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# hw9_print - STAT416 Probability — Spring 2011 Assignments...

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Unformatted text preview: STAT416: Probability — Spring 2011 Assignments 1 Homework #9 (Solution) April 13, 2011 1 Total possible points is 80 1. Ex. 7.1 (5 pts) . Consider f ( x ) = 1 x k , x ≥ 1. We want, 1 = Z ∞ 1 f ( x ) dx = Z ∞ 1 1 x k dx = x- k +1 (- k + 1) = 1 (1- k ) x k- 1 ∞ 1 = 1 1- k [0- 1] = 1 k- 1 ⇒ 1 k- 1 = 1 ⇒ k- 1 = 1 ⇒ k = 2 . If k = 1, then R ∞ 1 1 x dx = ln | x || ∞ 1 → ∞ . 2. Ex. 7.3 (20 pts) . (a) Let f ( x ) = c | x | (1 + x )(1- x ) ,- 1 ≤ x ≤ 1. Z 1- 1 c | x | (1- x 2 ) dx = c Z- 1 ( x 3- x ) dx + c Z 1 ( x- x 3 ) dx = c " x 4 4- x 2 2- 1 + x 2 2- x 4 4 1 # = c- 1 4 + 1 2 + 1 2- 1 4 = 1 2 c = 1 ⇒ c = 2 . (b) f ( x ) = 2 | x | ( 1- x 2 ) ,- 1 ≤ x ≤ 1. f ( x ) = 2( x 3- x ),- 1 ≤ x ≤ 0. 2( x- x 3 ), ≤ x ≤ 1. F ( x ) = P ( X ≤ x ) = R x- 1 2( t 3- t ) dt ,- 1 ≤ t ≤ x . 1 2 + R x 2( t- t 3 ) dt , ≤ t ≤ x . F ( x ) = 2( t 4 4- t 2 2 ) x- 1 ,- 1 ≤ t ≤ x ....
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hw9_print - STAT416 Probability — Spring 2011 Assignments...

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