Contents 3MLC-ACT...1.4 Curtate Future Lifetime Random Variable C1-11 1.5 Force of Mortality C1-13...

P-1Preface

Contents

Preface P-5 Syllabus Reference P-7 Flow Chart P-10 Chapter 1 Survival Distributions C1-1

1.1 Age-at-death Random Variables C1-1 1.2 Future Lifetime Random Variable C1-4 1.3 Actuarial Notation C1-6 1.4 Curtate Future Lifetime Random Variable C1-11 1.5 Force of Mortality C1-13

Exercise 1 C1-18 Solutions to Exercise 1 C1-23 Chapter 2 Life Tables C2-1

2.1 Life Table Functions C2-1 2.2 Fractional Age Assumptions C2-5 2.3 Select-and-Ultimate Tables C2-14 2.4 Moments of Future Lifetime Random Variables C2-19 2.5 Useful Shortcuts C2-26

Exercise 2 C2-28 Solutions to Exercise 2 C2-35 Chapter 3

Life Insurances C3-1 3.1 Continuous Life Insurances C3-2 3.2 Discrete Life Insurances C3-16 3.3 mthly Life Insurances C3-22 3.4 Relating Different Policies C3-24 3.5 Recursions C3-29 3.6 Relating Continuous, Discrete and mthly Insurance C3-35 3.7 Useful Shortcuts C3-38

Exercise 3 C3-40 Solutions to Exercise 3 C3-49 Chapter 4 Life Annuities C4-1

4.1 Continuous Life Annuities C4-1 4.2 Discrete Life Annuities (Due) C4-14 4.3 Discrete Life Annuities (Immediate) C4-21 4.4 mthly Life Annuities C4-25

© Actex 2012 | Johnny Li and Andrew Ng | SoA Exam MLC

Preface

P-2

4.5 Relating Different Policies C4-27 4.6 Recursions C4-31 4.7 Relating Continuous, Discrete and mthly Life Annuities C4-32 4.8 Useful Shortcuts C4-38

Exercise 4 C4-39 Solutions to Exercise 4 C4-48 Chapter 5 Premium Calculation C5-1

5.1 Traditional Insurance Policies C5-1 5.2 Benefit Premium and Equivalence Principle C5-3 5.3 Benefit Premiums for Special Policies C5-12 5.4 The Loss-at-issue Random Variable C5-18 5.5 Percentile Premium C5-26

Exercise 5 C5-30 Solutions to Exercise 5 C5-44 Chapter 6 Benefit Reserves C6-1

6.1 The Prospective Approach C6-2 6.2 The Recursive Approach: Basic Idea C6-17 6.3 The Recursive Approach: Further Applications C6-26

6.4 The Retrospective Approach C6-36 Exercise 6 C6-43 Solutions to Exercise 6 C6-62 Chapter 7 Insurance Models Including Expenses C7-1

7.1 Gross Premium C7-1 7.2 Gross Premium Reserve C7-5 7.3 Expense Reserve and FPT Reserve C7-12

7.4 Basis, Asset Share and Profit C7-19 Exercise 7 C7-31 Solutions to Exercise 7 C7-44 Chapter 8 Multiple Decrement Models: Theory C8-1

8.1 Multiple Decrement Table C8-1 8.2 Forces of Decrement C8-5 8.3 Associated Single Decrement C8-10

8.4 Discrete Jumps C8-23 Exercise 8 C8-28 Solutions to Exercise 8 C8-39


P-3Preface

Chapter 9 Multiple Decrement Models: Applications C9-1

9.1 Calculating Actuarial Present Values of Cash Flows C9-1 9.2 Cash Values C9-12 9.2 Calculating Asset Share under Multiple Decrement C9-17

Exercise 9 C9-22 Solutions to Exercise 9 C9-32 Chapter 10 Multiple State Models C10-1

10.1 Discrete-time Markov Chain C10-4 10.2 Continuous-time Markov Chain C10-13 10.3 Kolmogorov’s Forward Equations C10-19 10.4 Calculating Actuarial Present Value of Cash Flows C10-30 10.5 Calculating Reserves C10-36

Exercise 10 C10-41 Solutions to Exercise 10 C10-52 Chapter 11 Multiple Life Functions C11-1

11.1 Multiple Life Statuses C11-2 11.2 Insurances and Annuities C11-17 11.3 Dependent Life Models C11-29

Exercise 11 C11-36 Solutions to Exercise 11 C11-53 Chapter 12 Interest Rate Risk C12-1

12.1 Yield Curves C12-1 12.2 Interest Rate Scenario Models C12-9 12.3 Diversifiable and Non-Diversifiable Risks C12-12

Exercise 12 C12-17 Solutions to Exercise 12 C12-25 Chapter 13 Profit Testing C13-1

