normal form games with complete information part 2

iterated elimination of dominated strategies dominance solvability roadmap iterated elimination of dominated strategies dominance solvability strong and weak versions examples discussion rationality order of elimination non-existence

dominated strategies to sum up si’ is strictly dominated by si for player i if for all s-iє S-i, ui(si,s-i) > u i(si’,s-i) si’ is (weakly) dominated by si for player i if for all s-iє S-i, ui(si,s-i) ≥ u i(si’,s-i) and for some s-iє S-i, ui(si,s-i) > u i(si’,s-i) (and si is an undominated strategy for player i if it is not dominated by any other strategy)

example: prisoners’ dilemma 2 1 -1,-1 -6,0 0,-6 -3,-3 Not to confess

example: weakly dominated strategies 2 1 L R U 1,1 0,1 M 0,2 1,0 D 0,-1 0,0

best-response and question si is a best response to s-i if for all s’iє Si, ui(si,s-i) ≥ u i(si’,s-i) Can you relate the best-response concept with the concept of dominant strategy? R: si is a dominant strategy iff it is a best-response to every s-i

iterated elimination of dominated strategies (IEDS) If no rational player will play a dominated strategy, a rational player would not expect her opponents to play a dominated strategy Elimination of dominated strategies can therefore lead to a chain reaction that successively narrows down how a group of rational players will act. If there is a unique prediction that arises from IEDS, that is the IEDS solution and the game is said to be dominance solvable IEDS can be simultaneous or sequential

example of IEDS: weak 2 1 L M R U 1,1 0,1 1,2 D -1,-1 -1,0 0,0

example of IEDS: strong 2 1 L M R U 1,1 0,1 1,2 D -1,-1 -1,0 0,0

iterated elimination of dominated strategies (IEDS) A rational player never chooses a strictly dominated strategy But a rational player may choose a dominated strategy! Rationality is not enough to rule out weakly dominated stratgeies! IE(W)DS requires rationality + knowing that others are rational + … (rationality has to be common knowledge) + knowing others’ payoffs. Elimination of a strictly dominated strategy requires only rationality of the player.

discussion: order of elimination in IE(W)DS 2 1 L R U 0,0 0,1 D 1,0 Order of elimintion matters bc leads to different outcomes! Order of elimination matters in IEWDS when we remove dominated strategies in turn (ie, one player at a time). But not if we remove all players’ dominated strategies at the same time

discussion of IEDS: layers of rationality 2 1 L C R U 4,5 1,6 5,6 M 3,5 2,5 5,4 D 2,0 7,0 R dominada por C D dominada por M e U L dominada por C U dominada por M Note-se que 2 pode garantir um payoff 5 jogando L; só joga C pq tem a certeza q 1 é racional… 1 nãousa D pq 2 é racional e não usa R 2 não usa L pq sabe que 1 é racional e sabe que 1 sabe que ele prórpio é racional: 2 não usa R, logo 1 não usa D, log 2 não usa L 1 não usa U pq sabe que 2 é racional e sabe que 2 sabe que ele próprio é racional e sabe que 2 sabe que ele sabe que 2 é racional…. Uff… IE(W)DS requires rationality + knowing that others are rational + … (rationality has to be common knowledge) + knowing others’ payoffs. Elimination of a strictly dominated strategy requires only rationality of the player.

discussion of IEDS: non-existence 2 1 Heads Tails -1,+1 +1,-1

example 1 (homework) Each player i of n players selects a number xi between 0 and 100 simultaneously. Player i’s payoff is ui = xi-3y/5, where y denotes the average of all numbers chosen

example 1 (homework) Normal form representation Is there any dominated strategy? What numbers should be selected? All strategies are strictly dominated by 100 (the utility function is strictly increasing in xi) Everyone will select 100!

example 2 (homework) Two players (1 and 2) are bidding at a painting’s auction. Their valuations (represented by vi) (willingness to pay) are common knowledge and are such that v1 > v2. The auction format is as follows. Each player i (simultaneously) submits a non-negative bid xi in a closed envelope. The auctioneer opens the envelopes. The highest bidder wins the auction and pays the other bidder’s bid. In case of a tie, bidder 1 wins the auction and pays the bid.

example 2 (homework) Normal form representation Is there any dominated strategy? What bids should be selected?

