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## PowerPoint Slideshow about 'Auctions' - albert

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Strategic Situation

- You are bidding for an object in an auction.
- The object has a value to you of $20.
- How much should you bid?
- Depends on auction rules presumably

Review: Second Price Auctions

- Suppose that the auction is a second-price auction
- High bidder wins
- Pays second highest bid
- Sealed bids
- We showed (using dominance) that the best strategy was to bid your value.
- So bid $20 in this auction.

Review: English Auctions

- An English (or open outcry) auction is one where bidders shout bids publicly.
- Auction ends when there are no higher bids.
- Implemented as a “button auction” in Japan
- Implemented on eBay through proxy bidding.

What to Bid

- Again, suppose you value the object at $20.
- Dominance says to drop out when bid = value.
- The fact that bidding strategies are the same in the two auction forms means that they are strategically equivalent.

Revenues

- How much does the seller earn on the auction?
- Depends on the distribution of values.
- Suppose that there are 2 bidders and values are equally likely to be from $0 to $100.
- The seller earns an amount equal to the expected losing bid.

Order Statistics

- The seller is interested in the expected value of the lower of two draws from 0-100.
- This is called the second order statistic of the distribution.
- We will sometimes write this as E[Vk(n)] where the k denotes the order (highest, 2nd highest, etc.) of the draw and (n) denotes the number of draws.
- So we’re interested in E[V2(2)]

Order Statistics of Uniform Distributions

- There order statistics have simple regularity properties
- The mean of a uniform draw from 0-100 is 50.
- Note the mean could be written as E[V1(1)].

100

0

50

Key Observation

- With uniform distributions, the order statistics evenly divide the number line into n + 1 equal segments.
- Let’s try 3 draws:

1st

3rd

2nd

50

0

75

100

25

Generalizing

- So in general,
- E[Vk(n)] = 100* (n – k + 1)/(n + 1)
- So revenues in a second price or English auction in this setting are:
- E[V2(n)] = 100 * (n – 1)/(n + 1)
- As the number of bidders grows large, the seller’s revenues increase
- As the number of bidders grows unbounded, the seller earns all the surplus, i.e. 100!

First Price Auctions

- Now suppose you have a value of $20 and are competing with one other bidder in a first-price auction
- You don’t know the exact valuation of the other bidder.
- But you do know that it is randomly drawn from 0 to 100.
- How should you bid?

Setting Up the Problem

- As usual, you want to bid to maximize your expected payoff
- But now you need to make a projection about the strategy of the other bidder
- Presumably this strategy depends on the particular valuation the bidder has.
- Let b(v) be your projection for the bid of the other bidder when his valuation is v.

Bidder’s Problem

- Choose a bid, B, to maximize expected profits.
- E[Profit] = (20 – B) x Pr(B is the highest bid)
- What is Pr(B is the highest bid)?
- It is Pr(B > b(v))

Conjectures about b(v)

- Suppose that I believe that my rival’s strategy is to bid a constant fraction of his value
- Then b(v) = av
- Where a is some fraction
- I win whenever
- B >= av
- Or, equivalently
- v <= B/a
- So Pr(B > b(v)) becomes:
- Pr( v <= B/a) = B/100a

Bidder’s Problem Revisited

- So now I need to choose B to maximize
- E[Profit] = (20 – B)(B/100a)
- Optimize in the usual way:
- (1/100a) x (20 – 2B) = 0
- Or B = 10
- So I should bid 10 when my value is 20.

Other Values

- Suppose my value is V?
- E[Profit] = (V – B)(B/100a)
- Optimize in the usual way:
- (1/100a) x (V – 2B) = 0
- Or B = V/2
- So I should always bid half my value.

Equilibrium

- My rival is doing the same calculation as me.
- If he conjectures that I’m bidding ½ my value
- He should bid ½ his value (for the same reasons)
- Therefore, an equilibrium is where we each bid half our value.

Uncertainty about my Rival

- This equilibrium we calculated is a slight variation on our usual equilibrium notion
- Since I did not exactly know my rival’s payoffs in this game
- I best responded to my expectation of his strategy
- He did likewise

Bayes-Nash Equilibrium

- Mutual best responses in this setting are called Bayes-Nash Equilibrium.
- The Bayes part comes from the fact that I’m using Bayes rule to figure out my expectation of his strategy.

Comments

- In this setting, dominant strategies were not enough
- What to bid in a first-price auction depends on conjectures about how many rivals I have and how much they bid.
- Rationality requirements are correspondingly stronger.

Revenues

- How much does the seller make in this auction?
- Since the high bidder wins, the relevant order statistic is E[V1(2)] = 66.
- But since each bidder only bids half his value, my revenues are
- ½ x E[V1(2)] = 33
- Notice that these revenues are exactly the same as in the second price or English auctions.

Revenue Equivalence

- Two auction forms which yield the same expected revenues to the seller are said to be revenue equivalent
- Operationally, this means that the seller’s choice of auction forms was irrelevant.

More Rivals

- Suppose that I am bidding against n – 1 others, all of whom have valuations equally likely to be 0 to 100.
- Now what should I bid?
- Should I shade my bid more or less or the same?
- In the case of second-price and English auctions, it didn’t matter how many rivals I had, I always bid my value
- What about in the first-price auction?

Optimal Bidding

- Again, I conjecture that the others are bidding a fraction a of their value.
- E[Profit] = (V – B) x Pr(B is the high bid)
- To be the high bid means that I have to beat bidder 2.
- Pr( B >= b(v2)) = B/100a
- But I also have to now beat bidders 3 through n.

