Introduction to Game Theory

Introduction to Game Theory Part 3. Static games of incomplete information Chapter 2. Applications Ciclo Profissional 2 o Semestre / 2011 Graduação em Ciências Econômicas V. Filipe Martins-da-Rocha (FGV) Introduction to Game Theory October, 2011 1 / 36

Topics covered 1 Mixed strategies revisited 2 An auction 3 A double auction V. Filipe Martins-da-Rocha (FGV) Introduction to Game Theory October, 2011 2 / 36

Mixed strategies revisited Harsanyi (1973) suggested the following interpretation of mixed strategies Player j s mixed strategy represents player i s uncertainty about j s choice of a pure strategy Player j s choice in turn depends on the realization of a small amount of private information More precisely, a mixed-strategy Nash equilibrium in a game of complete information can (almost always) be interpreted as a pure-strategy Bayesian Nash equilibrium in a closely related game with a little bit of incomplete information The crucial feature of a mixed-strategy NE is not that player j chooses a strategy randomly, but rather that player i is uncertain about player j s choice V. Filipe Martins-da-Rocha (FGV) Introduction to Game Theory October, 2011 3 / 36

The Battle of Sexes Chris Pat Opera Fight Opera 2, 1 0, 0 Fight 0, 0 1, 2 The Battle of Sexes There are two pure-strategy Nash equilibria: (Opera,Opera) and (Fight,Fight) And a mixed-strategy Nash equilibrium in which Chris plays Opera with probability 2/3 and Pat plays Fight with probability 2/3 V. Filipe Martins-da-Rocha (FGV) Introduction to Game Theory October, 2011 4 / 36

The Battle of Sexes with incomplete information Suppose that Chris and Pat are not quite sure of each other s payoffs For instance, suppose that Chris s payoff if both attend the Opera is 2 + t c, where t c is privately known by Chris Pat s payoff if both attend the Fight is 2 + t p, where t p is privately known by Pat The parameters t c and t p are independent draws from a uniform distribution on [0, x], where x should be thought as small with respect to 2 All the other payoffs are the same V. Filipe Martins-da-Rocha (FGV) Introduction to Game Theory October, 2011 5 / 36

The Battle of Sexes with incomplete information The abstract static Bayesian games in normal-form is G = {A c, A p ; T c, T p ; p c, p p ; u c, u p } where the action spaces are A c = A p = {Opera, Fight} the type space are T c = T p = [0, x] the beliefs are p c (X t c ) = p p (X t p ) = λ(x)/x for all X [0, x], and the payoffs are as follows Chris Pat Opera Fight Opera 2 + t c, 1 0, 0 Fight 0, 0 1, 2 + t p The Battle of Sexes with incomplete information V. Filipe Martins-da-Rocha (FGV) Introduction to Game Theory October, 2011 6 / 36

The Battle of Sexes with incomplete information Fix two critical values c and p in [0, x] Consider the strategy profile s = (s c, s p) defined as follows Chris plays Opera if t c exceeds the critical value c and plays Fight otherwise, i.e., { s Opera if tc > c c(t c ) = Fight if t c c Pat plays Fight if t p exceeds the critical value p and plays Opera otherwise, i.e., { s Fight if tp > p p(t p ) = Opera if t p p Chris plays Opera with probability (x c)/x and Pat plays Fight with probability (x p)/x V. Filipe Martins-da-Rocha (FGV) Introduction to Game Theory October, 2011 7 / 36

The Battle of Sexes with incomplete information For a given value of x, we will determine values of c and p such that these strategies are a Bayesian Nash equilibrium Given Pat s strategy s p, Chris s expected payoff from playing Opera is u c (Opera, s p; t c ) = p [ x (2 + t c) + 1 p ] 0 = p x x (2 + t c) and from playing Fight, Chris s payoff is u c (Fight, s p; t c ) = p [ x 0 + 1 p ] 1 = 1 p x x Playing Opera is optimal (best response) if and only if t c x p 3 c V. Filipe Martins-da-Rocha (FGV) Introduction to Game Theory October, 2011 8 / 36

The Battle of Sexes with incomplete information Given Chris s strategy s c, Pat s expected payoff from playing Fight is [ u p (Fight, s c; t p ) = 1 c ] 0 + c x x (2 + t p) = c x (2 + t p) and from playing Opera, Pat s payoff is [ u p (Opera, s c; t p ) = 1 c ] 1 + c x x 0 = 1 c x Playing Fight is optimal if and only if t p x c 3 p V. Filipe Martins-da-Rocha (FGV) Introduction to Game Theory October, 2011 9 / 36

