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Preface
P-1
Contents
Preface
P-7
Syllabus Reference
P-10
Flow Chart
P-13
Chapter 0
Chapter 1
Some Factual Information
C0-1
0.1
0.2
0.3
0.4
0.5
C0-1
C0-3
C0-3
C0-4
C0-6
Traditional Life Insurance Contracts
Modern Life Insurance Contracts
Underwriting
Life Annuities
Pensions
Survival Distributions
1.1
1.2
1.3
1.4
1.5
Age-at-death Random Variables
Future Lifetime Random Variable
Actuarial Notation
Curtate Future Lifetime Random Variable
Force of Mortality
Exercise 1
Solutions to Exercise 1
Chapter 2
Life Tables
2.1
2.2
2.3
2.4
2.5
Life Table Functions
Fractional Age Assumptions
Select-and-Ultimate Tables
Moments of Future Lifetime Random Variables
Useful Shortcuts
Life Insurances
3.1
3.2
3.3
3.4
3.5
3.6
3.7
C1-21
C1-27
C2-1
C2-6
C2-18
C2-29
C2-39
C2-43
C2-52
C3-1
Continuous Life Insurances
Discrete Life Insurances
mthly Life Insurances
Relating Different Policies
Recursions
Relating Continuous, Discrete and mthly Insurance
Useful Shortcuts
Exercise 3
Solutions to Exercise 3
Actex Learning
C1-1
C1-4
C1-6
C1-10
C1-12
C2-1
Exercise 2
Solutions to Exercise 2
Chapter 3
C1-1
| Johnny Li and Andrew Ng | SoA Exam MLC
C3-2
C3-17
C3-26
C3-29
C3-36
C3-43
C3-46
C3-48
C3-61
Preface
P-2
Chapter 4
Life Annuities
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
Chapter 5
Continuous Life Annuities
Discrete Life Annuities (Due)
Discrete Life Annuities (Immediate)
mthly Life Annuities
Relating Different Policies
Recursions
Relating Continuous, Discrete and mthly Life Annuities
Useful Shortcuts
C4-47
C4-61
Premium Calculation
C5-1
Traditional Insurance Policies
Net Premium and Equivalence Principle
Net Premiums for Special Policies
The Loss-at-issue Random Variable
Percentile Premium and Profit
C5-1
C5-3
C5-12
C5-18
C5-27
Exercise 5
Solutions to Exercise 5
C5-38
C5-55
Net Premium Reserves
C6-1
6.1
6.2
6.3
6.4
The
The
The
The
Prospective Approach
Recursive Approach: Basic Idea
Recursive Approach: Further Applications
Retrospective Approach
Exercise 6
Solutions to Exercise 6
Chapter 7
C4-1
C4-18
C4-25
C4-29
C4-31
C4-35
C4-38
C4-44
Exercise 4
Solutions to Exercise 4
5.1
5.2
5.3
5.4
5.5
Chapter 6
C4-1
Insurance Models Including Expenses
7.1
7.2
7.3
7.4
Gross Premium
Gross Premium Reserve
Expense Reserve and Modified Reserve
Basis, Asset Share and Profit
Exercise 7
Solutions to Exercise 7
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| Johnny Li and Andrew Ng | SoA Exam MLC
C6-2
C6-15
C6-24
C6-33
C6-41
C6-65
C7-1
C7-1
C7-5
C7-13
C7-23
C7-39
C7-57
Preface
Chapter 8
P-3
Multiple Decrement Models: Theory
8.1
8.2
8.3
8.4
Multiple Decrement Table
Forces of Decrement
Associated Single Decrement
Discrete Jumps
Exercise 8
Solutions to Exercise 8
Chapter 9
Multiple Decrement Models: Applications
9.1
9.2
9.3
9.4
Calculating Actuarial Present Values of Cash Flows
Calculating Reserve and Profit
Cash Values
Calculating Asset Shares under Multiple Decrement
C8-1
C8-1
C8-5
C8-10
C8-23
C8-29
C8-40
C9-1
C9-1
C9-4
C9-17
C9-23
Exercise 9
Solutions to Exercise 9
C9-28
C9-40
Chapter 10 Multiple State Models
C10-1
10.1
10.2
10.3
10.4
10.5
Discrete-time Markov Chain
Continuous-time Markov Chain
Kolmogorov’s Forward Equations
Calculating Actuarial Present Value of Cash Flows
Calculating Reserves
Exercise 10
Solutions to Exercise 10
Chapter 11 Multiple Life Functions
11.1
11.2
11.3
Multiple Life Statuses
Insurances and Annuities
Dependent Life Models
Exercise 11
Solutions to Exercise 11
Chapter 12 Interest Rate Risk
12.1
12.2
12.3
Yield Curves
Interest Rate Scenario Models
Diversifiable and Non-Diversifiable Risks
Exercise 12
Solutions to Exercise 12
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| Johnny Li and Andrew Ng | SoA Exam MLC
