HW6Solution (1)

HW6Solution (1) - Econ 3190 Fall 2010 HW#6 Solution 1. f...

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Econ 3190 Fall 2010 HW#6 Solution 1. f (0) = 27 64 ; f (1) = 27 64 ; f (2) = 9 64 ; f (3) = 1 64 . E ( X ) = xf ( x ) = 0 f (0) + 1 f (1) + 2 f (2) + 3 f (3) = 27 64 0 + 27 64 1 + 9 64 2 + 1 64 4 = 3 4 2. E ( X ) = Z 1 xf ( x ) dx = Z 1 0 x 2 dx + Z 2 1 x (2 ± x ) dx = Z 1 0 x 2 dx + Z 2 1 2 x ± Z 2 1 x 2 dx = 1 3 x 3 j 1 0 + x 2 j 2 1 ± 1 3 x 3 j 2 1 = 1 3. Eg ( x ) = Z 1 g ( x ) f ( x ) dx = Z 1 0 e 2 x 3 e x dx = Z 1 0 e 1 3 x dx = ± 3 e 1 3 x j 1 0 = ± 3(0 ± 1) = 3 4. E ( X ) = X xf ( x ) = ± 2 & 0 : 3 + 3 & 0 : 2 + 5 & 0 : 5 = 2 : 5 1
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( X 2 ) = X x 2 f ( x ) = 4 & 0 : 3 + 9 & 0 : 2 + 25 & 0 : 5 = 15 : 5 V ar ( X ) = E ( X 2 ) ± E ( X ) 2 = 15 : 5 ± 2 : 5 2 = 9 : 25 or V ar ( X ) = E [( X ± E ( X )) 2 ] = ( ± 2 ± 2 : 5) 2 & 0 : 3 + (3 ± 2 : 5) 2 & 0 : 2 + (5 ± 2 : 5) 2 & 0 : 5 = 6 : 075 + 0 : 05 + 3 : 125 = 9 : 25 and X = p V ar ( X ) = p 9 : 25 = 3 : 04 5. E ( X ) = Z 1 0 x 2( x + 2) 5 dx = 1 5 Z 1 0 2 x 2 + 4 xdx = 1 5 ( 2 3 x 3 + 2 x 2 ) j 1 0 = 1 5 8 3 = 8 15 E ( X 2
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This note was uploaded on 04/06/2011 for the course ECON 3190 taught by Professor Hong during the Fall '07 term at Cornell.

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HW6Solution (1) - Econ 3190 Fall 2010 HW#6 Solution 1. f...

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