by yufeng guo - actuarial bookstore 3mlc-yg-12ssm-e... · 2011-11-07 · by yufeng guo . 12th...

http://actuary88.com

Deeper Understanding, Faster Calculation --SOA Exam MLC Insights & Shortcuts

by Yufeng Guo

12th Edition

Spring, 2012 This is a partial study guide and doesn’t cover everything in the MLC syllabus. This is best used as a second study guide. This electronic book is intended for individual buyer use for the sole purpose of preparing for Exam MLC. This book may NOT be resold to others or shared with others. No part of this publication may be reproduced for resale or multiple copy distribution without the express written permission of the author.

© 2011, 2012, 2013 Yufeng Guo

1 of 210

http://actuary88.com/�

Table of Contents Chapter 1 Passing Exam MLC the First Time..................................... 6

Origin of the COM method ............................................................................................. 6 Pitfalls of the Traditional Study Method ........................................................................ 6 How COM Works ........................................................................................................... 7 Why COM Works ........................................................................................................... 8 An example of the COM method .................................................................................... 9

Chapter 2 Eliminate calculation errors .............................................. 21 Solve 2 0ax bx c+ + = . ............................................................................................ 22 Linear Interpolation ...................................................................................................... 27 Calculate mean and variance of a discrete random variable ......................................... 31 Find the conditional mean and variance ....................................................................... 39 Find the present value of life insurance or annuity ....................................................... 41 How to eliminate errors in complex calculations ......................................................... 44

Chapter 3 Survival distribution ........................................................... 51 Age-at-death random variable of a newborn ................................................................ 51 What makes a good survival function ........................................................................... 51 Age-at-death random variable of an individual aged x ............................................... 52 Distribution function of ( )T x ...................................................................................... 52

Relationship between ( )T x and X ............................................................................. 52 Difficult symbol ............................................................................................................ 53 Curtate future lifetime of ( )x ....................................................................................... 54

Force of mortality (instantaneous death rate/mortality) - ( )xµ and ( )x tµ ................. 55 FAQ’s ............................................................................................................................ 56 Common problems and model solutions ....................................................................... 59

Type 1 From ( )S x to force of mortality xµ ........................................................ 59

Type 2 From ( )S x find ( )f t , ( )E T , ( )Var T ................................................... 60

Type 3 From ( )S x to t xp , t xq ............................................................................ 62 Type 4 From a series of t xq find xx t q .................................................................. 64

Type 5 What makes a good xµ ............................................................................. 65 Type 6 From ( )x tµ to t xp ................................................................................... 66 Type 7 Two independent lives, first death after m years but before n years ...... 67 Type 8 Two independent lives, second death after m years but before n years . 69 Type 9 Two independent lives, at least one will die in n years ........................... 70 Type 10 Multiple independent lives .................................................................... 73

Chapter 4 Life Table ............................................................................. 75 Basic concepts ............................................................................................................... 75 Other formulas .............................................................................................................. 76 Complete expectation of future lifetime ....................................................................... 76 Curtate expectation of life ............................................................................................. 76

2 of 210

Relationship between 0

xe and xe ................................................................................... 77

Relationship between 0

xe and xe under UDD ............................................................... 77 Less common life table functions ................................................................................. 77 Common problems and model solutions ....................................................................... 79

Type 1 Build a life table ........................................................................................ 79 Type 2 From the Life Table, find t xp and t xq ..................................................... 80 Type 3 Find mean and variance of the number of survivors ................................. 81 Type 4 Recursive formula xe ................................................................................ 82 Type 5 From xl find xµ ........................................................................................ 83 Type 6 Force of mortality doubles ........................................................................ 84

Chapter 5 UDD between integral ages ................................................ 86 Shortcut for UDD between [ ], 1x x + .......................................................................... 86 UDD shortcut examples .............................................................................................. 89

