Mechanism Design without Money

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Mechanism Design without Money. Lecture 2. Let’s start another way…. Everyone choose a number between zero and a hundred, and write it on the piece of paper you have, with your name. Then pass all the numbers to me. I will compute the average (hopefully correctly)

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### Mechanism Design without Money

Lecture 2

Let’s start another way…
• Everyone choose a number between zero and a hundred, and write it on the piece of paper you have, with your name.
• Then pass all the numbers to me.
• I will compute the average (hopefully correctly)
• The winner is the one who is closest to two thirds of the average.
Here is another game
• No dominant strategy for any player…
So how can we predict something?
• Imagine that the game is played many times.
• Imagine that at some point, the profile (R2, C2) is being played
• Then no player has incentive to move
Nash equilibrium
• The strategy profile is called a Nash equilibrium.
• Named after John Nash who proved existence in ‘51 (Nobel in ‘94). Original concept due to Von Neumann
• Formally: A strategy profile s = (s1, …sn) is a Nash equilibrium, if for every i and i we haveUi(s) ≥ Ui(s-i, i)
• So given that everyone else sticks, no player wants to move
Examples of Nash Equilibrium
• Suppose each player i has a dominant strategy di. Then (d1… dn) is a Nash equilibrium
Multiple Nash equilibria
• The battle of sexes
• Multiple Nash equilibria (see next slides)
Proof
• Claim: (Football, Football) is Nash equilibrium
• Proof:
• Consider AUalice(Football, Football) = 1 > Ualice (Play, Football) = 0
• Consider BUbob(Football, Football) = 2> Ubob (Football,play) = 0
Proof
• Claim: (Play,Play) is Nash equilibrium
• Proof:
• Consider A Ualice(Play, Play) = 2 > Ualice (Football, play) = 0
• Consider BUbob(Play, Play) = 1 > Ubob(Play, Football) = 0
Multiple Nash equilibria
• Different equilibria are good for different players.
• Can some equilibria be better than others for everyone?
What would be a good strategy?
• Well, you never want to put more than 66, right?
• But if everyone never puts more than 66, you never want to put more than 44, right?
• So everyone should play zero.
• There is a difference between rationality and common knowledge of rationality
What do you usually get?
• Politiken played this with 19,196 for 5000 krones
No (pure) Nash equilibrium
• But players can randomize…
Mixed Nash equilibrium
• Reminder – Si is the set of possible strategies of player i.
• Let pi : Si[0,1] be a probability distribution on strategies.
• Functions p1…pnare a (mixed) Nash equilibrium if for every player i and strategy i
Mixed Nash – properties (1)
• Claim: Every pure Nash is a mixed Nash. Proof: Let s1…snbe a pure Nash. Set pi(si)=1, pi(i) = 0 if i  si
Mixed Nash – properties (2)
• Claim: Let p1…pn be a Mixed Nash equilibrium. Then for every probability distribution qi on Si
• Proof is an exercise. Follows from linearity of expectation
Mixed Nash – properties (3)
• Claim: Let p1…pn be a Mixed Nash equilibrium. Then if pi(i) > 0, or equivalently i is in the support of pi then
• Proof is an exercise. Again follows from linearity of expectation.
Back to Rock Paper Scissors (RPS)
• Let p1(R)=p1(P)=p1(S)= 1/3 p2(R)=p2(P)=p2(S)= 1/3
• Then p1,p2 is a mixed Nash equilibrium for RPS, with utility 0 for both players.
• Note that property 2 and 3 hold:
• If player 1 switches to a distribution q1 she still gets expected utility 0
• For every pure strategy  in the support of p1, the expected utility for player 1 playing  is 0.
Existence of mixed Nash equilibrium
• Theorem (John Nash, 1951): In an N player game, if the strategy state is finite* a mixed equilibrium always exists
• Nobel prize in 1994
• The proof gives something a bit stronger
• The proof is based on Brower’s fixed point theorem
Brower’s fixed point theorem
• Brower Fixed point theorem: Let f be a continuous function from a compact set B Rn to itself. Then there exists xB with f(x)=x
• Examples:
• B = [0,1], f(x) = x2
• B=[0,1], f(x) = 1-x
• B=[0,1]2, f(x,y) = (x2, y2)
Brower’s fixed point theorem
• All the conditions are necessary
• Consider B = R, and f(x) = x+1
• Consider B = (0,1] and f(x) = x/2
• Consider B = the circle defined by x2+ y2=1, and f(x) is a rotation
Brower’s fixed point theorem
• Proof in 1D
• Let B = [a,b]
• If f(a) = a or f(b) = b we are done.
• Else define F(x) = f(x) – x
• Note:F(a) = f(a) – a > 0F(b) = f(b) – b < 0
• So there must be x such that F(x) = 0, or f(x)=x
Intuition for proving Nash given Brower
• Define an n dimensional function, which takes as input n strategies, and outputs n new strategies
• The Fi(s1…sn) is player i’s best response to s1…s-i…sn
• If F has a fixed point, it’s a Nash equilibrium
OK, so mixed Nash always exists
• But can we find it?
• Note it’s not NP complete or anything – it always exists
• For over 50 years, economists tried to come up with natural dynamics which would lead to a Nash equilibrium
• They always failed…
Finding special types of Nash
• Are there two Nash equilibria?
• Is there a Nash equlibrium where a player i gets utility more than k?
• Is there an equilibrium where player i has support larger than k?
• Is there an equilibrium where player i sometimes plays i?
• These are all NP complete!
So what can we say about finding just one Nash?
• Daskalakis Papadimitriou and Goldberg showed that finding a Nash is PPAD complete even for two player games
• Finding a Nash is just as hard as finding the fixed point
• And the proof we gave was not constructive…
• If there are two players, and we know the support of pi for each player, then finding the probabilities is just solving an LP
• Suppose you have a directed graph on 2n vertices (dente them 0, 1, … 2n-1), with the following properties:
• The in degree and the out degree of each vertex is at most 1
• Vertex 0 has out degree 1, and in degree 0
• You want to find a vertex x with in degree 1 and out degree 0
• Such an x always exists
• Finding it is PPAD complete
Back to the Prisoner’s Dilemma
• Both players confessing is a Nash equilibrium
• But it sux…
• Both players remaining silent is great
• But it’s not an equilibrium
How much can a Nash equilibrium Suck?
• Or in a more clean language:
• Consider a game G. The social welfare of a profile s is defined asWelfare (s) = iUi(s)
• Let O be the profile with maximal social welfare
• Let N be the Nash equilibrium profile with minimal social welfare (worst Nash solution)
• The Price of Anarchy of G is defined to beWelfare(O) / Welfare(N)
When is this notion meaningful?
• In the first game, PoA = 3. In the second, 100
• But it’s really the same game…
• PoA makes sense only when there is a real bound, based on the structure of the game
Back to the game we played…

