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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 6

CHAPTER 6 SOLUTIONS

6.1   SECTION 6.1

6.1

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6.2   SECTION 6.5

6.2 For Exercise 14.3, the values at k = 1, 2, 3 are 0.1000, 0.0994, and 0.1333, which are nearly constant. The Poisson distribution is recommended. For Exercise 14.5, the values at k = 1, 2, 3, 4 are 0.1405, 0.2149, 0.6923, and 1.3333, which is increasing. The geometric/negative binomial is recommended (although the pattern looks more quadratic than linear).

6.3 For the Poisson, λ > 0 and so it must be a = 0 and b > 0. For the binomial, m must be a positive integer and 0 < q < 1. This requires a < 0 and b > 0 provided −b/a is an integer ≥ 2. For the negative binomial, both r and β must be positive so a > 0 and b can be anything provided b/a > −1.

The pair a = −1 and b = 1.5 cannot work because the binomial is the only possibility but −b/a = 1.5, which is not an integer. For proof, let p0 be arbitrary. Then p1 = (−1 + 1.5/1)p = 0.5p and p2 = (−1 + 1.5/2)(0.5p) = −0.125p < 0.

6.3   SECTION 6.6

6.4

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The factors will be positive (and thus pk will be positive) provided p1 > 0, β > 0, r > −1, and r ≠ 0.

To see that the probabilities sum to a finite amount,

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The terms of the summand are the pf of the negative binomial ...

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