ISM_chapter6 - Chapter 6: Functions of Random Variables 6.1...

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121 Chapter 6: Functions of Random Variables 6.1 The distribution function of Y is = = y Y y y dt t y F 0 2 2 ) 1 ( 2 ) ( , 0 y 1. a. 2 2 1 2 1 2 1 2 1 1 ) ( ) ( 2 ) ( ) ( ) 1 2 ( ) ( ) ( 1 + + + + = = = = = u u u Y u U F Y P u Y P u U P u F . Thus, 1 1 , ) ( ) ( 2 1 1 1 = = u u F u f u U U . b. 2 2 1 2 1 2 1 1 2 1 2 ) ( ) ( 2 1 ) ( ) ( ) 2 1 ( ) ( ) ( 2 + + = = = = = = u u u Y u U F Y P u Y P u U P u F . Thus, 1 1 , ) ( ) ( 2 1 2 2 = = + u u F u f u U U . c. u u u F u Y P u Y P u U P u F Y U = = = = = 2 ) ( ) ( ) ( ) ( ) ( 2 3 3 Thus, 1 0 , 1 ) ( ) ( 1 3 3 = = u u F u f u U U . d. . 6 / 1 ) ( , 3 / 1 ) ( , 3 / 1 ) ( 3 2 1 = = = U E U E U E e. . 6 / 1 ) ( , 3 / 1 ) 2 1 ( , 3 / 1 ) 1 2 ( 2 = = = Y E Y E Y E 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 . 6.3 The distribution function for Y is > < = 5 . 1 1 5 . 1 1 2 / 1 1 0 2 / ) ( 2 y y y y y y F Y . a. ) ( ) ( ) 4 10 ( ) ( ) ( 10 4 10 4 + + = = = = u Y u U F Y P u Y P u U P u F . So, > < = + 11 1 11 6 6 4 ) ( 10 1 200 ) 4 ( 2 u u u u F u u U , and < = = + elsewhere 0 11 6 6 4 ) ( ) ( 10 1 100 4 u u u F u f u U U . b. E ( U ) = 5.583. c. E (10 Y – 4) = 10(23/24) – 4 = 5.583. 6.4 The distribution function of Y is 4 / 1 ) ( y Y e y F = , 0 y . a. 12 / ) 1 ( 3 1 3 1 1 ) ( ) ( ) 1 3 ( ) ( ) ( = = = + = = u u Y u U e F Y P u Y P u U P u F . Thus, 1 , ) ( ) ( 12 / ) 1 ( 12 1 = = u e u F u f u U U . b. E ( U ) = 13.
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122 Chapter 6: Functions of Random Variables Instructor’s Solutions Manual 6.5 The distribution function of Y is 4 / ) ( y y F Y = , 1 y 5. 2 3 4 1 2 3 2 3 2 ) ( ) ( ) 3 2 ( ) ( ) ( = = = + = = u u Y u U F Y P u Y P u U P u F . Differentiating, () 53 5 , ) ( ) ( 2 / 1 2 3 16 1 = = u u F u f u 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. a. ). ( ) ( ) ( ) ( ) ( ) ( 2 u F u F u Z u P u Z P u U P u F Z Z U = = = = The density function for U is then 0 ), ( ) ( ) ( ) ( ) ( 1 2 1 2 1 = + = = u u f u f u f u F u f Z u Z u Z u U U . Evaluating, we find 0 ) ( 2 / 2 / 1 2 1 = π u e u u f u U . b. U has a gamma distribution with α = 1/2 and β = 2 (recall that Γ (1/2) = π ).
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ISM_chapter6 - Chapter 6: Functions of Random Variables 6.1...

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