The Battle of Sexes with incomplete information Solving and simultaneously Yields p = c and p 2 + 3p x = 0 x p 3 c x c 3 p Solving the quadratic equation then shows that the probability that Chris plays Opera, namely (x c)/x, and the probability that Pat plays Fight, namely (x p)/x, both equal 1 3 + 9 + 4x 2x which approaches 2/3 as x approaches zero V. Filipe Martins-da-Rocha (FGV) Introduction to Game Theory October, 2011 10 / 36

The Battle of Sexes with incomplete information As the incomplete information disappears, the players behavior in this pure-strategy Bayesian Nash equilibrium of the incomplete information game approaches the players behavior in the mixed-strategy Nash equilibrium in the original game of complete information V. Filipe Martins-da-Rocha (FGV) Introduction to Game Theory October, 2011 11 / 36

An auction Consider the following first-price, sealed-bid auction There are two bidders, I = {1, 2} Bidder i has a valuation v i for the good If bidder i gets the good and pays the price p, then i s payoff is v i p The two players valuations are independently and uniformly distributed on [0, 1] The players simultaneously submit their non-negative bids The higher bidder wins the good and pays the price she bid; the other bidder gets and pays nothing In case of a tie, the winner is determined by a flip of a coin The bidders are risk-neutral and all this is common knowledge V. Filipe Martins-da-Rocha (FGV) Introduction to Game Theory October, 2011 12 / 36

An auction The static Bayesian game associated to this problem is defined by the action space A i = [0, ) the type space T i = [0, 1] Player i believes that v j is uniformly distributed on [0, 1] V B([0, 1]), p i ({v j V } v i ) = λ(v ) where λ is the Lebesgue measure on [0, 1] Abusing notations, p i ( v i ) is denoted p i ( ) since it is independent of v i Player i s (expected) payoff function is v i b i if b i > b j u i (b 1, b 2 ; v 1, v 2 ) = (v i b i )/2 if b i = b j 0 if b i < b j V. Filipe Martins-da-Rocha (FGV) Introduction to Game Theory October, 2011 13 / 36

An auction A strategy for player i is a function bi : v i b i (v i ) A profile ( b 1, b 2 ) is Bayesian Nash equilibrium if for each player i, for each valuation v i [0, 1], the value b i (v i ) belongs to { argmax (v i b i )p i {b i > b j } + 1 } 2 (v i b i )p i {b i = b j } : b i 0 Recall that p i {b i > b j } = λ{v j [0, 1] : b i > b j (v j )} and p i {b i = b j } = λ{v j [0, 1] : b i = b j (v j )} V. Filipe Martins-da-Rocha (FGV) Introduction to Game Theory October, 2011 14 / 36

An auction: existence of a linear equilibrium We propose to look for a linear equilibrium of the form where c i > 0 bi : v i a i + c i v i We are not restricting the players strategy spaces to include only linear strategies We allow players to choose arbitrary strategies but ask whether there is an equilibrium that is linear V. Filipe Martins-da-Rocha (FGV) Introduction to Game Theory October, 2011 15 / 36

An auction: existence of a linear equilibrium Suppose that player j adopts the linear strategy where bj = a j + c j Id Id : [0, 1] [0, 1] is defined by Id(v) = v For a given valuation v i, player i s best response solves where we recall that max (v i b i )λ{b i > a j + c j Id} b i 0 λ{b i > a j + c j Id} = λ{v j [0, 1] : b j > a j + c j v j } We have used the fact that λ{b i = b j } = 0 V. Filipe Martins-da-Rocha (FGV) Introduction to Game Theory October, 2011 16 / 36

An auction: existence Observe that the best reply b i (v i ) must satisfy If b i belongs to [a j, a j + c j ] then a j b i (v i ) a j + c j λ{b i > a j + c j Id} = λ[0, (b i a j )/c j ) = b i a j c j Player i s best response is therefore bi (v i ) = { (vi + a j )/2 if v i a j a j if v i < a j If 0 < a j < 1 then there are some values of v i such that v i < a j, in which case b i is not linear V. Filipe Martins-da-Rocha (FGV) Introduction to Game Theory October, 2011 17 / 36

