Chapter 7: Bayesian Games and Bayes–Nash Equilibrium

Date: Oct 2, 2026

In every game so far the players knew the whole game: who the players were, what each could do, and what each stood to gain. That is rarely true. A bidder does not know what the painting is worth to the other bidders. A firm does not know its rival’s costs. And in the seminar game, player 1 may not be sure that player 2 wants to meet at all; perhaps player 2 would prefer to avoid player 1.

Take that last case. Suppose player 1 believes that with probability \(3/4\) player 2 wants to meet, with the payoffs of Chapter 2, and that with probability \(1/4\) player 2 wants to avoid player 1 and prefers to be at whichever seminar player 1 is not. Player 2 knows which is true. What should player 1 do? The best choice depends on what player 2 will do, as always. But player 1 now has to think about what player 2 would do in each of the two cases, since both are possible from player 1’s point of view. And player 2, even knowing her own wishes, has to think about what player 1 expects her to do in the case that did not occur, because player 1’s choice depends on it. A strategy for player 2 must therefore say what she does in each case, including the one that she knows is false. This chapter develops that idea into a general model of games with private information and its equilibrium concept.

7.1 Types and beliefs

The device, due to Harsanyi, is to replace “player 1 does not know which game is being played” by “player 1 does not know which move chance made”. Everything a player knows privately is summarized in that player’s type, and the types are drawn at random before the game starts, each player observing only their own.

Definition 7.1 (Bayesian game). A Bayesian game consists of a set of players \(N=\{1,\dots,n\}\) ; for each player \(i\) a set \(A_i\) of actions and a set \(\Theta_i\) of types; a probability distribution \(p\) on \(\Theta=\Theta_1\times\dots\times\Theta_n\) , the common prior; and for each player \(i\) a payoff function \(u_i(a,\theta)\) depending on the action profile \(a=(a_1,\dots,a_n)\) and the type profile \(\theta=(\theta_1,\dots,\theta_n)\) . The game is played as follows: \(\theta\) is drawn according to \(p\) ; each player \(i\) observes \(\theta_i\) and nothing else; the players then choose actions simultaneously. All of this, including \(p\) , is common knowledge.

The game has private values if each \(u_i\) depends on \(\theta\) only through player \(i\) ’s own type, and interdependent values otherwise. Unknown costs and unknown tastes are private values; an unknown quality of a traded good, known to the seller and affecting the buyer’s payoff, is not.

A player who has observed their own type updates the prior. The belief of type \(\theta_i\) about the others’ types is the conditional distribution

\[ p(\theta_{-i}\mid\theta_i)=\frac{p(\theta_i,\theta_{-i})}{\sum_{\theta_{-i}'}p(\theta_i,\theta_{-i}')}, \]

defined whenever the denominator, the prior probability of the type \(\theta_i\) , is positive. This is the conditional probability of the others’ types given one’s own, which is what Bayes’ rule computes from the prior (math.probability-theory.bayes-rule). If the types are independent, \(p(\theta_{-i}\mid\theta_i)\) does not depend on \(\theta_i\) and every type of player \(i\) holds the same beliefs. If they are correlated, different types of the same player believe different things about the others, and those differences matter.

In the opening story \(\Theta_1\) has a single element, so player 1 has no private information; \(\Theta_2=\{\text{meet},\text{avoid}\}\) ; and the prior gives the two types probabilities \(3/4\) and \(1/4\) .

7.2 Strategies and equilibrium

Definition 7.2 (strategy and Bayes–Nash equilibrium). A (pure) strategy of player \(i\) in a Bayesian game is a function \(s_i:\Theta_i\to A_i\) , assigning an action to each type. A strategy profile \(s^\star\) is a Bayes–Nash equilibrium if for every player \(i\) and every type \(\theta_i\) of positive prior probability, the action \(s_i^\star(\theta_i)\) maximizes that type’s interim expected payoff:

\[ s_i^\star(\theta_i)\in\arg\max_{a_i\in A_i}\ \sum_{\theta_{-i}}p(\theta_{-i}\mid\theta_i)\;u_i\bigl(a_i,\,s_{-i}^\star(\theta_{-i}),\,\theta_i,\theta_{-i}\bigr). \]

For continuous type sets the sum is replaced by an integral against the conditional density.

In words: each type of each player chooses a best response to the strategies of the others, averaging over the others’ types with that type’s own beliefs. The expectation is over what the player does not know. The player’s own type is known and is not averaged over.

Why must a strategy specify an action for every type, when the player knows which type is real? Because the other players do not know it. Player 1’s best response is computed against the whole function \(s_2\) , weighting each of player 2’s types by its probability. The action of a type that did not occur is part of what the others are responding to.

Example 7.1: The seminar game with an uncertain partner

Player 1 chooses Algebra or Biology and gets \(2\) if both are at Algebra, \(1\) if both are at Biology, and \(0\) if they are apart. Player 2 is of type meet with probability \(3/4\) and then gets \(1\) if both are at Algebra, \(2\) if both are at Biology, and \(0\) if they are apart. With probability \(1/4\) she is of type avoid and then gets \(2\) if she is alone at Biology, \(1\) if she is alone at Algebra, and \(0\) if they are together. A strategy of player 2 is a pair (action of meet, action of avoid).

