Rubinstein2005-page46

Rubinstein2005-page46 - October 21, 2005 12:18 master Sheet...

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Unformatted text preview: October 21, 2005 12:18 master Sheet number 44 Page number 28 28 Lecture Three Proof: Define % by x % y if x = C ( { x , y } ) . Let us first verify that the relation % is a preference relation. • Completeness : Follows from the fact that C ( { x , y } ) is always well defined. • Transitivity : If x % y and y % z , then C ( { x , y } ) = x and C ( { y , z } ) = y . If C ( { x , z } ) 6= x then C ( { x , z } ) = z . By ∗ and C ( { x , z } ) = z , C ( { x , y , z } ) 6= x . By ∗ and C ( { x , y } ) = x , C ( { x , y , z } ) 6= y , and by ∗ and C ( { y , z } ) = y , C ( { x , y , z } ) 6= z . A contradiction to C ( { x , y , z } ) ∈ { x , y , z } . We still have to show that C ( B ) = C % ( B ) . Assume that C ( B ) = x and C % ( B ) 6= x . That is, there is y ∈ B so that y  x . By definition of % , this means C ( { x , y } ) = y , contradicting ∗ . What Is an Alternative Some of the cases where rationality is violated can be attributed to the incorrect specification of the space of alternatives. Consider thethe incorrect specification of the space of alternatives....
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This note was uploaded on 12/29/2011 for the course ECO 443 taught by Professor Aswa during the Fall '10 term at SUNY Stony Brook.

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