational advantages of a net premium policy value calculation (simple calculation, no expense assumptions required), with the advantages of the gross premium policy value calculation (more accurate assessment of prospective liability). To be effective, the FPT policy value should approximate reasonably closely the gross premium policy value. In this example we have the results summarized in Table S7.1. We see that at the end of the first, second and final years of the policy, the FPT policy value is fairly close to the gross premium policy value compared with the net premium policy value, even though at the start of the contract they are farther apart. In this example, the implicit allowance for initial expenses through the FPT method is Pg − 1 P = $451.56.

The actual initial expenses are 0.5Pg + 200 = $555.15, which is why the method is moderately effective at replicating the policy values after the first year.

7.19 (a) Note that the total premiums per year are $840 and the total renewal expenses per year are $60 plus 5% of premiums. The policy value at time 15 years is then (12) (12) 15V = 100 000 A55 + (60 − 0.95 × 840) ä55 = $12 548,

116

Solutions for Chapter 7 t

tV

g

tV

n

0 1 2 .. .

19

96.93 − 6.95 338.85

0.00 353.08 670.50

0.00 0.00 345.25

555.40

576.81

555.81

tV

FPT

Table S7.1 Summary of policy values for Exercise 7.18.

where i A(12) 55 = i(12) A55 = 0.24058 and ä(12) 55 = α(12) ä55 − β (12) = 15.5966.

Hence the cash value is $12 048 allowing for the $500 surrender charge.

t t +1 , (b) If death occurs in the time interval , for t = 0, 1, 2, . . ., then the 12 12 present value of the loss is 100 000 v (t+1)/12 − 738 ä(t+1)/12 = 100 000 v (t+1)/12 − 738

1 − v (t+1)/12 d (12)

738 738 = 100 000 + (12) v (t+1)/12 − (12) d d = 115 157 v (t+1)/12 − 15 157.

This present value is less than 12 048 if

v (t+1)/12 = exp{−δ (t + 1)/12} < 0.23624, which is equivalent to t > 353.88. So if t ≥ 354, then L15 is less than the cash value and as 354 months is 29 12 years, the required probability is 1 29.5 p55 = 2 (29 p55 + 30 p55 ) = 0.642291.

(c) Let S̃ be the paid-up sum insured. Then (12) 12 048 = S̃ A(12) 55 + 60 ä55 ⇒ S̃ = $46 189.

117

Solutions for Chapter 7 (d) Let S∗ be the revised sum insured under term insurance. Then 12 048 + 0.95 × 840 ä(12) = 60 ä(12) + S∗ A

(12)

55:10

55:10

1

55:10

,

where ä(12) = α(12) ä55:10 − β (12)(1 − 10 E55 ) = 7.8311 55:10

and A

(12)

1

55:10

i = (12) A 1 = 0.02527.

55:10 i

Hence the revised sum insured is S∗ = $705 374. (e) The original sum insured is $100 000 and the revised one is more than seven times this amount – a significant increase. A policyholder who knows that their health is compromised would have an incentive to make this change to acquire a much larger death benefit at no extra cost. This is an example of adverse selection, where the policyholder uses information not available to the insurer to gain access to benefits with much greater value than the insurer assumes. (Note that in the calculation above, we assume the same rates of mortality for the policyholder who converts as for the policyholder who does not convert.) If this conversion is allowed, the insurer would require substantial medical information, as if the policy were newly issued. (f) Typically, insurers use a conservative rate of interest in a valuation, resulting in higher reserves than they are likely to need, to provide a cushion in case experience is very much worse than expected. Policyholders who surrender their policies should not benefit from this extra level of security. Furthermore, the surrendering policyholders are creating additional costs and risks for the insurer, compared with the continuing policyholders, in general, including liquidity risk (having to sell supporting assets at short notice) and adverse selection risk (healthy policyholders surrender in higher numbers, leaving only the higher-cost policyholders in the continuing portfolio). Using a higher rate of interest in the cash value calculation gives smaller cash values than using a lower rate, which provides some protection for the insurer from the additional risks from surrendering policyholders.

7.20 (a) Let P be the annual premium. Using the equivalence principle, we have 0.96P ä[35]:20 − 0.11P = 100 000 A 1

[35]:20

so that P = $91.37.

+ 200

118

Solutions for Chapter 7 (b) The policy value just before the first premium is paid is zero since the premium is calculated using the equivalence principle and the policy value is calculated on the same basis as the premium. The policy value just after the first premium is paid is the premium received, $91.37, less the expenses paid, $200 and 0.15 × $91.37. Hence 0+V = 0 + 91.37 − 200 − 0.15 × 91.37 = −$122.33.

(c) The policy value just after the first premium is received is negative because the premium is insufficient to cover the initial expenses, as can be seen from the calculation in part (b). (d) The plot of the policy values is shown as Figure S7.1. It can be seen that the policy value first becomes positive at duration 3+ just after the fourth premium is paid; just before the fifth premium is paid the policy value is negative again, then becomes positive when the fifth premium is paid, and stays positive for the remainder of the contract.

400

300

Policy Valuue

200

100

0 0

2

4

6

8

10

12

14

16

18

20

-100

-200

Duration

Figure S7.1 Policy values for Exercise 7.20.

(e) The number of survivors at age 35 + t, t = 0, 1, . . . , k, is t p[35] N, and the

119

Solutions for Chapter 7

number of deaths between ages 35 + t and 35 + t + 1 is t |q[35] N. These are not expected numbers, they are actual numbers, according to the information from the question, with probabilities calculated on the premium/policy value basis. Hence, the accumulation of premiums paid up to, but not including, time k is P N 1.05k + 1.05k−1 p[35] + · · · + 1.05 k−1 p[35] = P N 1.05k

k−1

vt t p[35] ,

where v = 1/1.05

t=0

= P N 1.05k ä[35]:k .

Similarly, the accumulation of expenses is 200 N × 1.05k + 0.04 P N 1.05k ä[35]:k + 0.11 P N 1.05k , and the accumulation of death benefits is 100 000 N 1.05k−1 q[35] + 1.05k−2 1 |q[35] + · · · + k−1 |q[35] = 100 000 N 1.05k

k−1

vt+1 t |q[35]

t=0

= 100 000 N 1.05k A 1

[35]:k

.

Hence, the accumulated fund at time k is

1.05k N 0.96P ä[35]: k − 200 − 0.11P − 100 000 A 1

[35]: k

.

Since the number of survivors to time k is k p[35] N, the fund per survivor is

1.05k 0.96P ä[35]:k − 200 − 0.11P − 100 000 A 1 / k p[35] .

[35]: k

The equation for the premium, 0.96P ä[35]:20 − 0.11P = 100 000 A 1

[35]: 20

+ 200,

can be manipulated as follows, where we divide the functions into the first k years and the remaining 20 − k years.

We know that ä[35]:20 = ä[35]: k + k E[35] ä[35]+k: 20−k and A1

[35]:20

=A1

[35]: k

+ k E[35] A

1

[35]+k: 20−k

.

120

Solutions for Chapter 7 So we can rewrite the premium equation as 0.96P ä[35]: k + k E[35] ä[35]+k: 20−k − 0.11P

= 100 000 A 1 + k E[35] A 1 [35]: k

[35]+k: 20−k

+ 200.

We collect together all the first k years payments on the left-hand side, and the following 20 − k years on the right-hand side, giving 0.96 P ä[35]: k − 0.11P − 100 000 A 1

[35]: k

= 100 000 k E[35] A

1

[35]+k: 20−k

− 200 − 0.96P k E[35] ä[35]+k: 20−k

⇒ 1.05k 0.96 P ä[35]: k − 0.11P − 100 000 A 1

[35]: k

= 100 000 A

1

[35]+k: 20−k

− 200 /k p[35]

− 0.96P ä[35]+k: 20−k .

The left-hand side is the accumulated fund at time k per surviving policyholder, as derived above, and the right-hand side is kV .

7.21 (a) The premium, P, is calculated from P ā[40]:20 = 200 000 Ā[40]:20 , giving P = $6 020.40. (b) The policy value at duration 4, 4V , is calculated as 4V = 200 000 Ā44:16 − P ā44:16 = $26 131.42.

(c) The revised values needed for the calculation of the policy value are Ā44:16 = 0.463033 and ā44:16 = 11.00563.

The revised policy value is 4V = 200 000 × 0.463033 − 6 020.40 × 11.00563 = $26 348.41.

(d) The policy value has not changed by very much because, for the age range involved, 44 to 60, mortality is reasonably light. In particular, the change in the mortality basis changes 16 p44 from 0.9751 to 0.9723. Hence, the benefit on survival to the end of the term is still the most significant contribution to the EPV of the benefits.

Solutions for Chapter 7

121

(e) The revised values needed for the calculation of the policy value are Ā44:16 = 0.537698 and ā44:16 = 11.78720.

The revised policy value is 4V = 200 000 × 0.537698 − 6 020.40 × 11.78720 = $36 575.95.

(f) Since the benefit is very likely to be paid at the end of the term, the interest earned on the invested premiums matters to a considerable extent. A lower rate of interest, 4% rather than 5%, means that future premiums will accumulate at a lower rate and so more cash is needed now in order to pay for the benefit. (g) One advantage of using a proportionate paid-up sum insured is that it has an intuitive appeal which can easily be understood by the policyholder.

The policy values and EPV of the proportionate paid-up sum insured are shown in Figure S7.2. It can be seen that the EPV of the proportionate paid-up sum insured is less than the policy value for all durations between 0 and 20, with greater differences for the middle durations. Since t V represents the value of the investments the insurer should be holding at duration t, adopting the suggestion of a proportionate paid-up sum insured would give the insurer a small profit for each policy becoming paid-up, assuming that experience exactly follows the assumptions. It is generally considered reasonable for the insurer to retain a small profit, on average, as the policyholder has adjusted the terms of the contract.

7.22 (a) The equation for the annual premium, P, is 0.95 Pä[50] − 0.17 P = 90 + 10 ä[50] + 10 000 A[50] , giving P = $144.63. (b) For t = 1, 2, . . ., the policy value at time t, just before the premium then due, is t V = 10 000 A[50]+t + 10 ä[50]+t − 0.95 P ä[50]+t .

An excerpt from the resulting table of values is shown in part (c) below. (c) Let Prt denote the profit at the end of the year (t − 1, t), t = 1, 2, . . ., in respect of a policy in force at time t − 1. Then Pr1 = 1.055 (0.78 P − 100) − 10 000 q[50] − 1V p[50]

122

Solutions for Chapter 7 200000

Proportionate PUSI

180000

Policy Value

160000

Policy Valuue/PUSI P

140000 120000 100000 80000 60000 40000 20000 0 0

2

4

6

8

10 Duration

12

14

16

18

Figure S7.2 Policy values and the EPV of the proportionate paid-up sum insured for Exercise 7.21.

and for t = 2, 3, . . ., Prt = 1.055 ( t−1V + 0.95 P − 10) − 10 000 q[50]+t−1 − t V p[50]+t−1 .

For t = 1, 2, 3, . . . the bonus is 0.9 Prt /p[50]+t−1 .

We show an excerpt from the full table of calculations. t

tV

0 1 2 3 4 5 .. .

0.00 3.06 123.84 248.22 376.89 509.93 .. .

Dividend 0.00 0.12 1.18 2.26 3.39 4.55 .. .

20

123

Solutions for Chapter 7 (d) The EPV of the bonuses per policy issued can then be written as ∞

0.1 Prt vt t−1 p[50] = $29.26.

t=1

(e) The policy value on the premium basis at the end of the first year is very small ( 1V = $3.06). This is because most of the first year’s premium is used to pay the initial expenses. In these circumstances, and since it is the policyholder who has requested that the policy be surrendered, it would be reasonable for the insurer to offer no surrender value.

7.23 (a) For 0 < t < 10, d t V = δt t V + P − μ[40]+t (20 000 − t V ), dt and for 10 < t < 20, d t V = δt t V + P − μ[40]+t ( t V − t V ) dt = δt t V + P. Note that, because the death benefit for 10 < t < 20 is t V , the term μ40+t ( t V − t V ), which is the net cost of the death benefit in the second expression, is always zero.

The boundary conditions are lim t V = 60 000,

t→20−

lim t V = lim t V,

t→10+

t→10−

lim t V = 0.

t→0+

(b) With h = 0.05 Thiele’s differential equation gives us that, approximately, t+hV − t V

h

= δt t V + P − μ[40]+t (20 000 − t V )

for t = 0, h, 2h, . . . , 10 − h, and t+hV − t V

h

= δt t V + P ,

for t = 10, 10 + h, 10 + 2h, . . . , 20 − h. So we have recursive equations t+hV = t V + h δt t V + h P − h μ[40]+t (20 000 − t V )

for t = 0, h, 2h, . . . , 10 − h, and t+hV = t V + h δt t V + h P

124

Solutions for Chapter 7 for t = 10, 10 + h, 10 + 2h, . . . , 20 − h. Set 0V = 0, and using Solver in Excel, we set 20V = 60 000 for the target, and find that P = $1810.73. Note that we could also use a backward recursive approach by setting 20V = 60 000, with 0V = 0 for the target. (c) The graph of t V is shown in Figure S7.3.

60000

50000

Policy value, $

40000

30000

20000

10000

0 0

2

4

6

8

10 Duration

12

14

16

18

20

Figure S7.3 Policy values for Exercise 7.23.

7.24 First note that 0V = 0 since we are calculating policy values on the premium basis and the premium is calculated using the equivalence principle. Next, we have 1V = 500 000 A[50]+1:19 − P ä[50]+1:19 = $15 369.28

and 2V = 500 000 A52:18 − P ä52:18 = $31 415.28.

For t = 0.1, 0.2, . . . , 0.9 the required formula is t V = 500 000 1−t q[50]+t + 1V 1−t p[50]+t .

For t = 1.1, 1.2, . . . , 1.9 the required formula is t V = 500 000 2−t q[50]+t + 2V 2−t p[50]+t .

Solutions for Chapter 7 An excerpt from the table of values is shown below. t 0.0 0.1 0.2 .. .

1.9 2.0

tV

0.00 15 144.56 15 173.83 .. .

31 326.91 31 415.28

125

Solutions for Chapter 8

8.1 (a) For t > 0 we have

⎧ t

⎪ ⎪ ⎪

⎫ ⎪ ⎪ ⎪ ⎬ dr⎪ ⎪ ⎪ ⎩ ⎭ 0 ⎫ ⎧ t

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎨ (0.02 + 0.03) dr⎪ = exp ⎪ − ⎪ ⎪ ⎪ ⎪ ⎭ ⎩

⎨ 00 − t px = exp ⎪ ⎪ ⎪

01 02 + μx+r μx+r

0

= exp{−0.05t}.

So 5 p00 x = exp{−0.25} = 0.77880. (b) We require the probability that (x) is in State 0 at time 5, given that (x) is in either State 0 or State 1 at time 5, and we can write this conditional probability as 00 5 px 00 01 5 px + 5 px

.

Now 5 01 5 px =

00 01 11 r px μx+r 5−r px+r dr, 0

and

⎧ ⎫ ⎪ ⎪ 5−r ⎪ ⎪ ⎪ ⎪ 3 4 ⎪ ⎪ ⎨ ⎬ 11 12 μ ds − = exp −0.04(5 − r) . ⎪ 5−r px+r = exp ⎪ x+r+s ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ 0

126

127

Solutions for Chapter 8 Hence 5 01 5 px =

e−0.05r 0.02 e−0.04(5−r) dr

0 −0.2

5

= 0.02 e

e−0.01r dr

0

0.02 e−0.2 (1 − e−0.05 ) = 0.01 = 0.07986, and so the required probability is 0.77880/(0.77880+0.07986) = 0.90699.

8.2 In order for the benefit to be payable, the life must move from State 0 to State 1 at age x+r, where r < 14, and then the life must remain in State 1 continuously for at least one year. Hence the probability that the benefit is paid is 14 00 01 11 r px μx+r 1 px+r dr .

0 11 Further, r p00 x = exp{−0.05r} and 1 px+r = exp{−0.11}. Hence the required probability is

14

e−0.05r 0.02 e−0.11 dr = 0.02 e−0.11

0

14

e−0.05r dr

0

0.02 e−0.11 (1 − e−0.7 ) = 0.05 = 0.18039.

8.3 (a) The probability of remaining in State 1 for 15 years is ⎧ ⎫ ⎪ ⎪ 15 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ 11 10 12 p = exp + μ μ dr − ⎪ ⎪ 15 x x+r x+r ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ 0

⎫ ⎧ ⎪ ⎪ 15 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎨ (0.1 + 0.05) dr = exp ⎪ − ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎩ 0

= exp{−0.15 × 15} = 0.10540.

128

Solutions for Chapter 8 (b) We require the life to remain in State 1 until age x + r, where r < 15, to move to State 0 at that age, and to remain in State 0 until time 15. Hence the required probability is 15 11 10 00 r px μx+r 15−r px+r dr.

0 00 From part (a), we can see that r p11 x = exp{−0.15r} and, similarly, 15−r px+r = exp{−0.12(15 − r)}, giving the required probability as

15

e−0.15r 0.1 e−0.12(15−r) dr = 0.1 e−1.8

0

15

e−0.03r dr

0

0.1 e−1.8 (1 − e−0.45 ) = 0.03 = 0.19967.

8.4 (a) We have

⎧ t ⎫ ⎫ ⎧ n−t

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎨ ⎬ ⎨ 00 11 01 02 12 − (μx+s + μx+s ) ds⎪ exp ⎪ − μx+t+r dr⎪ t px n−t px+t = exp ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎩ ⎭ ⎩ 0 0 ⎫ ⎫ ⎧ n ⎧ t

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎬ ⎨ ⎨ 12 01 02 (μ + μ ) ds μ dr exp − = exp ⎪ − ⎪ ⎪ ⎪ x+s x+s x+r ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎭ ⎩ ⎩ t 0 ⎫ ⎫ ⎧ n ⎧ t

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎬ ⎨ ⎨ 01 02 μ ds μ ds exp = exp ⎪ − − ⎪ ⎪ ⎪ x+s x+s ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎭ ⎩ ⎩ 0

0

since μx02 = μx12 for all x. (b) We have n 01 n px =

00 01 11 t px μx+t n−t px+t dt 0

⎧ t ⎫ ⎧ n ⎫

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ 01 ⎨ ⎬ 01 02 exp ⎪ μ ds exp μ ds = − μ − dt ⎪ ⎪ ⎪ x+s x+t x+s ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ ⎩ ⎭ 0 0 0 ⎫ n ⎧ t ⎫ ⎧ n

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎨ ⎬ 01 ⎨ 02 01 μ ds exp μ ds = exp ⎪ − μx+t dt. − ⎪ ⎪ ⎪ x+s x+s ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎩ ⎭ ⎩ n

0

0

0

129

Solutions for Chapter 8 Noting that ⎧ t ⎫ ⎧ t ⎫

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ d ⎨ ⎬ ⎨ ⎬ 01 01 01 exp ⎪ μ ds exp μ ds − = −μ − , ⎪ ⎪ ⎪ x+s x+t x+s ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ dt ⎩ ⎭ ⎩ ⎭ 0

0

we get ⎧ t ⎫ n⎛ ⎫⎞ ⎜⎜

⎪ ⎪ ⎪ ⎟ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎜⎜⎜ d ⎨ ⎬⎟⎟⎟⎟ 01 ⎜⎜⎜− exp ⎪ ⎟⎟ dt μ ds − ⎪ x+s ⎪ ⎪ ⎪ ⎪ ⎝ dt ⎩ ⎭⎟⎠ ⎩ ⎭ 0 0 0 ⎫⎛ ⎫⎞ ⎧ n ⎧ n

⎪ ⎪ ⎪ ⎪ ⎜ ⎟ ⎪ ⎪ ⎪ ⎪ ⎜ ⎪ ⎪ ⎪ ⎪ ⎜ ⎬ ⎜⎜ ⎬⎟⎟⎟⎟ ⎨ ⎨ 02 01 ⎜ ⎟⎟ . μ ds μ ds 1 − exp = exp ⎪ − − ⎪ ⎪ ⎪ ⎜ x+s x+s ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎜⎝ ⎭⎟⎠ ⎩ ⎩ ⎧ n

⎪ ⎪ ⎪

⎨ 01 − n px = exp ⎪ ⎪ ⎪

⎬ 02 μx+s ds⎪ ⎪ ⎪

0

0

8.5 We have m

m

∞

∞

āixj =

j=0

e

−δt

ij t px dt =

j=0 0

⎛ m ⎞ ⎜⎜ i j ⎟⎟⎟ ⎜⎝ t px ⎟⎟⎟⎠ dt.

−δt ⎜ ⎜⎜

e

j=0

0

Now m

ij t px = 1,

j=0

so m

∞ āixj =

j=0

8.6 Ā02

79:1

e−δt dt =

0

1 . δ

01 12 00 02 = Ā01 79 − v 1 p79 Ā80 − v 1 p79 Ā80 = 0.0749.

8.7 Recall that t 01 t px =

00 01 11 r px μx+r t−r px+r dr.

0

Then ∞ ā01 x =

e−δt t p01 x dt

0

∞ =

−δt

t 00 01 11 r px μx+r t−r px+r dr dt

e 0

0

130

Solutions for Chapter 8 ⎞ ⎛ ∞ ⎜⎜ ∞ ⎟⎟⎟ ⎜⎜⎜ −δt 11 ⎜⎜⎜ e t−r px+r dt ⎟⎟⎟⎟⎟ r p00 = μ 01 dr ⎠ x x+r ⎝ r 0 ⎞ ⎛ ∞ ∞ ⎟⎟⎟ ⎜⎜⎜ ⎟ −δ r ⎜ −δ s 11 e ⎜⎜⎜⎜ e s px+r ds⎟⎟⎟⎟ r p00 μ 01 dr = ⎠ x x+r ⎝ 0

0

∞ =

01 11 e−δ r r p00 x μx+r āx+r dr.

0

8.8 (a) The EPV of the benefit is 5 100 000

01 02 e−δt t p00 60 μ60+t + μ60+t dt.

0

(b) First, ⎧ t ⎫ ⎫ ⎧ t

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ ⎬ ⎨ 00 01 02 + μ 0.025 ds μ ds − = exp − ⎪ ⎪ ⎪ t p60 = exp ⎪ 60+s 60+s ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ ⎭ ⎩ 0

0

= exp{−0.025t}, so that the EPV of the benefit is 5 100 000

e

−0.05t

e

−0.025t

5 0.025 dt = 2 500

0

e−0.075t dt

0

= 10 423.69.

8.9 Let P denote the annual premium. Then the EPV of premium income is P 1 + v p00 58 = 1.94762 P. Note that if the life is in State 0 at the start of a year and is in State 2 (i.e. dead) at the end of that year, then $100 000 will be the benefit amount at the year end since either (i) the life has died without making a transition to State 1, and so the benefit is a death benefit of $100 000, or (ii) the life moves from State 0 to State 1, and then to State 2, meaning that both the permanent disability benefit of $75 000 and the death benefit of $25 000 are payable. Thus, the EPV of the benefits is 01 00 02 01 2 100 000 p02 58 + 75 000 p58 v + p58 100 000 p59 + 75 000 p59 v

131

Solutions for Chapter 8 12 2 + p01 58 25 000 p59 v

= 993.08.

Equating this with the EPV of the premium income gives P = $509.90.

8.10 The monthly premium is $200 and the monthly sickness benefit is $3 000. Consider first the State 0 policy value at time 9 11 12 . We have V 9 11 12 giving 9 11 12

(0)

1

+ 200 1.05 12 = 3000 121 p01 + 150 000 121 p02 , 59 11 59 11 12

12

1 02 1 V (0) = v 12 3000 121 p01 + 150 000 p − 200 11 59 11 12 59 12

12

= −15.77.

Similarly, noting that the premium is not payable if the policyholder is in State 1, the State 1 policy value at time 9 11 12 is

1 (1) 12 12 3000 1 p11 V = v 11 + 150 000 1 p 11 9 11 12 12 59 12 59 12

12

= 3232.92.

The State 0 policy value at time 9 10 12 is

1 V (0) = v 12 9 1112 V (0) 121 p00 + 3000 + 9 1112 V (1) 9 10 59 10 12 12

1 12

p01 + 150 000 121 p02 59 10 59 10

1 12

p11 + 150 000 121 p12 59 10 59 10

12

12

− 200 = −29.26, and the State 1 policy value at time 9 10 12 is

1 V (1) = v 12 9 1112 V (0) 121 p10 + 3000 + 9 1112 V (1) 9 10 59 10 12 12

12

12

= 6 446.37.

8.11 (a) Using standard matrix multiplication (or the MMULT command in Excel) we get ⎛ ⎞ ⎜⎜⎜⎜0.253 0.396 0.351⎟⎟⎟⎟ ⎜⎜ ⎟ ⎜⎜⎜0.198 0.352 0.450⎟⎟⎟⎟⎟ .

3 Px = ⎜ ⎝ ⎠ 0.000 0.000 1.000 (b) The expected number of lives in State 2 at time 3 is 12 8 3 p02 x + 4 3 px = 4.608.

132

Solutions for Chapter 8 (c) If there are exactly two lives in State 2 at the end of three years, that is composed of, say, j lives who started in State 0 and 2 − j lives who started in State 1, where j could be 0, 1 or 2. The required probability is

2

j 8 4 02 j 02 2− j 12 2− j = 0.07594.

1 − 3 px 1 − 3 p12 3 px 3 px x j 2 − j j=0

8.12 (a) The Kolmogorov differential equations are as follows: d 00 01 10 00 01 03 t p = t px μx+t − t px μx+t + μx+t , dt x d 01 00 01 01 10 12 13 t p = t px μx+t − t px μx+t + μx+t + μx+t , dt x d 02 01 12 02 23 t p = t px μx+t − t px μx+t , dt x d 03 00 03 01 13 02 23 t p = t px μx+t + t px μx+t + t px μx+t . dt x (b) Summing the four identities in part (a) gives 3

d

dt j=0 (c) We have 3

0j t px = 1

j=0

0j t px = 0.

⎞ ⎛ 3 3

d 0j d ⎜⎜⎜⎜ 0 j ⎟⎟⎟⎟ ⎟ ⎜⎜⎝ ⇒ p = 0 ⇒ t px = 0. t x ⎟ ⎠ dt j=0 dt j=0

8.13 (a) The EPV of the care costs is 01 02 12 000 ā00 80 + 25 000 ā80 + 40 000 ā80

+ 5 000 v 5

00 02 01 12 02 22 5 p80 ā85 + 5 p80 ā85 + 5 p80 ā85

= 146 157.

The EPV of the additional costs is 03 10 000 Ā02 80 + 15 000 Ā80 = 11 017.

The EPV of the fees is 01 02 F ā00 80 + ā80 + ā80 = 6.4568F,

and by equating this with the EPV of the care costs and additional costs we obtain F = $24 342.38.

133

Solutions for Chapter 8 (b) Now the EPV of the fees is 01 02 X ā00 80 + 2X ā80 + 3X ā80 = 11.6851X,

and by equating this with the EPV of the care costs and additional costs we obtain X = $13 450.79. (c) In the following, F = 24 350. (i) For State 0, d (0) = δ t V (0) + (F − 12 000) tV dt 01 (1) 03 − t V (0) − μ80+t − μ80+t 15 000 − t V (0) . tV For State 1, d (1) 10 (0) = δ t V (1) + (F − 25 000) − μ80+t − t V (1) tV tV dt 12 13 10 000 + t V (2) − t V (1) − μ80+t 15 000 − t V (1) , − μ80+t and for State 2, noting that the care cost is now $45 000 per year, d (2) 23 = δ t V (2) + (F − 45 000) − μ80+t 15 000 − t V (2) . tV dt (ii) The policy values are 5V

(0)

01 02 02 03 = 12 000 ā00 85 + 25 000 ā85 + 45 000 ā85 + 10 000 Ā85 + 15 000 Ā85 01 02 − F ā00 85 + ā85 + ā85

= 900.05, 5V

(1)

11 12 12 13 = 12 000 ā10 85 + 25 000 ā85 + 45 000 ā85 + 10 000 Ā85 + 15 000 Ā85 11 12 − F ā10 85 + ā85 + ā85

= 14 397.68, 5V

(2)

23 22 = 45 000 ā22 85 + 15 000 Ā85 − F ā85

= 53 234.84. (iii) For State 0 d (0) 01 = δ t V (0) + (24 350 − 12 000) − μ80+t tV dt 03 15 000 − t V (0) , − μ80+t which gives 227.68 when t = 5.

tV

(1)

− t V (0)

134

Solutions for Chapter 8 For State 1 d (1) 10 (0) = δ t V (1) + (24 350 − 25 000) − μ80+t − t V (1) tV tV dt 12 (2) 13 − t V (0) − μ80+t 15 000 − t V (1) , − μ80+t tV which gives −64.42 when t = 5.

For State 2 d (2) 23 = δ t V (2) + (24 350 − 45 000) − μ80+t 15 000 − t V (2) , tV dt which gives −3019.43 when t = 5.

So, at time 5, the reserves are increasing in State 0, but decreasing in States 1 and 2.

For lives in State 0, the reserves should be building up, as the fee of $24 350 exceeds the annual cost of care in State 0 ($12 000), and more than compensates for the expected cost of immediate transition to State 1 or State 3.

For lives in States 1 and 2 the costs exceed the fee, so we can view the outgo from the care home’s point of view as an annuity (variable in the case of lives in State 1 as the amount will increase if transition to State 2 occurs) and a death benefit to a resident. So the reserve is being used up whether the resident survives or dies. The overall effect is a decrease in policy values in States 1 and 2.

8.14 (a) Let P be the total premium per year. Then the premium equation of value is P ā00

55:20

= 30 000 ā01

55:20

+ 100 000 Ā02

55:20

.

We need the following probabilities, annuity and insurance values, calculated using the tables in AMLCR Appendix D: 00 00 00 01 10 20 p55 = 10 p55 10 p65 + 10 p55 10 p65 = 0.28967, 01 00 01 01 11 20 p55 = 10 p55 10 p65 + 10 p55 10 p65 = 0.28807,

ā00

20 00 00 01 10 = ā00 20 p55 ā75 + 20 p55 ā75 = 9.7410, 55 − v

ā01

20 00 01 01 11 = ā01 20 p55 ā75 + 20 p55 ā75 = 1.3309, 55 − v

Ā02

20 00 02 01 12 = Ā02 20 p55 Ā75 + 20 p55 Ā75 = 0.24205.

55 − v

55:20 55:20 55:20

135

Solutions for Chapter 8

Inserting the relevant values into the premium equation of value we obtain P = $6 583.73. (b) The policy value for State 0 at time 10 is 10V

(0)

= 30 000 ā01

65:10

+ 100 000 Ā02

65:10

− P ā00

,

− P ā10

.

65:10

and the policy value for State 1 at time 10 is 10V

(1)

= 30 000 ā11

65:10

+ 100 000 Ā12

65:10

65:10

We have: 10 00 00 01 10 = ā00 10 p65 ā75 + 10 p65 ā75 65 − v 65:10 10 00 01 01 11 ā01 = ā01 10 p65 ā75 + 10 p65 ā75 65 − v 65:10 10 11 11 10 01 ā11 = ā11 10 p65 ā75 + 10 p65 ā75 65 − v 65:10

= 5.7976,

ā00

= 1.0950, = 6.5701,

10 11 10 10 00 = ā10 10 p65 ā75 + 10 p65 ā75 = 0.0290, 65 − v 65:10 10 00 02 01 12 Ā02 = Ā02 10 p65 Ā75 + 10 p65 Ā75 = 0.24429, 65 − v 65:10 10 11 12 10 02 Ā12 = Ā12 10 p65 Ā75 + 10 p65 Ā75 = 0.30178.

65 − v 65:10

ā10

We obtain 10V (0) = 19 110 and 10V (1) = 227 092. (c)

(i) A suitable model is shown in Figure S8.1.

Healthy, never claimed 0

-

Sick 1

Healthy, previously claimed 2

Q

Q Q

Q

Q Q s Q

?

Dead 3

+

.

Figure S8.1 Sickness–death model adapted for refund of premium benefit.

Using a tilde for intensities under this new model, we have the following identities: μ̃x01 = μx01 ,

μ̃x03 = μx02 ,

μ̃x12 = μx10 ,

136

Solutions for Chapter 8 μ̃x13 = μx12 ,

μ̃x21 = μx01 ,

μ̃x23 = μx02 .

Other relationships follow, and we use these below. Specifically, 00 02 00 t p̃55 + t p̃55 = t p55

01 01 t p̃55 = t p55 .

and

(ii) The refund is payable if the life remains continuously in State 0 for 20 years, the probability of which is ⎧ ⎫ ⎪ ⎪ 20 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ 00 01 03 p = exp + μ̃ μ̃ dt − ⎪ ⎪ 20 55 55+t 55+t ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ 0

⎫ ⎧ ⎪ ⎪ 20 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎨ 01 02 + μ μ dt = exp ⎪ − ⎪ 55+t 55+t ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎩ 0

b1 55c1 20c1 b2 55c2 20c2 e −1 + e −1 e e = exp − (a1 + a2 ) t + c1 c2

= 0.288199. (iii) Let P̂ denote the premium for the revised contract. Then the EPV of the premiums is 20

20 v

P̂

00 02 t p̃55 + t p̃55

t

0

00 vt t p00 55 dt = P̂ ā

dt = P̂

55:20

.

0

Similarly, the EPV of the death benefit is 20 vt

100 000

00 03 01 13 02 23 t p̃55 μ̃55+t + t p̃55 μ̃55+t + t p̃55 μ̃55+t

dt

0

20 = 100 000

vt

00 02 01 12 02 02 t p̃55 μ55+t + t p55 μ55+t + t p̃55 μ55+t

vt

00 02 01 12 t p55 μ55+t + t p55 μ55+t

0

20 = 100 000 0

= 100 000 Ā02

55:20

.

dt

dt

137

Solutions for Chapter 8 Lastly, the EPV of the sickness benefit is 20

20 v t p̃01 55 dt = 30 000

01 vt t p01 55 dt = 30 000 ā

t

30 000 0

55:20

.

0

The equation of value for the premium is then P̂ ā00

55:20

= 30 000 ā01

55:20

+ 100 000 Ā02

55:20

+ 10P̂ v 20 20 p00 55 ,

which gives P̂ = $7410. (iv) Similar arguments to those in part (iii) apply to valuing premiums and benefits to calculate policy values, and we omit the detail. All the ā and Ā notation below applies to the original model, and the tilde notation to the revised model.

The policy value for State 0 in the revised model is 10Ṽ

(0)

= 30 000 ā01

65:10

+ 100 000 Ā02

65:10

00 + 10P̂ v 10 10 p00 65 − P̂ ā

65:10

,

and as 10 p00 65 = 0.389662 (which is the same under both models), we get 10V (0) = 32 046.

The policy value for State 1 in the revised model is 10Ṽ

(1)

= 30 000 ā11

65:10

+ 100 000 Ā12

65:10

− P̂ ā10

65:10

= 227 068.

The policy value for State 2 in the revised model is 10Ṽ

(2)

= 30 000 ā01

65:10

+ 100 000 Ā02

65:10

− P̂ ā00

65:10

= 14 320.

(d) The introduction of the return of premium benefit would likely alter policyholders’ behaviour. Policyholders would be unlikely to claim small amounts of sickness benefit if the total of such amounts would be less than the return of premium benefit. So, we might expect the transition intensities from healthy to sick to reduce, most likely at older ages when policyholders who have not previously claimed would view the return of premium benefit as more valuable than potential sickness benefits. If policyholder behaviour did change, then those in State 0 would be less healthy overall than before the change, and this may result in an increase in transition intensities to the dead state. It would also affect the policyholders in the sick state, as they are now likely to be more sick on average as a result of fewer transfers from

138

Solutions for Chapter 8 the healthy state, leaving a higher proportion of long-term sick in the sick state, with a consequent increase in transition intensities to the dead state.

