Transcript Course Overview

Course administration
• Tirgul: will be held every only after exercise submission.
• Teaching Assistant: Liri Finkelstein,
[email protected]
• Exercises: four exercises. Each one is 8% of the final grade.
• Rest of grade: a final homework. Four hard exercises. To be
submitted by the end of the semester.
• Office hours: Flexible, email me ([email protected])
• Course website:
http://ie.technion.ac.il/~ronlavi/auction-theory/
– I will put all lecture slides there, plus different
announcements and supporting material.
What is Auction Theory?
• Auctions are becoming more and more popular
– The Internet only helps.
• We see different auction formats, different rules.
– Can we say anything about the difference, what will be the
result of one format vs. another.
– Examples: Ebay has proxies, different sites has different end
rules, etc.
• How to analyze this? We use Game Theory, that models
strategic decisions of players.
• A mathematical analysis, we will see one or two examples, but
the main focus is on the mathematical modeling and techniques.
Course Overview
1.
Basic game-theoretic notions, The basic auction model,
description and analysis of classic one-item auctions
2.
Revenue-maximizing auctions
3.
Auctions with interdependent values
4.
Auctions for multiple identical goods
5.
Combinatorial Auctions (if time permits)
No special knowledge is required, I only assume basic math. Skills
So don’t hesitate to stop me and ask any question whatsoever.
First topic: analysis of classic one-item
auction types
• Basic game-theory notions: normal-form games,
dominant strategies, games in incomplete information,
Nash equilibrium. Bayesian-Nash equilibrium.
• Analysis and comparison of four auction types:
– First-price, second-price, English, Dutch
– Efficiency and revenue.
A game (in complete information)
Player II’s actions
Action “L”
Action “T”
Player I’s
actions
Action “B”
Action “R”
Here’re the utilities
if player I plays T
and II plays L
5 , -3
If player I plays B and II plays R then player I gets 5 and II gets -3
Formalism (a “normal-form” game)
• A game is composed of:
– An action space Ai for each player i.
– A utility function ui : A1 x …x An -> R
• This modeling implies that players choose
actions simultaneously.
Example: Prisoner’s dilemma
• The story: Two partners wish to terminate the partnership. Each
one needs to hire a lawyer, to reach a settlement on how to split
their property. Total property is worth 10.
– Two types of lawyers: cheap costs 1, expensive costs 5.
– If both choose same type, property is split equally.
– Otherwise the one with the expensive lawyer gets everything.
“C”
• So, the game is:
“E”
“C”
4,4
-1 , 5
“E”
5 , -1
0,0
What will happen in this game?
DFN: Action A dominates action B (for player i) if for any
combination of actions of the other players, a-i, ui(A,a-i) > ui(B,a-i).
Action A is a dominant action if it dominates all other actions.
• Observation:
“E” is a dominant strategy.
• Assumption: Players indeed
play a dominant strategy if
they have one.
“C”
“E”
“C”
4,4
-1 , 5
“E”
5 , -1
0,0
A one-item auction
The story:
• A seller wishes to sell a good to n players. Player i will obtain
a value of vi > 0 from having the good. vi is known only to i.
• The seller can charge a payment pi from player i.
In this case player i’s utility is: vi – pi
A popular auction format (“English auction”):
• The seller starts from a very low price. All players that are
interested in buying the item in this price raise their hands.
• The seller gradually increases the price, and uninterested
buyers lower their hands.
• When only one hand remains up, the price stops. The
remaining player gets the item, and pays the last price.
How should we analyze this?
• Problem: this is not a normal-form game.
• So, let’s assume for a moment that each player drops exactly
when the price reaches his value:
– Then the player with the highest value wins, and pays the
second highest price.
• Therefore we get the following “sealed-bid” 2nd price auction:
– Each player submits a bid (integer number).
– The player with the highest bid gets the item, and pays the
second highest bid.
• Now, how should we analyze this?
Drawing the game (for two players)
• The strategy space of each player is composed of all integers.
• For fixed v1, v2, if player 1 plays b1 and player 2 plays b2,
and b2 > b1, then the utility of player 1 is 0, and of 2 is v2 – b1.
b2
.
.
.
b1
0, v2 – b1
Analysis for n=2
Claim: For fixed v1, v2, a dominant action for player i=1,2 is
to play bi=vi
Proof: Fix any b2
• If v1 > b2 then winning is better than losing for player 1.
Declaring b1=v1 will cause player 1 to win.
• If v1 < b2 then losing is better than winning for player 1.
Declaring b1=v1 will cause player 1 to lose.
