* Key Code Inside * ACTEX...2019/02/01 · (c) Bayesian Credibility (Nonlife Actuarial Models, Chapter 8.1–8.2) (d) Empirical implementation of Credibility ( Nonlife Actuarial Models

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Exam MAS-II Study Manual

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Spring 2019 Edition

Jeffrey S. Pai, Ph.D., ASA, ACIA

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Spring 2019 Edition

Jeffrey S. Pai, Ph.D., ASA, ACIA

ACTEX

ACTEX Learning | Learn Today. Lead Tomorrow.

Exam MAS-II Study Manual

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Contents

Contents iii

A Introduction to Credibility 1

1 Limited Fluctuation Credibility 5

1.1 Full Credibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.1.1 Full Credibility for Claim Frequency . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.1.2 Full Credibility for Claim Severity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.1.3 Full Credibility for Aggregate Loss . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.1.4 Full Credibility for Pure Premium . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.2 Partial Credibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.3 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

1.4 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

1.5 Problem Set Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2 Buhlmann and Buhlmann-Straub Credibility 29

2.1 Buhlmann Credibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

2.2 Buhlmann-Straub Credibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

2.3 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

2.4 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41


3 Bayesian Credibility 49

3.1 Bayesian Inference and Estimation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

3.2 Conjugate Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

3.3 Discrete Prior Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

3.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

3.5 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65


4 Non-parametric Empirical Bayes 71

4.1 Non-parametric Model in Buhlmann-Straub’s Case . . . . . . . . . . . . . . . . . . . . . . 71

4.2 Non-parametric Model in Buhlmann’s Case . . . . . . . . . . . . . . . . . . . . . . . . . . 75

4.3 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

iii

iv CONTENTS

4.4 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80


B Linear Mixed Models 87

5.0 Introduction and Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

5 Two-level Models for Clustered Data 95

5.1 Rat Pup Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

5.2 Model Specification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98

5.2.1 General Model Specification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98

5.2.2 Hierarchical Model Specification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102

5.3 Hypothesis Tests and Model Selection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

5.4 Interpretation, Residual Diagnostics, and Prediction . . . . . . . . . . . . . . . . . . . . . 115

5.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122

5.6 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123


6 Three-level Models for Clustered Data 129

6.1 Classroom Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130



6.4 Intraclass Correlation Coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142

6.5 Calculating Predicted Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

6.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144

6.7 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145


7 Models for Repeated-Measures Data: The Rat Brain Example 151

7.1 Rat Brain Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151



7.4 Interpretation and Residual Diagnostics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157

7.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160

7.6 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161


8 Random Coefficient Models for Longitudinal Data: The Autism Example 167

8.1 Autism Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167



8.4 Interpretation, Residual Diagnostics, and Predictions . . . . . . . . . . . . . . . . . . . . . 174

8.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181

CONTENTS v

8.6 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182


9 Models for Clustered Longitudinal Data: The Dental Veneer Example 185

9.1 Veneer Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186




9.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193

9.6 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194


10 Models for Data with Crossed Random Factors: The SAT Score Example 199

10.1 SAT Score Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200




10.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209

10.6 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210


C Bayesian Analysis and Markov Chain Monte Carlo 215

11 Bayesian Inference and Sampling 219

11.1 Model Components . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219

11.2 Posterior . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220

11.2.1 Grid Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221

11.2.2 Quadratic Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222

11.2.3 Markov Chain Monte Carlo . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224

11.3 Sampling from a Grid-approximate Posterior . . . . . . . . . . . . . . . . . . . . . . . . . 224

11.4 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228


12 Linear and Multivariate Regression Models 233

12.1 Gaussian (Normal) Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233

12.2 Simple Regression Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236

12.3 Multivariate Regression Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238

12.3.1 Polynomial Regression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238

12.3.2 Spurious Relationship . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239

12.3.3 Masked Relationship . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240

12.4 Categorical Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242

12.5 Ordinary Least Squares . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243

vi CONTENTS

12.6 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245


13 Overfitting, Regularization, and Information criteria 249

13.1 Overfitting and Underfitting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249

13.2 Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252

13.3 Information Criteria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253

13.4 Model Comparison and Model Averaging . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255

13.4.1 Model Comparison . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255

13.4.2 Model Averaging . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257

13.5 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259


14 Markov Chain Monte Carlo 263

14.1 Metropolis-Hastings Algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263

14.2 Gibbs Sampling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265

14.3 Hamiltonian Monte Carlo . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267

14.4 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273


15 Generalized Linear Models 277

15.1 Link Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277

15.2 Binomial Regression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278

15.3 Poisson Regression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287

15.4 Ordered Categorical Outcomes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291

15.5 Zero-inflated Outcome . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295

15.6 Over-dispersed Outcome . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 296

15.7 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299


16 Multilevel Models and Covariance 303

16.1 Varying Intercepts Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303

16.2 Multiple-Cluster Multilevel Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311

16.2.1 Varying Slopes - Random Intercepts . . . . . . . . . . . . . . . . . . . . . . . . . . . 314

16.2.2 Varying Slopes Multilevel Model with Logit Link . . . . . . . . . . . . . . . . . . . . 316

16.3 Continuous Categories and the Gaussian Process . . . . . . . . . . . . . . . . . . . . . . . 321

16.4 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323


D Statistical Learning 327

17 Supervised Learning 331

CONTENTS vii

17.1 K-Nearest Neighbors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331

17.2 Regression Trees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333

17.3 Classification Trees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 338

17.4 Advantages and Disadvantages of Trees . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343

17.5 Bagging, Random Forests, and Boosting . . . . . . . . . . . . . . . . . . . . . . . . . . . . 344

17.6 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 348


18 Unsupervised Learning 353

18.1 Principal Components Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353

18.2 K-means Clustering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 357

18.3 Hierarchical Clustering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 359

18.4 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 364


E Practice Set 369

F Solutions to Sample Questions 405

G Solutions to Fall 2018 Exam 415

Index 429

Preface

This study guide covers the material in the Casualty Actuarial Society (CAS) Exam Modern Actuarial

Statistics - II. To get the most current examination syllabus, start with http://www.casact.org and

browse under “SYLLABUS OF BASIC EDUCATION””.

This study guide consists of four parts, which cover the four topics listed in the syllabus. The topics, the

textbook references, and the weights are shown below.

