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ch2sol - Chapter 2 Transformations and Expectations 2.1 a...

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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, 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, 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, 1 0 15 y 1 2 (1 - y ) 2 dy = 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. a. f Y ( y ) = 1 2 y - 1 / 2 , 0 < y < 1. b. f Y ( y ) = ( n + m +1)! n ! m ! e - y ( n +1) (1 - e - y ) m , 0 < y < . c. f Y ( y ) = 1 σ 2 log y y e - (1 / 2)((log y ) ) 2 , 0 < y < . 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 -∞ f ( x ) dx = 0 -∞ 1 2 λe λx dx + 0 1 2 λe - λx dx = 1 2 + 1 2 = 1 .

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2-2 Solutions Manual for Statistical Inference b. Let X be a random variable with density f ( x ). P ( X < t ) = t -∞ 1 2 λe λx dx if t < 0 0 -∞ 1 2 λe λx dx + t 0 1 2 λe - λx dx if t 0 where, t -∞ 1 2 λe λx dx = 1 2 e λx t -∞ = 1 2 e λt and t 0 1 2 λe - λx dx = - 1 2 e - λx t 0 = - 1 2 e - λt + 1 2 . Therefore, P ( X < t ) = 1 2 e λt if t < 0 1 - 1 2 e - λt dx if t 0 c. P ( | X | < t ) = 0 for t < 0, and for t 0, P ( | X | < t ) = P ( - t < X < t ) = 0 - t 1 2 λe λx dx + t 0 1 2 λe - λx dx = 1 2 1 - e - λt + 1 2 - e - λt +1 = 1 - e - λt . 2.5 To apply Theorem 2.1.8. Let A 0 = { 0 } , A 1 = (0 , π 2 ), A 3 = ( π, 3 π 2 ) and A 4 = ( 3 π 2 , 2 π ). Then g i ( x ) = sin 2 ( x ) on A i for i = 1 , 2 , 3 , 4. Therefore g - 1 1 ( y ) = sin - 1 ( y ), g - 1 2 ( y ) = π - sin - 1 ( y ), g - 1 3 ( y ) = sin - 1 ( y ) + π and g - 1 4 ( y ) = 2 π - sin - 1 ( y ). Thus f Y ( y ) = 1 2 π 1 1 - y 1 2 y + 1 2 π - 1 1 - y 1 2 y + 1 2 π 1 1 - y 1 2 y + 1 2 π - 1 1 - y 1 2 y = 1 π y (1 - y ) , 0 y 1 To use the cdf given in (2.1.6) we have that x 1 = sin - 1 ( y ) and x 2 = π - sin - 1 ( y ). Then by differentiating (2.1.6) we obtain that f Y ( y ) = 2 f X (sin - 1 ( y ) d dy (sin - 1 ( y ) - 2 f X ( π - sin - 1 ( y ) d dy ( π - sin - 1 ( y ) = 2( 1 2 π 1 1 - y 1 2 y ) - 2( 1 2 π - 1 1 - y 1 2 y ) = 1 π y (1 - y ) 2.6 Theorem 2.1.8 can be used for all three parts. a. Let A 0 = { 0 } , A 1 = ( -∞ , 0) and A 2 = (0 , ). Then g 1 ( x ) = | x | 3 = - x 3 on A 1 and g 2 ( x ) = | x | 3 = x 3 on A 2 . Use Theorem 2.1.8 to obtain f Y ( y ) = 1 3 e - y 1 / 3 y - 2 / 3 , 0 < y < . b. Let A 0 = { 0 } , A 1 = ( - 1 , 0) and A 2 = (0 , 1). Then g 1 ( x ) = 1 - x 2 on A 1 and g 2 ( x ) = 1 - x 2 on A 2 . Use Theorem 2.1.8 to obtain f Y ( y ) = 3 8 (1 - y ) - 1 / 2 + 3 8 (1 - y ) 1 / 2 , 0 < y < 1 .
Second Edition 2-3 c. Let A 0 = { 0 } , A 1 = ( - 1 , 0) and A 2 = (0 , 1). Then g 1 ( x ) = 1 - x 2 on A 1 and g 2 ( x ) = 1 - x on A 2 . Use Theorem 2.1.8 to obtain f Y ( y ) = 3 16 (1 - 1 - y ) 2 1 1 - y + 3 8 (2 - y ) 2 , 0 < y < 1 . 2.7 Theorem 2.1.8 does not directly apply. a. Theorem 2.1.8 does not directly apply. Instead write P ( Y y ) = P ( X 2 y ) = P ( - y X y ) if | x | ≤ 1 P (1 X y ) if x 1 = y - y f X ( x ) dx if | x | ≤ 1 y 1 f X ( x ) dx if x 1 .

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ch2sol - Chapter 2 Transformations and Expectations 2.1 a...

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