Math 408, Actuarial Statistics I A.J. HildebrandVariance, covariance, and moment-generating functionsPractice problems — Solutions1. Suppose that the cost of maintaining a car is given by a random variable, X, with mean200 and variance 260. If a tax of 20% is introducted on all items associated with themaintenance 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) = 1.22Var(X) = 1.44 ·260 = 374.2. The profit for a new pro duct is given by Z = 3X − Y − 5, where X and Y are independentrandom 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 getVar(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 thatVar(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 varianceof the total benefit after these increases?Solution: We need to compute Var(X + 100 + 1.1Y ). Since adding constants does notchange 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) + 1.12Var(Y ) + 2 · 1.1 Cov(X, Y ).We are given that Var(X) = 5, 000, Var(Y ) = 10, 000, so the only remaining unknownquantity is Cov(X, Y ), which can be computed via the general formula for Var(X + Y ):Cov(X, Y ) =12(Var(X + Y ) − Var(X) − Var(Y )) =12(17, 000−5, 000−10, 000) = 1, 000.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, 5204. A company insures homes in three cities, J, K, L. The losses occurring in these cities areindependent. The moment-generating functions for the loss distributions of the citiesareMJ(t) = (1 − 2t)−3, MK(t) = (1 − 2t)−2.5, ML(t) = (1 − 2t)−4.5Let X represent the combined losses from the three cities. Calculate E(X3).Solution: Let J , K, L denote the losses from the three cities. Then X = J + K + L.1Math 408, Actuarial Statistics I A.J. HildebrandSince J , K, L are independent, the moment-generating function for their sum, X, isequal 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 getM0(t) = (−2)(−10)(1 − 2t)−11,M00(t) = (−2)2(−10)(−11)(1 − 2t)−12,M000(t) = (−2)3(−10)(−11)(−12)(1 − 2t)−13.Hence, E(X3) = M000X(0) = (−2)3(−10)(−11)(−12) = 10, 560.5. Given that E(X) = 5, E(X2) = 27.4, E(Y ) = 7, E(Y2) = 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 getE(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(X2) + 2.2E(XY ) + 1.2E(Y2)= 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 ) − 71.72To complete the calculation, it remains to find E(XY ). To this end we make use of thestill unused relation Var(X + Y ) = 8:8 = Var(X + Y ) = E((X + Y )2) − (E(X + Y ))2= E(X2) + 2E(XY ) + E(Y2) − (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 =
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