Chapter 6 The heart and soul of Actuarial Mathematics ................. 94 Fundamental law of life insurance ................................................................................ 94 Equivalence principle .................................................................................................. 100 n-year Term Insurance Model ..................................................................................... 102

n-year term fully discrete ........................................................................................ 102 n-year Term Fully Continuous ................................................................................ 106

Chapter 7 Fundamentals of life insurance and annuity .................. 110 Memorizing definitions ............................................................................................... 110

Term vs. whole life ................................................................................................. 110 Endowment ............................................................................................................. 111 Deferred Life Insurance .......................................................................................... 112

Memorizing symbols .................................................................................................. 112 Common problems and model solutions ..................................................................... 118

Type 1 Discrete life insurance, find E(Z), Var(Z) .............................................. 118 Type 2 Continuous life insurance, find E(Z), Var(Z) ......................................... 128

Free internet resources for learning life annuities ....................................................... 129 Chapter 8 Reserve ............................................................................... 130

What’s reserve all about? ............................................................................................ 130 What’s a benefit reserve? ............................................................................................ 131 How do I calculate reserve? ........................................................................................ 132 Retrospective and prospective method ....................................................................... 140 Steps to calculate reserve at time t for a fully discrete insurance .............................. 144 Recursive formulas for reserve ................................................................................... 157 Reserve for variable death benefit .............................................................................. 162 How to calculate reserve for a fully continuous insurance ......................................... 163

Chapter 9 Asset share ......................................................................... 167 Chapter 10 Expense loaded premium ................................................. 171 Chapter 11 Multiple decrement model ............................................... 175

Examples of multiple decrements ............................................................................... 175

3 of 210

Building a multiple decrement model ......................................................................... 176 The associated single decrement table ........................................................................ 178 Constant Force of Mortality or UDD under the Multiple Decrement Table .............. 181 Construct a multiple decrement table ......................................................................... 185 How to memorize formulas for two decrements ......................................................... 185 Common problems and model solutions ..................................................................... 187

Type 1 Find ( )f T J j= and ( )E T J j= ......................................................... 187

Type 2 ( )f J j T= and ( )Tf t ............................................................................... 189 Type 3 UDD in the multiple decrement table ..................................................... 191 Type 4 UDD in associated single decrement tables, 2 decrements ..................... 192 Type 5 UDD in associated single decrement tables, 3 decrements ..................... 194 Type 6 multiple decrements – some continuous and some discrete ................... 195 Type 7 PVDB under the two decrement table .................................................... 199

Chapter 18 Other commonly used shortcuts ...................................... 203 About the author .................................................................................... 210

4 of 210

This page is intentional blank.

5 of 210


Chapter 1 Passing Exam MLC the First Time Origin of the COM method When I became an actuary, I was in my late 30’s with a young family to support. I knew it was possible to need three or four sittings to pass Course 3 and that it could take five to eight years to achieve the ASA designation. I wanted to lessen this burden on my family. After passing Course 1, I set out to develop a study method that would help me pass Courses 2 and 3 in one sitting. By trial and error, I discovered the COM method: C = Calculation. Eliminate all calculation errors. O = Officially Released Exams. Master officially released

Exam M problems through reverse engineering.

M = Margin. Add some margin of error by studying more material than what you expect will be tested. Pitfalls of the Traditional Study Method Under the traditional method, students read a chapter from the required text, solve some practice problems in a corresponding study manual, and write down some notes. Then the student moves on to the next chapter and repeats the process until completion several months later. Next, candidates solve previous Course 3 and M exam problems. A zealous candidate might solve 1,000+ practice problems from sources like old SOA or CAS problems or original practice problems. This traditional study method is how we have all been trained to learn new information. However, the sad truth is that many candidates who use this method fail exams repeatedly. What goes wrong? The traditional study method has two major flaws: First, the traditional study method casts the net too wide. Candidates indiscriminately learn everything from the textbooks. After months of hard work, they walk into the exam room feeling confident only to discover that they know too much about what SOA doesn’t test and too little about what SOA does test. Second, the traditional study method encourages a candidate to solve vast numbers of practice problems without a thorough understanding and analysis of the critical concepts. Days and nights spent solving hundreds of old SOA/CAS problems and memorizing formulas with only a superficial understanding of the core concepts yield little fruit on exam day. Candidates are often shocked to discover that they did not pass the exam.