8.15 Equation (8.33) in AMLCR is āixj =

∞ m 0

j j −δt ir r j dt. t px μx+t āx+t e

r=0, r j

If State k is an absorbing state, then t pkk x = 1 for all t ≥ 0 and we have the following results: ∞ ākk x =

kk −δt dt = t px e

0

āik x =

∞

e−δt dt =

0

∞ m

δ

r=0, r k

∞

m

0

r=0, r k

0

ir rk −δt dt, t px μx+t e

r=0, r k

Āik x ir rk −δt ik . dt = δ āik t px μx+t e x ⇒ āx = δ

From Exercise 8.5 we have

m

āixj =

j=0

so

∞ m

1 ir rk −δt kk āx+t dt = t px μx+t e

0

Āik x =

1 , δ

m

āixj =

m

1 , δ

āixj − āik x =

j=0

j=0, j k

m

⇒ Āik x =1−δ

1 Āik − x δ δ āixj .

j=0, j k

8.16 Let P be the total premium per year. Then the equation of value is 2 e

P

−0.05t

2 00 t p50 dt = 60 000

0

e−0.05t t p01 50 dt.

0

We have 2

−0.05t

e 0

2 00 t p50 dt = 0

e−0.05t

2 −0.015t 1 −0.01t e + e dt 3 3

139

Solutions for Chapter 8 2 = 3

2 e

−0.065t

1 dt + 3

0

2

e−0.06t dt

0

= 1.87852, and as 2 01 00 02 t p50 = 1 − t p50 − t p50 = 3

e−0.01t − e−0.015t

we have 2

−0.05t

e

2 01 t p50 dt =

0

e−0.05t

2 −0.011t − e−0.015t dt e 3

0

⎛ 2 ⎞ 2 ⎜ ⎟⎟⎟ 2 ⎜⎜⎜⎜ ⎟ = ⎜⎜⎜ e−0.06t dt − e−0.065t dt ⎟⎟⎟⎟ ⎠ 3 ⎝ 0

0

= 0.00614.

Hence P = $195.99.

8.17 (a) Kolmogorov’s forward equations for this model are d 00 01 10 00 01 02 + μx+t μx+t , t p = t px μx+t − t px dt x d 01 00 01 01 10 12 + μx+t μx+t , t p = t px μx+t − t px dt x d 02 00 02 01 12 t p = t px μx+t + t px μx+t . dt x is the EPV of a benefit of 1 per year, payable continuously for a maxi(b) ā01 x:n mum of n years to a life who is now aged x and in State 0 (healthy), as long as the life is in State 1 (sick). We have n = ā01 x:n

vt t p01 x dt.

0

Ā02 is the EPV of a benefit of 1 payable on the death before time n years x:n of a life who is now aged x and in State 0. The benefit is payable should death occur from either live state. We have n Ā02 = x:n

vt 0

00 02 01 12 t px μx+t + t px μx+t

dt.

140

Solutions for Chapter 8 (c) The policy value in State 1 at time t is tV

(d)

(1)

= B ā11

40+t:20−t

+ S Ā12

40+t:20−t

− P ā10

40+t:20−t

.

(i) We use the approximations 00 00 t+h p40 ≈ t p40

01 02 10 + μ40+t 1 − h μ40+t + t p01 40 h μ40+t

01 01 t+h p40 ≈ t p40

10 12 01 + μ40+t 1 − h μ40+t + t p00 40 h μ40+t

and

01 with h = 0.1, 0 p00 40 = 1 and 0 p40 = 0. These give 00 0.1 p40 ≈ 1 − 0.1 (0.01074 + 0.00328) = 0.99860, 01 0.1 p40 ≈ 0.1 × 0.01074 = 0.00107, 00 0.2 p40 ≈ 0.99860(1 − 0.1 (0.01094 + 0.00330))

+ 0.00107 × 0.1 × 0.09003 = 0.99719, 01 0.2 p40 ≈ 0.00107 (1 − 0.1(0.09003 + 0.00719))

+ 0.99860 × 0.1 × 0.01094 = 0.00216. (ii) Thiele’s differential equations are d (0) 01 (1) 02 = δ t V (0) + P − μ40+t − t V (0) − μ40+t S − t V (0) tV tV dt and d (1) 10 (0) 12 = δ t V (1) − B − μ40+t − t V (1) − μ40+t S − t V (1) . tV tV dt (iii) Using the approximation d (i) 1 ≈ tV dt h

tV

(i)

− t−hV (i)

for i = 1, 2 we have tV

(0)

01 ≈ t−hV (0) + h δ t V (0) + P − μ40+t

tV

(1)

− t V (0)

tV

(0)

− t V (1)

02 − μ40+t S − t V (0)

and tV

(1)

10 ≈ t−hV (1) + h δ t V (1) − B − μ40+t

141

Solutions for Chapter 8 12 S − t V (1) , − μ40+t

with 20V (0) = 20V (1) = 0. Setting h = 0.1 these equations give 19.9V

(0)

02 = −h P − μ60 S = −427,

19.9V

(1)

12 = −h −B − μ60 S = 2594.

8.18 (a) Consider t+h p12 x , and split the time interval (0, t + h) as (0, t) and (t, t + h).

Then 12 11 12 12 22 t+h px = t px h px+t + t px h px+t

⎛ ⎜⎜

12 12 ⎜ ⎜⎜1 − = t p11 x h μx+t + t px ⎜ ⎝

j 2

⎞ ⎟

⎟⎟ 2j ⎟ ⎟⎠ + o(h) h px+t ⎟

⎛ ⎞

⎜⎜⎜ ⎟⎟⎟ 2 j 12 12 ⎜ μx+t ⎟⎟⎟⎠ + o(h). = t p11 ⎜⎝1 − h x h μx+t + t px ⎜ j 2

Then 1 h

12 12 t+h px − t px

12 12 = t p11 x μx+t − t px

2j μx+t +

j 2

o(h) . h

Dividing by h and letting h → 0 we get

d 12 2j 11 12 12 μx+t , t px = t px μx+t − t px dt j 2

or, as the only possible transitions out of State 2 are to State 3 or State 4, d 12 11 12 12 23 24 + μx+t μx+t . t p = t px μx+t − t px dt x (b) First note that as State 1 cannot be re-entered after it has been left, 11 11 t px = t px = exp{−(0.1 + 0.04 + 0.1) t} = exp{−0.24 t}.

Then the differential equation for t p12 x becomes d 12 −0.24t − 0.34 t p12 t p = 0.1 e x . dt x We write this as d 12 12 −0.24t , t p + 0.34 t px = 0.1 e dt x and note that, applying the product rule, d 0.34t 12 0.34t 12 0.34t d 12 e tp t px = 0.34 e t px + e dt dt x

142

Solutions for Chapter 8

d 12 12 p + 0.34 p = e0.34t t t x . dt x Hence d 0.34t 12 0.1t e . t px = 0.1 e dt Integrate over (0, r), noting that 0 p12 x = 0, giving 0.1r − 1, e0.34r r p12 x =e −0.24r − e−0.34r . and so r p12 x =e

Alternative solution The expression for t p12 x can be verified by checking that it satisfies the Kolmogorov differential equation, and the required boundary condition.

First, we require d 12 11 12 12 23 24 t p = t px μx+t − t px μx+t + μx+t dt x = 0.1e−0.24t − 0.34 t p12 x −0.24t as above. Assume that t p12 − e−0.34t as given. Then the left-hand x =e side of the Kolmogorov equation is

d −0.24t − e−0.34t = −0.24e−0.24t + 0.34e−0.34t e dt and the right-hand side is 0.1e−0.24t − 0.34 e−0.24t − e−0.34t = − 0.24e−0.24t + 0.34e−0.34t , so that the Kolmogorov equation is satisfied. Next, we see that the boundary −0 − e−0 = 0 is also satisfied. condition, 0 p12 x =e −0.24t Hence we have verified that t p12 − e−0.34t is the solution to the x =e Kolmogorov differential equation.

(c) The EPV of the benefit is ∞ 3 000

−δt

e 0

∞ 12 t p90 dt = 3 000 0

e−0.05t e−0.24t − e−0.34t dt

1 1 − = 3 000 = 2 652.52.

0.29 0.39

143

Solutions for Chapter 8 (d) First, t p22 x = exp{−(0.04 + 0.30) t} = exp{−0.34 t}. Then for y ≥ 90, 0.5 ā22 = y:0.5

e

−δt

−0.34t

e

0.5 dt =

0

e−0.39t dt =

1 − e−0.195 = 0.45.

0.39

0

(e) With a waiting period of one half-year, the EPV of the benefit is ∞

12 22 22 e−δt t p11 μ 90 90+t ā90+t − ā

90+t:0.5

dt

0

with

∞ ā22 90+t =

e

−δt

∞ 22 r p90+t dr =

0

e−0.39t dt = 2.56

0

(noting that this EPV doesn’t vary with age above age 90 due to the con12 stant transition intensities), and ā90+t:0.5 = 0.45. Hence, with μ90+t = 0.1 for all t, the EPV of the benefit is ∞ 0.1 (2.56 − 0.45)

e−δt e−0.24t dt = 2 182.59.

0

(f) One reason for having a waiting period is that a policy with a waiting period is less expensive than a policy with a waiting period. From an insurer’s perspective, lower premiums may attract a greater volume of business. A second reason for having a waiting period, similar to disability income insurance, is that the waiting period can eliminate claims that have a short duration. This can save the insurer the cost of administering these claims.

8.19 First, the policy value at time t if the policyholder is in State 0 is tV

(0)

= B ā01 − P ā00 , x+t:n−t x+t:n−t

and the policy value at time t if the policyholder is in State 1 is tV

(1)

= B ā11 − P ā10 . x+t:n−t x+t:n−t

The premium equation of value is P ā00 = B ä01 , and we can write this in x:n x:n terms of integrals as n 0=P

e 0

−δ s

n 00 s px ds − B 0

e−δ s s p01 x ds

144

Solutions for Chapter 8 t

−δ s

=P

e

n

e−δ s s p00 x ds

00 s px ds + P t

0

t −B

−δ s

e

n 01 s px ds − B

e−δ s s p01 x ds

t

0

n−t = P ā00 +P x:t

e−δ (s+t) s+t p00 x ds

0

n−t −B − B ā01 x:t

e−δ (s+t) s+t p01 x ds,

0

so that n−t P ā00 − B ā01 =B x:t x:t

e−δ (s+t) s+t p01 x ds − P

0

n−t

e−δ (s+t) s+t p00 x ds.

0

The left-hand side of this equation is the EPV at issue of income minus the cost of insurance, so we want to show that the right-hand side is the EPV at issue of the net premium policy value, which is (0) (1) e−δt t p00 + t p01 . x tV x tV

Noting that 00 00 00 01 10 s+t px = t px s px+t + t px s px+t

and 01 00 01 01 11 s+t px = t px s px+t + t px s px+t ,

we have n−t

−δ (s+t)

e

B

n−t 01 s+t px ds − P

0

e−δ (s+t) s+t p00 x ds

0

n−t =B

e−δ (s+t)

00 01 01 11 t px s px+t + t px s px+t

ds

0

n−t −P 0

e−δ (s+t)

10 00 00 01 t px s px+t + t px s px+t

ds

Solutions for Chapter 8 ⎞ ⎛ n−t n−t ⎟⎟⎟ ⎜⎜⎜ ⎜ ⎟⎟ 01 = B e−δt ⎜⎜⎜⎜t p00 e−δ s s p01 e−δ s s p11 ⎟⎟⎠ x x+t ds + t px x+t ds⎟ ⎝ 0

145

0

⎞ ⎛ n−t n−t ⎟⎟⎟ ⎜⎜⎜ ⎟⎟ ⎜ 01 e−δ s s p00 e−δ s s p10 − P e−δt ⎜⎜⎜⎜t p00 ⎟⎟⎠ x x+t ds + t px x+t ds⎟ ⎝ 0

0

01 01 11 −δt 00 00 01 10 = B e−δt t p00 t px āx+t:n−t + t px āx+t:n−t x āx+t:n−t + t px āx+t:n−t − P e

= e−δt t p00 − P ā00 − P ā10 B ā01 + e−δt t p01 B ā11 x x x+t:n−t x+t:n−t x+t:n−t x+t:n−t (0) (1) = e−δt t p00 + e−δt t p01 x tV x tV ,

as required.

8.20 (a) First note that

⎧ t+s ⎫

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ v(t + s)/v(t) = exp ⎪ δ du − , ⎪ u ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ t

so that v(t + s)/v(t) represents the present value at time t of a unit amount payable at time t + s.

The left-hand side of the formula, t V (i) , is the EPV of all future net cash flows from the insurer from time t, given that the life is in State i at time t. Suppose that the first transition out of State i occurs in the time interval (t + s, t + s + ds), where ds is small, and is to State j, where j = 0, 1, . . . , m, j i; the probability that this happens is (approximately) ij ii s px+t μx+t+s ds.

Given that it does happen, there will be a net single payment at that time (i j) of amount St+s and the EPV of all future net cash flows from time t + s is ( j) t+sV . The EPV at time t of these future cash flows is v(t + s) (i j) St+s + t+sV ( j) v(t)

ij ii s px+t μx+t+s ds.

Summing, that is, integrating, over all possible values for s, s = 0 → ∞, and summing over all possible future states j gives the first term on the right-hand side.

The second term on the right-hand side is the EPV of a continuous stream (i) of payments at rate Bt+s while the life remains in State i.

146

Solutions for Chapter 8 (b) Following the approach in Section 7.4.1 of AMLCR, we change the variable of integration to r = s + t, so that tV

(i)

∞ m

v(r) (i j) = Sr + rV ( j) v(t) j=0, j i t

∞ + t

ii r px ii t px

ij μx+r dr

v(r) (i) r piix B dr . v(t) r t piix

Next, note that ⎡ ⎧ t ⎫⎤ ⎤ ⎡

⎪ ⎢ ⎥ m ⎪ ⎪ ⎪

⎪ ⎪ d ⎢⎢⎢ 1 ⎥⎥⎥ d ⎢⎢⎢⎢ ⎨ ⎬⎥⎥⎥⎥ ij ⎢⎢⎢exp ⎪ ⎥⎦ = ⎥⎥ ⎢⎣ + μ δ ds ⎪ s x+s ⎪ ⎪ ⎪ ⎪ dt v(t) t piix dt ⎣ ⎩ ⎭⎥⎦ j=0, j i 0 ⎫ ⎧ t ⎞ ⎛

⎪ m m ⎪ ⎪ ⎪

⎟ ⎜⎜⎜ ⎪ ⎪ ⎟ ⎬ ij ⎟ ij ⎟⎟⎟ exp ⎨ μx+t + μ δ ds = ⎜⎜⎜⎝δt + ⎪ ⎪ s x+s ⎪ ⎪ ⎠ ⎪ ⎪ ⎭ ⎩ j=0, j i j=0, j i 0 ⎞ ⎛ m

⎟⎟ 5 ⎜⎜⎜ ij ⎟ ⎟⎟⎠⎟ v(t) t piix . μx+t = ⎜⎜⎝⎜δt + j=0, j i

Finally, note that for any function of two variables, g(r, t), ⎡ ∞ ⎤ ∞ ⎢ ⎥⎥⎥ d ⎢⎢⎢⎢ d ⎢⎢⎢ g(r, t) dr⎥⎥⎥⎥⎥ = −g(t, t) + [g(t, r)] dr. ⎦ dt ⎣ dt t

t

Putting these pieces together, we have m ii

d (i) v(t) (i j) t px ij V = − μx+t St + t V ( j) t ii dt v(t) t px j=0, j i ⎞ ⎛ ∞

m m

⎟⎟ ⎜⎜⎜ v(r) (i j) ij ⎟ ⎟ ⎜ + ⎜⎜⎝δt + μx+t ⎟⎟⎠ Sr + rV ( j) v(t) j=0, j i

j=0, j i t

ii r px ii t px

ij μx+r dr

⎞ ∞ ⎛ m

⎟⎟ v(r) (i) r piix v(t) (i) t piix ⎜⎜⎜⎜ ij ⎟ Bt B + ⎜⎜⎝δt + μx+t ⎟⎟⎟⎠ dr − ii v(t) v(t) r t piix t px j=0, j i t ⎞ ⎛ m m

⎟⎟ ⎜⎜⎜ ij ⎟ ij ⎜ μx+t ⎟⎟⎟⎠ t V (i) − Bt(i) − St(i j) + t V ( j) μx+t = ⎜⎜⎝δt + j=0, j i

= δt t V (i) − Bt(i) −

j=0, j i

m

ij , St(i j) + t V ( j) − t V (i) μx+t

j=0, j i

which is Thiele’s differential equation.

147

Solutions for Chapter 8 8.21 (a) (i) We have

⎫ ⎧ ⎪ ⎪ 10 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ 00 01 02 + μ μ dt − ⎪ 10 p30 = exp ⎪ 30+t 30+t ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎩ 0 ⎫ ⎧ ⎪ ⎪ 10 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎨ −4 30+t − A + Bc dt = exp ⎪ −10 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎩ 0

−4

= exp{−10 } s gc (c −1) , 10

30

10

where

g = exp −B/log c = 0.999118.

s = exp{−A} = 0.999490, Hence 10 p00 30 = 0.979122.

(ii) The formula for this probability is 10 01 10 p30 =

00 01 10 p30 μ30+t dt, 0

which can be evaluated by numerical integration, giving 01 10 p30 = 0.020779.

(iii) This probability is 02 00 01 10 p30 = 1 − 10 p30 − 10 p30 = 0.000099.

(b) (i) The EPV for the premium, P per year payable continuously, is 10 P ā00 = P 30:10

vt t p00 30 dt.

0

The EPV of the death benefit is 100 000 Ā02

+ 200 000 Ā01

= 100 000

02 01 vt t p00 30 μ30+t + 2μ30+t dt.

30:10 10

30:10

0

The integrals can be evaluated using numerical integration, giving P = $206.28.

148

Solutions for Chapter 8 (ii) The policy value at time 5 (in State 0) is given by 5V

= 100 000 Ā02 + 200 000 Ā01 − P ā00

(0)

35:5

35:5

35:5

= $167.15.

8.22 Let P be the monthly premium. The premium equation is P ä00(12) = 50 000 Ā01

30:20

30:20

+ 75 000 Ā02

30:20

,

where 1 k v 12 k p00 , 12 30 12 k=0 239

ä00(12) = 30:20

20 Ā01 = 30:20

20 01 v t p00 30 μ30+t dt, t

Ā02 = 30:20

0

02 vt t p00 30 μ30+t dt.

0

To calculate t p00 30 , we have

⎧ t ⎫

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎨ 00 01 02 + μ μ ds − ⎪ t p30 = exp ⎪ 30+s 30+s ⎪ ⎪ ⎪ ⎪ ⎭ ⎩ 0 ⎫ ⎧ t

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎨ 01 1.05 μ ds = exp ⎪ − ⎪ 30+s ⎪ ⎪ ⎪ ⎪ ⎭ ⎩ 0

⎛ ⎫⎞1.05 ⎧ t

⎪ ⎪ ⎜⎜⎜ ⎟ ⎪ ⎪ ⎪ ⎪ ⎬⎟⎟⎟⎟ ⎨ ⎜⎜⎜ 01 ⎟⎟ μ ds = ⎜⎜exp ⎪ − ⎪ 30+s ⎪ ⎪ ⎪ ⎪ ⎝ ⎭⎟⎠ ⎩ 0

1.05

= st gc (c −1) ,

where s = exp{−A} and g = exp −B/log c . Using numerical integration to value the benefits, we find that 30

ä00(12) = 13.2535, 30:20

Ā01

30:20

t

= 0.08288

and

Ā02

30:20

= 0.00414,

which gives a monthly premium of P = $28.01.

8.23 (a) Kolmogorov’s equations, discretized, in this case give us 00 00 00 t+h p30 = t p30 − t p30 h

01 02 03 10 μ30+t + μ30+t + μ30+t + t p01 30 h μ30+t

01 01 01 t+h p30 = t p30 − t p30 h

10 12 13 01 μ30+t + μ30+t + μ30+t + t p00 30 h μ30+t .

and

Solutions for Chapter 8

149

1 01 and using the starting values 0 p00 Setting h = 12 30 = 1 and 0 p30 = 0, we can use these two (approximate) equations to calculate successively 00 01 00 01 00 h p30 , h p30 , 2h p30 , 2h p30 , . . . , 35 p30 .

(b) (i) The EPV of the premiums of P per year payable monthly is P k v 12 k p00 = 15.58544P. 12 30 12 k=0 419

P ä00(12) = 30:35

The EPV of the death and critical illness benefits is 35 02 03 01 12 13 100 000 vt t p00 30 μ30+t + μ30+t + t p30 μ30+t + μ30+t

dt

0

= $8971.30.

The EPV of the disability income, paid continuously, is 35 75 000 ā01 = 75 000 30:35

vt t p01 30 dt = $29 660.94.

0

Hence, the monthly premium, P/12, is given by P/12 = (8 971.30 + 29 660.94)/(12 × 15.58544) = $206.56. (ii) With a premium P per year payable continuously, Thiele’s differential equations for t V (0) and t V (1) are d (0) 01 (1) = δ t V (0) + P − μ30+t − t V (0) tV tV dt 02 03 + μ30+t 100 000 − t V (0) − μ30+t and d (1) 10 (0) = δ t V (1) − 75 000 − μ30+t − t V (1) tV tV dt 12 13 + μ30+t 100 000 − t V (1) . − μ30+t 1 Using Euler’s method with a step size h = 12 , we have the following (approximate) equations:

1 (0) (0) 01 (1) 1 V − V = − t V (0) δ t V (0) + P − μ30+t t tV t− 12 12

02 03 (0) − μ30+t + μ30+t 100 000 − t V ,

150

Solutions for Chapter 8 giving 1 V t− 12

(0)

= tV

(0)

1 01 δ t V (0) + P − μ30+t − 12

tV

(1)

− t V (0)

02 03 + μ30+t − μ30+t 100 000 − t V (0)

and tV

(1)

− t− 121 V

(1)

1 10 = δ t V (1) − 75 000 − μ30+t 12

tV

(0)

,

− t V (1)

12 13 + μ30+t 100 000 − t V (1) − μ30+t

,

giving 1 V t− 12

(1)

= tV

(1)

1 10 − δ t V (1) − 75 000 − μ30+t 12

tV

(0)

− t V (1)

12 13 + μ30+t 100 000 − t V (1) − μ30+t

.

The boundary conditions are 35V

(0)

= 35V (1) = 0.

These equations can be solved, in terms of the unknown P, succes1 2 1 , 35 − 12 , . . . , 12 , 0. Requiring 0V (0) to be sively for t = 35, 35 − 12 equal to 0 gives P = $2 498.07. (iii) The value of 10V (0) is calculated as part of the recursive scheme in part (ii), 10V (0) = $16 925.88.

8.24 We present two approaches. First, we consider versions of Thiele’s differential equation that are specific to the functions we require, and which require separate evaluation. Then we consider a particular policy with premiums and benefits which we vary and which are inputs we can change in a spreadsheet calculation.

Consider the state-dependent policy values for an annuity of 1 per year payable continuously in State 0, issued to (50), with no premiums or other benefits. We have tV

(0)

= ā00 50+t

and

tV

(1)

= ā10 50+t .

151

Solutions for Chapter 8 Applying Thiele’s differential equation, we have d (0) 01 10 00 02 00 = δ ā00 tV 50+t − 1 − μ50+t ā50+t − ā50+t + μ50+t ā50+t , dt d (1) 10 00 10 12 10 = δ ā10 tV 50+t − μ50+t ā50+t − ā50+t + μ50+t ā50+t . dt

(8.1) (8.2)

For the spreadsheet calculations, we set h = 0.01 and let ω = 110, and 10 assume ā00 110 = h and ā110 = 0. We then work backwards from age 110 to age 50, in steps of h = 0.01, iterating the two derivatives from equations (8.1) and (8.2), and using those, step by step, in the Euler equations

d (0) d (1) 00 00 10 10 and ā50+t−h = ā50+t − h . ā50+t−h = ā50+t − h tV tV dt dt Similarly, for an annuity of 1 per year payable continuously in State 1, issued to (50), with no premiums or other benefits, we have tV

(0)

= ā01 50+t

and

tV

(1)

= ā11 50+t .

Applying Thiele’s differential equation, we have d (0) 01 11 01 02 01 = δ ā01 tV 50+t − μ50+t ā50+t − ā50+t + μ50+t ā50+t , dt d (1) 10 01 11 12 11 = δ ā11 tV 50+t − 1 − μ50+t ā50+t − ā50+t + μ50+t ā50+t . dt

(8.3) (8.4)

11 For the spreadsheet calculations, set ā01 110 = 0 and ā110 = h. We then iterate backwards, as above, where in each iteration we calculate the derivatives from equations (8.3) and (8.4), and use those, step by step, in the Euler equations

d (0) d (1) 01 11 11 = ā − h V = ā − h V and ā . ā01 t t 50+t 50+t−h 50+t 50+t−h dt dt

Now consider a policy issued to (50) with a payment of 1 on each transition into State 1, with no premium or other benefits. Then tV

(0)

= Ā01 50+t

and

tV

(1)

= Ā11 50+t .

Even though Ā11 y is not included in the tables, we need it in order to calculate Ā01 .

Applying Thiele’s differential equation, we have y d (0) 01 11 01 02 01 = δ Ā01 tV 50+t − μ50+t 1 + Ā50+t − Ā50+t + μ50+t Ā50+t , dt d (1) 10 01 11 12 11 = δ Ā11 tV 50+t − μ50+t Ā50+t − Ā50+t + μ50+t Ā50+t . dt

(8.5) (8.6)

152

Solutions for Chapter 8 11 We set Ā01 110 = 0 and Ā110 = 0, and iterate backwards using the derivative values from equations (8.5) and (8.6), with the Euler equations

d (0) d (1) 01 01 11 11 Ā50+t−h = Ā50+t − h and Ā50+t−h = Ā50+t − h . tV tV dt dt

Similarly, for a policy issued to (50) with a payment of 1 on each transition into State 0, with no premium or other benefits, we have tV

(0)

= Ā00 50+t

and

tV

(1)

= Ā10 50+t .

Even though Ā00 y is not included in the tables, we need it in order to calcu01 late Āy . Applying Thiele’s differential equation, we have d (0) 01 10 00 02 00 = δ Ā00 tV 50+t − μ50+t Ā50+t − Ā50+t + μ50+t Ā50+t , dt d (1) 10 00 10 12 10 = δ Ā10 tV 50+t − μ50+t 1 + Ā50+t − Ā50+t + μ50+t Ā50+t . dt

(8.7) (8.8)

10 We set Ā00 110 = 0 and Ā110 = 0, and iterate backwards using the derivative values from equations (8.7) and (8.8), with the Euler equations

d (0) d (1) 00 00 10 10 Ā50+t−h = Ā50+t − h and Ā50+t−h = Ā50+t − h . tV tV dt dt

Finally, consider a policy that pays 1 on transition into State 2, with no premium or other benefits. We have tV

(0)

= Ā02 50+t

and

tV

(1)

= Ā12 50+t .

Applying Thiele’s differential equation, we have d (0) 01 12 02 02 02 = δ Ā02 tV 50+t − μ50+t Ā50+t − Ā50+t − μ50+t 1 − Ā50+t , dt d (1) 10 02 12 12 12 = δ Ā12 tV 50+t − μ50+t Ā50+t − Ā50+t − μ50+t 1 − Ā50+t . dt

(8.9) (8.10)

12 We set Ā02 110 = 1 and Ā110 = 1, and iterate backwards using the derivative values from equations (8.9) and (8.10), with the Euler equations

d (0) d (1) 12 12 02 02 Ā50+t−h = Ā50+t − h and Ā50+t−h = Ā50+t − h . tV tV dt dt

Table S8.1 shows selected values from the spreadsheet calculations.

For our second approach, consider a policy issued to (50) under which premiums are payable continuously at rate P per year while in State 0,

153

Solutions for Chapter 8 Table S8.1 Values from Euler calculation of Standard Sickness–Death Table functions. x

ā00 x

ā01 x

ā11 x

ā10 x

Ā01 x

Ā02 x

Ā10 x

Ā12 x

50 60 70 80

11.7447 8.3902 4.9612 2.2934

1.9623 2.6295 3.0171 2.7703

12.3923 10.2081 7.3744 4.6935

0.6664 0.1080 0.0133 0.0012

0.24142 0.36314 0.51441 0.66805

0.33123 0.46235 0.61074 0.75294

0.06547 0.01564 0.00339 0.00065

0.36286 0.49668 0.63955 0.77095

benefits are payable at rate B per year while in State 1, and a benefit of amount S(i j) is payable on each transition from State i to State j for i, j = 0, 1 (where i j), and a benefit of S is payable on death from either State 0 or State 1. Then tV

(0)

(01) 01 Ā50+t + S Ā02 = B ā01 50+t + S 50+t ∞ (10) −δ r 00 01 00 +S e r p50+t μ50+t+r Ā10 50+t+r dr − P ā50+t , (8.11) 0

where the integral term values benefits payable on any future transitions from State 1 to State 0. (The life is in State 0 at time t; if the first move to State 1 after time t takes place at time t + r, the benefits payable on transition from State 1 to State 0 can be valued at time t + r as S(10) Ā10 50+t+r .)

Similarly, tV

(1)

(10) 10 Ā50+t + S Ā12 = B ā11 50+t + S 50+t ∞ 10 01 10 + S(01) e−δ r r p11 50+t μ50+t+r Ā50+t+r dr − P ā50+t . (8.12) 0

Further, Thiele’s differential equation gives d (0) 01 02 = δ t V (0) + P − μ50+t S(01) + t V (1) − t V (0) − μ50+t S − t V (0) , tV dt (8.13) d (1) 10 12 = δ t V (1) − B − μ50+t S(10) + t V (0) − t V (1) − μ50+t S − t V (1) . tV dt (8.14) Setting P = −1 and all benefit amounts to equal 0, equations (8.11) and (8.12) give tV

(0)

= ā00 50+t

and

tV

(1)

= ā10 50+t ,

and we can numerically solve for these by constructing Euler equations

154

Solutions for Chapter 8 from (8.13) and (8.14) as before, with the same boundary conditions at age 110. The columns in the table below show combinations of P, B, S(i j) and S which result in the policy values being the desired ā and Ā functions.

P B S(01) S(10) S

−1 0 0 0 0

0 1 0 0 0

0 0 1 0 0

0 0 0 1 0

0 0 0 0 1

tV

(0)

ā00 50+t

ā01 50+t

Ā01 50+t

–

Ā02 50+t

tV

(1)

ā10 50+t

ā11 50+t

–

Ā10 50+t

Ā12 50+t

An advantage of this approach is that we can set up cells in our spreadsheet to contain values of the premium and benefits, and then we need only set up two columns in our spreadsheet to calculate policy values from Euler equations.

Solutions for Chapter 9

(1) (2) (1) (2) 9.1 As l51 = l50 − d50 − d50 , we get l50 = 1200. Similarly, l52 = l51 − d51 − d51 (1) gives d51 = 70. Then 01 2 p50 =

(1) (1) + d51 d50 150 = 0.125. = l50 1200

9.2 (a) Assuming a uniform distribution of decrements in the single decrement tables, we have 2 400 p01 = q∗(1) 1 − 12 q∗(2) 40 = 40 40 15 490 and p02 40 =

51 = q∗(2) 1 − 12 q∗(1) 40 40 .

15 490

Now write the first equation as q∗(1) 40 =

p01 40 1 − 12 q∗(2) 40

and substitute into the second equation to get ⎛ ⎛ ⎞⎞ ⎜⎜⎜ p01 ⎜⎜ ⎟⎟⎟⎟⎟⎟ 1 ∗(2) ⎜ 40 02 ⎟⎟⎟⎟ . p40 = q40 ⎜⎜⎝1 − ⎜⎜⎝ ∗(2) 1 2 1 − q ⎠⎠ 2 40 Multiplying throughout by 1 − 12 q∗(2) 40 leads to the quadratic equation q∗(2) 40

2

01 02 − q∗(2) 2 + p02 40 − p40 + 2 p40 = 0, 40

which gives q∗(2) 40 = 0.00357.

155

156

Solutions for Chapter 9 (b) As withdrawals take place at the start of the year, (2) (1) q∗(2) l40 − d40 40 = d40 ,

giving q∗(2) 40 = 0.00390.

9.3 The dependent rate of exit by decrement 1 is ∗(1) + 0.75 qx

1 − 0.75 q∗(1) × 0.8 × 0.25 q∗(1) x x+0.75 ,

since decrement 2 operates only at age x + 0.75, and, with probability 0.8, the life survives with respect to decrement 2 at that age. = t q∗(1) for Under UDD for decrement 1 in the single decrement model, t q∗(1) x x ∗(1) 0 ≤ t ≤ 1, and we can find 0.25 qx+0.75 from = 0.75 q∗(1) + 1 − 0.75 q∗(1) q∗(1) x x x

∗(1) 0.25 qx+0.75 ,

which gives − 0.75 q∗(1) − 0.75 q∗(1) q∗(1) q∗(1) x x x x ∗(1) = = 0.01299.

0.25 qx+0.75 = ∗(1) 1 − 0.75 q∗(1) 1 − 0.75 q x x Hence the dependent rate of exit by decrement 1 is 0.0475.

9.4 (a) We count the number of exits by the different decrements before age 60.8 under our assumptions. All 400 decrement-1 exits take place by age 60.8, as do all 640 of the decrement-3 exits. Under the uniform distribution, 80% of the 800 decrement-2 exits take place by age 60.8, i.e. 640. So the total number of exits is 1680 and hence 1 680 00 = 0.790.

0.8 p60 = 1 − 8 000 (b) Counting as in part (a), all 400 decrement-1 exits take place by age 60.4, 40% of the 800 decrement-2 exits take place, i.e. 320, and 80% of the 640 decrement-3 exits take place (since they are uniformly distributed over the first half-year), i.e. 512. So the total number of exits is 1 232 and so 1 232 00 = 0.846.

0.4 p60 = 1 − 8 000

(c) The probability that a life exits by decrement 3 given that the life exits in the first 0.4 years is just the number of decrement-3 exits in that time, divided by the total number of exits, i.e. 640/1 232 = 0.4156. (This is the 00 same as 0.4 p03 60 / 1 − 0.4 p60 .)

157

Solutions for Chapter 9 9.5 (a) We have ⎧ t ⎫ ⎫ ⎧ n−t

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ ⎬ ⎨ 00 11 01 02 12 − + μ μ dr μ ds exp − ⎪ ⎪ ⎪ t px n−t px+t = exp ⎪ x+s x+s x+t+r ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ ⎭ ⎩ 0 0 ⎫ ⎧ t ⎫ ⎧ n

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎨ ⎬ ⎨ 01 02 12 + μ μ dr = exp ⎪ μ ds − exp − ⎪ ⎪ ⎪ x+s x+s x+r ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎩ ⎭ ⎩ t 0 ⎫ ⎫ ⎧ t ⎧ n

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎬ ⎨ ⎨ 01 02 μ ds μ ds = exp ⎪ exp − − ⎪ ⎪ ⎪ x+s x+s ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎭ ⎩ ⎩ 0

0

since μx02 = μx12 for all x. (b) We have n 01 n px =

00 01 11 t px μx+t n−t px+t dt 0

⎧ t ⎫ ⎧ n ⎫

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ ⎨ ⎬ 01 01 02 exp ⎪ μ ds exp μ ds = − dt − μ ⎪ ⎪ ⎪ x+s x+t x+s ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ ⎩ ⎭ 0 0 0 ⎫ n ⎧ t ⎫ ⎧ n

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎨ ⎬ 01 ⎨ 02 01 μ ds exp μ ds = exp ⎪ − μx+t dt. − ⎪ ⎪ ⎪ x+s x+s ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎩ ⎭ ⎩ n

0

0

0

Noting that ⎧ t ⎫ ⎧ t ⎫

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ d ⎬ ⎨ ⎬ ⎨ 01 01 01 exp ⎪ μ ds exp μ ds = −μ − , − ⎪ ⎪ ⎪ x+s x+t x+s ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ dt ⎭ ⎩ ⎭ ⎩ 0

0

we get ⎧ t ⎧ n ⎫ n⎛ ⎫⎞

⎜⎜

⎪ ⎪ ⎪ ⎪ ⎟ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎜ d ⎨ ⎨ ⎬ ⎬⎟⎟⎟⎟ ⎜⎜⎜ 01 02 01 ⎟⎟ dt exp μ ds μ ds − − − ⎪ ⎪ ⎪ ⎜ n px = exp ⎪ x+s x+s ⎪ ⎪ ⎪ ⎪ ⎜⎝ dt ⎪ ⎪ ⎪ ⎪ ⎩ ⎭⎟⎠ ⎩ ⎭ 0 0 0 ⎫⎛ ⎫⎞ ⎧ n ⎧ n

⎪ ⎪ ⎪ ⎪ ⎜ ⎟ ⎪ ⎪ ⎪ ⎪ ⎜ ⎪ ⎪ ⎪ ⎪ ⎜ ⎬ ⎬⎟⎟⎟⎟ ⎨ ⎨ ⎜⎜⎜ 02 01 ⎟⎟ . μ ds μ ds 1 − exp − = exp ⎪ − ⎪ ⎪ ⎪ ⎜ x+s x+s ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎜⎝ ⎭⎟⎠ ⎩ ⎩ 0

0

9.6 UDD in the single decrement models implies that for j = 1, 2 ∗( j) j) = t q∗( t qx x

and

∗( j) 0 j ∗( j) t px μx+t = qx .