13.1 Profit Vector and Profit Signature C13-1 13.2 Profit Measures C13-8

Exercise 13 C13-12 Solutions to Exercise 13 C13-15


Preface

P-4

Chapter 14 Universal Life Insurance C14-1

14.1 Basic Policy Design C14-2 14.2 Cost of Insurance and Surrender Value C14-5 14.3 Other Policy Features C14-15 14.4 Projecting Account Values C14-17 14.5 Profit Testing C14-22

Exercise 14 C14-27 Solutions to Exercise 14 C14-37 Chapter 15 Pension Mathematics C15-1

15.1 The Salary Scale Function C15-1 15.2 Pension Plans C15-9

15.3 Setting the DC Contribution rate C15-10 Exercise 15 C1-16 Solutions to Exercise 15 C15-19 Chapter 0 Numerical Techniques C0-1

0.1 Numerical Integration C0-1 0.2 Euler’s Method C0-7

Exercise 1 C1-18 Solutions to Exercise 1 C1-23 Mock Test 1 T1-1

Solution T1-18 Mock Test 2 T2-1

Solution T2-18


P-5Preface

Preface

Thank you for choosing ACTEX. Starting in Spring 2012, Exam MLC will be based on a new syllabus, which is significantly different from the old one. Two new textbooks, Actuarial Mathematics for Life Contingent Risks (AMLCR) and Models for Quantifying Risk (MQR) are adopted. A number of new topics, for example, interest rate risk and universal life insurance, are added to the syllabus. Because of these major changes, ACTEX decided to bring you this study manual, which is written anew (rather than developed from old stuff) to fit the current syllabus. When you took other actuarial exams, you might rely on only one of the alternative textbooks. This approach is not going to work for Exam MLC, because not all topics in the syllabus are covered at the same level in each textbook. The exam syllabus says:

Not all topics listed in the learning objectives are covered at the same level in each source (textbook). The sources differ in their exposition, examples, and exercises. The selection of examination questions is based on coverage of the learning objectives but the selection of the examination questions is not necessarily neutral with respect to the sources. That is, the sources may not be equivalent in their coverage of all the specific items included on an examination question.

For instance, discrete-time Markov chains, which are examined in the new Exam MLC Sample Questions (#152), are discussed only in MQR. In contrast, profit measures and gain by source, which are also examined in the sample questions (#300), are mentioned only in AMLCR. Using a good study manual is therefore crucially important for Exam MLC. In this study manual, we have fully integrated the strengths of the two textbooks, making sure that all topics in the syllabus are explained and practiced in sufficient depth. For your reference, a detailed mapping between this study guide and the textbooks are provided on pages P-7 to P-9. We have paid special attention to the newly introduced topics. Five full-length (up to 186 pages) chapters are especially devoted to these topics. Instead of treating the new topics as “orphans”, we demonstrate, as far as possible, how they can be related to the old topics in an exam setting. This is very important for you, because, as stated explicitly in the exam syllabus, questions may cover multiple learning outcomes. We totally understand that you may be panic, because no past exam questions are based on the new topics. To help you best prepare for the exam, we have incorporated all Exam MLC sample questions that are related to the new topics into the manual. On top of that, plenty of original exam-type questions on the new topics are provided for you to practice. Besides the topics specified in the exam syllabus, you also need to know a range of numerical techniques in order to succeed. These techniques include, for example, Euler’s method, which is involved in Exam MLC Sample Question #299. We know that quite a few of you have not even heard of Euler’s method before, so we have prepared a special chapter (C0, appended to the end of the study manual) to teach you all numerical techniques required for this exam. Whenever a


Preface

P-6

numerical technique is used, we clearly point out which technique it is, letting you follow our examples and exercises more easily. Other distinguishing features of this study manual include:

− We use graphics extensively. Graphical illustrations are probably the most effective way to explain formulas involved in Exam MLC. The extensive use of graphics can also help you remember various concepts and equations.

− A sleek layout is used. The font size and spacing are chosen to let you feel more comfortable in reading. Important equations are displayed in eye-catching boxes.

− Rather than splitting the manual into tiny units, each of which tells you a couple of formulas only, we have carefully grouped the exam topics into 15 chapters. Such a grouping allows you to more easily identify the linkages between different concepts, which, as we mentioned earlier, are essential for your success.

− Instead of giving you a long list of formulas, we point out which formulas are the most important. Having read this study manual, you will be able to identify the formulas you must remember and the formulas that are just variants of the key ones.

− We do not want to overwhelm you with verbose explanations. Whenever possible, concepts and techniques are demonstrated with examples and integrated into the practice problems.