C10-4
C10-13
C10-19
C10-32
C10-39
C10-44
C10-63
C11-1
C11-2
C11-17
C11-31
C11-43
C11-63
C12-1
C12-1
C12-13
C12-17
C12-26
C12-37
Preface
P-4
Chapter 13 Profit Testing
13.1
13.2
13.3
C13-1
Profit Vector and Profit Signature
Profit Measures
Using Profit Test to Compute Premiums and Reserves
Exercise 13
Solutions to Exercise 13
Chapter 14 Universal Life Insurance
14.1
14.2
14.3
14.4
14.5
14.6
Basic Policy Design
Cost of Insurance and Surrender Value
Other Policy Features
Projecting Account Values
Profit Testing
Asset Shares for Universal Life Policies
Exercise 14
Solutions to Exercise 14
Chapter 15 Participating Insurance
15.1
15.2
C13-1
C13-12
C13-16
C13-24
C13-32
C14-1
C14-2
C14-5
C14-17
C14-21
C14-30
C14-40
C14-43
C14-53
C15-1
Dividends
Bonuses
C15-2
C15-10
Exercise 15
Solutions to Exercise 15
C15-24
C15-28
Chapter 16 Pension Mathematics
16.1
16.2
16.3
16.4
16.5
The Salary Scale Function
Pension Plans
Setting the DC Contribution Rate
DB Plans and Service Table
Funding of DB Plans
Exercise 16
Solutions to Exercise 16
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| Johnny Li and Andrew Ng | SoA Exam MLC
C16-1
C16-1
C16-12
C16-14
C16-19
C16-39
C16-45
C16-60
Preface
P-5
Appendix 1 Numerical Techniques
1.1
1.2
1.3
Numerical Integration
Euler’s Method
Solving Systems of ODEs with Euler’s Method
Appendix 2 Review of Probability
2.1
2.2
2.3
2.4
2.5
2.6
Probability Laws
Random Variables and Expectations
Special Univariate Probability Distributions
Joint Distribution
Conditional and Double Expectation
The Central Limit Theorem
A1-1
A1-1
A1-7
A1-12
A2-1
A2-1
A2-2
A2-6
A2-9
A2-10
A2-12
Exam MLC: General Information
T0-1
Mock Test 1
T1-1
Solution
T1-29
Mock Test 2
T2-1
Solution
T2-28
Mock Test 3
T3-1
Solution
T3-29
Mock Test 4
T4-1
Solution
T4-29
Mock Test 5
T5-1
Solution
T5-29
Mock Test 6
T6-1
Solution
T6-29
Mock Test 7
T7-1
Solution
T7-29
Mock Test 8
T8-1
Solution
Actex Learning
T8-28
| Johnny Li and Andrew Ng | SoA Exam MLC
Preface
P-6
Suggested Solutions to MLC May 2012
S-1
Suggested Solutions to MLC Nov 2012
S-17
Suggested Solutions to MLC May 2013
S-29
Suggested Solutions to MLC Nov 2013
S-45
Suggested Solutions to MLC April 2014
S-55
Suggested Solutions to MLC Oct 2014
S-69
Suggested Solutions to MLC April 2015
S-81
Suggested Solutions to MLC Oct 2015
S-97
Suggested Solutions to MLC May 2016
S-109
Suggested Solutions to MLC Oct 2016
S-123
Suggested Solutions to Sample Structural Questions
S-133
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| Johnny Li and Andrew Ng | SoA Exam MLC
Chapter 1: Survival Distributions
C1-1
Chapter 1
Survival Distributions
OBJECTIVES
1.
To define future lifetime random variables
2.
To specify survival functions for future lifetime random variables
3.
To define actuarial symbols for death and survival probabilities and
develop relationships between them
4.
To define the force of mortality
In Exam FM, you valued cash flows that are paid at some known future times. In Exam MLC,
by contrast, you are going to value cash flows that are paid at some unknown future times.
Specifically, the timings of the cash flows are dependent on the future lifetime of the underlying
individual. These cash flows are called life contingent cash flows, and the study of these cash
flows is called life contingencies.
It is obvious that an important part of life contingencies is the modeling of future lifetimes. In
this chapter, we are going to study how we can model future lifetimes as random variables. A
few simple probability concepts you learnt in Exam P will be used.
1. 1 Age-at-death Random Variable
Let us begin with the age-at-death random variable, which is denoted by T0. The definition of T0
can be easily seen from the diagram below.