158

Solutions for Chapter 9 Then for 0 < t ≤ 1, t 01 t px =

t 00 01 r px μx+r dr =

0

t =

∗(1) ∗(2) 01 μx+r dr r px r px

0

dr 1 − r q∗(2) q∗(1) x x

0

= q∗(1) x

t

1 − r q∗(2) dr x

0

t = t q∗(1) 1 − q∗(2) . x x 2 In particular,

1 ∗(2) ∗(1) q = q 1 − . p01 x x 2 x

So the assumption of UDD in the single decrement models does not give 01 01 01 01 t px = t px for 0 < t < 1, and t px = t px is the relationship that holds under UDD in the multiple decrement model. So UDD in the single decrement model implies that the decrement is not uniformly distributed in the multiple decrement model.

9.7 (a) The independent probabilities of death are q∗(1) 39 = 0.00049,

q∗(1) 40 = 0.00053,

q∗(1) 41 = 0.00057.

For x = 39, 40+ , 41, where 40+ indicates age 40 after resignations at that age, we calculate ∗(1) p01 1 − 12 q∗(2) x = qx x

and

∗(2) p02 1 − 12 q∗(1) , x = qx x

giving the following values: x

p01 x

p02 x

39 40+ 41

0.00047 0.00050 0.00053

0.08998 0.09997 0.10997

For these values of x we can then calculate dx( j) as lx p0x j for j = 1, 2, with (3) d40 = 0.2 l40 . We get the following values:

159

Solutions for Chapter 9 x

lx

dx(1)

dx(2)

39 40 40+ 41

100 000.0 90 955.1 72 764.1 65 453.2

47.1

8 997.8

36.4 35.0

7 274.5 7 197.8

dx(3) 18 191.0

(b) The probability that an employee aged 39 is still employed at age 42 is l42 /l39 , and from the above multiple decrement table we find l42 = 58 220.4, so the probability is 0.58220. (c) The probability that an employee aged 39 transfers between ages 41 and (2) 42 is d41 /l39 = 0.07198. (d) Let D be the annual deposit. Then the EPV of deposits is D 1 + v l40+ + v 2 l41 /l39 = 2.2349 D and, assuming that transfers take place mid-way through the year of age, the EPV of the grant is (2) (2) (2) + v 3/2 d40 + v 5/2 d41 10 000 v 1/2 d39 /l39 = 2 107.75.

Equating these EPVs gives D = $943.11.

9.8 (a) We solve ∗(1) 1 ∗(2) p01 = 0.1, 60 = q60 1 − 2 q60 ∗(2) 1 ∗(1) p02 = 0.06.

60 = q60 1 − 2 q60 ∗(2) ∗(2) By differencing these we obtain q∗(1) 60 − q60 = 0.04. Substituting for q60 in the first equation leads to

q∗(1) 60

2

− 2.04 q∗(1) 60 + 0.2 = 0,

∗(2) from which we obtain q∗(1) 60 = 0.103267, and hence q60 = 0.063267.

(b) We first find t p00 60 since we need this in part (c). We have ∗(1) ∗(2) 00 t p60 = t p60 t p60

= 1 − t q∗(1) 1 − t q∗(2) 60 60 = 1 − t q∗(1) 1 − t q∗(2) 60 60 , from which 0.8 p00 60 = 0.870955.

160

Solutions for Chapter 9 (c) The Kolmogorov forward differential equation for t p02 x is d 02 00 02 t p = t px μx+t . dt x

(8.15)

Under our assumptions t 02 t px =

t 00 02 r px μx+r dr =

0

∗(1) ∗(2) 02 μx+r dr r px r px

0

= q∗(2) x

t

1 − r q∗(1) dr, x

0

giving d 02 ∗(2) 1 − t q∗(1) . t p = qx x dt x

(8.16)

(d) Under our assumptions, the Kolmogorov forward differential equation (8.15) can also be written as d 02 ∗(1) 02 , 1 − t q∗(2) μx+t t p = 1 − t qx x dt x where we have used part (b) to write t p00 x , and equating this with (8.16) gives 02 .

1 − t q∗(2) μx+t q∗(2) 1 − t q∗(1) = 1 − t q∗(1) x x x x

This gives 02 = μx+t

q∗(2) x 1 − t q∗(2) x

02 and hence μ60.8 = 0.06664.

9.9 First, note that Āx , the EPV of a unit payment immediately on the death of a life now aged x, does not depend on whether this individual is still an employee, has withdrawn or has retired. This is because the intensity of mortality is the same from States 0, 1 and 2 in our model.

We calculate probabilities by breaking the probability up into the part before the specified age, the part relating to transition at the specified age, and the part after. For example, the probability of surviving in employment to just before

161

Solutions for Chapter 9 age 60 from age 40 is, say, 20−p00 40 , where ⎧ ⎫ ⎪ ⎪ 20 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ 00 02 03 − p = exp + μ μ dt − ⎪ ⎪ 20 40 40+t 40+t ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 0 ⎭ ⎫ ⎧ ⎪ ⎪ 20 ⎪ ⎪ ⎪ ⎪ ⎪ −20 μ 02 ⎪ ⎬ ⎨ μ dt e = exp ⎪ − ⎪ 40+t ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎩ 0

= 20 p40 e−20μ .

02

At exact age 60, 40% of the survivors retire, so the probability of surviving to just after age 60, 20+p00 40 say, can be written as 00 00 20+p40 = 0.6 20−p40 .

Between ages 60 and 61 the only cause of decrement is mortality. So, the probability of surviving in employment from age 40 to just before age 61 is 00 00 21−p40 = 20+p40 p60 .

Then at age 61, another 40% exit, so the probability of being in employment just after age 61 is 00 00 2 −20μ 02 , 21+p40 = 0.6 21−p40 = 0.6 21 p40 e

and so on.

Consider the benefit on death after retirement. Retirement can take place only at exact ages 60, 61, 62, . . . , 65. If the employee retires at age x, the EPV of the benefit from that age is 150 000 Āx .

The probability that an employee currently aged 40 will retire at 60 is 00 −20μ02 .

20− p40 × 0.4 = 0.4 20 p40 e

Hence the EPV at age 40 of the retirement benefit from age 60 is 150 000 Ā60 e−20δ 0.4 20 p40 e−20μ02 = 150 000 e−20μ 0.4 20 p40 e−20δ Ā60 .

02

The probability of retiring at age 61 is the product of the probabilities of the following events: (i) surviving in employment to age 60− ,

162

Solutions for Chapter 9 (ii) not retiring at age 60, (iii) surviving from age 60+ to age 61− , and (iv) retiring at age 61.

This probability is equal to 00 −20μ 02 (0.6 × 0.4).

20− p40 × 0.6 × 1 p60 × 0.4 = 21 p40 e

Continuing in this way, the probability that the 40-year-old employee retires at age 65 is 00 20− p40 × 0.6 × 1 p60 × 0.6 × 1 p61 × 0.6 × 1 p62 × 0.6 × 1 p63 × 0.6 × 1 p64

= 25 p40 e−20μ 0.65 .

02

Hence the EPV of the benefit payable on death after retirement is ⎧ ⎫ ⎛ 20+k ⎪ ⎪

⎪ ⎪ 4 ⎪ ⎪ 4 ⎜⎜⎜⎜ ⎪ ⎪ ⎨ ⎬ −(20+k)δ 03 Ā60+k 0.6k × 0.4 exp ⎪ μ dt 150 000 exp −20μ 02 ⎜⎜⎜⎜⎜ − e ⎪ 40+t ⎪ ⎪ ⎪ ⎪ ⎝ k=0 ⎪ ⎪ ⎩ 0 ⎭ ⎧ ⎫ ⎞ ⎪ ⎪ 25 ⎟⎟⎟ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎟⎟⎟ ⎨ ⎬ 03 −25δ Ā +0.65 exp ⎪ μ dt − e ⎟⎟⎟ ⎪ 65 40+t ⎪ ⎪ ⎪ ⎪ ⎠ ⎪ ⎪ ⎩ ⎭ 0 ⎛ ⎞ 4 ⎜⎜ ⎟⎟⎟ −20μ 02 ⎜ k 5 ⎜⎜⎝⎜0.4 0.6 k |Ā60 + 0.6 5 |Ā60 ⎟⎟⎠⎟ . = 150 000 20 E40 e 3

k=0

The EPV of the lump sum payable on death as an employee can be expressed as the sum of the EPV of any benefit payable before age 60, the EPV of any benefit payable between 60 and 61, and so on up to the value of any benefit payable between 64 and 65. As with the death after retirement benefit, we need to split the probabilities after age 60 into up to, at and after the year-end exits. Recalling that, in this example, the probability of surviving in employment between exact

163

Solutions for Chapter 9 age retirements is the ordinary survival probability 1 px , the EPV is ⎛ 20 1 ⎜⎜⎜ ⎜⎜⎜ −δt 00 03 −20δ 00 03 e−δt t p00 200 000 ⎜⎜ e t p40 μ40+t dt + e 20− p40 × 0.6 60+ μ60+t dt ⎝ 0 2 + e−21δ 20− p00 40 × 0.6 × 1 p60 ×

1

0 03 e−δt t p00 61+ μ61+t dt + · · ·

0

+ e

−24δ

1 00 5 20− p40 × 0.6 × 4 p60 ×

e

−δt

⎞ ⎟⎟⎟

⎟⎟ 00 03 ⎟⎟⎠ t p64+ μ64+t dt ⎟

0

which can be written more neatly as ⎛ 5 ⎞⎞ ⎛ ⎜⎜ ⎟⎟⎟⎟⎟⎟ ⎜⎜⎜ 02 ⎜ + 20 E40 e−20μ ⎜⎜⎜⎝ 0.6k k−1 |Ā 1 ⎟⎟⎟⎠⎟⎟⎟⎠ .

200 000 ⎜⎜⎜⎝Ā 1 40:20 60:1 k=1

9.10 Let t p19 denote the probability that a student, currently aged 19, will be alive at age 19 + t. Then, for example, 0.5 p19 = exp{−0.5 × 5 × 19 × 10 1 p19 = exp{−5 × 19 × 10

−5

−5

} = 0.999525,

} = 0.999050,

−5 3.5 p19 = 3 p19 exp{−0.5 × 5 × 22 × 10 } = 0.996456.

Let t p∗19 denote the probability that a student, currently aged 19, will be alive at age 19 + t and will still be at university. Then, for example, ∗ 0.5+ p19 = 0.85 0.5 p19 , ∗ 2 1+ p19 = 0.85 1 p19 , ∗ 2 2 2 2 3.5+ p19 = 0.85 × 0.9 × 0.95 × 0.98 3.5 p19 .

The EPV of the tuition fees is 10 000

7

1.02 2 v 2 k + p∗19 = $53 285.18. k

k

2

k=0

9.11 First note that

⎧ t ⎫

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ 00 01 02 03 + μ + μ μ ds − ⎪ t px = exp ⎪ x+s x+s x+s ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ 0

= exp{−0.0005((x + t)2 − x2 )} exp{−0.01t} st gc (c −1) , x

t

164

Solutions for Chapter 9

where s = exp{−A} and g = exp −B/log c . So t p00 x can be evaluated for any values of x and t. (a) (i) The required probability can be written as 2 01 2 p25 =

00 01 t p25 μ25+t dt = 0.050002, 0

evaluated by numerical integration. (ii) The required probability is 1 00 03 00 2 p25 p27 = 2 p25

00 03 t p27 μ25+t dt = 0.003234.

0

(iii) The required probability is 3 p00 25 = 0.887168. (b) Let L denote the levy payable by each individual in active service on the first and second anniversaries of joining. The EPV (at the time of joining) of the levies payable by an individual is 2 00 L (v 1 p00 25 + v 2 p25 ).

The EPV of the lump sum payment on transfer is 3 10 000 Ā01 = 10 000 25:3

01 vt t p00 25 μ25+t dt.

0

Equating these two EPVs gives L = $397.24. (c) We now expand the model to incorporate transitions after transfer. The model is now Active 0 ?

Resigned 2

HH

- Transfer 1 HH j H Dead 3

-

Dead 4

165

Solutions for Chapter 9 We require 3 04 3 p25 =

00 01 14 t p25 μ25+t 3−t p25+t dt.

0

Now, once the life has transferred to State 1, the survival probability is a two-state Makeham survival probability, with parameters 1.5A, 1.5B and c, so that

1.5B 28 11 . c − c25+t 3−t p25+t = exp − 1.5A(3 − t) + log c Numerical integration gives 3 p04 25 = 0.000586.

9.12 Let t p50 denote the probability that the policyholder, currently aged 50, will still be alive at age 50 + t. Then t p50 = exp

−At −

B c50 ct − 1 log c

.

Let t p00 50 denote the probability that the policyholder, currently aged 50, will still be alive at age 50 + t, and will not have lapsed the policy. Then ⎧ ⎪ p for 0 ≤ t < 1, ⎪ ⎪ ⎪ t 50 ⎨ 00 0.98 p for 1 ≤ t < 2, t p50 = ⎪ t 50 ⎪ ⎪ ⎪ ⎩ 0.982 p for 2 ≤ t ≤ 10. t 50 Let P denote the monthly premium. The EPV of the premiums is 12 P ä00(12) = P 50:10

119

k

v 12 k p00 50 = 84.608P .

12

k=0

For t = 0, 1, 2, . . ., let t |p0d 50 denote the probability that the policyholder dies from the active state in the year from age 50 + t to age 50 + t + 1. Then ⎧ ⎪ for t = 0, q50 ⎪ ⎪ ⎪ ⎨ 0d 0.98 1 |q50 for t = 1, t |p50 = ⎪ ⎪ ⎪ ⎪ ⎩ 0.982 |q for t = 2, 3, . . . . t 50 Now let A0d denote the EPV of a payment of 1 at the end of the year of death, 50:10 before lapsation, within 10 years. Then A0d

50:10

=

9

t=0

0d t+1 = 0.143964. t |p50 v

166

Solutions for Chapter 9 Under UDD, the continuous insurance EPV is i Ā0d = A0d = 0.147534.

50:10 δ 50:10 So the premium equation, for the monthly premium P, is 0.975 × 12P ä00 (12) = 100 000 Ā0d

50:10

50:10

+ 200,

giving P = $181.27.

(1) (1) (1) 9.13 (a) 3 p01 60 = d60 + d61 + d62 /l60 = 0.109.

(b) As l63 = 8 945 − 380 − 110 − 70 = 8 385, we get 00 2 p61 = l63 /l61 = 0.8850.

(c) The EPV at 5% per year is (3) (3) (3) 10 000 v d60 + v 2 d61 + v 3 d62 /l60 = 125.09.

(d) The EPV at 5% per year is 1 000 l60 + v l61 + v 2 l62 /l60 = 2 713.72. (e) Let μ 0 j denote the (assumed) constant transition intensity from State 0 to State j at age 62, for j = 1, 2, 3, and let μ 0• = μ 01 + μ 02 + μ 03 .

Then for 0 < t ≤ 1,

⎧ t ⎫

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ 00 01 02 03 μ +μ +μ ds⎪ − = exp{−μ 0•t}, t p62 = exp ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ 0

0• with p00 62 = exp{−μ } = 8 385/8 945 = 0.93740. Further,

1 p01 62 =

00 01 t p62 μ dt 0

1 =μ

01

exp{−μ 0•t} dt 0

μ 1 − exp{−μ 0• } μ 0• = 0.04248.

=

01

167

Solutions for Chapter 9 01 Hence μ 01 = 0.04387 and so q∗(1) 62 = 1 − exp{−μ } = 0.04292.

(f) Under the constant force assumption, 01 q∗(1) 62 = 0.1 = 1 − exp{−μ },

giving μ 01 = 0.10536. Using the same approach as in part (e), we get μ 02 = 0.01270 and μ 03 = 0.00808, so μ 0• = 0.12614. Then 1 p01 62 =

exp{−tμ 0• } μ 01 dt 0

0.10536 (1 − exp{−0.12614}) = 0.09899. = 0.12614 03 Similarly, p02 62 = 0.01193 and p63 = 0.00759, and so (1) d62 = 885.4,

(2) d62 = 106.7,

(3) d62 = 67.9.

To use the UDD assumption in the single decrement models, we first cal∗(3) culate q∗(2) 62 = 0.01262 and q62 = 0.00805, using the same approach as in part (e). Then

1 ∗(2) 1 ∗(2) ∗(3) ∗(1) ∗(3) q p01 = q + q q 1 − = 0.09897, + q 62 62 62 2 62 3 62 62 03 and similarly p02 62 = 0.01194 and p62 = 0.00760, giving (1) d62 = 885.3,

(2) d62 = 106.8,

(3) d62 = 68.0.

9.14 The EPV of the contributions is 10 000

34

v k k+ p00 30 .

k=0

For k = 0, 1, . . . , 29, 00 k c30 (ck −1) exp{−k μ 02 } k+ p30 = s g

where s = exp{−A}, and g = exp{−B/log c}, and for k = 30, 31, . . . , 34, 3 4 00 k c30 (ck −1) exp −30 μ 02 0.6k−29 . k+ p30 = s g The EPV is $125 489.33.

Solutions for Chapter 10

10.1 (a) 10 p60 × 10 p70 = 0.7802. (b) 10 p60 (1 − 10 p70 ) + 10 p70 (1 − 10 p60 ) + 10 p60 × 10 p70 = 0.9898. (c) 10 p60 (1 − 10 p70 ) + 10 p70 (1 − 10 p60 ) = 0.2096. (d) 1 − 10 p60 × 10 p70 = 0.2198. (e) (1 − 10 p60 )(1 − 10 p70 ) = 0.0102. (f) Note that 20 p60 = 0.94 × 0.83 = 0.7802 and similarly 20 p70 = 0.5976. We require 10 p60:70 − 20 p60:70 = 0.94 × 0.83 − 0.7802 × 0.5976 = 0.3140.

(g) 10 p60 + 10 p70 − 10 p60:70 − (20 p60 + 20 p70 − 20 p60:70 ) = 0.0782.

10.2 We want both lives to survive for two years, with probability 0.72 = 0.49, and then in the third year (x) dies but (y) does not. Note that pxy = 0.7 = Pr[Tx > 1 and Ty > 1], and py = 0.85 = Pr[Ty > 1] = Pr[Ty > 1 and Tx > 1] + Pr[Ty > 1 and Tx ≤ 1].

Consequently Pr[Ty > 1 and Tx ≤ 1] = 0.15, and the same holds for ages y + 2 and x+2 as the given survival probabilities are not age-dependent. The required probability is therefore 0.49 × 0.15 = 0.0735.

10.3 (a) ä50:60:10 = 7.9044 from the tables.

168

169

Solutions for Chapter 10 (b) ä50:60 = ä50 + ä60 − ä50:60 = 17.6587. (c) ä60|50 = ä50 − ä50:60 = 2.7546. i A50:60 = 0.32843. δ 1 = (Ā60:60 − v 10 10 p60:60 Ā70:70 ) (e) Ā 1 60:60:10 2 i (A60:60 − v 10 10 p60:60 A70:70 ) = 0.04242. = 2δ 1 i 1 1 = Ā60 − Ā60:60 = (A60 − A60:60 ) = 0.10837. (f) Ā60:602 = Ā60 − Ā60:60 2 δ 2 10.4 (a) The EPV of the annuity is (d) Ā50:60 =

20 000

∞

vt t p60 (1 − t p70 ) + t p70 (1 − t p60 ) + t p60:70

t=1

= 20 000 (a60 + a70 − a60:70 ) = 293 808.37. (b) The EPV of the annuity is 30 000

9

vt t p60:70 = ä60:70:10 = 225 329.46.

t=0

(c) The EPV of the annuity is 25 000

∞

vt t p60 (1 − t p70 ) = 25 000 (a60 − a60:70 )

t=1

= 92 052.87.

10.5 To show the result we make use of Tx + Ty = Txy + Txy ,

(10.1)

Āx + Āy = Āxy + Āxy .

(10.2)

We have Cov(v Txy , v Txy ) = E[v Txy v Txy ] − E[v Txy ] E[v Txy ] = E[v Txy +Txy ] − Āxy Āxy = E[v Tx +Ty ] − Āxy Āxy

using (10.1)

= E[v ] E[v ] − Āxy Āxy Tx

Ty

using independence

170

Solutions for Chapter 10 = Āx Āy − Āxy Āxy = Āx Āy − (Āx + Āy − Āxy ) Āxy

using (10.2)

= (Āx − Āxy ) (Āy − Āxy ).

10.6 Let P be the annual premium. Then the EPV of premiums is P ä60:60:10 = 7.8080P. The EPV of the annuity payments is 120 000

∞

vt t p60:60 + 2 × 70 000

t=10

∞

vt t p60 (1 − t p60 )

t=10

= 120 000 10 |ä60:60 + 140 000 (10 |ä60 − 10 |ä60:60 ) = 140 000 10 |ä60 − 20 000 10 |ä60:60 .

Now 10 |ä60 = 10 E60 ä70 = 6.9485 and 10 |ä60:60 = ä60:60 − ä60:60:10 = 5.4417,

leading to P = $110 650.

10.7 Let Y denote the present value random variable for the reversionary annuity.

Then ⎧ ⎪ ⎪ if Ty > Tx , ⎨ āTy − āTx Y =⎪ ⎪ ⎩ 0 if Ty ≤ Tx , ⎧ ⎪ ⎪ ⎨ āTy − āTx ⇒Y =⎪ ⎪ ⎩ ā − ā Ty Ty

if Ty > Tx , if Ty ≤ Tx .

Now, if Ty > Tx then Tx = Txy , and, similarly, if Ty ≤ Tx then Ty = Txy , so ⎧ ⎪ if Ty > Tx , ⎪ ⎨ āTy − āTxy Y =⎪ ⎪ ⎩ ā − ā if Ty ≤ Tx , Ty

Txy

giving Y = āTy − āTxy .

10.8 As P is the total premium each year, the EPV of premiums is

1 − Ā75:75 P ā75:75 = P = 7.2970 P. δ6%

171

Solutions for Chapter 10 The EPV of the death benefit is 1 100 0000 Ā75 − Ā75:75

since the benefit is payable on John’s death, but not if he dies first. As 1 1 Ā75:75 + Ā75:75 = Ā75:75 ,

the EPV of the death benefit is 100 0000 0.46570 − 12 (0.57481) = 17 829.50, and so P = $2 443.39.

10.9 (a) Since the lives are independent, we have t pxy = t px t py

= gc (c −1) gc (c −1) x

t

x

t

= g(c +c )(c −1) x

w

y

t

= gc (c −1) t

= t pw , where cw = cx + cy so that w=

log (cx + cy ) . log c

(b) The insurance function can be written as 1 Ax:y =

∞

1 v k+1 k pxy qx+k:y+k =

k=0

∞

1 v k+1 k pw qx+k:y+k .

k=0

Now 1 qx+k:y+k =

t px+k:y+k μx+k+t dt

1

0

1 =

x+k+t dt t pw+k B c 0

1 =

w+k+t t pw+k B c

0

= qw+k

cx . cw

cx dt cw

172

Solutions for Chapter 10 So cx k+1 v k pw qw+k cw k=0 ∞

1 = Ax:y

=

cx Aw cw

as required.

10.10 (a) Since the lives are independent, we have t pxy = t px t py

= st gc (c −1) st gc (c −1) x

t

y

t

= s2t g(c +c )(c −1) x

y

w

t

t

= s2t g2c (c −1) = t pw:w , where cw = (cx + cy )/2, or w=

log((cx + cy )/2) . log c

(b) We calculate the equivalent equal age as w=

log((1.12448.24 + 1.12460 )/2) = 56. log 1.124

(i) The EPV of the joint life annuity is 10 000 ä56:56 = 143 725. (ii) The EPV of the last survivor annuity is 10 000 ä56:56 = 10 000 (2 ä56 − ä56:56 ) = 173 163.

10.11 (a) Let FTxy denote the distribution function of Txy . Then 1 − FTxy (t) = Pr[Txy > t] = Pr[Tx > t and Ty > t].

By independence, Pr[Tx > t and Ty > t] = Pr[Tx > t] Pr[Ty > t] = t px t py .

Hence FTxy (t) = 1 − t px t py and so d FT (t) dt xy d d = − t px t py − t py t px dt dt

fTxy (t) =

173

Solutions for Chapter 10 = t pxy (μx+t + μy+t ) using formula (2.18) of AMLCR.

(b) Let fx and fy be the probability density functions of Tx and Ty respectively. As Tx and Ty are independent, the joint probability density function of (Tx , Ty ) is the product of fx and fy . $ % We have Āxy1 = E v Tx I(Tx < Ty ) , and writing this expectation in integral form we have ∞ ∞ 1 Āxy = vt fx (t) fy (s) I(t < s) ds dt t=0 s=0 ∞ ∞

=

vt fx (t) fy (s) ds dt t=0 s=t ∞

=

∞

v fx (t) t

t=0

(note change of lower integration limit)

fy (s) ds dt. s=t

Now fx (t) = t px μx+t , and ∞ fy (s) ds = Pr[Ty > t] = t py , t

so ∞ Āxy =

∞ t px μx+t t py dt =

1

0

t pxy μx+t dt.

0

$ % (c) We now have Āxy2 = E v Tx I(Tx > Ty ) . The argument in part (b) now yields ∞ t Āxy =

vt fx (t) fy (s) ds dt

2

t=0 s=0 ∞

=

t px (1 − t py ) μx+t dt 0

= Āx − Āxy1 .

Rearranging this result, we see that Āx = Āxy1 + Āxy2 .

174

Solutions for Chapter 10 The result follows as (x) must die either before or after (y). If (x) were to effect two contracts, one paying $1 on death before (y), the other paying $1 on death after (y), then the total payout under these contracts is $1 on the death of (x).

10.12 (a) Let P denote the annual premium. The EPV of the premiums is P ä50 = 17.0245 P. To find the EPV of the death benefit, we initially use letters as subscripts.

In an obvious notation, the EPV of the death benefit is 2 100 000 AB:M .

We have 1 2 AB = AB:M + AB:M

so that 2 1 AB:M = AB − AB:M .

Replacing B and M by age 50 and noting (by symmetry) that 1 A50:50 = 12 A50:50 ,

we see that the EPV of the death benefit is 100 000 A50 − 12 A50:50 = 100 000 0.18931 − 12 (0.24669) = 6 596.50 using the tables in Appendix D. Equating this EPV with the EPV of the premiums gives P = $387.47. (b) (i) If Bob is alive at time 10, but Mike is not, the policy value is 10V = 100 000 A60 − P ä60 = $23 253.

(ii) If both are alive at time 10, the policy value is 10V = 100 000

A60 − 12 A60:60 − P ä60 = $4 800.

10.13 (a) We can value the lump sum death benefits as $1000 payable on the death of each life plus an extra $9000 payable on the first death. The EPV of these benefits is 1000 (Ā65 + Ā60 ) + 9000 Ā65:60 = $5440.32.

Solutions for Chapter 10

175

(b) There is a reversionary annuity to each of the lives so that the EPV is 5 000 (ā65 − ā65:60 + ā60 − ā65:60 )

1 − Ā65 1 − Ā60 1 − Ā65:60 = 5 000 + −2 δ δ δ = $25 262.16. (c) Let P be the annual rate of premium. The equation for P is P ā65:60 = 5440.32 + 25 262.16 so that P = (5440.32 + 25 262.16)/((1 − Ā65:60 )/δ ), giving P = $2470.55. (d) Using the labelling of states in Figure 10.2 of AMLCR, the policy values required in parts (i) and (ii) are denoted 10V (0) and 10V (2) , respectively. (i) The policy value is given by 10V

(0)

= 5000 (ā75 + ā70 − 2 ā75:70 ) + 1000 (Ā75 + Ā70 ) + 9000 Ā75:70 − P ā75:70 .

(ii) The policy value is given by 10V

(2)

= 5000 ā70 + 1000 Ā70 .

(iii) Thiele’s differential equation at policy duration t in these two cases is (1)

(2)

d (0) 02 (2) = δ t V (0) + P − μ65+t:60+t + 10 000 − t V (0) tV tV dt 01 (1) + 10 000 − t V (0) , − μ65+t:60+t tV d (2) 23 = δ t V (2) − 5000 − μ60+t 1000 − t V (2) . tV dt

10.14 (a) (i) The total intensity out of State 0 to States 2 and 4 at age x is μx∗ , and the intensity out of State 1 to State 3 at age x is also μx∗ . So, regardless of whether (y) is alive or not, the intensity of death for (x) is the same, and so survival or death probabilities for (x) do not depend on whether or not (y) is alive.

176

Solutions for Chapter 10 (ii) The model contains a common shock transition which allows for simultaneous deaths, so the lives are not independent with respect to mortality. (b) In general, ⎧ t

⎪ ⎪ ⎪

⎨ 00 − t pxy = exp ⎪ ⎪ ⎪

01 02 04 + μx+s:y+s + μx+s:y+s μx+s:y+s

⎩ 0 ⎫ ⎧ t

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎨ ∗ ∗ μy+s + μx+s − 0.0005 ds⎪ = exp ⎪ − ⎪ ⎪ ⎪ ⎪ ⎭ ⎩

⎫ ⎪ ⎪ ⎪ ⎬ ds⎪ ⎪ ⎪ ⎭

0

= t py t px e−0.0005t , where t py and t px are calculated according to the Standard Ultimate Survival Model. Hence 00 20 p60:55 = 0.78292 × 0.87079 × 0.99005 = 0.67498.

(c) The EPV of the non-simultaneous death benefit is 200 000 Ā03 60:55 = 38 614. (d) Let P be the total premium each year. Then P ā00

60:55:20

04 02 = 200 000 Ā03 60:55 + 400 000 Ā60:55 + 20 000 ā60:55 ,

where ā00

60:55:20

20 00 00 = ā00 60:55 − v 20 p60:55 ā80:75 = 11.7078,

giving P = $6 949.38. (e) The policy values at time 20 are 20V

(0)

04 02 = 200 000 Ā03 80:75 + 400 000 Ā80:75 + 20 000 ā80:75 = 151 510,

20V

(1)

= 200 000 Ā13 80:75 = 121 528,

20V

(2)

22 = 200 000 Ā23 80:75 + 20 000 ā75 = 294 856.

(f) Thiele’s differential equations for t < 20 are d (0) 01 (1) 02 (2) = δ t V (0) + P − μ60+t:55+t − t V (0) − μ60+t:55+t − t V (0) tV tV tV dt 04 400 000 − t V (0) , − μ60+t:55+t d (1) 13 = δ t V (1) − μ60+t:55+t 200 000 − t V (1) , tV dt

177

Solutions for Chapter 10 d (2) 23 = δ t V (2) − 20 000 − μ60+t:55+t 200 000 − t V (2) . tV dt (g) Euler’s backward method gives

(0) (0) 01 (1) V = V − h δ t V (0) + P − μ60+t:55+t − t V (0) t−h t tV 02 (2) − t V (0) − μ60+t:55+t tV

04 − μ60+t:55+t 400 000 − t V (0) , t−hV

(1)

13 = t V (1) − h δ t V (1) − μ60+t:55+t 200 000 − t V (1) ,

t−hV

(2)

23 = t V (2) − h δ t V (2) − 20 000 − μ60+t:55+t 200 000 − t V (2) .

Using h = 0.25 and applying these equations twice starting from the values in part (e), we obtain 19.5V

(0)

= 146 391,

19.5V

(1)

= 119 792,

19.5V

(2)

= 296 663.

(h) If we used Figure 10.3 in AMLCR we could not value the higher benefit payable if deaths occur simultaneously. A function like Ā03 xy for Figure 10.3 can be used to value a unit benefit only when the process enters State 3 (both dead), whereas in this question the amount of benefit payable when both are dead depends on how the deaths occur (simultaneously or sequentially). t

10.15 (a) The discount factor, v m , is the present value of a unit amount payable at time t/m. The term t−1 pxy − mt pxy is the probability that (x) and (y) are both m alive at time (t − 1)/m but are not both alive at time t/m. Hence m

t

vm

t−1 m

pxy − mt pxy

t=1

is the EPV of a payment of 1 at the end of the (1/m)th of a year in which the first death of (x) and (y) occurs, provided this is within one year from now. (b) Using the result from part (a), we can write A(m) xy =

∞

k=0

v k k pxy

m

t=1

t

vm

t−1 m

px+k:y+k − mt px+k:y+k .

178

Solutions for Chapter 10 (c) Since the lives are independent, we can write the joint life probabilities in terms of single life probabilities, as follows: t−1 m

pxy − mt pxy = t−1 px t−1 py − mt px mt py m m = 1 − t−1 qx 1 − t−1 qy − 1 − mt qx 1 − mt qy m m t−1 t t = 1 − t−1 m qx 1 − m qy − 1 − m qx 1 − m qy

1 − 2t 1 = qx qy qx + qy + m m2

m + 1 − 2t 1 = (qx + qy − qx qy ) + qx qy m m2

m − 2t + 1 1 qx qy = (1 − pxy ) + m m2

(using UDD)

as required.

Using this formula, we have m

v

t m

t−1 m

pxy − mt pxy = (1 − pxy )

t=1

t m

vm

t=1

m

+ qx qy

m

t=1

v

t m

m − 2t + 1 . m2

Now t m

vm

t=1

So m

t

vm

t=1

t−1 m

m

= a(m) = 1

1−v iv = (m) . (m) i i

m

t m − 2t + 1 iv pxy − mt pxy = (1 − pxy ) (m) + qx qy vm , i m2 t=1

as required. (d) Compare the two parts of the expression in part (c) for m

t

vm

t−1 m

pxy − mt pxy .

t=1

Since, for most ages, qx and qy are likely to be small, 1 − pxy is likely to be much larger than qx qy . Hence, the first term is likely to be much larger than the second term. Ignoring the second term in the expression for A(m) xy in part (c), we have A(m) xy ≈

∞

iv v k k pxy (1 − px+k:y+k ) (m) i k=0

179

Solutions for Chapter 10 i k+1 = (m) v k pxy (1 − px+k:y+k ) i k=0 ∞

i = (m) Axy . i 10.16 (a) For the joint life process (in Figure 10.2, 10.3 or 10.4 of AMLCR) ⎧ t ⎫

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ 00 0• p = exp μ ds − . ⎪ ⎪ t x:y x+s:y+s ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ 0

Then writing vt = e−δt we have ⎧ t ⎫ ⎫ ⎧ t

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ ⎬ ⎨ −δt 0• 0• e−δt t p00 = e exp μ ds δ + μ ds − = exp − ⎪ ⎪ ⎪ ⎪ x:y x+s:y+s x+s:y+s ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ ⎭ ⎩ 0

0

and so we get

⎧ t ⎫

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ d −δt 00 ⎨ ⎬ 0• 0• e t px:y = − δ + μx+t:y+t exp ⎪ δ + μ ds − ⎪ x+s:y+s ⎪ ⎪ ⎪ ⎪ dt ⎩ ⎭ 0

0• e−δt t p00 = − δ + μx+t:y+t x:y .