Suppose player 1 chooses Algebra. The meet type’s best response is Algebra (\(1>0\) ). The avoid type’s best response is Biology (\(2>0\) ). So player 2’s strategy must be (Algebra, Biology). Is Algebra a best response to that? Player 1 finds player 2 at Algebra with probability \(3/4\) , so Algebra yields \(2\cdot\tfrac34=\tfrac32\) . Biology would find her there with probability \(1/4\) and yield \(1\cdot\tfrac14=\tfrac14\) . Yes.

Suppose player 1 chooses Biology. The meet type’s best response is Biology (\(2>0\) ) and the avoid type’s is Algebra (\(1>0\) ), so the strategy is (Biology, Algebra). Against it, Biology yields \(1\cdot\tfrac34=\tfrac34\) for player 1 and Algebra yields \(2\cdot\tfrac14=\tfrac12\) . Biology is a best response.

There are two Bayes–Nash equilibria in pure strategies:

\[ \bigl(\text{Algebra};\ (\text{Algebra},\text{Biology})\bigr)\qquad\text{and}\qquad\bigl(\text{Biology};\ (\text{Biology},\text{Algebra})\bigr). \]

In the first, player 1’s expected payoff is \(3/2\) and the meet and avoid types get \(1\) and \(2\) ; in the second, player 1’s is \(3/4\) and the two types get \(2\) and \(1\) . In both, the avoid type succeeds in avoiding, and player 1 goes where the more likely type will be.

A Bayesian game can also be analysed before the types are drawn, as an ordinary game in strategic form.

Proposition 7.3 (interim and ex ante optimality). Let every type of every player have positive prior probability, and let the type sets be finite. Define the ex ante payoff of a strategy profile by \(U_i(s)=\sum_{\theta}p(\theta)\,u_i(s_1(\theta_1),\dots,s_n(\theta_n),\theta)\) . Then \(s^\star\) is a Bayes–Nash equilibrium if and only if it is a Nash equilibrium of the game in strategic form whose strategies are the functions \(s_i:\Theta_i\to A_i\) and whose payoffs are the \(U_i\) .

Proof. Group the sum defining \(U_i\) by player \(i\) ’s type:

\[ U_i(s_i,s_{-i}^\star)=\sum_{\theta_i}p(\theta_i)\Bigl[\sum_{\theta_{-i}}p(\theta_{-i}\mid\theta_i)\,u_i\bigl(s_i(\theta_i),s_{-i}^\star(\theta_{-i}),\theta\bigr)\Bigr]. \]

This is the decomposition of an expectation into conditional expectations (math.probability-theory.tower-property). The bracket for a given \(\theta_i\) is that type’s interim payoff, and it depends on \(s_i\) only through the single action \(s_i(\theta_i)\) . Because the function \(s_i\) can be chosen separately at each type and every weight \(p(\theta_i)\) is positive, the sum is maximized over \(s_i\) exactly when each bracket is maximized over \(s_i(\theta_i)\) . The first statement is the Nash condition and the second is the Bayes–Nash condition.

Corollary 7.4 (existence). Every Bayesian game with finitely many players, actions, and types has a Bayes–Nash equilibrium, possibly in mixed strategies.

Proof. If some type has prior probability zero, remove it; this does not change the equilibrium conditions. The strategic form of Proposition 7.3 is then a finite game, with \(\prod_i|A_i|^{|\Theta_i|}\) profiles, so it has a mixed Nash equilibrium by Theorem 4.7, and by the proposition, applied to mixed strategies in the same way, that is a Bayes–Nash equilibrium.

In Example 7.1 the strategic form has two rows and four columns, and an exercise asks you to construct it and recover the two equilibria. In practice the interim form is nearly always easier to work with: fix the others’ strategies and solve one optimization problem per type.

7.3 A rival with unknown cost

Example 7.2: Cournot competition with private cost

The two firms of Example 2.3 face the price \(14-Q\) . Firm 1’s unit cost is \(2\) and is known to both. Firm 2’s unit cost is \(0\) or \(4\) , each with probability \(1/2\) ; firm 2 knows it and firm 1 does not. A strategy of firm 1 is a quantity \(q_1\) . A strategy of firm 2 is a pair \((q_L,q_H)\) , the quantities of the low-cost and high-cost types.

Firm 2. Each type knows its cost and faces the known quantity \(q_1\) , so each solves the problem of Chapter 2:

\[ q_L=\frac{14-0-q_1}{2},\qquad q_H=\frac{14-4-q_1}{2}. \]

Firm 1. It does not know which type it faces. Its expected profit is

\[ \tfrac12\,q_1(12-q_1-q_L)+\tfrac12\,q_1(12-q_1-q_H)=q_1\bigl(12-q_1-\bar q\bigr),\qquad \bar q=\tfrac12(q_L+q_H). \]

Because profit is linear in the rival’s quantity, only the expected quantity \(\bar q\) matters, and \(q_1=(12-\bar q)/2\) .