X

20

n

• Time to count the votes…

A

B

0

n

20

Y

• Choose 1 for AXB
• Choose 2 for AXYB
• Choose 3 for AYXB
• Choose 4 for AYB
Optimal solution
• Suppose there are 26 players.
• 10 go for AYXB, and 16 for AXYB
• AYXB players take 10 minutes each
• AXYB players take 40 minutes each
• Total time spend on the road is 16*40+10*20 = 840 minutes
• But is this a Nash equilibrium?
• No – if a player moves from AXYB to AYXB they save 18 minutes!
Nash equilibrium
• Utility of a player is minus the time spent on the road
• Claim: The following is a Nash equilibrium:
• 20 players take route AYXB
• 6 players take route AXYB
• Proof: For every player i and deviation si we need to show that Ui(Nash)≥Ui(Nash-i,si)
• Suppose player 1 chose AYXB
• U1(Nash) = -40
• U1(Nash-1,AXB) = U1(Nash-1,AXB) = U1(Nash-1,AXB) = -40
Proof continued
• Suppose player 21 chose AXYB
• U21(Nash) = -40
• U21(Nash-21,AXB) = U21 (Nash-21,AYB) = -41
• U21(Nash-21,AXYB) = -42
Equilibrium Analysis
• The total time spent on the road is 26*40 = 1040 minutes
• Worse than the optimal time of 858 minutes, but not much worse
• How does that compare to us?
Back to the simpler game

X

20

n

A

B

n

20

Y

• Choose 5 for AXB
• Choose 6 for AYB
Analysis of the simpler game
• Same optimal solution as for the game with XY:
• 13 players use AYB, and 13 use AXB
• Same time on the road, 26*33 = 858 minutes
Nash equilibria of the simple game
• Claim: 13 people use AXB and 13 use AYB is a Nash equilibrium of the simpler game
• Proof: Suppose player 1 uses AXB
• U1(Nash) = -33
• U1(Nash-1 , AYB) = -34
• Similarly, for a player who uses AYB
• Making 42nd street one way
• Simulations on road networks in various cities
Can we bound Price of Anarchy on the road?
• Yes, but we won’t finish this lesson.
• Let’s begin by formally defining a “road network” and showing that a pure Nash exists
• Based on “potential functions”
Routing games
• The problem has three ingredients:
• A graph G
• Demands: Each demand (commodity) is of the form: sj,tj meaning j want to move 1 unit from a vertex sj to a vertex tj
• Each edge has a cost function: a monotone continuous function from traffic to the real numbers
Routing game example

X

• G is given
• We want to route 1 unit from A to B, through AXB or through AYB
• The costs are given. AX=AY=0, XB=x, YB=1

