Transcript Auctions - Faculty Directory | Berkeley-Haas

Auctions
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
Two Draws
 Now suppose there are two draws.
 What are the first and second order statistics?
0
33
66
100
Key Observation
 With uniform distributions, the order statistics
evenly divide the number line into n + 1 equal
segments.
 Let’s try 3 draws:
3rd
0
25
2nd
50
1st
75
100
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 firstprice 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))
What is Pr(B > b(v))?
b(v)
B
I win
b-1(B) I lose
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:

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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
P
Demand curve
Q
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
Monopoly Problem
P
Marginal Revenue
P*
MC
Q
Q*
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
Auction/Monopoly Problem
P
100
50
0
1
1/2
Q = Pr of sale
Auction/Monopoly Problem
P
Q = 1 – F(p)
100
50
0
1
1/2
Q = Pr of sale
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
Auction/Monopoly Problem
P
Marginal Revenue
100
P*
0
Q*
1
Q = Pr of sale
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