HW4solution

HW4solution - Econ 3190 Fall 2011 HW #4 Solution 1. #3.1...

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Unformatted text preview: Econ 3190 Fall 2011 HW #4 Solution 1. #3.1 X): discrete; Y): continuous; M): continuous; N): discrete; P): discrete; Q): continuous 2. #3.5 To be a valid probability mass function (pmf) for a discrete random vari- able(drv), f ( x ) has to satisfy the following two conditions: (i) & f X ( x ) & 1 for all x 2 R : (ii) P x f X ( x ) = 1 : (a) f (0) + f (1) + f (2) + f (3) = c [(0 + 4) + (1 + 4) + (4 + 4) + (9 + 4)] = 30 c = 1 ) c = 1 30 also f ( x ) = 1 30 ( x 2 + 4) ¡ ; for x = 0 ; 1 ; 2 ; 3 (b) f (0) + f (1) + f (2) = c [( 2 )( 3 3 ) + ( 2 1 )( 3 2 ) + ( 2 2 )( 3 1 )] = c (1 + 6 + 3) = 1 ) c = 1 10 also f ( x ) = 1 10 ( 2 x )( 3 3 & x ) ¡ ; for x = 0 ; 1 ; 2 3. #3.7 (a) P ( X < 1 : 2) = R 1 xdx + R 1 : 2 1 (2 ¢ x ) dx = 1 2 x 2 j 1 +2(1 : 2 ¢ 1) ¢ 1 2 x 2 j 1 : 2 1 = : 68 (b) P (0 : 5 < X < 1) = R 1 : 5 xdx = 1 2 x 2 j 1 : 5 = 0 : 375 4. #3.13 F ( x ) = 8 > > > > > > < > > > > > > : for x < : 41 for & x < 1 : 41 + 0 : 37 = 0 : 78 for 1 & x < 2 : 78 + 0 : 16 = 0 : 94 for 2 & x < 3 : 94 + 0 : 05 = 0 : 99 for 3 & x < 4 1 for x ¡ 4...
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This note was uploaded on 11/30/2011 for the course ECON 3190 at Cornell.

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HW4solution - Econ 3190 Fall 2011 HW #4 Solution 1. #3.1...

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