Part A. Introduction to Credibility 10%

Tse, Nonlife Actuarial Models 6.1 - 6.3, 7.1 - 7.4, 8.1 - 8.2, 9.1, 9.2

Part B. Linear Mixed Models 25%

(i) West, Linear Mixed Models 1 - 8 (excluding 2.9.6), and Appendix B

(ii) Notes on Shrinkage

Part C. Bayesian Analysis and Markov Chain Monte Carlo 50%

(i) McElreath, Statistical Rethinking 2 - 6, 8, 9.2, 10 - 13

(ii) Ford

Part D. Statistical Learning

James, et al., An Introduction to Statistical Learning 2.2.3, 8, 10 15%

Please feel free to contact [email protected] if you see something that is incorrect or unclear.

ix

Part A

Introduction to Credibility

1

3

This is the first part of the MAS-II syllabus. We will cover all the eight learning objectives from Nonlife

Actuarial Models. There are 4 main topics in this part:

(a) Limited Fluctuation Credibility (Nonlife Actuarial Models, Chapter 6.1–6.3)

(b) Buhlmann and Buhlmann–Straub Credibility (Nonlife Actuarial Models, Chapter 7.1–7.4)

(c) Bayesian Credibility (Nonlife Actuarial Models, Chapter 8.1–8.2)

(d) Empirical implementation of Credibility (Nonlife Actuarial Models, Chapter 9.1–9.2)

Under each topic (chapter), there are examples and questions indicated by level of difficulty with “*” (core

question), “**” (advanced question), and “***” (mastery).

Important concepts and formulas are summarized in each chapter followed by a problem set. We suggest

you work on core questions first, and move on to advanced and mastery questions when you are ready to

challenge yourself with more difficult questions.

Chapter 1

Limited Fluctuation Credibility

Learning objective: Understand the basic framework of credibility and be familiar with limited

fluctuation credibility, including partial credibility and full credibility

The limited fluctuation credibility approach1 (also called the classical credibility approach) calcu-

lates the updated prediction of the loss measure as a weighted average of D and M , where D is based on

the recent claim experience and M is based on a rate specified in the manual, called manual rate. The

weight Z, 0 ≤ Z ≤ 1, assigned to D is called the credibility factor, and the updated prediction, denoted

by U , is given by

U = Z D + (1− Z) M. (1.1)

The followings are the notations and terms used in this chapter.

Claim frequency: Denote the number of claims by N .

Claim severity: Denote the amount of the ith claim by Xi. The average claim severity is the sample

mean of X1, · · · , XN , i.e. X = (∑N

i=1 Xi)/N .

Aggregate loss: Denote the aggregate loss by S where S = X1 + X2 + · · ·+ XN .

Pure premium: Denote the pure premium by P where P = S/E and E is the number of exposure units.

Example 1.1.* The 2017 manual rate equals 1000. The loss experience is such that the actual pure

premium for that year equals 1200. The credibility factor is Z = 0.33. Estimate the pure premium

underlying the rate in 2018. (Assume no change in the pure premium due to inflation.)

Solution. The observed rate is D = 1200, the manual rate is M = 1000, and the credibility factor is

Z = 0.33. Using equation (1.1), the updated premium rate in 2018 is

U = ZD + (1− Z)M = (0.33)(1200)+ (0.67)(1000) = 1066.

1Nonlife Actuarial Models, p. 172

5

6 CHAPTER 1. LIMITED FLUCTUATION CREDIBILITY

Questions using equation (1.1) usually require the calculation of the credibility factor, Z, which will be

discussed in the next section “partial credibility”. The following examples use equation (1.1) in various

ways.

Example 1.2.** You are given D = 240, Z = 0.4, and U = 264. Determine the manual rate, M , used in

the limited fluctuation approach.

Solution. Using equation (1.1), we have

U = Z D + (1− Z) M

264 = (0.4)(240)+ (1− 0.4)M =⇒ M = $280.

Example 1.3.** You are given D = 230, M = 280, and U = 265. Determine the credibility factor Z used

in the limited fluctuation approach.


U = Z D + (1− Z) M

265 = Z (230) + (1 − Z)(280) =⇒ Z = 0.3.

Example 1.4.*** You are given D = 230, Z = 0.4, and U = 266. If the loss is increased by 10%, calculate

the updated predicted premium using the same manual rate and the credibility factor.

Solution. Using equation (1.1) we have

U = Z D + (1− Z) M

266 = (0.4)(230)+ (1− 0.4)M =⇒ M = $290.

After the 10% increase, the loss per worker is D′ = (1.1)(230) = 253 and the updated predicted pure

premium for the company’s insurance is

U ′ = Z D′ + (1− Z) M = (0.4)(253)+ (1− 0.4)(290) = $275.2.

Note that D is increased by 23, and 40% (Z = 0.4) of which attributes to U . Thus

U ′ = U + (10%)(230)(0.4) = 266 + 9.2 = $275.2.

Recap: The updated prediction is U = Z D + (1− Z) M .

Example 1.1: Find U given Z, D, and M .

Example 1.2: Find M given Z, D, and U .

Example 1.3: Find Z given U , D, and M .

Example 1.4: Find M given Z, D, and U . Find U ′ given D′.

1.1 Full Credibility 7

Since 0 ≤ Z ≤ 1, we have min{D, M} ≤ U ≤ max{D, M}. The updated premium, U , is a weighted

average of D and M where the weight Z is on the claim experience D. The credibility factor Z determines

the relative importance of the data (D) in calculating the updated prediction.

Full credibility is achieved if Z = 1, in which case the prediction depends only on the data, i.e. U = D.

When Z < 1, the data have partial credibility and the prediction is the weighed average of D and M as

in equation (1.1).

1.1 Full Credibility

The credibility factor Z is a function of the sample size and Z increases as sample size increases. The

limited fluctuation approach determines the minimum data size required for the experience data to be

given full credibility.2 The minimum data size is called the standard for full credibility.

1.1.1 Full Credibility for Claim Frequency

Let N be the claim frequency random variable. Denote E(N ) = µN and Var(N ) = σ2N . If we assume that

N is normal distributed, then the probability of observing claim frequency within 100k% of the mean

can be written as3

Pr(µN − k µN ≤ N ≤ µN + k µN ) = 2Φ

(k µN

σN

)− 1, (1.2)

where Φ(·) is the df (distribution function) of the standard normal random variable.4 For example,

Φ(1.645) = 0.95 and Φ(1.96) = 0.975.

If the coverage probability in (1.2) is 1 − α, then

kµN

σN

= z1−α/2,

where zβ is the 100βth percentile of the standard normal distribution, i.e. Φ(zβ) = β.