(b) Apply formula (B.4) in Appendix B.2 of AMLCR, with f (u) = v u u p00 x:y and n = ∞, so that, using the result in part (a), f (0) = 1,

f (n) = 0,

f (0) = −(δ + μxy ),

f (n) = 0.

This gives ∞ ∞

1 k m − 1 m2 − 1 0• − v m k p00 ≈ v k k p00 − , δ + μx:y x:y x:y m m k=0 2m 12m k=0

i.e. ä(m) xy ≈ äxy −

m − 1 m2 − 1 0• − . δ + μx:y 2m 12m

10.17 First note that for each life t cx (ct −1) , t px = s g

where s = exp{−A} and g = exp −B/log c . (a) Since the lives are independent, we have 10 p30:40 = 10 p30 10 p40 = 0.886962.

180

Solutions for Chapter 10 (b) We can evaluate this probability as follows: 10 10 q30:40 =

t p30:40 μ30+t dt = 0.037257.

1

0

(c) We can evaluate this probability as follows: 10 10 q30:40 =

t p30 (1 − t p40 ) μ30+t dt = 10 q30 −10 q30:40 = 0.001505.

2

1

0

(d) Finally, 10 p30:40 = 10 p30 + 10 p40 − 10 p30:40 = 0.997005.

10.18 The probability that Jones dies before age 50 and before Smith is, in an obvious notation, 20 (J) (J) (S) t p30 μ30+t t p30 dt.

0

Note that (J) c30 (ct −1) , t p30 = g

where g = exp{−B/log c} and (S) c30 (ct −1) exp{−0.039221 t}. t p30 = g

The integral expression for the probability can be evaluated numerically to give 0.567376.

10.19 (a) The EPV is $100 000 ä(12) 65 , using Ryan’s mortality, which is $802 639. (b) Recall the joint and last survivor model, Figure 10.2 in AMLCR. We use that figure to reference the states involved in the annuity payments. The annuity calculated in part (a) corresponds to ä00(12) + ä01(12) , x:y x:y where x denotes Ryan and y denotes Lindsay. The new annuity is found by equating the value of the benefits with the EPV of the single life annuity in part (a), so the equation of value for an annual starting benefit of B per year is 802 693 = B ä00(12) + B ä01(12) + 0.6B ä02(12) . x:y x:y x:y

181

Solutions for Chapter 10

using the approach of Section 8.7 of AMLCR, We can calculate ä02(12) x:y summing the product of the appropriate probability and discount function 02 at each month end. For t p02 x:y we calculate recursively, starting from 0 px:y = 0, using 02 02 22 00 02 t+h pxy = t pxy h py+t + t px:y h px+t:y+t , 1 . Let m denote male mortality, f m denote female married where h = 12 mortality and f w denote female widowed mortality. Then fm 00 m t px:y = t px t py

and

fw 22 h py+t = h py+t .

Now h 02 h px+t:y+t =

fw 00 02 r px+t:y+t μx+t+r:y+t+r h−r py+t+r dr.

0

We can approximate this using the trapezium rule as h 02 fw 02 + h p00 μ hp x+t:y+t μx+t+h:y+t+h .

2 x+t:y+t y+t Alternatively, if we assume Ryan’s death occurs half-way through the interval from t to t + h we obtain the approximation fm fw 02 m h px+t:y+t ≈ h qx+t × h py+t × h py+t+ h .

2

2

2

We find that ä02(12) = 4.0307, so that the revised benefit while both partners x:y survive is $76 846. (c) The equation of value for a starting benefit of B per year is now + 100 000 ä01(12) + 0.6B ä02(12) 802 693 = B ä00(12) x:y x:y x:y which gives a revised starting benefit of $73 942.

10.20 Let P denote the monthly premium. Let superscript f denote the female survival model, and superscript m denote the male survival model. The equation of value for the premium is 0.97 × 12P ä(12)

24:28:25

= 100 000 Ā24:28 + 250,

where k 1 f m k p k p v 12 = 13.3266 28 12 24 12 12 k=0

299

ä(12)

24:28:25

=

182

Solutions for Chapter 10 and

∞ f f m vt t pm 28 t p24 μ28+t + μ24+t dt = 0.24846,

Ā24:28 = 0

from which we calculate that P = $161.78.

10.21 First note that we can evaluate t p00 28:27 as follows: ⎧ t ⎫

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ 00 01 02 03 p = exp + μ + μ μ dt − ⎪ ⎪ t 28:27 28+t:27+t 28+t:27+t 28+t:27+t ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ 0

28 t 27 t = s gc1 (c −1) st gc2 (c −1) exp{−(5 × 10−5 ) t},

t

where s = exp{−A},

g1 = exp −B/log c ,

g2 = exp −D/log c .

The formula for the annual premium, P, is ∞ = 500 000 P ä00 28:27:30

02 03 vt t p00 28:27 μ28+t:27+t + μ28+t:27+t dt, 0

where ä00

28:27:30

=

29

v k k p00 28:27 .

k=0

Evaluating the integral numerically leads to P = $4948.24.

Solutions for Chapter 11

11.1 At the end of 2018 the member’s age is 56.75 and so the expected earnings in 2019 are 75 000

0.75 s58 + 0.25 s57 s57.75 = 75 000 = $76 130. s56.75 0.75 s57 + 0.25 s56

11.2 (a) The member’s expected final average salary is 75 000

s56 + s57 + s58 + s59 = $185 265.

4 s34

(b) Since salaries increase half-way through the calendar year, the member’s salary from age 55 to age 56 is $100 000. The member’s salary from age 60 + t to age 60 + t + 1 for t = 0, 1, 2, 3, 4 is therefore $100 000 s60+t /s55 , and so the expected average salary aged 60.5 to 64.5 is 100 000 4 s55

1 1 2 s60 + s61 + s62 + s63 + 2 s64

= 111 000.

11.3 Let S denote Xiaoxiao’s salary aged 35 to 36. Under the assumed salary scale, her salary aged 64 to 65 is 1.0429 S, or 3.1187S, and her total salary over her career is

1.0430 − 1 S 1 + 1.04 + 1.042 + · · · + 1.0429 = S = 56.0849S. 0.04 Her pension is 2.5% of the total salary over her career, giving a replacement ratio of 0.025 × 56.0849S = 44.96%.

3.1187S 183

184

Solutions for Chapter 11

11.4 The actuarial liability is the EPV of the accrued benefit at the valuation date, which is ⎞ ⎛ 1 3 5 ⎜⎜ v 2 d62 + v 2 d63 + v 2 d64 ⎟⎟⎟ ⎟⎠ = 6 773.31.

20 000 × 30 × ⎜⎜⎝ l62 Similarly, the actuarial liability one year later is ⎞ ⎛ 1 3 ⎜⎜⎜ v 2 d63 + v 2 d64 ⎟⎟⎟ ⎟⎠ = 5 324.98. ⎜ 20 000 × 31 × ⎝ l63 Let Cd denote the normal contribution. Then the actuarial liability now plus the normal contribution provide benefits on mid-year exit and the actuarial liability in one year for survivors. So 1

6 773.31 + Cd = v 2

d62 l63 (30.5 × 20 000) + v (5 324.98), l62 l62

and hence Cd = 186.51.

11.5 The EPV of the benefit is 10 l63 11 l64 v5% B(63, 10) + v5% B(64, 11) l53 l53 10 l63 11 l64 B(63, 0) 1.0410 + v5% B(63, 0) 1.0411 (1.025) = v5% l53 l53 = 8 888.52.

11.6 At the valuation date, Quinn’s final average salary is $102 000, and, as her service is 10 years, her accrued pension is 0.015 × 10 × 102 000 = 15 300.

As there are no exits before age 65, the EPV of the accrued pension, which is the actuarial liability, is 30 ä65 − 11 15 300 v 30 ä(12) 24 = 46 345.

65 = 15 300 v

One year after the valuation date, Quinn’s final average salary is 104 000 (1 + 1.04)/2 = 106 080.

Her service is now 11 years, and so her accrued pension is 0.015 × 11 × 106 080 = 17 503.

Hence, the actuarial liability one year after the valuation date is 17 503 v 29 ä(12) 65 = 55 669.

185

Solutions for Chapter 11

Let Cr denote the normal contribution. As there are no mid-year exits, the normal contribution equation is 46 345 + Cr = 55 669 v ⇒ Cr = $6 674.

11.7 (a) The probability that the employee dies in service before age 60 is (d55 + d56 + · · · + d59 )/l55 = 0.01171. (b) The EPV of the death benefit is 200 000

4

1.06−(t+1/2)

t=0

d55+t = $2011.21. l55

(c) The EPV of the death benefit is now 2 × 85 000

4

1.06−(t+1/2)

t=0

s55+t d55+t = $1776.02. s54.5 l55

11.8 (a) The accumulation to time 35 years of 1% of salary over the first 20 years is 20 1.07t 1.0735−t dt = 400 × 20 × 1.0735 = $85 412.65.

0.01 × 40 000 0

The accumulation to time 35 years of 1% of salary over the last 15 years is 15 0.01 × 40 000

1.0720 1.04t 1.0715−t dt 0

15 = 400 × 1.07

35

1.04 1.07

t dt

0

= 400 × 1.0735

1 − (1.04/1.07)15 log(1.07/1.04)

= $52 148.63.

Hence the total accumulation at age 60 of 1% of salary is $137 561.28.

186

Solutions for Chapter 11 The employee’s projected salary earned between ages 59 and 60 is 1 40 000 × 1.07 × 1.04 20

14

1.04t dt 0

= 40 000 × 1.0720 × 1.0414

0.04 log 1.04

= $273 367.45.

Using a replacement ratio of 70%, the pension in the first year is 0.7 × 273367.45 = $191 357.21.

Hence the EPV (at age 60 and with i = 5%) of the pension benefit is 191 357.21 ä(12) + v 10 10 p60 ä(12) = $2 795 692.86.

70 10 Now, if 1% of salary accumulates to 137 561.28 at age 60, then a contribution rate of c% of salary accumulates to 137 561.28c. Setting c such that the accumulated contributions at age 60 are equal to the EPV of the pension benefit at age 60, we have 137 561.28c = 2 795 962.86 ⇒ c = 20.3 so that the contribution rate is 20.3%. (b) Now the accumulation to time 35 years of contributions of 20.3% of salary is 35 0.203 × 40 000

1.05t 1.0635−t dt 0

= 0.203 × 40 000 × 1.0635

1 − (1.05/1.06)35 log(1.06/1.05)

= $1 861 128.98.

Her projected salary earned between ages 59 and 60 is now 1 40 000 × 1.05

34

1.05t dt 0

= 40 000 × 1.05

34

= $215 344.55.

0.05 log 1.05

187

Solutions for Chapter 11

The expression for the EPV of the pension benefits is unchanged from part (a), but it is calculated with i = 4.5%. The EPV of a pension of $X per year is 15.4068X. Setting this EPV equal to the accumulation of contributions, we find that X = 120 798.93. Dividing X by the salary earned between ages 59 and 60 gives the replacement ratio as 56.1%.

11.9 (a) Let S be the salary aged 30 to 31. Then the total salary up to age 65 is S 1 + 1.05 + · · · + 1.0515 + 1.0515 1.04 + 1.042 + · · · + 1.0419

1.0516 − 1 1.0420 − 1.04 = S + 1.0515 0.05 0.04 = 83.485S. The average salary in the final three years is S 1.0515 1.0417 + 1.0418 + 1.0419 3 and the salary in the final year is 4.3800S.

= 4.2137S,

Under the final salary plan, the pension is 0.012 × 35 × 4.2137S = 1.7697S, and hence the replacement ratio is 1.7697S = 0.4041.

4.3800S Under the CAE plan, with accrual rate α, the pension is 83.485 α S and the replacement ratio is still 0.4041, so 83.485 α S = 1.7697S ⇒ α = 2.12%. (b) The revalued earnings for the first 16 years total S 1.022534 + 1.05 1.022533 + 1.052 1.022532 + · · · + 1.01515 1.022519

1.052 1.05 1.0515 + = 1.022534 S 1 + + · · · + 1.0225 1.02252 1.022515

(1 + j)16 − 1 1.05 − 1, = 1.022534 S where j = j 1.0225 = 41.914S, and the revalued earning for the last 19 years total 1.0515 S 1.04 1.022518 + 1.042 1.022517 + · · · + 1.0419

1.042 1.0518 1.04 15 18 + + ··· + = 1.05 × S × 1.04 × 1.025 1 + 1.0225 1.02252 1.022518

188

Solutions for Chapter 11

(1 + k)19 − 1 = 1.0515 × S × 1.04 × 1.02518 k

where k =

1.04 − 1, 1.0225

= 71.742S, so that the total revalued earnings are 113.656S. Proceeding as in part (a), we set 113.656 α S = 1.7697S, giving an accrual rate of α = 1.56%.

11.10 At age 61, the EPV of the death benefit is 5000 × 35

3

1

vt+ 2

t=0

s61+t d61+t = $2351.48. s60 l61

If the member survives to age 62, the EPV of the death benefit at that age is 5000 × 36

2

1

vt+ 2

t=0

s62+t d62+t = $2094.46. s60 l62

The EPV at age 61 of benefits payable due to a mid-year death is 1

5000 × 35.5 v 2

s61 d61 = $630.47. s60 l61

Hence the normal contribution, C, for the death in service benefit is such that 2351.48 + C = v

l62 × 2094.46 + 630.47, l61

giving C = $58.31.

11.11 (a) The annual rate of pension from age 65 is B, say, where B = 0.016 × 40 × 50 000 × 1.0539 = $214 552. (b) As we have a new member, the value at time 0 (i.e. the date of entry) of the accrued benefits is 0. If the member survives to age 26, the value at that time of the accrued benefits is 0.016 × SFin × v 39 39 p26 ä(12) 65 , where SFin = 50 000 × 1.0539 , so the value at time 0 is −40 0.016 × SFin × v 40 40 p25 ä(12) × 0.8 × 11 = 4589.03.

65 = 5363.80 × 1.06

As there are no exits other than by death (and there is no death benefit), the funding equation gives the contribution as this amount, which is 9.18% of salary.

Solutions for Chapter 11

189

(c) Assuming contributions are made mid-year, the accumulation of contributions to the retirement date is

39

0.12 × 50 000 1.0840 − 1.0540 1.05t 1.0839.5−t = 0.12 × 50 000 1 − (1.05/1.08) 1.080.5 t=0 = $3 052 123.

At the retirement date, the EPV of a pension of B per year is 11 B, and setting this equal to the accumulated amount of contributions gives B = $277 466. (d) The defined benefit plan offers a pension that is known in the sense that it is expressed in terms of her final salary. The amount of the pension does not depend on the financial performance of the underlying assets. If, for example, the underlying assets provide a lower level of accumulation than expected, the replacement ratio would be reduced under a defined contribution plan, but not under a defined benefit plan. (e) From the employer’s point of view, the defined contribution plan leads to a known level of contributions, which is desirable from the point of view of setting a company’s budget. By contrast, with a defined benefit plan, the employer’s contribution rate could vary. For example, if the underlying assets do not perform strongly, the employer may have to increase its contributions to the fund to ensure that benefits can be paid.

11.12 Let Bx denote the pension accrued up to age x. The pension is payable from age 65 without actuarial reduction, or at age x with the reduction factor applied. For retirement at age 55 we set (12) 10 (1 − 10k) B55 ä(12) 55 = v B55 ä65

since the reduced pension is payable 10 years early, and we are assuming no exit by any decrement before age 65. This identity yields k = 4.85%.

Similarly, for retirement at age 60 we set (12) 5 (1 − 5k) B60 ä(12) 60 = v B60 ä65 ,

which gives k = 5.80%.

11.13 (a) On retirement at age 60.5, the pension is 0.015 × 30.5 × FAS60.5 × (1 − 18 × 0.005) = $58 615.08,

190

Solutions for Chapter 11 where FAS60.5 = 100 000 z60.5 /s44 and zx = (sx−1 + sx−2 )/2. Her salary in the year prior to retirement is s59.5 = $141 837.57 100 000 s44 and so the replacement ratio is 41.3%.

On retirement at age 62 exact, the pension is 0.015 × 32 × FAS62 = $69 105.82, and her salary in the year prior to retirement is s61 = $145 036.92, 100 000 s44 and so the replacement ratio is 47.6%.

On retirement at age 65 exact, the pension is 0.015 × 35 × FAS65 = $79 040.71, where FAS65 = 100 000 z65 /s44 . Her salary in the year prior to retirement is s64 = $151 681.71 100 000 s44 and so the replacement ratio is 52.1%. (b) For retirement at age x, the EPV of the pension is , v x−45 x−45 p45 Bx ä(12) x where Bx is the pension calculated in part (a). With i = 0.05, values are as follows: x = 60.5: EPV = 383 700, x = 62: EPV = 406 686, x = 65: EPV = 372 321. (c) The EPV of the withdrawal benefit is 0.015 × 15 × 93 000 v 17 17 p45 ä(12) 62 = $123 143.

11.14 There is no actuarial liability for the active members aged 25 since they have no past service. For the active members aged 35 the actuarial liability is 3 × 10 × 300 v 25 25 p35 ār60 = 26 565.77,

Solutions for Chapter 11

191

where ār60 = ā5 + v 5 5 p60 ā65 = 13.0517.

For the active member aged 45 the actuarial liability is 15 × 300 v 15 15 p45 ār60 = 23 913.20; for the active member aged 55 the actuarial liability is 25 × 300 v 5 5 p55 ār60 = 72 241.24; for the deferred pensioner the actuarial liability is 7 × 300 v 25 25 p35 ār60 = 6198.68; and for the pensioner aged 75 the actuarial liability is 25 × 300 ā75 = 68 771.71.

Summing these, the total actuarial liability is $197 691.

To calculate the normal contribution for the three members aged 25, we have 0V = 0 and l26 25 l60 r ā = $1478.75. v 1V = 3 × 300 v l25 l25 60 This is the normal contribution for these three members.

For the three members aged 35, we have 0V = 26 565.77 and if they all survive to age 36, 1V = 3 × 11 × 300 v

24

r 24 p36 ā60

so that v

l36 11 1V = 0V, l35 10

and hence the funding equation for the contributions from these three members, C35 say, is 11 26 565.77, 26 565.77 + C35 = 10 so that C35 = 29 565.77/10 = $2656.58. The same argument gives the contributions for the members aged 45 and 55 as C45 =

1 23 913.20 = $1594.21 15

192

Solutions for Chapter 11 and 1 72 241.24 = $2889.65.

25 Hence the total of the normal contributions is $8619.

C55 =

11.15 (a) (i) For Giles, the final salary is FSG = 40 000 s64 /s35 = $101 245.72, and the actuarial liability (assuming no exit before age 65) is 0VG = v

30

× 0.02 × 5 × FSG × ä(12) 65 = $44 990.14.

For Faith, the final salary is FSF = 50 000 s64 /s60 = $53 071.18, and the actuarial liability is 0VF = v

5

× 0.02 × 30 × FSF × ä(12) 65 = $377 210.84.

Hence the total actuarial liability is $422 201. (ii) The funding equation for Giles’ contribution (CG ) is 0VG + CG = v

30

× 0.02 × 6 × FSG × ä(12) 65

(since we are ignoring pre-retirement mortality and other modes of exit) so that CG = 0VG /5 = $8998.03, which is 22.5% of salary.

The funding equation for Faith’s contribution (CF ) is 0VF + CF = v

5

× 0.02 × 31 × FSF × ä(12) 65

so that CF = 0VF /30 = $12 573.69, which is 25.1% of salary. (b) (i) For Giles, the final salary is now $38 000 and the actuarial liability is 0VG = v

30

× 0.02 × 5 × 38 000 × ä(12) 65 = $16 885.90.

For Faith, the final salary is now $47 000 and the actuarial liability is 0VF = v

5

× 0.02 × 30 × 47 000 × ä(12) 65 = $334 059.06.

Hence the total actuarial liability is $350 945. (ii) For Giles’ contribution, the funding equation is 0VG + CG = v

30

× 0.02 × 6 × 40 000 × ä(12) 65 =

24 0VG 19

5 so that CG = 19 0VG = $4443.66, which is 11.1% of salary.

For Faith’s contribution, the funding equation is 0VF + CF = v

5

× 0.02 × 31 × 50 000 × ä(12) 65 =

155 0VF 141

Solutions for Chapter 11

193

14 so that CF = 141 0VF = $33 168.98, which is 66.3% of salary.

(c) Under the PUC method, the projected impact of future salary increases on the accrued benefits is included in the actuarial liability. The normal contribution pays for the increase in benefit arising solely from the impact of additional service. Under the TUC method, future salary increases are not included in the actuarial liability, and the normal contribution must fund both the impact of the additional service and the impact of salary increases on the whole accrued benefit. Because the PUC method pre-funds the projected salary increases, the actuarial liability is always higher than for the TUC method, but the values must converge at the retirement age. We see that, in this case, the actuarial liability for Giles is much lower under the TUC method than under the PUC method, because he is a long way from retirement. Faith is close to retirement, so the difference is smaller.

To build up the higher early actuarial liability, the contributions under the PUC method start out higher than under the TUC method, as the benefit funded under the PUC method is based on the higher, projected final salary. Later, the contribution under the TUC method becomes higher; the TUC contributions pay for the additional year of accrued benefit each year, and also pay for the entire past accrued benefit to be adjusted for the projected one-year salary increase. The cost of increasing the accrued benefits in line with salaries becomes very high as the employee moves closer to retirement, as the accrued benefit becomes larger. In this case, Giles’ contribution under the TUC method is still substantially below that under the PUC method, as his accrued benefit is still fairly small. Faith’s accrued benefit is large, and the cost of paying for the salary increase in the TUC contribution is very significant.

11.16 (a) For retirement at age 65, the EPV is B(65, 0) + v p65 B(66, 1) + v 2 2 p65 B(67, 2) + v 3 3 p65 B(68, 3) + · · · , where for t = 1, 2, 3, . . .

B(65 + t, t) = (1.025 × 1.04)t B(65, 0) and so vt B(65 + t, t) = vit∗ B(65, 0), where 1 + i∗ = (1 + i)/(1.025 × 1.04). So the EPV is B(65, 0) + v p65 B(66, 1) + v 2 2 p65 B(67, 2) + v 3 3 p65 B(68, 3) + · · ·

194

Solutions for Chapter 11 = B(65, 0) 1 + vi∗ p65 + vi∗2 2 p65 + vi∗3 3 p65 + · · · = B(65, 0) (1 + vi∗ p65 aB66 ) = B(65, 0) aB65 , and so aB65 = 1 + vi∗ p65 aB66 = 28.4679.

For retirement at age 64.5, it is useful to first find the EPV for retirement at age 65.5, as follows: B(65.5, 0) + v p65.5 B(66.5, 1) + v 2 2 p65.5 B(67.5, 2) + v 3 3 p65.5 B(68.5, 3) + · · · = B(65.5, 0) 1 + vi∗ p65.5 + vi∗2 2 p65.5 + vi∗3 3 p65.5 + · · · = B(65.5, 0) aB65.5 .

Then for retirement at age 64.5, we use the fact that B(65.5, 0) = B(64.5, 0) and so for k = 1, 2, . . ., B(65.5+k, 0) = 1.025k B(65.5, 0) = 1.025k B(64.5, 0).

Then the EPV for retirement at age 64.5 is B(64.5, 0) + v p64.5 B(65.5, 1) + v 2 2 p64.5 B(66.5, 2) + v 3 3 p64.5 B(67.5, 3) + · · · = B(64.5, 0) + 1.04v p64.5 B(65.5, 0) + (1.04v)2 2 p64.5 B(66.5, 0) + (1.04v)3 3 p64.5 B(67.5, 0) + · · · = B(64.5, 0) 1 + 1.04v p64.5 + 1.025(1.04v)2 2 p64.5 +1.0252 (1.04v)3 3 p64.5 + · · ·

= B(64.5, 0) 1 + 1.04v p64.5 1 + vi∗ p65.5 + vi∗2 2 p65.5 + vi∗3 3 p65.5 + · · · = B(64.5, 0) 1 + 1.04v p64.5 aB65.5 = B(64.5, 0) aB64.5 , so that aB64.5 = 1 + 1.04v p64.5 aB65.5 = 28.4193.

For retirement at age 63.5, the EPV is B(63.5, 0) + v p63.5 B(64.5, 1) + v 2 2 p63.5 B(65.5, 2) + v 3 3 p63.5 B(66.5, 3) + · · ·

Solutions for Chapter 11

195

= B(63.5, 0) + 1.04v p63.5 B(64.5, 0) + (1.04v)2 2 p63.5 B(65.5, 0) + (1.04v)3 3 p63.5 B(66.5, 0) + · · ·

= B(63.5, 0) + 1.04v p63.5 B(64.5, 0) + 1.04v p64.5 B(65.5, 0)

+ (1.04v)2 2 p64.5 B(66.5, 0) + · · · = B(63.5, 0) + 1.04v p63.5 B(64.5, 0) aB64.5 = B(63.5, 0) 1 + 1.04v p63.5 aB64.5 = B(63.5, 0) aB63.5 so that aB63.5 = 1 + 1.04v p63.5 aB64.5 = 29.0078. (b) The actuarial value of the benefits is 1500 (1.04v)1/2 r63 aB63.5 + (1.04v)3/2 r64 aB64.5 + (1.04v)2 r65− aB65 l63 = 41 584.

11.17 With i = 0.05, the accrued liability at time 0 is

r60 (12) r64 (12) 25 r60− (12) ä + v 25.5 ä + · · · + v 29.5 ä 0V = 5 × 350 v l35 60 l35 60.5 l35 64.5

r65− (12) + v 30 ä l35 65 = 2628.39, and this is the actuarial liability.

The accrued liability at time 1 if the member is alive then is

r60− (12) r60 (12) r64 (12) V = 6 × 350 v 24 ä + v 24.5 ä + · · · + v 28.5 ä 1 l36 60 l36 60.5 l36 64.5

r65− (12) + v 29 ä65 l36 so that v

l36 6 1V = 0V, l35 5

and as there is no mid-year retirement at age 35, the funding equation for the contribution C is 6 l36 0V + C = v 0V, 1V = l35 5

196

Solutions for Chapter 11 giving C=

1 0V = $525.68.

5

11.18 The accrued pension benefit at time 0 is 0.025 × 5 ×

175 000 = $4375.

5

With i = 0.05 this has EPV

r60− (12) r60 (12) r64 (12) V = 4375 v 25 ä + v 25.5 ä + · · · + v 29.5 ä 0 l35 60 l35 60.5 l35 64.5

r65− (12) + v 30 ä65 l35 = 6570.98, and this is the actuarial liability.

The accrued liability at time 1 if the member is alive then is based on the projected career average salary at age 36, (175 000 + 40 000)/6 = 35 833.33, giving a projected accrued pension at the year end of $5375. The expected present value at time 1 of this pension benefit is

24 r60− (12) 24.5 r60 (12) 28.5 r64 (12) 29 r65− (12) ä + v ä + ··· + v ä +v ä , 1V = 5375 v l36 60 l36 60.5 l36 64.5 l36 65 so that v

l36 5375 1V = 0V, l35 4375

and as there is no mid-year retirement at age 35, the funding equation for the contribution C is l36 5375 1V = 0V + C = v 0V, l35 4375 giving C = $1501.94.

11.19 (a) The accrued pension is 0.02 × 2 500 500 = 50 010 and so the actuarial liability is

1 r64 r65− (12) 50 010 v 2 ä(12) + v ä = 621 718. l64 64.5 l64 65 (b) If retirement takes place at age 64.5, David’s pension will be $51 810, and

Solutions for Chapter 11

197

if retirement takes place at age 65, his pension will be $53 610. So, if Cr denotes the normal contribution, we have 1

621 718 + Cr = 51 810 v 2

r64 (12) r65− (12) ä64.5 + 53 610 v ä , l64 l64 65

which gives Cr = $42 550. (c) There is no actuarial liability at the valuation date for a new member. If Rui is alive at age 31, the accrued pension will be 0.02 × 120 000 = $2 400, and the actuarial liability at that time will be

r60− (12) r60 (12) r61 (12) 2 400 v 29 ä60 + v 29.5 ä60.5 + v 30.5 ä l31 l31 l31 61.5 r64 (12) r65− (12) + · · · + v 33.5 ä64.5 + v 34 ä l31 l31 65 = 1 977.

Retirement cannot take place mid-year, so the normal contribution, Cr is Cr = 1 977 v

l31 = $1 702. l30

(d) We need to adjust the pension benefit to allow for revaluation of earnings.

For example, if retirement takes place at age 61.5, the accrued pension is 2% of the revalued earnings, that is 0.02 × 120 000 × 1.02530.5 = 2 400 × 1.02530.5 .

So the actuarial liability at time 1 if Rui is alive then is now

r60− (12) r60 (12) r61 (12) ä60 + (1.025v)29.5 ä60.5 + (1.025v)30.5 ä 2 400 (1.025v)29 l31 l31 l31 61.5

r64 (12) r65− (12) + · · · + (1.025v)33.5 ä64.5 + (1.025v)34 ä65 l31 l31 = 4 298, and so Cr = 4 298 v l31 /l30 = $3 699.

11.20 Let S denote the member’s salary from age 35 to age 36. Then the EPV of the member’s contributions is ⎛ ⎞ 29 29

⎜⎜⎜ l35+t+ 21 l35+t+ 12 ⎟⎟⎟ 1 s35+t 1 s35+t t+ t+ ⎜ ⎟⎟⎟ = 0.7557S. S ⎜⎜⎝0.04 v 2 + 0.01 v 2 ⎠ s l s l 35 35 35 35 t=0 t=15

198

Solutions for Chapter 11 Define zy = (sy−3 + sy−2 + sy−1 )/3. Then the EPV of the retirement benefit is S v 25

29

z35+t+ 12 r35+t t+ 12 (12) 1 z60 r60− 25 (12) ä60 + S ä vt+ 2 s35 l35 60 s35 l35 60 35+t+ 12 t=25

+ S v 30

z65 r65− 30 (12) ä s35 l35 60 65

= 2.3807S. If the employer contributes a multiple m of the member’s contribution, then 0.7557S (1 + m) = 2.3807S ⇒ m = 2.15.

Solutions for Chapter 12

12.1 (a) The spot rate for time t is calculated as

−1/t P(t) yt = − 1.

100 Thus, for example,

y3 =

84.45 100

−1/3 − 1 = 0.05795.

The other values are y1 = 0.05988, y2 = 0.05881, y4 = 0.05754 and y5 = 0.05701. (b) The one-year forward rate at time 0 is just y1 . For all other values of t, the one-year forward rate is calculated as P(t) − 1.

P(t + 1) Thus, for example, the one-year forward rate for t = 4 is 79.95 − 1 = 0.05489.

75.79 Similar calculations give the one-year forward rates for t = 1, 2 and 3 as 0.05774, 0.05625 and 0.05629. (c) The EPV of the five-year term annuity-due is ⎞ ⎛ 4

⎟⎟⎟ ⎜⎜⎜ P(t) t ⎜ 0.99 ⎟⎟⎟⎠ = $4395.73.

1000 ⎜⎜⎝1 + 100 t=1 199

200

Solutions for Chapter 12

12.2 Let v(t) be the discount factor for time t, t = 1, 2, 3, . . .. Values can be calculated recursively from v(t) =

v(t − 1) , 1 + f (t − 1, t)

with initial value v(0) = 1.

The EPV of the annuity is then 10 000

∞

v(t) t p60 = 155 413.

t=0

12.3 (a) Let Z denote the present value of the benefit payment under a randomly selected policy. Then, treating the sum insured as a random variable, S, which is equally likely to be 10 000 or 100 000, we have E[Z | S] = S Ā 1

75:5

and E[Z] = E[E[Z | S]] = 12 × 10 000 Ā 1

+ 12 × 100 000 Ā 1

75:5

= 12

10 + 10 4

5

Ā

75:5

1

75:5

= $5286.49 since Ā 1

75:5

Similarly,

= 0.096118. $ % E Z 2 = 12 108 + 1010

2

Ā 1

75:5

,

where the superscript 2 indicates that the term insurance function is evaluated using an effective interest rate of 1.062 − 1 = 0.1236 per year. As 2 1 Ā = 0.083041, we have E[Z 2 ] = 419 354 543 and hence the standard 75:5 deviation of Z is $19 784. (b) Consider a policy with sum insured S. The variance of the present value of the benefit is

2 2 2 1 1 Ā S − Ā .

75:5

75:5

There are 50 policies for which S = 10 000 and 50 for which S = 100 000.

Solutions for Chapter 12

201

By the independence of the policyholders, the variance of the present value of benefits from the portfolio is

2 = 37 269 931 272, 50 108 + 1010 2 Ā 1 − Ā 1 75:5

75:5

and hence the standard deviation is $193 054. (c) Now let Xi denote the present value of the benefit payment under the ith policy in a portfolio of size N in which half the policies have sum insured 10 000 and the other half have sum insured 100 000. For the case N = 100, 0 $1 % V Ni=1 Xi 193 054 = = 1930.54, N 100 while in the case N = 100 000, 0 $1 % √ V Ni=1 Xi 193 054 1 000 × 193 054 = = = 61.05. √ N 100 000 100 1 000 This result follows since the variance in the case N = 100 000 is 1000 times the variance in the case N = 100. For the risk to be fully diversifiable, we require 0 $1 % V Ni=1 Xi = 0, lim N→∞ N and we can see that this will be the case. If we set N = 100n, 0 $1 % V Ni=1 Xi 193 054 = √ N 100 n and this quantity goes to 0 as n → ∞ (and hence as N → ∞).

12.4 (a) For a portfolio of N insurance policies the coefficient of variation (CVN ) is 0 $1 % V Ni=1 Xi $1 % .

E Ni=1 Xi For a diversifiable risk we know that 0 $1 % V Ni=1 Xi = 0. lim N→∞ N

202

Solutions for Chapter 12 $1 % As E[Xi ] = μ for i = 1, 2, . . . , N, we have E Ni=1 Xi = Nμ, where μ > 0, so that 0 $1 0 $1 % % N V Ni=1 Xi 1 V i=1 Xi = .

CVN = Nμ μ N Taking limits of both sides, we have lim CVN =

N→∞

(b)

0 $1 % V Ni=1 Xi

1 lim μ N→∞

N

= 0.

(i) We treat the parameter c of Makeham’s law as a random variable, which we denote c. Then Pr[c = 1.124] = 0.75 = 1 − Pr[c = 1.114].

Let Z denote the present value of the benefit from an individual policy.

Then E[Z] = E[E[Z | c]] = 0.75 × 100 000 A 1

65:15

+ 0.25 × 100 000 A∗ 1

65:15

,

where the unstarred function is calculated using the Standard Ultimate Survival Model and the starred function is calculated using Makeham’s law with c = 1.114. We find that A1

65:15

= 0.11642

and

A∗ 1

65:15

= 0.06410,

giving E[Z] = 105 × 0.10334.

Next, V[Z] = E[V[Z | c]] + V[E[Z | c]].

Now

V[Z | c = 1.124] = 100 0002 2 A 1

65:15

= 10

10

− A1

2

65:15

0.07202 − 0.116422

= 1010 × 0.05846, where the superscript 2 indicates calculation using an effective interest rate of 1.062 − 1 = 0.1236 per year. Similarly, V[Z | c = 1.114] = 1010 × 0.03570,

203

Solutions for Chapter 12 giving E[V[Z | c]] = 0.75 × 1010 × 0.05846 + 0.25 × 1010 × 0.03570 = 1010 × 0.05277.

Finally,

V[E[Z | c]] = 0.75 × 105 A 1

65:15

2

+ 0.25 × 105 A∗ 1

2

65:15

− E[Z 2 ]

= 1010 × 0.00051, giving V[Z] = 1010 × 0.05328. Thus, the coefficient of variation of Z is √ √ V[Z] 0.05328 = = 2.2337.

