TABLE OF CONTENTS - VOLUME 1
INTRODUCTORY COMMENTS
MODELING
SECTION 1 - PROBABILITY REVIEW
PROBLEM SET 1
LM-1
LM-9
SECTION 2 - REVIEW OF RANDOM VARIABLES - PART I
PROBLEM SET 2
LM-19
LM-29
SECTION 3 - REVIEW OF RANDOM VARIABLES - PART II
PROBLEM SET 3
LM-35
LM-43
SECTION 4 - REVIEW OF RANDOM VARIABLES - PART III
PROBLEM SET 4
LM-51
LM-59
SECTION 5 - PARAMETRIC DISTRIBUTIONS AND TRANSFORMATIONS
PROBLEM SET 5
LM-63
LM-71
SECTION 6 - DISTRIBUTION TAIL BEHAVIOR
PROBLEM SET 6
LM-75
LM-79
SECTION 7 - MIXTURE OF TWO DISTRIBUTIONS
PROBLEM SET 7
LM-79
LM-87
SECTION 8 - MIXTURE OF DISTRIBUTIONS
PROBLEM SET 8
LM-93
LM-99
SECTION 9 - CONTINUOUS MIXTURES
PROBLEM SET 9
LM-107
LM-113
SECTION 10 - POLICY LIMITS
AND THE LIMITED LOSS RANDOM VARIABLE
PROBLEM SET 10
LM-121
LM-125
SECTION 11 - POLICY DEDUCTIBLE (1),
THE COST PER LOSS RANDOM VARIABLE
PROBLEM SET 11
LM-127
LM-133
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING
SECTION 12 - POLICY DEDUCTIBLE (2),
THE COST PER PAYMENT RANDOM VARIABLE
PROBLEM SET 12
LM-143
LM-149
SECTION 13 - POLICY DEDUCTIBLES APPLIED TO THE UNIFORM,
EXPONENTIAL AND PARETO DISTRIBUTIONS
PROBLEM SET 13
LM-161
LM-169
SECTION 14 - COMBINED LIMIT AND DEDUCTIBLE
PROBLEM SET 14
LM-173
LM-179
SECTION 15 - ADDITIONAL POLICY ADJUSTMENTS
PROBLEM SET 15
LM-191
LM-195
SECTION 16 - MODELS FOR THE NUMBER OF CLAIMS
AND THE ²Á Á ³ and ²Á Á ³ CLASSES
PROBLEM SET 16
LM-199
LM-207
SECTION 17 - MODELS FOR THE AGGREGATE LOSS,
COMPOUND DISTRIBUTIONS (1)
PROBLEM SET 17
LM-219
LM-223
SECTION 18 - COMPOUND DISTRIBUTIONS (2)
PROBLEM SET 18
LM-245
LM-251
SECTION 19 - MORE PROPERTIES OF THE AGGREGATE
LOSS RANDOM VARIABLE
PROBLEM SET 19
LM-265
LM-269
SECTION 20 - STOP LOSS INSURANCE
PROBLEM SET 20
LM-281
LM-287
SECTION 21 - RISK MEASURES
PROBLEM SET 21
LM-293
LM-297
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODEL ESTIMATION
SECTION 1 - REVIEW OF MATHEMATICAL STATISTICS (1)
ESTIMATORS
PROBLEM SET 1
ME-1
ME-7
SECTION 2 - REVIEW OF MATHEMATICAL STATISTICS (2)
CONFIDENCE INTERVALS AND HYPOTHESIS TESTS
PROBLEM SET 2
ME-11
ME-17
SECTION 3 - NON-PARAMETRIC EMPIRICAL POINT ESTIMATION
PROBLEM SET 3
ME-23
ME-31
SECTION 4 - KERNEL SMOOTHING ESTIMATORS
PROBLEM SET 4
ME-37
ME-51
SECTION 5 - EMPIRICAL ESTIMATION FROM GROUPED DATA
PROBLEM SET 5
ME-61
ME-67
SECTION 6 - ESTIMATION FROM CENSORED AND TRUNCATED DATA
PROBLEM SET 6
ME-75
ME-85
SECTION 7 - PROPERTIES OF SURVIVAL PROBABILITY ESTIMATORS
PROBLEM SET 7
ME-101
ME-109
SECTION 8 - MOMENT AND PERCENTILE MATCHING
PROBLEM SET 8
ME-123
ME-135
SECTION 9 - MAXIMUM LIKELIHOOD ESTIMATION
PROBLEM SET 9
ME-151
ME-161
SECTION 10 - MAXIMUM LIKELIHOOD ESTIMATION
FOR THE EXPONENTIAL DISTRIBUTION
PROBLEM SET 10
ME-173
ME-179
SECTION 11 - MAXIMUM LIKELIHOOD ESTIMATION
FOR PARETO AND WEIBULL DISTRIBUTIONS
PROBLEM SET 11
ME-185
ME-195
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODEL ESTIMATION
SECTION 12 - MAXIMUM LIKELIHOOD ESTIMATION
FOR DISTRIBUTIONS IN THE EXAM C TABLE
PROBLEM SET 12
ME-201
ME-211
SECTION 13 - PROPERTIES OF MAXIMUM LIKELIHOOD ESTIMATORS
PROBLEM SET 13
ME-219
ME-223
SECTION 14 - GRAPHICAL EVALUATION OF ESTIMATED MODELS
PROBLEM SET 14
ME-233
ME-237
SECTION 15 - HYPOTHESIS TESTS FOR FITTED MODELS
PROBLEM SET 15
ME-243
ME-255
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
INTRODUCTORY COMMENTS
This study guide is designed to help in the preparation for the Society of Actuaries Exam C and
Casualty Actuarial Society Exam 4. The exam covers the topics of modeling, model estimation,
construction and selection, credibility, simulation and risk measures.
The study manual is divided into two volumes. The first volume consists of a summary of notes,
illustrative examples and problem sets with detailed solutions on the modeling and model
estimation topics. The second volume consists of notes examples and problem sets on the
credibility, simulation and risk measures topics, as well as 14 practice exams.
The practice exams all have 35 questions. The level of difficulty of the practice exams has been
designed to be similar to that of the past 4-hour exams. Some of the questions in the problem sets
are taken from the relevant topics on SOA/CAS exams that have been released prior to 2009 but
the practice exam questions are not from old SOA exams.
I have attempted to be thorough in the coverage of the topics upon which the exam is based. I
have been, perhaps, more thorough than necessary on a couple of topics, such as maximum
likelihood estimation, Bayesian credibility and applying simulation to hypothesis testing.
Because of the time constraint on the exam, a crucial aspect of exam taking is the ability to work
quickly. I believe that working through many problems and examples is a good way to build up
the speed at which you work. It can also be worthwhile to work through problems that have been
done before, as this helps to reinforce familiarity, understanding and confidence. Working many
problems will also help in being able to more quickly identify topic and question types. I have
attempted, wherever possible, to emphasize shortcuts and efficient and systematic ways of setting
up solutions. There are also occasional comments on interpretation of the language used in some
exam questions. While the focus of the study guide is on exam preparation, from time to time
there will be comments on underlying theory in places that I feel those comments may provide
useful insight into a topic.
The notes and examples are divided into sections anywhere from 4 to 14 pages, with suggested
time frames for covering the material. There are over 330 examples in the notes and over 800
exercises in the problem sets, all with detailed solutions. The 14 practice exams have 35 questions
each, also with detailed solutions. Some of the examples and exercises are taken from previous
SOA/CAS exams. Questions in the problem sets that have come from previous SOA/CAS exams
are identified as such. Some of the problem set exercises are more in depth than actual exam
questions, but the practice exam questions have been created in an attempt to replicate the level of
depth and difficulty of actual exam questions. In total there are aver 1600
examples/problems/sample exam questions with detailed solutions. ACTEX gratefully
acknowledges the SOA and CAS for allowing the use of their exam problems in this study guide.
I suggest that you work through the study guide by studying a section of notes and then
attempting the exercises in the problem set that follows that section. My suggested order for
covering topics is
(1) modeling (includes risk measures), (2) model estimation , (Volume 1) ,
(3) credibility theory , and (4) simulation , (Volume 2).
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
It has been my intention to make this study guide self-contained and comprehensive for all Exam
C topics, but there are occasional references to the Loss Models reference book (3rd edition)
listed in the SOA/CAS catalog. While the ability to derive formulas used on the exam is usually
not the focus of an exam question, it is useful in enhancing the understanding of the material and
may be helpful in memorizing formulas. There may be an occasional reference in the review
notes to a derivation, but you are encouraged to review the official reference material for more
detail on formula derivations. In order for the review notes in this study guide to be most
effective, you should have some background at the junior or senior college level in probability
and statistics. It will be assumed that you are reasonably familiar with differential and integral
calculus. The prerequisite concepts to modeling and model estimation are reviewed in this study
guide. The study guide begins with a detailed review of probability distribution concepts such as
distribution function, hazard rate, expectation and variance.
Of the various calculators that are allowed for use on the exam, I am most familiar with the
BA II PLUS. It has several easily accessible memories. The TI-30X IIS has the advantage of a
multi-line display. Both have the functionality needed for the exam.
There is a set of tables that has been provided with the exam in past sittings. These tables consist
of some detailed description of a number of probability distributions along with tables for the
standard normal and chi-squared distributions. The tables can be downloaded from the SOA
website www.soa.org .
If you have any questions, comments, criticisms or compliments regarding this study guide,
please contact the publisher ACTEX, or you may contact me directly at the address below. I
apologize in advance for any errors, typographical or otherwise, that you might find, and it would
be greatly appreciated if you would bring them to my attention. ACTEX will be maintaining a
website for errata that can be accessed from www.actexmadriver.com .
It is my sincere hope that you find this study guide helpful and useful in your preparation for the
exam. I wish you the best of luck on the exam.
Samuel A. Broverman
Department of Statistics
University of Toronto
© ACTEX 2009
April, 2009
www.sambroverman.com
E-mail: sam@utstat.toronto.edu
or
2brove@rogers.com
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING
MODELING SECTION 1 - PROBABILITY REVIEW
LM-1
MODELING SECTION 1 - PROBABILITY REVIEW
Basic Probability, Conditional Probability and Independence
Exam C applies probability and statistical methods to various aspects of loss modeling and model
estimation. A good background in probability and statistics is necessary to fully understand
models and the modeling that is done. In this section of the study guide, we will review
fundamental probability rules. The suggested time frame for this section (not including exercises)
is two hours.
LM-1.1 Basic Probability Concepts
Sample point and probability space
A sample point is the simple outcome of a random experiment. The probability space (also called
sample space) is the collection of all possible sample points related to a specified experiment.
When the experiment is performed, one of the sample points will be the outcome. An experiment
could be observing the loss that occurs on an automobile insurance policy during the course of
one year, or observing the number of claims arriving at an insurance office in one week. The
probability space is the "full set" of possible outcomes of the experiment. In the case of the
automobile insurance policy, it would be the range of possible loss amounts that could occur
during the year, and in the case of the insurance office weekly number of claims, the probability
space would be the set of integers ¸Á Á Á ÀÀÀ¹.
Event
Any collection of sample points, or any subset of the probability space is referred to as an event.
We say "event ( has occurred" if the experimental outcome was one of the sample points in (.
Union of events ( and )
( r ) denotes the union of events ( and ), and consists of all sample points that are in either (
or ) .
(r)
Union of events ( Á ( Á ÀÀÀÁ (
( r ( r Ä r ( ~ r ( denotes the union of the events ( Á ( Á ÀÀÀÁ ( , and consists of all
~
sample points that are in at least one of the ( 's. This definition can be extended to the union of
infinitely many events.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-2
MODELING SECTION 1 - PROBABLITY REVIEW
Intersection of events ( Á ( Á ÀÀÀÁ (
( q ( q Ä q ( ~ q ( denotes the intersection of the events ( Á ( Á ÀÀÀÁ ( , and consists
~
of all sample points that are simultaneously in all of the ( 's.
(q)
Mutually exclusive events ( Á ( Á ÀÀÀÁ (
Two events are mutually exclusive if they have no sample points in common, or equivalently, if
they have empty intersection. Events ( Á ( Á ÀÀÀÁ ( are mutually exclusive if ( q ( ~ J
for all £ , where J denotes the empty set with no sample points. Mutually exclusive events
cannot occur simultaneously.
Exhaustive events ) Á ) Á ÀÀÀÁ )
If ) r ) r Ä r ) ~ : , the entire probability space, then the events ) Á ) Á ÀÀÀÁ ) are
referred to as exhaustive events.
Complement of event (
The complement of event ( consists of all sample points in the probability space that are not in
c
(. The complement is denoted (Á — (Á (Z or ( and is equal to ¸% ¢ % ¤ (¹. When the
underlying random experiment is performed, to say that the complement of ( has occurred is the
same as saying that ( has not occurred.
Subevent (or subset) ( of event )
If event ) contains all the sample points in event (, then ( is a subevent of ) , denoted ( ‰ ) .
The occurrence of event ( implies that event ) has occurred.
Partition of event (
Events * Á * Á ÀÀÀÁ * form a partition of event ( if ( ~ r * and the * 's are mutually
~
exclusive.
DeMorgan's Laws
(i) ²( r )³Z ~ (Z q ) Z , to say that ( r ) has not occurred is to say that ( has not occurred
and ) has not occurred ; this rule generalizes to any number of events;
Z
4 r ( 5 ~ ²( r ( r Ä r ( ³Z ~ (Z q (Z q Ä q (Z ~ q (Z
~
~
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 1 - PROBABILITY REVIEW
LM-3
(ii) ²( q )³Z ~ (Z r ) Z , to say that ( q ) has not occurred is to say that either ( has not
occurred or ) has not occurred (or both have not occurred) ; this rule generalizes to any
Z
number of events, 4 q ( 5 ~ ²( q ( q Ä q ( ³Z ~ (Z r (Z r Ä r (Z ~ r (Z
~
~
Indicator function for event (
The function 0( ²%³ ~ D
if %(
if %¤(
is the indicator function for event (, where % denotes a sample
point. 0( ²%³ is 1 if event ( has occurred.
Some important rules concerning probability are given below.
(i) 7 ´:µ ~ if : is the entire probability space (when the underlying experiment is
performed, some outcome must occur with probability 1).
(ii) 7 ´Jµ ~ (the probability of no face turning up when we toss a die is 0).
(iii) If events ( Á ( Á ÀÀÀÁ ( are mutually exclusive (also called disjoint) then
7 ´ r ( µ ~ 7 ´( r ( r Ä r ( µ ~ 7 ´( µ b 7 ´( µ b Ä b 7 ´( µ ~ 7 ´( µ .
~
~
This extends to infinitely many mutually exclusive events.
(iv) For any event (,  7 ´(µ  .
(1.1)
(v) If ( ‰ ) then 7 ´(µ  7 ´)µ .
(vi) For any events (, ) and * , 7 ´( r )µ ~ 7 ´(µ b 7 ´)µ c 7 ´( q )µ .
(1.2)
(vii) For any event (, 7 ´(Z µ ~ c 7 ´(µ .
(1.3)
(viii) For any events ( and ) , 7 ´(µ ~ 7 ´( q )µ b 7 ´( q ) Z µ
(1.4)
(ix) For exhaustive events ) Á ) Á ÀÀÀÁ ) , 7 ´ r ) µ ~ .
~
If ) Á ) Á ÀÀÀÁ ) are exhaustive and mutually exclusive, they form a partition of the
entire probability space, and for any event (,
7 ´(µ ~ 7 ´( q ) µ b 7 ´( q ) µ b Ä b 7 ´( q ) µ ~ 7 ´( q ) µ .
(1.5)
(1.6)
~
(x) The words "percentage" and "proportion" are used as alternatives to "probability".
As an example, if we are told that the percentage or proportion of a group of people that are of a
certain type is 20%, this is generally interpreted to mean that a randomly chosen person from
the group has a 20% probability of being of that type. This is the "long-run frequency"
interpretation of probability. As another example, suppose that we are tossing a fair die. In the
long-run frequency interpretation of probability, to say that the probability of tossing a 1 is is
the same as saying that if we repeatedly toss the coin, the proportion of tosses that are 1's will
approach .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-4
MODELING SECTION 1 - PROBABLITY REVIEW
LM-1.2 Conditional Probability and Independence of Events
Conditional probability arises throughout the Exam C material. It is important to be familiar and
comfortable with the definitions and rules of conditional probability.
Conditional probability of event ( given event )
If 7 ²)³ € , then 7 ²(O)³ ~
7 ²(q)³
7 ²)³
is the conditional probability that event ( occurs
given that event ) has occurred. By rewriting the equation we get 7 ²( q )³ ~ 7 ²(O)³ h 7 ²)³.
Partition of a Probability Space
Events ) Á ) Á ÀÀÀÁ ) are said to form a partition of a probability space : if
(i) ) r ) r Ä r ) ~ : and (ii) ) q ) ~ J for any pair with £ .
A partition is a disjoint collection of events which combines to be the full probability space.
A simple example of a partition is any event ) and its complement ) Z .
If ( is any event in probability space : and ¸) Á ) Á ÀÀÀÁ ) ¹ is a partition of probability space : ,
then 7 ²(³ ~ 7 ²( q ) ³ b 7 ²( q ) ³ b Ä b 7 ²( q ) ³ .
A special case of this rule is 7 ²(³ ~ 7 ²( q )³ b 7 ²( q ) Z ³ for any two events ( and ) .
Bayes rule and Bayes Theorem
For any events ( and ) with 7 ²(³ € , 7 ²)O(³ ~
7 ²(O)³h7 ²)³
7 ²(³
.
(1.7)
If ) Á ) Á ÀÀÀÁ ) form a partition of the entire sample space : , then
7 ²) O(³ ~
7 ²(O) ³h7 ²) ³
7 ²(O) ³h7 ²) ³
for each ~ Á Á ÀÀÀÁ .
(1.8)
~
The values of 7 ²) ³ are called prior probabilities, and the value of 7 ²) O(³ is called a
posterior probability. Variations on this rule are very important in Bayesian credibility.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 1 - PROBABILITY REVIEW
LM-5
Independent events ( and )
If events ( and ) satisfy the relationship 7 ²( q )³ ~ 7 ²(³ h 7 ²)³ , then the events are
said to be independent or stochastically independent or statistically independent. The
independence of (non-empty) events ( and ) is equivalent to 7 ²(O)³ ~ 7 ²(³ or
7 ²)O(³ ~ 7 ²)³ .
Mutually independent events ( Á ( Á ÀÀÀÁ (
The events are mutually independent if
(i) for any ( and ( , 7 ²( q ( ³ ~ 7 ²( ³ d 7 ²( ³ , and
(ii) for any ( , ( and ( , 7 ²( q ( q ( ³ ~ 7 ²( ³ d 7 ²( ³ d 7 ²( ³ ,
and so on for any subcollection of the events, including all events:
7 ²( q ( q Ä q ( ³ ~ 7 ²( ³ d 7 ²( ³ d Ä d 7 ²( ³ ~ 7 ²( ³ .
(1.9)
~
Here are some rules concerning conditional probability and independence. These
can be verified in a fairly straightforward way from the definitions given above.
(i) 7 ²( r )³ ~ 7 ²(³ b 7 ²)³ c 7 ²( q )³ for any events ( and )
(1.10)
(ii) 7 ²( q )³ ~ 7 ²)O(³ h 7 ²(³ ~ 7 ²(O)³ h 7 ²)³ for any events ( and )
(1.11)
(iii) If ) Á ) Á ÀÀÀÁ ) form a partition of the sample space : , then for any event (,
~
~
7 ²(³ ~ 7 ²( q ) ³ ~ 7 ²(O) ³ h 7 ²) ³ ;
(1.12)
as a special case, for any events ( and ) , we have
7 ²(³ ~ 7 ²( q )³ b 7 ²( q ) Z ³ ~ 7 ²(O)³ h 7 ²)³ b 7 ²(O)Z ³ h 7 ²)Z ³
(1.13)
(iv) If 7 ²( q ( q Ä q (c ³ € , then
7 ²( q ( q Ä q ( ³ ~ 7 ²( ³ h 7 ²( O( ³ h 7 ²( O( q ( ³Ä7 ²( O( q ( q Ä q (c ³
(v) 7 ²(Z ³ ~ c 7 ²(³ and 7 ²(Z O)³ ~ c 7 ²(O)³
(vi) if ( ‰ ) then 7 ²(O)³ ~
7 ²(q)³
7 ²)³
(1.14)
7 ²(³
~ 7 ²)³ , and 7 ²)O(³ ~ (vii) if ( and ) are independent events then (Z and ) are independent events,
( and ) Z are independent events, and (Z and ) Z are independent events
(viii) since 7 ²J³ ~ 7 ²J q (³ ~ ~ 7 ²J³ h 7 ²(³ for any event (, it follows that J is
independent of any event (
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-6
MODELING SECTION 1 - PROBABLITY REVIEW
Example LM1-1: Suppose a fair six-sided die is tossed. We define the following events:
( ~ "the number tossed is  " ~ ¸Á Á ¹ , ) ~ "the number tossed is even" ~ ¸Á Á ¹ ,
* ~ "the number tossed is a or a " ~ ¸Á ¹ ,
+ ~ "the number tossed doesn't start with the letters 'f' or 't'" ~ ¸Á ¹ .
The conditional probability of ( given ) is
7 ²¸ÁÁ¹q¸ÁÁ
¹³
7 ²¸ÁÁ
¹³
7 ²¸¹³
°
~ 7 ²¸ÁÁ
¹³ ~ ° ~ .
Events ( and ) are not independent, since ~ 7 ²( q )³ £ 7 ²(³ h 7 ²)³ ~ h ~ ,
or alternatively, events ( and ) are not independent since 7 ²(O)³ £ 7 ²(³.
7 ²(O*³ ~ £ ~ 7 ²(³, so that ( and * are not independent.
7 ²(O)³ ~
7 ²)O*³ ~ ~ 7 ²)³ , so that ) and * are independent
(alternatively, 7 ²) q *³ ~ 7 ²¸¹³ ~ ~ h ~ 7 ²)³ h 7 ²*³ ).
It is not difficult to check that both ( and ) are independent of +.
IMPORTANT NOTE: The following manipulation of event probabilities arises from time to
time: 7 (() ~ 7 ((O) ) h 7 ²)³ b 7 ((O) Z ) h 7 ²) Z ³ . If we know the conditional
probabilities for event ( given some other event ) and its complement ) Z , and if we know the
(unconditional) probability of event ), then we can find the probability of event (. One of the
important aspects of applying this relationship is the determination of the appropriate events (
and ) .
Example LM1-2: Urn I contains 2 white and 2 black balls and Urn II contains 3 white and 2
black balls. An Urn is chosen at random, and a ball is randomly selected from that Urn. Find the
probability that the ball chosen is white.
Solution: Let ) be the event that Urn I is chosen and ) Z is the event that Urn II is chosen. The
implicit assumption is that both Urns are equally likely to be chosen (this is the meaning of "an
Urn is chosen at random"). Therefore, 7 ²)³ ~ and 7 ²) Z ³ ~ . Let ( be the event that the
ball chosen in white. If we know that Urn I was chosen, then there is probability of choosing a
white ball (2 white out of 4 balls, it is assumed that each ball has the same chance of being
chosen); this can be described as 7 ²(O)³ ~ .
In a similar way, if Urn II is chosen, then 7 ²(O) Z ³ ~ (3 white out of 5 balls). We can now
apply the relationship described prior to this example.
7 ²( q )³ ~ 7 ²(O)³ h 7 ²)³ ~ ² ³² ³ ~ , and
7 ²( q ) Z ³ ~ 7 ²(O) Z ³ h 7 ²) Z ³ ~ ² ³² ³ ~ . Finally,
7 ²(³ ~ 7 ²( q )³ b 7 ²( q ) Z ³ ~ b ~ .
The order of calculations can be summarized in the following table
(
)
1. 7 ²( q )³ ~ 7 ²(O)³ h 7 ²)³
)Z
2. 7 ²( q ) Z ³ ~ 7 ²(O) Z ³ h 7 ²) Z ³
3. 7 ²(³ ~ 7 ²( q )³ b 7 ²( q ) Z ³
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 1 - PROBABILITY REVIEW
LM-7
Example LM1-3: Urn I contains 2 white and 2 black balls and Urn II contains 3 white and 2
black balls. One ball is chosen at random from Urn I and transferred to Urn II, and then a ball is
chosen at random from Urn II. The ball chosen from Urn II is observed to be white. Find the
probability that the ball transferred from Urn I to Urn II was white.
Solution: Let ) denote the event that the ball transferred from Urn I to Urn II was white and let
( denote the event that the ball chosen from Urn II is white. We are asked to find 7 ²)O(³ .
From the simple nature of the situation (and the usual assumption of uniformity in such a
situation, meaning all balls are equally likely to be chosen from Urn I in the first step), we have
7 ²)³ ~ (2 of the 4 balls in Urn I are white), and by implication, it follows that 7 ´) Z µ ~ .
If the ball transferred is white, then Urn II has 4 white and 2 black balls, and the probability of
choosing a white ball out of Urn II is ; this is 7 ²(O)³ ~ .
If the ball transferred is black, then Urn II has 3 white and 3 black balls, and the probability of
choosing a white ball out of Urn II is ; this is 7 ²(O) Z ³ ~ .
All of the information needed has been identified. We do calculations in the following order:
1.
7 ´( q )µ ~ 7 ´(O)µ h 7 ´)µ ~ ² ³² ³ ~ 7 ´( q ) Z µ ~ 7 ´(O) Z µ h 7 ´) Z µ ~ ² ³² ³ ~ 2.
7 ´(µ ~ 7 ´( q )µ b 7 ´( q ) Z µ ~ b ~ 7 ´)q(µ
°
7 ´)O(µ ~ 7 ´(µ ~ ° ~ .
3.
4.
Example LM1-4: Three dice have the following probabilities of throwing a "six": Á Á Á
respectively. One of the dice is chosen at random and thrown (each is equally likely to be
chosen). A "six" appeared. What is the probability that the die chosen was the first one?
Solution: The event " a 6 is thrown" is denoted by "6"
7 ´die O"
"µ ~
7 ´²die ³q²"
"³µ
7 ´"
" µ
7 ´"
"Odie µh7 ´die µ
7 ´"
"µ
h ~ 7 ´"
"µ .
But 7 ´"
"µ ~ 7 ´²"
"³ q ²die ³µ b 7 ´²"
"³ q ²die ³µ b 7 ´²"
"³ q ²die ³µ
~ 7 ´"
"Odie µ h 7 ´die µ b 7 ´"
"Odie µ h 7 ´die µ b 7 ´"
"Odie µ h 7 ´die µ
~ h
bh
© ACTEX 2009
bh
~
~
bb
h h S 7 ´die O"
"µ ~ 7 ´"
"µ ~ ²bb³h
~ bb .
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-8
© ACTEX 2009
MODELING SECTION 1 - PROBABLITY REVIEW
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LOSS MODELS - PROBLEM SET 1
LM-9
MODELING - PROBLEM SET 1
Review of Probability - Section 1
1. A survey of 1000 people determines that 80% like walking and 60% like biking, and
all like at least one of the two activities. How many people in the survey like biking but not
walking?
A) 0
B) .1
C) .2
D) .3
E) .4
2. A life insurer classifies insurance applicants according to the following attributes:
4 - the applicant is male
/ - the applicant is a homeowner
Out of a large number of applicants the insurer has identified the following information:
40% of applicants are male, 40% of applicants are homeowners and
20% of applicants are female homeowners.
Find the percentage of applicants who are male and do not own a home.
A) .1
B) .2
C) .3
D) .4
E) .5
3. Let (Á )Á * and + be events such that ) ~ (Z Á * q + ~ J, and
7 ´(µ ~ , 7 ´)µ ~ , 7 ´*O(µ ~ , 7 ´*O)µ ~ , 7 ´+O(µ ~ , 7 ´+O)µ ~ Calculate 7 ´* r +µ .
A) B) C) D) E) 4. You are given that 7 ´(µ ~ À and 7 ´( r )µ ~ À .
Actuary 1 assumes that ( and ) are independent and calculates 7 ´)µ based on that assumption.
Actuary 2 assumes that ( and ) mutually exclusive and calculates 7 ´)µ based on that
assumption. Find the absolute difference between the two calculations.
A) 0
B) .05
C) .10
D) .15
E) .20
5. A test for a disease correctly diagnoses a diseased person as having the disease with
probability .85. The test incorrectly diagnoses someone without the disease as having the disease
with a probability of .10. If 1% of the people in a population have the disease, what is the chance
that a person from this population who tests positive for the disease actually has the disease?
A) À B) À
C) À D) À E) À
6. Two bowls each contain 5 black and 5 white balls. A ball is chosen at random from bowl 1
and put into bowl 2. A ball is then chosen at random from bowl 2 and put into bowl 1. Find the
probability that bowl 1 still has 5 black and 5 white balls.
A) B) C) D) E) © ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-10
MODELING - PROBLEM SET 1
7. People passing by a city intersection are asked for the month in which they were born. It is
assumed that the population is uniformly divided by birth month, so that any randomly passing
person has an equally likely chance of being born in any particular month. Find the minimum
number of people needed so that the probability that no two people have the same birth month is
less than .5 .
A) 2
B) 3
C) 4
D) 5
E) 6
8. In a T-maze, a laboratory rat is given the choice of going to the left and getting food or going
to the right and receiving a mild electric shock. Assume that before any conditioning (in trial
number 1) rats are equally likely to go the left or to the right. After having received food on a
particular trial, the probability of going to the left and right become .6 and .4, respectively on the
following trial. However, after receiving a shock on a particular trial, the probabilities of going to
the left and right on the next trial are .8 and .2, respectively. What is the probability that the
animal will turn left on trial number 2?
A) .1
B) .3
C) .5
D) .7
E) .9
9. In the game show "Let's Make a Deal", a contestant is presented with 3 doors. There is a prize
behind one of the doors, and the host of the show knows which one. When the contestant makes
a choice of door, at least one of the other doors will not have a prize, and the host will open a
door (one not chosen by the contestant) with no prize. The contestant is given the option to
change his choice after the host shows the door without a prize. If the contestant switches doors,
what is the probability that he gets the door with the prize?
A) B) C) D) E) 10. A supplier of a testing device for a type of component claims that the device is highly
reliable, with 7 ´(O)µ ~ 7 ´(Z O) Z µ ~ À , where
( ~ device indicates component is faulty, and ) ~ component is faulty .
You plan to use the testing device on a large batch of components of which 5% are faulty.
Find the probability that the component is faulty given that the testing device indicates that the
component is faulty .
A) 0
B) .05
C) .15
D) .25
E) .50
11. An insurer classifies flood hazard based on geographical areas, with hazard categorized as
low, medium and high. The probability of a flood occurring in a year in each of the three areas is
Area Hazard
low
medium
high
Prob. of Flood
.001
.02
.25
The insurer's portfolio of policies consists of a large number of policies with 80% low hazard
policies, 18% medium hazard policies and 2% high hazard policies. Suppose that a policy had a
flood claim during a year. Find the probability that it is a high hazard policy.
A) .50
B) .53
C) .56
D) .59
E) .62
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LOSS MODELS - PROBLEM SET 1
LM-11
12. One of the questions asked by an insurer on an application to purchase a life insurance policy
is whether or not the applicant is a smoker. The insurer knows that the proportion of smokers in
the general population is .30, and assumes that this represents the proportion of applicants who
are smokers. The insurer has also obtained information regarding the honesty of applicants:
40% of applicants that are smokers say that they are non-smokers on their applications,
none of the applicants who are non-smokers lie on their applications.
What proportion of applicants who say they are non-smokers are actually non-smokers?
A) B) C) D) E) 13. When sent a questionnaire, 50% of the recipients respond immediately. Of those who do not
respond immediately, 40% respond when sent a follow-up letter. If the questionnaire is sent to 4
persons and a follow-up letter is sent to any of the 4 who do not respond immediately, what is the
probability that at least 3 never respond?
A) ²À³ b ²À³ ²À³
B) ²À³ ²À³
C) ²À³ b ²À³ ²À³
D) À²À³²À³ b ²À³
E) ²À³ b ²À³ ²À³
14. A fair coin is tossed. If a head occurs, 1 fair die is rolled; if a tail occurs, 2 fair dice are
rolled. If @ is the total on the die or dice, then 7 ´@ ~ µ ~
A) B) C) D) E) 15. In Canada's national 6-49 lottery, a ticket has 6 numbers each from 1 to 49, with no repeats.
Find the probability of matching exactly 4 of the 6 winning numbers if the winning numbers are
all randomly chosen.
A) .00095
B) .00097
C) .00099
D) .00101
E) .00103
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-12
MODELING - PROBLEM SET 1
MODELING - PROBLEM SET 1 SOLUTIONS
1. Let ( ~ "like walking" and ) ~ "like biking" . We use the interpretation that "percentage"
and "proportion" are taken to mean "probability".
We are given 7 ²(³ ~ À Á 7 ²)³ ~ À
and 7 ²( r )³ ~ .
From the diagram below we can see that since ( r ) ~ ( r ²) q (Z ³ we have
7 ²( r )³ ~ 7 ²(³ b 7 ²(Z q )³ S 7 ²(Z q )³ ~ À is the proportion of people who like biking
but (and) not walking . In a similar way we get 7 ²( q ) Z ³ ~ À .
An algebraic approach is the following. Using the rule 7 ²( r )³ ~ 7 ²(³ b 7 ²)³c7 ²( q )³,
we get ~ À b À
c 7 ²( q )³ S 7 ²( q )³ ~ À . Then, using the rule
7 ²)³ ~ 7 ²) q (³ b 7 ²) q (Z ³ , we get 7 ²) q (Z ³ ~ À
c À ~ À . Answer: C
2. 7 ´4 µ ~ À Á 7 ´4 Z µ ~ À
Á 7 ´/µ ~ À Á 7 ´/ Z µ ~ À
Á 7 ´4 Z q /µ ~ À Á
We wish to find 7 ´4 q / Z µ . From probability rules, we have
À
~ 7 ´/ Z µ ~ 7 ´4 Z q / Z µ b 7 ´4 q / Z µ , and
À
~ 7 ´4 Z µ ~ 7 ´4 Z q /µ b 7 ´4 Z q / Z µ ~ À b 7 ´4 Z q / Z µ .
Thus, 7 ´4 Z q / Z µ ~ À and then 7 ´4 q / Z µ ~ À . The following diagram identifies the
component probabilities.
The calculations above can also be summarized in the following table. The events across the top
of the table categorize individuals as male (4 ) or female (4 Z ), and the events down the left side
of the table categorize individuals as homeowners (/ ) or non-homeowners (/ Z ).
7 ²/³ ~ À
given
7 ²/ Z ³ ~ c À ~ À
Answer: B
© ACTEX 2009
7 ²4 ³ ~ À , given
7 ²4 Z ³ ~ c À ~ À
7 ²4 q /³
«
7 ²4 Z q /³ ~ À , given
Z
~ 7 ²/³ c 7 ²4 q /³ ~ À c À ~ À
®
7 ²4 q / Z ³ ~ 7 ²4 ³ c 7 ²4 q /³ ~ À c À ~ À
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LOSS MODELS - PROBLEM SET 1
LM-13
3. Since * and + have empty intersection, 7 ´* r +µ ~ 7 ´*µ b 7 ´+µ À
Also, since ( and ) are "exhaustive" events (since they are complementary events, their union is
the entire sample space, with a combined probability of
7 ´( r )µ ~ 7 ´(µ b 7 ´)µ ~ ).
We use the rule 7 ´*µ ~ 7 ´* q (µ b 7 ´* q (Z µ , and the rule 7 ´*O(µ ~
7 ´*µ ~ 7 ´*O(µ h 7 ´(µ b 7 ´*O(Z µ h 7 ´(Z µ ~ h b h ~ and
Z
Z
7 ´+µ ~ 7 ´+O(µ h 7 ´(µ b 7 ´+O( µ h 7 ´( µ ~ h b h ~ À
Then, 7 ´* r +µ ~ 7 ´*µ b 7 ´+µ ~ À
7 ´(q*µ
7 ´(µ
to get
Answer: C.
4. Actuary 1: Since ( and ) are independent, so are (Z and ) Z .
7 ´(Z q ) Z µ ~ c 7 ´( r )µ ~ À .
But À ~ 7 ´(Z q ) Z µ ~ 7 ´(Z µ h 7 ´) Z µ ~ ²À³7 ´)Z µ S 7 ´)Z µ ~ À
S 7 ´)µ ~ À .
Actuary 2: À ~ 7 ´( r )µ ~ 7 ´(µ b 7 ´)µ ~ À b 7 ´)µ S 7 ´)µ ~ À .
Absolute difference is OÀ c ÀO ~ À .
Answer: E
5. We define the following events: + - a person has the disease ,
; 7 - a person tests positive for the disease. We are given 7 ´; 7 O+µ ~ À and
7 ´; 7 O+Z µ ~ À and 7 ´+µ ~ À . We wish to find 7 ´+O; 7 µ .
Using the formulation for conditional probability we have 7 ´+O; 7 µ ~
7 ´+q; 7 µ
7 ´; 7 µ
.
But 7 ´+ q ; 7 µ ~ 7 ´; 7 O+µ h 7 ´+µ ~ ²À³²À³ ~ À , and
7 ´+Z q ; 7 µ ~ 7 ´; 7 O+Z µ h 7 ´+Z µ ~ ²À³²À³ ~ À . Then,
7 ´; 7 µ ~ 7 ´+ q ; 7 µ b 7 ´+Z q ; 7 µ ~ À S 7 ´+O; 7 µ ~ À
À ~ À .
The following table summarizes the calculations.
7 ´+µ ~ À , given
¬
7 ´+Z µ ~ c 7 ´+µ ~ À
®
®
7 ´+ q ; 7 µ
7 ´+Z q ; 7 µ
~ 7 ´; 7 O+µ h 7 ´+µ ~ À
~ 7 ´; 7 O+Z µ h 7 ´+Z µ ~ À
®
7 ´; 7 µ ~ 7 ´+ q ; 7 µ b 7 ´+Z q ; 7 µ ~ À
®
7 ´+q; 7 µ
À
7 ´+O; 7 µ ~ 7 ´; 7 µ ~ À ~ À .
Answer: B
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-14
MODELING - PROBLEM SET 1
6. Let * be the event that bowl 1 has 5 black balls after the exchange.
Let ) be the event that the ball chosen from bowl 1 is black, and
let ) be the event that the ball chosen from bowl 2 is black.
Event * is the disjoint union of ) q ) and )Z q )Z (black-black or
white-white picks), so that 7 ´*µ ~ 7 ´) q ) µ b 7 ´)Z q )Z µ .
The black-black combination has probability ² ³² ³ ,
since there is a chance of picking black from bowl 1, and then (with 6 black in bowl 2, which
now has 11 balls) is the probability of picking black from bowl 2. This is
7 ´) q ) µ ~ 7 ´) O) µ h 7 ´) µ ~ ² ³² ³ .
In a similar way, the white-white combination has probability ² ³² ³ .
Then 7 ´*µ ~ ² ³² ³ b ² ³² ³ ~ .
Answer: C
7. ( ~ event that second person has different birth month from the first.
7 ²( ³ ~ ~ À
À
( ~ event that third person has different birth month from first and second.
Then, the probability that all three have different birthdays is
7 ´( q ( µ ~ 7 ´( O( µ h 7 ²( ³ ~ ² ³² ³ ~ À
.
( ~ event that fourth person has different birth month from first three.
Then, the probability that all four have different birthdays is
7 ´( q ( q ( µ ~ 7 ´( O( q ( µ h 7 ´( q ( µ
~ 7 ´( O( q ( µ h 7 ´( O( µ h 7 ²( ³ ~ ² ³² ³² ³ ~ À .
( ~ event that fifth person has different birth month from first four.
Then, the probability that all five have different birthdays is
7 ´( q ( q ( q ( µ ~ 7 ´( O( q ( q ( µ h 7 ´( q ( q ( µ
~ 7 ´( O( q ( q ( µ h 7 ´( O( q ( µ h 7 ´( O( µ h 7 ²( ³
~ ² ³² ³² Answer: D
³² ³ ~ À .
8. 3 ~ turn left on trial 1, 9 ~ turn right on trial 1, 3 ~ turn left on trial 2 .
We are given that 7 ´3µ ~ 7 ´9µ ~ À .
7 ´3µ ~ 7 ´3 q 3µ b 7 ´3 q 9µ since 3Á 9 form a partition .
7 ´3O3µ ~ À
(if the rat turns left on trial 1 then it gets food and has a .6 chance of turning left
on trial 2). Then 7 ´3 q 3µ ~ 7 ´3O3µ h 7 ´3µ ~ ²À
³²À³ ~ À .
In a similar way, 7 ´3 q 9µ ~ 7 ´3O9µ h 7 ´9µ ~ ²À³²À³ ~ À .
Then, 7 ´3µ ~ À b À ~ À .
Answer: D
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LOSS MODELS - PROBLEM SET 1
LM-15
9. We define the events ( ~ prize door is chosen after contestant switches doors ,
) ~ prize door is initial one chosen by contestant . Then 7 ´)µ ~ , since each door is equally
likely to hold the prize initially. To find 7 ´(µ we use the Law of Total Probability.
7 ´(µ ~ 7 ´(O)µ h 7 ´)µ b 7 ´(O) Z µ h 7 ´) Z µ ~ ²³² ³ b ²³² ³ ~ .
If the prize door is initially chosen, then after switching, the door chosen is not the prize door, so
that 7 ´(O)µ ~ . If the prize door is not initially chosen, then since the host shows the other
non-prize door, after switching the contestant definitely has the prize door, so that 7 ´(O) Z µ ~ .
Answer: E
10. We are given 7 ´)µ ~ À . We can calculate entries in the following table in the order
indicated.
(
(Z
)
7 ´)µ ~ À
(given)
7 ´(O)µ ~ À (given)
1. 7 ´( q )µ ~ 7 ´(O)µ h 7 ´)µ ~ À
7 ´(Z O) Z µ ~ À (given)
)Z
7 ´) Z µ
~ c 7 ´)µ
~ À
3. 7 ´( q ) Z µ
~ 7 ´) Z µ c 7 ´(Z q ) Z µ
~ À c À ~ À
2. 7 ´(Z q ) Z µ
~ 7 ´(Z O) Z µ h 7 ´) Z µ
~ À ~ À
5. 7 ´)O(µ ~
4. 7 ´(µ ~ 7 ´( q )µ b 7 ´( q ) Z µ ~ À
7 ´)q(µ
7 ´(µ
~ À
À ~ À .
Answer: E
11. This is a classical Bayesian probability situation. Let * denote the event that a flood claim
occurred. We wish to find 7 ²/O*³ .
We can summarize the information in the following table, with the order of calculations
indicated.
3 Á 7 ²3³ ~ À
4 Á 7 ²4 ³ ~ À
/ Á 7 ²/³ ~ À
(given)
(given)
(given)
*
7 ²*O3³ ~ À
(given)
7 ²*O4 ³ ~ À
(given)
1. 7 ²* q 3³
~ 7 ²*O3³ h 7 ²3³
~ À
2. 7 ²* q 4 ³
~ 7 ²*O4 ³ h 7 ²4 ³
~ À
7 ²*O/³ ~ À
(given)
3. 7 ²* q /³
~ 7 ²*O/³ h 7 ²/³
~ À
4. 7 ²*³ ~ 7 ²* q 3³ b 7 ²* q 4 ³ b 7 ²* q /³ ~ À .
5. 7 ²/O*³ ~
© ACTEX 2009
7 ²/q*³
7 ²*³
À
~ À
~ À .
Answer: B
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-16
MODELING - PROBLEM SET 1
12. We identify the following events:
: - the applicant is a smoker,
5 : - the applicant is a non-smoker ~ : Z
+: - the applicant declares to be a smoker on the application
+5 - the applicant declares to be non-smoker on the application ~ +: Z .
The information we are given is 7 ´:µ ~ À Á 7 ´5 :µ ~ À Á 7 ´+5 O:µ ~ À Á 7 ´+:O5 :µ ~ .
7 ´5 :q+5 µ
.
7 ´+5 µ
7 ´+5 q:µ
7 ´+5 q:µ
We calculate À ~ 7 ´+5 O:µ ~ 7 ´:µ ~
S 7 ´+5 q :µ ~
À
7 ´+:q5 :µ
7 ´+:q5 :µ
and ~ 7 ´+:O5 :µ ~ 7 ´5 :µ ~
S 7 ´+: q 5 :µ ~ .
À
We wish to find 7 ´5 :O+5 µ ~
À ,
Using the rule 7 ´(µ ~ 7 ´( q )µ b 7 ´( q ) Z µ, and noting that +: ~ +5 Z and : ~ 5 : Z
we have 7 ´+: q :µ ~ 7 ´:µ c 7 ´+5 q :µ ~ À c À ~ À , and
7 ´+5 q 5 :µ ~ 7 ´5 :µ c 7 ´+: q 5 :µ ~ À c ~ À , and
7 ´+5 µ ~ 7 ´+5 q 5 :µ b 7 ´+5 q :µ ~ À b À ~ À .
7 ´5 :q+5 µ
7 ´+5 µ
À
~ À
~ .
These calculations can be summarized in the order indicated in the following table.
7 ²:³ Á À
¬
1. 7 ²5 :³ ~ c 7 ²:³ ~ À
given
®
Then, 7 ´5 :O+5 µ ~
6. +: «
7 ²+:³
~ 7 ²+: q :³
b 7 ²+: q 5 :³
~ À b ~ À
®
7. +5
7 ²+5 ³
~ c 7 ²+:³
~ c À
~ À
5. 7 ²+: q :³
~ 7 ²:³ c 7 ²+5 q :³
~ À c À ~ À
­
4. 7 ²+5 O:³ ~ À
given
7 ²+5 q :³
~ 7 ²+5 O:³ h 7 ²:³
~ ²À³²À³ ~ À
Then, 8. 7 ´5 :O+5 µ ~
7 ´5 :q+5 µ
7 ´+5 µ
À
~ À
~ .
2. 7 ²+:O5 :³ ~ , given
7 ²+: q 5 :³
~ 7 ²+:O5 :³ h 7 ²5 :³
~ ²³²À³ ~ 3. 7 ²+5 q 5 :³ ~
~ 7 ²5 :³ c 7 ²+: q 5 :³
~ À c ~ À
Answer: D
13. The probability that an individual will not respond to either the questionnaire or the
follow-up letter is ²À³²À
³ ~ À . The probability that all 4 will not respond to either the
questionnaire or the follow-up letter is ²À³ .
7 ´ don't respondµ ~ 7 ´ response on 1st round, no additional responses on 2nd roundµ
b 7 ´no responses on 1st round, 1 response on 2nd roundµ
~ ´²À³ ²À
³ µ b ´²À³ ²À
³ ²À³µ ~ ²À³ ²À³ . Then,
7 ´at least 3 don't respondµ ~ ²À³ b ²À³ ²À³ .
Answer: A
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LOSS MODELS - PROBLEM SET 1
LM-17
14. If 1 fair die is rolled, the probability of rolling a 6 is , and if 2 fair dice are rolled, the
probability of rolling a 6 is (of the 36 possible rolls from a pair of dice, the rolls 1-5,
2-4, 3-3, 4-2 and 5-1 result in a total of 6), Since the coin is fair, the probability of rolling a head
or tail is .5. Thus, the probability that @ ~ is ²À³² ³ b ²À³² ³ ~ À
Answer: C
15. Suppose you have bought a lottery ticket. There are 2 3 ~ ways of picking 4 numbers
from the 6 numbers on your ticket. Suppose we look at one of those subsets of 4 numbers from
your ticket. In order for the winning ticket number to match exactly those 4 of your 6 numbers,
the other 2 winning ticket numbers must come from the 43 numbers between 1 and 49 that are not
3 d
numbers on your ticket. There are 2 ~ d ~ ways of doing that, and since there are 15
subsets of 4 numbers on your ticket, there are d ~ Á ways in which the winning
ticket numbers match exactly 3 of your ticket numbers. Since there are a total of 13,983,816 ways
of picking 6 out of 49 numbers, your chance of matching exactly of the winning numbers is
13,545
Answer: B
ÁÁ
~ À
.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-18
© ACTEX 2009
MODELING - PROBLEM SET 1
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 2 - REVIEW OF RANDOM VARIABLES PART I
LM-19
MODELING SECTION 2 - REVIEW OF RANDOM VARIABLES -PART I
Probability, Density and Distribution Functions
This section relates to Chapter 2 (3rd edition) of "Loss Models". The suggested time frame for
covering this section is two hours. A brief review of some basic calculus relationships is
presented first.
LM-2.1 Calculus Review
Natural logarithm and exponential functions
²%³ ~ ²%³ is the logarithm to the base ;
²³ ~ Á ²³ ~ Á ~ Á
²& ³ ~ & Á ²%³ ~ % Á ²& ³ ~ & h ²³ Á
²& h '³ ~ ²&³ b ²'³ Á ² &' ³ ~ ²&³ c ²'³ Á
% ' ~ %b' Á ²% ³$ ~ %$ .
Differentiation
²%b³c ²%³
For the function ²%³ Á Z ²%³ ~ % ~ lim
Â
¦
Product rule: %
´²%³²%³µ ~ Z ²%³²%³ b ²%³Z ²%³ Â
Z
Z
²%³
²%³
Quotient rule: %
´ ²%³ µ ~ ²%³ ²%³c²%³
Â
´²%³µ
Z ²%³
%
Chain rule: %
´²%³µ ~ ²%³ Á %
´²%³µ ~ ´²%³µc h Z ²%³ Á %
~ % h ²³
(2.1)
(2.2)
(2.3)
(2.4)
(2.5)
Integration
b
%
% % ~ %
% % ~ b Á % ~ h ´ b %µ b b
Á
(2.6)
b
²³
b%
Integration by parts : "²!³ #²!³ ~ "²³#²³ c "²³#²³ c #²!³ "²!³ for definite integrals,
and " # ~ "# c # " for indefinite integrals (this is derived by integrating both sides of the
product rule) ; note that #²!³ ~ #Z ²!³ ! and "²!³ ~ "Z ²!³ ! Â
%
% ²!³ !
~ ²%³ Á %
% ²!³ ! ~ c ²%³ Á
²%³
% ²%³ ²!³!
~ ²²%³³ h Z ²%³ c ²²%³³ h Z ²%³ Á
B % c% % ~ [
b if € and is an integer ‚ .
© ACTEX 2009
(2.7)
(2.8)
(2.9)
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-20
MODELING SECTION 2 - REVIEW OF RANDOM VARIABLES PART I
The word "model" used in the context of a loss model, usually refers to the distribution of a loss
random variable. Random variables are the basic components used in actuarial modeling. In this
section we review the definitions and illustrate the variety of random variables that we will
encounter in the Exam C material.
A random variable is a numerical quantity that is related to the outcome of some random
experiment on a probability space. For the most part, the random variables we will encounter are
the numerical outcomes of some loss related event such as the dollar amount of claims in one year
from an auto insurance policy, or the number of tornados that touch down in Kansas in a one year
period.
LM-2.2 Discrete Random Variable
The random variable ? is discrete and is said to have a discrete distribution if it can take on
values only from a finite or countable infinite sequence (usually the integers or some subset of the
integers). As an example, consider the following two random variables related to successive
tosses of a coin:
? ~ if the first head occurs on an even-numbered toss, ? ~ if the first head occurs on an
odd-numbered toss;
@ ~ , where is the number of the toss on which the first head occurs.
Both ? and @ are discrete random variables, where ? can take on only the values or , and @
can take on any positive integer value.
Probability function of a discrete random variable
The probability function (pf) of a discrete random variable is usually denoted ²%³ ²or ²%³³, and
is equal to 7 ²? ~ %³. As its name suggests, the probability function describes the probability of
individual outcomes occurring.
The probability function must satisfy
(i)  ²%³  for all % , and (ii) ²%³ ~ .
(2.10)
%
For the random variable ? above, the probability function is ²³ ~ Á ²³ ~ ,
and for @ it is ²³ ~ for ~ Á Á Á ÀÀÀ .
An event ( is a subset of the set of all possible outcomes of ? , and the probability of event (
occurring is 7 ´(µ ~ ²%³ .
%(
For @ above, 7 ´@ is evenµ ~ 7 ´@ ~ Á Á Á ÀÀÀµ ~ b b b Ä ~ ~ 7 ²? ~ ³ .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 2 - REVIEW OF RANDOM VARIABLES PART I
LM-21
LM-2.3 Continuous Random Variable
A continuous random variable usually can assume numerical values from an interval of real
numbers, perhaps the whole set of real numbers. As an example, the length of time between
successive streetcar arrivals at a particular (in service) streetcar stop could be regarded as a
continuous random variable (assuming that time measurement can be made perfectly accurate).
Probability density function
A continuous random variable ? has a probability density function (pdf) denoted ²%³ or ? ²%³
(or sometimes denoted ²%³ ), which is a continuous function (except possibly at a finite or
countably infinite number of points). For a continuous random variable, we do not describe
probability at single points. We describe probability in terms of intervals. In the streetcar
example, we would not define the probability that the next street car will arrive in exactly 1.23
minutes, but rather we would define a probability such as the probability that the streetcar will
arrive between 1 and 1.5 minutes from now.
Probabilities related to ? are found by integrating the density function over an interval.
7 ´?  ²Á ³µ ~ 7 ´  ?  µ is defined to be equal to ²%³ % .
B
²%³ must satisfy (i) ²%³ ‚ for all % , and (ii) cB ²%³ % ~ .
(2.11)
Often, the region of non-zero density is a finite interval, and ²%³ ~ outside that interval. If
²%³ is continuous except at a finite number of points, then probabilities are defined and
calculated as if ²%³ was continuous everywhere (the discontinuities are ignored).
% for %
For example, suppose that ? has density function ²%³ ~ D
, elsewhere
. Then
B
satisfies the requirements for a density function, since cB ²%³ % ~ % % ~ .
Then, for example 7 ´À  ?  µ ~ À % % ~ % c ~ À . This is illustrated in the shaded
À
are in the graph below.
For a continuous random variable ? ,
7 ´  ?  µ ~ 7 ´  ?  µ ~ 7 ´  ?  µ ~ 7 ´  ?  µ , so that when
calculating the probability for a continuous random variable on an interval, it is irrelevant
whether or not the endpoints are included. For a continuous random variable, 7 ´? ~ µ ~ ;
non-zero probabilities only exist over an interval, not at a single point.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-22
MODELING SECTION 2 - REVIEW OF RANDOM VARIABLES PART I
LM-2.4 Mixed Distribution
A random variable may have some points with non-zero probability mass and with a continuous
pdf elsewhere. Such a distribution may be referred to as a mixed distribution, but the general
notion of mixtures of distributions will be covered later. The sum of the probabilities at the
discrete points of probability plus the integral of the density function on the continuous region for
? must be 1. For example, suppose that ? has probability of À at ? ~ , and ? is a continuous
random variable on the interval ²Á ³ with density function ²%³ ~ % for  %  , and ? has
no density or probability elsewhere. This satisfies the requirements for a random variable since
the total probability is
7 ´? ~ µ b ²%³ % ~ À b % % ~ À b À ~ .
À
Then, 7 ´  ?  Àµ ~ % % ~ ÀÁ and
7 ´  ?  Àµ ~ 7 ´? ~ µ b 7 ´  ?  Àµ ~ À b À ~ À
. Notice that for this random
variable 7 ´  ?  Àµ £ 7 ´  ?  Àµ because there is a probability mass at ? ~ .
LM-2.5 Cumulative Distribution Function, Survival Function and Hazard Function
Given a random variable ? , the cumulative distribution function of ? (also called the
distribution function, or cdf) is - ²%³ ~ 7 ´?  %µ (also denoted -? ²%³ ).
The cdf - ²%³ is the "left-tail" probability, or the probability to the left of and including %.
The survival function is the complement of the distribution function,
:²%³ ~ c - ²%³ ~ 7 ´? € %µ .
The event ? € % is referred to as a "tail" or right tail of the distribution.
For any cdf 7 ´  ?  µ ~ - ²³ c - ²³ Á lim - ²%³ ~ Á lim - ²%³ ~ .
%¦B
%¦cB
(2.12)
(2.13)
For a discrete random variable with probability function ²%³, - ²%³ ~ ²$³ , and
$%
in this case - ²%³ is a "step function" (see Example LM2-1 below); it has a jump (or step
increase) at each point with non-zero probability, while remaining constant until the next jump.
Note that for a discrete random variable, - ²%³ includes the probability at the point % as well as
the total probabilities of all the points to the left of %.
If ? has a continuous distribution with density function ²%³, then
%
B
(2.14)
- ²%³ ~ 7 ²?  %³ ~ cB ²!³ ! and :²%³ ~ 7 ²? € %³ ~ % ²!³ !
and - ²%³ is a continuous, differentiable, non-decreasing function such that
Z
Z
% - ²%³ ~ - ²%³ ~ c : ²%³ ~ ²%³ . Also, for a continuous random variable, the hazard rate
²%³
²%³
or failure rate is ²%³ ~ c- ²%³ ~ :²%³ ~ %
:²%³ .
If ? is continuous and ? ‚ , then the survival function satisfies :²³ ~ and
%
%
:²%³ ~ c ²!³ ! À The cumulative hazard function is /²%³ ~ ²!³ ! À
(2.15)
(2.16)
If ? has a mixed distribution with some discrete points and some continuous regions, then - ²%³
is continuous except at the points of non-zero probability mass, where - ²%³ will have a jump.
The region of positive probability of a random variable is called the support of the random
variable.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 2 - REVIEW OF RANDOM VARIABLES PART I
LM-23
LM-2.6 Examples of Distribution Functions
The following examples illustrate the variety of distribution functions that can arise from random
variables. The support of a random variable is the set of points over which there is positive
probability or density.
Example LM2-1: Finite Discrete Random Variable (finite support)
> ~ number turning up when tossing one fair die, so > has probability function
> ²$³ ~ 7 ´> ~ $µ ~ for $ ~ Á Á Á Á Á .
if $  if  $  if  $  -> ²$³ ~ 7 ´>  $µ ~ H if  $  if  $  if  $  if $ ‚ The graph of the cdf is a step-function that increases at each point of probability by the amount of
probability at that point (all 6 points have probability in this example). Since the support of >
is finite (the support is the set of integers from 1 to 6), -> ²$³ reaches 1 at the largest point
> ~ (and stays at 1 for all $ ‚ ).
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-24
MODELING SECTION 2 - REVIEW OF RANDOM VARIABLES PART I
Example LM2-2: Infinite Discrete Random Variable (infinite support)
? ~ number of successive independent tosses of a fair coin until the first head turns up.
? can be any integer ‚ 1, and the probability function of ? is ? ²%³ ~ % .
%
The cdf is -? ²%³ ~ ~ c % for % ~ Á Á Á ÀÀÀ .
~
The graph of the cdf is a step-function that increases at each point of probability by the amount of
probability at that point. Since the support of ? is infinite (the support is the set of
integers ‚ ) -? ²%³ never reaches 1, but approaches 1 as a limit as %SB. The graph of -? ²%³
is
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 2 - REVIEW OF RANDOM VARIABLES PART I
LM-25
Example LM2-3: Continuous Random Variable on a Finite Interval
@ is a continuous random variable on the interval ²Á ³ with density function
& for &
@ ²&³ ~ D
, elsewhere
.
if &  Then -@ ²&³ ~ F & if  &  .
if & ‚
Example LM2-4: Continuous Random Variable on an Infinite Interval
< is a continuous random variable on the interval ²Á B³ with density function
"c" for "€
< ²"³ ~ D
, for "
.
for "
Then -< ²"³ ~ D
c²b"³c" , for "€
.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-26
MODELING SECTION 2 - REVIEW OF RANDOM VARIABLES PART I
Example LM2-5: Mixed Random Variable
A has a mixed distribution on the interval ´Á ³. A has probability of À at A ~ , and A has
density function A ²'³ ~ ' for  '  , and A has no density or probability elsewhere.
Then,
if '  À if ' ~ -A ²'³ ~
.
À b ' if  '  if ' ‚ H
LM-2.7 The Empirical Distribution For a Data Sample
Given a sample of observations from a random variable ? , say % Á % Á ÀÀÀÁ % , the empirical
distribution for the sample is the discrete random variable which takes on the values
% Á % Á ÀÀÀÁ % with a probability of for each point.
If % ~ for distinct %'s ( repeated values), then the probability function of the empirical
distribution at is ²³ ~ À
For instance, suppose we have the following sample of size ~ :
% ~ Á % ~ Á % ~ Á % ~ Á % ~ Á %
~ .
The empirical distribution has probability function
²³ ~ Á ²³ ~ Á ²³ ~ Á ²
³ ~ Á ²³ ~ .
Concepts related to random samples will be considered in detail later in the study guide.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 2 - REVIEW OF RANDOM VARIABLES PART I
LM-27
LM-2.8 Gamma Function, Incomplete Gamma Function
and Incomplete Beta Function
Many of the continuous distributions described in the Exam C Tables make reference to the
gamma function and the incomplete gamma function. The definitions of these functions are
B
• gamma function: !²³ ~ !c c! ! for € %
• incomplete gamma function: !²Â %³ ~ !²³ h !c c! ! for € Á % € (2.17)
Some important points to note about these functions are the following:
• if is an integer and ‚ 1, then !²³ ~ ² c ³[
(2.18)
• !² b ³ ~ h !²³ and !² b ³ ~ ² b c ³² b c ³Ä h h !²³
for any € and integer ‚ (2.19)
!²b³
B
• % c% % ~ b for ‚ and € (use substitution " ~ %)
(2.20)
!²c³
B
• % c°% % ~ c for € and € (use substitution " ~ % )
(2.21)
Some of the table distributions make reference to the incomplete beta function, which is defined
as follows:
!²b³ % c
- ²Á  %³ ~
² c !³c ! for  %  , Á € .
(2.22)
!
!²³!²³
References to the gamma function have been rare and the incomplete functions have not been
referred to on the released exams. It is useful to remember the integral relationship
B % c% % ~ !²b³
, particularly in the case in which is a non-negative integer.
b
B
[
In that case, we get % c% % ~ b
, which can occasionally simplify integral
relationships. This relationship is embedded in the definition of the gamma distribution in the
c c!°
Exam C Table. The pdf of the gamma distribution with parameters and is ²!³ ~ ! !²³ ,
B c c!°
defined on the interval ! € . This means that ! ! ~ , which can be reformulated
B
as !c c!° % ~ B % c% % ~ !²b³
.
b
!²³ À If we let ~
!²³
and ~ b , we get the relationship
(2.23)
Looking at the various continuous distributions in the Exam C table gives some hints at
calculating a number of integral forms. For instance, the pdf of the beta distribution with
!²b³
parameters Á Á ~ is ²%³ ~ !²³!²³ h %c ² c %³c for  %  .
!²b³
Therefore, h %c ² c %³ % ~ , from which we get
!²³!²³
%c ²
!²³!²³
c %³c % ~ !²b³ .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-28
© ACTEX 2009
MODELING SECTION 2 - REVIEW OF RANDOM VARIABLES PART I
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LOSS MODELS - PROBLEM SET 2
LM-29
MODELING - PROBLEM SET 2
Review of Random Variables I - Section 2
1. Let ? be a discrete random variable with probability function
7 ´? ~ %µ ~ % for % ~ Á Á Á ÀÀÀ What is the probability that ? is even?
A) B) C) D) E) 2. For a certain discrete random variable on the non-negative integers, the probability function
satisfies the relationships
7 ²³ ~ 7 ²³ and 7 ² b ³ ~ h 7 ²³ for ~ Á Á Á ÀÀÀ Find 7 ²³.
A) B) c C) ² b ³c
D) c
E) ² c ³c
3. Let ? be a continuous random variable with density function
²%³ ~ D
A) À
%²c%³ for %
otherwise
B) À
. Calculate 7 ´ O? c O € µ .
C) À
D) À
E) À
4. Let ? be a random variable with distribution function
- ²%³ ~
A) for %  for  %  b % for  %  . Calculate 7 ´  ?  µ .
%
b for  %  for % ‚ B) C) D) E) H
%
5. Let ? Á ? and ? be three independent continuous random variables each with density
function ²%³ ~ D
jc% for %j
otherwise
.
What is the probability that exactly 2 of the 3 random variables exceeds 1?
A) c j
B) c j
C) ²j c ³² c j³
D) ² c j³ ²j c ³
© ACTEX 2009
E) ² c j³ ²j c ³
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-30
MODELING - PROBLEM SET 2
6. Let ? Á ? and ? be three independent, identically distributed random variables each with
density function ²%³ ~ D
Find 7 ´@ € µ .
A) B) % for %
otherwise
. Let @ ~ %¸? Á ? Á ? ¹.
C) D) E) c%
7. Let the distribution function of ? for % € be - ²%³ ~ c % [
What is the density function of ? for % € ?
A) c%
c%
B) % c%
C) % .
~
c%
D) % c c%
c%
E) % b c%
8. Let ? have the density function ²%³ ~ %
for  %  , and ²%³ ~ , otherwise. If
7 ´? € µ ~ , find the value of .
A) B) ² ³°
C) ² ³°
D) °
E) 9. A large wooden floor is laid with strips 2 inches wide and with negligible space between
strips. A uniform circular disk of diameter 2.25 inches is dropped at random on the floor. What
is the probability that the disk touches three of the wooden strips?
A) j
B) C) D) E) 10. If ? has a continuous uniform distribution on the interval from 0 to 10, then what is
7 ´? b ? € µ ?
A) B) C) D) E) 11. (CAS) For a loss distribution where % ‚ , you are given:
'
i) The hazard rate function: ²%³ ~ %
, for % ‚ ii) A value of the distribution function: - ²³ ~ À
Calculate ' .
A) 2
B) 3
C) 4
D) 5
E) 6
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LOSS MODELS - PROBLEM SET 2
LM-31
12. (CAS) A pizza delivery company has purchased an automobile liability policy for its delivery
drivers from the same insurance company for the past five years. The number of claims filed by
the pizza delivery company as the result of at-fault accidents caused by its drivers is shown
below:
Year
Claims
2002
4
2001
1
2000
3
1999
2
1998
15
Calculate the skewness of the empirical distribution of the number of claims per year.
A) Less than 0.50
B) At least 0.50, but less than 0.75
C) At least 0.75, but less than 1.00
D) At least 1.00, but less than 1.25
E) At least 1.25
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SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-32
MODELING - PROBLEM SET 2
MODELING - PROBLEM SET 2 SOLUTIONS
1. 7 ´? is evenµ ~ 7 ´? ~ µ b 7 ´? ~ µ b 7 ´? ~ µ b Ä
~ h ´ b b b ĵ ~ h c ~ . Answer: A
2. 7 ²³ ~ 7 ²³ ~ 7 ²³ Á 7 ²³ ~ h 7 ²³ ~ [
h 7 ²³ , . . .
7 ²³ ~ ²c³[
h 7 ²³ . The probability function must satisfy the requirement
B
B
7 ²³ ~ so that 7 ²³ b h 7 ²³ ~ 7 ²³² b ³ ~ ~
~
²c³[
(this uses the series expansion for % at % ~ ). Then, 7 ²³ ~ b
.
Answer: C
°
3. 7 ´ c? c c  µ ~ 7 ´ c  ? c  µ ~ 7 ´  ?  µ ~ ° %² c %³ %
~ À
S 7 ´ c? c c  µ ~ c 7 ´ c? c c ‚ µ ~ À .
Answer: C
4. 7 ´  ?  µ ~ 7 ´?  µ c 7 ´?  µ ~ - ²³ c limc - ²%³ ~ c ~ À
%¦
Answer: E
5. 7 ´?  µ ~ ²j c %³ % ~ j c Á 7 ´? € µ ~ c 7 ´?  µ ~ c j .
With 3 independent random variables, ? Á ? and ? , there are 3 ways in which exactly 2 of the
? 's exceed 1 (either ? Á ? or ? Á ? or ? Á ? ). Each way has probability
²7 ´? € µ³ h 7 ´?  µ ~ ² c j³ ²j c ³ for a total probability of
h ² c j³ ²j c ³ . Answer: E
6. 7 ´@ € µ ~ c 7 ´@  µ ~ c 7 ´²?  ³ q ²?  ³ q ²?  ³µ
°
Answer: E
~ c ²7 ´?  µ³ ~ c ´ % %µ ~ c ² ³ ~ À
c c%
7. ²%³ ~ - Z ²%³ ~ c %
~
c%
h ´ b
© ACTEX 2009
%c
~
b
% c%
b
c% c%
[
% c%
µ
c
~ c% h < % c%
=
[
~
c% %
~
À
Answer: C
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LOSS MODELS - PROBLEM SET 2
LM-33
8. Since ²%³ ~ if % € , and since 7 ´? € µ ~ , we must conclude that € .
Then, 7 ´? € µ ~ ²%³ % ~ %
% ~ c ~ , or equivalently, ~ .
Answer: E
9. Let us focus on the left-most point on the disk. Consider two adjacent strips on the floor.
Let the interval ´Á µ represent the distance as we move across the left strip from left to right. If
is between and À, then the disk lies within the two strips.
If is between À and , the disk will lie on 3 strips (the first two and the next one to the right).
Since any point between and is equally likely as the left most point on the disk (i.e.
uniformly distributed between and ) it follows that the probability that the disk will touch three
strips is À
~ . Answer: D
10. Since the density function for ? is ²%³ ~ for  %  , we can regard ? as
being positive. Then
7 ´? b ? € µ ~ 7 ´? c ? b € µ ~ 7 ´²? c ³²? c ³ € µ
~ 7 ´? € µ b 7 ´?  µ (since ²! c ³²! c ³ € if either both ! c Á ! c € or both ! c Á ! c  ) ~ ~ ~ . Answer: E
11. The survival function :²&³ for a random variable can be formulated in terms of the hazard
&
rate function: :²&³ ~ %´ c cB ²%³ %µ . In this question,
:²³ ~ c - ²³ ~ À
~ %´ c ' %µ ~ %´ c ' ² ³µ .
%
Taking natural log of both sides of the equation results in
solving for ' results in ' ~ .
Answer: A
c ' ² ³ ~ ²À
³ , and
12. The empirical distribution is a five point random variable with probability function
7 ´? ~ µ ~ 7 ´? ~ µ ~ 7 ´? ~ µ ~ 7 ´? ~ µ ~ 7 ´? ~ µ ~ À (each observation in an
empirical distribution of a data set is given the same probability as the others).
The skewness is
,´²?c,²?³³ µ
²= ´?µ³°
À
For the empirical distribution, ,´?µ ~ ²À³² b b b b ³ ~ Á
,´? µ ~ ²À³² b b b b ³ ~ Á
,´²? c ³ µ ~ ²À³´² c ³ b ² c ³ b ² c ³ b ² c ³ b ²³ µ ~ Á
= ´?µ ~ ,´? µ c ²,´?µ³ ~ c ~ À
Skewness ~ Answer: E
° ~ À
À
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SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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© ACTEX 2009
MODELING - PROBLEM SET 2
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 3 - REVIEW OF RANDOM VARIABLES PART II
LM-35
MODELING SECTION 3
REVIEW OF RANDOM VARIABLES - PART II
This section relates to Chapter 3 (3.1. - 3.3) of "Loss Models". The mean and variance of a
random variable are two fundamental distribution parameters. In this section we review those and
some other important distribution parameters. Chapter 3 of Loss models introduces deductibles
and policy limits. These topics will be considered in detail later in the study guide.
The suggested time frame for this section is 2 hours.
LM-3.1 Expected Value and Other Moments of a Random Variable
For a random variable ? , the expected value is denoted ,´?µ , or ? or . The expected
value of ? is also called the expectation of ? , or the mean of ? . The expected value is the
"average" over the range of values that ? can be, or the "center" of the distribution.
For a discrete random variable, the expected value of ? is % h ²%³ ,
(3.1)
all %
where the sum is taken over all points % at which ? has non-zero probability. For instance, if ?
is the result of one toss of a fair die, then ,´?µ ~ h b h b Ä b h ~ À
B
For a continuous random variable, the expected value is cB % h ²%³ % .
(3.2)
Although the integral is written with lower limit c B and upper limit B, the interval of
integration is the interval of support (non-zero-density) for ? . For instance, if ²%³ ~ % for
 %  , then the mean is ,´?µ ~ % h % % ~ .
If is a function, then ,´²?³µ is equal to ²%³ h ²%³ if ? is a discrete random variable,
%
B
and it is equal to cB ²%³ h ²%³ % if ? is a continuous random variable. ,´²?³µ is the
"average" value of ²?³ based on the possible outcomes of random variable ? .
The mean of a random variable ? might not exist, it might be b B or c B. For example, the
continuous random variable ? with
for %‚
pdf ²%³ ~ D %
, otherwise
B
has expected value % h % % ~ b B .
For any constants Á and and functions and ,
,´ ²?³ b ²?³ b µ ~ ,´ ²?³µ b ,´ ²?³µ b If ? is a non-negative random variable (defined on ´Á B³ or ²Á B³ ) then
B
B
,´?µ ~ ´ c - ²%³µ % ~ :²%³ %
This relationship is valid for any random variable, discrete, continuous or with a mixed
distribution. Using the example on the previous page, if ²%³ ~ % for  %  , then
% for %
- ²%³ ~ D
, for %€
(3.3)
(3.4)
B
Á and ´ c - ²%³µ % ~ ´ c % µ % ~ ~ ,´?µ .
It tends to be more awkward to apply this rule to discrete random variables.
Jensen's Inequality states that if is a function such the ZZ ²%³ ‚ on the probability space for
? , then ,´²?³µ ‚ ²,´?µ³ . For example, ,´? µ ‚ ²,´?µ³ .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-36
MODELING SECTION 3 - REVIEW OF RANDOM VARIABLES PART II
Moments of a random variable
If ‚ is an integer, then the -th (raw) moment of ? is ,´? µ (sometimes denoted Z ) and
B
is % h ²%³ in the discrete case, and cB % h ²%³ % in the continuous case.
(3.5)
all %
If the mean of ? is , then the -th central moment of ? (about the mean ) is defined to be
,´²? c ³ µ and may be denoted .
For instance, the 3rd central moment of the fair die toss random variable ? is
,´²? c ³ µ ~ ² c ³ h b ² c ³ h b Ä b ²
c ³ h ~ .
The 2nd central moment of the continuous random variable ? pdf ²%³ ~ % for  %  is ²% c ³ h % % ~ .
Variance of ?
The variance of ? is denoted = ´?µ , = ´?µ , ?
or . It is defined to be equal to
= ´?µ ~ ,´²?c? ³ µ ~ ,´? µ c ? ~ ,´? µ c ²,´?µ³ .
(3.6)
The representation of = ´?µ as ,´? µ c ²,´?µ³ is often the most convenient one to use. The
variance is the 2nd central moment of ? ; ~ = ´?µ ~ Z c .
Variance is a measure of the "dispersion" of ? about the mean. As the expected value of
²?c? ³ , the variance is the average squared deviation of ? from its mean ? .
A large variance indicates significant levels of probability or density for points far from ,´?µ.
The variance is always ‚ (the variance of ? is equal to only if ? has a discrete distribution
with a single point and probability at that point; in other words, not random at all).
prob. .5
has mean ,´< µ ~ and variance
prob. .5
prob. .5
= ´< µ ~ . The random variable > ~ D
has the same mean as < , ,´> µ ~ ,
prob. .5
but has variance = ´> µ ~ . The higher variance of > is reflected in the further dispersion of
the outcomes of > from the mean 5 as compared to < .
The random variable < ~ D
If and are constants, then = ´? b µ ~ = ´?µ .
Here is a useful shortcut for finding the variance of a 2-point discrete random variable.
If ? is the two-point random variable ? ~ F
then = ´?µ ~ ² c ³ d d ² c ³ .
Prob. Prob. c (3.7)
Standard deviation of ?
The standard deviation of the random variable ? is the square root of the variance, and is denoted
? ~ j = ´?µ .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 3 - REVIEW OF RANDOM VARIABLES PART II
Coefficient of variation
The coefficient of variation of ? is ? ~
?
j= ´?µ
,´?µ
Skewness and kurtosis
,´²?c³ µ
LM-37
.
(3.8)
,´²?c³ µ
The skewness of ? is
, and the kurtosis is
.
(3.9)
Skewness measures the symmetry of a random variable; skewness of 0 indicates a distribution
which is symmetric around its mean. The fair dies toss random variable has skewness of 0.
Kurtosis is a measure of the "peakedness" of a distribution. Higher kurtosis suggests that more of
the variance is due to less frequent large deviations, rather than more frequent smaller deviations.
There have been infrequent references to skewness and kurtosis on Exam C.
LM-3.2 Moment generating function of random variable ?
The moment generating function of ? (mgf) is denoted 4? ²!³ Á ? ²!³ Á 4 ²!³ or ²!³, and it
is defined to be 4? ²!³ ~ ,´!? µ , which is either !% ²%³ if ? is discrete,
%
B
or cB !% ²%³ % if ? is continuous. It is always true that 4? ²³ ~ .
(3.10)
The moment generating function of ? might not exist for all real numbers, but usually exists on
some interval of real numbers. The function ´4? ²!³µ is called the cumulant generating
function. The function 4? ² !³ ~ ,´!? µ is called the probability generating function and
may be denoted 7? ²!³ ~ ,´!? µ ; the probability generating function is usually used in the
case of a discrete random variable.
Some properties of moment and probability generating functions
Suppose that for the random variable ? , the moment generating function 4? ²!³ exists in
an interval containing the point ! ~ . Then
²³
Z
! 4? ²!³c!~ ~ 4? ²³ ~ ,´? µ ~ , the -th moment of ? , and
4?Z ²³
´4
²!³µ
~
~ ,´?µ , and !
~ = ´?µ .
c
?
´4? ²!³µc
!
4
²³
?
!~
!~
(3.11)
(3.12)
If ? and ? are random variables, and 4? ²!³ ~ 4? ²!³ for all values of ! in an
interval containing ! ~ , then ? and ? have identical probability distributions.
If ? Á ? Á ÀÀÀÁ ? are independent random variables and : ~ ? then
~
4: ²!³ ~ 4? ²!³ h 4? ²!³ h h h 4? ²!³ ~ 4? ²!³ .
(3.13)
~
If ? has a discrete non-negative integer distribution with ~ 7 ´? ~ µ , then the
B
probability generating function is 7? ²!³ ~ b h ! b h ! b Ä ~ h ! ,
and 7? ²³ ~ .
© ACTEX 2009
(3.14)
~
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-38
MODELING SECTION 3 - REVIEW OF RANDOM VARIABLES PART II
LM-3.3 Percentiles and Quantiles of a distribution
If   , then the -th percentile of the distribution of ? is the number which
satisfies the following inequalities:
- ´c µ ~ 7 ´?  µ   7 ´?  µ ~ - ´ µ .
(3.15)
For a continuous random variable, it is sufficient to find the for which 7 ´?  µ ~ . If
~ À , the 50-th percentile of a distribution is referred to as the median of the distribution,
it is the point 4 for which 7 ´?  4 µ ~ À . The median 4 is the 50% probability point, half
of the distribution probability is to the left of 4 and half is to the right. The word "quantile" is a
general term for the proportion or percent of a distribution below a certain given point and is
often used interchangeably with "percentile".
LM-3.4 The mode of a distribution
The mode is any point at which the probability or density function ²%³ is maximized. For the
fair die toss random variable, each of % ~ Á Á Á Á or would satisfy the requirements of
being a mode, since the probability at each point is , which is the maximum probability of any
individual point. For the continuous random variable with ²%³ ~ % for  %  , strictly
speaking, there is no mode since the upper bound of the density of 2 is never reached. The mode
could be described as occurring at 1 as a limit.
Example LM3-1: Let ? equal the number of tosses of a fair die until the first "1" appears.
Find ,´?µ.
Solution: ? is a discrete random variable that can take on any integer value ‚ . The
probability that the first 1 appears on the %-th toss is ²%³ ~ ² ³%c ² ³ for % ‚ (% c tosses that are not followed by a 1). This is the probability function of ? . Then
B
B
~
~
,´?µ ~ h ²³ ~ h ² ³c ² ³ ~ ² ³´ b ² ³ b ² ³ b ĵ .
We use the general increasing geometric series relation b b b Ä ~ ²c³
,
so that ,´?µ ~ ² ³ h ²c ³ ~ .
Example LM3-2: The moment generating function of ? is c!
for !  , where € .
Find = ´?µ.
Solution: = ´?µ ~ ,´? µ c ²,´?µ³ . ,´?µ ~ 4?Z ²³ ~ ²c!³
~ ,
c
!~
and ,´? µ ~ 4?ZZ ²³ ~ ²c!³
~ S = ´?µ ~ c ² ³ ~ .
c
!~
Alternatively, 4? ²!³ ~ ² c! ³ ~ c ² c !³ S !
´4? ²!³µ ~ c!
and !
~ À
´4? ²!³µ ~ ²c!³ so that = ´?µ ~ ! ´4? ²!³µc
!~
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 3 - REVIEW OF RANDOM VARIABLES PART II
LM-39
Example LM3-3: Given that the density function of ? is ²%³ ~ c% , for % € , and elsewhere, find the -th moment of ? , where is a non-negative integer (assuming that € ).
B
Solution: The -th moment of ? is ,´? µ ~ % h c% % . Applying integration by
parts, this can be written as
%~B
B
B
B % ² c c% ³ ~ c % c% c
c c %c c% % ~ %c c% % À
%~
Repeatedly applying integration by parts results in ,´? µ ~ [ . It is worthwhile noting the
general form of the integral that appears in this example; if ‚ is an integer and € , then
B
by repeated applications of integration by parts, we have ! c! ! ~ [
b , so that in this
B c%
B c%
[
[
example % % ~ % % ~ h b ~ .
An alternative solution uses the moment generating functionÀ
B
B
4? ²!³ ~ ,´!? µ ~ !% c% % ~ c²c!³% % ~ c!
(which will be valid for !  ).
Z
Then 4?Z ²!³ ~ ²c!³
so that ,´?µ ~ 4? ²³ ~ ,
²³
²³
and 4? ²!³ ~ ²c!³
so that ,´? µ ~ 4? ²³ ~ .
²³
²³
[
[
It can be shown by induction on that 4? ²!³ ~ ²c!³
b so that ,´? µ ~ 4? ²³ ~ .
Example LM3-4: The continuous random variable ? has pdf ²%³ ~ h cO%O for
c B  %  B . Find the 87.5-th percentile of the distribution.
Solution: The 87.5-th percentile is the number for which
À ~ 7 ´?  µ ~ cB ²%³ % ~ cB h cO%O % .
Note that this distribution is symmetric about , since ² c %³ ~ ²%³, so the mean and median
are both . Thus, € , and so
cO%O
cB
% ~ cB h cO%O % b h cO%O % ~ À b h c% %
h
~ À b ² c c ³ ~ À S ~ c ²À³ ~ .
Example LM3-5: ? is the outcome after 1 toss of a fair die. Find the 25-th percentile of ? .
Solution: The distribution function of ? was found in Example LM2-1 above. We see that
- ²c ³ ~ 7 ´?  µ ~  À  ~ 7 ´?  µ ~ - ²³ . Any number other than 2 will not
satisfy one side of the inequalities. The 25-th percentile is 2. Note that 2 would also be the 20-th
and the 30-th percentile; 2 would be any percentile from 16.7-th to 33.3-rd.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-40
MODELING SECTION 3 - REVIEW OF RANDOM VARIABLES PART II
LM-3.5 Normal Distribution and Normal Approximation
The standard normal distribution, ? — 5 ²Á ³, has a mean of and variance of 1. A table of
probabilities for the standard normal distribution is provided on the exam. The density function
is ²%³ ~ j h c% ° for c B  %  B . The density function has the following graph.
The shaded area is 7 ´?  %µ , which is denoted )²%³ .