Analysis for n>2
• For n > 2 players the game and the claim are similar.
Claim: For fixed v1, …,vn the dominant action of player i is to
play bi=vi
Proof: Fix any b-i , and let x = max j  i bj
• If vi > x then winning is better than losing for player i.
Declaring bi=vi will cause player i to win.
• If vi < x then losing is better than winning for player i.
Declaring bi=vi will cause player i to lose.
A game in incomplete information
• For each tuple of vi’s we get a new game: the strategy spaces
are all the same, but the utility functions may be different.
• A player knows his own type (the vi), but not the other types.
• Thus a player do not fully know the matrix of the game.
• A strategy si : vi -> Ai determines the player’s action, as a
function of his type.
• A strategy si is dominant if the action si(vi) is dominant for
every vi and v-i
THM: si(vi) = vi is a dominant strategy in the 2nd price auction.
Proof: Immediate from the previous claim.
Few more definitions and remarks
• A mechanism is essentially a game of incomplete information
• A mechanism is direct-revelation if the bidders are supposed
to declare their type (like the 2nd price auction).
• A direct-revelation mechanism is truthful if, in equilibrium,
players declare their true types (like the 2nd price auction).
Checkpoint
By now we know:
• Normal-form games and dominant actions.
• Mechanisms with incomplete information and dominant
strategies.
• The 2nd price auction (description, its truthfulness)
1st price auction
• What about the natural 1st price rule: The highest bidder
wins, and pays his bid?
• Observation: There is no dominant strategy in this game (if
a player wins with a bid b he would have preferred to say
slightly less than b).
Reminder: continuous probability
• Suppose X is a random variable that takes values in the
interval [a,b]. F() is its cumulative distribution function
if Pr(X < ) = F() for all  in [a,b].
• f(x)=F’(x) is called a probability distribution function.
• Remark: In contrast to discrete probability, here there is no positive
probability for a specific point, only to an interval.
• Example: The uniform distribution over [a,b] is F(x)=(x-a)/(b-a)
– F(a)=0, F(b)=1, F((a+b)/2)=1/2
• The expectation of X is E(X) = a f(x)·x·dx
– The interpretation (by the central limit theorem): the average of
many samples from f() will approach the expectation.
– If X is uniformly distributed over [a,b] then E(X)=(a+b)/2
b
Nash Equilibrium (in complete information)
• In the “coordination game”, players gain a positive utility if and
only if they play the same action.
“A”
“B”
“A”
2,2
DFN
• Action ai is best response to a-i if
0,0
for any other action a’i of player i,
“B”
ui(ai, a-i) > ui(a’i, a-i)
• The actions (a1,…,an) are in Nash equilibrium
ai is best response to a-i for all players i.
• What are the Nash equilibria in the coordination game?
0,0
1,1
Bayesian-Nash Equilibrium
(in incomplete information)
• Since players do not know the matrix of the game, we extend
the notion of Nash to Incomplete settings in several ways.
• Assumption: vi is drawn from a probability distribution fi, and
these distributions are known to all (“common priors”).
DFN: The strategies s1,…, sn are in Bayesian-Nash equilibrium
if for any i, vi, ai : Ev-i[ui(si(vi),s-i(v-i)] > Ev-i[ui(ai,s-i(v-i)]
Remark: One could think of two other definitions: (I) where the
expectation is on vi as well (this is called ex-ante equilibrium),
and (II) the “for any” quantifier is also on v-i, and no
expectation is involved (this is called ex-post equilibrium).
Analysis of the 1st price auction
• Clearly, in the 1st price auction a bidder needs to “shade
down” his value. The question is by how much?
• We will find a Bayesian-Nash equilibrium under the
following simplifying assumptions:
– vi is independently drawn from the uniform distribution
on [0,1] (Pr[vi < y] = y).
– si(vi) = xi·vi (the bidder shades down his value by a
constant fraction)
• We want to find strategies si(vi) that form a Bayesian-Nash
equilibrium.
• We will “guess” that si(vi) = x · vi and we will find the
appropriate x that indeed satisfies the equilibrium properties.
• Given that all players j  i play sj(vj), player i chooses a bid b
to maximize her expected utility:
ui(b) = (vi – b) · Pr[maxji (x · vj ) < b]
• We have:
– Pr[maxji (x · vj ) < b] = ji Pr[x · vj < b]
– Pr[x·vj < b] = Pr[vj < b / x] = min (b/x , 1)
• If b/x > 1 then player i wins with probability 1, but this b
clearly does not maximize the utility since by taking b’ < b
such that b’/x > 1 we increase the utility. Thus b/x < 1.