E[Z] 0.10334 (ii) Let Zn denote the total present value of benefits from a portfolio of n policies, and let Z be as in part (i) above, so that Z ≡ Z1 . Then E[Zn ] = n E[Z] = 105 × 0.10334 n and V[Zn ] = E[V[Zn | c]] + V[E[Zn | c]].

Now E[V[Zn | c]] = n E[V[Z | c]] = 1010 × 0.05277n and V[E[Zn | c]] = V[n E[Z | c]] = 1010 × 0.00051n2 , giving V[Zn ] = 1010 0.05277n + 0.00051n2 .

Hence for n = 10 000 we have E[Zn ] = 109 × 0.10334 and V[Zn ] = 1014 × 5.18584, so that the coefficient of variation in this case is 0.2204. (iii) For a general number of policies, n, we have √ 0

V[Zn ] lim = lim 1010 0.05277 n−1 + 0.00051 n→∞ n→∞ n √ = 105 × 0.00051 > 0,

204

Solutions for Chapter 12 which means that the mortality risk is not fully diversifiable. The limiting value of the coefficient of variation is ! √

1010 0.05277n + 0.00051n2 V[Zn ] = lim lim n→∞ E[Zn ] n→∞ 105 × 0.10334 n √ 105 × 0.00051 = 105 × 0.10334 = 0.2192.

12.5 (a) Let P denote the net annual premium. Then P ā40:25 = 100 000 Ā40:25 at 7%, giving P = $1608.13. (b) The mean rate of interest is 0.5 × 5% + 0.25 × 7% + 0.25 × 11% = 7%.

As this is the interest rate used to calculate P in part (a), the EPV of the net future loss is 0. (c) The modal interest rate is 5% and the EPV of the net future loss is 100 000 Ā40:25 − 1 608.13 ā40:25 at 5%, giving the EPV as $7325.40. (d) The present value (PV ) of the net future loss given i = i is 1 − v min(T40 ,25) δ

P min(T40 ,25) P = 100 000 + − . v δ δ

PV = 100 000 v min(T40 ,25) − P

Thus,

and

P $ min(T40 ,25) % P E[PV | i = i] = 100 000 + − E v δ δ

P P Ā40:25 − , = 100 000 + δ δ

P 2 $ min(T40 ,25) % V v V[PV | i = i] = 100 000 + δ

2 P 2 2 = 100 000 + Ā40:25 − Ā40:25 , δ

Solutions for Chapter 12

205

where 2 A indicates calculation using an effective interest rate of (1 + i)2 − 1 per year.

Then E[PV ] = 0.5 E[PV | i = 0.05] + 0.25 E[PV | i = 0.07] + 0.25 E[PV | i = 0.11] = $2129.80.

Note from part (b) that E[PV | i = 0.07] = 0.

Next,

$ % $ % V[PV ] = E V[PV | i + V E[PV | i] ,

where $ % E V[PV | i ] = 0.5 V[PV | i = 0.05] + 0.25 V[PV | i = 0.07] + 0.25 V[PV | i = 0.11] = 40 371 859, and $ % V E[PV | i ] = 0.5 E[PV | i = 0.05]2 + 0.25 E[PV | i = 0.11]2 − E[PV ]2 = 31 693 915 (using the fact that E[PV | i = 0.07] = 0), which leads to V[PV ] = 72 065 775, and hence the standard deviation is $8489.16. (e) It is common to use a deterministic approach to premium calculation. Using the expected interest rate, 7%, and using the equivalence principle to calculate the premium, generates a high risk (50%) of significant losses ($7325.40) from interest rate risk alone.

This risk is not compensated by the upside potential, allowing for the possibility of profits if i = 11%. Using the mean rate of interest in the deterministic premium calculation generates an expected loss of $2129.80 on each policy.

The significant uncertainty about the profitability of the contract is demonstrated by the variance of the present value of the net future loss. We may

206

Solutions for Chapter 12 compare the overall standard deviation, which allows for mortality and interest rate risk, with the standard deviation that would apply if the interest rate were fixed at, say, 7%. The overall standard deviation of present value of losses is $8489.16 from part (d) above. If the interest rate of 7% could be locked in, leaving only mortality risk, the standard deviation would be $6468.27.

12.6 (a) Intuitively, pandemic risk is a non-diversifiable risk, since all lives in a portfolio will be affected if a single life is affected. To illustrate that it is not a diversifiable risk, consider a portfolio consisting of n one-year term insurance policies, each with sum insured S, issued to independent lives who are the same age. Let q denote the normal mortality rate, and let q̂ = 1.25q. Let X denote the total amount of claims in the portfolio in the next year. Let Q be a random variable denoting the mortality rate with Pr[Q = q] = 0.99 = 1 − Pr[Q = q̂].

Then E[X] = E[E[X | Q]] = 0.99 S n q + 0.01 S n q̂, and V[X] = E[V[X | Q]] + V[E[X | Q]], where E[V[X | Q]] = 0.99 S2 n q (1 − q) + 0.01 S2 n q̂ (1 − q̂) and V[E[X | Q]] = 0.99 S2 n2 q2 + 0.01 S2 n2 q̂2 − (0.99 S n q + 0.01 S n q̂)2 .

The risk is fully diversifiable if lim

n→∞

√

V[X] = 0, n

and as V[X] = 0.99 S2 n−1 q (1 − q) + 0.01 S2 n−1 q̂ (1 − q̂) n2 + 0.99 S2 q2 + 0.01 S2 q̂2 − (0.99 S q + 0.01 S q̂)2 , we see that √ 0 V[X] = 0.99 S2 q2 + 0.01 S2 q̂2 − (0.99 S q + 0.01 S q̂)2 > 0, lim n→∞ n meaning that the risk is not fully diversifiable.

207

Solutions for Chapter 12

(b) The actuary would be able to calculate the distribution of the present value of future loss from the portfolio assuming no pandemic risk. To assess the impact of pandemic risk, the most suitable approach would be to simulate the losses assuming an appropriate model for the incidence of pandemic risk for each year that the portfolio is in force. The simulations could be used to re-evaluate the moments of the loss distribution, to quantify the impact of pandemic risk on this portfolio.

In addition to considering the impact of pandemic risk on the mean and variance of the future losses, the actuary might try stress testing the portfolio, to ensure that if a pandemic arose at any point in the near future, the insurer would have sufficient capital available to meet the extra costs.

Stress testing involves assuming a deterministic path which is adverse, to assess the most important vulnerabilities of a portfolio.

12.7 (a) Let P denote the net premium, and let v(t) = (1 + yt )−t for t = 1, 2, 3, . . ., with v(0) = 1. Then the equation of value is ⎞ ⎛ 14 14

⎜⎜⎜ ⎟⎟⎟ P v(t) t p45 = 100 000 ⎜⎜⎝⎜ v(t + 1) t |q45 + v(15) 15 p45 ⎟⎟⎠⎟ , t=0

t=0

which gives P = $4207.77. (b) Now let P̂ denote the net premium. Then the equation of value is, using standard actuarial notation, P̂ ä45:15 = 100 000 A45:15 at an effective interest rate of i = y15 per year, giving P̂ = $4319.50. The change in the interest rate basis has resulted in a 2.7% increase in the premium. (c) The policy value is ⎞ ⎛ 11 11

⎟⎟⎟ ⎜⎜⎜ v(t + 4) v(t + 3) v(15) ⎟⎟ − P ⎜ ⎜⎝ 3V = 100 000 ⎜ t |q48 + 12 p48 ⎟ t p48 . ⎠ v(3) v(3) v(3) t=0 t=0 Thus, 3V = $13 548.

12.8 (a) We have Pr[T50 ≤ t] = 1 − t p50 = 1 − gc (c −1) , x

t

where g = exp{−B/log c}. Using the inverse transform method, if u is a

208

Solutions for Chapter 12 random drawing from the U(0, 1) distribution, we find the simulated value of future lifetime, t, by setting u = 1 − gc (c −1) . x

t

Algebraic manipulation gives

log (1 − u) t = log 1 + 50 ÷ log c c log g

log (1 − u) ÷ log c. = log 1 − B c50 /log c If t > 15, no death benefit is payable and so the simulated value of the future loss is −550 ā15 .

If t ≤ 15, the simulated value of the future loss is 200 000 vt − 550 āt .

For example, with u = 0.013, the simulated value of future lifetime is t = 7.513 and the simulated value of future loss is $135 162. (b) The answers will depend on the random uniform variates used. (c) Let x̄ and s2 denote the sample mean and variance from 1000 simulations of the future loss. If the true mean and variance are denoted by μ and σ 2 , then the sample mean is approximately normally distributed with mean μ and variance σ 2 /1000. We approximate σ 2 by s2 , to get an estimated 90% confidence interval estimate for μ of s x̄ ± 1.645 √ .

1000 For example, if for one set of 1000 simulated values for the loss, we have x̄ = −795 and s = 25 267, then the 90% confidence interval for μ would be 25 267 −795 ± 1.645 √ = (−2109, 519).

1000 (d) The true value of the mean future loss is 200 000 Ā 1

50:15

− 550 ā50:15 = −$184.07.

This value does lie in the confidence interval calculated in part (c). (e) The answer here will depend on your simulations. See part (f) below.

209

Solutions for Chapter 12

(f) By the nature of a 90% confidence interval, we would expect that 10% of the sets of simulations would produce a confidence interval that does not contain the (true) mean future loss. Thus, the answer to part (e) above is expected to be 2. (g) Under this interest rate model, E[I] = 0.05, which is the deterministic rate used in the calculations in part (a). To simulate values of the interest rate, we can simulate from the N(0.0485, 0.02412 ) distribution in Excel.

Suppose a simulated value is y. Then the simulated value of 1 + I is ey . For each simulated value of I the calculation of the simulated value of future loss proceeds exactly as in part (a), using the same future lifetimes as in part (a). So, for the jth simulation, j = 1, 2, . . . , 1000, the procedure is as follows.

1. Generate a simulated future lifetime, as in part (a), t j , say.

2. Generate a simulated N(0.0485, 0.02412 ) variate, y j .

3. Set the interest rate for the simulation as i j = exp{y j } − 1.

4. Calculate the present value of the future loss at this simulated rate of interest as ⎧ ⎪ ⎪ if t j ≤ 15, ⎨ 200 000 (1 + i j )−t j − 550 āt j Lj = ⎪ ⎪ ⎩ −550 ā if t > 15.

15

j

Note that allowing for interest rate uncertainty, in addition to mortality uncertainty, does not increase the standard deviation of the future loss by very much. This is because term insurance is not very sensitive to the interest rate assumption. However, in this question there is no uncertainty about the parameters of the survival model, so that the mortality risk is diversifiable whereas the interest rate risk is non-diversifiable. Hence, for a large portfolio of identical policies the interest rate risk will be relatively more significant.

Solutions for Chapter 13

13.1 The surplus emerging at the end of the tenth policy year, per policy in force at the start of that year, is the accumulation of the reserve at the start of the year plus the premium (P) minus expenses, less the expected cost of the death benefit and expenses, less the expected cost of setting up the year-end reserve. Thus, the emerging surplus is 1.06 (9V + 0.985 P) − 150 060 q49 − 10V p49 = 15.36, since q49 = 0.0019, P = $270 and 9V = 10V = 300.

13.2 (a) Following the same logic as in the previous solution, the surplus emerging at the end of the tenth policy year, per policy in force at the start of that year, is 1.05 (9V + 0.975 P) − 250 040 q54 − 10V p54 = 773.86, where P = $8400 is the annual premium, and q54 = 0.0024. (b) As there is certain to be a claim (on death or maturity), the surplus emerging at the end of the twentieth policy year, per policy in force at the start of that year, is 1.05 (19V + 0.975 P) − 250 040 = 2172.10.

13.3 (a) Let P denote the annual premium. The surplus emerging at the end of the first policy year, per policy in force at the start of that year, is 1.05 P − 1V p45 = 580.70, since p45 = 0.99923 and P = $26 100. Note that there are no expenses in the first year, as initial expenses are deemed to be paid at time 0.

210

Solutions for Chapter 13

211

The surplus emerging at the end of the second policy year, per policy in force at the start of that year, is 1.05 (1V + P − 25) − 2V p46 = 688.11, since p46 = 0.99916.

As the initial expenses are $5220, NPV(2) = −5220 + 580.70 v + 688.11 v 2 p45 = −4067.02, where the risk discount rate is 10% per year. (b) At time 29 years, the annuity payment is 50 000 (1.029 ). Hence, the surplus emerging at the end of the 30th policy year, per policy in force at the start of that year, is 1.05 29V − 40 − 50 000 (1.029 ) − 30V p74 = 3503.74, where p74 = 0.98356 and the total expenses at the start of the year are 25 + 15 = 40.

13.4 The emerging profit in the 10th policy year is 1

1.06 (9V + 0.93 × 12 000) − 106 q59 (1.06 2 ) − 0.05 p59 85 000 − 0.95 p59 10V = 1911.81, 1

where the factor 1.06 2 allows for payment of the death benefit mid-year, and the probabilities of surrender and being in force at time 10 are 0.05 p59 and 0.95 p59 respectively. Then Π10 = 9 p[50] × 1911.81 = 1185.08.

13.5 The reserve required at time 4 to zeroize the negative emerging surplus at time 5 is 4V

Z

= 217/1.05 = 206.67.

This reserve cannot be established from the emerging surplus of −55 at time 4, so the reserve required at time 3 is 3V

Z

= 55/1.05 + 206.67 px+3 /1.05 = 246.06

since a loss of 55 emerges for each policy in force at time 3 and a reserve of 4V Z

212

Solutions for Chapter 13

is required only for policies still in force at time 4. This reserve cannot be established from the emerging surplus of 94 at time 3, but this surplus does contribute to the reserve. Hence, the reserve required at time 2 is 2V

Z

= −94/1.05 + 246.06 px+2 /1.05 = 141.30.

This amount is required in respect of policies in force at time 2, and so the emerging surplus at time 2 is 229 − 141.30 px+1 = 89.68.

Hence, the revised profit vector is (−310, 436, 89.68, 0, 0, 0) .

13.6 (a) We calculate survival probabilities t px for t = 1, 2, 3 and 4 from t px =

t−1 2

(1 − qx+r ).

r=0

The tth element of the profit signature is then calculated as Πt = Prt × t−1 px , for t = 1, 2, . . . , 5, with Π0 = Pr0 . Thus, the profit signature is (−360.98, 149.66, 14.62, 268.43, 377.66, 388.29) . (b) The NPV using a risk discount rate of 10% per year is 5

1.1−t Πt = 487.88.

t=0

(c) The NPV using a risk discount rate of 15% per year is 5

1.15−t Πt = 365.69.

t=0

(d) A higher risk discount rate generates a lower NPV. However, both the 10% rate and the 15% rate indicate a positive NPV for this profit vector. (e) The IRR, say j per year, satisfies 5

t=0

The solution is j = 42.7%.

(1 + j)−t Πt = 0.

213

Solutions for Chapter 13

13.7 (a) We show the profit test table, and explain below how the entries are calculated. t (1) 0 1 2 3 4

(2)

Pt (3)

Et (4)

It (5)

EBt (6)

EVt (7)

Prt (8)

330 330 330

1100 1100 1100 1100

330 0 22 22 22

88.00 112.64 112.64 112.64

800.48 900.54 1000.60 1200.72

327.36 327.03 326.70 0.00

−330.00 60.16 293.07 193.34 319.92

t−1V

In column (1) of the table, t refers to the cash flow date.

The first row deals with the initial expenses, assumed incurred at time t = 0.

We assume that the insurer does not establish a reserve for the policy until the end of the first year, so that the initial expenses represent the only outgo at inception.

Column (2) is for the reserve brought forward from the previous year; the reserve is set at t−1V = $330 for t = 2, 3, 4.

Column (3) is the premium, Pt = $1100, collected at the start of each year.

Column (4) shows the renewal expenses, Et .

Column (5) shows the interest earned through the year, It = 0.08 (t−1V + Pt − Et ).

Column (6) is the expected cost of the death benefit, $100 000, and claim expenses, $60, for a policy in force at the start of the year; that is, for t = 1, 2, 3, 4, EBt = q60+t−1 100 060.

Column (7) is the expected cost of the year-end reserve for policies which are continuing, given that the policy is in force at the start of the year; that is, for t = 1, 2, 3, EVt = p60+t−1 330.

Column (8) is the profit vector: Pr0 = −330

214

Solutions for Chapter 13 and for t = 1, 2, 3, 4, Prt = t−1V + Pt − Et + It − EBt − EVt .

(b) We have Π0 = Pr0 , and for t = 1, 2, 3 and 4, Πt = t−1 p60 Prt .

So Π = (−330.00, 60.16, 290.73, 190.07, 311.36) . (c) The NPV, using a risk discount rate of 12% per year, is 4

1.12−t Πt = $288.64.

t=0

(d) The EPV of premium income, at 12% per year interest, is 1100

3

1.12−t t p60 = 3698.36

t=0

and so the profit margin is 288.64 = 7.8%.

3698.36 (e) The discounted payback period is the least integer t such that the partial NPV up to time t, denoted NPVt , is greater than or equal to 0, that is, where NPVt =

t

1.12−k Πk ≥ 0.

k=0

In this case we have t

Πt

NPVt

0 1 2 3 4

−330.00 60.16 290.73 190.07 311.36

−330.00 −276.29 −44.52 90.76 288.64

We see that the partial NPV changes sign when t = 3, so the discounted payback period is three years.

215

Solutions for Chapter 13 (f) To calculate the IRR we find j such that 4

(1 + j)−t Πt = 0.

t=0

The solution is j = 41.9%, which is less than 50% per year. (g) As the IRR exceeds the hurdle rate of 15% per year, the contract is satisfactory.

13.8 (a) The profit test table is shown below.

(1) t 0 1 2 3

(2) t−1V 5 40 80

(3) Pt

(4) Et

75 75 75

5 2 2 2

(5) It 5.46 7.91 10.71

(6) EDBt 33.98 37.92 42.34

(7) EW Bt

(8) ESBt

1.87 3.74

(9) EVt

(10) Prt

5.00 37.87 75.71

−10.00 9.74 3.55 1.88

119.49

Column (1) refers to the cash flow date.

The first row of the table deals with the initial expenses that are incurred at time 0, and the reserve 0V which is established at that time, leading to a total outgo of 10 immediately before the first premium payment.

Column (2) shows the reserve brought forward at the start of the year.

Column (3) shows the premium, Pt = $75, payable at the start of each year.

Column (4) shows the renewal expenses of $2 at the start of each year.

Column (5) shows the interest for the year, calculated as It = 0.07 (t−1V + Pt − Et ) .

Column (6) shows the expected cost of the death benefit at the year end, calculated as EDBt = 10 000 q60+t−1 .

Column (7) shows the expected withdrawal benefit, calculated for t = 1, 2 as EW Bt = 0.05 p60+t−1 (0.5 × t × P).

216

Solutions for Chapter 13 Column (8) shows the expected survival benefit for the final policy year, calculated as ESB3 = 120 p62 .

Column (9) shows the expected cost of establishing the reserve at the year end, calculated as EVt = p60+t−1 t V .

Column (10) shows the profit vector; in terms of the column entries, we have Prt = (2) + (3) − (4) + (5) − (6) − (7) − (8) − (9).

In terms of the column headers, we have Pr0 = −E0 − EV0 , and for t = 1, 2, 3 Prt = t−1V + Pt − Et + It − EDBt − EW Bt − ESBt − EVt .

(b) We have Π0 = Pr0 , and for t = 1, 2, 3 , Πt = t−1 p60 Prt , so Π = (−10.00, 9.74, 3.36, 1.69) . (c) The NPV, using a risk discount rate of 10% per year, is 3

1.1−t Πt = $2.89.

t=0

(d) We work backwards through the policy years. First we calculate 2V Z from 1.07 (2V Z + 75 − 2) − EDB3 − ESB3 = 0, giving 2V Z = 78.24. Next, 1.07 (1V Z + 75 − 2) − EDB2 − EW B2 − 0.95 p61 2V Z = 0, giving 1V Z = 35.13. Finally, 1.07 (0V Z + 75 − 2) − EDB1 − EW B1 − 0.95 p60 1V Z = 0, giving 0V Z = −8.41, which means that a reserve is not required at time 0, and so 0V Z = 0. (e) The revised profit vector is the same as the revised profit signature, namely (−5.00, 9.00, 0.00, 0.00) and hence the revised NPV is 3.18.

Solutions for Chapter 13

217

(f) The total, undiscounted profit under the original approach is larger than the total undiscounted profit under the zeroized approach (4.74 compared with 4.00). However, under the zeroized approach the profit is released sooner, and, given that the hurdle rate is quite high, the present value of the profit under the zeroized approach is larger than the present value under the original reserve schedule.

Note that we assume that reserves earn only 7%, compared with the hurdle rate of 10%, meaning that holding higher reserves than required will reduce the emerging profit. If the values were the same, holding extra capital would make no difference to the NPV. 13.9 (a) First, calculate the reserves and cash values for the year, denoting the start of the year as time t − 1. Then t −1V = 100 000 A40 (4%) = 17 758.80, t V = 100 000 A41 (4%) = 18 426.14,

CVt = 100 000 A41 (5%) = 12 665.17.

Next, project the expected surplus emerging at the year end (time t), per policy in force at the start of the year (time t − 1): Prt = 1.06 (t −1V − 20) − 0.0004 × 100 000 − 0.9996 × 0.08 CVt − 0.9996 × 0.92 t V = 805.05.

The policyholder’s share is 80%, or $644.04, per policy in force at time t − 1.

The dividend is only payable to policyholders who are still in force at time t, so the projected dividend per policy in force is 644.07 = 700.32.

0.9996 × 0.92 (b) The projected surplus at time t, per policy in force at time t − 1, ignoring surrenders, is Prt = 1.06 (t −1V − 20) − 0.0004 × 100 000 − 0.9996 × t V = 344.35.

The dividend is 80% of this surplus, shared between the projected in-force policyholders. It is not proposed that surrender rates are changed, nor that surrendering policyholders share in the profit distribution. This means that the denominator in the dividend calculation is the same as in part (a), so the

218

Solutions for Chapter 13 revised projected dividend is 0.8 × 344.35 = 299.56.

0.9996 × 0.92 Comments: (i) If profit sharing has been marketed specifically as sharing in investment profits, it is reasonable not to include surrender profits in the dividend calculations. (ii) On the other hand, if profit sharing is presented as an 80/20 split of profits between policyholders and shareholders, it would seem unreasonable to omit surrender profits. (iii) In either case, it would be more equitable to include surrendering policyholders in the profit distribution, as they have contributed to the investment profits for the policy year.

13.10 Assume that the insurer holds no reserves for this contract.

We show here an excerpt from the profit test table, and explain the entries in more detail below. t (1)

Pt (2)

Et (3)

It (4)

EBt (5)

Prt (6)

100 100 100 .. .

100

550 0 5 5 .. .

5

0.49 0.46 0.46 .. .

0.46

31.76 32.04 32.31 .. .

93.18

−550.00 68.73 63.43 63.15 .. .

2.28

0 1 12 2 12 3 12

.. .

10

Column (1) shows the time at which the cash flows are valued; the first row accounts for the initial expenses, incurred at time t = 0.

All other rows accumulate the cash flows to the end of each month (i.e. time t).

For each row, the cash flows are determined assuming that the policy is in force 1 ). at the start of the month (i.e. at time t − 12 Column (2) shows the premium income at the start of each month.

Column (3) shows the expenses incurred; the initial expenses in the first row

Solutions for Chapter 13

219

include all first-month expenses, which is why there are no further expenses in the second row (which shows cash flows during the first month of the contract).

Column (4) shows the interest earned during the month on the beginning-month cash flows, i.e. (1.061/12 − 1)(Pt − Et ).

Column (5) shows the expected benefit outgo for the month for a policy in force at the beginning of the month, accumulated to the end of the month. We assume that, on average, the death benefit is paid half-way through the month of death, so that EBt = 200 000 121 q55+t− 121 (1.061/24 ) .

Column (6) shows the profit vector, which is the expected profit at time t for each 1 policy in force at time t − 12 , where Prt = Pt − Et + It − EBt .

Mortality probabilities are calculated from Makeham’s formula.

13.11 There are some preliminary calculations to be carried out to establish the net premium and the net premium reserves. Let Pn denote the net premium. Recall that for net premium policy values, the net premium is always calculated on the policy value basis, and is not affected by changes in the gross premium. Then Pn ä55:10 = 100 000 A55:20 .

We have ä55:10 = 7.71822 and A55:20 = 0.33236, giving Pn = $4306.23.

Next, we can calculate the net premium reserves recursively as 0V = 0, and for t = 0, 1, 2, . . . , 9, (t V + Pn ) (1.06) = 100 000 q55+t + p55+t t+1V and for t = 10, 11, 12, . . . , 18, after premiums cease, t V (1.06) = 100 000 q55+t + p55+t t+1V .

The approach to calculating the premium is to set up a spreadsheet calculation with an arbitrary, but realistic, guess at the gross premium entered in a particular cell. The net premium calculated above would be a good starting value. The NPV, and hence the profit margin, are calculated using this value of the premium and then the premium is adjusted until the profit margin is 15%. Solver in Excel will do this automatically.

Given the premium P we set up the profit test table as usual. The first four rows

220

Solutions for Chapter 13

and last two rows of the profit test are shown here, using the net premium in place of the gross premium.

Profit test using first guess premium. Rounded for presentation. t (1) 0 1 2 3 .. .

19 20

(2)

Pt (3)

Et (4)

It (5)

EBt (6)

EVt (7)

Prt (8)

0 4 374 9 000 .. .

89 078 94 340

4306 4306 4306 .. .

0 0

300 0 108 108 .. .

0 0

323 643 990 .. .

6681 7075

199 221 246 .. .

1 466 100 000

4 365 8 980 13 858 .. .

92 956 0

−300.00 64.59 14.47 83.86 .. .

1336.17 1415.09

t−1V

As usual, the first row accounts for the initial expenses and each row from t = 1 to t = 20 considers the cash flows for that year, accumulated to the year end, assuming the policy is in force at the start of the year.

Column (1) is the time to which the year’s cash flows are accumulated.

Column (2) shows the reserve brought forward in the tth year, t−1V , calculated from the above net premium recursion.

Column (3) shows the premium, Pt = P, for t = 1, 2, . . . , 10, and Pt = 0 thereafter.

Column (4) shows the expenses, Et .

Column (5) shows the interest earned during the year on the start-year cash flows, It = 0.075 (t−1V + Pt − Et ) .

Column (6) shows the expected cost of benefits at year end, given that the policy is in force at the start of the year, EBt . For t = 1, 2, . . . , 19, this is the expected death benefit, and in the final year, if the policyholder survives to the start of the year, the full benefit is payable at the year end on death or survival, so EBt = 100 000 q55+t−1 for t = 1, 2, . . . , 19 and EB20 = 100 000.

Column (7) shows the expected cost of the reserve for continuing policies, EVt = p55+t−1 t V .

221

Solutions for Chapter 13 Column (8) shows the profit vector Prt = t−1V + Pt − Et + It − EBt − EVt .

The net present value of the policy using a risk discount rate of 12% per year is Pr0 +

20

1.12−t t−1 p55 Prt .

t=1

To obtain the profit margin, we divide this by the EPV of the premiums, i.e. by P

9

1.12−t t p55 .

t=0

Using Solver in Excel, with a target profit margin of 15%, we find that the required value of P is $4553.75.

The updated profit test using this premium is shown below.

Profit test using the correct premium. t (1) 0 1 2 3 .. .

19 20

(2)

Pt (3)

Et (4)

It (5)

EBt (6)

EVt (7)

Prt (8)

0 4 374 9 000 .. .

89 078 94 340

4554 4554 4554 .. .

0 0

300 0 114 114 .. .

0 0

342 661 1008 .. .

6681 7075

199 221 246 .. .

1 466 100 000

4 365 8 980 13 858 .. .

92 956 0

−300.00 330.68 273.91 343.29 .. .

1336.17 1415.09

t−1V

13.12 The approach to this question is virtually identical to the previous one. There are changes to the detail of the calculations, most notably that the time unit for the profit test is a month rather than a year.

Once again we start with calculations relating to the net premium reserve. Now let Pn denote the total net premium in a year. Then Pn ä(12)

55:1 0

= 100 000 Ā55:20 .

We have ä(12) = 7.50515 and Ā55:20 = 0.33414, resulting in Pn = $4452.16 or 55:10 $371.01 per month.

222

Solutions for Chapter 13

Next, we calculate the net premium reserves recursively. The death benefit is payable immediately on death, and we approximate the cost of this by assuming that it is payable in the middle of the month of death, which means that the end of month cost has an extra half-month accumulation. We have 0V = 0, and 1 2 for t = 0, 12 , 12 , . . . , 119 12 , 1 1

n 1 q55+t + 1 V 1 p55+t , t V + P /12 (1.06 12 ) = 100 000 (1.06 24 ) 12 t+ 12 12 121 238 and for t = 120 12 , 12 , . . . , 12 we have 1

1

1 q55+t + 1 V 1 p55+t . t V (1.06 12 ) = 100 000 (1.06 24 ) 12 t+ 12 12

As in the previous question, the approach to calculating the premium is to set up a spreadsheet calculation with an arbitrary, but realistic, guess at the gross premium to calculate the net present value of the policy, and hence the profit margin, under this monthly premium. Once the spreadsheet is complete, we can use Solver to adjust the premium until it gives the required 15% profit margin.

We show the profit test details for the first three months and the last two months, based on the premium which generates a 15% profit margin.

Profit test using the correct premium. t (1) 0 1/12 2/12 3/12 .. .

239/12 240/12

(2)

Pt (3)

Et (4)

It (5)

EBt (6)

EVt (7)

Prt (8)

0 357 716 .. .

99 035 99 516

394.27 394.27 394.27 .. .

0 0

300 0 9.86 9.86 .. .

0 0

2.38 4.48 6.65 .. .

598.66 601.57

15.89 16.03 16.17 .. .

144.50 100 000.44

356.94 715.53 1 075.79 .. .

99 372.59 0

−300.00 23.82 14.33 14.75 .. .

116.52 117.08

t−1/12V

As usual, the first row accounts for the initial expenses and each row from t = 1/12 to t = 20 considers the cash flows for that month, accumulated to the month end, assuming the policy is in force at the start of the month.

Column (1) is the time to which the month’s cash flows are accumulated.

Column (2) shows the reserve brought forward, t−1/12V , calculated from the above net premium recursion.

Column (3) shows the monthly premium, Pt , which equals P/12, for t = 1/12, 2/12, . . . , 10, and 0 thereafter.

223

Solutions for Chapter 13 Column (4) shows the expenses, Et .

Column (5) shows the interest earned during the month on the start-of-month cash flows, It = (t−1/12V + Pt − Et ) × 1.0751/12 − 1 .

Column (6) shows the expected cost of benefits at month end, given that the policy is in force at the start of the month. We assume that, on average, death benefits are paid half-way through the month of death. If the policyholder survives to the start of the final month, the full benefit is payable at the month end on survival, so 1 , . . . , 19 11 EBt = 100 000 121 q55+t− 121 (1.0751/24 ) for t = 12 12

and EB20 = 100 000 121 q74 1112 (1.0751/24 ) + 100 000 121 p74 11 .

12 Column (7) shows the expected cost of the reserve for continuing policies, EVt = 121 p55+t− 121 t V .

Column (8) shows the profit vector Prt = t−1/12V + Pt − Et + It − EBt − EVt .

The net present value of the policy using a risk discount rate of 12% per year is Pr0 +

240

1.12−k/12 k−1 p55 Pr k .

12

12

k=1

To obtain the profit margin, we divide this by the EPV of the premiums, i.e.

P

119

1.12−k/12 k p55 .

12

k=0

The value of P which gives a profit margin of 15% is $4731.22, or $394.27 per month.

13.13 (a) Let P denote the single premium. We set P equal to the EPV of the expenses and death benefit at an effective interest rate of 4% per year.

At each policy anniversary, the expenses incurred are 0.03P if at least one life is alive. The probability that at least one of the lives is alive at time t is 1 minus the probability that both are dead at time t, i.e.

1 − (1 − t p50 )(1 − t p50 ) = 2 t p50 − t p50:50 .

224

Solutions for Chapter 13 So the EPV of the premium related expenses is 0.03P

9

vt (2 t p50 − t p50:50 ) = 0.25304 P .

t=0

For the EPV of the death benefit, consider three possible situations for each policy year t to t+1: • 100 000 is payable at time t+1 if both lives are alive at time t and exactly one of them dies during the year; • 300 000 is payable at time t+1 if both lives are alive at time t and both die during the year; and • 200 000 is payable at time t+1 if exactly one life is alive at time t and that life dies during the year (and note that this situation can apply only from the second year, as both lives are alive at time t = 0).

Then the EPV of the death benefit is 100 000

9

vt+1 t p50:50 (2 p50+t q50+t ) + 300 000

t=0

+ 200 000

9

vt+1 t p50:50 (q50+t )2

t=0

9

vt+1 2 t p50 (1 − t p50 ) q50+t

t=1

= 3122.55.

The single premium is then P = $4180.35. (b) The reserves can be calculated recursively. Let t V (1) denote the reserve if exactly one life is alive at time t, and let t V (0) denote the reserve if both lives are alive at time t.

The situation when just one life is alive at time t is simpler; we have tV

(1)

− 0.03P 1.04 = 200 000 q50+t + t+1V (1) p50+t .

This can be calculated recursively, using 10V (1) = 0, or by using standard functions as tV

(1)

= 200 000 A 1

50+t:10−t

+ 0.03Pä50+t:10−t .

225

Solutions for Chapter 13 If both lives are alive at time t we have tV

(0)

− 0.03P 1.04 = 100 000 + t+1V (1) (2 p50+t q50+t ) + 300 000 q250+t + t+1V (0) p250+t ,

since if exactly one of the lives dies aged 50 + t, a death benefit of 100 000 and a reserve of t+1V (1) are required at the year end; if both lives die, a death benefit of 300 000 is required, and if both lives survive, a reserve of t+1V (0) is required. The recursion can be used in conjunction with values of t V (1) , backwards, using the fact that 10V (0) = 0. This should generate 0V = P, which is the reserve immediately after receipt of the single premium. (c) We calculate the profit signature by evaluating first two separate profit vectors; one assuming that only one life is alive at the start of the policy year, denoted Pr(1) , and the other assuming that both lives are alive at the start of the policy year, denoted Pr(0) . We then multiply these by the appropriate probabilities to obtain the overall profit signature.

We show here an excerpt from the profit test table for calculating the emerging profit at time t in the case that both lives are alive at time t − 1. t (1) 0 1 2 3 .. .

(2)

Et (3)

It (4)

EBt (5)

EVt(0) (6)

EVt(1) (7)

Prt(0) (8)

4180.35 3975.37 3737.66 .. .

62.71 0.00 65.21 67.82 .. .

334.43 312.81 293.59 .. .

241.85 266.39 293.96 .. .

3965.77 3727.71 3452.69 .. .

9.52 9.86 10.09 .. .

−62.71 297.64 219.01 206.69 .. .

t−1V

(0)

We account separately for the initial expenses as a time-0 outgo.

Column (2) shows the reserve brought forward in the tth year, t−1V (0) , calculated from the above net premium recursion, given that both lives are alive at time t − 1. Note that we include the single premium in the first year’s cash flows in column (2), as 0V (0) , for convenience.

Column (3) shows the expenses in the tth year, Et = 0.015P (1.04t−1 ).

Column (4) shows the interest earned during the year on the start-year cash flows, It = 0.08 (t−1V (0) − Et ).

Column (5) shows the expected cost of benefits at year end, given that both

226

Solutions for Chapter 13 lives are alive at the start of the year, EBt . If one of the lives dies, and the other survives, the benefit is $100 000. If both lives die, the benefit is $300 000, so EBt = (2 p50+t−1 q50+t−1 ) 100 000 + (q50+t−1 )2 300 000.

Column (6) shows the expected cost of the reserve for continuing policies, for the case when both lives survive to the year end EVt(0) = (p50+t−1 )2 t V (0) .

Column (7) shows the expected cost of the reserve for continuing policies, for the case when exactly one life survives to the year end EVt(1) = (p50+t−1 )(q50+t−1 ) t V (1) .