6 of 210


After failing the exam, many candidates decide to solve even more practice problems, leading to repeated failures of Exam M. How COM Works The COM method fixes the two major flaws of the traditional study method. First, candidates find out what concepts are really important for the exam through reverse engineering. After identifying these core concepts, candidates then learn these concepts through solving the officially released Exam M problems. By repeatedly testing themselves with Exam M problems under exam conditions, candidates perfect their command of the core concepts essential for Exam M. Calculation

• Eliminate all calculation errors. • Develop error free calculation processes • Maximize the use of calculators

Officially released M problems (including Sample M problems and additional sample problems) from the SOA website are the core for your study plan. Master Exam M problems through reverse engineering:

• Never suggest there aren’t enough practice problems. The officially released Exam M problems are sufficient for anyone to pass Exam M.

• Resist the temptation to solve 1,000+ old SOA/CAS problems. Don’t bother with

old SOA problems. SOA is innovative and consistently develops new ways of measuring your command of the information.

• Resist the temptation to solve problems from a textbook. Textbook problems tend

to be more theoretical; exam problems are more practical.

• If you do want to solve additional practice problems (such as problems from the CAS website or from a seminar), remember to master the officially released Exam M problems first. Only after you have mastered these problems should you consider solving other practice problems. However, keep in mind that the officially-released Exam M problems are the best practice problems.

• Work problems in an exam like-condition. Then work and rework problems from

Exam M until you can solve all the problems 100% right, 100% of the time.

• Do not rely on academic solutions. They look nice on paper but are too lengthy and complex to use under exam conditions.

7 of 210


• Read Exam M problems and solutions first. Use reverse engineering to identify

what concepts are tested; then use your textbook and study manuals to learn the tested concepts.

• Build 3-minute solution scripts.

Margin -- Learn additional information in the event that there are subjects tested which do not appear within the officially released SOA problems. This step should be completed last, assuming you have achieved all of the previous study goals and that you have additional time.

• There are many topics included in the texts that weren’t tested in the SOA officially released practice questions. There is a fair chance some of this material will appear in the official examination.

• If you have the time, prepare for the unexpected.

• Take a little time to go over the textbooks and study the core concepts that are

listed in the syllabus but were not tested in the officially released questions.

• If you are short on time, skip building margin. Time Allocation Spend 95% of your study time on perfecting your calculation skills and on mastering the officially released M problems. Spend 5% of your time studying extra material. Why COM Works Officially released Exam M questions demonstrate the depth of knowledge that SOA expects a passing candidate to have. If you can solve the officially released Exam M problems under exam-like conditions with 100% accuracy, you should have the necessary aptitude to pass Exam M. With the additional preparation of concepts not included in the officially released questions, you have added some margin in case SOA decides to test new concepts. Reverse engineering -- First analyze exams and then, study Under this method, you analyze what concepts are tested in Exam M. Then you learn these concepts from the textbooks and study manuals.

1. Get the most recent SOA M exam and its solutions. Start from Problem 1 in the exam. Read the solution. Identify what concepts are involved in Problem 1. Exam problems are generally arranged from simple concepts to complex concepts. Problem 1 most likely contains the easiest concepts.

8 of 210


2. Learn the concepts involved in Problem 1 from the textbooks and study manuals

you have. Build a 3-minute solution script to this problem. 3. Work on Problem 2. Do the same: read the solution to the problem, identify what

concepts are tested, look up these concepts from your textbooks and study manuals. Learn these concepts. Build a 3-minute solution script to Problem 2.

4. Repeat this process till you have analyzed all the test problems, identified all the

concepts involved, and built 3-minute solution scripts to all the problems.

