Rubinstein2005-page48

Rubinstein2005-page48 - cannot be chosen without y while y...

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October 21, 2005 12:18 master Sheet number 46 Page number 30 30 Lecture Three Figure 3.2 Violation of the weak axiom. Proposition: Assume that C is a choice function with a domain that includes at least all subsets of size not greater than 3. Assume that C satisFes WA. Then, there is a preference % so that C = C % . Proof: DeFne x % y if x C ( { x , y } ) . We will now show that the relation is a preference: Completeness : ±ollows from C ( { x , y } ) 6=∅ . Transitivity :I f x % y and y % z then x C ( { x , y } ) and y C ( { y , z } ) .I f x / C ( { x , z } ) , then C ( { x , z } ) ={ z } .B yW A , x / C ( { x , y , z } ) (by WA x cannot be revealed to be as good as z because z was chosen without x from { x , z } ). Similarly, y / C ( { x , y , z } ) (by WA, y cannot be chosen without x while x C ( { x , y } ) ). And also, z / C ( { x , y , z } )
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Unformatted text preview: cannot be chosen without y while y C ( { y , z } )). This contradicts the nonempti-ness of C ( { x , y , z } ) . It remains to be shown that C ( B ) = C % ( B ) . Assume that x C ( B ) and x / C % ( B ) . That is, there is y B so that y is strictly better than x , or in other words, C ( { x , y } ) = { y } , thus contradicting WA. Assume that x C % ( B ) and x / C ( B ) . Let y C ( B ) . By WA x / C ( { x , y } ) and thus C ( { x , y } ) = { y } , and therefore y x , contradicting x C % ( B ) ....
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