Solving. Averaging firm 2’s two conditions gives \(\bar q=(12-q_1)/2\) . Together with \(q_1=(12-\bar q)/2\) this is the system of Example 2.3, so \(q_1=4\) and \(\bar q=4\) . Then

\[ q_1^\star=4,\qquad q_L^\star=\frac{14-4}{2}=5,\qquad q_H^\star=\frac{10-4}{2}=3 . \]

If the cost is low, total output is \(9\) , the price is \(5\) , firm 1 earns \(4\cdot 3=12\) and firm 2 earns \(5\cdot 5=25\) . If it is high, output is \(7\) , the price is \(7\) , firm 1 earns \(4\cdot 5=20\) and firm 2 earns \(3\cdot 3=9\) . Firm 1’s expected profit is \(16\) .

Compare what would happen if the cost were common knowledge. With cost \(0\) the Cournot quantities would be \(q_1=10/3\) and \(q_2=16/3\) ; with cost \(4\) they would be \(14/3\) and \(8/3\) . Under private information firm 1 cannot adjust to the rival’s cost and produces \(4\) in both cases, too much against the strong rival and too little against the weak one. The low-cost type, facing a firm 1 that has not cut back, produces \(5\) , less than \(16/3\) ; the high-cost type, facing a firm 1 that has not expanded, produces \(3\) , more than \(8/3\) . Each type’s output is shaped by what firm 1 thinks the other type would do. That is the mark of a Bayesian game.

7.4 A continuum of types

When types are real numbers, equilibrium strategies are functions on an interval, and the search is usually narrowed by a structural observation: a type that likes an action more should be more willing to take it.

Example 7.3: Who will pay for the public good?

Two housemates each decide whether to call a plumber. If at least one calls, the tap is fixed and each gets a benefit of \(1\) . Calling costs housemate \(i\) an amount \(c_i\) in time and trouble, known only to \(i\) ; the two costs are independent and uniformly distributed on \([0,2]\) . A housemate who calls gets \(1-c_i\) ; one who does not gets \(1\) if the other calls and \(0\) otherwise.

Best responses are thresholds. Fix any strategy of housemate \(j\) and let \(z\) be the probability that \(j\) calls. Housemate \(i\) with cost \(c_i\) gets \(1-c_i\) from calling and \(z\) from not calling, so calling is optimal exactly when \(c_i\le 1-z\) . Whatever the other does, the best response is to call when the cost is below some threshold.

Equilibrium. So look for an equilibrium in which each housemate calls exactly when \(c_i\le c^\star\) . Then the other calls with probability \(z=c^\star/2\) , since the cost is uniform on \([0,2]\) , and the best-response threshold is \(1-c^\star/2\) . Consistency requires

\[ c^\star=1-\frac{c^\star}{2},\qquad\text{so}\qquad c^\star=\frac23 . \]

Each housemate calls with probability \(1/3\) . By the first step every best response is a threshold strategy, so every equilibrium is a pair of thresholds \((c_1^\star,c_2^\star)\) with \(c_1^\star=1-c_2^\star/2\) and \(c_2^\star=1-c_1^\star/2\) , and that system has the single solution \((\tfrac23,\tfrac23)\) . The equilibrium is unique.

Outcome. Nobody calls with probability \((2/3)^2=4/9\) , exactly one calls with probability \(4/9\) , and both call, wastefully, with probability \(1/9\) . The tap is fixed with probability \(5/9\) . Yet fixing it is always worth doing, since the joint benefit is \(2\) and the smaller cost is never above \(2\) . A housemate with cost \(0.9\) does not call, although calling would give \(0.1\) for sure, because not calling gives \(1\) with probability \(1/3\) . Each is hoping the other will do it, as in the volunteer’s dilemma of Chapter 3; the difference is that here nobody randomizes. Every type has a strict preference, except the threshold type, and the uncertainty each housemate faces comes entirely from not knowing the other’s cost.

7.5 Where mixed strategies come from

That last remark suggests a new reading of mixed equilibria. In a mixed equilibrium each player is indifferent, and it was unclear in Chapter 3 why an indifferent player should randomize with exactly the right probabilities. Suppose instead that each player has a little private information, some small consideration that tilts the payoffs one way or the other and that the opponent cannot see. Then each type plays a pure strategy, and what looks like randomization to the opponent is ignorance of the type.

Example 7.4: Purifying the mixed equilibrium of the seminar game

In the seminar game of Chapter 2 the mixed equilibrium has each player going to their favourite seminar with probability \(2/3\) . Now perturb the game. Player 1’s payoff when both are at Algebra is \(2+t_1\) , and player 2’s payoff when both are at Biology is \(2+t_2\) , where \(t_1\) and \(t_2\) are independent, uniformly distributed on \([0,x]\) , and each is known only to its owner. All other payoffs are as before. For small \(x\) this is a slight uncertainty about how keen the other person is on their favourite subject.