This is an excerpt from the Exam C Table provided at the exam. The entries in the table are
probabilities )²%³ ~ 7 ´?  %µ . The 95-th percentile of A is 1.645 (sometimes denoted 'À )
since )²À
³ ~ À (the shaded region to the left of % ~ À
in the graph above). We
use the symmetry of the distribution to find )²%³ for negative values of %. For instance,
)² c ³ ~ 7 ´A  c µ ~ 7 ´A ‚ µ ~ c )²³ since the two regions have the same area
(probability). Notice also that 7 ´ c À
 A  À
µ ~ À , since
7 ´A € À
µ ~ c )²À
³ ~ À, and this area is deleted from both ends of the curve.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 3 - REVIEW OF RANDOM VARIABLES PART II
LM-41
The general form of the normal distribution has mean and variance . This is a continuous
distribution with a "bell-shaped" density function similar to that of the standard normal, but
symmetric around the mean . The median and the mode are also .
Given any normal random variable > — 5 ²Á ³, it is possible to find 7 ´  >  µ by first
> c
"standardizing" , ? ~ and then
c
7 ´  >  µ ~ 7 ´ > c


c
µ
c
c
~ )² ³ c )² ³ .
(3.16)
As an example, suppose that > has a normal distribution with mean 1 and variance 4. Then
7 ´>  Àµ ~ 7 ´ >jc  Àc
j µ ~ 7 ´?  Àµ ~ )²À³ ~ À .
The 95-th percentile of > is , where
c
c
7 ´>  µ ~ À S 7 ´ >jc  c
j µ ~ )² j ³ ~ À S j ~ À
S ~ À .
The moment generating function of > is 4> ²!³ ~ %<! b ! =.
(3.17)
LM-3.6 Approximating a Distribution Using a Normal Distribution
Given a random variable with mean and variance , probabilities related to the distribution of
? are sometimes approximated by assuming the distribution of ? is approximately 5 ²Á ³. If
? is discrete and integer-valued then an "integer correction" should be applied; the probability
7 ´  ?  µ is approximated by assuming that ? is normal and then finding the probability
7 ´ c  ?  b µ . This is also sometimes referred to as the "correction for
discontinuity". The integer correction should be used on an exam question when the normal
approximation is applied to an integer distribution.
:c,´:µ
If the random variable : is (approximately) normal, then the transformed variable j
has a
= ´:´
standard normal distribution; a standard normal random variable has mean 0 and variance 1.
As an example, the 95-th percentile of : , say À Á can be found by solving the expression
:c,´:µ
= ´:µ
7 ´:  À µ ~ 7 < j
c,´:µ
c,´:µ
À
À
 j
~ À
(3.18)
= ~ À , so that j
= ´:´
= ´:´
is the 95-th percentile of the standard normal distribution (found from the standard normal table
which is part of the Exam C exam tables). Thus, if ,´:µ and = ´:µ are known, then using the
normal approximation, the approximate percentiles of : can be found.
If @ Á @ Á ÀÀÀÁ @ are independent and identically distributed random variables with common mean
,´@ µ and common variance = ´@ µ, then the Central Limit Theorem of probability states that
: ~ @ has a distribution which is approximately normal with mean ,´@ µ and variance
~
= ´@ ´ . As gets larger, : approaches a normal distribution. In practice, a value of 30 or so is
regarded as "large enough" for the normal distribution to be a reasonable approximation to the
distribution of : . This is a justification for using the normal approximation in some circumstances.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-42
MODELING SECTION 3 - REVIEW OF RANDOM VARIABLES PART II
Example LM3-6: ? Á ? Á ÀÀÀÁ ? are independent random variables each uniformly distributed
on the interval ²Á ³. Use the normal approximation to find the 95-th percentile of : ~ ? .
~
~
~
~
~
Solution: It is always true that , < ? = ~ ,´? µ , so that ,´:µ ~ ,´?µ ~ .
If ? Á ? Á ÀÀÀÁ ? are independent, then = < ? = ~ = ´? µ , so that
= ´:µ ~ = ´?µ ~ ² ³ ~ À .
The 95-th percentile must satisfy the relationship 7 ´:  µ ~ À .
Standardizing the probability results in
7 ´:  µ ~ 7 ´ j:c  jc µ ~ )² jc ³ ~ À S jc ~ À
S ~ À .
À
© ACTEX 2009
À
À
À
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 3
LM-43
MODELING - PROBLEM SET 3
Review of Random Variables II - Section 3
1. If ? is a random variable with density function ²%³ ~ D
then ,´?µ ~
A) B) D) C) Àc% bÀc% for %€
,
, elsewhere
E) 2. Let ? be a continuous random variable with density function
²%³ ~ D
%²b%³
for %
.
, otherwise
A) B) C) Find , ´ ?
µÀ
D) E) 3. Let ? be a continuous random variable with density function ²%³ ~ D
Find ,´ b? c ,´?µb µ .
B) A) C)
D) %
for %
, otherwise
.
E) 4. Let @ be a continuous random variable with cumulative distribution function
- ²&³ ~ D
A) - ²À³
for &
cc ²&c³ otherwise
B) c j 2
, where is a constant. Find the 75-th percentile of @ .
C) b j 2
D) c 2j 2
E) b j 2
5. Let ? be a continuous random variable with density function ²%³ ~ D
If the median
A)
of this distribution is ,then B) C) D)
c% for %€
, otherwise
~
E) 6. Let ? be a continuous random variable with density function
²%³ ~ D
A) %²c%³
for %
, otherwise
B) © ACTEX 2009
C) . What is the mode of ? ?
D) E) SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-44
MODELING - PROBLEM SET 3
7. A system made up of 7 components with independent, identically distributed lifetimes will
operate until any of 1 of the system's components fails. If the lifetime ? of each component has
density function
²%³ ~ D
%
for %
, otherwise
,
what is the expected lifetime until failure of the system?
A) À
B) À
C) À
D) À
E) À
8. Let ? have the density function ²%³ ~ D
%
for %
, otherwise
.
For what value of is the variance of ? equal to ?
A) B) C) D) E) 9. Two players put one dollar into a pot. They decide to throw a pair of dice alternately. The
first one who throws a total of 5 on both dice wins the pot. How much should the player who
starts add to the pot to make this a fair game?
A) B) C) D) E) 10. If ²%³ ~ ² b ³% for  %  , find the moment generating function of ? .
A)
D)
! ²
b
!b! ³
!
! ²
b
!b! ³
!
B)
b !
! ²
c
!b! ³
!
! ²
c
!b! ³
E)
!
C)
c !
! ²
b
!b! ³
!
c !
11. If the moment generating function for the random variable ? is 4? ²!³ ~ b!
,
find the third moment of ? about the point % ~ .
A) B) C) D) c E) c !
12. Let ? be a random variable with moment generating function 4 ²!³ ~ ² b
³
for c B  %  B . Find the variance of ? .
A) B) C) D) E) 13. If the mean and variance of random variable ? are 2 and 8, find the first three terms in the
Taylor series expansion of the moment generating function of ? about the point ! ~ .
A) ! b ! B) b ! b !
C) b ! b !
D) b ! b !
E) b ! b !
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 3
LM-45
14. Let ? be a random variable with a continuous uniform distribution on the interval ²Á ³
where € . If ,´?µ ~ h = ´?µ , then ~
A) B) C) j
D) E) 15. A student received a grade of 80 in a math final where the mean grade was 72 and the
standard deviation was . In the statistics final, he received a 90, where the mean grade was 80
and the standard deviation was 15. If the standardized scores (i.e., the scores adjusted to a mean
of 0 and standard deviation of 1) were the same in each case, find .
A) B) C) D) E) 16. If ? has a standard normal distribution and @ ~ ? , what is the -th moment of @ ?
A) B) C) °
D) °
E) if ~ c and ²c³²c³Äh if ~ 17. The random variable ? has an exponential distribution with mean °. It is found that
4? ² c ³ ~ À . Find .
A) B) C) D) E) 18. Let ? be a continuous random variable with density function
²%³ ~ j c% ° for c B  %  B . Calculate ,´?O? ‚ µ .
A) B) j
2
C) D) k E) 19. ? has a lognormal distribution with parameters with a mean of and variance of c .
Find 7 ´?  µ .
A) À
B) À
C) À
D) À
E) À
20. If ? has a normal distribution with mean 1 and variance 4, then 7 ´? c ?  µ ~ ?
A) À
B) À
C) À
D) À
E) À
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-46
MODELING - PROBLEM SET 3
21. (SOA) For watches produced by a certain manufacturer:
(i)
Lifetimes follow a single-parameter Pareto distribution with € and ~ .
(ii) The expected lifetime of a watch is 8 years.
Calculate the probability that the lifetime of a watch is at least 6 years.
A) 0.44
B) 0.50
C) 0.56
D) 0.61
E) 0.67
22. The claim amount random variable ) has the following distribution function
|
%




%°Á  %  À
% ~ .
- ²%³ ~ }


 %  
 ²% b Á ³°
Á % ‚ ~
What is ,´)µ b j= ²)³ ?
A) 2400
B) 2450
C) 2500
D) 2550
E) 2600
23. Given ,´? “ @ ~ &µ ~ & and = ´? “ @ ~ &µ ~ ,
where @ has an exponential distribution with a mean of , what is = ´?µ?
A) B) 1
C) D) 2
E) 3
24. ? has a gamma distribution with mean 8 and skewness 1. Find the variance of ? .
A) 4 B) 8 C) 16 D) 32 E) 64
25. For a random variable ? such that its df has - ²³ ~ ,
,´²? c ³ µ  ²,´?µ c ³ .
A) True
B) False
26. A random variable has pdf ²%³ ~ % for  %  .
Find the 75-th percentile of the distribution, À .
A) .750
B) .777
C) .833
D) .866
E) .902
27. (SOA) You are given:
(i) A and A are independent 5 ²Á ³ random variables.
(ii) Á Á Á Á Á are constants such that not both and are 0.
(iii) @ ~ b A b A and ? ~ b A b A
Determine ,²@ O?³ .
A) B) b ² b ³²? c ³
C) b ² b ³²? c ³
D) b ´² b ³°² b ³µ?
E) b ´² b ³°² b ³µ²? c ³
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 3
LM-47
MODELING - PROBLEM SET 3 SOLUTIONS
B
B
1. ,´?µ ~ cB % h ²%³ % ~ ²À%c% b À%c% ³ %
~ ² c À%c% c Àc% c À%c% c Àc% ³c
%~B
%~
~ À .
The integrals were found by integration by parts. Note that we could also have used
B % c% % ~ [
if is an integer ‚ , and € .
Answer: A
b
2. , ´ ?
µ ~ % h %² b %³ % ~ .
Answer: B
3. ,´?µ ~ % h % % ~ . Then, ? c ,´?µ ~ ? c , which is negative for
 %  and is positive for  %  . Thus, b? c ,´?µb ~ c ? if  ?  and b? c ,´?µb ~ ? c if  %  . Then,
°
,´ b? c ,´?µb µ ~ ² c %³ h % % b ° ²% c ³ h % % ~ .
Answer: C
4. Let us denote the 75-th percentile of @ by . Thus, - ²³ ~ À, so that
c c ²c³ ~ À . Solving this equation for results in c ²c³ ~ À ,
or equivalently, ² c ³ ~ S ~ b j ~ b j . Answer: E
°
5. c% % ~ c c° ~
S ~ .
Answer: D
6. The mode is the point at which ²%³ is maximized. Z ²%³ ~ c % b ² c %³ ~ c % .
Setting Z ²%³ ~ results in % ~ .
Since ZZ ²³ ~ c  , that point is a relative maximum.
Answer: E
7. Let ; be the time until failure for the system. In order for the system to not fail by
time ! € , it must be the case that none of the components have failed by time !. For
B
a given component, with time until failure of > , 7 ´> € !µ ~ ! % % ~ ! . Thus,
7 ´; € !µ ~ 7 ´²> € !³ q ²> € !³ q Ä q ²> € !³µ
~ 7 ´> € !µ h 7 ´> € !µÄ7 ´> € !µ ~ ! (because of independence of the > 's).
The cumulative distribution function for ; is
-; ²!³ ~ 7 ´;  !µ ~ c 7 ´; € !µ ~ c ! , so the density function for ; is ; ²!³ ~ !
.
B
The expected value of ; is then ,´; µ ~ ! h ! ~
~ À .
!
Alternatively, once the cdf of ; is known, since the region of density for ; is ! € ,
B
B
the expected value of ; is ,´; µ ~ b ´ c -; ²!³µ ! ~ b ! ! ~ b .
Answer: D
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-48
MODELING - PROBLEM SET 3
%
8. ,´?µ ~ % h %
% ~ Á ,´? µ ~ % h % ~ S = ´?µ ~ c ² ³ ~ ~ S ~ .
Answer: B
9. Player 1 throws the dice on throws 1, 3, 5, . . . and the probability that player wins on throw
b is ² ³ h for ~ Á Á Á Á ÀÀÀ (there is a probability of throwing a total of 5 on any
one throw of the pair of dice). The probability that player 1 wins the pot is
b ² ³ h b ² ³ h b Ä ~ h c² ³ ~ À
Player 2 throws the dice on throws 2, 4, 6, . . The probability that player 2 wins the pot on
throw is ² ³c h for ~ Á Á Á ÀÀÀ and the probability that player 2 wins is
h b ² ³ h b ² ³ h b Ä ~ h h c² ³ ~ ~ c À
If player 1 puts b dollars into the pot, then his expected gain is h c ² b ³ h À
and player 2' expected gain is ² b ³ h c h In order for the two players to have the same expected gain, we must have
h c ² b ³ h ~ , so that ~ Answer: C
10. Since ²%³ % ~ , it follows that ² b ³ h ~ , so that ~ , and ²%³ ~ % .
Then, 4? ²!³ ~ ,´!? µ ~ !% h % %. Applying integration by
parts, we have
!%
!% %~
!%
!% h % % ~ % ² ³ ~ % c
c %
%
!
!
!
%~
%~
!
!%
!
%!%
!%
~ ! c %
! ² ! ³ ~ ! c < ! c%~ c ! %=
!
!
²! c³
! ²
c
!b! ³
~ ! c ~
c !
.
! b
!
!
Answer: E
11. ,´²? c ³ µ ~ ,´? µ c ,´
? µ b ,´?µ c ,´µ
²³
²³
~ 4? ²³ c 4? ²³ b 4?Z ²³ c .
²³
²³
Z
4?Z ²!³ ~ c ²b!³
S 4? ²³ ~ c , 4? ²!³ ~ ²b!³ S 4? ²³ ~ Á
4? ²!³ ~ c ²b!³
S 4? ²³ ~ c . Then, ,´²? c ³ µ ~ c . Answer: D
²³
²³
12. = ´?µ ~ ,´? µ c ²,´?µ³ , and ,´?µ ~ 4 Z ²³ Á ,´? µ ~ 4 ZZ ²³ .
!
!
!
!
!
!
b b ZZ
4 Z ²!³ ~ ² b
³ h Á 4 ²!³ ~ h ² ³ h ² ³ b ² ³ h .
Then, 4 Z ²³ ~ and 4 ZZ ²³ ~ b ~ , so that = ´?µ ~ c ~ .
Alternatively, = ´?µ ~ !
4 ²!³c
!
!~
. In this case,
!
4 ²!³ ~ ² b
³ ~ ´² b ³ c µ ,
²b! ³²! ³c²! ³²! ³
!
so that !
4 ²!³ ~ b
,
! , and ! 4 ²!³ ~
²b! ³
and then !
4 ²!³c
© ACTEX 2009
!~
~
²³²³c²³²³
²³
~.
Answer: A
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 3
LM-49
!
13. 4 ²!³ ~ ,´!? µ ~ ,´ b !? b ! ?
b ĵ ~ ,´µ b ! h ,´?µ b h ,´? µ b Ä
,´µ ~ Á ,´?µ is given as , and = ´?µ ~ ,´? µ c ²,´?µ³ is given to be ,
so that ,´? µ ~ . Then the first 3 terms of the expansion of 4 ²!³ are b ! b ! .
Answer: B
²c³
14. ,´?µ ~ b
and = ´?µ ~
, so that
b
~
h
²c³
S c ~ S ~ Á S ~ (since € ).
Answer: B
15. The standardized statistics score is c
~ . The standardized math score is
c
~ ~ S ~ .
Answer: B
16. The -th moment of @ is ,´@ µ ~ ,´? µ ~ 4? ²³ ~ Answer: D
°
(since ~ and ~ ).
17. 4? ²!³ ~ c!°
S 4? ² c ³ ~ c²c
³ ~ b ~ b ~ À S ~ .
Answer: D
18. ? has a 5 ²Á ³ distribution, so that the density function of the conditional distribution is
²%³
²%³
²% O ? € ³ ~ 7 ´?€µ ~ ° ~ ²%³À The conditional expectation is
%~B
B % h ²% O ? € ³ % ~ B % h c% ° % ~ c c% ° c
~
j
j
%~
j
~ k Answer: D
19. A lognormal random variable with parameters and has mean
,´?µ ~ b ~ , and variance = ´?µ ~ ² c ³b ~ c .
Then, b ~ ²b ³ ~ S c ~ c S ~ Á ~ .
> ~ ²?³ — 5 ²Á ³ ~ 5 ²Á ³ S 7 ´?  µ ~ 7 ´²?³  µ ~ 7 ´>  µ À
> c,´> µ
Then, A ~ j
~ >c has a standard normal distribution, so that
= ´> µ
7 ´>  µ ~ 7 ´ >c  c
µ ~ 7 ´A  Àµ ~ À
(from the table of the standard normal
distribution) .
Answer: B
20. Since ? — 5 ²Á ³, A ~ ?c
has a standard normal distribution. The probability
in question can be written as
7 ´? c ?  µ ~ 7 ´? c ? b  µ ~ 7 ´²? c ³  µ ~ 7 ´ c  ? c  µ
~ 7 ´ c À  ?c
 Àµ ~ 7 ´ c À  A  Àµ ~ )²À³ c ´ c )²À³µ ~ À
(from the standard normal table).
Answer: D
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-50
MODELING - PROBLEM SET 3
21. ,´?µ ~ c ~ c ~ S ~ S 7 ´? ‚ µ ~ c - ²
³ ~ ² ³ ~ À .
Answer: A
22. The pdf of ) is ²%³ ~ for %  and ²%³ ~ for % ‚ ,
it is ²%³ ~ À for  %  , and it is ²%³ ~ À
for  %  .
There is a point mass of probability with ²%³ ~ À at % ~ () has a mixed
distribution).
,´)µ ~ % h ²À³% b ²³²À³ + % h ²À
³% ~ ,
,´) µ ~ % h ²À³% b ² ³²À³ b % h ²À
³% ~ Á Á = ´)µ ~ ,´) µ c ²,´)µ³ = Á Á . ,´)µ b j= ´)µ ~ .
Answer: B.
23. = ´?µ ~ = ´,´? “ @ µµ b ,´= ´? “ @ µµ ~ = ´@ µ b ,´µ ~ = ´@ µ b ~ h b ~ , since the variance of an exponential random variable is the square of the
mean, thus, = ´@ µ ~ .
Answer: E.
,´²?c,´?µ³ µ
24. ,´?µ ~ , Skewness ~ ²,´²?c,´?µ³ µ³° ~
~
(b³²b³c ²b³²³b²³
´ µ°
,´? µc,´? µh,´?µb²,´?µ³ c²,´?µ³
´,´? µc²,´?µ³ µ°
~ ° ~ S ~ .
,´?µ ~ ~ ~ S ~ S = ´?µ ~ ~ À
Answer: C
25. Jensen's inequality states that if ? is a random variable and ²%³ is a function such that
on the region for which ? has positive probability or density, then
Z ²%³ ‚ ,´²?³µ ‚ ²,´?µ³ . Since - ²³ ~ 7 ´?  µ ~ , it follows that the regions of non-zero
density of ? is ? ‚ . With function ²%³ ~ ²% c ³ , we have Z ²%³ ~ ²% c ³ ‚ (for any %). From Jensen's inequality it follows that
Answer: B
,´²? c ³ µ ~ ,´²?³µ ‚ ²,´?µ³ ~ ²,´?µ c ³ .
26. À ~ 7 ´?  À µ ~ À %% ~ ²À ³ S À ~ jÀ ~ À
.
Answer: D
27. This problem can be solved by eliminating answers based on a careful choice of the constant
values. Suppose that ~ . Then ? ~ b A , or equivalently, A ~ ²? c ³° . Then,
,²@ O?³ ~ ,´ b A b A O?µ ~ ,´ b A b A OA ~ ²? c ³°µ ~ b ´²? c ³°µ
(since A does not appear in ? , it follows that ,´A µ ~ ).
The only answer consistent with this expectation is E, since with ~ Answer E becomes
b ´²³°² ³µ²? c ³ ~ b ´²? c ³°µ .
There is an alternative solution to this problem. If < and > are any two normal random
variables with means < and > , and variances < and >
and covariance < > ,
?@
then it is true that ,´@ O? ~ %µ ~ @ b h ²% c ? ³ .
?
In this example, ? and @ are normal with ? ~ Á @ ~ Á ?
~ b ,
and ?@ ~ *#´ b A b A Á b A b A µ ~ b .
b
Then, ,´@ O?µ ~ b b h ²? c ³ .
Answer: E
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 4 - REVIEW OF RANDOM VARIABLES PART III
LM-51
MODELING SECTION 4 REVIEW OF RANDOM VARIABLES - PART III
Joint, Marginal and Conditional Distributions
The suggested time frame for this section is 2-3 hours.
LM-4.1 Joint Distribution of Random Variables ? and @
A joint distribution of two random variables has a probability function or probability density
function ²%Á &³ that is a function of two variables (sometimes denoted ?Á@ ²%Á &³). It is defined
over a two-dimensional region. For joint distributions of continuous random variables ? and @ ,
the region of probability (the probability space) is usually a rectangle or triangle in the %-& plane.
If ? and @ are discrete random variables, then ²%Á &³ ~ 7 ´²? ~ %³ q ²@ ~ &³µ is the joint
probability function, and it must satisfy
(i)  ²%Á &³  and (ii) ²%Á &³ ~ .
(4.1)
% &
If ? and @ are continuous random variables, then ²%Á &³ must satisfy
B
B
(i) ²%Á &³ ‚ and (ii) cB cB ²%Á &³ & % ~ .
(4.2)
It is possible to have a joint distribution in which one variable is discrete and one is continuous,
or either has a mixed distribution. The joint distribution of two random variables can be extended
to a joint distribution of any number of random variables.
If ( is a subset of two-dimensional space, then 7 ´²?Á @ ³  (µ is the summation (discrete case)
or double integral (continuous case) of ²%Á &³ over the region (.
Example LM4-1: ? and @ are discrete random variables which are jointly distributed
with the following probability function ²%Á &³ (in table form):
?
c
From this table we see, for example, that
@
7 ´? ~ Á @ ~ c µ ~ ²Á c ³ ~ .
c
Find (i) 7 ´? b @ ~ µ , (ii) 7 ´? ~ µ and (iii) 7 ´?  @ µ .
Solution: (i) We identify the ²%Á &³-points for which ? b @ ~ , and the probability is the sum
of ²%Á &³ over those points. The only %Á & combinations that sum to 1 are the points ²Á ³ and
²Á ³ . Therefore, 7 ´? b @ ~ µ ~ ²Á ³ b ²Á ³ ~ b ~ À
(ii) We identify the ²%Á &³-points for which ? ~ . These are ²Á c ³ and ²Á ³ (we omit
²Á ³ since there is no probability at that point). 7 ´? ~ µ ~ ²Á c ³ b ²Á ³ ~ b ~ (iii) The ²%Á &³-points satisfying ?  @ are ² c Á ³ Á ² c Á ³ and ²Á ³ .
Then 7 ´?  @ µ ~ ² c Á ³ b ² c Á ³ b ²Á ³ ~ b b ~ À
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-52
MODELING SECTION 4 - REVIEW OF RANDOM VARIABLES PART III
Example LM4-2: Suppose that ²%Á &³ ~ 2²% b & ³ is the density function for the joint
distribution of the continuous random variables ? and @ defined over the unit square bounded by
the points ²Á ³ Á ²Á ³ Á ²Á ³ and ²Á ³, find 2 . Find 7 ´? b @ ‚ µ .
Solution: In order for ²%Á &³ to be a properly defined joint density, the (double) integral of the
density function over the region of density must be 1, so that
~ 2²% b & ³ & % ~ 2 h ¬ 2 ~ ¬ ²%Á &³ ~ ²% b & ³ for  %  and  &  .
In order to find the probability 7 ´? b @ ‚ µ, we identify the two dimensional region
representing ? b @ ‚ . This is generally found by drawing the boundary line for the
inequality, which is % b & ~ (or & ~ c %) in this case, and then determining which side of
the line is represented in the inequality. We can see that % b & ‚ is equivalent to & ‚ c % .
This is the shaded region in the graph below.
The probability 7 ´? b @ ‚ µ is found by integrating the joint density over the two-dimensional
region. It is possible to represent two-variable integrals in either order of integration. In some
cases one order of integration is more convenient than the other. In this case there is not much
advantage of one direction of integration over the other.
&~
7 ´? b @ ‚ µ ~ c% ²% b & ³ & % ~ ²% & b & c
~ ²% b c % ² c %³ c ² c %³ ³ % ~ À
Reversing the order of integration, we have % ‚ c & , so that
7 ´? b @ ‚ µ ~ c& ²% b & ³ % & ~ À
&~c%
³ %
Expectation of a function of jointly distributed random variables
If ²%Á &³ is a function of two variables, and ? and @ are jointly distributed random variables,
then the expected value of ²?Á @ ³ is defined to be
,´²?Á @ ³µ ~ ²%Á &³ h ²%Á &³ in the discrete case, and
% &
B
B
,´²?Á @ ³µ ~ cB cB ²%Á &³ h ²%Á &³ & % in the continuous case.
(4.3)
Example LM4-3: ? and @ are discrete random variables which are jointly distributed
with the following probability function ²%Á &³ (from Example LM4-1):
?
c
@
c
Find ,´? h @ µ.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 4 - REVIEW OF RANDOM VARIABLES PART III
LM-53
Solution: ,´?@ µ ~ %& h ²%Á &³ ~ ² c ³²³² ³ b ² c ³²³² ³ b ² c ³² c ³² ³
% &
b ²³²³² ³ b ²³²³²³ b ²³² c ³² ³ b ²³²³² ³ b ²³²³² ³ b ²³² c ³² ³
Note that the summation is taken over all pairs ²%Á &³ in the joint distribution.
~ .
Example L4-4: Suppose that ²%Á &³ ~ ²% b & ³ is the density function for the joint
distribution of the continuous random variables ? and @ defined over the unit square defined on
the region  %  and  &  . Find ,´? b @ µ .
Solution:
,´? b @ µ ~ ²% b & ³ h ²%Á &³ & % ~ ²% b & ³² ³²% b & ³ & %
~ ²À% b % b À³ % ~ .
LM-4.2 Marginal distribution of ? found from a joint distribution of ? and @
If ? and @ have a joint distribution with joint density or probability function ²%Á &³, then the
marginal distribution of ? has a probability function or density function denoted ? ²%³, which
is equal to ? ²%³ ~ ²%Á &³ in the discrete case, and is equal to
&
B
? ²%³ ~ cB ²%Á &³ & in the continuous case.
(4.4)
The density function for the marginal distribution of @ is found in a similar way, @ ²&³ is equal
B
to either @ ²&³ ~ ²%Á &³ or @ ²&³ ~ cB ²%Á &³ % .
%
For instance, ? ²³ ~ ²Á &³ in the discrete case. What we are doing is "adding up" the
&
probability for all points whose %-value is 1 to get the overall probability the ? is 1.
Marginal probability functions and marginal density functions must satisfy all the requirements of
probability and density functions. A marginal probability function must sum to 1 over all points
of probability and a marginal density function must integrate to 1. The marginal distribution of ?
describes the behavior of ? alone without reference to @ (and same for marginal of @ ).
Example LM4-5: Find the marginal distributions of ? and @ for the joint distribution in
Example LM4-1 .
Solution: The joint distribution was given as
?
c
@
c
To find the marginal probability function for ? , we first note that ? can be c Á or .
We wish to find ? ² c ³ ~ 7 ´? ~ c µ Á ? ²³ and ? ²³ .
As noted above, to find ? ²%³ we sum over the other variable @ :
? ² c ³ ~ ² c Á &³ ~ ² c Á c ³ b ² c Á ³ b ² c Á ³ ~ b b ~ .
all &
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-54
MODELING SECTION 4 - REVIEW OF RANDOM VARIABLES PART III
Example LM4-5 continued
In a similar way we get ? ²³ ~ b b ~ and ? ²³ ~ b b ~ .
In Example LM4-1 we saw that 7 ´? ~ µ ~ . What we were finding was the marginal
probability ? ²³ that was just found here . Note also that
? ²%³ ~ ? ² c ³ b ? ²³ b ? ²³ ~ b b ~ . This verifies that ? ²%³ satisfies
all %
the requirements of a probability function. The marginal probability function of @ is found in the
same way, except that we sum over % (across each row in the table above).
@ ² c ³ ~ b b ~ , @ ²³ ~ b b ~ and @ ²³ ~ b b ~ .
Example LM4-6: Find the marginal distributions of ? and @ for the joint distribution in
Example LM4-2 .
Solution: The joint density function is ²%Á &³ ~ ²% b & ³ for  %  and  &  .
The marginal density function of ? is found by integrating out the other variable &.
? ²%³ ~ all & ²%Á &³ & ~ ²%Á &³ & ~ ²% b & ³ & ~ % b for  %  .
We can verify that this is a proper density function by checking that ? ²%³ % ~ .
In a similar way, @ ²&³ ~ & b for  &  .
LM-4.3 Independence of random variables ? and @
Random variables ? and @ with density functions ? ²%³ and @ ²&³ are said to be independent
(or stochastically independent) if the probability space is rectangular (  %  Á  &  ,
where the endpoints can be infinite) and if the joint density function is of the form
(4.5)
²%Á &³ ~ ? ²%³ h @ ²&³ .
If ? and @ are independent, then for any functions and ,
,´²?³ h ²@ ³µ ~ ,´²?³µ h ,´²@ ³µ , and in particular, ,´? h @ µ ~ ,´?µ h ,´@ µ.
(4.6)
For the discrete joint distribution in Example LM4-1 we can see that ? and @ are not
independent, because, for instance, ² c Á c ³ ~ £ h ~ ? ² c ³ h @ ² c ³ . For the
continuous joint distribution of Example LM4-2, we see that
²%Á &³ ~ ²% b & ³ £ ² % b ³² & b ³ ~ ? ²%³ h @ ²&³ , so ? and @ are not
independent.
Example LM4-7: Suppose that ? and @ are independent continuous random variables with the
following density functions: ? ²%³ ~ for  %  and @ ²&³ ~ & for  &  .
Find 7 ´@  ?µ.
Solution: Since ? and @ are independent, the density function of the joint distribution of ? and
@ is ²%Á &³ ~ ? ²%³ h @ ²&³ ~ & , and is defined on the rectangle created by the intervals for
%
? and @ , which, in this case, is the unit square. 7 ´@  ?µ ~ & & % ~ .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 4 - REVIEW OF RANDOM VARIABLES PART III
LM-55
LM-4.4 Conditional distribution of @ given ? ~ %
Conditional distributions are very important in the loss models and credibility material and must
be understood well..
The way in which a conditional distribution is defined follows the basic definition of conditional
probability, 7 ´(O)µ ~
7 ´(q)µ
7 ´)µ
. In fact, given discrete joint distribution, this is exactly how a
conditional distribution is defined. Example LM4-1 described a discrete joint distribution of ?
and @ , and then Example LM4-5 showed how to formulate the marginal distributions of ? and
@ . We now wish to formulate a conditional distribution. For instance, for the joint distribution of
Example LM4-1, suppose we wish to describe the conditional distribution of ? given @ ~ .
What we are trying to describe are conditional probabilities of the form 7 ´? ~ %O@ ~ µ.
We find these conditional probabilities in the usual way that conditional probability is defined.
7 ´? ~ c O@ ~ µ ~
7 ´²?~c³q²@ ~³µ
7 ´@ ~µ
.
The denominator is the marginal probability that @ ~ , @ ²³ ~ . The numerator is the joint
probability ² c Á ³ ~ , which is found in the joint probability table. Then,
²cÁ³
°
7 ´? ~ c O@ ~ µ ~ ²³ ~ ° ~ . We would denote this conditional probability
@
²Á³
°
?O@ ² c O@ ~ ³ . In a similar way, we can get ?O@ ²O@ ~ ³ ~ ²³ ~ ° ~ , and
@
²Á³
°
?O@ ²O@ ~ ³ ~ ²³ ~ ° ~ . This completely describes the conditional distribution of
@
? given @ ~ . As with any discrete distribution, probabilities must add to 1, and this is the case
for this conditional distribution, since
?O@ ² c O@ ~ ³ b ?O@ ²O@ ~ ³ b ?O@ ²O@ ~ ³ ~ b b ~ .
A conditional distribution satisfies all the same properties of any distribution. We can find a
conditional mean, a conditional variance, etc. For instance, the conditional mean of ? given
@ ~ in the example we have just been considering is
,´?O@ ~ µ ~ ' % h ?O@ ²%O@ ~ ³
%
~ ² c ³?O@ ² c O@ ~ ³ b ²³?O@ ²O@ ~ ³ b ²³?O@ ²O@ ~ ³
~ ² c ³² ³ b ²³² ³ b ²³² ³ ~ À
We can find the second conditional moment in a similar way,
,´? O@ ~ µ ~ ' % h ?O@ ²%O@ ~ ³ . Then the conditional variance would be
%
= ´?O@ ~ µ ~ ,´? O@ ~ µ c ²,´?O@ ~ µ³ .
The expression for conditional probability that was used above in the discrete case was
²%Á&³
?O@ ²%O@ ~ &³ ~ ²&³ . This can be applied to find a conditional distribution of @ given
@
²%Á&³
? ~ % also, so that we define @ O? ²&O? ~ %³ ~ ²%³ .
?
© ACTEX 2009
(4.7)
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-56
MODELING SECTION 4 - REVIEW OF RANDOM VARIABLES PART III
We also apply this same algebraic form to define the conditional density in the continuous case,
with ²%Á &³ being the joint density and ? ²%³ being the marginal density. In the continuous case,
the conditional mean of @ given ? ~ % would be
,´@ O? ~ %µ ~ & h @ O? ²&O? ~ %³ & ,
(4.8)
where the integral is taken over the appropriate interval for the conditional distribution of @
given ? ~ %. The conditional density/probability is also written as @ O? ²&O%³ , or ²&O%³ .
If ? and @ are independent random variables, then @ O? ²&O? ~ %³ ~ @ ²&³ and
?O@ ²%O@ ~ &³ ~ ? ²%³ , which indicates that the density of @ does not depend on ? and viceversa. The conditional density function must satisfy the usual requirement of a density function,
B
cB
@ O? ²&O%³ & ~ . Note also that if the marginal density of ? is known, ? ²%³, and the
conditional density of @ given ? ~ % is also known, @ O? ²&O? ~ %³ , then the joint density of
? and @ can be formulated as ²%Á &³ ~ @ O? ²&O? ~ %³ h ? ²%³ À
(4.9)
Example LM4-8: Find the conditional distribution of @ given ? ~ c for the joint
distribution of Example LM4-1. Find the conditional expectation of @ given ? ~ c .
Solution: The marginal probability function for ? was found in Example LM4-5, where it was
found that ? ² c ³ ~ . The conditional probability function of @ given ? ~ c is
²cÁ&³
²cÁ&³
° . Then,
²cÁc³
°
²cÁ³
°
@ O? ² c O? ~ c ³ ~
~ ° ~ Á @ O? ²O? ~ c ³ ~ ° ~ °
°
²cÁ³
°
and @ O? ²O? ~ c ³ ~ ° ~ ° ~ .
,´@ O? ~ c µ ~ & h @ O? ²&O? ~ c ³ ~ ² c ³² ³ b ²³² ³ b ²³² ³ ~ c .
@ O? ²&O? ~ c ³ ~ ²c³ ~
?
all &
~ Á
Example LM4-9: Find the conditional density and conditional expectation and conditional
variance of ? given @ ~ À for the joint distribution of Example LM4-2.
Solution: ?O@ ²%O@ ~ À³ ~
²%ÁÀ³
@ ²À³
~
²% bÀ ³
À b The conditional expectation is
~
²% bÀ³
À
.
²% bÀ³
,´?O@ ~ Àµ ~ % h ?O@ ²%O@ ~ À³ % ~ % h À
% ~ À
.
The conditional second moment of ? given @ ~ À is
²% bÀ³
,´? O@ ~ Àµ ~ % h ?O@ ²%O@ ~ À³ % ~ % h À
% ~ À .
The conditional variance is
= ´?O@ ~ Àµ ~ ,´? O@ ~ Àµ c ²,´?O@ ~ Àµ³ ~ À c ²À
³ ~ À .
We can construct the joint density ²%Á &³ from knowing the conditional density @ O? ²&O%³ and
the marginal density ? ²%³ using the relationship ²%Á &³ ~ ²&O%³ h ? ²%³ . When doing this,
care must be taken to ensure that proper two-dimensional region is being formulated for the joint
distribution.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 4 - REVIEW OF RANDOM VARIABLES PART III
LM-57
Example LM4-10: Suppose that ? has a continuous distribution with pdf ? ²%³ ~ %
on the interval ²Á ³ , and ? ²%³ ~ elsewhere. Suppose that @ is a continuous random
variable such that the conditional distribution of @ given ? ~ % is uniform on the interval ²Á %³.
Find the mean and variance of (the marginal distribution of) @ .
Solution: We find the unconditional (marginal) distribution of @ . We are given ? ²%³ ~ %
for  %  , and @ O? ²&O? ~ %³ ~ % for  &  % .
Then, ²%Á &³ ~ ²&O%³ h ? ²%³ ~ % h % ~ for  &  %  .
The unconditional (marginal) distribution of @ has pdf.
B
@ ²&³ ~ cB ²%Á &³ % ~ & % ~ ² c &³ for  &  (and @ ²&³ is elsewhere). Then ,´@ µ ~ & h ² c &³ & ~ , ,´@ µ ~ & h ² c &³ & ~ ,
and = ´@ µ ~ ,´@ µ c ²,´@ µ³ ~
c ² ³
~
.
The idea applied in Example LM4-10 is quite important in the continuous Bayesian credibility
approach that will be covered later in the study guide.
LM-4.5 Covariance Between Random Variables ? and @
If random variables ? and @ are jointly distributed with joint density/probability function
²%Á &³, the covariance between ? and @ is
* #´?Á @ µ ~ ,´²? c ,´?µ³²@ c ,´@ µ³µ
~ ,´²? c ? ³²@ c @ ³µ ~ ,´?@ µ c ,´?µ h ,´@ µÀ
(4.10)
Note that *#´?Á ?µ ~ = ´?µ .
If ? and @ are independent, then ,´? h @ µ ~ ,´?µ h ,´@ µ and *#´?Á @ µ ~ .
(4.11)
For constants Á Á Á Á Á and random variables ?Á @ Á A and > ,
*#´? b @ b Á A b > b µ
~ *#´?Á Aµ b *#´?Á > µ b *#´@ Á Aµ b *#´@ Á > µ
(4.12)
An important application of the covariance is in finding the variance of the sum of ? and @ .
Suppose that , and are constants. Then
= ´? b @ b µ ~ = ´?µ b = ´@ µ b *#´?Á @ µ .
(4.13)
If ? and @ are independent, then = ´? b @ µ ~ = ´?µ b = ´@ µ .
~
~
If ? Á ? Á ÀÀÀÁ ? are independent, then = ´ ? µ ~ = ´? µ .
(4.14)
If ? Á ? Á ÀÀÀÁ ? are independent, and if : ~ ? , then the moment generating
~
function of : is 4: ²!³ ~ 4? ²!³ .
(4.15)
~
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-58
MODELING SECTION 4 - REVIEW OF RANDOM VARIABLES PART III
LM-4.6 Coefficient of correlation between random variables ? and @
The coefficient of correlation between random variables ? and @ is defined to be
* #´?Á@ µ
²?Á @ ³ ~ ?Á@ ~ ,
? @
where ? and @ are the standard deviations of ? and @ respectively. Note that
c  ?Á@  always.
(4.16)
Example LM4-11: Find *#´?Á @ µ for the jointly distributed discrete random variables in
Example LM4-1 above.
Solution: *#´?Á @ µ ~ ,´?@ µ c ,´?µ h ,´@ µ. In Example LM4-3 it was found that
,´?@ µ ~ . The marginal probability function for ? is 7 ´? ~ µ ~ b b ~ ,
7 ´? ~ µ ~ and 7 ´? ~ c µ ~ , and the mean of ? is
,´?µ ~ ²³² ³ b ²³² ³ b ² c ³² ³ ~ .
In a similar way, the probability function of @ is found to be 7 ´@ ~ µ ~ Á 7 ´@ ~ µ ~ Á
and 7 ´@ ~ c µ ~ Á with a mean of ,´@ µ ~ c .
Then, *#´?Á @ µ ~ c ² ³² c ³ ~ .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 4
LM-59
MODELING - PROBLEM SET 4
Review of Random Variables III - Section 4
1. Let ? and @ be discrete random variables with joint probability function
²%Á &³ ~ F
A) %bc&
for %~Á and &~Á
Á otherwise
B) C) . Calculate ,´ ?
@ µ.
D) E) 2. Let ? and @ be continuous random variables with joint cumulative distribution function
- ²%Á &³ ~ ²%& c % & c %& ³ for  %  and  &  .
Determine 7 ´? € µ .
A) B) C) D) ²& c & ³
E) ²
& c & ³
3. Let ? and @ be discrete random variables with joint probabilities given by
?
b b @
b b Let the parameters and satisfy the usual assumption associated with a joint probability
distribution and the additional constraints c À   À and
  À . If ? and @ are independent, then ² Á ³ ~
A) ²Á ³ B) ² Á ³ C) ² c Á ³ D) ² c Á ³ E) ² Á ³
4. Let ? and @ be continuous random variables with joint density function
for %&
²%Á &³ ~ D
, otherwise
.
Determine the density function of the conditional distribution of @ given ? ~ %, where
 %  .
A) c%
for %  &  B) ² c %³ for %  &  C) for %  &  D) & for %  &  E) c&
for %  &  5. A wheel is spun with the numbers 1, 2 and 3 appearing with equal probability of each. If the
number 1 appears, the player gets a score of 1.0; if the number 2 appears, the player gets a score
of 2.0; if the number 3 appears, the player gets a score of ? , where ? is a normal random
variable with mean 3 and standard deviation 1. IF > represents the player's score on 1 spin of
the wheel, then what is 7 ´>  Àµ?
A) À
B) À
C) À
D) À
E) À
6. Let ? and @ be continuous random variables with joint density function
²%Á &³ ~ D
%b& for % Á &
Á otherwise
.
What is the marginal density function for ?Á where nonzero?
A) & b © ACTEX 2009
B) %
C) %
D) %b%
E) % b SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-60
MODELING - PROBLEM SET 4
7. Let ? and @ be discrete random variables with joint probability function
²%Á &³ ~ D
²%b³²&b³
for %~ÁÁ Â &~ÁÁ
Á otherwise
What is ,´@ O? ~ µ?
A) B) C) D)
&b
.
E)
& b&
8. Let ? and @ be continuous random variables with joint density function
²%Á &³ ~ D
Note that
A) % for %&
.
Á otherwise
,´?µ ~ and ,´@ µ
B) C) ~ . What is *#´?Á @ µ ?
D) E) 9. The distribution of Smith's future lifetime is ? , an exponential random variable with mean ,
and the distribution of Brown's future lifetime is @ , an exponential random variable with mean .
Smith and Brown have future lifetimes that are independent of one another. Find the probability
that Smith outlives Brown.
c
c
A) b
B) b
C) D) E) 10. In reviewing some data on smoking (? , number of packages of cigarettes smoked per year),
income ²@ , in thousands per year³ and health (A , number of visits to the family physician per
year) for a sample of males, it is found that
,´?µ ~ Á = ´?µ ~ Á ,´@ µ ~ Á = ´@ µ ~ Á ,´Aµ ~ Á = ´Aµ ~ ,
and *#²?Á @ ³ ~ c Á *#²?Á A³ ~ À (covariances).
Dr. N.A. Ively, a young statistician, attempts to describe the variable A in terms of ? and @ by
the relation A ~ ? b @ , where is a constant to be determined . Dr. Ively's methodology for
determining is to find the value of for which *#²?Á A³ remains equal to 2.5 when A is
replaced by ? b @ . What value of does Dr. Ively find?
A) 2.00
B) 2.25
C) 2.50
D) c 2.00
E) c 2.25
11. In order to simplify an actuarial analysis Actuary A uses an aggregate distribution
: ~ ? b Ä b ?5 , where 5 has a Poisson distribution with mean 10 and ? ~ À for all .
Actuary A’s work is criticized because the actual severity distribution is given by
7 ²@ ~ ³ ~ ²@ ~ ³ ~ À, for all , where the @ ’s are independent.
Actuary A counters this criticism by claiming that the correlation coefficient between : and
: i ~ @ b Ä b @5 is high.
Calculate the correlation coefficient between : and : i .
(A) 0.75
(B) 0.80
(C) 0.85
(D) 0.90
(E) 0.95
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 4
LM-61
MODELING - PROBLEM SET 4 SOLUTIONS
? h ²? Á @ ³ ~ h b h
1. ,´ ?
µ
~
@
@
%~ &~ b h b h ~ Answer: D
2. -? ²³ ~ 7 ´?  µ ~ lim - ²Á &³ ~ - ²Á ³ ~ ~ ,
&¦B
so that 7 ´? € µ ~ c 7 ´?  µ ~ À
Answer: C
3. Since the total probability must be 1, we have b ~ .
The marginal distributions of ? and @ have
7 ´? ~ µ ~ 7 ´? ~ µ ~ 7 ´@ ~ µ ~ 7 ´@ ~ µ ~ b ~ À Then, because
of independence, 7 ´? ~ Á @ ~ µ ~ 7 ´? ~ µ h 7 ´@ ~ µ ~ ~ b .
Solving the two equations in and ( b ~ and b ~ ) results in
~ , ~ .
Answer: B
4. The region of joint density is the triangular region above the line & ~ % and below the
horizontal line & ~ for  %  À The conditional density of & given ? ~ % is
²%Á&³
²& O ? ~ %³ ~ ²%³ Á where ? ²%³ is the marginal density function of %.
?
B
? ²%³ ~ cB ²%Á &³ & ~ % & ~ ² c %³ Á so that ²& O ? ~ %³ ~ ²c%³
~ c%
and the region of density for the conditional distribution of @ given ? ~ % is %  &  À
It is true in general that if a joint distribution is uniform (has constant density in a region) then
any conditional (though not necessarily marginal) distribution will be uniform on it restricted
region of probability - the conditional distribution of @ given ? ~ % is uniform on the interval
%  &  , with constant density c%
.
Answer: A
5. Let 5 denote the number that appears on the wheel, so that
7 ´5 ~ µ ~ 7 ´5 ~ µ ~ 7 ´5 ~ µ ~ . Then, conditioning over 5 ,
7 ´>  Àµ ~ 7 ´>  ÀO5 ~ µ h 7 ´5 ~ µ b µ7 ´>  ÀO5 ~ µ h 7 ´5 ~ µ
b 7 ´>  ÀO5 ~ µ h 7 ´5 ~ µ .
If 5 ~ then > ~ , so that 7 ´>  ÀO5 ~ µ ~ Á and
if 5 ~ then > ~ , so that 7 ´>  ÀO5 ~ µ ~ .
If 5 ~ then > — 5 ²Á ³ so that
7 ´>  ÀO5 ~ µ ~ 7 ´ >c  Àc
O5 ~ µ ~ 7 ´A  c Àµ ~ À
(A has a standard normal distribution - the probability is found from the table).
Then, 7 ´>  Àµ ~ h b h b ²À³ h ~ À .
Answer: C
6. ? ²%³ ~ ²% b &³ & ~ % b .
© ACTEX 2009
Answer: E
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-62
MODELING - PROBLEM SET 4
B
7. ? ²³ ~ 7 ´? ~ µ ~ ²Á &³ ~ ²Á ³ b ²Á ³ b ²Á ³ ~ .
&~cB
²Á³
°
Then we have conditional probabilities 7 ´@ ~ O? ~ µ ~ 7 ´?~µ ~ ° ~ Á
and similarly, 7 ´@ ~ O? ~ µ ~ and 7 ´@ ~ O? ~ µ ~ .
Then, ,´@ O? ~ µ ~ h b h b h ~ À
Answer: C
8. *#´?Á @ µ ~ ,´?@ µ c ,´?µ h ,´@ µ
The region of probability is the triangle
above the line & ~ % in the unit square
%,&.
&
,´?@ µ ~ %& h % % & ~ S *#´?Á @ µ ~ c h ~ .
Alternatively,
,´?@ µ ~ % %& h % & % ~ .
Answer: A
B B
9. 7 ´@  ?µ ~ & ? ²%³@ ²&³ % & (since ? and @ are independent, the joint density
function of ? and @ is the product of the two separate density functions).
The density function of ? is c%° , and of @ is c%° , so that
B B
B
7 ´@  ?µ ~ & c%° c&° % & ~ c&° c&° & ~ b
~ b .
Answer: A
10. *#²?Á ? b @ ³ ~ *#²?Á ?³ b *#²?Á @ ³ ~ = ´?µ b *#²?Á @ ³
~ c . This is set equal to *#²?Á A³ ~ À , so that
c ~ À S ~ À .
Answer: B
11. The covariance between : and : i is ,´:: i µ c ,´:µ h ,´: i µ .
,´:µ ~ ,´5 µ h ,´?µ ~ ²³²À³ ~ Á
,´: i µ ~ ,´5 µ h ,´@ µ ~ ²³²À³ ~ .
We use the double expectation rule ,´: h : i µ ~ ,´,´: h : i O5 µµ .
,´: h : i O5 µ ~ ,´²? b Ä b ?5 ³²@ b Ä b @5 ³O5 µ
~ ,´À5 ²@ b Ä b @5 ³O5 µ ~ ²À5 ³²5 h ,´@ µ³ ~ À5 h ²À³ ~ À5 .
Then ,´: h : i µ ~ ,´À5 µ ~ À,´5 µ ~ À²= ´5 µ b ²,´5 µ³ ³
~ À² b ³ ~ À À
*#´:Á : i µ ~ ,´: h : i µ c ,´:µ h ,´: i µ ~ À c ²³²³ ~ À .
= ´:µ ~ = ´À5 µ ~ À= ´5 µ ~ À Á
= ´: i µ ~ = ´5 µ²,´@ µ³ b ,´5 µ= ´@ µ ~ ,´@ µ (since 5 is Poisson)
~ ²³´² ³²À³ b ² ³²À³µ ~ .
Then the correlation coefficient between : and : i is
*#´:Á: i µ
À
j= ´:µh= ´: i µ ~ j²À³²³ ~ À .
© ACTEX 2009
Answer: E
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 5 - PARAMETRIC DISTRIBUTIONS AND TRANSFORMATIONS
LM-63
MODELING SECTION 5
PARAMETRIC DISTRIBUTIONS AND TRANSFORMATIONS
The material in this section relates to Sections 4.2.1-4.2.2, 5.2.1-5.2.3 and 5.3 of "Loss Models".
The suggested time frame for this section is 2 hours. There is been very infrequent reference to
this topic on the exam. Later topics do not depend on this material, and it can be postponed and
covered at a later time.
LM-5.1 Parametric Distributions (Loss Models Sections 4.2.1-4.2.2)
For any random variable, the mean and variance (skewness, etc.) are "parameters" of the
distribution (some of these might be infinite) that can be calculated if the form of the distribution
is known, or they can be estimated when a sample of data is available. There is also the notion of
a parametric distribution, which means that the random variable ? has a pdf (or cdf) which is
formulated in terms of parameters. In this definition, "parametric distribution" refers to a
collection of distributions based on the set of all possible values of the parameters. Some
examples are as follows.
° if %
(i) The uniform distribution on the interval ´Á µ has pdf ²%³ ~ D
otherwise
.
This is a parametric distribution with parameter (the mean is , and other distribution quantities
such as variance, skewness, etc. are formulated in terms of the parameter ).
(ii) The normal distribution has parameters (mean) and (standard deviation) has pdf
°
c²%c³
²%³ ~ j
for c B  %  B.
(iii) The Poisson distribution with parameter € is a discrete, non-negative integer-valued
c %
random variable with pf ²%³ ~ 7 ´? ~ %µ ~ %[ for % ~ Á Á Á ÀÀÀ
(iv) The exponential distribution with parameter € has pdf ²%³ ~ c%° for % € and
cdf - ²%³ ~ c c%° .
There are a large number of parametric distributions in the Tables for Exam C (taken from an
Appendix of the "Loss Models" book). Information on the pdf, cdf, moments, etc. is given there.
Scale distribution and scale parameter
A continuous parametric distribution is a scale distribution if ? is a member of the set of
distributions whenever ? is a member and € is a constant. is a scale parameter if the
corresponding parameter for ? is . It is possible for a distribution to be a scale distribution
and yet not have a scale parameter (the lognormal distribution is an example).
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-64
MODELING SECTION 5 - PARAMETRIC DISTRIBUTIONS AND TRANSFORMATIONS
Example LM5-1: The exponential distribution has a pdf of the form ²%Â ³ ~ c%° , or
equivalently, cdf of the form - ²% ³ ~ c c%° , € .
Show that this is a scale distribution and that is the scale parameter.
Solution: If @ ~ ? and € , then 7 ´@  &µ ~ 7 ´?  &°µ ~ c c²&°³° ~ c c&° .
@ has the cdf of an exponential distribution with parameter . Therefore the exponential family
is a scale family and the exponential parameter is a scale parameter.
The Exam C table of distributions has a number of distributions which involve the parameter .
The table has been arranged so that always is a scale parameter. Any distribution that has the
parameter in the table is a scale distribution. The only continuous distributions in the table that
does not have a parameter are the lognormal and the log-! distributions.
The Pareto distribution is another example of a scale distribution. Suppose that ? has a Pareto
distribution with parameters and , and suppose that € is a constant. Let @ ~ ? . Then
@ also has a Pareto distribution with parameters and ( is the scale parameter, is
unchanged).
LM-5.2 Families of Distributions (5.3)
Section 5.3 of "Loss Models" outlines how families of probability distributions can be organized.
In additional, it is shown that there is a very general 4-parameter distribution, the Transformed
Beta that has most of the other distributions in the Exam C Table as limits or special cases. There
has not been any reference to this on released exams, so it may not be necessary to devote a lot of
time to it.
A couple of simple observations that can be made are that the exponential distribution is a special
case of the gamma distribution and it is also a special case of the Weibull distribution.
Suppose ? has a gamma distribution with parameters and .
c c%°
The pdf of ? is ²%³ ~ % !²³ .
c%°
If ~ , the pdf becomes ²%³ ~ , which is the pdf of the exponential distribution with
parameter .
Suppose ? has a Weibull distribution with parameters and .
If ~ , the pdf
parameter .
(%/) c(%°)
.
%
c%°
becomes ²%³ ~ The pdf of ? is ²%³ ~
, which is the pdf of the exponential distribution with
There are some other examples in Section 5.3 of "Loss Models" in which limits of pdf's are taken
as a parameter goes to 0 or B, and the limiting distribution is of a recognizable form.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 5 - PARAMETRIC DISTRIBUTIONS AND TRANSFORMATIONS
LM-65
LM-5.2.1 The Linear Exponential Family
A random variable with probability function or a density function that can be written in the form
²%Â ³ ~
²%³ ²³%
²³
, where ²%³ does not depend on is said to be from the exponential
family. Also, the region on which the random variable is defined cannot depend on . Some
examples of random variables that are exponential family members are:
c %
c ² ³%
(i) Poisson with parameter : ²%Â ³ ~ %[ ~ %[
,
²%³ ~ %[ , ²³ ~ and ²³ ~ .
(ii) Binomial with trials and parameter :
²%Â ³
²%³ ~
c%
3 %
~ 2
~ % c % ² c ³
% ² c ³
c
2
3
% Á ²³ ~ c Á ²³ ~ ² c ³
2 3²
²
³
~
³
2
3 ² c ³ %
% ²c³c
,
c%°
(iii) Exponential with parameter : ²%Â ³ ~ , ²%³ ~ Á ²³ ~ Á ²³ ~ The textbook shows that the normal distribution is a member of the linear exponential family if
the mean is the parameter, and the gamma distribution is also a member of the linear exponential
family.
We have the following general rules for a member of the linear exponential family:
Z ²³
,²?³ ~ ²³ ~ Z ²³h²³
and
Z ²³
= ²?³ ~ Z ²³ .
(5.1)
Example LM5-2: Use Equation 5.1 to find the mean and variance of a binomial random
variable.
Solution: For a binomial with trials and probability of success, the parameter for the linear
exponential family representation is ~ . ²%Â ³ ~
where ²%³ ~
Z ²³ ~ ² c ³
Z ²³
²³h²³
~ ² c ³
and Z ²³ ~ 2
3
% Á ²³
cc
²c³cc
2
3 ² c ³ %
% ²c³c
,
Á ²³ ~ ² c ³c .
b c
so that
~ ² b ³h²c³c ~ ~ ²³ (mean of the binomial).
c
Z ²³
Then Z ²³ ~ so that Z ²³ ~ b ~ ² c ³ (variance of the binomial).
c
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-66
MODELING SECTION 5 - PARAMETRIC DISTRIBUTIONS AND TRANSFORMATIONS
LM-5.3 A Comment on the Pareto (two-parameter) and
the Single Parameter Pareto
The Exam C table gives information on the Pareto distribution, which has parameters and ,
and pdf ²%³ ~ ²%b³b for % € . The Single Parameter Pareto distribution is also defined
in the table, with pdf ²&³ ~ &
b for & € .
These two distributions are essentially the same. They are related through the relation
@ ~ ? b . The Single Parameter Pareto is just the Pareto shifted to the right by a distance of ;
if & ~ % b , then & € is equivalent to % € .
Note that the mean of the Pareto is ,´?µ ~ c
, and the mean of the Single Parameter Pareto is
,´@ µ ~ c , which is equal to ,´?µ b (as should be the case if @ ~ ? b ).
If a Pareto distribution is mentioned in the Exam C context, it refers to the two-parameter version;
that is the default meaning of Pareto. The Single Parameter Pareto distribution will be explicitly
named as such if that is the version being intended.
Both versions of the Pareto distribution arise in the estimation topics to be covered later in the
Exam C material. In the case of the Single Parameter Pareto, the value of would be given, so
only the one parameter is estimated. In the case of the default (two parameter) Pareto
distribution, both and would be estimated.
LM-5.4 The Distribution of a Transformed Continuous Random Variable
The Loss Models book presents a few ways of constructing new continuous distributions from
existing ones. If ? is a continuous random variable with cdf -? ²%³ and pdf ? ²%³ and if @ is
constructed as a function or transformation of ? , then for some transformations it may be
possible to formulate -@ ²&³ and @ ²&³ in terms of -? ²%³ and ? ²%³ .
There is a general rule that can be applied if @ ~ ²?³ and is an invertible function. An
invertible function is one for which we can reformulate the original function to write ? in terms
of @ , say ? ~ ²@ ³. An example of an invertible function is & ~ ²%³ ~ % b , which can
&c
be reformulated as % ~ . Another example is & ~ ²%³ ~ % for % € , which has inverse
% ~ &° (positive square root).
In general, will be an invertible function if it is strictly increasing or strictly decreasing on the
region for which it is being used. That is why ²%³ ~ % for % € was invertible; it is strictly
increasing for % € (but not for all real %).
Suppose that ²&³ is the inverse function of ²%³, so that % ~ ²&³ ~ ²²%³³ ~ % and
& ~ ²%³ ~ ²²&³³ . For instance, if ²%³ ~ % , then ²&³ ~ ²&³ is the inverse of ,
since ²²%³³ ~ ²% ³ ~ % and ²²&³³ ~ & ~ & .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 5 - PARAMETRIC DISTRIBUTIONS AND TRANSFORMATIONS
LM-67
Under these assumptions, we have the following relationship:
if the pdf of ? is ? ²%³, and if @ ~ ²?³ , then the pdf of @ is
@ ²&³ ~ ? ²²&³³ h O Z ²&³O
(5.2)
(the absolute value ensures that the pdf of @ is non-negative).
Also -@ ²&³ ~ -? ²²&³³ if is increasing, and -@ ²&³ ~ c -? ²²&³³ if is decreasing.
It is also important to determine the region of probability for the transformed variable @ .
If the region of probability for ? is the interval ²Á ³ , then the region of probability for @ will
be the interval ²²³ Á ²³³ if is an increasing function, and ²²³ Á ²³³ if is a decreasing
function.
Example LM5-3: ? has the pdf ? ²%³ ~ % for  %  .
@ is defined by the transformation @ ~ c ²?³ . Find the pdf of @ .
Solution: The inverse of the function ²%³ ~ c ²%³ is ²&³ ~ c& .
@ ²&³ ~ ? ²²&³³ h O Z ²&³O ~ c& h O c c& O ~ c& defined on the region  &  B .
? is defined on the interval ²Á ³ and ²%³ ~ c ²%³ is a decreasing function, so @ is defined
on the interval ²²³ Á ²³³ . We see that ²³ ~ c ²³ ~ , and to find
²³ ~ c ²³ , we take the limit of ²!³ ~ c ²!³ as ! S . This limit is b B.
The region of probability for @ is  &  B .
We now summarize some of the more typical transformations that can arise.
The pdf and cdf of ? are denoted ? ²%³ and -? ²%³, with similar notation for the transformed
variable @ .
Constant multiple transformation (5.2.1)
If @ ~ ? ~ ²?³ , where € is a constant, then ? ~ @ ~ ²@ ³ is the inverse
&
transformation. Then @ ²&³ ~ ? ²²&³³ h O Z ²&³O ~ ? ² ³ h .
(5.3)
A scale family can be created using the constant multiple transformation. If ? is a continuous
random variable with ? € , then the family of random variables ¸? ¢ € ¹ is a parametric
family with scale parameter .
Power Transformation (5.2.2)
The general form of the power transformation is @ ~ ²?³ ~ ? ° .
Then % ~ ²&³ ~ & is the inverse function of .
(5.4)
If € , then @ ²&³ ~ & c ? ²& ³ and -@ ²&³ ~ -? ²& ³ .
@ is called a "transformed distribution" of ? .
(5.5)
If  , then
i
i
@ ²&³ ~ c & c ? ²& ³ ~ i &c c ? ²&c ³ and -@ ²&³ ~ c -? ²& ³ ,
where i ~ c . If ~ c then @ is called an inverse distribution of ? , and
if  but is not c then @ is called an inverse transformed distribution of ? .
© ACTEX 2009
(5.6)
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-68
MODELING SECTION 5 - PARAMETRIC DISTRIBUTIONS AND TRANSFORMATIONS
For example, using the exponential distribution with mean ~ as the base distribution ? for
this transformation, with € , we have
@ ²&³ ~ ? ²²&³³ h O Z ²&³O ~ c²&³° h & c
~
c& °
h &
c
²&°° ³ c²&°
~
&
° ³
.
This is the pdf of a Weibull random variable with parameters and ~ ° .
Note that the "" in the Weibull distribution is not the same numerical value as the ~ in the
original exponential distribution.
As another example, suppose that we use the Pareto with parameters and as the base
distribution. If @ ~ ? c ~ ²?³ (so ~ c in the power transformation), we have
? ~ ²@ ³ ~ @ . This time we will use the relationship -@ ²&³ ~ -? ²²&³³ to determine the
distribution of @ . Since ²%³ ~ % is a decreasing function, we get
&
-@ ²&³ ~ c -? ² & ³ ~ ( b
³ ~ ( &b ³ .
&
This is the cdf of the inverse Pareto with parameters and .
Note that the "" parameter from the original Pareto distribution has been inverted in the
transformation to the inverse Pareto, but the parameter has been maintained.
Using the transformation @ ~ ? c is how we get inverse transformations in general. We must
be careful to identify the effect on the original distribution parameters and how they relate to the
parameters in the transformed distribution. Each of the following distributions and their inverses
involve a parameter labeled (in the Exam C table) and another parameter (except for the
exponential distribution): exponential, Pareto, loglogistic, paralogistic, gamma, and Weibull.
For each of these, when the transformation @ ~ ? c is applied, the distribution of @ is the
inverse of the original distribution, and the "" in the inverse distribution is numerically equal to
,where is the value from the original distribution, and the other parameters are unchanged.
For instance, if ? has a gamma distribution with parameters ~ and ~ , then @ ~ ? c
has an inverse gamma distribution with ~ and ~ . A couple of additional points to note
are:
- in the case of the Pareto which has parameters and , and the inverse Pareto which has and
, ~ when constructing the inverse distribution, and
- if ? has a loglogistic distribution with parameters and , then @ ~ ? c has a loglogistic
distribution with parameters and (the inverse of loglogistic is also loglogistic).
Example LM5-4: For the random variable ? with pdf ²%³ ~ % for  %  (and 0
elsewhere), find the pdf of (i) ? ( € ) , (ii) ? c , and (iii) ? ° ( € ) .
&
&
&
&
Solution: (i) @ ²&³ ~ h ? ² ³ ~ h ~ for   ,
&
or equivalently, @ ²&³ ~ for
(ii) @ ²&³ ~ &c h ? ²&c ³ ~ &
&.
h & ~ & for  &  ,
or equivalently, for  &  B . Note that @ has a single parameter Pareto distribution with
~ and ~ .
(iii) @ ²&³ ~ & c h ? ²& ³ ~ & c h & ~ & c for  &  .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 5 - PARAMETRIC DISTRIBUTIONS AND TRANSFORMATIONS
LM-69
Example LM5-5: Find expressions for the pdf of the transformed gamma and the inverse
gamma by applying the definition of transforming and inverting a distribution. Apply the
transformations to the base gamma distribution with ~ .
²%³ c%
Solution: ? is the gamma random variable, with pdf ²%³ ~ %h!²³ .
²& ³ c&
Transformed gamma: @ ~ ? ° ; @ ²&³ ~ & c ? ²& ³ ~ & c h & h!²³ ~
²&c ³ c°&
Inverse gamma: @ ~ ? c ; @ ²&³ ~ &c ? ²&c ³ ~ &c h &c h!²³ ~
This is the inverse gamma with ~ .
& c c&
!²³
&c c c°&
!²³
.
.
Exponential transformation (5.2.3)
The exponential transformation is @ ~ ? ~ ²?³ , so that ? ~ ²@ ³ ~ ²@ ³ .
Then -@ ²&³ ~ -? ² &³ , @ ²&³ ~ & h ? ² &³ .
(5.7)
For this transformation, if ? is normal with mean and variance , then @ ~ ? has a
lognormal distribution.
Sums of certain random variables
Suppose that ? Á ? Á ÀÀÀÁ ? are independent random variables and @ ~ ?
~
distribution of ?
distribution of @
Bernoulli )²Á ³
binomial )²Á ³
binomial )² Á ³
binomial )²' Á ³
Poisson Poisson '
geometric negative binomial ~ Á negative binomial , negative binomial ' Á normal 5 ² Á ³
5 ²' Á ' ³
exponential with mean gamma with ~ , gamma with , gamma with ' , Chi-square with df
Chi-square with ' df
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-70
© ACTEX 2009
MODELING SECTION 5 - PARAMETRIC DISTRIBUTIONS AND TRANSFORMATIONS
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 5
LM-71
MODELING - PROBLEM SET 5
Parametric Distributions and Transformations - Section 5
"ÎB
1. 0\ ÐBÑ œ / B# , B  ! . ] œ )\ . Find 0] ÐCÑ .
)ÎB
)ÎB
# )ÎB
)ÎB
A) / B#
B) )/B#
C) ) /B#
D) /)B#
# )ÎB
E) ) )/# B#
2. \ has a uniform distribution on the interval Ð!ß -Ñ . ] œ #\ . Find the distribution of ] .
A) uniform on Ð!ß #- Ñ
B) uniform on Ð!ß #-Ñ
C) uniform on Ð-ß #-Ñ
D) uniform on Ð-ß $-Ñ
E) None of A, B, C or D is correct
3. \ has a uniform distribution on the interval Ð!ß -Ñ . ] œ \ "Î# . Find the distribution of ] .
#C
#C
A) 0] ÐCÑ œ - for !  C  ÈB) 0] ÐCÑ œ - for !  C  - #
C) 0] ÐCÑ œ - for !  C  ÈÈC
E) 0] ÐCÑ œ
for !  C  ÈC#
C#
D) 0] ÐCÑ œ - for !  C  - #
-
4. \ has a uniform distribution on the interval Ð!ß -Ñ . ] œ /\ . Find the distribution of ] .
68 C
A) 0] ÐCÑ œ - ß !  68 C  - , "  C  /"
B) 0] ÐCÑ œ - 68
C ß !  68 C  - , "  C  /
"
C) 0] ÐCÑ œ -C
ß !  68 C  - , "  C  /C
D) 0] ÐCÑ œ /- ß !  68 C  - , "  C  68 -
D) 0] ÐCÑ œ -/"C ß !  68 C  - , "  C  68 -
5. \ has a Weibull distribution with parameters 7 and ).
] œ 1Ð\Ñ has an exponential distribution with mean )7 Þ
Find the transformation 1Ð\Ñ.
6. \ has a Pareto distribution with parameters α and ). ] œ 68Ð \)
) Ñ.
Find the distribution of ] .
7. \ has an exponential distribution with mean ) . ] œ /\ . Find the distribution of ] .
A) Weibull
B) Inverse Weibull
C) Exponential
D) Inverse Exponential
E) Single Parameter Pareto
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-72
MODELING - PROBLEM SET 5
8.. (CAS) Claim severities are modeled using a continuous distribution and inflation impacts
claims uniformly at an annual rate of 3. Which of the following are true statements regarding the
distribution of claim severities after the effect of inflation?
1. An Exponential distribution will have scale parameter Ð"  3Ñ)
2. A 2-parameter Pareto distribution will have scale parameters Ð"  3Ñα and Ð"  3Ñ)
3. A Paralogistic distribution will have a scale parameter )ÎÐ"  3Ñ
A) 1 only
B) 3 only
C) 1 and 2 only
D) 2 and 3 only
E) 1, 2 and 3
9. (CAS Nov 05) Claim size, \ , follows a Pareto distribution with parameters α and ).
A transformed distribution, ] , is created such that ] œ \ "Î7 .
Which of the following is the probability density function of ] ?
7 )C7 "
A) ÐC)Ñ7 "
α)α 7 C7 "
B) ÐC7 )Ñα"
)
C) ÐC)α
αÑ)"
α7 ÐCÎ)Ñ7
D) CÒ"ÐCÎ)Ñ7 Óα"
α
α)
E) ÐC7 
αÑα"
10. (CAS May 06) The aggregate losses of Eiffel Auto Insurance are denoted in euro currency
and follow a Lognormal distribution with . œ ) and 5 œ #. Given that 1 euro œ 1.3 dollars,
which set of lognormal parameters describes the distribution of Eiffel's losses in dollars?
A) . œ 'Þ"&, 5 œ #Þ#'
B) . œ (Þ(%, 5 œ #Þ!!
C) . œ )Þ!!, 5 œ #Þ'!
D) . œ )Þ#', 5 œ #Þ!!
E) . œ "!Þ%!, 5 œ #Þ'!
11. (CAS May 06) The following information is available regarding the random variables \
and ] :
• \ follows a Pareto distribution with α œ # and ) œ "!!
• ] œ 68Ò"  Ð\Î)ÑÓ
Calculate the variance of ] .
A) Less than 0.1
B) At least 0.1, but less than 0.2
C) At least 0.2, but less than 0.3
D) At least 0.3, but less than 0.4
E) At least 0.4
12. (CAS May 06) Calculate the skewness of a Pareto distribution with α œ % and ) œ "ß !!! .
A) Less than 2
B) At least 2, but less than 4
C) At least 4, but less than 6
D) At least 6, but less than 8
E) At least 8
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 5
LM-73
MODELING - PROBLEM SET 5 SOLUTIONS
"ÎÐBÎ)Ñ
)ÎB
1. 0] ÐCÑ œ ") † 0\ Ð B) Ñ œ )" † /ÐBÎ)Ñ# œ )/B# . Answer: B
C
C
"
2. 0\ ÐBÑ œ "- for !  B  - . 0] ÐCÑ œ "# † 0\ Ð # Ñ œ #for !  #  - ,
"
or equivalently, 0] ÐCÑ œ #for !  C  #- . ] is uniform on Ð!ß #-Ñ .
Answer: B
3. 0] ÐCÑ œ #C † 0\ ÐC# Ñ œ - for !  C#  - , or equivalently, !  C  È- . Answer: A
#C
"
4. 0] ÐCÑ œ C" † 0\ Ð68 CÑ œ -C
ß !  68 C  - , or equivalently, !  C  /- Þ Answer: C
7
5. 0] ÐCÑ œ 0\ Ð5ÐCÑÑ † 5 w ÐCÑÞ
0\ ÐBÑ œ
7 B7 " /ÐBÎ)Ñ
Þ
)7
We are given that 0] ÐCÑ œ )"7 /CÎ) .
By inspection, it appears that C œ B7 œ 1ÐBÑ may be the proper transformation.
"
With this transformation B œ 5ÐCÑ œ C"Î7 , 5 w ÐCÑ œ 7" C" 7 ß
7
7 ÐC"Î7 Ñ7 " /ÐC
"Î7 Î)Ñ7
"
and 0] ÐCÑ œ
† 7" C 7 " œ )"7 † /CÎ) .
)7
This is the pdf of an exponential random variable with mean )7 .
7
6. 0] ÐCÑ œ 0\ Ð5ÐCÑÑ † 5 w ÐCÑ .
C
w
C
In this case, C œ 1ÐBÑ œ 68Ð B)
) Ñ p B œ )Ð/  "Ñ œ 5ÐCÑ ß and 5 ÐCÑ œ ) / .
α
For the Pareto distribution, 0\ ÐBÑ œ ÐBα))Ñα" Þ
α)α
C
 αC
Then 0] ÐCÑ œ Ò)Ð/C "Ñ
.
)Óα" † )/ œ α/
This is the pdf of an exponential distribution with mean α" .
"
α
B
7. 0] ÐCÑ œ C" † 0\ Ð68 CÑ œ C" † ") † /Ð68 CÑÎ) œ C" † )" † C"Î
) œ Cα" for C œ /  " , where
α œ ") . ] has the pdf of a single parameter Pareto distribution with ) œ "Þ Answer: E
8. To say that random variable \ has scale parameter ) means that if - is a constant and
] œ -\ , then ] has the same distributional form as \ with ) replaced by - ). In this case, after
inflation the loss is ] œ Ð"  3Ñ\ , and so will have the same distributional type with scale
parameter Ð"  3Ñ) .
The distributions in the Exam C Table have been formulated so that the parameter ) is a scale
parameter (the only continuous distribution in the table that do not use the parameter ) is the
lognormal, all others have scale parameter )).
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-74
MODELING - PROBLEM SET 5
8. continued
1. The exponential distribution has scale parameter ), so after inflation ] œ Ð"  3Ñ\ will be
exponential with parameter Ð"  3Ñ) . True
2. The Pareto has scale parameter ), so after inflation ] œ Ð"  3Ñ\ will have a Pareto
distribution with scale parameter Ð"  3Ñ) (the Pareto distribution also has parameter α, which is
not a scale parameter). False
3. The Paralogistic distribution has scale parameter ), ] œ Ð"  3Ñ\ will have a Paralogistic
distribution with scale parameter Ð"  3Ñ) (the Paralogistic distribution also has parameter α,
which is not a scale parameter). False
Answer: A
9. We use the following transformation rule. If \ has pdf 0\ ÐBÑ and if ] œ 2Ð\Ñ , then we
find the inverse transformation \ œ 5Ð] Ñ. The pdf of ] is 0] ÐCÑ œ 0\ Ð5ÐCÑÑ † l5 w ÐCÑl .
α
In this example, 0\ ÐBÑ œ ÐBα))Ñα" (Pareto).
The transformation is ] œ 2Ð\Ñ œ \ "Î7 . The inverse transformation is \ œ ] 7 œ 5Ð] Ñ .
α
α)α 7 C7 "
α)
7 "
The pdf of ] is 0] ÐCÑ œ 0\ ÐC7 Ñ † 7 C7 " œ ÐC7 
œ ÐC7 )Ñα" . Answer: B
)Ñα" † 7 C
" #
10. The mean of a lognormal distribution with parameters . and 5 is /. # 5
#
and the second moment is /#.#5 .
If the loss measured in euros is \ , then IÒ\Ó œ /"! and IÒ\ # Ó œ /#% .
The loss measured in dollars is ] œ "Þ$\ , with mean IÒ] Ó œ IÒ"Þ$\Ó œ "Þ$/"! and
with second moment IÒ] # Ó œ IÒÐ"Þ$\Ñ# Ó œ "Þ'*/#% .
w " w#
w
w#
If ] is lognormal with parameters .w and 5 w , then IÒ] Ó œ /.  # 5 and IÒ] # Ó œ /#. #5 .
Therefore, .w  "# 5 w # œ 68Ð"Þ$/"! Ñ œ 68 "Þ$  "! , and
#.w  #5 w # œ 68 "Þ'*  #% . Solving this system of two equations in .w and 5 w results in
Answer: D
.w œ )Þ#' and 5 w œ #Þ!! .
11. ] œ 68Ò"  \
) Ó œ 1Ð\Ñ . We can find the pdf of ] from the relationship
0] ÐCÑ œ 0\ Ð5ÐCÑÑ † l5 w ÐCÑl , where 5ÐCÑ is the inverse function to 1ÐBÑ.
\
]
]
From ] œ 68Ò"  \
) Ó , we get / œ "  ) , and then \ œ )Ð/  "Ñ œ 5Ð] Ñ is the inverse
function. Therefore, 0] ÐCÑ œ 0\ Ð)Ð/C  "ÑÑ † Ð)/C Ñ .
α
The pdf of the Pareto distribution is 0\ ÐBÑ œ ÐBα))Ñα" .
α
α)
C
 αC
Then, 0] ÐCÑ œ Ð)Ð/C "Ñ
.
)Ñα" † Ð)/ Ñ œ α/
This is the pdf of an exponential distribution with mean α" , so the variance is α"# .
We are given α œ # , so the variance of ] is .25 . Answer: C
IÒ\ $ Ó$IÒ\ # Ó†IÒ\Ó#ÐIÒ\ÓÑ$
.
ÒIÒ\ # ÓÐIÒ\ÓÑ# Ó$Î#
)
For the Pareto distribution, we have IÒ\Ó œ α"
œ "!!!
$ ,
#
#
#)
"!!!
') $
'‚"!!!$
$
IÒ\ # Ó œ Ðα#ÑÐ
.
α"Ñ œ $ , and IÒ\ Ó œ Ðα$ÑÐα#ÑÐα"Ñ œ
'
"
#
$
#
$
" 
"!!! $Ð"!!! Î$ÑÐ"!!!Î$Ñ#Ð"!!!Î$Ñ
The skewness is
œ Ò " Ð$" Ñ##(
$Î# œ (Þ!( .
Ò"!!!# Î$  Ð"!!!Î$Ñ# Ó$Î#
$
$ Ó
12. The skewness is
IÒÐ\IÒ\ÓÑ$ Ó
ÐZ +<Ò\ÓÑ$Î#
œ
Answer: D
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 6 - DISTRIBUTION TAIL BEHAVIOR
LM-75
MODELING SECTION 6 - DISTRIBUTION TAIL BEHAVIOR
The material in this section relates to Section 3.4 of "Loss Models". The suggested time frame for
this section is 1 hour. There is been very infrequent reference to this topic on the exam. Later
topics do not depend on this material, and it can be postponed and covered at a later time.
LM-6.1 Measuring Tail Weight Using Existence of Moments
A right tail of the distribution for random variable ? is an interval of the form ²%Á B³ , with
B
probability 7 ´? € %µ ~ :? ²%³ ~ c -? ²%³ ~ % ? ²!³ !
(:? is the survival function of ? ). A random variable with a lot of probability in the right tails
is said to have "heavy right tails", or just heavy tails. Heavy tails are characteristic of a random
variable that has relatively high probability for large numerical outcomes. The opposite would be
true for a light-tailed distribution. There are various ways of classifying tails as heavy or light.
One classification considers the moments of ? . Under this classification, distributions for which
,´? µ is finite for all € indicate a light right tail, and distributions for which ,´? µ is
infinite for above a certain value indicate a heavy right tail. Any distribution on ²Á B³ whose
pdf is proportional (or asymptotically proportional) to % will have heavy right tails because
,´? µ will be infinite for ‚ c . Any distribution ²Á B³ whose pdf is proportional to
% c% , with € , will have a light right tail since
B % h c% %  B if € and € .
Heavy right-tailed distributions from the Exam C table of distributions based on this existence of
moments classification are: Pareto, inverse Pareto, loglogistic, paralogistic, inverse paralogistic,
inverse gamma (and inverse exponential) and inverse Weibull.
Light right-tailed distributions from the Exam C table of distributions based on this existence of
moments classification are: gamma (and exponential), Weibull, normal, lognormal, and inverse
Gaussian.
LM-6.2 Comparing the Tail Weights of Two Distributions
The tail weights of two distributions can be compared by taking the limit of the ration of their
survival functions. Suppose that ? and @ are two continuous random variables. Then
lim
:? ²%³
%¦B :@ ²%³
²%³
~ lim ? ²%³
%¦B @
(this follows from l'Hospital's rule).
: ²%³
(6.1)
²%³
lim ?
~ lim ? ²%³ ~ %¦B :@ ²%³
%¦B @
We define the relative tail weights of ? and @ as follows:
Suppose that the limit is
• if ~ Á we say that ? has a lighter right tail than @
• if   BÁ we say that ? and @ have similar (proportional) right tails
• if ~ B, we say that ? has a heavier right tail than @
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-76
MODELING SECTION 6 - DISTRIBUTION TAIL BEHAVIOR
Example LM6-1: Compare the tail weights of the inverse Pareto distribution and the inverse
gamma distribution (assume the inverse gamma distribution has € ).
²%³
c
%h!²³
h!²³
b
i °%
%
% h
Solution: # 7 ! ²%³ ~ ²%b
³ b h ²i °%³ hci °% ~ ²i ³ h ²%b³ b À
# i
% b
% b
lim °% ~ ~ ; ²%b
h %c S B as %SB (since € ).
³ b ~ ² %b ³
%¦B
The inverse Pareto has a heavier tail than the inverse gamma with € .
LM-6.3 Measuring Tail Weight Using Hazard Rate and Mean Residual Lifetime
The hazard rate (the hazard rate is the force of mortality in survival analysis terminology, and it
is also called failure rate) is
²%³
- Z ²%³
: Z ²%³
²%³ ~ :²%³ ~ c- ²%³ ~ c %
:²%³ ~ c :²%³ .
(6.2)
%
If ? ‚ (a non-negative random variable), then :²%³ ~ c ²!³ ! .
In Exam C, the notation ; ²%³ is the continuous random variable representing time until death of
someone now alive at age %. The expected value of ; ²&³ is the expected time until death for
B
someone at age ²&³ which is ,´; ²%³µ ~ ° % ~ ! % ! , called the complete expectation of life.
If ? represents the random variable of age at death, then ,´; ²%³µ ~ ,´? c %O? € %µ . What
this indicates is that a newborn must survive to age %, and then we measure the time until death
from age % for someone still alive. ° % is also referred to as the mean residual lifetime (given
survival to %), because it measures the average additional number of years until death from age %
given that an individual has survived to age %.
B
:? ²"³ "
: ²%b!³
B : ²%b!³
You may recall from Exam M that ! % ~ :? ²%³ , so that ° % ~ :? ²%³ ! ~ % : ²%³
.