Confirm the coverage probability:

2Φ

(k µN

σN

)− 1 = 2Φ(z1−α/2) − 1 = 2(1 − α/2) − 1 = 1 − α.

Example 1.5.*5 Suppose the claim frequency of a risk group is normal distributed with mean µN = 420

and variance σ2N = 521. Find the probability that the observed number of claims is within 10% of the true

mean.

2Nonlife Actuarial Models, p. 1733Nonlife Actuarial Models, p. 1744MAS-II Tables5Nonlife Actuarial Models, Example 6.2


Solution. Using equation (1.2), the probability that the observed number of claims is within 10% of the

true mean is

Pr(µN − k µN ≤ N ≤ µN + k µN) = 2Φ

(k µN

σN

)− 1 = 2Φ

(10%(420)√

521

)− 1

= 2Φ (1.84)− 1 = 2(0.9671)− 1 = 93.42%.

Example 1.6.** (continued) The probability that the observed number of claims is within 100k% of the

true mean is 90%. Determine k.


0.9 = Pr(µN − k µN ≤ N ≤ µN + k µN) = 2Φ

(k(420)√

521

)− 1 =⇒ Φ

(k(420)√

521

)= 0.95.

From MAS-II Tables, we have Φ(1.645) = 0.95. Therefore,

k(420)√521

= 1.645 =⇒ k = 0.0894.

Example 1.7.*** A block of health insurance policies consists of two independent groups. Suppose

the claim number of the first group is normal distributed with mean 420 and variance 521, and the claim

number of the second group is normal distributed with mean 360 and variance 1,000. Find the probability

that the combined number of claims is within 5% of the true mean of the given risk block.

Solution. The claim number of the risk block is the sum of that of the two groups, i.e. N = N1 + N2,

and N is normal distributed with mean µN = 420 + 360 = 780 and variance σ2N = 521 + 1, 000 = (39)2.

Using equation (1.2), the probability that the observed number of claims is within 5% of the true mean is


(k µN

σN

)− 1 = 2Φ

(5%(780)

39

)− 1

= 2Φ(1)− 1 = 2(0.8413)− 1 = 68.26%.

Recap: The probability of observing claim frequency within k of the mean is


(k µN

σN

)− 1.

Example 1.5: Calculate the coverage probability given k, µN , and σN .

Example 1.6: Calculate k given µN , σN , and coverage probability.

Example 1.7: Calculate the coverage probability for a heterogeneous risk group.

The focus of this section is to find the minimum data size required to give full credibility. It is often

assumed that the claim frequency is distributed as a Poisson variable with mean λN , which is large enough

to apply the normal approximation. The followings are the key steps.


Requirement: The probability that the observed number of claims is within k of the true mean is at least

1 − α:

Pr(µN − k µN ≤ N ≤ µN + k µN) = Pr

(−k

µN

σN

≤ N − µN

σN

≤ kµN

σN

)≥ 1− α.

Poisson Assumption: If N is Poisson distributed with parameter λN

, then µN = σ2N = λ

Nand

kµN

σN

= kλN√λN

= k√

λN.

Normal Approximation: If the number of claims is large enough, we can use normal approximation

such that

k√

λN = z1−α/2 or λN =(z1−α/2

k

)2

. (1.3)

Thus, full credibility for claim frequency is attributed to the data if λN≥(

z1−α/2

k

)2.

Standard for full credibility: Define

λF =(z1−α/2

k

)2

, (1.4)

which is the standard for full credibility for claim frequency. Therefore, full credibility is attained

if the observed number of claims is greater than λF .

Example 1.8.*6 Suppose the claim frequency of a risk group is Poisson distributed with mean 850, and

that the normal approximation can be used for the Poisson. Find the probability that the observed number

of claims is within 10% of the true mean.

Solution. Using equation (1.3), we have z1−α/2 = k√

λN = (10%)√

850 = 2.92. Since Φ(2.92) = 0.9982 =

1 − α/2, we have α = 0.0036. The coverage probability is 1 − α = 99.64%.

Example 1.9.** (continued) The probability that the observed number of claims is within 100k% of the

true mean is 100(1− α)%. Determine k if α = 10%.

Solution. For α = 10%, we have 1− α/2 = 0.95 and z0.95 = Φ−1(0.95) = 1.645. Using equation (1.3), we

have k =z1−α/2√

λN

=1.645√

850= 5.64%.

Example 1.10.*** The claim frequency is Poisson distributed with mean 850, and the normal approxi-

mation can be used for the Poisson. Suppose the coverage probability remains unchanged.

What is the percentage change in k if the Poisson mean increased by 20%?

Solution. Using equation (1.3), we have z1−α/2 = k√

λN

=√

850k. After the 20% increase, the Poisson

mean is λ′N

= (1.2)(850) = 1020. Using equation (1.3) again, we have z1−α/2 = k′√

λ′N

=√

1020k′. Since

the coverage probability remains unchanged, z1−α/2 remains the same and√

850k =√

1020k′. Thus, the

percentage change in k is

k′

k− 1 =

√850√1020

− 1 = −8.7%,

6Nonlife Actuarial Models, Example 6.3


Example 1.11.*7 If an insurance company requires a coverage probability of 99% for the number of claims

to be within 5% of the true expected claim frequency, how many claims in the recent period are required

for full credibility?

Solution. We have α = 1 − 0.99 = 0.01 and thus z1−α/2 = z0.995 = Φ−1(0.995) = 2.576. Using equation

(1.4) and k = 0.05, we have

λF =(z1−α/2

k

)2

=

(2.576

0.05

)2

= 2, 655. (rounded up)

Example 1.12.** (continued) Recent experience of a workers compensation insurance has established the

mean accident rate as 0.045 and the standard for full credibility of claims as 1,200. For a group with a

similar risk profile, what is the minimum number of exposure units (workers) required for full credibility?

Solution. Since the standard for full credibility is 1,200 claims, the same standard base on exposure is

1, 200/0.045 = 26, 667 workers.

Recap: We assume that the claim frequency is Poisson distributed.

Example 1.8: Find the coverage probability given λN

and k.

Example 1.9: Find k given λN and α.

Example 1.10: Find the percentage change in k for a change in λN .

Example 1.11: Find λF, the minimum number of claims required for full credibility.

Example 1.12: Find the minimum number of exposure units required for full credibility.

1.1.2 Full Credibility for Claim Severity

Assume that X1, X2, · · · , XN are N independent, identically distributed random variables with Xi repre-

senting the size of the ith claim. All claim sizes have the same mean, µX , and the same variance, σ2X . Full

credibility is attributed to X if the probability of X being within k of the true mean of claim loss µX is at

lease 1 − α:

Pr(µX − k µX ≤ X ≤ µX + k µX) = 2Φ

(k µX

σX/√

N

)− 1 ≥ 1 − α.