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Chapter 1: Survival Distributions
C1-2
Death occurs
Age
T0
0
The age-at-death random variable can take any value within [0, ∞). Sometimes, we assume that
no individual can live beyond a certain very high age. We call that age the limiting age, and
denote it by ω. If a limiting age is assumed, then T0 can only take a value within [0, ω].
We regard T0 as a continuous random variable, because it can, in principle, take any value on
the interval [0, ∞) if there is no limiting age or [0, ω] if a limiting age is assumed. Of course, to
model T0, we need a probability distribution. The following notation is used throughout this
study guide (and in the examination).
− F0(t) = Pr(T0 ≤ t) is the (cumulative) distribution function of T0.
− f0(t) =
d
F0 (t ) is the probability density function of T0. For a small interval Δt, the product
dt
f0(t)Δt is the (approximate) probability that the age at death is in between t and t + Δt.
In life contingencies, we often need to calculate the probability that an individual will survive to
a certain age. This motivates us to define the survival function:
S0(t) = Pr(T0 > t) = 1 – F0(t).
Note that the subscript “0” indicates that these functions are specified for the age-at-death
random variable (or equivalently, the future lifetime of a person age 0 now).
Not all functions can be regarded as survival functions. A survival function must satisfy the
following requirements:
1. S0(0) = 1. This means every individual can live at least 0 years.
2. S0(ω ) = 0 or lim S 0 (t ) = 0. This means that every individual must die eventually.
t →∞
3. S0(t) is monotonically decreasing. This means that, for example, the probability of surviving
to age 80 cannot be greater than that of surviving to age 70.
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Chapter 1: Survival Distributions
C1-3
Summing up, f0(t), F0(t) and S0(t) are related to one another as follows.
F O R M U L A
Relations between f0(t), F0(t) and S0(t)
f 0 (t ) =
d
d
F0 (t ) = − S0 (t ) ,
dt
dt
∞
t
t
0
S 0 (t ) = ∫ f 0 (u )du = 1 − ∫ f 0 (u )du = 1 − F0 (t ) ,
Pr(a < T0 ≤ b) =
∫
b
f 0 (u )du = F0 (b) − F0 (a ) = S 0 (a ) − S 0 (b) .
a
(1.1)
(1.2)
(1.3)
Note that because T0 is a continuous random variable, Pr(T0 = c) = 0 for any constant c. Now, let
us consider the following example.
Example 1.1
[Structural Question]
You are given that S0(t) = 1 – t/100 for 0 ≤ t ≤ 100.
(a) Verify that S0(t) is a valid survival function.
(b) Find expressions for F0(t) and f0(t).
(c) Calculate the probability that T0 is greater than 30 and smaller than 60.
Solution
(a) First, we have S0(0) = 1 – 0/100 = 1.
Second, we have S0(100) = 1 – 100/100 = 0.
Third, the first derivative of S0(t) is −1/100, indicating that S0(t) is non-increasing.
Hence, S0(t) is a valid survival function.
(b) We have F0(t) = 1 – S0(t) = t/100, for 0 ≤ t ≤ 100.
Also, we have and f0(t) =
d
F0(t) = 1/100, for 0 ≤ t ≤ 100.
dt
(c) Pr(30 < T0 < 60) = S0(30) – S0(60) = (1 – 30/100) – (1 – 60/100) = 0.3.
[ END ]
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Chapter 1: Survival Distributions
C1-4
1. 2 Future Lifetime Random Variable
Consider an individual who is age x now. Throughout this text, we use (x) to represent such an
individual. Instead of the entire lifetime of (x), we are often more interested in the future
lifetime of (x). We use Tx to denote the future lifetime random variable for (x). The definition of
Tx can be easily seen from the diagram below.
0
x
x + Tx
Now
Death
occurs
0
Tx
Age
Time from now
[Note: For brevity, we may only display the portion starting from age x (i.e., time 0) in future
illustrations.]
If there is no limiting age, Tx can take any value within [0, ∞). If a limiting age is assumed, then
Tx can only take a value within [0, ω − x]. We have to subtract x because the individual has
attained age x at time 0 already.
We let Sx(t) be the survival function for the future lifetime random variable. The subscript “x”
here indicates that the survival function is defined for a life who is age x now. It is important to
understand that when modeling the future lifetime of (x), we always know that the individual is
alive at age x. Thus, we may evaluate Sx(t) as a conditional probability:
S x (t ) = Pr(Tx > t ) = Pr(T0 > x + t | T0 > x)
=
Pr(T0 > x + t ∩ T0 > x) Pr(T0 > x + t ) S0 ( x + t )
.
