Dr. Hackney STA Solutions pg 16

Dr. Hackney STA Solutions pg 16 - a. f Y ( y ) = 1 2 y-1 /...

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Chapter 2 Transformations and Expectations 2.1 a. f x ( x ) = 42 x 5 (1 - x ), 0 < x < 1; y = x 3 = g ( x ), monotone, and Y = (0 , 1). Use Theorem 2.1.5. f Y ( y ) = f x ( g - 1 ( y )) ± ± ± ± d dy g - 1 ( y ) ± ± ± ± = f x ( y 1 / 3 ) d dy ( y 1 / 3 ) = 42 y 5 / 3 (1 - y 1 / 3 )( 1 3 y - 2 / 3 ) = 14 y (1 - y 1 / 3 ) = 14 y - 14 y 4 / 3 , 0 < y < 1 . To check the integral, Z 1 0 (14 y - 14 y 4 / 3 ) dy = 7 y 2 - 14 y 7 / 3 7 / 3 ± ± ± ± 1 0 = 7 y 2 - 6 y 7 / 3 ± ± ± 1 0 = 1 - 0 = 1 . b. f x ( x ) = 7 e - 7 x , 0 < x < , y = 4 x + 3, monotone, and Y = (3 , ). Use Theorem 2.1.5. f Y ( y ) = f x ( y - 3 4 ) ± ± ± ± d dy ( y - 3 4 ) ± ± ± ± = 7 e - (7 / 4)( y - 3) ± ± ± ± 1 4 ± ± ± ± = 7 4 e - (7 / 4)( y - 3) , 3 < y < . To check the integral, Z 3 7 4 e - (7 / 4)( y - 3) dy = - e - (7 / 4)( y - 3) ± ± ± 3 = 0 - ( - 1) = 1 . c. F Y ( y ) = P (0 X y ) = F X ( y ). Then f Y ( y ) = 1 2 y f X ( y ). Therefore f Y ( y ) = 1 2 y 30( y ) 2 (1 - y ) 2 = 15 y 1 2 (1 - y ) 2 , 0 < y < 1 . To check the integral, Z 1 0 15 y 1 2 (1 - y ) 2 dy = Z 1 0 (15 y 1 2 - 30 y + 15 y 3 2 ) dy = 15( 2 3 ) - 30( 1 2 ) + 15( 2 5 ) = 1 . 2.2 In all three cases, Theorem 2.1.5 is applicable and yields the following answers.
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Unformatted text preview: a. f Y ( y ) = 1 2 y-1 / 2 , 0 &lt; y &lt; 1. b. f Y ( y ) = ( n + m +1)! n ! m ! e-y ( n +1) (1-e-y ) m , 0 &lt; y &lt; . c. f Y ( y ) = 1 2 log y y e-(1 / 2)((log y ) / ) 2 , 0 &lt; y &lt; . 2.3 P ( Y = y ) = P ( X X +1 = y ) = P ( X = y 1-y ) = 1 3 ( 2 3 ) y/ (1-y ) , where y = 0 , 1 2 , 2 3 , 3 4 ,. .. , x x +1 ,. .. . 2.4 a. f ( x ) is a pdf since it is positive and Z - f ( x ) dx = Z- 1 2 e x dx + Z 1 2 e-x dx = 1 2 + 1 2 = 1 ....
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This note was uploaded on 02/03/2012 for the course STA 1014 taught by Professor Dr.hackney during the Spring '12 term at UNF.

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