?
?
?
It is natural to describe mean residual lifetime in terms of an age at death random variable ? as
we just have done. Algebraically, we can define the mean residual lifetime for any non-negative
random variable in the same way. We might see the notation ²%³ instead of ° % . Mean residual
lifetime will be an important concept that arises again when we consider policy deductibles a
little later on in the study guide.
Tail weight of a continuous distribution can be classified by the behavior of the hazard rate and
also by the behavior of the mean residual lifetime. Distributions with increasing hazard rate
functions have a light tail and those with decreasing hazard rate functions have a heavy tail. The
following is a summary of some relationships involving tail weight, hazard rate, survival function
and mean residual lifetime of a random variable.
Light right tail corresponds to the following conditions
²%b&³
²%³
is a decreasing function of % for all values of & ‚ ¬ the hazard rate ²%³ is an increasing function of %
¬ ²%³ (mean residual lifetime) is a decreasing function of %
¬
j= ´?µ
,´?µ
 (coefficient of variation of ? is  )
(reverse implications are not true, in general).
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 6 - DISTRIBUTION TAIL BEHAVIOR
? has an increasing hazard rate ¯
:²%b&³
:²%³
LM-77
is a decreasing function of %.
(6.3)
Examples of such a distribution are gamma with € and Weibull with € . Note that the
exponential distribution has a constant hazard rate, but all moments exist, so that it is considered a
light right-tailed distribution using the existence of moments criterion.
Heavy right tail corresponds to the following conditions
²%b&³
²%³
is an increasing function of % for all values of & ‚ ¬ the hazard rate ²%³ is a decreasing function of %
¬ ²%³ is an increasing function
¬
j= ´?µ
,´?µ
‚ (coefficient of variation of ? is ‚ )
(reverse implications are not true, in general).
? has a decreasing hazard rate ¯
:²%b&³
:²%³
is an increasing function of %.
(6.4)
Examples of such a distribution are Pareto, inverse Pareto, inverse gamma, inverse Weibull,
gamma with  and Weibull with  .
Note that all moments of the gamma distribution exist even if   , so that when the
existence of moments is used as the measure of the right tail weight, the gamma always has light
right tails. On the other hand, if the behavior of the hazard rate is used as the measure of right-tail
behavior then the gamma has a light right-tail if € and a heavy right tail if  (the
concept of heavy/light right-tail becomes somewhat vague in this case, and it may be more
meaningful when comparing the relative tail weights of two distributions). A similar comment
applies to the Weibull distribution. All moments of the Weibull distribution exist, but the hazard
rate is increasing when € , the hazard rate is constant when ~ , and the hazard rate is
decreasing when  .
The following are some additional relationships involving hazard rate and mean residual lifetime.
• The mean residual lifetime is ²%³ ~
²³
%B ²!c%³ ²!³ !
:²%³
and :²%³ ~ ²%³ %´ c ²!³
! µ for % ‚ .
•
~
%B :²!³ !
:²%³
B :²%b&³
~ :²%³
%
& ,
(6.5)
lim ²%³ ~ lim ²%³
.
%¦B
(6.6)
%¦B
²³
% • :²%³ ~ ²%³ h %´ c ²!³
!µ
(6.7)
:²%³
• The equilibrium distribution of the random variable ? has pdf ²%³ ~ ,´?µ .
(6.8)
The hazard rate of the equilibrium distribution is ²%³ ~ ²%³ , where ²%³ is the mean
residual lifetime of ? .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-78
MODELING SECTION 6 - DISTRIBUTION TAIL BEHAVIOR
Example LM6-2: ? has mean residual lifetime ²%³ ~ % b for % ‚ .
Find :²%³ Á ²%³ and ²%³. Determine the tail behavior of ? by considering the moments of ? ,
the behavior of the hazard rate, and the behavior of the mean residual lifetime. Find the pdf of the
equilibrium distribution.
²³
%
B
Solution: :²%³ ~
h %´ c !µ ~ h %´ c %µ
~
%b
²%³
h %´ c ²% b ³µ ~
Then ²%³ ~ c : Z ²%³ ~
²!³
²%b³ .
²%b³
%b
, and ²%³ ~
²%³
:²%³
%b
~
%b
.
%
The first moment of ? is ,´?µ ~ % h ²%³ % ~ ²%b³
% ~ (the integral can be found by using the substitution " ~ % b ).
B
B
%
The second moment of ? is ,´? µ ~ % h ²%³ % ~ ²%b³
% ~ B.
Using the moment criterion for tail weight indicates that ? has a heavy tail.
B
B
The hazard rate is decreasing and the mean residual lifetime is increasing, which also is an
indication of a heavy tail.
:²%³
The equilibrium distribution has pdf ²%³ ~ ,´?µ ~ ²%b³
for % ‚ .
Example LM6-3: Suppose that ? has a Weibull distribution with parameters and , and pdf
²%°³ c²%°³
²%³ ~
. Find the hazard rate, and determine the behavior of the mean residual
%
lifetime of ? .
Solution: The survival function for this distribution is :²%³ ~ c²%°³ . The hazard rate is
²%³
c
²%³ ~ :²%³ ~ % . If € then ²%³ is an increasing function of % , and therefore ?
has a decreasing mean residual lifetime À If  , the distribution has a decreasing hazard rate
and an increasing mean residual lifetime.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 6
LM-79
MODELING - PROBLEM SET 6
Distribution Tail Behavior - Section 6
1. Using the criterion of existence of moments, determine which of the following distributions
have heavy tails.
I. Normal distribution with mean , variance .
II. Lognormal distribution with parameters and .
III. Single parameter Pareto.
A) I only
B) II only
C) III only
D) All but I
E) All but II
°
2. You are given that ? has pdf ²%³ ~ b% for  %  B .
How many of the following distributions have a lighter right tail than ? ?
I. Pareto with ~ II. Pareto with € III. Paralogistic with ~ A) 0
B) 1
IV. Inverse paralogistic with € C) 2
D) 3
E) 4
3. ? has pdf ²%³ ~ %c% , % € .
:²%b&³
(a) Find :²%³ and determine whether or not :²%³
is an increasing function of %.
(b) Find the hazard rate and determine whether it is increasing or decreasing.
(c) Use the fact that the standard normal distribution is symmetric around the origin and
B
that cB j
B
j
h c% ° % ~ to show that ,´?µ ~ c% % ~ .
(d) Find the pdf of the equilibrium distribution for ? .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-80
MODELING - PROBLEM SET 6
MODELING - PROBLEM SET 6 SOLUTIONS
1. I. The moment generating function of the normal is 4? ²!³ ~ !b ! .
Each successive derivative exists and is finite when ! ~ . All moments exist, so the tail is not
heavy.
II. From the Exam C table, the -th moment of the lognormal is ,´? µ ~ b finite for every . All moments exist, so the tail is not heavy.
, which is
III. From the Exam C table, the -th moment of the single parameter Pareto with parameters and is ,´? µ ~ c , which exists only for  . The tail is heavy.
²%³
°
2. I. ²%³ ~ b% h
0
²%³
°
²%³
°
²%³
°
²%b³
Answer: C
S as % S B . Same right tail weight.
²%b³b
²%b³
c
II. ²%³ ~ b% h ~ S B as % S B .
h b% h ²% b ³
00
? has heavier right tail weight.
III. ²%³ ~ b% h [ b ²%°³µ ~ h
000
%[b²%°³ µ b
IV. ²%³ ~ b% h
0=
²%°³ has heavier right tail weight.
B
B
[b²%°³µ
b%
S as % S B . Same right tail weight.
%[b²%°³ µ b
~ h
S B as % S B since € . ?
b%
Answer: C
3.(a) :²%³ ~ % ²!³ ! ~ % !c! ! ~ c% (the antiderivative of !c! is c c! ).
:²%b&³
:²%³
c²%b&³
~ c% ~ c%&c& , which is a decreasing function of % for any & € .
²%³
c%
(b) The hazard rate is ²%³ ~ :²%³ ~ %c% ~ % , which is an increasing function of %.
:²%b&³
This is also implied by (a), since :²%³ decreasing is equivalent to ²%³ increasing.
(c) Since the standard normal density is symmetric around the origin, it follows that
B
B h c% ° % ~ cB
h c% ° % ~ ,
j
j
j
~
k .
Then, with the change of variable % ~ !j , the integral
B
B
becomes h c% ° % ~ j h c! ! ~ k ,
B
and then h c% ° % ~
j
B
so that c! ! ~ .
j
B
B
,´?µ ~ :²%³ % ~ c% % ~ .
:²%³
c%
(d) The pdf of the equilibrium distribution is ²%³ ~ ,´?µ ~ j
.
°
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 7 - MIXTURE OF TWO DISTRIBUTIONS
LM-81
MODELING SECTION 7 - MIXTURE OF TWO DISTRIBUTIONS
The material in this section relates to Section 4.2.3 of "Loss Models". The suggested time frame
for this section is 2 hours.
LM-7.1 Mixture of Two Distributions
We begin with a formal algebraic definition of how a mixture of two distributions is constructed,
and later we will look at how a mixture distribution is described by general reasoning.
Given random variables ? and ? , with pdf's or pf's ? ²%³ and ? ²%³, and given
  , we construct the random variable @ with pdf
@ ²&³ ~ ? ²&³ b ² c ³? ²&³ .
(7.1)
This is called a mixture of two distributions or a two point mixture of the distributions of ? and
? .
The two-point mixture random variable @ can also be defined in terms of the cdf,
-@ ²&³ ~ -? ²&³ b ² c ³-? ²&³
(7.2)
? and ? are the component distributions of the mixture, and the factors and c are
referred to as mixing weights. It is important to understand that we are not adding ? and
² c ³? , @ is not a sum of random variables. @ is defined in terms of a pad (or cdf) that is a
weighted average of the pdf's (or cdf's) of ? and ? . We are adding ? and ² c ³?
to get @ .
Example LM7-1: As a simple illustration of a mixture distribution, consider two bowls. Bowl A
has 5 balls with the number 1 on them and 5 balls with the number 2 on them, and bowl B has 3
balls with the number 1 and 7 balls with the number 2. Let ? denote the number on a ball
randomly chosen from bowl A, and let ? denote the number on a ball randomly chosen from
bowl B. The probability functions of ? and ? are
? ²³ ~ ? ²³ ~ À and ? ²³ ~ À , ? ²³ ~ À .
Suppose we create the mixture distribution with mixing weights ~ À and c ~ À .
The mixture distribution @ has probability function
@ ²³ ~ ²À³²À³ b ²À³²À³ ~ À , @ ²³ ~ ²À³²À³ b ²À³²À³ ~ À
.
Note that the outcomes of the mixture distribution @ come from the possible outcomes of the
component distributions ? and ? .
An alternative way of looking at this mixture distribution is by means of conditioning on a
"parameter". This will be important when we look at continuous mixing. The parameter approach
to describe the mixture distribution in Example 7-1 is as follows.
Suppose that a fair coin is tossed. If the toss is a head, a ball is chosen at random from bowl A
and if the toss is a tail, a ball is chosen at random from bowl B. We define the random variable A
to be the number on the ball. We will see that A has the same distribution as the mixture
distribution labeled @ above.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-82
MODELING SECTION 7 - MIXTURE OF TWO DISTRIBUTIONS
The random variable A can be interpreted as follows. Consider the 2-point random variable #,
for which # ~ Bowl A if the coin toss is a head, and # ~ Bowl B if the toss is a tail. Then
7 ²# ~ (³ ~ 7 ²# ~ )³ ~ À (since the coin is fair). # is used to indicate which bowl the coin
will be chosen from depending on the outcome of the coin toss.
If the toss is a head, the bowl is A, and then A has the ? distribution for the number on the ball,
so ? ²³ ~ 7 ²A ~ O# ~ (³ ~ À and ? ²³ ~ 7 ²A ~ O# ~ (³ ~ À .
In a similar way, if the toss is a tail, the bowl is B, and then A has the ? distribution for the
number on the ball, so ? ²³ ~ 7 ²A ~ O# ~ )³ ~ À and ? ²³ ~ 7 ²A ~ O# ~ )³ ~ À.
We are describing A as a combination of two conditional distributions based on the parameter #.
To find the overall, or unconditional distribution of A , we use some basic rules of probability.
Since # must be A or B, we can think of bowl B as the "complement" of bowl A, and then
7 ²A ~ ³ ~ 7 ²A ~ O# ~ (³ h 7 ²# ~ (³ b 7 ²A ~ O# ~ )³ h 7 ²# ~ )³
~ ²À³²À³ b ²À³²À³ ~ À . In a similar way, 7 ²A ~ ³ ~ À
.
We have used the rule 7 ²*³ ~ 7 ²*O+³ h 7 ²+³ b 7 ²*O+Z ³ h 7 ²+Z ³ .
This shows that the distribution of A is the same as the mixture distribution @ in Example
LM7-1. The mixing weights for the two bowls are the probabilities of the coin indicating bowl A
or bowl B.
Language used on exam questions that identifies a mixture distribution
There is some typical language that is used in exam questions that indicates a mixture of
distributions is being considered. This language will be illustrated using the distributions
involved in the bowl example.
Suppose that we are told that there are two types of individuals. Type A individuals have a
mortality probability of .5 (and survival probability of .5) in the coming year, and Type B
individuals have a mortality probability of .3 in the coming year. In a large group of these
individuals, 50% are Type A and 50% are Type B. An individual is chosen at random from the
group. We want to find this individual's mortality probability.
We can use the usual rules of conditional probability to formulate this probability, just as above:
7 ²dying this year³ ~ 7 ²dying q Type A³ b 7 ²dying q Type B³
~ 7 ²dyingOType A³ h 7 ²Type A³ b 7 ²dying q Type B³ h 7 ²Type B³
~ ²À³²À³ b ²À³²À³ ~ À .
This is exactly @ ²³ ~ ? ²³ h b ? ²³² c ³ where "1" corresponds to the event of
dying within the year.
Note that the phrase "50% are Type A" is interpreted as meaning that if an individual is chosen
from the large group, the probability of being Type A is .5. This is language that is often used in
exam questions to indicate a mixture.
Mixture of a discrete and a continuous distribution
In Section 2.4 of the study guide we looked at "mixed distributions". In that section, a mixed
distribution referred to a random variable that was continuous on part of its probability space and
also had one or more discrete points. The concept of mixed distribution just introduced in this
section can be used to describe the Section 2.4 type of mixed distribution. The following example
uses the example from Section 2.4 to illustrate this
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 7 - MIXTURE OF TWO DISTRIBUTIONS
LM-83
Example LM7-2: Suppose that ? has probability of À at ? ~ , and ? is a continuous random
variable on the interval ²Á ³ with density function ²%³ ~ % for  %  , and ? has no
density or probability elsewhere. Show that this random variable can be described as a mixture of
two distributions with appropriate definitions for component distributions ? , ? and mixing
weights and c .
Solution: Let ? ~ be a constant (not actually a random variable, sometimes called a
degenerate random variable), so that ? ²³ ~ 7 ²? ~ ³ ~ and ? ²%³ ~ for % £ .
Let ? be continuous on ²Á ³ with pdf ? ²%³ ~ % .
With mixing weights of ~ À and c ~ À, using the definition of the mixture of two
distributions, we have the mixed random variable @ satisfying
@ ²³ ~ h ? ²³ b ² c ³ h ? ²³ ~ ²À³²À³ b ~ À , and
@ ²%³ ~ h ? ²%³ b ² c ³ h ? ²%³ ~ ²À³²³ b ²À³²%³ ~ % for  %  .
We should be a little careful about the situation in Example LM7-2. When we are mixing discrete
probability points with a continuous density, for any particular discrete point we always assign a
probability of 0 at that point for any continuous component distribution. For instance, in example
LM7-2, if ? was at the single point .4, then ? ²À³ ~ and ? ²%³ ~ elsewhere, and for the
mixture distribution, @ ²À³ ~ ²À³²³ ~ À (we do not add À? ²À³ ).
Some important relationships for mixture distributions
We have already seen the defining relationships
@ ²&³ ~ ? ²&³ b ² c ³? ²&³
and
-@ ²&³ ~ -? ²&³ b ² c ³-? ²&³ À
We can interpret these relationships by saying that the pdf and cdf of @ are weighted averages of
the component pdf's and cdf's. This weighted average interpretation can be applied to a number of
other distribution related quantities.
• if * is any event related to @ , then 7 ²*³ ~ h 7? ²*³ b ² c ³ h 7? ²*³;
(7.3)
7? ²*³ is the event probability based on random variable ? , and the same for ?
• if is any function (that doesn't involve the parameters of @ ), then
,´²@ ³µ ~ h ,´²? ³µ b ² c ³ h ,´²? ³µ
(7.4)
The justification for this relationship follows from the form of the mixed density;
,´²@ ³µ ~ ²&³ @ ²&³ & ~ ²&³ h ´? ²&³ b ² c ³? ²&³µ &
~ ²&³ h ? ²&³ & b ² c ³ ²&³ h ? ²&³ & ~ ,´²? ³µ b ² c ³,´²? ³µ
Some of the important examples of these weighted average relationships are:
Interval Probability; the event * is  @  7 ²  @  ³ ~ h 7 ²  ?  ³ b ² c ³ h 7 ²  ?  ³
(7.5)
-th moment of @ ; the function is ²@ ³ ~ @ ,´@ µ ~ ,´? µ b ² c ³,´? µ
(7.6)
Power function; the function is ²@ ³ ~ @ , where is a constant
,´ @ µ ~ ,´ ? µ b ² c ³,´ ? µ ;
(7.7)
a special case of this is the moment generating function, 4@ ²!³ ~ ,´!@ µ
4@ ²!³ ~ 4? ²!³ b ² c ³4? ²!³ .
(7.8)
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-84
MODELING SECTION 7 - MIXTURE OF TWO DISTRIBUTIONS
LM-7.2 Formulating a Mixture Distribution as a Combination of
Conditional Distributions
The relationships above can also be formulated as applications of probability rules involving
conditional probability and conditional expectation.
(i) For any random variable @ and event +,
@ ²&³ ~ ²&O+³ h 7 ²+³ b ²&O+Z ³ h 7 ²+Z ³
(7.9)
(ii) For any events * and +,
7 ´* µ ~ 7 ´* q +µ b 7 ´* q +Z µ ~ 7 ´* O+µ h 7 ´+µ b 7 ´* O+Z µ h 7 ´+Z µ .
(7.10)
(iii) For any random variable @ and any event +,
,´@ µ ~ ,´@ O+µ h 7 ´+µ b ,´@ O+Z µ h 7 ´+Z µ
(7.11)
We can define the random variable # to be 1 or 2, indicating whether or not event + has
occurred, so # ~ – + has occurred, and # ~ – +Z has occurred.
We can think of ? as the conditional distribution of @ given # ~ (event +) and we can think
of ? as the conditional distribution of @ given # ~ (event +Z ), so that
²&O+³ ~ ²&O# ~ ³ ~ ? ²&³ and ²&O+Z ³ ~ ²&O# ~ ³ ~ ? ²&³ .
The cdf, pdf, expectation and moment generating function of @ are adaptations of the conditional
probability rules above. For instance, the mean of the mixed distribution is the "mixture of the
means" of the component distributions.
,´@ µ ~ ,´@ O# ~ µ h 7 ´# ~ µ b ,´@ O# ~ µ h 7 ´# ~ µ ~ ,´? µ h b ,´? µ h ² c ³ .
Example LM7-3: A collection of insurance policies consists of two types. 25% of policies are
Type 1 and 75% of policies are Type 2. For a policy of Type 1, the loss amount per year follows
an exponential distribution with mean 200, and for a policy of Type 2, the loss amount per year
follows a Pareto distribution with parameters ~ and ~ . For a policy chosen at
random from the entire collection of both types of policies, find the probability that the annual
loss will be less than 100, and find the average loss.
Solution: The two types of losses are the random variables ? and ? .
? has an exponential distribution with mean 100, so it has cdf -? ²%³ ~ c c%° .
? has cdf -? ²%³ ~ c 4 %b
5 ~ c 4 %b 5 .
The mixing weights are the proportions of policies of each type, so that ~ À (the proportion
of Type 1 policies) and c ~ À . The distribution @ of the randomly chosen policy loss is
the mixture of ? and ? .
The cdf of @ is
-@ ²&³ ~ -? ²&³ b ² c ³-? ²&³ ~ ²À³² c c&° ³ b ²À³² c 4 &b
5 ³ .
Then, -@ ²³ ~ À
.
The mean of @ is ²À³,´? µ b ²À³,´? µ ~ ²À³²³ b ²À³² c
³ ~ .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 7 - MIXTURE OF TWO DISTRIBUTIONS
LM-85
LM-7.3 The Variance of a Mixed Distribution IS (usually) NOT the Weighted
Average of the Variances
For a mixture of two random variables, the weighted average pattern applies to most quantities
for the mixed distribution. One important exception to this rule is the formulation of the variance
of the mixture. If @ is any random variable, then we can formulate the variance of @ as
= ´@ µ ~ ,´@ µ c ²,´@ µ³ . If @ is the mixture of ? and ? with mixing weights and c ,
then we know that ,´@ µ ~ ,´? µ b ² c ³,´? µ and ,´@ µ ~ ,´? µ b ² c ³,´? µ.
It is tempting to guess that = ´@ µ is the weighted average of = ´? µ and = ´? µ .
This is not true in general (we will see a special case later for which it is true).
Example LM7-3 continued: Find the variance of the loss on the policy chosen at random, and
compare it to the weighted average of the variances of the two component loss distributions.
Solution: The mean of @ is ²À³,´? µ b ²À³,´? µ ~ ²À³²³ b ²À³² c
³ ~ ,
and the second moment of @ is
d
²À³,´? µ b ²À³,´? µ ~ ²À³² d ³ b ²À³² ²c³²c³
³ ~ Á , so
= ´@ µ ~ ,´@ µ c ²,´@ µ³ ~ Á .
= ´? µ ~ Á (variance of an exponential distribution is the square of the mean).
d
= ´? µ ~ ²c³²c³
c ²³ ~ Á .
The weighted average of the two variances is ²À³²Á ³ b ²À³²Á ³ ~ Á .
Weighted average does not apply to percentiles in a mixture distribution
In Example LM7-3, the 50-th percentile of ? is , the solution of the equation
-? ² ³ ~ c c ° ~ À, so that ~ À
.
The 50-th percentile of ? is , the solution of -? ² ³ ~ c 4 b 5 ~ À,
so that ~ À .
The weighted average of the two 50-th percentiles is ²À³²À
³ b ²À³²À³ ~ À
.
The cdf of the mixed distribution @ is -@ ²&³ ~ ²À³-? ²&³ b ²À³-? ²&³ , so that
-@ ²À
³ ~ À , so 73.64 is not the 50-th percentile of @ .
To find the 50-th percentile, say , of the mixed distribution in Example LM7-3, we must solve
the equation À ~ -@ ²³ ~ ²À³-? ²³ b ²À³-? ²³
~ ²À³² c c° ³ b ²À³² c 4 b
5 . There is no algebraic solution, and would have
to be found by a numerical approximation method. The 50-th percentile turns out to be
about 65.7.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-86
© ACTEX 2009
MODELING SECTION 7 - MIXTURE OF TWO DISTRIBUTIONS
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 7
LM-87
MODELING - PROBLEM SET 7
Two Point Mixtures - Section 7
1. A portfolio of insurance policies is divided into low and high risk policies. 80% of the policies
are low risk and the rest are high risk. The annual number of claims on a low risk policy has a
Poisson distribution with a mean of .25 and the annual number of claims on a high risk policy has
a Poisson distribution with a mean of 2. A policy is randomly chosen from the portfolio.
(a) Find the mean and variance of the number of annual claims for the policy.
(b) Find the probability that the policy has at most 1 claim in the coming year.
2. The distribution of a loss, ? , is a two-point mixture:
(i) With probability 0.8, ? has a two-parameter Pareto distribution with ~ and ~ .
(ii) With probability 0.2, ? has a two-parameter Pareto distribution with ~ and ~ .
Calculate 7 ²?  ³ .
A) 0.76
B) 0.79
C) 0.82
D) 0.85
E) 0.88
3. (SOA) The random variable 5 has a mixed distribution:
(i) With probability , 5 has a binomial distribution with ~ À and ~ .
(ii) With probability c , 5 has a binomial distribution with ~ À and ~ .
Which of the following is a correct expression for Prob²5 ~ ³?
A) À B) À b À C) À b À
D) À c À
E) À c À
4. @ is a mixture of two exponential distributions, @ ²&³ ~ c% b c%° .
The random variable A defined by the equation A ~ @ .
A is a mixture of two exponentials. The means of those two exponential distributions are
A) 1 and 3
B) 1 and 6
C) 2 and 3
D) 2 and 6
E) 3 and 6
5. ? has a uniform distribution on the interval ²Á ³ and ? has a uniform distribution on
the interval ²Á ³. @ is defined as a mixture of ? and ? with mixing weights of .5 for each
mixture component. Find the pdf, cdf and median (50-th percentile) of @ .
6. A population of people aged 50 consists of twice as many non-smokers as smokers. Nonsmokers at age 50 have a mortality probability of .1 and smokers at age 50 have a mortality
probability of .2 . Two 50-year old individuals are chosen at random from the population.
(a) Find the probability that at least one of them dies before age 51.
(b) Suppose that the mortality probabilities for smokers and non-smokers remain the same at age
51. Find the mortality probability of a randomly chosen survivor at age 51 in this population.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-88
MODELING - PROBLEM SET 7
7. @ is the mixture of an exponential random variable with mean and mixing weight , and an
exponential distribution with mean 2 and mixing weight .
Find the pdf, cdf, mean, variance and 90-th percentile of @ .
8. Which of the following statements are true?
I. A mixture of two different exponentials with the mixture having a mean 2 has a heavier right
tail than a single exponential distribution with mean 2.
II. If @ ²&³ ~ h ? ²&³ b h ? ²&³ , where   and   , then
@ ²³ ~ h ? ²³ b h ? ²³ .
9. You are given two independent estimates of an unknown quantity, :
(i) Estimate A: , (( ) ~ 1000 and (( ) ~ 400
(ii) Estimate B: , () ) ~ 1200 and () ) ~ 200
Estimate C is a weighted average of the two estimates A and B, such that:
* ~ $ h ( b (1c$) h )
Determine the value of $ that minimizes (* ).
A) 0
B) 1/5
C) 1/4
D) 1/3
E) 1/2
10. (SOA) You are given the claim count data for which the sample mean is roughly equal to the
sample variance. Thus you would like to use a claim count model that has its mean equal to its
variance. An obvious choice is the Poisson distribution. Determine which of the following models
may also be appropriate.
A) A mixture of two binomial distributions with different means.
B) A mixture of two Poisson distributions with different means.
C) A mixture of two negative binomial distributions with different means.
D) None of A, B, or C
E) All of A, B, and C
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 7
LM-89
MODELING - PROBLEM SET 7 SOLUTIONS
1.(a) @ is the mixture of two Poisson distributions.
,´@ µ ~ ²À³,´?3 µ b ²À³,´?/ µ ~ ²À³²À³ b ²À³²³ ~ À
.
To find the second moment of @ , we recall that for a Poisson distribution with mean , the
variance is also . Since = ´?µ ~ ,´? µ c ²,´?µ³ , it follows that
,´? µ ~ = ´?µ b ²,´?µ³ ~ b for a Poisson distribution.
Then, ,´@ µ ~ ²À³,´?3 µ b ²À³,´?/
µ ~ ²À³²À b À ³ b ²À³² b ³ ~ À .
Then = ´@ µ ~ À c ²À
³ ~ À .
(b) 7 ²@  ³ ~ ²À³7 ²?  ³ b ²À³7 ²?  ³
~ ²À³´cÀ b ÀcÀ µ b ²À³´c b c µ ~ À
.
2. The probability is a mixture of the probabilities for the two components.
7 ²?  ³ ~ ²À³7 ²?  ³ b ²À³7 ²?  ³ , where ? and ? are the two Pareto
distributions. 7 ²?  ³ ~ c ² b
³ ~ À , and
7 ²?  ³ ~ c ² b
³ ~ À .
7 ²?  ³ ~ ²À³²À³ b ²À³²À³ ~ À .
Answer: A
3. 7 ²5 ~ ³ ~ 7 ²5 ~ ³ b ² c ³7 ²5 ~ ³ ~ ²À³ b ² c ³
²À³ ~ À c À .
c
3 We have used the binomial probabilities 2 .
Answer: E
² c ³
c%°
4. @ ²&³ ~ c% b h . -@ ²&³ ~ ² c c% ³ b ² c c%° ³ .
-A ²'³ ~ -@ ² ' ³ ~ ² c c'° ³ b ² c c'°
³ .
A is a mixture of two exponentials with means 2 and 6.
Answer: D
5. @ ²&³ ~ ²À³? ²&³ b ²À³? ²&³
²À³²À³ b ²À³²À³ ~ À if  &  ~F
.
²À³²À³ ~ À
if  &  -@ ²&³ ~ ²À³-? ²&³ b ²À³-? ²&³
²À³²À%³ b ²À³²À%³ ~ À% if  &  ~F
.
²À³²³ b ²À³²À%³ ~ À b À% if  &  The median of @ , , must satisfy the equation - @ ²³ ~ À .
We see that at & ~ , -@ ²³ ~ À . Therefore,  , so that
-@ ²³ ~ À ~ À . Therefore, ~ À
.
Note that the mean of @ is ²À³²³ b ²À³²³ ~ .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-90
MODELING - PROBLEM SET 7
6. For a randomly chosen individual at age 50, the mortality probability is the mixture of the
mortality probabilities for non-smokers and smokers. This is ~ ² ³²À³ b ² ³²À³ ~ À
The probability that both of two independent 50-year old individuals survive the year is
² c ³ ~ À À The probability at least one of them dies by age 51 is c À ~ À .
Suppose that there are 1000 non-smokers and 500 smokers at age 50. The expected number of
surviving non-smokers at age 51 is ²À³ ~ , and the number of surviving smokers is
²³²À³ ~ . A randomly chosen survivor at age 51 has a chance of being a non-smoker
and a chance of being a smoker. The mortality probability at age 51 for the randomly chosen
survivor is ² ³²À³ b ² ³²À³ ~ À .
7. @ ²&³ ~ ² ³²c& ³ b ² ³² c&° ³ .
-@ ²&³ ~ ² ³² c c& ³ b ² ³² c c&° ³ .
,´@ µ ~ ² ³²³ b ² ³²³ ~ .
,´@ µ ~ ² ³² d ³ b ² ³² d ³ ~ S = ´@ µ ~ c ² ³ ~ .
The 90-th percentile of @ is , which must satisfy the equation
-@ ²³ ~ ² ³² c c ³ b ² ³² c c° ³ ~ À .
If we let ~ c° , then this becomes a quadratic equation in ,
² ³² c ³ b ² ³² c ³ ~ À , or equivalently, b c À ~ .
Solving for results in ~ À or ~ c À . We ignore the negative root, since ~ c°
must be positive. Then ~ c ~ c À ~ À is the 90-th percentile of @ .
8. I. If @ ²&³ ~ h ? ²&³ b h ? ²&³ ~ h c&° b h c&° ,
and b ~ . Then   (or vice-versa) .
For an exponential distribution with mean 2, we have A ²&³ ~ c&°
@ ²&³
A ²&³
~
h c&° b h c&°
c&°
&² c ³
&² c ³
~ b
.
Since   , it follows that
  . Therefore, c € @ ²&³
&² c ³
&² c ³
so that lim ²&³ ~ lim b ~ b B . True.
&¦B A
&¦B II. Suppose that @ ²&³ ~ c& b c&° . Then :@ ²&³ ~ c& b c&° Á
so that @ is mixture of exponentials with means of 1 and 3, with equal mixing weights of .
@ ²³ ~
B :²&³ &
:²³
c
b c°
~ c b c° À
? ²³ ~ Á ? ²³ ~ S ? ²³ b ? ²³ ~ . False.
9. Since the two estimates are independent,
= ´* µ ~ = ´$ h ( b ² c $³ h ) µ ~ $ = ´( µ b ² c $³ = ´) µ
~ $ ² ³ b ² c $³ ² ³ ~ Á $ c Á $ b Á .
This is a quadratic expression in $ with positive coefficient of $ . The minimum can be found
by differentiating with respect to $ and setting equal to 0:
Á $ c Á ~ S $ ~ is the value of $ that minimizes = ´ µ ,
and therefore also minimizes ²* ³ ~ j= ²* ³ .
Answer: B
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 7
LM-91
10. If @ is a mixture of ? and ? with mixing weights and c , we can define the
parameter # ~ ¸Á ¹, with 7 ´# ~ µ ~ Á 7 ´# ~ µ ~ c .
Then ,´@ µ ~ ,´ ,´@ O#µ µ ~ ,´@ O# ~ µ h 7 ´# ~ µ b ,´@ O# ~ µ h 7 ´# ~ µ
~ ,´? µ h b ,´? µ h ² c ³
(this is the usual way the mean of a finite mixture is formulated).
= ´@ µ ~ ,´ = ´@ O#µ µ b = ´ ,´@ O#µ µ , where
,´ = ´@ O#µ µ ~ = ´? µ h b = ´? µ h ² c ³ and
= ´ ,´@ O#µ µ ~ ²,´? µ c ,´? µ³ h ² c ³ .
The last equality follows from the fact that if A is a two-point random variable
" prob. A ~D
, then = ´Aµ ~ ²" c #³ h ² c ³; ,´@ O#µ is
# prob. c ,´? µ prob. a two-point random variable ,´@ O#µ ~ D
.
,´? µ prob. c Therefore = ´@ µ ~ = ´? µ h b = ´? µ h ² c ³ b ²,´? µ c ,´? µ³ h ² c ³ .
B) If ? Á ? are Poisson random variables then ,´? µ ~ = ´? µ and ,´? µ ~ = ´? µ ,
so that ,´@ µ ~ ,´? µ h b ,´? µ h ² c ³ ~ = ´? µ h b = ´? µ h ² c ³ .
It follows that
= ´@ µ ~ = ´? µ h b = ´? µ h ² c ³ b ²,´? µ c ,´? µ³ h ² c ³ € ,´@ µ .
C) For a negative binomial random variable with parameters and , the mean is and the
variance is ² b ³, so the variance is larger than the mean. If ? and ? have negative
binomial distributions, the ,´? µ  = ´? µ and ,´? µ  = ´? µ .
Therefore,
,´@ µ ~ ,´? µ h b ,´? µ h ² c ³  = ´? µ h b = ´? µ h ² c ³ , and
= ´? µ h b = ´? µ h ² c ³
 = ´? µ h b = ´? µ h ² c ³ b ²,´? µ c ,´? µ³ h ² c ³ ~ = ´@ µ ;
therefore ,´@ µ  = ´@ µ .
A) For a binomial random variable with parameters and , the mean is and the variance is
² c ³, which is smaller than the mean. Therefore
,´@ µ ~ ,´? µ h b ,´? µ h ² c ³ € = ´? µ h b = ´? µ h ² c ³ , and it is
possible that when we add ²,´? µ c ,´? µ³ h ² c ³ to the right side, we get
approximate equality. So it is possible that ,´@ µ ~ = ´@ µ for a mixture of binomials.
Answer: A
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-92
© ACTEX 2009
MODELING - PROBLEM SET 7
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 8 - MIXTURE OF DISTRIBUTIONS
LM-93
MODELING SECTION 8 - MIXTURE OF DISTRIBUTIONS
The material in this section relates to Section 4.2.3 and 5.2.6 of "Loss Models". The suggested
time frame for this section is 2 hours.
LM-8.1 Mixture of Distributions
A mixture of two distributions can be generalized in a straightforward way.
Given random variables ? Á ? Á ÀÀÀÁ ? and non-negative numbers such that
b b Ä b ~ , suppose that the random variable @ has pdf
@ ²&³ ~ h ? ²&³ b h ? ²&³ b Ä b h ? ²&³ ~ h ? ²&³ .
(8.1)
~
The 's are the "weights". The cdf will be the weighted average
-@ ²&³ ~ -? ²&³ b -? ²&³ b Ä b -? ²&³ ~ h -? ²&³ .
(8.2)
~
@ is called an -point mixture of the ? 's. The probabilities, moments and the moment
generating function of @ can be formulated in the same "weighted-average" way:
7 ²  @  ³ ~ h 7 ²  ?  ³ b h 7 ²  ?  ³
b Ä b h 7 ²  ?  ³
(8.3)
,´@ µ ~ h ,´? µ b h ,´? µ b Ä b h ,´? µ ~ h ,´? µ Á
(8.4)
,´ @ µ ~ ,´ ? µ b ,´ ? µ b Ä b ,´ ? µ
(8.5)
~
4@ ²!³ ~ h 4? ²!³ b h 4? ²!³ b Ä b h 4? ²!³ ~ h 4? ²!³ .
(8.6)
~
As in the case of a mixture of two distributions, for any mixture of distributions, the
variance of the mixed distribution IS NOT, in general, the weighted average of the
variances of the component distributions. To find = ´@ µ, we use ,´@ µ c ²,´@ µ³ .
Also as in the case of a mixture of two distributions, an alternative way of looking at the mixture
of distributions is to define an -point random variable # with probability function
7 ´# ~ µ ~ for ~ Á Á ÀÀÀÁ . Then for each from 1 to we think of ? as being the
conditional distribution of @ given # ~ . In this interpretation, for each ~ Á Á ÀÀÀÁ we
have
-? ²&³ ~ -@ O# ²&O# ~ ³ and ? ²&³ ~ @ O# ²&O# ~ ³ .
(8.7)
~
~
Then -@ ²&³ ~ h -? ²&³ ~ - ²&O# ~ ³ h 7 ´# ~ µ
and
~
~
@ ²&³ ~ h ? ²&³ ~ ²&O# ~ ³ h 7 ´# ~ µ .
© ACTEX 2009
(8.8)
(8.9)
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 8 - MIXTURE OF DISTRIBUTIONS
LM-94
The distribution function and density function of @ can be conditioned over the possible values
of #. The distribution of # is called the mixing (or prior) distribution.
The indication that a situation involves a finite (usually two-point or three-point) mixture of
distributions is usually the use of language of the form
"30% of a population are Type 1 and 70% are Type 2" , or
"with probability .3, @ has distribution 1, with probability .3, @ has distribution 2, and
with probability .4, @ distribution 3".
The percentages, proportions or probabilities are the mixing weights, and the distributions types
are the ? , ? , etc.
Example LM8-1: ? Á ? and ? have exponential distributions with parameter (mean) equal
to 1, 2 and 3, respectively. A mixture of these distributions is created with cdf
-@ ²&³ ~ -? ²&³ . Find the pdf, mean and variance of @ .
~
Solution: @ ²&³ ~ ? ²&³ ~ c&° ~ ´c& b c&° b c&° µ ,
~
~
,´@ µ ~ ,´? µ ~ ´ b b µ ~ , ,´@ µ ~ ,´? µ ~ ´ b b µ ~ Á
~
= ´@ µ ~ c ~ .
Note that = ´? µ ~ ´ b b µ
~
~
~ £ = ´@ µ , so in this example, the variance of @
is not the weighted average of the variances of the component distributions.
Variable component distribution
Suppose that there is a large collection of distributions with cdf's - ²%³Á - ²%³Á À À À, which will
be called component distributions, and suppose that a mixture consists of 2 of them, where 2 is
allowed to vary. In this situation, 2 becomes a parameter also. A distribution constructed this
way is called a variable-component mixture distribution. An example would be the set of all
exponential distributions comprising the component distributions. The variable component
distributions would be all possible mixtures of finite numbers of exponential random variables.
LM-8.2 Two Important Rules of Probability
Mixing of distributions can be considered from an alternative point of view that uses some
important rules of probability involving conditional expectation and conditional variance.
If @ and > are any random variables, then
,´@ µ ~ ,´,´@ O> µµ and
(8.10)
= ´@ µ ~ ,´= ´@ O> µµ b = ´,´@ O> µµ .
(8.11)
These rules are illustrated in the following example.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 8 - MIXTURE OF DISTRIBUTIONS
LM-95
Example LM8-2: Suppose that > is the 3-point discrete uniform random variable ¸Á Á ¹ ,
with 7 ´> ~ µ ~ 7 ´> ~ µ ~ 7 ´> ~ µ ~ , and suppose that the conditional distribution of
@ given > ~ $ is exponential with mean $ .
Find the (unconditional) mean and (unconditional) variance of @ À
c&°$
Solution: The conditional pdf of @ given > ~ $ is @ O> ²&O> ~ $³ ~ $
.
Also, we are given ,´@ O> ~ $µ ~ $ , so that ,´@ O> µ ~ H if > ~ , prob. if > ~ , prob. .
if > ~ , prob. We see that ,´@ O> µ is a 3-point random variableÀ
Let us use the notation A ~ ,´@ O> µ for this 3-point random variable.
Then ,´Aµ ~ ,´ ,´@ O> µ µ ~ ²³² ³ b ²³² ³ b ²³² ³ ~ .
This is ,´@ µ according to the rule cited above..
Also, as a random variable, A ~ ,´@ O> µ has a variance:
= ´Aµ ~ = ´ ,´@ O> µ µ ~ ,´A µ c ²,´Aµ³ ~ ,´²,´@ O> µ³ µ c ²,´ ,´@ O> µ µ³
~ ´² ³² ³ b ² ³² ³ b ² ³² ³µ c ~ .
The variance of an exponential random variable is the square of the mean, so since the
conditional distribution of @ given > is exponential with mean $, the conditional variance of @
given > ~ $ is = ´@ O> ~ $µ ~ $ . As with the conditional mean of @ O> , the
conditional variance of @ O> is a random variable dependent on the outcome of > .
= ´@ O$ ~ µ ~ if > ~ , prob. = ´@ O> µ ~ H = ´@ O> ~ µ ~ if > ~ , prob. .
= ´@ O> ~ µ ~ if > ~ , prob. = ´@ O> µ is a three-point random variable. Let us use the notation < ~ = ´@ O> µ .
Then ,´< µ ~ ,´ = ´@ O> µ µ
~ ²= ´@ O$ ~ µ³² ³ b ²= ´@ O$ ~ µ³² ³ b ²= ´@ O$ ~ µ³² ³
~ ²³² ³ b ²³² ³ b ²³² ³ ~ .
Using the variance rule cited above, we have
= ´@ µ ~ ,´ = ´@ O> µ µ b = ´ ,´@ O> µ µ ~ ,´< µ b = ´Aµ ~ b ~ .
Example LM8-2 represents the mixing distribution situation presented in Example LM8-1 in an
alternative, but equivalent conditional distribution form. The three random variables ? , ? and
? are the conditional random variables @ O> ~ , @ O> ~ and @ O> ~ .
In Example LM8-2 we found ,´@ µ ~ ,´? µ ~ ´ b b µ ~ , which can then be
written as ,´@ µ ~
~
,´@ O> ~ µ
~
~ ,´@ O> ~ µ h 7 ´> ~ µ b ,´@ O> ~ µ h 7 ´> ~ µ b ,´@ O> ~ µ h 7 ´> ~ µ
~ ,´ ,´@ O> µ µ .
This is the form that has been applied in example LM8-2.
Note that > plays the role of # in the comments made before Example LM8-1.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-96
MODELING SECTION 8 - MIXTURE OF DISTRIBUTIONS
The expectation rule ,´@ µ ~ ,´ ,´@ O> µ µ can be applied to find the expected value of a
function of @ : ,´²@ ³µ ~ ,´ ,´²@ ³O> µ µ . Applying this rule to find ,´@ µ for the
distributions in Example LM8-2, we have
,´@ µ ~ ,´ ,´@ O> ~ $µ µ ~ ,´ > µ ~ ´² ³² ³ b ² ³² ³ b ² ³² ³µ ~ .
Then, = ´@ µ ~ ,´@ µ c ²,´@ µ³ ~ c ~ .
One more comment on the variance of a mixed distribution
We have mentioned several times that the variance of a mixed distribution is not, in general,
equal to the weighted average of the variances of the component distributions. The conditioning
approach we have just looked at provides a little more insight into this. In Example LM8-2, we
say that = ´@ O> µ was a 3-point random variable, and we found its mean to be
,´ = ´@ O> µ µ ~ ²= ´@ O$ ~ µ³² ³ b ²= ´@ O$ ~ µ³² ³ b ²= ´@ O$ ~ µ³² ³ ~ .
This is the weighted average of the variances of the component distributions in the mixture.
From the probability rule for variance of @ by conditioning over > , we know that
= ´@ µ ~ ,´ = ´@ O> µ µ b = ´ ,´@ O> µ µ .
Therefore, ,´ = ´@ O> µ µ, the weighted average of the variance of the component distributions
is only part of = ´@ µÀ The full variance of @ must also include = ´ ,´@ O> µ µ , which we saw
was in Example LM8-2.
There is a special case in which the variance of the mixed distribution is equal to the weighted
average of the variances of the component distributions. That case occurs if = ´ ,´@ O> µ µ ~ .
What this means is that ,´@ O> µ is the same numerical value for all possible value of > ~ $.
As an example to illustrate this, suppose we consider a mixture of an exponential distribution
with mean 100 and a Pareto distribution with ~ and ~ , and with mixing weights of
for the exponential distribution and for the Pareto distribution.
The mean of the Pareto is c
~ . Suppose that we define the random variable > ~ ¸Á ¹
to have probabilities 7 ²> ~ ³ ~ and 7 ²> ~ ³ ~ . Then suppose that we define @
conditionally as @ O> ~ is exponential with mean 100, and @ O> ~ is Pareto with ~ ,
~ . Then = ´@ µ ~ ,´ = ´@ O> µ µ b = ´ ,´@ O> µ µ .
Since ,´@ O> ~ µ ~ ,´@ O> ~ µ ~ , it follows that = ´ ,´@ O> µ µ ~ .
Then = ´@ µ ~ ,´ = ´@ O> µ µ ~ = ´@ O> ~ µ h ² ³ b = ´@ O> ~ µ h ² ³
is the weighted average of the variances of the component distributions.
Questions on mixtures of distributions have appeared frequently on many of the released exams.
This is an important topic with which to be familiar. The conditioning approach used to describe
a mixed distribution is particularly important in the case of continuous mixing which will be
considered in the next section. Also, the two important probability rules discussed in this
subsection have a wide range of application. In particular, they are important in analyzing
compound distributions that we will look at a little later in this study guide.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 8 - MIXTURE OF DISTRIBUTIONS
LM-97
LM-8.3 Spliced distributions
A spliced model is a model which is defined in a "piecewise" way on a series of disjoint intervals.
The intervals will usually be adjacent. We consider two ways of defining a spliced model.
Version 1 of a Spliced Distribution
Suppose that ²%³ is a pdf on the interval ´c Á ³ , for ~ Á Á ÀÀÀÁ so that c
²%³ % ~ .
Suppose also that Á Á ÀÀÀÁ are numbers that satisfy the relationships
(i)   for all , and (ii) ~ .
~
We define the pdf ²%³ to be ²%³ ~ h ²%³ for c  %  .
(8.12)
This will be a properly defined pdf, and the random variable ? with this pdf has a spliced
distribution.
Version 2 of a Spliced Distribution
Suppose we have the following conditions satisfied:
(i) for each ~ Á Á ÀÀÀÁ we have a non-negative function ²%³ on the interval ´c Á ³ ,
(ii) for each ~ Á Á ÀÀÀÁ we have a positive number ,
(iii) c
²%³ % ~ .
(8.13)
~
The function ²%³ defined to be ²%³ ~ h ²%³ on the interval ´c Á ³ is a pdf, and has a
spliced distribution.
Example LM8-3: A spliced distribution is defined to have the following density function.
²%³ ~ F
À
h c%
%
.
%
Find the mean of the spliced distribution.
Solution: This appears to follow Version 2 described above, with ²%³ ~ À and ~ on the
interval ²Á ³, and with ²%³ ~ c% and ~ on the interval ²Á B³ .
B ²%³ % ~ S À % b B c% % ~ À b c ~ S ~ À .
B
B
,´?µ ~ % ²%³ % ~ %²À³ % b % h À h c% % ~ b ~ À .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-98
© ACTEX 2009
MODELING SECTION 8 - MIXTURE OF DISTRIBUTIONS
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 8
LM-99
MODELING - PROBLEM SET 8
Point Mixtures - Section 8
1. A carnival gambling game involves spinning a wheel and then tossing a coin. The wheel lands
on one of three colors, red, white or blue. There is a 1/2 chance that the wheel lands on red, and
there is a 3/8 chance of white and a 1/8 chance of blue. A coin of the color indicated by the wheel
is then tossed. Red coins have a 50% chance of tossing a head, white coins have a 3/4 chance of
tossing a head, and blue coins have a 7/8 chance of tossing a head. If the game player tosses a
head, she wins $100, if she does not toss a head she wins 0.
(a) Find the cost to play the game so that the carnival wins an average of $1 per play of the game.
(b) Let the amount won on one play of the game be @ . Find the variance of @ each of the
following ways: (i) ,´@ µ c ²,´@ µ³ , and (ii) ,´ = ´@ Ocolorµ µ b = ´ ,´@ Ocolorµ µ .
2. Annual claim counts for each policyholder are independent and follow a common Negative
Binomial distribution. A priori, the parameters for this distribution are ² Á ³ ~ ²Á ³ or
² Á ³ ~ ²Á ³ or ² Á ³ ~ ²Á ³. Each parameter set is considered equally likely. A policy file
is sampled at random.
(a) Find the probability of at least one claim in the coming year.
(b) Find the variance of the number of claims for the policy in the coming year.
3. A portfolio of insurance policies consists of two types of policies. Losses on Type 1 policies
have a Pareto distribution with parameters ~ Á ~ . Losses on policies of Type 2 have
an Inverse Pareto distribution with parameters ~ Á ~ . The policies are evenly divided
between the two types. A policy is chosen at random from the portfolio.
Show that the distribution of the loss on the randomly chosen policy is a Pareto distribution and
identify the parameters.
4. An insurer selects risks from a population that consists of three independent groups.
• The claims generation process for each group is Poisson.
• The first group consists of 50% of the population. These individuals are expected to generate
one claim per year.
• The second group consists of 35% of the population. These individuals are expected to generate
two claims per year.
• Individuals in the third group are expected to generate three claims per year.
An individual is chosen at random from the population, and the individual is observed until a
claim occurs.
(a) The number of full years with no claims until the first claim year is denoted @ . Find ,´@ µ.
(b) Find the expected number of full years with no claims until the first claim year for each
separate group. Find the weighted average of those 3 numbers.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-100
MODELING - PROBLEM SET 8
5. @ is a mixture of ? and ? with mixing weights and c .
@ is a mixture of ? and ? with mixing weights and c .
A is a mixture of @ and @ with mixing weights and c .
Show that A is a mixture of ? Á ? Á ? and ? , and find the mixing weights.
6. ? has a uniform distribution on the interval ´Á µ for ~ Á Á Á ÀÀÀ
@ is a mixture of ? Á ? Á ? Á ÀÀÀ with mixing weights ~ Á ~ Á ~ Á À À À
Find the mean and variance of @ and -@ ²³.
7. The random variable ? has the uniform distribution on the interval ² c Á ³ for ‚ .
Determine the distribution and density functions for a variable mixture of ? 's.
8. A spliced distribution is defined to have the following density function.
h ²%³  %  .
²%³ ~ F
h ²%³  %  ²%³ is the density function of a uniform random variable on the interval ²Á ³ , ~ À, and
²%³ is the density function of the uniform distribution on the interval ´Á ³. Find the
variance of the spliced distribution.
9. ? has a uniform distribution on the interval ´Á µ.
? has a single parameter Pareto distribution with ~ and ~ .
? has a single parameter Pareto distribution with ~ and ~ .
@ is a mixture of ? and ? with mixing weight applied to ? .
h ? ²%³
%
A has a spliced distribution with pdf A ²%³ ~ F
.
² c ³ h @ ²%³  %
A ²%³ is a continuous function and the mean of A is . Find and .
10. You are given ,´?O@ ~ &µ ~ & and = ´?O@ ~ &µ ~ , where @ has an exponential
distribution with a mean of . Find = ´?µ.
A) B) C) D) E) 11. Random variable @ is defined as the mixture of three uniform random variables:
Uniform on ´Á µ , Uniform on ´Á µ , Uniform on ´Á µ .
The median of @ is and the mean of @ is 1.2 .
Find the variance of @ .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 8
LM-101
12. ? is a mixture of Poisson distributions with means 1, 2 and 3.
The mean of ? is 2.1 and the variance of ? is 2.59. Find 7 ²? ~ ³.
13. (SOA) The repair costs for boats in a marina have the following characteristics:
Number
Probability that Mean of repair cost Variance of repair
Boat Type
of boats
repair is needed given a repair
cost given a repair
Power Boats
100
0.3
300
10,000
Sailboats
300
0.1
1000
400,000
Luxury Yachts 50
0.6
5000
2,000,000
At most one repair is required per boat each year. The marina budgets an amount, @ , equal to the
aggregate mean repair costs plus the standard deviation of the aggregate repair costs.
Calculate @ .
A) 200,000
B) 210,000
C) 220,000
D) 230,000
E) 240,000
14. (SOA) An actuary for a medical device manufacturer initially models the failure time for a
particular device with an exponential distribution with mean 4 years.
This distribution is replaced with a spliced model whose density function:
(i) is uniform over ´ Á µ
(ii) is proportional to the initial modeled density function after 3 years
(iii) is continuous
Calculate the probability of failure in the first 3 years under the revised distribution.
A) 0.43
B) 0.45
C) 0.47
D) 0.49
E) 0.51
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-102
MODELING - PROBLEM SET 8
MODELING - PROBLEM SET 8 SOLUTIONS
1. The amounts won for each coin type are the component distributions:
prob. prob.
Red coin: ?9 ~ F
, White coin: ?> ~ F
prob. prob.
Blue coin: ?) ~ F
prob. ,
.
prob. 7 ²Red³ ~ , 7 ²White³ ~ , 7 ²Blue³ ~ .
@ ~ amount won is a mixture of ?9 Á ?> and ?) , with the mixing weights above.
Then 7 ²@ ~ ³
~ 7 ²?9 ~ ³ h 7 ²Red³ b 7 ²?> ~ ³ h 7 ²White³ b 7 ²?) ~ ³ h 7 ²Blue³
~ ² ³² ³ b ² ³² ³ b ² ³² ³ ~ , and 7 ²@ ~ ³ ~ .
(a) The expected amount won is ² ³ ~ À
on a play of the game, so the carnival
should charge the player À
per play.
(b)(i) ,´@ µ ~ ,´?9 µ h 7 ²Red³ b ,´?>
µ h 7 ²White³ b ,´?) µ h 7 ²Blue³
~ ² ³² ³² ³ b ² ³² ³² ³ b ² ³² ³² ³ ~ Á À ,
or alternatively, ,´@ µ ~ ² ³² ³ ~ Á À .
= ´@ µ ~ Á À c ²
À
³ ~ Á À .
Red, prob. ² ³² ³ ~ (ii) = ´@ Ocolorµ ~ H ² ³² ³ ~ ² ³² ³ ~ À
White, prob. .
Blue, prob. ,´ = ´@ Ocolorµ µ ~ ²³² ³ b ²³² ³ b ²À³² ³ ~ À .
² ³ ~ ,´@ Ocolorµ ~ H ² ³ ~ Red, prob. White, prob. .
² ³ ~ À Blue, prob. = ´ ,´@ Ocolorµ µ
~ ² ³² ³ b ² ³² ³ b ²À ³² ³ c ´²³² ³ b ²³² ³ b ²À³² ³µ
~ À .
Then ,´ = ´@ Ocolorµ µ b = ´ ,´@ Ocolorµ µ ~ À .
2. For a negative binomial distribution ? with parameters and ,
7 ²? ‚ ³ ~ c 7 ²? ~ ³ ~ c ² b ³c . The mean of the negative binomial is , and the
variance is ² b ³. @ is the mixture of three negative binomial random variables, with
mixing weights of for each component.
(a) 7 ²@ ‚ ³ ~ c 7 ²@ ~ ³
~ c ´7 ²? ~ ³ h b 7 ²? ~ ³ h b 7 ²? ~ ³ h µ
~ c h ´ b
b ²b³
b ²b³ µ ~ À
.
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SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 8
LM-103
2.(b) = ´@ µ ~ ,´ = ´@ Ocomponentµ µ b = ´ ,´@ Ocomponentµ µ .
= ´? µ ~ = ´@ Ocomponentµ ~ H = ´? µ ~ = ´? µ ~ prob. prob. prob. S ,´ = ´@ Ocomponentµ µ ~ ² h b h b h ³ ~ .
,´? µ ~ ,´@ Ocomponentµ ~ H ,´? µ ~ ,´? µ ~ prob. prob. S = ´ ,´@ Ocomponentµ µ ~ .
prob. = ´@ µ ~ b~ À
3. If @ is the loss on the randomly chosen policy, then @ is the mixture of the Pareto with
~ Á ~ and the Inverse Pareto with ~ Á ~ .
The mixing weights are for each component.
The pdf of @ is
Ábh%
h
hh%
@ ²&³ ~ ²À³ h ²%b³
b ²À³ h ²%b³ ~
²%b³ ~ ²%b³ .
This is the pdf for a Pareto distribution with ~ Á ~ .
4.(a) There are 3 component distributions in this mixture > . They are
? - Poisson mean 1, mixing weight .5,
? - Poisson mean 2, mixing weight .35, and
? - Poisson 3, mixing weight .15.
> is the number of claims occurring in one year for the randomly chosen individual
We first consider the probability of no claims occurring in a one year period.
7 ²> ~ ³ ~ 7 ²? ~ ³ h 7 ²Type 1³ b 7 ²? ~ ³ h 7 ²Type 2³ b 7 ²? ~ ³ h 7 ²Type 3³
~ ²c ³²À³ b ²c ³²À³ b ²c ³²À³ ~ À.
If @ is the number of full years with no claim until the first claim year, then
7 ²@ ~ ³ ~ ²À³ h ²À
³ .
This is a geometric distribution with ~ b
~ À
, so that ~ À ~ ,´@ µ .
(b) Group 1. 7 ²? ~ ³ ~ c .
7 ² years with no claims until 1st claim year³ ~ c h ² c c ³ .
#years with no claims until 1st claim year has geometric distribution with b
~ c c , so
~ À ~ ,´#years with no claims until 1st claim yearµ .
Group 2: 7 ²? ~ ³ ~ c .
7 ² years with no claims until 1st claim year³ ~ c h ² c c ³ .
#years with no claims until 1st claim year has geometric distribution with b
~ c c , so
~ À
~ ,´#years with no claims until 1st claim yearµ .
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SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-104
MODELING - PROBLEM SET 8
4. continued
Group 3: 7 ²? ~ ³ ~ c .
7 ² years with no claims until 1st claim year³ ~ c h ² c c ³ .
#years with no claims until 1st claim year has geometric distribution with b
~ c c , so
~ À
~ ,´#years with no claims until 1st claim yearµ .
Weighted average is ²À³²À³ b ²À
³²À³ b ²À
³²À³ ~ À
.
5. A ²'³ ~ @ ²'³ b ² c ³@ ²'³
~ ´? ²'³ b ² c ³? ²'³µ b ² c ³´? ²'³ b ² c ³? ²'³µ
~ ? ²'³ b ² c ³? ²'³ b ² c ³? ²'³ b ² c ³² c ³? ²'³ .
² ³
c
6. For each , ,´? µ ~ b
and ,´? µ ~ ~ .
~
B
B
B B
,´@ µ ~ ,´? µ ~ h c ~ ~ ² ³ ~ h ´ b b ² ³ b ĵ
~
~
~
~
~ h c ~ À
B
B
B
,´@ µ ~ ,´? µ ~ h ~ h ² ³
~
~
~
This sum is infinite, so ,´@ µ ~ B, and therefore = ´@ µ ~ B .
B
B
B
-@ ²³ ~ -? ²³ ~ h h ~ ~ h ´ b b ² ³ b ĵ ~ À
~
~
~
7. Suppose the mixture is based on ? Á ? Á ÀÀÀÁ ?2 with mixing weights Á Á ÀÀÀÁ 2 ,
where   . Then ²%³ ~ for c  %  , because ? ²%³ ~ for all % not in
c
interval ² c Á ³ . - ²%³ ~ b h ²% c b ³ for c  %  , because
~
-? ²%³ ~ for  and -? ²%³ ~ for € when c  %  .
À
 %  .
h ²%³  %  ²%³ % ~ S ²À³ % b À % ~ S ~ À
.
,´?µ ~ % ²%³ % ~ À% % b À
% % ~ b ~ À
,´? µ ~ % ²%³ % ~ À% % b À
% %
~ À b Á ~ Á À
= ´?µ ~ Á À c ²³ ~ À À
8. ²%³ ~ F
9. The pdf of ? is ? ²%³ ~ for  %  .
The pdf of ? is ? ²%³ ~ % for % € ,
and the pdf of ? is ? ²%³ ~ % for % € .
The pdf of @ is @ ²%³ ~ h % b ² c ³ h % for % € .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 8
LM-105
9. continued
Since A ²%³ is continuous, it must be true that A ²³ is the same from both sides of % ~ .
Therefore, h ? ²³ ~ ² c ³ h @ ²³ .
This equation becomes ~ ² c ³ h ´ h b ² c ³ h µ ~ ² c ³² c ³ .
Solving for in terms of , we have ~ c
c À
B
B
The mean of A is % h A ²%³ % ~ % h h ? ²%³ % b ² c ³ h % h @ ²%³ %
~ h ,´? µ b ² c ³ h ,´@ µ ~ h ,´? µ b ² c ³ h ´ ,´? µ b ² c ³,´? µ µ .
B
The mean of ? is . The mean of ? is % h % % ~ ,
B
and the mean of ? is % h % ~ .
%
The mean of A is ,´Aµ ~ h b ² c ³´ b ² c ³µ ~ b ² c ³² b
³ ~ À
c
c
Substituting ~ c
c , we get b h ² c ³ ~ c ~ , so that ~ ,
and ~ .
10. = ´?µ ~ = ´,´?O@ µµ b ,´= ´?O@ µµ
~ = ´@ µ b ,´µ ~ = ´@ µ b ~ h b ~ (the variance of an exponential random variable is the square of the mean). Answer: E
11. We will denote the mixing weights as for the uniform ´Á µ and for the uniform ´Á µ.
Then the mixing weight for uniform ´Á µ is ~ c c .
The cdf's of the mixing components are
- ²%³ ~ % for  %  for uniform ´Á µ ,
- ²%³ ~ % for  %  for uniform ´Á µ , and
- ²%³ ~ % for  %  for uniform ´Á µ .
The cdf for @ is -@ ²%³ ~ - ²%³ b - ²%³ b ² c c ³- ²%³
~ % b d % b ² c c ³ d % ~ ² b b ³ d % .
Then -@ ² ³ ~ ² b b ³ d ~ À .
This equation can be written as b ~ À .
The means of the mixing components are
À for uniform ´Á µ , for uniform ´Á µ , for uniform ´Á µ .
The mean of @ is d À b d b ² c c ³ d ~ c À c ~ À .
This equation can be written as À b ~ À .
Solving the equations for and results in ~ À and ~ À ,
so c c ~ À (these are the mixing weights).
c
The 2nd moment of a uniform distribution on ´Á µ is ²c³
, so the 2nd moments of the mixing
distributions are for uniform ´Á µ . for uniform ´Á µ , for uniform ´Á µ .
The 2nd moment of @ is the mixture of these 2nd moments, so
,´@ µ ~ À d b À d b À d ~ .
The variance of @ is ,´@ µ c ²,´@ µ³ ~ c ²À³ ~ .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-106
MODELING - PROBLEM SET 8
12. With mixing weight applied to the Poisson with mean 1 and mixing weight applied to the
Poisson with mean 2, the mean of ? is ,²?³ ~ b b ² c c ³ ~ À .
The second moment of ? is ,²? ³ ~ b b ² c c ³ ~ À b À ~ À .
We get the two equations b ~ À and b ~ .
Solving these two equations results in ~ À and ~ À .
Then 7 ²? ~ ³ ~ Àc b Àc b Àc ~ À
.
13. For each boat type we find the mean and the variance of the repair cost. The mean of the
aggregate repair costs is the sum of the mean repair costs for the 450 boats, and assuming
independence of boat repair costs for all 450 boats, the variance of the aggregate cost is the sum
of the variances for the 450 boats.
For each type of boat, the repair cost is a mixture of 0 (if no repair is needed) and ? (repair cost
variable for boat if a repair is needed). The mean repair cost for boat is
,´? µ d prob. repaid is needed for boat , and the second moment of the repair cost for boat is
,´? µ d prob. repaid is needed for boat (note that ,´? µ ~ = ´? µ b ²,´? µ³ ).
The variance of the repaid cost for boat is the second moment minus the square of the first
moment.
Power boats: Mean repair cost for one boat ~ ²À³ ~ ,
second moment of repair cost for one boat ~ ´Á b µ²À³ ~ Á .
Variance of repair cost for one power boat ~ Á c ~ Á .
Sailboats: Mean repair cost for one boat ~ ²À³ ~ ,
second moment of repair cost for one boat ~ ´Á b µ²À³ ~ Á .
Variance of repair cost for one power boat ~ Á c ~ Á .
Luxury Yachts: Mean repair cost for one boat ~ ²À
³ ~ ,
second moment of repair cost for one boat ~ ´Á Á b µ²À
³ ~ Á Á .
Variance of repair cost for one power boat ~ Á Á c ~ Á Á .
The mean of the aggregate repair cost is ²³ b ²³ b ²³ ~ Á ,
and the variance is ²Á ³ b ²Á ³ b ²Á Á ³ ~ Á Á .
The amount budgeted by the marina is Á b jÁ Á ~ Á . Answer: B
%
.
cÀ%
h ²À
³ %€
Since the spliced model is continuous, it must the case that ~ ²ÀcÀ ³
(since the density is continuous at % ~ ) , which can be written as ~ À .
B
It also must be true that ²%³ % ~ , so that
% b B h ²ÀcÀ% ³ % ~ b cÀ ~ .
Then, b ~ S ~ , and 7 ´?  µ ~ % ~ ~ À . Answer: A
14. For the spliced model, ²%³ ~ F
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 9 - CONTINUOUS MIXTURES
LM-107
MODELING SECTION 9 - CONTINUOUS MIXTURES
The material in this section relates to Loss Models, Section 5.2
The suggested time for this section is 2 hours.
LM-9.1 Continuous Mixtures and Conditional Distributions
In the -point mixture distribution @ considered in the previous section, we can define the point random variable # ~ ¸Á Á ÀÀÀÁ ¹ with probability function 7 ´# ~ µ ~ , the mixing
weight for ? . Then we can think of the distribution of # as the conditioning distribution, also
called the mixing distribution, so that given # ~ , we have @ ~ ? , or equivalently
@ O# ²&O# ~ ³ ~ ? ²&³ À
Then, the unconditional distribution of @ is the mixture with pdf
@ ²&³ ~ h ? ²&³ ~ @ O# ²&O# ~ ³ h 7 ´# ~ µ .
(9.1)
~
~
It is possible to have a continuous mixing distribution, and the pdf is similar in form to the one in
Equation 9.1. Suppose that # is a continuous random variable with pdf # ²³, and suppose that
the distribution of @ depends upon (or is conditional upon) the parameter , with conditional pdf
(or pf for the discrete case of @ ³ @ O# ²&O³. Then the unconditional (also called marginal)
distribution of @ has pdf
@ ²&³ ~ @ O# ²&O³ h # ²³ .
(9.2)
This is a continuous version of the summation in Equation 9.1 for the pdf of the -point mixture.
Another way of looking at this is from the joint distribution point of view. We can think of @ and
# as having a joint distribution with joint density function ²&Á ³. If we are given the
conditional pdf of @ given #, @ O# ²&O³ , and if we are given the (marginal) pdf of #, # ²³ ,
then from joint distribution relationships, we know that the joint density can be formulated as
²&Á ³ ~ @ O# ²&O³ h # ²³ . Then, we can find the marginal pdf of @ by integrating out the
variable , so that @ ²&³ ²&Á ³ ~ @ O# ²&O³ h # ²³ , as above.
The weighted average relationships that we have for @ in the finite mixture case are valid in the
continuous mixing case. We replace summation with integration:
(i) the expected value of @ is ,´@ µ ~ ,´@ O# ~ µ h # ²³ ,
(9.3)
(ii) the -th moment of @ is ,´@ µ ~ ,´@ O# ~ µ h # ²³ ,
(9.4)
(iii) the cdf of @ is -@ ²&³ ~ -@ O# ²&O³ h # ²³ ,
(9.5)
(iv) the moment generating function of @ is 4@ ²!³ ~ 4@ ´!O# ~ µ h # ²³ ,
(9.6)
iv) the probability generating function of @ is 7@ ²!³ ~ 7@ ´!O# ~ µ h # ²³ ,
(9.6)
(vi) interval probability is 7 ´  @  µ ~ 7 ´  @  O# ~ µ h # ²³ .
(9.7)
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-108
MODELING SECTION 9 - CONTINUOUS MIXTURES
These representations can be extended to any probability so that if ( is any event, then we have
7@ ²(³ ~ 7@ O# ²(O³ # ²³ (9.8)
The integrals in each case have limits that are the same as the limits of the distribution of #. Keep
in mind that in the continuous mixture situation, the mixing random variable # is continuous, but
@ could be discrete or continuous.
In Subsection 8.2 of the previous section we reviewed two important probability rules. Those
rules can be applied to a continuous mixture distribution.
The mean of the mixed distribution is ,´@ µ ~ & h @ ²&³ & (if @ is continuous).
Using the "double expectation rule" of Subsection 8.2, we also have
,´@ µ ~ ,´ ,´@ O#µ µ .
(9.9)
Recall the general rule for finding the expectation of some function of a random variable;
for random variable A , ,´²A³µ ~ ²'³ h A ²'³ ' (if A is continuous).
Using # for A , and ,´@ O# ~ µ for ²³, we get
,´@ µ ~ ,´ ,´@ O#µ µ ~ ,´@ O# ~ µ h # ²³ .
This is Equation 9.3 above. This is also how we get relationships such as Equations 9.4 to 9.7
above. For instance ,´@ µ ~ ,´ ,´@ O#µ µ ~ ,´@ O# ~ µ h # ²³ ,
4@ ²!³ ~ ,´!@ µ ~ ,´ ,´!@ O#µ µ ~ ,´ 4@ O# ²!³ µ
The variance of @ can be expressed in the form
= ´@ µ ~ ,´@ µ c ²,´@ µ³ ~ ,´ ,´@ O#µ µ c ²,´@ µ³ .
(9.10)
The following variance rule given in Section 8 can be used as an alternative (but algebraically
equivalent) way of finding = ´@ µ ¢
= ´@ µ ~ ,´= ´@ O#µµ b = ´,´@ O#µµ .
(9.10)
Note that = ´?µ is not equal to = ´@ Oµ h # ²³ .
There are a couple of ways that a continuous mixture situation can be identified from the
language that is used to describe a situation. If a problem specifies a random variable @ that
depends on a parameter # and then goes on to describe the distribution of #, that indicates that #
is the mixing distribution. Sometimes a problem states that @ has a conditional distribution given
some variable #, and then describes that conditional distribution as well as the distribution of #.
Example LM9-1: Suppose that the conditional distribution of @ given $ ~ is Poisson with
parameter (and mean) . Suppose that the distribution of $ is exponential with parameter (and
mean) ~ . Find the pf of the unconditional distribution of @ .
c &
Solution: The conditional probability function of @ is @ O$ ²&O³ ~ &[ for & ~ Á Á Á ÀÀÀ
(@ is a discrete random variable) and the pdf of $ is $ ²³ ~ c for € . Then,
B
B c &
B c &
@ ²&³ ~ @ O$ ²&O³ h $ ²³ ~ h c ~ ~ ² ³&b for
&[
&[
& ~ Á Á Á ÀÀÀ The unconditional distribution of @ is geometric with parameter (mean) ~ .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 9 - CONTINUOUS MIXTURES
LM-109
The following identity is useful in simplifying exponential integrals of the type in Example
LM9-1:
B
[
if is an integer with ‚ , and if € , then ! c! ! ~ b
.
(9.11)
Applying this to the integral in Example LM9-1, we get
c &
&[
B ~ h B & c ~ h &b
~ ² ³&b ; in this integral the variable of
&[
&[
&[
integration is instead of !, is replaced by the integer & , and ~ .
Finding the mean and variance of the mixed random variable @
For a mixture random variable @ , it may be possible to find the mean and variance of @ without
finding @ ²&³. This can be illustrated using the distributions in Example LM9-1.
In Example LM9-1, the conditional expectation of @ given $ is ,´@ O$µ ~ $ , since @ is
Poisson with parameter $. Then, we have ,´@ µ ~ ,´,´@ O$µµ ~ ,´$µ ~ , since $ has an
exponential distribution with mean 1.
To find = ´@ µ we use the relationship = ´@ µ ~ ,´ = ´@ O$µ µ b = ´ ,´@ O$µ µ .
Since @ O$ is Poisson with parameter $, we have = ´@ O$µ ~ $ and (as before) ,´@ O$µ ~ $ À
Then ,´ = ´@ O$µ µ ~ ,´$µ ~ , since $ has a mean of 1, and
= ´ ,´@ O$µ µ ~ = ´$µ ~ ($ has an exponential distribution with mean 1, so the variance is
the square of the mean). Then, = ´@ µ ~ b ~ . This approach does not describe the actual
distribution of @ , but it does give us the mean and variance.
This approach to finding the mean and variance of the mixed distribution is particularly useful
when ,´@ O$µ and = ´@ O$µ have recognizable parametric forms. As another example, suppose
that @ O$ has an exponential distribution with mean $, and $ has an exponential distribution with
mean . Then ,´@ O$µ ~ $ and ,´@ O$µ ~ $ , from which we get
,´@ µ ~ ,´ ,´@ O$µ µ ~ ,´$µ ~ , and
,´@ µ ~ ,´ ,´@ O$µ µ ~ ,´$ µ ~ ,´$ µ ~ ² h ³ ~ .
Then = ´@ µ ~ c ~ . This does not tell us about the pdf of @ .
A generalization of Example LM9-1 shows that if the conditional distribution of @ given
$ ~ is Poisson with parameter , and if $ has a gamma distribution with parameters and , then the unconditional distribution of @ is negative binomial with parameters ~ and ~ . This particular relationship has come up a number of times on exam questions.
To verify this, we must show that
B
7 ´@ ~ µ ~ 7 ´@ ~ O$ ~ µ h $ ²³ ²°³ hc°
!²b³
B c ~ [ h !²³ ~ [h!²³ h ²b³b
(the last equality follows from the definition of the gamma function).
Another combination of @ and # is outlined in the following example.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-110
MODELING SECTION 9 - CONTINUOUS MIXTURES
Example LM9-2: Suppose that @ O$ has an exponential distribution with mean $, and $ has an
inverse gamma distribution with parameters and . Find the unconditional distribution of @ .
c °
Solution: @ O$ ²&O³ ~ c&° and $ ²³ ~ +1hh!() .
c °
B
B
B c²b&³°
@ ²&³ ~ @ O$ ²&O³ h $ ²³ ~ c&° h +1hh!() ~ !() h + c²b&³°
h!²b³
h!²b³
B ² b&³b
~
h h + ~
~ b , since
b
b
²b&³
h!²³
!(b)
c²b&³°
h + ~ (this is
b
B ²b&³
!(b)
²b&³
h!²³
²b&³
the integral of the pdf of the inverse gamma with
parameters b & and b ). Therefore, @ ²&³ ~ ²b&³
b , which is the pdf of the Pareto
distribution with parameters and .
Another continuous mixture that might arise is the combination of @ O normal with unknown
mean and known variance , along with having a normal distribution with known mean and
known variance . The unconditional distribution of @ is also normal with mean and variance
b . It is possible to show this results using moment generating functions:
4@ ²!³ ~ ,´!@ µ ~ ,´ ,´!@ Oµ µ ~ ,´ !b ! µ ~ ! ,´! µ ~ ! h 4 ²!³
~ ! h !b ! ~ !b ²b ³! , which is the mgf of a normal distribution with mean and
variance b .
One other recognizable mixture combination is considered in Problem Set 9, where the problem
asks you to show that the combination of @ O$ having an inverse exponential distribution and $
having a gamma distribution results in @ having an inverse Pareto distribution.
Summary of some recognizable continuous mixture combinations
Conditional distribution
of @ O$
Distribution of $
Mixed Distribution of @
Poisson,mean $
Gamma, , Negative binomial ~ , ~ Exponential, mean $
Inverse gamma, , Pareto with same , Inverse Exponential
Gamma, , Normal, mean $, var. Normal, mean , var. Inverse Pareto, ~ , same Normal, mean , var. b Example LM9-3: A carnival game based on a random integer generator has the following rules.
The player pays $7 to play. The random integer generated is 5 , where 5 has a Poisson
distribution with a mean of , and has a gamma distribution with a mean of 2 and a variance
of 2. The player receives a payment of $5 . The game operator chooses 7 so that the operator
has an expected gain of $1 each time the game is played. Find the probability that a player has a
net gain on one play of the game.
Solution: 5 is a continuous mixture of a Poisson over a gamma distribution, so 5 has a
negative binomial distribution with parameters ~ and ~ . We are given that
,´µ ~ ~ and = ´µ ~ ~ , so that ~ ~ and ~ ~ .
The expected payout on one play of the game is
,´5 µ ~ = ´5 µ b ²,´5 µ³ ~ ² b ³ b ² ³ ~ .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 9 - CONTINUOUS MIXTURES
LM-111
Example LM9-3 continued
The expected gain by the game operator for one play of the game is 7 c ~ , so that 7 ~ .
A player has a net gain if 5 € , which is equivalent to 5 ‚ . À
7 ²5 ‚ ³ ~ c 7 ²5 ~ Á Á Á ³
²³
²³²³
~ c @ ²b³
b ²b³b b ²b³b b ²b³b A ~ .
Example LM9-4: A restaurant supply company has a large number of machines that produce
chopsticks for restaurants. The length of a particular chopstick produced by a randomly chosen
machine has a normal distribution with a mean of centimeters and a standard deviation of .2
centimeters, where has a normal distribution with a mean of 20 centimeters and a standard
deviation of centimeters. Over a long period of time, it is found that 95% of all chopsticks
produced are between 19.3 and 20.7 centimeters long. Find .
Solution: The length of a chopstick produced on a randomly chosen machine is a continuous
mixture of a normal over a normal distribution. The mean and variance is and À b .
The 95% interval centered at 20 is f À
jÀ b which we are told is ²ÀÁ À³.
Therefore, À
jÀ b ~ À , so ~ À , and ~ À
.
LM-9.2 Some Other Related Expected Values
We have seen that for a mixed distribution @ over parameter distribution $, the moment
generating function of @ is 4@ ²!³ ~ ,´!@ µ ~ ,´ 4@ O$ ²!O$³ µ ~ ,´ ,´!@ O$µ µ ,
and the probability generating function is 7@ ²!³ ~ ,´@ ! µ ~ ,´ 7@ O$ ²!O$³ µ ~ ,´ ,´@ ! O$µ µ .
We see that ,´!@ O$µ is the moment generating function of the conditional distribution of @
given $, and ,´@ ! O$µ is the probability generating function of the conditional distribution of @
given $. Knowing the moment or probability generating functions of the mixture and conditional
distributions may allow for an easier calculation of the overall moment or probability generating
function of @ .
Example LM9-5: A carnival offers the following gambling game. A player randomly chooses a
coin from a large collection of coins and tosses the coin 10 times. The player receives $2 , where
2 is the number of heads that were tossed. For a randomly chosen coin, the probability of tossing
a head is , where is uniformly distributed between .2 and .7, Find the expected amount that the
carnival will pay out on a play of the game.
Solution: 2 has a binomial distribution with parameters ~ and . The expected payout
is ,´2 µ . This is the probability generating function for 2 , which for the binomial is
72 ²'³ ~ ,´' 2 µ ~ ´ b ²' c ³µ . Since ~ , with ' ~ this becomes ´ b µ ,
which still depends on . This is the probability generating function for the conditional
distribution of 2 given , so ,´2 Oµ ~ ² b ³ . Using the double expectation rule, we
have ,´2 µ ~ ,´ ,´2 Oµ µ ~ ,´² b ³ µ . The pdf of the uniform distribution on ´ÀÁ Àµ is
²³ ~ À
~ . Using the uniform distribution for , we have
À
,´² b ³ µ ~ À ² b ³ h ~ À
is the expected payout on the game.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-112
MODELING SECTION 9 - CONTINUOUS MIXTURES
LM-9.3 Frailty Models
A frailty model is a particular type of continuous mixture which is based on a function ²%³ that
is used as a "base" hazard rate function. The conditional hazard rate of ? given is
?O ²%O³ ~ ²%³ , where $ is the random variable for the parameter .
%
We define the function (²%³ to be (²%³ ~ ²!³ ! .
Then the conditional survival function of ? given is
%
%
:?O$ ²%O³ ~ %´ c ?O ²%O³ µ ~ c ²!³ ! ~ c(²%³ .
The marginal survival function is :? ²%³ ~ ,´c$(²%³ µ ~ 4$ ´ c (²%³µ (moment generating
function of $).
For example, if ²%³ ~ then (²%³ ~ % , and :?O$ ²%O³ ~ c% , so that the conditional
distribution of ? given is exponential with mean .
Example LM9-6: A frailty model has ²%³ ~ b%
, % € , and $ has a uniform distribution
between 1 and 2. Find the marginal survival function :? ²%³.
% Solution: (²%³ ~ b!
! ~ ² b %³ , and :?O$ ²%O³ ~ c ²b%³ ~ ²b%³
.
$ ²³ ~ for   .
~
c
Then :? ²%³ ~ ,´:?O$ ²%O$³µ ~ ²b%³
h ~ ²b%³ h²b%³ c
~
%
~ ²b%³
h ´ b%
c ²b%³
µ
~
,
for
%
€
.
²b%³ ²b%³
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 9
LM-113
MODELING - PROBLEM SET 9
Continuous Mixtures - Section 9
1. has a uniform distribution on the interval ²Á ³ and
the conditional distribution of ? given is uniform on the interval ²Á ³ .
Find the following probabilities.
(a) 7 ´ € O? ~ µ
(b) 7 ´ € O? € µ
2. A model for the lifetime of a tire assumes that a randomly chosen tire
has a lifetime that is normally distributed with a mean of $ miles and a
standard deviation of miles. According to the model, $ is normally
distributed with a mean of 80,000 and a standard deviation of 10,000.
You are given that the 95-th percentile of the lifetime of a randomly chosen
tire is 116,783 miles. Find the probability that a randomly selected tire has
a lifetime of at most 100,000 miles.
3. ? has a binomial distribution with a mean of and a variance of ² c ³ , and has a
beta distribution with parameters ~ Á ~ (the pdf of is ²³ ~ ² c ³ for   ³.
Find the variance of the unconditional distribution of ? .
A) 400
B) 410
C) 420
D) 430
E) 440
4. Show that the continuous mixture of an inverse exponential distribution over a gamma
distribution has an inverse Pareto distribution. Specifically, if has a gamma distribution with
parameter and , and if the conditional distribution of ?O is inverse exponential with
parameter (as defined in the table of distributions), then the unconditional distribution of ? is
inverse Pareto with ~ and the same value of .
5. ? has a normal distribution with a mean of $ and variance of 1. $ has a normal distribution
with a mean of 1 and variance of 1. Find the mean and variance of ? .
6. ? has a uniform distribution on the interval ´Á j$µ , and $ has a uniform distribution on the
interval ´Á µ . Find the pdf and variance of the unconditional distribution of ? .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-114
MODELING - PROBLEM SET 9
7. (SOA) The length of time, in years, that a person will remember an actuarial statistic is
modeled by an exponential distribution with mean °@ . In a certain population, @ has a gamma
distribution with ~ ~ . Calculate the probability that a person drawn at random from this
population will remember an actuarial statistic less than ° year.
