Deeper Understanding, Faster Calculation --Exam C Insights & Shortcuts
by Yufeng Guo
The Missing Manual
This electronic book is intended for individual buyer use for the sole purpose of preparing for Exam C. This book can NOT be resold to others or shared with others. No part of this publication may be reproduced for resale or multiple copy distribution without the express written permission of the author.
© Yufeng Guo
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Table of Contents Introduction 4 Chapter 1 Doing calculations 100% correct 100% of the time.. 5
6 strategies for improving calculation accuracy ............................................................. 5 6 powerful calculator shortcuts....................................................................................... 6
#1 Solve 2 0ax bx c+ + = . .................................................................................... 6 #2 Keep track of your calculation...................................................................... 10 #3 Calculate mean and variance of a discrete random variable......................... 21 #4 Calculate the sample variance....................................................................... 29 #5 Find the conditional mean and conditional variance .................................... 30 #6 Do the least squares regression ..................................................................... 36 #7 Do linear interpolation .................................................................................. 46
Chapter 2 Maximum likelihood estimator ......................................... 52 Basic idea ...................................................................................................................... 52 General procedure to calculate the maximum likelihood estimator ............................. 53 Fisher Information ........................................................................................................ 58 The Cramer-Rao theorem ............................................................................................. 62 Delta method................................................................................................................. 66
Chapter 3 Kernel smoothing................................................................ 75 Essence of kernel smoothing ........................................................................................ 75 Uniform kernel.............................................................................................................. 77 Triangular kernel........................................................................................................... 82 Gamma kernel ............................................................................................................... 90
Chapter 4 Bootstrap.............................................................................. 95 Essence of bootstrapping .............................................................................................. 95 Recommended supplemental reading ........................................................................... 96
Chapter 5 Bühlmann credibility model ............................................ 102 Trouble with black-box formulas................................................................................ 102 Rating challenges facing insurers ............................................................................... 102 3 preliminary concepts for deriving the Bühlmann premium formula ....................... 106
Preliminary concept #1 Double expectation ....................................................... 106 Preliminary concept #2 Total variance formula.................................................. 108 Preliminary concept #3 Linear least squares regression ..................................... 111
Derivation of Bühlmann’s Credibility Formula.......................................................... 112 Summary of how to derive the Bühlmann credibility premium formulas .................. 117 Special case................................................................................................................. 122 How to tackle Bühlmann credibility problems ........................................................... 123 An example illustrating how to calculate the Bühlmann credibility premium ........... 123 Shortcut ....................................................................................................................... 126 Practice problems........................................................................................................ 126
Chapter 6 Bühlmann-Straub credibility model ............................... 148 Context of the Bühlmann-Straub credibility model.................................................... 148 Assumptions of the Bühlmann-Straub credibility model............................................ 149 Summary of the Bühlmann-Straub credibility model................................................. 154
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General Bühlmann-Straub credibility model (more realistic) .................................... 155 How to tackle the Bühlmann-Straub premium problem............................................. 158
Chapter 7 Empirical Bayes estimate for the Bühlmann model...... 168 Empirical Bayes estimate for the Bühlmann model ................................................... 168
Summary of the estimation process for the empirical Bayes estimate for the Bühlmann model..................................................................................................... 170
Empirical Bayes estimate for the Bühlmann-Straub model........................................ 173 Semi-parametric Bayes estimate................................................................................. 182
Chapter 8 Limited fluctuation credibility ........................................ 187 General credibility model for the aggregate loss of r insureds ................................. 188 Key interim formula: credibility for the aggregate loss............................................. 190 Final formula you need to memorize .......................................................................... 191 Special case................................................................................................................. 192
Chapter 9 Bayesian estimate ......................................................... 202 Intuitive review of Bayes’ Theorem ........................................................................... 202 How to calculate the discrete posterior probability .................................................... 206 Framework for calculating the discrete posterior probability..................................... 208 How to calculate the continuous posterior probability ............................................... 213 Framework for calculating discrete-prior Bayesian premiums................................... 219 Calculate Bayesian premiums when the prior probability is continuous.................... 251 Poisson-gamma model ................................................................................................ 260 Binomial-beta model................................................................................................... 264
Chapter 10 Claim payment per payment ........................................... 268 Chapter 11 LER (loss elimination ratio)............................................. 274 Chapter 12 Find E(Y-M)+.................................................................... 276 About the author .................................................................................... 284
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Introduction This manual is intended to be a missing manual. It skips what other manuals explain well. It focuses on what other manuals don’t explain or don’t explain well. This way, you get your money’s worth. Chapter 1 teaches you how to do manual calculation quickly and accurately. If you studied hard but failed Exam C repeatedly, chances are that you are “concept strong, calculation weak.” The calculator techniques will improve our calculation accuracy. Chapter 2 focuses on the variance of a maximum likelihood estimator (MLE), a difficult topic for many. Chapter 3 explains the essence of kernel smoothing and teaches you how to derive complex kernel smoothing formulas for ( )yk x and ( )yK x . You shouldn’t have any trouble memorizing complex kernel smoothing formulas after this chapter. Many candidates don’t know the essence of bootstrap. Chapter 4 is about bootstrap. Chapter 5 explains the core theory behind the Bühlmann credibility model. Chapter 6 compares and contrasts the Bühlmann-Straub credibility models with the Bühlmann credibility model. Many candidates are afraid of empirical Bayes’ estimate problems. The formulas are just too hard to remember. Chapter 7 will relieve your pain. Many candidates find that there are just too many limited fluctuation credibility formulas to memorize. To address this, Chapter 8 gives you a unified formula. Chapter 9 presents a framework for quickly calculating the posterior probability (discrete or continuous) and the posterior mean (discrete or continuous). Many candidates can recite Bayes’ theorem but can’t solve related problem in the exam condition. Their calculation is long, tedious, and prone to errors. This chapter will drastically improve your calculation efficiency. Chapter 10 is about claim payment per payment. Chapter 11 is about loss elimination ratio. Chapter 12 is about how to quickly calculate ( )E Y M
+− .
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Chapter 5 Bühlmann credibility model
Trouble with black-box formulas The Bühlmann credibility premium formula is tested over and over in Course 4 and Exam C. However, many candidates don’t have a good understanding of the inner workings of the Bühlmann credibility premium model. They just memorize a series of black-box formulas:
nZn k
=+
,( )
( )E Var X
kVarθ
θ
θ
µ θ
=
, and ( )1P Z Z Xµ≈ − +
Rote memorization of a formula without fully grasping the concepts is tedious, difficult, and prone to errors. Additionally, a memorized formula will not yield the needed understanding to grapple with difficult problems.
In this chapter, we’re going to dig deep into Bühlmann’s credibility premium formula and gain a crystal clear understanding of the concepts.
Rating challenges facing insurers Let’s start with a simple example to illustrate one major challenge an insurance company faces when determining premium rates. Imagine you are the founder and the actuary of an auto insurance company. Your company’s specialty is to provide auto insurance for taxi drivers. Before you open your business, there are half of dozen insurance companies in your area that offer auto insurance to taxi drivers. The world has been going on fine for many years without your start up. It can continue going on without your start up. So it’s tough for you to get customers. Finally, you take out a big portion of your saving account and buy TV advertising, which brings in your first three customers: Adam, Bob, and Colleen. Since your corporate office is your garage and you have only one employee (you), you decide that three customers is good enough for you to start your business. When you open your business at 0t = , you sell three auto insurance policies to Adam, Bob, and Colleen. The contract of your insurance policy says that the premium rate is guaranteed for only two years. Once the two-year guarantee period is over, you have the right to set the renewal premium, which can be higher than the guaranteed initial premium. When you set your premium rate at 0t = , you notice that Adam, Bob, and Colleen are similar in many ways. They are all taxicab drivers. They work at the same taxi company in the same city. They are all 35 years old. They all graduated from the same high school.
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They are all careful drivers. Therefore, at 0t = you treat Adam, Bob, and Colleen as identical risks and charge the same premium for the first two years. To actually set the initial premium for the first two years, you decide to buy a rate book from a consulting firm. This consulting firm is well-known in the industry. Each year it publishes a rate manual that lists the average claim cost of a taxi driver by city, by mileage and by several other criteria. Based on this rate manual, you estimate that Adam, Bob, and Colleen may each incur $4 claim cost per year. So at 0t = , you charge Adam, Bob, and Colleen $4 each. This premium rate is guaranteed for two years. During the 2-year guaranteed period, Adam, Bob, and Colleen have incurred the following claims:
Year 1 Claim
Year 2 Claim
Total Claim Average claim per insured per year
Adam $0 $0 $0 $0 / 2 = $0 Bob $1 $7 $8 $8 / 2 = $4 Colleen $4 $9 $13 $13 / 2 =$6.5 Grand Total $21
Average claim per person per year (for the 3-person group): $21 / (3 × 2) = $3.5
Now the two-year guarantee period is over. You need to determine the renewal premium rate for Adam, Bob, and Colleen respectively for the third year. Once you have determined the premium rates, you will need to file these rates with the insurance department of the state where you do business (called domicile state). Question: How do you determine the renewal premium rate for the third year for Adam, Bob, and Colleen respectively? One simple approach is to charge Adam, Bob, and Colleen a uniform rate (i.e. the group premium rate). After all, Adam, Bob, and Colleen are similar risks; they form a homogeneous group. As such, they should pay a uniform group premium rate, even though their actual claim patterns for the past two years are different. You can continue charging them the old rate of $4 per insured per year. However, since the average claim cost for the past two years is $3.50 per insured per year, you can charge them $3.50 per person for year three. Under the uniform group rate of $3.50, Bob and Colleen will probably underpay their premiums; their actual average annual claim for the past two years exceeds this group premium rate. Adam, on the other hand, may overpay his premiums; his average annual claim for the past two years is below the group premium rate. When you charge each policyholder the uniform group premium rate, low-risk policyholders will overpay their premiums and the high-risk policyholders will underpay their premiums. Your business as whole, however, will collect just enough premiums to pay the claim costs.
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However, in the real world, most likely you won’t be able to charge Adam, Bob, and Colleen a uniform rate of $3.50. Any of your customers can easily shop around, compare premium rates, and buy an insurance policy elsewhere with a better rate. For example, Adam can easily find another insurer who sells a similar insurance policy for less than your $3.50 group rate. Additionally, the commissioner of your state insurance department is unlikely to approve your uniform rate. The department will want to see that your low risk customers pay lower premiums. Key points to remember: Under the classical theory of insurance, people with similar risks form a homogeneous group to share the risk. Members of a homogeneous group are photocopies of each other. The claim random variable for each member is independent identically distributed with a common density function ( )Xf x . The uniform pure premium rate is ( )E X . Each member
of the homogeneous group should pay ( )E X .