We assume that the sample size N is sufficiently large so that X is approximately normal distributed with

mean µX and variance σ2X/N .

If the coverage probability is 1 − α, then

kµX

σX/√

N= z1−α/2.



For the coverage probability to be larger than 1 − α, we want

kµX

σX/√

N≥ z1−α/2.

To satisfy the condition for given α, k, µX , and σX , we must have

N ≥(z1−α/2

k

)2(

σX

µX

)2

= λF C2X ,

where CX = σX/µX is the coefficient of variation of X . See Table 1.2 for C2X for some commonly used

severity distributions.

The standard for full credibility for claim severity is λF C2X . That is, full credibility is attained if

N ≥ λF C2X .

Example 1.13.*8 What is the standard for full credibility for claim severity with α = 0.01 and k = 0.05,

given that the mean and variance estimates of the severity are 1,000 and 2,000,000, respectively?

Solution. For α = 0.01, we have z1−α/2 = z0.995 = Φ−1(0.995) = 2.576. And for k = 0.05, we have

λF =(z1−α/2

k

)2

=

(2.576

0.05

)2

= 2, 655.

The standard for full credibility for claim severity is

λF

(σX

µX

)2

= (2, 655)2, 000, 000

(1, 000)2= 5310.

Example 1.14.** The standard for full credibility for claim frequency for a risk group is 234. If the claim

severity follows an exponential distribution with mean 1,000. Determine standard for full credibility for

claim severity for the same risk group.

Solution. The standard for full credibility for claim frequency is 234, i.e. λF = 234. For the given

exponential distribution, we have µX = 1, 000 and σ2X = 1, 000, 000. Thus, the standard for full credibility

for claim severity is

λF

(σ2

X

µ2X

)= (234)

1, 000, 000

(1, 000)2= 234.

Note that if X is exponential distributed, then CX = 1.

Example 1.15.*** Claim severity follows gamma distribution with parameters a = 50 and β = 20 (the

mean is 1,000). The classical credibility approach is adopted to predict mean severity, with minimum

coverage probability of 92% to within 1% of the mean severity. Determine standard for full credibility for

claim severity.



Solution. For the given gamma distribution, calculate

µX = aβ = (50)(20) = 1, 000,

σ2X = aβ2 = (50)(20)2 = 20, 000.

For 92% coverage probability, we have α = 1 − 0.92 = 0.08 and z1−α/2 = z0.96 = Φ−1(0.96) = 1.75. For

k = 0.01, the standard for full credibility for claim severity is

λF

(σ2

X

µ2X

)=

(1.75

0.01

)2 20, 000

(1, 000)2= 613.

Note that if X is gamma distributed with parameters a and β, then C2X = 1/a and λF C2

X = λF /a.

Recap: The standard for full credibility for claim severity is λF C2X .

Example 1.13: Find the standard given α, k, µX , and σX .

Example 1.14: Find the standard given λF and that X is exponential distributed.

Example 1.15: Find the standard given α, k, and that X is gamma distributed.

1.1.3 Full Credibility for Aggregate Loss

Denote the aggregate loss by S where

S = X1 + X2 + · · ·+ XN

.

Denote E(S) = µS and Var(S) = σ2S. Assume that N and X1, X2, · · · , X

Nare mutually independent. We

have

µS = (µN)(µX)

σ2S = (µN)(σ2

X) + (µ2X)(σ2

N).

Assume that N is Poisson distributed with mean λN, then µN = σ2

N = λN, and therefore

µS = (λN)(µX)

σ2S = (λ

N)(σ2

X + µ2X).

Full credibility is attributed to S if the probability of S being within k of the true mean µS is at least

1 − α.

If we assume that S is normal distributed, the coverage probability is

Pr(µS − k µS ≤ S ≤ µS + k µS) = 2Φ

(k µS

σS

)− 1.

If the coverage probability is 1 − α, then

kµS

σS

= z1−α/2.


For the coverage probability to be larger than 1 − α, we must have

kµS

σS

≥ z1−α/2.

To satisfy the above condition for given α, k, µX , and σX , we must have

λN≥(z1−α/2

k

)2(

µ2X + σ2

X

µ2X

)= λF

(E(X2)

µ2X

)= λF

(1 + C2

X

).

The standard for full credibility for aggregate loss is λF

(1 + C2

X

).

Note that λF

(1 + C2

X

)= λF + λF C2

X . Therefore, the standard for full credibility for aggregate loss is the

sum of the standard for claim frequency and the standard for claim severity.

Example 1.16.* You are given the following:

(i) The number of claims follows a Poisson distribution.

(ii) Claims sizes follow a Poisson distribution with mean 4.

(iii) The number of claims and claim sizes are independent.

The full credibility standard has been selected so that actual claim costs (aggregate loss) will be within

10% of expected claim costs 95% of the time using a normal approximation. Using the methods of limited

fluctuation credibility, determine the expected number of claims required for full credibility for aggregate

loss.

Solution. The coverage probability is 0.95. Thus α = 1 − 0.95 = 0.05, and z1−α/2 = z0.975 = 1.96. The

expected number of claims required for full credibility is

λF (1 + C2X ) =

(z0.975

k

)2(

1 +σ2

X

µ2X

)=

(1.96

0.10

)2(1 +

4

42

)= 481. (rounded up)

Example 1.17.* The standard for full credibility for claim frequency for a risk group is 112. The standard

for full credibility for claim severity for the risk group is 424. Determine standard for full credibility for

aggregate loss for the same risk group.

Solution. We are given λF = 112 and λF C2X = 424. The standard for full credibility for aggregate loss is

λF (1 + C2X ) = λF + λF C2

X = 112 + 424 = 536.

Example 1.18.*** Claim severity follows inverse gamma distribution with parameters α = 3 and θ =

2, 000. The classical credibility approach is adopted to predict mean aggregate loss, with minimum coverage

probability of 95% to within 10% of the mean aggregate loss. Determine standard for full credibility for

aggregate loss.

Solution. Look for the inverse gamma distribution in MAS-II Tables, we have

µX =θ

α − 1=

2, 000

3− 1= 1, 000,

E(X2) =θ2

(α − 1)(α− 2)=

(2, 000)2

(3− 1)(3− 2)= 2, 000, 000.