A) 0.125
B) 0.250
C) 0.500
D) 0.750
E) 0.875
8. (SOA) You are given:
(i) For 8 ~ Á ? Á ? Á ÀÀÀÁ ? are independent, identically distributed Bernoulli random
variables. with parameter .
(ii) : ~ ? b ? b Ä b ?
(iii) The prior distribution of 8 is beta with ~ Á ~ Á and ~ .
Determine the smallest value of such that the mean of the marginal distribution of : is
greater than or equal to 50.
A) 1082
B) 2164
C) 3246
D) 4950
E) 5000
9. (SOA) Bob is a carnival operator of a game in which a player receives a prize worth > ~ 5
if the player has 5 successes, 5 ~ Á Á Á Á ÀÀÀ Bob models the probability of success for a
player as follows:
(i) 5 has a Poisson distribution with mean $.
(ii) $ has a uniform distribution on the interval (0, 4).
Calculate ,´> µ.
A) 5
B) 7
C) 9
D) 11
E) 13
10. The number of claims in one exposure period follows a Bernoulli distribution with mean .
The density function of is assumed to be ²³ ~ Á   .
Determine the expected number of claims. Hint: ~ À
A) B) C) D) E) ²c
c³
11. (SOA) An actuary for an automobile insurance company determines that the distribution of
the annual number of claims for an insured chosen at random is modeled by the negative binomial
distribution with mean 0.2 and variance 0.4.
The number of claims for each individual insured has a Poisson distribution and the means of
these Poisson distributions are gamma distributed over the population of insureds.
Calculate the variance of this gamma distribution.
A) 0.20
B) 0.25
C) 0.30
D) 0.35
E) 0.40
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 9
LM-115
12. (SOA) A claim count distribution can be expressed as a mixed Poisson distribution. The
mean of the Poisson distribution is uniformly distributed over the interval [0, 5].
Calculate the probability that there are 2 or more claims.
A) 0.61
B) 0.66
C) 0.71
D) 0.76
E) 0.81
13. (SOA) The scores on the final exam in Ms. B's Latin class have a normal distribution with
mean and standard deviation equal to 8. is a random variable with a normal distribution with
mean equal to 75 and standard deviation equal to 6. Each year, Ms. B chooses a student at
random and pays the student 1 times the student's score. However, if the student fails the exam
(score  65), then there is no payment. Calculate the conditional probability that the payment is
less than 90 given that there is a payment.
A) 0.77
B) 0.85
C) 0.88
D) 0.92
E) 1.00
14. (SOA May 07) You are given:
(i) Conditional on 8 ~ , the random variables ? Á ? Á ÀÀÀÁ ? are independent and follow a
Bernoulli distribution with parameter .
(ii) : ~ ? b ? b Ä b ?
(iii) The distribution of 8 is beta with ~ , ~ , and ~ .
Determine the variance of the marginal distribution of : .
A) 1.00
B) 1.99
C) 9.09
D) 18.18
E) 25.25
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-116
MODELING - PROBLEM SET 9
MODELING - PROBLEM SET 9 SOLUTIONS
1.(a) 7 ´ € O? ~ µ ~
7 ´?~q²€³µ
7 ´?~µ
.
7 ´? ~ q ² € ³µ ~ ²Á ³ ~ ²O³ ²³ ~ h ~ À
7 ´? ~ µ ~ ²Á ³ ~ ²O³ ²³ ~ h ~ À
7 ´ € O? ~ µ ~
7 ´?~q²€³µ
7 ´?~µ
(b) 7 ´ € O? € µ ~
À ²³
~ À ²³ ~ À
7 ´²?€³q²€³µ
7 ´?€µ
À
7 ´²? € ³ q ² € ³µ ~
% ~ ² c ³² ³ ~ À c À ²³
7 ´? € µ ~ ²%Á ³ % ~ ² c ³² ³ ²%Á ³
~ À c À ²³
7 ´ € O? € µ ~
7 ´²?€³q²€³µ
7 ´?€µ
ÀcÀ ²³
~ ÀcÀ ²³ ² ~ À³ À
2. $ has a normal distribution with mean 80,000 and standard deviation 10,000
? is the tire lifetime. We are told that the conditional distribution of ? given $
is normal with a mean of $ and a standard deviation of .
? is a continuous mixture over $. The unconditional distribution of ? is normal
with a mean of 80,000 and a standard deviation of jÁ b .
The 95-th percentile of ? is Á b À
jÁ b ~ Á .
It follows that jÁ b ~ Á .
?cÁ
Then, 7 ²?  Á ³ ~ 7 ² 22,360

100,000cÁ
)
22,360
~ )²À³ ~ À .
3. = ´?µ ~ ,´ = ´?Oµ µ b = ´ ,´?Oµ µ ~ ,´² c ³µ b = ´µ
~ ,´µ c ,´ µ b = ´µ
~ ,´µ c ,´ µ b Á ,´ µ c Á ²,´µ³
~ ,´µ b Á ,´ µ c Á ²,´µ³ .
The beta distribution with parameters and is in the Exam C table, and
²b³
,´µ ~ b
~ , and ,´ µ ~ ²b³²bb³ ~ .
Then, = ´?µ ~ ²À
³ b Á ²À³ c Á ²À
³ ~ . Answer: C
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 9
LM-117
c c°
4. The pdf of is ²³ ~ h!h() and the pdf of ?O is ²%O³ ~
Then the joint density of ? and is
²%Á ³ ~
c°%
% h
c hc°
h!()
~
hc´ b % µ
% h h!()
~
% µ
c°´ b%
c°%
% .
c°´ % µ
b%
h
% h h!()
.
%
B
B B
Then ? ²%³ ~ ²%Á ³ ~ %hh h!() ~ % hh!() h c°´ b% µ À
%
B h c° b% ~ !² b ³ h ² % ³b , so that
b%
? ²%³ ~
% b
!²b³h² b%
³
% h h!()
%c
~ ²
b%³b .
This is the pdf of the inverse Pareto with parameters ~ and .
5. The conditional distribution of ? given $ has ,´?O$µ ~ $ and = ´?O$µ ~ .
We are given that ,´$µ ~ = ´$µ ~ .
? is a continuous mixture with mean ,´?µ ~ ,´ ,´?O$µ µ ~ ,´$µ ~ and
= ´?µ ~ = ´ ,´?O$µ µ b ,´ = ´?O$µ µ ~ = ´$µ b ,´µ ~ b ~ .
The continuous mixture of a normal "over" a normal distribution is also normal, so the
unconditional distribution of ? is normal with mean 1 and variance 2.
6. The uniform distribution on ´Á j$µ has pdf ?O$ ²%O³ ~ j
or equivalently, ‚ % . Also, the pdf of $ is $ ²³ ~ .
for  %  j ,
Then ? ²%³ ~ ?O$ ²%O³ h $ ²³ ~ % j h 1 d ~ ² c %³ for  %  .
The mean of % is ,´?µ ~ % h ² c %³ % ~ .
j$
Alternatively, since ,´?O$µ ~ , we have
j$
j
j
,´?µ ~ ,´ ,´?O$µ µ ~ ,´ µ ~ h $ ²³ ~ ~ .
The 2nd moment of ? is ,´? µ ~ % h ² c %³ % ~ .
The variance of ? is = ´?µ ~ ,´? µ c ²,´?µ³ ~ c ~ À
Alternatively, since the variance of a uniform distribution is the interval length squared over 12,
we have = ´?O$µ ~ , so that
j$
$
= ´?µ ~ ,´ = ´?O$µ µ b = ´ ,´?O$µ µ ~ ,´ µ b = ´ µ .
$
,´ µ ~ h ,´$µ ~ h ~ .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-118
MODELING - PROBLEM SET 9
7. Let ; denote the time that the person remembers the statistic. Then the conditional pdf of ;
given @ ~ & is &c&! , and the conditional cdf is c c!& .
&c&°
The pdf of @ is h[ ~ h &c&° .
7 ´;  µ ~ c 7 ´; ‚ µ and
B
ÀB
7 ´; ‚ µ ~ 7 ´; ‚ O@ ~ &µ h @ ²&³ & ~ c&° h h &c&° &
B
~ h ´ &c& &µ ~ B
( &c& & ~ since this is the integral of a gamma pdf with ~ Á ~ ).
Then, 7 ´;  µ ~ c ~ .
Note also that
B
B
7 ´; ‚ µ ~ 7 ´; ‚ O@ ~ &µ h @ ²&³ & ~ c&° h @ ²&³ & ~ 4@ ² c ³
(moment generating function of @ ). The mgf of a gamma distribution with parameters and is
4 ²³ ~ ² c ³c , so 4 ² c ³ ~ ´ c ² c ³µc ~ .
Answer: D
8. The Bernoulli distribution is a 0,1 distribution, with 7 ²? ~ ³ ~ and 7 ²? ~ ³ ~ c .
: has a binomial distribution with parameters and .
Using the double expectation rule, we have ,´: µ ~ ,´ ,´: O8µ µ .
We know that ,´: O8 ~ µ ~ , since : has a binomial distribution with parameters and
. Then, ,´: µ ~ ,´8µ ~ ,´8µ . The pdf of 8 is
!²b³
!²³ !²³
h c ² c ³c ~ [[
[ h ² c ³ ~ ² c ³ .
The expected value of is
h ² c ³ ~ ´ c ² c ³µ² c ³ ~ ´ ² c ³ c ² c ³ µ ~ ´ c µ ~ .
Then, ,´: µ ~ ,´8µ ~ .
In order for this to be at least 50, we must have ‚ .
Answer: E
9. We can use the double expectation rule to find ,´> µ ~ ,´5 µ .
,´> µ ~ ,´5 µ ~ ,´ ,´5 O$µ µ .
The inside expectation is ,´5 O$µ . This is the probability generating function of 5 evaluated at
! ~ , given that we know the value of $; 75 ²!³ ~ ,´!5 µ, so ,´5 µ ~ 75 ²³ ,
Since 5 is Poisson with mean $, the probability generating function is 75 ²!³ ~ $²!c³ .
Therefore, ,´5 O$µ ~ $²c³ ~ $ . Then, ,´> µ ~ ,´$ µ . Since $ has a uniform
distribution on the interval ²Á ³, we have
,´> µ ~ ,´$ µ ~ h ² ³ ~ c ~ À .
If we had not made the observation that ,´5 O$µ is the probability generating function of 5
B
c$ evaluated at ! ~ , then we would have to find ,´5 O$µ directly as h [$ , which
~
B
²$³
becomes c$ h [
~
~ c$ h $ ~ $ (the summation is the Taylor expansion for $ ).
We would then proceed as before.
© ACTEX 2009
Answer: E
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 9
LM-119
10. If ? is Bernoulli given , then the expected value of ? is ,´?Oµ ~ ³ .
From the hint, we have ,´µ ~ h ²³ ~ h ~ , and
,´?µ ~ ,´,´?Oµµ ~ ,´µ ~ À
Answer: D
11. If the conditional distribution of @ given $ ~ is Poisson with parameter , and if $ has a
gamma distribution with parameters and , then the unconditional distribution of @ is negative
binomial with parameters ~ and ~ . The mean of the negative binomial is h , and the
variance is h h ² b ³. We are given that the negative binomial distribution has mean
À ~ , and variance À ~ ² b ³ .
Therefore, b ~ À
À ~ , so that ~ , and then ~ À .
Since ~ and ~ from the gamma distribution, it follows that
~ À and ~ . Referring to the table of distributions. we see that the variance of the gamma
distribution ? with parameters and is ,´? µ c ²,´?µ³ ~ ² b ³ c ´µ .
In this case, the variance is ²À³²À³ c ²À³ ~ À .
Answer: A
12. The distribution of 5 given $ is Poisson with mean $, where $ has a uniform distribution on
´Á µ , with pdf ²³ ~ À for   .
Then, 7 ´5 ‚ µ ~ c 7 ´5 ~ Á µ . We find 7 ´5 ~ µ by conditioning over $.
7 ´5 ~ µ ~ 7 ´5 ~ Oµ h ²³ ~ c h ²À³ ~ ²À³² c c ³ ~ À
.
We find 7 ´5 ~ µ in a similar way.
7 ´5 ~ µ ~ 7 ´5 ~ Oµ h ²³ ~ c h ²À³ ~ ²À³´ c c c c c
~
~
µ ~ ²À³´ c c c c b µ ~ À .
Then, 7 ´5 ‚ µ ~ c ²À
b À³ ~ À
.
Answer: A
13. ? represents the randomly chosen student's score. We wish to find 7 ´?  O? € µ .
We are given that ? has a normal distribution with mean and standard deviation 8, and has a
normal distribution with mean 75 and standard deviation 6. Therefore ? is a continuous mixture
of a normal distribution with a normal mixing distribution. The resulting unconditional
distribution of ? is normal with mean 75 and variance b ~ . Then
7 ´
?µ
7 ´?€
µ . We standardize each probability.
c
7 ´? € µ ~ 7 ´ ?c
€ µ ~ c )² c ³ ~ )²³ ~ À , and
?c
c
7 ´
 ?  µ ~ 7 ´ c
  µ ~ )²À³ c )² c ³
7 ´?  O? € µ ~
~ À c À ~ À.
Then, 7 ´?  O? € µ ~
© ACTEX 2009
À
À
~ À .
Answer: D
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-120
MODELING - PROBLEM SET 9
14. = ´: µ ~ = ´ ,´: O8µ µ b ,´ = ´: O8µ µ .
The conditional distribution of : given 8 is binomial with ~ and 8.
,´: O8µ ~ 8 and = ´: O8µ ~ 8² c 8³ .
From the Exam C table of distributions, we have
²b³
,´8µ ~ b
~ b
~ and ,´8 µ ~ ²b³²bb³ ~ ²³²³
.
Then, = ´ ,´: O8µ µ ~ = ´8µ ~ = ´8µ ~ ´ ²³²³
c µ ~ À
and
,´ = ´: O8µ µ ~ ,´8² c 8³µ ~ ²,´8µ c ,´8 µ³ ~ ² c ²³²³
³ ~ À .
Therefore, = ´: µ ~ À b À ~ À .
© ACTEX 2009
Answer: B
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 10 - POLICY LIMITS
LM-121
MODELING SECTION 10 - POLICY LIMITS AND
THE LIMITED LOSS RANDOM VARIABLE
The material in this section relates to Loss Models, Section 8.4
The suggested time for this section is 1 hour.
For a loss random variable ? , the amount paid by the insurer may be subject to some
adjustments. The actual loss amount, prior to any modifications is referred to as the ground up
loss, ?. After adjustments are applied to the amount of the loss, the result is the amount paid by
the insurer (which may be 0). When considering the amount paid by the insurer, we can consider
the cost per loss, which includes the 0 amounts paid by the insurer that can result from some
losses (if a loss is below the deductible for instance). We can also consider the cost per
payment, in which we only consider the non-zero payments of the insurer for losses above a
deductible (this would be a conditional distribution, given that the insurer actually makes a
payment).
The distribution of loss amount or of the cost to the insurer may be referred to as the severity
distribution. The distribution of the number of losses, or amounts paid per unit time is referred
to as the frequency distribution. In this section we consider policy limits applied to a loss
amount. The ground up loss random variable is generally denoted ? , and it is usually assumed
that ? ‚ .
LM-10.1 Policy Limit " and the Limited Loss Random Variable
The policy limit is the maximum amount paid by the insurance policy for a single loss. If there
are no other adjustments on the policy, then the insurer pays the full loss amount ? for losses up
to amount ", and the insurer pays the limit amount " for losses that are above amount ". The
amount paid by the insurer is denoted ? w " , and is referred to as the limited loss random
variable. In this situation, the insurer is subject to pay a maximum covered loss of ".
The limited loss random variable may also be referred to as being right-censored at limit ". Data
points are right-censored at limit " if any data value above " is recorded as ". If an insurer had
data based on insurance payments on a limited loss insurance policy with limit ", then the
recorded payments would be " for any loss above ". The data would not show that actual loss
amounts that were above ", we would only know that for a limit payment the loss was greater
than ".
Limited loss random variable with policy limit "
Amount paid by insurer ~ ? w " ~ F
© ACTEX 2009
?
"
if ?  "
~ ²?Á "³
if ? € "
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-122
MODELING SECTION 10 - POLICY LIMITS
The Loss Models book also identifies the density function of the limited loss random variable.
Exam questions have generally not focused on this density function. The following examples and
notes summarize the density function as well as expected amount paid by the insurer, but based
on past exam experience, the emphasis has been on the expected values and variances. The same
will be true for the other main policy adjustment, which is the policy deductible.
When a loss occurs, the amount paid by the insurer is the cost per loss, and the expected amount
paid by the insurer per loss is the expected cost per loss. When the only policy adjustment is a
policy limit of amount ", the cost per loss has a maximum value of " which is reached if ? € ",
and the probability of this happening is 7 ´? € "µ ~ c -? ²"³ ~ :? ²"³ . In the following
comments, the cost per loss ? w " will be denoted @ . The pdf and cdf of @ can be expressed in
terms of the distribution of ? .
? ²&³
&"
The pdf of @ is @ ²&³ ~ F c -? ²"³ & ~ " À
(10.1)
&€"
The cdf of @ is -@ ²&³ ~ D
-? ²&³ &  "
.
&€"
(10.2)
Note that @ has a discrete point of probability at @ ~ " , so if ? is continuous, @ will be
continuous for @  ", and @ has a discrete point of probability at ".
LM-10.2 Limited Expected Value
We are often interested in ,´? w "µ , which is called the limited expected value.
The limited expected value of ? with limit " is denoted ,´? w "µ À
For a continuous loss variable ? , this is
"
,´²?Á "³µ ~ ,´? w "µ ~ cB % h ²%³ % b " h ´ c - ²"³µ
(10.3)
and for a discrete ? this is ,´? w "µ ~ % h ²% ³ b " h ´ c - ²"³µ .
(10.4)
In both the continuous and discrete cases it can be shown that
"
,´? w "µ ~ c cB - ²%³ % b ´ c - ²%³µ % .
(10.5)
% "
If ? is a non-negative random variable (? ‚ ), then
"
"
,´? w "µ ~ ´ c - ²%³µ % ~ :²%³ % .
(10.6)
Almost all loss and survival random variables that are considered on Exam C are ‚ (the
normal distribution is the main exception). Equation 10.6 can be a convenient way of finding a
limited expected value as opposed to Equation 10.3 which may involve an integration by parts.
The variance of the cost per loss is
= ´? w "µ ~ ,´²? w "³ µ c ²,´? w "µ³ ,
"
where ,´²? w "³ µ ~ % h ²%³ % b " h ´ c -? ²"³µ .
© ACTEX 2009
(10.7)
(10.8)
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 10 - POLICY LIMITS
LM-123
In general, the -th moment of the limited loss random variable ? w " is (assuming ? ‚ )
"
,´²? w "³ µ ~ % h ²%³ % b " h ´ c - ²"³µ (continuous)
(10.9)
or ,´²? w "³ µ ~ % h ²% ³ b " h ´ c - ²"³µ (discrete).
(10.10)
% "
Usually the random variable ? is non-negative (for instance, when ? represents a loss or a time
until death), and the lower integration limits of c B become 0. We will generally assume
? ‚ unless indicated otherwise.
Limited loss random variable with limit " ²? ‚ ³
"
,´? w "µ ~ % h ²%³ % b " h ´ c -? ²"³µ (continuous ? )
,´? w "µ ~ % h ²% ³ b " h ´ c -? ²"³µ (discrete ? )
% "
"
"
,´? w "µ ~ ´ c -? ²%³µ % ~ :? ²%³ %
(discrete or continuous)
"
,´²? w "³ µ ~ % h ²%³ % b " h ´ c - ²"³µ (continuous)
The Exam C table provides the limited expected value for many of the distributions listed.
The temporary expected lifetimes found in life contingencies are limited expected values for ; ²&³
and 2²&³, the complete and curtate future lifetime random variables of someone at age & .
° &¢O ~ ,´; ²&³ w µ ~ ! & ! ~ 7 ´; ²&³ € !µ ! ~ ´ c -; ²&³ ²!³µ !
(10.11)
and
&¢O ~ ,´2²&³ w µ ~ & .
(10.12)
~
Example LM10-1: Suppose that the ground up loss ? has a uniform distribution on the interval
² Á ³. There is a policy limit of 50. Find the pdf of the limited loss random variable.
Solution: The pdf of ? is ? ²%³ ~ À for  %  . For losses above 50, the amount
paid by the insurer is 50, so 7 ´@ ~ µ ~ 7 ´? ‚ µ ~ À , where @ denotes the cost per loss.
For losses below 50, @ ~ ? , so @ ²&³ ~ ? ²&³ ~ À for  &  .
À  &  The pdf of @ is @ ²&³ ~ D
. @ has a mixed distribution this is continuous on
À
& ~
the interval  &  , and @ has a single point of probability at & ~ .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-124
MODELING SECTION 10 - POLICY LIMITS
Example LM10-2: Find the pdf, cdf and expected value and variance of the cost per loss when
there is a policy limit of " in each of the following cases.
(a) ? is exponential with mean .
(b) ? has a uniform distribution on ²Á ³ , and "  .
c&°
&"
? ²&³
&"
Solution: (a) @ ²&³ ~ F c -? ²"³ & ~ " ~ F c"°
&~" À
&€"
&€"
Note that @ ²"³ ~ c"° is the probability 7 ²@ ~ "³ , but for &  "
@ ²&³ ~ c&° is a density.
-@ ²&³ ~ D
-? ²&³ &  "
c c&°
~D
&€"
&"
.
&€"
"
"
,´@ µ ~ ,´? w "µ ~ ´ c -? ²%³µ % ~ c%° % ~ ² c c"° ³
(the Exam C table of distributions gives limited expected value formulations).
"
Note that we could use the formulation ,´@ µ ~ % ²%³ % b " h ´ c -? ²"³µ ,
but this would require an integration by parts.
We formulate the variance of @ as = ´@ µ ~ ,´@ µ c ²,´@ µ³ .
"
,´@ µ ~ ,´²? w "³ µ ~ % h ²%³ % b " h ´ c -? ²"³µ
"
~ % h c%° % b " h c"° .
The antiderivative of % h c%° is found by integration by parts to be
c % c%° c %c%° c c%° , so that
"
,´@ µ ~ % h c%° % b " h c"°
~ ² c % c%° c %c%° c c%° ³c
%~"
%~
b " h c"°
~ c "c"° c c"° .
Then = ´@ µ ~ c "c"° c c"° c ´² c c"° ³µ .
&"
&~" .
&€"
The same comment regarding @ ²"³ ~ c " being a probability and @ ²&³ being a density
apply here as in part (a).
? ²&³
&"
(b) @ ²&³ ~ F c -? ²"³ & ~ " ~ F c "
&€"
-@ ²&³ ~ D
&
-? ²&³ &  "
~D
&€"
&"
.
&€"
"
"
,´@ µ ~ ,´? w "µ ~ ´ c -? ²%³µ % ~ ´ c % µ % ~ " c " .
"
,´@ µ ~ % h ²%³ % b " h ´ c -? ²"³µ
"
~ % h % b " h ´ c " µ ~ " b " h ´ c " µ ~ " c "
À
" "
"
= ´@ µ ~ ,´@ µ c ²,´@ µ³ ~ " c "
c ²" c ³ ~ c .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 10
LM-125
MODELING - PROBLEM SET 10
Policy Limits - Section 10
1. Under a group insurance policy, an insurer agrees to pay 100% of the medical bills incurred
during the year by employees of a small company, up to a maximum total of one million dollars.
The total amount of bills incurred, ? , has probability density function
²%³ ~ H
%²c%³
,
%
, where % is measured in millions.
otherwise
Calculate the total amount, in millions of dollars, the insurer would expect to pay under this
policy.
A) 0.120
B) 0.301
C) 0.935
D) 2.338
E) 3.495
,
2. Claim amounts follow a Weibull distribution with ~ and unknown . An insurer sets a
policy limit of 100 and finds that 50% of the claims are below the policy limit. After a uniform
inflation adjustment of 10% on all claims amounts, find the percentage of the claims that will be
below the policy limit of 100.
A) 40%
B) 42%
C) 44%
D) 46%
E) 48%
3. A loss distribution is uniformly distributed on the interval from 0 to 100 .
Two insurance policies are being considered to cover part of the loss.
Insurance policy 1 insures 80% of the loss.
Insurance policy 2 covers the loss up to a maximum insurance payment of 3  .
Both policies have the same expected payment by the insurer. Find the ratio
= ´insurer payment under policy 2µ
= ´insurer payment under policy 1µ
A) 1.5
B) 1.2
C) .9
(nearest .1).
D) .6
E) .3
4. (SOA) The unlimited severity distribution for claim amounts under an auto liability insurance
policy is given by the cumulative distribution: - ²%³ ~ c ÀcÀ% c ÀcÀ% Á % ‚ The insurance policy pays amounts up to a limit of 1000 per claim.
Calculate the expected payment under this policy for one claim.
A) 57
B) 108
C) 166
D) 205
E) 240
5. ? and @ are random losses with joint density function
%
²%Á &³ ~ Á
for  %  and  &  .
An insurance policy on the losses pays the total of the two losses to a maximum payment of 100.
Find the expected payment the insurer will make on this policy (nearest 1).
A) 90
B) 92
C) 94
D) 96
E) 98
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-126
MODELING - PROBLEM SET 10
MODELING - PROBLEM SET 10 SOLUTIONS
1. The insurer will pay 3 ~ D
% if %  (million)
.
if % € (million)
The expected payment by the insurer will be
%²c%³
%²c%³
,´3µ ~ % h ²%³ % b h ²%³ % ~ % h % b %
~ b ~ ~ À.
Answer: C
2. -? ²³ ~ c c²°³ ~ À S ~ À
c²°³
After 10% inflation, @ ~ À? Á -@ ²³ ~ -? ² ~ À
.
À ³ ~ c Note that because of the scale properties of the Weibull distribution, @ also has a Weibull
distribution with the same ~ , but with ~ . Answer: C
3. Expected payment under policy 2 is
3 &²À³& b 3 h 7 ´? € 3µ ~ À3 b ²3³² c3 ³ ~ 3 c À3 .
This is equal to the expected payment under policy 1, which is
²À³,´?µ ~ ²À³²³ ~ À Solving 3 c À3 ~ results in
3 ~ À Á À . We discard 144.72 as a limit since it is larger than the maximum loss
amount. Thus, 3 ~ À .
The variance of insurer payment under policy 1 is
= ´À?µ ~ À
= ´?µ ~ ²À
³² ³ ~ À .
Under policy 2,
À
,´(insurer payment) µ ~ & ²À³& b ²À³ h 7 ´? € Àµ
~ À b ²À³ ´ cÀ
µ ~ À
, and
= ´insurer paymentµ ~ À
c ²³ ~ À
.
= ´insurer payment under policy 2µ
= ´insurer payment under policy 1µ
4. ,´? w µ ~ c
À
~ À
~ À
.
´ c - ²%³µ % ~ c
Answer: D
²ÀcÀ% b ÀcÀ% ³ %
c
~ ²À³² c
À ³ b ²À³² À ³ ~ À .
Answer: C
5. The maximum payment on the policy occurs when ? b @ ‚ . The expected payment is
9 ²% b &³ %
& % b 9 ²³ %
& %
Á
Á
where 9 and 9 are the regions in the square ¸²%Á &³O  %  Á  &  ¹
represented in the graph below: 9 ~ ¸²%Á &³O% b &  ¹ , 9 ~ ¸²%Á &³O% b & ‚ ¹ .
c%
The expected payment is ²% b &³ %
&% b c% ²³ %
& %
~
Á
%
Á ´% ² c %³ b %² c %³ µ % b © ACTEX 2009
Á
% ~ b À š À Answer: B
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 11 - POLICY DEDUCTIBLE (1), COST PER LOSS
LM-127
MODELING SECTION 11 - POLICY DEDUCTIBLE (1),
THE COST PER LOSS RANDOM VARIABLE
The material in this section relates to Loss Models, Section 8.2
The suggested time for this section is 2 hours.
LM-11.1 Ordinary Policy Deductible and Cost Per Loss
The insurer pays when the ground up loss amount ? is above the deductible . In that case, the
insurer pays the difference between the loss amount and . For loss amounts below the
deductible the insurer pays 0. This is referred to as an ordinary deductible , often just referred
to as policy deductible . The amount paid by the insurer is called the cost per loss (or the left
censored and shifted variable), and is denoted @3 or ²? c ³b .
Policy deductible , cost per loss
Amount paid by insurer ~ cost per loss
~ left-censored and shifted random variable
@3 ~ ²? c ³b ~ F
?c
if ?  ~ %²? c Á ³
if ? € (11.1)
The usual interpretation of the cost per loss variable is that ? represents a loss and is the
deductible applied to the insurance payment. No insurance payment is made for losses below the
deductible , but for losses that are larger than the deductible, the insurance pays ?c , the loss
amount in excess of the deductible.
Note that ²? w ³ b ²? c ³b ~ ? .
(11.2)
For a loss below , the limited loss variable ? w pays the full amount, and the cost per loss
variable with deductible pays 0. For a loss above , the limited loss pays and the cost per loss
pays ? c , for a total of ? . Therefore, for any loss amount, the limited loss with limit and the
cost per loss with deductible combine to pay the full loss amount ? .
It follows that
²? c ³b ~ ? c ²? w ³ .
(11.3)
Therefore two insurance policies, one with limit and one with deductible amount , would
combine to pay the entire loss no matter what the size of the loss is.
The loss models book identifies the pdf of the cost per loss random variable. There has been very
little reference on exam questions to pdf's of random variables related to policy limits and policy
deductibles.
The pdf and cdf of @3 , the cost per loss, are
@3 ²&³ ~ D
© ACTEX 2009
-? ²³
&~
- ²³
&~
, -@3 ²&³ ~ D ?
.
? ²& b ³ & € -? ²& b ³ & € SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-128
MODELING SECTION 11 - POLICY DEDUCTIBLE (1), COST PER LOSS
Expected cost per loss
When a loss occurs, the average amount paid by the insurance is ,´@3 µ ~ ,´²? c ³b µ. This is
the expected cost per loss. Note that some losses (those below ) result in a payment of 0 by the
insurer.
Expected Cost Per Loss With Deductible ²? c ³b ~ %²? c Á ³ ~ F
?c
if ?  ~ ? c ²? w ³
if ? ‚ Expected cost per loss ~ ,* 3 ~ ,´²? c ³b µ ~ ,´@3 µ
B
~ ²% c ³ h ²%³ % (continuous ? )
~ ²% c ³ h ²% ³ (discrete ? )
(11.5)
It is possible to show that for any loss variable ? € (continuous or discrete)
B
B
,´²? c ³b µ ~ ,´?µ c ,´? w µ ~ ´ c - ²%³µ % ~ :²%³ %
(11.6)
(11.4)
% €
B
We know that for ? ‚ , the mean can be formulated as ,´?µ ~ ´ c - ²%³µ % .
Recall that the limited expected loss can be formulated as ,´? w µ ~ ´ c - ²%³µ %.
B
Then, ,´?µ ~ ´ c - ²%³µ % b ´ c - ²%³µ % ~ ,´? w µ b ,´²? c ³b µ .
This is consistent with the earlier observation that ? ~ ²? w ³ b ²? c ³b .
Example LM11-1: Find the pdf, cdf and expected value of the cost per loss when there is an
ordinary deductible of in each of the following cases.
(a) ? is exponential with mean .
(b) ? has a uniform distribution on ²Á ³ , where  .
c c°
&~
- ²³
&~
Solution: (a) @3 ²&³ ~ D ?
~ D c²&b³°
.
? ²& b ³ & € h
&
€
-@ 3 ²&³ ~ D
-? ²³
&~
c c°
~D
-? ²& b ³ & € c c²&b³°
&~
.
&€
B
B
The formulation ,´@3 µ ~ ²% c ³ h ²%³ % ~ ²% c ³ h h c%° %
requires integration by parts. That can be avoided in this case by using
B
B
,´@3 µ ~ ,´²? c ³b µ ~ ´ c -? ²%³µ % ~ c%° % ~ c° .
Alternatively, ,´@3 µ ~ ,´²? c ³b µ ~ ,´?µ c ,´? w µ .
From the Exam C table or from Example LM10-2 of the previous section, we have
,´? w µ ~ ² c c° ³ , so that ,´²? c ³b µ ~ c ² c c° ³ ~ c° .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 11 - POLICY DEDUCTIBLE (1), COST PER LOSS
Example LM11-1 continued
(b) @3 ²&³ ~ D
-? ²³
&~
~ D ? ²& b ³ & € - ²³
&~
-@3 ²&³ ~ D ?
~ H &b
-? ²& b ³ & € &~
& c
LM-129
.
&~
& c .
&‚c ²c³
,´@3 µ ~ ²% c ³ h ? ²%³ % ~ ²% c ³ h % ~ (no integration parts needed for this distribution). Alternatively
²c³
,´@3 µ ~ ´ c -? ²%³µ % ~ ´ c % µ % ~ c c c
.
~
B
Also, ,´@3 µ ~ ,´²? c ³b µ ~ ,´?µ c ,´? w µ .
From Example LM10-2 we have ,´? w µ ~ c , so that
²c³
b ,´@3 µ ~ ,´²? c ³b µ ~ ,´?µ c ,´? w µ ~ c ² c ³ ~ c
~ .
Variance of Cost Per Loss @3 With Deductible The variance of the cost per loss is
= ´²? c ³b µ ~ = ´@3 µ ~ ,´@3 µ c ²,´@3 µ³ ~ ,´²? c ³b µ c ²,´²? c ³b µ³ .
(11.7)
There are a couple of ways in which we can find this variance.
For many of the distributions in the Exam C table we are given formulations for ,´? µ and
,´²? w ³ µ (the -th moment and the -th limited moment). We can use these formulations to
find ,´@3 µ ~ ,´²? c ³b µ ~ ,´?µ c ,´? w µ .
It can also be shown that
,´@3 µ ~ ,´²? c ³b µ ~ ,´? µ c ,´²? w ³ µ c ´ ,´?µ c ,´? w µ µ ,
(11.8)
which makes use of the formulations in the Exam C table of distributions.
Note that it is not true that ,´²? c ³b µ ~ ,´? µ c ,´²? w ³ µ .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-130
MODELING SECTION 11 - POLICY DEDUCTIBLE (1), COST PER LOSS
Example LM11-2: Find the variance of the cost per loss when there is an ordinary deductible of
in each of the following cases.
(a) ? is exponential with mean .
(b) ? has a uniform distribution on ²Á ³ , where  .
Solution: (a) In Example LM11-1 we have seen that ,´@3 µ ~ c° .
B
From Equation 11.8, we have ,´@3 µ ~ ²% c ³ h h c%° % .
Using the change of variable ' ~ % c , this integral becomes
B ' h h c²'b³° ' ~ c° h B ' h h c'° ' .
B c'°
But ' h h ' is the second moment of an exponential random variable with mean ,
which is . Therefore, ,´@3 µ ~ c° .
The variance of @3 is ,´@3 µ c ²,´@3 µ³ ~ c° c ²c° ³ ~ c° ² c c° ³ .
In the Exam C table for the exponential distribution, the formulation for ,´²? w ³ µ involves the
partial gamma function, which would require integration by parts. We can avoid the integration
by parts using the change of variable approach shown above. A little later when we consider the
cost per payment random variable,, we will see that for some distributions (the exponential is
one), we can avoid integration altogether if we are familiar with some additional relationships.
(b) From Example LM11-2 we have ,´@3 µ ~
,´@3 µ ~ ²% c ³ h % ~
Loss elimination ratio (LER)Á
²c³
²c³
.
. Then = ´@3 µ ~
²c³
²c³ c ´ µ
.
,´?wµ
,´?µ
The loss elimination ratio is the average fraction of ,´?µ that is saved by the insurer if a
deductible of is imposed. ,´?µ is the expected loss with no deductible, and ,´?µ c ,´? w µ
is the expected payment with a deductible of . Therefore, the average amount by which
expected cost per loss is reduced when the deductible is included is ,´? w µ . Therefore, the
average fraction of ,´?µ by which the expected payment is reduced when the deductible is
included is
,´?wµ
,´?µ
© ACTEX 2009
.
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 11 - POLICY DEDUCTIBLE (1), COST PER LOSS
LM-131
LM-11.2 Modeling Bonus Payments
There have been a few Exam questions that relate to a bonus payment made if losses are below a
certain level. The following comments discuss this concept in a general way.
Suppose that ? is a loss random variable. Suppose that a risk manager receives a bonus equal to
the amount by which the loss is less than some specified limit ". If the loss is above ", there is no
bonus. We can formulate the bonus as
Bonus ~ H
"c?
?"
.
?€"
An important observation is that the bonus is equal to
and the expected bonus is
,´Bonusµ ~ " c ,´? w "µ
(11.9)
Bonus ~ " c ²? w "³ ,
(11.10)
If the bonus is equal to the fraction of the amount by which the loss is less than ", then
Bonus ~ ´" c ²? w "³µ .
The expected bonus is ,´Bonusµ ~ ´" c ,´? w "µ µ .
Example LM11-3: A risk manager receives a bonus of 25% of the amount by which the loss is
below 150. The loss is uniformly distributed between 0 and 200. Find the expected value and
variance of the bonus received by the risk manager.
²À³² c ?³  ?  .
Solution: The bonus is Bonus ~ H
? € The pdf of the ground up loss ? is ? ²%³ ~ ~ À for  %  .
The expected bonus is ²À³ ² c %³²À³ % ~ À
.
Note that this can also be formulated as ,´bonusµ ~ À² c ,´? w µ³ ~ À
.
The second moment of the bonus is ,´bonus µ ~ ²À³ ² c %³ ²À³ % ~ À
.
The variance of the bonus is À
c ²À
³ ~ À .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-132
MODELING SECTION 11 - POLICY DEDUCTIBLE (1), COST PER LOSS
LM-11.3 Franchise Deductible
A variation on the concept of a deductible amount is the franchise deductible . As with the
ordinary deductible, for a loss of amount ?  , the insurer pays nothing. Unlike the ordinary
deductible, for a loss of amount ? € , the insurer pays the full loss amount (not just the loss
amount that is in excess of the deductible.
Franchise deductible Amount paid by insurer ~ F
?
if ?  if ? € Expected cost per loss
B
~ % ? ²%³ % ~ ,´²? c ³b µ b ´ c -? ²³µ
(11.11)
(11.12)
Example LM11-4: For franchise deductible find the expected cost per loss in each of the
following cases.
(a) ? is exponential with mean .
(b) ? has a uniform distribution on ²Á ³ , where  .
Solution: (a) If is a franchise deductible, then the expected cost per loss is
,´²? c ³b µ b ´ c -? ²³µ ~ c° b c° ~ ² b ³c° À
(b) If is a franchise deductible, then the expected cost per loss is
²c³
b ´ c µ ~ c
.
There has been little direct reference to franchise deductible on the exam. However, the concept is
closely related to Conditional Tail Expectation that will be considered later on under the Risk
Measures topic.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 11
LM-133
MODELING - PROBLEM SET 11
Policy Deductible (1), Cost Per Loss - Section 11
1. A company has annual losses that can be described by the continuous random variable ? , with
density function ²%³ . The company wishes to obtain insurance coverage that covers annual
losses above a deductible. The company is trying to choose between deductible amounts and
, where  . With deductible the expected annual losses that would not be covered by
insurance is , , and with deductible the expected annual losses that would not be covered by
insurance is , . Which of the following is the correct expression for , c , ?
A) % h ²%³ %
B) ² c ³´- ² ³ c - ² ³µ
C) % h ²%³ % b ² c ³´- ² ³ c - ² ³µ
D) % h ²%³ % b ² c ³
E) % h ²%³ % b ² c ³ c ´ - ² ³ c - ² ³µ
À%  %  .
otherwise
Find the mean and variance of the left-censored and shifted variable ²? c ³b .
2. The continuous random variable ? has pdf ²%³ ~ D
3. An insurer will pay the amount of a loss in excess of a deductible amount . Suppose that the
loss amount has a continuous uniform distribution between 0 and * € . When a loss occurs, let
the expected payout on the policy be ²³. Find Z ²³.
A) *
B) c *
C) *
b
D) *
c
E) c *
4. You are given:
(i) Losses follow an exponential distribution with the same mean in all years.
(ii) The loss elimination ratio this year is 70%.
(iii) The ordinary deductible for the coming year is 4/3 of the current deductible.
Compute the loss elimination ratio for the coming year.
A) 70%
B) 75%
C) 80%
D) 85%
E) 90%
5. Losses have an exponential distribution with a mean of 1,000. There is a deductible of 500.
The insurer wants to double the loss elimination ratio. Determine the new deductible that
achieves this.
A) 219
B) 693
C) 1,046
D) 1,193
E) 1,546
6. (SOA) A claim severity distribution is exponential with mean 1000. An insurance company
will pay the amount of each claim in excess of a deductible of 100. Calculate the variance of the
amount paid by the insurance company for one claim, including the possibility that the amount
paid is 0.
A) 810,000
B) 860,000
C) 900,000
D) 990,000
E) 1,000,000
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-134
MODELING - PROBLEM SET 11
?
. For random loss ? :
?c ? ‚
7 ´? ~ µ ~ 7 ´? ~ µ ~ À and ,´²? c ³b µ ~ .
Determine .
A) 4.5
B) 5.0
C) 5.5
D) 6.0
E) 6.5
7. (SOA) ²? c ³b ~ F
8. (SOA) A random loss is uniformly distributed over (0 , 80). Two types of insurance are
available.
Type
Premium
ordinary
Insurer's expected claim
deductible 10
plus 14.6
Complete
Insurer's expected claim
times ² b ³
The two premiums are equal. Determine .
A) 0.07
B) 0.09
C) 0.11
D) 0.13
E) 0.15
9. (SOA) The random variable for a loss, ? , has the following characteristics:
%
- ²%³ ,²? w %³
À
À
À
À
Calculate the mean excess loss for a deductible of 100.
A) 250
B) 300
C) 350
D) 400
E) 450
10. (SOA) The graph of the density function for losses is:
Calculate the loss elimination ratio for an ordinary deductible of 20.
A) 0.20
B) 0.24
C) 0.28
D) 0.32
E) 0.36
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 11
LM-135
"
11. Which of the following is a correct expression for % h ²%³ % for  "?
A) ,´? w "µ c ,´? w µ
B) ,´? w "µ c ,´? w µ c ²" c ³´- ²"³ c - ²³µ
C) ,´? w "µ c ,´? w µ b ²" c ³´- ²"³ c - ²³µ
D) ²,´? w "µ c "´ c - ²"³µ³ c ²,´? w µ c ´ c - ²³µ³
E) ²,´? w "µ b "´ c - ²"³µ³ c ²,´? w µ b ´ c - ²³µ³
12. (SOA) Total hospital claims for a health plan were previously modeled by a two-parameter
Pareto distribution with ~ 2 and ~ 500. The health plan begins to provide financial
incentives to physicians by paying a bonus of 50% of the amount by which total hospital claims
are less than 500. No bonus is paid if total claims exceed 500. Total hospital claims for the health
plan are now modeled by a new Pareto distribution with ~ 2 and ~ 2 . The expected claims
plus the expected bonus under the revised model equals expected claims under the previous
model. Calculate 2 .
A) 250
B) 300
C) 350
D) 400
E) 450
(SOA) Use the following information for Questions 13 and 14.
An insurer has excess-of-loss reinsurance on auto insurance. You are given:
(i) Total expected losses in the year 2001 are 10,000,000.
(ii) In the year 2001 individual losses have a Pareto distribution with
2
- (%) ~ 1 c ² %b
³ , %€0
(iii) Reinsurance will pay the excess of each loss over 3000.
(iv) Each year, the reinsurer is paid a ceded premium, *year , equal to 110% of the expected losses
covered by the reinsurance.
(v) Individual losses increase 5% each year due to inflation.
(vi) The frequency distribution does not change.
13. Calculate *2001 .
A) 2,200,000
B) 3,300,000
14. Calculate *2002 /*2001 .
A) 1.04
B) 1.05
C) 4,400,000
C) 1.06
D) 5,500,000
D) 1.07
E)
6,600,000
E) 1.08
15. (CAS) Let ? be the random variable representing aggregate losses for an insured. ? follows
a gamma distribution with a mean of $1 million and coefficient of variation of 1. An insurance
policy pays for aggregate losses that exceed twice the expected value of ? .
Calculate the expected loss for this policy.
A) Less than $100,000
B) At least $100,000, but less than $200,000
C) At least $200,000, but less than $300,000
) At least $300,000, but less than $400,000
E) At least $400,000
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-136
MODELING - PROBLEM SET 11
16. (CAS) High-Roller Insurance Company insures the cost of injuries to employees of ACME
Dynamite Manufacturing , Inc.
• 30% of injuries are fatal and the rest are "Permanent Total" (PT). There are no other injury
types.
• Fatal injuries follow a log-logistic distribution with ~ and ~ .
• PT injuries follow a log-logistic distribution with ~ and ~ .
• There is a $750 deductible per policy.
Calculate the probability that an injury will result in a claim to High-Roller.
A) Less than 30%
B) At least 30%, but less than 35%
C) At least 35%, but less than 40%
E) At least 40%, but less than 45%
E) At least 45%
17. (CAS May 2005) An insurance company offers two types of polices. Type 8 and type 9 .
Type 8 has non deductible, but a policy limit of 3,000-. Type 9 has no limit, but an ordinary
deductible of . Losses follow a Pareto distribution with ~ Á and ~ . Calculate the
deductible , such that both policies have the same expected cost per loss.
A) Less than 50
B) At least 50, but less than 100
C) At least 100, but less than 150
D) At least 150, but less than 200
E) 200 or more
18. (SOA) For an insurance:
À%  %  elsewhere
(ii) The insurance has an ordinary deductible of 4 per loss.
(iii) @ 7 is the claim payment per payment random variable.
Calculate ,´@ 7 µ .
A) 2.9
B) 3.0
C) 3.2
D) 3.3
E) 3.4
(i) Losses have a density function ? ²%³ ~ F
19. (SOA) Insurance agent Hunt N. Quotum will receive no annual bonus if the ratio of incurred
losses to earned premiums for his book of business is 60% or more for the year. If the
ratio is less than 60%, Hunt’s bonus will be a percentage of his earned premium equal to
15% of the difference between his ratio and 60%. Hunt’s annual earned premium is
800,000. Incurred losses are distributed according to the Pareto distribution, with ~ Á and ~ .
Calculate the expected value of Hunt’s bonus.
A) 13,000
B) 17,000
C) 24,000
D) 29,000
E) 35,000
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 11
LM-137
20. (CAS Nov 2005) In year 2005, claim amounts have the following Pareto distribution:
- ²%³ ~ c 4 %b
5
The annual inflation rate is 8%. A franchise deductible of 300 will be implemented in 2006.
Calculate the loss elimination ratio of the franchise deductible.
A) Less than 0.15
B) At least 0.15, but less than 0.20
C) At least 0.20, but less than 0.25
D) At least 0.25, but less than 0.30
E) At least 0.30
21. (SOA) The annual number of doctor visits for each individual in a family of 4 has a
geometric distribution with mean 1.5. The annual numbers of visits for the family members are
mutually independent. An insurance pays 100 per doctor visit beginning with the 4-th visit per
family. Calculate the expected payments per year for this family.
A) 320
B) 323
C) 326
D) 329
E) 332
22. (SOA May 07) You are given:
(i) The frequency distribution for the number of losses for a policy with no deductible is negative
binomial with ~ and ~ .
(ii) Loss amounts for this policy follow the Weibull distribution with ~ and ~ À.
Determine the expected number of payments when a deductible of 200 is applied.
A) Less than 5
B) At least 5, but less than 7
C) At least 7, but less than 9
D) At least 9, but less than 11
E) At least 11
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-138
MODELING - PROBLEM SET 11
MODELING - PROBLEM SET 11 SOLUTIONS
B
1. , ~ % h ²%³ % b h ²%³ % ~ % h ²%³ % b ´ c - ² ³µ ,
and similarly, , ~ % h ²%³ % b ´ c - ² ³µ . Answer: E
2. ,´²? c ³b µ ~ ²% c ³²À%³ % ~ ,
or alternatively, ,´²? c ³b µ ~ ´ c - ²%³µ % ~ ´ c À% µ % ~ ,
or alternatively, ,´²? c ³b µ ~ ,´?µ c ,´? w µ ~ c ~ À
,´²? c ³b µ ~ ²% c ³ ²À%³ % ~ , and
= ´²? c ³b µ ~ ,´²? c ³b µ c ²,´²? c ³b µ³ ~ c ² ³ ~ ~ À À
*
3. ²³ ~ ²% c ³ h
*
% ~
²*c³
*
S Z ²³ ~
*
c.
Answer: D
4. Suppose that the exponential mean is ,´?µ ~ in all years.
From the distribution table we have ,´? w µ ~ ´ c c° µ .
With deductible this year, the loss elimination ratio this year is
,´?wµ
´cc° µ
À ~ ,´?µ ~
~ c c° . It follows that c° ~ À .
In the coming year the deductible is ° but the distribution of ? remains the same, so that the
loss elimination ratio in the coming year is
,´?w²°³µ
,´?µ
~
´cc° µ
~ c c° ~ c ²c° ³° ~ c ²À³° ~ À .
Answer: C
,´?wµ
5. With a basic loss random variable ? and deductible , the loss elimination ratio is ,´?µ À
The mean of ? is ,´?µ ~ , and ,´? w µ ~ ² c c° ³ ~ À (this is found
in the table of distributions)À With deductible 500 the loss elimination ratio is À
~ À À In
order for the loss elimination ratio to double we must have
,´?wµ
,´?µ
~
²cc° ³
~ À À Solving for results in ~ Á À
Answer: E
6. We are asked to find the variance of the amount paid per loss, @3 ~ ²? c ³b , which is
,´@3 µ c ²,´@3 µ³ À For an exponential distribution loss random variable ? with mean , we
know that with an ordinary deductible of , the cost per payment random variable @7 (which is
the conditional distribution of ? c O? € ) is also exponential with mean . We also know
,´²?c³ µ
that ,´@7 µ ~ ,´? c O? € µ ~ 7 ²?€³b , and
,´²?c³ µ
,´@7 µ ~ ,´²? c ³ O? € µ ~ 7 ²?€³b .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 11
LM-139
6. continued
Therefore, since ,´@7 µ ~ and ,´@7 µ ~ (since @7 is exponential with mean ), it follows
that ,´@3 µ ~ ,´²? c ³b µ ~ ,´@7 µ h 7 ²? € ³ ~ c° , and
,´@3 µ ~ ,´²? c ³b µ ~ ,´@7 µ h 7 ²? € ³ ~ c° .
With ~ and ~ , we get
= ´@3 µ ~ ,´@3 µ c ²,´@3 µ³ ~ ² ³c° c ²c° ³ ~ Á .
We can also find ,´@3 µ ~ ,´²?c³b µ in a couple of other ways:
B
B
(i) ²% c ³ h ? ²%³ % ~ ²% c ³ h c%° % , or
B
B
c%°
(ii) ´ c -? ²%³µ % ~ % .
Integral (ii) is quite simple, integral (i) can be simplified with the substitution ' ~ % c .
B
Also, ,´@3 µ ~ ,´²?c³b µ ~ ²% c ³ h c%° %
can be simplified using the substitution ' ~ % c .
Answer: D
7. Since all answers are ‚ , we make that assumption. Then
?  , prob. .5
²? c ³b ~ F
S ,´²? c ³b µ ~ ² c ³²À³ ~ S ~ À
c ? ‚ , prob. .5
Alternatively, we have ,´²? c ³b µ ~ ,´?µ ~ ÀÁ ,´²? c ³b µ ~ À Á ,´²? c ³b µ ~ .
Since ? has no density between 3 and 12, ,´²? c ³b µ is a linear function of for   .
c
Therefore, ,´²? c ³b µ ~ À c c
²À c ³ Á and if ,´²? c ³b µ ~ , then
c
À c c
²À c ³ ~ Á so that ~ .
Answer: D
8. ²% c ³² ³ % b À
~ À ~ ² b ³ S ~ À .
9. Mean excess loss with deductible of is
Answer: D
,´?µc,´?wµ
c- ²³
We are given that - ²³ ~ , so that all losses are  1000 .
Therefore ,´?µ ~ ,´? w µ ~ . Then,
,´?µc,´?wµ
c- ²³
~ c
cÀ ~ .
Answer: B
,´?wµ
10. The loss elimination ratio is ,´?µ , where is the deductible.
À
 %  We see that ²%³ ~ H À² c %³  %  .
otherwise
,´?µ ~ % ²%³ % ~ À% b À²% c % ³ % ~ À
.
,´? w µ ~ % ²%³ % b 7 ´? € µ ~ À% b ² c 7 ´?  µ³
~ À% b ´ c À %µ ~ .
The loss elimination ratio with a deductible of 20 is À
~ À .
Answer: E
"
11. ,´? w "µ ~ % h ²%³ % b "´ c - ²"³µ Á ,´? w µ ~ % h ²%³ % b ´ c - ²³µ À
"
,´? w "µ c ,´? w µ ~ % h ²%³ % b "´ c - ²"³µ c ´ c - ²³µ À
" % h ²%³ % ~ ²,´? w "µ c "´ c - ²"³µ³ c ²,´? w µ c ´ c - ²³µ³ . Answer: D
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-140
MODELING - PROBLEM SET 11
12. We denote the bonus by ) and the revised hospital claims by ? . Then,
?
?  ) ~ %¸ c ?Á ¹ ~ h ´ c F
µ ~ h ´ c ? w µ
? € Since ? has a Pareto distribution with parameters ~ and ~ 2 , we have ,´?µ ~ ~ 2 ,
2
c
and ,´? w µ ~ c
´ c ² b
µ ~ ´ c ² b
³
³µ ~ 2´ c ² b2 ³µ .
Then, ,´? b )µ ~ ,´?µ b c ,´? w µ
2
~ 2 b c h 2´ c ² b2
³µ .
The expected claims of the previous model (Pareto with ~ and ~ ) is 500.
2
2
Then, 2 b c h 2´ c ² b2
³µ ~ S h 2 b b2
~ S 2 b 2 ~ Á b 2 S 2 ~ Á S 2 ~ . Answer: C
13. The individual loss is Pareto with ~ Á ~ . The expected individual loss is
,´?µ ~ ~ . Since total expected losses are 10,000,000 for year 2001, it follows that the
ÁÁ
expected number of losses is
~ .
Expected payment per individual loss by reinsurer is
,´²? c ³b µ ~ ,´?µ c ,´? w µ ~ c ´ c ² b
³µ ~ .
Ceded premium per policy is ²À³ ~ . Total ceded premium for 5000 policies is
²³²³ ~ Á Á ~ * .
Answer: C
14. Expected loss per policy in 2002 is ,´À?µ ~ ²À³²³ ~ À
Expected reinsurer payment per policy in 2002 is
,´²À? c ³b µ ~ ,´À?µ c ,´À? w µ ~ À²,´?µ c ,´? w À µ³ .
Since ,´? w À µ ~ ´ c ² À b
³µ ~ À , the expected reinsurer
payment per policy in 2002 is ˫ c ˵ ~ .
The ceded premium in 2002 is ²³²À³²
³ ~ Á Á ~ * .
ÁÁ
* °* ~ ÁÁ ~ À .
Answer: E
15. We will use units of $1,000,000 . The mean and variance of the gamma distribution with
= ´?µ
parameters and are and . The square of the coefficient of variation is ²,´?µ³ .
We are given that ~ and ²
³ ~ . Therefore ~ and ~ . Since ~ , ? has an
exponential distribution with mean ~ . The insurance pays for losses in excess of 2 (twice the
expected value); this is ²? c ³b . The expected loss for the policy is
B
B
,´²? c ³b µ ~ ´ c -? ²%³µ % ~ c% % ~ c ~ À (in millions),
for an expected insurance loss of 135,335 .
Answer: B
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 11
LM-141
16. The cost ? of an injury is a mixture of the two log-logistic distributions with mixing weight
.3 for "Fatal" injury cost @ and weight .7 for "Permanent Total" injury cost @ . An injury will
results in a claim if the loss is over the deductible of 750. This probability is
7 ´? € µ ~ ²À³7 ´@ € µ b ²À³7 ´@ € µ
²°³
²°
³
~ ²À³´ c b²°³ µ b ²À³´ c b²°
³ µ ~ À .
²%°³
We use the log-logistic distribution function - ²%³ ~ b²%°³ from the table of distributions.
Answer: B
17. The expected cost per loss for a Type Q policy is
,´? w µ ~ c ´ c ² b ³ µ ~ (,´? w %µ is given in the distribution table for
the Pareto). The expected cost per loss for a Type R policy is
,´²? c ³b µ ~ ,´?µ c ,´? w µ ~ c c c ´ c ² b ³ µ .
If this is also 840, then c c c ´ c ² b ³ µ ~ , so that ² b ³ ~ À , and
then ~ À .
Answer: D
,´²?c³ µ
18. ,´@ 7 µ ~ ,´? c O? € µ ~ 7 ²?€³b À
,´²? c ³b µ ~ ²% c ³²À%³ % ~ À , 7 ²? € ³ ~ À% % ~ À .
!
Alternatively, we can find - ²!³ ~ À% % ~ À! ,
and ,´²? c ³b µ ~ ´ c - ²!³µ ! ~ ² c À! ³ ! ~ À .
,´@ 7 µ ~ À
Answer: E
À ~ À .
3
3
19. The ratio of losses to premium is Á
. If Á
 .6, then the bonus is
3
²À³²À
c Á
³²Á ³ ~ ²À³²Á c 3³ . This bonus can be written
Á c 3 3  Á , which is equal to
3 ‚ Á ) ~ ²À³ h ´Á c ²3 w Á ³µ .
,´)µ ~ ²À³ h ´Á c ,²3 w Á ³µ À
Since 3 has a Pareto distribution, using the table of distributions for Exam C, we have
Á
Á
,´3 w Á µ ~
h ´ c ² ÁbÁ ³ µ ~ Á .
as ) ~ ²À³F
Then ,´)µ ~ ²À³²Á c Á ³ ~ Á .
© ACTEX 2009
Answer: E
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-142
MODELING - PROBLEM SET 11
20. The claim amount random variable in 2005, say ? , has a Pareto distribution with ~ and
~ . The claim amount random variable in 2006 is @ ~ À? . Since the Pareto is a scale
distribution, @ will also have a Pareto distribution, but with ~ and Z ~ À ~ .
Expected cost per loss after deductible
The loss elimination ratio is c Expected cost per loss without deductible .
Z
In 2006, the denominator is ,´@ µ ~ c ~ c
~ À
The numerator is the expected cost per loss with a franchise deductible of 300.
For a franchise deductible , if a loss is above , then the insurer pays the full loss @ (note
@ c ). The expected cost per loss with a franchise deductible is equal to
,´²@ c ³b µ b h ´ c -@ ²³µ .
Note that it is always true that ,´²@ c ³b µ ~ ,´@ µ c ,´@ w µ .
20. continued
Since @ is Pareto with ~ and Z ~ , from the table of distributions, we get
,´@ w µ ~ h ´ c ² b
³ µ ~ .
The expected cost per loss with the franchise deductible of 300 is
,´²@ c ³b µ b h ´ c -@ ²³µ ~ ,´@ µ c ,´@ w ´ b h ´ c -@ ²³µ
~ c b ² b
³ ~ .
The loss elimination ratio is c ~ À
.
Answer: B
21. The number of visits by family members in a year is @ ~ ? b ? b ? b ? , where each
? has a geometric distribution with mean ~ À. The sum of 4 independent geometric
distributions with the same is negative binomial with ~ and ~ À. The number of
visits resulting in insurance payments is ²@ c ³b ~ @ c ²@ w ³ , so the expected number of
visits resulting in insurance payments is ,´@ µ c ,´@ w µ ~ ²³²À³ c ,´@ w µ .
,´@ w µ ~ 7 ²@ ~ ³ b 7 ²@ ~ ³ b 7 ² ‚ ³ .
8 continued
For the negative binomial with ~ and ~ À, we have
²³²À³
7 ²@ ~ ³ ~ ²À³
~ À
Á 7 ²@ ~ ³ ~ ²À³b ~ À
Á
²³²³²À³
7 ²@ ~ ³ ~ ²À³b ~ À
Á
and 7 ²@ ‚ ³ ~ c 7 ²@ ~ Á Á ³ ~ À .
,´@ w µ ~ ²À
³ b ²À
³ b ²À³ ~ À .
The expected number of visits resulting in insurance payments is
,´@ µ c ,´@ w µ ~ c À ~ À .
With a cost per insured visit of 100, the total expected insurance payment for the year is 329.
Answer: D
22. The expected number of losses from the negative binomial frequency distribution
is ~ . The probability that a loss is above the deductible is
À
7 ²? € ³ ~ c²°³ ~ c²°³ ~ À .
Out of the 15 losses that occur on average, the expected number that are above the deductible of
200 (and result in an insurance payment being made) is ²À³ ~ À .
Answer: C
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 12 - POLICY DEDUCTIBLE (2), COST PER PAYMENT
LM-143
MODELING SECTION 12 - POLICY DEDUCTIBLE (2),
THE COST PER PAYMENT RANDOM VARIABLE
The material in this section relates to Loss Models, Section 8.2
The suggested time for this section is 2 hours.
LM-12.1 Ordinary Policy Deductible and Cost Per Payment
In Section 11, with an ordinary deductible , we considered the cost per loss, which includes the
0 amounts paid by the insurer that result from losses below a deductible. When an insurance
policy has a deductible, the insurer might only be interested in looking at the losses that resulted
in a payment, and the actual amount paid by the insurance. The insurer might ignore the losses
below , since they do not result in an insurance payment being made (the insurer might not be
aware of the losses below the deductible since the policyholder might not contact the insurer if no
insurance payment will be made).
For example, suppose a policy has a deductible of 1000, and the following eight losses occurred:
2500 , 800 , 1200 , 3000 , 500 , 5000 , 3500 , 250 . The actual amounts paid by the insurer would
be 1500 , 0 , 200 , 2000 , 0 , 4000 , 2500 , 0 . The losses below 1000 would likely not even be
reported to the insurer, so the insurer's payment record would only show that 5 payments were
made, and might not show how many other losses there were below the deductible of 1000. The
insurer's record would only show payments of 1500 , 200 , 2000 , 4000 , 2500 , with no other
information about losses below 1000. What the insurer has is information about the distribution
of ? c , the amount paid by the insurer. Since we only have data for those losses that are
? € , we really only have information about the conditional distribution of ? c given that ? € . This is what is meant by the cost per payment random variable, @7 .
For losses that are above the deductible, the cost per payment is equal numerically to the cost per
loss, it is the amount paid by the insurer. But there is a meaningful difference between "cost per
loss" and "cost per payment", since not all losses result in a payment. The cost per payment @7
is the conditional distribution of the cost per loss @3 given that ? € ;
@7 ~ @3 O? € ~ ²? c ³b O? € ~ ²? c ³O? € .
(12.1)
Example LM12-1: As an illustration of the cost per payment random variable and its distinction
from the cost per loss, let us suppose that ? is a discrete 6 point random variable based on the
toss of a fair die. The probability function is
?¢
²%³ ¢
Suppose we apply a deductible of 2. The cost per loss random variable will be
@3 ~ ²? c ³b ¢
probability ¢
The cost per payment random variable @7 is the conditional distribution of ²? c ³b given that
? € . Probabilities for @7 are found using the usual definition of conditional probability. Note
that @7 will not be 0, it will only be 1, 2, 3 or 4.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-144
MODELING SECTION 12 - POLICY DEDUCTIBLE (2), COST PER PAYMENT
Example LM12-1 continued
°
For instance, 7 ²@7 ~ ³ ~ 7 ²@3 ~ O? € ³ ~ 7 ²? ~ O? € ³ ~ °
~ .
Because of the uniform probability in this example, the other probabilities for @7 are also each
@7 ~ @3 O? € ~ ²? c ³b O? € ¢
probability ¢
:
It is always the case that cost per payment @7 is the € part of cost per loss @3 , with conditional
probabilities.
This "cost per payment" random variable @7 is also called the excess loss random variable, and
it may be referred to as the left truncated and shifted variable.
@3 and @7 do not have the same pdf and cdf. The pdf and cdf of the cost per payment @7 are the
conditional density and distribution function of the cost per loss @3 given that ? € . Note that if
the payment is & € after deductible is applied, then the original loss is % ~ & b . The
denominator in the conditional density is the probability of the conditional event,
7 ²? € ³ ~ c -? ²³,
pdf and cdf of @3 ¢ @3 ²&³ ~ D
-? ²³
&~
- ²³
&~
, -@3 ²&³ ~ D ?
? ²& b ³ & € -? ²& b ³ & € ²&b³
?
pdf and cdf of @7 ¢ @7 ²&³ ~ cfor & € , -@7 ²&³ ~
? ²³
-? ²&b³c-? ²³
c-? ²³
(12.2)
for & € (12.3)
The pdf and cdf of cost per loss and cost per payment have rarely appeared on exam questions.
LM-12.2 Expected Cost Per Payment
Most of the Exam questions related to policy limits and deductibles involve determination of the
expected insurer payment, and some refer to variance. When there is a policy deductible, we
must be careful to distinguish between the expected cost per loss and the expected cost per
payment. The Loss Models book also identifies the density functions of the cost per loss and cost
per payment random variables under the various policy adjustments. Exam C questions have
generally not focused on those density functions.
We have seen that with a deductible of amount applied to a loss variable ? , the cost per
payment random variable is the conditional distribution of ²? c ³b given than ? € . The
expected cost per payment is the conditional expectation ,´? c O? € µ .
(12.4)
The expected cost per payment is also referred to as the mean excess loss, or the mean residual
loss or mean residual lifetime, and it may be denoted ²³ or ? ²³. You should be prepared
to see this terminology and notation. Exam questions might use the terminology "payment per
payment" instead of "cost per payment"
The expected cost per payment is the average amount paid by the insurer given that the insurer
had to make a payment. The algebraic representation of expected cost per payment is summarized
below.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 12 - POLICY DEDUCTIBLE (2), COST PER PAYMENT
LM-145
Expected Cost Per Payment With Deductible Expected cost per payment ~ ,* 7
~ ,´? c O? € µ ~ ,´@7 µ ~ ²³
~
~
~
,´@3 µ
7 ²?€³
~
,´²?c³b µ
c-? ²³
B ²%c³h ²%³ %
c- ²³
²% c³h²% ³
% €
c- ²³
~
,´?µc,´?wµ
c- ²³
~
B ²%c³h ²%³ %
B ²%³ %
(continuous ? )
(12.5)
²% c³h²% ³
~
% €
(discrete ? )
²% ³
(12.6)
% €
It is possible to show that for any loss variable ? € (continuous or discrete)
,´? c O? € µ ~
B ´c- ²%³µ %
c- ²³
~
B :²%³ %
:²³
Example LM12-2: The continuous random variable ? has pdf ²%³ ~ D
(12.7)
À%
 %  .
otherwise
Find
(i) the mean and variance of ? w ,
(ii) the mean and variance of the left-censored and shifted variable ²? c ³b ,
(iii) the mean and variance of the excess loss random variable ? c given ? € ,
Solution:
(i) The limited expected value with limit " ~ is
,´²?Á ³µ ~ ,´? w µ ~ % ²À%³ % b ´ c - ²³µ
~ b ´ c Àµ ~ ~ À ,
or alternatively, ,´? w µ ~ ´ c - ²%³µ % ~ ´ c À% µ % ~ À
,´²? w ³ µ ~ % ²À%³ % b ´ c - ²³µ ~ À b ´ c Àµ ~ À
= ´? w µ ~ ,´²? w ³ µ c ²,´? w µ³ ~ À c ² ³ ~ ~ À .
(ii) ,´²? c ³b µ ~ ²% c ³²À%³ % ~ ,
or alternatively, ,´²? c ³b µ ~ ´ c - ²%³µ % ~ ´ c À% µ % ~ ,
or alternatively, ,´²? c ³b µ ~ ,´?µ c ,´? w µ ~ c ~ À
,´²? c ³b µ ~ ²% c ³ ²À%³ % ~ , and
= ´²? c ³b µ ~ ,´²? c ³b µ c ²,´²? c ³b µ³ ~ c ² ³ ~ ~ À À
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-146
MODELING SECTION 12 - POLICY DEDUCTIBLE (2), COST PER PAYMENT
Example LM12-2 continued
,´²?c³b µ
°
c- ²³ ~ cÀ
,´²?c³b µ
°
c- ²³ ~ cÀ ~ .
c ² ³ ~ ~ ÀÀ
(iii) ²³ ~ ,´? c O? € µ ~
,´²? c ³ O? € µ ~
= ´? c O? € µ ~
~ .
Example LM12-3: Find the pdf, cdf and expected value of the cost per payment random
variable when there is a policy deductible of in each of the following cases.
(a) ? is exponential with mean .
(b) ? has a uniform distribution on ²Á ³ , where  .
²&b³
?
Solution: (a) @ 7 ²&³ ~ c~
? ²³
-@7 ²&³ ~
,´@7 µ ~
c²&b³°
c°
~ c&° ,
-? ²&b³c-? ²³
²cc²&b³° ³c²cc° ³
~
c-? ²³
c°
,´@3 µ
,´?µc,´?wµ
c²cc° ³
~
~
c-? ²³
c-? ²³
c°
~ c c&° ,
c°
~ c° ~ .
Note that the cost per payment random variable @7 has the same distribution as ? (this is unique
to the exponential distribution, due to the "lack-of-memory" property it satisfies).
²&b³
?
(b) @ 7 ²&³ ~ c~ ~ c
.
c ? ²³
-@7 ²&³ ~
&b c
c cc c
c -? ²&b³c-? ²³
c-? ²³
,´@ µ
~
&
~ c , for  &  c .
,´@7 µ ~ c- 3²³ ~
~ c
2 .
?
Note that the cost per payment random variable has a uniform distribution on the
interval ² Á c ³.
Example LM12-4: A jewelry store has obtained two separate insurance policies that together
provide full coverage. You are given:
(i) The average ground-up loss is 11,100.
(ii) Policy A has an ordinary deductible of 5,000 with no policy limit.
(iii) Under policy A, the expected amount paid per loss is 6,500.
(iv) Under policy A, the expected amount paid per payment is 10,000.
(v) Policy B has no deductible and a policy limit of 5,000.
Given that a loss has occurred, determine the probability that the payment under policy B is
5,000.
Solution: ? is the ground-up loss random variable. The average ground-up loss is given to be
11,100, so that ,´?µ ~ Á . With a deductible of 5,000 on policy A, the expected amount
paid per loss on policy A is given to be 6,500, so that ,´?µ c ,´? w µ ~ Á , and the
expected amount paid per payment is given to be 10,000 so that
,´?µc,´?wµ
c-? ²³
~ Á .
From this given information we get ,´? w µ ~ Á c Á ~ Á , and
-? ²³ ~ c
,´?µc,´?wµ
Á
Á
~ c Á ~ À .
Under Policy B with a limit of 5,000, the payment will be 5,000 if the ground-up loss is at least
5000, and the probability of that is 7 ´? ‚ µ ~ c -? ²³ ~ À
.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 12 - POLICY DEDUCTIBLE (2), COST PER PAYMENT
LM-147
The mean residual lifetime
It was noted above that the expected cost per payment is also called the mean excess loss, and is
also called the mean residual lifetime. In life contingencies, the complete whole life expectation
for someone at age ²'³ is
B
B
B ²'b!³
'B ²$³ $ ~ ' ´c-? ²$³µ $ .
° ' ~ ! ' ! ~ !
~
²'³
²'³
c- ²'³
(12.8)
?
In the final expression on the right, ? is age at death. We see that complete whole life
expectation, or the mean residual lifetime is the same algebraically as a mean excess loss or
expected cost per payment, if we interpret ? as age at death instead of size of loss. In other
words, ° ' is the expected number of years that ²'³ will live beyond age ' , given alive at age ' .
Therefore,
,´?µc,´?wµ
,´²?c³b µ
,´? c O? € µ ~ ,´²? c ³b O? € µ ~
~
~ ²³ ~ ° ~
B ²!c³ ? ²!³ !
c-? ²³
~
B ´c-? ²!³µ !
c-? ²³
~
c- ²³
B :? ²!³ !
:? ²³ .
c- ²³
(12.9)
LM-12.3 Variance of Cost Per Payment With Deductible The variance of the cost per payment is
= ´? c O? € µ ~ = ´@7 µ ~ ,´@7 µ c ²,´@7 µ³
~ ,´²? c ³ O? € µ c ²,´? c O? € µ³
(12.10)
,´²?c³ µ
,µ@ µ
3
It was noted earlier in this section that ,´@7 µ ~ ,´? c O? € µ ~ c- ²³b ~ 7 ²?€³
À
?
,´²?c³ µ
,µ@ µ
3
It is also true that ,´@7 µ ~ ,´²? c ³ O? € µ ~ c- ²³b ~ 7 ²?€³
.
?
(12.11)
We can find ,´@3 µ ~ ,´²? c ³b µ a couple of ways:
,´@3 µ ~ ,´²? c ³b µ
B
~ ²% c ³ h ? ²%³ %
~ ,´? µ c ,´²? w ³ µ c ²,´?µ c ,´? w µ³ .
(12.13)
This last expression allows us to make use of the limited loss moment formulations that are given
in the Exam C table. Actually, for the exponential, uniform and Pareto distributions, a little later
we will see a shortcut for finding ,´@3 µ that avoids any integration.
Example LM12-5: The ground up loss ? has a lognormal distribution with parameters ~ .
and ~ À
, and there is an ordinary policy deductible of . Find the mean and variance of @3
and @7 .
Solution: ,´²? c ³b µ cannot be found directly by integration. We use
,´²? c ³b µ ~ ,´?µ c ,´? w µ and the Exam C table formulation for ,´? w µ for the
lognormal distribution.
,´? w µ ~ b h )²
c c ³ b ´ c )² c ³µ
.
With the given parameter values, this becomes
,´? w µ ~ À
h )² c À³ b ´ c )² c À³µ ~ À .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-148
MODELING SECTION 12 - POLICY DEDUCTIBLE (2), COST PER PAYMENT
Example LM12-5 continued
For the lognormal ? , we have ,´?µ ~ b ~ À
, so that
,´@3 µ ~ ,´²? c ³b µ ~ À
c À ~ À .
,´@ µ
À
3
The mean of @7 is ,´@7 µ ~ c- ²³
~
~ À
~ À
.
?
c)² c ³
To find = ´@3 µ we first find
,´@3 µ ~ ,´²? c ³b µ ~ ,´? µ c ,´²? w ³ µ c ´ ,´?µ c ,´? w µ µ .
For the lognormal distribution ? , from the Table we have ,´? µ ~ b ~ Á , and
c c ,´²? w ³ µ ~ b h )²
³ b ´ c )²
~ À
h )² c À³ b ´ c )² c À³µ ~ À .
c ³µ
Then, ,´@3 µ ~ Á À c À c ´À
c Àµ ~ Á À ,
and = ´@3 µ ~ Á À c À ~ À À
,´@ µ
Á
We find ,´@7 µ from ,´@7 µ ~ c- 3²³ ~
~ Á ,
?
c)² c ³
and then = ´@7 µ ~ Á c ²À
³ ~ .
LM-12.4 Franchise Deductible The franchise deductible was introduced in Section 11. With franchise deductible applied to
if ?  loss variable ? , the amount paid by the insurer is F
.
? if ? € The expected cost per loss is ,*3 ~ ,´²? c ³b µ b ´ c -? ²³µ .
The expected cost per payment is the conditional expectation,
3
,* 3
,* 7 ~ 7 ,*
²?€³ ~ c-? ²³ ~
,´²?c³b µb´c-? ²³µ
c-? ²³
,´²?c³ µ
~ c- ²³b b ?
(12.14)
Note the expected cost per payment is just larger than for an ordinary deductible.
Example LM12-6: For franchise deductible find the expected value of (i) the cost per loss
random variable and (ii) the cost per payment random variable in each of the following cases.
(a) ? is exponential with mean .
(b) ? has a uniform distribution on ²Á ³ , where  .
Solution: (a)(i) For franchise deductible the expected cost per loss is
,´²? c ³b µ b ´ c -? ²³µ ~ c° b c° ~ ² b ³c° À
,´²?c³ µ
(ii) The expected cost per payment is c- ²³b b ~ b ?
(b)(i) The expected cost per loss is
(ii) The expected cost per payment
© ACTEX 2009
²c³
b ´ c µ ~ c
b
is c
2 b ~ 2 À
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 12
LM-149
MODELING - PROBLEM SET 12
Policy Deductible (2), Cost Per Payment - Section 12
1. ? has a uniform distribution on the interval ´Á µ. An ordinary deductible of is applied.
(a) Find if ,´@3 µ ~ .
(b) Find if ,´@7 µ ~ .
2. You are given the following information about a Pareto loss distribution:
• €
• expected cost per loss with deductible is 1105
• expected cost per payment with deductible is 1778
• loss elimination ratio for deductible is .2633
Find the expected cost per payment if the deductible is doubled to .
A) Less than 1500
B) at least 1500 but less than 1700
C) at least 1700 but less than 1900
D) at least 1900 but less than 2100 E) at least 2100
3. Ground up losses are modeled according to an exponential distributions with a mean of 100.
An insurer considers the following two polices.
Policy 1 has no limit but has a deductible of 20.
Policy 2 has no limit but has a deductible of 50.
For each of the two policies the insurer calculates the coefficient of variation for the cost per
payment, say (coefficient of variation for Policy 1) and . By what percentage of is below ?
A) 40%
B) 30%
C) 0%
D) c 30%
E) c 40%
4. A jewelry store has obtained two separate insurance policies that together provide full
coverage. You are given:
(i) The average ground-up loss is 11,100.
(ii) Policy A has an ordinary deductible of 5,000 with no policy limit.
(iii) Under policy A, the expected amount paid per loss is 6,500.
(iv) Under policy A, the expected cost per payment is 10,000.
(v) Policy B has no deductible and a policy limit of 5,000.
Given that a loss less than or equal to 5,000 has occurred, what is the expected cost per payment
under policy B?
A) Less than 2,500
B) At least 2,500, but less than 3,000
C) At least 3,000, but less than 3,500
D) At least 3,500, but less than 4,000
E) At least 4,000
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-150
MODELING - PROBLEM SET 12
5. A loss random variable ? has the following characteristics. There is a 90% chance of no loss
occurring, ? ~ , and there is a 10% that a positive loss occurs. If a positive loss occurs, it is
uniformly distributed between 1000 and 5000. An insurance policy on this loss has an ordinary
deductible of 2000 applied. Find the expected cost per loss (including when ? ~ ) and the
expected cost per payment for this policy. Find the variance of the cost per loss.
6. (CAS) Auto liability losses for a group of insureds (Group R) follow a Pareto distribution with
~ and ~ . Losses from a second group (Group S) follow a Pareto distribution with
~ and ~ . Group R has an ordinary deductible of 500, Group S has a franchise
deductible of 200. Calculate the amount that the expected cost per payment for Group S exceeds
that for Group R.
A) Less than 350
B) At least 350, but less than 650
C) At least 650, but less than 950
D) At least 950, but less than 1,250
E) At least 1,250
7. A loss random variable is exponentially distributed with mean . A franchise deductible is
chosen to be applied to the loss so that the expected cost per loss is 75% of the expected cost per
payment. If the deductible is doubled to , what is the new expected cost per loss as a
percentage of the new expected cost per payment?
A) 12.5%
B) 25%
C) 37.5%
D) 50%
E) 56.25%
8. Losses in 2003 follow a two-parameter Pareto distribution with ~ and ~ .
Losses in 2004 are uniformly 20% higher than in 2003. An insurance covers each loss subject to
an ordinary deductible of 10. Calculate the Loss Elimination Ratio in 2004.
A) 5/9
B) 5/8
C) 2/3
D) 3/4
E) 4/5
9. (CAS May 2005) Well-Traveled Insurance Company sells a travel insurance policy that
reimburses travelers for any expense incurred for a planned vacation that is cancelled because of
airline bankruptcies. Individual claims follow a Pareto distribution with ~ and ~ .
Because of financial difficulties in the airline industry, Well-Traveled imposes a limit of $1,000
on each claim. If a policyholder's planned vacation is cancelled due to airline bankruptcies and he
or she has incurred more than $1,000 in expenses, what is the expected non-reimbursed amount
of the claim?
A) Less than $500
B) At least $500, but less than $1,000
C) At least $1,000, but less than $1,500
D) At least $1,500, but less than $2,000
E) $2,000 or more
10. An insurer notices that for a particular class of policies, whenever the claim amount is over
1000, the average amount by which the claim exceeds 1000 is 500. The insurer assumes that the
claim amount distribution has a uniform distribution on the interval ´ Á µ , where € .
Find the value of that is consistent with the observation of the insurer.
A) 1500
B) 2000
C) 2500
D) 4000
E) 5000
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 12
LM-151
11. ? is a mixture of ? and ? with mixing weights and so that
²%³ ~ h ? ²%³ b h ? ²%³ .
,*3 ~ expected cost per loss,
,*7 ~ expected cost per payment and
3,9 ~ loss elimination ratio.
Which of the following statements are true?
I. With deductible , ,*3? ~ h ,*3 b h ,*3 .
II. With deductible , ,*7? ~ h ,*7 b h ,*7 .
III. With deductible , 3,9? ~ h 3,9 b h 3,9 .
A) I only
B) I and II only
C) I and III only
D) All
E) All but I
12. (SOA) The random variables ? and @ have joint density function
&
²%Á &³ ~ c%c ,  %  B Á  &  B .
Determine the mean residual life function for the marginal distribution of ? evaluated at % ~ .
A) 1/4
B) 1/2
C) 1
D) 2
E) 4
13. (SOA) You are given the following:
- The random variable ? follows a Pareto distribution with parameters ~ Á ~ .
- The mean residual life function, ? ²³ is defined to be ,´? c O? ‚ µ.
(a) Determine the range of ? ²³ over its domain ´Á B³.
²³
(b) @ ~ À? . Determine the range of the function @ ²³ over its domain ´Á B³.
?
(c) A ~ ²?Á ³ . Determine the range of A ²³ over its domain ´Á µ.
14. Suppose that ? has a uniform distribution on the interval ´Á µ. With a deductible of ,
the mean excess loss is ²³ (this is the expected cost per payment). Find the deductible Z that
results in a mean excess loss which is one-half as big (² Z ³ ~ ²³).
A) c B) b C) c D) c E) b 15. A particular survival distribution has a mean residual lifetime at age % of c% , % ‚ À
Find the survival function of %, :²%³.
A) %c
%
B) c
%
C) %bc
%
D) c%
c
E) c%
cc%
16. (SOA) For an insurance:
(i) Losses can be 100, 200 or 300 with respective probabilities 0.2 , 0.2 , and 0.6.
(ii) The insurance has an ordinary deductible of 150 per loss.
(iii) @ 7 is the claim payment per payment random variable.
Calculate = ²@ 7 ³ .
A) 1500
B) 1875
C) 2250
D) 2625
E) 3000
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-152
MODELING - PROBLEM SET 12
17. When an ordinary deductible is applied to a loss random variable ? , we can formulate the
distribution of @7 , the cost per payment random variable. @7 is the conditional distribution of
? c given that ? € .
For each of the following distributions for ? , and for ordinary deductible € ,
we wish to investigate the relative tail weights of @7 and ? .
I. exponential with mean € II. Weibull with parameters € and € III. Pareto with parameters and For which of these distributions does @7 have lighter right tails than ? ?
18. (SOA May 07) The loss severity random variable ? follows the exponential distribution
with mean 10,000. Determine the coefficient of variation of the excess loss variable
@ ~ %²? c Á ³ .
