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Student Solutions Manual to Accompany Loss Models: From Data to Decisions, Fourth Edition by Gordon E. Willmot, Harry H. Panjer, Stuart A. Klugman

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CHAPTER 3

CHAPTER 3 SOLUTIONS

3.1   SECTION 3.1

3.1 image

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

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For Model 2, σ2 = 4,000,000 − 1,0002 = 3,000,000, σ = 1,732.05. μ3 and μ4 are both infinite so the skewness and kurtosis are not defined.

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

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

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

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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 ...

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