Look for an equilibrium in which each player goes to their favourite seminar exactly when their own \(t_i\) is at least a threshold \(k\) . Then player 2 goes to Biology with probability \((x-k)/x\) and to Algebra with probability \(k/x\) . Player 1 of type \(t_1\) gets \((2+t_1)\,k/x\) from Algebra and \(1\cdot(x-k)/x\) from Biology, and prefers Algebra when

\[ (2+t_1)k\ge x-k\quad\Longleftrightarrow\quad t_1\ge\frac xk-3 . \]

So player 1’s best response is a threshold strategy with threshold \(x/k-3\) , and by symmetry the same holds for player 2. Consistency requires \(k=x/k-3\) , that is,

\[ k^2+3k-x=0,\qquad k=\frac{-3+\sqrt{9+4x}}{2}. \]

Each player goes to their favourite seminar with probability \(1-k/x\) . For \(x=7/4\) the threshold is \(k=1/2\) and that probability is \(1-\tfrac27=\tfrac57\approx 0.714\) . For \(x=13/16\) it is \(k=1/4\) and the probability is \(\tfrac{9}{13}\approx 0.692\) . As \(x\to 0\) , \(\sqrt{9+4x}\approx 3+\tfrac23x\) , so \(k\approx x/3\) and the probability tends to \(2/3\) .

In the perturbed game every type except the threshold type has a strict best response, and no one randomizes. Yet as the perturbation vanishes, the distribution of each player’s action converges to the mixed equilibrium of the unperturbed game. The mixed equilibrium describes what an opponent can predict about a player whose exact payoffs are not known. This result, in its general form, is Harsanyi’s purification theorem, and it removes much of the awkwardness of the indifference condition: the probabilities are properties of beliefs, and beliefs can be precise even when no individual tosses a coin.

7.6 When the informed side knows what the good is worth

With private values, a player’s uncertainty concerns only what others will do. With interdependent values it concerns the player’s own payoff, and then an opponent’s willingness to act is itself information.

Example 7.5: A market that closes

A seller owns a used car whose quality \(\theta\) is uniformly distributed on \([0,1]\) . The seller knows \(\theta\) and values the car at \(\theta\) . A buyer does not know \(\theta\) and would value the car at \(\tfrac32\theta\) : whatever the quality, the car is worth half as much again to the buyer as to the seller, so trade would always create a gain. The buyer names a price \(p\) , and at the same time the seller names the lowest price \(r\) she will accept; if \(p\ge r\) the car is sold at the price \(p\) .

The seller. Type \(\theta\) gains \(p-\theta\) from a sale and \(0\) otherwise, so she wants to sell exactly when \(p\ge\theta\) . Naming \(r=\theta\) achieves this whatever \(p\) is, and no other \(r\) does better against any \(p\) ; it is weakly dominant.

The buyer. Against \(r(\theta)=\theta\) , an offer \(p\in[0,1]\) is accepted exactly by the types \(\theta\le p\) . The buyer’s expected payoff is

\[ \int_0^p\Bigl(\tfrac32\theta-p\Bigr)\,d\theta=\tfrac34p^2-p^2=-\tfrac14p^2 . \]

Every positive offer loses money on average, and the buyer’s best response is \(p=0\) . In the equilibrium there is no trade, although trade is efficient for every single car.

The reason is in the limits of the integral. The buyer gets the car only when \(\theta\le p\) , and conditional on that event the average quality is \(p/2\) , worth \(\tfrac34p\) to the buyer, less than the price. An offer attracts exactly the sellers whose cars are worth less than it. If the buyer valued the car at \(k\theta\) , the same calculation gives \(\bigl(\tfrac k2-1\bigr)p^2\) : the market operates only when \(k\ge 2\) . This is adverse selection, and it previews the winner’s curse of Chapter 9: winning the object is news about its value.

Common mistake. Averaging over one’s own type, or treating the opponent as an “average type”. In Example 7.2 it is wrong to say that firm 2 “expects its cost to be \(2\) ” and produces \(4\) : firm 2 knows its cost, and each of its types solves its own problem, giving \(5\) and \(3\) . It is firm 1 that averages, and what it averages is the rival’s quantity under the rival’s strategy, weighted by firm 1’s beliefs. A second mistake arises when types are correlated: each type must use its own conditional distribution \(p(\theta_{-i}\mid\theta_i)\) , not the prior. A firm that learns its own cost is high in an industry where costs move together should conclude that its rival’s cost is probably high as well. A third is to specify an action only for the type that “really” occurred; the equilibrium conditions of the other players cannot even be written down without the actions of the other types.

The games of this chapter were static: types were drawn, and then everyone moved once, simultaneously. Nothing a player did could reveal anything, because nobody observed it in time. When moves are sequential, an informed player’s early action is seen by those who move later, and they will try to read the type from it. The informed player knows this and may act to be read in a particular way. The next chapter is about that interplay.

