The Encyclopedia of Economic Principals

Chapter 27

Auction Formats, Values, and Revenue Equivalence

Who bids, what they know and how they pay decide what an auction earns.

Four of the chapter's worked examples, made interactive: an extra bidder against an optimal reserve, the winner's curse, revenue equivalence between first and second price, and affiliated signals.

Every example here is a constructed teaching example: it uses the hypothetical numbers of the chapter's worked examples, plus a few values added for comparison and labelled as such in each panel. Nothing here measures a real market, firm or household.

Demonstration 1 of 4

One more bidder beats the optimal reserve

Is it better to set the optimal reserve or to find one more bidder?

The Bulow-Klemperer theorem says an extra serious bidder in a plain auction raises more expected revenue than the best reserve with one bidder fewer. Recruiting costs can still reverse the net ranking.

Equation, written in LaTeX: \phi(v)=2v-1

Equation, written in LaTeX: 2(0.5)(0.5)=0.50.

Equation, written in LaTeX: 0.5+\frac{0.5}{3}=\frac23.

Equation, written in LaTeX: \frac{2}{3+1}=\frac12.

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Values are independent and uniform on 0 to 1. phi(v) is the virtual value; the optimal reserve for two bidders is 0.5. The third bidder costs c to recruit.

Predict first. What is the largest recruiting cost at which the extra bidder still beats the optimal reserve?

Your prediction

Choose an example

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Figure: One more bidder beats the optimal reserve. Two bars: the two-bidder auction earns 0.417 and the three-bidder auction 0.500 after a recruiting cost of 0.00.
Reserve in the two-bidder auction: 0.5 (optimal), Cost of recruiting the third bidder: 0
Constructed example: the chapter's hypothetical uniform sale (reserve 0.5, revenues 5/12, 1/3 and 1/2, recruiting cost 0.12); a recruiting cost of 0.05 is added.

Calculated values

Two-bidder revenue
0.417
Three-bidder gross revenue
0.500
Recruiting cost
0.00
Three-bidder net revenue
0.500
Better format
three bidders
Break-even recruiting cost
0.083

With reserve 0.5, one value clears it with probability 2(0.5)(0.5) = 0.50 and pays 0.5, giving 0.25; both clear it with probability 0.25 and the lower averages 2/3, giving 0.167. Revenue is 0.25 + 0.167 = 0.417. A third bidder with no reserve pays the expected second-highest of three, 2/(3 + 1) = 0.500; net of a 0.00 recruiting cost that is 0.500 - 0.00 = 0.500. The difference is 0.500 - 0.417 = 0.083, so the three-bidder auction earns more.

Worked steps

  1. One above reserve: 0.50 x 0.5 = 0.25
  2. Both above: 0.25 x 2/3 = 0.167
  3. Two-bidder revenue = 0.25 + 0.167 = 0.417
  4. Three bidders: 2 / 4 = 0.500, net 0.500 - 0.00 = 0.500
  5. Difference = 0.500 - 0.417 = 0.083

Use the idea

Spend effort widening the bidder pool before fine-tuning reserves, but count what recruitment costs.

Where the conclusion applies

Independent private values from a regular distribution, symmetric risk-neutral bidders and a second-price or equivalent auction.

Check your understanding: With a recruiting cost of 0.05, does the three-bidder auction still win?
0.50 - 0.05 = 0.45 > 5/12 = 0.417, yes; the break-even cost is 1/12 = 0.083.

Chapter 27 source: section "Bulow-Klemperer theorem".

Demonstration 2 of 4

The winner's curse grows with competition

Why does the winner of a common-value auction tend to overpay, and how does that change with more bidders?

Winning tells a bidder that its estimate was the highest, which is news that it was probably too high. Bidding the raw estimate builds that selection into the price.

Equation, written in LaTeX: B_{\max}=V+\max_i\varepsilon_i.

Equation, written in LaTeX: E[\max_i\varepsilon_i\mid n=2]=\frac{-20(1)+0(3)+20(5)}{9}=\frac{80}{9}\approx\$8.89.

Equation, written in LaTeX: E[\max_i\varepsilon_i\mid n=3]=\frac{-20(1)+0(7)+20(19)}{27}=\frac{360}{27}\approx\$13.33.

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The lease is worth V, $100 on average. Each bidder sees V plus an error of -e, 0 or +e, equally likely, and naively bids that signal. n is the number of bidders.

Predict first. Does adding a bidder raise or lower the naive winner's expected loss?

Your prediction

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Figure: The winner's curse grows with competition. Probability bars for the largest of 2 errors of size 20, and the expected naive winning bid rising with the number of bidders; at 2 it is $108.89.
Number of naive bidders: 2, Error size: $20
Constructed example: the chapter's hypothetical leases (mean value $100, errors of $20, two and three bidders); one and four bidders and errors of $10 and $30 are added, computed the same way.

Calculated values

Expected largest error
$8.89
Expected naive winning bid
$108.89
Expected winner payoff
-$8.89
Chance the largest error is +e
5/9

With 2 naive bidders and errors of -20, 0 or +20, the largest error is -20 in 1 of 9 cases, 0 in 3 and +20 in 5. So E = (-20(1) + 0(3) + 20(5)) / 9 = 80 / 9 = $8.89. The winner bids $108.89 on average for a lease worth $100, a payoff of -$8.89.