− We write the practice problems and the mock exams in a similar format as the released exam and sample questions. This will enable you to comprehend questions more quickly in the real exam.

On page P-10, you will find a flow chart showing how different chapters of this manual are connected to one another. You should first study Chapters 1 to 10 in order. Chapters 1 to 4 will build you a solid foundation, while the Chapters 5 to 10 will get you to the core of the exam. You should then study Chapters 11 to 15 in any order you wish. Immediately after reading a chapter, do the practice problems we provide for that chapter. Make sure that you understand every single practice problem. Finally, work on the mock exams. Before you begin your study, please download the exam syllabus from the following hyperlink:

http://www.soa.org/files/pdf/edu-2012-spring-mlc-syllabus.pdf

On page 2 of the pdf file, you will find a link to Exam MLC Tables, which are frequently used in the exam. You should keep a copy of the tables, as we are going to refer to them from time to time. On the same page, you will also find links to Exam MLC Sample Questions and the supplementary notes for AMLCR. If you find a possible error in this manual, please let us know at the “Customer Feedback” link on the ACTEX homepage (www.actexmadriver.com). Any confirmed errata will be posted on the ACTEX website under the “Errata & Updates” link. Enjoy your study!


P-7Preface

Syllabus Reference

Our Manual AMLCR + Supplement MQR

Chapter 1: Survival Distributions

1.1 2.1, 2.2 5.1 1.2 2.2 5.3, 6.3 1.3 2.4 5.1 1.4 2.6, SN1.1 5.3 1.5 2.3 5.1

Chapter 2: Life Tables

2.1 3.1, 3.2 6.1, 6.2 2.2 3.3 6.6 2.3 3.7, 3.8, 3.9 5.4, 6.7 2.4 2.5, 2.6 6.3 2.5 5.2

Chapter 3: Life Insurances

3.1 4.4.1, 4.4.5, 4.4.7, 4.6 7.3, 7.4 3.2 4.4.2, 4.4.6, 4.4.7, 4.6 7.1, 7.2, 7.4 3.3 4.4.3 7.3.4 3.4 4.4.8, 4.5 7.1.3 3.5 4.4.4 3.6 4.5 7.3.3, 7.5.1, 7.5.2 3.7 7.3.5

Chapter 4: Life Annuities

4.1 5.5 8.1.3, 8.2.3, 8.3.3, 8.5 4.2 5.4.1, 5.4.2, 5.9, 5.10 8.1.2, 8.2.2, 8.3.2, 8.5 4.3 5.4.3, 5.4.4 8.1.1, 8.2.1, 8.3.1, 8.5 4.4 5.6 8.4 4.5 5.8 Scattered in Chapter 8 4.6 5.11.1 4.7 5.11.2, 5.11.3 8.4.4 4.8


Preface

P-8


Chapter 5: Premium Calculation

5.1 6.1, 6.2 5.2 6.5 9.1, 9.2, 9.4, 9.5 5.3 5.4 6.4 9.2 5.5 9.3

Chapter 6: Benefit Reserves

6.1 7.1, 7.3.1, 7.8 10.1.1, 10.1.3, 10.1.4, 10.1.5, 10.3.3, 10.4

6.2 7.3.3 10.2, 11.3 6.3 7.4 11.2 6.4 7.7, SN2.2 10.1.2, 10.1.6

Chapter 7: Insurance Models Including Expenses

7.1 6.6 9.6 7.2 7.3.2, 7.5 10.3, 11.4 7.3 SN2.3 11.1.2 7.4 6.7, 7.3.4, 7.3.5

Chapter 8: Multiple Decrement Models: Theory

8.1 SN3.2 13.1.1 8.2 8.8, SN3.4 13.1.2, 13.2, 13.3 8.3 8.8, SN3.3, SN3.5 13.4 8.4 8.10

Chapter 9: Multiple Decrement Models: Applications

9.1 14.1 9.2 7.6 14.3 9.3 14.2

Chapter 10: Multiple State Models

10.1 3.1, 7.6.1, 8.6 10.2 8.2, 8.3 3.2, 5.5, 14.4 10.3 8.4, 8.5 14.4.3 10.4 8.6 3.3, 7.6.2, 8.6, 14.4.2 10.5 8.7 14.4.5