Exercises

Players 1 and 2 have types in \(\{a,b\}\) and \(\{c,d\}\) with prior \(p(a,c)=0.4\) , \(p(a,d)=0.1\) , \(p(b,c)=0.2\) , \(p(b,d)=0.3\) . Compute the beliefs of each of the four types about the other player’s type, and say whether the types are independent.

Answer

Type \(a\) has prior probability \(0.5\) and believes \(c\) with probability \(0.4/0.5=0.8\) . Type \(b\) has probability \(0.5\) and believes \(c\) with probability \(0.2/0.5=0.4\) . Type \(c\) has probability \(0.6\) and believes \(a\) with probability \(0.4/0.6=2/3\) . Type \(d\) has probability \(0.4\) and believes \(a\) with probability \(0.1/0.4=1/4\) . The types are not independent: independence would require \(p(a,c)=0.5\cdot 0.6=0.3\) , and it is \(0.4\) .
In Example 7.1 let the probability of the meet type be \(\pi\) instead of \(3/4\) . Find all pure-strategy Bayes–Nash equilibria as a function of \(\pi\) , and list them for \(\pi=1/2\) .

Solution

Player 2’s types have strict best responses to each pure action of player 1, so there are only two candidates.

Candidate 1: player 1 chooses Algebra. Then meet chooses Algebra and avoid chooses Biology. Against (Algebra, Biology), Algebra yields \(2\pi\) for player 1 and Biology yields \(1\cdot(1-\pi)\) , since only the avoid type is at Biology. Algebra is a best response when \(2\pi\ge 1-\pi\) , that is, \(\pi\ge 1/3\) .

Candidate 2: player 1 chooses Biology. Then meet chooses Biology and avoid chooses Algebra. Against (Biology, Algebra), Biology yields \(\pi\) and Algebra yields \(2(1-\pi)\) . Biology is a best response when \(\pi\ge 2(1-\pi)\) , that is, \(\pi\ge 2/3\) .

So for \(\pi<1/3\) there is no equilibrium in pure strategies; for \(1/3\le\pi<2/3\) the only one is (Algebra; (Algebra, Biology)); for \(\pi\ge 2/3\) both candidates are equilibria. At \(\pi=1/2\) there is exactly one, the first. When the avoid type is likely enough, player 1 cannot coordinate on Biology, because the avoid type would then be at Algebra, where player 1’s payoff from meeting is higher.
Construct the ex ante strategic form of Example 7.1: a table with two rows for player 1 and four columns for player 2’s strategies, each cell holding the two ex ante expected payoffs. Find its pure Nash equilibria and confirm that they are the two Bayes–Nash equilibria found in the example.

Answer

Write player 2’s strategies as (meet, avoid). Player 1’s ex ante payoffs: row Algebra gives \(2,\ 1.5,\ 0.5,\ 0\) against \((A,A),(A,B),(B,A),(B,B)\) ; row Biology gives \(0,\ 0.25,\ 0.75,\ 1\) . Player 2’s ex ante payoffs: against Algebra, \(0.75,\ 1.25,\ 0,\ 0.5\) ; against Biology, \(0.25,\ 0,\ 1.75,\ 1.5\) . The cells with both payoffs maximal are (Algebra, \((A,B)\) ) with payoffs \((1.5,1.25)\) and (Biology, \((B,A)\) ) with payoffs \((0.75,1.75)\) , the two equilibria of the example.
In a Bayesian game with finite type sets, prove that whether the action \(s_i(\theta_i)\) is optimal for type \(\theta_i\) does not depend on the actions that \(s_i\) assigns to player \(i\) ’s other types. Then explain why the actions of those other types nevertheless matter for the equilibrium.

Hint

Write type \(\theta_i\) ’s interim payoff from Definition 7.2. It contains \(s_{-i}\) and the action taken by type \(\theta_i\) , and nothing about the actions of player \(i\) ’s other types. Those actions matter because they enter the other players’ interim payoffs, which average over all of player \(i\) ’s types, and so they shape the \(s_{-i}\) that type \(\theta_i\) faces.
Two firms face the price \(14-Q\) . Each firm’s unit cost is \(1\) or \(3\) with probability \(1/2\) each, independently, and each firm knows only its own cost. Find the symmetric Bayes–Nash equilibrium, giving the quantity of each type, and compute the expected total output.

Solution

Look for a symmetric equilibrium in which a firm of cost \(c\) produces \(q(c)\) , and let \(m=\tfrac12q(1)+\tfrac12q(3)\) be the expected quantity of a firm. A firm of cost \(c\) does not know the rival’s cost and maximizes its expected profit

\[ \tfrac12\,q\bigl(14-c-q-q(1)\bigr)+\tfrac12\,q\bigl(14-c-q-q(3)\bigr)=q(14-c-q-m), \]

a concave quadratic in \(q\) maximized at \(q=(14-c-m)/2\) . Hence

\[ q(1)=\frac{13-m}{2},\qquad q(3)=\frac{11-m}{2}. \]

Averaging, \(m=(12-m)/2\) , so \(m=4\) . Therefore \(q(1)=4.5\) and \(q(3)=3.5\) : a firm produces more when its cost is low. Each type’s quantity is positive, as the derivation assumed, and each is the unique best response to the rival’s strategy, so this is a Bayes–Nash equilibrium.

