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Mechanism Design without Money

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

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  1. Mechanism Design without Money Lecture 2

  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.

  3. Here is another game • No dominant strategy for any player…

  4. 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

  5. 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

  6. Examples of Nash Equilibrium • Suppose each player i has a dominant strategy di. Then (d1… dn) is a Nash equilibrium

  7. Multiple Nash equilibria • The battle of sexes • Multiple Nash equilibria (see next slides)

  8. 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

  9. 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

  10. Multiple Nash equilibria • Different equilibria are good for different players. • Can some equilibria be better than others for everyone?

  11. Yes - coordination games

  12. 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

  13. What do you usually get? • Politiken played this with 19,196 for 5000 krones

  14. No (pure) Nash equilibrium • But players can randomize…

  15. 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

  16. 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

  17. 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

  18. 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.

  19. 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.

  20. 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

  21. 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)

  22. 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

  23. 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

  24. 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

  25. 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…

  26. 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!

  27. 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?

  28. 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

  29. Questions about mixed Nash?

  30. 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

  31. 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)

  32. 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

  33. 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

  34. 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!

  35. 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

  36. Proof continued • Suppose player 21 chose AXYB • U21(Nash) = -40 • U21(Nash-21,AXB) = U21 (Nash-21,AYB) = -41 • U21(Nash-21,AXYB) = -42

  37. 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?

  38. Back to the simpler game X 20 n • Let’s count the votes A B n 20 Y • Choose 5 for AXB • Choose 6 for AYB

  39. 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

  40. 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

  41. 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

  42. 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”

  43. 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

  44. 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

  45. 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

  46. Two ways to compute social welfare

  47. Nash flow • Theorem – in a Nash flow, for every demand j, all paths from sjto tj have the same cost

  48. 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

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

  50. 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

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