A) 1.0
B) 3.0
C) 6.3
D) 9.0
E) 39.2
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 12
LM-153
MODELING - PROBLEM SET 12 SOLUTIONS
1.(a) ,´@3 µ ~ ,´²? c ³b µ ~ ²% c ³ h ²À³ % ~
(b) ,´@7 µ ~
,´@3 µ
7 ²?€³
~
²c³
c
,² ³
~
c
²c³
~ S ~ .
~ S ~ .
2. From the Exam C table, for the Pareto distribution, - ²%³ ~ c ² %b
³ , ,´?µ ~ c
and ,´? w %µ ~ c
´ c ² %b ³c µ .
Then ,*3 ~ ,´?µ c ,´? w µ ~ ² c
³² b ³c ~ ,
,*7 ~
,´?µc,´?wµ
c- ²³
~ b
c ~ .
,´?wµ
c
3,9 ~ ,´?µ ~ c ² b
~ À
.
³
Then, ,´? w µ ~ À
,´?µ S ,´?µ h ² c À
³ ~ S ,´?µ ~ c
~ .
b
Then, b
~ ² c ³,² c ³ ~ ~ À
.
Then, ~ ² c
³² b ³c ~ ²³²À
³c S ~ À .
b
Then, ~ b
c ~ À S b ~ , and b ~ À
S ~ and ~ .
With deductible of ~ , the expected cost per payment is
,*7 ~
,´?µc,´?wµ
c- ²³
b
~ b
~ .
c ~
Àc
Answer: D
3. If ? is exponential with parameter and @7 is the cost per payment when there is a deductible
of , then @7 also has an exponential distribution with parameter .
This is true since if the ground up loss ? has an exponential distribution with parameter ,
then the pdf of ? is ? ²%³ ~ c%° , and the cdf of ? is -? ²%³ ~ c c%° À
²&b³
c²&b³°
c&°
?
The pdf of @7 will be @7 ²&³ ~ c~ c²c
,
c° ³ ~ ? ²³
which is also exponential with parameter . Therefore for both deductibles the cost per
payment made has an exponential distribution with mean 100 and therefore both policies have the
same coefficient of variation.
Answer: C
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-154
MODELING - PROBLEM SET 12
4. Policy B pays the ground up loss ? up to a maximum payment of 5000. Policy B pays the
limited loss @ ~ ? w , which has a mixed distribution with density function
for &  ²&³
) ²&³ ~ F ?
c -? ²³ for & ~ The conditional distribution of amount paid under policy B given that a loss ? is less than or
²%³
equal to 5000 has pdf )O? ²%³ ~ - ?²³ for %  (and )O? ²%³ ~ if
?
% € , since it is impossible for ? to be greater than 5000 if we are given that ? is less than
or equal to 5000).
The expected payment under policy B given that a loss less than or equal to 5000 has occurred is
% h )O? ²%³ % ~ % h
? ²%³
-? ²³
% ~
%h? ²%³ %
-? ²³
In order to compete this calculation, we need the values of
%
.
h ? ²%³ % and -? ²³ .
We are given ,´?µ ~ Á (average ground up loss). Policy A has a deductible of 5000,
and we are given that the expected amount paid per loss under policy A is 6500.
Therefore, ,´²? c ³b µ ~ ,´?µ c ,´? w µ ~ , and it follows that
,´? w µ ~ .
We are also given that the expected cost per payment for policy A is 10,000.
Therefore, ,´@7 µ ~
,´?µc,´?wµ
7 ´?€µ
It follows that -? ²³ ~ À .
~ c-
~ Á À
? ²³
We now use the relationship ,´? w µ ~ % h ? ²%³ % b ²³´ c -? ²³µ
to get ~ % h ? ²%³ % b ²³´ c Àµ , so that % h ? ²%³ % ~ .
Finally, the conditional expectation we are trying to find (the expected cost per payment under
policy B given that a loss less than or equal to 5000 has occurred) is
%h? ²%³ %
-? ²³
~ À ~ À
Answer: D
5. The distribution of ? is a mixture of a discrete point at ? ~ , with mixing weight
7 ²? ~ ³ ~ À, and a continuous uniform distribution on then interval ´Á µ with
mixing weight .1. The pdf of ? on ´Á µ is ? ²%³ ~ ²À³²À³ ~ À
for  %  .
if ?  .
? c if ? € ,´@3 µ ~ ²% c ³ h ? ²%³ % ~ ²% c ³ h ²À³ % ~ À À
,´@3 µ ~ ²% c ³ h ²À³ % ~ Á À
= ´@3 µ ~ Á c ²À³ ~ Á À .
The cost per loss variable is @3 ~ ²? c ³b ~ D
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 12
LM-155
,´?µc,´?wµ
c-? ²³
6. The expected cost per payment for Group R is
À
For the Pareto distribution, ,´?µ ~ c
~ c ~ ,
c
c
,´? w %µ ~ c ´ c ² %b ³ µ S ,´? w µ ~ µ ~ À
c ´ c ² b ³
-? ²%³ ~ c ² %b
³ S -? ²³ ~ c ² b ³ ~ À
À
The expected cost per payment for Group R is
c
cÀ
~ Á À
For a ground up loss @ and franchise deductible , the expected cost per loss is
,´@ µ c ,´@ w µ b ´ c -@ ²³µ , and the expected cost per payment is
,´@ µc,´@ wµb´c-@ ²³µ
c-@ ²³
~
,´@ µc,´@ wµ
c-@ ²³
b.
The expected cost per payment for Group S is
,´@ µc,´@ wµ
c-@ ²³
b ~
c
µ
c c c ´c² b ³
² b ³
b ~ Á À
The amount by which the expected cost per payment for Group S exceeds that for Group R is
c ~ À
Answer: C
7. With either an ordinary deductible or a franchise deductible, the relationship between ,*3
,*3
and ,*7 is ,*7 ~ c²³ , or ,*3 ~ ´ c -? ²³µ h ,*7 .
?
We are given that ,*3 ~ À d ,*7 .
Therefore, since ,*3 ~ ´ c - ²³µ h ,*7 , it follows that
- ²³ ~ À ~ c ° S c° ~ À .
Then, if is doubled, c° ~ ²c° ³ ~ À
, and
,*3 ~ ´ c - ²³µ h ,*7 ~ ²À
³,*7 .
Answer: E
8. Using the Pareto ? from 2003, the loss in 2004 is À? . The loss elimination ratio in 2004
À,´?w À
µ
,´?w À
µ
,´À?wµ
~
~
,´À?µ
À,´?µ
,´?µ .
c
µ ~ À .
À µ ~ ² c ³´ c ² b ³
À
with an ordinary deductible of 10 is
,´?µ ~ c
~ , and ,´? w
The LER in 2004 is À
~ À
.
Answer: B
9. For a loss ? and number , it is always true that ,´?µ ~ ,´? w µ b ,´²? c ³b µ .
This indicates that the total loss can be separated into two parts, the part of the loss paid on a
policy with limit , and the part of the loss paid on a policy with deductible . Therefore, for a
policy with a limit of , the expected amount not reimbursed when a claim occurs is
,´²? c ³b µ , which is the expected cost per loss for a policy with a deductible of .
We are asked to find the expected non-reimbursed cost of the claim if the claim is over 1000.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-156
MODELING - PROBLEM SET 12
9. continued
The expected amount not reimbursed, given that the claim is greater than 1000, is the conditional
,´²?c³ µ
expectation ,´²? c ³b O? € µ ~ c- ²³b .
?
For this problem, ,´²? c ³b µ ~ ,´?µ c ,´? w µ À
The total loss is ,´?µ ~ c
~ c
~ . For the Pareto distribution, the expected payment
c
on a policy with limit " is ,´? w "µ ~ c
´ c ² "b
µ , and the distribution function is
³
- ²%³ ~ c ² %b
³ .
In this case, ,´? w µ ~ c
´ c ² b
³c µ ~ and -? ²³ ~ c ² b
³ ~ À
Therefore, the amount not reimbursed by the policy with limit 1000 is
,´²?c³b µ
c-? ²³
~
,´?µc,´?wµ
c-? ²³
~
c c ~ .
Answer: D
10. If the claim amount ? is uniform on the interval ´ Á µ , then the conditional
²%³
°
density ²%O? € ³ ~ 7 ´?€µ ~ ²c³° ~ c
, for  %  .
This conditional density is uniform on the interval ´ Á c µ , and has a mean of
c
À In order for this to be 500, we must have c
~ , so that ~ .
Answer: B
B
11. I. ,*3? ~ ,´?µ c ,´? w µ ~ :? ²%³ %
B
B
B
~ ´ h :? ²%³ b h :? ²%³µ % ~ h :? ²%³ % b b h :? ²%³ %
~ h ,*3 b h ,*3 . True.
,´?µc,´?wµ
,*3?
h,*3 b h,*3
~ h: ²³b
~ h:
:²³
?
h:? ²³
? ²³b h:? ²³
,*3
,*3
h:? ²³b h:? ²³ b h h:? ²³b h:? ²³ .
II. ,*7? ~
~ h
However, h :? ²³ b h :? ²³ is not necessarily equal to :? ²³ , so that
,*3
,*3
h: ²³b h: ²³ £ : ²³ ~ ,*7 , and similarly for ,*7 .
?
?
?
Therefore, in general ,*7? £ h ,*7 b h ,*7 À False.
,´?wµ
h,´? wµb h,´? wµ
h,´? µb h,´? µ
,´? wµ
h,´? wµ
h,´? µb h,´? µ b h h,´? µb h,´? µ
III. 3,9? ~ ,´?µ ~
~ h
.
Since it will generally be the case that ,´? µ £ h ,´? µ b h ,´? µ (and the same for ? ),
,´? wµ
h,´? µb h,´? µ
£ 3,9 (and the same for ? ), so that
3,9? £ h 3,9 b h 3,9 À False.
© ACTEX 2009
Answer: A
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 12
LM-157
12. The mean residual lifetime is the mean excess loss. The marginal distribution of ? has pdf
&
B c%c & ~ c% , so ? has an exponential distribution with mean . Because of the lack
of memory property of the exponential distribution, the mean excess loss for any deductible is
the same as the original exponential distribution loss mean, .
Alternatively, ? ²³ ~
,´?µc,´?wµ
c- ²³
~
B ´c- ²%³µ %
c- ²³
~
B c% %
c
c
~ c ~
À
Answer: B
c
13. For the Pareto distribution, ,´?µ ~ c
Á ,´? w µ ~ c
´ c ² b
µÁ
³
- ²³ ~ c ² b ³ À
(a) ? ²³ ~
,´?µc,´?wµ
c- ²³
~
c²³´c² b
³µ
² b ³
~ b . Range is ´Á B³.
&
À (b) -@ ²&³ ~ 7 ´@  &µ ~ 7 ´À?  &µ ~ -? ² À ³ ~ c ² & b ³ ~ c ² &bÀ
³ .
À
This shows that @ has a Pareto distribution with parameters À ~ and ~ .
c²³´c² b
³µ
~ b .
² b ³
b
b , which has a range of ²Á Àµ (
Therefore, @ ²³ ~
²³
Then @ ²³ ~
?
the ratio is 1.1).
(c) A ²³ ~
,´Aµc,´Awµ
c-A ²³
SB, the ratio goes to 1, and at ~ ,
À ,´Aµ ~ ,´? w µ ~ ²³´ c ² b
³µ ~ À .
For  Á ,´A w µ ~ ,´? w µ ~ ²³´ c ² b
³µ ,
and -A ²³ ~ -? ²³ ~ c ² b
³ . Therefore,
A ²³ ~
Àc²³´c² b
³µ
² b ³
~ b c À
² b
³ . On the interval   ,
the maximum and minimum are either at the endpoints, ~ Á or at a critical point.
The critical points occur where c ²
À
³²³² b
³² ³ ~ Á so that ~ À
The values of A ²³ at these points are A ²³ ~ À Á A ²³ ~ Á A ²³ ~ .
The min and max of A ²³ are 0 and 150 . The range is ´Á µ .
14. The mean excess loss with deductible is
²³ ~
,´?µc,´?wµ
c- ²³
~
cb c and with deductible Z it is ²Z ³ ~
Answer: E
²c ³²c ³
~ c ,
c Z
c ~ ² c ³ S Z ~
~
15. The mean residual lifetime of ? with deductible % is
We are given that this is c% ~
c c% ~
%B :²!³ !
:²%³
B
:²%³h´c:²%³µc´% :²!³ !µh: Z ²%³
Z
´: ²%³µ
© ACTEX 2009
%B :²!³ !
:²%³
b .
.
. If we differentiate both sides with respect to % we get
: Z ²%³
~ c c :²%³ h
B :²!³ !
%
:²%³
: Z ²%³
~ c c :²%³ h c% .
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-158
MODELING - PROBLEM SET 12
15. continued
: Z ²%³
Therefore, :²%³ ~ c c% . From the chain rule of calculus we have
: Z ²%³
%
´:²%³µ ~ :²%³ ~ c % , and therefore ´:²%³µ ~ % c % b (antiderivative of
%
% c ), and then :²%³ ~ %c b . Since :²³ ~ for any survival function, it follows that
%
:²³ ~ c b ~ , so that ~ , and :²%³ ~ %bc .
Answer: C
16. @ 7 is the conditional random variable @ c given the @ € .
The variance of @ 7 is ,´²@ c ³ O@ € µ c ²,´@ c O@ € µ³ .
,´²@ c³ µ
²À³b²À³b²À
³
~ .
ÀbÀ
,´²@ c³b µ
²À³b ²À³b ²À
³
~
~ Á 7 ²@ €³
ÀbÀ
,´@ c O@ € µ ~ 7 ²@ €³b ~
,´²@ c ³ O@ € µ ~
7
= ´@ µ ~ Á c ²³ ~ À
.
Answer: B
:@ ²%³
%¦B :? ²%³
17. We say that @ has a lighter right tail than ? is lim
~.
I. If ? has an exponential distribution with mean , then @7 also has an exponential distribution
with mean . Therefore, ? and @7 have proportional tail weights. @7 does not have a lighter right
tail than ? .
II. If ? has a Weibull distribution, then :? ²%³ ~ c²%°³ .
The survival function for @7 is
:@7 ²%³ ~ 7 ²@7 € %³ ~ 7 ²? c € %O? € ³ ~ 7 ²? € b %O? € ³
7 ²?€b%³
c´²b%³°µ
~ 7 ²?€³ ~ c²°³
:@7 ²%³
:? ²%³
.
c´²b%³°µ
c²°³
c´²b%³°µ
c ´²b%³ c
~ ,c²%°³ ~ c²°
³ c²%°³ ~ Since € , it follows that ² b %³ c c % S B as % S B .
Then
c% µ
.
:@ ²%³
Therefore, lim : 7²%³ ~ , so @7 does have a lighter right tail than ? .
%¦B ?
III. If ? is Pareto with parameters and , then :? ²%³ ~ ² %b
³ .
Also, in for this ? , the distribution of @7 is also Pareto, but with parameters and b . Therefore, the survival function for @7 is :@7 ²%³ ~ ² %bb
b ³ .
Then
and
:@7 ²%³
%b b b :? ²%³ ~ ² %bb ³ ,² %b ³ ~ ² ³ d ² %bb ³ ,
:@ ²%³
lim : 7²%³ ~ ² b
³ . Since this is between and B, ? and @7
%¦B ?
have
proportional right tail weights.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 12
LM-159
18. The coefficient of variation of @ is
j= ²@ ³
,²@ ³
.
@ ~ ? c ²? w ³ , so ,²@ ³ ~ ,²?³ c ,²? w ³ .
For an exponential random variable > with mean , ,²> w "³ ~ ² c c"° ³ .
It follows that ,²? w ³ ~ Á ² c cÁ°Á ³ ~ .
Then, ,²@ ³ ~ Á c ~ .
= ²@ ³ ~ ,²@ ³ c ´,²@ ³µ .
B
,²@ ³ ~ ²% c ³ h Á
c%°Á % .
Using the change of variable ' ~ % c Á , this integral becomes
B ' h c²'bÁ³°Á ' ~ c h B ' h c'°Á '
Á
Á
~ c h ² d Á ³ ~ Á Á (the integral is the 2nd moment of an exponential random
variable with mean 10,000).
Then, = ²@ ³ ~ Á Á c ²³ ~ Á Á .
jÁÁ
The coefficient of variation of @ is
~ À
.
Note that we can use the following approach to get ,²@ ³ .
If > has an exponential distribution with mean , then he conditional distribution of > c given that > € also has an exponential distribution with mean .
Therefore, ,´²> c ³ O> € µ ~ . But it is also true that
,´²> c³ µ
,´²> c ³ O> € µ ~ 7 ²> €³b , so that
,´²> c ³b µ ~ ,´²> c ³ O> € µ h 7 ²> € ³ ~ ² ³²c° ³ .
Applying this to ? , and @ ~ ²? c ³b , we see that
,²@ ³ ~ ,´²? c ³b µ ~ ,´²? c ³ O? € µ h 7 ²? € ³
~ ² d Á ³²cÁ°Á ³ ~ Á Á , as before.
Answer: C
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-160
© ACTEX 2009
MODELING - PROBLEM SET 12
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 13 - THE UNIFORM, EXPONENTIAL AND PARETO
LM-161
MODELING SECTION 13 - POLICY DEDUCTIBLES APPLIED TO
THE UNIFORM, EXPONENTIAL AND PARETO DISTRIBUTIONS
The material in this section relates to Loss Models, Section 8.2
The suggested time for this section is 2 hours.
The moments of @3 and @7
Suppose that is an ordinary deductible applied to a loss random variable ? . For many of the
distributions in the Exam C Table, we are given ,´? w µ , so that it is fairly mechanical to
find the expected cost per loss, ,´@3 µ ~ ,´²? c ³b µ ~ ,´?µ c ,´? w µ .
(13.1)
Finding ,´@3 µ ~ ,´²? c ³b µ may not be quite so straightforward, depending on the
distribution of ? . We have seen in Section 12 that
(13.2)
,´²? c ³b µ ~ ,´? µ c ,´²? w ³ µ c ´ ,´?µ c ,´? w µ µ ,
so this is a way to find ,´@3 µ ~ ,´²? c ³b µ .
Note that it is not true that ,´²? c ³b µ ~ ,´? µ c ,´²? w ³ µ .
The variance of the cost per loss is
= ´@3 µ ~ = ´²? c ³b µ ~ ,´²? c ³b µ c ²,´²? c ³b µ³ ~ ,´@3 µ c ²,´@3 µ³ .
(13.3)
The cost per payment random variable @7 is the conditional distribution of @3 or ? c (payment amount) given that ? € (given that a payment is made). We have seen that the
expected value of the cost per payment (also called the mean excess loss or the mean residual
lifetime) is
,´²?c³ µ
,´@ µ
,´@7 µ ~ ,´? c O? € µ ~ c- ²³b ~ c- 3²³ ~
?
?
Expected cost per loss
7 ²?€³
.
(13.4)
It is also true that the second moment of the cost per payment can be formulated in terms of the
second moment of the cost per loss:
,´²?c³ µ
,´@ µ
,´@7 µ ~ ,´²? c ³ O? € µ ~ c- ²³b ~ c- 3²³ .
?
?
(13.5)
Rewriting these expressions, we get
(13.6)
,´@3 µ ~ ,´²? c ³b µ ~ ,´? c O? € µ h ´ c -? ²³µ ~ ,´@7 µ h ´ c -? ²³µ ,
and
(13.7)
,´@3 µ ~ ,´²? c ³b µ ~ ,´²? c ³ O? € µ h ´ c -? ²³µ ~ ,´@7 µ h ´ c -? ²³µ .
These can be useful simplifying relationships (particularly the second moment relationship) for
some specific loss distributions when we are trying to find the variance of the cost per loss, @3 .
The loss distributions for which this is a useful approach are the uniform, exponential and
Pareto.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-162
MODELING SECTION 13 - THE UNIFORM, EXPONENTIAL AND PARETO
The Distribution of Cost Per Payment @7 for the Uniform, Exponential
and Pareto Distributions
If ? is a loss random variable and is an ordinary deductible, we have seen in Section 12 that
²&b³
?
the pdf of @7 , the cost per payment random variable is @7 ²&³ ~ cfor & € .
? ²³
We apply this to the uniform, exponential and Pareto distributions.
(13.8)
Uniform Distribution on ´Á µ
Suppose that ? has a uniform distribution on ´Á µ and an ordinary deductible is applied.
The pdf of ? is ? ²!³ ~ for  !  , and the cdf is -? ²!³ ~ ! for  !  .
The mean of ? is , the second moment is , and the variance is .
The pdf of @7 , cost per payment is
²&b³
°
?
@7 ²&³ ~ c~
~ c
for  & b  .
c ? ²³
The inequality  & b  is the same as  &  c (actually it is c  &  , but & cannot be negative).
(13.9)
We see that the pdf of @7 is the pdf of a uniform distribution on the interval ´Á c µ ,
and therefore, @7 has that distribution.
²c³
Then ,´@7 µ ~ c
,
and
,´@
µ
~
,
²c³
and the variance of @7 is = ´@7 µ ~ .
(13.10)
From Equation 13.6, we get the expected cost per loss
²c³
,´@3 µ ~ ,´@7 µ h ´ c -? ²³µ ~ c
.
h ´ c µ ~
(13.11)
From Equation 13.7, we get the second moment of the cost per loss,
,´@3 µ ~ ,´@7 µ h ´ c -? ²³µ ~
²c³
²c³
h ´ c µ ~ .
(13.12)
These relationships are relatively easy to get directly from the definitions
²c³
,´@3 µ ~ ²! c ³ ? ²!³ ! ~ ²! c ³ h ! ~
,
and
,´@3 µ
²c³
~ ²! c ³ ? ²!³ ! ~ ²! c ³ h ! ~ .
We would then get = ´@3 µ ~ ,´@3 µ c ²,´@3 µ³ ~
© ACTEX 2009
²c³
²c³
c ´ µ .
(13.13)
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 13 - THE UNIFORM, EXPONENTIAL AND PARETO
LM-163
Exponential distribution
Suppose that ? has an exponential distribution with mean .
The pdf of ? is ? ²%³ ~ c%° for % € , and the cdf is -? ²%³ ~ c c%° for % € .
The mean of ? is , the second moment is , and the variance is .
The pdf of @7 , cost per payment is
²&b³
?
@7 ²&³ ~ c~
? ²³
c²&b³°
c°
~ c&° for & € .
(13.14)
This is the pdf of the exponential distribution with mean . Therefore, @7 has an exponential
distribution with mean . This is a consequence of a special property of the exponential
distribution called the "lack of memory" property. For an exponential random variable ? , the
lack of memory property can be stated as
7 ²? € & b O? € ³ ~ 7 ²? c € &O? € ³ ~ 7 ²? € &³ .
The main consequence for our purposes is that the distribution of the cost per payment
@7 ~ ? c O? € , is the same as the original exponential distribution of ? ,
exponential with mean .
Therefore, ,´@7 µ ~ , ,´@7 µ ~ , and = ´@7 µ ~ .
(13.15)
From Equation 13.6, we get the expected cost per loss
,´@3 µ ~ ,´@7 µ h ´ c -? ²³µ ~ c° .
(13.16)
From Equation 13.7, we get the second moment of the cost per loss,
,´@3 µ ~ ,´@7 µ h ´ c -? ²³µ ~ c° .
(13.17)
We then get = ´@3 µ ~ ,´@3 µ c ²,´@3 µ³ ~ c° c ´c° µ
~ c° h ² c c° ³ .
(13.18)
Pareto distribution
Suppose that ? has a Pareto distribution with parameters and , and suppose that € (if ~ , then ,´?µ ~ B). The pdf of ? is ? ²%³ ~ ²%b³b for % € ,
and the cdf is -? ²%³ ~ c ² %b
³ for % € .
The mean of ? is ,´?µ ~ c
, the second moment is ,´? µ ~ ²c³²
c³ ,
and the variance is = ´?µ ~ ²c³ ²c³ .
The pdf of @7 , cost per payment is
²&b³
²b³
?
@7 ²&³ ~ c~ ²&bb
³b ,² b ³ ~ ²&bb³b for & € .
? ²³
(13.19)
If we define Z ~ b , then the pdf of @7 is @7 ²&³ ~ ²&bZ ³b for & € ,
(13.20)
Z ² ³
We see that the pdf of @7 is the same as the pdf of a Pareto distribution with parameters and Z ~ b , and therefore, @7 has that distribution.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-164
MODELING SECTION 13 - THE UNIFORM, EXPONENTIAL AND PARETO
²b³
b
Therefore, ,´@7 µ ~ c and the second moment is ,´@7 µ ~ ²c³²c³ .
From Equation 13.6, we get the expected cost per loss
,´@3 µ ~ ,´@7 µ h ´ c -? ²³µ ~ ² b
c ³ h ² b ³ .
(13.21)
(13.22)
From Equation 13.7, we get the second moment of the cost per loss,
²b³
,´@3 µ ~ ,´@7 µ h ´ c -? ²³µ ~ ²c³²c³ h ² b
³ .
(13.23)
We would then get = ´@3 µ ~ ,´@3 µ c ²,´@3 µ³ .
Single parameter Pareto distribution
Suppose that ? has a single parameter Pareto distribution with parameters and , and suppose
that € (if ~ , then ,´?µ ~ B). The pdf of ? is ? ²%³ ~ %
b for % € ,
and the cdf is -? ²%³ ~ c ² % ³ for % € .
The mean of ? is ,´?µ ~ ,
the
second
moment
is
,´?
µ
~
c
²c³ .
We consider two cases; either  , or ‚ .
Case 1:  Since ? is defined on the interval % € , it must be true that % € (so there is really no
deductible). Also, -? ²³ ~ , since no losses are below . Therefore there is no conditioning
that occurs, and @7 ~ @3 ~ ? c . Then
,´@7 µ ~ ,´@3 µ ~ ,´? c µ ~ ,´?µ c ~ c
c ,
(13.24)
and = ´@7 µ ~ = ´@3 µ ~ = ´? c µ ~ = ´?µ .
(13.25)
Case 2: € The pdf of @7 , cost per payment is
²&b³
?
@7 ²&³ ~ c~ ²&b³
b ,² ³ ~ ²&b³b for & € .
? ²³
(13.26)
This is the pdf for a (two parameter) Pareto distribution with parameters and ~ ,
so that @7 has that distribution.
Therefore, ,´@7 µ ~ c
, and the second moment is ,´@7 µ ~ ²c³²
c³ .
From Equation 13.6, we get the expected cost per loss
,´@3 µ ~ ,´@7 µ h ´ c -? ²³µ ~ ² c
³ h ² ³ .
(13.22)
From Equation 13.7, we get the second moment of the cost per loss,
,´@3 µ ~ ,´@7 µ h ´ c -? ²³µ ~ ²c³²
c³ h ² ³ .
(13.23)
We would then get = ´@3 µ ~ ,´@3 µ c ²,´@3 µ³ .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 13 - THE UNIFORM, EXPONENTIAL AND PARETO
LM-165
Example LM13-1: For each of the distributions (a), (b) and (c) below for ground up loss random
variable ? , find the expected value and variance of the cost per loss @3 and the cost per payment
@7 .
(a) ? has a uniform distribution on the interval ² Á ³, and there is an ordinary policy
deductible of 20.
(b) ? has an exponential distribution with a mean of 100 and there is an ordinary deductible of
20.
(c) ? has a Pareto distribution with parameters ~ and ~ , and there is an ordinary
policy deductible of .
Solution: In each case, we will find the mean and variance of @7 first.
(a) From the comments earlier in this section, we know that @7 has a uniform distribution on the
interval ²Á c ³ ~ ²Á ³ . Then,
,´@7 µ ~ ~ Á ,´@7 µ ~ ~ Á , and = ´@7 µ ~ ~ .
In the case of the uniform distribution, it is straightforward to find ,´@3 µ and ,´@3 µ
directly.
,´@3 µ ~ ,´²? c ³b µ ~ ²% c ³²À³ % ~ and ,´@3 µ ~ ,´²? c ³b µ ~ ²% c ³ ²À³ % ~ .
Then, = ´@3 µ ~ c ~ .
We have a few other ways to find ,´@3 µ:
(i) ,´@3 µ ~ ,´@7 µ h ´ c -? ²³µ ~ ²³² c À³ ~ .
(ii) ,´@3 µ ~ ´ c -? ²!³µ ! ~ ´ c À!µ ! ~ .
(iii) ,´@3 µ ~ ,´²? c ³b µ ~ ,´?µ c ,´? w µ , where ,´?µ ~ ~ ,
and ,´? w µ ~ % h ²À³ % b ´ c -? ²³µ³ ~ .
From the comments preceding this example, we have an alternative way to find ,´@3 µ.
,´@3 µ ~ ,´@7 µ h ´ c -? ²³µ ~ ²Á ³² c À³ ~ .
Actually, we also have the relationship
,´@3 µ ~ ,´²? c ³b µ ~ ,´? µ c ,´²? w ³ µ c ²³´ ,´?µ c ,´? w µ µ
~ c c ´ c ²³µ ~ , but this would be less convenient to apply.
Notice that once we identified that the distribution of @7 was uniform on ²Á ³, the
calculations to find ,´@3 µ and ,´@3 µ could be done quickly.
(b) Since ? is exponential with mean 100, @7 is also exponential with mean 100 (for any
deductible ³À Then ,´@7 µ ~ Á ,´@7 µ ~ d ~ Á and = ´@7 µ ~ ~ Á .
As in part (a), we can find ,´@3 µ directly.
,´@3 µ ~ ,´²? c ³b µ ~ ²% c ³ h ²ÀcÀ% ³ % .
This integral can be simplified using integration by parts, or we can apply the change of variable
' ~ % c . Integration by parts is not difficult, but also it is not necessary.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-166
MODELING SECTION 13 - THE UNIFORM, EXPONENTIAL AND PARETO
Example LM13-1 continued
Using the change of variable, the integral becomes
B ' h ²ÀcÀ²'b³ ³ ' ~ cÀ h B ' h ²ÀcÀ' ³ ' ~ cÀ ~ À .
We have a few other ways to find ,´@3 µ:
(i) ,´@3 µ ~ ,´@7 µ h ´ c -? ²³µ ~ ²³²c° ³ ~ À .
B
B
(ii) ,´²? c ³b µ ~ ´ c -? ²!³µ ! ~ cÀ! ! ~ cÀ ~ À.
(iii) ,´@3 µ ~ ,´²? c ³b µ ~ ,´?µ c ,´? w µ . For the exponential distribution, the Exam
C table gives ,´? w µ ~ ² c c° ³ . Then,
,´@3 µ ~ c ² c c° ³ ~ cÀ ~ À .
B
To find = ´@3 µ , we first find ,´@3 µ ~ ,´²? c ³b µ ~ ²% c ³ ²ÀcÀ% ³ % .
Again, we can apply integration by parts. We can also apply the same substitution that
we used to find ,´@3 µ , ' ~ % c . The integral becomes
B ' h ²ÀcÀ²'b³ ³ ' ~ cÀ h B ' h ²ÀcÀ' ³ '
~ cÀ h ² d ³ ~ Á cÀ ~ Á À
.
(the last integral is the second moment of an exponential distribution with mean 100,
which is then multiplied by cÀ ).
An easier alternative is to use
,´@3 µ ~ ,´@7 µ h ´ c -? ²³µ ~ ²Á ³²c° ³ ~ Á À
Then = ´@3 µ ~ Á cÀ c ²cÀ ³ ~ Á cÀ ² c cÀ ³ ~ À .
h
(c) The pdf of ? is ? ²%³ ~ ²%b³
for % € ,
the cdf of ? is -? ²%³ ~ c ² %b
³ for % € and ,´?µ ~ c
~ À
Again from the comments preceding this example, we know that if ? has a Pareto distribution
with parameters ~ and ~ , and a deductible ~ is applied, then @7 has a Pareto
distribution with parameters ~ and Z ~ b ~ .
²b³
Then ,´@7 µ ~ b
c ~ À
and ,´@7 µ ~ ²c³²c³ ~ Á À , and
= ´@7 µ ~ Á À c ²
À
³ ~ Á .
B
h² ³
We can find ,´@3 µ directly from ,´@3 µ ~ ,´²? c ³b µ ~ ²% c ³ h ²%b³ %
This integral is awkward (the change of variable ' ~ % c would help a little, but integration
by parts would still be needed).
As in parts (a) and (b), we have other approaches to find ,´@3 µ.
(i) ,´@3 µ ~ ,´@7 µ h ´ c -? ²³µ ~ ²
À
³² b
³ ~ À .
B
B
(ii) ,´²? c ³b µ ~ ´ c -? ²!³µ ! ~ ² ³ ! ~ À .
!b
(iii) ,´@3 µ ~ ,´²? c ³b µ ~ ,´?µ c ,´? w µ . For the Pareto distribution, the Exam C
table gives ,´? w µ ~ c
h ´ c ² b
³ µ . Then,
,´@3 µ ~ c h ´ c ² b
³ µ ~ À .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 13 - THE UNIFORM, EXPONENTIAL AND PARETO
LM-167
B
h
To find = ´@3 µ , we first find ,´@3 µ ~ ,´²? c ³b µ ~ ²% c ³ h ²%b³
% .
This integral is awkward. However, we know that @7 has a Pareto distribution with ~ and
~ , and we know that
h² ³
,´@3 µ ~ ,´@7 µ h ´ c -? ²³µ ~ ²c³²c³ h ² b
³ ~ Á .
Then, = ´@3 µ ~ Á c ²À³ ~ Á .
An alternative way to find ,´@3 µ is
,´@3 µ ~ ,´²? c ³b µ ~ ,´? µ c ,´²? w ³ µ c ´ ,´?µ c ,´? w µ µ,
but this approach gets complicated because it involves the incomplete beta function needed in the
formulation of ,´²? w ³ µ for the Pareto distribution.
The main theme throughout this section is that there is a shortcut available for finding ,´@3 µ
when there is an ordinary deductible if ? has an exponential or Pareto distribution. The
shortcut is to use the relationship ,´@3 µ ~ ,´@7 µ h ´ c -? ²³µ . The shortcut also involves
recognizing that @7 is exponential with mean if ? is, and @7 is Pareto with parameters and
Z ~ b is ? is Pareto with parameters and .
The shortcut also applies to the uniform distribution on ´Á µ, but because the uniform
distribution is so simple, the shortcut is not really necessary.
The shortcut also applies to finding ,´@3 µ ~ ,´@7 µ h ´ c -? ²³µ , but we can use
,´@3 µ ~ ,´?µ c ,´? w µ fairly easily, because ,´? w µ is given in the Exam C table.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-168
© ACTEX 2009
MODELING SECTION 13 - THE UNIFORM, EXPONENTIAL AND PARETO
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 13
LM-169
MODELING - PROBLEM SET 13
Deductibles Applied to Uniform, Exponential and Pareto - Section 13
1. ? has a uniform distribution on ´Á µ.
(a) Find the ordinary deductible so that ,´@7 µ ~ ,´?µ .
(b) Find the ordinary deductible so that = ´@7 µ ~
(c) Find the ordinary deductible so that ,´@3 µ ~
= ´?µ .
,´?µ .
2. Ground up losses are modeled according to an exponential distributions with a mean of 100.
An insurer considers the following two polices.
Policy 1 has no limit but has a deductible of 20.
Policy 2 has no limit but has a deductible of 50.
For each of the two policies the insurer calculates the coefficient of variation for the cost per
payment, say (coefficient of variation for Policy 1) and . By what percentage of is below ?
A) 40%
B) 30%
C) 0%
D) c 30%
E) c 40%
3. (CAS) Auto liability losses for a group of insureds (Group R) follow a Pareto distribution with
~ and ~ . Losses from a second group (Group S) follow a Pareto distribution with
~ and ~ . Group R has an ordinary deductible of 500, Group S has a franchise
deductible of 200. Calculate the amount that the expected cost per payment for Group S exceeds
that for Group R.
A) Less than 350
B) At least 350, but less than 650
C) At least 650, but less than 950
D) At least 950, but less than 1,250
E) At least 1,250
4. (CAS) Claim sizes this year are described by a 2-parameter Pareto distribution with
parameters ~ Á and ~ . What is the expected claim size per loss next year after 20%
inflation and the introduction of a $100 deductible?
A) Less than 490
B) At least 490, but less than 500
C) At least 500, but less than 510
D) At least 510, but less than 520
E) At least 520
5. ? has a single parameter Pareto distribution with ~ and ~ . An ordinary
deductible of 120 is applied to ? . Find the expected value and the variance of the cost per loss.
6. (SOA) A loss, ? , follows a 2-parameter Pareto distribution with ~ and unspecified
parameter .
You are given: , :? c O? € ; ~ , :? c O? € ;
Calculate , :? c O? € ; .
A) 150
B) 175
C) 200
D) 225
E) 250
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-170
MODELING - PROBLEM SET 13
MODELING - PROBLEM SET 13 SOLUTIONS
1. ,´?µ ~ and = ´?µ ~ .
²c³
,´@7 µ ~ c
and
=
´@
µ
~
7
.
²c³
,´@3 µ ~ .
(a) If ,´@7 µ ~ ,´?µ , then c
~ h , so that c ~ , and then ~ .
²c³
(b) = ´@7 µ ~ = ´?µ , then
~ h , so that ² c ³ ~ ,
and then ~ ² c j ³ ~ À .
²c³
(c) If ,´@3 µ ~ ,´?µ , then
~
h
,
so
that
²
c
³
~
,
and then ~ ² c j ³ ~ À .
2. If ? is exponential with parameter and @7 is the cost per payment made when there is a
deductible of and no policy limit, then @7 is also exponential with parameter .
Therefore for both deductibles the cost per payment made has an exponential distribution with
mean 100 and therefore both policies have the same coefficient of variation.
This is Problem #3 from Problem Set 12, but it is repeated here to illustrate the use of the shortcut
described in Section 13.
Answer: C
3. Since ?9 has a Pareto distribution with parameters ~ and ~ , the cost per
payment with an ordinary deductible of 500 for Group R also has a Pareto distribution with
~ and Z ~ b ~ À The expected cost per payment for Group R is
c ~ .
When a franchise deductible is applied to ? , the expected cost per payment is larger than the
expected cost per payment for an ordinary deductible of amount . Therefore, for Group S which
has ?: as Pareto with ~ and ~ , the expected cost per payment with an ordinary
deductible of 200 would be b
~ . For a franchise deductible of 200, the expected
c
cost per payment would be b ~ À
The amount by which the expected cost per payment for Group S exceeds that for Group R is
c ~ À
This is Problem #6 from Problem Set 12, but it is repeated here to illustrate the use of the shortcut
described in Section 13.
Answer: C
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 13
LM-171
4. This year the loss random variable ? has a Pareto distribution with ~ and ~ .
The claim size random variable next year will be A ~ À? . Since the Pareto is a scale
distribution and is the scale parameter, A also has a Pareto distribution with ~ and
Z ~ À²³ ~ . The cdf of A is -A ²!³ ~ c ² !b
³ .
A way in which we can find the expected cost per loss next year, is to use the relationship
,´@3 µ ~ ,´@7 µ h ´ c -A ²³µ . With a deductible of 100 applied to A , @7 has a Pareto
distribution with ~ and ZZ ~ b ~ À
Then ,´@3 µ ~ Answer: D
c h ² b ³ ~ À .
5. The cdf of ? is -? ²!³ ~ c ² ! ³ for ! € .
Since the deductible is greater than , the cost per payment variable is two parameter Pareto with
~ and ~ ~ . Then,
,´@3 µ ~ ,´@7 µ h ´ c -? ²³µ ~ c
h ² ³ ~ À , and
² ³
,´@3 µ ~ ,´@7 µ h ´ c -? ²³µ ~ ²c³²c³
h ² ³ ~ À .
= ´@3 µ ~ À c ²À³ ~ .
6. From the table of distributions, we have - ²%³ ~ c 4 %b
5 ~ c 4 %b 5 and
c
,´?µ ~ c
~ , and ,´? w µ ~ c
h < c 4 b
5
since ~ .
We use the relationship ,´? c O? € µ ~
,´? c O? € µ ~
,´²?c³b µ
c- ²³
~
,´²?c³b µ
c- ²³
~
,´?µc,´?wµ
c- ²³
= ~ < c 4 b
5= ,
,´?µc,´?wµ
c- ²³
~
.
c<c4 b
5=
4 b
5
Á
which simplifies to b .
In a similar way, ,´? c O? € µ ~ b .
We are given that b ~ ² b ³ , from which we get ~ .
Then (as above), ,´? c O? € µ ~ b ~ .
A quicker solution than this uses the following property of the Pareto distribution.
If ? is Pareto with parameters and , and if is an ordinary deductible, then the distribution if
@7 , the cost per payment (conditional distribution of @ c given @ € ) is also Pareto with the
same , but with Z ~ b . Therefore, ,´@ c O@ € µ ~ b
c . Applying this property to
b
this question, we see that , :? c O? € ; ~
, , :? c O? € ; ~ b , and
c
c
b
b
, :? c O? € ; ~ b
c . Then, c ~ h c , from which we get ~ ,
and therefore, b
Answer: B
c ~ .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-172
© ACTEX 2009
MODELING - PROBLEM SET 13
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 14 - COMBINED LIMIT AND DEDUCTIBLE
LM-173
MODELING SECTION 14 - COMBINED LIMIT AND DEDUCTIBLE
The material in this section relates to Loss Models, Section 8.5
The suggested time for this section is 1-2 hours.+
LM-14.1 Maximum Covered Loss " in Combination With Policy Deductible  "
It is possible that a policy has both a policy limit and a deductible applied to determine the
amount paid by the insurance. We define the maximum covered loss to be the amount " above
which no additional benefits are paid. With a deductible and maximum covered loss ", the
amount paid by the insurer is defined as follows.
Maximum covered loss " in combination with policy deductible  "
Cost per lossÁ @3
Amount paid by insurer ~ H ? c "c
?
 ?  " ~ ²? w "³ c ²? w ³ (14.1)
?€"
We see that the maximum amount payable by the insurer, which is called the policy limit, is
" c . In this situation, it is assumed that we first check to see if ? € ", the maximum covered
loss. If the loss is ?  ", we apply the deductible to ? and the insurance payment is ? c . If
the loss is ? € ", we apply the deductible to the maximum covered loss " and the insurance
payment is " c (the policy limit).
Note that the amount paid by the insurer in this case can be formulated as the difference of two
limited loss random variables, * ! 3 ~ @3 ~ ²? w "³ c ²? w ³ .
Whenever there is a deductible , we can consider the cost per payment @7 , which is the
conditional distribution of the amount pad by the insurer given that the loss is above the
deductible. Expectations for @3 and @7 would relate in the same way as when there was only a
,´@ µ
,´@ µ
3
3
deductible, ,´@7 µ ~ 7 ²?€³
and ,´@7 µ ~ 7 ²?€³
.
(14.2)
As a simple example, if a policy has a deductible of 100 (so ~ ) and maximum covered
loss of 1000 (so " ~ ), then the amount paid by the insurer is
?  ?
c
 ?  ~ ²? w ³ c ²? w ³ .
H
" c ~ ? € If the policy was described as having a deductible of 100 and a policy limit of 1000, then
~ and " c ~ , so that " ~ . The maximum covered loss " ~ ,
corresponds to a policy limit (maximum paid by the insurer) of " c ~ .
Expected values related to cost per loss and cost per payment are summarized in the box below.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-174
MODELING SECTION 14 - COMBINED LIMIT AND DEDUCTIBLE
Deductible , maximum covered loss ", policy limit " c The expected cost per loss is:
,´@3 µ ~ ,* 3 ~ ,´? w "µ c ,´? w µ
"
"
~ ²% c ³ h ? ²%³ % b ²" c ³´ c -? ²"³µ ~ ´ c -? ²%³µ %
(14.3)
The expected cost per payment is:
,´@ µ
,* 3
,´@7 µ ~ c~ c- 3²³ ~
? ²³
?
,´?w"µc,´?wµ
c-? ²³
(14.4)
Second moment of cost per loss is
"
,´@3 µ ~ ²% c ³ h ? ²%³ % b ²" c ³ ´ c -? ²"³µ
~ 4,´²? w "³ µ c ,´²? w ³ µ5 c h 4,´? w "µ c ,´? w µ5
(14.5)
Second moment of cost per payment is
,´@ µ
,´@7 µ ~ c- 3²³
?
(14.6)
Example LM14-1: You are given the following:
- Losses follow a uniform distribution on the interval from 0 to 50,000.
- There is a maximum covered loss of 25,000 per loss and a deductible of 5,000 per loss.
- The insurer applies the maximum covered loss prior to applying the deductible
Determine the expected payment per payment made.
A) Less than 15,000
B) At least 15,000 but less than 17,000
C) At least 17,000 but less than 19,000
D) At least 19,000 but less than 21,000
E) At least 21,000
Solution: ? — unif ´ Á Á µ . The "expected payment per payment made" is the language
that was used to describe the expected cost per payment. @7 is the cost per payment, and we are
trying to find ,´@7 µ. The maximum covered loss is " ~ Á and the policy deductible is
~ Á (the policy limit is 20,000). The uniform distribution is quite simple, and any of the
standard formulations for ,´? w µ can be found fairly quickly. The integral of c - ²%³ is a
straightforward one that can be useful for other distributions, particularly the exponential and
Pareto.
,´@ µ
3
,´@7 µ ~ c- ²³
~
~
´c
%
Á µ %
c Á
,´?wÁµc,´?wµ
c- ²³
~
Á
À
~ Á .
~
Á ´c- ²%³µ %
c- ²³
If we had been asked for the variance of cost per loss or cost per payment, we would first find
,´@ µ
3
,´@3 µ and then ,´@7 µ ~ c- ²³
. For the uniform distribution, using ²%³ ~ Á
, we
Á
would get ,´@3 µ ~ Á ²% c ³ h Á % b ²Á c Á ³ h ´ c - ²Á ³µ.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 14 - COMBINED LIMIT AND DEDUCTIBLE
LM-175
Example LM14-2: The distribution of the loss random variable ? is exponential with mean .
With deductible and maximum covered loss " € , find
(a) the expected value and variance of the cost per loss, and
(b) the expected cost per payment,
Solution: (a) Expected cost per loss is ,´@3 µ ~ ,´? w "µ c ,´? w µ .
From Table C we know that ,´? w "µ ~ ² c c"° ³ for an exponential variable ? .
Therefore, ,´@3 µ ~ ² c c"° ³ c ² c c° ³ ~ ²c° c c"° ³ .
We can also find ,´@3 µ from more basic formulations.
"
"
,´@3 µ ~ ,´? w "µ c ,´? w µ ~ ´ c - ²!³µ ! ~ c!° ! ~ ´c° c c"° µ .
We can also use the formulation
"
"
,´? w "µ ~ ! h ²!³ ! b "´ c - ²"³µ ~ ! h c!° ! b "c!° ,
but this will require integration by parts.
To find the variance of @3 , we can find
,´@3 µ ~ 4,´²? w "³ µ c ,´²? w ³ µ5 c h 4,´? w "µ c ,´? w µ5 .
Table C formulations for the exponential distribution shows that ,´²? w "³ µ involves the partial
Gamma function for € . We can find the expectations from more basic principles, but
integration by parts will be needed.
"
,´²? w "³ µ ~ % c%° % b " ´ c - ²"³µ
%
%
%
From integration by parts, we get % % % ~ % c %
b .
Therefore,
,´²? w "³ µ ~ c % c%° c % c%° c c%° c
%~"
%~
b " c"°
~ c c"° ´" b µ ,
and in a similar way, ,´²? w ³ µ ~ c c° ´ b µ .
Then = ´@3 µ ~ ,´@3 µ c ²,´@3 µ³
~ ,´²? w "³ µ c ,´²? w ³ µ c ,´? w "µ b ,´? w µ c ´,²? w "³ c ,²? w ³µ
~ c c"° ´" b µ c 4 c c° ´ b µ5
c ´ c c"° µ b ´ c c° µ c 4´c° c c"° µ5
~ c° ´ b µ c c"° ´" b µ c ´c° c c"° µ c 4´c° c c"° µ5 .
(b) expected cost per payment is
,´@ µ
3
,´@7 µ ~ c- ²³
~
© ACTEX 2009
,´?w"µc,´?wµ
c- ²³
~
´c° cc"° µ
c°
~ ´ c c²"c³° µ À
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-176
MODELING SECTION 14 - COMBINED LIMIT AND DEDUCTIBLE
LM-4.2 PDF and CDF of @3 and @7
There has been little or no reference to the pdf's or cdf's of @3 or @7 on exam questions. We will
summarize them briefly.
The pdf and cdf of the cost per loss @3 are
-? ²³
&~
-? ²³
&~
@3 ²&³ ~ H ? ²& b ³  &  " c , -@3 ²&³ ~ H -? ²& b ³  &  " c c -? ²"³ & ~ " c & ‚"c
(14.7)
The pdf and cdf of the cost per payment @7 are
@7 ²&³ ~ H
? ²&b³
c-? ²³
c-? ²"³
c-? ²³
& "c
& ~"c
, -@7 ²&³ ~ H
-? ²&b³c-? ²³
c-? ²³
& "c
& ‚"c
(14.8)
Example LM14-1 continued: Refer to Example LM14-1 a couple of pages earlier. Formulate
the pdf of @7 and find ,´@7 µ using that pdf.
Solution:
@7 ²&³ ~
H
²&b³
? ²&b³
c-? ²³
?
~ c~ cÁ
~ Á
? ²³
Á
c-? ²"³
c-? ²³
~ c-? ²³ ~ c Á
~ Á
?
Á
c- ²Á³
Then ,´@7 µ ~
Á &
h
Á
c Á
Á
Á
& b ²Á ³² Á ³
0  &  Á (density)
& ~ Á (a point of probability)
& € Á ~ Á .
Example LM14-3: ? has an exponential distribution with mean . Find the pdf, cdf and
expected value of (i) @3 and (ii) @7 when there is a policy deductible of and maximum
covered loss " (limit " c ) .
Solution: (i) The pdf, cdf and expected value of @3 are
c c° & ~ -? ²³
&~
@3 ²&³ ~ H ? ²& b ³  &  " c ~ H c²&b³°  &  " c c -? ²"³ & ~ " c c"°
& ~"c
-? ²³
&~
c c°
-@3 ²&³ ~ H -? ²& b ³  &  " c ~ H c c²&b³°
& €"c
&~
& "c ,
& €"c
"
,´@3 µ ~ ,´? w "µ c ,´? w µ ~ ´ c -? ²%³µ % ~ ²c° c c"° ³
(as in Example LM14-1) .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 14 - COMBINED LIMIT AND DEDUCTIBLE
LM-177
Example LM14-3 continued
(ii) The pdf, cdf and expected value of @7 are
@7 ²&³ ~ H
-@7 ²&³ ~ H
? ²&b³
c-? ²³
c-? ²"³
c-? ²³
& "c
& ~"c
-? ²&b³c-? ²³
c-? ²³
,´@ µ
c&°
~ H c²"c³°
& "c
& ‚"c
,´?w"µc,´?wµ
,´@7 µ ~ c- 3²³ ~
c-? ²³
?
(as in Example LM14-1) .
~
~H
& "c
& ~"c
c c&°
²c° cc"° ³
c°
.
& "c
.
& ‚"c
~ ² c c²"c³° ³
LM-4.3 A Few Comments
Exam questions on policy limits and deductibles will often require some juggling of the
following main factors and relationships.
B
B
Mean of ? is ,´?µ ~ ´ c -? ²!³µ ! ~ % h ? ²%³ %
Limited expected value is ,´? w µ ~ ´ c -? ²!³µ !
B
Expected cost per loss with deductible is ,´?µ c ,´? w µ ~ ´ c -? ²!³µ !
Expected cost per loss with deductible and maximum covered loss " (limit " c ) is
"
,´? w "µ c ,´? w µ ~ ´ c -? ²!³µ !
Expected cost per payment with deductible is
,´?µc,´?wµ
c-? ²³
Expected cost per payment with deductible and max. covered loss " is
,´?w"µc,´?wµ
c-? ²³
,´?wµ
Loss elimination ratio is ,´?µ
Expected cost per loss with franchise deductible is
B
,´?µ c ,´? w µ b ´ c -? ²³µ ~ ´ c - ²!³µ ! b ´ c -? ²³µ
Expected cost per payment with franchise deductible is
© ACTEX 2009
,´?µc,´?wµ
c-? ²³
b
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-178
MODELING SECTION 14 - COMBINED LIMIT AND DEDUCTIBLE
LM-4.4 Graphical Representation of ,´@3 µ
The graphical representation of ,´@3 µ is not a part of the exam syllabus for Exam C, but it
provides an alternative way of looking at the expectations with various policy adjustments. It
also may be helpful as an approach to stop-loss insurance which we look at a little later.
The graphical representation of ,´@3 µ is based on its formulation as an integral of c - ²%³,
and can be represented as a shaded region under the graph of c - ²%³.
Policy Limit: Suppose that - ²%³ is the cdf of the ground up loss ? , and the insurance policy
has a limit of ". Then the expected cost per loss (the limited expected value) can be written in
"
the form ,´? w "µ ~ ´ c - ²%³µ %. ,´? w "µ is the shaded region the graph below.
Policy Deductible: Suppose that the policy has a deductible of . Then the expected cost per
loss can be written in the form
B
,´²? c ³b µ ~ ´ c - ²%³µ % ~ ,´?µ c ,´? w µ .
,´²? c ³b µ is the shaded region in graph below.
Combination Maximum Covered Loss " and Deductible (Policy Limit " c )
"
In this case, the expected cost per loss is ,´? w "µ c ,´? w µ ~ ´ c - ²%³µ % .
,´? w "µ c ,´? w µ is the shaded region in the graph below.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 14
LM-179
MODELING - PROBLEM SET 14
Combined Limit and Deductible - Section 14
Problems 1 to 2 relate to a loss random variable ? with a uniform distribution on the interval
´Á µ.
1. Find the variance of the insurer cost per loss if there is a deductible of amount 100 and a
maximum covered loss of amount 500 (nearest 1000).
A) 20,000
B) 21,000
C) 22,000
D) 23,000
E) 24,000
2. With a deductible of , the mean excess loss is ²³. Find the deductible Z that results in a
mean excess loss which is one-half as big (² Z ³ ~ ²³).
A) c B) b C) c D) c E) b 3. The distribution of the loss random variable ? is exponential with parameter . With
deductible and maximum covered loss " € , find the cdf and pdf of each of the two random
variables (a) cost per loss, and (b) cost per payment.
4. The ground up loss random variable is ? . Two insurance policies are being compared. Policy
( has a franchise deductible of amount but no policy limit. Policy ) has an ordinary
deductible of amount and a maximum covered loss of amount " € . @( and @)
are the cost per payment on Policies A and B respectively. Find ,´@( µ c ,´@) µ.
A)
,´?w"µc,´?wµ
c-? ²³
D)
,´?w"µc,´?wµ
c-? ²³
b
B)
E)
,´?µc,´?w"µ
c-? ²³
b
C)
,´?µc,´?wµ
c-? ²³
b
,´?µc,´?w"µ
c-? ²³
5. The ground-up loss for a particular insured has an exponential distribution. If an insurance
policy on the loss has a deductible of and a maximum covered loss of , then the
expected cost per loss is À , and the expected cost per payment is À . Find the mean
ground-up loss.
A) 500
B) 1000
C) 1500
D) 2000
E) 2500
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-180
MODELING - PROBLEM SET 14
6. Losses follow a uniform distribution on the interval from 0 to 50,000. There is a maximum
covered loss of 25,000 and a deductible of 5000. The insurer applies the maximum covered loss
prior to applying the deductible. Determine the expected cost per payment made by the insurer.
A) Less than 15,000
B) At least 15,000, but less than 17,000
C) At least 17,000, but less than 19,000
D) At least 19,000, but less than 21,000
E) At least 21,000
7. (SOA) Michael is a professional stuntman who performs dangerous motorcycle jumps at
extreme sports events around the world. The annual cost of repairs to his motorcycle is modeled
by a two parameter Pareto distribution with ~ and ~ . An insurance reimburses
Michael's motorcycle repair costs subject to the following provisions:
(i) Michael pays an annual ordinary deductible of 1000 each year.
(ii) Michael pays 20% of repair costs between 1000 and 6000 each year.
(iii) Michael pays 100% of the annual repair costs above 6000 until Michael has paid 10,000 in
out-of-pocket repair costs for each year.
(iv) Michael pays 10% of the remaining repair costs each year.
Calculate the expected annual insurance reimbursement.
A) 2300
B) 2500
C) 2700
D) 2900
E) 3100
8. An insurance policy on the loss ? has an ordinary deductible of 40. The policy also has the
following adjustments. If the loss is between 40 and 60, the insurance policy pays the amount of
the loss above 40. If the loss is between 60 and 80, the insurance pays 20 plus 75% of the loss
above 60. If the loss is above 80, the insurance pays 35. The distribution of ? is uniform on the
interval ²Á ³ , find the expected cost per loss .
(a) Express the cost per loss random variable as a combination of ? and ? w "
factors for appropriate values of ".
(b) Find expressions for the distributions (density and probabilities) of @3 and @7 .
Use @7 ²&³ to find the expected cost per payment.
,´@ µ
3
(It should be equal to c- ²³
where ,´@3 µ was found in part (a).)
?
9. (CAS) The severity distribution function of claims data for automobile property damage
coverage for Le Behemoth Insurance Company is given by an exponential distribution, - ²%³.
c%
- ²%³ ~ c %² ³
To improve the profitability of this portfolio of policies, Le Behemoth institutes the following
policy modifications:
i) It imposes a per-claim deductible of 500.
ii) It imposes a per-claim maximum covered loss of 25,000.
Previously, there was no deductible and no limit.
Calculate the average savings per (old) claim if the new deductible and policy limit had been in
place.
A) 490
B) 500
C) 510
D) 520
E) 530
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 14
LM-181
10. A loss random variable has density function ²%³ ~ À% for  %  .
(a) Find the expected cost per loss if a policy limit of 80 is applied to the loss.
(b) Formulate the density function of the left-censored and shifted random variable for
an ordinary deductible of 40, and calculate the expected cost per loss.
(c) Formulate the distribution function, -@7 ²&³, for the excess-loss random variable for
B
an ordinary deductible of 40. Calculate ,´@7 µ ~ ´ c -@7 ²&³µ & and verify that it is the
,´@ µ
3
same as the mean excess loss (expected cost per payment), ,´? c O? € µ ~ c- ²³
.
?
(d) Calculate the expected cost per loss if a maximum covered loss of 80 is combined
with a franchise deductible of 40. Show the region representing this expected cost per loss in the
following graph.
11. (CAS): You are given the following:
- Losses follow a lognormal distribution with parameters ~ Á ~ .
- For each loss less than or equal to 50,000, the insurer makes no payment.
- For each loss greater than 50,000, the insurer pays the entire amount of the loss up to the
maximum covered loss of 100,000.
Determine the expected amount paid per loss.
12. The ground up loss random variable ? for a health insurance policy in 2006 is modeled with
an exponential distribution with mean 1000. An insurance policy pays the loss above an ordinary
deductible of 100, with a maximum annual payment of 500. The ground up loss random variable
is expected to be 5% larger in 2007, but the insurance in 2007 has the same deductible and
maximum payment as in 2006. Find the percentage increase in the expected cost per payment
from 2006 to 2007.
13. (SOA) Annual prescription drug costs are modeled by a two-parameter Pareto distribution
with ~ and ~ . A prescription drug plan pays annual drug costs for an insured
member subject to the following provisions:
(i) The insured pays 100% of costs up to the ordinary annual deductible of 250.
(ii) The insured then pays 25% of the costs between 250 and 2250.
(iii) The insured pays 100% of the costs above 2250 until the insured has paid 3600 in total.
(iv) The insured then pays 5% of the remaining costs.
Determine the expected annual plan payment.
A) 1120
B) 1140
C) 1160
D) 1180
E) 1200
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-182
MODELING - PROBLEM SET 14
14. (SOA) For a special investment product, you are given:
(i) All deposits are credited with 75% of the annual equity index return, subject to a minimum
guaranteed crediting rate of 3%.
(ii) The annual equity index return is normally distributed with a mean of 8% and a standard
deviation of 16%.
(iii) For a random variable ? which has a normal distribution with mean and standard
deviation , you are given the following limited expected values:
,´? w %µ
~ %
~ %
~ %
c À%
À%
~ %
c À%
c À%
~ %
~ %
,´? w %µ
~ %
À%
c À%
~ %
À%
c À%
Calculate the expected annual crediting rate.
A) 8.9%
B) 9.4%
C) 10.7%
D) 11.0%
© ACTEX 2009
E) 11.6%
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 14
LM-183
MODELING - PROBLEM SET 14 SOLUTIONS
1. = ´@ µ ~ ,´²? w "³ µ c ,´²? w ³ µ c ,´? w "µ b ,´? w µ
c <,²? w "³ c ,²? w ³ =
" %
"
"
,´? w "µ ~ % b "´ c µ ~ " c ~ Á ,´? w µ ~ c ~ "
,´²? w "³ µ ~ % % b " ´ c " µ ~ " c " ~ Á Á
c ,´²? w ³ µ ~ ~ Á S = ´@ µ ~ Á c Á c ²³²³ b ²³²³ c ² c ³ ~ Á À
Answer: D
2. The mean excess loss with deductible is
²³ ~
,´?µc,´?wµ
c- ²³
~
cb c and with deductible Z it is ²Z ³ ~
Answer: E
²c ³²c ³
~ c ,
c Z
c ~ ² c ³ S Z ~
~
b .
3. (a) @3 ~ if ?  (prob. c c° ), and @3 ~ " c if ? € " (prob. c"° ) .
c c°
& ~ (%  ³
i
-@3 ²&³ ~ 7 ´@  &µ ~ H c c²&b³°  &  " c (  %  ")
& € " c (% ‚ ")
c c° & ~ (%  ³
c²&b³°
 &  " c (  %  ")
@3 ²&³ ~
c"°
& ~ " c (% ‚ ")
& €"c
H
²³ -@7 ²&³ ~ H
@7 ²&³ ~
H
© ACTEX 2009
-? ²&b³c-? ²³
c-? ²³
~c
? ²&b³
c-? ²³
c-? ²"³
c-? ²³
& ~ (%  ³
c&°
 &  " c (  %  " )
& ‚ " c (% ‚ ")
~ c&°
0  &  " c (density)
~ c²"c³°
& ~ " c (a point of probability)
& €"c
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-184
MODELING - PROBLEM SET 14
4. For the franchise deductible with no policy limit @( ~ F
and @( ²&³ ~ F ? ²&³
c-? ²³
&
?€
S ,´@( µ ~
B % ? ²%³ %
c-? ²%³
undefined
?
?
,
?€
Á and the numerator is
B % ? ²%³ % ~ B % ? ²%³ % c % ? ²%³ %
~ ,´?µ c ´,´? w µ c ² c -? ²%³³µ ~ ,´?µ c ,´? w µ b ´ c -? ²³µ ,
so that ,´@( µ ~
,´?µc,´?wµ
c-? ²³
b.
undefined
For the ordinary deductible with maximum covered loss ", @) ~ H ? c "c
and
@) ²&³ ~ F
? ²&b³
c-? ²³
c-? ²"³
c-? ²³
& "c
& ~"c
S ,´@) µ ~
?
?"
?‚"
,´?w"µc,´?wµ
c-? ²³
(this expectation is in the notes in a general form for a payment that has coinsurance factor and
inflation rate ). Then ,´@( µ c ,´@) µ ~
,´?µc,´?w"µ
c-? ²³
b.
Answer: B
5. Expected insurer payment per loss ~ ,´? w "µ c ,´? w µ
"
"
~ ´ c - ²!³µ ! ~ c!° ! ~ ´c° c c"° µ S ´c° c c° µ ~ À À
Expected insurer cost per payment made
~
,´?w"µc,´?wµ
c- ²³
~
´c° cc"° µ
c°
~ ´ c c²"c³° µ ~ ° ´c° c c"° µ
S ° ´c° c c° µ ~ À .
Then ° ~ À
À ~ À
S ~ ~ ,´?µ .
Answer: D
6. The distribution of cost per payment can be formulated in terms of the original loss
distribution ? , after taking into account the maximum covered loss and deductible.
If @ is the random variable of amount paid by the insurer on a loss of amount ? , then
Á
?  @ ~ H ? c  ?  Á ,
Á Á  ?  Á where ? has a uniform distribution on ´ Á Á µ À
,´@ µ
3
,´@7 µ ~ c- ²³
~
²%c³ ? ²%³ %bÁh7 ´?‚Áµ
.
7 ´?‚µ
Since ? has a uniform distribution on the interval ´ Á Á µ, the density function of is
? ²%³ ~ À , and 7 ´? ‚ Á µ ~ À Á 7 ´? ‚ µ ~ À À
The expectation becomes
Á ²%c³ ²À³ %bÁh²À³
²À³
~
bÁ
À
~ Á À
Á
%
%
Alternatively, using - ²?³ ~ Á
, we get ,´@3 µ ~ Á ´ c Á
µ % ~ Á ,´@3 µ
Á
and then ,´@7 µ ~ c- ²³ ~ À ~ Á À Answer: B
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 14
LM-185
7. ? is the annual repair cost. The insurance pays À²? c ³ up to a payment of
À²
c ³ ~ , which occurs if repair costs are 6000, and Michael pays 2000 if repair
costs are 6000. Michael pays all of the next 8000 of repair costs, so if costs are 14,000, Michael
pays 10,000.
If costs are above 14,000, insurance pays À²? c Á ³ .
The insurance pays À´²? c ³b c ²?c
³b µ b À²? c Á ³b .
~ À´²? w ³ c ²? w ³µ b À´? c ²? w Á ³µ .
The expected insurance payment is
À´,´? w µ c ,´? w µµ b À´,´?µ c ,´? w Á µµ .
c
For the Pareto distribution, ,´? w µ ~ c
h ´ c ² b
µ and ,´?µ ~ c
.
³
The expected insurance payment is À´ c µ b À´ c µ ~ . Answer: C
8. (a) Amount paid by insurer (cost per loss) is
? c @3 ~ H
b À²? c ³
²? c ³b ~ F
? c ?   ?   ?  ? € ?  ? € is valid for @3 up to ? ~ .
If we subtract À²? c ³b , we get
?   ?  ²? c ³b c À²? c ³b ~ H ? c ²? c ³ c À²? c ³ ~ b À²? c ³ ? € If we subtract À²? c ³b , we get
²? c ³b c À²? c ³b c À²? c ³b
?  ? c  ?  ~
b À²? c ³
 ?  b À²? c ³ c À²? c ³ ~ ? € H
This is the cost per payment made by the insurance.
This is
²? c ³b c À²? c ³b c À²? c ³b
~ ? c ²? w ³ c À´? c ²? w ³µ c À´? c ²? w ³µ
~ À²? w ³ b À²? w ³ c ²? w ³
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-186
MODELING - PROBLEM SET 14
8.(b) Expected cost per payment ~ ,´@7 µ ~ & @7 ²&³ &
@3 's distribution is @3 ²³ ~ À Á @3 ²&³ ~ À for  &  ,
@3 ²&³ ~ À
À ~ for  &  , and @3 ²³ ~ À .
@7 's distribution is the conditional distribution of @3 given that @3 € .
We know 7 ´@3 € µ ~ À
, so
À
@7 ²&³ ~ À
À
for  &  , @7 ²&³ ~ À
²³ for  &  , and @7 ²³ ~ À
Then ,´@7 µ ~ &² ³ & b &² ³ & b ²³² ³ ~ À b À b À
~ À .
,´@3 µ
Note that ,´@7 µ ~ c- ²³
~ À
À
~ À .
9. The original loss distribution is exponential with a mean of ~ Á .
The expected cost per claim after the deductible and limit are imposed is
,´? w Á µ c ,´? w µ . From the Exam C Tables we see that for the
exponential distribution with parameter , ,´? w µ ~ ² c c° ³ .
Therefore, ,´? w Á µ ~ Á ² c cÁ°Á ³ ~ Á À and
,´? w µ ~ Á ² c c°Á ³ ~ À , so that
,´? w Á µ c ,´? w µ ~ Á À . This is the expected cost per claim after the
deductible and limit are imposed. The average savings per claim is
Á c Á À ~ À.
Answer: C
10. (a) ,´? w µ ~ % ²%³ % b ´ c - ²³µ ~ %²À%³% b À% %
~ b ² ³ ~ À .
or c - ²%³ ~ c À%
¬ ,´? w µ ~ ´ c - ²%³µ% ~ ´ c À% µ % ~ À
(b) @3 ~ H
? c @3 ²&³ ~ H
?  À
? € - ²³ ~ À
& ~ Á %  ²& b ³ ~ À²& b ³  &  Á  %  ,´@3 µ ~ & ²À³²& b ³ & ~ À .
or ,´@3 µ ~ ´ c - ²%³µ % ~ ´ c À% µ % ~ À .
7 ´@ &µ
(c) -@7 ²y³ ~ 7 ´@7  &µ ~ 7 ´@3  &O? € µ ~ c-3²³ ~
À´²&b³ c µ
~ À
ÀcÀ² ³
À ²& b &³  &
,´@7 µ ~ ´ c À
À ²& b &³µ & ~ À
~
7 ´?&bµ
c- ²³
 .
,´@ µ
3
From (b) ,´@3 µ ~ À , so ,´? c O? € µ ~ c- ²³
~ À
À ~ À .
?
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 14
LM-187
10.(d) ,´@3 µ ~ % ²%³ % b ´ c - ²³µ ~ % ²À%³ % b ² ³ ~ 58.67 .
11. In this case, @3 is 0 if the loss is less than 50,000 , @3 ~ ? if the loss is between 50,000 and
100,000, and @3 ~ Á if ? € Á (this is a franchise deductible of 50,000 with a
maximum covered loss and policy limit of 100,000).
Á
,´@3 µ ~ Á % ²%³ % b Á ´ c - ²Á ³µ
Á
Á
~ % ²%³ % c % ²%³ % b Á ´ c - ²Á ³µ
~ ,´? w Á µ c ²,´? w Á µ c Á ² c - ´Á µ³³
(this follows from the relationship ,´? w µ ~ % ²%³ % b ´ c - ²³µ , so that
% ²%³ % ~ ,´? w µ c ´ c - ²³µ , and therefore
Á % ²%³ % b Á ´ c - ²Á ³µ ~ ,´? w Á µ and
Á % ²%³ % ~ ,´? w Á µ c Á ´ c - ²Á ³µ ).
Note that this policy has a combination of a maximum covered loss of 100,000 and a franchise
deductible of 50,000 .
From the distribution tables, for the lognormal distribution, we have
" c c ,´²? w "³ µ ~ %² b ³ h )4
5 b " ´ c - ²"³µ . Therefore,
with ~ Á ~ and ~ ,
,´? w Á µ ~ %² b ³ h )4
where - ²Á ³ ~ )4
À
Á c c
5 b Á ´ c - ²Á ³µ ,
Á c 5
, so that
,´? w Á µ ~ h )²À³ b Á ´ c )²À³µ ~ Á À
Also, ,´? w Á µ c Á ² c - ´Á µ³
Á c c
~ %² b ³ h )4
5 ~ À h )² c À³ ~ Á À
The insurer's expected annual payment is Á c Á ~ Á .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-188
MODELING - PROBLEM SET 14
12. In 2006 the deductible is ~ and the maximum covered loss is " ~ (policy limit
of
" c ~ ). The expected cost per loss in 2006 is ,´? w "µ c ,´? w µ ,
and the expected cost per payment is
,´?w"µc,´?wµ
c-? ²³
.
For the exponential distribution with mean , we have
-? ²%³ ~ c c%° and
,´? w %µ ~ ² c c%° ³ .
The expected cost per payment in 2006 is
,´?w
µc,´?wµ
c-? ²³
~
²cc
° ³c²cc° ³
c²cc° ³
~ ² c c° ³ À
In 2007 the loss random variable is @ ~ À? . The exponential distribution is a scale
distribution, which means that a constant multiple is also exponential with a scaled mean.
Therefore, @ has an exponential distribution with mean 1050.
The expected cost per payment in 2007 (same deductible and policy limit as 2006)
,´@ w
µc,´@ wµ
c-@ ²³
~
²cc
° ³c²cc° ³
c²cc° ³
~ ² c c° ³ À
The ratio of expected cost per payment in 2007 to that of 2006 is
²cc° ³
²cc° ³
~ À , an increase of 1.1% from 2006 to 2007.
An alternative way to find the expected cost per payment in 2007 is as follows.
If @ ~ ² b ³? and a maximum covered loss of " is applied to @ , then it is possible to
"
formulate the expected cost per loss for @ in terms of ? : ,´@ w "µ ~ ² b ³,´? w b
µ.
&
Also, the distribution function of @ is -@ ²&³ ~ -? ² b ³ .
If there is a policy deductible of and a maximum covered loss of " applied to @ , then
the expected cost per loss is
"
,´@ w "µ c ,´@ w µ ~ ² b ³²,´? w b
µ c ,´? w b
µ³ .
The expected cost per payment is
,´@ w"µc,´@ wµ
c-@ ²³
~
"
²b³²,´?w b
µc,´?w b
µ³
c-? ² b ³
.
Applying this to exponential random variable ? with ~ ÀÁ ~ Á ~ and
" ~ , we get an expected cost per payment in 2007 of
²À³²,´?w À
µc,´?w À
µ³
c-? ² À ³
~
²À³´²cc
° ³c²cc° ³µ
c²cc° ³
.
This reduces to the same expression as the first approach.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 14
LM-189
13. If the annual drug cost is 2250, then the insured pays b ²À³² c ³ ~ .
Since 3600 is 2850 greater than 750, the insured would reach a payment of 3600 when annual
drug cost reaches b ~ . The plan pays 95% of cost above 5350.
Suppose that ? is the annual drug cost. The plan pays the following amounts
?  À²? c ³
 ?  @ ~H
 ?  b À²? c ³ ? € We can build up an expression for @ in the following way
?  ²À³´²? w ³ c ²? w ³µ will pay H À²? c ³  ?  .
? € Then À´? c ²? w ³µ pays À²? c ³ for ? € .
Therefore @ ~ ²À³´²? w ³ c ²? w ³µ b À´? c ²? w ³µ .
Then the expected annual payment is
,´@ µ ~ ²À³´,²? w ³ c ,²? w ³µ b ²À³´,²?³ c ,²? w ³µ
~ ²À³´² c ³´ c ² b ³µ c ² c ³´ c ² b ³µ µ
b ²À³´ ² c ³ c ² c ³´ c ² b ³µ ~ . Answer: C
14. The annual crediting rate is 75% of the annual index return ? , subject to a minimum
guaranteed rate of 3%. The annual credit rating @ is
À?  ?
?
@ ~F
~F
~bF
À? À? € À? ? € À? c ? € ~ b ²À? c ³b ~ b À? c ²À? w ³À
Since ? is normal with mean 8% and standard deviation 16%, using the values in the table, we
have the expected annual credit rate is
b À,´?µ c ,´À? w µ ~ b c À,´? w µ ~ c À² c À³ ~ À%.
Answer: B
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-190
© ACTEX 2009
MODELING - PROBLEM SET 14
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 15 - ADDITIONAL POLICY ADJUSTMENTS
LM-191
MODELING SECTION 15 - ADDITIONAL POLICY ADJUSTMENTS
The material in this section relates to Loss Models, Section 8.3 and 8.5
The suggested time for this section is 1-2 hours.
LM-15.1 Coinsurance factor
A coinsurance factor is a number ,   . The usual way in which a coinsurance factor is
applied is to first apply any policy limit and deductible on the insurance policy, and then the
insurer pays proportion of that amount.
Maximum covered loss " in combination with policy deductible  " and coinsurance
factor (policy limit ²" c ³)
If the policy has deductible and limit maximum covered loss " and a coinsurance factor , then
?
the amount paid by the insurer (the cost per loss) is H ²? c ³  ?  " .
(15.1)
²" c ³ ? € "
The maximum amount paid by the insurer (the policy limit) is ²" c ³ .
With coinsurance, the amount paid by the insurer is multiplied by the amount that would have
been paid by the insurer without coinsurance. The expected values of @3 and @7 after
coinsurance are equal to multiplied by those expected values before coinsurance, and the
variances after coinsurance are equal to multiplied by the variances before coinsurance.
Expected cost per loss
"
,* 3 ~ ,´@3 µ ~ ²,´? w "µ c ,´? w µ³ ~ ´ c -? ²%³µ % ,
Expected cost per payment
,´@ µ
,* 7 ~ ,´@7 µ ~ c- 3²³ ~
?
²,´?w"µc,´?wµ³
c-? ²³
.
(15.2)
(15.3)
LM-15.2 Inflation
A factor than can arise when forecasting future losses and insurer payments is inflation. An
inflation rate , or growth factor ² b ³ might be applied to the current loss distribution in
order to model the loss distribution for the next period. The "inflated" loss variable next period
will be ² b ³? , and insurer payments next period are made based on the inflated loss random
variable for the next period. The reason that this may be useful is that some aspects of the
behavior or distribution of the inflated loss random variable may be similar to the original
(current) loss random variable. For instance, if the current loss random variable is uniformly
distributed on the interval ² Á ³, and if the inflation factor between now and next period is
10%, then general reasoning suggests that the loss random variable next period will also be
uniformly distributed, but on the interval ² Á ³. Also, most of the distributions in the Exam
C table are scale distributions, and if a scale distribution is multiplied by a constant, the resulting
random variable is of the same type with an adjusted scale parameter.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-192
MODELING SECTION 15 - ADDITIONAL POLICY ADJUSTMENTS
For instance if ? has a Pareto distribution with parameters and , then ² b ³? has a Pareto
distribution with parameters and ² b ³.
If ? is the current loss random variable and ? ~ ² b ³? is the loss random variable after
inflation. We see that ,´? µ ~ ² b ³,´?µ , = ´? µ ~ ² b ³ = ´?µ
(15.4)
(mean and variance).
The following relationships also help to identify and analyze the after-inflation loss random
variable:
-? ²³ ~ 7 ´?  µ ~ 7 ´?  b
µ ~ -? ² b
³ (distribution function),
(15.5)
%
%
? ²%³ ~ -?Z ²%³ ~ b
h -?Z ² b
³ ~ b
h ? ² b
³ (density function),
4? ²³ ~ ,´? µ ~ ,´²b³? µ ~ 4? ´² b ³µ (moment generating function).
(15.6)
(15.7)
A ground up loss random variable ? will usually be defined on either the interval ² Á B³ , or
the interval ² Á µ , where  B . The after-inflation random variable ? ~ ² b ³? will
be defined on either ² Á B³ , or ² Á ² b ³µ , respectively.
Example LM15-1: Using the pdf of the loss random variable, find the distribution of the afterinflation loss random variable when inflation is 10% for the following pre-inflation loss random
variables:
(i) continuous uniform distribution on the interval ² Á ³ ;
(ii) exponential distribution with mean 1000 ;
(iii) Pareto distribution with parameters and .
Solution: With an inflation factor of , the after-inflation loss random variable ? will have pdf
%
? ²%³ ~ b
h ? ² b
³.
(i) If ? has a uniform distribution on the interval ² Á ³ , then the pdf of ? is
? ²%³ ~ for  %  .
With an inflation factor of 10%, the after-inflation loss random variable ? will have pdf
? ²%³ ~ À
h ~ for  %  ²³²À³.
This is the pdf of a uniform distribution on the interval ² Á ³ .
(ii) The pdf of the exponential distribution with mean 1000 is ? ²%³ ~ ÀcÀ% for % € .
The pdf of ? is ? ²%³ ~ À
h ²À³cÀ²%°À³ ~ h c%° for % € À
This is the pdf for the exponential distribution with mean 1100. Another way to see this is that
since the exponential distribution is a scale distribution with scale parameter (the mean), À?
will also be exponential with mean À.
(iii) The pdf of ? is ? ²%³ ~ ²%b³b for % € .
´²b³µ
The pdf of ? is ? ²%³ ~ b
h ² % ~
b
´%b²b³µb for % € .
b b ³
This is the pdf for the Pareto distribution with parameters and ² b ³ .
Another way to see this is that since the Pareto distribution is a scale distribution with scale
parameter , À? will also be Pareto with mean the same and Z ~ ² b ³ .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 15 - ADDITIONAL POLICY ADJUSTMENTS
LM-193
A limited loss expected value formula for inflation
Care must be taken when combining inflation with the policy adjustments of deductible and
policy limit. It is important to determine what deductible and policy limit are being applied to the
after-inflation loss variable; they may be the same as those applied to the current loss variable, or
they might not be the same (for instance, they may be subject to the same inflation factor).
The formulations of after-inflation expected cost per loss and expected cost per payment can
make use of the following relationship. Suppose that € and ? is a loss random variable.
Then ,´? w "µ ~ h ,´? w " µ .
(15.8)
Suppose that ? is the before-inflation random variable and there is a policy deductible and
maximum covered loss of ". Then the expected cost per loss (before inflation) is
,´? w "µ c ,´? w µ . If the same deductible and limit apply after inflation at rate , then the
expected cost per loss after inflation is
"
3
,´² b ³? w "µ c ,´² b ³? w µ ~ ² b ³ h 2,´? w b
µ c ,´? w b
µ (15.9)
and 7 ´² b ³? € µ ~ 7 ´? € b
µ ~ c -? ² b
³
( b replaces in this formulation).
(15.10)
Keep in mind that if ? has a scale distribution, then we may be able to use Table C formulations
to find ,´² b ³? w µ . For instance, if ? is exponential with mean 100 and inflation of 10%
is applied, and then a policy limit of 200 is imposed after inflation, the limited expected value
would be ,´À? w µ ~ ,´A w µ , where A has an exponential distribution with mean
110. Using the Table formulation for limited expected value, we see that
,´A w µ ~ ² c c° ³ . We could also have used the formulation from the previous
paragraph, ,´À? w µ ~ À h ,´? w À µ . Since ? is exponential with mean 100, this
c °
becomes À h ²³² c À
³ ~ ² c c° ³ .
LM-15.3 Policy Deductible in Combination with Maximum Covered Loss " and
Coinsurance Factor and Inflation Factor We summarize expressions for the expected value of the after-inflation random variables @3 (cost
per loss) and of @7 (cost per payment) when there is a maximum covered loss ", a policy
deductible , a coinsurance factor and inflation factor . Those results assume that the same
policy deductible and the same maximum covered loss " are applicable to both before and
after-inflation losses. The objective of those results is to express the after inflation ,´@3 µ and
,´@7 µ in terms of the before-inflation ground up loss random variable ? . Note that for the after
inflation @3 , 7 ´@3 € µ ~ 7 ´² b ³? € µ ~ 7 ´? € b
µ ~ c -? ² b
³.
(15.11)
Expected cost per loss ~ ,* 3 ~ ,´@3 µ
"
~ ² b ³4,´? w b
µ c ,´? w b
µ5 Á
(15.12)
Expected cost per payment ~ ,* 7 ~ ,´@7 µ ~ ,´@3 O@3 € µ ~ ,´@3 O? € b
µ
~
,* 3
7 ´@3 €µ
© ACTEX 2009
~
,´@3 µ
c-? ² b
³
~
"
µc,´?w b
µ5
²b³4,´?w b
c-? ² b
³
À
(15.13)
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-194
MODELING SECTION 15 - ADDITIONAL POLICY ADJUSTMENTS
The second moment of @3 is
" ,´@3 µ ~ ² b ³ 4,´²? w b
³ µ c ,´²? w b
³ µ5
"
c ² b ³ h 4,´? w b
µ c ,´? w b
µ5 ,
and the second moment of @7 is ,´@7 µ ~
,´@3 µ
c-? ² b
³
(15.14)
À
(15.15)
Note that since the after-inflation loss is ? ~ ² b ³? , it follows that ? € is true if
"
? € b
(similarly for "). That is how the factors b
and b
arise in formulating afterinflation expectations in terms of the before-inflation loss random variable ? . The amount
@3 (cost per loss) paid by the insurer can be expressed in terms of either ? or ?