Each firm produces \(4\) on average, so the expected total output is \(8\) , the same as in the complete-information game with cost \(2\) for both firms. Realized output ranges from \(7\) , when both costs are high, to \(9\) , when both are low.
In Example 7.2 let firm 2’s cost be \(c_L\) with probability \(\pi\) and \(c_H\) with probability \(1-\pi\) , and let the demand intercept and firm 1’s cost be \(a\) and \(c_1\) . Derive \(q_1^\star=(a-2c_1+\bar c)/3\) with \(\bar c=\pi c_L+(1-\pi)c_H\) , and the quantities of both types of firm 2, assuming all are positive.

Answer

The types of firm 2 produce \(q(c)=(a-c-q_1)/2\) , so \(\bar q=(a-\bar c-q_1)/2\) , and firm 1 produces \(q_1=(a-c_1-\bar q)/2\) . Solving, \(q_1^\star=(a-2c_1+\bar c)/3\) , and then \(q_L^\star=(a-c_L-q_1^\star)/2\) and \(q_H^\star=(a-c_H-q_1^\star)/2\) . They differ by \((c_H-c_L)/2\) .
In the public-good game of Example 7.3 let the costs be uniform on \([0,1]\) instead of \([0,2]\) . Find the symmetric equilibrium threshold with two housemates. Then find it with three housemates, where the benefit \(1\) is received if at least one of the three calls.

Solution

Two housemates. If the other calls exactly when the cost is at most \(c^\star\) , the other calls with probability \(c^\star\) , the cost now being uniform on \([0,1]\) . A housemate with cost \(c_i\) gets \(1-c_i\) from calling and \(c^\star\) from not calling, so calls when \(c_i\le 1-c^\star\) . Consistency requires \(c^\star=1-c^\star\) , so \(c^\star=1/2\) . The tap is fixed with probability \(1-\tfrac14=\tfrac34\) .

Three housemates. If each of the other two calls with probability \(c^\star\) , nobody else calls with probability \((1-c^\star)^2\) . Calling gives \(1-c_i\) ; not calling gives \(1-(1-c^\star)^2\) . Calling is optimal when \(c_i\le(1-c^\star)^2\) : the cost must be below the probability of being the only possible caller. Consistency requires \(c^\star=(1-c^\star)^2\) , that is, \((c^\star)^2-3c^\star+1=0\) , whose root in \([0,1]\) is \(c^\star=(3-\sqrt5)/2\approx 0.382\) .

Each housemate is less willing to call when there are three. The probability that nobody calls is \((1-c^\star)^3=c^\star(1-c^\star)\approx 0.236\) , slightly below the \(0.25\) of the two-housemate case, so here the extra housemate helps a little, unlike in the volunteer’s dilemma with a common known cost.
A student writes for Example 7.2: “Firm 2’s expected cost is \(2\) , the same as firm 1’s, so by symmetry both firms produce \(4\) .” Say exactly which player is entitled to take an expectation over firm 2’s cost and which is not, and show that \(q_2=4\) is not a best response for either type of firm 2 against \(q_1=4\) .

Hint

Firm 1 is uncertain about firm 2’s cost and averages over it. Firm 2 knows its cost. Against \(q_1=4\) , the low-cost type’s profit \(q(14-4-q)\) is maximized at \(q=5\) and the high-cost type’s profit \(q(10-4-q)\) at \(q=3\) ; at \(q=4\) the two profits are \(24<25\) and \(8<9\) .
Two bidders have values \(v_1,v_2\) independent and uniform on \([0,1]\) . Each submits a bid; the higher bid wins and pays its bid. Suppose bidder 2 bids \(v_2/2\) . Write bidder 1’s expected payoff from a bid \(b\in[0,\tfrac12]\) when the value is \(v_1\) , maximize it, and conclude that bidding half of one’s value is a Bayes–Nash equilibrium.

Answer

A bid \(b\le\tfrac12\) wins when \(v_2/2<b\) , which has probability \(2b\) , so the expected payoff is \((v_1-b)\cdot 2b\) . Its derivative \(2v_1-4b\) vanishes at \(b=v_1/2\) , and the function is concave, so \(b=v_1/2\) is the best response; a bid above \(\tfrac12\) wins no more often than \(\tfrac12\) and pays more. By symmetry the profile in which each bids half the value is a Bayes–Nash equilibrium.
Two firms simultaneously decide whether to enter a market. Firm \(i\) ’s entry cost \(\theta_i\) is private, and the costs are independent and uniform on \([0,1]\) . An entrant earns a gross profit of \(1\) if it enters alone and \(1/4\) if both enter; a firm that stays out gets \(0\) . Find the symmetric equilibrium threshold and the probability that the market is served by nobody.

