Midterm-2-S-431-2011-F

Midterm-2-S-431-2011-F - b (3 ,p ). Thus, p 3 = 0 . 125 and...

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SOLUTIONS TO MIDTERM TEST # 2 – ACTSC 431/831, FALL 2011 1. It follows from E ( N ) = k =1 kp k = k =1 k ( a + b k ) p k - 1 that E ( N ) = a + b 1 - a . 2. The loss elimination ratio in the current year is LER X (20) = E ( X 20) E ( X ) = 0 . 51775 . 3. (a) Let I j = 1 if X j > 30 and 0 if X j 30. Notice that N P = I 1 + ... + I N L . We have E ( I j ) = Pr { X j > 30 } = e - 0 . 6 . Thus, E ( N P ) = E ( N L ) E ( I j ) = 16 . 4643 . (b) We have V ar ( N P ) = E ( N L ) V ar ( I j ) + V ar ( N L )[ E ( I j )] 2 , where V ar ( I j ) = e - 0 . 6 (1 - e - 0 . 6 ) . Hence, V ar ( N P ) = 106 . 82. (c) We have Cov ( N P , N P * ) = E ( N P N P * ) - E ( N P ) E ( N P * ) , where E ( N P * ) = E ( N L ) - E ( N P ) , E ( N P N P * ) = E ( N P N L ) - E [( N P ) 2 ] , E [( N P ) 2 ] = V ar ( N P ) + [ E ( N P )] 2 , E ( N P N L ) = E ( N L E ( N P | N L )) = E ( N L N L e - 0 . 6 ) = e - 0 . 6 E [( N L ) 2 ] . Therefore, Cov ( N P , N P * ) = 74 . 289. 4. (a) Clearly, N is neither a Poisson random variable nor a binomial random variable. Suppose that N has a binomial distribution
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Unformatted text preview: b (3 ,p ). Thus, p 3 = 0 . 125 and p = 0 . 5. But Pr { N = 1 } = 0 . 5 6 = 3 . 5 . 5 2 . Hence, N is not a binomial random variable. Therefore, N is not in an ( a,b, 0) class since the class contains only Poisson, binomial, and negative binomial distributions. (b) The probability of Pr { S = 9 } = 0 . 25 Pr { M 1 + M 2 = 9 } + 0 . 125 Pr { M 1 + M 2 + M 3 = 9 } = 0 . 25 . 02 + 0 . 125 . 064 = 0 . 013. 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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