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## PowerPoint Slideshow about 'Mechanism Design without Money' - kiayada-madden

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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 xB 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
- Who knows about LP’s?

But what’s PPAD?

- 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?

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

Braess paradox

- By adding a road, we made the situation worse
- The paradox exists in road networks
- 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

Road example

- 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

Nash in road example

- 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

Braess’ paradox

- 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

Road example

- 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

Nash in road example

- 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

Braess’ paradox

- 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)

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