In reality, however, there’s no such thing as a homogeneous group. No two policyholders, however similar, have exactly the same risks. If you as an insurer charge everybody a uniform group rate, then low-risk policyholders will leave and buy insurance elsewhere.
To stay in business, you have no choice but to charge individualized premium rates that are proportional to policyholders’ risks.
Now let’s come back to our simple case. We know that uniform rating won’t work in the real world. We’ll want to set up a mathematical model to calculate the fair renewal premium rate for Adam, Bob, and Colleen respectively. Our model should reflect the following observations and intuition:
• Adam, Bob, and Colleen are largely similar risks. We’ll need to treat them as a rating group. This way, our renewal rates for Adam, Bob, and Colleen are somewhat related.
• On the other hand, we need to differentiate between Adam, Bob, and Colleen. We
might want to treat Adam, Bob, and Colleen as potentially different sub-risks within a largely similar rate group. This way, our model will produce different renewal rates. We hope the renewal rate calculated from our model will agree with our intuition that Adam deserves the lowest renewal rate, Bob a higher rate, and Colleen the highest rate.
• To reflect the idea that Adam, Bob, and Colleen are different sub-risks within a
largely similar rate group, we may want to divide the largely similar rate group into four sub-risks (or more sub-risks if you like): super preferred, preferred, standard, and sub-standard. So the rate group actually consists of four sub-risks. Adam or Bob or Colleen can be any one of the four sub-risks.
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• Here comes a critical point: we don’t know who belongs to which sub-risk. We don’t know whether Adam is a super-preferred sub-risk, or a preferred sub-risk, a standard sub-risk, or a sub-standard sub-risk. Nor do we know to which sub-risk Bob or Colleen belongs. This is so even if we have Adam’s two-year claim data. Judged from his 2-year claim history, Adam seems to be a super preferred or at least a preferred sub-risk. However, a bad driver can have no accidents for a while due to good luck; a good driver can have several big accidents in a row due to bad luck. So we really can’t say for sure that Adam is indeed a better risk. All we know that Adam’s sub-risk class is a random variable consisting of 4 possible values: super preferred, preferred, standard; and substandard.
To visualize that Adam’s sub-risk class is a random variable, think about rolling a 4-sided die. One side of the die is marked with the letters “SP” (super preferred); another side is marked with “PF” (preferred); the third side is marked with “STD” (standard); and the fourth side is marked with “SUB” (substandard). To determine Adam belongs to which sub-class, we’ll roll the die. If the result is “SP,” then we’ll assign Adam to the super preferred class. If the result is “PF,” we’ll assign him to the preferred class. And so on and so forth. Similarly, we can roll the die and randomly assign Bob or Colleen to one of the four sub-classes: SP, PF, STD, and SUB.
Now we are ready to come up with a model to calculate the renewal premium rate: Let random variable j tX represent the claim cost incurred in year t by the j -th insured, where 1, 2,...,t n= , and 1n + and j =1,2,…, and m . Here in our example, 2n = (we have two years of claim data) and m =1,2,3 (corresponding to Adam, Bob, and Colleen).
For any j =1,2,…, and m , 1jX , 2jX ,…, j nX , and 1j nX + are identically distributed with a
common density function ( ), ,Xf x θΘ , a common mean ( )j tE Xµ = , and a common
variance ( )2j tVar Xσ = . What we are saying here is that all policyholders j =1,2,…,
and m have identical mean claim µ and identical claim variance 2σ .
θ is a realization of Θ . Θ is a random variable (or a vector of random variables) representing the presence of multiple sub-risks. 1jX , 2jX ,…, j nX , and 1j nX + , which represent the claim costs incurred by the same policyholder, belong to the same sub risk class θ .
However, θ is unknown to us. All we know is that θ is a random realization of Θ . Here in our example, { }SP, PF, STD, SUBΘ = . When we say that θ is a realization of Θ , we mean that with probability 1p , SPθ = ; with probability 2p , PFθ = ; with probability
3p , STDθ = ; with probability ( )4 1 2 31p p p p= − + + , SUBθ = .
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Because 1jX , 2jX ,…, j nX , and 1j nX + are claims generated from the same (unknown) sub-risk class, we assume that given θ , 1jX , 2jX ,…, j nX , and 1j nX + are independent
identically distributed. That is, 1jX θ , 2jX θ , …, j nX θ , 1j nX θ+ are independent
identically distributed with a common conditional mean ( ) ( )j tE X θ µ θ= and a
common conditional variance ( )j tVar X θ .
We have observed 1jX , 2jX ,…, j nX . Our goal is to estimate 1j nX + , the claim cost in
year 1n + by the j -th insured, using his prior n -year average claim cost1
1 n
j j tt
X Xn =
= ∑ .
The estimated value of 1j nX + is the pure renewal premium for year 1n + . Bühlmann’s
approach is to use ja Z X+ to approximate 1j nX + subject to the condition that
( ) 2
1j j nE a Z X X ++ − is minimized.
The final result:
( )1j ja Z X Z Z Xµ+ = − + ,
nZn k
=+
,( )( )
( )( )
j t j t
j t
E Var X E Var Xk
VarVar E X
θ θ
θθ
θ θ
µ θθ
= =
( ) ( ) ( )j t j tE X E E X Eθ θµ θ µ θ = = = .
Next, we’ll derive the above formulas. However, before we derive the Bühlmann premium formulas, let’s go over some preliminary concepts.
3 preliminary concepts for deriving the Bühlmann premium formula
Preliminary concept #1 Double expectation
( ) ( )E X E E Xθ θ =
If X is discrete, ( ) ( ) ( ) ( )all
E X E E X p E Xθθ
θ θ θ = = ∑ .
If X is continuous, ( ) ( ) ( ) ( )E X E E X E X f dθ θ θ θ θ+∞
Θ−∞
= = ∫
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I’ll explain the double expectation theorem assuming X is discrete. However, the same logic applies when X is continuous. Let’s use a simple example to understand the meaning behind the above formula. A class has 6 boys and 4 girls. These 10 students take a final. The average score of the 6 boys is 80; the average score of the 4 girls is 85. What’s the average score of the whole class? This is an elementary level math problem. The average score of the whole class is:
( ) ( )6 80 4 85Total score 820Average score 82# of students 10 10
+= = = =
Now let’s rearrange the above equation:
( ) ( )6 4Average score 80 8510 10
= +
If we express the above calculation using the double expectation theorem, then we have:
( ) ( ) ( ) ( )GenderE Score E E Score Gender P Gender E Score Gender = = ∑
( ) ( ) ( ) ( )P boy E score boy P girl E score girl= +
( ) ( )6 480 85 8210 10
= + =
So instead of directly calculating the average score for the whole class, we first break down the whole class into two groups based on gender. We then calculate the average score of these two groups: boys and girls. Next, we calculate the weighted average of these two group averages. This weighted average is the average of the whole class. If you understand this formula, you have understood the essence of the double expectation theorem. The Double Expectation Theorem in plain English: Instead of directly calculating the mean of the whole population, you first break down the population into several groups based on one standard (such as gender). You calculate the mean of each group. Next, you calculate the mean of all the group means. This is the mean of the whole population.
Problem A group of 20 graduate students (12 with non-math major and 8 with math major) have a total GRE score of 12,940. The GRE score distribution by major is as follows:
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Total GRE scores of 12 non-math major 7,740 Total GRE scores of 8 math major 5,200 Total GRE score 12,940 Find the average GRE score twice. First time, do not use the double expectation theorem. The second time, use the double expectation theorem. Show that you get the same result. Solution (1) Find the mean without using the double expectation theorem. The average GRE score for 20 graduate students is:
Total score 12,940Average score 647# of students 20
= = =
(2) Find the mean using the double expectation theorem.
( ) ( ) ( ) ( )MajorE GRE E E GRE Major P Major E GRE Major = = ∑
( ) ( ) ( ) ( )P non math E GRE non math P math E GRE math= − − +
12 7,740 8 5, 200 64720 12 20 8
= + =
You can see the two methods produce an identical result.
Preliminary concept #2 Total variance formula
( ) ( ) ( )Y YVar X E Var X Y Var E X Y = +
Proof.
( ) ( ) ( )2 2Var X E X E X= −
Put the double expectation theorem into use:
( ) ( )YE X E E X Y = , ( ) ( )2 2YE X E E X Y =
However, ( ) ( ) ( )2 2E X Y Var X Y E X Y= + .
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⇒ ( ) ( ) ( ) ( ) ( ) ( ){ }22 2 2
Y YVar X E X E X E Var X Y E X Y E E X Y = − = + −
( ) ( ) ( )( ){ }22
Y Y YE Var X Y E E X Y E E X Y = + −
( ) ( )Y YE Var X Y Var E X Y = +
If X is the lost amount of a policyholder and Y is the risk class of the policyholder, then ( ) ( ) ( )Y YVar X E Var X Y Var E X Y = + means that the total variance of the loss
consists of two components:
• ( )YE Var X Y , the average variance by risk class
• ( )YVar E X Y , the variance of the average loss by risk class.
( )YE Var X Y is called the expected value of process variance. ( )YVar E X Y is called the variance of hypothetical mean.
( ) ( ) ( )Total variance expected process variance variance of hypothetical mean
Y YVar X E Var X Y Var E X Y = + ����� ��������� �����������
Next, let’s look at a comprehensive example using double expectation and total variance. Example. The number of claims, N , incurred by a policyholder has the following distribution:
( ) ( ) ( )33! 1! 3 !
nnP n p pn n
−= −−
.
P is uniformly distributed over [0, 1]. Find ( )E N and ( )Var N .
Solution If p is constant, N has a binomial distribution with mean and variance:
( ) 3E N p= , ( ) ( )3 1Var N p p= −
However, p is also a random variable. So we cannot directly use the above formula.
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To find ( )E N , we divide N into different groups by p , just as we divided the class into boys and girls. The only difference is that this time we have an infinite number of groups ( p is a continuous random variable). Let’s consider a small group [ ],p p dp+
Each value of p is a separate group. For each group, we will calculate its mean. Then we will find the weighted average mean of all the groups, with weight being the probability of each group’s p value. The result should be ( )E N .
( ) ( ) ( ) ( )11 1
2
00 0
3 332 2P P
p pE N E E N p E N p f p dp p dp p
= =
= = = = = ∫ ∫
Please note that p is uniform over [0, 1]. Consequently, ( ) 1Pf p = .
Alternatively, ( ) ( ) [ ] ( ) 1 33 3 32 2P PE N E E N p E p E P = = = = =
Next, we’ll calculate ( )Var N . One method is to calculate ( )Var N from scratch using
the standard formula ( ) ( ) ( )2 2Var N E N E N= − . We’ll use the double expectation
theorem to calculate ( )2E N and ( )E N .
