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A1-S-431F08 - SOLUTIONS TO ASSIGNMENT 1 ACTSC 431/831 FALL...

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SOLUTIONS TO ASSIGNMENT 1 – ACTSC 431/831, FALL 2008 1. (a) F Y ( y ) = Pr { Y y } = Pr { Y y | N = 0 } Pr { N = 0 } + Pr { Y y | 1 N 5 } Pr { 1 N 5 } + Pr { Y y | N > 5 } Pr { N > 5 } = 0 , y < 0 0 . 3 + 0 . 5(1 - e - y/ 100 ) + 0 . 2(1 - e - y/ 200 ) , y 0 = 0 , y < 0 , 1 - 0 . 5 e - y/ 100 - 0 . 2 e - y/ 200 , y 0 . [5 marks] (b) The probability is Pr { Y > 500 } = 1 - F Y (500) = 0 . 5 e - 5 + 0 . 2 e - 2 . 5 = 0 . 02. [5 marks] (c) The probability Pr { 0 < Y < 300 } = F (300 - ) - F (0) = 0 . 63 . [5 marks] (d) The mean of the total loss is E ( Y ) = 90 . [5 marks] (e) Since 0 = Pr { Y < 0 } < 0 . 2 < Pr { Y 0 } = 0 . 3, the 20th percentile of Y is 0. [5 marks] (f) Since Pr { Y 0 } = 0 . 3 < 0 . 8 and Pr { Y y } = F Y ( y ) is continuous and strictly increasing in y > 0, the 80th percentile of Y is the root of equation F Y ( y ) = 0 . 8, or equivalently, is the root of the following equation: 0 . 5 e - y/ 100 + 0 . 2 e - y/ 200 = 0 . 2 . Solving the equation by substitution e - y/ 200 = x gives x = 0 . 463325 and hence the 80th percentile of Y is - 100 ln 0 . 463325 = 153 . 87. [5 marks] 2. (a) Note that Y L [0 , 70]. Thus, F Y L ( y ) = Pr { Y L y } = 0 , y < 0; Pr { X - 80 y } , 0 y < 40; Pr { 40 + 0 . 75( X - 120) y } , 40 y < 70; 1 , 70 y ; = 0 , y < 0; y 200 + 0 . 4
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