Dr. Hackney STA Solutions pg 18

# Dr. Hackney STA Solutions pg 18 - -log(1-y b(i lim x...

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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 - p 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 = ² R y - y f X ( x ) dx if | x | ≤ 1 R y 1 f X ( x ) dx if x 1 . Diﬀerentiation gives f y ( y ) = ² 2 9 1 y if y 1 1 9 + 1 9 1 y if y 1 . b. If the sets B 1 ,B 2 ,. .. ,B K are a partition of the range of Y , we can write f Y ( y ) = X k f Y ( y ) I ( y B k ) and do the transformation on each of the B k . So this says that we can apply Theorem 2.1.8 on each of the B k and add up the pieces. For A 1 = ( - 1 , 1) and A 2 = (1 , 2) the calculations are identical to those in part (a). (Note that on A 1 we are essentially using Example 2.1.7). 2.8 For each function we check the conditions of Theorem 1.5.3. a.(i) lim x 0 F ( x ) = 1 - e - 0 = 0, lim x →-∞ F ( x ) = 1 - e -∞ = 1. (ii) 1 - e - x is increasing in x . (iii) 1 - e - x is continuous. (iv) F - 1 x ( y ) =
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Unformatted text preview: -log(1-y ). b.(i) lim x →-∞ F ( x ) = e-∞ / 2 = 0, lim x →∞ F ( x ) = 1-( e 1-∞ / 2) = 1. (ii) e-x/ 2 is increasing, 1 / 2 is nondecreasing, 1-( e 1-x / 2) is increasing. (iii) For continuity we only need check x = 0 and x = 1, and lim x → F ( x ) = 1 / 2, lim x → 1 F ( x ) = 1 / 2, so F is continuous. (iv) F-1 X ( y ) = ± log(2 y ) ≤ y < 1 2 ≤ y < 1, 1-log(2(1-y )) 1 2 ≤ y < 1 c.(i) lim x →-∞ F ( x ) = e-∞ / 4 = 0, lim x →∞ F ( x ) = 1-e-∞ / 4 = 1. (ii) e-x / 4 and 1-e-x / 4 are both increasing in x . (iii) lim x ↓ F ( x ) = 1-e-/ 4 = 3 4 = F (0), so F is right-continuous. (iv) F-1 X ( y ) = ± log(4 y ) ≤ y < 1 4-log(4(1-y )) 1 4 ≤ y < 1...
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