Worked steps

  1. Counts out of 9: 1, 3, 5
  2. E = (-20 + 0 + 100) / 9 = 80 / 9 = 8.89
  3. Winning bid = 100 + 8.89 = 108.89
  4. Winner payoff = -8.89

Use the idea

Shade bids in common-value auctions, and shade more when competition is stronger.

Where the conclusion applies

Independent, equally likely errors, naive bidders and a common value. With private values the curse is absent.

Check your understanding: With n = 4 and e = $20, what is the expected naive overpayment?
Counts out of 81 are 1, 15 and 65; E = (-20 + 1,300) / 81 = 1,280 / 81 = $15.80.

Chapter 27 source: section "The Winner's Curse in Common-Value Auctions".

Demonstration 3 of 4

Same expected revenue, different realized payments

If first-price and second-price auctions earn the same on average, why do their payments differ?

Revenue equivalence holds in expectation: both formats give the object to the highest value and leave the lowest type nothing. Individual draws differ, and an entry fee changes the lowest type's utility, so equivalence fails.

Equation, written in LaTeX: E[\min(v_1,v_2)]=\frac{\$100}{3}\approx\$33.33.

Equation, written in LaTeX: b(v)=\frac{v}{2}.

Equation, written in LaTeX: \frac{\$100}{3}+\$10=\frac{\$130}{3}\approx\$43.33.

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Two risk-neutral bidders with values uniform from $0 to $100. b(v) is the first-price bid. The fee is a nonrefundable entry charge in the second-price auction only.

Predict first. For which realized pairs does the first-price auction collect more than the second-price auction?

Your prediction

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Figure: Same expected revenue, different realized payments. Bars of realized receipts: second price $50.00, first price $40.00, with expected revenue lines at $33.33 and $33.33.
Higher realized value: $80, Lower realized value: $50, Second-price entry fee: $0
Constructed example: the chapter's hypothetical auction (values uniform to $100, draw $80 and $50, fee $5); high values of $60 and $100 and a low value of $30 are added.

Calculated values

Second-price receipts
$50.00
First-price payment
$40.00
Expected revenue, second price
$33.33
Expected revenue, first price
$33.33
More in this draw
second price

With values 80 and 50, the second-price winner pays the lower value 50. The first-price winner bids 80 / 2 = 40. In this draw second price collects more. Before values are known, both formats expect 100 / 3 = 33.33.

Worked steps

  1. Second price pays 50
  2. First price pays 80 / 2 = 40
  3. Expected first price = (1/2)(2 x 100 / 3) = 33.33
  4. Expected second price = 100 / 3 = 33.33

Use the idea

Choose between standard formats on grounds other than expected revenue (risk, collusion, simplicity) unless the theorem's conditions fail.

Where the conclusion applies

Independent private values, risk-neutral symmetric bidders, the same allocation and a zero payoff for the lowest type.

Check your understanding: With values $100 and $30, which format collects more in that draw?
Second price pays 30; first price pays 100 / 2 = 50, so first price, though both expect 33.33.

Chapter 27 source: section "Revenue-equivalence theorem".

Demonstration 4 of 4

When a rival's signal is news

When does seeing what a rival knows change what an object is worth to you?

Affiliated signals move together: a high signal makes a rival's high signal more likely, so seeing the rival stay in an ascending auction is good news about value.

Equation, written in LaTeX: \Pr(X_2=H\mid X_1=H)=\frac{0.40}{0.50}=0.80

Equation, written in LaTeX: 0.40(0.40)\geq0.10(0.10).

Equation, written in LaTeX: 0.8(\$100)+0.2(\$75)=\$95\text{ million}.

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Each bidder sees a high (H) or low (L) signal. The tract is worth $50, $75 or $100 million with 0, 1 or 2 high signals. a is the probability on each diagonal cell.

Predict first. At what diagonal mass does the rival's signal stop carrying information?

Your prediction

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Figure: When a rival's signal is news. Joint probability grid with 0.40 on the diagonal and 0.10 off it, and the expected tract value of 95.0 million for a bidder with a high signal.
P(H,H) = P(L,L): 0.4
Constructed example: the chapter's hypothetical tract (joint probabilities 0.40 and 0.10, values $50, $75 and $100 million, counterfactual 0.10); diagonal masses of 0.30 and 0.25 are added.

Calculated values

P(rival H | own H)
0.80
P(rival H | own L)
0.20
Expected value given own H
$95.0m
Affiliation check
0.1600 >= 0.0100
Signals
affiliated

P(rival H | own H) = 0.40 / 0.50 = 0.80. A bidder with signal H expects 0.80 x 100 + 0.20 x 75 = 95.0 million. The check 0.40 x 0.40 = 0.1600 >= 0.10 x 0.10 = 0.0100 says the signals are affiliated.

Worked steps

  1. P(H | H) = 0.40 / 0.50 = 0.80
  2. Value = 0.80 x 100 + 0.20 x 75 = 95.0
  3. Check: 0.1600 >= 0.0100

Use the idea

In common-value sales, formats that reveal rivals' information (ascending auctions, public appraisals) tend to raise prices when signals are affiliated.

Where the conclusion applies

Two binary signals, a symmetric joint distribution and values that depend only on the number of high signals.

Check your understanding: With a = 0.30, what is the expected value for a bidder with signal H?
P(H | H) = 0.30 / 0.50 = 0.60; value = 0.6 x 100 + 0.4 x 75 = $90 million; check 0.09 >= 0.04 holds.

Chapter 27 source: section "Affiliation".