Solution

Suppose the rival enters exactly when its cost is at most \(\theta^\star\) , which has probability \(\theta^\star\) . A firm with cost \(\theta_i\) that enters earns the gross profit \(1\) with probability \(1-\theta^\star\) and \(\tfrac14\) with probability \(\theta^\star\) , so its expected payoff from entering is

\[ (1-\theta^\star)\cdot 1+\theta^\star\cdot\tfrac14-\theta_i=1-\tfrac34\theta^\star-\theta_i , \]

against \(0\) from staying out. It enters when \(\theta_i\le 1-\tfrac34\theta^\star\) . The best response is a threshold strategy whatever the rival does, since the expected gross profit does not depend on \(\theta_i\) . In a symmetric equilibrium \(\theta^\star=1-\tfrac34\theta^\star\) , so \(\theta^\star=4/7\) .

Each firm enters with probability \(4/7\) . The market is served by nobody with probability \((3/7)^2=9/49\) , by both with probability \(16/49\) , and by exactly one with probability \(24/49\) . A firm with cost \(0.6\) stays out although it would earn \(0.4\) as a monopolist, because the chance of facing a competitor is too high.
In Example 7.4, verify the two numerical cases by substituting \(k=1/2\) and \(k=1/4\) into \(k^2+3k-x=0\) , and compute the equilibrium probability of going to one’s favourite seminar for \(x=1\) . Explain in one sentence what a player of type exactly \(t_i=k\) does and why it does not matter.

Answer

\((\tfrac12)^2+\tfrac32=\tfrac74\) and \((\tfrac14)^2+\tfrac34=\tfrac{13}{16}\) . For \(x=1\) , \(k=(\sqrt{13}-3)/2\approx 0.303\) and the probability is \(1-k\approx 0.697\) . The type \(t_i=k\) is indifferent and may do either; a single type has probability zero under a continuous distribution, so the choice affects nobody’s expected payoff.
In Example 7.5 let the buyer value a car of quality \(\theta\) at \(\theta+\tfrac12\) . Find the buyer’s optimal offer, the set of qualities traded, and the buyer’s expected payoff. What fraction of the possible gains from trade is realized?

Solution

The seller’s weakly dominant strategy is still to accept exactly the prices \(p\ge\theta\) . An offer \(p\in[0,1]\) is accepted by the types \(\theta\le p\) , and the buyer’s expected payoff is

\[ \int_0^p\Bigl(\theta+\tfrac12-p\Bigr)\,d\theta=\tfrac12p^2+\tfrac12p-p^2=\tfrac12p-\tfrac12p^2 . \]

This is concave with derivative \(\tfrac12-p\) , so the optimal offer is \(p=1/2\) . An offer above \(1\) is accepted by everyone and is worse than \(p=1\) , which already gives \(0\) . The cars traded are those with \(\theta\le\tfrac12\) , and the buyer’s expected payoff is \(1/8\) .

Every trade creates a gain of \(\tfrac12\) , so the possible gains from trade are \(\tfrac12\) in expectation, and those realized are \(\tfrac12\cdot\tfrac12=\tfrac14\) : one half. The market does not close as in Example 7.5, because the buyer’s premium does not shrink with quality, but the better half of the cars is not traded. The realized gain is split as \(\tfrac18\) to the buyer and, on average, \(\int_0^{1/2}(\tfrac12-\theta)\,d\theta=\tfrac18\) to the seller.
In Example 7.3, prove that against any strategy of the other housemate, measurable but otherwise arbitrary, a housemate’s set of best responses consists of threshold strategies, up to the action of the single threshold type.

Hint

Let \(z\) be the probability, under the other’s strategy and the distribution of the other’s cost, that the other calls. Compare \(1-c_i\) with \(z\) . The comparison depends on the other’s strategy only through the single number \(z\) , and the set of costs for which calling is optimal is \(\{c_i\le 1-z\}\) .
A state is good or bad with probability \(1/2\) each. Each of two investors privately observes a signal that equals the true state with probability \(3/4\) , the two signals being independent given the state. Each then chooses Invest or Not. Not gives \(0\) . Invest gives \(1\) if the state is good and the other investor also invests, and \(-1\) otherwise. Show that “invest exactly when the signal is good” is a Bayes–Nash equilibrium, and that it stops being one if the signal accuracy is \(2/3\) .

Solution

Let the accuracy be \(\alpha\) , and suppose the other investor invests exactly when her signal is good.

Beliefs. The states are equally likely, so after a good signal the probability of the good state is \(\alpha/(\alpha+(1-\alpha))=\alpha\) , and after a bad signal it is \(1-\alpha\) . Given the state, the other’s signal is good with probability \(\alpha\) if the state is good.

A good signal. Investing pays \(1\) when the state is good and the other’s signal is good, an event of probability \(\alpha\cdot\alpha\) given my signal, and \(-1\) otherwise. With \(\alpha=\tfrac34\) this is \(\tfrac{9}{16}-\tfrac{7}{16}=\tfrac18>0\) , so investing beats not investing.