For 95% coverage probability, we have z0.975 = Φ−1(0.975) = 1.96. For k = 0.1, the standard for full

credibility for aggregate loss is

λFE(X2)

µ2X

=

(1.96

0.1

)2 2, 000, 000

(1, 000)2= 769. (rounded up)

See Table 1.2 for 1 + C2X = (α − 1)/(α− 2) = 2.

Recap: The standard for full credibility for aggregate loss is λF + λF C2X .

Example 1.16: Calculate the standard given α, k, µX , and σX .

Example 1.17: Calculate the standard given λF and λF C2X .

Example 1.18: Calculate the standard given α, k, and that X is inverse gamma distributed.

See Table 1.2 for other commonly used severity distributions.

1.1.4 Full Credibility for Pure Premium

Let E be the number of exposure units of the risk group, the pure premium per exposure unit is defined

as P = S/E. Denote E(P ) = µP and Var(P ) = σ2P . Full credibility is attributed if the probability of P

being within 100k% of the true mean E(P ) is at lease 1 − α. Since µP /σP = µS/σS, the standard for full

credibility for pure premium is the same as that for aggregate loss.

Example 1.19.* The number of claims has a Poisson distribution. Claim sizes have a two-parameter

Pareto distribution with parameters θ = 5.5 and α = 3. The number of claims and claim sizes are

independent. The observed pure premium should be within 4% of the expected pure premium 95% of the

time. Determine the expected number of claims needed for full credibility for pure premium.

Solution. Look for the Pareto distribution in MAS-II Tables, we have

µX =θ

α − 1,

E(X2) =2θ2

(α − 1)(α − 2),

1 + C2X =

E(X2)

µ2X

=2(α − 1)

α − 2= 4.

We have 1− 0.95 = 0.05 and thus z1−0.05/2 = z0.975 = 1.96. The expected number of claims needed for full

credibility for pure premium is

λF (1 + C2X ) =

(1.96

0.04

)2

(4) = 9, 604.

1.2 Partial Credibility 15

Example 1.20.**9 A block of accident insurance policies has mean claim frequency of 0.03 per policy.

Claim-frequency distribution is assumed to be Poisson. If the claim-severity distribution is lognormal

distributed with µ = 5 and σ = 1, calculate the number of policies required to attain full credibility for

pure premium, with α = 0.02 and k = 0.05.

Solution. Look for the lognormal distribution in MAS-II Tables, we have

µX = eµ+σ2/2

and

E(X2) = e2µ+2σ2.

Thus

1 + C2X =

E(X2)

µ2X

= eσ2= e.

Now z1−α/2 = z0.99 = Φ−1(0.99) = 2.326, so that the standard for full credibility for pure premium requires

a minimum expected claim frequency of

λFE(X2)

µ2X

=

(2.326

0.05

)2

(e) = 5, 883.

Hence, the minimum number of policies for full credibility for pure premium is

5, 883

0.03= 196, 100.

Recap: The standard for full credibility for pure premium is λF + λF C2X .

Example 1.19: Calculate the standard for given α, k, and that X is Pareto distributed.

Example 1.20: Calculate the standard given α, k, and that X is lognormal distributed.

See Table 1.2 for other commonly used severity distributions.

1.2 Partial Credibility

When the full credibility is not attained, a value of Z < 1 has to be determined and the updated prediction

is U = Z D + (1− Z) M .

For the claim frequency, we require that the probability of (Z×N ) lying within interval (ZµN −kµN , ZµN +

kµN) is 1 − α for a given k:

Pr(ZµN − k µN ≤ Z N ≤ ZµN + k µN ) = 1 − α

=⇒ Pr

(−k µN

ZσN

≤ N − µN

σN

≤ k µN

ZσN

)= 1 − α.



Assuming Poisson claim-frequency distribution with mean λN

and applying normal approximation, we

have

k µN

ZσN

= z1−α/2 =⇒ k λN

Z√

λN= z1−α/2 =⇒ Z =

(k

z1−α/2

)√λN =

√λ

N

λF.

The partial credibility factors for claim severity, aggregate loss, and pure premium can be derived in a

similar way.

The partial credibility factors for claim frequency, claim severity, aggregate loss, and pure premium are

summarized below.

Claim Frequency: Z =

√λN

λF

Claim Severity: Z =

√N

λF C2X

Aggregate Loss and Pure Premium: Z =

√λN

λF + λF C2X

.

It is also called the square-root rule. Within the square root, the denominator is the standard for full

credibility of the corresponding risk measure, and the numerator, λN or N , is observed from data. Note

that if the ratio is greater than 1 then the full credibility is attained and Z = 1.

Example 1.21.*10 A block of insurance policies had 896 claims this period with mean loss of 45 and

variance of loss of 5,067. Full credibility is based on a coverage probability of 98% for a range of within

10% deviation from the true mean. The mean frequency of claims is 0.09 per policy and the block has

18,600 policies. Calculate Z for the claim frequency for the next period.

Solution. The expected claim frequency is λN = (18, 600)(0.09) = 1, 674. Since z1−α/2 = Φ−1(0.99) =

2.326 and k = 10%, the full-credibility standard for claim frequency is λF = (2.326/0.1)2 = 542. Since

λN

> λF , full credibility is attained for claim frequency and Z = 1.

Example 1.22.* (continued) A block of insurance policies had 896 claims this period with mean loss of

45 and variance of loss of 5,067. Full credibility is based on a coverage probability of 98% for a range of

within 10% deviation from the true mean. The mean frequency of claims is 0.09 per policy and the block

has 18,600 policies. Calculate Z for the claim severity for the next period.

Solution. The standard for full credibility for claim severity is

λF C2X =

(2.326

0.1

)2 (5, 067

452

)= 1, 357.

The block had N = 896 claims this period. Therefore, the partial credibility factor is

Z =

√N

λF C2X

=

√896

1, 357= 0.813.


1.3 Summary 17

Example 1.23.** Claim severity has mean 342 and standard deviation 408. An insurance company has

75,000 insurance policies. Using the classical credibility approach with coverage probability of 95% to

within 5% of the aggregate loss, determine the credibility factor Z if the average claim per policy is 4%.

Solution. For 95% coverage probability, α = 1−0.95 = 0.05 and thus z1−α/2 = z0.975 = Φ−1(0.975) = 1.96.

We are given µX = 342, σX = 408. For k = 0.05, the standard for full credibility for aggregate loss is

λF (1 + C2X ) =

(1.96

0.05

)2(1 +

(408)2

(342)2

)= 3, 724.