Á
if ?  b
Á
if ?  @3 ~ H ´? c µ if  ?  " ~ H ´² b ³? c µ if  ?  " .
b
b
"
if ? ‚ "
²" c ³
if ? ‚ b
²" c ³
(15.16)
Note also that if the after-inflation policy deductible or limit are, say Z and "Z c Z (instead of
being equal to the before-inflation values and " c ), then the expressions for after-inflation
,´@3 µ and ,´@7 µ remain the same, except that "Z replaces " and Z replaces .
Example LM15-2: You are given the following:
- Losses follow a distribution (Prior to the application of any deductible) with cumulative
distribution function and limited expected values as follows:
Loss Size (%)
- ²%³
,²? w %³
Á À
Á Á À
Á Á À
Á B
À
Á - There is a deductible of 15,000 per loss and no policy limit.
The amount paid by the insurer per payment made is @7 .
(a) Determine the expected value of @7 .
(b) After several years of inflation, all losses have increased in size by 50%, but the deductible
has remained the same. Determine the expected value of @7 (after-inflation).
Solution: (a) @3 ~ insurer's payment if loss is ? , so that
@3 ~ if ?  Á and @3 ~ ? c Á if ? ‚ Á .
@7 is the amount paid given that a payment is made.
With a deductible of and no policy limit, the conditional expectation of payment amount given
that a payment is made is the mean excess loss
,´@7 µ ~
,´?µc,´?wµ
c- ²³
~
,´?wBµc,´?wÁµ
c- ²Á³
~ Á .
(b) With deductible and inflation rate and no policy limit,
,´@7 µ ~ ,´AO? € b
µ~
²À³²Ác
Á³
²b³4,´?µc,´?w b
µ5
c- ´ b
µ
~
²À³²,´?µc,´?w Á
À µ³
c- ² Á
³
À
~
~ Á .
cÀ
Note the expected cost per loss grows from ,´?µ c ,´? w Á µ ~ Á before inflation,
Á
to ,´Aµ ~ ²À³4,´?µ c ,´? w À µ5 ~ Á after inflation, increasing over 50%.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODEL - PROBLEM SET 15
LM-195
MODELING - PROBLEM SET 15
Additional Policy Adjustments - Section 15
1. ? has a uniform distribution on the interval ´Á µ. With an inflation rate of ~ À, a
deductible of ~ , and a maximum covered loss of " ~ (before application of
deductible), find the expected cost per payment after inflation.
A) 28
B) C) D) E) 2. A group dental policy has a negative binomial claim count distribution with mean 300 and
variance 800. Ground-up severity is given by the following table:
Severity Probability
40
0.25
80
0.25
120
0.25
200
0.25
You expect severity to increase 50% with no change in frequency. You decide to impose a per
claim deductible of 100. Calculate the expected total claim payment after these changes.
A) Less than 18,000
B) At least 18,000, but less than 20,000
C) At least 20,000, but less than 22,000
D) At least 22,000, but less than 24,000
E) At least 24,000
²%°³  %   %
An insurance pays 80% of the amount of the loss in excess of an ordinary deductible of 20,
subject to a maximum payment of 60 per loss. Calculate the conditional expected claim payment,
given that a payment has been made.
A) 37
B) 39
C) 43
D) 47
E) 49
3. (SOA) Loss amounts have the distribution function - ²%³ ~ F
4. (SOA) A risk has a loss amount which has a Poisson distribution with mean 3. An insurance
covers the risk with an ordinary deductible of 2. An alternative insurance replaces the deductible
with coinsurance of , which is the proportion of the loss paid by the insurance, so that the
expected insurance cost remains the same. Calculate .
A) 0.22
B) 0.27
C) 0.32
D) 0.37
E) 0.42
5. (SOA) In 2005 a risk has a two-parameter Pareto distribution with ~ and ~ À
In 2006 losses inflate by 20%. An insurance on the risk has a deductible of 600 in each year.
7 , the premium in year , equals 1.2 times expected claims. The risk is reinsured with a
deductible that stays the same in each year. 9 , the reinsurance premium in year , equals 1.1
times the expected reinsured claims. 9 °7 ~ À .
Calculate 9
°7
.
A) 0.46
B) 0.52
C) 0.55
D) 0.58
E) 0.66
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-196
MODELING - PROBLEM SET 15
MODELING - PROBLEM SET 15 SOLUTIONS
1. ,´@ µ ~
~
"
²b³4,´?w b
µc,´?w b
µ5
c- ² b
³
²À³4,´?w
Àµc,´?wÀµ5
c- ²À³
~
The cdf for the uniform distribution on
,´@ µ ~
À
!
²À³À ´c µ!
À
c ~
²À³4,´?w À
µc,´?w À
µ5
c- ² À
³
À
²À³À ´c- ²!³µ!
.
c- ²À³
!
´Á µ is - ²!³ ~ , so
that
~ À À Answer: C
2. After the 50% increase, severity is
À
À
À
À
Expected total claim is ,´5 µ h ²,´?µ c ,´? w µ³ , where ? is the severity after the 50%
increase. We are given ,´5 µ ~ . From the severity (after increase) we get
,´?µ ~ ²
b b b ³²À³ ~ , and
,´? w µ ~ ²
³²À³ b ²³²À³ ~ .
The expected total claim is ²³²
c ³ ~ Á .
Answer: D
3. The maximum payment of 60 is triggered by a loss of 95 (95 minus 20 is 75, and 80% of 75 is
the maximum payment of 60).
?  The cost per loss is @3 ~ H À²? c ³  ?  ~ À´²? w ³ c ²? w ³µ .
? € The expected cost per loss is
,´@3 µ ~ ²À³²,´? w µ c ,´? w µ³ ~ À ´ c - ²%³µ %
~ À ´ c À% µ % ~ À² c À³ ~ À .
,´@ µ
À
À
3
The expected cost per payment is ,´@7 µ ~ 7 ²?€³
~ c²³ ~ À
~ À .
Answer: B
4. The expected cost of the original insurance is ,´²? c ³b µ ~ ,´?µ c ,´? w µ , where ?
has a Poisson distribution with mean ,´?µ ~ . Then
,´? w µ ~ 7 ²? ~ ³ b 7 ²? ‚ ³ ~ 7 ²? ~ ³ b ´ c 7 ²? ~ Á ³µ
~ c b ´ c ²c b c ³µ ~ c c .
The expected insurance cost is c ² c c ³ ~ À .
The alternative insurance has expected cost ,´?µ ~ .
In order for this to be the same expected cost as the original insurance, we must have
~ À
Answer: E
~ À .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODEL - PROBLEM SET 15
LM-197
5. Suppose that ? is the risk in 2005. Expected claims in 2005 is
,´²? c ³b µ ~ ,´?µ c ,´? w µ ~ c c c ´ c ² b0 ³µ ~ .
The premium in 2005 is 7 ~ ²À³²³ ~ .
In 2006, the risk is À? . This will be Pareto with ~ and ~ À²³ ~ (the Pareto is a scale distribution with scale parameter ).
With deductible 600 in 2006, the premium in 2006 is
7
~ ²À³² c ³² b
³ ~ .
Suppose that the deductible for the reinsurance is . Then the reinsurance based on risk ? has a
deductible of b . The reinsurance premium in 2005 is
9 ~ ²À³² c ³² bb ³ .
We are given that ²À³² c ³² bb ³, ~ À , so that ~ .
With a reinsurance deductible of 2400 in 2006, we have
9
~ ²À³² c ³² bb
³ ~ , and 9
°7
~ À . Answer: D
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-198
© ACTEX 2009
MODELING - PROBLEM SET 15
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 16 - MODELS FOR THE NUMBER OF CLAIMS, THE ²Á Á ³ CLASS
LM-199
MODELING SECTION 16
MODELS FOR THE NUMBER OF CLAIMS AND THE ²Á Á ³ CLASS
The material in this section relates to Loss Models, Sections 6.1-6.6 and 6.7.
The suggested time for this section is 2 hours.
The last several sections have looked at the distribution of loss amount and adjustments that can
be applied to the loss amount. The loss amount or the amount paid by the insurer may be referred
to as the severity distribution. An insurer may have a portfolio of a large number of policies and
we may also be interested in the number of losses that occur in some particular period of time.
The distribution of the number of losses in a unit time is referred to as the frequency
distribution.
Over a fixed period of time such as one month or one year, the total amount of all losses is the
aggregate loss, and the total of all amounts paid is the aggregate payment for the period. There
are two main models used to represent aggregate losses or payments.
Collective risk model: This model assumes that the number of losses (or payments) 5 in the
period is a discrete integer-valued random variable, and the amount of each loss (or payment) is a
random variable coming from the distribution of ? (? may have a continuous, discrete, or mixed
distribution). For each loss (or payment) amount ? Á
for ~ Á Á ÀÀÀÁ 5 , ? has the distribution of ? . The aggregate loss (payment) for the period is
the random variable : ~ ? b ? b Ä b ?5 .
(16.1)
: is a random sum because 5 , the number of terms in the sum is a random variable, and each ?
is also a random variable. Such a random sum is also referred to as a compound distribution.
For our purposes, for the most part, it will be assumed that 5 Á ? Á ? Á ÀÀÀÁ ?5 are mutually
independent. Compound distributions will be considered in more detail in a later section
Individual risk model: In this model, the portfolio is made up of a fixed number, say , of risks
(usually insurance policies). For ~ Á Á ÀÀÀÁ , ? is the random variable representing the loss
(or payment) generated by risk in a particular period. The ? 's do not necessarily have the same
distribution, but are usually assumed to be mutually independent. The aggregate loss (payment)
in the period is : ~ ? b ? b Ä b ? .
(16.2)
Notice that is not random.
Chapters 4 and 5 of the Loss Models book look at the various distributions and variations on
those distributions that can be considered for ? , the loss amount (also called the severity) random
variable. Chapter 6 considers models for the claim number random variable, which is often
denoted 5 . Chapter 9 of the book analyzes aggregate claim distributions. In this section of the
notes we are reviewing parts of Chapter 6 of the Loss Models book on models for the number of
claims.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 16 - MODELS FOR THE NUMBER OF CLAIMS, THE ²Á Á ³ CLASS
LM-200
LM-16.1 Probability Generating Function of a Discrete Random Variable
The number of claims in a specific period of time (such as one month) is modeled as a discrete
non-negative integer-valued random variable 5 , with probability function
(16.3)
7 ´5 ~ µ ~ Á ~ Á Á Á ÀÀÀ
B
The probability generating function of 5 is 75 ²!³ ~ ,´!5 µ ~ h ! .
~
The probability generating function can be used to find certain expected values related to 5 as
well as the probability function values, 7 ²5 ~ ³ ~ .
Successive derivatives of 75 ²!³ evaluated at ! ~ have the following pattern.
²³
²³
75 ²³ ~ Á 75Z ²³ ~ Á 75ZZ ²³ ~ Á 75 ²³ ~ Á ÀÀÀ Á 75 ²³ ~ [ h Á ÀÀÀ
(16.4)
Successive derivatives of 75 ²!³ evaluated at ! ~ have a patter involving expectations related
to 5 .
75Z ²³ ~ ,´5 µ Á 75ZZ ²³ ~ ,´5 ²5 c ³µ ~ ,´5 µ c ,´5 µ ,
²³
75 ²³ ~ ,´5 ²5 c ³²5 c ³µ , and in general,
²³
75 ²³ ~
!
75 ²!³c
!~
~ ,´5 ²5 c ³Ä²5 c b ³µ
(16.5)
If 5 Á 5 Á ÀÀÀÁ 5 are independent discrete non-negative integer-valued random variables, and
~
~
4 ~ 5 , then 74 ²!³ ~ 75 ²!³ .
(16.6)
A minor point to note is that the probability generating function is closely related to the moment
generating function. 75 ²!³ ~ ,´!5 µ ~ ,´² !³5 µ ~ 45 ² !³ .
Example LM16-1: 5 has the following distribution.
7 ´5 ~ µ ~ ~ À Á 7 ´5 ~ µ ~ ~ À Á 7 ´5 ~ µ ~ ~ À Á 7 ´5 ~ µ ~ ~ À .
Find the probability generating function of 5 and use it to find the mean and variance of 5 .
Solution:
B
75 ²!³ ~ h ! ~ ²À³ h ! b ²À³ h ! b ²À³ h ! b ²À³ h ! ~ À b À! b À! b À! .
~
7 Z ²!³ ~ À b À
! b À
! S 7 Z ²³ ~ À ~ ,´5 µ ~ h ~ d À b d À b d À b d À
~
7 ZZ ²!³ ~ À
b À! S 7 ZZ ²³ ~ À ~ ,´5 µ c ,´5 µ ~ ,´5 µ c À S ,´5 µ ~ À .
Finally, = ´5 µ ~ ,´5 µ c ²,´5 µ³ ~ À c ²À³ ~ À .
The distributions summarized here are the Poisson, Negative Binomial and Binomial. They are
all discrete, non-negative integer-valued random variables. Their descriptions are also in the
Exam C Tables, so it is not necessary to memorize the probability function, mean, variance or
probability generating functions. Also, note that there has been little reference to probability
generating functions on Exam C questions.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 16 - MODELS FOR THE NUMBER OF CLAIMS, THE ²Á Á ³ CLASS
LM-201
LM-16.2 Poisson Distribution
The Poisson distribution is described in terms of a single positive parameter, often denoted
€ . The probability function of the Poisson random variable 5 with parameter is
c ~ 7 ´5 ~ µ ~ [ for ‚ .
c
c
For instance, if ~ , then ~ c Á ~ c Á ~ [
~ c Á ~ Á À À À
The mean and variance of 5 are both equal to the parameter , ,´5 µ ~ = ´5 µ ~ .
(16.7)
(16.8)
The probability generating function of 5 is 75 ²!³ ~ ²!c³ .
!
The moment generating function of 5 is 45 ²!³ ~ ² c³ ~ 75 ²! ³ À
If 5 Á 5 Á ÀÀÀÁ 5 are independent Poisson random variables with parameters Á Á ÀÀÀÁ ,
respectively, then @ ~ 5 has a Poisson distribution with parameter ~ b b Ä b .
~
The Poisson distribution is often used as a model for the number of events per unit time
Suppose that 5 has a Poisson distribution with parameter , and 5 represents the overall number
of events in a certain time period. Suppose also that each time an event occurs it can be any one
of distinct types of events, and suppose that given an event occurs, the probability that it is of
type is ( b b Ä b ~ ). Then for each ~ Á Á ÀÀÀÁ , the number of events of type per period, say 5 , has a Poisson distribution with parameter . For instance, suppose that the
number of cars arriving for service at a service facility in one week has a Poisson distribution
with a mean of 20. Suppose that each car is classified as either domestic or foreign. Suppose also
that each time a car arrives for service there is a 75% chance that it domestic and a 25% chance
that it is foreign. Then 5 , the number of domestic cars arriving per week for service has a
Poisson distribution with a mean of ²³²À³ ~ , and 5 , the number of foreign cars arriving
per week is Poisson with a mean of ²³²À³ ~ . Furthermore, 5 and 5 are independent.
Example LM16-2: 5 has a Poisson distribution with a mean of 3.
Find ,´²5 c ³b µ and = ´5 w µ À
Solution:
7 ²5 ~ ³ ~ c
7 ²5 ~ ³ ~ c
5 w~
.
c
7 ²5 ~ ³ ~ [ ~ Àc
7 ²5 ‚ ³ ~ c Àc
,´5 w µ ~ c b ²³²Àc ³ b ²³² c Àc ³ ~ c Àc ~ À .
,´²5 w ³ µ ~ c b ² ³²Àc ³ b ² ³² c Àc ³ ~ c Àc ~ À
.
= ´5 w µ ~ À
c ²À³ ~ À .
H
,´²? c ³b µ ~ ,´?µ c ,´? w µ ~ c ´c b ² c c ³µ ~ À .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-202
MODELING SECTION 16 - MODELS FOR THE NUMBER OF CLAIMS, THE ²Á Á ³ CLASS
LM-16.3 Negative Binomial Distribution
The negative binomial distribution is defined in terms of two parameters, and , both of which
are € . The probability function of the negative binomial random variable 5 with parameters
€ and € is ( and do not have to be integers, but is often an integer)
bc 2 3 2 3
~ 7 ´5 ~ µ ~ 4
, ~ Á Á Á ÀÀÀ
(16.9)
5
b
b
%
[
The factor 2 3 is a generalization of the binomial quantity 2 3 ~ [ ²c³[
,
2 % 3 ~
²%³²%c³Ä²%cb³
[
!²%b³
(16.10)
~ !²b³!²%cb³ for integer ‚ , and any real % .
%
%
Note that it is always true for any % that 2 3 ~ and 2 3 ~ % .
(16.11)
The mean and variance of 5 are ,´5 µ ~ and = ´5 µ ~ ² b ³;
note that ,´5 µ  = ´5 µ .
(16.12)
The probability generating function of 5 is 75 ²!³ ~ ´c ²!c³µ
.
(16.13)
The moment generating function of 5 is 45 ²!³ ~ ´c ²! c³µ .
(16.14)
If ~ , this distribution is referred to as the geometric distribution,
and ~ ²b ³b .
(16.15)
If 5 Á 5 Á ÀÀÀÁ 5 are independent Negative Binomial random variables with a common parameter
and with -parameters Á Á ÀÀÀÁ , respectively, then @ ~ 5 has a Negative Binomial
distribution with parameters and ~ b b Ä b .
~
Here is an example of negative binomial probability calculations for the negative binomial 5
with ~ and ~ :
bc 32 3 2 3
7 ´5 ~ µ ~ 2
~ ²³² ³²³ ~ 2 3 ~ ,
b
b
b
bc
3 2 3
7 ´5 ~ µ ~ 2 32 b
~ ²³² ³² ³ ~ 2 b 3 2 b 3 ~
b
bc
3 2 3
and 7 ´5 ~ µ ~ 2 32 b
~ ²³² ³² ³ ~ .
b
If is an integer, then the negative binomial random variable 5 can be interpreted as follows.
Suppose that an experiment ends in either failure or success, and the probability of success for a
particular trial of the experiment is . Suppose further that the experiment is performed repeatedly
(independent trials) until the -th success occurs. If 5 is the number of failures until the -th
c
success occurs, then 5 has a negative binomial distribution with parameters and ~ .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 16 - MODELS FOR THE NUMBER OF CLAIMS, THE ²Á Á ³ CLASS
LM-203
Mixture of the Poisson distribution over the gamma distribution
This was considered in Section 9 earlier in these notes under the topic "Continuous Mixing
Distributions" on Section 5.2.4 of the Loss Models book. In Example LM9-2 of Section 9 we
considered this particular mixture.
If the conditional distribution of 5 given $ ~ is Poisson with parameter
(and mean) , and if $ has a gamma distribution with parameters and (as defined in the Exam C Table of distributions), then the unconditional
distribution of 5 is negative binomial with parameters ~ and ~ .
Example LM16-3: The conditional distribution of @ given $ ~ is Poisson with parameter
(and mean) . The distribution of $ is exponential with parameter (and mean) ~ . Find the pf
of the unconditional distribution of @ .
Solution: Note first that the exponential distribution is a special case of the gamma distribution
with parameter ~ and exponential parameter . Then, using the principle in the previous
paragraph, the unconditional distribution of @ is negative binomial with ~ ~ and
~ ~ . If ~ , the negative binomial is the geometric distribution.
The probability function is ~ ²b ³b ~ b
.
LM-16.4 Binomial Distribution
The probability function of the binomial distribution with parameters (an integer ‚ ) and (with   ) is ~ 7 ´5 ~ µ ~ 2 3 ² c ³c , ~ Á Á ÀÀÀÁ .
(16.16)
If an experiment is performed times, independent of one another, and is the probability of a
"successful event" on a particular trial, then is the probability of exactly successes in the trials.
The mean and variance of 5 are ,´5 µ ~ and = ´5 µ ~ ² c ³ ;
note that ,´5 µ € = ´5 µ .
(16.17)
The probability generating function is 75 ²!³ ~ ´ b ²! c ³µ .
(16.18)
The moment generating function is 45 ²!³ ~ ´ b ²! c ³µ .
(16.19)
Example LM16-4: Smith and Jones each write the same multiple choice test. The test has 5
questions, and each question has 5 answers (exactly one of which is right). Smith and Jones are
not very well prepared for the test and they answer the questions randomly. Find the probability
that they both get the same number of answers correct.
Solution: Let ? be the number of answers that Smith gets correct. Then ? has a binomial
distribution with ~ , ~ À , and the probability function is 7 ´? ~ µ ~ 2 3²À³ ²À³c À
?¢
²%³ ¢ À
© ACTEX 2009
À
À
À
À
À
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-204
MODELING SECTION 16 - MODELS FOR THE NUMBER OF CLAIMS, THE ²Á Á ³ CLASS
LM16-4 continued
The number of answers Jones gets correct, say @ , has the same distribution. Then,
7 ´? ~ @ µ ~ 7 ´²? ~ ³ q ²@ ~ ³µ b 7 ´²? ~ ³ q ²@ ~ ³µ b Ä b 7 ´²? ~ ³ q ²@ ~ ³µ
~ 7 ´? ~ µ h 7 ´@ ~ µ b 7 ´? ~ µ h 7 ´@ ~ µ b Ä b 7 ´? ~ µ h 7 ´@ ~ µ
(this follows from independence of ? and @ ), which is equal to
²À
³ b ²À
³ b ²À³ b ²À³ b ²À
³ b ²À³ ~ À .
LM-16.5 The ²Á Á ³ Class of Discrete Distributions
The three discrete distributions that have been considered (Poisson, Negative Binomial and
Binomial) are members of a class of discrete non-negative integer-valued distributions called the
²Á Á ³ class. A discrete non-negative integer-valued random variable with probability function
Á ~ Á Á Á Á ÀÀÀ is a member of the ²Á Á ³ class if there are constants and such that for
all ~ Á Á Á ÀÀÀ, the probability function satisfies the relationship ~ b .
(16.20)
c
²c ³°[
The Poisson with parameter has ~ ²c c ³°²c³[ ~ , so that ~ Á ~ À
c
(16.21)
²c³
For the Negative Binomial distribution with parameters and , we have ~ b Á ~ b
(16.22)
(recall that the Geometric distribution with parameter is a special case off the Negative
Binomial with ~ )À
For the Binomial distribution with parameters and , we have ~ c c Á ~
²b³
c
À
(16.23)
Example LM16-5: You are given that 5 is a member of the ²Á Á ³ class of distributions, and
you are given that ~ À and ~ . Find 7 ´5 ~ µ and find the mean of 5 .
Solution: Since   and € , it follows that 5 must have a negative binomial
²c³
distribution with ~ b ~ À S ~ and ~ b ~ S ~ .
bc
3 2 3
Then 7 ´5 ~ µ ~ 2 32 b
~ , and ,´5 µ ~ ~ d ~ À
b
Example LM16-6: You are given that 5 is a member of the ²Á Á ³ class of distributions, and
~ À
Á ~ À
and ~ À
. Determine the distribution of 5 À
À
Solution: ~ b ~ À
À
~ and ~ b ~ À
~ À .
Solving the to equations for and , we get ~ À and ~ c À .
The only ²Á Á ³ class distribution with  is the binomial, so that
~ c À ~ c c S ~ À and ~ À ~
© ACTEX 2009
²b³
c
~
À²b³
À
S ~.
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 16 - MODELS FOR THE NUMBER OF CLAIMS, THE ²Á Á ³ CLASS
LM-205
LM-16.6 The ²Á Á ³ Class of Discrete Distributions
If 5 is a discrete non-negative integer-valued random variable with probability function , it is
possible to create a related random variable 5 i that has i ~ 7 ²5 i ~ ³ equal to any number in
ci
the interval  i  , and i ~ 7 ²5 i ~ ³ ~ c h .
5 i is said to have a zero-modified distribution, and in the particular case that i is chosen to be
0, we say that 5 i has a zero-truncated distribution .
In the Loss models book, this is applied to the ²Á Á ³ class of distributions, while still
maintaining the recursive relationship ~ b for ~ Á Á ÀÀÀ . Remember that ²Á Á ³
c
really means that the distribution is either Poisson, Negative Binomial, or Binomial. These
variations on the ²Á Á ³ class are referred to as the ²Á Á ³ class of distributions.
Zero Truncation
The probabilities in a zero-truncated distribution are denoted ; , and we set ; ~ .
In this case, the other probabilities are defined to be ; ~ c
for ~ Á Á Á ÀÀÀ
( and are from the distribution being truncated).
,´5 µ
The mean of the zero-truncated distribution is ,´5; µ ~ c
75 ²!³c
;
and the probability generating function is 75 ²!³ ~ c
,
(16.24)
(16.25)
(16.26)
where 75 ²!³ is the probability generating function of the distribution being truncated.
Zero Modification
The probabilities in a zero-modified distribution are denoted 4 . To construct the zero-modified
distribution, the first step is to choose the value of 4 that the distribution will have (it can be
any number between 0 and 1). Once 4 has been chosen, the other probabilities are
c4
4 ~ c h for ~ Á Á Á ÀÀÀ
(16.27)
Note that a zero-truncated distribution is really a zero-modified distribution with 4 ~ .
c4
The mean of the zero-modified distribution is ,´54 µ ~ c h ,´5 µ
c4
and the probability generating function is 754 ²!³ ~ 4 b c h ´75 ²!³ c µ À
(16.28)
(16.29)
Note that the zero-modified distribution 54 is a mixture of the single point 0 with a mixing
weight (probability at 0) of 4 , and the zero-truncated distribution 5; with mixing weight
² c 4 ³ .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-206
MODELING SECTION 16 - MODELS FOR THE NUMBER OF CLAIMS, THE ²Á Á ³ CLASS
The extended truncated negative binomial (ETNB) distribution
This variation on a negative binomial distribution allows the parameter to be € c . This
distribution has ; ~ , and for ~ Á Á ÀÀÀ the ; 's follows the ²Á Á ³ negative binomial
;
²c³
recursion relationship ; ~ b , with ~ b Á ~ b .
c
(16.28)
A special case of this occurs with ~ , and is called the logarithmic distribution, which has
²
³
b
probabilities ; ~ ²b
³ for ~ Á Á ÀÀÀ .
(16.29)
Example LM16-7: A Poisson random variable has parameter ~ .
(a) Find Á and and the mean and variance for the zero-truncated version of this
distribution.
(b) Find Á and and the mean for the zero-modified version of this distribution
with 4 ~ À.
Solution: (a) The truncated version must have ; ~ , and then ; ~ c
~ À for
‚ . For the original Poisson distribution, ~ c ~ c ~ À
Á ~ c h [
~ À
,
À
À
and ~ c h [ ~ À , so that ; ~ cÀ
~ À , and ; ~ cÀ
~ À .
;
Since ~ c for ‚ , it follows that
B
B
B
B
,´5; µ ~ h ; ~ h ; ~ h c
~ c
h h ~
~
~
~
B
,´5 µ
~ c
h h ~ c ~ cÀ
~ À .
~
Note that this is the same as ,´5 O5 € µ .
B
,´5 µ
To find = ´5; µ we need ,´5; µ ~ h ; ~ c . Since 5 has a Poisson distribution
~
with mean 1, it follows that ,´5 µ ~ = ´5 µ b ²,´5 µ³ ~ b ~ , so that
,´5; µ ~ cÀ
~ À
. Then = ´5; µ ~ À
c ²À³ ~ À
.
c4
À
(b) The modified version has 4 ~ À, and then ; ~ c h ~ cÀ
h ~ À .
4
4
Then, ~ ²À³²À
³ ~ À Á ~ À . Following the same reasoning as in (a),
B
B
B
4
4
B
c
c
,´5 4 µ ~ h 4 ~ h 4 ~ c h ~ c h h ~
~
~
~
c4
B
c4
~ c h h ~ c h ,´5 µ ~ ²À³,´5 µ ~ À .
~
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 16
LM-207
MODELING - PROBLEM SET 16
Models for the Number of Claims - Section 16
1. Given (   ) , 5 has a distribution which is geometric with 7 ´5 ~ µ ~ . If is
continuously uniformly distributed on the interval ´ÀÁ Àµ , what is ,´5 µ?
A) 0.
B) .
C) .
D) .
E) .
2. A population is equally divided into two classes of drivers. The number of accidents per
individual driver is Poisson for all drivers. For a driver selected at random from Class I, the
expected number of accidents is uniformly distributed over (0.2 , 1.0). For a driver selected at
random from Class II, the expected number of accidents is uniformly distributed over (0.4 , 2.0).
For a driver selected at random from this population, determine the probability of zero accidents.
A) 0.41
B) 0.42
C) 0.43
D) 0.44
E) 0.45
3. The distribution of the number of claims, 5 , given $, is Poisson with parameter $. The
distribution of $ is exponential with mean 1. Determine 7 ´5 ~ µ .
A) 0.00
B) 0.05
C) 0.20
D) 0.35
E) 0.50
4. (CAS) The Independent Insurance Company insures 25 risks, each with a 4% probability of
loss. The probabilities of loss are independent. On average, how often would 4 or more risks
have losses in the same year?
A) Once in 13 years
B) Once in 17 years
C) Once in 39 years
D) Once in 60 years
E) Once in 72 years
5. (CAS) A new actuarial student analyzed the claim frequencies of a group of drivers and
concluded that they were distributed according to a negative binomial distribution and that the
two parameters, and , were equal. An experienced actuary reviewed the analysis and pointed
out the following:
"Yes, it is a negative binomial distribution. The parameter is fine, but the value of the parameter is wrong. Your parameters indicate that of the drivers should be claim-free,
but in fact, of them are claim-free."
Based on this information, calculate the variance of the corrected negative binomial distribution.
A) 0.50
B) 1.00
C) 1.50
D) 2.00
E) 2.50
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-208
MODELING - PROBLEM SET 16
6. (CAS) A pharmaceutical company must decide how many experiments to run in order to
maximize its profits.
• The company will receive a grant of $1 million if one or more of its experiments is successful.
• Each experiment costs $2,900.
• Each experiment has a 2% probability of success, independent of the other experiments.
• All experiments are run simultaneously.
• Fixed expenses are $500,000.
• Ignore investment income.
The company performs the number of experiments that maximizes its expected profit. Determine
the company's expected profit before it starts the experiments.
A) 77,818
B) 77,829
C) 77,840
D) 77,851
E) 77,862
7. (CAS May 05) A service guarantee covers 20 television sets. Each year, each set has a 5%
chance of failing. These probabilities are independent. If a set fails, it is replaced with a new set at
the end of the year of failure. This new set is included under the service guarantee. Calculate the
probability of no more than 1 failure in the first two years.
A) Less than 40.5%
B) At least 40.5%, but less than 41.0%
C) At least 41.0%, but less than 41.5%
D) At least 41.5%, but less than 42.0%
E) 42.0% or more
8. (CAS May 05) Which of the following are true regarding sums of random variables?
1. The sum of two independent negative binomial distributions with parameters ² Á ³ and
² Á ³ is negative binomial if an only if ~ .
2. The sum of two independent binomial distributions with parameters ² Á ³ and ² Á ³ is
binomial if and only if ~ .
3. The sum of two independent Poisson distributions with parameters and is Poisson if and
only if ~ .
A) None are true
B) 1. only
C) 2. only
D) 3. only
E) 1. and 3. only
9. (CAS May 05) You are given a negative binomial distribution with ~ À and ~ .
For what value of does take on its largest value?
A) Less than 7
B) 7
C) 8
D) 9
E) 10 or more
10. (CAS May 05) Longterm Insurance Company insures 1000,000 drivers who have each been
driving for at least 5 years. Each driver gets violations at a Poisson rate of 0.5 per year. Currently,
drivers with 1 or more violations in the past three years pay a premium of 1000. Drivers with 0
violations in the past 3 years pay 850. Your marketing department wants to change the pricing so
that drivers with 2 or more accidents in the past five years pay 1,000 and drivers with zero or one
violation sin the past 5 years pay ? . Find ? so that the total premium revenue for your firm
remains constant when this change is made.
A) Less than 900
B) At least 900, but less than 925
C) At least 925, but less than $950
D) At least $950, but less than $975
E) 975 or more
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 16
LM-209
11. (SOA) The distribution of accidents for 84 randomly selected policies is as follows:
Number of Accidents
Number of Policies
0
32
1
26
2
12
3
7
4
4
5
2
6
1
Total
84
Which of the following models best represents these data?
A) Negative binomial
B) Discrete uniform
C) Poisson
D) Binomial
E) Either Poisson or Binomial
12. (CAS May 06) Total claim counts generated from a portfolio of 1,000 policies follows a
Negative Binomial distribution with parameters ~ and ~ À .
Calculate the variance in total claim counts if the portfolio increases to 2,000 policies.
A) Less than 1.0
B) At least 1.0, but less than 1.5
C) At least 1.5, but less than 2.0
D) At least 2.0, but less than 2.5
E) At least 2.5
13. 5 has a geometric distribution with a mean of 2. Describe the probability functions, the
mean and the variance of the zero-truncated distribution and the zero-modified distribution with
4 ~ .
14.(CAS) Vehicles arrive at the Bun-and-Run drive-thru at a Poisson rate of 20 per hour. On
average, 30% of these vehicles are trucks. Calculate the probability that at least 3 trucks arrive
between noon and 1:00 PM.
A) Less than 0.80
B) At least 0.80, but less than 0.85
C) At least 0.85, but less than 0.90
D) At least 0.90, but less than 0.95
E) At least 0.95
15.(CAS) Coins are tossed into a fountain according to a Poisson process with a rate of one every
three minutes. The coin denominations are independently distributed as follows:
Probability
Coin Denomination
Penny
0.5
Nickel
0.2
Dime
0.2
Quarter
0.1
Calculate the probability that the fourth dime is tossed into the fountain in the first two hours.
A) Less than 0.89
B) At least 0.89, but less than 0.92
C) At least 0.92, but less than 0.95
D) At least 0.95, but less than 0.98
E) At least 0.98
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-210
MODELING - PROBLEM SET 16
16. (CAS) You are given:
• Claims are reported at a Poisson rate of 5 per year.
• The probability that a claim will settle for less than $100,000 is 0.9.
What is the probability that no claim of $100,000 or more is reported for the next 3 years?
A) 20.59%
B) 22.31%
C) 59.06%
D) 60.65%
E) 74.08%
17. (CAS) XYZ Insurance Company introduces a new policy and starts a sales contest for 1000
of its agents. Each makes a sale of the new product at a Poisson rate of 1 per week. Once an
agent has made 4 sales, he gets paid a bonus of $1000. The contest ends after 3 weeks. Assuming
0% interest, what is the cost of the contest?
A) $18,988
B) $57,681
C) $168,031
D) $184,737
E) $352,768
18. (CAS May 2005) For Broward County, Florida, hurricane season is 24 weeks long. It is
assumed that the time between hurricanes is exponentially distributed with a mean of 6 weeks. It
is also assumed that 30% of all hurricanes will hit Broward County. Calculate the probability that
in any given hurricane season, Broward County will be hit by more than 1 hurricane.
A) Less than 15%
B) At least 15%, but less than 20%
C) At least 20%, but less than 25%
D) At least 25%, but less than 30%
E) 30% or more
19. (SOA) ? is a discrete random variable with a probability function which is a member of the
²Á Á ³ class of distributions. You are given:
(i) 7 ²? ~ ³ ~ 7 ²? ~ ³ ~ À
(ii) 7 ²? ~ ³ ~ À .
Calculate 7 ²? ~ ³ .
A) 0.120
B) 0.125
C) 0.130
D) 0.135
E) 0.140
20. (SOA) For a discrete probability distribution, you are given the recursion relation
²³ ~ h ²c³,
~ Á Á Ã À
Determine ²³.
A) 0.07
B) 0.08
C) 0.09
D) 0.10
E) 0.11
21. (SOA) A discrete probability distribution has the following properties:
(i) ~ 2 b 3c for ~ Á Á ÀÀÀ (ii) ~ À
Calculate .
A) 0.06
B) 0.13
C) 0.29
D) 0.35
E) 0.40
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 16
LM-211
22. (SOA) An actuary has created a compound claims frequency model with the following
properties:
(i) The primary distribution is the negative binomial with probability generating function
7 ²'³ ~ ´ c ²' c ³µc .
(ii) The secondary distribution is the Poisson with probability generating function
7 ²'³ ~ ²!c³ .
(iii) The probability of no claims equals 0.067.
Calculate .
A) 0.1
B) 0.4
C) 1.6
D) 2.7
E) 3.1
23. 8 has a beta Á Á distribution ( ~ , 8 is distributed on the interval ²Á ³ ).
The conditional distribution of @ given 8 ~ has probability function
7 ²@ ~ O8 ~ ³ ~ ²b³b .
You are given that the unconditional mean and variance of @ are
,²@ ³ ~ À
and = ²@ ³ ~ À .
Find the values of and .
24. The distribution of 5 given 7 ~ is binomial with parameters and . If 7 is
continuously uniformly distributed on the interval ´Á µ , what is = ´5 µ ?
A) c
B) b
C) c
D) b
E) 25. An insurer is combining two independent blocks of insurance. The aggregate claims in both
blocks are represented by compound Poisson distributions. With 5 ²³ , 5 ²³ denoting the
number of claims in block 1 and block 2, respectively, and 5 ~ 5 ²³ b 5 ²³ , you are given
7 ´5 ²³ ~ µ ~ ÀÁ 7 ´5 ²³ ~ µ ~ À
and 7 ´5 ~ µ ~ À . What is 7 ´5 ~ µ?
A) .02
B) .04
C) .06
D) .08
E) .10
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-212
MODELING - PROBLEM SET 16
MODELING - PROBLEM SET 16 SOLUTIONS
c
1. ,´5 µ ~ ,´,´5 “ µµ = , ´ µ = ~ ²À³ h ´²³ c
(À
À
µ ~ À .
À
À c
h ² À
³ = ²À³ h ² c ³ À
À
Answer: D.
2. 7 ´5 ~ µ ~ 7 ´5 ~ OClass Iµ h 7 ´Class Iµ b 7 ´5 ~ OClass IIµ h 7 ´Class IIµ
7 ´5 ~ OClass Iµ ~ À 7 ´5 ~ OClass IÁ µ h 0 ²³ ~ À c h À
~ À
,
c
7 ´5 ~ OClass IIµ ~ À 7 ´5 ~ OClass IÁ µ h 00 ²³ ~ À h ~ À .
7 ´5 ~ µ ~ ²À
³²À³ b ²À³²À³ ~ À .
Answer: E
À
3. The pdf of the exponential random variable with mean 1 is ²³ ~ c .
B
B
7 ´5 ~ µ ~ 7 ´5 ~ Oµ h ²³ ~ c h c ~ .
Alternatively, the exponential distribution is a special case of the gamma distribution with ~ .
The exponential distribution with mean 1 is a gamma distribution with ~ and ~ . If 5
given $ is Poisson with parameter $, and $ has a gamma distribution with parameters and ,
then the unconditional distribution of 5 is negative binomial with ~ and ~ . Therefore,
5 has a negative binomial distribution with ~ and ~ . Since ~ , the negative
binomial becomes a geometric distribution with ~ , and 7 ´5 ~ µ ~ ~ b ~ .
Answer: E
4. The number of risks 5 that have a loss in a particular year has a binomial distribution with
~ trials and a probability of "success" (loss) occurring of ~ À for any particular risk.
The probability that 4 or more risks have losses in a given year is 7 ´5 ‚ µ , which is equal to
c 7 ´5 ~ or or or µ . For the binomial distribution, 7 ´5 ~ µ ~ 2 3 ² c ³c .
In this question, 7 ´5 ~ µ ~ 2 3²À³ ²À
³ ~ À
,
7 ´5 ~ µ ~ 2 3²À³ ²À
³ ~ À Á 7 ´5 ~ µ ~ 2 3²À³ ²À
³ ~ À Á and
7 ´5 ~ µ ~ 2 3²À³ ²À
³ ~ À
.
Then, 7 ´5 ‚ µ ~ c ²À
b À b À b À
³ ~ À
.
There is a .0165 probability of 4 or more risks having losses in a given year. Since À
~ (approximately), this probability can be restated by saying that on average, in one of every 60
years there will be 4 or more risks that have losses.
Answer: D
5. Let the random variable 5 denote the number of claims that a driver has. To say that of the
drivers are claim-free is the same as saying that the probability of a driver being claim-free
is , or equivalently 7 ´5 ~ µ ~ . For the negative binomial distribution with parameters and , 7 ´5 ~ µ ~ ²b ³ .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 16
LM-213
5. continued
According to the actuarial student, ~ and 7 ´5 ~ µ ~ ²b ³ ~ ²b³
~ . From this
equation we can see that ~ . According to the experienced actuary, this value of ~ is
correct, but of the drivers are claim-free, so that 7 ´5 ~ µ ~ ²b ³ ~ ²b ³ ~ .
Solving this equation for results in ~ . Then, based on the experienced actuary's analysis,
the variance of the corrected negative binomial distribution is ² b ³ ~ h h ² b ³ ~ .
Answer: C
6. If the company runs experiments its costs will be Á b Á ,
and its revenue will be either (if there are no successes) or Á Á if there is at least one
success. Let ? be defined to be 0 if there are no successes in the trials, and ? ~ if there is at
least one success in the trials. The profit based on experiments is
if ? ~ c Á c Profit ~ F
.
Á Á c Á c Á if ? ~ The probability the ? ~ is the probability of 0 successes in trials, which is ²À³ , and the
probability that ? ~ is c ²À³ . The expected profit based on trials is
² c Á c ³²À³ b ²Á Á c Á c Á ³´ c ²À³ µ
~ Á Á ´ c ²À³ µ c Á c Á .
We wish to find the value of that maximizes expected profit.
For the function ²%³ ~ Á Á ´ c ²À³% µ c Á c Á % , we have
Z ²%³ ~ c Á Á ²À³% ²À³ c . The critical point occurs where
Z ²%³ ~ c Á Á ²À³% ²À³ c ~ . Solving for % results in
c
% ~ ´ ÁÁ
²À³ µ°²À³ ~ À .
At ~ the expected profit is 77,818 .
At ~ the expected profit is 77,794 .
The maximum expected profit is 77,818, based on 96 experiments.
Answer: A
7. The probability of at most one failure within the first two years can be written as
7 ´0 failures in 1st year q 0 failures in 2nd yearµ
b 7 ´1 failure in 1st year q 0 failures in 2nd yearµ
b 7 ´0 failures in 1st year q 1 failure in 2nd yearµ .
Because of independence, this becomes
7 ´0 failures in 1st yearµ d 7 ´0 failures in 2nd yearµ
b 7 ´1 failure in 1st yearµ d 7 ´0 failures in 2nd yearµ
b 7 ´0 failures in 1st yearµ d 7 ´1 failure in 2nd yearµ .
The number of failures in a year has a binomial distribution with ~ and ~ À .
The probability above becomes
´2 3²À³ ²À³ µ d ´2 3²À³ ²À³ µ
b ´2 3²À³ ²À³ µ d ´2 3²À³ ²À³ µ
b ´2 3²À³ ²À³ µ d ´2 3²À³ ²À³ µ ~ À .
© ACTEX 2009
Answer: A
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-214
MODELING - PROBLEM SET 16
8. I. False. The sum of independent negative binomials with ~ is also negative binomial
with ~ b .
II. False. The sum of independent binomials with ~ is also binomial with ~ b .
III. False. The sum of independent Poissons is always Poisson with mean b . Answer: A
9. The negative binomial distribution is a member of the ²Á Á ³ class, with
²c³
²Àc³²³
~ b and ~ b . In this case, ~ and ~
~ .
The definition of ²Á Á ³ class is that ~ b . In this case that is ~ b .
-1
-1
As long as this is € , € c . This is true if € , which is equivalent to
 À . Therefore, for ~ , ~ b ~ b ~ À so that € .
But for ~ , ~ b ~ b ~ À so that  . The values will continue to
decrease from there. Therefore, the maximum value of occurs at . Answer: B
10. For a particular driver, the number of violations in the past 3 years has a Poisson distribution
with a mean of d À ~ À . The probability of a driver having 0 violations in the past 3 years
is cÀ ~ À , and the probability of having 1 or more violations in the past 3 years is
c À ~ À
. Of the 100,000 insured drivers, we expect 22,313 of them to have had
no violations in the past 3 years, and we expect 77,687 of them to have at least 1 violation in the
past 3 years. The total premium currently being collected is
²³²Á ³ b ²³²Á ³ ~ Á Á .
After the change in pricing, we consider the number of violations in 5 years, which has a Poisson
distribution with a mean of 2.5. The probability of having zero or one violation in the past 5 years
is cÀ b cÀ ²À³ ~ À, so we expect Á insured drivers to have zero or one
violation in the past 5 years, and we expect the rest, 71,270 drivers, to have had 2 or more
violations in the past 5 years. The total premium collected after the pricing change is
Á ? b Á ²³ ~ Á Á b Á ? . In order for this to be the same as
premium revenue of 96,653,050 collected before the pricing change, we must have
Á Á b Á ? ~ Á Á , so that ? ~ À .
Answer: A
b²³²³b²³²³b²³²³b²³²³b²³²
³
c
11. The estimated mean is ? ~
~ À
À
The estimated (unbiased) variance is
c
c
~ ²? c ?³ ~ ´ ? c ? µ
~
~
´
²³ b ² ³ b ² ³ b ² ³ b ² ³ b ²
³ c ²À
³ µ
~
~ À .
The negative binomial distribution has variance larger than the mean, the Poisson has variance
equal to mean and the binomial has variance less than mean. Since the estimated variance is
significantly larger than the mean, this suggests the negative binomial as the best model.
Answer: A
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 16
LM-215
12. Based on 1,000 policies the variance of claim count is ²³²À³²À³ ~ À .
If another 1,000 (independent) policies are added, the variance will be doubled to À .
This situation is the same as finding the variance of > b < , where > and < are independent
each with the same variance. The number of claims from the first group of 1000 policies is >
and the number of claims from the second group of 1000 policies is < À Answer: D
13. The probability function of the geometric distribution with mean ,´5 µ ~ ~ is of the
form ~ ²b ³b ~ ²³b ~ ² ³² ³ , so that ~ . The variance of the geometric
distribution is = ´5 µ ~ ² b ³ ~ ²³ ~ (since ~ for a geometric distribution).
Note also that ,´5 µ ~ = ´5 µ b ²,´5 µ³ ~ b ~ .
The zero-truncated distribution has probability function
; ~ c
~ °
~ ~ ² ³² ³ .
,´5 µ
The mean of the zero-truncated distribution is ,´5; µ ~ c ~ °
~.
,´5 µ
The second moment of 5; is ,´5; µ ~ c ~ °
~ , and the variance is
= ´5; µ ~ c ~ À
The zero-modified distribution has probability function
c4
c 4 ~ c h ~ °
h ~ for ~ Á Á Á ÀÀÀ
c4
The mean of the zero-truncated distribution is ,´5; µ ~ c h ,´5 µ ~ h ~ .
c4
The second moment of 54 is ,´54
µ ~ c h ,´5 µ ~ h ~ and the variance is = ´54 µ ~ c
²
³
~
À
14. The number of trucks arriving in an hour, say 5 , will have a Poisson distribution with a
mean of ²³²À³ ~ . The probability of at least 3 trucks arriving in the one hour period
between noon and 1:00 PM is
c
c
c
7 ´5 ‚ µ ~ c 7 ´5 ~ or or µ ~ c ´ [h
b [h
b [h
µ ~ c À
~ À .
Answer: D
15. The number of coins tossed into the fountain follows a Poisson process at a rate of 20 per
hour (one every three minutes). The number of dimes tossed into the fountain also follows a
Poisson process, with a rate of À d ~ per hour. The number of dimes tossed into the
fountain in the first two hours has a Poisson distribution with a mean of d ~ . Let 2
denote the number of dimes tossed into the fountain in the first two hours. Then
7 ´4-th dime is tossed into the fountain in the first two hoursµ ~ 7 ´2 ‚ µ
c c c c ~ c 7 ´2  µ ~ c 7 ´2 ~ Á Á or µ ~ c ² [h b [h b [h b [h ³
~ c À ~ À
.
Answer: D
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-216
MODELING - PROBLEM SET 16
16. The number of claims in 3 years is Poisson with a mean of d ~ .
Since the probability of a claim being less than 100,000 is .9, the probability of a claim being
over 100,000 is .1. Then the number of claims 5 over 100,000 in 3 years has a Poisson
distribution with a mean of ²³²À³ ~ À . The probability that the number of claims in 3 years
over 100,000 is 0 is 7 ´5 ~ µ ~ cÀ ~ À .
Answer: B
17. We find the expected cost for one agent and multiply by 1000 agents.
The number of sales in 3 weeks for an agent has a Poisson distribution with mean 3 (Poisson
average of 1 per week for 3 weeks). The bonus paid to an agent is either 0 (if sales are 0,1,2 or
3 in 3 weeks) or 1000 (if sales are 4 or more). The probability of an agent getting no bonus is
c
c
c
7 ´0,1,2 or sales in 3 weeksµ ~ c b [h b [h b [h ~ À
.
The probability of an agent getting a bonus is c À
~ À
.
The expected bonus for an agent is ²À
³ ~ À
.
The expected bonus for 1000 agents is 352,768 .
Answer: E
18. If the time between events has an exponential distribution with mean , then the number of
events per unit time follows a Poisson process with mean per unit of time. In this case, the unit
of time is a week, and we are told that the time between Florida hurricanes is exponentially
distributed with a mean of ~ weeks. The number of Florida hurricanes follows a Poisson
process with a mean of hurricanes per week. Since each hurricane has a 30% chance of
hitting Broward County, the number of hurricanes hitting Broward County follows a Poisson
process with a mean rate of ²À³² ³ ~ À hurricanes per week. The number of hurricanes hitting
Broward County in the 24 week hurricane season, say 5 , has a Poisson distribution with a mean
of ²À³ ~ À . The probability of more than one hurricane hitting Broward County during a
hurricane season is 7 ´5 € µ ~ c 7 ´5 ~ Á µ ~ c ´cÀ b cÀ h ²À³µ ~ À.
Answer: E
19. As a member of the ²Á Á ³ class of distributions, the probability function ~ 7 ²? ~ ³
must satisfy ~ b for ~ Á Á ÀÀÀ
c
From the given probabilities, we have ~ À
À ~ ~ b , and
À
~ À ~ À ~ b . Solving the two equations for and results in
~ À and ~ À . Then ~ À
~ b ~ À b À
S ~ 7 ²? ~ ³ ~ À .
Answer: B
20. This question can be solved by referring to the ²Á Á ³ class of distributions. A discrete non
negative integer-valued random variable with probability function ²³ ~ Á ~ Á Á Á Á ÀÀÀ
is a member of the ²Á Á ³ class is there are constants and such that for all
~ Á Á Á ÀÀÀ, the probability function satisfies the relationship ~ b .
The Poisson with parameter has
c
~
²c ³°[
c
² c ³°²c³[
~
c
, so that ~ Á ~ À
Therefore, the distribution in this problem is Poisson with ~ , and
²³ ~ c h [ ~ À À
© ACTEX 2009
Answer: C
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 16
LM-217
21. 19. The ²Á Á ³ class satisfies the relationship ~ b .
c
The given distribution is in the ²Á Á ³ class with ~ ~ .
There are three ²Á Á ³ distributions. The Poisson has ~ , so the distribution cannot be
²b³
Poisson. The binomial has ~ c c and ~ c , and since ‚ , and cannot be
equal, so the distribution in this problem is not binomial. The negative binomial has
²c³
~ b and ~ b , so the distribution is negative binomial with ~ to make ~ À
For the negative binomial with ~ , ~ ²b ³ ~ ²b ³ ~ À . Therefore b ~ j ,
jc
and ~ b ~ j ~ À .
Answer: C
22. If 2 is a discrete non-negative integer random variable with probability generating function
72 ²'³ , then 7 ´2 ~ µ ~ 72 ²³ .
Suppose that 5 is the primary distribution (negative binomial) and 4 is the secondary (Poisson),
and let 2 denote the compound claims frequency model. Then the probability generating
function of 2 is 72 ²'³ ~ 75 ²74 ²'³³³ ~ ´ c ²²'c³ c ³µc .
Then, À
~ 72 ²³ ~ ´ c ²²c³ c ³µc . Solving this equation for results in ~ À .
Answer: E
23. The conditional distribution of @ given 8 has a geometric distribution with mean 8
and ,²@ O8 ~ ³ ~ , = ²@ O8 ~ ³ ~ ² b ³ À
²b³d
The first and second moments of 8 are ,²8³ ~ b
and ,²8 ³ ~ ²bb³d²b³ À
@ has a continuous mixture distribution. The unconditional mean of @ is
,²@ ³ ~ ,´,²@ O8³µ ~ ,´8µ ~ b
~ À
.
The unconditional variance of @ is
= ²@ ³ ~ = ´,²@ O8³µ b ,´= ²@ O8³µ ~ = ²8³ b ,´8² b 8³µ
~ ,²8 ³ c ´,²8³µ b ,²8³ b ,²8 ³ ~ ,²8 ³ b ,²8³ c ´,²8³µ
²b³d
²b³
~ ²bb³d²b³ b b
c ² b
³ ~ ²bb³ d À
b À
c ²À
³
À²b³
b
~ ²bb³ b À ~ À , and then bb
~
.
b
From the two equations b
~ À
and bb
~ ,
we get ~ À
b À
and b ~ b b .
Solving these equations results in ~ and ~ .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-218
MODELING - PROBLEM SET 16
24. = ´5 µ ~ = ´,´5 “ 7 µµ b ,´= ´5 “ 7 µµ . But
,´5 “ 7 µ ~ h 7 , and = ´5 “ 7 µ ~ h 7 h ² c 7 ³ .
Thus, = ´,´5 “ 7 µµ ~ = ´ h 7 µ ~ h = ´7 µ ~ (the variance of the uniform
distribution on ´Á µ is
²c³
), and
,´= ´5 “ 7 µµ ~ ,´ h 7 c h 7 µ ~ h ²,´7 µ c ,´7 µ³ ~ h ² c ³ ~ (the variance of 7 is ~ ,´7 µ c ²,´7 µ³ ¦ ,´7 µ ~ b ² ³ ~ ³ .
b
Thus, = ´5 µ ~ .
Answer: E.
25. Let and be the mean values of 5 ²³ and 5 ²³ , respectively.
Then, c ~ À and ~ c ²À³ ~ À , h c ~ À
, and
² b ³ hc c
~ À . Thus,
hc hc b hc hc b hc hc
=
²À³ h²À³h À
bh²À³h²À³h²À
³b h²À³²À
³
~ À .
This results in a quadratic in : À c À
b À
~ ¦ ~ ÀÁ À .
Noting that À h cÀ ~ À
, but À h cÀ ~ À
, we ignore the root À .
Thus, with ~ À and ~ À , 7 ´5 ~ µ ~ ² b ³ h c c ~ 3.9 h cÀ ~ À .
Answer: D.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 17 - THE AGGREGATE LOSS, COMPOUND DISTRIBUTIONS (1)
LM-219
MODELING SECTION 17 - MODELS FOR THE AGGREGATE LOSS,
COMPOUND DISTRIBUTIONS (1)
The material in this section relates to Loss Models, Sections 9.1-9.3
The suggested time for this section is 2 hours.
LM-17.1 The Collective or Compound Model for Aggregate Claims
in One Period of time
Aggregate claims or aggregate losses refers to the total of all losses that occur in a specified
period of time. The components of the compound model for aggregate claims are as follows.
(i) 5 , the number of claims or claim count random variable whose distribution is called the
claim count distribution or frequency distribution. 5 is a discrete, non-negative integer
random variable representing the number of claims or losses or payments that occur in the period.
The probability function of 5 is 7 ´5 ~ µ ~ .
(ii) ? , the single or individual loss random variable whose distribution is called the severity
distribution. Each time a loss or claim occurs, its amount is assumed to follow the distribution
of ? . ? can be continuous or discrete, or can have a mixed distribution, but is generally
assumed to be non-negative. The pf or pdf will be denoted ? ²%³ and the cdf is
-? ²%³ ~ 7 ´?  %µ. ? might also refer to the amount paid after any modifications (such as
policy deductible, or policy limit) based on insurance coverage.
(iii) If 5 claims occur in the period, the amounts will be ? Á ? Á ÀÀÀÁ ?5 , all from the
distribution of ? . It is assumed that 5 Á ? Á ? Á ÀÀÀÁ ?5 are mutually independent random
variables.
(iv) : ~ ? b ? b Ä b ?5 is the aggregate loss per period.
(17.1)
: is a random sum, and has a compound distribution. The terminology "compound
distribution" refers to the combination of 5 (the random variable representing the number of
losses in one period of time) with ? (the random variable representing the size of each individual
loss).
The mean and variance of a compound distribution
To find the mean and variance of : , we use the conditioning expectation rules that were reviewed
earlier in Section 8.2:
If < and > are any random variables, then
,´< µ ~ ,´ ,´< O> µ µ and
(17.2)
= ´< µ ~ = ´ ,´< O> µ µ b ,´ = ´< O> µ µ .
(17.3)
We apply these rules with < ~ : and > ~ 5 (the frequency). Applying Equation 17.2, we get
the mean of the compound distribution
,´:µ ~ ,´ ,´:O5 µ µ ~ ,´ 5 h ,´?µ µ ~ ,´5 µ h ,´?µ
(17.4)
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-220
MODELING SECTION 17 - THE AGGREGATE LOSS, COMPOUND DISTRIBUTIONS (1)
The key point in Equation 17.4 is that ,´:O5 µ ~ 5 h ,´?µ . The reason that this is true can be
seen in the following way. Suppose we are given that 5 ~ , so that there are 3 claims, and
therefore : ~ ? b ? b ? . We see that ,´:µ ~ ,´? b ? b ? µ ~ h ,´?µ .
This relationship would work no matter what the actual value of 5 is, so for an 5 , if we are
given the value of 5 , we know that there are 5 claims, each with mean ,´?µ, so that
,´:O5 µ ~ 5 h ,´?µ . But then we note that ,´?µ is a number, so that 5 h ,´?µ is the constant
,´?µ multiplied by 5 . Therefore ,´5 h ,´?µ µ ~ ,´5 µ h ,´?µ (whenever we multiply a
random variable A by a constant, say , we know that ,´Aµ ~ h ,´Aµ ; in this situation, 5 is
the random variable instead of A , and ,´?µ is the number ).
Similar reasoning is used in applying Equation 17.3. First, we apply the first part of (17.4). We
have seen that ,´:O5 µ ~ 5 h ,´?µ , where 5 is a random variable and ,´?µ is a number. Then
we use the rule of probability = ´Aµ ~ h = ´Aµ if A is a random variable and is a
number. With 5 as A , and ,´?µ as , we get
= ´ ,´:O5 µ µ ~ = ´ 5 h ,´?µ µ ~ = ´5 µ h ²,´?µ³
(17.5)
Now we apply the second part of (17.4). Similar reasoning to that above shows that
= ´:O5 µ ~ 5 h = ´?µ . Again, = ´?µ is a number, so
,´ = ´:O5 µ µ ~ ,´ 5 h = ´?µ µ ~ ,´5 µ h = ´?µ
(17.6)
Z
Notation used in the Loss Models book denotes raw moments as ,´A µ ~ A
, and central
moments as ,´²A c ,´Aµ³ µ ~ A so that = ´Aµ ~ A . Using this notation, we can express
the mean of : as
,´:µ ~ ,´5 µ h ,´?µ ~ Z: ~ Z5 Z? .
(17.7)
We can express the variance of : as
Z
Z
= ´:µ ~ ,´5 µ h = ´?µ b = ´5 µ h ²,´?µ³ ~ : ~ 5
? b 5 ²? ³ À
It is also possible to show that the 3rd central moment of : is
,´²: c ,´:µ³ µ ~ : ~ Z5 ? b 5 Z? ? b 5 ²Z? ³ À
²17.8³
²17.9³
The probability generating function and moment generating function of :
An application of the double expectation rule allows us to formulate the probability generating
function (pgf) of : in terms of the pgf's of 5 and ? .
(17.10)
7: ²³ ~ ,´: µ ~ 75 ² 7? ²³ ³ .
We have seen that for a random variable A , the moment generating function is closely related to
the pgf, 4A ²!³ ~ ,´!A µ ~ ,´²! ³A µ ~ 7A ²! ³ .
It follows that the mgf of : can be formulated as
(17.11)
4: ²!³ ~ 7: ²! ³ ~ 75 ² 7? ²! ³ ³ ~ 75 ²4? ²!³³ ~ 45 ² 4? ²!³ ³ .
The pgf and mgf of : have rarely come up on exam questions.
Example LM17-1: The number of losses per week 5 has the following probability function:
~ ²5 ~ ³ ~ À Á ~ À and ~ À . The size of each loss is uniformly distributed on
the interval ²Á ³. The number of losses and loss sizes are mutually independent. Find the
mean and variance of the aggregate loss for one week. Suppose that a policy deductible of 50 is
applied to each loss. Find the expected aggregate insurance payment for one week.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 17 - THE AGGREGATE LOSS, COMPOUND DISTRIBUTIONS (1)
LM-221
Solution: For the frequency distribution we have
,´5 µ ~ ²³²À³ b ²³²À³ ~ Á ,´5 µ ~ ² ³²À³ b ² ³²À³ ~ À ,
so that = ´5 µ ~ À c ~ À À
For the severity distribution we have ,´?µ ~ and = ´?µ ~ ~ .
For the aggregate claim distribution, we have
,´:µ ~ ,´5 µ h ,´?µ ~ ²³²³ ~ , and
= ´:µ ~ ,´5 µ h = ´?µ b = ´5 µ h ²,´?µ³ ~ ²³² ³ b ²À³²³ ~ À
If a deductible of 50 is applied to each loss, then the amount paid by the insurer for a particular
loss ? is ²? c ³b , with mean ,´²? c ³b µ ~ ²% c ³²À³ % ~ À .
The expected aggregate payment would be ,´5 µ h ,´²? c ³b µ ~ À
Example LM17-2: The number of claims per day 5 has a geometric distribution with mean 2.
The size of each claim has an exponential distribution with mean 1000. The number of losses and
loss sizes are mutually independent. Find the mean and variance of the aggregate loss for one day.
Solution: ,´5 µ ~ ~ and = ´5 µ ~ ² b ³ ~ ²³²³ ~ .
,´?µ ~ ~ Á = ´?µ ~ ~ .
,´:µ ~ ,´5 µ h ,´?µ ~ d ~ .
= ´:µ ~ ,´5 µ d = ´?µ b = ´5 µ d ²,´?µ³ ~ ²³² ³ b ²
³²³ ~ d .
LM-17.2 The Compound Poisson Distribution
A compound distribution that appears frequently on exam questions is the compound Poisson
distribution : . The frequency distribution is Poisson, with mean . Then
,´:µ ~ ,´5 µ h ,´?µ ~ ,´?µ , and
(17.12)
= ´:µ ~ ,´5 µ h = ´?µ b = ´5 µ h ²,´?µ³ ~ h ,´? µ .
(17.13)
Example LM17-3: When an individual is admitted to the hospital, the hospital charges have the
following characteristics:
Mean
Standard Deviation
(i)
Charges
Room
1000
500
Other
500
300
(ii) the covariance between an individual's Room Charges and Other Charges is 100,000.
An insurer issues a policy that reimburses 100% for Room Charges and 80% for Other Charges.
The number of hospital admissions has a Poisson distribution with parameter 4. Determine the
variance of the insurer's payout for the policy.
5
Solution: The insurer's payout is : ~ ? , where 5 is Poisson with parameter 4, and
~
? ~ 9 b À( (room plus other charges) is the total hospital charge for one admission.
= ´:µ ~ ,´5 µ h = ´?µ b = ´5 µ h ²,´?µ³ .
Since 5 is Poisson, ,´5 µ ~ = ´5 µ ~ .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-222
MODELING SECTION 17 - THE AGGREGATE LOSS, COMPOUND DISTRIBUTIONS (1)
Example LM17-3 continued
,´?µ ~ ,´9µ b À,´(µ ~ b ²À³²³ ~ , and
= ´?µ ~ = ´9µ b À
= ´(µ b ²À³*#´9Á (µ
~ b ²À
³² ³ b ²À³²Á ³ ~ Á .
Then, = ´:µ ~ Á Á .
LM-17.3 The Normal Approximation to a Compound Distribution
Many of the aggregate loss problems on the exam involve identifying frequency 5 and severity
? , and then finding ,´:µ and = ´:µ, and then applying the normal approximation to : to
compute probabilities.
Example LM17-4: A claim amount distribution is normal with mean 100 and variance 9. The
distribution of the number of claims, 5 , is:
7 ´5 ~ µ
À
À
À
À
Determine the probability that aggregate claims exceed 100 using the normal approximation to
the aggregate claim random variable : .
Solution: ,´5 µ ~ À and = ´5 µ ~ ,´5 µ c ²,´5 µ³ ~ À c ²À³ ~ À .
,´:µ ~ ,´5 µ h ,´?µ ~ Á = ´:µ ~ ,´5 µ h = ´?µ b = ´5 µ h ²,´?µ³ ~ Á À .
Applying the normal approximation to : , we get
7 ´: € µ ~ 7 ´ j :c € jc µ ~ 7 ´A € À
µ ~ À
(from the normal distribution
ÁÀ
ÁÀ
table in the exam tables).
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 17
LM-223
MODELING - PROBLEM SET 17
Compound Distributions (1) - Section 17
1. An auto insurer has analyzed claim experience and has chosen as a model for aggregate daily
claims a compound Poisson random variable with an average of 20 claims per day. The insurer
will pay the full amount of damage when a claim is made. and the claim is modeled to be
exponentially distributed with a mean claim amount of $1000. The company considers offering
a policy with a deductible amount of $250 (i.e., if damage occurs, the company will pay the
amount of damage that is in excess of $250). The company charges an aggregate premium
(divided equally among all policy holders) that is one standard deviation larger than the
expected aggregate claim. What percentage reduction in premium results if all policyholders
switch from full coverage to the coverage with $250 deductible.
A) 10%
B) 15%
C) 20%
D) 25%
E) 30%
2. In a group insurance scheme, the claim per member has a normal distribution with a mean of
and a variance of . Claims by members of the group are independent. The insurer has
determined that for a group of 100 members, a premium of 7 per member would result in a
probability of .95 that aggregate claims would be less than total premium collected. The insurer
has also determined that if the group were of size , then the same premium of 7 per member
would give a probability of .99 that aggregate claims would be less than total premium
collected. What is ?
A) 120
B) 140
C) 160
D) 180
E) 200
3. After a study of workdays missed in a year due to illness per employee in a particular
company, the following information has been determined:
Female
Male
Mean
4
7
Variance
30
20
The number of workdays missed in a year by a given employee has been determined to be
independent of the number of workdays missed by any other employee. In a randomly chosen
department of 5 employees, the number of female employees is (discretely) uniformly
distributed between and 5 . If ? is the number of workdays missed per year in a randomly
selected department with 20 employees, what is ,´?µ b j= ´?µ ?
A) 131
B) 135
C) 139
D) 143
E) 144
4. Aggregate claims : follow a compound distribution with a Poisson frequency distribution 5
with mean 1, and a geometric severity distribution ? with probability function
7 ²? ~ ³ ~ ²b ³b , for ~ Á Á Á ÀÀÀ
It is found that ,´:O: € µ ~ c c° .
Determine ,²?³ .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-224
MODELING - PROBLEM SET 17
5. : has a compound distribution with frequency 5 and severity ?
and : has a compound distribution with frequency 5 and severity ? .
5 is from the ²Á Á ³ class of distributions.
? has an exponential distribution with mean , and the mean and variance of :
are 72 and 2268 .
? has a uniform distribution on the interval ²Á ³ (same value of as ? ), and the mean and
variance of : are 36 and 351 .
Find 7 ²5 ~ ³ .
6. (SOA) For an insurance portfolio:
(i) The number of claims has the probability distribution
0
0.1
1
0.4
2
0.3
3
0.2
(i) Each claim amount has a Poisson distribution with mean 3; and
(ii) The number of claims and claim amounts are mutually independent.
Calculate the variance of aggregate claims.
A) 4.8
B) 6.4
C) 8.0
D) 10.2
E) 12.4
7. (SOA) For an insurance portfolio:
(i) the number of claims has the probability distribution
¢
²³ ¢ À
À
À
À
(ii) each claim amount has a Poisson distribution with mean 4; and
(iii) the number of claims and claim amounts are mutually independent.
Determine the variance of aggregate claims.
A) 8
B) 12
C) 16
D) 20
E) 24
8. (SOA) An insurer issues a portfolio of 100 automobile insurance policies. Of these 100
policies, one-half have a deductible of 10 and the other half have a deductible of 0. The
insurance policy pays the amount of damage in excess of the deductible subject to a maximum of
125 per accident. Assume:
(i) the number of automobile accidents per year per policy has a Poisson distribution with mean
0.03: and
(ii) given that an accident occurs, the amount of vehicle damage has the distribution:
%¢
²%³ ¢ °
°
°
Compute the total amount of claims the insurer expects to pay in a single year.
A) 270
B) 275
C) 280
D) 285
E) 290
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 17
LM-225
9. (SOA) Let : be the aggregate claims for a collection of insurance policies. You are given:
. is the premium with a relative security loading of ( is the proportion by which premium is
above expected claims);
: has a compound Poisson distribution with Poisson parameter and severity ? ; and
:
9~ .
(the loss ratio).
Which of the following is an expression for = ´9µ?
,´? µ
A) ,´?µ h b
,´? µ
,´? µ
²,´? µ³
B) ²,´?µ³ h ²b
³
D) ²,´?µ³ h
²b³ C) ²,´?µ³ h ²b
³
²,´? µ³
E) ²,´?µ³ h
²b³ 10. (SOA) For : ~ ? b ? b Ä b ?5 ;
(i) ? Á ? Á ÀÀÀ each has an exponential distribution with mean ;
(ii) the random variables 5 Á ? Á ? Á ÀÀÀ are mutually independent;
(iii) 5 has a Poisson distribution with mean 1.0; and
(iv) 4: ²À³ ~ À .
Determine .
A) .50
B) .52
C) .54
D) .56
E) .58
11. (SOA) A company has a machine that occasionally breaks down. An insurer offers a
warranty for this machine. The number of breakdowns and their costs are independent. The
number of breakdowns each year is given by the following distribution:
Probability
# of breakdowns
0
50%
1
20%
2
20%
3
10%
The cost of each breakdown is given by the following distribution:
Probability
Cost
1,000
50%
2,000
10%
3,000
10%
5,000
30%
To reduce costs, the insurer imposes a per claim deductible of 1,000. Compute the standard
deviation of the insurer's losses for this year.
A) 1,359
B) 2,280
C) 2,919
D) 3,092
E) 3,434
5
12. (SOA) For aggregate claims : ~ ? , you are given:
~
(i) ? has distribution ? ²³ ~ Á ? ²³ ~ c .
(ii) $ is a Poisson random variable with parameter ;
(iii) given $ ~ , 5 is Poisson with parameter ;
(iv) the number of claims and claim amounts are mutually independent; and
(v) = ´:µ ~ .
Determine .
A) B) C) D) E) © ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-226
MODELING - PROBLEM SET 17
5
13. (SOA) For aggregate claims : ~ ? , you are given:
~
(i) the conditional distribution of 5 , given $, is Poisson with parameter $;
(ii) $ has a gamma distribution with ~ and ~ À;
(iii) ? Á ? Á ? Á ÀÀÀ are identically distributed with ? ²³ ~ ? ²³ ~ À ;
(iv) 5 Á ? Á ? Á ÀÀÀ are mutually independent.
Determine = ´:µ.
A) 4.4
B) 4.5
C) 4.6
D) 4.7
E) 4.8
14. (SOA) You are the producer for the television show Actuarial Idol. Each year, 1000 actuarial
clubs audition for the show. The probability of a club being accepted is 0.20. The number of
members of an accepted club has a distribution with mean 20 and variance 20. Club acceptances
and the numbers of club members are mutually independent. Your annual budget for persons
appearing on the show equals 10 times the expected number of persons plus 10 times the standard
deviation of the number of person. Calculate your annual budget for persons appearing on the
show.
A) 42,600
B) 44,200
C) 45,800
D) 47,400
E) 49,000
15. (CAS) Daily claim counts are modeled by the negative binomial distribution with mean 8 and
variance 15. Severities have a mean 100 and variance 40,000. Severities are independent of each
other and of the number of claims. Let be the standard deviation of a day's aggregate losses.
On a certain day, 13 claims occurred, but you have no knowledge of their severities. Let Z be
the standard deviation of that day's aggregate losses, given that 13 claims occurred.
Calculate Z c .
A) Less than c À%
B) At least c À%, but less than 0
C) 0
D) More than 0, but less than 7.5%
E) At least 7.5%
16. (SOA) The number of auto vandalism claims reported per month at Sunny Daze Insurance
Company (SDIC) has mean 110 and variance 750. Individual losses have mean 1101 and
standard deviation 70. The number of claims and the amounts of individual losses are
independent. Using the normal approximation, calculate the probability that SDIC’s aggregate
auto vandalism losses reported for a month will be less than 100,000.
A) 0.24
B) 0.31
C) 0.36
D) 0.39
E) 0.49
17. (SOA) For an aggregate loss distribution : :
(i) The number of claims has a negative binomial distribution with ~ and ~ .
(ii) The claim amounts are uniformly distributed on the interval ²Á ³.
(iii) The number of claims and claim amounts are mutually independent.
Using the normal approximation for aggregate losses, calculate the premium such that the
probability that aggregate losses will exceed the premium is 5%.
A) 500
B) 520
C) 540
D) 560
E) 580
© ACTEX 2009
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MODELING - PROBLEM SET 17
LM-227
18. (CAS) You are asked to price a Workers' Compensation policy for a large employer. The
employer wants to buy a policy from your company with an aggregate limit of 150% of total
expected loss. You know the distribution for aggregate claims is Lognormal. You are also
provided with the following:
Standard Deviation
Mean
Number of Claims
50
12
Amount of individual loss
4,500
3,000
Calculate the probability that the aggregate loss will exceed the aggregate limit.
A) Less than 3.5%
B) At least 3.5%, but less than 4.5%
C) At least 4.5%, but less than 5.5%
D) At least 5.5%, but less than 6.5%
E) At least 6.5%
19. (SOA) For aggregate losses, : :
(i) The number of losses has a negative binomial distribution with mean 3 and variance 3.6.
(ii) The common distribution of the independent individual loss amounts is uniform from 0 to 20.
Calculate the 95-th percentile of the distribution of : as approximated by the normal distribution.
A) 61
B) 63
C) 65
D) 67
E) 69
20. (CAS) The mean annual number of claims is 103 for a group of 10,000 insureds. The
individual losses have an observed mean and standard deviation of 6,382 and 1,781, respectively.
The standard deviation of the aggregate claims is 22,874. Calculate the variance of the annual
number of claims.
A) 1.47
B) 2.17
C) 4.82
D) 21.73
E) 47.23
21. (CAS) An insurance policy provides full coverage for the aggregate losses of the Widget
Factory. The number of claims for the Widget Factory follows a negative binomial distribution
with mean 25 and coefficient of variation 1.2. The severity distribution is given by a lognormal
distribution with mean 10,000 and coefficient of variation 3. To control losses, the insurer
proposes that the Widget Factory pay 20% of the cost of each loss. Calculate the reduction in the
95-th percentile of the normal approximation of the insurer's loss.
A) Less than 5%
B) At least 5%, but less than 15%
C) At least 15%, but less than 25%
D) At least 25%, but less than 35%
E) At least 35%
22. (SOA) For an insurance:
(i) The number of losses per year has a Poisson distribution with ~ .
(ii) Loss amounts are uniformly distributed on ²Á ³ .
(iii) Loss amounts and the number of losses are mutually independent.
(iv) There is an ordinary deductible of 4 per loss.
Calculate the variance of aggregate payments in a year.
A) 36
B) 48
C) 72
D) 96
E) 120
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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MODELING - PROBLEM SET 17
23. (SOA) In a CCRC, residents start each month in one of the following three states:
Independent Living (State #1), Temporary in a Health Center (State #2) or
Permanently in a Health Center (State #3). Transitions between states occur at the of the month.
If a resident receives a physical therapy, the number of sessions that the resident receives in a
month has a geometric distribution with a mean which depends on the state in which the resident
begins the month. The numbers of sessions are independent. The number in each state at the
beginning of a given month, the probability of needing physical therapy in the month, and the
mean number of sessions received for residents receiving therapy are displayed in the following
table:
State #
Number in State
Probability of
Mean number
needing therapy
of visits
1
400
0.2
2
2
300
0.5
15
3
200
0.3
9
Using the normal approximation for the aggregate distribution, calculate the probability that more
than 3000 physical therapy session will be required for the given month.
A) 0.21
B) 0.27
C) 0.34
D) 0.42
E) 0.50
24. (CAS May 05) Annual losses for the New Widget Factory can be modeled using a Poisson
frequency model with mean of 100 and an exponential severity model with mean of $10,000. An
insurance company agrees to provide coverage for that portion of any individual loss that exceeds
$25,000. Calculate the standard deviation of the insurer's annual aggregate claim payments.
A) Less than $36,000
B) At least $36,000, but less than $37,000
C) At least $37,000, but less than $38,000
D) At least $38,000, but less than $39,000
E) $39,000 or more
25. (CAS May 05) An insurance company has two independent portfolios. In Portfolio A, claims
occur with a Poisson frequency of 2 per week and severities are distributed as a Pareto with mean
1,000 and standard deviation 2,000. In Portfolio B, claims occur with a Poisson frequency of 1
per week and severities are distributed as a lognormal with mean 2,000 and standard deviation
4,000.
Determine the standard deviation of the combined losses for the next week.
A) Less than 5,500
B) At least 5,500, but less than 5,600
C) At least 5,600, but less than 5,700
D) At least $5,700, but less than $5,800
E) 5,800 or more
26. (CAS Nov 05) The number of accidents reported to a local insurance adjusting office is a
Poisson process with parameter ~ claims per hour. The number of claimants associated with
each reported accident follows a negative binomial distribution with parameters ~ and
~ À. If the adjusting office opens at 8:00a.m., calculate the variance in the distribution of the
number of claimants before noon.
A) 9
B) 16
C) 47
D) 108
E) 189
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 17
LM-229
27. (CAS Nov 05) On January 1, 2005, Dreamland Insurance sold 10,000 insurance policies that
pay $100 for each day in 2005 that a policyholder is in the hospital. The following assumptions
were used in pricing the policies:
• The probability that given policyholder will be hospitalized during the year is 0.05. No
policyholder will be hospitalized more than one time during the year.
• If a policyholder is hospitalized, the number of days spent in the hospital follows a lognormal
distribution with ~ À and ~ À.
Using the normal approximation, calculate the premium per policy such that there is a 90%
probability that the total premiums will exceed total losses.
A) Less than 21.20
B) At least 21.20, but less than 21.50
C) At least 21.50, but less than 21.80
D) At least 21.80, but less than 22.10
E) At least 22.10
28. (CAS May 06) The following information is known about a consumer electronics store:
• The number of people who make some type of purchase follows a Poisson distribution with a
mean of 100 per day.
• The number of televisions bought by a purchasing customer follows a Negative Binomial
distribution with parameters ~ À and ~ À .
Using the normal approximation, calculate the minimum number of televisions the store must
have in inventory at the beginning of each day to ensure that the probability of its inventory being
depleted during the day is no more than 1.0%.
A) Fewer than 138
B) At least 138, but fewer than 143
C) At least 143, but fewer than 148
D) At least 148, but fewer than 153
E) At least 153
29. (SOA) You are given:
Number of Claims
Probability
°
Claim Size
Probability
°
°
°
°
°
Claim sizes are independent. Determine the variance of the aggregate loss.
A) 4,050
B) 8,100
C) 10,500
D) 12,510
E) 15,612
°
30. (SOA) The number of annual losses has a Poisson distribution with a mean of 5. The size of
each loss has a two-parameter Pareto distribution with ~ and ~ À. An insurance for
the losses has an ordinary deductible of 5 per loss.
Calculate the expected value of the aggregate annual payments for this insurance.
A) 8
B) 13
C) 18
D) 23
E) 28
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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MODELING - PROBLEM SET 17
31. (SOA) Two types of insurance claims are made to an insurance company. For each type, the
number of claims follows a Poisson distribution and the amount of each claim is uniformly
distributed as follows:
Type of Claim
Poisson Parameter Range of Each Claim
for Number of Claims
Amount
I
12
²Á ³
II
4
²Á ³
The numbers of claims of the two types are independent and the claim amounts and claim
numbers are independent. Calculate the normal approximation to the probability that the total of
claim amounts exceeds 18.
A) 0.37
B) 0.39
C) 0.41
D) 0.43
E) 0.45
32. (SOA) Computer maintenance costs for a department are modeled as follows:
(i) The distribution of the number of maintenance calls each machine will need in a year
is Poisson with mean 3.
(ii) The cost for a maintenance call has mean 80 and standard deviation 200.
(iii) The number of maintenance calls and the costs of the maintenance calls are all
mutually independent.
The department must buy a maintenance contract to cover repairs if there is at least a 10%
probability that aggregate maintenance costs in a given year will exceed 120% of the expected
costs. Using the normal approximation for the distribution of the aggregate maintenance costs,
calculate the minimum number of computers needed to avoid purchasing a maintenance
contract.
A) 80
B) 90
C) 100
D) 110
E) 120
33. (SOA) Aggregate losses for a portfolio of policies are modeled as follows:
(i) The number of losses before any coverage modifications follows a Poisson distribution with
mean .
(ii) The severity distribution of each loss before coverage modifications is uniformly distributed
between and .
The insurer would like to model the impact of imposing an ordinary deductible , (   ), on
each loss and reimbursing only a percentage, (   ), of each loss in excess of the
deductible. It is assumed that the coverage modifications will not affect the loss distribution. The
insurer models its claims with modified frequency and severity distributions. The modified claim
amount is uniformly distributed on the interval ´ Á ² c ³ µ . Determine the mean of the
modified frequency distribution.
A) B) C) D) c
E) c
34. (SOA) At the beginning of each round of a game of chance the player pays 12.5. The player
then rolls one die with outcome N. The player then rolls N dice and wins an amount equal to the
total of the numbers showing on the N dice. All dice have 6 sides and are fair. Using the normal
approximation, calculate the probability that a player starting with 15,000 will have at least
15,000 after 1000 rounds.
A) 0.01
B) 0.04
C) 0.06
D) 0.09
E) 0.12
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 17
LM-231
35. (SOA) A towing company provides all towing services to members of the City Automobile
Club.
You are given:
(i)
Towing Distance
Towing Cost
Frequency
0-9.99 miles
80
50%
10-29.99 miles
100
40%
30+ miles
160
10%
(ii) The automobile owner must pay 10% of the cost and the remainder is paid by the City
Automobile Club.
(iii) The number of towings has a Poisson distribution with mean of 1000 per year.
(iv) The number of towings and costs of individual towings are all mutually independent.
Using the normal approximation for the distribution of aggregate towing costs, calculate the
probability that the City Automobile Club pays more than 90,000 in any given year.
A) 3%
B) 10%
C) 50%
D) 90%
E) 97%
36. (SOA) The number of claims in a period has a geometric distribution with mean 4. The
amount of each claim ? follows 7 ²? ~ %³ ~ À Á % ~ Á Á Á .
The number of claims and the claim amounts are independent. : is the aggregate claim amount in
the period. Calculate -: ²³.
A) 0.27
B) 0.29
C) 0.31
D) 0.33
E) 0.35
37. (CAS) Frequency of losses follows a binomial distribution with parameters ~ and
~ À. Severity follows a Pareto distribution with parameters ~ and ~ . Calculate
the standard deviation of the aggregate losses.
