Hello, and welcome to this tutorial on Exam C, problem 032. This question falls under the learning objective "Credibility". Here is a view of the problem. Please complete this problem and select one of the five answers. If incorrect, you can click the gray X to try again. Here is a solution to the problem.
In this question we are asked to estimate the number of claims within the next year, given two years of claim totals. The distribution for number of claims is Poisson distributed with parameter lambda. Lambda is gamma distributed with parameters alpha=1 and theta=1.2. We will go through two ways of solving this problem. The first we will use is a shortcut for the Poisson/Gamma conjugate prior. The second method will be completing the problem without the shortcut.
A conjugate prior results when the prior and the posterior distributions are from the same family. In this problem, we can use the posterior distribution to estimate the number of claims for year 3, if we know what it is. The alignment of Poisson and Gamma distributions in this problem is a very common combination, and it's a good idea to memorize the shortcut we are about to use. For a Poisson distribution with a gamma-distributed lambda parameter, we can reparametrize to get the posterior distribution. The posterior distribution is gamma distributed for this combination, with the following parameters. Alpha is equal to the prior alpha plus the number of experienced claims, which is denoted here with x. It is given that there were three claims so far, so alpha equals 4. For the posterior theta, I think it is easier to first calculate 1/theta. 1/theta is equal to 1/(prior theta) plus the number of observations in the sample, which in this question is sample years. Then to find theta, we take the reciprocal, calculating a value of 1/2.8333. Now we have the parameters of the posterior distribution, so we can quickly solve for the answer. Finding the expected value of a gamma distribution from the tables, we can calculate the expected number of claims next year; 1.4118. This is answer choice D.
We can also do this problem without the shortcut. We can calculate the sample mean first; there were three claims in two years, so the average number of claims per year was 1.5. Now we must calculate the hypothetical mean, the variance of the hypothetical mean, and the expected process variance. The number of claims is Poisson distributed; we can ascertain from the inventory of discrete distributions that the mean and variance are equal. So the hypothetical mean, mu, and the expected process variance, v, are both equal to the expected value of lambda, which is 1.2. To calculate the variance of the expectation of lambda, we use the variance of a gamma function. This is not given in the tables, but is commonly used; it is another good thing to memorize. The variance is alpha times theta-squared. Our value of a is equal to 1.44.
We are now able to calculate the value of Buehlmann's k. v/a equals 5/6. The credibility factor, Z, that we can assign to our sample, is equal to n/(n+k). As shown, this value equals 12/17. Weighting the sample mean with the credibility factor, and weighting the hypothetical mean with the credibility complement, we solve for the expected number of claims, 1.4118, which matches the value we got using the shortcut.
That's all for this tutorial. Thanks for watching!