Probability of Winning

- So now my chance of winning is
- B/100a x B/100a x …B/100a
- For n – 1 times.
- Or equivalently
- Pr(B is the highest) = [B/100a]n-1

Bidder 1’s optimization

- Choose B to maximize expected profits
- E[Profit] = (V – B) x Pr(B is highest)
- E[Profit] = (V – B) x [B/100a]n-1
- E[Profit] = (1/100a)n-1 x (V – B) x [B]n-1
- Optimizing in the usual way:
- (1/100a)n-1 x ((n-1)V – nB) [B]n-2 = 0
- So the optimal bid is
- B = V x (n-1)/n

Equilibrium

- I bid a proportion of my value
- But that proportion is (n-1)/n
- As I’m competing against more rivals, I shade my bid less.
- Since all my rivals are making the same calculation, in equilibrium everyone bids a fraction (n-1)/n of their value.

Revenues

- How much does the seller make in this auction?
- The relevant order statistic is E[V1(n)] = 100* n/(n + 1)
- But eveyone shades by (n-1)/n so
- Revenues = (n-1)/n x E[V1(n)]
- Revenues = 100 x (n-1)/(n+1)

Comments

- Revenues are increasing in the number of bidders
- As that number grows arbitrarily large, the seller gets all the surplus, i.e. 100!
- How does this compare to the English or Second-Price auction?

Comparing Revenues

- First-price:
- R = (n-1)/n x E[V1(n)]
- R = 100 x (n-1)/(n+1)
- Second-price:
- R = E[V2(n)]
- R = 100 x (n-1)/(n+1)
- The auctions still yield the same expected revenues.

Revenue Equivalence Theorem

- In fact, revenue equivalence holds quite generally
- Consider any auction which:
- Allocates the object to the highest bidder
- Gives any bidder the option of paying zero
- Then if bidders know their values
- Values are uncorrelated
- Values are drawn from the same distribution
- Then all such auctions are revenue equivalent!

Implications

- This means that we can determine the revenues quickly and easily for all sorts of auctions
- Consider an all-pay auction
- Bidders submit cash payments to the seller (bribes)
- The bidder submitting the highest bribe gets the object
- The seller keeps all the bribe money
- This auction auction yields the same revenues as an English auction.

Other Strange Auction forms

- Suppose that all bidders submit bribes to the auctioneer
- The object is awarded to the person paying the highest bribe
- And the seller gives back the bribe of the winner, but keeps all the others
- This is also revenue equivalent.

Optimal Auctions

- Revenue equivalence says that the form of the auction does not affect how much money the seller makes.
- But there are other tools the seller has to make money.

One Bidder Auctions

- Suppose that the seller is running an auction that attracts only one bidder.
- What should he do?
- If he goes with the usual auction forms, he’ll make nothing since the second highest valuation for the object is zero.

Monopoly

- Since the seller is a monopoly provider of the good, maybe some tricks from monopoly theory might help.
- Suppose a monopolist faced a linear demand curve and could only charge a single price
- What price should he charge?

Monopoly Problem

- The monopolist should choose p to maximize profits
- Profits = P x Q(P) – C(Q(P))
- Or equivalently, the monopolist could choose Q to maximize profits
- Profits = P(Q) x Q – C(Q)
- P(Q) is the inverse demand function
- Optimizing in the usual way, we have:
- MR = MC

Back to Auctions

- What is the demand curve faced by a seller in a one bidder auction?
- One can think of the “quantity” as the probability of making a sale at a given price.
- So if the seller asks for $100, he will make no sales.
- If he asks for $0, he will sell with probability = 1
- If he asks $50, he will sell with probability .5

Demand Curve

- So the demand curve is just the probability of making a sale
- Pr(V > P)
- If we denote by F(p) the probability that V <=p, then
- Q = 1 – F(p)
- But we need the inverse demand curve to do the monopoly problem the usual way.
- P = F-1(1 – Q)

Auction/Monopoly Problem

- Now we’re in a position to do the optimization.
- The seller should choose a reserve price to maximize his expected profits
- E[Profits] = p x (1 – F)
- Equivalently, the auctioneer chooses a quantity to maximize
- E[Profits] = F-1(1 – Q) x Q

Optimization

- As usual the optimal quantity is where MR = MC
- But MC is zero in this case
- So the optimal quantity is where MR = 0

So what is Marginal Revenue?

- Revenue = F-1(1 – Q) x Q
- Marginal Revenue = F-1(1 – Q) – Q/f(F-1(1 – Q))
- where f(p) is (approximately) the probability that v = p
- Now substitute back:
- P – (1 – F(p))/f(p) = 0

Uniform Case

- In the case where valuations are evenly distributed from 0 to 100
- F(p) = p/100
- f(p) = 1/100
- So
- P – (1 – P) = 0
- Or
- P = 50!

Recipe for Optimal Auctions

- The seller maximizes his revenue in an auction by:
- Step 1: Choosing any auction form satisfying the revenue equivalence principle
- Step 2: Placing a reserve price equal to the optimal reserve in a one bidder auction
- Key point 1: The optimal reserve price is independent of the number of bidders.
- Key point 2: The optimal reserve price is NEVER zero.

Conclusions

- Optimal bidding depends on the rules of the auction
- In English and second price auctions, bid your value
- In first-price auctions, shade your bid below your value
- The amount to shade depends on the competition
- More competition = less shading

More Conclusions

- As an auctioneer, the rules of the auction do not affect revenues much
- However reserve prices do matter
- The optimal reserve solves the monopoly problem for a one bidder auction

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