0

x

A

B

0

1

Y

How much do you pay?
• Suppose we have a flow on the graph
• Each edge now has a cost – the function evaluated on the flow
• Each path has a cost – the sum of costs of all edges in the paths
• Each demand has a cost. If for every pj the demand passes xj on it, the cost is
Nash flow
• Theorem – in a Nash flow, for every demand j, all paths from sjto tj have the same cost
Marginal cost
• The cost of an edge is fecost(fe)
• The marginal cost of an edge is(fecost(fe) )’ = cost(fe) + fecost’(fe)
• Marginal cost of a path is the sum of marginal costs of its edges
• Theorem: In an optimal flow, all paths from sjto tj have the same marginal cost

Theorem – A flow f is optimal in G, if and only if it is Nash with respect to the marginal cost

Using the theorem
• We know how to find optimal flows (greedy algorithm works)
• Can we use this to get Nash flows?
• For every e, we want a function gesuch that ge’ = ce
• Then we can find the optimal flows according to the cost function ge
• Using the theorem it’s a Nash flow for the cost ce
Questions?
• Feedback
• Office hours
• 50 people want to get from A to B
• There are two roads, each one has two segments. One takes an hour, and the other one takes the number of people on it

1 hour

N minutes

A

B

1 hour

N minutes

• In the Nash equilibrium, 25 people would take each route, for a travel time of 85 minutes

1 hour

N minutes

A

B

1 hour

N minutes

• Now suppose someone adds an extra road which takes no time at all. Travel time goes to 100 minutes

1 hour

N minutes

A

B

Free

1 hour

N minutes

Multiple Nash equilibria
• The battle of sexes
• See multiple equilibria on the board
• Note different equilibria are better for some players
No (Pure) Nash equilibrium
• We will get back to this – players can randomize
Back to the Prisoner’s Dilemma
• Both players confessing is a Nash equilibrium
• But it sux…
How much can a Nash equilibrium Suck?
• Or in a more clean language:
• Consider a game G. The social welfare of a profile s is defined asWelfare (s) = iUi(s)
• Let O be the profile with maximal social welfare
• Let N be the Nash equilibrium profile with minimal social welfare (worst Nash solution)
• The Price of Anarchy of G is defined to beWelfare(O) / Welfare(N)
Price of Anarchy
• Suppose a 100 people want to get from BIU to Jerusalem
• The train takes two hours
• Driving the car takes 1 hour+1 minute for every other driver
• How many people will drive to Jerusalem?
The train game
• Each player has two strategies – Car and Train.
• The utility of each player is 0 for taking the train (regardless of the number of passengers)
• The utility of taking the car is 60 – the number of car drivers.
• The Nash equilibrium is that 60 drivers take the car and 40 take the train. Social welfare = 0
• The optimal solution is that 30 people take the car, for a social welfare of 30*30 = 900
• Think about the new entrance to Tel Aviv
• Rigorous treatment of Price of Anarchy later
Questions?
• Feedback
• Office hours
• 50 people want to get from A to B
• There are two roads, each one has two segments. One takes an hour, and the other one takes the number of people on it

1 hour

N minutes

A

B

1 hour

N minutes

• In the Nash equilibrium, 25 people would take each route, for a travel time of 85 minutes

1 hour

N minutes

A

B

1 hour

N minutes

• Now suppose someone adds an extra road which takes no time at all. Travel time goes to 100 minutes

1 hour

N minutes

A

B

Free

1 hour

N minutes

Feedback points
• Lectures online: look in http://u.cs.biu.ac.il/~avinatan/
• Slides are numbered
• Algorithmic versus game theoretic focus – my initial plan was to give a few lectures of background, but will try to give juice today
• Relevant book chapters – 17,18 (In Algorithmic Game Theory). Focus on 18.3
A scheduling problem
• I can’t do Tuesday next week.
• Two options:
• Will ask someone to fill me in
• Will move to a different time
• I prefer the second, but it depends on you
• Two stage vote – first we find a good time, and then you vote if you want a filler or not
A game

X

20

n

• You need to get from A to B
• Travelling on AX or YB takes 20 minutes
• Travelling on AY or XB takes n minutes, where there are n travellers
• XY takes no time

A

B

0

n

20

Y

• Choose 1 for AXB
• Choose 2 for AXYB
• Choose 3 for AYXB
• Choose 4 for AYB
A simpler game

X

20

n

• You need to get from A to B
• Travelling on AX or YB takes 20 minutes
• Travelling on AY or XB takes n minutes, where there are n travellers

A

B

n

20

Y

• Choose 5 for AXB
• Choose 6 for AYB
Reminder
• Nash equilibrium: A strategy profile s = (s1, …,sn) is a Nash equilibrium, if for every i and i we haveUi(s) ≥ Ui(s-i, i)