A bad signal. The success event has probability \((1-\alpha)\alpha=\tfrac{3}{16}\) , so investing pays \(\tfrac{3}{16}-\tfrac{13}{16}=-\tfrac58<0\) , and not investing is better.

Both types are best-responding, so “invest exactly when the signal is good” is a Bayes–Nash equilibrium for \(\alpha=\tfrac34\) .

With \(\alpha=\tfrac23\) , a good signal gives success probability \(\tfrac49\) and an expected payoff of \(\tfrac49-\tfrac59=-\tfrac19<0\) . Even the optimistic type prefers not to invest, so the profile is not an equilibrium; the only equilibrium is that nobody invests. The investor needs to be confident both about the state and about the other’s signal, and the product of the two probabilities must exceed one half.
Describe Example 7.1 as an extensive game in which Nature moves first. How many decision nodes and information sets does each player have, how many pure strategies, and why does the game have no proper subgames?

Hint

Let Nature choose the type, then player 2, then player 1. Player 2 has two decision nodes, each its own information set, since she knows her type. Player 1 has four decision nodes in one information set, knowing neither the type nor her action. Player 2 has \(4\) pure strategies and player 1 has \(2\) . A subgame would have to begin at a node that is alone in its information set and whose successors are not tied by information sets to nodes outside; player 1’s information set spans both branches from Nature’s move.
In the stag hunt of Chapter 2, suppose that with probability \(\mu\) hunter 2 is a type for whom Hare is strictly dominant, and otherwise has the usual payoffs; hunter 1 has the usual payoffs and no private information. Find all \(\mu\) for which there is a Bayes–Nash equilibrium in which hunter 1 and the usual type of hunter 2 choose Stag.

Answer

Against a hunter 2 who chooses Stag with probability \(1-\mu\) , Stag gives hunter 1 the payoff \(4(1-\mu)\) and Hare gives \(3(1-\mu)+2\mu\) . Stag is a best response when \(4(1-\mu)\ge 3-\mu\) , that is, \(\mu\le 1/3\) . The usual type of hunter 2 faces Stag for sure and replies with Stag. So the equilibrium exists exactly for \(\mu\le 1/3\) .
Show that a game in strategic form is a Bayesian game in which each \(\Theta_i\) has one element, and that its Bayes–Nash equilibria are its Nash equilibria. Then, using Chapter 2, compute the complete-information Cournot quantities for each of firm 2’s costs in Example 7.2 and verify the comparison made in the text.

Hint

With one type each, the conditional distribution in Definition 7.2 is trivial and the interim condition is the Nash condition. For the comparison use \(q_1=(a-2c_1+c_2)/3\) and \(q_2=(a-2c_2+c_1)/3\) from Chapter 2 with \(c_2=0\) and with \(c_2=4\) .
In Example 7.2, compute firm 1’s expected profit if firm 2’s cost were made public before quantities are chosen, and compare it with \(16\) . Then decide, for each type of firm 2 separately, whether that type would prefer its cost to be public, and whether firm 2 would prefer publicity before learning its cost.

Solution

Public cost. With \(c_2=0\) the Cournot quantities are \(q_1=10/3\) and \(q_2=16/3\) , the price is \(14-\tfrac{26}{3}=\tfrac{16}{3}\) , firm 1 earns \(\tfrac{10}{3}\cdot\tfrac{10}{3}=\tfrac{100}{9}\) and firm 2 earns \((\tfrac{16}{3})^2=\tfrac{256}{9}\) . With \(c_2=4\) the quantities are \(14/3\) and \(8/3\) , the price is \(\tfrac{20}{3}\) , firm 1 earns \((\tfrac{14}{3})^2=\tfrac{196}{9}\) and firm 2 earns \((\tfrac83)^2=\tfrac{64}{9}\) .

Firm 1. Its expected profit with public costs is \(\tfrac12\cdot\tfrac{100}{9}+\tfrac12\cdot\tfrac{196}{9}=148/9\approx 16.44\) , against \(16\) under private information. Firm 1 gains \(4/9\) from knowing the cost, because it can tailor its quantity to the rival it actually faces.

Firm 2, type by type. The low-cost type earns \(\tfrac{256}{9}\approx 28.44\) with a public cost and \(25\) with a private one: it wants its strength known, so that firm 1 cuts back. The high-cost type earns \(\tfrac{64}{9}\approx 7.11\) with a public cost and \(9\) with a private one: it wants to hide its weakness.

Firm 2 before learning its cost. The expected profit is \(\tfrac12\cdot\tfrac{256}{9}+\tfrac12\cdot\tfrac{64}{9}=160/9\approx 17.78\) with public costs and \(\tfrac12(25+9)=17\) without. Ex ante, firm 2 prefers publicity. But once the cost is learned the two types disagree, and a firm that could choose whether to reveal would have to reckon with what silence itself reveals. That is the subject of Chapter 8.