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slide33-nov21 - ECON 401 Auctions Revenue Siyang Xiong Rice...

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ECON 401: Auctions: Revenue Siyang Xiong Rice University November 21, 2011 Xiong (Rice University) ECON 401 November 21, 2011 1 / 17

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revenue equivalence Suppose there are n bidders, and bidders°values v i are i.i.d. distributed on [ 0 , 1 ] following the cdf F . ° / . De±nition An auction is standard if the rules of the auction dictate that the person who bids the highest amount is awarded the object. Xiong (Rice University) ECON 401 November 21, 2011 2 / 17
revenue equivalence Theorem Suppose that values are independently and identically distributed. Then any symmetric and increasing equilibrium of any standard auction, such that the expected payment of a bidder with value zero is zero, yields the same expected revenue to the seller. Consider any symmetric and increasing equilibrium of some standard auction A . symmetric: every bidder take the same bidding strategy b . ° / ; increasing + the standard auction: the bidder with the highest value wins the auction. Xiong (Rice University) ECON 401 November 21, 2011 3 / 17

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revenue equivalence: proof consider bidder 1 who observes v 1 2 [ 0 , 1 ] . Given other bidders taking the strategy b . ° / , how should bidder 1 bid? bidder 1 could pretend he has value v 0 1 2 [ 0 , 1 ] and bids b ° v 0 1 ± , and he gets the expected payo/: G ° v 0 1 ± ± v 1 ² m A ° v 0 1 ± m A ° v 0 1 ± is the expected payment of bidder 1, if he bids b ° v 0 1 ± ; ² depends on the rules of the auction A ; G ° v 0 1 ± is the probability of bidder 1 winning, if he bids b ° v 0 1 ± ; G ° v 0 1 ± D Pr °² . v 2 ,..., v n / : b . v 2 / ³ b ° v 0 1 ± , b . v 3 / ³ b ° v 0 1 ± ,..., b . v n / ³ b ° v 0 1 ±³ D Pr °² v 2 : v 2 ³ v 0 1 ³± ± ... ± Pr °² v n : v n ³ v 0 1 ³± , because b . k / is incr D F 2 ° v 0 1 ± ± F 3 ° v 0 1 ± ± ... ± F n ° v 0 1 ± D ´ F ° v 0 1 ±µ n ² 1 Xiong (Rice University) ECON 401 November 21, 2011 4 / 17
revenue equivalence: proof To summarize, bidder 1 who observes v 1 2 [ 0 , 1 ] solves the following maximization problem: max v 0 1 2 [ 0 , 1 ] G ° v 0 1 ± ± v 1 ² m A ° v 0 1 ± (1) since b .

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