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Unformatted text preview: Second Edition 77 We know that x and 2 X = i ( x i x ) 2 /n maximizes A ; the question is whether given Y , Y , and , does x , 2 X maximize B ? Let us first fix 2 X and look for X , that maximizes B . We have log B X  2 X i ( y i Y ) Y X ( x i X ) ! Y X set = 0 X i ( y i Y ) = Y X ( x i X ) . Similarly do the same procedure for L (  y ) L ( , y  x ) This implies i ( x i X ) = X Y i ( y i Y ). The solutions X and Y therefore must satisfy both equations. If i ( y i Y ) 6 = 0 or i ( x i X ) 6 = 0, we will get = 1 / , so we need i ( y i Y ) = 0 and i ( x i X ) = 0. This implies X = x and Y = y . ( 2 log B 2 X < 0. Therefore it is maximum). To get 2 X take log B 2 X X i Y 2 X ( x i X ) ( y i Y ) Y X ( x i X ) set...
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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.
 Spring '12
 Dr.Hackney
 Statistics

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