5. Work on the next official Exam M. Do the same: read the solution to the problem, identify what concepts are tested, look up these concepts from your textbooks and study manuals. Learn these concepts. Build a 3-minute solution script to each of the tested problems.

Why Reverse Engineering?

• If SOA thinks a concept is important, most likely it tested this concept in Exam M.

• If SOA doesn’t think that a concept is important, most likely it didn’t test it in

Exam M.

• Concepts tested on Exam M are the bare minimum knowledge you have to learn.

• Concepts not tested in Exam M are less important and should command less of your time.

• You can add some margin by learning concepts that were not tested in Exam M,

but this should be done only after you have first mastered Exam M problems. An example of the COM method

Typical problem -- Sample M problem # 66 For a select-and-ultimate mortality table with a 3-year select period: (i)

x [ ]xq [ ] 1+xq [ ] 2+xq 3+xq 3+x 60 0.09 0.11 0.13 0.15 63 61 0.10 0.12 0.14 0.16 64 62 0.11 0.13 0.15 0.17 65 63 0.12 0.14 0.16 0.18 66 64 0.13 0.15 0.17 0.19 67

9 of 210


(ii) White was a newly selected life on 01/01/2000. (iii) White’s age on 01/01/2001 is 61. (iv) P is the probability on 01/01/2001 that White will be alive on 01/01/2006. Calculate P . How to reverse engineer this problem and build a 3-minute solution script: Step 1 – Identify key concepts involved: Select & Ultimate Mortality Table Select period Variable x , xq Step 2 – Understand the above key concepts. Translate the textbook definitions of these key concepts into your own words. Select

• If John is selected at age 50, this simply means that John bought an insurance policy at age 50; the insurance company selected John as a customer when John was 50 years old.

• Similarly, if John was selected on 1/1/2000, then John bought an insurance policy

on 1/1/2000; the insurance company selected John as a customer on 1/1/2000.

• The word “select” implies that not everyone wanting a life insurance policy can actually get one.

• Life insurance companies are picky about their customers. For example, they do

not want to sell a life policy to someone on his death bed. If someone on his death bed is allowed to buy a life insurance policy, he can purchase a big policy (such as a policy that pays $1,000,000 death benefits if the insured dies). The insured about to die will pay only a small initial premium (such as $1,000). If he dies in a few days then the insurance company, having collected only $1,000 premium, has to pay $1,000,000 to the insured’s family. If a life insurance company makes a lot of “deathbed sales,” it will soon go bankrupt.

Selection process

• When someone applies for an insurance policy, the insurance company often pays a doctor to give the applicant a free medical exam. Then the insurance company analyzes the exam results to decide if this applicant is a good risk, and how much premium to charge him. This process is called underwriting.

Age and death rates x , [ ]x , xq , [ ]xq

10 of 210


• Once a customer’s application for a life policy is approved, the insurance

company needs to estimate the likelihood that this customer will die in each of the future years and charge a premium accordingly.

• If the customer is more likely to die (e.g. he is a pilot), he will be charged a higher

premium.

• The likelihood that someone who went through medical exam is going to die next year is called [ ]xq or xq where x represents the age of the customer. We can simply call [ ]xq or xq as the death rate for the next year. Here [ ]xq is the probability that [ ]x dies between x and 1+x ; xq is the probability that x will die between ages x and 1+x .

• [ ]x and x refer to two people both aged x . The difference is that [ ]x has gone

through a medical exam and was approved to buy a life policy. In contrast, x merely represents someone aged x who is randomly chosen from the general population.

• So [ ]x and x mean that we have two people both aged x , but [ ]x is less likely to

die than x . So [ ] xx qq < and [ ]x should pay less premium than x .

• [ ]xq represents the probability of someone aged x , having gone through medical underwriting, who will die before reaching age 1+x .

• xq represents the probability of someone aged x randomly chosen from the

general population who will die before reaching age 1+x .

