ct010s08 - UNSW ACTL1001 Actuarial Studies and Commerce...

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Unformatted text preview: UNSW ACTL1001 Actuarial Studies and Commerce Sample Solutions Exercises 8 Exercise 1 In this case 1. For the individual risks with random loss L E [L] = 150; 000 0:05 + 75; 000 0:05 + 0 0:9 = 11; 250 V ar [L] = (150; 000 + (75; 000 11; 250)2 2 11; 250) 0:05 0:05 2 + (0 11; 250) 0:9 = 1; 279; 687; 500 or a standard deviation of 35,772.7 2. See following table (note that there are faster ways of doing this by using binomial). There are 27 possibilities for the combinations of possible losses for the 3 risks. The expected value is 11,250 (as is expected) and the variance is 426,562,500 or a standard deviation of 20,653.4. 1 Con…rming that a reduction in variance has resulted from risk pooling. Loss 1 150000 150000 150000 75000 150000 150000 150000 75000 75000 0 150000 150000 75000 75000 75000 0 0 150000 75000 75000 0 0 0 75000 0 0 0 Loss 2 150000 150000 75000 150000 150000 75000 0 150000 75000 150000 75000 0 150000 75000 0 150000 75000 0 75000 0 150000 75000 0 0 75000 0 0 Loss 3 150000 75000 150000 150000 0 75000 150000 75000 150000 150000 0 75000 0 75000 150000 75000 150000 0 0 75000 0 75000 150000 0 0 75000 0 Averag Prob 1 Prob 2 Prob 3 Prob 150000 0.05 0.05 0.05 0.0001 125000 0.05 0.05 0.05 0.0001 125000 0.05 0.05 0.05 0.0001 125000 0.05 0.05 0.05 0.0001 100000 0.05 0.05 0.9 0.0023 100000 0.05 0.05 0.05 0.0001 100000 0.05 0.9 0.05 0.0023 100000 0.05 0.05 0.05 0.0001 100000 0.05 0.05 0.05 0.0001 100000 0.9 0.05 0.05 0.0023 75000 0.05 0.05 0.9 0.0023 75000 0.05 0.9 0.05 0.0023 75000 0.05 0.05 0.9 0.0023 75000 0.05 0.05 0.05 0.0001 75000 0.05 0.9 0.05 0.0023 75000 0.9 0.05 0.05 0.0023 75000 0.9 0.05 0.05 0.0023 50000 0.05 0.9 0.9 0.0405 50000 0.05 0.05 0.9 0.0023 50000 0.05 0.9 0.05 0.0023 50000 0.9 0.05 0.9 0.0405 50000 0.9 0.05 0.05 0.0023 50000 0.9 0.9 0.05 0.0405 25000 0.05 0.9 0.9 0.0405 25000 0.9 0.05 0.9 0.0405 25000 0.9 0.9 0.05 0.0405 0 0.9 0.9 0.9 0.729 1 E(AveL) var(AveL) 18.75 2406445 15.63 1617383 15.63 1617383 15.63 1617383 225.00 17722266 12.50 984570 225.00 17722266 12.50 984570 12.50 984570 225.00 17722266 168.75 9144141 168.75 9144141 168.75 9144141 9.38 508008 168.75 9144141 168.75 9144141 168.75 9144141 2025.00 60813281 112.50 3378516 112.50 3378516 2025.00 60813281 112.50 3378516 2025.00 60813281 1012.50 7657031 1012.50 7657031 1012.50 7657031 0.00 92264063 11250 426,562,500 20653.4 We can summarise the distribution of the average loss as follows: Average 150000.0 125000.0 100000.0 75000.0 50000.0 25000.0 0.0 Prob ln(1+W)E(U(W)) 0.000125 18.75 0 0 0.000375 46.875 10.127 0.003798 0.007125 712.5 10.82 0.077091 0.013625 1021.875 11.225 0.152944 0.12825 6412.5 11.513 1.476534 0.1215 3037.5 11.736 1.425933 0.729 0 11.918 8.688512 11.82481 1 11250 2 3. Before pooling, the expected utility will be E (U ) = ln (1 + 150; 000 + ln (1 + 150; 000 + ln (1 + 150; 000 = 11:2878 150; 000) 0:05 75; 000) 0:05 0) 0:9 After pooling, the expected utility will be E (U ) = ln (1 + 150; 000 150; 000) 0:000125 + ln (1 + 150; 000 125; 000) :000375 + ln (1 + 150; 000 100; 000) 0:007125 + : : : + ln (1 + 150; 000 0) 0:729 = 11:8248 Note that if we replaced the average loss with its long run average as the cost charged to each individual, as would be the case in an insurance company, we would get