Practice Questions 4 Answer

Practice Questions 4 Answer - -100) , 100 < X 2 , 100 1...

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ANSWERS TO PRACTICE QUESTIONS 4 – ACTSC 431/831, FALL 2011 1. Use the pgf of X to prove that X is infinitely divisible. 2. (a) Let v = Pr { X j 50 } . Use P N P * ( z ) = P N L (1 + v ( z - 1)) to prove that N P * is infinitely divisible. (b) i. V ar ( N P ) = 2108 . 05. ii. Cov ( N P , N P * ) = 4803 . 95. 3. E ( Y P ) = R d yf ( y ) dy 1 - F ( d ) = 148 . 36 . 4. (a) F Y L ( y ) = 0 , y < 0 0 . 875 , 0 y 100 1 - 125 , 000 y 3 , y > 100 and F Y P ( y ) = 0 , y 100 , 1 - 1 , 000 , 000 y 3 , y > 100 . Furthermore, E ( Y L ) = 18 . 75 , and E ( Y P ) = E ( Y L ) 1 - F ( d ) = 150 . (b) The loss elimination ratio for the current year is LER X (180) = E ( X 180) E ( X ) , = 0 . 9743 . (c) (i) The loss elimination ratio for the next year is LER 1 . 03 X (180) = E (1 . 03 X 180) E (1 . 03 X ) = 0 . 9727 . (ii) To keep the same loss elimination ratio for the next year as that for current year, the deductible of d for the next year should be d = 1 . 03 × 180 = 185 . 4. (d) (i) We have d = 100, α = 0 . 85, and α ( u - d ) = 1 , 700. Then u = 2 , 100, Y L = 0 , X 100 0 . 85( X
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Unformatted text preview: -100) , 100 < X 2 , 100 1 , 700 , X > 2 , 100 Let Y P be the per-payment random variable. Note that Since Y P = Y L | X > d . Thus, the cdf of Y P is F Y P ( y ) = , y , 1- 85 y +85 3 , < y < 1 , 700 , 1 y 1 , 700 . (ii) We have E ( Y L ) = Z 2 , 100 100 . 85( x-100) f ( x ) dx + 1 , 700 Z 2 , 100 f ( x ) dx = 5 . 30045 , E [( Y L ) 2 ] = Z 2 , 100 100 [0 . 85( x-100)] 2 f ( x ) dx + 1 , 700 2 Z 2 , 100 f ( x ) dx = 819 . 161 . Thus, E ( Y P ) = 5 . 30045 / (1-F ( d )) = 42 . 4036 and E ( Y P ) 2 = 819 . 161 / (1-F ( d )) = 6553 . 29 . Therefore, V ar ( Y P ) = 4755 . 22 . 1...
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This note was uploaded on 02/08/2012 for the course ACTSC 431 taught by Professor Laundriualt during the Fall '09 term at Waterloo.

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