CPS 196.
2
Bayesian games and their
use in auctions
Vincent Conitzer
conitzer@cs.duke.edu
What is mechanism design?
• In mechanism design, we get to design the game (or
mechanism)
– e.g. the rules of the auction, marketplace, election, …
• Goal is to obtain good outcomes when agents
behave strategically (game-theoretically)
• Mechanism design often considered part of game
theory
• Sometimes called “inverse game theory”
– In game theory the game is given and we have to figure
out how to act
– In mechanism design we know how we would like the
agents to act and have to figure out the game
• 2007 Nobel Prize in Economics!
Example: (single-item) auctions
• Sealed-bid auction: every bidder submits bid in a
sealed envelope
• First-price sealed-bid auction: highest bid wins, pays
amount of own bid
• Second-price sealed-bid auction: highest bid wins,
pays amount of second-highest bid
bid 1: $10
first-price: bid 1 wins, pays $10
bid 2: $5 second-price: bid 1 wins, pays $5
bid 3: $1
0
Which auction generates more revenue?
• Each bid depends on
– bidder’s true valuation for the item (utility = valuation - payment),
– bidder’s beliefs over what others will bid (→ game theory),
– and... the auction mechanism used
• In a first-price auction, it does not make sense to bid your true
valuation
– Even if you win, your utility will be 0…
• In a second-price auction, (we will see later that) it always
makes sense to bid your true valuation
bid 1: $10
a likely a likely outcome
outcome for for the second-
the first-price bid 1: $5 price mechanism bid 2: $5
mechanism bid 2: $4
bid 3: $1 bid 3: $1
0 0
Are there other auctions that perform better? How do we know when we have found the best one?
Bidding truthfully is optimal in
the Vickrey auction!
• What should a bidder with value v bid?
Option 1: Win
the item at price
b, get utility v - b Would like to win if
b = highest bid and only if v - b > 0 –
among other but bidding truthfully
bidders accomplishes this!
Option 2: Lose
the item, get
utility 0
We say the Vickrey
0 auction is strategy-proof
Collusion in the Vickrey auction
• Example: two colluding bidders
v1 = first colluder’s
true valuation
v2 = second price colluder 1 would pay when
colluder’s true colluders bid truthfully
valuation gains to be distributed among colluders
b = highest bid
price colluder 1 would pay if
among other bidders
colluder 2 does not bid
0
Bayesian games
• In a Bayesian game a player’s utility depends on that player’s
type as well as the actions taken in the game
– Notation: θi is player i’s type, drawn according to some distribution from
set of types Θi
– Each player knows/learns its own type, not those of the others, before
choosing action
• Pure strategy si is a mapping from Θi to Ai (where Ai is i’s set of actions)
– In general players can also receive signals about other players’
utilities; we will not go into this
L R L R
row player U 4 6 column player U 4 6
type 1 (prob. 0.5) D 2 4 type 1 (prob. 0.5) D 4 6
L R L R
row player U 2 4 column player U 2 2
type 2 (prob. 0.5) D 4 2 type 2 (prob. 0.5) D 4 2
Converting Bayesian games to normal form
L R L R
row player U 4 6 column player U 4 6
type 1 (prob. 0.5) D 2 4 type 1 (prob. 0.5) D 4 6
L R L R
row player U 2 4 column player U 2 2
type 2 (prob. 0.5) D 4 2 type 2 (prob. 0.5) D 4 2
type 1: L type 1: L type 1: R type 1: R
type 2: L type 2: R type 2: L type 2: R
type 1: U
type 2: U
3, 3 4, 3 4, 4 5, 4
type 1: U exponential
type 2: D
4, 3.5 4, 3 4, 4.5 4, 4 blowup in size
type 1: D
type 2: U
2, 3.5 3, 3 3, 4.5 4, 4
type 1: D
type 2: D
3, 4 3, 3 3, 5 3, 4
Bayes-Nash equilibrium
• A profile of strategies is a Bayes-Nash
equilibrium if it is a Nash equilibrium for the
normal form of the game
– Minor caveat: each type should have >0
probability
• Alternative definition: for every i, for every type
θi, for every alternative action ai, we must
have:
Σθ-i P(θ-i) ui(θi, σi(θi), σ-i(θ-i)) ≥
Σθ-i P(θ-i) ui(θi, ai, σ-i(θ-i))
First-price sealed-bid auction BNE
• Suppose every bidder (independently) draws a
valuation from [0, 1]
• What is a Bayes-Nash equilibrium for this?
• Say a bidder with value vi bids vi(n-1)/n
• Claim: this is an equilibrium!