Column (8) shows the profit vector for the case when both lives are alive at the start of the year Prt(0) = t−1V (0) − Et + It − EBt − EVt(0) − EVt(1) .

Next, we show an excerpt from the profit test to determine the profit emerging at time t when exactly one life survives to time t − 1. This is only possible for t = 2, 3, . . . , 10, as we know that both lives are alive at the start of the policy.

(2)

Et (3)

It (4)

EBt (5)

EVt(1) (6)

Prt(1) (7)

3943.26 3709.30 .. .

65.21 67.82 .. .

310.24 291.32 .. .

266.21 293.75 .. .

3704.36 3433.50 .. .

217.73 205.55 .. .

t (1)

t−1V

2 3 .. .

(1)

The entries in this table are similar to the Pr(0) case above. The reserve brought forward in Column (2) is now the reserve for the one-survivor case. EBt is the expected benefit cost in the tth year, given that only one life survived to the start of the year. Hence EBt = 200 000 q50+t−1 and EVt(1) is the expected cost of establishing the year-end reserve in the case that the life survives, that is EVt(1) = p50+t−1 t V (1) .

227

Solutions for Chapter 13

For the profit signature, we multiply the profit vector elements by the appropriate survival probabilities up to time t −1, to give the unconditional expected emerging profits at time t. For t = 0, 1, we know that both lives are alive at time t = 0, so Π0 = Pr(0) 0

and Π1 = Pr(0) 1 .

For t = 2, 3, . . . , 10, we multiply Prt(0) by the probability that both lives are alive at time t − 1, and we multiply Prt(1) by the probability that exactly one life is alive at time t − 1, giving Πt = t−1 p50:50 Prt(0) + 2 t−1 p50 t−1 q50 Prt(1) .

Some selected values from the profit signature are Π2 = 219.01,

Π4 = 192.75,

Π6 = 159.29,

Π8 = 116.68,

Π10 = 62.52.

13.14 (a) Let P be the annual premium. The EPV of premium payments is P ä60:60:5 = 4.5628P and the EPV of the annuity and expenses is 300 + 10 200

20

vt (1 − t p60 ) t p60 = 8362.63

t=1

giving P = $1832.79. (b) First, we calculate the reserves. Let t V (1) denote the reserve at time t if just the husband is alive and let t V (0) denote the reserve if both the husband and the wife are alive.

If the wife has died by time t, then the reserve must be sufficient to pay the annuity plus annuity expenses from time t + 1 to time 20, inclusive, so tV

(1)

= 10 200 a60+t:20−t .

The values can be determined from tables or recursively, backwards, given (1) = 0 and 20V tV

(1)

1.04 = p60+t 10 200 + t+1V (1) .

Given the values for t V (1) , we can also calculate t V (0) recursively: for t = 0 (P − 300) 1.04 = p60:60 1V (0) + p60 q60 10200 + 1V (1) ,

228

Solutions for Chapter 13 for t = 1, 2, 3, 4, tV

(0)

+ P 1.04 = p60+t:60+t t+1V (0) + p60+t q60+t 10 200 + t+1V (1) ,

and, as premiums cease after five years, for t = 5, 6, . . . , 19, tV

(0)

1.04 = p60+t:60+t t+1V (0) + p60+t q60+t 10 200 + t+1V (1) .

The recursion can work backwards, as 20V (0) = 0, or forwards, using 0V (0) = 0, because the premium and reserve are calculated on the same basis.

We calculate two profit vectors each year (after the first). Prt(0) is the expected profit emerging at time t, conditional on both partners surviving to time t − 1.

Prt(1) is the expected emerging profit at time t conditional on the husband surviving and the wife having died at time t −1. The profit signature is calculated by multiplying each of these by the appropriate probability, and summing, to give the expected emerging profit at time t, conditional only on both partners being alive at the start of the policy.

We show here an excerpt from the profit test table for Pr(0) . t (1) 0 1 2 3 .. .

t−1V

(0)

(2)

Pt (3)

0 1144.36 2623.61 .. .

1832.79 1832.79 1832.79 .. .

It (4)

EBt (5)

EVt(0) (6)

EVt(1) (7)

Prt(0) (8)

109.97 178.63 267.38 .. .

34.54 38.53 43.00 .. .

1136.59 2603.75 4105.50 .. .

422.96 453.95 486.15 .. .

−300.00 348.66 59.54 89.13 .. .

We account separately for the initial expenses as a time 0 outgo; this is captured as Pr(0) 0 in the table. The benefit expenses are accounted for with the benefit costs.

Column (2) shows the reserve brought forward in the tth year, t−1V (0) , calculated from the above net premium recursion, given that both lives are alive at time t − 1.

Column (3) shows the premiums, which are Pt = 1832.79 for the first five years, with Pt = 0 thereafter.

Column (4) shows the interest earned during the year on the start-year cash flows, It = 0.06 (t−1V (0) + Pt ).

Column (5) shows the expected cost of benefits at year end, given that both

229

Solutions for Chapter 13

partners are alive at the start of the year, EBt . If the wife dies and the husband survives, the year-end benefit, including expenses, is $10 200, so EBt = 10 200 p60+t−1 q50+t−1 .

Column (6) shows the expected cost of the reserve for continuing policies, for the case when both lives survive to the year end: EVt(0) = (p60+t−1 )2 t V (0) .

Column (7) shows the expected cost of the reserve for continuing policies, for the case when only the husband survives to the year end: EVt(1) = (p60+t−1 )(q60+t−1 ) t V (1) .

Column (8) shows the profit vector for the case when both partners are alive at the start of the year: Prt(0) = t−1V (0) + Pt + It − EBt − EVt(0) − EVt(1) .

For the case when the husband is alive at the start of a policy year and the wife is not, the calculations are similar, but for an alive–dead model. We show an excerpt from the profit test table below.

t (1)

t−1V

(1)

(2)

It (3)

EBt (4)

EVt(1) (5)

Prt(1) (6)

2 3 4 .. .

124 890 120 180 115 319 .. .

7493 7211 6919 .. .

10 161 10 157 10 152 .. .

119 725 114 831 109 780 .. .

2497.80 2403.60 2306.38 .. .

Combining the profit vectors with the appropriate probabilities, we obtain the profit signature. We have Π0 = −300,

Π1 = Pr(0) 1

and for t = 2, 3, . . . , 20, Πt = t−1 p60:60 Prt(0) + t−1 q60 t−1 p60 Prt(1) .

230

Solutions for Chapter 13 The net present value is 20

1.15−t Πt = $779.26.

t=0

13.15 (a) Let P denote the single premium. Then, noting that the return of premium applies on death before age 66, P = 20 020 (1.06−5 ) 5 p60 a65 +

5

1.06−(t+1) (1.05t+1 P) t |q60 + 275,

t=0

where a65 is calculated at 5%, giving P = $192 805.84. (b) We calculate reserves recursively. We have, for the first year, (0V + P − 275)(1.06) = q60 P (1.05) + p60 1V .

For t = 1, 2, 3, 4, we have t+1 + p60+t t+1V ; t V (1.06) = q60+t P (1.05)

for t = 5 we have 5V (1.05) = q65 P (1.05)

6

+ p65 (20 020 + 6V ) ;

and for t = 6, 7, . . . we have t V (1.05) = p60+t (20 020 + t+1V ) .

The recursions can be applied forwards (premiums and reserves are calculated using the same assumptions, so 0V = 0), or backwards, using an appropriately high assumption for the maximum lifetime, or we can set 6V = 20 020 a66 at 5% effective interest, and use recursions for the reserves for earlier years. (c) (i) Note that the mortality basis for the profit test is different from the premium and reserve basis.

We show an excerpt from the profit test table, and explain each column in more detail below.

231

Solutions for Chapter 13 t (1)

t−1V + Pt − Et

(2)

It (3)

EBt (4)

EVt (5)

Prt (6)

Πt (7)

0 1 2 3 .. .

12 13 14 .. .

−275 192 806 204 088 216 346 .. .

219 787 213 207 206 516 .. .

15 424 16 327 17 308 .. .

13 187 12 792 12 391 .. .

688 806 945 .. .

19 786 19 758 19 726 .. .

203 394 215 526 228 379 .. .

210 719 203 815 196 796 .. .

−275 4 148 4 083 4 330 .. .

2 469 2 427 2 385 .. .

−275 4 148 4 069 4 298 .. .

2 303 2 237 2 170 .. .

Column (2): the first row accounts for the initial expenses. The second row accounts for the single premium. The other entries in this column are the reserve values from the calculations in part (b).

Column (3): the interest earned at 8% (before vesting) or 6% (after vesting) on the Column (2) cash flow, for t = 1, 2, 3, . . ..

Column (4): EBt is the expected cost of the year-end benefits, given that the contract is in force at the start of the year. For t = 1, 2, 3, 4, 5 this is the expected cost of the return of premium with interest. For t = 6, we allow for the return of premium, if the policyholder dies during the year, and for the annuity payment and expenses, if the policyholder survives. For t = 7, 8, 9, . . . the expected benefit is the expected cost of the annuity. That is ⎧ ⎪ for t = 1, 2, 3, 4, 5, P (1.05t ) q60+t−1 ⎪ ⎪ ⎪ ⎨ t EBt = ⎪ P (1.05 ) q60+t−1 + 20 020 p60+t−1 for t = 6, ⎪ ⎪ ⎪ ⎩ 20 020 p for t = 7, 8, . . ..

60+t−1 Column (5): EVt is the expected cost of the year-end reserve for continuing contracts, so for t = 1, 2, 3, . . .

EVt = t V p60+t−1 .

Column (6): Prt is the profit emerging at time t, conditional on the contract being in force at time t − 1 (for t = 1, 2, . . .), so ⎧ ⎪ −275 for t = 0, ⎪ ⎪ ⎪ ⎨ Prt = ⎪ for t = 1, P + It − EBt − EVt ⎪ ⎪ ⎪ ⎩ V + I − EB − EV for t = 2, 3, 4, . . .. t−1 t t t

232

Solutions for Chapter 13 Column (7): the profit signature, Πt , is the profit emerging at time t, per policy issued, so Π0 = Pr0 and Πt = t−1 p60 Prt for t = 1, 2, 3, . . .. (ii) Using a risk discount rate of 10% per year, the net present value of the contract is ω−60

1.1−t Πt = 28 551.36 t=0

(where ω is a suitable limiting age, e.g. 145), and dividing this by the single premium gives the profit margin as 14.8%.

13.16 We have a survival model, as shown in Figure 13.2 of AMLCR, with two parameterizations, one for profit testing and one for the calculation of the gross premium and the reserves. The difference between the parameterizations is that the former allows for withdrawals, whereas the latter does not.

Let t pixj∗ denote the probability that a life in State i at age x will be in State j at age x + t given the parameterizations allowing for withdrawals and let t pixj denote the corresponding probability when withdrawals are not included. The following probabilities are useful in our calculations.

For 0 ≤ t ≤ 10,

⎧ t ⎫

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ 00∗ 01 02 03 + μ + μ μ ds = exp{−0.035t}, − ⎪ t p55 = exp ⎪ 55+s 55+s 55+s ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ 0 ⎧ t ⎫

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ 00 01 02 + μ μ ds − = exp{−0.025t}, ⎪ t px = exp ⎪ x+s x+s ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ 0

11∗ 11 t px = exp{−0.03t} = t px , t 01∗ 00∗ 01 11∗ s p55 μ55+s t−s p55+s ds t p55 = 0

t exp{−0.035s} 0.01 exp{−0.03(t − s)} ds

= 0

= 2 (exp{−0.03t} − exp{−0.035t}), t 01 00 01 11 s p55 μ55+s t−s p55+s ds t p55 = 0

233

Solutions for Chapter 13 t =

exp{−0.025s} 0.01 exp{−0.03(t − s)} ds 0

= 2 (exp{−0.025t} − exp{−0.03t}), 12∗ 12 t px = 1 − exp{−0.03t} = t px .

Further, for 55 ≤ x ≤ 64 11 12 , 1 12

12 1 p , p12∗ x = 1 − exp{−0.03/12} = 12 x

5 (1 − exp{−0.035/12}) − 2(exp{−0.03/12} − exp{−0.035/12}), 7 02 1 p x = 1 − 3 exp{−0.025/12} + 2 exp{−0.03/12}, 12

1 12

p02∗ x =

2 (1 − exp{−0.035/12}), 7 01∗ 1 p x = 2 (exp{−0.03/12} − exp{−0.035/12}), 12 1 12

p03∗ x =

1 12

p01 x = 2 (exp{−0.025/12} − exp{−0.03/12}).

(a) Let P denote the monthly premium calculated on the net premium policy value basis, so that withdrawals are ignored. Then P

119

t

t 12

00 10 12 = 50 000 p00 10 p55 v 55 v

t=0

+ 50 000

119

t 12

t+1 01 01 12 1 p p00 v 12 t t + t p55 1 p 55 12 55+ 12 12 55+

+ 100 000

t 12

02 1 p p00 v 12 .

55 12 55+ t

119

12

12

t=0

t+1

12

t=0

Using the formulae we have developed, we can calculate that 119

t 12

t 12 = 85.13 p00 55 v

t=0

and

119

t

t 12

12 = 3.65, p01 55 v

t=0

giving P = $452.00. (b) Let t V (0) and t V (1) denote the net premium policy values at policy duration t years given that the policyholder is healthy and disabled, respectively. If t is an exact number of months, then the policy value is calculated before payment of a premium and after payment of any benefits due at that time.

234

Solutions for Chapter 13 Then 10V

(0)

= 10V (1) = 0

10 1 and we can calculate the policy values recursively for t = 9 11 12 , 9 12 , . . . , 12 , 0 from these starting values using the formulae tV

(0)

(0) (1) 1 V 1 V + P 1.05 12 = 121 p00 + 121 p01 55+t t+ 12 55+t 50 000 + t+ 12 1

+ 100 000 121 p02 55+t and tV

(1)

1

(1) 1 V 1.05 12 = 121 p11 + 50 000 121 p12 55+t t+ 12 55+t .

Policy values for a selection of durations are shown in Table S13.1.

Table S13.1 Net premium policy values. t years 0

tV

(0)

0.00 279.32 560.40 .. .

15 237.52 15 613.44 15 991.75

1 12 2 12

.. .

3 11 12 4 1 4 12

(1)

t years

− 10 301.49 10 244.19 .. .

7 234.67 7 157.17 7 079.16

7 11 12 8 1 8 12 .. .

9 10 12 9 11 12 10

tV

tV

(0)

tV

36 252.19 36 761.39 37 273.82 .. .

48 818.44 49 407.35 0.00

(1)

2 876.14 2 769.93 2 663.02 .. .

247.86 124.34 0.00

(c) Let P denote the monthly gross premium. Then, using the equivalence principle, 0.95 P

119

t

t 12

00 10 12 = 50 000 p00 10 p55 v 55 v

t=0

+ 50 000

119

t 12

t+1 01 01 12 t p 1 p 1 p p00 + v 12 t t 55 12 55+ 12 55 12 55+

+ 100 000

t 12

02 1 p p00 v 12 + 1000 55 12 55+ t

119

12

12

t=0

t+1

12

t=0

where the rate of interest is now 5.25% per year. Hence, we now have 119

t

t 12

12 = 84.26 p00 55 v

t=0

giving P = $484.27.

and

119

t

t 12

t=0

12 = 3.59, p01 55 v

235

Solutions for Chapter 13

(d) The emerging surplus at the end of each month is calculating in two parts: first we assume the life is healthy at the start of the month and then we assume the life is disabled at the start of the month. Parts of the calculation are shown in Tables S13.2 and S13.3.

Table S13.2 Emerging surplus assuming the life is healthy at the start of the month. t years (1)

P

Et

It

Dis.

Dis.

Death

With’l

Healthy

(2)

(3)

(4)

(5)

ben.

(6)

res.

(7)

ben.

(8)

ben.

(9)

res.

(10)

(11)

484.27 484.27 . . .

484.27 484.27 484.27

1024.21 0.00 24.21 . . .

224.21 224.21 224.21

278.50 558.77 . . .

48 676.26 49 263.46 49 854.38

−1024.21 33.15 9.29 . . .

93.24 94.27 95.30

0 1 12 2 12

. . .

9 10 12 9 11 12 10

Prt(0)

V (0)

1 t− 12

0.00 279.32 . . .

48 233.24 48 818.44 49 407.35

2.74 4.18 . . .

275.32 278.63 281.96

41.55 41.55 . . .

41.55 41.55 41.55

8.56 8.51 . . .

0.21 0.10 0.00

124.92 124.92 . . .

124.92 124.92 124.92

0.32 0.51 . . .

32.43 32.82 33.21

The key to the columns in Table S13.2 is as follows.

1 (1) denotes the time interval from t − 12 to t, measured in years, except that t = 0 denotes time 0.

(2) denotes the reserve held at the start of the time interval for a life who is 1 healthy at that time, t− 121 V (0) . No reserve is required for t = 0 or t = 12 (0) since 0V = 0.

(3) denotes the gross premium, $484.27, payable at the start of the month.

(4) denotes the expenses payable in the time interval. The entry for t = 0 includes all the initial expenses, so that 1024.21 = 1000 + 0.05 × P = 1000 + 0.05 × 484.27.

1 The entry for t = 12 is zero since all the initial expenses have been assigned to the row for t = 0. For other rows the expense is 0.05P, which is assumed to be incurred at the start of the month.

(5) denotes the interest earned during the month at the assumed rate of 7% per year effective, so that 1

It = (1.07 12 − 1) t− 121 V (0) + P − Et .

(6) is the expected disability benefit payable at the end of the month,

236

Solutions for Chapter 13 50 000 121 p01∗ 55+t , in respect of a life who was healthy at the start of the month but disabled at the end of the month.

(7) is the expected cost of setting up the required reserve at the end of the month for a life who was healthy at the start of the month but disabled at (1) the end of the month. This expected cost is 121 p01∗ 55+t t V .

(8) is the expected death benefit payable at the end of the month for a life who was healthy at the start of the month. This expected cost is 100 000 121 p02∗ 55+t .

(9) is the expected cost of the withdrawal benefit. This is 1 12

p03∗ 55+t

1 V t− 12

(0)

+ P × 0.8.

(10) denotes the expected cost of setting up the reserve required at the start of the following month for a life who remains healthy throughout the month.

(0) This expected cost is 121 p00∗ 55+t t V .

(11) denotes the expected surplus emerging at the end of the month in respect of a policyholder who was healthy at the start of the month, so that Prt(0) = (2) + (3) – (4) + (5) – (6) – (7) – (8) – (9) – (10).

Table S13.3 Emerging surplus assuming the life is disabled at the start of the month. t years (1) 1 12 2 12 3 12

.. .

10 9 12 11 9 12 10

(1)

(2)

(3)

Death ben.

(4)

0.00 10 031.49 10 244.19 .. .

370.58 247.86 124.34

0.00 58.25 57.92 .. .

2.10 1.40 0.70

0.00 124.84 124.84 .. .

124.84 124.84 124.84

1 V t− 12

It

Disabled res.

(5)

Prt(1)

0.00 10 218.61 10 161.08 .. .

247.24 124.03 0.00

0.00 16.28 16.19 .. .

0.59 0.39 0.20

(6)

The key to the columns in Table S13.3 is as follows.

1 to t, measured in years.

(1) denotes the time interval from t − 12

Solutions for Chapter 13

237

(2) denotes the reserve held at the start of the time interval for a life who is disabled at that time, t− 121 V (1) . No cash flows are included for the first 1 month, corresponding to t = 12 , since the life is healthy when the policy is issued.

(3) denotes the interest earned during the month, at the rate of 7% per year effective, on the reserve held at the beginning of the month.

(4) is the expected death benefit payable at the end of the month for a life who was disabled at the start of the month. This expected cost is 50 000 121 p12∗ 55+t .

(5) denotes the expected cost of setting up the reserve required at the start of the following month for a life who remains disabled throughout the (1) month. This expected cost is 121 p11∗ 55+t t V .

(6) denotes the expected surplus emerging at the end of the month in respect of a policyholder who was disabled at the start of the month, so that Prt(1) = (2) + (3) – (4) – (5). (e) The profit signature vector, Π = (Π0 , Π 121 , . . . , Π9 1112 , Π10 ) , is calculated as Π0 = Pr(0) 0 1 2 and for t = 12 , 12 , . . . , 10, (1) 01∗ 1 p Πt = Prt(0) t− 121 p00∗ 55 + Prt t− 12 55 .

(1) 01∗ 1 Values of Prt(0) , t− 121 p00∗ 55 , Prt , t− 12 p55 and Πt for selected values of t are shown in Table S13.4.

The internal rate of return is the rate of interest, r, per year such that 120

Π k (1 + r)− 12 = 0. k

12

k=0

This gives an internal rate of return of 32.7% per year. (f) The net present value, NPV, is given by NPV =

120

Π k 1.15− 12 = $992.29. k

12

k=0

238

Solutions for Chapter 13 Table S13.4 Calculation of the profit signature. t years

Prt(0)

00∗ 1 p t− 12 55

Prt(1)

01∗ 1 p t− 12 55

Πt

0

−1024.21 33.15 9.29 .. .

34.82 35.48 36.13 .. .

71.38 72.27 73.16 .. .

93.24 94.27 95.30

1.00 1.00 0.9971 .. .

0.8744 0.8719 0.8694 .. .

0.7602 0.7580 0.7558 .. .

0.7109 0.7088 0.7067

− 0.00 16.28 .. .

11.55 11.43 11.31 .. .

4.71 4.54 4.38 .. .

0.59 0.39 0.20

0.00 0.00 0.0008 .. .

0.0338 0.0345 0.0351 .. .

0.0607 0.0612 0.0617 .. .

0.0710 0.0714 0.0719

−1024.21 33.15 9.27 .. .

30.84 31.33 31.81 .. .

54.55 55.06 55.56 .. .

66.33 66.85 67.37

1 12 2 12

.. .

3 11 12 4 1 4 12 .. .

7 11 12 8 1 8 12 .. .

9 10 12 9 11 12 10

The profit margin, i.e. the NPV as a percentage of the EPV of gross premiums, is NPV = 3.84%, 119

k 00∗ − 12 k p P 55 1.15 12

k=0

and the NPV as a percentage of the acquisition costs is NPV = 97.0%.

0.05 P + 1 000 The discounted payback period is (m/12) years, where m is the smallest integer such that m

k Π k (1 + r)− 12 ≥ 0.

12

k=0

This gives a discounted payback period of five years and five months.

Solutions for Chapter 14

14.1 Let AV t and ASt denote the account value and asset share, respectively, at the end of year t. Recall that the asset share at time t is the accumulated income minus outgo, expressed per policy in force at time t.

We have AV4 = 30 and AS4 = 20, so AV5 = 1.06 (AV4 + 20 − 2 − 7) = 43.46 and AS5 =

1.08 (AS4 + 20 − 2) − 0.001 × 1000 − 0.05 (AV5 − 20) = 40.96.

1 − 0.001 − 0.05

14.2 Let CVt and AVt denote the cash (surrender) value and the account value, respectively, at the end of the tth month; CoIt is the cost of insurance in the tth month.

The effective monthly rate of credited interest is 0.054/12 = 0.0045. This is also the CoI interest rate. We have AV11 = CV11 + 500 = 1700.00, AV13 = CV13 + 125 = 1927.94, CoI12 = 0.002 × 50 000/1.0045 = 99.55, CoI13 = 0.003 × 50 000/1.0045 = 149.33, AV12 = 1.0045 (AV11 + 300(1 − W ) − 99.55 − 10) = 1898.96 − 301.35W, 239

240

Solutions for Chapter 14 AV13 = 1.0045 (AV12 + 300 × 0.85 − 149.33 − 10) = 2003.60 − 302.71W. Then 1927.94 = 2003.60 − 302.71W ⇒ W = 0.25.

14.3 First, we calculate the CoI and account values for the first two years as CoI1 = 0.0054 × 200 000/1.06 = 1018.87, CoI2 = 0.0060 × 200 000/1.06 = 1132.08, AV1 = 1.06 (5000 − 1018.87 − 100) = 4114.00, AV2 = 1.06 (4114.00 + 5000 − 1132.08 − 100) = 8354.84.

So the cash value at the end of the second year is expected to be CV2 = 0.93 × 8354.84 = 7770.00 .

Next, calculate the probability of surrender in the second year, say 1 |1 q(w) 60 . (w) 1 |1 q60 = (1 − 0.0034) (1 − 0.06) (1 − 0.0038) 0.06 = 0.055995.

Then the EPV of the second-year surrender benefit is 2 0.055995 × 7770.00 v7% = 380.01.

14.4 From the question, using the notation of AMLCR, we have the following information: FA = 100 000,

γt = 1.95,

ECt = 0.004 AVt −1 + 25,

q∗x+t −1 = 0.005,

AVt −1 = 49 500,

vq = v4% ,

itc = 5%.

So, if the ADB is based on the face value death benefit, we have AVt f = AVt−1 − ECt − q∗x+t −1 vq FA − AVt f ) (1 + itc ) ⇒ AVt f = 51 236.04 + 0.005048AVt f 51 236.04 = 51 496.00. ⇒ AVt f = 0.994952 So ADBtf = 100 000 − 51 496.00 = 48504.00, and CoItf = 0.005 v4% (ADBtf ) = 233.19.

Solutions for Chapter 14

241

If the ADB is based on the corridor factor, then

AVtc = AVt−1 − ECt − q∗x+t −1 vq (γt − 1) AVtc (1 + itc ) ⇒ AVtc = 51740.85 − 0.004796AVtc 51740.85 = 51493.90. ⇒ AVtc = 1.004796 So ADBtc = (0.95)(51 493.90) = 48 919.20, and CoItc = 0.005 v4% (ADBtc ) = 235.19.

The ADB is the greater of the two values calculated, so the corridor factor applies here, and we have AVt = 51 493.90, ADBt = 48 919.20 and CoIt = 235.19.

14.5 The account value immediately after premiums and deductions at time 5 is AV5+ = 2029 + 200 − 400 − 60 = 1769.

The EPV of the guaranteed term insurance is 100 000 A 1

45:20

= 100 000 × 0.02391 = 2391.29.

So the no lapse guarantee reserve is 2391.41 − 1769.00 = $622.29.

14.6 First, calculate AVt f from AVt f = 200 000 + (25 000 × 0.98) − 0.03 vq (500 000 − AVt f ) (1 + ic ), and note that here vq (1 + ic ) = 1. Then AVt f =

(200 000 + (25 000 × 0.98) − 0.03 vq 500 000)(1 + ic ) = 227552, 1 − 0.03

leading to an additional death benefit of $272 448.

Now calculate AVtc from AVtc = 200 000 + (25 000 × 0.98) − 0.03 vq (2.5 − 1) AVtc (1 + ic ), leading to AVtc =

(200 000 + (25 000 × 0.98))(1 + ic ) = 225 574.

1 + 0.03 × 1.5

The additional death benefit in this case is 1.5 AVtc = $338 361, and as this is the larger of the two additional death benefits calculated, this is the additional death benefit.

242

Solutions for Chapter 14

14.7 To prevent lapse, we want AV5 > 0. So if we consider the equation AV5 = (AV4 + P4 − EC4 − CoI4 )(1 + ic ), then by setting AV5 = 0 we see that we require a premium greater than P, where 0 = 5000 + 0.95P − 50 − 0.01 × 106 /1.045.

The minimum premium is therefore $4862.50.

14.8 (a) For years t = 1, 2, 3, the cost of insurance is CoIt = 150 000 q35+t−1 /1.04.

The values are 333.17, 354.81 and 377.88 for t = 1, 2, 3, respectively. We use these to project the account values as AV1 = 1.04 (4800 − 960 − 333.17) = 3647.10, AV2 = 1.04 (AV1 + 4800 − 240 − 354.81) = 8166.38, AV3 = 1.04 (AV2 + 4800 − 240 − 377.88) = 12 842.44. (b) The profit vector calculations are as follows.

(1) t

(2) AVt−1

(3) Pt

(4) Et

(5) It

(6) EDBt

(7) ESBt

(8) EAVt

(9) Prt

1 2 3

0.00 3647.10 8166.38

4800 4800 4800

1200 336 336

288.00 648.89 1010.43

231.22 238.00 245.01

114.54 1 331.28 12 823.18

3277.47 6523.31 0.00

264.78 667.40 572.63

Columns (1)–(3) are self-explanatory. Column (4) shows expenses, Et , being 25% of premium in year 1, and 7% in years 2 and 3.

Column (5), It , shows interest, calculated as It = 0.08 (AVt−1 + Pt − Et ).

Column (6) shows the expected death benefit, EDBt , calculated as 0.0015 (150 000 + AVt + 500), where the mortality rate does not vary with age and 500 is the expense amount on death.

Column (7) shows the expected cost of surrender. For example, in year 1 this is 0.1 × 0.9985 (AV1 − 2500),

Solutions for Chapter 14

243

as 10% of surviving policyholders surrender, paying a surrender charge of 2500.

Column (8) shows the expected account value, EAVt , required at the end of the year. For year 1 this is 0.9 × 0.9985 AV1 .

Column (9) shows the profit vector with Prt = AVt−1 + Pt − Et + It − EDBt − ESBt − EAVt , with the understanding that AV0 = 0. (c) Allowing for mortality and surrenders, the probabilities that a policy is in force at the start of the second and third policy years are respectively 0.8987 and 0.7178. Hence the profit signature is Π = (−800.00, 240.70, 599.76, 411.05) and the NPV at 10% is Π0 +

3

vt Πt = 245.21.

t=1

(d) From a first-principles calculation, the EPV of the premiums is 11 569 and so the profit margin is 245.21/11 569 = 2.12%.

14.9 As this is a Type B policy, the additional death benefit is constant, at $50 000.

The account value at time t = 11 months, AV11 , is 860 + 500 = 1 360. So

1 12 AV12 = AV11 + 300 − (0.2)(300) − 20 − (0.002)(v4% )(50 000) (1.0045) = 1 486.99.

Similarly,

1 12 AV13 = AV12 + 300 − (0.15)(300) − 10 − (0.003)(v4% )(50 000) (1.0045) = 1 589.60.

Deducting the second-year surrender penalty of 125 gives the cash value as 1 464.60.

244

Solutions for Chapter 14

14.10 (a) First, project the account values and the death benefit, assuming the policy remains in force for four years. In general, AVt = (AVt −1 + P − ECt − CoIt )(1 + itc ) and DBt = max(12 000, 1.5 AVt ).

The calculations are set out in the following table. t

AVt −1

P

ECt

CoIt

itc

AVt

DBt

1 2 3 4

0.00 2 969.20 6 094.78 9 437.40

3 000 3 000 3 000 3 000

120.00 111.88 124.38 137.75

25 25 25 25

4.0% 4.5% 5.5% 5.5%

2 969.20 6 094.78 9 437.40 12 949.75

12 000.00 12 000.00 14 156.09 19 424.63

Next, use AVt and DBt from this projection to profit test the insurer’s cash flows. At time t = 0, we have the pre-premium expenses of $200, giving Pr0 = −200. For t = 1, 2, 3, 4, the emerging profit at the year end for a policy in force at the start of the year is Prt = (AVt −1 + P − Et ) (1 + it ) − q65+t −1 DBt − p65+t −1 AVt .

The full profit vector calculation is tabulated below. t

AVt −1

P

Et

It

EDBt

EAVt

Prt

0 1 2 3 4

0.00 2 969.20 6 094.78 9 437.40

3 000 3 000 3 000 3 000

200.00 50.00 51.00 52.02 53.06

132.75 325.50 587.78 804.98

70.98 79.42 104.88 161.16

2 951.64 6 054.44 9 367.47 12 842.31

−200.00 60.14 109.84 158.18 185.84

The profit signature, Π, is the vector of emerging profits per policy issued, where Πt = t −1 p65 Prt for t = 1, 2, 3, 4, and Π0 = Pr0 . So Π = (−200.00, 60.14, 109.19, 156.20, 182.16) and NPV = Π0 +

4

t Πt v8% = 207.19.

t=1

(b) Given that the policyholder dies in the first year, the profit signature is Π = (−200, −8 917.25, 0, 0, 0) .

Solutions for Chapter 14

245

Note that the profit signature and profit vector are the same when there is no uncertainty about the year of death.

We see that there is no impact on Pr0 , as the value was not dependent on survival in any case. The second term is from the year 1 cash flows, given that the death benefit is paid in the first year: Π1 = 1.045 (3000 − 50) − 12 000 = −8 917.25.

The NPV is −$8 456.71. (c) The calculations are similar to those in part (a), but assuming the policyholder survives the term. For example Π1 = 1.045 (3000 − 50) − AV1 = 113.55, Π2 = 1.055 (AV1 + 3000 − 51) − AV2 = 148.92.

The profit signature, given survival to the end of the term, is Π = (−200, 113.55, 148.92, 193.14, 239.57) , and the NPV is $362.23. (d) (i) The calculations are similar to those in part (c), but assuming the policyholder survives to time 2, and then receives the full account value, AV2 , and there are no further cash flows. Using part (c), we have Π = (−200, 113.55, 148.92, 0, 0) , and the NPV is $32.82. (ii) Assuming the policyholder survives to time 2, and then receives 90% of the account value, we have Π2 = 1.055 (AV1 + 3 000 − 51) − 0.9 AV2 = 758.40, giving Π = (−200, 113.55, 758.40, 0, 0) , and the NPV is $555.35. (e) The profit margin, denoted pm, for policyholders who remain in force

246

Solutions for Chapter 14 throughout the contract is NPV given survival to the contract end PV Premiums NPV 362.23 = 3.38%. = = 3 000 ä4 8% 10 731.29

pm =

The profit margin for a policyholder who surrenders at time 2 is NPV given surrender at time 2 PV Premiums NPV NPV = = .

3 000 (1 + v8% ) 5 777.78

pm =

Setting this profit margin equal to 3.38% gives NPV = 0.0338 × 5 777.78 = 195.02, 1 k and so, since we also have NPV = 2k=0 Πk v8% , 2 , 195.02 = −200 + 113.55 v8% + Π2 v8%

giving Π2 = 338.12. Then, if α represents the surrender penalty, 338.12 = 1.055 (2 969.20 + 3 000 − 51) − (1 − α) AV2 , giving α = 3.1%. So a cash value of 96.9% of the account value would generate the 3.38% profit margin required from policyholders surrendering during the second policy year. (f) We see from part (a) that, ignoring surrenders, there will be a profit for the insurer, at the 8% risk discount rate, of $207.19 per policy. (The profit margin is 1.95%.) However, part (b) shows that early death is quite costly, as we would expect from a policy with a term insurance component. On the other hand, survivals generate a small profit in NPV terms, of $362.23 each. The mix of more severe losses from death benefit claims, which are assumed to be relatively rare, and smaller gains from survivals, which are more common, gives the average result found in part (a).

If the policyholder surrenders at time 2, we see from part (d) that the cash value determination is significant. If there is no surrender penalty, so policyholders take the full account value on early exit, the NPV is very small, at $32.82; the profit margin (which is a useful measure as it adjusts for fewer premiums) is 0.6% in this case, compared with 1.95% overall.

If the cash value is reduced to 90% of the account value, the NPV is much

Solutions for Chapter 14

247

stronger, at $555.35, representing a profit margin of 9.61%. At this level of penalty, the insurer gains a NPV greater than that earned from the full term survivors in part (c). It is not ideal to have surrenders generating significantly higher returns than those who stay – it creates perverse incentives for insurers to stimulate higher withdrawal rates, and indicates a lack of equitable treatment for leavers compared with stayers.

From part (e) we see that a cash value of 96.9% of the account value for the time 2 surrenders would generate the same profit margin for leaving policyholders as for those who stay throughout the term. This would appear to be more equitable. The NPV generated by the surrendering policyholders is less than for those staying for the full term, but is slightly more than the overall expected NPV, which allows for mortality but not for surrenders.

Solutions for Chapter 15

15.1 Note that the values in Table 15.10 of AMLCR have, conveniently, been ordered by size. (a) There are 54 values in Table 15.10 of AMLCR which are larger than 10.

