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Unformatted text preview: Solutions to Assignment 3 ACTSC 433/833, Winter 2012 1. (a) S (60) = 1 10 (0 + 0 + 0 + 1 10 + 4 10 + 1 + 1 + 1 + 1) = 0 . 45. (b) S (60) = 1 10 (0 + 0 + 2 7 30 + 11 30 + 14 30 + 23 30 + 23 30 + 25 30 + 29 30 + 1) = 0 . 486667. 2. (a) S (220) = 1 5 0 + (190+75 220) 2 2 75 2 + 1 (220 241+75) 2 2 75 2 + 1 + 1 = 0 . 58416. (b) S (220) = 1 5 0 + (190+100 220) 2 2 100 2 + 1 (220 241+100) 2 2 100 2 + 1 + 1 = 0 . 58659. 3. (a) We have L ( ) = n e n i =1 x i . Then, it follows from ln L = 0 that = n n i =1 X i . (b) Note that S = n i =1 X i has the gamma distribution with pdf f S ( y ) = n y n 1 e y ( n ) , y > 0. Thus, E ( ) = n R 1 y f S ( y ) dy = n n 1 as n and E ( 2 ) = n 2 Z 1 y 2 f S ( y ) dy = n 2 ( n 1)( n 2) 2 . Hence, V ar ( ) 0 as n . Thus, is consistent for ....
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This note was uploaded on 03/07/2012 for the course ACTSC 433 taught by Professor Johnny during the Winter '09 term at Waterloo.
 Winter '09
 johnny

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