=
=
Pr(T0 > x)
Pr(T0 > x)
S0 ( x)
The third step above follows from the equation Pr( A | B) =
Exam P.
Actex Learning
Pr( A ∩ B)
, which you learnt in
Pr( B)
| Johnny Li and Andrew Ng | SoA Exam MLC
Chapter 1: Survival Distributions
C1-5
F O R M U L A
Survival Function for the Future Lifetime Random Variable
S x (t ) =
S0 ( x + t )
S0 ( x)
(1.4)
With Sx(t), we can obtain Fx(t) and fx(t) by using
Fx(t) = 1 – Sx(t) and fx(t) =
d
Fx (t ) ,
dt
respectively.
Example 1.2
[Structural Question]
You are given that S0(t) = 1 – t/100 for 0 ≤ t ≤ 100.
(a) Find expressions for S10(t), F10(t) and f10(t).
(b) Calculate the probability that an individual age 10 now can survive to age 25.
(c) Calculate the probability that an individual age 10 now will die within 15 years.
Solution
(a) In this part, we are asked to calculate functions for an individual age 10 now (i.e., x = 10).
Here, ω = 100 and therefore these functions are defined for 0 ≤ t ≤ 90 only.
First, we have S10 (t ) =
S0 (10 + t ) 1 − (10 + t ) /100
t
=
= 1 − , for 0 ≤ t ≤ 90.
S0 (10)
1 − 10 /100
90
Second, we have F10(t) = 1 – S10(t) = t/90, for 0 ≤ t ≤ 90.
Finally, we have f10 (t ) =
d
1
F10 (t ) =
.
dt
90
(b) The probability that an individual age 10 now can survive to age 25 is given by
Pr(T10 > 15) = S10(15) = 1 −
15
5
= .
90 6
(c) The probability that an individual age 10 now will die within 15 years is given by
Pr(T10 ≤ 15) = F10(15) = 1 − S10(15) =
1
.
6
[ END ]
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Chapter 1: Survival Distributions
C1-6
1. 3 Actuarial Notation
For convenience, we have designated actuarial notation for various types of death and survival
probabilities.
Notation 1: t px
We use t px to denote the probability that a life age x now survives to t years from now. By
definition, we have
t px
= Pr(Tx > t) = Sx(t).
When t = 1, we can omit the subscript on the left-hand-side; that is, we write 1px as px.
Notation 2: t qx
We use t qx to denote the probability that a life age x now dies before attaining age x + t. By
definition, we have
t qx
= Pr(Tx ≤ t) = Fx(t).
When t = 1, we can omit the subscript on the left-hand-side; that is, we write 1qx as qx.
Notation 3: t|u qx
We use t |u qx to denote the probability that a life age x now dies between ages x + t and x + t + u.
By definition, we have
t|u qx
= Pr(t < Tx ≤ t + u) = Fx(t + u) − Fx(t) = Sx(t) − Sx(t + u).
When u = 1, we can omit the subscript u; that is, we write t |1 qx as t | qx.
Note that when we describe survival distributions, “p” always means a survival probability,
while “q” always means a death probability. The “|” between t and u means that the death
probability is deferred by t years. We read “t | u” as “t deferred u”. It is important to remember
the meanings of these three actuarial symbols. Let us study the following example.
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Chapter 7: Insurance Models Including Expenses
Chapter 7
C7-1
Insurance Models Including Expenses
OBJECTIVES
1.
To calculate gross premium and gross premium reserve
2.
To calculate expense reserve and FPT and other modified reserves
3.
To calculate asset share, annual profit and gain
This is an important chapter and serves as a first check point of the basics of life contingencies.
In this chapter we discuss how expenses are incorporated into the calculation of premiums and
reserves. We are going to assume that the expenses are incurred with certainty.
7. 1 Gross Premium
Four Types of Expenses
Recall that net premiums are calculated under a basis that does not allow for expenses. The net
premium is used to pay for the cost of insurance. However, when a policy is sold, various
expenses are incurred by the insurer. It is important to understand the characteristics of these
expenses because they have a significant effect on premiums and reserves.
Initial / acquisition expenses
Initial expenses are expenses incurred when the policy is issued. This typically includes the
cost of underwriting and other costs of acquisition, including clerical expenses (setting up
records on computer systems), medical check-up fees, commissions and advertising expenses.
Some of these costs are fixed, while some of them vary with the amount of insurance or size of
the premium.
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Chapter 7: Insurance Models Including Expenses
C7-2
Maintenance / renewal expenses
After the policy is issued, there are continuing administrative expenses and commission. For
example, there is money spent on collecting premium payments and keeping track of all the
accounting records. Such expenses are normally incurred each time a premium is payable.