( ) ( ) ( ) ( )1
2 2 2
0PE N E E N p E N p f p dp = = ∫
( ) ( ) ( ) ( ) ( )22 2 23 3 1 6 3E N p E N p Var N p p p p p p= + = + − = +
( ) ( ) ( ) ( )11 1
2 2 2 3 2
00 0
3 76 3 22 2
E N E N p f p dp p p dp p p = = + = + = ∫ ∫
⇒ ( ) ( ) ( )2
2 2 7 3 52 2 4
Var N E N E N = − = − =
Alternatively, you can use the following formula to calculate the variance:
( ) ( ) ( )p PVar N E Var N p Var E N p = +
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Because N p is binomial with parameter 3 and p , we have:
( ) 3E N p p= , ( ) ( )3 1Var N p p p= −
( ) ( ) ( )23 1 3 3p p pE Var N p E p p E p p = − = −
( ) ( ) ( ) ( )2 23 3 3 3p p p pE p E p E p E p= − = −
( ) ( ) ( )3 9P PVar E N p Var p Var p = =
⇒ ( ) ( ) ( )p PVar N E Var N p Var E N p = + ( ) ( ) ( )23 3 9p pE p E p Var p = − +
Applying the general formula:
If X is uniform over [ , ]a b , then ( )2
a bE X += , ( ) ( )2
12b a
Var X−
=
We have:
( ) 0 1 12 2
E P += = , ( ) ( )21 0 1
12 12Var P
−= =
( ) ( ) ( )2
2 2 1 1 42 12 12
E P E P Var P = + = + =
⇒ ( ) ( ) ( ) ( ) ( ) ( )23 3 9p P p pVar N E Var N p Var E N p E p E p Var p = + = − +
1 4 1 53 3 92 12 12 4
= − + =
Preliminary concept #3 Linear least squares regression In a regression analysis, you try to fit a line (or a function) through a set of points. With least squares regression, you get a better fit by minimizing the distance squared of each point to the fitted line. Let’s say you want to find out how a person’s income level affects how much life insurance he buys. Let X represent income. Let Y represent the amount of life insurance this person buys. You have collected some data pairs of ( ),X Y from a group of consumers. You suspect there’s a linear relationship between X and Y . You want to predict Y using the function a bX+ , where a and b are constant. With least squares regression, you want to minimize the following:
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( )2Q E a bX Y = + −
Next, we’ll derive a and b .
( ) ( )2 2Q E a bX Y E a bX Ya a a
δ δ δδ δ δ
= + − = + −
( ) ( ) ( )2 2E a bX Y a bE X E Y= + − = + −
Setting 0Qa
δδ
= . ⇒ ( ) ( ) 0a bE X E Y+ − = ( Equation I )
( ) ( )2 2Q E a bX Y E a bX Yb b b
δ δ δδ δ δ
= + − = + −
( ) ( ) ( ) ( )22 2E a bX Y X aE X bE X E X Y = + − = + −
Setting 0Qb
δδ
= . ⇒ ( ) ( ) ( )2 0aE X bE X E X Y+ − = (Equation II )
(Equation II) - (Equation I) × ( )E X :
⇒ ( ) ( ) ( ) ( ) ( )2 2b E X E X E X Y E X E Y − = −
However, ( ) ( ) ( )2 2E X E X Var X− = , ( ) ( ) ( ) ( ),E X Y E X E Y Cov X Y− = .
⇒( )( )
,Cov X Yb
Var X= , ( ) ( )a E Y bE X= −
Derivation of Bühlmann’s Credibility Formula Now I’m ready to give you a quick proof of the Bühlmann credibility formula. To simplify notations, I’m going to fix on one particular insured (such as Adam) and change the symbol j tX to tX . Remember, our goal is to estimate 1nX + , the individualized
premium rate for year 1n + , using a Z X+ . Z is the credibility factor assigned to the
mean of past claims ( )1 21 ... nX X X Xn
= + + + . We’ll want to find a and Z that
minimize the following:
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( ) 2
1nE a Z X X ++ −
Please note that 1X , 2X ,…, nX , and 1nX + are claims incurred by the same policyholder (whose risk class θ is unknown to us) during year 1, 2, …, n , and 1n + .
Applying the formula developed in preliminary concept #3, we have:
( )( )
1, nCov X Xz
Var X+
=
( ) ( ) ( )1 1 2 1 1 2 11 1, ... , ... ,n n n n nCov X X Cov X X X X Cov X X X Xn n+ + +
= + + + = + + +
( ) ( ) ( )1 1 2 1 11 , , ... ,n n n nCov X X Cov X X Cov X Xn + + += + + +
One common mistake is to assume that 1X , 2X ,…, nX , 1nX + are independent identically distributed. If indeed 1X , 2X ,…, nX , 1nX + are independent identically distributed, we would have
( ) ( ) ( )1 1 2 1 1, , ... , 0n n n nCov X X Cov X X Cov X X+ + += = = =
⇒( )
( )1,
0nCov X X
ZVar X
+= =
The result 0Z = simply doesn’t make sense. What went wrong is the assumption that 1X , 2X ,…, nX , 1nX + are independent identically distributed. The correct statement is
that 1X , 2X ,…, nX , and 1nX + are identically distributed with a common density function
( ),f x θ , where θ is unknown to us. Or stated differently, 1X , 2X ,…, nX , and 1nX + are independent identically distributed given risk class θ . In other words, if we fix the sub-class variable at θ , then all the claims incurred by the policyholder who belongs to sub-class θ are independent identically distributed. Mathematically, this means that 1X θ , 2X θ ,…, nX θ , and
1nX θ+ are independent identically distributed. Here is an intuitive way to see why iX and jX have non-zero covariance. iX and jXrepresent the claim amount incurred at time i and j by the policyholder whose sub-class
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θ is unknown to us. So iX and jX are controlled by the same risk-class factor θ . If θ is a low risk, then iX and jX both tend to be small. On the other hand, if θ is a high risk, then iX and jX both tend to be big. So iX and jX are correlated and have a non-zero variance. Next, let’s derive the formula:
( ) ( ) ( ) ( ) ( ),i j i j i jCov X X E X X E X E X Varθ µ θ= − = where i j≠ .
Using the double expectation theorem, we have ( ) ( )i j i jE X X E E X Xθ θ = .
Because iX θ and jX θ are independent identically distributed with a common
conditional mean ( )µ θ , we have:
( ) ( ) ( ) ( ) ( ) ( ) 2i j i jE X X E X E Xθ θ θ µ θ µ θ µ θ= = × =
( ) ( ) ( ) ( ) 2i j i jE E X X E E X E X Eθ θ θθ θ θ µ θ = =
⇒ ( ) ( ) ( ){ } ( )22
,i jCov X X E E Varθ θ θµ θ µ θ µ θ= − =
⇒ ( ) ( )1 1 2 11, ... ,n n nCov X X Cov X X X Xn+ += + + +
( ) ( ) ( )1 1 2 1 11 , , ... ,n n n nCov X X Cov X X Cov X Xn + + += + + +
( ){ } ( )1 nVar Varn θ θµ θ µ θ= =
Next, we’ll calculate ( )Var X .
( ) ( ) ( )1 2 1 22
1 1... ...n nVar X Var X X X Var X X Xn n
= + + + = + + +
Once again, we have to be careful here. One temptation is to write:
( ) ( ) ( ) ( )1 2 1 2... ...n nVar X X X Var X Var X Var X+ + + = + + + Wrong! This is wrong because 1X , 2X ,…, nX are not independent. Instead, 1X θ ,
2X θ ,…, nX θ are independent. So we have to include covariance among 1X ,
2X ,…, nX . The correct expression is:
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( )1 2 ... nVar X X X+ + +
( ) ( ) ( )1 2 ... nVar X Var X Var X= + + +
( ) ( ) ( )1 2 1 3 12 , 2 , ... 2 ,n nCov X X Cov X X Cov X X−+ + + +
So we have n variance terms. Though 1X , 2X ,…, nX are not independent, they have a common mean ( )E Xµ = and common variance ( )Var X .
⇒ ( ) ( ) ( ) ( )1 2 ... nVar X Var X Var X nVar X+ + + = .
Next, let’s look at the covariance terms:
( ) ( ) ( )1 2 1 3 12 , 2 , ... 2 ,n nCov X X Cov X X Cov X X−+ + + .
Out of 1X , 2X ,…, nX , if you take out any two items iX and jX where i j≠ , you’ll get
a covariance ( ) ( ),i jCov X X Varθ µ θ= . Since there are ( )2 12n
n nC
−= ways of taking
out two items iX and jX where i j≠ , the sum of the covariance terms becomes:
( ) ( ) ( )1 2 1 3 12 , 2 , ... 2 ,n nCov X X Cov X X Cov X X−+ + +
( ){ } ( ) ( ) ( ) ( )2 12 2 1 12nVar C Var n n n n Varθ θµ θ µ θ µ θ= = × − = −
⇒ ( ) ( )1 22
1 ... nVar X Var X X Xn
= + + +
( ) ( ) ( ){ 1 22
1 ... nVar X Var X Var Xn
= + + +
( ) ( ) ( ) }1 2 1 3 12 , 2 , ... 2 ,n nCov X X Cov X X Cov X X−+ + + +
( ) ( ) ( ){ } ( ) ( ) ( ){ }2
1 11 1nVar X n n Var Var X n Varn nθ θµ θ µ θ= + − = + −
( ) ( ) ( )Var X Var
Varn
θθ
µ θµ θ
− = +
Using the total variance formula, we have:
( ) ( ) ( )Var X E Var X Var E Xθ θθ θ = +
⇒ ( ) ( ) ( )Var X Var E Var Xθ θµ θ θ − =
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⇒ ( ) ( ) ( )1Var X Var E Var Xnθ θµ θ θ = +
Finally, we have:
( )( )
( )( )
( )
( ) ( )1,
1nCov X X Var Var
ZVar X Var X Var E Var X
n
θ
θ θ
µ θ µ θ
µ θ θ
+ = = = +
( )( )
nE Var X
nVarθ
θ
θ
µ θ
= +
Let ( )
( )E Var X
kVarθ
θ
θ
µ θ
=
. Then ( )
( )Var nZ
n kVar Xθ µ θ = =
+
Next, we need to find ( ) ( )1na E X Z E X+= − . Remember, 1X , 2X ,…, nX , though not
independent, have a common mean ( )E X µ= and a common variance ( )Var X .
