Complete the table in Example 3.2.2 for x = 0,...,6. Determine the expected value and the variance of N, X and S. Example 3.2.2 (Application of the convolution formula) Let Pr[N=j-1]=j/10 for j=1,2,3,4, and let p(1) = 0.4, p(2) = 0.6. By using (3.10), F(x) can be calculated as follows: X p0 (x) p*(x) p*(x) p*(x) 012345 0 : 1 Ix Pr[N=n] 0.1 0.4 0.6 0.16 0.48 0.064 0.36 0.288 0.432 : +Ix 0.4 +1x +1x 0.2 0.3 = f(x) 0.1000 0.0800 0.1680 0.1696 F(x) 0.1000 0.1800 0.3480 0.5176 : The probabilities Pr[N = n] in the bottom row are multiplied by the numbers in a higher row. Then, the sum of these results is put in the corresponding row in the column f(x). For example: 0.2 x 0.6+0.3 x 0.16 0.168. V Note that if we attempt convolution in case of arbitrary discrete claim sizes rather than integer-valued ones such as here, the number of possible values and the re- quired number of computations increase exponentially.