• We get ui(b) = (vi – b) · bn-1 / xn-1
• We want to find the b that maximizes the utility, so we take the
derivative:
u’(b) = [(n-1) · vi · bn-2 – n · bn-1]/ xn-1 = 0  b=[(n-1)/n] · vi
THM: A Bayesian-Nash equilibrium of the 1st price auction is
when every player i bids si(vi) =[(n-1)/n] · vi
Remarks
• 1st price auction is equivalent to a descending (“Dutch”) auction:
the auctioneer gradually lowers the price, the first to accept wins,
for this price.
• A comparison to second price:
– No dominant strategies, so less obvious how to play.
– In both auctions, in equilibrium, the bidder with the highest
value gets the item (the “efficient” outcome).
– 2nd price may look bad - the winner’s price may be much lower
than his bid. An extreme once happened in a New-Zealand governmentauction: One firm bid NZ$100,000 for a license, and paid the second-highest
price of only NZ$6 (http://www.economicprincipals.com/issues/06.05.21.html).
– But which auction indeed raises more revenue?
1st price
• We have n random variables X1,…,Xn with
– The probability density function is f(x)=1 (for 0<x<1)
– The cumulative distribution function is F(x)=x
– Let Y1(n) = maxi Xi . Y1(n) is a random variable [with c.d.f.
F1(n)(x) and p.d.f. f1(n)(x)].
1st price
• We have n random variables X1,…,Xn with
– The probability density function is f(x)=1 (for 0<x<1)
– The cumulative distribution function is F(x)=x
– Let Y1(n) = maxi Xi . Y1(n) is a random variable [with c.d.f.
F1(n)(x) and p.d.f. f1(n)(x)].
What are its statistical properties?
1st price
• We have n random variables X1,…,Xn with
– The probability density function is f(x)=1 (for 0<x<1)
– The cumulative distribution function is F(x)=x
– Let Y1(n) = maxi Xi . Y1(n) is a random variable [with c.d.f.
F1(n)(x) and p.d.f. f1(n)(x)].
• F1(n)(x)=[F(x)]n
1st price
• We have n random variables X1,…,Xn with
– The probability density function is f(x)=1 (for 0<x<1)
– The cumulative distribution function is F(x)=x
– Let Y1(n) = maxi Xi . Y1(n) is a random variable [with c.d.f.
F1(n)(x) and p.d.f. f1(n)(x)].
• F1(n)(x)=[F(x)]n => f1(n)(x) = n [F(x)]n-1 f(x)=nxn-1
1st price
• We have n random variables X1,…,Xn with
– The probability density function is f(x)=1 (for 0<x<1)
– The cumulative distribution function is F(x)=x
– Let Y1(n) = maxi Xi . Y1(n) is a random variable [with c.d.f.
F1(n)(x) and p.d.f. f1(n)(x)].
• F1(n)(x)=[F(x)]n => f1(n)(x) = n [F(x)]n-1 f(x)=nxn-1
• E[Y1
(n)]
= 0 f(x)xdx = 0
1
1
nxndx
=
[n/(n+1)]xn+1
|0
1
1st price
• We have n random variables X1,…,Xn with
– The probability density function is f(x)=1 (for 0<x<1)
– The cumulative distribution function is F(x)=x
– Let Y1(n) = maxi Xi . Y1(n) is a random variable [with c.d.f.
F1(n)(x) and p.d.f. f1(n)(x)].
• F1(n)(x)=[F(x)]n => f1(n)(x) = n [F(x)]n-1 f(x)=nxn-1
• E[Y1
(n)]
= 0 f(x)xdx = 0
1
1
nxndx
=
[n/(n+1)]xn+1
1
|0 = n/(n+1)
1st price
• We have n random variables X1,…,Xn with
– The probability density function is f(x)=1 (for 0<x<1)
– The cumulative distribution function is F(x)=x
– Let Y1(n) = maxi Xi . Y1(n) is a random variable [with c.d.f.
F1(n)(x) and p.d.f. f1(n)(x)].
• F1(n)(x)=[F(x)]n => f1(n)(x) = n [F(x)]n-1 f(x)=nxn-1
• E[Y1
(n)]
= 0 f(x)xdx = 0
1
1
nxndx
=
[n/(n+1)]xn+1
1
|0 = n/(n+1)
• E[Revenue of 1st price] = E[s(Y1(n))] = E[ ((n-1)/n) Y1(n) ]
1st price
• We have n random variables X1,…,Xn with
– The probability density function is f(x)=1 (for 0<x<1)
– The cumulative distribution function is F(x)=x
– Let Y1(n) = maxi Xi . Y1(n) is a random variable [with c.d.f.
