Question

A seller wishes to auction a single unit of an indivisible object. She can do so in one of the following two ways (i) through a first-price auction to n bidders, (ii) through a second-price auction with optimally chosen reserve price to n - 1 bidders. Each bidder's valuation is private information and is an independent draw from the uniform distribution on [0, 1]. The seller values the object at 0.

(a) Which of the two alternatives would you recommend to the seller?

(b) Show that for a large number of bidders, the seller would be indifferent between the two alternatives.

Answer 1

The seller can prefer the first option because the indivisible object can be sell by n bid auction.

• A single indivisible object is for sale.

• There is a fixed number n of bidders i = 1, . . . , n. •

The bidders submit sealed bids. So bids are simultaneous and independent. The bid of bidder _i is denoted b_i .

• Each bidder i has a valuation for the object, denoted by v_i . He enjoys vi if and only if he wins the object.

• The highest bidder wins the object. If there are two or more bids tied at the top, the winner is chosen randomly. All top bidders win with positive probability.

• The price paid depends on whether we are in a secondprice or a first-price auction.

Only the winning bidder pays. The price paid for the object is denoted by p.

• Bidders are risk-neutral. If i wins the object and the price is p, his utility is v_i ? p. All non-winners have a utility of 0. If there is uncertainty about who wins, expected payoffs guide the bidders’ behavior.

Second-Price, Known Valuations

• Typically, each bidder will know his own vi and not observe the valuations of other bidders. But to begin with we look at the case in which all bidders know the valuations of all other bidders as well.

• This simplifies things and will be useful later when we look at the case in which i does not observe the other bidders’ valuations.

• Fix (v₁, . . . , v_n), known to all. The only thing that matters for i in deciding how much to bid is the largest bid of all other bidders. This is because if he wins (becausehis bid is highest), this bid will be the price he pays. • For each i, let mi be the largest bid among all bidders but i.

That is m_i = max   {b_j}

• Remember, bids are simultaneous and independent. So, given any possible configuration of the bids of others, how should i set bi? We can proceed on a case-by-case basis as follows.

1. Suppose m_i > vi.

(a) Bidding b_i ? v_i makes i lose the auction. He gets a payoff of zero.

(b) Bidding v_i < b_i < m_i makes i lose the auction. He gets a payoff of zero.

(c) Bidding b_i = m_i makes i’s bid tied as the top bid with one or more other bids. So i wins with positive probability, and pays a price p = m_i . If he wins he gets v_i ? p < 0. If he does not win he gets zero. So, in this case, his expected payoff is negative.

(d) Bidding b_i > m_i makes i win the auction for sure. So, in this case he gets v_i ? p = v_i ? m_i < 0.

• From cases (1a) through (1d) we can conclude that if m_i > v_i , then there is no choice of bi that is better than setting b_i = v_i . Some other choices are just as good for bidder i, but no choice is better.

2. Suppose m_i = v_i .

(a) Bidding b_i < v_i makes i lose the auction. He gets a payoff of zero.

(b) Bidding b_i = v_i makes i’s bid tied as the top bid with one or more other bids. So i wins with positive probability, and pays a price p = m_i = v_i . If he wins he gets v_i ? p = 0. If he does not win he gets zero. So, in this case, his expected payoff is zero.

(c) Bidding b_i > v_i makes i win the auction for sure. His payoff is v_i ? p. Since p = m_i = v_i , his payoff is zero.

• From cases (2a) through (2c) we can conclude that if m_i = v_i , then there is no choice of bi that is better than setting b_i = v_i . Bidder i gets a payoff of zero, regardless of his bid bi .

3. Suppose m_i < v_i .

(a) Bidding b_i < m_i makes i lose the auction. He gets a payoff of zero.

(b) Bidding b_i = m_i makes i’s bid tied as the top bid with one or more other bids. So i wins with positive probability, and pays a price p = m_i . If he wins he gets v_i ? m_i > 0. If he does not win he gets zero. So, in this case, his expected payoff positive but below v_i ? m_i .

(c) Bidding m_i < b_i < v_i makes i win the auction for sure. His payoff is v_i ? p. Since p = mi , his payoff is v_i ? m_i > 0.

(d) Bidding b_i = v_i makes i win the auction for sure. His payoff is v_i ? p. Since p = m_i , his payoff is v_i ? m_i > 0.

(e) Bidding b_i > v_i makes i win the auction for sure. His payoff is v_i ? p. Since p = m_i , his payoff is v_i ? m_i > 0.

• From cases (3a) through (3e) we can conclude that if m_i < v_i , then there is no choice of b_i that is better than setting b_i = v_i . Some other choices are just as good for bidder i, but no choice is better.

• Putting together all the cases, (1a) through (3e) we can conclude that, regardless of mi , there is no choice of b_i that is better for i than setting b_i = v_i . Depending on m_i , some choices may be worse. But no choice is ever better.

• Another way to say this is that bidding b_i = v_i weakly dominates all other strategies.

• Remember that it is okay to rule out weakly dominated strategies (not okay in general to do this iteratively). They are not robust to the possibility of “mistakes.”

• Also, we know that after deleting weakly dominated strategies, at least one Nash equilibrium will still be there.

• Hence we can conclude that the unique Nash equilibrium not involving weakly dominated strategies is for each bidder _i to bid b_i = v_i .

• Notice one appealing property of the Nash equilibrium we have found. The bidder with the largest vi wins the auction (if there are two or more valuations tied at the top, then one of the top valuers wins for sure).

• An auction with the property that (one of) the highest valuers wins the object is called an efficient auction.