⇒ ( ) ( ) ( )1 2 1 21 1... ...n nE X E X X X E X X Xn n
= + + + = + + + ( )1 n
nµ µ= =
⇒ ( )1nE X µ+ =
⇒ ( ) ( ) ( )1 1na E X Z E X Z Zµ µ µ+= − = − = −
⇒ ( ) ( )1 1a Z X Z Z X Z X Zµ µ+ = − + = + − , where nzn k
=+
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Summary of how to derive the Bühlmann credibility premium formulas
( )( )
1, nCov X XZ
Var X+
= , ( )1a Z µ= −
( ) ( ) ( )1, ,n i jCov X X Cov X X Var VEθ µ θ+ = = = , where i j≠
( ) ( )1 22
... nVar X X XVar X
n+ + +
=
( ) ( ) ( ) ( )1 2 ... 1 ,n i jVar X X X nVar X n n Cov X X+ + + = + −
( ) ( ) ( )1nVar X n n Varθ µ θ= + −
( ) ( ){ } ( )2n Var X Var n Varθ θµ θ µ θ= − +
( ) ( )2nE Var X n Varθ θθ µ θ = +
⇒ ( ) ( ) ( ) ( )1 22
... 1 1nVar X X XVar X Var E Var X VE EV
n n nθ θµ θ θ+ + +
= = + = +
⇒( )
( )( )
( )( )
( ) ( ){ }1,
1nCov X X Var Var
ZVar X Var X E Var X nVar
n θ
µ θ µ θ
θ µ θ
+ = = = +
( )( )
n nn kE Var X
nVarθ θ
µ θ
= =+ +
,
Or ( )
( )( )
( )1,
1nCov X X Var VE nZ EVVar X Var X VE EV n
n VE
µ θ+ = = = =+ +
⇒ ( )1P a Z X Z Z Xµ= + = − +
Let’s look at the final formula:
( )Renewal global meanpremium
risk-specificsample mean
1P Z X Z µ= + −����� ������� �����
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Here P is the renewal premium rate during year 1n + for a policyholder whose sub-risk is unknown to us. X is the sample mean of the claims incurred by the same policyholder (hence the same sub-risk class) during year 1, 2, …, n . µ is the mean claim cost of all the sub-risks combined. If we apply this formula to set the renewal premium rate for Adam for Year 3, then the formula becomes:
( ) , ,
Renewal global meanpremiumrisk-specificsample mean
1AdamAdam Adam Bob ColleenP Z X Z µ= + −������� ��������� �����������
At first, the above formula may seem counter-intuitive. If we are interested only in Adam’s claim cost in Year 3, why not set Adam’s renewal premium for Year 3 equal to his prior two-year average claim X (so P X≈ )? Why do we need to drag in µ , the global average, which includes the claim costs incurred by Bob and Colleen? Actually, it’s blessing that the renewal premium formula includes µ . X varies widely based on your sample size. However, the state insurance departments generally want the renewal premium to be stable and responsive to the past claim data. If your renewal premium P is set to X , then P will fluctuate wildly depending on the sample size. Then you’ll have a difficult time getting your renewal rates approved by state insurance departments. In addition, you may have 0P X≈ = ; this is the case for Adam. You’ll provide free insurance to the policyholder who has not incurred any claim yet. This certainly doesn’t make any sense. By including the global mean µ , the renewal premium ( )1P Z Z Xµ= − + is stabilized.
At the same time, P is still responsive to X . Since Adam Bob
X X< , the renewal premium formula ( )1P Z Z Xµ= − + will produce <Adam BobP P .
There are other ways to derive the Bühlmann credibility formula. For example, instead of
minimizing ( ) 2
1nE a Z X X ++ − , we can minimize
( )2
E a Z X µ θ + −
Please note that ( ) ( )E Xµ θ θ= is a random variable. In our taxi driver insurance
example, ( )µ θ has four possible values:
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( )E X SP , ( )E X PF , ( )E X STD , and ( )E X SUB
The idea behind ( )2
E a Z X µ θ + − is this. If we know that a policyholder belongs to
sub-risk θ , then we can set our renewal premium for year 1n + equal to his conditional mean claim cost ( ) ( ) ( ) ( ) ( )1 1 2 ...n nE X E X E X E Xµ θ θ θ θ θ+= = = = = . However, we
don’t know θ . As a result, we list all the possible values of ( )µ θ and find the least mean
squared errors estimator of ( )µ θ by minimizing ( )2
E a Z Xθ µ θ + − .
Then using Preliminary Concept #3, we have:
( )( ),Cov X
ZVar X
µ θ =
( ),Cov X µ θ
( ) ( )1 21 ... ,nCov X X Xn
µ θ = + + + ( ) ( )1 2
1 ... ,nCov X X Xn
µ θ= + + +
( ) ( ) ( ){ }1 21 , , ... ,nCov X Cov X Cov Xn
µ θ µ θ µ θ= + + +
For 1, 2,...,i n= , we have:
( ) ( ) ( ) ( ),i i iCov X E X E X Eµ θ µ θ µ θ= −
( ) ( ){ }i iE X E E Xθµ θ µ θ θ =
For a fixed θ , ( )µ θ is a constant. Hence ( ) ( ) ( ) 2i iE X E Xµ θ θ µ θ θ µ θ = =
⇒ ( ) ( ){ } ( ) 2i iE X E E X Eθ θµ θ µ θ θ µ θ = =
( ) ( ) ( ) ( ) ( ){ }2
i iE X E E E X E Eθ θ θµ θ θ µ θ µ θ = =
⇒ ( ) ( ) ( ){ } ( )22
,iCov X E E Varθ θ θµ θ µ θ µ θ µ θ= − =
⇒ ( ) ( ) ( ) ( ), 1 , 2 ,, , ... ,X X X nCov X Cov X Cov X nVarθ θ θ θµ θ µ θ µ θ µ θ+ + + =
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( ),Cov X µ θ
( ) ( ) ( ){ } ( )1 21 , , ... ,nCov X Cov X Cov X Varn θµ θ µ θ µ θ µ θ= + + + =
( )Var X is the same whether ( ) 2
1nE a Z X X ++ − or ( )2
,XE a Z Xθ µ θ + − is to be
minimized:
( ) ( ) ( ){ }1Var X E Var X nVarn θ θθ µ θ = +
One again, we get:
( )( )
( )( ) ( )
( )
1, nCov X X Var n nZn kE Var XVar X Var X
nVar
θ
θ
µ θ
θ
µ θ
+ = = = =+ +
( ) ( ) ( )1 1na E X Z E X Z Zµ µ µ+= − = − = −
( ) ( )1 1a Z X Z Z X Z X Zµ µ+ = − + = + − ,
There’s a third approach to deriving Bühlmann’s credibility formula. Instead of minimizing
( ) 2
1nE a Z X X ++ − or ( )2
E a Z X µ θ + −
we can minimize ( ) 2
1 1 2, ,...,n nE a Z X E X X X X+ + − .
Here 1 1 2, ,...,n nX X X X+ represents the claim cost at year 1n + of the policyholder who
incurred claims 1 2, ,..., nX X X in year 1,2,…, n . The notation 1 1 2, ,...,n nX X X X+
emphasizes that the claim amounts 1 2, ,..., nX X X , 1nX + are from the same sub-class θ .This condition must hold for the Bühlmann credibility formula to be valid. For example, if 1nX + comes from sub class 1θ and 1 2, ,..., nX X X from sub-class 2θ , then the Bühlmann credibility formula will not hold true. However, the requirement that the claim amounts 1 2, ,..., nX X X , 1nX + are from the same sub-class θ shouldn’t bother us at all. At the very beginning when we presented the Bühlmann credibility formula, we already used 1 2, ,..., nX X X , 1nX + to refer to the claims incurred by the same policyholder whose sub-risk is θ . As a result,
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( ) ( )1 1 2, ,...,n nE X X X X E X θ µ θ+ = =
So ( ) ( )2 2
1 1 2, ,...,n nE a Z X E X X X X E a Z X µ θ+ + − = + −
Key Points We can derive the Bühlmann credibility formula by minimizing any of the following three terms:
( ) 2
1nE a Z X X ++ − , ( )2
E a Z X µ θ + − , ( ) 2
1 1 2, ,...,n nE a Z X E X X X X+ + − .
The Bühlmann credibility premium is the least squares linear estimator of any of the following three terms:
1nX + , the claim amount in year 1n + incurred by the policyholder who has
1 2, ,..., nX X X claims in year 1, 2,.., n .
( )µ θ , the mean claim amount of the sub-class θ that has generated 1 2, ,..., nX X X
( )1 1 2, ,...,n nE X X X X+ , the Bayes’ posterior estimate of the mean claim in year 1n +
given we have observed that the same policyholder has 1 2, ,..., nX X X claim costs in years 1, 2,.., n respectively.
Even though we have derived the Bühlmann credibility formula assuming X is the claim cost, the Bühlmann credibility formula works if X is any other quantity such as loss ratio, the aggregate loss amount, or the number of claims. Popularity of the Bühlmann credibility formula The Bühlmann credibility formula is popular due to its simplicity. The renewal premium is the weighted average of the uniform group rate and the sample mean of the past claims. The renewal premium is easy to calculate and easy to explain to clients.
In contrast, Bayesian premiums (the posterior means) are often difficult to calculate, requiring knowledge of prior distributions and involving complex integrations.
Next, let’s derive a special case of the Bühlmann credibility formula. This special case is presented in Loss Models.
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Special case If ( )iE X µ= , ( ) 2
iVar X σ= , and for i j≠ ( ) 2,i jCov X X ρσ= where correlation coefficient ρ satisfies 1 1ρ− < < , determine the Bühlmann credibility premium.
Once again, the credibility premium is a Z X+ .
( )( )
1, nCov X XZ
Var X+
= , ( ) ( ) ( ) 21, ,n i jCov X X Var Cov X Xθ µ θ ρσ+ = = =
( ) ( )1 22
... nVar X X XVar X
n+ + +
=
( ) ( ) ( ) ( ) ( )2 21 2 ... 1 , 1n i jVar X X X nVar X n n Cov X X n n nσ ρσ+ + + = + − = + −
⇒( )
( ) ( ) ( )2
1
2 22
,1 1 11
nCov X X nZnVar X n n n
n
ρσ ρρσ ρσ
+= = =
+ − + −
( ) ( )( )
( )1
1 11 1 1 1
na Zn n
ρ µρµ µρ ρ
−= − = − = + − + −
The Bühlmann credibility premium is
( ) ( )( )
( )( )
1
1 11
1 1 1 1 1 1
n
ii
nZ X Z X Xn n n n
ρ µ ρ µρ ρµρ ρ ρ ρ ρ ρ=
− −+ − = + = +
+ − + − + − + −∑
You don’t need to memorize the Bühlmann credibility premium formula for this special case. If you understand how to derive the general Bühlmann credibility premium formula, you can derive the special case formula any time by setting ( ) 2,i jCov X X ρσ= .
