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HW 2 Solutions

# HW 2 Solutions - HW 2 Solutions 2 3 6.2 The distribution...

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HW 2 Solutions 6.2 The distribution function of Y is - - = = y Y y dt t y F 1 3 2 ) 1 )( 2 / 1 ( ) 2 / 3 ( ) ( , –1 ≤ y ≤ 1. a. ) 1 18 / ( ) 3 / ( ) 3 / ( ) 3 ( ) ( ) ( 3 2 1 1 1 - = = = = = u u F u Y P u Y P u U P u F Y U . Thus, 3 3 , 18 / ) ( ) ( 2 1 1 - = = u u u F u f U U . b. ] ) 3 ( 1 [ ) 3 ( 1 ) 3 ( ) 3 ( ) ( ) ( 3 2 1 2 2 u u F u Y P u Y P u U P u F Y U - - = - - = - = - = = . Thus, 4 2 , ) 3 ( ) ( ) ( 2 2 3 2 2 - = = u u u F u f U U . c. 2 / 3 2 3 ) ( ) ( ) ( ) ( ) ( ) ( 3 u u F u F u Y u P u Y P u U P u F Y Y U = - - = - = = = . Thus, 1 0 , ) ( ) ( 2 3 3 3 = = u u u F u f U U . 6.6 Refer to Ex. 5.10 ad 5.78. Define ) ( ) ( ) ( ) ( 2 1 2 1 u Y Y P u Y Y P u U P u F U + = - = = . a. For u ≤ 0, 0 ) ( ) ( ) ( 2 1 = - = = u Y Y P u U P u F U . For 0 ≤ u < 1, ∫ ∫ = = - = = + u u y y U u dy dy u Y Y P u U P u F 0 2 2 2 1 2 1 2 / 1 ) ( ) ( ) ( 2 2 . For 1 ≤ u ≤ 2, ∫ ∫ - + - - = - = - = = u u y U u dy dy u Y Y P u U P u F 2 0 2 2 2 1 2 1 2 / ) 2 ( 1 1 1 ) ( ) ( ) ( 2 . Thus, - < = = elsewhere 0 2 1 2 1 0 ) ( ) ( y u u u u F u f U U . b. E ( U ) = 1. 6.7 Let F Z ( z ) and f Z ( z ) denote the standard normal distribution and density functions respectively.

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