• [ ] 1+xq refers to someone who is now 1+x years old and who went medical exam last year at age x . [ ] 1+xq represents the likelihood that this person will die between age 1+x and 2+x .

• 1+xq refers to someone who is now 1+x years old and who was randomly chosen

from the general population last year at age x . Here 1+xq represents the likelihood that this person will die between age 1+x and 2+x .

Select period

• If after n years, the effect of the medical exam wears off and [ ] nxnx qq ++ = , then n is called the select period. After n years, there are no differences between [ ] nx + and nx + .

11 of 210


• So if nm ≥ , we’ll no longer use [ ] mx + and [ ] mxq + ; instead we’ll use mx + and

mxq + Select table, ultimate table, select & ultimate table

• Select table - A table listing the year-by-year death rate for [ ]x . In a select table, the effect of the medical exam has not worn off yet.

• Ultimate table - A table listing the year-by-year death rate for x . In the ultimate

table, the effect of the medical exam has worn off. • Select & ultimate table – If you merge a select table and an ultimate table, you’ll

get a select & ultimate table. In this table, the medical exam is effective for a number of years and then wears off.

Step 3 – Design a 3-minute script for looking up rates Look up rates in a select table

• To look up a death rate, first identify the issue age --- how old the insured was when he first bought the policy. This is the age of the insured when the insurance company issued the policy.

• Next, identify how many years elapsed after the issue of the policy (this is called

policy year or duration).

• Example #1. Mary is 62 years old now. She bought her insurance when she was 60. What’s her death rate from age 62 to 63?

x (issue age) [ ]xq [ ] 1+xq [ ] 2+xq 60 0.09 0.11 0.13 61 0.10 0.12 0.14 62 0.11 0.13 0.15 63 0.12 0.14 0.16 64 0.13 0.15 0.17

We have [ ] [ ]60=x (issue age). So we’ll use the row where 60=x . The policy year or duration is 3. Policy Year 1 runs from age 60 to 61; Policy Year 2 runs from age 61 to 62; and Policy Year 3 runs from age 62 to 63. So

12 of 210


among the three columns of q values, we’ll need to use the third column of q values – this is the [ ] 2+xq column. This gives us [ ] [ ] 13.02602 == ++ qq x

• Example #2. John is 62 year old. He bought his insurance when he was 61. What’s his death rate from age 62 to 63? Use the same table above. Solution: We have [ ] [ ]61=x (issue age); we’ll use the second row.

The policy year is 2. Policy Year 1 runs from age 61 to 62; Policy Year 2 runs from age 62 to 63. This gives us [ ] [ ] 12.01611 == ++ qq x

Mary and John are both 62 years old today. However, Mary’s probability to die next year is 0.13, while John’s probability to die next year is 0.12. So under the select table, people with the same age today have different chances of dying each year. The death rates depend, among other things, on the issue age and the duration (i.e. policy year).

Look up rates in an ultimate table

• Identify the insured’s actual age (called attained age). All the people with the same attained age have the same death rate.

• Example #1. Mary is 62 years old now. She bought her insurance when she was

60. What’s her death rate from age 62 to 63? Solution: we have 62=x (attained age); 15.062 =q

x (attained age) xq 60 0.13 61 0.14 62 0.15 63 0.16 64 0.17

• Example #2. John is 62 years old now. He bought his insurance when he was 61.

What’s his death rate from age 62 to 63? Solution: we have 62=x (remember that x is the actual age, not issue age). So 15.062 =q

Look up rates in a select and ultimate table

• Within the select period, look up the death rate by issue age and by Policy Year. After the select period, look up rates only by the attained age.

