Estimation, Evaluation, and Selection of Actuarial Models

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78 CHAPTER 4. MODEL EVALUATION AND SELECTION The total is χ 2 = 1.4034. With four degrees of freedom (6 rows minus 1 minus 1 estimated parameter) the critical value for a test at a 5% significance level is 9.4877 (this can be obtained with the Excel function CHIINV(.05,4)) and the p-value is 0.8436 (from CHIDIST(1.4034,4)). The exponential model is a good fit. For Data Set B censored at 1,000, the first interval is from 0—150 and the last interval is from 1,000—∞. Unlike the previous two tests, the censored observations can be used for this test. Range ˆp Expected Observed χ 2 0—150 0.1885 3.771 4 0.0139 150—250 0.1055 2.110 3 0.3754 250—500 0.2076 4.152 4 0.0055 500—1,000 0.2500 5.000 4 0.2000 1,000—∞ 0.2484 4.968 5 0.0002 The total is χ 2 =0.5951. With three degrees of freedom (5 rows minus 1 minus 1 estimated parameter) the critical value for a test at a 5% significance level is 7.8147 and the p-value is 0.8976. The exponential model is a good fit. For Data Set C the groups are already in place. The results are given below. Range ˆp Expected Observed χ 2 7,500—17,500 0.2023 25.889 42 10.026 17,500—32,500 0.2293 29.356 29 0.004 32,500—67,500 0.3107 39.765 28 3.481 67,500—125,000 0.1874 23.993 17 2.038 125,000—300,000 0.0689 8.824 9 0.003 300,000—∞ 0.0013 0.172 3 46.360 The test statistic is χ 2 =61.913. There are four degrees of freedom for a critical value of 9.488. The p-value is about 10 −12 . There is clear evidence that the exponential model is not appropriate. A more accurate test would combine the last two groups (because the expected count in the last group is less than 1). The group from 125,000 to infinity has an expected count of 8.997 and an observed count of 12 for a contribution of 1.002. The test statistic is now 16.552 and with three degrees of freedom the p-value is 0.00087. The test continues to reject the exponential model. ¤ Exercise 82 Repeat the above example for the Weibull model. There is one important point to note about the tests in this Section. Suppose the sample size were to double, but sampled values weren’t much different (imagine each number showing up twice instead of once). For the Kolmogorov-Smirnov test, the test statistic would be unchanged, but the critical value would be smaller. For the Anderson-Darling and chi-square tests, the test statistic would double while the critical value would be unchanged. As a result, for larger sample sizes, it is more likely that the null hypothesis (and thus the proposed model) will be rejected. This should not be surprising. We know that the null hypothesis is false (it is extremely unlikely that a simple distribution using a few parameters can explain the complex behavior that produced the observations) and with a large enough sample size we will have convincing evidence of that truth. When using these tests we must remember that although all our models are wrong, some may be useful.
4.6. SELECTING A MODEL 79 Exercise 83 (*) 150 policyholders were observed from the time they arranged a viatical settlement until their death. No observations were censored. There were 21 deaths in the first year, 27 deaths in the second year, 39 deaths in the third year, and 63 deaths in the fourth year. The survival model t(t +1) S(t) =1− , 0 ≤ t ≤ 4 20 is being considered. At a 5% significance level, conduct the chi-square goodness-of-fit test. Exercise 84 (*) Each day, for 365 days, the number of claims is recorded. The results were 50 days with no claims, 122 days with 1 claim, 101 days with 2 claims, 92 days with 3 claims, and no days with 4 or more claims. For a Poisson model determine the maximum likelihood estimate of λ and then perform the chi-square goodness-of-fit testata2.5%significance level. Exercise 85 (*) During a one-year period, the number of accidents per day was distributed as follows: No. of accidents Days 0 209 1 111 2 33 3 7 4 3 5 2 Test the hypothesis that the data are from a Poisson distribution with mean 0.6 using the maximum number of groups such that each group has at least five expected observations. Use a significance level of 5%. 4.6 Selecting a model 4.6.1 Introduction Almost all of the tools are in place for choosing a model. Before outlining a recommended approach, two important concepts must be introduced. The first is parsimony. The principle of parsimony states that unless there is considerable evidence to do otherwise, a simpler model is preferred. The reason is that a complex model may do a great job of matching the data, but that is no guarantee the model matches the population from which the observations were sampled. For example, given any set of 10 (x, y) pairs with unique x-values, there will always be a polynomial of degree 9 or less that goes through all 10 points. But if these points were a random sample, it is highly unlikely that the population values all lie on that polynomial. However, there may be a straight line that comes close to the sampled points as well as the other points in the population. This matches the spirit of most hypothesis tests. That is, do not reject the null hypothesis (and thus claim a more complex description of the population holds) unless there is strong evidence to do so. The second concept doesn’t have a name. It states that if you try enough models, one will look good, even if it isn’t. Suppose I have 900 models at my disposal. For most data sets, it is likely that one of them will fit well, but this does not help us learn about the population. Thus, in selecting models, there are two things to keep in mind:
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