P-9Preface


Chapter 11: Multiple Life Functions

11.1 8.9.1 12.1, 12.2, 12.3, 12.6 11.2 8.9.4 12.4 11.3 8.9.2, 8.9.3 12.5, 12.7

Chapter 12: Interest Rate Risk

12.1 10.1 − 10.3 15.1, 15.3, 15.4 12.2 15.2 12.3 6.8, 10.4

Chapter 13: Profit Testing

13.1 11.1, 11.2, 11.4 13.2 11.3

Chapter 14: Universal Life Insurances

14.1 SN4.1 11.5, 16.1 14.2 SN4.3, SN4.5 14.3 SN4.1 14.4 SN4.2, SN4.3 14.5 SN Example 4.1 − 4.3

Chapter 15: Pension Mathematics

15.1 9.3 14.5.1 15.2 9.1, 9.2 15.2 9.4

Chapter 0: Numerical Techniques

0.1 8.6 0.2 7.5.2 14.4.3


PrefaceP-10

Flow Chart

1. Survival Distributions


2. Life Tables

3. Life Insurances

4. Life Annuities

6. Benefit Reserves

12. Interest Rate Risk

9. Multiple Decrement Models: Applications

8. Multiple Decrement Models: Theory

13. Profit Testing

14. Universal Life Insurance

10. Multiple State Models

11. Multiple Life Functions

0. Numerical Techniques

5. Premium Calculation

7. Insurance Models Including Expenses

15. Pension Mathematics

Sample Pages from Chapter 10

Chapter 10 contains 72 pages. We are showing pages 4 to 10.



C10-4

The Disability Income Model

A multiple state model models the transition rates between states. Transition to any state can be

reversible or irreversible. Consider the following disability income model:

Model 3: The Disability Income Model In this model, the insured can switch between state 0 and state 1. However, transition to state 2

is irreversible. We can change the above to a permanent disability model by disallowing

transition from state 1 back to state 0.

In this chapter we want to analyze models with structures similar to Model 3. We shall first

discuss a simple version, where state changes can only occurs at the end of each period. Then

we shall discuss the more general case when state changes can only occur any time.

In this section we discuss discrete-time Markov chain, a model for which transitions can only

occur at the end of each period. Such a model is useful in modeling health status, bonus malus

system and credit rating in group health and non-life insurances. Let the state at time t be Yt. Let

the state space, which is the set of all possible states, be E. For example, in the disability income

model above, we have E = {0, 1, 2} and Y0 = 0. As time progresses, Y switches between 0 and 1

and finally get struck in state 2. The following figure shows one of the possible paths of Y.

0. Healthy

1. Disabled

2. Dead

10. 1 Discrete-time Markov Chain


C10-5Chapter 10: Multiple State Models

Transition Probability Matrix and the Markov Property

For a discrete-time Markov chain, we assume that when the process is in state i at time t, the

probability that it would be in state j at time t + 1 is ijtp . That is, Pr(Yt+1 = j | Yt = i) = ij

tp . It is

customary to place all probabilities into a transition probability matrix:

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

nnt

nt

nt

nt

ntttt

ntttt

ntttt

t

pppp

pppppppppppp

L

MOMMM

L

L

L

210

2222120

1121110

0020100

P

The sum of the probabilities on each row is 1. Moreover, we assume for any t ≥ 0 and i, j ∈ E.

The Markov Property

Pr(Yt+1 = j | Yt = i, Yt–1 = it–1, Yt–2 = it–2, … ) = Pr(Yt+1 = j | Yt = i)

This above says that the conditional distribution of any future state Yt+1, given the past states {Y0,

Y1, …, Yt–1} and the present state Yt, depends only on the present state Yt but not the past states.

Loosely speaking, “Markov” means given the present, you can forget about the past!

If Pt is constant with t, the Markov chain is said to be homogeneous. If Pt is time-dependent, the

Markov chain is said to be inhomogeneous.

time t

Y t

2

1

0

0 1 2 3 4 5 6 7 8 9

0 1 2 3 t t − 1 t + 1 …

P0 P1 P2 Pt−1 Pt

Y1 Y0 Y2 Y3 Yt−1 Yt Yt+1



C10-6

In many actuarial applications, there would be one or more states that cannot be left once it is

entered (examples include withdrawal, death, and bankruptcy). If you look at the following 3-

state transition probability matrix:

10021.06.03.012.01.07.00

210

then state 2 cannot be left once it is entered. Such a state is called an absorbing state.

For an absorbing state, all elements in the row, except the one in the main diagonal †, are 0. The

one in the main diagonal is 1.

Kevin and Kira are modeling the future lifetime of (60).

(i) Kevin uses a double decrement model:

Age )(τxl )1(

xd )2(xd

60 1000 120 80 61 800 160 80 62 560 − −

(ii) Kira uses a non-homogeneous Markov model:

(a) The states are 0 (alive), 1 (death due to cause 1), 2 (death due to cause 2).

(b) P60 is the transition matrix from age 60 to 61; P61 is the transition matrix from age 61 to

62.

(iii) The two models produce equal probabilities of decrement.

Calculate P61.

