Midterm-1-S-431-2011-F

# Midterm-1-S-431-2011-F - N is F N k = Pr N ≤ k = ∑ k j...

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SOLUTIONS TO MIDTERM TEST # 1– ACTSC 431/831, FALL 2011 1. (a) Pr { Y L > 60 } = 1 - Pr { Y L 60 } = 1 - Pr { 50 + 0 . 8( X - 50) 60 } = 0 . 375 . (b) Pr { Y P 40 } = Pr { Y L 40 | X > 20 } = Pr { X 40 | X > 20 } = 0 . 25 . (c) E ( Y L ) = R 50 20 x × 1 100 dx + R 80 50 (50 + 0 . 8( x - 50)) × 1 100 dx + R 100 80 74 × 1 100 dx = 43 . 9 . 2. (a) The mean of the loss is E ( Y ) = 0 . 4 × 10 + 0 . 6 × 10 2 - 1 = 10. (b) The probability is Pr { Y > E ( Y ) } = Pr { Y > 10 } = 0 . 4 e - 10 10 + 0 . 6 ± 10 10+10 ² 2 = 0 . 297 . 3. (a) The mgf of N is given by M N ( t ) = E ( e tN ) = E ( E ( e tN | Θ)) = E ( e Θ( e t - 1) ) = M Θ ( e t - 1) = (1 - ( e t - 1)) - 1 , which means that N has the negative binomial distribution NB (1 , 1) or the geometric distribution with probability function Pr { N = k } = 1 2 ( 1 2 ) k ,k = 0 , 1 , 2 ,.... (b) By (a), we know that for any k = 0 , 1 , 2 ,.... , the distribution function of
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Unformatted text preview: N is F N ( k ) = Pr { N ≤ k } = ∑ k j =0 1 2 ( 1 2 ) j = 1-( 1 2 ) k +1 . Obviously, F N (4) = 0 . 96875 > . 95 and F N (3) = 0 . 9375 < . 95. Hence, V aR . 95 ( N ) = 4. 4. We have f Θ | X ( θ | 10) ∝ f X | Θ (10 | θ ) f Θ ( θ ) ∝ e-(10-θ ) 2 2 × 4 e-( θ-8) 2 2 × 16 ∝ e-5 θ 2-96 θ 2 × 16 ∝ e-( θ-96 10 ) 2 2 × 16 5 . Hence, the conditional distribution of Θ, given X = 10, is a normal distribution N ( 96 10 , 16 5 ) = N (9 . 6 , 3 . 2). 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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