F1(n)(x) and p.d.f. f1(n)(x)].
• F1(n)(x)=[F(x)]n => f1(n)(x) = n [F(x)]n-1 f(x)=nxn-1
• E[Y1
(n)]
= 0 f(x)xdx = 0
1
1
nxndx
=
[n/(n+1)]xn+1
1
|0 = n/(n+1)
• E[Revenue of 1st price] = E[s(Y1(n))] = E[ ((n-1)/n) Y1(n) ] =
= [(n-1)/n]E[Y1(n)] = (n-1)/(n+1)
2nd price
• Let Y2(n) be the second highest of X1,…,Xn (with c.d.f. F2(n)(x))
2nd price
• Let Y2(n) be the second highest of X1,…,Xn (with c.d.f. F2(n)(x))
• F2(n)(x) = ??
2nd price
• Let Y2(n) be the second highest of X1,…,Xn (with c.d.f. F2(n)(x))
• F2(n)(x) = F(x)n + n F(x)n-1(1-F(x))
2nd price
• Let Y2(n) be the second highest of X1,…,Xn (with c.d.f. F2(n)(x))
• F2(n)(x) = F(x)n + n F(x)n-1(1-F(x)) = F(x)n + n F(x)n-1-nF(x)n =
= n F(x)n-1 - (n-1) F(x)n
2nd price
• Let Y2(n) be the second highest of X1,…,Xn (with c.d.f. F2(n)(x))
• F2(n)(x) = F(x)n + n F(x)n-1(1-F(x)) = F(x)n + n F(x)n-1-nF(x)n =
= n F(x)n-1 - (n-1) F(x)n = n F1(n-1)(x) – (n-1) F1(n)(x)
2nd price
• Let Y2(n) be the second highest of X1,…,Xn (with c.d.f. F2(n)(x))
• F2(n)(x) = F(x)n + n F(x)n-1(1-F(x)) = F(x)n + n F(x)n-1-nF(x)n =
= n F(x)n-1 - (n-1) F(x)n = n F1(n-1)(x) – (n-1) F1(n)(x)
=> E[Y2(n)] = n E[Y1(n-1)] – (n-1) E[Y1(n)]
2nd price
• Let Y2(n) be the second highest of X1,…,Xn (with c.d.f. F2(n)(x))
• F2(n)(x) = F(x)n + n F(x)n-1(1-F(x)) = F(x)n + n F(x)n-1-nF(x)n =
= n F(x)n-1 - (n-1) F(x)n = n F1(n-1)(x) – (n-1) F1(n)(x)
=> E[Y2(n)] = n E[Y1(n-1)] – (n-1) E[Y1(n)]
= n(n-1)/n – (n-1)n/(n+1) = (n-1)(1 – n/(n+1))=
= (n-1)/(n+1)
2nd price
• Let Y2(n) be the second highest of X1,…,Xn (with c.d.f. F2(n)(x))
• F2(n)(x) = F(x)n + n F(x)n-1(1-F(x)) = F(x)n + n F(x)n-1-nF(x)n =
= n F(x)n-1 - (n-1) F(x)n = n F1(n-1)(x) – (n-1) F1(n)(x)
=> E[Y2(n)] = n E[Y1(n-1)] – (n-1) E[Y1(n)]
= n(n-1)/n – (n-1)n/(n+1) = (n-1)(1 – n/(n+1))=
= (n-1)/(n+1)
• E[Revenue of 2nd price] = E[Y2(n)] = (n-1)/(n+1)
2nd price
• Let Y2(n) be the second highest of X1,…,Xn (with c.d.f. F2(n)(x))
• F2(n)(x) = F(x)n + n F(x)n-1(1-F(x)) = F(x)n + n F(x)n-1-nF(x)n =
= n F(x)n-1 - (n-1) F(x)n = n F1(n-1)(x) – (n-1) F1(n)(x)
=> E[Y2(n)] = n E[Y1(n-1)] – (n-1) E[Y1(n)]
= n(n-1)/n – (n-1)n/(n+1) = (n-1)(1 – n/(n+1))=
= (n-1)/(n+1)
• E[Revenue of 2nd price] = E[Y2(n)] = (n-1)/(n+1)
• Thus E[Revenue of 1st price] = E[Revenue of 2nd price] !!
Questions
• Is this by accident?
• Can we design an auction with a higher revenue?