Hence, the estimate of Pr[L0 > 10] is 54/1000 = 0.054. (b) The variance of the estimate in part (a) is p (1 − p)/1000, where p = Pr[L0 > 10]. This can be estimated by 0.054 (1 − 0.054)/1000 = 5.108 × 10−5 .

The 99.5% point of the N(0, 1) distribution is 2.5758. Hence, an approximate 99% confidence interval for p is !

0.054 ± 2.5758 5.108 × 10−5 , which is (0.036, 0.072). (c) Since there are 1000 simulated values of L0 , the estimate of the 99% point of the distribution is between the 10th and 11th largest values. We may estimate Q0.99 (L0 ) by taking the mid-point between these values, giving an estimate of (17.357 + 17.248)/2 = $17.30.

Note that there are other ways to determine an appropriate estimate between the 10th and 11th largest values. The smoothed empirical estimation of the p-quantile from an ordered sample of n values uses interpolation to 248

Solutions for Chapter 15

249

find the p(n + 1)th value, which in this case would be the 990.99th value, which is 17.36. (d) The estimate of CTE0.99 (L0 ) is the average of the ten largest simulated values of L0 . Hence CTE0.99 (L0 ) = (26.140 + 24.709 + · · · + 17.774 + 17.357)/10 = $21.46.

15.2 (a) The fund value at the end of the year is F1 = 97R, where R ∼ LN(0.09, 0.18).

Hence Pr[F1 < 100] = Pr[97R < 100] = Pr[log 97 + log R < log 100] & ' log R − 0.09 log 100 − log 97 − 0.09 < = Pr 0.18 0.18 = Pr[Z < −0.3308], where Z ∼ N(0, 1) = 0.37040. (b) The mean of F1 is E[F1 ] = 97 E[R] = 97 exp{0.09 + 0.5 × 0.182 } = $107.87. (c) For a standard normal random variable, Z, the lower 5% point is −1.64485. Hence ' & log R − 0.09 < −1.64485 = 0.05 Pr 0.18 ⇒ Pr[R < exp{−0.18 × 1.64485 + 0.09}] = 0.05 ⇒ Pr[R < 0.81377] = 0.05. (d) From part (c) we know that Pr[R > 0.81377] = 0.95. Hence Pr[97R > 97 × 0.81377] = 0.95 ⇒ Pr[100 − 97R < 100 − 97 × 0.81377] = 0.95 ⇒ Pr[max(100 − 97R, 0) < 100 − 97 × 0.81377] = 0.95 since 100 − 97 × 0.81377 > 0 ⇒ Pr[L0 < (100 − 97 × 0.81377) e−0.05 − 0.5] = 0.95 ⇒ Q0.95 (L0 ) = (100 − 97 × 0.81377) e−0.05 − 0.5 ⇒ Q0.95 (L0 ) = $19.54.

250

Solutions for Chapter 15 (e) (i) Let f denote the probability density function of R. The random variable L0 has the following distribution: ⎧ ⎪ ⎪ if R > 100/97, ⎨ −0.5 L0 = ⎪ ⎪ ⎩ (100 − 97R) e−0.05 − 0.5 if R ≤ 100/97.

Hence 100/97

(100 − 97x) e−0.05 f (x) dx

E[L0 ] = −0.5 + 0

−0.05

= −0.5 + 100 e

Pr[R < 100/97] − 97 e

−0.05

100/97

x f (x) dx 0

= $3.46. (ii) CTE0.95 (L0 ) is E[L0 | L0 > Q0.95 (L0 )] (as Q0.95 (L0 ) lies in the continuous part of the distribution of L0 ). Note first that Pr[L0 > Q0.95 (L0 )] ≡ Pr[R < 0.81377] = 0.05.

Then CTE0.95 (L0 ) = E[L0 + 0.5 | L0 > Q0.95 (L0 )] − 0.5 = e−0.05 E[(L0 + 0.5) e0.05 | L0 > Q0.95 (L0 )] − 0.5 = e−0.05 E[(L0 + 0.5) e0.05 | R < 0.81377] − 0.5 = e−0.05 E[max(100 − 97R, 0) | R < 0.81377] − 0.5 = e−0.05 E[100 − 97R | R < 0.81377] − 0.5 since 100 − 97R > 0 for R < 0.81377 0.81377

97e−0.05 −0.05 = 100 e − 0.5 − x f (x) dx Pr[R < 0.81377] 0

= $24.83. (f) The values are simulated using random N(0, 1) numbers to generate 100 different values for R, say R1 , R2 , . . . , R100 . Use these to generate 100 values for F1 and 100 values for L0 . (a) If n is the number of simulated values of F1 less than 100, then the estimate of Pr[F1 < 100] is n/100.

Solutions for Chapter 15

251

(b) The estimate of E[F1 ] is the average of the 100 simulated values of F1 . (c), (d) The 5th percentile of the distribution of R and the 95th percentile of the distribution of L0 can be found using spreadsheet functions. Note that the Excel PERCENTILE function uses a smoothed approach, so that the estimated 5th percentile is interpolated between the 5th and 6th values. (e)(i) The estimate of E[L0 ] is the average of the 100 simulated values of L0 . (e)(ii) The estimate of CTE0.95 (L0 ) is the average of the five largest simulated values of L0 .

15.3 Let P denote the annual premium. We use the following notation: It is the return on investments in the tth year.

Et is the total expense incurred at time t − 1 per policy in force at the start of the tth year. All the expenses at time 0 are included in E0 rather than E1 , so that E0 = 0.1P + 100; E1 = 0; Et = 0.06P for t = 2, 3, 4, 5.

Prt is the profit emerging at the end of the tth year per policy in force at the start of the year. Πt is t−1 p50 Prt . tV

is the reserve required at time t for a policy still in force at that time.

The profit, Prt , is calculated as follows: Pr0 = −E0 , Prt = ( t−1V + P − Et )(1 + It ) − 10 000 q50+t−1 − t V p50+t−1 for t = 2, 3, 4, Pr5 = ( 4V + P − E5 )(1 + I5 ) − 10 000 q54 − 20 000 p54 . (a) In this part, we have It = 0.08 for t = 1, 2, . . . , 5. The EPV of future profit, or NPV, as a function of P, is given by NPV = Pr0 +

5

vt t−1 p50 Prt ,

t=1

where v = 1/1.1. Requiring NPV = P/3 gives P = $3739.59.

252

Solutions for Chapter 15 5 , where (b) For this part, we first generate 500 simulated values of {1 + It }t=1 each (1 + It ) is independently generated from the LN(0.07, 0.13) distribution. For each set of simulated values of the investment returns, we calcu5 late {Prt }t=0 using P = $3740.

(i) We are interested in the probability of a loss in the final year, given that the contract is in force at the start of that year – that is, the probability that Pr5 is negative. Let N denote the number of simulations (out of 500) for which Pr5 is negative. Then N ∼ B(500, p), where p = Pr[Pr5 < 0], so that an estimate of p is p̂ = N/500. The standard ! error of this estimate is p (1 − p)/500, which can be estimated by ! p̂ (1 − p̂)/500.

In our projections, we find that 263 simulated values of Pr5 have a negative value, out of 500 simulations, so the estimate of p is 263/500 = 0.526, and an approximate 95% confidence interval for p is 0 p̂ ± 1.96 p̂ (1 − p̂)/500 , which is (0.482, 0.570). (ii) The formula for Pr5 is Pr5 = (1 + I5 )(15 000 + 0.94 × 3740) − 10 000 × 0.001797 − 20 000 × 0.998203.

Hence, the probability that Pr5 is negative can be calculated as

log 1.0792 − 0.07 Pr [1 + I5 < 1.0792] = Φ = 0.519.

0.13 The 95% confidence interval calculated in part (i) contains the true value of p, as we might expect given the high level of confidence. (iii) For each of the 500 simulations, we can calculate the NPV using a risk discount rate of 10%. In our calculations, 254 simulations have an NPV greater than 3740/3 and so meet the profit objective. So the estimated probability that the profit objective will be met is 254/500 = 0.508, and an estimated 95% confidence interval for this probability is 0 0.508 ± 1.96 0.508 (1 − 0.508)/500 , which is (0.464, 0.552).

253

Solutions for Chapter 15

15.4 (a) We assume the policy stays in force until at least the start of the final year.

Let Ft− denote the fund value at time t, t = 0, 1, . . . , 4, just before payment of the premium and deduction of the management charge. Then the management charge at time t, MCt , is 0.03(Ft− + 100). Ft− can be calculated recursively as follows: F0− = 0, and for t = 1, 2, 3, 4, Ft− = 1.08 × (1 − 0.03)(Ft−1− + 100).

The table of projected fund values and management charges is as follows.

We also show the implied additional death benefit (ADBt ) for the insurer, which is the difference between the death benefit ($500) and the end-year fund, if positive. t

Ft−1 + P before charges

MCt

Ft

ADBt

1 2 3 4 5

100.00 204.76 314.51 429.48 549.92

3.00 6.14 9.44 12.88 16.50

104.76 214.51 329.48 449.92 576.10

395.24 285.49 170.52 50.08 0.00

(b) The profit test table is shown below. t

MCt

Et

It

EBt

Prt

Πt

1 2 3 4 5

3.00 6.14 9.44 12.88 16.50

2.00 4.10 6.29 8.59 11.00

0.06 0.12 0.19 0.26 0.33

0.79 0.80 0.55 0.19 0.00

0.27 1.37 2.79 4.37 5.83

0.27 1.37 2.77 4.33 5.76

Note that we are told that the insurer does not establish reserves for this contract.

MCt is from the policyholder fund projection in part (a), and is an item of income for the insurer.

Et denotes the insurer’s expenses incurred at the start of the tth year.

254

Solutions for Chapter 15 It is the interest earned on the insurer’s funds, It = 0.06 (MCt − Et ).

EBt is the expected cost of benefits. Note that the insurer only pays for the excess over the policyholder’s funds, which is ADBt from part (a). There is no projected liability for the maturity benefit, as in the projection in part (a), F5 > 500, so the guaranteed minimum payment has no projected cost.

Prt is the emerging profit at time t for a policy in force at time t − 1, so Prt = MCt − Et + It − EBt .

The profit signature is Πt = t−1 p60 Prt . (c) The calculations in parts (a) and (b) do not show any negative cash flows for the insurer and so it may appear that it is unnecessary to set up reserves for this contract. However, these calculations are based on assumptions, most notably concerning fund growth, which may or may not be realized.

It would be prudent for the insurer to carry out a stochastic analysis, possibly using simulation, to determine the extent of any liabilities arising from different scenarios and then to decide if reserves are required to cover these liabilities. (d) Suppose, for simplicity, that interest rate risk is the only source of future uncertainty when the policy is issued. In practice all assumptions – expenses, mortality, growth of the insurer’s fund – are uncertain, but uncertainty about future interest rates is likely to have the greatest impact. We would choose an appropriate stochastic model for future interest rates, possibly the independent lognormal model. We would then generate a large number of future interest rate scenarios and, for each scenario, calculate the EPV of the future loss, after paying any initial expenses. Let Li denote the EPV of the future loss calculated from the ith scenario. The 99% quantile reserve required at time 0 is then the 99th percentile of the distribution of L as given by the simulated values {Li }, provided this is positive, and zero otherwise. The 99% CTE reserve required at time 0 is E[L | L > Q0.99 (L)], which would be estimated by taking the average of the simulated values {Li } which are greater than the (estimated) value of the 99% quantile reserve.

Solutions for Chapter 15

255

(e) (i) There will be a payment under the guarantee if the fund at the end of the year is less than $500. Let the accumulation factor for the fifth year be denoted 1 + i5 , so that 1 + i5 ∼ LN(0.09, 0.18).

The fund just before the start of the year is $485, so the fund at the end of the year will be $(1 + i5 ) 0.97 (485 + 100). Hence, the required probability is Pr[1 + i5 < 500/(0.97 × 585)] = 0.114. (ii) The insurer’s liability at the end of year 5 is the same, max(500 − F5− , 0), whether the policyholder survives or dies in the year. The management charge at the start of the year is given by MC4 = 0.03(485 + 100) = $17.55, the insurer’s expenses at the start of the year are 0.02 (485 + 100) = $11.70, and F5− is given by F5− = 0.97 (485 + 100)(1 + i5 ).

Hence, the present value of the future loss at time 4, L4 , is given by L4 = max(500 − F5− , 0)/1.06 + 11.70 − 17.55.

We know from part (i) that there is a probability of 0.114 (> 0.01) that 500 − F5− will be positive. Hence, the 99% point of the distribution of L4 , Q0.99 (L4 ), will be the 99% point of the random variable (500 − 0.97 (485 + 100)(1 + i5 ))/1.06 − 5.85.

This is equal to 500 − 0.97 (485 + 100) e0.09+0.18z0.01 /1.06 − 5.85, where z0.01 (= −2.3263) is the 1% point of the N(0, 1) distribution, so that e0.09+0.18z0.01 is the 1% point of the distribution of 1 + i5 . This gives Q0.99 (L4 ) = 500 − 0.97(485 + 100) e0.09−0.18×2.3263 /1.06 − 5.85 = $80.50.

256

Solutions for Chapter 15

15.5 (a) First project the policyholder’s fund assuming the policy remains in force throughout. Let At denote the tth allocated premium (paid at time t − 1), let Ft− denote the fund at time t before the deduction of the management charge, let Ft denote the fund at time t after the deduction of the management charge, and let MCt denote the management charge deducted at time t for a policy in force at time t − 1.

Then for the first premium we have A1 = 0.25 × 0.95 × 750 = 178.13, and for t = 2, . . . , 5, At = 1.025 × 0.95 × 750 = 730.31.

For t = 1, 2, . . . , 5 Ft− = (At + Ft−1 ) 1.065,

MCt = 0.01 Ft− , and Ft = 0.99 Ft− .

This gives the table of projected fund values shown below. t

Ft−1

At

MCt

Ft

1 2 3 4 5

0.00 187.81 968.02 1 790.64 2 657.96

178.13 730.31 730.31 730.31 730.31

1.90 9.78 18.09 26.85 36.09

187.81 968.02 1 790.64 2 657.96 3 572.43

The profit emerging at time 0, Pr0 , arises only from the insurer’s initial expenses, so that Pr0 = −(150 + 0.1P) = −$225.

In each of the five years, assuming the contract is in force at the start of the tth year, the insurer receives income from the unallocated premium, say UPt = 750 − At at the start of the year, and from the management charge at the end of the year, as well as from interest earned during the year at 5.5%. Outgo comprises the incurred expenses, the expected cost of death benefits and the expected cost of the maturity benefit. So, the profit table is as follows:

257

Solutions for Chapter 15 t (1)

UPt (2)

Et (3)

It (4)

MCt (5)

EBt (6)

Prt (7)

Πt (8)

0 1 2 3 4 5

0.00 571.88 19.69 19.69 19.69 19.69

225.00 0.00 83.75 83.75 83.75 83.75

31.45 −3.52 −3.52 −3.52 −3.52

1.90 9.78 18.09 26.85 36.09

3.40 2.70 1.78 0.56 356.60

−225.00 601.83 −60.51 −51.27 −41.29 −388.10

−225.00 601.83 −54.40 −41.43 −29.98 −281.34

The expenses at time 0 have been included in the calculation of Pr0 and so are not included in the calculation of Pr1 .

The cost to the insurer of the death benefit is the cost of the extra amount, if any, above the bid value of the units; there is a cost to the insurer in the final year from maturities of 10% of the bid value of the units. These costs are all included in the EBt column.

For t ≥ 1, Prt is the expected profit emerging at time t given that the contract is in force at time t − 1. The profit signature element for time t, Πt , is the expected profit emerging at time t per policy issued. We must allow for the probability that the policyholder has not died or surrendered before time t − 1 to find Πt from Prt , so Π0 = Pr0 , Πt = t−1 p50 (0.9t−1 ) Prt for t = 1, 2, 3, 4, and Π5 = 4 p50 (0.93 ) Pr5 .

The EPV of the profit is 5

vt Πt = $42.30

t=0

and the EPV of the premiums is P

5

vt 0.9min(t−1,3) t−1 p50 = $2704.75,

t=1

where v is calculated at the risk discount rate, so that v = 1/1.085. Hence, the profit margin is 42.30/2704.75 = 0.0156 or 1.56%.

258

Solutions for Chapter 15 (b) (i) The deterministic profit test in part (a) shows negative cash flows for the insurer at the end of years 2, 3, 4 and 5. Reserves should be established to meet these future liabilities. (ii) The effect of introducing reserves is shown in the revised profit test table below. t (1)

UPt (2)

t−1V

(3)

Et (4)

It (5)

MCt (6)

EBt (7)

EVt (8)

0 1 2 3 4 5

571.88 19.69 19.69 19.69 19.69

0 400 400 400 375

225.00 0.00 83.75 83.75 83.75 83.75

31.45 18.48 18.48 18.48 17.10

1.90 9.78 18.09 26.85 36.09

3.40 2.70 1.78 0.56 356.60

359.56 359.52 359.47 374.39 0.00

Prt −225.00 242.26 1.97 11.25 6.32 7.52

To understand these calculations, consider, for example, the effect of the reserve of $400 required at the start of the fourth year. This has to be set up at the end of the third year, for policies which are continuing at that time. The expected cost, given that the policy is in force at the start of the third year, is EV3 in the table above, where EV3 = 3V × p52 × 0.9, allowing for survival and for the probability that the policyholder does not surrender the contract.

The effect on Pr3 is to reduce it by EV3 . The effect of this reserve on Pr4 is to increase it by 1.055 × 400, since it is assumed the reserve will earn interest at 5.5% throughout the fourth year.

The overall effect of introducing reserves is to reduce the profit margin to 0.0051 or 0.51%. The negative cash flows in years 2 to 5 have been eliminated by setting aside, as a reserve, some of the profit that emerged at time 1. Note that Pr1 reduces from $601.83 in part (a) to $242.26 in this table. The reserve is assumed to earn interest at 5.5% per year rather than the (higher) risk discount rate of 8.5% per year.

This reduces the insurer’s NPV and hence the profit margin.

5 (c) Let {zt }t=1 denote the random standard normal deviates given in the ques5 tion. The simulated rates of return in successive years, {1 + it }t=1 , are calculated as

Solutions for Chapter 15

259

1 + it = exp{0.07 + 0.2zt } for t = 1, 2, . . . , 5. These rates of return are then used in place of the fixed rate 1.055 for projecting the growth of the policyholder’s fund and hence the calculation of the management charge, the death benefit and the cost to the insurer of the maturity benefit.

With these changes, the calculations follow as in part (b)(ii).

The revised profit margin is −0.0143 or −1.43%.

Solutions for Chapter 16

16.1 (a) Under the risk neutral measure the probability of an upward movement is 1.04 − 0.8 = 0.5333 1.25 − 0.8 and the probability of a downward movement is 0.4667. (b) The option price is the expected present value under the risk neutral probability measure. The option pays $10 if the stock price rises twice or falls twice, and the risk neutral expected present value is therefore 10 1.04−2 0.53332 + 0.46672 = 4.6433.

16.2 Formulae (16.8) and (16.9) in AMLCR give the expressions we require to answer this question, namely c(t) = St Φ(d1 (t)) − K e−r(T −t) Φ(d2 (t)), log(St /K) + (r + σ 2 /2)(T − t) , √ σ T −t √ d2 (t) = d1 (t) − σ T − t. d1 (t) =

Then d d d c(t) = St Φ(d1 (t)) + Φ(d1 (t)) − Ke−r(T −t) Φ(d2 (t)) dSt dSt dSt d d d1 (t) + Φ(d1 (t)) − Ke−r(T −t) φ (d2 (t)) d2 (t), = St φ (d1 (t)) dSt dSt √ where φ (x) = exp{−x2 /2}/ 2π is the standard normal density function.

260

Solutions for Chapter 16

261

Now d 1 d d1 (t) = d2 (t) = √ dSt dSt St σ T − t so that St φ (d1 (t))

φ (d1 (t)) d . d1 (t) = √ dSt σ T −t

Next, φ (d2 (t)) =

exp{−d2 (t)2 /2} √ 2π

√ 2 exp{− d1 (t) − σ T − t /2} = √ 2π √ σ2 (T − t) , = φ (d1 (t)) exp d1 (t) σ T − t − 2 and as √ σ2 σ2 (T − t) = log(St /K) + (r + σ 2 /2)(T − t) − (T − t) d1 (t) σ T − t − 2 2 = log(St /K) + r(T − t), we have φ (d2 (t)) = φ (d1 (t)) Thus Ke−r(T −t) φ (d2 (t))

St exp{r(T − t)}.

K

φ (d1 (t)) d d2 (t) = √ dSt σ T −t

and hence d c(t) = Φ(d1 (t)). dSt 16.3 (a) Under the Q-measure, Sn = S0 ur d n−r for r = 0, 1, 2, . . . , n (i.e. r upward stock price movements and n − r downward ones) with probability

n (1 − q)r qn−r , r where q=

u − er . u−d

262

Solutions for Chapter 16 Thus EQ [Sn ] =

n

r=0

= S0

S0 ur d n−r

n

n r=0

r

n (1 − q)r qn−r r

(u (1 − q))r (d q)n−r

= S0 (u (1 − q) + d q)n r n u − er e −d +d = S0 u u−d u−d rn = S0 e .

(Note that the second line is simply the binomial expansion of the third.) (b) Under the Q-measure in the Black–Scholes–Merton model, Sn /S0 has a √ lognormal distribution with parameters (r − σ 2 /2)n and σ n. We know that if X ∼ LN(μ, σ ) then E[X] = exp{μ + σ 2 /2} (see Appendix A in AMLCR). From this it follows that E[Sn /S0 ] = exp{(r − σ 2 /2)n + σ 2 n/2} = exp{rn}, i.e. E[Sn ] = S0 ern .

16.4 (a) Consider an investment of $a in zero-coupon bonds and $100b (i.e. b units) in the security. To construct the replicating portfolio at time t = 0 we require a e0.06 + 110 b = 20 (i.e. the payoff is replicated if the security price goes up) and a e0.06 + 90 b = 0 (i.e. the payoff is replicated if the security price goes down). Differencing these identities gives b = 1, and hence a = −90 e−0.06 = −84.76.

Thus, the replicating portfolio comprises one unit of the security and −$84.76 in zero-coupon bonds, and hence the option price is $100−$84.76 = $15.24.

Solutions for Chapter 16

263

(b) Consider the situation at time t = 1 if the security price has increased to 110. At time 2 the security price will be either 110 × 1.1 = 121, if the price moves up again, or 110×0.9 = 99, if the price moves down in the next time unit. If $au is invested in zero-coupon bonds and bu units of the security are purchased, then we require au e0.06 + 121 bu = 20, au e0.06 + 99 bu = 0, since a security price of S2 = 99 is less than S0 , giving a payoff of $0 under the option. Differencing these identities gives (121 − 99) bu = 20, so that bu = 10 11 and hence au = −99 bu e−0.06 = −84.76.

Hence the replicating portfolio at time t = 1 has value 110 bu +au = $15.24 if the security price goes up in the first time period.

Now consider the situation at time t = 1 if the security price has decreased to 90. At time t = 2 the security price will be either 99 or 81. Thus, the security price will be below S0 = 100, and hence the option will be worth 0 at time t = 2. The replicating portfolio at time t = 1 therefore consists of $0 in zero-coupon bonds and 0 units of the security. (You can also see this by setting up equations.)

To find the replicating portfolio at time t = 0 we consider an investment of $a in zero-coupon bonds and $100 b in the security. Then we require a e0.06 + 110 b = 15.24, a e0.06 + 90 b = 0, so that b = 15.24/20 = 0.7621 and hence a = −90 b e−0.06 = −64.59.

The option price is thus 100 × 0.7621 − 64.59 = $11.61.

264

Solutions for Chapter 16

16.5 (a) Under the P-measure the payoff under Option A is 3 2 1

with probability with probability with probability

0.52 = 0.25, 2 × 0.52 = 0.5, 0.52 = 0.25.

Hence the EPV of the payoff is e−2×0.04879 (3 × 0.25 + 2 × 0.5 + 1 × 0.25) = 2 e−0.09758 = 1.81.

Under the P-measure the payoff under Option B is 1 2 3

with probability with probability with probability

0.52 = 0.25, 2 × 0.52 = 0.5, 0.52 = 0.25.

Hence the EPV of the payoff is e−2×0.04879 (1 × 0.25 + 2 × 0.5 + 3 × 0.25) = 2 e−0.09758 = 1.81. (b) We can price the options using the same approach as calculating the expected present values in part (a), except that we replace the true probability of up and down moves with the artificial Q-measure values. From equation (16.2) in AMLCR we see that the Q-measure ‘probability’ of an up movement in each time period is er − d = 0.4, u−d where u is the proportionate increase in the security price on an up move, which is 1.2 in this case, and d is the proportionate decrease on a down move, 0.95 in this case. The option prices are then Option A: e−2r 3 × 0.42 + 2 × (2 × 0.4 × 0.6) + 1 × 0.62 = 1.633, Option B: e−2r 1 × 0.42 + 2 × (2 × 0.4 × 0.6) + 3 × 0.62 = 1.995. (c) We price options using the principle of replication. As the payoffs differ under the two options, the values of the replicating portfolios must differ, firstly at time t = 1, then at time t = 0.

Intuitively, Option A is less expensive because its payoff moves in the same direction as the stock, and the upside move of the stock is bigger, relatively, than the downside move; when we hedge, we benefit from the extra return on the upside. The starting replicating portfolios for the two options are as

265

Solutions for Chapter 16

follows, where a is the value of the risk free asset, and b is the value of the security, at time t = 0: Option A: a = −2.18 and

b = 3.81,

Option B: a = 5.81 and

b = −3.81.

16.6 (a) Under the risk neutral measure the probability of an upward movement is 2 1.1 − 0.8 = 1.25 − 0.8 3 and the probability of a downward movement is 1/3. (b) Consider the situation at time t = 1 if the security price has increased to 125. If $au is invested in zero-coupon bonds and bu units of the security are purchased, then we require 1.1 au + 156.25 bu = 1, 1.1 au + 100 bu = 2, giving bu = −1/56.25 = −0.01778 and au = (2 − 100bu )/1.1 = 3.4343.

Thus, the replicating portfolio has value Pu = au + 125 bu = 1.2121.

Next, if at time t = 1 the security price has decreased to 80, and if $ad is invested in zero-coupon bonds and bd units of the security are purchased, then we require 1.1 ad + 100 bd = 2, 1.1 ad + 64 bd = 0, giving bd = 2/36 = 0.05556 and ad = −64bd /1.1 = −3.2323.

Thus, the replicating portfolio has value Pd = ad + 80 bd = 1.2121, which is the same as Pu .

Thus, the amount required at time 1 is 1.2121 regardless of the movement in the security price in the first time period. We can replicate this by investing $1.2121/1.1 in cash, so D0 = $1.1019.

266

Solutions for Chapter 16 (c) The hedging strategy at time t = 0 is to hold $1.1019 of the risk free asset only, i.e. do not hold any of the security.

The hedging strategy at time t = 1 if S1 = 125 is to hold $3.4343 in the risk free asset and −0.01778 units (i.e. a short position) in the security. On the other hand, if S1 = 80, the hedge portfolio at time t = 1 comprises 0.05556 units of the security, and −$3.2323 (a short position) in the risk free asset. (d) The hedging strategy at time t = 0 is unusual in that it involves holding the risk free asset only. Although different replicating portfolios are required at time t = 1 depending on the value of S1 , these replicating portfolios have the same value at time t = 1, so the time t = 1 liability is certain. We always hedge a certain liability with the risk free investment.

16.7 (a) The put-call parity formula (see equation (16.11) in AMLCR) is c(t) + K e−r(T −t) = p(t) + St .

We have t = 0, T = 12 , r = 0.07, S0 = 400, K = 420 and c(0) = 41. The price of the put option is thus p(0) = c(0) + K e−r/2 − S0 = 41 + 420 e−0.035 − 400 = $46.55. (b) The implied volatility is the value of σ implied by the price of the European call option. Inserting the parameter values from part (a) into formula (16.8) in AMLCR we have c(0) = 41 = 400 Φ(d1 (0)) − 420 e−0.035 Φ(d2 (0)), where d1 (0) = and d2 (0) = d1 (0) − σ

log(4/4.2) + 12 (0.07 + σ 2 /2) ! σ 1/2

!

1/2.

This can only be solved numerically for σ . Using Solver in Excel we find that σ = 38.6%. (c) The delta of the option at time t is d c(t) , dS(t)

267

Solutions for Chapter 16

which, from Exercise 16.2, is Φ(d1 (t)). Thus, with t = 0 and σ as in part (b), we have d1 (0) =

log(4/4.2) + 12 (0.07 + 0.3862 /2) = 0.08595 !

0.386 1/2

and Φ(0.08595) = 0.5342, so the delta of the option is 53.42%. (d) Formula (16.8) in AMLCR tells us that 41 = 400 Φ(d1 (0)) − 420 e−0.035 Φ(d2 (0)) and the right-hand side gives the self-financing replicating portfolio at time 0 as 400 Φ(d1 (0)) = $213.6981 −420 e

−0.035

Φ(d2 (0)) = −$172.6982

in stock, in bonds.

The total holding for 10 000 units of the call option is then $2 136 981 in stock, i.e. 5342.45 units of stock, and −$1 726 982 in bonds, i.e. a short holding of 17 270 bonds with face value $100.

Solutions for Chapter 17

17.1 (a) The single premium is P = $100 and the value of the GMMB is ( ) + −10r 11 π(0) = 0.959 10 p60 EQ S e , 0.85P − P(1 − m) 10 0 where the term 0.959 allows for withdrawals at the end of years 1 to 9, and m = 0.02 so that the term (1 − m)11 allows for both the front-end expense loading and the annual management charge. We can rewrite this as &

+ ' 0.85 −10r − S e . π(0) = 0.959 10 p60 P (1 − m)11 EQ 10 0 (1 − m)11 Using formula (16.10) in AMLCR, with K = 0.85/(1 − m)11 , t = 0 and T = 10, we have &

+ ' 0.85 0.85 e−10r −10r − S Φ (−d2 (0)) − S0 Φ(−d1 (0)), EQ e = 10 0 11 (1 − m) (1 − m)11 where S0 = 1, d1 (0) =

log (1 − m)11 /0.85 + 0.04 + 0.22 /2 10 = 0.85427 √ 0.2 10

and

√ d2 (0) = d1 (0) − 0.2 10 = 0.22182.

Hence

& EQ 0

−10r

e

0.85 − S10 (1 − m)11

and π(0) = $4.61.

268

+ ' = 0.09685

269

Solutions for Chapter 17 (b) The value of annual risk premiums of c is m = 0.0204.

1−m Equating this to π(0) gives c = 0.68 or 0.68% of the fund. c ä60:10 ∗ = 6.7562c, where i∗ = i

(c) At time 2 years we have S2 = 0.95. The value of the GMMB is &

+ ' 0.85 7 11 Q −8r − S10 0.95 8 p62 P (1 − m) E0 e (1 − m)11 and

&

−8r EQ e 2

0.85 − S10 (1 − m)11

+ ' =

0.85 e−8r Φ (−d2 (2)) − S2 Φ(−d1 (2)), (1 − m)11

where d1 (2) = and

log (1 − m)11 S2 /0.85 + 0.04 + 0.22 /2 8 = 0.65230 √ 0.2 8 √ d2 (2) = d1 (0) − 0.2 8 = 0.08661.

Hence

& EQ 0

−8r

e

0.85 − S10 (1 − m)11

+ ' = 0.11456

and the value of the GMMB is $6.08.

17.2 (a) The death benefit payable at time t, conditional on death in the previous month, is max (P, 1.05 Ft ), where P = $10 000 is the single premium, and Ft = P e−0.03t St (with S0 = 1) since there is a management charge of 3% per year, deducted daily. We can write this death benefit as 0.03t + −0.03t e − St .

1.05 Ft + max(P − 1.05 Ft , 0) = 1.05 Ft + 1.05 P e 1.05 As the fund will provide Ft at time t, we need to price the additional benefit payable on death, of 0.03t + e 0.05 Ft + 1.05 P e−0.03t − St 1.05 at time t. To value this benefit, use equation (17.3) in AMLCR, where in this example &

0.03t + ' Q −rt −0.03t e v(0, t) = E0 e − St 0.05 Ft + 1.05 P e .

1.05

270

Solutions for Chapter 17 Consider the first part: $ % " " −rt # −rt # −0.03t P St e−rt = 0.05 e−0.03t P EQ = EQ EQ 0 0.05 Ft e 0 0.05 e 0 St e rt and we know that EQ 0 [St ] = S0 e , from the risk neutral property of the Q-measure, so that " −rt # EQ 0.05 Ft = 0.05 P e−0.03t .

0 e

The price at time 0 of this component of the death benefit is P

120

0.05 e−0.03t t−1 | 1 q60 = 24.1436.

12 12

t=1

The price of the second component of the death benefit, conditional on the benefit being paid at time t, is & 0.03t + ' −rt −0.03t e − S 1.05 P e e , EQ t 0 1.05 which can be found as

0.03t e e−rt Φ (−d2 (0, t)) − S0 Φ(−d1 (0, t)) , 1.05 P e−0.03t 1.05 where d1 (0, t) =

log(1.05 e−0.03t ) + (r + σ 2 /2) t , √ σ t

where r = 0.04 and σ = 0.25, and

√ d2 (0, t) = d1 (0, t) − σ t .

The price of this second component of the death benefit is then P

120

e−rt Φ (−d2 (0, t)) − 1.05 e−0.03t S0 Φ(−d1 (0, t))

| q60

t−1 1 12 12

t=1

= 83.6097.

Hence the price of the death benefit at issue is 24.1436 + 83.6097 = $107.75. (b) The value of the risk premium deductible continuously from the fund (as part of the m = 3% per year management charge), at a rate c per year, is P c ā60:10 δ =m = 8.4465 c P. Equating this to the benefit cost of $107.75 gives c = 0.128% of the fund.

271 √ 17.3 (a) Under the P-measure, S10 /S0 ∼ LN(10μ, σ 10), where μ = 0.08 and σ = 0.25. Setting S0 = 1 we have Solutions for Chapter 17

F10 = 0.9710 × 100 000 × S10 as there is a 3% management charge at the start of each year. The GMMB matures in the money if F10 < 100 000. The required probability is thus $ % " # Pr 0.9710 S10 < 1 = Pr log S10 < −10 log 0.97 & ' −10 log 0.97 − 10μ = Pr Z < , where Z ∼ N(0, 1) √ σ 10 = Φ(−0.6266) = 0.26545. √ (b) Under the Q-measure, S10 /S0 ∼ LN(10 (r −σ 2 /2), σ 10), where r = 0.04.

Proceeding exactly as in part (a), the required probability is ⎤ ⎡ ⎢⎢⎢ −10 log 0.97 − 10 r − σ 2 /2 ⎥⎥⎥ ⎥⎥⎥ = Φ (0.2746) = 0.60819. ⎢ Pr ⎢⎢⎣Z < √ ⎦ σ 10 (c) The EPV of the option payoff is ( ) + EP e−10r 100 000 − 0.9710 × 100 000 S10 , which can be written as

( ) + 100 000 × 0.9710 × e−10r EP 0.97−10 − S10 .

Let f denote the probability density function of S10 under the P-measure (we still have S0 =1). Then P

(

E

0.97

−10

− S10

+

)

0.97 −10

=

0.97−10 − x f (x) dx

0

= 0.97−10 Pr[S10 ≤ 0.97−10 ] −

0.97 −10

x f (x) dx.

0

From Appendix A of AMLCR we can write 0.97 −10

10 σ 2 x f (x) dx = exp 10 μ + 2

0

log 0.97−10 − 10 μ − 10 σ 2 Φ √ σ 10

= exp{1.1125} Φ (−1.4172) = 0.23791,

272

Solutions for Chapter 17 and from part (a) we know that Pr[S10 ≤ 0.97−10 ] = 0.26545. Hence the EPV of the option payoff under the P-measure is $6033. (d) The price of the option is % $ −10r 10 max 100 000 − 0.97 × 100 000 S , 0 EQ e 10 0 which can be written as

( ) + −10r −10 0.9710 × 100 000 EQ − S e .