Next, let’s turn our attention toward how to solve the Bühlmann credibility problem on the exam.
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How to tackle Bühlmann credibility problems Step 1 Divide the policyholders into sub-classes 1θ , 2θ , 3θ …
Step 2 For each sub-class θ , calculate the average claim cost (or loss ration, aggregate claim, etc) ( ) ( )E Xµ θ θ= ; calculate the variance of the claim
cost ( )Var X θ .
Step 3 Calculate EV= ( )E Var Xθ θ , the average variance for all sub-classes
combined. Calculate VE= ( )Var E Xθ θ , the variance of the average claim for all sub-classes combined.
Step 4 Calculate EVkVE
= , nZn k
=+
Step 5 Calculate ( )E E Xθµ θ = , the average claim cost for all sub-classes combined. This is the uniform group premium rate you would charge under the classical theory of insurance.
Step 6 Calculate the sample claim of the past data 1
1 n
ii
X Xn =
= ∑ .
Step 7 Calculate the Bühlmann credibility premium ( )1Z X Z µ+ − . This is the weighted average of the sample mean and the uniform group rate.
An example illustrating how to calculate the Bühlmann credibility premium (Nov 2003 #23) You are given: Two risks have the following severity distributions:
Amount of Claim Probability of Claim
Amount for Risk 1 Probability of Claim Amount for Risk 2
250 0.5 0.7 2,500 0.3 0.2 60,000 0.2 0.1
Risk 1 is twice as likely to be observed as Risk 2.
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A claim of 250 is observed. Determine the Bühlmann credibility estimate of the second claim amount from the same risk. Solution This is a typical problem for Exam C. Here policyholders are from two risk classes. Even though the problem doesn’t say that Risk 1 and Risk 2 are two sub-risks of a similar bigger risk group (i.e. homogeneous group), we should assume so. Otherwise, the Bühlmann credibility formula won’t work. Remember the Bühlmann credibility premium is the weighted average of the uniform group rate µ and the risk specific sample mean X . If Risk 1 and Risk 2 are not sub-risks of a homogeneous group, then the uniform group rate µ doesn’t exist; we have no way of calculating ( )1Z X Z µ+ − .
The problem says that a claim of 250 is observed. This means that a policyholder of an unknown sub-class has incurred a claim of 1X =$250. Since Risk 1 is twice as likely as
Risk 2, the $250 claim has 23
chance of coming from Risk 1 and 13
chance of from Risk
2. The question asks us to estimate the next claim amount 2X incurred by the same policyholder. Amount of Claim Probability of Claim
Amount for Risk 1 Probability of Claim Amount for Risk 2
250 0.5 0.7 2,500 0.3 0.2 60,000 0.2 0.1
Let X represent the dollar amount of a claim randomly chosen.
( )risk 1E X = 250(0.5) + 2,500(0.3) + 60,000(0.2) = 12,875
( )risk 2E X = 250(0.7) + 2,500(0.2) + 60,000(0.1) = 6,675 The uniform group rate is:
( ) ( ) ( ) ( ) ( )from risk 1 risk 1 from risk 2 risk 2E X P X E X P X E Xµ = = +
( ) ( )2 112,875 6,675 10,808.333 3
= + =
The variance of the conditional mean is:
( ) ( ) ( ) ( )2 2 2from risk 1 risk 1 from risk 2 risk 2VE P X E X P X E X µ = + −
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( ) ( )2 2 22 112,875 6,675 10,808.33 8,542, 222.223 3
= + − =
( )2 risk 1E X =2502(0.5) + 2,5002(0.3)+ 60,0002(0.2)=721,906,250
( )2 risk 2E X =2502(0.7 )+ 2,5002(0.2) + 60,0002(0.1) =361,293,750
( ) ( ) ( )2 2 2risk 1 risk 1 risk 1 721,906, 250 12,875 556,140,625Var X E X E X= − = − =
( ) ( ) ( )2 2 2risk 2 risk 2 risk 2 361, 293,750 6,675 316,738,125Var X E X E X= − = − =
The average conditional variance is:
( ) ( ) ( ) ( )from risk 1 risk 1 from risk 2 risk 1EV P X Var X P X Var X= +
( ) ( )2 2556,140,625 316,738,125 476,339,791.673 3
= + =
⇒476,339,791.67 55.768,542, 222.22
EVkVE
= = =
1 1.76%1 55.76
nZn k
= = =+ +
,
( ) ( ) ( )1 1.76% 250 1 1.76% 10,808.33 10,622.50P Z X Z µ= + − = + − =
Next, I want to emphasize an important point. In the Bühlmann credibility premium formula, what matters is the
( )1 21 ... nX X X Xn
= + + + , not the individual claims data 1X , 2X ,…, nX .
For example, for 3n = , ( ) ( )1 2 3, , 0,3,6X X X = , ( ) ( )1 2 3, , 1,7,1X X X = , and
( ) ( )1 2 3, , 2,2,5X X X = have the same 3X = and will produce the same renewal premium
( ) ( )1 3 1P Z X Z Z Zµ µ= + − = + − .
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Shortcut We can rewrite the Bühlmann credibility premium formula as
( ) 11
n
ii
k Xn k k nXP Z X Z X
n k n k n k n k
µµµ µ =
++
= + − = + = =+ + + +
∑
We can interpret EVkVE
= as the number of samples taken out of the global mean µ .
Imagine we have two urns, A and B. A contains an infinite number of identical balls with each ball marked with the number µ . B contains an infinite number of identical balls with each ball marked with the number X . You take out k balls from Urn A and n balls from Urn B.
Then the average value per ball is: 1
n
ii
k Xk nXP
n k n k
µµ =
++
= =+ +
∑
This is the renewal premium just for year 1n + .
k n
1
n
ii
k Xk nXP
n k n k
µµ =
++
= =+ +
∑
A B
Practice problems Q1 You are an actuary on group health insurance pricing. You want to use the Bühlmann credibility premium formula ( )1P Z X Z µ= + − to set the renewal premium rate for a policy. One day the vice president of your company stops by. He has a Ph.D. degree in statistics and is widely regarded as an expert on the central limit theorem. He asks you to throw the formula ( )1P Z X Z µ= + − into the trash can and focus on µ .
“All we care about is µ . As long as we charge each policyholder µ , we’ll be okay,” the vice president says. “The fundamental concept of insurance is that many people form a group to share the risk. If we charge µ , the law of large numbers will work its magic and we’ll be able to collect enough premiums to pay our guarantees.”
µ X
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Comment on the vice president’s remarks. Solution According to the law of large numbers, for a homogeneous group of policyholders, we can set the premium rate equal to the average claim cost ( )E Xµ = . Some policyholders
will suffer losses greater than ( )E X , while others will suffer losses less than ( )E X .However, on average, insurance companies will have collected just enough premiums to offset the loss. As long as each policyholder pays µ , then the insurer will be solvent. However, in practice, insurance companies can’t charge X µ≈ . Members of a so-called homogenous risk group are really different risks. Policyholders of different risks can shop around and compare premium rates. If any policyholder believes that his premium is too high, he can terminate his policy and buy cheaper insurance elsewhere. If an insurer charges µ to similar yet different risks, good risks will stop doing business with the insurer and buy cheaper insurance elsewhere; only bad risks will remain in the insurer’s book of business. As more and more good risks leave the insurer’s book of business, the actual expected claim cost will exceed the original average premium rate µ .Then the insurer has to increase µ , causing more policyholders to terminate their policies. Gradually, the insurer’s customer base will shrink and the insurer will go bankrupt.
Q2 Compare and contrast the classical theory of insurance and the credibility theory of insurance. Solution
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The classical theory vs. the credibility theory: Classical theory of
insurance Credibility
Theory Is there a homogeneous group?
Yes. This is the foundation of insurance. Identical risks form a homogeneous group to share risks.
Each member of a seemingly homogeneous group belongs to a sub-class. The insurer doesn’t know who belongs to which sub-class.
Are claim random variables X of different members of a group independent identically distributed?
Yes. Since each member of a homogeneous group has identical risk, each member’s claim random variable is independent identically distributed at all times.
No. Since members of a similar risk group are actually of different sub-risk classes, only claims incurred by the same sub-class are independent identically distributed.
What’s the fair premium rate?
The fair premium is ( )E X µ= , where X is
the random loss variable of any policyholder in the homogeneous group. Every member of a homogeneous group needs to pay µ , the uniform group pure premium rate.
The fair premium is ( ) ( )E X θ µ θΘ = = , which
is the mean claim cost of the sub-class θ . Every member of the same sub-class θneeds to pay ( )µ θ ,
Q3 One day you visited your college statistics professor. He asked what you were doing in your job. You told him that you used the Bühlmann credibility premium formula to set the renewal premium for group health insurance policies. The Bühlmann credibility theory was new to the professor. After listening to your explanation of the formula
( )1P Z X Z µ= + − , he looked puzzled. He told you that for 20 years he had been telling
his students that X is the unbiased estimator of ( )E X . “I don’t get it. Why don’t you
just set P X≈ ?”
Explain why it’s not a good idea to set P X≈ .
Solution Your stats professor is perfectly correct in saying that the sample mean is an unbiased estimator of the population mean. If the number of observations n is large (so we have observed 1X , 2X , …, nX claims), for any policyholder, setting his renewal premium equal to his prior average mean claim is a good idea.
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In reality, however, it’s hard to implement the idea P X≈ . Often you, as an insurer, have to set the renewal premium with limited data (so n may be small). For a small n ,X may not be a good estimate of ( )E X . In addition, we may have a weird situation
where 0X = . In our taxi driver insurance example, if you use P X≈ to set the renewal premium for Adam, you’ll get P = 0. This clearly doesn’t make any sense. Q4 Nov 2005 #26 For each policyholder, losses 1X , 2X , …, nX , conditional on Θ , are independent identically distributed with mean
( ) ( )jE Xµ θ θ= Θ = , 1, 2,...,j n=
and variance
( ) ( )jv Var Xθ θ= Θ = , 1, 2,...,j n=
You are given: • The Bühlmann credibility factor assigned for estimating 5X based on 1X , 2X ,
3X , 4X is 0.4Z =
• The expected value of the process variance is known to be 8. Calculate ( ),i jCov X X , i j≠ .
Solution
( )( )
( )( )
1,1
nCov X X Var VEZVar X Var X VE EV
n
µ θ+ = = =+
We are told that 4n = (we have four years of claim data), 0.4Z = , and 0.8VE = .
⇒ 0.4 8 24
VE VEVEVE
= =++
, 1.33VE = .
