HW4solution

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

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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) 0 ° 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) ± 0 ; for x = 0 ; 1 ; 2 ; 3 (b) f (0) + f (1) + f (2) = c [( 2 0 )( 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 ) ± 0 ; for x = 0 ; 1 ; 2 3. #3.7 (a) P ( X < 1 : 2) = R 1 0 xdx + R 1 : 2 1 (2 ² x ) dx = 1 2 x 2 j 1 0 +2(1 : 2 ² 1) ² 1 2 x 2 j 1 : 2 1 = 0 : 68 (b) P (0 : 5 < X < 1) = R 1 0 : 5 xdx = 1 2 x 2 j 1 0 : 5 = 0 : 375 4. #3.13 F ( x ) = 8 > > > > > > < > > > > > > : 0 for x < 0 0 : 41 for 0 ° x < 1 0 : 41 + 0 : 37 = 0 : 78 for 1 ° x < 2 0 : 78 + 0 : 16 = 0 : 94 for 2 ° x < 3 0 : 94 + 0 : 05 = 0 : 99 for 3 ° x < 4 1 for x ± 4 5. #3.14 Convert 12 mins in terms of hours, which is 0.2 (a) P ( X < 0 : 2) = F (0 : 2) = 1 ² e ° 8 ± 0 : 2 = 1 ² e ° 1 : 6 = 0 : 798 (b) f ( x ) = ° 0 x < 0 8 e ° 8 x x ± 0 P ( X < 0 : 2) = R 0 : 2 0 8 e ° 8 x dx = 8 R 0 : 2 0 e ° 8 x dx = ² e ° 8 x j 0 : 2 0 = 1 ² e ° 1 : 6 = 0 : 798 1

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6. #3.21 Need to verify f ( x ) is a valid pdf, and derive F ( x ) : (a) k ± 0; R 1 0 k p xdx = k R 1 0 x 1 2 dx = k 2 3 x 3 2 j
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