more homework: exercises 1 and 2

normal form games with complete information part 3

roadmap Nash equilibrium definition interpretations examples relation between Nash equilibrium and IEDS references Sec 1.1.C and 1.2.A, B of Gibbons Sec 2.6-2.8 of Osborne Ch 5 of Dutta

definition of Nash equilibrium (recall) strategy si* is a best response to s-i*є S-i if ui(si*,s-i*) ≥ ui(si,s-i*) for all si A vector of strategies s*=(s1*,s2*,…,sN*) is a Nash equilibrium if ui(si*,s-i*) ≥ u i(si,s-i*) for all si and for all i

definition of Nash equilibrium or si* solves Max ui(s1*,..., si-1*, si, s-i+1*,..., sn*) subject to si є Si therefore, a Nash equilibrium is a profile of strategies, such that each player’s strategy is the best for her, given that the other players are playing equilibrium strategies

definition of Nash equilibrium So, Each player is playing a best response against a conjecture Conjectures are correct

interpretations of Nash equilibrium play prescription preplay communication rational introspection focal point trial and error Play prescr: if the vector is proposed, then no-one wants to deviate Preplay comm: players could meet before the game and coordinate on a NE RI: each player asks himselfwhat the outcome will be; only NE appear to make sense Focal: NE is a focal point (Schelling) bc it stands out TE: NE is reached by trial and error… (!)

prisoners’ dilemma 2 1 Not to confess Confess -1,-1 -6,0 0,-6 -3,-3

coordination game problem: uniqueness 2 1 Book launch Movie 2,1 0,0 1,2

matching pennies problem: existence 2 1 Heads Tails -1,+1 +1,-1

IE(W)DS and Nash equilibria 2 1 L M R U 1,1 0,1 1,2 D -1,-1 -1,0 0,0

IE(S)DS and Nash equilibria 2 1 L M R U 1,1 0,1 1,2 D -1,-1 -1,0 0,0

discussion: order of elimination, IE(W)DS and Nash equilibria 2 1 L R U 0,0 0,1 D 1,0 When more than a NE exists, some may be eliminated by IEWDS

to sum up Any IE(W)DS solution (i.e., when the game is dominance solvable) has to be a Nash equilibrium; there may be Nash equilibria that are not IE(W)DS solutions Any IE(S)DS solution is a Nash equilibrium (but the strategies that survive IESDS don’t have to be a Nash equilibrium); all Nash equilibria survive IE(S)DS IEWDS solvable implies NE; the reverse does not hold bc (1) IEWDS eliminates some NE when more than one NE exists and (2) the game may not be dominance solvable (eg, battle of sexes)

to sum up A dominant strategy equilibrium must be a Nash equilibrium; the reverse implication does not hold

normal form games with complete information part 4

roadmap solving a maximization problem Cournot model Bertrand models (with differentiated and homogeneous products) contributing to a public good references Sec 1.2.B of Gibbons and Sec 3.1 and 3.2 of Osborne Ch 6 of Dutta

Cournot model of duopoly a product is produced by only two firms: firm 1 and firm 2. The quantities are denoted by q1 and q2, respectively. Each firm chooses the quantity without knowing the other firm has chosen the market price is P(Q)=a-Q, where a is a constant number and Q=q1+q2 the cost to firm i of producing quantity qi is Ci (qi)=cqi

Cournot model of duopoly how to find a Nash equilibrium: find a quantity pair (q1*, q2*) such that q1* solves: Max u1(q1, q2*)=q1[a-(q1+q2*)-c] subject to 0 ≤ q1 ≤ +∞ and q2* solves Max u2(q1*, q2)=q2[a-(q1*+q2)-c] subject to 0 ≤ q2 ≤ +∞

concave function a function f(x) is concave if, for any x and y, f(tx + (1- t)y) ≥ t f(x) + (1- t) f(y), for all 0 ≤ t ≤1

convex function a function f(x) is convex if, for any x and y, f(tx + (1- t)y) ≤ t f(x) + (1- t) f(y), for all 0 ≤ t ≤1

concavity and convexity a (differentiable) function f(x) is concave iff f’’(x) ≤ 0 a (differentiable) function f(x) is convex iff f’’(x) ≥ 0 A function f(x) is concave iff -f (x) is convex

concavity and convexity example 1: f(x) = -3x2 + 6x - 4, f’(x) = -6x + 6, f’’(x) = -6 < 0. Hence, f(x) is concave example 2: f(x) = ex, f’(x)= f’’(x) = ex > 0. Hence, f (x) is convex example 3: f(x,y) = 2xy + 4y - x2 + 3y2 For any fixed y, f (x, y) is concave in x because f’x(x,y) = 2y - 2x and f’’(x,y) = -2 < 0 For any fixed x, f (x, y) is convex in y because f’y(x,y) = 2x + 4 + 6y and f’’y(x,y) = 6 > 0

maximum and minimum a maximum of f(x) is a point x’ such that f(x’) ≥ f(x) for any x a minimum of f(x) is a point x* such that f(x*) ≤ f(x) for any x