So ( ) ( ) ( )1, , 1.33i j nCov X X Cov X X Var VEµ θ+= = = =
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Q5 Nov 2005 #19 For a portfolio of independent risks, the number of claims for each risk in a year follows a Poisson distribution with means given in the following table:
Class Mean # of claims per risk # of risks 1 1 900 2 10 903 20 10
You observe x claims in Year 1 for a randomly selected risk. The Bühlmann credibility estimate of the number of claims for the same risk in Year 2 is 11.983. Determine x .
Solution The problem states that x claims in Year 1 have been observed for a randomly selected risk. The wording “a randomly selected risk” is needed because in order for the Bühlmann credibility formula to work, the risk class must be unknown to us. If we already know the risk class, we can calculate the expected number of claims in Year 2; we don’t need to estimate any more. Please also pay attention to the wording “the Bühlmann credibility estimate of the number of claims for the same risk in Year 2 is …” In order for the Bühlmann credibility formula to work, the renewal premium (or the expected number of claims in this problem) in year 1n + and the prior n year claims 1X , 2X , …, nX must refer to the same (unknown) risk class. And now back to the problem. Let Y represent the number of claims incurred in a year by a randomly chosen class. Since Y θ has is Poisson random variable,
( ) ( )E Y Var Yθ θ= .
Class θΘ =
Mean # of claims per risk ( ) ( )E Y Var Yθ θ=
( )P θΘ = # of risks
1 1 90% 9002 10 9% 903 20 1% 10
Total 100% 1,000
The global mean (using the double expectation theorem):
( ) ( ) ( ) ( ) ( ) ( ) ( )1 90% 10 9% 20 1% 2E Y E E Y P E Yθθ
µ θ θ θ = = = = + + = ∑
The average conditional variance:
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( ) ( ) ( ) ( ) ( )1 90% 10 9% 20 1% 2EV P Var Yθ
θ θ= = + + =∑
The variance of conditional means:
( ) ( ) ( ){ }22VE Var E Y E E Y E E Yθ θ θθ θ θ = = −
( ) ( ) ( )2 2 2 21 90% 10 9% 20 1% 2 9.9= + + − =
Alternatively, ( ) ( ) ( ){ }2VE Var E Y E E Y E E Yθ θ θθ θ θ = = −
( ) ( ) ( ) ( ) ( ) ( )2 2 21 2 90% 10 2 9% 20 2 1% 9.9= − + − + − =
⇒2
9.9EVkVE
= =
( )1
2 29.9 11.98321
9.9
n
ii
Y k xnY kPn k n k
µµ =
+ ++= = = =
+ + +
∑, ⇒ 14x =
Q6 Nov 2005 #7 For a portfolio of policies, you are given:
• The annual claim amount on a policy has probability density function
( ) 2
2xf x θθ
= , 0 x θ< <
• The prior distribution of θ has density function: ( ) 34π θ θ= , 0 1θ< <
• A randomly selected policy had claim amount 0.1 in Year 1. Determine the Bühlmann credibility estimate of the claim amount for the selected risk in Year 2. Solution The conditional mean is:
( ) ( )2
32 2 2 0
0 0 0
2 2 2 1 23 3
x xE X x f x dx x dx dx xθ θ θ
θθ θ θ
θ θ θ = = = = = ∫ ∫ ∫
Please don’t write:
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( ) ( )0
E X x f x dθ
θ θ θ= ∫ Wrong!
( )E X θ is the expected value of X if we fix the random variable θΘ = . So θ should
be treated as a constant and 0dθ = . The correct calculation is to integrate ( )x f x θregarding x , not θ .
Conditional variance: ( ) ( ) ( )2 2Var X E X E Xθ θ θ= −
( ) ( )3
2 2 2 4 22 2 2 0
0 0 0
2 2 2 1 14 2
x xE X x f x dx x dx dx xθ θ θ
θθ θ θ
θ θ θ = = = = = ∫ ∫ ∫
⇒ ( ) ( ) ( )2
2 2 2 21 2 12 3 18
Var X E X E Xθ θ θ θ θ θ = − = − =
The global mean: ( ) ( ) ( )2 23 3
E X E E X E Eθ θµ θ θ θ = = = =
The expected conditional variance: ( ) ( )2 21 118 18
EV E Var X E Eθ θθ θ θ = = =
The variance of the conditional mean: ( ) ( )22 2
3 3VE Var E X Var Varθ θθ θ θ = = =
( ) ( ) ( )1 1 1
3 4
0 0 0
44 45
E d d dθ θ π θ θ θ θ θ θ θ= = = =∫ ∫ ∫
( ) ( ) ( )1 1 1
2 2 2 3 5
0 0 0
44 46
E d d dθ θ π θ θ θ θ θ θ θ= = = =∫ ∫ ∫
( ) ( ) ( )2
2 2 4 4 26 5 75
Var E Eθ θ θ = − = − =
⇒ ( )21 1 418 18 6
EV E θ = =
, ( )2 22 2 2
3 3 75VE Var θ = =
( )2 2 4 83 3 5 15
Eµ θ = = =
, 2
1 418 6 3.1252 23 75
EVkVE
= = =
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The above fraction is complex. We don’t want to bother expressing k in a neat fraction; trying to expressing k in a neat fraction is prone to errors.
⇒ 1 1
83.125 0.115 0.428
1 1 3.125
n
ii
k Xk XP
n k k
µµ=
++ + = = = =+ + +
∑
Alternative method of calculation. This method is more complex.
( ) ( )2
32 2 2 0
0 0 0
2 2 2 1 23 3
x xE X x f x dx x dx dx xθ θ θ
θθ θ θ
θ θ θ = = = = = ∫ ∫ ∫ (as before)
( ) ( ) ( ) ( )11 1
3 5
00 0
2 2 4 843 3 5 15
E X E E X E X d dθµ θ θ π θ θ θ θ θ θ = = = = = = ∫ ∫
( ) ( ) ( ){ }22VE Var E X E E X E E Xθ θ θθ θ θ = = −
( ) ( ) ( ) ( )21 1
2 2
0 0
23
E E X E X d dθθ θ
θ θ π θ θ θ π θ θ= =
= = ∫ ∫
21 13 5
0 0
2 16 16 1 843 9 9 6 27
d dθ θ
θ θ θ θ θ= =
= = = = ∫ ∫
⇒ ( ) ( ){ }2
22 8 8 0.0118527 15
VE VE E X E E Xθ θθ θ = − = − =
( )EV E Var X θ = , ( ) ( ) ( )2 2Var X E X E Xθ θ θ= −
( ) ( )3
2 2 2 4 22 2 2 0
0 0 0
2 2 2 1 14 2
x xE X x f x dx x dx dx xθ θ θ
θθ θ θ
θ θ θ = = = = = ∫ ∫ ∫
⇒ ( ) ( ) ( )2
2 2 2 21 2 12 3 18
Var X E X E Xθ θ θ θ θ θ = − = − =
( ) ( ) ( )1 1
3 2
0 0
1 4 1418 18 6
EV E Var X Var X d dθ π θ θ θ θ θ θ = = = = ∫ ∫
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4 118 6 3.1250.01185
EVkVE
= = =
⇒ 1 1
83.125 0.115 0.428
1 1 3.125
n
ii
k Xk XP
n k k
µµ=
++ + = = = =+ + +
∑
Q7 May 2005, #20 For a particular policy, the conditional probability of the annual number of claims given
θΦ = , and the probability distribution of Φ are as follows:
# of claims 0 1 2 Probability 2θ θ 1 3θ−
Two claims are observed in Year 1. Calculate the Bühlmann credibility estimate of the number of claims in Year 2. Solution Let X represent the annual number of claims.
X θ 0 1 2
Probability 2θ θ 1 3θ−
( ) ( ) ( ) ( )0 2 1 2 1 3 2 5E X θ θ θ θ θ= + + − = −
( ) ( ) ( ) ( )2 2 2 20 2 1 2 1 3 4 11E X θ θ θ θ θ= + + − = −
( ) ( )2 24 11 2 5 9 25Var X θ θ θ θ θ= − − − = −
( ) [ ] ( )2 5 2 5E E X E Eθ θµ θ θ θ = = − = −
( ) ( ) ( ) ( )2 5 5 25VE Var E X V V Varθ θ θθ θ θ θ = = − = =
( ) ( ) ( ) ( )2 29 25 9 25EV E Var X E E Eθ θθ θ θ θ θ = = − = −
( ) ( ) ( )0.05 0.8 0.3 0.2 0.1E θ = + =
( ) ( ) ( )2 2 20.05 0.8 0.3 0.2 0.02E θ = + =
θ 0.05 0.30Probability 0.80 0.20
θ 0.05 0.30Probability 0.80 0.20
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( ) 20.02 0.1 0.01Var θ = − =
( ) ( )2 5 2 5 0.1 1.5Eµ θ= − = − =
( ) ( )25 25 0.01 0.25VE Var θ= = =
( ) ( ) ( ) ( )29 25 9 0.1 25 0.02 0.4EV E Eθ θ= − = − =
0.4 1.60.25
EVkVE
= = = ⇒( )1 1 1.6 1.5 2
1.691 1 1.6
n
ii
k Xk XP
n k k
µµ=
+ ++= = = =
+ + +
∑
Q8 May 2005, #17 You are given:
• The annual number of claims on a given policy has a geometric distribution with parameter β
• The prior distribution of β has the Pareto density function
( )( ) 11 α
απ ββ +=
+, 0 β< < ∞
where α is a known constant greater than 2. A randomly selected policy has x claims in Year 1. Calculate the Bühlmann credibility estimate of the number of claims for the selected policy in Year 2. Solution Let X represent the annual number of claims on a randomly selected policy. Here the risk factor is β . The conditional random variable X β has geometric distribution. If you look up “Tables for Exam C/4,” you’ll find geometric random variable N with parameter β has mean and variance as follows:
( )E N β= , ( ) ( )1Var N β β= +
Applying the above mean and variance formula, we have: The conditional mean: ( )E X β β=
The conditional variance: ( ) ( )1Var X β β β= +
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( ) ( ) ( ) ( )21EV E Var X E E Eβ β β ββ β β β β = = + = + . Typically, we write ( )Eβ β
as ( )E β and ( )2Eβ β as ( )2E β . So
( ) ( ) ( ) ( )21EV E Var X E E Eβ ββ β β β β = = + = +
( ) ( )VE V E X Vβ β β = =
The global mean is: ( ) ( )E E X Eβµ β β = =
We are told that the prior distribution ofβ has the Pareto density function
( )( ) 11 α
απ ββ +=
+, 0 β< < ∞
Here the phrase “prior distribution” refers to the fact that we know ( )π β prior to our
observation of x claims in Year 1. In other words, ( )π β hasn’t incorporated our observation of x claims in Year 1 yet. Please note that the prior distribution, not the posterior distribution, is used in Bühlmann’s credibility estimate. Frankly, I think SOA’s emphasis that ( )π β is prior (as opposed to posterior) distribution is unnecessary and really meant to scare exam candidates. When we talk about density function, we always refer to prior distribution. So there’s never a need to say “prior distribution.” If we want to refer to a distribution that has incorporated our recent observations, at that time we say “posterior distribution.” Back to the problem. … We are told that β has Pareto distribution. Is it a one-parameter Pareto or two-parameter Pareto? Many candidates have trouble knowing which one to use. Here is a simple rule: To decide whether to use one-parameter Pareto or two-parameter Pareto, look at your random variable X . If X is greater than a positive number, then use single-parameter Pareto. If X is greater than zero, then use two-parameter Pareto:
If X > a positive constant θ , then use single-parameter Pareto ( ) 1f xx
α
α
αθ+= ;
If 0X > , then use two-parameter Pareto ( )( ) 1f xx
α
α
αθθ +=
+.