An auction: existence Can we find a Nash equilibrium where a j 1 or a j 0? Assume that a j 0, in this case player i s best response is bi (v i ) = a j 2 + 1 2 v i The function b i takes the form a i + c i Id where a i = a j /2 and c i = 1/2 This implies that ((1/2) Id, (1/2) Id) is a Nash Bayesian equilibrium V. Filipe Martins-da-Rocha (FGV) Introduction to Game Theory October, 2011 18 / 36

An auction: uniqueness We propose to prove that there is a unique symmetric Bayesian Nash equilibrium which is the linear equilibrium already derived A BNE is called symmetric if the players strategies are identical We propose to prove that there is a single function b such that ( b, b) is a BNE Since players valuations typically will be different, their bids typically will be different, even if both use the same strategy V. Filipe Martins-da-Rocha (FGV) Introduction to Game Theory October, 2011 19 / 36

An auction: uniqueness Suppose that player j adopts a strategy b and assume that b is strictly increasing and differentiable For a given value of v i, player i s optimal bid solves 1 max (v i b i )λ{b i > b} b i 0 If b i Im( b) then {b i > b} = [ 0, b ) 1 (b i ) implying that b 1 ( b i (v i )) + (v i b i (v i )) [ b 1 ] ( bi (v i )) = 0 1 Observe that λ{b i = b} = 0 V. Filipe Martins-da-Rocha (FGV) Introduction to Game Theory October, 2011 20 / 36

An auction: uniqueness In order to get a symmetric BNE we need to have b i = b The first-order condition is then b 1 ( b(v i )) + (v i b(v [ b 1 ] i )) ( b(vi )) = 0 Since we have b 1 ( b(v i )) = v i and [ b 1 ] ( b(vi )) = 1 ] [ b (vi ) Then the function b must satisfy v i [ b] (vi ) + b(v i ) = v i V. Filipe Martins-da-Rocha (FGV) Introduction to Game Theory October, 2011 21 / 36

An auction: uniqueness Observe that [ Id b] (vi ) = v i [ b] (vi ) + b(v i ) This leads to [ Id b (1/2) Id 2] = 0 Therefore there exists a constant k such that v i [0, 1], v i b(vi ) = 1 2 v2 i + k We need a boundary condition to determine k V. Filipe Martins-da-Rocha (FGV) Introduction to Game Theory October, 2011 22 / 36

An auction: uniqueness A player s action should be individually rational: No player should bid more than his valuation Thus, we require b Id This implies k = 0 and b = 1 2 Id V. Filipe Martins-da-Rocha (FGV) Introduction to Game Theory October, 2011 23 / 36

A double auction We consider the case in which a buyer and a seller each have private information about their valuations The seller names an asking price p s The buyer simultaneously names an offer price p b If p b p s then trade occurs at price p = (p b + p s )/2 If p b < p s then no trade occurs V. Filipe Martins-da-Rocha (FGV) Introduction to Game Theory October, 2011 24 / 36

A double auction The buyer s valuation for the seller s good is v b, the seller s valuation is v s These valuations are private information and are drawn from independent uniform distribution on [0, 1] If the buyer gets the good for price p then his utility is v b p; if there is no trade the buyer s utility is zero If the seller sells the good for price p then his utility is p v s ; if there is no trade the seller s utility is zero A strategy for the buyer is a function p b : v b p b (v b ) specifying the price the buyer will offer for each of his possible valuation A strategy for the seller is a function p s : v s p s (v s ) specifying the price the seller will demand for each of his possible valuation V. Filipe Martins-da-Rocha (FGV) Introduction to Game Theory October, 2011 25 / 36

A double auction A profile of strategies ( p b, p s ) is a BNE if the two following conditions hold For each v b [0, 1], the price p b (v b ) solves [ max v b p ] b + p s (v s ) λ(dv s ) p b 0 {p b p s} 2 where {p b p s } = {v s [0, 1] : p b p s (v s )} For each v s [0, 1], the price p s (v s ) solves [ ] ps + p b (v b ) max v s λ(dv b ) p s 0 {p s p b } 2 where {p s p b } = {v b [0, 1] : p s p b (v b )} V. Filipe Martins-da-Rocha (FGV) Introduction to Game Theory October, 2011 26 / 36

