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Exam MAS-II Study Manual
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Spring 2019 Edition
Jeffrey S. Pai, Ph.D., ASA, ACIA
ACTEX
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
2.5 Problem Set Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
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
3.6 Problem Set Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
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
4.5 Problem Set Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
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
5.7 Problem Set Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126
6 Three-level Models for Clustered Data 129
6.1 Classroom Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130
6.2 Model Specification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132
6.3 Hypothesis Tests and Model Selection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136
6.4 Intraclass Correlation Coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142
6.5 Calculating Predicted Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143
6.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144
6.7 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145
6.8 Problem Set Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148
7 Models for Repeated-Measures Data: The Rat Brain Example 151
7.1 Rat Brain Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151
7.2 Model Specification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153
7.3 Hypothesis Tests and Model Selection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155
7.4 Interpretation and Residual Diagnostics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157
7.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160
7.6 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161
7.7 Problem Set Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164
8 Random Coefficient Models for Longitudinal Data: The Autism Example 167
8.1 Autism Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167
8.2 Model Specification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170
8.3 Hypothesis Tests and Model Selection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171
8.4 Interpretation, Residual Diagnostics, and Predictions . . . . . . . . . . . . . . . . . . . . . 174
8.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181
CONTENTS v
8.6 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182
8.7 Problem Set Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184
9 Models for Clustered Longitudinal Data: The Dental Veneer Example 185
9.1 Veneer Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186
9.2 Model Specification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186
9.3 Hypothesis Tests and Model Selection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187
9.4 Interpretation and Residual Diagnostics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190
9.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193
9.6 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194
9.7 Problem Set Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197
10 Models for Data with Crossed Random Factors: The SAT Score Example 199
10.1 SAT Score Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200
10.2 Model Specification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203
10.3 Hypothesis Tests and Model Selection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203
10.4 Interpretation and Residual Diagnostics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206
10.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209
10.6 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210
10.7 Problem Set Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212
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
11.5 Problem Set Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231
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
12.7 Problem Set Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247
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
13.6 Problem Set Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261
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
14.5 Problem Set Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275
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
15.8 Problem Set Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301
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
16.5 Problem Set Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325
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
17.7 Problem Set Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 351
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
18.5 Problem Set Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 367
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.
Solution. Using equation (1.1), we have
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
8 CHAPTER 1. LIMITED FLUCTUATION CREDIBILITY
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.
Solution. Using equation (1.2), we have
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
Pr(µN − k µN ≤ N ≤ µN + k µN) = 2Φ
(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
Pr(µN − k µN ≤ N ≤ µN + k µN) = 2Φ
(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.
1.1 Full Credibility 9
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
10 CHAPTER 1. LIMITED FLUCTUATION CREDIBILITY
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.
7Nonlife Actuarial Models, Example 6.5
1.1 Full Credibility 11
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.
8Nonlife Actuarial Models, Example 6.7
12 CHAPTER 1. LIMITED FLUCTUATION CREDIBILITY
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.
1.1 Full Credibility 13
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.
14 CHAPTER 1. LIMITED FLUCTUATION CREDIBILITY
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.
9Nonlife Actuarial Models, Example 6.10
16 CHAPTER 1. LIMITED FLUCTUATION CREDIBILITY
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.
10Nonlife Actuarial Models, Example 6.11
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.
18 CHAPTER 1. LIMITED FLUCTUATION CREDIBILITY
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?
20 CHAPTER 1. LIMITED FLUCTUATION 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:
(i) The number of claims follows a Poisson distribution.
(ii) The coefficient of variation of the claim size distribution is 4.
(iii) The number of claims and claim sizes are independent.
(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.
(iii) The number of claims and claim sizes are independent.
(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.
Question 1.16. **(SOA/CAS Exam C 2003 Fall #03) You are given:
(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.
(iii) The number of claims and claim sizes are independent.
(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.
22 CHAPTER 1. LIMITED FLUCTUATION CREDIBILITY
Question 1.17. **(SOA/CAS Exam C 2005 Fall #35) You are given:
(i) The number of claims follows a Poisson distribution.
(ii) Claim sizes follow a gamma distribution with parameters α (unknown) and θ = 10,000.
(iii) The number of claims and claim sizes are independent.
(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:
(i) The number of claims follows a Poisson distribution.
(ii) Claim sizes follow an inverse gamma distribution with parameters α = 3 and θ = 10,000.
(iii) The number of claims and claim sizes are independent.
(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:
(i) The number of claims follows a Poisson distribution.
(ii) Claim sizes follow an inverse Gaussian distribution with parameters µ = 10 and θ = 40.
(iii) The number of claims and claim sizes are independent.
(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:
(i) The number of claims follows a Poisson distribution.
(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.
24 CHAPTER 1. LIMITED FLUCTUATION CREDIBILITY
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.
1.5 Problem Set Solutions 25
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
26 CHAPTER 1. LIMITED FLUCTUATION CREDIBILITY
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.
Note: The standard for full credibility for the aggregate loss is
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.
1.5 Problem Set Solutions 27
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
Note: The standard for full credibility for the aggregate loss is
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.
Note: The standard for full credibility for the aggregate loss is
λF (1 + C2X ) =
(z1−α/2
k
)2 E(X2)
[E(X)]2
=(z1−α/2
k
)2
eσ2(if X is lognormal distributed)