(A) ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

00.100000.1008.012.000.1

(B) ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

00.10008.016.056.008.012.080.0

(C) ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

00.100000.1008.016.076.0

(D) ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

00.100000.1010.020.070.0

(E) ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

00.100000.1012.028.060.0

† The main diagonal of a square matrix is the diagonal which runs from the top left corner to the bottom right corner.

Example 10.1 [Exam M 2005 Nov #4]



Solution

Obviously, both states 1 and 2 are absorbing. This means (B) must be wrong. For a life age (61),

the probability that he/she would still be alive after one year is

7.0800/560)(61 ==τp .

The probability that he/she would die due to cause 1 within the year is

2.0800/160)1(61 ==q .

This means (D) is the correct answer.

[ END ]

The following is another actuarial application of homogeneous Markov chain. An insurance

company classifies its insureds based on each insured’s credit rating, as one of Preferred,

Standard or Poor. Individual transition between classes is modeled by the following transition

matrix:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

75.0 25.0 00.005.0 80.0 15.001.0 04.0 95.0

Poor

StandardPreferred

Poor StandardPreferred

Very frequently we are not only interested in transition probabilities for a single period. For

example, we may want to find the probability that an insured who is “Standard” at time 0 would

end up in “Poor” after 2 years. This is called a 2-step transition probability.

One way to calculate such probabilities is to list out all possible paths:

Path Probability Standard → Preferred → Poor 0.15 × 0.01 = 0.0015 Standard → Standard → Poor 0.8 × 0.05 = 0.04

Standard → Poor → Poor 0.05 × 0.75 = 0.0375 Sum 0.079

Similarly, we can calculate the probability that an insured who is “Standard” at time 0 would

end up in “Preferred” after 2 years:



C10-8

Path Probability Standard → Preferred → Preferred 0.15 × 0.95 = 0.1425 Standard → Standard → Preferred 0.8 × 0.15 = 0.12

Standard → Poor → Preferred 0.05 × 0 = 0 Sum 0.2625

What is the probability for the insured to be in the state of “Standard” after 2 years? It is simply

1 − 0.079 − 0.2625 = 0.6585.

Actually there is a very efficient way to calculate such probabilities if you know some

elementary matrix algebra:

In case you have forgotten about matrix multiplication, read the following short example:

If ⎥⎦

⎤⎢⎣

⎡−−

=571

301A ,

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−=3421

12B , then

⎥⎦

⎤⎢⎣

⎡−

−=⎥

⎦

⎤⎢⎣

⎡−+−+−−−+−+−

+−+−+−+=

30111010

)3)(5()2)(7()1)(1()4)(5()1)(7()2)(1()3)(3()2)(0()1)(1()4)(3()1)(0()2)(1(

AB .

In this chapter we would frequently calculate the product of two square matrices. Remember

that for two square matrices A and B, the products AB and BA can be different!

Now let us look at

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

575.03875.00375.0079.06585.02625.0019.00725.09085.0

75.025.0005.08.015.001.004.095.0

75.025.0005.08.015.001.004.095.0

.

What can you find from the matrix on the right? The 2-step transition probabilities are all in the

second row of the matrix product P × P = P2. If we want to find

Pr(Yt+2 = j | Yt = i) = ijtp2 ,

then we can simply look at the ij-th element of the matrix P2. So, P2 is the 2-step transition

matrix. More generally, if we want to find

Pr(Yx+s = j | Yx = i) = ijxs p ,

we only need to look up the ij-th element in sPx = PxPx+1 ... Px+s−1.



This result can be thought of as a generalization of the relation

121 −+++= sxxxxxs ppppp K .

The above implies

The order of multiplication on the right hand side of the CK equation matters!

For a multiple state model with three states, Healthy (0), Disabled (1), and Dead (2):

(i) For k = 0, 1:

7.000 =+kxp , 2.001 =+kxp , 1.010 =+kxp , 25.012 =+kxp

(ii) There are 100 lives at the start, all Healthy. Their future states are independent.

Calculate the variance of the number of the original 100 lives who die within the first two years.

(A) 11 (B) 14 (C) 17 (D) 20 (E) 23

Example 10.2 [Sample #151]

x x + 1 x + 2 x + 3 x + s x + s − 1 x + s + 1 …

Px Px+1 Px+2 Px+s−1 Px+s

Yx+1 Yx Yx+2 Yx+3 Yx+s−1 Yx+s Yx+s+1

sPx

F O R M U L AThe Chapman-Kolmogorov (CK) Equation

txsxtxst ++ = PPP



C10-10

Solution

We are given the following information:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡== +

???25.0?1.0?2.07.0

1xx PP .

Since state 2 is an absorbing state, the last row must be [0 0 1]. By using the property that each

row of a transition probability matrix sums to 1, we get:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡== +

10025.065.01.01.02.07.0

1xx PP .

