Loss Models: From Data to Decisions, 4e Student Solutions Manual

Stuart A. Klugman, Harry H. Panjer, Gordon E. Willmot

• Éditeur: Wiley
• Année: 2014
• ISBN:
  9781118472026
• Édition: 4
• Sujet: Economía
• Sous-sujet: Estadísticas para los negocios y la economía

Student Solutions Manual to Accompany Loss Models: From Data to Decisions, Fourth Edition. This volume is organised around the principle that much of actuarial science consists of the construction and analysis of mathematical models which describe the process by which funds flow into and out of an insurance system.

CHAPTER 1

INTRODUCTION

The solutions presented in this manual reflect the authors’ best attempt to provide insights and answers. While we have done our best to be complete and accurate, errors may occur and there may be more elegant solutions. Errata will be posted at the ftp site dedicated to the text and solutions manual: ftp://ftp.wiley.com/public/sci_tech_med/loss_models/

Should you find errors or would like to provide improved solutions, please send your comments to Stuart Klugman at [email protected].

CHAPTER 2

CHAPTER 2 SOLUTIONS

2.1 SECTION 2.2

2.1

2.2 The requested plots follow. The triangular spike at zero in the density function for Model 4 indicates the 0.7 of discrete probability at zero.

2.3 f′(x) = 4(1 + x²)^–3 – 24x²(l + x²)^–4. Setting the derivative equal to zero and multiplying by (1 + x²)⁴ give the equation 4(1 + x²) – 24x² = 0. This is equivalent to x² = 1/5. The only positive solution is the mode of

.

2.4 The survival function can be recovered as

Taking logarithms gives

and thus A = 0.2009.

2.5 The ratio is

From observation or two applications of L’Hôpital’s rule, we see that the limit is infinity.

CHAPTER 3

CHAPTER 3 SOLUTIONS

3.1 SECTION 3.1

3.1

3.2 For Model 1, σ² = 3,333.33 – 50² = 833.33, σ = 28.8675.

For Model 2, σ² = 4,000,000 – 1,000² = 3,000,000, σ = 1,732.05.

and

are both infinite so the skewness and kurtosis are not defined.

For Model 3, σ² = 2.25 – .93² = 1.3851, σ = 1.1769.

For Model 4, σ² = 6,000,000,000 – 30,000² = 5,100,000,000, σ = 71,414.

For Model 5, σ² = 2,395.83 – 43.75² = 481.77, σ = 21.95.

3.3 The Standard deviation is the mean times the coefficient, of Variation, or 4, and so the variance is 16. From (3.3) the second raw moment is 16 + 2² = 20. The third central moment is (using Exercise 3.1) 136 – 3(20)(2) + 2(2)³ = 32. The skewness is the third central moment divided by the cube of the Standard deviation, or 32/4³ = 1/2.

3.4 For a gamma distribution the mean is αθ. The second raw moment is α(α + 1)θ², and so the variance is αθ². The coefficient of Variation is

/αθ = α^–1/2 = 1. Therefore α = 1. The third raw moment is α(α + 1)(α + 2)θ³ = 6θ³. From Exercise 3.1, the third central moment is 6θ³ – 3(2θ²)θ + 2θ³ = 2θ³ and the skewness is 2θ³/(θ²)^3/2 = 2.

3.5 For Model 1,

For Model 2,...