The University of New South Wales ACTL 3003/5106 Insurance Risk Models Final Exam|Session 2, 2014 SOLUTIONS 1 Question 1 Denote by the aggregate loss S = 15X i=1 IiBi + 15X i=1 Ii+15Bi+15: where Iiis the indicator of making a claim. q1 := Pr(I1 = 1) = = Pr(I15 = 1) = 0:005; q2 := Pr(I16 = 1) = = Pr(I30 = 1) = 0:01; andB1; B2; ; B30 are i.i.d. Gamma distributed random variables, i.e. Gamma (; ) with = 2 and = 0:002. We rst calculate the expectation and variance of S: E[S] = 15X i=1 E[Bi]Pr(Ii = 1) + 15X i=1 E[Bi+15]Pr(Ii+15 = 1) = 15 2 0:002 0:005 + 15 2 0:002 0:01 = 225; and Var[S] = 15X i=1 q1(1 q1)(E[Bi])2 + q1Var(Bi) + 15X i=1 q2(1 q2)(E[Bi+15])2 + q2Var(Bi+15) = 15 0:005 (1 0:005) 2 0:002 2 + 0:005 2 0:0022 +15 0:01 (1 0:01) 2 0:002 2 + 0:01 2 0:0022 = 335625: With the retained proportion , the net premium collected by the insurer is (1 + 1)E[S] (1 + 2)(1 )E[S] = (1 2 + + 2)E[S]: So the probability is calculated as follows Pr ( S (1 2 + + 2)E[S]) = Pr S E[S]p Var[S] [(1 2)= + 2]E[S]p Var[S] ! Pr Z [(1 2)= + 2]E[S]p Var[S] ! : 2 To make the above probability no less than 0:4325 (use the standard normal table), we should set [(1 2)= + 2]E[S]p Var[S] 0:17 Therefore, 2 1 2 0:17 p Var[S]=E[S] = 0:2163: Question 2 a. For column II, the maximum value is 8.5, which is not the minimum value in its row. So 8.5 is not a saddle point. For column I, the maximum value is maxf5:5; Y g = ( 5:5; if Y < 5:5; Y; if Y > 5:5: However, the rst case 5:5 is not the minimum value in its row. For the second case, only when 5:5 < Y < 6:5, Y is the minimum value of its row and hence a saddle point. For column III, the maximum value is maxf6:5; Xg = ( 6:5; if X < 5:5; X; if X > 5:5: However, the second case X > 5:5 is not the minimum value in its row. For the second case, only when Y > 6:5, 6:5 is the minimum value of its row and hence a saddle point. So the event that there exists two saddle points can be denoted by f5:5 < Y < 6:5g \ (fX > 5:5g \ fY > 6:5g) = ;; Therefore, its probability is zero. b. Similarly, the event that there exists at least one saddle points can be denoted by f5:5 < Y < 6:5g [ (fX > 5:5g \ fY > 6:5g): 3 Therefore, the probability of this event can be calculated as Pr