0.97 10 0

Now

( ) + −10r −10 EQ − S e = 0.97−10 e−10r Φ(−d2 (0)) − S0 Φ(−d1 (0)), 0.97 10 0

where d1 (0) = and

Thus

log 0.9710 + r + σ 2 /2 10 = 0.51597 √ σ 10

√ d2 (0) = d1 (0) − σ 10 = −0.27460. ( ) + −10r −10 EQ − S e = 0.24991 0.97 10 0

and the price of the option is $18 429. (e) Using the P-measure ignores the fact that the guarantee risk is non-diversifiable. For diversifiable risk, the P-measure expectation will be close to the true cost, provided enough contracts are sold for the diversification benefit from the central limit theorem. For non-diversifiable risk, the central limit theorem does not apply, and the P-measure expectation may be a long way from the true cost. The use of the Q-measure indicates that what we are valuing is not an expected value in the conventional sense, but the cost of replicating the option payoff. This is achieved by taking the EPV under the Q-measure. (f) We start by simulating 1000 values from the standard normal distribution.

In Excel, this can be done using the random number generation tool. Let zi be the ith such random number. We then calculate our ith simulated value of S10 as 3 √ 4 S10,i = 100 000 exp 10 (r − σ 2 /2) + zi σ 10 .

Solutions for Chapter 17

273

The simulated fund value at time 10 is then F10,i = 0.9710 S10,i and the simulated payoff is hi (10) = max(100 000 − F10,i , 0).

The estimate of the price of the option is then the average of the present values (at the risk free rate) of the simulated payoffs, 1 −0.04×10 e hi (10).

1000 i=1 1000

The result will depend on the random numbers used; our calculations give an estimate of $18 385, which is close to the true price calculated in part (d).

17.4 (a) The policyholder’s fund value at time t years is Ft = 100 × 0.97 × 0.99t St , where S0 = 1. As F5 = 110 we have S5 = 1.1/(0.97 × 0.995 ) = 1.19246.

The value at time 5 of the original guarantee is $ % −5r (100 − F10 )+ EQ 5 e &

+ ' 1 10 Q −5r = 100 × 0.97 × 0.99 E5 e − S10 0.97 × 0.9910 and & EQ 5

+ ' 1 − S10 e 0.97 × 0.9910 e−5r = Φ (−d2 (5, 10)) − S5 Φ (−d1 (5, 10)) , 0.97 × 0.9910 −5r

where d1 (5, 10) =

log(S5 × 0.97 × 0.9910 ) + (r + σ 2 /2) 5 = 0.93432 √ σ 5

since r = 0.05 and σ = 0.18, and

√ d2 (5, 10) = d1 (5, 10) − σ 5 = 0.53183.

Thus the value of the original guarantee is 100 e−5r Φ (−0.53183) − 0.995 × 1.1 Φ(−0.93432) = $4.85.

274

Solutions for Chapter 17 Under the reset guarantee, the guarantee changes to 110. The value at time 5 of this guarantee is $ % −10r (110 − F15 )+ EQ 5 e &

+ ' 1.1 −5r − S e = 100 × 0.97 × 0.9915 EQ 15 5 0.97 × 0.9915 and & EQ 5

+ ' 1.1 − S15 e 0.97 × 0.9915 1.1 e−10r Φ (−d2 (5, 15)) − S5 Φ (−d1 (5, 15)) , = 0.97 × 0.9915 −5r

where d1 (5, 15) = and

log S5 × 0.97 × 0.9915 /1.1 + (r + σ 2 /2)10 = 0.98645 √ σ 10 √ d2 (5, 15) = d1 (5, 15) − σ 10 = 0.41724.

Thus the value of the reset guarantee is 100 1.1 e−10r Φ (−0.41724) − 0.9910 × 1.1 Φ (−0.98645) = $6.46. (b) The threshold value has to be found by numerical methods. We calculate the values of the original and reset guarantees as in part (a), except we set the value of S5 to be (1 + x)/(0.97 × 0.995 ). For a given value of x, we can calculate the value of the guarantees. Solving numerically, for example using Excel Solver, we determine that x = 0.03433, so that F5 = 103.43 and each guarantee has value $6.07.

17.5 (a) The single premium is P = $100 000 and the value of the GMMB is % $ −5r (100 000 − F5 )+ ) , π(0) = 5 p60 EQ 0 e where F5 = 100 000 (1 − 0.0025)60 S5 , with S0 = 1. Now % $ −5r (100 000 − F5 )+ ) EQ 0 e $ % −5r = 100 000 × 0.997560 EQ (0.9975−60 − S5 )+ 0 e

Solutions for Chapter 17

275

and $ % −5r (0.9975−60 − S5 )+ EQ 0 e = 0.9975−60 e−5r Φ(−d2 (0)) − S0 Φ(−d1 (0)), where log(0.997560 ) + (r + σ 2 /2) 5 = 0.44679 √ σ 5

d1 (0) =

since r = 0.05 and σ = 0.2, and

√ d2 (0) = d1 (0) − σ 5 = −0.00042.

Thus π(0) = 5 p60 100 000 e−5r Φ(0.00042) − 0.997560 Φ(−0.44679) = 0.97874 × 10 769.16 = 10 540.21.

The value of risk premiums of c per month deducted from the fund is cP

59

0.9975t t/12 p60 = 55.26545c P,

t=0

and setting this equal to π(0) gives c = 0.19% of the fund. (b) (i) We illustrate the calculation for time 2 months, i.e. t = 16 years. At this time, the cost of the option is $ 5 % Q −4 6 r (100 000 − F5 )+ , 4 5 p60 16 E1/6 e 6

which can be written as $ 5 % e−4 6 r (0.9975−60 − S5 )+ .

100 000 × 0.997560 4 5 p60 16 EQ 1/6 6

Now $ 5 % e−4 6 r (0.9975−60 − S5 )+ EQ 1/6 = 0.9975−60 e−4 6 r Φ(−d2 ( 16 , 5)) − S1/6 Φ(−d1 ( 16 , 5)), 5

where d1 16 , 5 =

log(S1/6 × 0.997560 ) + (r + σ 2 /2) 4 56 = 0.35264 0 σ 4 56

276

Solutions for Chapter 17 and d2 16 , 5 = d1 16 , 5 − σ

0 4 56 = −0.08706.

Thus, the cost of the option is 100 000 4 5 p60 16 e−4 6 r Φ(0.08706) − 0.997560 S1/6 Φ(−0.35264) , 5

6

so that the stock part of the hedge is 100 000 4 5 p60 16 × 0.997560 S1/6 Φ(−0.35264) = −$29 528 6

and the bond part of the hedge is 100 000 4 5 p60 16 e−4 6 r Φ(0.08706) = $41 120.

5

6

1 Thus, the cost of the option at time t = 12 is

−29 528 + 41 120 = $11 592.

1 The hedge at time t = 12 comprised −29 737 of stock and 41 668 in 1 bonds. At time t = 6 this has value

−29 737

S 16 S 121

r

+ 41 668 e 12 = $11 701.

The rebalancing cost at time t = 16 is therefore 11 592 − 11 701 = −$109.

If H(t) denotes the hedge rebalancing cost at time t years (so that H 16 = −109), then the present value of the hedge rebalancing costs is 60

t 1.05− 12 H 12t = −$1092.35. t=1

(ii) Again, we illustrate the calculation using time t = 2 months as an example. The fund at time 2 months is F2 = 100 000 S2 × 0.99752 = $96 261.88.

The management cost is 0.0025 × F2 = $240.65, the expenses are 0.00065 × F2 = $62.57,

Solutions for Chapter 17

277

and, from part (i), the hedge rebalancing cost is −$109.16. Thus, the emerging profit is Pr2 = 240.65 − 62.57 + 109.16 = $287.24.

The profit margin is calculated as 60

1.1− 12 12t p60 Prt /100 000 = −1.23%. t

t=0

(iii) In part (a) we calculated that the initial hedge cost converts to a monthly outgo of 0.19% of the fund. The incurred renewal expenses are 0.065% of the fund, so the monthly cost of these two items is 0.255% of the fund. As the monthly management fees are 0.25% of the fund, we would not expect the contract to be profitable.

Solutions for Chapter 18

18.1 (a) Measuring in years, the first four death times are t(1) = 27, t(2) = 34, t(3) = 68 and t(4) = 70, meaning that Ŝ(69) = Ŝ(68). As r1 = 9, r2 = 7 and r3 = 5, and as the death times are unique, we have Ŝ(t(3) ) = Ŝ(69) = p̂(1) p̂(2) p̂(3) =

8 6 4 = 0.6095.

9 7 5

From Greenwood’s formula, we estimate the standard deviation of Ŝ(69) as 6 8 7 dj 1 1 1 = Ŝ(69) + + = 0.1805.

Ŝ(69) r (r − d ) 72 42 20 j j j:t ≤69 j ( j)

(b) The Nelson–Aalen estimate of H(69) is dj 1 1 1 Ĥ(69) = = + + = 0.4540, r 9 7 5 j:t ≤69 j ( j)

giving Ŝ(69) = exp{−0.4540} = 0.6351.

The estimated standard deviation of Ŝ(69) is 6 8 7 d j (r j − d j ) 8 6 4 Ŝ(69) = Ŝ(69) + 3 + 3 = 0.1562.

3 3 9 7 5 rj j:t ≤69 ( j)

18.2 (a) We have Ŝ(t(6) ) = Ŝ(t(5) ) p̂(6) where p̂(6) = 1 − d6 /r6 = 1 − 1/r6 . So 1−

0.85 1 = 0.9714, = r6 0.875

giving r6 = 35.

278

279

Solutions for Chapter 18

(b) Similarly, Ŝ(t(7) ) = Ŝ(t(6) ) p̂(7) gives r7 = 40. If there are n new entrants between times t(6) and t(7) then 35 − 1 + n − 2 = 40 gives n = 8, since we have one death at time t(6) and two exits between times t(6) and t(7) .

18.3 (a) The complete table is shown below. j

t( j)

rj

0 1 2 3 4 5

0 22.0 31.5 50.7 51.3 53.0

500 496 485 449 448 447

dj

1 1 1 1 1

p̂( j)

Ŝ(t), t( j) ≤ t < t( j+1)

cj

1.00000 0.99798 0.99593 0.99371 0.99149 0.98927

4 10 35 0 0 0

0.99798 0.99794 0.99777 0.99777 0.99776

We have used r1 = r0 − c0 and for j = 1, 2, 3, 4, r j+1 = r j − d j − c j , and we have calculated p̂( j) = 1 − d j /r j for j = 1, 2, . . . , 5. The survival probabilities were calculated as Ŝ(t) =

2

p̂( j) ,

j:t( j) ≤t

so that, for example, Ŝ(50) = p̂(1) p̂(2) = 0.99593. (b) In the original study, there were 35 exits between times t(2) and t(3) , corresponding to ages 31.5 to 50.7 (the ages at death are recorded in the question). We now add 100 lives at age 50, none of whom exit before age 51.5.

Hence c2 = 35 − 100 = −65. We add two further rows to the table in part (a) to allow for the deaths at ages 51.5 and 51.8.

The calculations then proceed as in part (a) and the complete table is shown below.

280

Solutions for Chapter 18 j

t( j)

rj

dj

p̂( j)

0 1 2 3 4 5 6 7

0 22.0 31.5 50.7 51.3 51.5 51.8 53.0

500 496 485 549 548 547 546 545

1 1 1 1 1 1 1

0.99798 0.99794 0.99818 0.99818 0.99817 0.99817 0.99817

Ŝ(t), t( j) ≤ t < t( j+1)

cj

1.00000 0.99798 0.99593 0.99411 0.99230 0.99048 0.98867 0.98686

4 10 −65 0 0 0 0 0

0 18.4 (a) The 99% linear confidence interval is Ŝ(80) ± 2.576 V [Ŝ(80)], which is (−0.1461, 0.7461), and we adjust this to (0, 0.7461). (b) The approximate 99% confidence interval for log(−log S(80)) is 9 2 1 V[Ŝ(80)] , log(−log Ŝ(80)) ± 2.576 Ŝ(80) log Ŝ(80) which is (−1.0496, 1.4208) = (gL , gU ). The 99% log-confidence interval is then (exp{− exp{gU }}, exp{− exp{gL }}) = (0.0159, 0.7046).

18.5 (a) The 95% linear confidence interval is Ĥ(30) ± 1.96 SD[Ĥ(30)], which is (−0.0009, 0.0089), and we adjust this to (0, 0.0089). (b) Using the transformation g(H) = log H so that 1 d g(H) = , dH H we get V[g(Ĥ(30))] ≈

1 V[Ĥ(30)] .

Ĥ(30)2

Then a 95% confidence interval for log H(30) is log Ĥ(30) ± 1.96

SD[Ĥ(30)] , Ĥ(30)

which is (−6.7464, −4.2965). Exponentiating gives the 95% log-confidence interval as (0.0012, 0.0136).

281

Solutions for Chapter 18 18.6 (a) The estimates of the (assumed constant) transition intensities are 01 = μ58+t

150 , 785

02 μ58+t =

200 , 785

03 μ58+t =

20 .

785

Hence we estimate 1 p00 58 as ⎧ ⎫ ⎪ ⎪ 1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ 01 02 03 exp ⎪ + μ + μ μ dt − = exp{−370/785} = 0.6242. ⎪ 58+t 58+t 58+t ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 0 ⎭ (b) From part (a) we see that for 0 ≤ t ≤ 1, t p00 58 = exp{−370 t/785}. Then 0.5 02 0.5 p58 =

200 00 02 1 − exp{−370 (0.5)/785} = 0.1135. t p58 μ58+t dt = 370

0 01 to represent the constant transition intensity over the year of (c) We take μ58 01 age from 58 to 59. Then the approximate variance of μ̂58 is 150/7852 .

01 Taking a log transformation, the approximate variance of log μ̂58 is

7852 150 1 01 V[ μ̂ ] = = 58 01 2 2 2 150 150 785 (μ̂58 ) 1

01 and so a 95% confidence interval for log μ58 is 8 1 01 log μ̂58 , ± 1.96 150

which is (−1.8151, −1.4950). Exponentiating gives the 95% log-confidence interval as (0.1628, 0.2242).

18.7 (a) For (i), interval censoring occurred since we know only that the disease symptom appeared between ages 0 and 45.

For (ii), the lives are subject to right censoring and to Type 1 censoring, since they survived to the predetermined end of the study period without the symptoms appearing.

For (iii), the lives are subject to interval censoring since all we know for each life is that the symptoms first appeared between two particular ages.

For (iv), the life is subject to random censoring as the date of death (and hence exit from the study) was unknown when the life joined the study.

For (v), random censoring is present as the exit times of these lives were

282

Solutions for Chapter 18 not predictable at the start of the study. Right censoring is also present as the symptoms had not appeared by the times at which these individuals left the study. (b) Let the random variable T denote the age at which a disease symptom first appeared. The contributions to the likelihood are as follows.

For (i), Pr[T < 45].

For (ii), the contributions are Pr[T > 60 | T > 30] and Pr[T > 72 | T > 30].

For (iii), the contributions are Pr[40 < T < 41 | T > 35] and Pr[55 < T < 56 | T > 35].

For (iv), the contribution is Pr[T > 65.2 | T > 41].

For (v), the contributions are Pr[T > 55.1 | T > 36], Pr[T > 47.4 | T > 42] and Pr[T > 62.9 | T > 44].

18.8 (a) For the Gompertz model, survival probabilities from birth are given by t p0 = exp

−B t c −1 . log c

The likelihood function is L(B, c) =

5 2

t j p0 μt j

j=1

and the log-likelihood function, denoted l(B, c), is l(B, c) = log L(B, c) =

5

log t j p0 + log μt j

j=1

=−

5

5

B (ct j −1) + log B + t j log c . log c j=1 j=1

(b) Differentiate the log-likelihood with respect to the parameters B and c, then set the derivatives to zero to obtain the equations satisfied by the maximum

Solutions for Chapter 18

283

likelihood estimates B̂ and ĉ as 5 −1 t j (ĉ −1) + , log ĉ j=1 B̂

5 5 t j ĉt j ĉt j − 1 1 − tj . + 0 = −B̂ ĉ log ĉ ĉ (log ĉ)2 ĉ j=1 j=1 5

0=

(18.1)

(18.2)

Inserting the given value ĉ = 1.0936 into equation (18.1) gives B̂ = 0.00082 to five decimal places. Inserting both these (rounded) values into equation (18.2) gives the value of the right-hand side as −0.09, which is close enough to 0, allowing for rounding. (c) Let t1 , t2 , . . . , t5 be the original five ages at death and let t j = 65 for j = 6, 7, . . . , 10. Then the log-likelihood is now l(B, c) = −

5

B (ct j −1) + log B + t j log c , log c j=1 j=1

10

and the equations satisfied by the maximum likelihood estimates are 5 −1 t j (ĉ −1) + , log ĉ j=1 B̂

10 5 t j ĉt j ĉt j − 1 1 0 = −B̂ − tj . + ĉ log ĉ ĉ (log ĉ)2 ĉ j=1 j=1 10

0=

(18.3)

(18.4)

Inserting the given value ĉ = 1.0568 into equation (18.3) gives B̂ = 0.00109 to five decimal places. Inserting both these (rounded) values into equation (18.4) gives the value of the right-hand side as −0.12, which is close enough to 0, allowing for rounding. (d) As the parameter estimates in part (c) are based on a larger amount of information, we would prefer these to the estimates in part (b).

18.9 (a) We use the fact that exp{−z} = 1 − z +

z 2 z3 − + ··· > 1 − z 2 6

provided 0 < z ≤ 1. For t ≥ t(1) this gives ⎧ ⎫

⎪ ⎪ ⎪ 2 ⎪ ⎪ dj ⎨ dj ⎪ ⎬ ŜNA (t) = exp ⎪ 1− = ŜKM (t) . − > ⎪ ⎪ ⎪ ⎪ rj ⎩ j:t( j) ≤t r j ⎪ ⎭ j:t( j) ≤t

284

Solutions for Chapter 18 For 0 < t < t(1) , ŜNA (t) = ŜKM (t) = 1, so that ŜNA (t) ≥ ŜKM (t) for all t > 0. (b) Using the definition of Ĥ(t) we have ⎤ ⎡ ⎢⎢⎢ d ⎥⎥⎥ j⎥ ⎢ ⎥⎥ V[Ĥ(t)] = V ⎢⎢⎢⎣ r ⎥⎦ j:t( j) ≤t j ⎡ ⎤ ⎢⎢⎢ ⎥⎥⎥ ⎢ = V ⎢⎢⎢⎣ 1 − p̂( j) ⎥⎥⎥⎥⎦ j:t( j) ≤t

V[ p̂( j) ] by independence = j:t( j) ≤t

=

p( j) (1 − p( j) )

j:t( j) ≤t

rj

dj dj ≈ 1− r j r2j j:t( j) ≤t (r j − d j ) d j = . r3j j:t( j) ≤t

18.10 (a) We can calculate the following values: j

t( j)

rj

1 2 3 4 5 6 7 8

12.8 35.3 35.6 40.6 43.3 48.0 48.8 53.8

18 15 14 13 12 10 9 8

(i) As there is a single death at each death time, the estimate is

7 2 1 Ŝ10 (40) = 1− = 0.55407. rj j=1

285

Solutions for Chapter 18 (ii) From Greenwood’s formula, SD[Ŝ10 (40)] = Ŝ10 (40)

6 : 7 7

j=1

(b)

1 r j (r j − 1)

= 0.12697.

(i) Note that f10 (x) =

f0 (x + 10) 1 −x/θ = e , S0 (10) θ

and S10 (x) = e−x/θ . Let x j = t j − 10 for j = 1, 2, . . . , 18. Then the likelihood function is ⎧ ⎫ ⎪ ⎪ 18 18 ⎪ 2 ⎪ ⎪ ⎨ 1 ⎪ ⎬ 1 −x j /θ 1 e = exp x , L(θ ) = − ⎪ ⎪ j ⎪ ⎪ δj ⎪ ⎪ θ θ ⎩ ⎭ θ 15 j=1 j=1 since 15 lives die, and the log-likelihood is −1 x j − 15 log θ . θ j=1 18

l(θ ) = Then

18 ∂ l(θ ) 1 15 = 2 , xj − ∂θ θ θ j=1

and hence the maximum likelihood estimate of θ is 1 871 − 180 = 46.0667. xj = 15 j=1 15 18

θ̂ =

(ii) The asymptotic variance of the maximum likelihood estimate is ' −1 & 2 ∂ l(θ ) , −E ∂θ2 where 15 ∂ 2 l(θ ) −2 = 3 xj + 2 . ∂θ2 θ j=1 θ 18

Inserting the parameter estimate, our estimate of the asymptotic variance is 11.89442 . (iii) The maximum likelihood estimate of S10 (40) is e−40/θ̂ = 0.4197.

286

Solutions for Chapter 18 (iv) With g(θ ) = e−40/θ we have g (θ ) = (40/θ 2 ) e−40/θ and so by the delta method,

2 40 −40/θ V[g(θ̂ )] ≈ 2 e V[θ̂ ] = 0.094092 . θ

18.11 (a) For Life 1, the contribution is 11 −(μx12 +μx13 ) .

1 px = e

For Life 2, the contribution is 00 01 11 12 22 0.25 px μx 0.5 px+0.25 μx 0.25 px+0.75

= μx01 μx12 e−0.25(μx +μx ) e−0.5(μx +μx ) e−0.25μx .

01

03

12

13

23

For Life 3, the contribution is 11 12 22 23 12 23 −0.5(μx12 +μx13 ) −0.3μx23 . e 0.5 px μx 0.3 px+0.5 μx = μx μx e

(b) (i) The likelihood function is L = μx01

163

μx03

125

μx12

154

μx13

42

× e−(μx +μx )T e−(μx +μx )T e−μx T 01

02

(0)

12

13

(1)

23

μx23

25

(2)

and the log-likelihood is log L =163 log μx01 + 125 log μx03 + 154 log μx12 + 42 log μx13 + 25 log μx23 − μx01 + μx02 T (0) − μx12 + μx13 T (1) − μx23 T (2) .

We differentiate the log-likelihood with respect to each parameter and set the partial derivative to equal 0. So, for example, ∂ 163 log L = 01 − T (0) , 01 ∂ μx μx gives μ̂x01 = 163/T (0) = 0.0361.

Similarly, ∂2 −163 log L = 01 2 , ∂ (μx01 )2 (μx ) and so V[μx01 ] ≈

& −E

∂2 log L ∂ (μx01 )2

' −1 =

We calculate V[μx01 ] as 163/(T (0) )2 = 0.00282 .

(μx01 )2 .

E[D01 ]

287

Solutions for Chapter 18 The complete list of numerical solutions is as follows.

Parameter

Estimate

Standard error

μx01 μx03 μx12 μx13 μx23

0.0361 0.0276 0.4303 0.1174 0.3592

0.0028 0.0025 0.0347 0.0181 0.0718

(ii) There is much more data to estimate transition intensities from State 0 than from State 1, both in terms of number of transitions and in terms of waiting times. We would expect to be able to estimate with greater accuracy if we have a larger volume of data, so we would expect our estimates for State 0 to be more accurate. In statistical terms, this means that we would expect a smaller standard deviation of the estimate.

18.12 (a) The likelihood function is of the form 2 f0 (t j ) 2 S0 (t j ) L(γ, θ ) = .

S0 (10) S0 (10) j:δ j =1

j:δ j =0

We set up cells in Excel to represent the contribution to the log-likelihood for each life using trial values of γ and θ contained in individual cells. We sum the individual contributions to the log-likelihood to obtain a value of the log-likelihood, then we use Excel Solver to maximize the value in the cell containing the value of the log-likelihood by changing the values in the cells containing γ and θ . We obtain γ̂ = 3.383 and θ̂ = 58.57. (b) The plots are shown in Figure S18.1. For most ages the Nelson–Aalen estimate is less than the Weibull estimate, but the two plots are reasonably close up to about age 70 where we have most data. Note that the Weibull cumulative hazard increases with age whereas the Nelson–Aalen estimate of the cumulative hazard is constant after the final death time.

288

Solutions for Chapter 18

Figure S18.1 Weibull (solid line) and Nelson–Aalen (dotted line) cumulative hazard functions.

Solutions for Chapter 19

19.1 (a) First, lm(40, Y ) = α40 + β40 KY gives −6.83 = α40 − 3.5 β40 .

Second, lm(40, Y + 1) = α40 + β40 KY +1 = α40 + β40 (KY + c + σk ZY ) gives E[lm(40, Y + 1) = α40 + β40 (KY + c) = α40 − 4β40 = −6.845 as E[ZY ] = 0. Solving these two equations gives α40 = −6.725 and β40 = 0.03. (b) We know that 1 − ϕ (m) (40, Y + 1) ∼ LN(β40 c, β40 σk ). So 3 4 1 − E[ϕ (m) (40, Y + 1)] = exp β40 c + 0.5(β40 σk )2 , giving E[ϕ (m) (40, Y + 1)] = 1.311%.

Let m be the median. Then Pr[ϕ (m) (40, Y + 1) ≤ m] = 0.5 or, equivalently, Pr[1 − ϕ (m) (40, Y + 1) ≥ 1 − m] = 0.5.

289

290

Solutions for Chapter 19 Let X = log(1 − ϕ (m) (40, Y + 1)). Then X ∼ N(β40 c, (β40 σk )2 ), and so 0.5 = Pr[1 − ϕ (m) (40, Y + 1) ≥ 1 − m] = Pr[X ≥ log(1 − m)] & ' log(1 − m) − β40 c = Pr Z ≥ , where Z ∼ N(0, 1) β40 σk ⇒

log(1 − m) − β40 c =0 β40 σk

⇒ log(1 − m) − β40 c = 0 ⇒ m = 1.489%.

19.2 (a) From 1

mx =

r px μx+r dr

0

,

1 r px dr 0

the assumption that μx+r = μ for 0 ≤ r < 1 gives mx = μ, and px = e−μ .

So, if m(x, t) = 0.025 then

q(x, t) 1 − e−0.025 logit q(x, t) = log = −3.676353. = log 1 − q(x, t) e−0.025 (b) Under the UDD assumption, mx =

qx , 1 − 12 qx

qx =

mx .

1 + 12 mx

which gives

So q(x, t) = 0.02469 and hence logit q(x, t) = −3.676301.

19.3 (a) We calculate m(70, 0) as m(70, 0) = exp{α70 + β70 K0 } = 0.017249. (b) The conditions m(70, 1) > m(70, 0) and lm(70, 1) > lm(70, 0) are equivalent,

Solutions for Chapter 19

291

so lm(70, 1) > lm(70, 0) ⇔ α70 + β70 K1 > α70 + β70 K0 ⇔ β70 (K0 + c + σk Z1 ) > β70 K0 ⇔ Z1 > −c/σk , and Z1 ∼ N(0, 1). Hence Pr[m(70, 1) > m(70, 0)] = Pr[Z1 > 0.2667] = 0.3949.

Note that it was convenient here to write lm(70, 0) as α70 + β70 K0 rather than as the numerical value of log 0.017249. We take the same approach in answering parts (c) and (d). (c) The conditions m(71, 1) < m(70, 0) and lm(71, 1) < lm(70, 0) are equivalent so lm(71, 1) < lm(70, 0) ⇔ α71 + β71 K1 < α70 + β70 K0 ⇔ 0.1 + 0.03 (K0 + c + σk Z1 ) < 0.03 K0 −0.1 − 0.03 c ⇔ Z1 < .

0.03 σk Hence Pr[m(71, 1) < m(70, 0)] = Pr[Z1 < −1.9556] = 0.0253. (d) Extending the arguments in part (b), lm(70, 1) > lm(70, 0) ⇔ α70 + β70 K1 + ε70,1 > α70 + β70 K0 ⇔ β70 (K0 + c + σk Z1 ) + ε70,1 > β70 K0 ⇔ β70 σk Z1 + ε70,1 > −β70 c.

On the left-hand side of the previous line we have a combination of two independent normal random variables – call this sum X. Then X is normally distributed, and as each variable has mean 0, we have E[X] = 0. Further, V[X] = (β70 σk )2 V[Z1 ] + V[ε70,1 ] = 0.002925.

Hence Pr[m(70, 1) > m(70, 0)] = Pr[X > 0.2219] = 0.4122.

292

Solutions for Chapter 19

19.4 (a) As x̄ = 60, lq(65, Y + 1) = KY(1)+1 + 5 KY(2)+1 , where KY(1)+1 = KY(1) + c(1) + σk1 ZY(1) = −3.04 + 0.05 ZY(1) and similarly KY(2)+1 = KY(2) + c(2) + σk2 ZY(2) = 0.021 + 0.005 ZY(2) .

Hence lq(65, Y + 1) = −2.935 + 0.05 ZY(1) + 0.005 ZY(2) , giving E[lq(65, Y + 1)] = −2.935 and V[lq(65, Y + 1)] = 0.052 + 0.0252 + 2(0.3)(0.05)(0.025) = 0.062252 . (b) Let xq be the q-quantile of p(65, Y + 1). Then Pr[p(65, Y + 1) ≤ xq ] = q ⇒ Pr[q(65, Y + 1) > 1 − xq ] = q &

' 1 − xq ⇒ Pr lq(65, Y + 1) > log =q xq ' & log((1 − xq )/xq ) + 2.935 = q. ⇒ Pr Z > 0.06225 Now Pr[Z > 0] = 0.5 gives 1 − x0.5 = e−2.935 , x0.5 so that the median, x0.5 , is 0.94955, and Pr[Z > −1.645] = 0.95 gives 1 − x0.95 = e−2.935+1.645(0.06225) , x0.95 so that the 0.95-quantile, x0.95 , is 0.95424.

19.5 The times series models are KY(1)+1 = −5.04 + 0.03 ZY(1)+1 , KY(2)+1 = 0.01 + 0.001 ZY(2)+1 , KY(3)+1 = 0.0001 + 0.0006 ZY(3)+1 .

With x̄ = 60, s2x = 140 and GY −64 = −0.05 we get lq(65, Y + 1) = −4.6015 + 0.03 ZY(1)+1 + 0.005 ZY(2)+1 − 0.069 ZY(3)+1 .

Solutions for Chapter 19

293

(a) As E[ZY(i)+1 ] = 0 for i = 1, 2, 3, we have E[lq(65, Y + 1)] = −4.6015.

As only Zt(1) and Zt(2) are correlated, V[lq(65, Y + 1)] = 0.032 + 0.0052 + 0.0692 + 2(0.5)(0.03)(0.005) = 0.076392 . (b) Let xq be the q-quantile of p(65, Y + 1). Then Pr[p(65, Y + 1) ≤ xq ] = q ⇒ Pr[q(65, Y + 1) > 1 − xq ] = q &

' 1 − xq ⇒ Pr lq(65, Y + 1) > log =q xq ' & log((1 − xq )/xq ) + 4.6015 = q. ⇒ Pr Z > 0.07639 Now Pr[Z > 0] = 0.5 gives 1 − x0.5 = e−4.6015 , x0.5 so that the median, x0.5 , is 0.99006, and Pr[Z > −1.645] = 0.95 gives 1 − x0.95 = e−4.6015+1.645(0.07639) , x0.95 so that the 0.95-quantile, x0.95 , is 0.99123.

19.6 We assume p(x, t) = exp{−m(x, t)} with m(x, t) = exp{αx + βx Kt } = exp{αx + βx (Kt−1 + c + σk Zt )} = exp{αx + βx (Kt−2 + c + σk Zt−1 ) + βx c + βx σk Zt } = exp{αx + βx Kt−2 + 2 βx c + βx σk (Zt−1 + Zt )} = m(x, t − 2) exp{2 βx c + βx σk (Zt−1 + Zt )}, and this generalizes to

⎧ ⎫ ⎪ ⎪ s−1 ⎪ ⎪

⎪ ⎪ ⎨ ⎬ c + β σ Z m(x, t) = m(x, t − s) exp ⎪ s β . ⎪ x x k t− j ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ j=0

1 As s−1 j=0 Zt− j is normally distributed with mean 0, the median of this sum is 0, and hence the median of m(x, t) is m(x, t − s) exp{s βx c}. Thus, the median of p(x, t) is p(x, t − s)exp{s βx c} .

294

Solutions for Chapter 19

19.7 (a) (i) We have m(50, Y + 2) = exp{α50 + β50 KY +2 }, so lm(50, Y + 2) = α50 + β50 KY +2 = α50 + β50 (KY +1 + c + σk ZY +1 ) = α50 + β50 (c + σk ZY +1 ) + β50 (KY + c + σk ZY ) = α50 + β50 (2c + KY ) + β50 σk (ZY +1 + ZY ).

As Zt ∼ N(0, 1) for all t, we have that lm(50, Y +2) is normally distributed with E[lm(50, Y + 2)] = α50 + β50 (2c + KY ) = −6.864 and as ZY and ZY +1 are independent, 2 σk2 = 0.33942 .

V[lm(50, Y + 2)] = 2 β50

Then m(50, Y + 2) is lognormally distributed with 4 3 E[m(50, Y + 2)] = exp −6.864 + 12 0.33942 = 0.0011067 and V[m(50, Y + 2)] = E[m(50, Y + 2)]2 exp{0.33942 } − 1 = 0.00038672 . (ii) Let zq be the q-quantile of the N(0, 1) distribution. Then the q-quantile of lm(50, Y + 2) is −6.864 + 0.3394 zq , and of m(50, Y + 2) is exp{−6.864 + 0.3394 zq }. As z0.5 = 0 the median of m(50, Y + 2) is exp{−6.864} = 0.001045, and as z0.75 = 0.6745, the 75% quantile of m(50, Y + 2) is exp{−6.864 + 0.6745 × 0.3394} = 0.001313. (b) Under the constant force assumption, q(50, Y + 2) = 1 − exp{−m(50, Y + 2)}.

Under this assumption, the median of q(50, Y + 2) is 1 − exp{−0.001045} = 0.001044, and the 75% quantile of q(50, Y + 2) is 1 − exp{−0.001313} = 0.001313.

Solutions for Chapter 19

295

(c) Under the assumption of UDD, q(50, Y + 2) =

m(50, Y + 2) .

1 + 12 m(50, Y + 2)

Under this assumption, the median of q(50, Y + 2) is 0.001045 = 0.001044, 1 + 12 0.001045 and the 75% quantile of q(50, Y + 2) is 0.001313 = 0.001313.

1 + 12 0.001313 19.8 We calculate m(x, t) = exp{αx + βx Kt } for t = 1, 2, . . . , 40 and x = 70 + t, where αx = −11 + 0.1x, ⎧ ⎪ 0.03 for x = 70, 71, . . . , 79, ⎪ ⎪ ⎪ ⎨ βx = ⎪ 0.01 for x = 80, 81, . . . , 89, ⎪ ⎪ ⎪ ⎩ 0 for x = 90, 91, . . . , 110, and Kt = Kt−1 − 0.4 + 1.5 Zt for t = 1, 2, . . . , 40 with K0 = −2. Simulating Z1 , Z2 , . . . , Z40 using the random number generator in Excel allows us to calculate values for K1 , K2 , . . . , K40 , and hence for values of m(70 + t, t). We can repeat this process to obtain the 1 000 paths. (a) Under the constant force assumption we calculate survival probabilities for the jth path ( j = 1, 2, . . . , 1 000) as p j (x, t) = exp{−m j (x, t)}. (b) For the jth path, we calculate r-year survival probabilities for r = 2, 3, . . . , 41 as r−1 2 p = p j (70 + t, t). r 70 t=0

The EPV of the annuity due at 5% per year is then 1+

41

1.05−r r p70 ,

r=1

where we assume that 42 p70 = 0. (c) Your answer will depend on your random drawings from the N(0, 1) distribution. The authors obtained a mean of 11.08 and a variance of 0.1622 .

The histogram is displayed in Figure S19.1.

296

Solutions for Chapter 19

Figure S19.1 Histogram of simulated annuity values for Exercise 19.8.