The average claim is λN = (75, 000)(4%) = 3, 000 and the credibility factor is

Z =

√λN

λF (1 + C2X)

=

√3, 000

3, 724= 0.898.

Recap: Partial credibility factors:

Example 1.21: Calculate Z for the claim frequency: Z =

√λ

N

λF

Example 1.22: Calculate Z for the claim severity: Z =

√N

λF C2X

Example 1.23: Calculate Z for the aggregate loss: Z =

√λN

λF + λF C2X

1.3 Summary

The main topic of this chapter is to calculate the updated prediction

U = Z D + (1− Z) M.

where D is based on the recent claim experience, M is based on a rate specified in the manual, and the

weight Z is the credibility factor.

When the sample size is large enough, we assign full credibility Z = 1, and thus U = D. If not, we apply

the square-root rule for partial credibility. Table 1.1 summarizes the formulas of full- and partial-credibility

factors for the four risk measures.

Note that

(i) The standard for full credibility for aggregate loss and that for pure premium are the same.

(ii) The standard for full credibility for aggregate loss is the sum of the standard for claim frequency and

the standard for claim severity.

(iii) The partial credibility is also called the square-root rule. Within the square root, the denominator

is the standard for full credibility of the corresponding risk measure.


Table 1.1: Summary of standards for full-credibility and partial credibility factor Z

Standard for Partial-credibility

Loss measures full credibility factor Z

Claim frequency λF Z =

√λN

λF

Claim severity λF C2X Z =

√N

λF C2X

Aggregate loss and λF + λF C2X Z =

√λN

λF + λF C2XPure premium

Note that λF =(z1−α/2

k

)2

, CX =σX

µX

, and 1 + C2X =

E(X2)

µ2X

.

Table 1.2 shows C2X and 1 + C2

X for some commonly used severity distributions. Since the coefficient of

variation is scale invariant, the scale parameter θ has no effect in calculating CX .

Table 1.2: CX for some commonly used severity distributions

X C2X 1 + C2

X

(Two-parameter) Pareto (α, θ)α

α − 2

2(α − 1)

α − 2

Single-parameter Pareto (α, θ)1

α(α − 2)

(α − 1)2

α(α − 2)

Gamma (α, θ)1

α

α + 1

α

Inverse Gamma (α, θ)1

α − 2

α − 1

α − 2

Inverse Gaussian (µ, λ)µ

λ

λ + µ

λ

Lognormal (µ, σ) eσ2 − 1 eσ2

Uniform in (0, θ) 1/3 4/3

1.4 Problem Set 19

1.4 Problem Set

Question 1.1. * The standard for full credibility for claim frequency for a risk group is 256 and the

standard is based on a coverage probability of 90% and an accuracy parameter k. If the required accuracy

parameter is doubled, what is the new standard for full credibility?

Question 1.2. * (CAS Exam 4B 1986 Spring #34) Using the limited fluctuation credibility approach,

X is the number of claims needed for full credibility for claim frequency. The estimate is to be within

k = 5% of the expected value with a 90% probability. Let Y be the similar number using 10% rather than

5%. What is the ratio of X to Y ?

Question 1.3. * Assume the claim severity has a mean of 594 and a standard deviation of 462. A

sample of 441 claims are observed. What is the probability that the sample mean is within 5% of the true

mean?

Question 1.4. * The standard for full credibility for claim severity for distribution A is 1000 claims for a

given coverage probability and k. Claim distribution B has the same standard deviation as distribution A,

but a mean that is twice as large as A’s. Given the same coverage probability and k, what is the standard

for full credibility for claim severity for distribution B?

Question 1.5. ** Assume the claim severity has a mean of 336 and a standard deviation of 532. A sample

of 361 claims are observed. Within what percentage of the true mean will the sample mean be observed

with a probability of 98%?

Question 1.6. *** (CAS Exam 4B 1987 Spring #35) The number of claims for a company’s major line of

business is Poisson distributed and during the past year, the following claim size distribution was observed:

Claim Size Interval Frequency

0 − 400 20

400− 800 240

800− 1200 320

1200− 1600 210

1600− 2000 100

2000− 2400 60

2400− 2800 30

2800− 3200 10

3200− 3600 10

Total 1000

The mean of this claim size distribution is 1216 and the standard deviation is√

362, 944. You need to select

the number of claims needed to ensure that the observed aggregate loss is within 8% of the expected

value 90% of the time. Using the limited fluctuation credibility approach, how many claims are needed for

full credibility?


Question 1.7. ** (SOA/CAS Exam C 2006 Fall #30) A company has determined that the limited

fluctuation full credibility standard for claim frequency is 2000 claims if the total number of claims is to be

within 3% of the true value with probability 1 − α. The number of claims follows a Poisson distribution.

The standard is changed so that the total cost of claims is to be within 5% of the true value with probability

1 − α. Also, the claim severity has probability density function:

f(x) =1

10, 000, 0 ≤ x ≤ 10, 000

Using limited fluctuation credibility, determine the expected number of claims necessary to obtain full

credibility for aggregate loss under the new standard.

Question 1.8. *** (CAS Exam 4B 1992 Spring #1) You are given the following information:

(i) A standard for full credibility of 1,000 claims has been selected so that the actual total loss costs

(aggregate loss) would be within 10% of the expected total loss costs 95% of the time.

(ii) The number of claims follows a Poisson distribution, and is independent of the severity distribution.

Using the limited fluctuation credibility approach and a normal approximation, determine the coefficient

of variation of the severity distribution underlying the full credibility standard.

Question 1.9. ** Given the following information, what is the minimum number of policies that will be

given full credibility for pure premium?

(i) Mean claim frequency is 0.05 claims per policy. (Assume a Poisson distribution.)

(ii) Mean claim severity is $1,000.

(iii) Variance of the claim severity is $3 million.

(iv) Full credibility is defined as having a 98% probability of being within plus or minus 5% of the true

pure premium.

Question 1.10. ** Using the limited fluctuation credibility approach, X is the number of claims needed

for full credibility for pure premium. The estimate is to be within 5% of the expected value with a 90%

probability. Let Y be the similar number using 10% rather than 5%. Determine X/Y .

Question 1.11. * (CAS 4B 1999 Spring #18) You are given the following:


(ii) The coefficient of variation of the claim size distribution is 4.


(iv) 800 expected claims are needed for full credibility for aggregate loss.

(v) The full credibility standard has been selected so that the observed aggregate loss will be within

100k% of the expected value with probability 1 − α. A normal approximation is used.