By the CK equation,

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡== +

100???22.0??

10025.065.01.01.02.07.0

10025.065.01.01.02.07.0

12 xxx PPP ,

where 0.22 = 0.7 × 0.1 + 0.2 × 0.25 + 0.1 × 1.

For a single life, the probability that he would die within the first two years is 0.22. If there are

100 independent lives, we let 2D0 as the number of deaths within the first two years. Then

2D0 ~ B(100, 0.22)

(B(n, p) here means a binomial distribution with number of trials n and probability of success p)

and hence the variance is

Var(2D0) = 100 × 0.22 × (1 − 0.22) = 17.

So the correct answer is (C).

[ END ]

Calculating APV of Cash Flows

There are mainly two kinds of cash flows associated with discrete-time Markov chain:

(1) Cash flow arising from being in a particular state.

(2) Cash flow arising from a transition from one state to another.

It is easy to calculate APV of cash flows under a discrete-time Markov chain model. Study the

following example:

Sample Pages from Chapter 14

Chapter 14 contains 46 pages. We are showing pages 2 to 11.

Chapter 14: Universal Life Insurance


C14-2

Any interest that the account value earns is credited back to the account value and invested, so

essentially, the policyholder’s money is earning money. You may therefore view a universal life

policy as a mixture of life insurance and an investment product.

Let us begin with a description of how the account value is accumulated.

We define the following notation:

Pt Premium for the tth time period, paid at time t − 1

ECt The expense charge (also called MER, for Management Expense Rate) for the tth time

period, deducted from the account value at time t – 1

CoIt The cost of insurance for the tth time period, deducted from the account value at time t – 1

AVt The account value at time t, before premium and deductions cti The credited rate of interest for the tth time period (i.e., from time t – 1 to time t)

The accumulation of the account value over the tth time period (i.e., from time t – 1 to time t) is

illustrated in the following diagram:

14. 1 Basic Policy Design

Account Value (AVt−1)

Premium (Pt)

Expense Charge (ECt)

Cost of Insurance

(CoIt)

Account Value (AVt)

Accumulate at the credited rate of interest

( cti )

Add

Deduct

Time t − 1 Time t


C14-3Chapter 14: Universal Life Insurance

Keep in mind that the premium (Pt), expense charge (ECt) and the cost of insurance (CoIt) are

for the tth time period and are deducted or paid at time t – 1.Assuming the policy is still in force

at time t, we have the following relation:

Note:

− For a basic universal life policy, there is no separate investment account. Policyholders’

funds are merged to the insurer’s general funds. The credited interest rate is declared by the

insurer and is based on the investment performance on the insurer’s assets.

− The amount of premium Pt is at the policyholder’s discretion.

− The expense charge ECt is determined by the insurer.

− The length of each time step depends on the frequency of premium payments. In the exam,

the accumulation of account value may be calculated at yearly or monthly time steps.

For a basic universal life policy, you are given:

(i) The account value at t = 10 (before premium payment and deductions) is $10,000.

(ii) The policyholder pays premiums at t = 10 and t = 11. Each premium is $500.

(iii) The expense charge is 1% of each premium.

(iv) The costs of insurance deducted at t = 10 and t = 11 are $20 and $15, respectively.

(v) The credited interest rate over the period from t = 10 to t = 12 is 2% per annum effective.

Assume the policy is still in force at t = 12.

(a) Calculate the account value at t = 12 if the length of each time step is one year.

(b) Calculate the account value at t = 12 if the length of each time step is one month.

Example 14.1

F O R M U L A

Account Value Accumulation

AVt = (AVt−1 + Pt − ECt − CoIt)(1 + cti )



C14-4

Solution

(a) Here, the length of each time step is one year. At t = 11, the account value is given by

AV11 = (AV10 + P11 – EC11 – CoI11)(1 + 11ci ) = (10000 + 500 – 5 – 20)(1 + 0.02) = 10684.5.

At t = 12, the account value is given by

AV12 = (AV11 + P12 – EC12 – CoI12)(1 + 12ci ) = (10684.5 + 500 – 5 – 15)(1 + 0.02) = 11387.8.

(b) Here, the length of each time step is one month. The credited interest rates are calculated as

follows:

112

11 121 1 (1 0.02)c ci i+ = + = +


AV11 = (AV10 + P11 – EC11 – CoI11)(1 + 11ci ) = (10000 + 500 – 5 – 20)

112(1 0.02)+ = 10492.3.


AV12 = (AV11 + P12 – EC12 – CoI12)(1 + 12ci ) = (10492.3 + 500 – 5 – 15)

112(1 0.02)+ = 10990.4.

[ END ]

You may wonder how the cost of insurance is calculated. We are going to address this issue in

the next section.