A) Less than 7000
B) At least 7000, but less than 7500
C) At least 7500, but less than
8000
D) At least 8000, but less than 8500
E) At least 8500
38. (CAS) You are given the following information for a group of policyholders:
• The frequency distribution is negative binomial with ~ and ~ .
• The severity distribution is Pareto with ~ and ~ .
Calculate the variance of the number of payments if a $500 deductible is introduced.
A) Less than 30
B) At least 30, but less than 40
C) At least 40, but less than 50
D) At least 50, but less than 60
E) At least 60
39. (CAS) You are given:
• Annual frequency follows a Poisson distribution with mean 0.3.
• Severity follows a normal distribution with - ²Á ³ ~ À
.
Calculate that probability that there is at least one loss greater than 100,000 in a year.
A) Less than 11%
B) At least 11%, but less than 13%
C) At least 13%, but less than 15%
D) At least 15%, but less than 17%
E) At least 17%
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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MODELING - PROBLEM SET 17
40. (CAS May 2005) An insurance company pays claims at a Poisson rate of 2,000 per year.
Claims are divided into three categories: "minor", "major", and "severe", with payment amounts
of $1,000, $5,000, and $10,000, respectively. The proportion of "minor" claims is 50%. The total
expected claim payments per year is $7,000,000. What proportion of claims are "severe"?
A) Less than 11%
B) At least 11%, but less than 12%
C) At least 12%, but less than 13%
D) At least 13%, but less than 14%
E) 14% or more
41. (SOA May 07) You are given:
(i) Aggregate losses follow a compound model.
(ii) The claim count random variable has mean 100 and standard deviation 25.
(iii) The single-loss random variable has mean 20,000 and standard deviation 5000.
Determine the normal approximation to the probability that aggregate claims exceed 150% of
expected losses.
A) 0.023
B) 0.056
C) 0.079
D) 0.092
E) 0.159
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 17
LM-233
MODELING - PROBLEM SET 17 SOLUTIONS
1. Let : denote the aggregate daily claim random variable with no deductible, and : Z the
corresponding random variable with the deductible. Then ,´:µ ~ h ~ h ~ Á and = ´:µ ~ h ~ h Á Á ~ Á Á (for an exponential distribution with
mean ~ , the variance is c ~ so that ~ ). Also, ,´: Z µ ~ h Z and
= ´: Z µ ~ h Z . The claim distribution with deductible is ? Z ,
where 7 ´? Z ~ µ ~ 7 ´?  µ ~ c c²À³h ~ c cÀ ~ À , and the pdf of
? Z is ²%³ ~ À h cÀ²%b³ ~ cÀ h ²À³ h cÀ% (? Z is ? c for ? ‚ ).
B
B
Then, Z ~ ,´? Z µ ~ h ²À³ b % h ²%³% ~ cÀ h % h ²%³% ~ cÀ h ~ ²À³ h ~ À (where ²%³ is the pdf of ? ) .
Alternatively,
B
B
B
Z ~ ,´? Z µ ~ ²% c ³ h ? ²%³ % ~ ´ c -? ²%³µ % ~ cÀ% % ~ cÀ .
B
B
In a similar way, Z ~ ,´²? Z ³ µ ~ h ²À³ b % h ²%³% ~ cÀ h % h ²%³%
~ cÀ h ~ ²À³ h Á Á ~ Á Á .
Thus, ,´: Z µ ~ h ²À³ ~ Á and = ´ : Z µ ~ h ²Á Á ³ ~ Á Á .
The (daily) premium with no deductible is Á b jÁ Á ~ Á , and the
premium with the $250 deductible is Á b jÁ Á ~ Á . The reduction is
Á
Á , which is Á ~ À
, or À
% .
Answer: C.
2. 7 ~ b ²À
³ ¦ 7 ~ b À
.
À
7 ~ b j²À
³ ¦ 7 ~ b À
j . Thus,
j = À
¦ ~ .
Answer: E
3. Let 5- denote the number of females in the department. Then ,´5- µ ~ and
= ´5- µ ~ c ~ À
. The mean of ? can be found from
,´?µ ~ ,´,´? “ 5- µµ . If the department has 5- female employees, then there are
c 5- male employees and ,´? “ 5- µ ~ h 5- b h ² c 5- ³ .
Thus, ,´?µ ~ ,´ c h 5- µ ~ c h ~ .
The variance of ? can be found from = ´?µ ~ = ´,´? “ 5- µµ b ,´= ´? “ 5- µµ .
But = ´,´? “ 5- µµ ~ = ´ c h 5- µ ~ h = ´5- µ ~ .
With 5- female employees and c 5- male employees, the variance of the total number of
workdays missed in a year for the entire department will be h 5- b h ² c 5- ³ (the sum
of the variances for the individuals, since number of workdays missed is independent from one
employee to another), thus, ,´= ´? “ 5- µµ ~ ,´ b h 5- µ ~ .
Then, = ´?µ ~ Á and ,´?µ b j= ´?µ ~ À .
Answer: C.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-234
MODELING - PROBLEM SET 17
4. ,²?³ ~ .
: is a non-negative integer-valued random variable with probability function
7 ²: ~ ³ ~ .
B
B
7 ²:~³
,²:³
,´:O: € µ ~ h 7 ²: ~ O: € ³ ~ h c7 ²:~³ ~ c7 ²:~³ À
~
~
The severity distribution is geometric with 7 ²? ~ ³ ~
,²:³ ~ ,²5 ³ h ,²?³ ~ ²³² ³ ~ .
b
.
B
7 ²: ~ ³ ~ ~ 7 ²: ~ O5 ~ ³ h 7 ²5 ~ ³
~
B
B ² ³ c ²³
b
~ 7 ²? ~ q ? ~ q Ä q ? ~ O5 ~ ³ h 7 ²5 ~ ³ ~ ~
c
~
h
°²b ³
~
c °²b ³
,²:³
~
[
.
À
,´:O: € µ ~ c7 ²:~³ ~ cc°²b³ ~ c c° ~ ccÀ°²bÀ³
.
We see that ~ À .
5. ,²: ³ ~ ,²5 ³ d ,²? ³ ~ ,²5 ³ d ~ (Eq. 1)
= ²: ³ ~ ,²5 ³ d = ²? ³ b = ²5 ³ d ´,²? ³µ
~ ´,²5 ³ b = ²5 ³µ d ~ (Eq. 2)
,²: ³ ~ ,²5 ³ d ,²? ³ ~ ,²5 ³ d ~ (Eq. 3)
= ²: ³ ~ ,²5 ³ d = ²? ³ b = ²5 ³ d ´,²? ³µ
~ ,²5 ³ d b = ²5 ³ d ~ (Eq. 4)
From Equations 2 and 4, we get d = ²5 ³ d ~ so that = ²5 ³ d ~ , and then from equation 2 we have
,²5 ³ d ~ . Now from Equation 1, we get
,²5 ³d
~ ,²5 ³d ~ ~ .
From this we get ,²5 ³ ~ and = ²5 ³ ~ .
Since 5 is in the ²Á Á ³ class, it must be either Poisson, Negative Binomial
or Binomial. Binomial is the only one of these with = ²5 ³  ,²5 ³ ,
so 5 is binomial. If the parameters of 5 are and , then
~ and ² c ³ ~ , so that c ~ , and ~ and ~ .
Then 7 ²5 ~ ³ ~ 4 5 ² c ³
c ² ³
~ À .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 17
LM-235
6. = ´:µ ~ ,´5 µ h = ´?µ b = ´5 µ h ²,´?µ³ .
,´5 µ ~ ²À³ b ²³²À³ b ²³²À³ ~ À
, ,´5 µ ~ ²À³ b ² ³²À³ b ² ³²À³ ~ À ,
and = ´5 µ ~ À c ²À
³ ~ À .
,´?µ ~ = ´?µ ~ .
Then, = ´:µ ~ ²À
³²³ b ²À³² ³ ~ À
.
Answer: E
7. = ´:µ ~ = ´5 µ²,´?µ³ b ,´5 µ= ´?µ À
,´5 µ ~ ²³²À³ b ²³²À³ b ²³²À³ b ²³²À³ ~ Á
= ´5 µ ~ ,´5 µ c ²,´5 µ³ ~ c ~ .
,´?µ ~ = ´?µ ~ S = ´:µ ~ ²³² ³ b ²³²³ ~ .
Answer: D
8. Let > denote the annual aggregate claim from one policy with deductible 0. Then > has a
compound Poisson distribution, with Poisson parameter .01, and claim-per-accident distribution
? . Then ,´> µ ~ ²À³´ ² b b ³µ ~ À .
Let = denote the annual aggregate claim from one policy with deductible 0. Then = has a
compound Poisson distribution, with Poisson parameter .01, and claim-per-accident distribution
? . Then ,´= µ ~ ²À³´ ² b b ³µ ~ À Since there are 50 policies of each type, the
expected annual claim on the insurer is ´À b Àµ ~ .
Answer: B
9. For the compound Poisson distribution : with relative security loading , ,´:µ ~ ,´?µ Á
= ´:µ ~ ,´? µ , and . ~ ² b ³,´:µ ~ ² b ³,´?µ .
:
= ´ .
µ~
. =
,´? µ
´:µ ~ ´²b³,´?µµ h ²,´? µ³ ~ ²,´?µ³ h
.
²b³ Answer: D
10. For the compound Poisson distribution : with Poisson parameter and severity random
variable ? , the moment generating function of : is 4: ²!³ ~ ´4? ²!³cµ . Since ? has an
exponential distribution with mean , 4? ²!³ ~ c! . In this case,
À ~ 4: ²À³ ~ %´ c
c µ S ~ À .
Answer: B
11. After the deductible is imposed on a breakdown, the cost per breakdown ? has distribution
Cost
Probability
0
50%
1,000
10%
2,000
10%
4,000
30%
The insurer's losses in a year has a compound distribution with frequency 5 as indicated in the
question and severity ? given above. The variance of the insurer's total losses for the year is
,´5 µ h = ´?µ b = ´5 µ h ²,´?µ³ À
,´5 µ ~ ²³²À³ b ²³²À³ b ²³²À³ b ²³²À³ ~ À Á
,´5 µ ~ ² ³²À³ b ² ³²À³ b ² ³²À³ b ² ³²À³ ~ ÀÁ
= ´5 µ ~ À c ²À³ ~ À À
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-236
MODELING - PROBLEM SET 17
11. continued
,´?µ ~ ²³²À³ b ²³²À³ b ²³²À³ b ²³²À³ ~ Á Á
,´? µ ~ ² ³²À³ b ² ³²À³ b ² ³²À³ b ² ³²À³ ~ Á Á Á
= ´?µ ~ Á Á c ²Á ³ ~ Á Á À
Then the variance of the insurer's losses for the year is
²À³²Á Á ³ b ²À³²Á ³ ~ Á Á À
The standard deviation is jÁ Á ~ Á À
Answer: B
12. = ´:µ ~ ,´5 µ h = ´?µ b = ´5 µ h ²,´?µ³ .
,´?µ ~ c Á = ´?µ ~ ,´? µ c ²,´?µ³ ~ c c ² c ³ ~ c .
,´5 µ ~ ,´,´5 O$µµ ~ ,´$µ ~ Á
= ´5 µ ~ ,´= ´5 O$µµ b = ´,´5 O$µµ ~ ,´$µ b = ´$µ ~ b ~ .
= ´:µ ~ h ² c ³ b h ² c ³ ~ . The resulting quadratic equation in has roots
~ Á . Since   , it follows that ~ .
Answer: E
13. When the conditional distribution of 5 given $ is Poisson, and the distribution of $ is
gamma with parameters and , the unconditional distribution of 5 is negative binomial with
~ and ~ . Thus, 5 has a negative binomial distribution with ~ and ~ À ,
so that ,´5 µ ~ ~ À and = ´5 µ ~ ² b ³ ~ À . From the information about ?
we have ,´?µ ~ Á = ´?µ ~ . Then,
Answer: B
= ´:µ ~ ,´5 µ h = ´?µ b = ´5 µ h ²,´?µ³ ~ À .
14. The number of people appearing on the show has a compound distribution with frequency 5
that is binomial with ~ and ~ À, and severity ? that has mean ,´?µ ~ and
variance = ´?µ ~ . The mean and variance of 5 are ,´5 µ ~ ²³²À³ ~ , and
= ´5 µ ~ ²³²À³²À³ ~ . The mean of the number appearing on the show in one year is
,´:µ ~ ,´5 µ h ,´?µ ~ ²³²À³²³ ~ , and the variance is
= ´:µ ~ ,´5 µ h = ´?µ b = ´5 µ h ²,´?µ³ ~ ²³²³ b ²
³²³ ~ Á .
The standard deviation of : is j
Á ~ .
The annual budget is ² b ³ ~ Á .
Answer: A
15. The aggregate loss for the day, say : , has a compound distribution with frequency 5 and
severity ? . The variance of : is
~ = ´:µ ~ ,´5 µ h = ´?µ b = ´5 µ h ²,´?µ³ ~ ²³²Á ³ b ²³² ³ ~ Á .
² Z ³ ~ = ´:O5 ~ µ ~ h = ´?µ ~ ²³²Á ³ ~ Á .
Then, ~ jÁ ~ À
and Z ~ jÁ ~ À ,
so that Z c ~ c À ( c À%).
Answer: B
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 17
LM-237
16. Aggregate losses per month : follow a compound distribution with frequency 5 (number of
claims) with ,´5 µ ~ and = ´5 µ ~ , and with severity (individual claim size) @ with
,´@ µ ~ and j= ´@ µ ~ . Then, ,´:µ ~ ,´5 µ h ,´@ µ ~ Á , and
= ´:µ ~ ,´5 µ h = ´@ µ b = ´5 µ h ²,´@ µ³ ~ Á Á ~ Á .
Using the normal approximation, we have
:cÁ
ÁcÁ
7 ´:  Á µ ~ 7 ´ Á

µ ~ 7 ´A  c Àµ ~ c )²À³ ~ À .
Á
(A has a standard normal distribution)
Answer: A
17. : has a compound distribution with ,´5 µ ~ ~ and = ´5 µ ~ ² b ³ ~ ,
and ,´?µ ~ and = ´?µ ~ ~ À .
,´:µ ~ ,´5 µ h ,´?µ ~ and = ´:µ ~ ,´5 µ h = ´?µ b = ´5 µ h ²,´?µ³ ~ Á .
We wish to find so that 7 ²: € ³ ~ À using the normal approximation.
c,´:µ
~ À
S ~ À
jÁ b ~ .
Answer: D
j= ´:µ
18. The aggregate loss : has a compound distribution. The expected aggregate loss is
,´:µ ~ ,´5 µ h ,´?µ ~ d Á ~ Á , and the variance of the aggregate loss is
= ´:µ ~ ,´5 µ h = ´?µ b = ´5 µ h ²,´?µ³
~ ²³²Á ³ b ² ³²Á ³ ~ Á Á Á À
The aggregate limit on the policy is À d ,´:µ ~ À d Á ~ Á .
Using the lognormal distribution for : , we have ,´:µ ~ Á ~ b , and
,´: µ ~ = ´:µ b ²,´:µ³ ~ Á Á Á ~ b À
Then b ~ À and b ~ À ,
so that ~ À and ~ À
, and ~ ÀÀ
Then 7 ´: € Á µ ~ c 7 ´:  Á µ ~ c )²
~ c À ~ À .
Answer: B
²Á³cÀ
³
À
~ c )²À³
19. : has a compound distribution with negative binomial frequency distribution 5 and uniform
severity distribution @ . Then ,´:µ ~ ,´5 µ h ,´@ µ ~ ²³² b
³ ~ and
= ´:µ ~ ,´5 µ h = ´@ µ b = ´5 µ h ²,´@ µ³ ~ ²³² ³ b ²À
³²³ ~ (we have used the fact that the variance of the uniform distribution on the interval ²Á ³
is
²c³
). Using the normal approximation to : , the 95-th percentile of : is , where
:c
c
c
c
7 ´:  µ ~ 7 ´ j
 j
µ ~ )² j
³ ~ À . Therefore, j
~ À
,
so that ~ À .
Answer: C
20. This is an implied compound distribution : with frequency 5 and severity ? .
We are given ,´5 µ ~ , ,´?µ ~ Á , = ´?µ ~ Á and = ´:µ ~ Á .
We use the relationship = ´:µ ~ ,´5 µ h = ´?µ b = ´5 µ h ²,´?µ³ to get
Á ~ ²³² ³ b = ´5 µ h ²
³ , from which we get = ´5 µ ~ À .
Answer: C
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-238
MODELING - PROBLEM SET 17
21. The frequency distribution has mean ,´5 µ ~ ~ and variance = ´5 µ ~ ² b ³.
= ´5 µ
The square of the coefficient of variation of frequency is ²,´5 µ³ ~
Since ~ , it follows that the variance of 5 is 900À
²b ³
² ³
~ À ~ À .
= ´?µ
In a similar way, we can find the variance of the severity, from ²,´?µ³ ~ , and since
,´?µ ~ Á it follows that = ´?µ ~ Á .
The aggregate loss : has a compound distribution with mean
,´:µ ~ ,´5 µ h ,´?µ ~ ²³²Á ³ ~ Á , and variance
= ´:µ ~ ,´5 µ h = ´?µ b = ´5 µ h ²,´?µ³
~ ²³²Á ³ b ²³²Á ³ ~ ²³²Á ³ .
The 95-th percentile of insurance losses before the 20% copayment is * , where
*cÁ
7 ´:  *µ ~ À, and using the normal approximation, we get j
~ À
,
²Á ³
so that * ~ Á À
After the copayment, the insurance loss is 80% of what it would have been, so : Z ~ À: ,
and ,´: Z µ ~ ²À³,´:µ ~ Á , and = ´: Z µ ~ ²À³ = ´:µ ~ ²À³ ²³²Á ³À
Using the normal approximation, the 95-th percentile of : Z is
* Z ~ À
d ²À³j²Á ³ b Á ~ Á .
This problem could have been solved much more quickly with the following observation. The 95*c,´:µ
th percentile of : is * , which is the solution of j
~ À
, so that
= ´:µ
* ~ ,´:µ b À
j= ´:µ . When we apply the 20% copayment, the insurer loss becomes
: Z ~ À: , with ,´: Z µ ~ ²À³,´:µ and j= ´: Z µ ~ ²À³j= ´:µ , so that the 95-th percentile
of : Z is * Z ~ ,´: Z µ b À
j= ´: Z µ ~ ²À³* . The reduction in the percentile is 20%.
Answer: C
22. The frequency is Poisson with ~ .
The severity (amount paid per loss) is @ ~ ²? c ³b ,
where ? has a uniform distribution on ²Á ³ . The aggregate annual payment, : , had a
compound Poisson distribution. = ´:µ ~ h ,´@ µ . The pdf of ? is .1 on ²Á ³ .
,´@ µ ~ ²% c ³ ²À³ % ~ À .
Then, = ´:µ ~ ²À³ ~ .
Notice that we can also regard : as a compound Poisson distribution with modified frequency 4
which is Poisson with mean Z ~ ²À
³ ~ , the expected number of payments, and modified
severity > , the cost per payment. With a deductible of 4 on the uniform (0,10), the cost per
payment > has a uniform distribution, but on (0,6) . The variance of : is
Answer: C
Z ,´> µ ~ ²
³²³ ~ .
23. For a resident in state 1, let 5 be the number of therapy visits needed in a month.
5 is a mixture of the constant 0 with probability .8 (no therapy needed) and geometric ? with
mean 2 with probability .2. The first and second moments of a geometric distribution with mean
are and b . The first and second moments of ? are and 10. The first and second
moments of 5 are ²À³²³ ~ À and ²À³²³ ~ ,
so the variance of 5 is c ²À³ ~ À .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 17
LM-239
23. continued
An alternative calculation for the variance of @ is based on the following rule if @ is a mixture of
0 with probability c and > with probability The mean of @ is ,´> µ , and the variance of
@ is ,´> µ c ²,´> µ³ ~ = ´> µ b ² c ³²,´> µ³ .
Using this rule with @ ~ 5 and > geometric with mean 2,
= ´5 µ ~ ²À³²³² b ³ b ²À³²À³²³ ~ À
In a similar way, if 5 is the number of therapy visits needed in a month for a resident in state 2,
then the mean of 5 is ²À³²³ ~ À , and the second moment of 5 is ²À³²
³ ~ À,
and the variance of 5 is À .
Again, in a similar way, the mean and second moment of 5 are ²À³²³ ~ À and
²À³²³ ~ À , and the variance of 5 is À .
The mean number of visits for all residents in all states is
²³²À³ b ²³²À³ b ²³²À³ ~ , and because of independence of residents, the
variance of the number of visits needed for all residents in all states is
²³²À³ b ²³²
À³ b ²³²À³ ~ Á .
Applying the normal approximation to 5 (an integer), the total number of visits by all resident,
we get 7 ´5 € µ ~ 7 ´ 5jcÀ € cÀ
µ ~ c )² cÀ
³
j
j
~ c )²À³ ~ À .
Á
Á
Á
Answer: D
24. The insurance applies an ordinary deductible of 25,000 to each loss @ . The severity @ is
exponential with mean 10,000. The cost per loss is ²@ c Á ³b , and the 2nd moment of the
cost per loss is ,´²@ c Á ³b µ . This can be found from the integral
B
Á
²& c Á ³ Á
c&°Á & , which can be found by using the substitution
' ~ & c Á . The integral becomes
B ' c²'bÁ³°Á ' ~ cÀ d (2nd moment of an exponential with mean 10,000)
Á
~ cÀ d ² d Á ³ .
We can also find ,´²@ c Á ³b µ by using the fact that
,´²@ cÁ³ µ
,´²@ c Á ³b O@ € Á µ ~ c- ²Á³b , so that
@
,´²@ c Á ³b µ ~ ,´²@ c Á ³b O@ € Á µ d ´ c -@ ²Á ³µ
~ ,´²@ c Á ³b O@ € Á µ d cÀ .
We then use the lack of memory property of the exponential distribution, which gives the result
that ,´²@ c Á ³b O@ € Á µ is the same as ,´@ µ ~ d Á , and then
,´²@ c Á ³b µ ~ d Á d cÀ .
Since the frequency is Poisson, the variance of the annual aggregate claim payments is
,´5 µ d ²2nd moment of payment per claim³ ~ d d Á d cÀ , and
the standard deviation is j d d Á d cÀ ~ Á . Answer: E
25. Since the portfolios have compound Poisson distributions for aggregate losses, the variance
of aggregate losses for portfolio A is
= ´:( µ ~ ,´5( µ h ,´?( µ ~ ,´5( µ h ²= ´?( µ b ²,´?( µ³ ³ ~ ² b ³ .
Similarly,
= ´:) µ ~ ,´5) µ h ,´?) µ ~ ,´5) µ h ²= ´?) µ b ²,´?) µ³ ³ ~ ² b ³ .
Since the portfolios are independent, we have
= ´:( b :) µ ~ = ´:( µ b = ´:) µ ~ Á Á b Á Á ~ Á Á .
The standard deviation is jÁ Á ~ Á .
Answer: A
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-240
MODELING - PROBLEM SET 17
26. The number of accidents reported by noon has a Poisson distribution with a mean of 12 (3
accidents per hour, on average, for 4 hours from 8AM to noon). This is the frequency 5 . The
severity ? for an accident is the number of claimants associated with the accident. ? has a
negative binomial distribution with ~ and ~ À .
The total number of claimants before noon is : ~ ? b ? b Ä b ?5 , a compound
distribution (actually, a compound Poisson distribution).
The mean and the variance of 5 are ,´5 µ ~ = ´5 µ ~ , and the mean and variance of ?
are ,´?µ ~ ~ À and = ´?µ ~ ² b ³ ~ À .
Then = ´:µ ~ ,´5 µ h = ´?µ b = ´5 µ h ²,´?µ³ ~ . Answer: D
27. For a single policy, the number of hospitalizations in the year, 5 , is a Bernoulli random
variable with 7 ´5 ~ µ ~ À and 7 ´5 ~ µ ~ À . When a hospitalization takes place, the
cost is @ ~ ? , where ? has a lognormal distribution with ~ À and ~ À .
The total costÁ :Á in 2005 for a single policy is a compound distribution based on frequency 5
and severity @ . The mean and variance of 5 are ,´5 µ ~ À Á = ´5 µ ~ ²À³²À³ ~ À .
The mean and second moment of @ are
,´@ µ ~ ,´?µ ~ b ~ Àb ²À ³ ~ ,´@ µ ~ ,´? µ ~ Á b ~ Á Àb²À ³ ~ Á .
The variance of @ is ,´@ µ c ²,´@ µ³ ~ Á .
The mean and variance of : are ,´:µ ~ ,´5 µ h ,´@ µ ~ and
= ´:µ ~ ,´5 µ h = ´@ µ b = ´5 µ h ²,´@ µ³ ~ Á .
The aggregate cost for 10,000 policies is > ~ : b : b Ä b :Á .
The mean and variance of > are ,´> µ ~ Á ,´:µ ~ Á ²³ and
= ´> µ ~ Á = ´:µ ~ Á ²Á ³ (we assume that the policies are independent, so
the variance of the sum does not involve any covariances).
We apply the normal approximation to find the fund size - which satisfies the probability
> c,´> µ
- cÁ²³
7 ´>  - µ ~ À . The probability can be written as 7 ´ j
 j
µ ~ À .
= ´> µ
Á²Á
³
- cÁ²³
Then j
is the 90-th percentile of the standard normal distribution, which, from the
Á²Á
³
table, is À . Solving for - results in - ~ Á .
Á
This is the premium for 10,000 policies. The premium per policy is Á ~ À
.
This solution interprets the question as meaning that the policy will pay $100 for each day for a
hospitalization that occurs in 2005 (so that if the hospitalization occurs on Dec. 31, 2005, all days
are still covered, even though the stay may carry over to 2006). Answer: C
28. The number of televisions purchased in a day has a compound distribution with Poisson
frequency 5 with a mean of 100, and Negative Binomial severity ? with parameters ~ À
and ~ À . The total number of television purchased in one day, say : , has mean
,´:µ ~ ,´5 µ h ,´?µ ~ ²À³²À³ ~ and variance
= ´:µ ~ ,´5 µ h = ´?µ b = ´5 µ h ²,´?µ³ ~ ²À³²À³²À³ b ´À µ ~ .
We wish to find the number of televisions so that,using the normal approximation,
7 ´: € µ ~ À . This probability can be written as
7 ´ :c
€ c
µ ~ À , so that c
is the 99-th percentile of the standard normal
j
j
j
distribution. Therefore, c
j ~ À
and ~ .
© ACTEX 2009
Answer: E.
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 17
LM-241
29. We formulate the variance of : by conditioning on 5
= ´:µ ~ = ´,´:O5 µµ b ,´= ´:O5 µµ .
We work from the inside out for each of the two expression in the sum.
First, if 5 ~ then ,´:O5 ~ µ ~ ,
if 5 ~ then ,´:O5 ~ µ ~ ²³² ³ b ²³² ³ ~ , and
if 5 ~ then ,´:O5 ~ µ ~ ´²³² ³ b ²³² ³µ ~ .
prob. °
Therefore, ,´:O5 µ ~ H prob. ° , and
prob. °
= ´,´:O5 µµ ~ ´²³ ² ³ b ² ³ ² ³ b ²³ ² ³µ c ´²³ ² ³ b ² ³² ³ b ²³² ³µ
~ À
If 5 ~ then = ´:O5 µ ~ (no claims),
Á0
if 5 ~ then = ´:O5 ~ µ ~ ² c ³ ² ³² ³ ~ , and
if 5 ~ then = ´:O5 ~ µ ~ ´² c ³ ² ³² ³µ ~ Á (2 independent claims
results in the variance of each claim being counted).
prob. °
Á0
Therefore = ´:O5 µ ~ H prob. ° , and
Á prob. °
Á0
,´= ´:O5 µµ ~ ²³² ³ b ² ³² ³ b ²Á ³² ³ ~ .
Then, = ´:µ ~ b ~ Á .
Answer: B
30. The expected cost per loss with a deductible of 5 per loss is
,´²? c ³b µ ~ ,´?µ c ,´? w µ ~ c
c ² c
³´ c ² b ³c µ
À
~ Àc
h ² b
³ ~ À
.
The expected aggregate payment is ,´5 µ h ,´²? c ³b µ ~ ²³²À
³ ~ À .
Answer: C
31. The aggregate claim distribution is the sum of two compound Poisson distributions.
:0 has mean and variance ,´:0 µ ~ ²³²À³ ~ , and = ´:0 µ ~ ²³² ³ ~ .
We use the rule for a compound Poisson distribution that = ´:µ ~ ,´5 µ h ,´? µ, and for the
c
uniform distribution on ²Á ³ the second moment is ²c³
.
:00 has mean ,´:00 µ ~ ²³²À³ ~ , and variance = ´:0 µ ~ ²³² d
³ ~ À .
Since the two distributions are independent, the variance of > ~ :0 b :00 is
b À ~ À.
The mean of the sum is 16. Using the normal approximation to the sum > ,
> c
c
7 ´> € µ ~ 7 ´ j
€ j
µ ~ 7 ´A € Àµ ~ c )²À³ ~ À .
Answer: A
À
© ACTEX 2009
À
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-242
MODELING - PROBLEM SET 17
32. The annual maintenance cost for one computer, say > , has a compound Poisson distribution
with a Poisson frequency distribution 5 with a mean of 3 and severity distribution ? with mean
80 and variance 40,000 . Then ,´> µ ~ ,´5 µ h ,´?µ ~ ²³²³ ~ and
= ´> µ ~ ,´5 µ h = ´?µ b = ´5 µ h ²,´5 µ³ ~ ²³²Á ³ b ²³²³ ~ Á .
With computers, the annual maintenance cost will be ; ~ > b Ä b > , with mean
,´; µ ~ and variance = ´; µ ~ Á .
The probability that annual maintenance costs exceed 120% of expected costs is
7 ´; € À,´; µ µ ~ 7 ´; € µ .
Applying the normal approximation, this probability is
; c
7 ´ j
€ c
µ ~ c )²Àj³ À
j
Á
Á
A maintenance contract will be avoided if this probability is less than 10%.
The probability is less than 10% if )²Àj³ € À .
From the normal table, the 90-th percentile of the standard normal distribution is 1.282 , so that
the maintenance contract will be avoided if Àj € À , or equivalently,
if € À .
Answer: C
33. After the deductible is imposed, a payment will only be made if the loss is above the
deductible. The probability that a loss ? will result in a payment being made is
7 ´? € µ ~ c
. The expected number of losses that will result in a payment being made is
(total expected number of losses) d (probability a loss results in a payment) ~ h c
.
Answer: D
34. Suppose : is the amount won in a single play of the game. : has a compound distribution
with "frequency" 5 (number of dice tossed) and "severity" @ (number turning up on an
individual die). 5 is an integer from 1 to 6, with
c
7 ²5 ~ ³ ~ for ~ Á Á ÀÀÀÁ À ,´5 µ ~ , = ´5 µ ~ ~ .
@ has the same distribution as 5 , since @ is the outcome of a single die toss.
Then ,´:µ ~ ,´5 µ h ,´@ µ ~ ² ³² ³ ~ , and
= ´:µ ~ ,´5 µ h = ´@ µ b = ´5 µ h ²,´@ µ³ ~ ² ³² ³ b ² ³² ³ ~ À .
? ~ : is the total amount won in 1000 rounds of the game.
~
Then ,´?µ ~ ,´:µ ~ Á and since the rounds are independent of one another,
= ´:µ ~ = ´?µ ~ Á À.
Since the cost is 12.5 for each play of the game, the cost for 1000 plays is 12,500. In order for
the player to have at least as much as he started with after 1000 plays of the game, he must win a
total of at least 12,500 . We use the normal approximation with continuity correction:
?cÁ
ÁÀcÁ
7 ´? ‚ Á Àµ ~ 7 ´ j
‚ j
µ
ÁÀ
ÁÀ
~ 7 ´A ‚ Àµ ~ c )²À³ ~ c À ~ À .
© ACTEX 2009
Answer: E
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 17
LM-243
35. The annual cost to the Club : has a compound Poisson distribution with Poisson mean
prob. .5
prob. .4 (90% of the towing cost).
~ , and 3-point severity distribution ? ~ H prob. .1
Then ,´:µ ~ ,´?µ ~ ´²³²À³ b ²³²À³ b ²³²À³µ ~ Á , and
= ´:µ ~ ,´? µ (valid only for compound Poisson) ~ Á Á .
Applying the normal approximation with continuity correction to : (since ? is an integer),
we have
:c
Á
ÁÀc
Á
7 ´: € Á Àµ ~ 7 ´ j
€ j
µ ~ c )²À³ ~ À . Answer: B
ÁÁ
ÁÁ
36. : has a compound distribution with frequency 5 and severity ? .
There are a couple of approaches that can be taken to solve this problem.
B
The first approach uses the relationship -: ²&³ ~ 7 ´:  &O5 ~ µ h 7 ´5 ~ µ ,
~
where 5 is the frequency distribution which is geometric with mean ~ in this case.
The probability function of 5 is 7 ´5 ~ µ ~ b Á ~ Á Á Á ÀÀÀ
Since ? must be at least 1, 7 ´:  O5 ~ µ ~ for ‚ (if there are 4 or more claims, :
must be at least 4). Therefore,
B
~
~
-: ²³ ~ 7 ´:  O5 ~ µ h 7 ´5 ~ µ ~ 7 ´:  O5 ~ µ h 7 ´5 ~ µ
7 ´:  O5 ~ µ ~ since if there are no claims, : must be 0.
7 ´:  O5 ~ µ ~ since 3 of the 4 claim amounts are  3 .
7 ´:  O5 ~ µ ~ since each pair ? Á ? of claims has ² ³² ³ ~ probability of
occurring, and 3 of the 16 pairs results in :  (the 1,1 pair, the 1,2 pair and the 2,1 pair).
7 ´:  O5 ~ µ ~ since if there are 3 claims, the only way the total is  is if all three are
for amount 1 each; this probability is ² ³² ³² ³À
Then -: ²³ ~ 7 ´:  O5 ~ µ h 7 ´5 ~ µ
~
~ ²³² ³ b ² ³² ³ b ² ³² ³ b ² ³² ³ ~ À
.
An alternative combinatorial approach is to find
-: ²³ ~ 7 ´: ~ µ b 7 ´: ~ µ b 7 ´: ~ µ b 7 ´: ~ µ .
7 ´: ~ µ ~ 7 ´5 ~ µ ~ , since : ~ only if 5 ~ .
7 ´: ~ µ ~ 7 ´5 ~ µ h 7 ´? ~ µ ~ ² ³² ³ ~ (one claim of amount 1 is the only
combination which results in : ~ ).
7 ´: ~ µ ~ 7 ´5 ~ µ h 7 ´? ~ µ b 7 ´5 ~ µ h ²7 ´? ~ µ³ ~ ² ³² ³ b ² ³² ³ ~ (one claim of amount 2, or two claims each of amount 1)À
7 ´: ~ µ ~ 7 ´5 ~ µ h 7 ´? ~ µ b 7 ´5 ~ µ h 7 ´? ~ µ h 7 ´? ~ µ
b 7 ´5 ~ µ h ²7 ´? ~ µ³ ~ ² ³² ³ b ² ³²³² ³² ³ b ² ³² ³ ~ .
Then -: ²³ ~ b b b ~ ~ À
.
© ACTEX 2009
Answer: E
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-244
MODELING - PROBLEM SET 17
37. Aggregate loss : has a compound distribution : with ,´5 µ ~ ²³²À³ ~ Á
= ´5 µ ~ ²³²À³²À³ ~ for frequency, and ,´?µ ~ c
~ ,
² ³
,´? µ ~ ²c³²c³ ~ Á and = ´?µ ~ Á c ²³ ~ Á for severity.
= ´?µ ~ ,´5 µ h = ´?µ b = ´5 µ h ²,´?µ³ ~ Á Á and
j= ´?µ ~ .
Answer: D
38. The probability that a payment is above the deductible of 500 is
c - ²³ ~ ² b
³ ~ À
. The negative binomial distribution satisfies the "infinite
divisibility" property, which implies that the number of losses that are above the deductible, say
4 , is also negative binomial with the same , and with "new " ~ "old " d .64 ~ À
.
Then = ´4 µ ~ ²³²À
³²À
³ ~ À .
Answer: A
39. The Poisson distribution also satisfies the infinite divisibility property. The number of losses
greater than 100,000, say 4 , has a Poisson distribution with
"new " ~ "old " d 7 ²? € Á ³ ~ ²À³²À³ ~ À .
Probability of at least one loss greater than 100,000 in a year is
Answer: B
7 ²4 ‚ ³ ~ c 7 ²4 ~ ³ ~ c cÀ ~ À .
40. Let us suppose that the proportion of major claims is . Then the proportion of severe claims
must be c À c ~ À c (since the three proportions must add to 1).
We expect ²À³ ~ minor claims per year, major claims per year,
and ²À c ³ ~ c severe claims per year.
The total expected claim payments per year will be
Á ²Á ³ b Á ²Á ³ b ² c ³²Á ³ ~ Á Á c Á Á .
We are told that this total is 7,000,000 . Therefore Á Á c Á Á ~ Á Á from which it follows that ~ À. The proportion of severe claims is À c ~ À . Answer: A
41. Aggregate losses will be represented by the random variable : .
We wish to find 7 ´: € À,²:³µ , using the normal approximation.
:c,²:³
À,²:³c,²:³
À,²:³
This will be 7 ´ j
€ j
µ ~ c )² j
³.
= ²:³
= ²:³
= ²:³
For the compound distribution, we have
,²:³ ~ ,²5 ³ h ,²?³ ~ ²³²Á ³ ~ Á Á (5 is the claim count, and ? is the single-loss).
We also have
= ²:³ ~ ,²5 ³ h = ²?³ b = ²5 ³ h ´,²?³µ ~ ²³² ³ b ² ³²Á ³
~ À d .
À,²:³
³
= ²:³
The probability is c )² j
~ c À
~ À À
© ACTEX 2009
À²ÁÁ³
³
Àd
~ c )² j
~ c )²À³
Answer: A
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 18 - COMPOUND DISTRIBUTIONS (2)
LM-245
MODELING SECTION 18 - COMPOUND DISTRIBUTIONS (2)
The material in this section relates to Loss Models, Sections 9.3 - 9.6.
The suggested time for this section is 2-3 hours.
LM-18.1 The Convolution Method for Finding the Distribution
of a Sum of Random Variables
Given a collection of random variables ? Á ? Á ÀÀÀÁ ? there are a number of ways of finding the
distribution of the sum > ~ ? b ? b Ä b ? . If the ? 's are independent of one another,
than we might be able to use the moment generating function relationship 4> ²!³ ~ 4? ²!³
~
or the probability generating function relationship 7> ²³ ~ 7? ²³ to identify the distribution
~
of > . This approach has not come up on the exam in recent years.
An alternative approach to find the distribution of a sum of random variables is the method of
convolution. In this method, we would first find the distribution of ? b ? . Then we would us
that to find the distribution of ²? b ? ³ b ? , and so on. This method can be applied whether
or not the ? 's are independent, but it is often the case that they are assumed to be independent.
Convolution applied to discrete random variables
To apply the method of convolution to discrete integer-valued random variables, we apply a
combinatorial approach. For instance, to find 7 ²? b ? ~ ³ , we look at all ? Á ? pairs that
add up to , and add up those probabilities:
7 ²? b ? ~ ³ ~ 7 ²? ~ q ? ~ c ³ .
(18.1)
! If ? and ? are independent, then the sum on the right in Equation 18.1 becomes
7 ²? ~ ³ h 7 ²? ~ c ³ .
(18.2)
! This approach can also be applied to find the distribution function of ? b ? it the pf of ? and
the cdf of ? are known. If we let A ~ ? b ? , then
-A ²³ ~ 7 ²? b ?  ³ ~ 7 ²? ~ q ?  c ³ .
(18.3)
! Assuming independence, 18.3 becomes
-A ²³ ~ ! ? ²³ h -? ² c ³ .
(18.4)
If ? and ? are ‚ , then Equation 18.2 becomes 7 ²? b ? ~ ³ ~ ²³ h ² c ³
~
and Equation 18.4 becomes 7 ²? b ?  ³ ~ ²³ h - ² c ³ .
~
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-246
MODELING SECTION 18 - COMPOUND DISTRIBUTIONS (2)
Example LM18-1: Suppose that ? and @ are independent discrete integer-valued random
variables with ? uniformly distributed on the integers to , and @ having the following
probability function: @ ²³ ~ À , @ ²³ ~ À , @ ²³ ~ À . Let A ~ ? b @ . Find
7 ´A ~ µ .
Solution: Using the fact that ? ²%³ ~ À for % ~ Á Á Á Á , and the convolution method for
independent discrete random variables, we have A ²³ ~ ? ²³@ ² c ³
~
~ ²À³²³ b ²À³²À³ b ²À³²³ b ²À³²À³ b ²À³²À³ ~ À
Convolution applied to continuous random variables
To apply the method of convolution to continuous random variables, we apply an integral version
of the summations in Equations 18.1 to 18.4. Suppose that ? and @ are non-negative random
variables with a joint pdf ²%Á &³. If A ~ ? b @ , then the pdf of A is
'
A ²'³ ~ ²!Á ' c !³ !.
(18.5)
'
If ? and @ are independent, then this becomes ? ²!³ h @ ²' c !³ !
(18.6)
'
and the cdf of A is -A ²'³ ~ ? ²!³ h -@ ²' c !³ ! .
(18.7)
Example LM18-2: ? and ? are independent exponential random variables each with a mean
of 1. Find 7 ´? b ?  µ.
Solution: Using the convolution method, the density function of @ ~ ? b ? is
&
&
@ ²&³ ~ ? ²!³ h ? ²& c !³ ! ~ c! h c²&c!³ ! ~ &c& , so that
7 ´? b ?  µ ~ 7 ´@  µ ~ &c& & ~ ´ c &c& c c& µc
&~
&~
~ c c
(the last integral required integration by parts).
Note that the sum of independent exponentials with the same mean has a gamma distribution.
Since ? and ? are independent each with a mean of 1, @ has a gamma distribution with ~ and ~ À
Convolution may arise in the context of compound distributions. If ? Á ? Á ÀÀÀÁ ? are
independent random variables that all have the same pdf or pf, then we may see the following
notation used to described the distribution of ? b ? b Ä b ? , called the -fold convolution
of ? :
7 ² ?  '³ ~ -?i ²'³
~
(18.8)
%
i²c³
In the case of continuous ? , we have -?i ²'³ ~ -?
²% c !³ h ? ²!³ !
(18.9)
(18.10)
%
and the pdf is ?i ²%³ ~ i²c³ ²% c &³ h ? ²&³ & .
%
i²c³
If ? is discrete and integer-valued, then -?i ²%³ ~ -?
%
~
i²c³
?i ²%³ ~ ?
~
© ACTEX 2009
²% c ³ h ? ²³ is the cdf, and
²% c ³ h ? ²³ is the pf for % ~ Á Á Á ÀÀÀ.
(18.11)
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 18 - COMPOUND DISTRIBUTIONS (2)
LM-247
LM-18.2 Describing the Distribution of : by Conditioning on 5
Suppose that : has a compound distribution with frequency 5 and severity ? , so that
: ~ ? b ? b Ä b ?5 . The cdf of : can be expressed using the following rule of
probability. If ( is any event and U is any random variable, then
7 ²(³ ~ ,´ 7 ²(O< ³ µ .
(18.12)
We can apply this rule to formulate an expression for the cdf and pdf of : by conditioning over
the frequency random variable 5 . ( is the event that :  &, and < is 5 .
B
-: ²&³ ~ 7 ²:  &³ ~ ,´ 7 ²:  &O5 ³ µ ~ 7 ²:  &O5 ~ ³ h 7 ²5 ~ ³
~
B
~ -?i ²&³
~
h 7 ²5 ~ ³
(because if 5 ~ , then : ~ ? b ? b Ä b ? ).
(18.13)
B
This relationship also applies to the pdf and of, so that : ²&³ ~ ?i ²&³ h 7 ²5 ~ ³ . (18.14)
~
Example LM18-3: For aggregate claim : , you are given:
B
(i) : ²%³ ~ ?i ²%³2 b 3²À
³ ²À³ ; and
~
%
? ²%³
À
À
À
Determine = ´:µ.
(ii)
B
3
Solution: From : ²%³ ~ h ?i ²%³ it follows that 7 ´5 ~ µ ~ ~ 2 b
²À
³ ²À³ À
~
This is the probability function for the negative binomial distribution with parameters ~ and
~ . Thus, ,´5 µ ~ ~ , and = ´5 µ ~ ² b ³ ~ .
From the distribution of ? we have ,´?µ ~ À and = ´?µ ~ ,´? µ c ²,´?µ³ ~ À
.
= ´:µ ~ ,´5 µ h = ´?µ b = ´5 µ h ²,´?µ³ ~ À .
Example LM18-4: A claim amount distribution is normal with mean 100 and variance 9. The
distribution of the number of claims, 5 , is:
7 ´5 ~ µ ~ À
À
À
À
Determine the probability that aggregate claims exceed 100.
~
~
Solution: 7 ´: € µ ~ 7 ²: € O5 ~ ³ h 7 ²5 ~ ³ ~ ´ c -?i ²³µ h .
-?i ²³ ~ , -?i ²³ ~ 7 ´?  µ ~ À since ? — 5 ²Á ³ .
-?i ²³ ~ 7 ´? b ?  µ , where ? b ? — 5 ²Á ³ (sum of independent normal
c
random variables) S -?i ²³ ~ 7 ´ ? b?
 c
µ ~ 7 ´A  c Àµ ~ , where
j
j
A — 5 ²Á ³ (after standardizing ? b ? ). In a similar way, -?i ²³ ~ .
Then 7 ´: € µ ~ ² c ³ b ² c À³ b ² c ³ b ² c ³ ~ À .
This is like Example LM17-4, where the normal approximation gave a probability of .46.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-248
MODELING SECTION 18 - COMPOUND DISTRIBUTIONS (2)
LM-18.3 The Combinatiorial Approach to Describing the Distribution of :
When Severity ? is an Integer Random Variable
If the severity ? is an integer random variable, then : must also be integer-valued, since : is
always a sum of the form ? b ? b Ä b ?5 . We can describe the distribution of : by
carefully considering combinations of 5 and ? . This approach is an application of the
conditioning approach described in the previous section. We can find 7 ²: ~ ³ this way:
B
B
7 ²: ~ ³ ~ 7 ²: ~ O5 ~ ³ h ~ 7 ²? ~ ? ~ Ä ~ ? ~ ³ h B
~
~
~ ´7 ²? ~ ³µ h ~ 75 ´7 ²? ~ ³µ,
(18.15)
~
where 75 is the probability generating function of ? .
The following example illustrates the combinatorial nature of this approach.
Example LM18-5: For a compound distribution : , the frequency distribution 5 has
probability function ~ À Á ~ À Á ~ À Á ~ À , and the severity distribution ? has
probability function 7 ²? ~ ³ ~ À Á 7 ²? ~ ³ ~ À Á 7 ²? ~ ³ ~ À .
Find ,´:µ Á = ´:µ Á and 7 ²: ~ ³Á 7 ²: ~ ³ and 7 ²: ~ ³ .
Solution: ,´5 µ ~ À and = ´5 µ ~ À , and ,´?µ ~ ²³²À³² b ²³²À³ b ²³²À³ ~ À Á
,´? µ ~ ²³²À³ b ²³²À³ b ²³²À³ ~ À , and = ´4 µ ~ À c ²À³ ~ À
.
,´:µ ~ ,´5 µ h ,´4 µ ~ ²À³²À³ ~ À and
= ´:µ ~ ,´5 µ h = ´4 µ b = ´5 µ h ²,´4 µ³ ~ ²À³²À
³ b ²À³²À³ ~ À À
To find the probability function of : , we can use
B
7 ²: ~ ³ ~ 7 ²? b Ä b ? ~ ³ h 7 ²5 ~ ³ . Note that 5  , so the summation only
~
goes up to an upper limit of ~ .
7 ²: ~ ³ ~ 7 ²? b Ä b ? ~ ³ h . Since any ? must be at least 1, the only way to
~
have ? b Ä b ? ~ is to have ~ . Therefore, 7 ²: ~ ³ ~ ~ À .
7 ²: ~ ³ ~ 7 ²? b Ä b ? ~ ³ h . Since any ? must be at least 1, the only way to
~
have ? b Ä b ? ~ is to have ~ and ? ~ .
Therefore, 7 ²: ~ ³ ~ 7 ²? ~ ³ h ~ ²À³²À³ ~ À .
7 ²: ~ ³ ~ 7 ²? b Ä b ? ~ ³ h .
~
The only combinations of ? and 5 that result in : ~ are
(i) 5 ~ Á ? ~ and (ii) 5 ~ Á ? ~ ? ~ .
It follows from independence of the ? 's that
7 ²? ~ ? ~ ³ ~ 7 ²? ~ ³ h 7 ²? ~ ³ ~ ²À³ .
Then 7 ²: ~ ³ ~ 7 ²? ~ ³ h b 7 ²? b ? ~ ³ h .
~ ²À³²À³ b ²À³²À³²À³ ~ À .
Calculation of additional probabilities for : would involve identifying the appropriate
combinations of ? and 5 .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 18 - COMPOUND DISTRIBUTIONS (2)
LM-249
LM-18.4 The Recursive Method for Finding the Distribution of : When 5 is in
the ²Á Á ³ or ²Á Á ³ Class and ? is Integer-Valued and ‚ Recall that 5 is in the ²Á Á ³ class if 5 is Poisson, negative binomial, or binomial. The ²Á Á ³
class consists of the zero-modified and zero-truncated random variables from the ²Á Á ³ class.
The distribution of : if 5 is in the ²Á Á ³ class
Suppose that the severity is a non-negative integer-valued random variable with probability
function 7 ²? ~ %³ ~ ? ²%³ . Suppose that 5 is in the ²Á Á ³ class, with parameter values and . We will denote the probability function of 5 by 7 ²5 ~ ³ ~ , and we will use the
notation 7 ²: ~ ³ ~ : ²³ to denote the probability function of : . It is possible to show that
the probability function for : can be defined recursively in the following way.
%
´ c²b³ µh? ²%³b ²b h³h? ²³h: ²%c³
~
7 ²: ~ %³ ~ : ²%³ ~
%
ch? ²³
(18.16)
Note that if the severity has a maximum possible value of ", then the sum in the numerator stops
at the minimum of " and %,
¸%Á"¹
.
~
The distribution of : if 5 is in the ²Á Á ³ class
This is just a special case of the formulation in Equation 18.16. If 5 is in the ²Á Á ³ class, then
~ b ~ b , so that c ² b ³ ~ . The first term in the numerator of 18.16
becomes 0, and we get
%
²b h³h? ²³h: ²%c³
7 ²: ~ %³ ~ : ²%³ ~
~
%
ch? ²³
(18.17)
Note further, that if ? is strictly positive, ? € , then 18.17 becomes
%
: ²%³ ~ ² b % h ³ h ? ²³ h : ²% c ³ .
(18.18)
~
In order to apply any of these recursive calculations, we need a starting value for : ²³ .
In general, as pointed out in the section, : ²³ ~ 75 ´7 ²? ~ ³µ ~ 75 ´? ²³µ . If it is also true
that 7 ²? ~ ³ ~ , then : ²³ ~ 75 ²³ ~ ~ 7 ²5 ~ ³ - this is how we get the starting
value. Then we apply the recursion in a step-by-step way. Using Equation 18.18 to illustrate this,
we get
: ²³ ~ ² b h ³ h ? ²³ h : ² c ³ ~ ² b ³ h ? ²³ h : ²³ .
(18.19)
~
Then to get : ²³ we have
: ²³ ~ ² b h ³ h ? ²³ h : ² c ³
~
~ ² b ³ h ? ²³ h : ²³ b ² b h ³ h ? ²³ h : ²³ .
(18.20)
We continue in this way to get successive probability values for : . Each time the sum adds
another term, but note the upper end of the summation may eventually be limited.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-250
MODELING SECTION 18 - COMPOUND DISTRIBUTIONS (2)
The distribution of : if 5 is Poisson
If 5 has a Poisson distribution with mean , then the ²Á Á ³ parameters for 5 are ~ and
~ . Equation 18.18 becomes
%
: ²%³ ~ h h ? ²³ h : ²% c ³ .
(18.20)
%
~
Example LM18-6: The distribution of aggregate claims has a compound Poisson distribution.
The first three values of the distribution are
%¢
-: ²%³ ¢
Individual claim amounts can only have positive integral values. Determine -: ²³.
Solution: Since claim amounts are positive integers, ? ²³ ~ , and thus, with Poisson
parameter , : ²³ ~ -: ²³ ~ 75 ²³ ~ 7 ²5 ~ ³ ~ c ~ S ~ .
From the given information, : ²³ ~ -: ²³ c -: ²³ ~ , and : ²³ ~ -: ²³ c -: ²³ ~ .
Then using the recursive relationship when 5 is Poisson, we get
~ : ²³ ~ ´ h ? ²³ h : ²³µ S ? ²³ ~ , and
~ : ²³ ~ ´ h ? ²³ h : ²³ b h ? ²³ h : ²³µ S ? ²³ ~ (and ? ²%³ ~ for % £ ).
Then : ²³ ~ ´ h ? ²³ h : ²³ b h ? ²³ h : ²³ b h ? ²³ h : ²³µ ~ , and
: ²³ ~ ´ h ? ²³ h : ²³µ ~ . Finally, -: ²³ ~ : ²%³ ~ .
%~
The frequency 5 itself may have a compound distribution of the form
5 ~ 2 b 2 b Ä b 21 , where 2 and 1 are both non-negative integer random variables.
5 has a compound distribution with "frequency" 1 and "severity" 2 .
In that case, the pgf of 5 would be 75 ²!³ ~ 71 ²72 ²!³³. If both 1 and 2 are from the ²Á Á ³
or ²Á Á ³ class, then we note that 7: ²!³ ~ 75 ²7? ²!³³ ~ 71 ²72 ²7? ²!³³³ .
We note that 72 ²7? ²!³³ is the pgf of a compound distribution, say > , and we can find the
distribution of > by the recursive method since 2 is an ²Á ³-type random variable. Then, since
1 is also an ²Á ³ type variable, 71 ²7> ²!³³ ~ 71 ²72 ²7? ²!³³³ ~ 7: ²!³ is the pgf of the
compound distribution : with "frequency" 2 and "severity" > , and we can apply recursion
again to find the distribution of : . What we have done is express : as a compound distribution in
two ways:
(i) original frequency 5 (which itself is compound with 1 and 2 ) and original severity ? , and
(ii) new frequency 1 and new severity > (where > is a compound distribution with 2 and ? ).
The topics covered in this section have appeared on the exam rarely in the past, particularly since
a major exam restructuring in 2000.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 18
LM-251
MODELING - PROBLEM SET 18
Compound Distributions (2) - Section 18
1. In a compound negative binomial distribution all claims are of the same amount * € .
For % € , -: ²%³ ~ -5 ²&³ , where & ~
A) * h %
B) * h %
C) %°*
D) %°* E) % b *
2. For each of the following frequency (5 ) and severity (4 ) pairs of integer-valued
distributions, find : ²³ and : ²³, where : ~ 4 b 4 b Ä b 45 .
(i) 5 ~ Poisson with parameter , 4 ~ geometric distribution with parameter ;
also find the probability generating function of : , 7: ²'³ .
(ii) 5 ~ geometric with parameter , 4 ~ Poisson with parameter ;
also find the probability generating function of : , 7: ²'³ .
(iii) 5 ~ binomial with Á , 4 ~ discrete uniform distribution on integers Á ÀÀÀÁ .
3. A compound distribution : has frequency distribution 5 which is Poisson with a mean of .
The severity distribution ? is a discrete random variable, with
7 ´? ~ µ ~ for ~ Á Á Á ÀÀÀ
Find 7 ´: ~ µ two ways
(i) Using a combinatorial approach by considering all combinations of 5 and ? which
result in : ~ .
(ii) Using the recursive approach that applies if the frequency is in the ²Á Á ³ class.
4. For compound distribution : , suppose that 5 has a Poisson distribution with ~ , and ? ,
the severity distribution, has a Bernoulli distribution (binomial with ~ ) with ~ À. Find
,´:µ, = ´:µ , 7: ²!³ and : ²³Á : ²³Á : ²³Á and : ²³ for the compound Poisson
distribution : .
Find the : probabilities three different ways:
(i) using the combinatorial approach,
(ii) using the recursive approach, and
(iii) identify the probability generating function of :À
5. : has a compound distribution.
The frequency 5 is Poisson with a mean of 1.
The severity random variable ? has a distribution for which ? c has a Poisson distribution
with mean 1.
Frequency and severity are independent, and severity amounts are independent of one another.
7 ~ 7 ²:  ³ exactly and
8 ~ 7 ²:  ³ using the normal approximation to : (with continuity correction).
Find 8°7 .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-252
MODELING - PROBLEM SET 18
6. (SOA) : ~ ? b ? b ?
distributed as follows:
%
²%³
À
À
À
À
You are given -: ²³ ~ À .
where ? Á ? and ? are independent random variables
²%³
c
Determine .
²%³
À
À
À
À
7. (SOA) The random variables ? Á ? Á ? Á and ? are independent with probability density
functions ²%³Á ²%³Á ²%³Á and ²%³Á respectively. Your are given:
%
²%³
²%³
²%³
²%³
0
0.5
0.4
0.4
0.3
1
0.5
0.3
0.3
0.3
2
0.3
0.2
0.4
3
0.1
What is 7 ¸? b ? b ? b ? ~ ¹ ?
A) .14
B) .15
C) .16
D) .17
E) .18
8. (SOA) Aggregate claims : ~ ? b ? b ? , where ? Á ? and ? are mutually
independent random variables with probability functions as follows:
%
²%³
²%³
²%³
À
À
À
À
À
À
À
À
À
À
À
À
À c À
You are given -: ²³ ~ À
. Determine .
A) 0.0
B) 0.1
C) 0.2
D) 0.3
E) 0.4
%
9. (SOA) The distribution function of ? is - ²%³ ~ H À b À%  %  .
%‚
What is - i ²À³ ?
A) .52275
B) .58575
C) .62575
D) .67875
E) .73125
10. (SOA) For aggregate claims : , you are given:
B
c ²³
(i) : ²%³ ~ ?i ²%³ h
; and
~
%¢
? ²%³ ¢
Determine = ´:µ.
A) 20.5
B) 85.0
(ii)
© ACTEX 2009
[
À
À
C) 144.5
À
D) 165.0
E) 205.0
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 18
LM-253
11. (SOA) Aggregate losses are modeled as follows:
(i) The number of losses has a Poisson distribution with ~ .
(ii) The amount of each loss has a Burr (Burr Type XII, Singh-Maddala) distribution
with ~ , ~ , and ~ .
(iii) The number of losses and the amounts of the losses are mutually independent.
Calculate the variance of aggregate losses.
A) 12
B) 14
C) 16
D) 18
E) 20
12. (CAS May 06) Prior to the application of any deductible, aggregate claim counts during 2005
followed a Poisson distribution with ~ . Similarly, individual claim sizes followed a Pareto
distribution with ~ and ~ . Annual severity inflation is 10%. If all policies have a
$250 ordinary deductible in 2005 and 2006, calculate the expected increase in the number of
claims that will exceed the deductible in 2006.
A) Fewer than 0.41 claims
B) At least 0.41, but fewer than 0.45
C) At least 0.45, but fewer than 0.49
D) At least 0.49, but fewer than 0.53
E) At least 0.53
13. (CAS) An insurance portfolio produces 5 claims with the following distribution:
7 ²5 ~ ³
À
À
À
Individual claim amounts have the following distribution:
%
? ²%³
À
À
À
Individual claim amounts and claim counts are independent. Calculate the probability that the
ratio of aggregate claim amounts to expected aggregate claim amounts will exceed 4.
A) Less than 3%
B) At least 3%, but less than 7%
C) At least 7%, but less than 11%
D) At least 11%, but less than 15%
E) At least 15%
14. (SOA) For an insured portfolio, you are given:
(i) the number of claims has a geometric distribution with ~ ;
(ii) individual claim amounts can take on values 3, 4 or 5, with equal probability;
(iii) the number of claims and claim amounts are independent; and
(iv) the premium charged equals expected aggregate claims plus the variance of aggregate
claims.
Determine the exact probability that aggregate claims exceeds the premium.
A) 0.01
B) 0.03
C) 0.05
D) 0.07
E) 0.09
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-254
MODELING - PROBLEM SET 18
15. (SOA) For a collective risk model the number of losses, ? , has a Poisson distribution with
~ . The common distribution of the individual losses has the following characteristics:
(i) ,´?µ ~ (ii) ,´? w µ ~ (iii) 7 ²? € ³ ~ À
(iv) ,´? O? € µ ~ An insurance covers aggregate losses subject to an ordinary deductible of 30 per loss. Calculate
the variance of the aggregate payments of the insurance.
A) 54,000
B) 67,500
C) 81,000
D) 94,500
E) 108,000
16. (SOA) You are given:
(i) : has a compound Poisson distribution with ~ ; and
(ii) individual claim amounts % are distributed as follows:
%¢
? ²%³ ¢
À
À
Determine : ²³ .
A) 0.05
B) 0.07
C) 0.10
D) 0.15
E) 0.21
17. (SOA) The number of accidents follows a Poisson distribution with mean 12. Each accident
generates 1, 2, or 3 claimants with probabilities , , , respectively.
3 6
Calculate the variance in the total number of claimants.
A) 20
B) 25
C) 30
D) 35
E) 40
18. (SOA) Aggregate claims : has a compound Poisson distribution with discrete individual
claim amount distribution: ? ²³ ~ , ? ²³ ~ .
Also, : ²³ ~ : ²³ b : ²³ . Determine = ´:µ .
A) 76
B) 78
C) 80
D) 82
E) 84
19. (SOA) A compound Poisson distribution has ~ and claim amount distribution as
follows:
%
²%³
À
À
À
Calculate the probability that aggregate claims will be exactly 600.
A) 0.022
B) 0.038
C) 0.049
D) 0.060
E) 0.070
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 18
LM-255
20. (SOA May 07) Annual aggregate losses for a dental policy follow the compound Poisson
distribution
with ~ . The distribution of individual losses is:
Loss
Probability
1
0.4
2
0.3
3
0.2
4
0.1
Calculate the probability that aggregate losses in one year do not exceed 3.
A) Less than 0.20
B) At least 0.20, but less than 0.40
C) At least 0.40, but less than 0.60
D) At least 0.60, but less than 0.80
E) At least 0.80
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-256
MODELING - PROBLEM SET 18
MODELING - PROBLEM SET 18 SOLUTIONS
1. If ? is constant at the value * , then : ~ * h 5 so that
-: ²%³ ~ 7 ´:  %µ ~ 7 ´* h 5  %µ ~ 7 ´5  %°*µ ~ -5 ²%°*³ .
Answer: C.
c 2.(i) 5 is Poisson with parameter , so that ~ 7 ´5 ~ µ ~ [ , and 75 ²&³ ~ ²&c³ .
4 is geometric with parameter , so that 4 ²³ ~ 7 ´4 ~ µ ~ ²b ³b , and
74 ²'³ ~ ´ c ²' c ³µc .
: ²³ ~ 75 ² ³ ~ ² c³ ~ ² b c³ ~ c °²b ³
Since the frequency distribution is Poisson with parameter , we have
: ²³ ~ h 4 ²³ h : ² c ³ for ~ Á Á ÀÀÀ .
~
Then, : ²³ ~ h 4 ²³ h : ² c ³ ~ h ´ h 4 ²³ h : ²³µ ~ h ²b ³ h c °²b ³ .
~
²'c³
7: ²'³ ~ 75 ´74 ²'³µ ~ ²74 ²'³c³ ~ %´²´ c ²' c ³µc c µ³ ~ %´ c ²'c³ µ
c (ii) ~ ²b ³b and 4 ²³ ~ [ .
: ²³ ~ 75 ²? ²³³ ~ ´ c ² c ³µc ~ ´ c ²c c ³µc .
Since the primary distribution is geometric, it is in the ²Á Á ³ class, with ~ b and ~ ,
: ²³ ~ 7 ´: ~ µ ~ c
h ² b ³4 ²³ h : ² c ³ , ~ Á Á Á ÀÀÀ
~
and
Then, : ²³ ~ c
h ² b ³4 ²³ h : ² c ³ ~
h h ? ²³ h : ²³
c
c
b h
~
~
c b hc
h b h c h h ´ c ²c c ³µc .
7: ²'³ ~ 75 ´74 ²'³µ ~ ´ c ²74 ²'³ c ³µc ~ ´ c ²²'c³ c ³µc .
(iii) ~ 2 3 ² c ³c and 75 ²'³ ~ ´ b ²' c ³µ , 4 ²³ ~ for ~ Á Á ÀÀÀÁ .
: ²³ ~ 75 ² ³ ~ 75 ²³ ~ ² c ³ .
B
: ²³ ~ 7 ´: ~ µ ~ 7 ´4 b Ä b 4 ~ µ h 7 ´5 ~ µ
~
~ 7 ´4 b Ä b 4 ~ µ h 7 ´5 ~ µ
~
~ ²³ h 7 ´5 ~ µ b 7 ´4 ~ µ h 7 ´5 ~ µ b ²³ h 7 ´  5  µ ~ h 2 3 ² c ³c
(if 5 ~ , then there are no terms in : , and
if 5 ‚ Á then : ~ 4 b 4 b Ä b 45 ‚ 5 ‚ so that : cannot be equal to 1).
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 18
LM-257
3. (i) 7 ´: ~ µ ~ 7 ´5 ~ µ h 7 ´? ~ µ b 7 ´5 ~ µ h h 7 ´? ~ µ h 7 ´? ~ µ
b 7 ´5 ~ µ h ²7 ´? ~ µ³
c
c
c
~ c h b h h ² ³² ³ b h ² ³ ~ (ii) : ²k³ ~ 7 ´: ~ µ ~ c
h ² b ³? ²³ h : ² c ³ , ~ Á Á Á ÀÀÀ
~
For the Poisson ²Á Á ³ dist. ~ and ~ , so
~ 7 ´: ~ µ ~ h ? ²³ h : ² c ³ , ~ Á Á Á ÀÀÀ
~
We start with : ²³ ~ 7 ´: ~ µ ~ c (from (a) above)
and ? ²³ ~ 7 ´? ~ µ ~ ~ Á Á ÀÀÀ
Then : ²³ ~ ? ²³ h : ²³ ~ c ,
²³²³
: ²³ ~ h ? ²³ h : ²³ b h ? ²³ h : ²³ ~ h h c b h c ~ c ,
²³²³
²³²³
: ²³ ~ h ? ²³ h : ²³ b h ? ²³ h : ²³ b h ? ²³ h : ²³
c
~ h h c b h h c b h c ~ ~ : ²³ ~ 7 ´: ~ µ
4. For the Poisson random variable 5 , we have ,´5 µ ~ = ´5 µ ~ ~ and
the probability generating function is 75 ²³ ~ ²c³ .
For the Bernoulli random variable ? , we have ,´?µ ~ ~ ²³²À³ ~ À ,
= ´?µ ~ ² c ³ ~ ²³²À³²À³ ~ À . ? is a two-point random variable with
7 ²? ~ ³ ~ ? ²³ ~ c ~ À and 7 ²? ~ ³ ~ ? ²³ ~ À ,
so that 7? ²!³ ~ À! b À! ~ À²! b ³ is the pgf of ? .
,´:µ ~ ,´5 µ h ,´?µ ~ ²³²À³ ~ À .
= ´:µ ~ ,´5 µ h = ´?µ b = ´5 µ h ²,´?µ³ ~ ²³²À³ b ²³²À³ ~ À .
The pgf of : is 7: ²!³ ~ 75 ´7? ²!³µ ~ ²7? ²!³c³ ~ 1´À²!b³c³ ~ À²!c³ .
B
(i) 7 ²: ~ ³ ~ 7 ²? b Ä b ? ~ ³ h 7 ²5 ~ ³.
~
B
B
c
: ²³ ~ 7 ²: ~ O5 ~ ³ h ~ 7 ²? b Ä b ? ~ ³ h [ .
~
~
The only way we can have ? b Ä b ? ~ is if ? ~ ? ~ Ä ~ ? ~ ;
the probability of this is ²À³²À³Ä²À³ ~ ²À³ . Therefore,
B
B c
&
: ²³ ~ ²À³ h [ ~ c h À ~ cÀ (we have used the Taylor series [ ~ & ).
~
~
B
B
c
: ²³ ~ 7 ²: ~ O5 ~ ³ h ~ 7 ²? b Ä b ? ~ ³ h [ .
~
~
The only way we can have ? b Ä b ? ~ is if one of the ? 's is 1 and the rest are 0.
There are ways this can happen, and each one has probability ²À³ .
B
B
B
c
c
c
: ²³ ~ ²À³ h [ ~ ²À³ h ²c³[
~ ²À³ ²À³c h ²c³[
~
B
~ ²À³ ²À³ h
~
© ACTEX 2009
c
[
~
~ ²À³
c
~
À
h ~ À
cÀ
.
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-258
MODELING - PROBLEM SET 18
4. continued
We can continue, but this is a somewhat tedious and lengthy process.
(ii) Since 5 has a Poisson distribution, which is of the ²Á Á ³ class, we can use the recursive
relationship the compound Poisson distribution, : ²³ ~ h ? ²³ h : ² c ³.
: ²³ ~ 75 ²? ²³³ ~ : ²³ ~
: ²³ ~
c²c ³
~
c²cÀ³
~
cÀ
~
.
h h ? ²³ h : ² c ³ for ‚ .
~
h h ? ²³ h : ² c ³ ~ ? ²³ h : ²³. ~ ²À³²cÀ ³ ~ ÀcÀ .
~
: ²³ ~ h h ? ²³ h : ² c ³ ~ h ²? ²³ h : ²³ b ? ²³ h : ²³³
~
²À³
~ h ²À³²ÀcÀ ³ b ²³²cÀ ³ ~ h cÀ .
Note that since ? ²³ ~ for ‚ , we have
~
~
h ? ²³ h : ² c ³ ~ h ? ²³ h : ² c ³ ~ ? ²³ h : ² c ³ ,
and therefore, : ²³ ~ h h ? ²³ h : ² c ³ ~ h ²À³: ² c ³ .
~
(iii) The probability generating function of : was found above to be 7: ²!³ ~ À²!c³ .
This is the pgf of the Poisson distribution with ~ À . Therefore, : must have a Poisson
distribution with mean À . Then : ²³ ~ cÀ , : ²³ ~
: ²³ ~
cÀ
²À³
[
~
²À³
²À³
h cÀ , and : ²³ ~ [ h cÀ
cÀ ²À³
[
~ ÀcÀ ,
5. 7 ²: ~ ³ ~ 7 ²5 ~ ³ ~ c .
7 ²: ~ ³ ~ 7 ²5 ~ ³ d 7 ²? ~ ³ ~ 7 ²5 ~ ³ d 7 ²? c ~ ³ ~ c d c ~ c À
7 ²: ~ ³ ~ 7 ²5 ~ ³ d 7 ²? ~ ³ b 7 ²5 ~ ³ d ´7 ²? ~ ³µ
~ 7 ²5 ~ ³ d 7 ²? c ~ ³ b 7 ²5 ~ ³ d ´7 ²? c ~ ³µ
c
c
~ c d c b [ d ²c ³ ~ c b .
7 ²: ~ ³ ~ 7 ²5 ~ ³ d 7 ²? ~ ³ b 7 ²5 ~ ³ d 7 ²? ~ ³ d 7 ²? ~ ³ d b 7 ²5 ~ ³ d ´7 ²? ~ ³µ
~ 7 ²5 ~ ³ d 7 ²? c ~ ³ b 7 ²5 ~ ³ d 7 ²? c ~ ³ d 7 ²? c ~ ³ d
b 7 ²5 ~ ³ d ´7 ²? c ~ ³µ
c
c
c
~ c d [ b [ d c d c d b [ d ´c µ
c
c
~ b c b .
The exact probability 7 ²:  ³ is 7 ²: ~ ³ b 7 ²: ~ ³ b 7 ²: ~ ³ b 7 ²: ~ ³
c
c
c
7 ~ c b c b c b b b c b c
c
c
~ c b b b ~ À .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 18
LM-259
5. continued
The mean and variance of 5 are both 1, and the mean of ? is 2 and the variance of ? is 1.
The mean and variance of : are ,²:³ ~ ,²5 ³ h ,²?³ ~ ²³²³ ~ and
= ²:³ ~ ,²5 ³ d ,²? ³ ~ ,²?³ d ´= ²?³ b ´,²?³µ µ ~ ²³´ b µ ~ À
Applying the normal approximation (with continuity correction) to 7 ²:  ³ results in
:c,²:³
= ²:³
8 ~ 7 ²:  À³ ~ 7 ² j
(this is )²À
³ ) .
8°7 ~
À
À
Àc,²:³
 j
³ ~ )² Àc
j ³ ~ )²À
³ ~ À
= ²:³
~ À .
6. We first find - ²%³ ¢ - ²³ ~ À Á - ²³ ~ À Á - ²%³ ~ for % ‚ À
Now we find - i- ²%³ by the convolution method ¢ - i- ²³ ~ - ²³ h ²³ ~ À Á
- i- ²³ ~ - ²³ h ²³ b - ²³ h ²³ ~ À Á
- i- ²³ ~ - ²³ h ²³ b - ²³ h ²³ b - ²³ h ²³ ~ À Á
- i- ²³ ~ - ²³ h ²³ b - ²³ h ²³ b - ²³ h ²³ b - ²³ h ²³ ~ Á
and - i- ²³ ~ for ‚ .
Then -: ²³ ~ ²- i- ³i- ²³ ~ - i- ²³ h ²³ b - i- ²³ h ²³ b - i- ²³ h ²³
b - i- ²³ ²³ b - i- ²³ ²³ ~ b b À b À² c ³ b ~ À
S ~ À .
7. i ²%³ is the pf of ? b ? .
%
²%³
i ²%³
À
À
À
À
c
À
i i ²%³
À
À
À
i i i ²%³
À
À
À
Answer: D.
8. The maximum that : can be is 7 (when ? ~ Á ? ~ and ? ~ ).
À ~ c -: ²³ ~ : ²³ b : ²
³ b : ²³ . The combinations leading to : equal to 5, 6 or 7
are
:
?
?
?
Prob.
²À
³²À c ³²À³
²À
³²À c ³²À³
²À³²À c ³²À³
²À³²À c ³²À³
À ~ 7 ´: ‚ µ ~ ²³²À
³²À c ³²À³ b ²³²À³²À c ³²À³ ~ À c S ~ À .
It is possible to solve using the standard convolution method, first finding the distribution of
? b ? , and then adding ? .
%
- ²%³
²%³
-? b? ²%³
²%³ -? b? b? ²%³ ~ -: ²%³
À
À
À
À
À
À
À
À b À
b À
À c b À
From -: ²³ ~ b À ~ À
, we get ~ À
Answer: D
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-260
MODELING - PROBLEM SET 18
9. ? has a mixed distribution with probability function ²³ ~ À , ²³ ~ À
and density ²%³ ~ À for  %  .
We must use a combination of the discrete and continuous approach to find the
convolution (note that - ²À c %³ ~ for  %  À, and
- ²À c %³ ~ À b À²À c %³ ~ À c À% for À  %  ) :
- i ²À³ ~ ²³ h - ²À³ b ²%³ h - ²À c %³% b ²³ h - ²À³
À
= ²À³ h ²³ b ²À³²1³% b ²À³²À c À%³% b ²À³²À³ ~ À
. Answer: D
À
10. Aggregate claims : for a compound distribution has probability function
B
c ²³
: ²%³ ~ ?i ²%³ h 7 ´5 ~ µ . Therefore, 7 ´5 ~ µ ~
, which is the
[
~
probability function for the Poisson distribution with ~ .
Therefore, : has a compound Poisson distribution, and
= ´:µ ~ ,´? µ ~ ²³´² ³²À³ b ² ³²À³ b ² ³²À³µ ~ .
Answer: D
11. Aggregate loss : is a compound Poisson distribution with Poisson parameter ~ .
= ´:µ ~ ,´? µ ~ ,´? µ , where ? is the severity. From the table of distributions, we see
that ,´? µ ~
!²b ³ !²c ³
!²³
= ´:µ ~ ²³ ~ .
~
!²b ³ !²c ³
!²³
~
²³²³
~ À
Answer: A
12. The expected number of claims in 2005 is 14 and the probability that a claim will exceed the
deductible of 250 in 2005 is 7 ²? € ³ ~ ² b
³ ~ À in 2005 .
We have used the Pareto cdf - ²%³ ~ c ² %b
³ , with ~ and ~ for 2005.
The expected number of claims exceeding the deductible in 2005 is ²À³ ~ À
.
With inflation of 10%, the severity in 2006 is À? . The Pareto is a scale distribution, with scale
parameter , so the severity in 2006 is Pareto with ~ and ~ .
The expected number of claims in 2006 is still 14, but the probability of exceeding the deductible
of 250 in 2006 is ² b
³ ~ À , so the expected number of claims exceeding the
deductible in 2005 is ²À³ ~ À . The increase in the expected number of claims that will
exceed the deductible in 2006 is À c À
~ À
. Answer: A
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 18
LM-261
13. Aggregate claim amount : has a compound distribution with expected value
,´:µ ~ ,´5 µ h ,´?µ ~ ²À b d À³² d À b d À³ ~ À .
:
The ratio of aggregate claims to expected aggregate claims is À
. We wish to find
:
7 ´ À
€ µ ~ 7 ´: € Àµ . Since each claim is either 0 or a multiple of 10, we see that the only
way that : can be greater than 20 is if there are 5 ~ claims, and at least one of them is of size
20. If there are 2 claims, there are three combinations that result in : € À. These are
(i) ? ~ and ? ~ (prob. À d À ~ À),
(ii) ? ~ and ? ~ (prob. À d À ~ À) , and
(iii) ? ~ and ? ~ (prob. À d À ~ À) .
The total probability that two claims total more than 20.8 is À b À b À ~ À .
Then 7 ´: € Àµ ~ 7 ´²5 ~ ³ q ²? b ? € ³µ ~ ²À³²À b À b À³ ~ À ~ % .
Answer: A
14. ,´:µ ~ ,´5 µ h ,´?µ ~ h ~ ,
= ´:µ ~ ,´5 µ h = ´?µ b = ´5 µ h ²,´?µ³ ~ h b h ²³ ~ .
Premium ~ b ~ ~ .
B
7 ´: € µ ~ c 7 ´:  µ (: is integer-valued). 7 ´:  µ ~ -?i ²³ h 7 ´5 ~ µ .
~
-?i ²³ ~ Á -?i ²³ ~ Á -?i ²³ ~ (6 of the nine combinations of 2 claims add to  8).
-?i ²³ ~ for ‚ (each claim is at least amount 3, so 3 or more claims must total at least 9).
For the geometric distribution with ~ , 7 ´5 ~ µ ~ ²b ³b ~ b
.
Then 7 ´:  µ ~ ²³² ³ b ²³² ³ b ² ³² ³ b ~ À
S 7 ´: € µ ~ À .
Answer: B
15. Let @ denote the amount covered per loss. Then @ ~ ²? c ³b .
Then the aggregate payment : has a compound Poisson distribution with Poisson parameter
~ and severity @ . The variance of : is = ´:µ ~ h ,´@ µ ~ h ,´²? c ³b µ .
We can write ,´²? c ³b µ ~ ,´²? c ³b O? € µ h 7 ´? € µ .
Also, ,´²? c ³b O? € µ ~ ,´²? c ³ O? € µ ~ ,´? c ? b O? € µ
~ ,´? O? € µ c ,´?O? € µ b ~ ,´? O? € µ c ,´? c O? € µ c b ~ c ,´? c O? € µ c ~ c ,´? c O? € µ .
But ,´? c O? € µ ~
,´?µc,´?wµ
7 ²?€³
~ c
À ~ , so that
,´²? c ³b O? € µ ~ c ²
³ ~ , and ,´²? c ³b µ ~ ²³²À³ ~ À
Finally, = ´:µ ~ ²³ ~ Á .
Answer: B
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-262
MODELING - PROBLEM SET 18
B
16. There are two ways to proceed. We can use : ²%³ ~ ?i ²%³ h 7 ´5 ~ µ , where
~
?i ²%³ ~ 7 ´? b ? b Ä b ? ~ %µ is the probability function for the sum of independent
losses from the severity distribution ? . Then, ?i ²³ ~ ?i ²³ ~ , since if there are 0 or 1
claim, the total claim is at most 2. To find ?i ²³ for other values of , we must consider
combinations of claim amounts that add to 4. ?i ²³ ~ ²À
³²À
³ ~ À
since the only way that 2
claims can add to 4 is if both claims are of amount 2. ?i ²³ ~ ²À³²À³²À
³ ~ À since total 4
from 3 claims must be b b , b b or b b . ?i ²³ ~ ²À³ ~ À
since all 4
claims must be amount 1 each.
For ‚ , ?i ²³ ~ since 5 or more claims must add to at least 5.
From the Poisson distribution, we have 7 ´5 ~ µ ~ c [ , so that
B
: ²³ ~ ?i ²³ h 7 ´5 ~ µ
~
~ b b ²À
³c h [ b ²À³c h [ b ²À
³c h [ b b b Ä ~ À .
The alternative approach is to use the recursive method (valid when the frequency distribution is
Poisson, binomial or negative binomial). For the compound Poisson, we have the starting value
of : ²³ ~ 7 ´5 ~ µ ~ c ~ À , and the recursive relationship
B
: ²%³ ~ % &? ²&³ h : ²% c &³ , % ~ Á Á Á ÀÀÀ
&~
Then, : ²³ ~ ´ h ? ²³ h : ²³µ ~ ²³²À³²c ³ ~ À ,
: ²³ ~ ´²À³²À³ b ²À
³²À³µ ~ À Á
: ²³ ~ ´²À³²À³ b ²À
³²À³ b µ ~ À Á
: ²³ ~ ´²À³²À³ b ²À
³²À³ b µ ~ À .
Answer: D
17. This problem involves the compound Poisson distribution. The frequency is 5 , which is
Poisson with mean 12, and the severity is ? , which is 1, 2 or 3. Total number of claimants : has
a compound Poisson distribution.
= ´:µ ~ ,´5 µ h = ´?µ b = ´5 µ h ²,´?µ³ .
,´5 µ ~ = ´5 µ ~ , and ,´?µ ~ Á = ´?µ ~ ,´? µ c ²,´?µ³ ~ . = ´:µ ~ À
Alternatively, for the compound Poisson distribution,
= ´:µ ~ ,´5 µ h ,´? µ ~ ²³² Answer: E
³ ~ .
18. For the compound Poisson : , = ´:µ ~ ,´5 µ h ,´? µ ~ ² ³.
From the other information, we can use the recursive relationship for the compound Poisson,
B
: ²%³ ~ &? ²&³ h : ²% c &³ , % ~ Á Á Á ÀÀÀ
%
&~
: ²³ ~ ´? ²³: ²³ b b ? ²³: ²³ b µ
~ : ²³ b : ²³ ~ : ²³ b : ²³ .
~ ´ : ²³ b ² ³: ²³µ
It follows that ~ , and then = ´:µ ~ ²³² ³ ~ .
© ACTEX 2009
Answer: A
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 18
LM-263
19. We consider all combinations of losses that can total 600. This will occur with the following
combinations:
c
5 ~ Á ? ~ Á ? ~ , probability is h ²À³²À
³ ~ À ,
c
5 ~ Á ? ~ Á ? ~ , probability is h ²À
³²À³ ~ À ,
c
5 ~ Á ? ~ ? ~ Ä ~ ?