13 of 210


• Example #1. Mary is 62 years old now. She bought her insurance when she was

60. Identify her death rate by each policy year.

x [ ]xq [ ] 1+xq [ ] 2+xq 3+xq 3+x 60 0.09 0.11 0.13 0.15 63 61 0.10 0.12 0.14 0.16 64 62 0.11 0.13 0.15 0.17 65 63 0.12 0.14 0.16 0.18 66 64 0.13 0.15 0.17 0.19 67

Solution: Mary’s death rates by policy year are listed below in color. The rates in red are select rates; the rates in blue are ultimate rates.

x [ ]xq [ ] 1+xq [ ] 2+xq 3+xq 3+x 60 0.09 0.11 0.13 0.15 63 61 0.10 0.12 0.14 0.16 64 62 0.11 0.13 0.15 0.17 65 63 0.12 0.14 0.16 0.18 66 64 0.13 0.15 0.17 0.19 67

. • Example #2. John is 62 years old now. He bought his insurance when he was 61.

Identify his death rate by each policy year. Solution: John’s death rates by policy year are listed below in colors. The rates in red are select rates; the rates in blue are ultimate rates.

x [ ]xq [ ] 1+xq [ ] 2+xq 3+xq 3+x 60 0.09 0.11 0.13 0.15 63 61 0.10 0.12 0.14 0.16 64 62 0.11 0.13 0.15 0.17 65 63 0.12 0.14 0.16 0.18 66 64 0.13 0.15 0.17 0.19 67

Finally, let’s solve the sample problem. Repeat of the problem For a select-and-ultimate mortality table with a 3-year select period: (i)

x [ ]xq [ ] 1+xq [ ] 2+xq 3+xq 3+x 60 0.09 0.11 0.13 0.15 63 61 0.10 0.12 0.14 0.16 64 62 0.11 0.13 0.15 0.17 65 63 0.12 0.14 0.16 0.18 66 64 0.13 0.15 0.17 0.19 67

14 of 210


(ii) White was a newly selected life on 01/01/2000. (iii) White’s age on 01/01/2001 is 61. (iv) P is the probability on 01/01/2001 that White will be alive on 01/01/2006. Calculate P . Solution Issue age = [ ] 60=x . x [ ]xq [ ] [ ] 1601 ++ = qq x [ ] [ ] 2602 ++ = qq x 3+xq 3+x 60 0.09 0.11 0.13 0.15 63 61 0.10 0.12 0.14 0.16 64 62 0.11 0.13 0.15 0.17 65 63 0.12 0.14 0.16 0.18 66 64 0.13 0.15 0.17 0.19 67 Policy Year From age To age Death rate Survival rate 2 61 62 0.11 0.89 3 62 63 0.13 0.87 4 63 64 0.15 0.85 5 64 65 0.16 0.84 6 65 66 0.17 0.83 So P =0.89(0.87)(0.85)(0.84)(0.83)=0.4589 Let’s see why this script is powerful … (1) Designing a 3-minute solution script forces us to thoroughly analyze the key concepts involved:

• Select, Ultimate, Select & Ultimate Tables • Select period • Variables x , [ ]x , xq , [ ]xq

(2) Designing a 3-minute solution script forces us to understand the key concepts not only in mathematical terms but also in a business sense. For example, we know that selection means going through a medical exam; we know that after a period of time, the effect of medical underwriting wears off. We not only know the math, but we also know what’s going on in the real world. To solve problems fast, we need to understand what’s going on in the real world. Knowing how concepts are actually used in the real world transforms abstract math

15 of 210


symbols and complex equations into concrete objects, enabling us to quickly solve an exam problem. (3) We developed the following step-by-step sequence on how to look up death rates:

