HW7solution - Econ 3190 Fall 2011 HW #7 Solution 1. #4.2 By

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Unformatted text preview: Econ 3190 Fall 2011 HW #7 Solution 1. #4.2 By de&nition, given X is a DRV, we know E ( X ) = P X x & f ( x ) From the pmf, we have f (0) = C 3 & (1 = 4) & (3 = 4) 3 = 27 = 64 ; f (1) = C 1 3 & (1 = 4) 1 & (3 = 4) 2 = 27 = 64 ; f (2) = C 2 3 & (1 = 4) 2 & (3 = 4) 1 = 9 = 64 ; f (3) = C 3 3 & (1 = 4) 3 & (3 = 4) = 1 = 64 ) E ( X ) = P X x & f ( x ) = 0 & f (0)+1 & f (1)+2 & f (2)+3 & f (3) = 48 = 64 = 3 = 4 2. #4.12 By de&nition, given X is a CRV, we know E ( X ) = R 1 &1 x & f ( x ) dx: ) E ( X ) = R 1 x & 2(1 x ) dx = 2 R 1 x x 2 dx = 2( x 2 = 2 x 3 = 3) j 1 = 1 = 3 3. #4.34 (a) To &nd the standard deviation of X; we need to compute the mean of X &rst. E ( X ) = P X x & f ( x ) = 2 & : 3 + 3 & : 2 + 5 & : 5 = 2 : 5 E ( X 2 ) = P X x 2 & f ( x ) = ( 2) 2 & : 3 + 3 2 & : 2 + 5 2 & : 5 = 15 : 5 ) & 2 X = E ( X 2 ) [ E ( X )] 2 = 15 : 5 6 : 25 = 9 : 25 ) The standard deivation & X = p & 2 X = p 9 : 25 = 3 : 04 4. #4.43 By de&nition, given X is a CRV, we know E ( X ) = R 1 &1 x & f ( x ) dx = R 1 x & 1 4 e & x= 4 dx = R 1 xde & x= 4 = [( xe & x= 4 ) j 1 R 1 e & x= 4 dx ] = [0 ( 4 e & x= 4 ) j 1 ] = 4...
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This note was uploaded on 11/30/2011 for the course ECON 3190 at Cornell University (Engineering School).

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HW7solution - Econ 3190 Fall 2011 HW #7 Solution 1. #4.2 By

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