~ , probability is [
h ²À³
~ À .
Total probability is À
.
Answer: D
20. : has a compound Poisson distribution.
7 ²:  ³ ~ 7 ²: ~ ³ b 7 ²: ~ ³ b 7 ²: ~ ³ b 7 ²: ~ ³ .
7 ²: ~ ³ ~ 7 ²5 ~ ³ ~ c .
c 7 ²: ~ ³ ~ 7 ²5 ~ ³ h 7 ²? ~ ³ ~ [ h ²À³ ~ Àc .
7 ²: ~ ³ ~ 7 ²5 ~ ³ h 7 ²? ~ ³ b 7 ²5 ~ ³ h ´7 ²? ~ ³µ
c c ~ [ h ²À³ b [ h ²À³ ~ À
c .
7 ²: ~ ³ ~ 7 ²5 ~ ³ h 7 ²? ~ ³ b 7 ²5 ~ ³ h 7 ²? ~ ³7 ²? ~ ³
b 7 ²5 ~ ³ h ´7 ²? ~ ³µ
c c c ~ [ h ²À³ b [ h ²À³²À³ b [ h ²À³ ~ À
c .
Aggregate losses do not exceed 3 if :  .
Answer: B
7 ²:  ³ ~ c ´ b À b À
b À
µ ~ À .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-264
© ACTEX 2009
MODELING - PROBLEM SET 18
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 19 - MORE PROPERTIES OF THE AGGREGATE LOSS
LM-265
MODELING SECTION 19 - MORE PROPERTIES OF THE AGGREGATE
LOSS RANDOM VARIABLE
The material in this section relates to Loss Models, Sections 9.4, 9.6, 9.7 and 9.11.2.
The suggested time for this section is less than 1 hour.
LM-19.1 Modification of the Severity and Frequency Distributions for :
Suppose for a compound distribution : , the frequency distribution is 5 . If a modification (such
as deductible, coinsurance, etc.) is applied to ? , and we wish to find the aggregate amount paid
by the insurer, then we can model the aggregate payment in two ways:
(i) find @ , the cost per loss (which might be 0 if there is a deductible), and then the aggregate
amount paid is : i ~ @ b @ b Ä b @5 , where 5 has the original frequency distribution, or
(ii) find A , the cost per payment, and find the distribution of 5 i , the number of payments made,
and then : i ~ A b A b Ä b A5 i is the aggregate amount paid.
5
In this second case, 5 i ~ 0 , where each 0 is the indicator random variable with
0 ~ F
~
if loss does not result in a payment
.
if loss results in a payment
Suppose that there is a deductible applied to each individual loss. Then
7 ´0 ~ µ ~ 7 ´?  µ ~ c Á 7 ´0 ~ µ ~ 7 ´? € µ ~ Á . Then 5 i has a compound
distribution, where the "frequency" distribution is 5 and the "severity" distribution is Bernoulli.
The probability generating function of 5 i is 75 i ²!³ ~ 75 ²70 ²!³³ ~ 75 ´ b ²! c ³µ .
(i) If 5 is Poisson with parameter , then 5 i is Poisson with parameter .
(ii) If 5 is binomial with parameters and 8, then 5 i is binomial with parameters and 8 .
(iii) If 5 is negative binomial with parameters and , then 5 i is negative binomial with
parameters and .
Example LM19-1: A portfolio of risks has a Poisson frequency distribution with parameter
~ . The severity distribution has a ground up loss random variable ? which has a
continuous uniform distribution on the interval ²Á ³. A deductible of per loss is imposed.
Describe the aggregate claim distribution in the two ways described above. Show that the two
ways of describing the aggregate claims result in the same means and variances.
Solution: (i) The cost per loss is the random variable @3 ~ ²? c ³b .
%
The distribution of @ has mean ,´@3 µ ~ ,´²? c ³b µ ~ ² c ³ % ~ À . The second
moment of @ is ²% c ³ ²À³ % ~
. The number of losses 5 based on the original
frequency distribution is Poisson with a mean of 10. The aggregate claim distribution : Z is
compound Poisson with severity @ , so that
: Z ~ @ b @ b Ä b @5 . The mean and variance of : Z are
,´: Z µ ~ ,´5 µ h ,´@3 µ ~ ²³²À³ ~ , and = ´: Z µ ~ ,´@3 µ ~ ²³² ³ ~ .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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MODELING SECTION 19 - MORE PROPERTIES OF THE AGGREGATE LOSS
Example LM19-1 continued
(ii) The cost per payment is the conditional distribution of @ given that @ € ; this is the excess
loss random variable @7 with deductible ~ . Since the original severity was uniform, the
distribution of the cost per payment with deductible Á @7 , is also uniform on the interval ²Á ³
(in Section 13, there was summary of policy adjustments to certain severity distributions; the
uniform was one of them, and it was shown that for a uniform severity on ²Á ³, with deductible
, the cost per loss random variable is uniform on ²Á c ³). The probability that a loss results
in a payment is .8 (the probability that ? € ). According to the comments above, since the
original frequency 5 was Poisson with mean 10, the modified frequency 5 i (the number of
losses that result in a claim payment) is also Poisson, with mean ²³²À³ ~ . Therefore, the
aggregate claim payment : ZZ has a compound distribution with modified frequency 5 i that is
Poisson with mean 8, and modified severity @7 , which is uniform on ²Á ³. The mean and
variance of : ZZ are
,´: ZZ µ ~ ,´5 µ h ,´@7 µ ~ ²³²³ ~ , and = ´: Z µ ~ ²³,´@7 µ ,
ZZ
where ,´@7 µ ~ & ² ³ & ~
. Therefore, = ´: µ ~ ²³² ³ ~ .
There have been just a few exam questions on this topic over several years.
LM-19.2 The Sum of Independent Compound Poisson Distributions
If : Á : Á ÀÀÀÁ : are mutually independent compound Poisson random variables with Poisson
parameters Á Á ÀÀÀÁ and claim amount d.f.'s - ²%³Á - ²%³Á ÀÀÀÁ - ²%³ respectively, then
: ~ : b : b Ä b : has a compound Poisson distributionÀ : will have Poisson parameter
~ and the severity will be a mixture of the severities of the compound Poisson
~
components. The d.f. of the combined severity is - ²%³ ~ - ²%³ and the pdf (pf) of the
~
combined severity is ²%³ ~ ²%³ . This is Theorem 9.7 in the Loss Models book.
~
Example LM19-2: : and : have independent compound Poisson distributions with ~ and ~ . The severities ? and ? the following distributions:
7 ²? ~ ³ ~ , ~ Á Á and 7 ²? ~ ³ ~ and 7 ²? ~ ³ ~ .
Find the severity of the compound Poisson distribution : ~ : b : .
Solution: The severity of : will be a mixture of ? and ? with mixing weights b ~ and . The probability function for the severity of ? is
7 ²? ~ ³ ~ h 7 ²? ~ ³ b h 7 ²? ~ ³ ~ h b h ~ .
7 ²? ~ 2³ ~ h 7 ²? ~ 2³ b h 7 ²? ~ 2³ ~ h b h ~ .
7 ²? ~ ³ ~ h 7 ²? ~ ³ b h 7 ²? ~ ³ ~ h b h ~ .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 19 - MORE PROPERTIES OF THE AGGREGATE LOSS
LM-267
LM-19.3 Discretization of a Continuous Distribution
If the random variable ? has a continuous distribution with pdf ? ²%³, it is possible to
approximate ? with a discrete integer-valued distribution. For each integer , we define the
b random variable ? to have probability at of 7 ²? ~ ³ ~ ? ²%³ % .
c ? can be used as a discrete approximation to ? . In a compound distribution context, if 5 is in
the ²Á Á ³ or ²Á Á ³ class, then the recursive approach can be applied to determine the
distribution of : .
A variation on this method would be define the discretized variable ? at multiples of some
quantity instead of at all integers. For instance, if ~ , ? would be defined at multiples
of 100.
The discretization method has not appeared on exam questions in recent years.
LM-19.4 The Individual Risk Model
The individual risk model for a short term is based on the aggregate claims from a portfolio of
policies for a "single period" (on year or one month, or one week). This portfolio consists of a
specific number of policies, say , and each policy has a claim amount random
variable for the single period, ? for policy . The aggregate claim for the period is
: ~ ? b ? b Ä b ? ~ ? .
(19.1)
~
It is usually assumed the claim amounts are independent of one another, so that
,´:µ ~ ,´? µ and = ´:µ ~ = ´? µ .
~
(19.2)
~
The distribution of ? may be defined as a mixture of 0 with probability c and loss amount
) € with probability . ? ~ ) if a claim occurs from policy , and the probability of a
claim occurring from policy is . Using rules involving mixtures of random variables, we have
,´? µ ~ ² c ³²³ b h ,´) µ ~ h ,´) µ , and
(19.3)
,´? µ ~ ² c ³² ³ b h ,´) µ ~ h ,´) µ , so that
(19.4)
(19.5)
= ´? µ ~ h ,´) µ c ² h ,´) µ³ .
Using the fact that = ´) µ ~ ,´) µ c ²,´) µ³ , with a little algebra, we get
= ´? µ ~ h = ´) µ b ² c ³²,´) µ³ .
(19.6)
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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MODELING SECTION 19 - MORE PROPERTIES OF THE AGGREGATE LOSS
Example LM19-3: A portfolio of independent insurance policies has three classes of policies:
Probability
of Claim
Number
Claim
Class
in Class
per Policy
Amount
1
1000
.01
1
2
2000
.02
1
3
500
.04
2
The insurer calculates the variance of the aggregate claims random variable. The insurer
changes his assumptions regarding the claims and now supposes that the claim amount is also a
random variable and the claim amount listed in the table above is the expected claim amount for
each of the policies, and the variance of the claim amount per policy is . The insurer
recalculates the variance of the aggregate claims and finds that it is 67% larger than the initial
calculation. What is the value of ?
Solution: The initial calculation is based on = ´) µ ~ for each ) , so for each ? we have
= ´? µ ~ ² c ³²,´) µ³ . The initial calculation results in a variance of the aggregate
claims of
h h ²À³²À³ b h h ²À³²À³ b h h ²À³²À
³ ~ À .
The revised calculation is has = ´) µ ~ for each , so the variance of the aggregate claim is
À b ´ h ²À³ b h ²À³ b h ²À³µ h ~ À b ~ ²À³²À
³ ~ À . Thus, ~ À .
Note that each Class individually can be analyzed from the point of view of a compound
distribution. For Class 1, the number of claims, say 5 , can be modeled as binomial with
~ and ~ À, so ,´5 µ ~ and = ´5 µ ~ À . Using the mean claim amount of 1
for ,´) µ and 1.2 for = ´) µ, the variance : (aggregate claim from Class 1 policies) is
,´5 µ h = ´) µ b = ´) µ h ²,´? µ³ ~ ²³²À b ²À³²³ ~ À .
Example LM19-2 continued
In a similar way, we get = ´: µ ~ ²³²À³ b ²À³²³ ~ À, and
= ´: µ ~ ²³²À³ b ²À³²³ ~ À .
The variance of the combination of Classes 1, 2 and 3 is À b À b À ~ À.
Questions on the Individual Model used to come up frequently on exams prior to 2000, but have
rarely appeared since then. Problem Set 19 has several exercises on this topic.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 19
LM-269
MODELING - PROBLEM SET 19
More on Aggregate Loss Distributions - Section 19
1. The exponential distribution with mean 1 is being used as the model for a loss distribution. An
actuary attempts to "discretize" the distribution by assigning a probability to b for
~ Á Á Á ÀÀÀ . The probability assigned to b is 7 ´  ?  b µ , where ? is the
exponential random variable with mean 1. Find the mean of the discretized distribution.
A) 1.00
B) 1.02
C) 1.04
D) 1.06
E) 1.08
2. The frequency distribution is Poisson with a mean of 20 and the severity distribution is
exponential with a mean of 100. A deductible of 20 is imposed on each individual loss. Find the
mean and variance of the aggregate payment made by the insurer.
3. The frequency distribution 5 has a negative binomial distribution with ~ and ~ .
The severity distribution ? is Pareto with parameters ~ and ~ . A deductible of 100
is applied to every individual loss. Find the mean of the aggregate payment.
4. A company has 1500 employees. The insurer for the company's health plan has noticed that
the employees can be divided into three groups with regard to likelihood and amount of claim.
Claim Amount
Distribution
Probability
Group
of Claim
Mean
Variance
1
.02
100
100
2
.01
200
200
3
.01
500
1000
= ´:µ
The insurer charges ,´:µ b , where : is the aggregate claim random variable, and
calculates the premium to be 4,454.40 . The insurer later notices that the mean claim amount for
group 3 should actually have been 400, and recalculates the premium to be 4,076.20 . How
many employees are in group 1?
A) 50
B) 100
C) 200
D) 400
E) 800
5. An insurer has a portfolio of 500 policies described below:
Probability
Claim
Number in
Category
Category
of Claim
Distribution
~ , ~ 1
.01
2
c .04
~ 3 , ~ 1
The insurer will charge a premium of ,´:µ b j= ´:µ , where : is the aggregate claim
random variable. What is the maximum possible premium that will be charged?
A) 74.25
B) 74.5
C) 74.75
D) 75.0
E) 75.25
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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MODELING - PROBLEM SET 19
6. (SOA) The number of claims, 5 , made on an insurance portfolio follows the following
distribution:
7 (5 ~ )
0
.7
2
.2
3
.1
If a claim occurs, the benefit is 0 or 10 with probability 0.8 and 0.2, respectively. The number of
claims and the benefit for each claim are independent. Calculate the probability that aggregate
benefits will exceed expected benefits by more than 2 standard deviations.
A) .02
B) .05
C) .07
D) .09
E) .12
7. (SOA) You own a fancy light bulb factory. Your workforce is a bit clumsy — they keep
dropping boxes of light bulbs. The boxes have varying numbers of light bulbs in them, and when
dropped, the entire box is destroyed.
You are given:
Expected number of boxes dropped per month:
50
Variance of the number of boxes dropped per month:
100
Expected value per box:
200
Variance of the value per box:
400
You pay your employees a bonus if the value of light bulbs destroyed in a month is less than
8000.
Assuming independence and using the normal approximation, calculate the probability that you
will pay your employees a bonus next month.
A) 0.16
B) 0.19
C) 0.23
D) 0.27
E) 0.31
8. (SOA) A dam is proposed for a river which is currently used for salmon breeding. You have
modeled:
(i) For each hour the dam is opened the number of salmon that will pass through and reach the
breeding grounds has a distribution with mean 100 and variance 900.
(ii) The number of eggs released by each salmon has a distribution with mean of 5 and
variance of 5.
(iii) The number of salmon going through the dam each hour it is open and the numbers of eggs
released by the salmon are independent.
Using the normal approximation for the aggregate number of eggs released, determine the least
number of whole hours the dam should be left open so the probability that 10,000 eggs will be
released is greater than 95%.
A) 20
B) 23
C) 26
D) 29
E) 32
9. (SOA) In a clinic, physicians volunteer their time on a daily basis to provide care to those who
are not eligible to obtain care otherwise. The number of physicians who volunteer in any day is
uniformly distributed on the integers 1 through 5. The number of patients that can be served by a
given physician has a Poisson distribution with mean 30. Determine the probability that 120 or
more patients can be served in a day at the clinic, using the normal approximation with continuity
correction.
A) 1 c )(0.68)
B) 1 c )(0.72)
C) 1 c )(0.93)
D) 1 c )(3.13)
E) 1 c )(3.16)
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 19
LM-271
10. (SOA) For an individual over 65:
(i) The number of pharmacy claims is a Poisson random variable with mean 25.
(ii) The amount of each pharmacy claim is uniformly distributed between 5 and 95.
(iii) The amounts of the claims and the number of claims are mutually independent.
Determine the probability that aggregate claims for this individual will exceed 2000 using the
normal approximation.
A) 1 c )(1.33) B) 1 c )(1.66)
C) 1 c )(2.33) D) 1 c )(2.66) E) 1 c )(3.33)
11.(SOA) A company has 50 employees whose dental expenses are mutually independent. For
each employee, the company reimburses 100% of dental expenses in excess of a $100 deductible.
The dental expense for each employee is distributed as follows:
Expense
Probability
0
0.20
50
0.30
200
0.30
500
0.10
1,000
0.10
Determine, by normal approximation, the 95-th percentile of the cost to the company.
A) 8,000
B) 9,000
C) 10,000
D) 11,000
E) 12,000
12. (SOA) An insurer provides life insurance for the following group of independent lives:
NumberDeath
Probability
of Lives
Benefit
of Death
100
1
0.01
200
2
0.02
300
3
0.03
The insurer purchases reinsurance with a retention of 2 on each life. The reinsurer charges a
premium of / equal to its expected claims plus the standard deviation of its claims. The insurer
charges a premium . equal to expected retained claims plus the standard deviation of retained
claims plus / . Determine ..
A) 44
B) 46
C) 70
D) 94
E) 96
13. (SOA) Two portfolios of independent insurance policies have the following characteristics:
Portfolio A:
Class NumberProb. of Claim
Claim Amount
in Class
per Policy
1
2000
0.05
1
2
500
0.10
2
Portfolio B:
Class NumberProb. of Claim
Claim Amount Distribution
in Class
per Policy
Mean Variance
1
2000
0.05
1
1
2
500
0.10
2
4
The aggregate claims in the portfolios are denoted by :( and :) , respectively.
= ´: µ
Determine = ´:) µ .
(
A) 1.0
B) 1.1
© ACTEX 2009
C) 1.2
D) 2.0
E) 2.1
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-272
MODELING - PROBLEM SET 19
14. (SOA) A group medical insurance policy cover the medical expenses incurred by 100,000
mutually independent lives. The annual loss amount ? incurred by each life is distributed as
follows:
%¢
Á 7 ²? ~ %³ ¢
À
À
À
À
À
À
The policy pays 80% of the annual losses for each life. The premium is equal to the 95-th
percentile of the normal distribution which approximates the distribution of total claims.
Determine the security loading.
A) 1,213,000
B) 1,356,000
C) 1,446,000
D) 1,516,000
E) 1,624,000
15. (SOA) The policies of a building insurance company are classified according to the location
of the building insured:
Number of
Claim
Claim Amount
Policies in Region
Probability
Region
A
20
300
0.01
B
10
500
0.02
C
5
600
0.03
D
15
500
0.02
E
18
100
0.01
Using the normal approximation, relative security loadings are computed for each region such
that the probability that the total claims for the region do not exceed premium collected from
policies in that region is 0.95. Which region pays the largest relative security loading?
A) A
B) B
C) C
D) D
E) E
16. (SOA) An insurer has the following portfolio of policies:
Benefit
Number
Probability
Class Amount
Covered
of a Claim
1
1
400
0.02
2
10
100
0.02
The insurer reinsures the amount in excess of 9 (9 € ) per policy at a cost of 0.025 per unit of
coverage. The insurer wants to minimize the probability, as determined by the normal
approximation, that retained claims plus cost of reinsurance exceeds 34. Determine 9 .
A) 1.5
B) 2.0
C) 2.5
D) 3.0
E) 3.5
17. (SOA) An insurance company is selling policies to individuals with independent future
lifetimes and identical mortality profiles. For each individual, the probability of death by all
causes is 0.10 and the probability of death due to accident is .01. Each insurance policy pays the
following benefits:
10 for accidental death, 1 for non-accidental death .
The company wishes to have at least 95% confidence that premiums with a relative security
loading of 0.20 are adequate to cover claims. Using the normal approximation, determine the
minimum number of policies that must be sold.
A) 1793
B) 1975
C) 2043
D) 2545
E) 2804
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 19
LM-273
18. (CAS May 06) Between 9am and 3pm Big National Bank employs 2 tellers to service
customer transactions. The time it takes Teller X to complete a transaction follows an exponential
distribution with a mean of 10 minutes. Transaction times for Teller Y follow an exponential
distribution with a mean of 15 minutes. Both Teller X and Teller Y are continuously busy while
the bank is open. On average, every third customer transaction is a deposit and the amount of the
deposit follows a Pareto distribution with parameters ~ and ~ . Each transaction
that involves a deposit of at least $7500 is handled by the branch manager. Calculate the total
deposits made through the tellers each day.
A) Less than $31,000
B) At least $31,000, but less than $32,500
C) At least $32,500, but less than $35,000
D) At least $35,000, but less than $37,500
E) At least $37,500
19. (SOA) Each life within a group medical expense policy has loss amounts which follow a
compound Poisson process with ~ À
. Given a loss, the probability that it is for Disease 1
is . Loss amount distributions have the following parameters:
Mean per loss
Standard Deviation per loss
Disease 1
5
50
Other diseases
10
20
Premiums for a group of 100 independent lives are set at a level such that the probability (using
the normal approximation to the distribution for aggregate losses) that aggregate losses for the
group will exceed aggregate premiums for the group is 0.24.
A vaccine which will eliminate Disease 1 and costs 0.15 per person has been discovered.
Define:
( ~ the aggregate premium assuming that no one obtains the vaccine, and
) ~ the aggregate premium assuming that everyone obtains the vaccine and the cost of the
vaccine is a covered loss.
Calculate (°) .
A) 0.94
B) 0.97
C) 1.00
D) 1.03
E) 1.06
20. (SOA) A company insures a fleet of vehicles. Aggregate losses have a compound Poisson
distribution. The expected number of losses is 20. Loss amounts, regardless of vehicle type, have
exponential distribution with ~ . In order to reduce the cost of the insurance, two
modifications are to be made:
(i) A certain type of vehicle will not be insured. It is estimated that this will reduce loss
frequency by 20%.
(ii) A deductible of 100 per loss will be imposed.
Calculate the expected aggregate amount paid by the insurer after the modifications.
A) 1600
B) 1940
C) 2520
D) 3200
E) 3880
21. An insurer combines two independent lines of insurance, both of which have compound
Poisson distributions. : , : and : denote the aggregate claim random variables for line 1, line
2 and the two lines combined, respectively, and ? ²³ Á ? ²³ and ? denote the claim amount
random variables for line 1, line 2 and the two lines combined. Given ,´:µ ~ ,
= ´:µ ~ , ,´: µ ~ , ,´? ²³ µ ~ , = ´? ²³ µ ~ , and = ´? ²³ µ ~ , what is ?
A) 1
B) 2
C) 3
D) 4
E) 5
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-274
MODELING - PROBLEM SET 19
MODELING - PROBLEM SET 19 SOLUTIONS
1. @ is the discretized distribution. 7 ´@ ~ b µ ~ 7 ´  ?  b µ ~ c c cc for
~ Á Á Á ÀÀÀ
B
B
B
,´@ µ ~ ² b ³²c c cc ³ ~ ²c c cc ³ b ²c c cc ³ .
~
B
²
~
c
c
B
~
cc
c
³ ~ ² c ³ b ²
c
B
c
~
c
c ³bÄ~.
c
c ~ c Á cc ~ c ²c ³
²c ³
~
~
B
c
c
c
S ²c c cc ³ ~ ²c
c ³ c ²cc ³ ~ cc .
~
c
,´@ µ ~ c
c b ~ À .
Answer: E
2. : ~ @ b Ä b @5 , where 5 is Poisson with mean 10 and @ ~ ²? c ³b .
,´:µ ~ ,´5 µ h ,´@ µ ~ ²³,´²? c ³b µ .
Since ? is exponential,
B
B
,´²? c ³b µ ~ ´ c -? ²!³µ ! ~ c!° ! ~ cÀ ~ À .
Then, ,´:µ ~ ²³²À³ ~ À .
,´²? c ³b µ ~ ,´²? c ³ O? € µ h 7 ²? € ³ .
For the exponential distribution, we know that the cost per payment
? c O? € also has an exponential distribution with the same mean of 100.
Therefore, the second moment is d ~ Á , so that
,´²? c ³b µ ~ Á h 7 ²? € ³ ~ Á c° ~ Á À
Then, since 5 is Poisson, = ´:µ ~ h ,´@3 µ ~ ²³²
Á À
³ ~ Á .
An alternative approach would be to let 5 i be the number of losses above 20, and let
@7 be the cost per payment. Then : ~ @7 b Ä b @7 5 i .
Since 5 is Poisson, 5 i is also Poisson with mean
7 ²? € ³ ~ ²³c° ~ À
.
The cost per payment is exponential with mean 100, so ,´@7 µ ~ d ~ Á and = ´:µ ~ ²À³²Á ³ ~ Á .
3. The aggregate payment is : ~ @ b Ä b @5 , where @ ~ ²? c ³b .
If we define 5 i to be the number of losses above the deductible, then
: ~ @7 b @7 b Ä b @7 5 i , where @7 is the cost per payment.
5 i is also negative binomial, with ~ and
Z ~ h 7 ²? € ³ ~ h ² b
³ ~ , so ,´5 µ ~ ²³² ³ ~ ² ³ .
Since ? is Pareto, @7 is also Pareto with ~ and Z ~ b ~ so ,´@7 µ ~ c
~ .
Then, ,´:µ ~ ,´5 i µ h ,´@7 µ ~ ² ³²³ ~ .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 19
LM-275
4. Let and denote the numbers of employees in groups 1 and 2, respectively, so that there
are c c employees in group 3. With the original information,
,´:µ ~ h ²À³²³ b h ²À³²³ b ² c c ³ h ²À³²³
~ c c , and
= ´:µ ~ h ´²À³²À³² ³ b ²À³²³µ b h ´²À³(À³² ³ b ²À³²³µ
+² c c ³ h ´²À³²À³² ³ b ²À³²³µ
~ Á Á c c .
ÁÁc c
Thus, c c b
~ Á À , or equivalently,
Á À c À c À ~ Á À (Equation 1).
With the group 3 mean claim amount changed to 400, these calculations become
,´:µ ~ h ²À³²³ b h ²À³²³ b ² c c ³ h ²À³²³
~ c c , and
= ´:µ ~ h ´²À³²À³² ³ b ²À³²³µ b h ´²À³(À³² ³ b ²À³²³µ
+² c c ³ h ´²À³²À³² ³ b ²À³²³µ
~ Á Á c c .
ÁÁc
c
Thus, c c b
~ Á À , or equivalently,
Á À c À
c À
~ Á À (Equation 2).
Solving equations 1 and 2 results in ~ À
Note that the number in group 3 could be found directly by noting that the decrease from 500 to
400 in mean claim amount for group 3 results in a reduction in ,´:µ of and a reduction in
= ´:µ of ²À³²À³² c ³ ~ . The reduction in the premium will be
À ~ À, so that ~ . This will now slightly simplify the two equations in and .
Answer: E.
5. ,´:µ ~ h ²À³²³ b ² c ³ h ²À³²³ ~ c À , and
= ´:µ ~ h ´²À³²À³² ³ b ²À³²³µ b ² c ³ h ´²À³²À
³² ³ b ²À³²³µ
~ À b À
.
The premium is ² ³ ~ c À b jÀ b À
, where   . The
maximum of occurs at either ~ or ~ or at a critical point of .
Z ² ³ ~ c À b j À
. Setting Z ² ³ ~ and solving results in ~ À .
ÀbÀ
Calculating ²³ ~ À , ²
À³ ~ À , and ²³ ~ À .
Answer: A.
6. We are given the distribution of 5 (claims number) and ? (claim amount).
The mean and variance of 5 and ? are found as follows.
,´5 µ ~ ²³²À³ b ²³²À³ b ²³²À³ ~ À Á
,´5 µ ~ ² ³²À³ b ² ³²À³ b ² ³²À³ ~ À Á = ´5 µ ~ ,´5 µ c ²,´5 µ³ ~ À.
,´?µ ~ ²³²À³ b ²³²À³ ~ Á ,´? µ ~ ² ³²À³ b ² ³²À³ ~ Á
= ´?µ ~ ,´? µ c ²,´?µ³ ~ .
The mean and variance of the aggregate benefit : are
,´:µ ~ ,´5 µ h ,´?µ ~ À Á = ´:µ ~ = ´5 µ h ²,´?µ³ b ,´5 µ h = ´?µ ~ À
The standard deviation of : is j
À ~ À .
We are asked to find 7 ´: c ,´:µ € j= ´:µ µ ~ 7 ´: € Àµ .
The event ": € À" will occur only if at least one claim of amount 10 occurs.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
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MODELING - PROBLEM SET 19
6. continued
7 ´at least one claim of amount 10]
~ 7 ´at least one claim of amount 10 q ²5 ~ ³]
b 7 ´at least one claim of amount 10 q ²5 ~ ³]
~ 7 ´at least one claim of amount 10O5 ~ ] h 7 ´5 ~ µ b
b 7 ´at least one claim of amount 10O5 ~ ] h 7 ´5 ~ µ
~ ´ c ²À³ µ h ²À³ b ´ c ²À³ µ h ²À³ ~ À .
Note that if there are 5 ~ claims, then in order for neither claim to be 10, both claims must be
0, and the probability of that is ²À³ .
Therefore, 7 ´at least one claim of amount 10O5 ~ ] ~ c ²À³ .
A similar comment applies to 7 ´at least one claim of amount 10O5 ~ ] ~ c ²À³
Alternatively, since event ": € À" will occur only if at least one claim of amount 10 occurs,
and since non-zero claims are all 10,
7 ´: € Àµ ~ c 7 ´:  Àµ ~ c 7 ´: ~ µ (the only way : is less than 9.4 is if : is 0).
7 ´: ~ µ ~ 7 ´: ~ q 5 ~ µ b 7 ´: ~ q 5 ~ µ b 7 ´: ~ q 5 ~ µ
~ 7 ´: ~ O5 ~ µ h 7 ´5 ~ µ b 7 ´: ~ O5 ~ µ h 7 ´5 ~ µ
b 7 ´: ~ O5 ~ µ h 7 ´5 ~ µ ~ ²³²À³ b ²À³ ²À³ b ²À³ ²À³ ~ À .
7 ´: € Àµ ~ c À ~ À .
Answer: E
7. ,´5 µ ~ Á = ´5 µ ~ Á ,´@ µ ~ Á = ´@ µ ~ À
Total value destroyed in one month is : ~ @ b Ä b @5 (compound distribution).
,´:µ ~ ,´5 µ h ,´@ µ ~ Á = ´:µ ~ = ´5 µ h ²,´@ µ³ b ,´5 µ h = ´@ µ ~ Á Á .
Using the normal approximation,
:cÁ
ÁcÁ
7 ´:  µ ~ 7 ´ j
 j
µ ~ 7 ´A  c Àµ
ÁÁ
~ c )²³ ~ c À ~ À
.
ÁÁ
Answer: A
8. This problem involves a compound distribution. : is the number of eggs released in one
hour and has a compound distribution. 5 is the number of salmon passing through in one hour,
and ? is the number of eggs released by a salmon.
We are given ,´5 µ ~ Á = ´5 µ ~ Á ,´?µ ~ = ´?µ ~ .
Then, = ´:µ ~ ,´5 µ h ,´?µ ~ , and
= ´:µ ~ ,´5 µ h = ´?µ b = ´5 µ h ²,´?µ³ ~ Á .
The total number of eggs released in hours is ; ~ : b : b Ä b : , with mean
,´; µ ~ h ,´:µ ~ and with variance = ´; µ ~ h = ´:µ ~ Á .
Using the normal approximation, the probability that (at least) 10,000 eggs will be released is
7 ´; ‚ Á µ ~ 7 ´; ‚ Á Àµ (integer correction for the discrete distribution of ; ).
Standardizing the probability, we get
ÁÀc
; c
7 ´; ‚ Á Àµ ~ 7 ´ j
‚ j
µ ~ 7 ´A ‚ ÁÀc
µ,
j
Á
Á
Á
where A has a standard normal distribution.
We continue with trial and error on the possible values of given in the answers.
~ S 7 ´A ‚ µ ~ c )²³ ~ À ,
~ S 7 ´A ‚ c À
µ ~ )²À
³ ~ ²À³²À³ b ²À
³²À³ ~ À ‚ À .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 19
LM-277
8. continued
Therefore, 23 hours is the number of hours required so that the probability that 10,000 eggs will
be released is greater than 95%. Note that in this solution the integer correction was used because
; was an integer-valued random variable. Because of the size of the numbers involved, it was
irrelevant to use the integer correction. There would have been no change in the value of c À
(at the 2nd decimal) in the ~ case if the integer correction had not been used.
Answer: B
9. We wish to find 7 ´: ‚ µ , where : is the number of patients that can be served in a day.
: has a compound distribution: : ~ 4 b 4 b Ä b 45 , where 4 is the number of
patients served by physician , and 5 is the number of physicians volunteering. We are told that
for each , 4 has a Poisson distribution with mean 30, so that ,´4 µ ~ = ´4 µ ~ . We are
also told that 5 has a uniform distribution on the integers from 1 to 5, so that
²c³ bÄb²c³
,´5 µ ~ bbbb
~ and = ´5 µ ~ ,´²5 c ³ µ ~
~.
The mean and variance of : are then ,´:µ ~ ,´5 µ h ,´4 µ ~ ²³²³ ~ and
= ´:µ ~ ,´5 µ h = ´4 µ b = ´5 µ h ²,´4 µ³ ~ ²³²³ b ²³²³ ~ .
Since : is a sum of integers (each 4 ) is an integer, we are asked to apply the continuity
correction to find 7 ´: ‚ µ when using the normal approximation. This means that we find
7 ´: ‚ Àµ (the continuity correction means that if a discrete random variable + is being
approximated by a continuous random variable * , then if is an integer, we use
7 ´+ ~ µ š 7 ´ c À  *  b Àµ , and then 7 ´+ ‚ µ š 7 ´* ‚ c Àµ ).
Applying the normal approximation to : , we get
:c,²:³
= ²:³
7 ´: ‚ µ š 7 ´: ‚ Àµ ~ 7 ´ j
‚ Àc
µ ~ 7 ´A ‚ À
µ ~ c )²À
³.
j
Answer: A
10. The aggregate claim : has a compound Poisson distribution with
~ , and ? uniformly distributed between 5 and 95. Then ,´?µ ~ b
~ ²c³
and = ´?µ ~ ~ .
,´:µ ~ ,´?µ ~ ²³²³ ~ Á
= ´:µ ~ ,´? µ ~ ´= ´?µ b ²,´?µ³ µ ~ ²³´
b ²³ µ ~ Á .
Applying the normal approximation to : we get
:c,²:³
= ²:³
7 ´: € µ ~ 7 ´ j
€ c
µ ~ 7 ´A € À
µ ~ c )²À
³ . Answer: D
j
Á
11. Let ? be the amount paid for one member of the group. The distribution of ? (after
deductible) is
?¢
Prob: À
À
À
À
,´?µ ~ Á = ´?µ ~ ,´? µ c ²,´?µ³ ~ Á c Á ~ Á À
: ~ ? S ,´:µ ~ ,´?µ ~ Á = ´:µ ~ = ´:µ ~ Á Á .
~
The 95-th percentile of : is , where 7 ´:  µ ~ À S 7 ´ j:c  jc µ ~ À .
ÁÁ
ÁÁ
Using the normal approximation, jc ~ À
S ~ Á À Answer: D
ÁÁ
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-278
MODELING - PROBLEM SET 19
12. The reinsurer covers a benefit of 1 for each of the third group of 300 lives. The expected
claim on the reinsurer is ,´:9 µ ~ ²³²À³ ~ , and the variance of the claims paid by the
reinsurer is = ´:9 µ ~ ² ³²À³²À³ ~ À . The reinsurer's premium is
/ ~ b jÀ ~ À . The ceding insurer's expected retained claim is
,´:* µ ~ ²³²À³ b ²³²À³ b ²³²À³ ~ , and the variance of retained claims
by the ceding insurer is
= ´:* µ ~ ²³²À³²À³ b ² ³²À³²À³ b ² ³²À³²À³ ~ À À
The ceding insurer's premium is . ~ / b b jÀ ~ . Answer: B
13. The variance individual policy claim is ² c ³ b . The policies in Portfolio A
all have ~ . Portfolio B policies have the same 's as those in Portfolio A, but Portfolio B
policies have € .
= ´:( µ ~ ² ³²À³²À³ b ² ³²À³²À³ ~ .
= ´:) µ ~ ´² ³²À³²À³ b ²³²À³µ b ´² ³²À³²À³ b ²³²À³µ ~ .
Answer: E
14. The security loading (not relative security loading in this case) is * , where
:c,´:µ
7 ´:  ,´:µ b *µ ~ À , or equivalently 7 ´ j
 j *
µ ~ À , so that
= ´:µ
= ´:µ
* ~ À
j= ´:µ , where is the aggregate payment random variable. For one life, the
expected loss amount is ,´?µ ~ Á and = ´?µ ~ ,´? µ c ²,´?µ³ ~ ÀÁ .
Since the insurer pays 80% of each loss, the insurer pays @ ~ À? on loss ? , with
,´@ µ ~ À,´?µ ~ , and = ´@ µ ~ À
= ´?µ ~ Á Á . For the portfolio of
100,000 policies, j= ´:µ ~ jÁ ²Á Á ³ ~ Á . The security loading is
* ~ À
²Á ³ ~ Á Á . Note that the premium would be ,´:µ b * . Answer: A
15. The relative security loading for a region is , where 7 ´:  ² b ³,´:µ µ ~ À and : is
the claim random variable for the region. . This is equivalent to
:c,´:µ
,´:µ
 j
µ ~ À , which, under the normal approximation
= ´:µ
= ´:µ
,´:µ
j
j= ´:µ =1.645 , or equivalently, ~ ,´:µ h À
= ´:µ À
7´j
becomes
For the five regions we have
( ~ ²³²³²À³
h À
j²³² ³²À³²À³ ~ À Á
) ~
h À
j²³² ³²À³²À³ ~ À Á
* ~
+ ~
, ~
²³²³²À³
j
²
³²³²À³ h À
²
³² ³²À³²À³ ~ À Á
j
²³²³²À³ h À
²³² ³²À³²À³ ~ À Á
j
²³²³²À³ h À
²³² ³²À³²À³ ~ À
.
© ACTEX 2009
Answer: E
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 19
LM-279
16. : ~ retained claims, * ~ cost of reinsurance.
* ~ ²³² c 9³²À³ ~ À² c 9³ Á ,´:µ ~ ²³²³²À³ b ²³²9³²À³ ~ b 9
= ´:µ ~ ²³² ³²À³²À³ b ²³²9 ³²À³²À³ ~ À b À
9 .
7 ´: b * € µ ~ 7 ´: b c À9 € µ ~ 7 ´: € b À9µ
:c,´:µ
= ´:µ
~ 7´j
bÀ9c²b9³
€ j
µ.
ÀbÀ
9 This probability is minimized if
bÀ9c²b9³
jÀbÀ
9 bÀ9
~ j bÀ9 ~
is maximized.
ÀbÀ
9
²À³jb9 At this point, we can substitute in the five possible answers to see which gives the maximum
value, or we can use standard calculus maximizing methods. Since 9 € € , this is
equivalent to maximizing ´
bÀ9
µ
²À³jb9 ²bÀ9³
~ ²À
³²b9 ³ . The critical points occur where
² b 9 ³² b À9³ c ² b À9³ ²9³ ~ , so that b 9 c ² b À9³²9³ ~ , or
equivalently, 9 ~ .
Answer: B
17. The probability of death by non-accident is À c À ~ À . For a single policy with loss
random variable ? ,. we have ,´?µ ~ ²³²À³ b ²³²À³ ~ À ,
and = ´?µ ~ ² ³²À³ b ² ³²À³ c ²À³ ~ À . For a portfolio of policies, with
aggregate loss : , ,´:µ ~ À , = ´:µ ~ À . With relative security loading .2, the
premium received is À,´:µ ~ À . In order for this premium to provide at least 95%
probability of covering claims, we must have 7 ´:  Àµ ‚ À .
:c,´:µ
This is equivalent to 7 ´ j
 ÀcÀ
jÀ µ ‚ . . Applying the normal approximation
= ´:µ
to : , we have jÀ03 ‚ À
, or equivalently, ‚ . Answer: B
À
18. The number of transactions completed per hour by Teller ? has a Poisson distribution with
a mean of 6 (since transaction time is exponential with a mean of 1/6 hour), and the number of
transactions completed per hour by Teller Y has a Poisson distribution with a mean of 4. The
total number of transactions completed per hour for both tellers combined has a Poisson
distribution with a mean of 10. Since, on average, every third customer makes a deposit, the
average number of deposits per hour is 10/3 . Each deposit amount has a Pareto distribution, and
the probability of a deposit being  7500 is c ² b
³ ~ À
.
The expected number of deposits per hour made through the Tellers is ² ³²À
³ ~ À
.
The average amount of a deposit handled by a teller is
%h ²%³
,´?O?  µ ~ % h ²%O?  ³ % ~ % .
7 ²?³
,´? w µ ~ % h ²%³ % b h 7 ²? € ³ , so that
% h ²%³ % ~ ,´? w µ c h 7 ²? € ³
~ c ´ c ² b ³ µ c ² b ³ ~ .
Then, the expected total deposits made through the tellers in an hour is
² ³²À
³ h ,´?O?  µ ~ ² ³²À
³ h À
~ .
The expected total per day (6 hours) is d ~ Á . Answer: B
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-280
MODELING - PROBLEM SET 19
19. The losses can be separated into two independent Poisson processes, one for Disease 1, and
one for other diseases. The Poisson rate for Disease 1 is ² ³²À
³ ~ À , and the rate for other
diseases is .15 . The severity for Disease 1 is ? , with mean 5 and variance 2500 and ? for other
diseases with mean 10 and variance 400. The aggregate loss for one life is the combination of the
compound distributions, : being losses for Disease 1, and : being losses for other diseases.
,´: µ ~ ²À³²³ ~ À , = ´: µ ~ ²À³²³ ~ À (the second moment of ? is 2525),
and ,´: µ ~ ²À³²³ ~ À and = ´: µ ~ ²À³²³ ~ .
Then : , aggregate losses for one life, has mean ,´:µ ~ À and variance = ´:µ ~ À .
Aggregate losses for 100 independent lives, say > , has mean 155 and variance 10,025.
Applying the normal approximation to > , the aggregate premium is (, where
> c
(c
(c
À ~ 7 ´> € (µ ~ 7 ´ j
€ j
µ ~ c )² j
³ , so that j(c ~ À
Á
Á
Á
Á
and then ( ~ À . This is the premium if non obtains the vaccine.
If everyone obtains the vaccine, then : losses are eliminated, and aggregate losses < for the 100
lives has mean ,´< µ ~ ²À b À³ ~ and = ´< µ ~ ²³ ~ .
Then À ~ 7 ´< € )µ ~ c )² )c
³ , so that )c
~ À
and ) ~ À .
j
j
(°) ~ À . Answer: C
20. This problem involves the Compound Poisson distribution and application of a deductible to
an individual loss. After the modification to the types of vehicle that will be insured, the loss
frequency will be Poisson with ,´5 µ ~ ²À³²³ ~ expected losses. Before the application of
the deductible, the amount paid by the insurer for a loss is ? , the full loss amount, which has an
exponential distribution with ~ . After the application of the deductible, the amount paid
if ?  by the insurer per loss is @ ~ %²? c Á ³ ~ F
.
? c if ? € The expected amount paid by the insurer per loss is
B
B
,´@ µ ~ ²% c ³ h ? ²%³ % ~ ²% c ³ h ²À³c%° % .
This expectation can also be represented as
B
B
,´@ µ ~ ´ c -? ²%³µ % ~ c%° % ~ c° ~ À
%
%
(-? ²%³ ~ ; ²!³ ! ~ ²À³c!° ! ~ c c!° ).
The expected aggregate loss paid by the insurer after the modifications is
,´5 µ h ,´@ µ ~ ²
³²À³ ~ À
.
Answer: B
21. ,´:µ ~ ,´: µ b ,´: µ ~ ,´: µ b h ,´? ²³ µ ¦ ~ b h ¦ ~ .
= ´:µ ~ = ´: µ b = ´: µ ~ = ´: µ b h ,´²? ²³ ³ µ .
But, ,´²? ²³ ³ µ ~ = ´? ²³ µ b ²,´? ²³ µ³ ~ b ~ ¦ ~ = ´: µ b h ¦ = ´: µ ~ ~ h ,´²? ²³ ³ µ . Thus,
= ´: µ
,´: µ
,´²? ²³ ³ µ
~ ~ ,´? ²³ µ
²³
²³ ²³ Since = ´? µ ~ ,´²? ³ µ c ²,´? µ³ ~ , we get a quadratic equation in ,´? ²³ µ :
²,´? ²³ µ³ c ,´? ²³ µ b ~ , or equivalently, ,´? ²³ µ ~ .
,´: µ
Thus, ~ ,´? ²³
~ ~ .
µ
© ACTEX 2009
~
Answer: B.
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 20 - STOP-LOSS INSURANCE
LM-281
MODELING SECTION 20 - STOP LOSS INSURANCE
The material in this section relates to Loss Models, Sections 9.3.
The suggested time for this section is less than 2 hours.
When a deductible is applied to aggregate losses (not individual losses), the insurance payment
will be the aggregate loss in excess of the deductible. If aggregate losses are : for a period, and
the deductible for the period is , then the stop-loss insurance payment is
4 %¸: c Á ¹ ~ ²: c ³b ~ : c ²: w ³ ~ F
: c
if :  .
if : € (20.1)
This is algebraically identical to ordinary deductible covered earlier in Section 11 of this study
guide.
The expected value of stop-loss insurance paid is the net stop-loss premium, which is equal to
,´²: c ³b µ . This can be formulated various ways.
B
(20.2)
,´²: c ³b µ ~ ²& c ³ h : ²&³ & if : is continuous,
B
or ² c ³ h : ²³ if : is discrete and integer valued.
(20.3)
~b
B
We also have the formulations ,´²: c ³b µ ~ ,´:µ c ,´: w µ ~ ´ c -: ²%³µ % . (20.4)
Equation 20.4 is valid for any non-negative distribution of : , continuous or discrete. For most
exam questions, the formulation ,´²: c ³b µ ~ ,´:µ c ,´: w µ is usually quite efficient to
use.
Insurance with a deductible was considered earlier in Section 11 of this study guide, and we are
considering the same idea here, except that the random variable to which the deductible is applied
is the aggregate loss : .
The following illustration will provide some insight into the mechanics of applying a deductible
to a loss. Suppose that : has the following discrete distribution:
&¢
À
: ²&³ ¢
À
À
À
À
À
À
-: ²&³ ¢
À
À
À
À
À
À
c -: ²&³ ¢
À
À
À
À
À
The mean can be found from
,´:µ ~ & h : ²&³ ~ ²³²À³ b ²³²À³ b ²³²À³ b ²³²À³ b ²
À³²À³ b ²³²À³ ~ À .
all &
B
The mean is also equal to ,´:µ ~ ´ c -: ² ³µ . In the case that : has a discrete
distribution, the distribution function and its complement are step-functions.
The graph of -: ² ³ is on the next page.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-282
MODELING SECTION 20 - STOP-LOSS INSURANCE
Note that -: ²³ ~ and -: ²&³ ~ for & ‚ .
The graph of the function c -: ² ³ is given below.
Note that c -: ²³ ~ .
B
Therefore, in this case ,´:µ ~ ´ c -: ² ³µ ~ ´ c -: ² ³µ .
Since c -: ² ³ is a step function, ´ c -: ² ³µ is the area under the curve, which
becomes the area of a series of rectangles, the area of the shaded region above:
´ c -: ² ³µ ~ ²³²À³ b ²³²À
³ b ²³²À³ b ²À³²À³ b ²À³²À³ ~ À .
Suppose that we wish to apply a deductible of ~ to create the stop loss insurance ²: c ³b
for this example. We can find ,´²: c ³b µ from ,´²: c ³b µ ~ ,´:µ c ,´: w µ , where
: w~F
:
if :  ~F
if : € if : ~ , : ²³ ~ À
.
if : ‚ , 7 ²: € ³ ~ À
Then, ,´: w µ ~ ²³ h : ²³ b ²³ h 7 ²: € ³ ~ À, and
,´²: c ³b µ ~ ,´:µ c ,´: w µ ~ À c À ~ À .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 20 - STOP-LOSS INSURANCE
LM-283
B
We can also use the relation ,´²: c ³b µ ~ ´ c -: ²&³µ & .
With a deductible of 1, this becomes ,´²: c ³b µ ~ ´ c -: ²&³µ & . The integral is the
area of the following shaded region, again a series of rectangles.
,´²: c ³b µ ~ ´ c -: ²&³µ & ~ ²³²À
³ b ²³²À³ b ²À³²À³ b ²À³²À³ ~ À .
Notice that ,´²: c ³b µ ~ ´ c -: ²&³µ &
~ ´ c -: ²&³µ & c ´ c -: ²&³µ & ~ ,´:µ c ´ c -: ²³µ .
This is true because c -: ²&³ ~ c -: ²³ ~ À is constant for  &  .
From the graphical point of view, ,´²: c ³b µ is found by subtracting the area of the first
rectangle on the left of the graph of c -: ²&³
Now suppose that we wish to apply a deductible of ~ to create the stop loss insurance
²: c ³b for this example. Again, to find ,´²: c ³b µ we can use the relationship
,´²: c ³b µ ~ ,´:µ c ,´: w µ ,
where : w ~ F : ~ , : ²³ ~ À
: ~ , : ²³ ~ À
.
: ‚ , 7 ²: € ³ ~ À
Then ,´: w µ ~ ²³²À³ b ²³²À³ b ²³²À
³ ~ À , and ,´²: c ³b µ ~ À c À ~ À .
B
We can also use the relation ,´²: c ³b µ ~ ´ c -: ²%³µ % . In this case, this becomes
,´²: c ³b µ ~ ´ c -: ²%³µ % . The integral is the area of the shaded region on the next
page, again a series of rectangles.
,´²: c ³b µ ~ ´ c -: ²%³µ % ~ ²³²À³ b ²À³²À³ b ²À³²À³ ~ À .
Notice that ,´²: c ³b µ ~ ´ c -: ²%³µ %
~ ´ c -: ²%³µ % c ´ c -: ²%³µ % ~ ,´²: c ³b µ c ´ c -: ²³µ .
This is true because c -: ²%³ ~ c -: ²³ ~ À
is constant for  %  .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-284
MODELING SECTION 20 - STOP-LOSS INSURANCE
Now suppose that we wish to apply a deductible of ~ À to create the stop loss insurance
²: c À³b for this example. We can use ,´²: c À³b µ ~ ,´:µ c ,´: w Àµ ,
and ,´: w Àµ ~ ²³²À³ b ²³²À³ b ²³²À³ b ²À³²À³ ~ À , so that
,´²: c À³b µ ~ À c À ~ À .
B
We can also use the relation ,´²: c À³b µ ~ À ´ c -: ²%³µ % . In this case, this becomes
,´²: c À³b µ ~ À ´ c -: ²%³µ % . The integral is the area of the following shaded region,
again a series of rectangles. One of the rectangles is a part of the rectangle whose base runs from
% ~ to % ~ .
,´²: c À³b µ ~ À ´ c -: ²%³µ % ~ ²À³²À³ b ²À³²À³ b ²À³²À³ ~ À .
Notice that ,´²: c À³b µ ~ À ´ c -: ²%³µ %
À
~ ´ c -: ²%³µ % c ´ c -: ²%³µ % ~ ,´²: c ³b µ c ²À³² c -: ²³³ .
This is true because c -: ²%³ ~ c -: ²³ ~ À is constant for  %  .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 20 - STOP-LOSS INSURANCE
LM-285
The general point being made by the example above using the graphical approach can be
described in the following way. Suppose that : has a discrete distribution, and suppose that and are two successive values of : (so that  , and there are no values of : between and
). Since : has a discrete distribution, -: ²%³ is a step function which steps up at each successive
point of probability of : . Then for any number % between and (actually, for  %  ), we
have -: ²%³ ~ -: ²³ since the next step after is at and there is no probability between and . This can be seen in the following diagram.
Therefore, it is also true that if  %  then c -: ²%³ ~ c -: ²³ .
Now suppose that is a deductible such that   , and suppose we know the expected
stop loss payment when the deductible is (that is, we know ,´²: c ³b µ ).
B
We wish to find ,´²: c ³b µ . We use the relationship ,´²: c ³b µ ~ ´ c -: ²%³µ % .
This can be written in the form
B
B
,´²: c ³b µ ~ ´ c -: ²%³µ % ~ ´ c -: ²%³µ % c ´ c -: ²%³µ % .
(20.5)
B
We know that ,´²: c ³b µ ~ ´ c -: ²%³µ % .
Also, since c -: ²%³ ~ c -: ²³ for  %  , we have
´ c -: ²%³µ % ~ ´ c -: ²³µ % ~ ² c ³´ c -: ²³µ
(this is the integral of a constant over the interval from to ) .
Therefore, ,´²: c ³b µ ~ ,´²: c ³b µ c ² c ³´ c -: ²³µ .
(20.6)
This relationship is valid for any between the successive probability points and .
In particular, ,´²: c ³b µ ~ ,´²: c ³b µ c ² c ³´ c -: ²³µ if and are successive points
of probability for : .
The most typical illustration of this relationship occurs when : is integer-valued. If that is so,
then for successive integer deductibles we have
(20.7)
,´²: c ³b µ ~ ,´:µ c ´ c -: ²³µ ,
(20.8)
,´²: c ³b µ ~ ,´²: c ³b µ c ´ c -: ²³µ Á ÀÀÀ
, :´: c ² b ³µb ; ~ ,´²: c ³b µ c ´ c -: ²³µ , for any integer ‚ .
(20.9)
This has come up regularly on the exam.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-286
MODELING SECTION 20 - STOP-LOSS INSURANCE
Some additional properties of stop-loss insurance are as follows.
B
(i) If : is continuous for : € , then ,´²: c ³b µ ~ ²% c ³ h : ²%³ % .
(ii) If : is discrete, then ,´²: c ³b µ ~ ²% c ³ h : ²%³ .
%€
(iii) If 7 ´  :  µ ~ and   , then
c
,´²: c ³b µ ~ ,´²: c ³b µ c ² c ³´ c -: ²³µ ~ c
c h ,´²: c ³b µ b c h ,´²: c ³b µ.
This is linear interpolation between the deductibles of and , and it is valid if : cannot assume
values (has no density) between and .
if :  (iv) If  , then ²: c ³b c ²: c ³b ~ H : c if  :  .
c if : € The maximum paid under this modified stop-loss arrangement is c .
This is a combination of maximum covered loss and policy deductible that was considered in
Section 14 of the study guide.
Example LM20-1: Under a stop-loss insurance arrangement, the insurance amount paid is 80%
of the excess of aggregate claims above 20, subject to a maximum payment of 5. All claim
amounts are non-negative integers. You are given:
Deductible:
Net stop-loss premium:
À À À À À
À
Determine the expected amount paid by the insurer.
Solution: The maximum payment is triggered if 80% of claims above 20 is 5, or equivalently, if
aggregate claims is above 26.25 (for then the excess over 20 is 6.25 or more, and 80% of 6.25
is 5). The payment made by the insurer can be expressed as
²À³´²: c ³b c ²: c À³b µ , and the expected amount paid by the insurer will be
²À³4,´²: c ³b µ c ,´²: c À³b µ5. We are given ,´²: c ³b µ ~ À .
Since all claim amounts are integer-valued, 7 ´
 :  µ ~ (: must be an integer).
Using the linear interpolation relationship (iii) above,
Àc
,´²: c À³b µ ~ c
À
c
h ,´²: c ³b µ b c
h ,´²: c ³b µ
~ ²À³²À
³ b ²À³²À
³ ~ À
.
The expected amount paid by the insurer is ²À³²À c À
³ ~ À .
Example LM20-2: For aggregate claims, : , you are given:
(i) : can assume values that are multiples of 10
(ii) ,´²: c ³b µ ~ À
(iii) ,´²: c ³b µ ~ À
Find -: ²³.
Solution: Since : ~ and : ~ are successive points of probability,
, 4´: c ²³µb 5 ~ ,´²: c ³b µ c h ´ c -: ²³µ
S À ~ À
c ²³´ c -: ²³µ S -: ²³ ~ À
.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 20
LM-287
MODELING - PROBLEM SET 20
Stop-Loss Insurance - Section 20
1. Annual aggregate losses : follow a compound distribution with annual frequency 5
and severity ? (the usual assumption of independence of 5 and the ? 's applies).
The probability function of 5 is uniform on the integers from 0 to .
? has a uniform distribution on the integers from 1 to .
Annual stop loss insurance on aggregate losses has a deductible of 2. The insurer collects a
premium equal to the sum of the mean and standard deviation of the stop loss. Find the stop loss
premium.
2. (SOA) An aggregate claim distribution has the following characteristics:
7 ´: ~ µ ~ for ~ Á Á ÀÀÀÁ . A stop-loss insurance with deductible amount has an
expected insurance payment of 1.5 . Find .
A) 1.75
B) 2.00
C) 2.25
D) 2.50
E) 2.75
3. (SOA): For aggregate claims, :À you are given ¢
(i) : takes on only positive integer values;
(ii) ,´:µ ~ (iii) ,´²: c ³b µ ~ (iv) ,´²: c ³b µ ~ Determine : ²³.
4. (SOA) Aggregate claims have a compound Poisson distribution with
~ , ? ²³ ~ , ? ²³ ~ . Determine ,´²: c ³b µ À
A) 3.05
B) 3.07
C) 3.09
D) 3.11
E) 3.13
5. (SOA) For an aggregate claim distribution : , the amount paid by a reinsurance policy is the
following random variable:
if :  : c if  :  0~
À:
if  :  if : ‚ Which of the following correctly expresses ,´0µ in terms of stop-loss expectations?
A) ,´²: c ³b µ c ,´²: c ³b µ c ,´²: c ³b µ
B) ,´²: c ³b µ c À,´²: c ³b µ c ,´²: c ³b µ
C) ,´²: c ³b µ c ,´²: c ³b µ c À,´²: c ³b µ
D) ,´²: c ³b µ c À,´²: c ³b µ c À,´²: c ³b µ
E) ,´²: c ³b µ c À,´²: c ³b µ c À,´²: c ³b µ
H
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-288
MODELING - PROBLEM SET 20
6. (SOA) For a certain company, losses follow a Poisson frequency distribution with mean 2 per
year, and the amount of a loss is 1, 2, or 3, each with probability 1/3. Loss amounts are
independent of the number of losses, and of each other. An insurance policy covers all losses in a
year, subject to an annual aggregate deductible of 2. Calculate the expected claim payments for
this insurance policy.
A) 2.00
B) 2.36
C) 2.45
D) 2.81
E) 2.96
7. (SOA) Prescription drug losses, S, are modeled assuming the number of claims has a geometric
distribution with mean 4, and the amount of each prescription is 40.
Calculate ,´²: c ³b µ .
A) 60
B) 82
C) 92
D) 114
E) 146
8. (SOA) WidgetsRUs owns two factories. It buys insurance to protect itself against major repair
costs. Profit equals revenues, less the sum of insurance premiums, retained major repair costs, and
all other expenses. WidgetsRUs will pay a dividend equal to the profit, if it is positive.
You are given:
(i)
Combined revenue for the two factories is 3.
(ii) Major repair costs at the factories are independent.
(iii) The distribution of major repair costs for each factory is
Prob²³
À
À
À
À
(iv) At each factory, the insurance policy pays the major repair costs in excess of that
factory’s ordinary deductible of 1. The insurance premium is 110% of the expected claims.
(v) All other expenses are 15% of revenues.
Calculate the expected dividend.
A) 0.43
B) 0.47
C) 0.51
D) 0.55
E) 0.59
9. (SOA) For a stop-loss insurance on a three person group:
(i) Loss amounts are independent.
(ii) The distribution of loss amount for each person is:
Loss Amount Probability
0
0.4
1
0.3
2
0.2
3
0.1
(iii) The stop-loss insurance has a deductible of 1 for the group.
Calculate the net stop-loss premium.
A) 2.00
B) 2.03
C) 2.06
D) 2.09
© ACTEX 2009
E) 2.12
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 20
LM-289
10. (SOA) For a collective risk model:
(i) The number of losses has a Poisson distribution with ~ .
(ii) The common distribution of the individual losses is:
%
? ²%³
À
À
An insurance covers aggregate losses subject to a deductible of 3. Calculate the expected
aggregate payments of the insurance.
A) 0.74
B) 0.79
C) 0.84
D) 0.89
E) 0.94
11. (SOA) In a given week, the number of projects that require you to work overtime has a
geometric distribution with ~ . For each project, the distribution of the number of overtime
hours in the week is the following:
%
²%³
5
0.2
10
0.3
20
0.5
The number of projects and the number of overtime hours are independent. You will get paid for
overtime hours in excess of 15 hours in the week. Calculate the expected number of overtime
hours for which you will get paid in the week.
A) 18.5
B) 18.8
C) 22.1
D) 26.2
E) 28.0
12. (SOA) A compound Poisson claim distribution has ~ and individual claims amounts
distributed as follows: %
? ²%³
À
À
where € The expected cost of an aggregate stop-loss insurance subject to a deductible of 5 is 28.03.
Calculate .
A) 6
B) 7
C) 8
D) 9
E) 10
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-290
MODELING - PROBLEM SET 20
MODELING - PROBLEM SET 20 SOLUTIONS
1. ,´5 µ ~ Á = ´5 µ ~ , ,´?µ ~ Á = ´?µ ~ .
,´:µ ~ ,´5 µ h ,´?µ ~ .
The stop loss insurance pays ²: c ³b ~ : c ²: w ³ .
,´: w µ ~ 7 ²: ~ ³ b 7 ²: € ³ .
7 ²: ~ ³ ~ 7 ²5 ~ ³ ~ À Á 7 ²: ~ ³ ~ 7 ²5 ~ ³ h 7 ²? ~ ³ ~ ²À³²À³ ~ À .
7 ²: € ³ ~ c À c À ~ À
À
,´: w µ ~ À b ²À
³ ~ À
À
,´²: c ³b µ ~ c À
~ À .
,´²: c ³b µ ~ ,´: µ c ,´²: w ³ µ c ²³´,²:³ c ,²: w ³µ À
= ´:µ ~ ,´5 µ h = ´?µ b = ´5 µ h ²,´?µ³ ~ ~ ,´: µ c ²,´:µ³ .
,´: µ ~ b ~ À
,´²: w ³ µ ~ 7 ²: ~ ³ b 7 ²: € ³ ~ À b ²À
³ ~ À .
,´²: c ³b µ ~ c ²À³ c ²
c À
³ ~ À
À
= ´²: c ³b µ ~ À
c ²À³ ~ À .
The premium for the stop loss insurance is b jÀ ~ À .
2. If  then
,´²: c ³b µ ~ ´² c ³ b ² c ³ b ² c ³ b ² c ³ b ²
c ³µ h ~ À S ~ À ,
which contradicts the assumption that  . If   , then
,´²: c ³b µ ~ ´² c ³ b ² c ³ b ² c ³ b ²
c ³µ h ~ À S ~ À .
3. From (iv) it follows that 7 ´: ‚ µ ~ , so that : has probability only at % ~ Á Á ,
so that : ²³ b : ²³ b : ²³ ~ .
From (ii) we get ,´:µ ~ h : ²³ ~ : ²³ b : ²³ b : ²³ ~ .
~
From (iii) we get ,´²: c ³b µ ~ ² c ³ h : ²³ ~ : ²³ ~ .
~
From the three equations, we can solve for : ²³, which is .
Actually, we do not need (iv). From (i) we have : ‚ , so that : ²³ ~ .
Then ,´²: c ³b ~ ,´:µ c ´ c -: ²³µ ~ c ~ .
Then ~ ,´²: c ³b ~ ,´²: c ³b µ c ´ c -: ²³µ ~ c ´ c -: ²³µ S -: ²³ ~ .
But ~ -: ²³ ~ : ²³ b : ²³ ~ b : ²³ .
4. When : is integer valued, then for any integer ‚ we have
,´²: c c ³b µ ~ ,´²: c ³b µ c ´ c -: ²³µ .
²: c ³b ~ : S ,´²: c ³b µ ~ ,´:µ ~ ,´?µ ~ ²³² ³ ~ .
-: ²³ ~ : ²³ ~ 7 ´5 ~ µ ~ c ~ c ~ À
S ,´²: c ³b µ ~ ,´²: c ³b µ c ´ c -: ²³µ ~ c ² c À³ ~ À .
-: ²³ ~ -: ²³ b : ²³ . : ²³ ~ ? ²³ h 7 ´5 ~ µ ~ ² ³²c h ³ ~ À
S -: ²³ ~ À b À ~ À À
Then ,´²: c ³b µ ~ ,´²: c ³b µ c ´ c -: ²³µ ~ À c ² c À³ ~ À . C
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 20
LM-291
5. 0 is the same as ²: c ³b for  :  .
Since ²: c ³b pays : c for : € , we see that À²: c ³b pays À: c for
: € . Thus, ²: c ³b c À²: c ³b pays : c for  :  , and pays
: c c ²À: c ³ ~ À: for : € . To limit the reinsurance coverage to a maximum
of ~ ²À³²³ , note that À²: c ³b pays À: c for : € .
Thus, ²: c ³b c À²: c ³b c À²: c ³b pays À: c ²À: c ³ ~ for : € .
Answer: D
6. We wish to find ,´²: c ³b µ ~ ,´:µ c ,´: w µ ~ ,´:µ c ´ ²³ b ² c - ²³³µ .
Since the claim amounts are integers, : is also an integer.
7 ²? ~ ³ ~ 7 ²? ~ ³ ~ 7 ²? ~ ³ ~ . ,´:µ ~ ,´5 µ h ,´?µ ~ ²³²³ ~ .
: ²³ ~ 7 ²: ~ ³ ~ 7 ²5 ~ ³ ~ c ~ À .
: ²³ ~ 7 ²: ~ ³ ~ 7 ²5 ~ ³ h 7 ²? ~ ³ ~ ²c h ³² ³ ~ À .
-: ²³ ~ : ²³ b : ²³ ~ À
S ,´²: c ³b µ ~ c ´À b ² c À c À³µ ~ À
.
Answer: B
7. ,´²: c ³b µ ~ ,´:µ c ,´: w µ ~ ,´:µ c ²³ c ²³ c ´ c - ²³µ
~ c ² ³ c ² ³ c ´ c c c µ ~ À
.
Answer: C
8. Expected dividend ~ Revenue c Insurance Payments c Expected retained costs .
Revenue ~ 3 .
Expected major repair cost per factory ~ ²³²À³ b ²³²À³ b ²³²À³ ~ .
Expected retained major repair cost per factory ~ À b À b À ~ À
.
Expected reinsurance payment per factory ~ c À
~ À .
Insurance premium per factory ~ ²À³²À³ ~ À .
Insurance payments for both factories ~ ²À³ ~ À .
Revenue c Insurance Payments c Other Costs ~ c À c À ~ À
Dividend ~ Revenue c Insurance Payments c Retained Costs (if ‚ ).
Fac. 1 Claim Fac. 2 Claim Retained Claim
Dividend
Prob.
0
0
0
1.67
.16
0
‚1
1
.67
.24
‚1
0
1
.67
.24
‚1
‚1
2
0
.36
Expected dividend ~ ²À
³²À
³ b ²À
³²À b À³ ~ À . Answer: E
9. We use the formulation ,´²: c ³b µ ~ ,´:µ c ,´: w µ ,
7 ²: ~ ³
where : w ~ F
.
7 ²: € ³
,´:µ ~ ,´? µ b ,´? µ b ,´? µ ~ h ,´?µ ~ h ´²³²À³ b ²³²À³ b ²³²À³µ ~ À
7 ´: ~ µ ~ 7 ´²? ~ ³ q ²? ~ ³ q ²? ~ ³µ ,
and by independence of the ? 's this becomes
7 ´? ~ µ h 7 ´? ~ µ h 7 ´? ~ µ ~ ²À³ ~ À
. Then,
,´²: c ³b µ ~ ,´:µ c ,´: w µ ~ ,´:µ c 7 ²: € ³ ~ c ´ c À
µ ~ À
. Answer: C
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-292
MODELING - PROBLEM SET 20
10. The amount paid is a stop-loss insurance with a deductible of 3 applied to aggregate losses.
The aggregate loss has a compound Poisson distribution with an integer-valued severity.
Therefore, : is integer-valued. ,´:µ ~ ,´5 µ h ,´?µ ~ ´²³²À
³ b ²³²À³µ ~ À .
,´²:c³b µ ~ ,´:µ c ,´: w µ .
: is integer valued with probability values 7 ²: ~ ³ ~ 7 ²5 ~ ³ ~ c ~ À Á
c
7 ²: ~ ³ ~ 7 ²5 ~ ³ h 7 ²? ~ ³ ~ [h h ²À
³ ~ À
Á
7 ²: ~ ³ ~ 7 ²5 ~ ³ h 7 ²? ~ ³ b 7 ²5 ~ ³ h ²7 ²? ~ ³³ ~ À ,
and 7 ²: ‚ ³ ~ c 7 ²: ~ Á Á ³ ~ À
.
prob. .1353
prob. .1624
Then, : w ~
prob. .2057
prob. .4966
so that ,´: w µ ~ ²³²À
³ b ²³²À³ b ²³²À
³ ~ À
.
Answer: A
,´²: c ³b µ ~ À c À
~ À
.
H
11. This is a stop-loss problem where : is the aggregate number of overtime hours worked in the
week and the deductible is 15. : has a compound distribution with frequency 5 that is geometric
with mean 2 and severity ? that is 5 (prob. .2), 10 (prob. .3) or 20 (prob. .5). We wish to find
,´²: c ³b µ ~ ,´:µ c ,´: w µ . The mean of : is
,´:µ ~ ,´5 µ h ,´?µ ~ ´²À³ b ²À³ b ²À³µ ~ .
Note that : must be a multiple of 5, with
7 ²: ~ ³ ~ 7 ´5 ~ µ ~ b
~ À (the only way that : ~ is if 5 ~ ),
7 ´: ~ µ ~ 7 ´5 ~ µ h 7 ´? ~ µ ~ ² ³²À³ ~ À , and
7 ´: ~ µ ~ ²7 ´5 ~ µ h 7 ´? ~ µ b 7 ´5 ~ µ h ²7 ´? ~ µ³ ³
~ ² ³²À³ b ² ³²À³ ~ À
.
Then, 7 ²: ‚ ³ ~ c 7 ²: ~ Á Á ³ ~ À .
: ~ , prob. À
: ~ , prob. À
,
: w ~
: ~ , prob. À
: ‚ , prob. .5497
so ,´: w µ ~ ²À³ b ²À
³ b ²À³ ~ À .
Then, ,´²: c ³b µ ~ c À ~ À .
Answer: B
H
12. The minimum claim amount is 5 if a claim occurs. The minimum value of : is 0, which
occurs if there are no claims,. The next possible value of : is À
The stop-loss insurance with deductible 5 pays ²: c ³b ~ : c ²: w ³ ,
:~
where : w ~ F
.
:‚
,´:µ ~ ,´5 µ h ,´?µ ~ ²³´²³²À
³ b ²À³µ ~ b .
,´: w µ ~ ´ c 7 ²: ~ ³µ ~ ´ c 7 ²5 ~ ³µ ~ ´ c c µ ~ À
.
We are given that À ~ ,´²: c ³b µ ~ ,´:µ c ,´: w µ ~ b c À
.
Solving for results in ~ . Answer: D
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 21 - RISK MEASURES
LM-293
MODELING SECTION 21 - RISK MEASURES
The material in this section relates to Section 3.5 of the text. The suggested time frame for this
section is 1 hour.
A "risk measure" is a variation on a premium calculation for a loss. In the notation of the text, the
risk measure for loss random variable 3 is ²3³. As the name suggests, a risk measure is a way
of assessing the risk of a potential loss random variable. The risk measure for loss 3 could be
²3³ ~ ,´3µ or ² b ³,´3µ (expected value principle), which is the average size of the loss.
Another measure of risk could involve the variability of 3, which is usually related to the
variance of 3; for instance ²3³ ~ ,´3µ b j= ´3µ (standard deviation principle) . Other risk
measures look at "worst-case" scenarios. The Value-at-Risk (VaR) and Tail-Value-at-Risk
(TVaR, also called Conditional Tail Expectation CTE) measures are of this type.
LM-21.1 Value at Risk
For a continuous loss random variable 3, if   , the Value-at-Risk at the 100% level
is denoted VaR ²3³. VaR , also denoted is defined to be the % percentile (or
quantile) of the distribution of 3;
7 ²3  VaR ³ ~ -3 ²VaR ³ ~ -3 ² ³ ~ .
(21.1)
For a continuous random variable, the value of would usually be unique.
Example LM21-1: Find À for each of the following loss random variables 3:
(i) exponential with mean ,
(ii) Pareto with parameters ~ and ~ .
Solution:
(i) -3 ²!³ ~ c c!° ~ À S ! ~ c ²À³ ~ À ~ À .
(ii) -3 ²!³ ~ ² !b ³ ~ c ² !b
³ ~ À S ! ~ j c ~ À ~ À .
À
LM121.2 Tail-Value-at-Risk
The Value at Risk measure gives percentiles of the loss distribution, but does not indicate how
large the losses could be above a particular percentile. The Tail-Value-at-Risk measure does
consider losses above a percentile.
For a continuous loss random variable 3, if   , the Tail-Value-at-Risk (TVaR) at
confidence level is
TVaR ~ ,´3O3 € µÁ
(21.2)
where is the % Value at Risk. This is also the expected cost per payment with a
franchise deductible of .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-294
MODELING SECTION 21 - RISK MEASURES
TVaR is related to the cost per payment formulations that were considered earlier in this
Modeling Unit of the study guide. We review a few of the relationships developed there and look
at the connection to TVaR.
We recall that for a continuous loss random variable 3 with an ordinary deductible , the
expected cost per loss (ECL) is
B
B
,´²3 c ³b µ ~ ²! c ³ 3 ²!³ ! ~ ´ c -3 ²!³µ ! ~ ,´3µ c ,´3 w µ ,
and the expected cost per payment is
,´²3c³ µ
,´3 c O3 € µ ~ c- ²³b ~
3
,´3µc,´3wµ
c-3 ²³
.
For a franchise deductible , the expected cost per loss is
B ! 3 ²!³ ! ~ B ²! c ³ 3 ²!³ ! b h ´ c -3 ²³µ ~ ,´²3 c ³b µ b h ´ c -3 ²³µ .
The expected cost per payment for a franchise deductible is
,*3
,´3O3 € µ ~ c~
3 ²³
~ ,´3 c O3 € µ b ~
,´²3c³b µbh´c-3 ²³µ
c-3 ²³
,´3µc,´3wµ
b.
c-3 ²³
TVaR ~ ,´3O3 € µ is an expected cost per payment for a franchise deductible of which becomes
TVaR ~ ,´3 c O3 € µ b VaR ~
,´3µc,´3w µ
c-3 ² ³
since -3 ² ³ ~ (this is the definition of ).
b ~
,´3µc,´3w µ
c
b For a continuous random variable, the TVaR for loss random variable 3 is equal to the
VaR (which is ) plus the expected cost per payment with an ordinary deductible of ²
,´3µc,´3w µ
³.
c
Example LM21-2: Find TVaRÀ for each of the following loss random variables 3:
(i) exponential with mean ,
(ii) Pareto with parameters ~ and ~ .
Solution:
(i) From Example LM21-1(i), we have À ~ c ²À³ .
For the exponential distribution with mean , we have ,´3 w "µ ~ ² c c"° ³ .
Since c cÀ ° ~ À , it follows that
,´3µc,´3w µ
c²ccÀ ° ³
À
TVaRÀ ~
b À ~
b À ~ b À .
cÀ
cÀ
Note that from the lack of memory property of the exponential distribution (with mean ), we
,´²3c³ µ
have ,´3 c O3 € µ ~ ,´²3 c ³b µ ~ c- ²³b ~
3
,´3µc,´3wµ
c-3 ²³
~.
(ii) From Example LM21-1(ii), we have À ~ À .
À
Then TVaRÀ ~ ,´3 c À O3 € À µ b À ~ b
b À ~ À .
c
(we have used the following property of the Pareto distribution:
,´? c O? € µ ~ b
c ( is the Pareto parameter here).
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING SECTION 21 - RISK MEASURES
LM-295
TVaR for Some Particular Distributions
In Section 13 of the Modeling Unit we saw some simplified representations for expected cost per
payment with ordinary deductible for the uniform, Pareto and exponential distributions.
• Uniform ²Á ³: ,´? c O? € µ ~ c
,
• Pareto , : ,´? c O? € µ ~ b
c ( is the Pareto parameter here),
• Exponential : ,´? c O? € µ ~ .
For these and some other loss distributions we can get convenient representations of the TVaR.
Uniform ²Á ³ ¢ For   , ~ ~ .
TVaR ~ ,´3 c 8 O3 € µ b ~
c
b ~
b
²b³
~ b ~
À
Pareto Á (using instead of in the Exam C table parametrization) ¢
For   , ~ 8 satisfies the equation ~ c ² b ³ , so c ~ ² b ³ .
TVaR ~
b
c
b ~
b
c À
Exponential ¢ For   , ~ satisfies the equation ~ c c ° .
TVaR ~ b ~ ´ c ² c ³µ .
Lognormal , : is the solution of the equation ~ )²
c
³
.
cc From the Exam C Table, ,´? w µ ~ h )²
³ b ´ c )² c
³µ .
cc If ~ , this becomes ,´? w µ ~ b h )²
³ b ´ c )² c ³µ
c
From the definition of is follows that )² ³ ~ , so
cc ,´? w µ ~ b h )²
³ b ´ c µ .
b .
Then, ,´²3 c ³b µ ~ ,´3µ c ,´3 w µ
cc ³ c ´ c µ
cc
~ b h ´ c )²
³µ c ´ c µ , and
,´3µc,´3w µ
cc b TVaR ~
b
~
³µ
c
c h ´ c )²
~ b c b h )²
.
c
Normal , ¢ is the solution of the equation ~ )² ³ .
c
c! °
TVaR ~ b c
(the pdf of the standard normal
h ² ³ , where ²!³ ~ j h distribution). The derivation of the TVaR for the normal distribution is in the textbook.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-296
MODELING SECTION 21 - RISK MEASURES
Example LM21-3: Find TVaRÀ for each of the following loss distributions.
(a) Uniform distribution on ²Á ³ .
(b) Pareto distribution with ~ and ~ .
(c) Exponential distribution with mean 1000.
(d) Lognormal distribution with ~ and ~ .
(e) Normal distribution with mean 1000 and standard deviation 100.
Solution: (a) TVaRÀ ~
²b³
~
²À³²³
~ .
(b) We find À from À ~ c ² b ³ S À ~ À .
Then
TVaRÀ ~
bÀ
c
À
b À ~
b
À
c
b À ~ À
.
(c) TVaRÀ ~ b À ~ ´ c ² c À³µ ~ À
.
(d) À is the solution of the equation À ~ )²
S À ~ À
.
À c
³
S
À c
~ À
b b cc TVaRÀ ~ c h ´ c )²
³µ ~ cÀ
h ´ c )² À
cc
³µ
~ ²Á ³´ c )² c À³µ ~ .
(e) À is the solution of the equation À ~ )²
TVaRÀ ~ b
© ACTEX 2009
cÀ
À c
³
h ²À³ ~ b Á h
S
j
À c
h
~ À .
c²À³ °
~ À .
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 21
LM-297
MODELLING - PROBLEM SET 21
Risk Measures - Section 21
1. Find VaRÀ and TVaRÀ for each of the following loss distributions.
(a) Exponential distribution with a mean of 10,000
(b) Pareto distribution with mean 10,000 and standard deviation of 12,247.
(c) Lognormal with mean 10,000 and standard deviation 5000.
(d) Normal with mean 10,000 and standard deviation 5000.
2. The distribution of ? is a mixture of two continuous random variables. The mixing weight is
for random variable ? and the mixing weight is c for random variable ? , where   Suppose that   , and Var is the -th percentile of the mixed distribution ? .
(a) ,*7?- Á and ,*7?- Á denote the expected cost per payment for random variables ? and
? with franchise deductible , respectively. Show that TVaR for the mixture distribution is the
weighted average of ,*7?- ÁVar and ,*7?- ÁVar using the mixing weights and c .
(b) Show that the following statement regarding the mixed random variable ? false. Use ? and
? as exponential random variables with means 1 and 2 and with mixing weights .5 and .5 to
construct an example to show that the statement is false. The statement is:
"The VaR of ? is the weighted average of the VaR of ? and the VaR of ? using the mixing
weights and c ."
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-298
MODELING - PROBLEM SET 21
MODELLING - PROBLEM SET 21 SOLUTIONS
1.(a) À ~ c cVaRÀ °Á S VaRÀ ~ Á .
TVaRÀ ~ b VaRÀ ~ Á b Á ~ Á .
(b) ,´3µ ~ c
~ Á ,
,´3 µ ~ ² c³²
c³ ~ = ´3µ b ²,´3µ³ ~ Á b ²Á ³ ~ d .
² c³² c³ ,² c ³
~
² c³
c
~ À S ~ and ~ Á .
Á
À ~ c ² VaR bÁ ³
S VaRÀ ~ Á À À
À
ÁbÁÀ
bVaR
TVaRÀ ~ c b VaR ~
b Á À ~ Á .
c
(c) ,´3µ ~ b ~ Á Á ,´3 µ ~ b ~ Á b ~ d .
b ~ Á ~ À Á b ~ ² d ³ ~ À
.
~ À Á ~ À Á ~ À .
VarÀ c
À ~ )²
³ ~ )² VarjÀ cÀ ³ S VarjÀ cÀ ~ À
À
À
S VarÀ ~ À S VarÀ ~ Á .
b Var cc Á
TVarÀ ~ c h ´ c )²
³µ ~ cÀ h ´ c )²
Á
~ cÀ h ´ c )²À
c À³µ ~ Á .
(d) À ~ )²
VarÀ cÁ
TVarÀ ~ b
VarÀ cÁ
³S
Var c
c h ² ³
VarÀ c
c ³µ
~ À
S VarÀ ~ Á .
~ Á b cÀ
h ²À
³ ~ Á .
2.(a) TVaR ~ ,*7 - for a franchise deductible of 8 for the mixture distribution.
,´²?cVaR ³ µ
b
This is
b VaR .
c
B
But, ,´²? c ³b µ ~ ´ c -? ²!³µ !
B
B
~ h ´ c -? ²!³µ ! b ² c ³ h ´ c -? ²!³µ !
~ h ,´²? c ³b µ b ² c ³ h ,´²? c ³b µ
Therefore
,´²?cVaR ³ µ
,´²? cVaR ³ µ
b
b
TVaR?Á8
~
~h
b ² c ³ h
c
c
~ h ,*7? Á8 b ² c ³ h ,*7? Á8
,´²? cVaR ³b µ
c
,´²?cVaR ³ µ
b
TVaR ~
b VaR
c
~ h ,*7?- ÁVaR b ² c ³ h ,*7?- ÁVaR b h VaR b ² c ³ h VaR
~ h ´,*7?- ÁVaR b VaR µ b ² c ³ h ´,*7?- ÁVaR b VaR µ
This is the weighted average of the expected cost per loss for franchise deductible VaR for ?
and ? .
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
MODELING - PROBLEM SET 21
LM-299
(b) We will use the 50-th percentile, but another would work just as well.
The 50-th percentile of ? is , where c ~ À, so that ~ À
.
The 50-th percentile of ? is , where c ° ~ À, so that ~ À
.
With mixing weights of .5 each, we get À²À
³ b À²À
³ ~ À .
The cdf of the mixture distribution is -? ²%³ ~ À² c c% ³ b À² c c%° ³ .
-? ²À³ ~ À² c cÀ ³ b À² c cÀ° ³ ~ À
.
Therefore, À is not the 50th percentile of the mixed distribution.
© ACTEX 2009
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models
LM-300
© ACTEX 2009
MODELING - PROBLEM SET 21
SOA Exam C/CAS Exam 4 - Construction and Evaluation of Actuarial Models