In this problem, the Pareto random variable 0β > . So we should use the two-parameter Pareto formula in “Tables for Exam C/4.”
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( ) ( )( ) ( )!
1 2 ...
kk kE X
kθ
α α α=
− − −.
Please note that the denominator has k items.
⇒ ( )1
E X θα
=−
, ( ) ( )( )2
2 21 2
E X θα α
=− −
( ) ( ) ( ) ( )( ) ( ) ( )
22 22 2
221 2 1 1 2
Var X E X E X θ θ αθα α α α α
= − = − = − − − − −
Since the two-parameter Pareto is frequently tested in Exam C, you might want to memorize the following formulas:
( )1
E X θα
=−
, ( ) ( )( )2
2 21 2
E X θα α
=− −
, ( )( ) ( )
2
21 2Var X αθ
α α=
− −
β is a two-parameter Pareto random variable with pdf ( )( ) 11 α
απ ββ +=
+. So the two
parameters are α and 1θ = . So we have:
( ) 11
E βα
=−
, ( ) ( )( )2 2
1 2E β
α α=
− −, ( )
( ) ( )21 2Var αβ
α α=
− −
⇒ ( ) ( ) ( )( ) ( )( )2 1 2
1 1 2 1 2EV E E αβ β
α α α α α= + = + =
− − − − −
( )( ) ( )21 2
VE V αβα α
= =− −
, ( ) 11
Eµ βα
= =−
( ) ( )
( ) ( )2
1 21
1 2
EVkVE
αα α
ααα α
− −= = = −
− −
⇒( )
( )1 1
11 111 1 1
n
ii
k X xk X xPn k k
µ αµ αα α
=
+ − ++ +−= = = =+ + + −
∑
137/284
Q9 May 2005 #11 You are given:
• The number of claims in a year for a selected risk follows Poisson distribution with mean λ
• The severity of claims for the selected risk follows exponential distribution with mean θ
• The number of claims is independent of the severity of claims. • The prior distribution of λ is exponential with mean 1. • The prior distribution of θ is Poisson with mean 1. • A priori, λ and θ are independent.
Using the Bühlmann credibility for aggregate losses, determine k .
Solution Let N represent the annual number of claims for a randomly selected risk. Let X represent the loss dollar amount per loss incident. Let S represent the aggregate annual claim dollar amount incurred by a risk.
Then 1 21
...N
i Ni
S X X X X=
= = + + +∑ .
N is a Poisson random variable with mean λ . So ( )!
n
Nf n en
λ λλ −= ( 0,1, 2,...N = ).
Here λ is an exponential random variable with pdf ( )f e λλ −= . We have ( ) 1E λ = ,
( ) 21 1Var λ = = , and ( ) ( ) ( )2 2 21 1 2E E Varλ λ λ= + = + =
1X , 2X ,…, NX are independent identically distributed with a common pdf
( ) 1 xXf x e θθ
θ−= . Here θ is a Poisson random variable with pdf ( ) 1 1
!f eθ
θ−= . We
have ( ) ( ) 1E Varθ θ= = . Hence ( ) ( ) ( )2 2 21 1 2E E Varθ θ θ= + = + = .
Here the risk parameters are ( ),λ θ .
( ) ( ) ( )E S E N E X= , ( ) ( ) ( ) ( ) ( )2Var S E N Var X Var N E X= +
To remember that you need to use ( )2E X , not ( )E X , in the ( )Var S formula, please
note that ( )Var S is dollar squared. If you use ( ) ( )Var N E X , you’ll get dollar, not dollar
squared. As a result, you need to use ( ) ( )2Var N E X .
For a fixed pair of ( ),λ θ , the conditional mean is:
138/284
( ) ( ) ( ),E S E N E Xλ θ λ θ λθ = =
The conditional variance is: ( ) ( ) ( ) ( ) ( )2,Var S E N Var X Var N E Xλ θ λ θ λ θ = +
2 2 22λθ λθ λθ= + =
Please note that N λ is a Poisson random variable with mean λ ; X θ is an exponential random variable with mean θ .
( ){ } ( )2, ,, 2EV E Var S Eλ θ λ θλ θ λθ= = .
Since λ and θ are independent, we have: ⇒ ( ) ( ) ( ) ( )( )2 2
, 2 2 2 1 2 4EV E E Eλ θ λθ λ θ= = = =
( ){ } ( ) ( ) ( ) 22, , , ,,VE Var E S Var E Eλ θ λ θ λ θ λ θλ θ λθ λθ λθ = = = −
( ) ( ) ( ) ( ) ( )2 2 2 2 2, , 2 2 4E E E Eλ θ λ θλθ λ θ λ θ = = = =
( ) ( ) ( ) ( ), 1 1 1E E Eλ θ λθ λ θ= = =
⇒ ( ) ( ) 22 2, , 4 1 3VE E Eλ θ λ θλθ λθ = − = − =
⇒43
EVkVE
= =
Q10 May 2005 #6 You are given:
• Claims are conditionally independent and identically Poisson distributed with mean Θ
• The prior distribution of Θ is:
( )2. 611
1F
θ Θ = − +
, 0θ >
Five claims are observed. Determine the Bühlmann credibility factor. Solution
139/284
Let X represent the number of claims. The risk factor is Θ . We are told that X θ is a Poisson random variable with mean θ .
The conditional mean is: ( )E X θ θΘ = =
The conditional variance is: ( )Var X θ θΘ = =
( ) ( )EV E Var X Eθ θ θ = = , ( ) ( )VE Var E X Varθ θ θ = =
⇒( )( )
EEVkVE Var
θθ
= =
To quickly calculate ( )E θ and ( )Var θ , you’ll need to recognize that:
cdf for a two-parameter Pareto random variable is: ( ) 1F xx
αθθ
= − + , where 0x > .
( )1
E X θα
=−
, ( ) ( )( )2
2 21 2
E X θα α
=− −
, ( )( ) ( )
2
21 2Var X αθ
α α=
− −
Here we are given that ( )2. 611
1F θ
θ = − +
. So θ is a two-parameter Pareto random
variable with parameters 1θ = and 2.6α = .
So ( ) 1 12.6 1 1.6
E X = =−
and ( )( ) ( ) ( )2 2
2.6 2.61.6 0.62.6 1 2.6 2
Var X = =− −
⇒( )( )
( )
( )
2
11.6 0.61.6 0.3692.6 2.6
1.6 0.6
EEVkVE Var
θθ
= = = = =
5 0.935 0.369
nZn k
= = =+ +
Q11 Nov 2004 #29 You are given:
• Claim counts follow a Poisson distribution with mean λ• Claim sizes follow a lognormal distribution with parameters µ and σ• Claim counts and claim amounts are independent. • The prior distribution has joint pdf:
140/284
( ), , 2f λ µ σ σ= , 0 1λ< < , 0 1µ< < , 0 1σ< <
Calculate Bühlmann’s k credibility for aggregate losses. Solution Let N represent the claim counts, iX the dollar amount of the i -th claim, and S the aggregate losses. N λ has Poisson distribution with mean of λ . ,iX µ σ has lognormal
distribution with parameters µ and σ . In addition, for 1i = to N , ,iX µ σ is independent identically distributed. The aggregate loss is:
1
N
ii
S X=
=∑
( ) ( ) ( )E S E N E X= , ( ) ( ) ( ) ( ) ( )2Var S E N Var X Var N E X= +
The risk parameters are λ , µ , and σ . If we fixλ , µ , and σ , then
( ) ( ) ( ) ( ), , , ,E S E N E X E Xλ µ σ λ µ σ λ µ σ= =
( ) ( ) ( ) ( ) ( )2, , , ,Var S E N Var X Var N E Xλ µ σ λ µ σ λ µ σ= +
( ) ( )2, ,Var X E Xλ µ σ λ µ σ= +
( )2,E Xλ µ σ =
From “Tables for Exam C/4,” we know the lognormal distribution has the following moments:
( ) 2 21exp2
kE X k kµ σ = +
⇒ ( ) 21exp2
E X µ σ = +
, ( ) ( )2 2 2 21exp 2 2 exp 2 22
E X µ σ µ σ = + × = +
⇒ ( ) ( ) 21, , , exp2
E S E Xλ µ σ λ µ σ λ µ σ = = +
141/284
( ) ( ) ( )2 2, , , exp 2 2Var S E Xλ µ σ λ µ σ λ µ σ = = +
The global mean is:
( ) 2, , , ,
1, , exp2
E E S Eλ µ σ λ µ σµ λ µ σ λ µ σ = = +
( )1 1 1
2
0 0 0
1exp , ,2
f d d dσ µ λ
λ µ σ λ µ θ λ µ σ= = =
= + ∫ ∫ ∫
1 1 12
0 0 0
1exp 22
d d dσ µ λ
λ µ σ σ λ µ σ= = =
= + ∫ ∫ ∫
21 1 1
0.5
0 0 0
2 e e d d dσ µ
σ µ λ
σ λ λ µ σ= = =
=
∫ ∫ ∫
21 1
0.5
0 0
1 22
e e d dσ µ
σ µ
σ µ σ= =
= ∫ ∫ ( ) 21
0.5
0
1 1 22
e e dσ
σ
σ σ=
= − ∫
Set 20.5 yσ = . Then d dyσ σ = .