maximum and minimum if a point x is a maximum or a minimum, then it satisfies the first order condition (FOC) f’(x) = 0 if f(x) is concave and x’ satisfies the FOC, then x’ is a maximum if f(x) is convex and x* satisfies the FOC, then x* is a minimum

maximum and minimum a maximum of f(x) in the domain [x1, x2] is a point x’ in [x1, x2] such that f(x’) ≥ f(x) for all x € [x1, x2] a minimum of f(x) in the domain [x1, x2] is a point x* in [x1, x2] such that f(x*) ≤ f(x) for all x € [x1, x2]

finding a maximum of a concave function in [x1, x2] find a maximum x’ of f(x) without constraints if x’ is in [x1, x2] , then x’ is also a maximum for the constrained problem otherwise, if f(x1) > f( x2) or x’ < x1, then x1 is a maximum if f(x1) < f( x2) or x’ > x2, then x2 is a maximum if f(x1) = f( x2), then any point in [x1, x2] is a maximum

finding a maximum of a concave function in [x1, x2] example 4: Max f(x) = -3x2 + 6x – 4 subject to -2 ≤ x ≤ 2 solution: x = 1 example 5: subject to x ≥ 2

finding a maximum of a concave function in [x1, x2]

Cournot model of duopoly solve Max u1(q1, q2*)=q1[a-(q1+q2*)-c] subject to 0 ≤ q1 ≤ +∞ FOC: a -2q1 – q2* - c = 0 q1 = (a - q2* - c)/2 Max u2(q1*, q2)=q2[a-(q1*+q2)-c] subject to 0 ≤ q2 ≤ +∞ FOC: a -2q2 – q1* - c = 0 q2 = (a – q1* - c)/2

Cournot model of duopoly (q1*, q2*) is a Nash equilibrium iff q1* = (a - q2* - c)/2 q2* = (a – q1* - c)/2 solving the two equations q1* = q2*= (a - c)/2

Cournot model of duopoly best response functions R1(q2) = (a – q2 - c)/2 if q2 < a-c 0 otherwise R2(q1) = (a – q1 - c)/2 if q1 < a-c

Cournot model of oligopoly solve Max ui(q1*,…, qi,…, qn*)=qi[a-(q1*+…+ qi+…+ qn*)-c] subject to 0 ≤ qi ≤ +∞ … FOC: a -2qi -(q1*+…+ qi-1+qi+1+…+ qn*)- c = 0 qi = (1) Since all firms are symmetric, in NE we will have qi*= qj* for all i, j. Substituting in (1), we obtain qi*= (a-c)/(n+1) for all i. Therefore, NE is (q1*,…, qn*)=((a-c)/(n+1),…, (a-c)/(n+1))

application: the problem of the commons n farmers in a village; each summer, all the farmers graze their goats on the village green gi: number of goats owned by farmer I c: cost of buying and caring for each goat v: value of a goat is v(G) per goat, where G = g1 + g2 +...+gn and v(G): v’(G) < 0 and v’’(G) < 0 there is a maximum number of goats that can be grazed on the green: v(G)>0 if G < Gmax, and v(G)=0 if G ≥ Gmax each spring, all the farmers simultaneously choose how many goats to own

application: the problem of the commons normal-form representation set of players {farmer 1,…, farmer n} sets of strategies Si = [0,Gmax), for i = 1, 2,…, n payoff functions ui (g1,...,gn ) = gi v(g1+…+ gn ) – c gi for i =1, 2,…, n

application: the problem of the commons computing the Nash equilibrium Max ui(g1*,…, gi-1*, gi, gi+1*,…, gn*)= gi v(g1*+…+ gi-1*+ gi + gi+1*+…+ gn*) - cgi subject to 0 ≤ gi ≤ Gmax for i = 1, 2,…, n FOC: v(g1+ g2*+…+ gn*) + g1v’(g1+ g2 *+…+ gn*) – c = 0 v(g1*+ g2+…+ gn*) + g2v’(g1*+ g2 +…+ gn*) – c = 0 … v(g1*+ g2*+…+ gn) + gnv’(g1*+ g2 *+…+ gn) – c = 0

application: the problem of the commons computing the Nash equilibrium Summing over all farmers’ FOC’s and dividing by n yields v(G*) + (1/n)G*v’(G*) – c = 0 where G*= g1*+ g2*+…+ gn*

application: the problem of the commons the social planner’s problem Max Gv(G) – Gc subject to 0 ≤ G ≤ Gmax FOC v(G) + Gv’(G) – c = 0 so, G**: v(G**) + G**v’(G**) – c = 0 G* vs G**?

to solve: all exercises except for ex. 3 and 7