Using the methods of limited fluctuation credibility determine the number of expected claims needed for

50% credibility for aggregate loss.

1.4 Problem Set 21

Question 1.12. * The 1984 pure premium underlying the rate equals 1000. The loss experience is such

that the actual pure premium for that year equals 1200 and the number of claims equals 600. If 5400

claims are needed for full credibility and the square root rule for partial credibility is used, estimate the

pure premium underlying the rate in 1985. (Assume no change in the pure premium due to inflation.)

Question 1.13. * (CAS 4B Exam 1986 Spring #35) You are in the process of revising rates.

(i) The premiums currently being used reflect a pure premium per insured of 100. The pure premium

experienced during the two year period used in the rate review averaged 130 per insured.

(ii) The average frequency during the two year review period was 250 claims per year.

(iii) Using a full credibility standard of 2500 claims and assigning partial credibility according to the

limited fluctuation credibility approach, what pure premium per insured should be reflected in the

new rates? (Assume that there is no inflation.)

Question 1.14. ** (SOA/CAS Exam C 2000 Spring #26) You are given:

(i) Claim counts follow a Poisson distribution.

(ii) Claim sizes follow a lognormal distribution with coefficient of variation 3.

(iii) Claim sizes and claim counts are independent.

(iv) The number of claims in the first year was 1000.

(v) The aggregate loss in the first year was 6.75 million.

(vi) The manual premium for the first year was 5.00 million.

(vii) The exposure in the second year is identical to the exposure in the first year.

(viii) The full credibility standard is to be within 5% of the expected aggregate loss 95% of the time.

Determine the limited fluctuation credibility for pure premium (in millions) for the second year.

Question 1.15. **(SOA/CAS Exam C 2003 Fall #03) You are given:

(i) The number of claims has a Poisson distribution.

(ii) Claim sizes have a Pareto distribution with parameters θ = 0.5 and α = 6.


(iv) The observed pure premium should be within 2% of the expected pure premium 90% of the time.

Determine the expected number of claims needed for full credibility for pure premium.


(i) The number of claims has a Poisson distribution.

(ii) Claim sizes have a Single-parameter Pareto distribution with parameters α = 6 and θ = 1000.


(iv) The observed pure premium should be within 2% of the expected pure premium 90% of the time.

Determine the expected number of claims needed for full credibility for pure premium.




(ii) Claim sizes follow a gamma distribution with parameters α (unknown) and θ = 10,000.


(iv) The full credibility standard has been selected so that actual aggregate losses will be within 10% of

expected aggregate losses 95% of the time.

Using limited fluctuation (classical) credibility, determine the expected number of claims required for full

credibility for aggregate loss.

Question 1.18. ***(CAS/SOA Exam C 2005 Fall #35) You are given:


(ii) Claim sizes follow an inverse gamma distribution with parameters α = 3 and θ = 10,000.


(iv) The full credibility standard has been selected so that actual aggregate losses will be within 10% of

expected aggregate losses 95% of the time.

Using limited fluctuation (classical) credibility, determine the expected number of claims required for full

credibility of aggregate loss.

Question 1.19. ***(CAS Exam 4B 1998 Spring #18) You are given the following:


(ii) Claim sizes follow an inverse Gaussian distribution with parameters µ = 10 and θ = 40.


(iv) The full credibility standard has been selected so that actual aggregate claim costs will be within 5%

of expected aggregate claim costs 95% of the time, using a normal approximation.

Using the methods of limited fluctuation credibility, determine the expected number of claims required for

full credibility for aggregate loss.

Question 1.20. ***(CAS Exam 4B 1999 Fall #2) You are given the following:


(ii) Claim sizes follow a lognormal distribution with parameters µ and σ.

(iii) The number of claims and the size of a claim are independent.

(iv) 13,000 claims are needed for full credibility for aggregate loss according to the limited fluctuation

credibility approach.

(v) The full credibility standard has been selected so that the actual aggregate claim costs will be within

5% of expected aggregate claim costs 90% of the time using a normal approximation.

Determine σ.

1.5 Problem Set Solutions 23

1.5 Problem Set Solutions

Question 1.1. We are given λF = 256. If the required accuracy parameter is doubled, then the new

standard for full credibility is

λ′F =

(z1−α/2

k′

)2

=(z1−α/2

2k

)2

=λF

4= 64.

Question 1.2. Using equation (1.4), we have

X =(z1−α/2

k

)2

=(z1−α/2

0.05

)2

.

The second standard has k = 0.10, and thus

Y =(z1−α/2

0.10

)2

.

Therefore the ratio is X/Y = (0.10/0.05)2 = 4.

Question 1.3. We are given µX = 594, σX = 462, and N = 441. The probability that the sample mean

is within k = 5% of the true mean is

Pr(µX − k µX ≤ X ≤ µX + k µX) = Pr

(− k µX

σX/√

N≤ X − µX

σX/√

N≤ k µX

σX/√

N

)

= 2Φ

(k µX

σX/√

N

)− 1 = 2Φ

((0.05)(594)

462/√

441

)− 1

= 2Φ(1.35)− 1 = (2)(0.9115)− 1 = 82.3%.

Question 1.4. For distribution A, we are given

λF (CA)2 = λF(σA)2

(µA)2= 1000.

For distribution B, we have

λF(σB)2

(µB)2= λF

(σA)2

(2µA)2=

1000

4= 250.

Question 1.5. We are given µX = 336, σX = 532, and N = 361. We have α = 1 − 0.98 = 0.02 and thus

z1−α/2 = z0.99 = Φ−1(0.99) = 2.326. Therefore,

z1−α/2 =k µX

σX/√

N=⇒ 2.326 =

k(336)

532/√

361=⇒ k = 19.4%.

Question 1.6. We have α = 1 − 0.9 = 0.1 and thus z1−α/2 = z0.95 = 1.645. We have µX = 1216,

σ2X = 362944, and k = 0.08. So the number of claims needed for full credibility is

λF (1 + C2X ) =

(1.645

0.08

)2 [1 +

362944

(1216)2

]= 527.


Question 1.7. Before the change we have

λF =(z1−α/2

0.03

)2

= 2, 000

After the change we have

λ′F =

(z1−α/2

0.05

)2

= (2, 000)(3/5)2.

The expected number of claims necessary to obtain full credibility for aggregate loss under the new standard

is

λ′F

(1 +

σ2X

µ2X

)= 2, 000(3/5)2

(1 +

1/12

(1/2)2

)= 960.