Suppose that a universal life insurance policy is in force at time t − 1. At time t, there are three

possibilities:

Possibility (1): The policy is still in force

At time t, the policy can still be in force. In this case, the process described in Section 14.1

repeats.

Possibility (2): The policy is surrendered

At time t, the policyholder can choose to surrender the policy. In this case, the policyholder will

be paid a surrender value. The total cash available to the policyholder on surrender is the

account value (AVt) minus the surrender charge (SCt), which is determined by the insurer. This

amount of money is called the cash value (CVt) of the contract. Mathematically,

CVt = max(AVt – SCt, 0)

14. 2 Cost of Insurance and Surrender Value

The policy is in force.

The policy is in force.

The policyholder has died. A death benefit is paid

The policy is surrendered. A surrender value is paid.

Time t Time t − 1



C14-6

Possibility (3): The policyholder has died

Death may occur between time t – 1 and time t. In this case, a death benefit must be paid at time

t. The amount of death benefit depends on the contract type.

Type A: Specified Amount

For a specified amount (Type A) contract, the total death benefit is level. The total death benefit

offered by Type A contract is called the Face Amount (FA) of the policy.

Type B: Specified Amount plus the Account Value

For a specified amount plus the account value (Type B) contract, the total death benefit is the

account value plus a fixed amount, which we denote by X.

For both types, the death benefit is subject to a corridor factor requirement, which stipulates the

minimum amount of insurance coverage required by the tax laws to prevent the contract from

becoming one of investment only (which has a less favorable tax status). The minimum total

death benefit is defined by the corridor factor (γt) times the account value (AVt) at death. Taking

the corridor factor requirement into account, the total death benefit at time t can be expressed as

follows.

The difference between the total death benefit and the account value is known as the Additional

Death Benefit (ADB). If the corridor factor requirement is satisfied, then the ADB for a

specified amount (Type A) contract, which equals FA – AVt, decreases as the account value

increases, but the ADB for a specified amount plus the account value (Type B) contract is a

F O R M U L A

Total Death Benefit

Specified Amount (Type A): max(FA, γtAVt)

Specified Amount plus the Account Value (Type B): max(AVt + X, γtAVt)



constant X. If the corridor factor requirement is not satisfied, then the ADB equals (γt − 1)AVt

for both types. Hence, we have the following expressions for the ADB at time t.

For a specified amount universal life insurance policy, you are given:

(i) The face amount of the policy is $100,000

(ii) The account value on December 31, 2011 was $50,000.

(iii) On January 1, 2012, a premium of $2000 was made. No other premiums were made in

2012.

(iv) The expense charge and the cost of insurance deducted on January 1, 2012 were $150 and

$200, respectively.

(v) The credited interest rate in 2012 was 6% per annum effective.

(vi) The corridor factor applicable in 2012 was 2.5.

(vii) The surrender charge applicable in 2012 was $10 per $1,000 face amount.

Calculate the following:

(a) The cash value of the policy on December 31, 2012.

(b) Suppose that death occurred in 2012 and that the death benefit was payable at the end of

the year of death. Calculate the amount of death benefit.

Example 14.2

F O R M U L A

Additional Death Benefit

Specified Amount (Type A):

ADBt = max(FA – AVt, (γt − 1)AVt))

Specified Amount plus the Account Value (Type B):

ADBt = max(X, (γt − 1)AVt))



C14-8

Solution

(a) First, we need to project the account value on December 31, 2012. For convenience, we

refer December 31, 2011 to t – 1 and December 31, 2012 to t. We have

AVt = (AVt−1 + Pt − ECt − CoIt)(1 + cti ) = (50000 + 2000 – 150 – 200)(1.06) = 54749.

The cash value of the policy on December 31, 2012 is given by

CVt = max(AVt – SCt, 0) = max(54749 – 1000, 0) = 53749.

(b) This is a Type A policy. The total death benefit is given by

max(FA, γtAVt) = max(100000, 2.5 × 54749) = 136872.5.

[ END ]

Suppose that at the time of death, the policyholder’s account value is AVt. The amount of AVt

will cover part of the total death benefit, and the shortfall (i.e., ADBt) will be topped up by the

insurer. It is now clear that the cost of insurance collected at the beginning of a time period is

used to support the expected ADB over that time period. In general, the cost of insurance can be

calculated by the following formula.

F O R M U L A

General Formula for the Cost of Insurance

CoIt = *tq × vq × ADBt,

where

− CoIt is the cost of insurance for the tth time period, deducted from the account value at time t – 1,

− *tq is the death probability (for the tth time period) used to calculate the cost of

insurance,

− vq is the discount factor for discounting the cost of insurance to time t − 1,

− ADBt is the additional death benefit at time t.



