Dr. Hackney STA Solutions pg 14

Dr. Hackney STA Solutions pg 14 - Second Edition -y 1-11...

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Second Edition 1-11 e. lim y →-∞ 1 - ± 1+ e - y = 0, lim y →∞ ± + 1 - ± 1+ e - y = 1, d dx ( 1 - ± 1+ e - y ) = (1 - ± ) e - y (1+ e - y ) 2 > 0 and d dx ( ± + 1 - ± 1+ e - y ) > 0, F Y ( y ) is continuous except on y = 0 where lim y 0 ( ± + 1 - ± 1+ e - y ) = F (0). Thus is F Y ( y ) right continuous. 1.48 If F ( · ) is a cdf, F ( x ) = P ( X x ). Hence lim x →∞ P ( X x ) = 0 and lim x →-∞ P ( X x ) = 1. F ( x ) is nondecreasing since the set { x : X x } is nondecreasing in x. Lastly, as x x 0 , P ( X x ) P ( X x 0 ), so F ( · ) is right-continuous. (This is merely a consequence of defining F ( x ) with “ ”.) 1.49 For every t , F X ( t ) F Y ( t ). Thus we have P ( X > t ) = 1 - P ( X t ) = 1 - F X ( t ) 1 - F Y ( t ) = 1 - P ( Y t ) = P ( Y > t ) . And for some t * , F X ( t * ) < F Y ( t * ). Then we have that P ( X > t * ) = 1 - P ( X t * ) = 1 - F X ( t * ) > 1 - F Y ( t * ) = 1 - P ( Y t * ) = P ( Y > t * ) . 1.50 Proof by induction. For
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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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