Course administration

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)

Action “T”

Action “B”

Action “R”Action “L”

Player I’s actions

Player II’s actions

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.

• So, the game is:

“C”

“E” 0 , 0

“C” “E”

4 , 4

5 , -1

-1 , 5

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 indeedplay a dominant strategy ifthey have one.

“C”

“E” 0 , 0

“C” “E”

4 , 4

5 , -1

-1 , 5

A one-item auctionThe 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.

.

.

.b1

b2

0, v2 – b1

Analysis for n=2Claim: 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 functionif 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) = ab 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

Nash Equilibrium (in complete information)

• In the “coordination game”, players gain a positive utility if and only if they play the same action.

DFN

• Action ai is best response to a-i iffor any other action a’i of player i,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?

1 , 1

2 , 2

0 , 0

0 , 0“A”

“B”

“A” “B”

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[maxji (x · vj ) < b]

• We have:

– Pr[maxji (x · vj ) < b] = ji 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 government-auction: 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?

http://www.economicprincipals.com/issues/06.05.21.html

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







What are its statistical properties?

1st price







• F1(n)(x)=[F(x)]n

1st price







• F1(n)(x)=[F(x)]n => f1

(n)(x) = n [F(x)]n-1 f(x)=nxn-1

1st price







• F1(n)(x)=[F(x)]n => f1

(n)(x) = n [F(x)]n-1 f(x)=nxn-1

• E[Y1(n)] = 0

1 f(x)xdx = 0

1 nxndx = [n/(n+1)]xn+1 |0

1

1st price





(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

1 f(x)xdx = 0

1 nxndx = [n/(n+1)]xn+1 |0

1 = n/(n+1)

1st price







• F1(n)(x)=[F(x)]n => f1

(n)(x) = n [F(x)]n-1 f(x)=nxn-1

• E[Y1(n)] = 0

1 f(x)xdx = 0

1 nxndx = [n/(n+1)]xn+1 |0

1 = n/(n+1)

• E[Revenue of 1st price] = E[s(Y1(n))] = E[ ((n-1)/n) Y1

(n) ]

1st price







• F1(n)(x)=[F(x)]n => f1

(n)(x) = n [F(x)]n-1 f(x)=nxn-1

• E[Y1(n)] = 0

1 f(x)xdx = 0

1 nxndx = [n/(n+1)]xn+1 |0

1 = 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


(n)(x))

• F2(n)(x) = ??

2nd price


(n)(x))

• F2(n)(x) = F(x)n + n F(x)n-1(1-F(x))

2nd price


(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


(n)(x))


= n F(x)n-1 - (n-1) F(x)n = n F1(n-1)(x) – (n-1) F1

(n)(x)

2nd price


(n)(x))


= 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


(n)(x))


= 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


(n)(x))


= 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


(n)(x))


= 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?

Documents

Course administration