Question #161983

Problem 2: We consider an auction with two bidders. Each bidder's valuation v is drawn as follows: 

• With probability the valuation v is distributed uniformly between 0 and 2.

 • With probability 3 the valuation v is distributed uniformly between a and 1. 

1. If we run a first price auction, what is the optimal bid function of a bidder with valuation v?

 2. What is the expected revenue of the seller? 

3. Suppose now that there is only one bidder and that the seller runs a second price auction with a reserve price. Calculate the optimal reserve price.


1
Expert's answer
2021-02-09T18:18:15-0500
SolutionSolution


1.Let BiB_i be agent i's bid and πs\pi's agent i's profit. If B_i\geq\ v_i,\ then\ \pi\leq0,\then assuming rationality BiViB_i\leq V_i

Thus π=0\pi=0

If you want to maximize your expected profit (hence your valuation of money is 

risk-neutral), then your maximum bid is V2B=0     B=v2V-2B=0\implies\ B=\frac{v}{2} where BB is the bid.


2. Here there are two options in place. One, a second-price auction, the seller is committed to collecting less money, because of charging the second-highest bid. On the other hand, in a first price auction, the bidders reduce their bids, in return reducing what the seller can collect.

Therefore, the seller's expected revenue is

=(n2n)(nn+2)=n2n+2Where n=2=222+2=04=0=(\frac{n-2}{n})(\frac{n}{n+2})=\frac{n-2}{n+2}\\ Where\ n=2\\ =\frac{2-2}{2+2}=\frac{0}{4}\\ =0

3.Due to the reserve price the item will be on the highest bidder, only if the highest bid is above rr, otherwise the item should not be sold. In the second price auction, if there is a winner, he pays the maximum of the second-place bid and the reserve price rr.

If the item is worth vv to the seller, then r=v with a probability of 2r2-r, the bidders' value is above rr and this item will be sold at a price rr. Given a probability rr, the bidders value will be below rr, and the seller will hold the item with a payoff of v=0v=0. Therefore, the seller expects revenue which is r(2r),r(2-r), this is maximized at r=12r=\frac{1}{2}


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