⇒ ( ) ( ) ( )( )21 0.5
0.5 0.5
0 0
1 1 1 1y
y
e e d e e dy e eσ
σ
µ σ σ= =
= − = − = − −∫ ∫
( ) ( ) ( ){ }22
, , , , , ,, , , , , ,VE Var E S E E S E E Sλ µ σ λ µ σ λ µ σλ µ σ λ µ σ λ µ σ = = −
22
, ,1exp2
E λ µ σ λ µ σ + ( )2 2
, , exp 2E λ µ σ λ µ σ = +
( ) ( )1 1 1
2 2
0 0 0
exp 2 , ,f d d dσ µ λ
λ µ σ λ µ θ λ µ σ= = =
= +∫ ∫ ∫
( )1 1 1
2 2
0 0 0
exp 2 2 d d dσ µ λ
λ µ σ σ λ µ σ= = =
= +∫ ∫ ∫
( ) ( )1 1 1
2 2
0 0 0
2 exp exp 2 d d dσ µ λ
σ σ µ λ λ µ σ= = =
= ∫ ∫ ∫ ( ) ( )1 1
2
0 0
1 2 exp exp 23
d dσ µ
σ σ µ µ σ= =
= ∫ ∫
( ) ( )1
2 2
0
1 12 exp 13 2
e dσ
σ σ σ=
= −∫ ( ) ( )1
2 2
0
1 1 exp3
e dσ
σ σ σ=
= − ∫
142/284
Set 2 yσ = . Then 2 d dyσ σ = .
( ) ( ) ( )2 1 1
2 2 2 2, ,
0 0
1 1 1 1exp 1 exp 12 3 3 2
y
y
E e d e e dyλ µ σσ
λ µ σ σ σ σ= =
+ = − = − ∫ ∫
( ) ( ) ( ) ( )2 21 1 11 1 1 13 2 6
e e e e = − − = − −
( ){ } ( ) ( )2 222 0.5
, , , , 1 1E E S e eλ µ σ λ µ σ µ = = − −
⇒ ( ) ( ){ }22
, , , ,, , , ,VE E E S E E Sλ µ σ λ µ σλ µ σ λ µ σ = −
( )( ) ( ) ( )222 0.51 1 1 1 1 0.58726
e e e e= − − − − − =
( ) ( )2, , , ,, , exp 2 2EV E Var S Eλ µ σ λ µ σλ µ σ λ µ σ = = +
( ) ( )1 1 1
2
0 0 0
exp 2 2 , ,f d d dσ µ λ
λ µ σ λ µ θ λ µ σ= = =
= +∫ ∫ ∫
( )1 1 1
2
0 0 0
exp 2 2 2 d d dσ µ λ
λ µ σ σ λ µ σ= = =
= +∫ ∫ ∫
( ) ( )1 1 1
2
0 0 0
2 exp 2 exp 2 d d dσ µ λ
σ σ µ λ λ µ σ= = =
= ∫ ∫ ∫ ( ) ( )1 1
2
0 0
1 2 exp 2 exp 22
d dσ µ
σ σ µ µ σ= =
= ∫ ∫
( ) ( ) ( ) ( ) ( )1
22 2 2 2 2
0
1 1 1 1 1 11 2 exp 2 1 1 1 5.1032 2 2 2 2 8
e d e e eσ
σ σ σ=
= − = − − = − = ∫
⇒5.103 8.69
0.5872EVkVE
= = =
Shortcut to avoid the hard-core integration seen above.
The joint pdf is ( ) ( ) ( ) ( ), , 2f a b cλ µ σ σ λ µ σ= = , where ( ) 1a λ = , ( ) 1b µ = , and
( ) 2c σ σ= . In addition, λ , µ , and σ lie in the cube 0 1λ< < , 0 1µ< < , 0 1σ< < .
143/284
Consequently, λ , µ , and σ are independent random variables with the following marginal pdf:
( ) 1fλ λ = , 0 1λ< < ;
( ) 1fµ µ = , 0 1µ< < ;
( ) 2fσ σ σ= , 0 1σ< < .
( ) ( ) ( ) ( )22 0.5, , , ,
1, , exp2
E E S E E E e E eµ σλ µ σ λ µ σµ λ µ σ λ µ σ λ = = + =
( ) 12
E λ = , ( )1
0
1E e e du eµ µ= = −∫ , ( ) ( )2 2 21 1
0.5 0.5 0.5 0.5
00
2 2 2 1E e e d e eσ σ σσ σ = = = − ∫
⇒ ( ) ( ) ( ) ( )( )20.5 0.51 1E E e E e e eµ σµ λ= = − −
( ) ( ) ( ) ( ) ( )22 2 2, , , ,, , exp 2 2EV E Var S E E E e E eµ σ
λ µ σ λ µ σλ µ σ λ µ σ λ = = + =
( ) ( )1
12 2 2 2
00
1 1 12 2
E e e du e eµ µ µ = = = − ∫
( ) ( )2 2 21 1
2 2 2 2
00
1 12 12 2
E e e d e eσ σ σσ σ = = = − ∫
⇒ ( ) ( ) ( ) ( ) ( ) ( )2 22 2 2 2 21 1 1 11 1 1 5.1032 2 2 8
EV E E e E e e e eµ σλ= = × − × − = − =
( ) ( ) ( ){ }22
, , , , , ,, , , , , ,VE Var E S E E S E E Sλ µ σ λ µ σ λ µ σλ µ σ λ µ σ λ µ σ = = −
( ) 2 2, , , ,E E Sλ µ σ λ µ σ µ = −
( ) ( ) ( ) ( )22
2 2 2 2 2, , , ,
1exp exp 22
E E E E e E eµ σλ µ σ λ µ σλ µ σ λ µ σ λ + = + =
( )1
2 2
0
13
E dλ λ λ= =∫ , ( ) ( )2 21 12
E e eµ = − , ( )2 2 21 1
00
2 1E e e d e eσ σ σσ σ = = = − ∫
( ) ( )( ) ( ) ( )2 222 2 0.5, ,
1, , 1 1 1 1 0.58726
VE E E S e e e eλ µ σ λ µ σ µ = − = − − − − − =
⇒5.103 8.69
0.5872EVkVE
= = =
144/284
Please note … The joint pdf ( ) ( ) ( ) ( ), ,f a b cλ µ σ λ µ σ= alone doesn’t guarantee that λ , µ , and σ are independent. The additional requirement for λ , µ , and σ to be independent is that λ , µ ,and σ lie in a cube A Bλ< < , C Dµ< < , E Fσ< < , where A,B,C,D,E, and F are constant. For example, say A Bλ< < , C Dµ< < , ( ) ( )e fλ σ λ< < . Even if
( ) ( ) ( ) ( ), ,f a b cλ µ σ λ µ σ= , then λ , µ , and σ are NOT independent.
Q12 Nov 2004 #25 You are given:
• A portfolio of independent risks is divided into two classes. • Each class contains the same number of risks. • For each risk in Class 1, the number of claims per year follows a Poisson
distribution with mean 5. • For each risk in Class 2, the number of claims per year follows a binomial
distribution with mean 8m = and 0.55q = .• A randomly selected risk has three claims in Year 1, r claims in Year 2, and four
claims in Year 3. The Bühlmann credibility estimate for the number of claims in Year 4 for this risk is 4.6019. Determine r .
Solution Risk ( )P Risk X Risk ( )E X Risk ( )Var X Risk#1 0.5 Poisson with mean 5 5 5 #2 0.5 Binomial 8m = and 0.55q = 8(0.55) 4.4= 8(0.55)0.45 1.98=
( )1 5 4.4 4.72
µ = + = , ( )1 5 1.98 3.492
EV = + =
( )2 25 4.4 0.5 0.09VE = − =
3.49 38.780.09
EVkVE
= = = , 1
n
ii
k XP
n k
µ=
+=
+
∑
⇒( ) ( )38.78 4.7 3 4
4.60193 38.78
r+ + +=
+, 3r =
145/284
Q13 Nov 2001 #23 You are given the following information on claim frequency of auto accidents for individual drivers:
Business Use Pleasure use Expected claims Claim variance Expected claims Claim variance
Rural 1.0 0.5 1.5 0.8 Urban 2.0 1.0 2.5 1.0 Total 1.8 1.06 2.3 1.12
You are given: • Each driver’s claim experience is independent of every other driver’s. • There are an equal number of business and pleasure use drivers.
Determine the Bühlmann credibility factor for a single driver. Solution The key to solving this problem is correctly identifying risk classes. There are four risk classes:
( ), , ,BR BU PR PUΘ =
BR=Business & Rural Use, BU=Business & Urban Use PR=Pleasure & Rural Use, PU=Pleasure & Urban Use Next, we need to calculate the probability of Rural Use and Urban Use.
Expected claims Rural 1.0 Urban 2.0 Total 1.8
( ) ( ) ( ) ( )1.0 2.0 1.8P R P U+ = ; ( ) ( ) 1P R P U+ =
Solving these two equations, we get: ( ) 0.2P R = , ( ) 0.8P U = .
Next, we list the probability for each class: Business Use 0.5 Pleasure use 0.5 Rural 0.2 P(BR)=0.2(0.5)=0.1 P(PR)=0.2(0.5)=0.1 Urban 0.8 P(BU)=0.8(0.5)=0.4 P(PU)=0.8(0.5)=0.4
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Let X represent the claim frequency of auto accidents of a randomly selected driver. θ ( )E X θ ( )Var X θ ( )P θ ( ) 2
E X θ BR 1.0 0.5 0.1 1.0 BU 2.0 1.0 0.4 4.0 PR 1.5 0.8 0.1 2.25 PU 2.5 1.0 0.4 6.25
( )E E Xθµ θ = = 1.0(0.1) + 2.0(0.4) + 1.5(0.1) + 2.5(0.4) = 2.05
( )EV E Var Xθ θ = = 0.5(0.1) + 1.0(0.4) + 0.8(0.1) + 1.0(0.4) = 0.93
( ) ( ) ( ){ }22VE Var E X E E X E E Xθ θ θθ θ θ = = −
( ) 2E E Xθ θ = 1.0(0.1) + 4.0(0.4) + 2.25(0.1) + 6.25(0.4) = 4.425
( ) ( ){ }22VE E E X E E Xθ θθ θ = − =4.425 – 2.052 = 0.2225
0.93 4.180.2225
EVkVE
= = =
The Bühlmann credibility factor for a single driver is:
1 0.1931 4.18
nZn k
= = =+ +
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About the author Yufeng Guo was born in central China. After receiving his Bachelor’s degree in physics at Zhengzhou University, he attended Beijing Law School and received his Masters of law. He was an attorney and law school lecturer in China before immigrating to the United States. He received his Masters of accounting at Indiana University. He has pursued a life actuarial career and passed exams 1, 2, 3, 4, 5, 6, and 7 in rapid succession after discovering a successful study strategy. Mr. Guo’s exam records are as follows: Fall 2002 Passed Course 1 Spring 2003 Passed Courses 2, 3 Fall 2003 Passed Course 4 Spring 2004 Passed Course 6 Fall 2004 Passed Course 5 Spring 2005 Passed Course 7
Mr. Guo currently teaches an online prep course for Exam P, FM, MFE, and MLC. For more information, visit http://actuary88.com/.
If you have any comments or suggestions, you can contact Mr. Guo at [email protected].
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