Note:

(1) The credited interest rate and the interest rate for calculating the cost of insurance are not

necessarily the same. For this reason, the notation vq is introduced.

(2) If premiums are paid annually, then the cost of insurance is just the single premium for a 1-

year term life insurance with sum insured equal to the ADB.

The actual calculation of the cost of insurance is a lot more complicated than it seems. This is

because of the following circular relationship:

The cost of insurance CoIt is a function of the additional death benefit ADBt, which is a function

of the account value AVt. However, AVt depends on CoIt, which is deducted at the beginning of

the time period. We now discuss how this problem can be solved.

First, let us consider a specified amount (Type A) policy. If the corridor requirement is satisfied,

then

ADBt = FA – AVt = FA − (AVt−1 + Pt − ECt − CoIt)(1 + cti ).

Let CoI ft be the cost of insurance when the corridor factor requirement is satisfied. We have

CoI ft = *

tq vq ADBt = *tq vq [FA − (AVt−1 + Pt − ECt − CoI f

t )(1 + cti )],

which implies

*1

*

(FA (AV EC )(1 ))CoI

1 (1 )

ct q t t t tf

t ct q t

q v P iq v i

−− + − +=

− +.

CoIt

ADBt AVt



C14-10

If the corridor factor requirement is not satisfied, then

ADBt = (γt – 1)AVt = (γt – 1) (AVt−1 + Pt − ECt − CoIt)(1 + cti ).

Let CoIct be the cost of insurance when the corridor factor requirement is not satisfied. We have

*1

*1

*

CoI ( 1)(AV EC CoI )(1 )

(1 )( 1)(AV EC ).

1 (1 )( 1)

c c ct t q t t t t t t

ct q t t t t t

ct q t t

q v P i

q v i Pq v i

γ

γγ

−

−

= − + − − +

+ − + −=

+ + −

Finally, the insurer should charge CoIt = max( CoI ft , CoIc

t ) for the cost of insurance.

We now consider a specified amount plus the account value (Type B) policy. If the corridor

factor requirement is satisfied, then the additional death benefit is ADBt = X. The cost of

insurance is simply

CoI ft = *

tq vq X.

The same as a Type A contract, the additional death benefit is ADBt = (γt – 1)AVt if the corridor

factor requirement is not satisfied. Hence, for Type B, policy, we also have

*1

*

(1 )( 1)(AV EC )CoI

1 (1 )( 1)

ct q t t t t tc

t ct q t t

q v i Pq v iγ

γ−+ − + −

=+ + −

.

F O R M U L A

Cost of Insurance for a Specified Amount (Type A) Policy

CoIt = max( CoI ft , CoIc

t )

where

*1

*

(FA (AV EC )(1 ))CoI

1 (1 )

ct q t t t tf

t ct q t

q v P iq v i

−− + − +=

− +,

*1

*

(1 )( 1)(AV EC )CoI

1 (1 )( 1)

ct q t t t t tc

t ct q t t

q v i Pq v iγ

γ−+ − + −

=+ + −

.



For a universal life insurance policy, you are given:

(i) The death benefit is $100,000 plus the account value at the end of the year of death. The

benefit is payable at the end of the year of death.

(ii) The account value on December 31, 2011 was $20,000.

(iii) On January 1, 2012, a premium of $1,000 was made. No other premiums were made in

2012.

(iv) The expense charge deducted on January 1, 2012 was $100.

(v) The credited interest rate in 2012 was 1% per annum effective.

(vi) The surrender charge applicable in 2012 was $10 per $1,000 face amount.

(vii) There is no corridor factor requirement for this policy.

(viii) In calculating the cost of insurance, it was assumed that the probability of death in 2012

was 0.001. The interest rate used was 3% per annum effective.

(iv) The policyholder is alive on December 31, 2012.

Calculate the account value on December 31, 2012.

Solution

For convenience, we refer December 31, 2011 to t – 1 and December 31, 2012 to t. The required

value, AVt, can be calculated with the following relation:

Example 14.3

F O R M U L A

Cost of Insurance for a Specified Amount plus Account Value (Type B) Policy

CoIt = max( CoI ft , CoIc

t )

where CoI ft = *

tq vq X and *

1*

(1 )( 1)(AV EC )CoI

1 (1 )( 1)

ct q t t t t tc

t ct q t t

q v i Pq v iγ

γ−+ − + −

=+ + −

.

Contents 3MLC-ACT...1.4 Curtate Future Lifetime Random Variable C1-11 1.5 Force of Mortality C1-13...

Documents

Transcript of Contents 3MLC-ACT...1.4 Curtate Future Lifetime Random Variable C1-11 1.5 Force of Mortality C1-13...