Note that the coefficient of variation is the same for any uniform distribution in (0, θ). So we can assume

a uniform (0,1) severity for this question. See Table 1.2 for 1 + C2X = 4/3.

Question 1.8. We have α = 1− 0.95 = 0.05 and thus z1−α/2 = z0.975 = Φ−1(0.975) = 1.96. We are given

k = 10% and thus λF = (z0.975/k)2 = (1.96/0.1)2 = 384.16. Using the equation

1, 000 = λF (1 + C2X) = (384.16)(1+ C2

X) =⇒ CX = 1.27.

Question 1.9. We have α = 1 − 0.98 = 0.02 and thus z1−α/2 = z0.99 = 2.326. The minimum number of

claims for full credibility for pure premium is

λF (1 + C2X ) =

(2.326

0.05

)2 [1 +

3, 000, 000

1, 000, 000

]= 8, 657.

The minimum number of policies is 8, 657/0.05 = 173, 140.

Question 1.10. We have

X =(z1−α/2

0.05

)2

(1 + C2X).

The second standard has k = 0.10 rather then k = 0.05, but otherwise it is the same. Thus,

Y =(z1−α/2

0.10

)2

(1 + C2X ).

Therefore the ratio is X/Y = 4.

Question 1.11. Since 800 expected claims are needed for full credibility for aggregate loss, we have

λF (1 + C2X) = 800. The credibility factor is Z = 0.5. Hence,

Z =

√λN

λF (1 + C2X)

=⇒ 0.5 =

√λN

800=⇒ λN = 200

Question 1.12. We are given that 5400 claims are required for full credibility but the actual number of

claims was only 600. Thus the credibility factor is Z =√

600/5400 = 1/3. The observed rate is 1200 so

the new rate will be

U = ZD + (1− Z)M = (1/3)(1200)+ (2/3)(1000) = 1066.67.


Question 1.13. The full credibility standard is given as 2500. The data is based on a total of λN =

(2)(250) = 500 claims, so its credibility is Z =√

500/2500 = 0.4472. For M = 100 and D = 130, the new

rate is U = Z D + (1 − Z)M = (0.4472)(130)+ (1 − 0.4472)(100) = 113.42.

Question 1.14. We have α = 1 − 0.95 = 0.05 and thus z1−α/2 = z0.975 = 1.96. The minimum number of

insureds needed for the aggregate loss to be fully credible is

λF (1 + C2X ) =

(1.96

0.05

)2

(1 + 32) = 15366.4,

Z =

√λN

λF (1 + C2X)

=

√1000

15366.4= 0.255,

U = Z D + (1 − Z) M = (0.255)(6.75)+ (0.745)(5) = 5.45.

Question 1.15. For Pareto distribution, we have

µX =θ

α − 1

E(X2) =2θ2

(α − 1)(α − 2)

1 + C2X =

E(X2)

µ2X

=2(α − 1)

α − 2=

2(6− 1)

6 − 2= 2.5

We have α = 1 − 0.9 = 0.1 and thus z1−α/2 = z0.95 = 1.645. The expected number of claims needed for

full credibility is

λF (1 + C2X ) =

(1.645

0.02

)2

(2.5) = 16913. (rounded up)

Note: The standard for full credibility for the aggregate loss is

λF (1 + C2X ) = λF

E(X2)

µ2X

= λF2(α− 1)

α − 2(if X is Pareto distributed)

Question 1.16. We have α = 1 − 0.9 = 0.1 and thus z1−α/2 = z0.95 = 1.645. For the single-parameter

Pareto distribution, we have

µX =αθ

α − 1

E(X2) =αθ2

α − 2

1 + C2X =

E[X2]

µ2X

=(α − 1)2

α(α − 2)

The expected number of claims needed for full credibility is

λF (1 + C2X) =

(z0.95

k

)2 (α − 1)2

α(α − 2)=

(1.645

0.02

)2 (6− 1)2

(6)(6− 2)= (6765)

25

24= 7047


Note: The standard for full credibility for the aggregate loss/pure premium is

λF

(1 + C2

X

)= λF

E(X2)

µ2X

= λF(α − 1)2

α(α − 2)(if X is single-parameter Pareto distributed)

Question 1.17. For the gamma distribution, we have

µX = αθ

σ2X = αθ2

1 + C2X = 1 +

σ2X

µ2X

=α + 1

α

We have α = 1− 0.95 = 0.05 and thus z1−α/2 = z0.975 = 1.96. The expected number of claims required for

full credibility for aggregate loss is

λF (1 + C2X) =

(z1−α/2

k

)2(

α + 1

α

)=

(1.96

0.1

)2(α + 1

α

).

The value cannot be determined since α is unknown.

Question 1.18. For the inverse gamma distribution we have

E(X) =θ

α − 1,

E(X2) =θ2

(α − 1)(α− 2).

We have α = 1 − 0.95 = 0.05 and thus z1−α/2 = z0.975 = Φ−1(0.975) = 1.96. The expected number of

claims required for full credibility is

λF (1 + C2X ) =

(z1−α/2

k

)2 E[X2]

(E[X ])2=

(1.96

0.1

)2 (α − 1

α − 2

)=

(1.96

0.1

)2

(2) = 769.


λF

(1 +

σ2X

µ2X

)= λF

(α − 1

α − 2

)(if X is inverse gamma distributed)

Question 1.19. We have α = 1 − 0.95 = 0.05 and thus z1−α/2 = z0.975 = Φ−1(0.975) = 1.96. For the

inverse Gaussian distribution, we have

µX = µ,

σ2X = µ3/θ,

C2X =

σ2X

µ2X

=µ

θ=

10

40= 2.5.


The full credibility standard in terms of claims is

λF (1 + C2X ) =

(z1−α/2

k

)2

(1 + 2.5) =

(1.96

0.05

)2

(3.5) = 5, 378


λF (1 + C2X ) =

(z1−α/2

k

)2 (1 +

µ

θ

)(if X is inverse Gaussian distributed)

Question 1.20. We have α = 1 − 0.9 = 0.1 and thus z1−α/2 = z0.95 = Φ−1(0.95) = 1.645. For the

lognormal distribution, we have

E(X) = eµ+σ2/2,

E(X2) = e2µ+2σ2.

Solving σ from

13, 000 = λFE(X2)

E(X)2=

(1.645

0.05

)2

eσ2

we have σ = 1.5.


λF (1 + C2X ) =

(z1−α/2

k

)2 E(X2)

[E(X)]2

=(z1−α/2

k

)2

eσ2(if X is lognormal distributed)

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