Chapter 3
Solutions

Section 3.1

1. 3.1
2. 3.2 For Model 1, , . , and , . , .

   For Model 2, , and . and are both infinite, so the skewness and kurtosis are not defined.

   For Model 3, and . , , , , , .

   For Model 4, and . , , . , , .

   For Model 5, and . , , . , , .

3. 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 . The third central moment is (using Exercise 3.1) . The skewness is the third central moment divided by the cube of the standard deviation, or .
4. 3.4 For a gamma distribution, the mean is . The second raw moment is , and so the variance is . The coefficient of variation is . Therefore . The third raw moment is . From Exercise 3.1, the third central moment is and the skewness is .
5. 3.5 For Model 1,

   For Model 2,

   For Model 3,

   For Model 4,

   The functions are straight lines for Models 1, 2, and 4. Model 1 has negative slope, Model 2 has positive slope, and Model 4 is horizontal.

6. 3.6 For a uniform distribution on the interval from 0 to w, the density function is . The mean residual life is

   The equation becomes

   with a solution of .

7. 3.7 From the definition,
8. 3.8
9. 3.9 For Model 1, from (3.8),

   and from (3.10),

   From (3.9),

   For Model 2, from (3.8),

   and from (3.10),

   From (3.9),

   For Model 3, from (3.8),

   and from (3.10),

   For Model 4, from (3.8),

   and from (3.10),

10. 3.10 For a discrete distribution (which all empirical distributions are), the mean residual life function is

    When d is equal to a possible value of X, the function cannot be continuous because there is a jump in the denominator but not in the numerator. For an exponential distribution, argue as in Exercise 3.7 to see that it is constant. For the Pareto distribution,

    which is increasing in d. Only the second statement is true.

11. 3.11 Application of the formula from the solution to Exercise 3.10 gives

    which cannot be correct. Recall that the numerator of the mean residual life is . However, when , the expected value is infinite and so is the mean residual life.

12. 3.12 The right truncated variable is defined as , given that . When , this variable is not defined. The kth moment is
13. 3.13 This is a single-parameter Pareto distribution with parameters and . The moments are and . The coefficient of variation is .
14. 3.14 . . . . Skewness is . Kurtosis is .
15. 3.15 The Pareto mean residual life function is

    and so .

16. 3.16 Sample mean: . Sample variance: . Sample third central moment: . Skewness coefficient: .

Section 3.2

1. 3.17 The pdf is . The mean is . The median is the solution to , which is 1.4142. The mode is the value at which the pdf is highest. Because the pdf is strictly decreasing, the mode is at its smallest value, 1.
2. 3.18 For Model 2, solve , and so , and the requested percentiles are 519.84 and 1419.95.

   For Model 4, the distribution function jumps from 0 to 0.7 at zero and so . For percentiles above 70, solve , and so and .

   For Model 5, the distribution function has two specifications. From to , it rises from 0.0 to 0.5, and so for percentiles at 50 or below, the equation to solve is for . For , the distribution function rises from 0.5 to 1.0, and so for percentiles from 50 to 100 the equation to solve is for . The requested percentiles are 50 and 65.

3. 3.19 The two percentiles imply

   Rearranging the equations and taking their ratio yields

   Taking logarithms of both sides gives for .

4. 3.20 The two percentiles imply

   Subtracting and then taking logarithms of both sides gives

   Dividing the second equation by the first gives

   Finally, taking logarithms of both sides gives for .

Section 3.3

1. 3.21 The sum has a gamma distribution with parameters and . Then . From the central limit theorem, the sum has an approximate normal distribution with mean and variance for a standard deviation of 1,000. The probability of exceeding 6,000 is .
2. 3.22 A single claim has mean and variance

   The sum of 100 claims has mean 480,000 and variance 9,216,000,000, which is a standard deviation of 96,000. The probability of exceeding 600,000 is approximately

3. 3.23 The mean of the gamma distribution is and the variance is . For 100 independent claims, the mean is 500,000 and the variance is 500,000,000 for a standard deviation of 22,360.68. The probability of total claims exceeding 525,000 is
4. 3.24 The sum of 2,500 contracts has an approximate normal distribution with mean and standard deviation . The answer is

Section 3.4

1. 3.25 While the Weibull distribution has all positive moments, for the inverse Weibull moments exist only for . Thus, by this criterion, the inverse Weibull distribution has a heavier tail. With regard to the ratio of density functions, it is (with the inverse Weibull in the numerator, and with asterisks (*) used to mark its parameters)

   The logarithm is

   The middle term goes to zero, so the issue is the limit of which is clearly infinite. With regard to the hazard rate, for the Weibull distribution we have

   which is clearly increasing when , constant when , and decreasing when . For the inverse Weibull,

   The derivative of the denominator is

   and the limiting value of this expression is . Therefore, in the limit, the denominator is increasing and thus the hazard rate is decreasing.

   Figure 3.1 displays a portion of the density function for Weibull and inverse Weibull distributions with the same mean and variance. The heavier tail of the inverse Weibull distribution is clear.

   Figure 3.1 The tails of a Weibull and an inverse Weibull distribution.

2. 3.26 Means:

   Second moments:

   Density functions:

   The gamma and lognormal densities are equal when , while the lognormal and Pareto densities are equal when . Numerical evaluation indicates that the ordering is as expected.

3. 3.27 For the Pareto distribution,

   Thus is clearly increasing and , indicating a heavy tail.

   The square of the coefficient of variation is

   which is greater than 1, also indicating a heavy tail.

4. 3.28

   This result assumes that . An application of L'Hôpital's rule shows that this is the same limit as . This limit must be zero, otherwise the integral defining will not converge.

5. 3.29
   1. For ,

      Thus,

   2. from (a) and (c).

   3. Using (b),
   4. Using (d), we have

      and so

      The derivative is equal to zero only at .

      Also, from (b) and (e), , and .

6. 3.30
   1. Integration by parts yields

      and hence , from which the result follows.

   2. It follows from (a) that

      which gives the result by addition of to both sides.

   3. Because , from (b),

      and the first result follows by division of both sides by . The inequality then follows from .

   4. Because , the result follows from the inequality in (c), with replaced by in the denominator.
   5. As in (c), it follows from (b) that

      that is,

      from which the first result follows by solving for . The inequality then follows from the first result, since .

Section 3.5

1. 3.31 Denote the risk measures by , , and . Note that and

   where here is the correlation coefficient, not the risk measure, and must be less than or equal to one. Therefore, . Thus,

   which establishes subadditivity.

   Because and , , which establishes positive homogeneity.

   Because and , , which establishes translation invariance.

   For the example given, note that for all possible pairs of outcomes. Also, , , , and . With , , which is greater than , violating monotonicity.

2. 3.32 The cdf is , and so the percentile solves and the solution is . Because the exponential...