Under the US commission system, usually the rate of commission is higher in the first year, and
thus the maintenance expenses are typically smaller than initial expenses. You will see that this
would create a serious problem in later sections.
Settlement / termination expenses
Upon the termination of the policy, there are some paperwork and costs to finalize and
disburse benefit payments. Occasionally, a claim investigation is required and incurs a cost.
Usually, settlement expenses are insignificant when compared to acquisition and renewal
expenses.
Other miscellaneous expenses
There are some other expenses that may or may not be directly related to the policies but are
loaded to the policies for pricing. One such example is investment expense, which is reflected in
pricing by a reduction in the assumed interest rate. Another example includes general expenses
spent on research, legal services, taxes, licenses and other fees.
Typically, we can summarize and aggregate all expenses in the following form:
Year
1st year
2nd year
M
Per policy
40
6
M
Per 1000 insurance
5
0.5
M
Percentage of premium
50%
15%
M
The expenses above are payable at the beginning of each policy year
Settlement expense: 20 per policy plus 0.4 per 1000 insurance
For example, for a policy with a death benefit of 5000, and an annual level premium of 800:
1st year expense (payable at issuance) is 40 + 5 × 5 + 0.5 × 800 = 465
2nd year expense (payable 1 year after issuance) is 6 + 0.5 × 5 + 0.15 × 800 = 128.5
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Chapter 7: Insurance Models Including Expenses
C7-3
Upon death, the settlement expense is 20 + 0.4 × 5 = 22.
In what follows, we need to compute the APV of expenses. The APV of settlement expenses is
obviously in the form of a term or endowment insurance. The APV of initial and renewal
expense are in the form of a life annuity. For example, for an n-year level annual premium
policy, with premium G, if we have the following expenses:
Year
1st year
Renewal years
e0 + c0 G
Per policy
e0
e1
e1 + c1G e1 + c1G
Percentage of premium
c0
c1
...
e1 + c1 G
time
0
1
2
...
n−1
n
then the APV of the all renewal expenses is (e1 + c1G )a x:n −1| .
The APV of initial and renewal expenses is
(e0 + c 0 G ) + (e1 + c1G )a x:n −1| = (e0 + c0 G ) + (e1 + c1G )(a&&x:n| − 1)
= (e0 − e1 ) + (c0 − c1 )G + (e1 + c1G )a&&x:n|
Expense Augmented Model
We can formulate the gross future loss random variable at issue that incorporates expense
payments by defining
g
0L
= PV of benefit outgo + PV of expenses − PV of gross premium income.
The superscript g denotes “gross.” Under the equivalence principle, the gross premium (or
expense-loaded premium) is the premium such that the mean of 0Lg is 0 under a specific set of
mortality, interest and expenses assumptions that form a premium basis. (We shall elaborate on
basis later in section 4.) That is,
APV of benefits + APV of expenses = APV of gross premium income.
Other premium principles (e.g. percentile premium, premium calculated based on a profit
margin) can also be used and they would be discussed in Chapters 12 and 13. In what follows
we usually use a level gross premium, though in the discussion on the general theory we permit
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Chapter 7: Insurance Models Including Expenses
C7-4
other premium payment patterns. The gross premium at time t is Gt, while the net premium at
time t is πt.
As in the previous chapters, usually one needs not write down the precise mathematical
formulation of 0Lg when the equivalence principle is applied. However, the form of 0Lg would
be needed if one computes the percentile or the variance of 0Lg, or invoke any other premium
principles. The good news is that such questions are rarely examined in Exam MLC.
Example 7.1
[Structural Question]
Consider a 4-year annual level premium special endowment insurance on (40). The death
benefit is 1000, payable at the end of the year of death if death occurs within 4 years. If the life
survives 4 years, 1200 is payable 4 years after the issuance of the policy.
Calculate the net level annual premium and level gross premium under the following basis:
(i)
Mortality: q40 = 0.08, q41 = 0.09, q42 = 0.10, q43 = 0.11.
(ii)
Expenses: Initial expenses and renewal expenses (payable at the beginning of each policy
year)
Year
1st year
Renewal years
Per policy
10
2
Percentage of premium
25%
5%
Settlement expense is 5, payable at the end of the year for the first 3 years. There is no
settlement expense in the fourth year.
(iii) Interest: 6% annual effective.
Solution
We first calculate the required insurance and annuity functions:
a&&40:4| = 1 +
0.92 0.92 × 0.91 0.92 × 0.91 × 0.90
+
+
= 3.2456659,
1.06
1.06 2
1.06 3
4E40
=
0.92 × 0.91 × 0.90 × 0.89
= 0.53117579,
1.06 4
1
A40
= A40:4| − 4 E 40 = 1 −
:4|
0.06
× 3.2456659 − 0.53117579 = 0.28510727,
1.06
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Chapter 7: Insurance Models Including Expenses
1
A40
=
:3|
C7-5
0.08 0.92 × 0.09 0.92 × 0.91 × 0.1
+
+
= 0.21945633 .