• Proof: suppose all others use this strategy
• For a bid b < (n-1)/n, the probability of winning is
(bn/(n-1))n-1, so the expected value is (vi-b)(bn/(n-1))n-1
• Derivative w.r.t. b is - (bn/(n-1))n-1 + (vi-b)(n-1)bn-2(n/(n-
1))n-1 which should equal zero
• Implies -b + (vi-b)(n-1) = 0, which solves to b = vi(n-1)/n
Analyzing the expected revenue of the first-price
and second-price (Vickrey) auctions
• First-price auction: probability of there not being a
bid higher than b is (bn/(n-1))n (for b < (n-1)/n)
– This is the cumulative density function of the highest bid
• Probability density function is the derivative, that is,
it is nbn-1(n/(n-1))n
• Expected value of highest bid is
n(n/(n-1))n∫(n-1)/nbndb = (n-1)/(n+1)
• Second-price auction: probability of there not being
two bids higher than b is bn + nbn-1(1-b)
– This is the cumulative density function of the second-highest bid
• Probability density function is the derivative, that is,
it is nbn-1 + n(n-1)bn-2(1-b) - nbn-1 = n(n-1)(bn-2 - bn-1)
• Expected value is (n-1) - n(n-1)/(n+1) = (n-1)/(n+1)
Revenue equivalence theorem
• Suppose valuations for the single item are drawn
i.i.d. from a continuous distribution over [L, H] (with
no “gaps”), and agents are risk-neutral
• Then, any two auction mechanisms that
– in equilibrium always allocate the item to the bidder with
the highest valuation, and
– give an agent with valuation L an expected utility of 0,
will lead to the same expected revenue for the
auctioneer
(As an aside) what if bidders are not risk-neutral?
• Behavior in second-price/English/Japanese does
not change, but behavior in first-price/Dutch does
• Risk averse: first price/Dutch will get higher
expected revenue than second
price/Japanese/English
• Risk seeking: second price/Japanese/English will
get higher expected revenue than first price/Dutch
(As an aside) interdependent valuations
• E.g. bidding on drilling rights for an oil field
• Each bidder i has its own geologists who do tests,
based on which the bidder assesses an expected
value vi of the field
• If you win, it is probably because the other bidders’
geologists’ tests turned out worse, and the oil field is
not actually worth as much as you thought
– The so-called winner’s curse
• Hence, bidding vi is no longer a dominant strategy in
the second-price auction
• In English and Japanese auctions, you can update
your valuation based on other agents’ bids, so no
longer equivalent to second-price
• In these settings, English (or Japanese) > second-
price > first-price/Dutch in terms of revenue
Expected-revenue maximizing
(“optimal”) auctions [Myerson 81]
• Vickrey auction does not maximize expected revenue
– E.g. with only one bidder, better off making a take-it-or-
leave-it offer (or equivalently setting a reserve price)
• Suppose agent i draws valuation from probability
density function fi (cumulative density Fi)
• Bidder’s virtual valuation ψ(vi)= vi - (1 - Fi(vi))/fi(vi)
– Under certain conditions, this is increasing; assume this
• The bidder with the highest virtual valuation (according
to his reported valuation) wins (unless all virtual
valuations are below 0, in which case nobody wins)
• Winner pays value of lowest bid that would have
made him win
• E.g. if all bidders draw uniformly from [0, 1], Myerson
auction = second-price auction with reserve price ½
Vickrey auction without a seller
v( )=2 v( )=4 v( )=3
pays 3
(money wasted!)
Can we redistribute the payment?
Idea: give everyone 1/n
of the payment
v( )=2 v( )=4 v( )=3
receives 1 pays 3 receives 1
receives 1
not strategy-proof
Bidding higher can increase your redistribution payment
Incentive compatible redistribution
[Bailey 97, Porter et al. 04, Cavallo 06]
Idea: give everyone 1/n of
second-highest other bid
v( )=2 v( )=4 v( )=3
receives 1 pays 3 receives 2/3
receives 2/3
2/3 wasted (22%)
Strategy-proof
Your redistribution does not depend on your bid;
incentives are the same as in Vickrey
Bailey-Cavallo mechanism…
• Bids: V1≥V2≥V3≥... ≥Vn≥0 R1 = V3/n
• First run Vickrey auction R2 = V3/n
• Payment is V2 R3 = V2/n
• First two bidders receive V3/n R4 = V2/n
• Remaining bidders receive V2/n ...
• Total redistributed: 2V3/n+(n- Rn-1= V2/n
2)V2/n Rn = V2/n
Is this the best possible?
Another redistribution mechanism
• Bids: V1≥V2≥V3≥V4≥... ≥Vn≥0
• First run Vickrey
R1 = V3/(n-2) - 2/[(n-2)(n-3)]V4
• Redistribution: R2 = V3/(n-2) - 2/[(n-2)(n-3)]V4
Receive 1/(n-2) * second- R3 = V2/(n-2) - 2/[(n-2)(n-3)]V4
highest other bid, R4 = V2/(n-2) - 2/[(n-2)(n-3)]V3
- 2/[(n-2)(n-3)] third-highest ...
other bid Rn-1= V2/(n-2) - 2/[(n-2)(n-3)]V3
• Total redistributed: Rn = V2/(n-2) - 2/[(n-2)(n-3)]V3
V2-6V4/[(n-2)(n-3)]
Idea pursued further in Guo & Conitzer 07 / Moulin 07