Dr. Hackney STA Solutions pg 75

Dr. Hackney STA Solutions pg 75 - > z | x ) P (...

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Second Edition 5-9 where we use the independence of X and Y . Since X and Y are identically distributed, P ( X > a ) = P ( Y > a ) = 1 - F X ( a ), so F Z 2 ( z ) = (1 - F X ( - z )) 2 - (1 - F X ( z )) 2 = 1 - 2 F X ( - z ) , since 1 - F X ( z ) = F X ( - z ). Differentiating and substituting gives f Z 2 ( z ) = d dz F Z 2 ( z ) = f X ( - z ) 1 z = 1 2 π e - z/ 2 z - 1 / 2 , the pdf of a χ 2 1 random variable. Alternatively, P ( Z 2 z ) = P ± [min( X,Y )] 2 z ² = P ( - z min( X,Y ) z ) = P ( - z X z,X Y ) + P ( - z Y z,Y X ) = P ( - z X z | X Y ) P ( X Y ) + P ( - z Y z | Y X ) P ( Y X ) = 1 2 P ( - z X z ) + 1 2 P ( - z Y z ) , using the facts that X and Y are independent, and P ( Y X ) = P ( X Y ) = 1 2 . Moreover, since X and Y are identically distributed P ( Z 2 z ) = P ( - z X z ) and f Z 2 ( z ) = d dz P ( - z X z ) = 1 2 π ( e - z/ 2 1 2 z - 1 / 2 + e - z/ 2 1 2 z - 1 / 2 ) = 1 2 π z - 1 / 2 e - z/ 2 , the pdf of a χ 2 1 . 5.23 P ( Z > z ) = X x =1 P ( Z > z | x ) P ( X = x ) = X x =1 P ( U 1 > z,. ..,U x
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Unformatted text preview: > z | x ) P ( X = x ) = X x =1 x Y i =1 P ( U i > z ) P ( X = x ) (by independence of the U i s) = X x =1 P ( U i > z ) x P ( X = x ) = X x =1 (1-z ) x 1 ( e-1) x ! = 1 ( e-1) X x =1 (1-z ) x x ! = e 1-z-1 e-1 < z < 1 . 5.24 Use f X ( x ) = 1 / , F X ( x ) = x/ , 0 < x < . Let Y = X ( n ) , Z = X (1) . Then, from Theorem 5.4.6, f Z,Y ( z,y ) = n ! 0!( n-2)!0! 1 1 z y-z n-2 1-y = n ( n-1) n ( y-z ) n-2 , < z < y < ....
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