1.06
1.06 2
1.06 3
Calculation of net level premium:
Let π be the net level premium.
1
πa&&40:4| = 1000 A40
+ 1200 4 E 40
:4|
π=
1
1000 A40
+ 1200 4 E 40
:4|
a&&40:4|
= 284.2308
Calculation of gross premium:
Let G be the level gross premium. The initial and renewal expenses add up to
10 + 0.25G + (2 + 0.05G )a 40:3| = 8 + 0.2G + (2 + 0.05G )a&&40:4| .
1
The APV of the settlement expense is 5A40
.
:3|
1
1
Ga&&40:4| = 1000 A40
+ 1200 4 E 40 + 5 A40
+ 8 + 0.2G + (2 + 0.05G )a&&40:4|
:4|
:3|
G=
1
1
1000 A40
+ 1200 4 E 40 + 5 A40
+ 8 + 2a&&40:4|
:3|
:4|
0.95a&&40:4| − 0.2
= 325.3494
[ END ]
The insurance company may charge a premium that is greater than the gross premium because
the insurer can only break even but not make a profit if the insured is charged a gross premium.
7. 2 Gross Premium Reserve
Consider an n-year term insurance policy which is still in force t years after issue. Define the
present value of gross future loss random variable by
g
tL
= Present value, at time t, of future benefits and expenses
− Present value at time t, of future gross premiums.
Death benefit:
bt − 1
Settlement expense:
Et − 1
Renewal / Initial expense: Rt − 1
bt
Et
Rt
bt + 1
Et + 1
Rt + 1
bn − 1
En − 1
Rn − 1
bn
En
time
0
Gross premiums:
…
t−1
t
t+1
Gt − 1
Gt
Gt + 1
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…
n−1
n
Gn − 1
| Johnny Li and Andrew Ng | SoA Exam MLC
Chapter 7: Insurance Models Including Expenses
C7-6
The present value, at time t, of expenses includes the renewal expense at time t because they are
associated with the premium then due at time t. The settlement expense at time t is not included
in the calculation of tLg because tLg is defined condition on Tx > t.
The gross premium reserve (or expense-augmented reserve) at time t is the conditional mean
g
tV
= E( tLg | Tx > t ),
where the expectation is calculated under a specific basis. As a result, gross premium reserve
can be calculated prospectively by
g
tV
= APV at time t of future benefits and expenses − APV at time t of future gross premiums.
This definition holds no matter if the policy is discrete, continuous or semi-continuous. The
gross premium above is the actual premium charged for the policy. The gross premium may be,
or may not be determined from the equivalence principle. If it is determined from the
equivalence principle, then 0Vg = 0. (Note, however, that when one compute the net premium
reserve, net premium is always used, so that 0V must be 0.)
Example 7.2
Consider the set up in Example 7.1. Calculate the gross premium reserve at the end of year 2
and year 3.
Solution
Death benefit:
1000
Survival benefit:
5
Settlement expense:
Renewal expenses:
2 + 0.05G
Initial expense: 10 + 0.25G
1000
5
2 + 0.05G
1000
5
2 + 0.05G
1000
1200
time
Gross premiums:
For 2Vg,
0
1
2
3
G
G
G
G
4
g
2V
Actex Learning | Johnny Li and Andrew Ng | SoA Exam MLC
Mock Test 4
T4-1
Exam MLC Mock Test 4
SECTION A − Multiple Choice
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| Johnny Li and Andrew Ng | SoA Exam MLC
Mock Test 4
T4-2
**BEGINNING OF EXAMINATION**
1.
For a fully discrete whole life annuity immediate of 10 per year issued to (37), you are
given:
(i)
Mortality follows the Illustrative Life Table.
(ii)
i = 0.06
(iii) Y is the present value random variable for the annuity immediate.
Calculate the standard deviation of Y.
(A) 21
(B) 22
(C) 23
(D) 24
(E) 25
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| Johnny Li and Andrew Ng | SoA Exam MLC
Mock Test 4
2.
T4-3
The present value random variable for an insurance policy on (x) is expressed as:
⎧ 0
⎪2v K x +1
⎪
Z = ⎨ K x +1
⎪v
⎪⎩2v K x +1
if K x < 5
if 5 ≤ K x < 10
if 10 ≤ K x < 15
thereafter
Which of the following is a correct expression for E(Z)?
