Estimation, Evaluation, and Selection of Actuarial Models

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The total of the last column is the test statistic of 2.85. With three degrees of freedom (four
groups less one less zero estimated parameters) the critical value is 7.81 and the null hypothesis of
a Poisson distribution cannot be rejected.
Exercise 86 These are discrete data from a discrete population, so the normal and gamma models
are not appropriate. There are two ways to distinguish among the three discrete options. One
is to look at successive values of kn k /n k−1 . They are 2.67, 2.33, 2.01, 1.67, 1.32, and 1.04. The
sequence is linear and decreasing, indicating a binomial distribution is appropriate. An alternative
is to compute the sample mean and variance. They are 2 and 1.494 respectively. The variance is
considerably less than the mean, indicating a binomial model.
Exercise 87 The various tests for the three data sets produce the following results. For Data
Set B truncated at 50, the estimates are ˆα =0.40982, ˆτ =1.24069, andˆθ =1, 642.31. For Data
Set B censored at 1,000 there is no maximum likelihood estimate. For Data Set C, the maximum
likelihood estimate is ˆα =4.50624, ˆτ =0.28154, andˆθ =71.6242.
B truncated at 50 B censored at 1,000
Criterion Exponential Weibull Trans gam Exponential Weibull Trans gam
K-S 0.1340 0.0887 0.0775 0.0991 0.0991 N/A
A-D 0.4292 0.1631 0.1649 0.1713 0.1712 N/A
χ 2 1.4034 0.3615 0.5169 0.5951 0.5947 N/A
p-value 0.8436 0.9481 0.7723 0.8976 0.7428 N/A
loglikelihood −146.063 −145.683 −145.661 −113.647 −113.647 N/A
SBC −147.535 −148.628 −150.078 −115.145 −116.643 N/A
C
Criterion Exponential Weibull Trans gam
K-S N/A N/A N/A
A-D N/A N/A N/A
χ 2 61.913 0.3698 0.3148
p-value 10 −12 0.9464 0.8544
loglikelihood −214.924 −202.077 −202.046
SBC −217.350 −206.929 −209.324
For Data Set B truncated at 50, there is no reason to use a three-parameter distribution. For
Data Set C, the transformed gamma distribution does not provide sufficient improvement to drop
the Weibull as the model of choice.
Exercise 88 The loglikelihood values for the two models are −385.9 for the Poisson and −382.4 for
the negative binomial. The test statistic is 2(−382.4 + 385.9) = 7.0. There is one degree of freedom
(two parameters minus one parameter) and so the critical value is 3.84. The null hypothesis is
rejected and so the data favors the negative binomial distribution.
Exercise 89 The penalty function subtracts ln(100)/2 =2.3 for each additional parameter. For
the five models, the penalized loglikelihoods are – generalized Pareto: −219.1 − 6.9 =−226.0,
Burr: −219.2 − 6.9 =−226.1, Pareto: −221.2 − 4.6 =−225.8, lognormal: −221.4 − 4.6 =−226.0,
and inverse exponential: −224.3 − 2.3 =−226.6. The largest value is for the Pareto distribution.