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Respond to the question: When one bidder has an advantage?

01/18/2000 05:09 PM by John G. Riley; An answer
Answer from John Riley The central issue is how bidding is affected if one of the bidders is allowed to match rather than bid himself.  (Since he gets to match, he may as well bid zero if invited to bid and wait to see if he wants to match.)  It is not clear that there is a general analytical answer to the question so numerical methods might be needed (The Li-Riley BIDCOMP program could be modified to do this.)  However there is one special case which is easy to analyse.  Suppose a bidders valuation v = a+xc where x is unifomly distributed on the interval [0,1]. To simplify the exposition a little bit we will consider the case in which a = 0 and c = 1.  Suppose that there are n bidders.  Suppose also that they bid linearly, that is b(v) = v/k, where k >1.  The maximum bid is therefore 1/k.   For any one of these bidders, the probability of winning witha bid of b is Pr{b(v) < b} = Pr{v < kb}=kb.   Then if buyer 1 bids b, and all other buyers adopt the above strategy, buyer 1's probability of winning is G(b) = (kb)^(n-1).   If his valuation is v, it follows that his expected payoff is

U(v,b) = (v-b)G(b).

Taking logs,

lnU = ln(v-b) + (n-1)lnb + constant terms.

Differentiating by b, it follows that buyer 1's best response is to bid b(v) = (n-1)v/n.  Exactly the same agument holds for each bidder.  Thus the equilibrium bidding strategy is b(v) = (n-1)v/n. It is  also not difficult to confirm that the expected high bid (hence expected revenue) is (n-1)/(n+1).

Next suppose that bidder n is allowed to match.  Let b be the high bid among the n-1 other buyers.  Bidder n will match if his valaution exceeds b.  Thus bidder n matches with probability Pr{vn < b} = b.  It follows that bidder 1 wins with a bid of b with probabilty H(b) = b(kb)^(n-2).   Arguing as before,

U(v,b) = (v-b)H(b).

Taking logs,

lnU = ln(v-b) + (n-1)lnb + constant terms.

Differentiating by b, it follows that buyer 1's best response is to bid b(v) = (n-1)v/n.  Thus with one biider matching, the bidding strategy is the same as if all n bidders were bidding.  Note also that if bidder 1 drops out completely, ther are n-1 bidders so the new equilibrium is b(v) = (n-2)v/(n-1).   Thus adding the nth bidder (whether or not he matches or bids like the other raises the bids of all players.

What about revenue?  It is not difficult to coinfirm that with bidder n matching, revenue is ((n-1)/n)^2

We thus have the following revenue results.

N bidders all bidding symmetrically  Revenue = (n-1)/(n+1)
N bidders with 1 of them matching    Revenue =  ((n-1)/n)^2
N-1 bidders all bidding symmetrically Revenue = (n-2)/n


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01/17/2000 01:23 PM by name withheld; Some more details of the problem
Background Developer (D) is developing a power plant and wishes to seek competitive bids for fuel supply. There are five (5) qualified bidders, Companies 1 - 5, all who are capable of supplying fuel. D will be required to [View full text and thread]

01/15/2000 11:30 AM by David K. Levine; The answer is complicated
What happens in auctions depends both on the circumstances of the auction and the rules. One possibility is that the bidders have private information relevant to the other bidders - for example some inside information about future [View full text and thread]

01/15/2000 10:41 AM by name withheld; What happens when one bidder has an advantage?
The background is that our company has offered to provide development funding to X in exchange for the right to match the best offer to supply them with fuel. Assume the development funding is adequate value for the option received. [View full text and thread]