(A) 25|Ax − 10|Ax
(B) 25|Ax − 10|Ax + 215|Ax
(C) 25ExAx − 10ExAx+10 + 15ExAx+15
(D) 25ExAx+5 − 10ExAx+10 + 15ExAx+15
(E)
5Ex(2Ax+5
− 5Ex+5Ax+10 + 15Ex+5Ax+15)
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| Johnny Li and Andrew Ng | SoA Exam MLC
Mock Test 4
T4-4
3.
For a special fully continuous whole life insurance on (44), you are given:
(i)
Premiums and benefits:
Premium Rate
Benefit
(ii)
μx =
First 20 years
3P
100,000
Thereafter
P
300,000
1
for x ≤ 84.
84 − x
(iii) δ = 0.06
Calculate P using the equivalence principle.
(A) 1,900
(B) 1,915
(C) 1,930
(D) 1,945
(E) 1,960
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| Johnny Li and Andrew Ng | SoA Exam MLC
Mock Test 4
4.
T4-5
You are given:
(i)
i = 0.05
(ii)
A60 = 0.560
(iii) A40 = 0.176
(iv)
20p40
= 0.75
( 3)
using the two-term Woolhouse approximation.
Calculate a 40
:20|
(A) 13.7
(B) 14.2
(C) 14.5
(D) 14.9
(E) 15.1
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| Johnny Li and Andrew Ng | SoA Exam MLC
Mock Test 4
T4-6
5.
For a multiple decrement model on (60), you are given:
(i)
μ 60(1)+t follows the Illustrative Life Table for t ≥ 0.
(ii)
μ 60(τ +) t = 2.5μ 60(1)+t for t ≥ 0.
Calculate the probability that failure of (60) occurs during the 10th year due to cause 1.
(A) 0.25
(B) 0.27
(C) 0.29
(D) 0.31
(E) 0.33
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| Johnny Li and Andrew Ng | SoA Exam MLC
Mock Test 4
19.
T4-21
For a defined benefit pension plan, you are given:
(i)
The retirement benefit is 15 per month of service.
(ii)
Retirement is allowed beginning at age 63.
(iii) There is one employee who is currently exact age 45 and was hired at exact age
40.
(iv) i = 0.07
(v)
The following data
x
63
64
65
q ′x( r )
0.2
0.3
1.0
a&&x(12)
10
9
8.5
Assume:
(i)
Employees retire only on their birthdays.
(ii)
There are no pre-retirement decrements.
Calculate the actuarial present value of the projected retirement benefit for this employee.
(A) 10,600
(B) 11,400
(C) 12,600
(D) 13,900
(E) 15,300
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| Johnny Li and Andrew Ng | SoA Exam MLC
Mock Test 4
T4-22
20.
Consider a universal life insurance sold to (40). The death benefit is $120,000 plus the
account value at the end of the year of death. You are given:
(i)
The following basis is used to calculate cost of insurance:
− Mortality follows the Illustrative Life Table.
− Interest rate is 5% per annum effective.
(ii)
For each premium paid, the expense charge is $50 plus 3% of the premium.
(iii)
There is no corridor requirement.
(iv)
The death benefit is payable at the end of the year of death.
(v)
The surrender penalties are as follows:
Year of surrender
Penalty
1
$3100
2
$2700
3
$2500
≥4
$900
Project the cash value at the end of the third year using the following assumptions:
− The policy remains in force for three years.
− Interest is credited to the policyholder’s account at 1% per annum.
− The policyholder pays a premium of $2000 at the beginning of each year.
(A) 0
(B) 770
(C) 1340
(D) 1770
(E) 2240
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Mock Test 4
T4-23
Exam MLC Mock Test 4
SECTION B − Written Answer
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| Johnny Li and Andrew Ng | SoA Exam MLC
Mock Test 4
T4-24
1.
(7 points) For this question, you are not required to show your steps.
(a)
(4 points) Identify the actuarial functions with the following integral or sum
representations. Verbally explain the meaning behind each of the actuarial functions
you use.
n
(i)
∫v
(ii)
1
2
3
⎞
1 ⎛⎜
n
m
m
m
⎟
p
v
p
v
p
v
...
p
v
+
+
+
+
x
n
x
1
2
3
x
x
⎟
m ⎜⎝ m
m
m
⎠
(iii)
t
0
∞
∑v
k = n +1
(iv)
(b)
∫
∞
0
p x dt
t
k
k
px
at| t p x μ x +t dt
(3 points) For each of the following actuarial functions, write down its integral or
summation form and explain verbally what it means.
(i)
n|
(ii)
a&&x:n|
Ax
(iii) ex
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| Johnny Li and Andrew Ng | SoA Exam MLC
Mock Test 4
2.