• If the table is select, we look up rates by issue age and policy year • If the table is ultimate, we look up rates by attained age • If the table is select & ultimate, we use select rates within the select period; we

use ultimate rates beyond the select period This step-by-step sequence has remarkable power. It avoids the need for us to rethink the key concepts involved in “looking up rates” problems. We have already spent a lot of time identifying and understanding the key concepts necessary to solve the problem. We definitely don’t want to waste our time in the exam going through this time-consuming process again. When we walk into the exam room and see a rate-looking-up problem, we activate this script and quickly find the required death rate in the correct row and the correct column in several seconds without thinking. Putting this script to work After designing a script for looking up death rates, we walk into the exam room and see the following problem: Sample M problem #73 For a select-and-ultimate table with a 2-year select period:

x [ ]xp [ ] 1+xp 2+xp 2+x 48 0.9865 0.9841 0.9713 50 49 0.9858 0.9831 0.9698 51 50 0.9849 0.9819 0.9682 52 51 0.9838 0.9803 0.9664 53

Keith and Clive are independent lives, both age 50. Keith was selected at age 45 and Clive was selected at age 50. Calculate the probability that exactly one will be alive at the end of three years. [A] Less than 0.115 [B] At least 0.115, but less than 0.125 [C] At least 0.125, but less than 0.135 [D] At least 0.135, but less than 0.145 [E] At least 0.145 Solution

16 of 210


We are very glad that we have a script ready for looking up rates problems. Without a script, we have to invent solutions on the spot. This problem has a new symbol xp and [ ]xp . You just need to know that p and q are complements; 1=+ qp . So xq and [ ]xq represent the likelihood that x and [ ]x die, respectively, next year; xp and [ ]xp represent the likelihood that x and [ ]x , respectively, will not die (i.e. will be alive) next year. Next, we get the big picture: P (3 years later only K or C alive but not both) = P (next 3 years K alive but C dead) + P (next 3 years C alive but K dead) Because K and C are independent, we have P (next 3 years K alive but C dead) = P (next 3 years K alive) P (next 3 years C dead) P (next 3 years C alive but K dead) = P (next 3 years C alive) P (next 3 years K dead) Next, we need to use the script to look up the right rates. The problem tells us that the select period is 2 years. Even if the problem doesn’t tell us this information, we can tell from the select & ultimate table that the select period is 2 years. We see that 2+xp is not written as [ ] 2+xp . So we know that after 2 years the selection effect wears off and [ ] 2+x is the same as

2+x . Keith had a medical exam at age 45. Five years have passed since then and he is now 50. The effect of his medical exam has worn off and we should look up his survival rates using the ultimate table: K’s survival rates for the next three years:

x [ ]xp [ ] 1+xp 2+xp 2+x 48 0.9865 0.9841 0.9713 50 49 0.9858 0.9831 0.9698 51 50 0.9849 0.9819 0.9682 52 51 0.9838 0.9803 0.9664 53

Please note that we don’t use 0.9664 at age 53. The value of 0.9664 represents K’s survival rate from age 53 to 54.

17 of 210


Let’s turn to Clive. Clive was selected just today. So for the first two years, we’ll look up his rates from the select table. For Year 3, we have to use the ultimate table: C’s survival rates for the next 3 years:

x [ ]xp [ ] 1+xp 2+xp 2+x 48 0.9865 0.9841 0.9713 50 49 0.9858 0.9831 0.9698 51 50 0.9849 0.9819 0.9682 52 51 0.9838 0.9803 0.9664 53

We see that both Keith and Clive have the same survival rate of 0.9682 when they move from age 52 to age 53. Why? At age 52, they both have passed the select period and their survival rates depend only on their attained age. Since they have the same attained age, their survival rates are always the same in the ultimate table. Now we are ready to calculate the final result: P (next 3 years K alive but C dead) = P (next 3 years K alive) P (next 3 years C dead) = P (next 3 years K alive) * [1- P (next 3 years C alive) ] =0.9713(0.9698)(0.9682) * (1-0.9849 * 0.9819 * 0.9682)= 5.807% P (next 3 years C alive but K dead) = P (next 3 years C alive) P (next 3 years K dead) = P (next 3 years C alive) * [ 1- P (next 3 years K alive) ] =0.9849 * 0.9819 * 0.96821 * (1 - 0.9713 * 0.9698 * 0.9682)= 8.238% P (next 3 years later only K or C alive but not both) = 5.807% + 8.238% = 14.045% So the answer is [D] At least 0.135, but less than 0.145 FAQ I just bought your manual and finished reading Chapter 1. What should I do next? Should I start reverse engineering official M problems? Or should I read your manual first and then start reverse engineering? Either way is fine. Do whatever makes sense to you. I started reverse engineering old M problems but got stuck in Problem #1. The reverse study method is a lot harder than the traditional study method. What should I do next?

