Math 408 Actuarial Statistics I A J Hildebrand Variance covariance and moment generating functions Practice problems Solutions 1 Suppose that the cost of maintaining a car is given by a random variable X with mean 200 and variance 260 If a tax of 20 is introducted on all items associated with the maintenance of the car what will the variance of the cost of maintaining a car be Solution The new cost is 1 2X so its variance is Var 1 2X 1 22 Var X 1 44 260 374 2 The profit for a new product is given by Z 3X Y 5 where X and Y are independent random variables with Var X 1 and Var Y 2 What is the variance of Z Solution Using the properties of a variance and independence we get Var Z Var 3X Y 5 Var 3X Y Var 3X Var Y 9 Var X Var Y 11 3 An insurance policy pays a total medical benefit consisting of a part paid to the surgeon X and a part paid to the hospital Y so that the total benefit is X Y Suppose that Var X 5 000 Var Y 10 000 and Var X Y 17 000 If X is increased by a flat amount of 100 and Y is increased by 10 what is the variance of the total benefit after these increases Solution We need to compute Var X 100 1 1Y Since adding constants does not change the variance this is the same as Var X 1 1Y which expands as follows Var X 1 1Y Var X Var 1 1Y 2 Cov X 1 1Y Var X 1 12 Var Y 2 1 1 Cov X Y We are given that Var X 5 000 Var Y 10 000 so the only remaining unknown quantity is Cov X Y which can be computed via the general formula for Var X Y Cov X Y 1 1 Var X Y Var X Var Y 17 000 5 000 10 000 1 000 2 2 Substituting this into the above formula we get the answer Var X 1 1Y 5 000 1 12 10 000 2 1 1 1 000 19 520 4 A company insures homes in three cities J K L The losses occurring in these cities are independent The moment generating functions for the loss distributions of the cities are MJ t 1 2t 3 MK t 1 2t 2 5 ML t 1 2t 4 5 Let X represent the combined losses from the three cities Calculate E X 3 Solution Let J K L denote the losses from the three cities Then X J K L 1 Math 408 Actuarial Statistics I A J Hildebrand Since J K L are independent the moment generating function for their sum X is equal to the product of the individual moment generating functions i e MX t MK t MJ t ML t 1 2t 3 2 5 4 5 1 2t 10 Differentiating this function we get M 0 t 2 10 1 2t 11 M 00 t 2 2 10 11 1 2t 12 M 000 t 2 3 10 11 12 1 2t 13 000 0 2 3 10 11 12 10 560 Hence E X 3 MX 5 Given that E X 5 E X 2 27 4 E Y 7 E Y 2 51 4 and Var X Y 8 find Cov X Y X 1 2Y Solution By definition Cov X Y X 1 2Y E X Y X 1 2Y E X Y E X 1 2Y Using the properties of expectation and the given data we get E X Y E X 1 2Y E X E Y E X 1 2E Y 5 7 5 1 2 7 160 8 E X Y X 1 2Y E X 2 2 2E XY 1 2E Y 2 27 4 2 2E XY 1 2 51 4 2 2E XY 89 08 Cov X Y X 1 2Y 2 2E XY 89 08 160 8 2 2E XY 71 72 To complete the calculation it remains to find E XY To this end we make use of the still unused relation Var X Y 8 8 Var X Y E X Y 2 E X Y 2 E X 2 2E XY E Y 2 E X E Y 2 27 4 2E XY 51 4 5 7 2 2E XY 65 2 so E XY 36 6 Substituting this above gives Cov X Y X 1 2Y 2 2 36 6 71 72 8 8 2
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