T4-25
(13 points)
(a)
(4 points) Explain in words the difference between Type A and Type B universal
life insurance policies.
(b)
(5 points) For a Type A universal life insurance policy with a face amount of
$60,000, you are given:
•
The account value on December 31, 2013 was $30,000.
•
On January 1, 2014, a premium of $2,500 was made. No other premiums
were made in 2014.
•
The total expense charge and the cost of insurance deducted on January 1,
2014 were $170 and $250, respectively.
•
The credited interest rate in 2014 was 8% per annum effective.
•
The corridor factor requirement is 2.5.
•
Death occurred in 2014 and a death benefit was payable at the end of
2014.
Compute the amount of the death benefit.
(c)
(4 points) Suppose that the universal life insurance policy in (b) is a Type B policy
with an additional death benefit of $60,000 instead. Recalculate the amount of the
death benefit.
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| Johnny Li and Andrew Ng | SoA Exam MLC
Mock Test 4
T4-26
3.
(14 points)
(a)
(9 points) For each of the following risks, determine if the risk is diversifiable.
Explain your answers verbally.
(i)
The uncertainty associated with future lifetimes of policyholders, provided that
the underlying mortality rates are fixed and known and that policyholders'
lifetimes are independent
(ii)
The uncertainty associated with death probabilities
(iii) The uncertainty associated with interest rates
(b)
(5 points) For pricing a 3-year life annuity-due of 1 on (x) to be issued on January 1,
2015, you are given:
= 0.95 − 0.03k, k = 1, 2
•
kpx
•
δ(k) is the force of interest for year k, k = 2014, 2015, 2016.
•
δ(2014) = 0.03
•
δ(k + 1) − δ(k) equals 0.01 with probability 70% and equals −0.01 with
probability 30%, for k = 2014, 2015.
•
Y is the expected present value of this annuity on January 1, 2015.
Calculate Y.
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| Johnny Li and Andrew Ng | SoA Exam MLC
Mock Test 4
4.
T4-27
(15 points) You are given:
•
i = 0.05
•
10|A40
•
2
•
a&&50 = 14.63606
= 0.178809, A40 = 0.20799, 10E40 = 0.590041
A40 = 0.06741, 2A50 = 0.12446
Let Y be the present value random variable for a 10-year deferred life annuity-due to (40)
of 1 per year.
Let Z1 be the present value random variable for a whole life insurance of 1 on (40) with
benefits at the end of the year of death.
Let Z2 be the present value random variable for a 10-year endowment insurance of 1 on
(40) with benefits at the end of the year of death.
(a)
(3 points) Express Y, Z1 and Z2 in terms of K40, v and d.
(b)
(2 points) Using your results in (a), show that Y = 21(Z2 – Z1).
(c)
(5 points)
(i)
Show that
Z1Z 2 = Z 32 + v10 Z 4
for some present value random variables Z3 and Z4. Identify Z3 and Z4.
(ii)
Hence, calculate E(Z1Z2).
(d) (5 points) Calculate the variance of Y.
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| Johnny Li and Andrew Ng | SoA Exam MLC
Mock Test 4
T4-28
5.
(7 points) Let x, y and m be positive integers. The lifetimes of (x) and (y) are independent
and follow the uniform distribution of death (UDD) fractional age assumption.
(a)
(4 points)
It is known that Ax( m ) =
assumption.
(i)
i
i
(m)
Ax under the specified fractional age
Using the given fact, or otherwise, show that under the specified fractional age
assumption,
a&&x( m ) = α (m)a&&x − β (m) ,
id
i − i (m)
where α (m) = ( m ) ( m ) and β (m) = ( m ) ( m ) .
i d
i d
(ii)
By considering the result from (a)(i), derive a formula for calculating a&&x( m:n)|
under the specified fractional age assumption. Express your answer in terms of
α(m), β(m), nEx and a&&x:n| .
(b)
(3 points) It is known that under the specified fractional age assumption,
a&&xy( m ) = α (m)a&&xy − β (m) −
1
2
2 ⎞ ∞ k +1
i
⎛
+ ⎟∑ v k pxy qx + k q y + k
⎜1 + −
i( m)d ( m) ⎝ m d (m) i ⎠ k =0
≈ α (m)a&&xy − β (m).
(i)
Explain why
∞
∑v
k =0
(ii)
k +1
k
p xy q x + k q y + k is typically small.
Rather than using the UDD assumption and the approximation above, one may
approximate a&&xy( m ) by using the Woolhouse formula. Write down the Woolhouse
formula for approximating a&&xy( m ) . Express your answer in terms of a&&xy , m, δ and
μxy.
**END OF EXAMINATION**
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| Johnny Li and Andrew Ng | SoA Exam MLC