18 of 210


If reverse engineering is too time consuming, throw it away. Use the traditional study sequence. Or you can use 60% (or some other combinations) traditional method (“learn first, solve problem second”) and 40% reverse engineering (“first solve SOA problems and next learn what’s necessary”). Do whatever makes sense to you. I like reverse engineering for two reasons: (1) I’m in the driver’s seat. Under the traditional study method, I’m spoon-fed by authors of textbooks and study manuals. They do too much thinking for me. In contrast, under reverse engineering, I have to figure things out myself. This forces me to really understand the core concepts. (2) Reverse engineering cuts a lot of the fluff. Under the traditional study method, you’ll spend weeks mastering a difficult concept only to find out that it was not tested in the exam. Under reverse engineering, however, you learn what really matters for the exam from day one. Whether you use reverse engineering or the traditional method, make sure you master the official M exam problems. What’s the risk of using the COM method? Major risk: passing the exam quickly with only a moderate amount of study time. You may have to change your vacation plans and go to the beach rather than studying with your friends for a second exam-sitting. Your approach makes sense. However, I still want to solve lots of old SOA problems. If I don’t solve lots of practice problems, I feel insecure. Is there anything wrong with me solving lots of practice problems? The central message of the COM method is this: “Look. You really don’t need to study that hard. Master the official exam problems and you’ll be fine.” However, not every candidate is comfortable with this minimalist approach. If you still want to solve lots of practice problems, do so. Does the COM method work for the type of essay problems seen for Courses 5, 6, 7 and 8? No, the COM method won’t work for essay exams because in essay problems, there are an unlimited number of ways to test a candidate’s knowledge. To pass Courses 5, 6, 7, and 8, you must read everything in the syllabus. On the other hand, Exams P, FM, M, and C are multiple-choice, formula-driven exams. In these exams, there are a finite number of core concepts and formulas to master. SOA tests the same core concepts and formulas over and over. As a result, if you can master the officially-released M problems, you have gained sufficient knowledge of the core concepts in Exam M and should be able to pass. If SOA finds out that many people are using the COM method, can’t they make the exam different to purposely render COM useless?

19 of 210


The COM method works even if SOA knows that many candidates are using it. SOA purposely releases M exams so candidates will know what’s really tested on the exam. SOA wants people to master the official exam M problems. SOA will be flattered to know that you have mastered the officially released Exam M problems prior to taking the exam.

20 of 210


About the author Yufeng Guo was born in central China. After receiving his Bachelor’s degree in physics at Zhengzhou University, he attended Beijing Law School and received his Masters of law. He was an attorney and law school lecturer in China before immigrating to the United States. He received his Masters of accounting at Indiana University. He has pursued a life actuarial career and passed exams 1, 2, 3, 4, 5, 6, and 7 in rapid succession after discovering a successful study strategy. Mr. Guo’s exam records are as follows: Fall 2002 Passed Course 1 Spring 2003 Passed Courses 2, 3 Fall 2003 Passed Course 4 Spring 2004 Passed Course 6 Fall 2004 Passed Course 5 Spring 2005 Passed Course 7 Mr. Guo currently teaches an online prep course for Exam P, FM, and MLC, and MFE. For more information, visit http://actuary88.com. If you have any comments or suggestions, you can contact Mr. Guo at [email protected].

210 of 210



mailto:[email protected]�

by yufeng guo - actuarial bookstore 3mlc-yg-12ssm-e... · 2011-11-07 · by yufeng guo . 12th...

Documents