A double auction A profile of strategies ( p b, p s ) is a BNE if the two following conditions hold For each v b [0, 1], the price p b (v b ) solves [ max v b p ] b + E[ p s p b p s ] λ{p b p s } p b 0 2 where we recall that E[ p s p b p s ] = 1 p s (v s )λ(dv s ) λ{p b p s } λ{p b p s} For each v s [0, 1], the price p s (v s ) solves [ ] ps + E[ p b p s p b ] max v s λ{p s p b } p s 0 2 V. Filipe Martins-da-Rocha (FGV) Introduction to Game Theory October, 2011 27 / 36

A double auction There are many BNE, we propose to exhibit one of them where trade occurs at a single price if it occurs at all For any x [0, 1], let the buyer s strategy be to offer x if vb x and to offer 0 otherwise let the seller s strategy be to demand x if vs x and to demand one otherwise This profile of strategies is a Bayesian Nash equilibrium V. Filipe Martins-da-Rocha (FGV) Introduction to Game Theory October, 2011 28 / 36

A double auction Trade would be efficient for all (v s, v b ) pairs such that v b v s, but does not occur in the two shaded regions V. Filipe Martins-da-Rocha (FGV) Introduction to Game Theory October, 2011 29 / 36

A double auction We propose to derive a linear Bayesian equilibrium Suppose the seller s strategy is p s : v s a s + c s v s with c s > 0 If the buyer s valuation is v b, his best reply p b (v b ) should solve [ max v b 1 { p b + a }] s + p b [pb a s ] + p b 0 2 2 c s The first order condition for which yields p b (v b ) = 2 3 v b + 1 3 a s Thus, if the seller plays a linear strategy, then the buyer s best response is also linear V. Filipe Martins-da-Rocha (FGV) Introduction to Game Theory October, 2011 30 / 36

A double auction Analogously, suppose the buyer s strategy is p b : v b a b + c b v b with c b > 0 If the seller s valuation is v s, his best reply p s (v s ) should solve [ 1 max {p s + 2a b + c b + [p s a b ] + } ] cb [p s a b ] + v s p s 0 2 2 The first order condition for which yields { 2 p s (v s ) = 3 v s + 1 3 (a b + c b ) if v s a b c b /2 a b if v s < a b c b /2 Thus, if the buyer plays a linear strategy, then the seller s best response may also be linear It will be the case if ab c b /2 c b V. Filipe Martins-da-Rocha (FGV) Introduction to Game Theory October, 2011 31 / 36

A double auction Assume a b c b /2 If the players linear strategies are to be best response to each other then we get c b = 2 3, c s = 2 a b = a s 3 and 3 a s = a b + c b 3 We obtain a b = c b /8 implying that the condition a b c b /2 is satisfied The linear strategies are then p b (v b ) = 2 3 v b + 1 12 and p s (v s ) = 2 3 v s + 1 4 V. Filipe Martins-da-Rocha (FGV) Introduction to Game Theory October, 2011 32 / 36

A double auction V. Filipe Martins-da-Rocha (FGV) Introduction to Game Theory October, 2011 33 / 36

A double auction Trade occurs if and only if the pair (v s, v b ) is such that p b (v b ) p s (v s ), i.e., iff v b v s + (1/4) V. Filipe Martins-da-Rocha (FGV) Introduction to Game Theory October, 2011 34 / 36

A double auction: comparing the solutions In both cases, the one-price and linear equilibria, the most valuable trade (v s = 0 and v b = 1) does occur The one-price equilibrium misses some valuable trades vs = 0 and v b = x ε where ε is small and achieves some trades that are worth next to nothing vs = x ε and v b = x + ε where ε is small The linear equilibrium, in contrast, misses all trades worth next to nothing but achieves all trades worth at least 1/4 V. Filipe Martins-da-Rocha (FGV) Introduction to Game Theory October, 2011 35 / 36

A double auction This suggests that the linear equilibrium may dominate the one-price equilibria, in terms of the expected gains the players receive One may wonder if it is possible to find other BNE for which the players might do even better Myerson and Satterthwaite (Jet 1983) show that, for the uniform valuation distributions, the linear equilibrium yields higher expected gains for the players than any other Bayesian Nash equilibria of the double auction This implies that there is no BNE of the double auction in which trade occurs if and only if it is efficient (i.e., if and only if v b v s ) V. Filipe Martins-da-Rocha (FGV) Introduction to Game Theory October, 2011 36 / 36