expected utility of ln (1 + 150; 000 11; 250) = ln (138; 751) = 11:8404 Exercise 2 We use the results from the lecture and text that E [A] = qL and V ar [A] = q ) L2 q (1 n to get Number 10 100 1000 10000 Expected value Variance 10 9990 10 999 10 99.9 10 9.99 Exercise 3 We have 3 1. The expected value for Class A is A = 1; 000 = 950 0:2 + 1; 000 0:7 + 500 0:1 The variance for Class A is = (1; 000 950)2 0:2 + (1; 000 2 A 2 0:7 950) + (500 950)2 = 22; 500 0:1 or a standard deviation of 150. The expected value for Class B is B = 900 = 850 0:2 + 900 0:7 + 400 0:1 The variance for Class B is = (900 850)2 0:2 + (900 2 0:7 2 2 B 0:1 850) + (400 850) = 22; 500 or a standard deviation of 150. Thus both have the same variance but B has a lower expected value. 2. The covariance between A and B is 2 3 (1; 000 950) (900 850) 0:2 cov (A; B ) = 4 + (1; 000 950) (900 850) 0:7 5 + (500 950) (400 850) 0:1 = 22; 500 and we note that the correlation will be +1 (i.e. perfect positive correlation). 4 Exercise 4 Results below 1. To calculate the probability of being on di¤erent discount levels for the …fth year we need to raise the transition probability matrix to the power of 5. Firstly the probability of no claims (and moving to the next higher level in a year) is e 0:1 = 0:904837 and the probability of one or more claims is 0.095163. The transition matrix will be NCD level 0 1 2 0 0.095163 0.904837 0 1 0.095163 0 0.904837 2 0 0.095163 0.904837 In Excel we can multiply matrices using the MMULT(,) function. To do this, select an area for the results the size of the resulting matrix ( in this case 3x3). Then enter =mmult(array1, array2) and then press CTRL, SHIFT, ENTER (not just ENTER) and the selected area (3X3) will have the resulting matrix in it. In this case array1 is just the transition matrix and array2 is the same matrix so we multiply the matrix by itself to get the square of the matrix and so on to the 5 th power in this case. The results are given in this case below 5 Transition matrix Prob of no claims 0.904837 NCD level 0 1 2 0 0.095163 0.904837 0 1 0.095163 0 0.904837 2 0 0.095163 0.904837 Year 2 0.095163 0.086107 0.818731 0.009056 0.172213 0.818731 0.009056 0.086107 0.904837 Year 3 0.01725 0.164019 0.818731 0.01725 0.086107 0.896643 0.009056 0.094301 0.896643 Year 4 0.01725 0.093521 0.889229 0.009836 0.100935 0.889229 0.009836 0.093521 0.896643 Year 5 0.010541 0.10023 0.889229 0.010541 0.093521 0.895938 0.009836 0.094227 0.895938 Power 10 0.009914 0.094219 0.895867 0.009909 0.094224 0.895867 0.009909 0.094219 0.895872 Power 20 0.009909 0.009909 0.009909 0.09422 0.895871 0.09422 0.895871 0.09422 0.895871 and the probability that a new policyholder will be on the maximum NCD level in the …fth year will be 0.889229. 2. Let i be the stable proportion in NCD discount level i: When the pro- 6 portions become stable we will have 1 + 0 = 0:095163 0 + 0:095163 1 1 = 0:904837 0 + 0:095163 2 2 = 0:904837 1 + 0:904837 2 0 + 2 =1 From the 2nd equation we have 0 = 0:095163 0:904837 1 and from the last equation we have 2 = 0:904837 0:095163 1 Substituting into the …rst equation then gives 0:095163 0:904837 so that 1 = 1 + 1 + 0:904837 0:095163 1 =1 1 = 0:0942196 10:613502 and therefore 0 = 0:0099092 and 2 = 0:89587124 and the long run proportion paying the full premium is 7 0 = 0:0099092: ...
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