(501B) Problem Set 5. Bayesian Games Suggested Solutions by Tibor Heumann

Dirk Bergemann Department of Economics Yale University

Microeconomic Theory (501b) Problem Set 5. Bayesian Games Suggested Solutions by Tibor Heumann

1. (Market for Lemons) Here I ask that you work out some of the details in perhaps the most famous of all information economics models. By contrast to discussion in class, we give a complete formulation of the game. A seller is privately informed of the value v of the good that she sells to a buyer. The buyer’s prior belief on v is uniformly distributed on [x, y] with 3 0 < x < y. The good is worth 2 v to the buyer. (a) Suppose the buyer proposes a price p and the seller either accepts or rejects p. If she accepts, the seller gets payoff p−v, and the buyer gets 3 2 v − p. If she rejects, the seller gets v, and the buyer gets nothing. Find the optimal offer that the buyer can make as a function of x and y. (b) Show that if the buyer and the seller are symmetrically informed (i.e. either both know v or neither party knows v), then trade takes place with probability 1. (c) Consider a simultaneous acceptance game in the model with private information as in part a, where a price p is announced and then the buyer and the seller simultaneously accept or reject trade at price p. The payoffs are as in part a. Find the p that maximizes the probability of trade.

[SOLUTION]

(a) We look for the subgame perfect equilibrium, thus we solve by back- ward induction. From the next point it should be clear why there are other Nash equilibrium which are not subgame perfect. If the buyer proposes price p, then the seller accepts if and only if v ≥ p (we assume that, if indifferent the seller accepts). Thus, the payoff of the buyer conditional on offering price p is:

3 {v ≥ p}· [ v − p|v ≥ p]. P E 2 That is, the expected utility of the buyer is the probability of the seller accepting the oﬀer times the expected utility conditional on

1 the seller accepting the oﬀer. Expanding the terms, we get:

3 (p − x) 3 {v ≥ p}· [ v − p|v ≥ p] = ( [ v|v ≤ p] − p) P E 2 y − x E 2 (p − x) 3 x + p = ( − p) y − x 2 2 p − x = (3(x + p) − 4p) 4(y − x) p − x = (3x − p). 4(y − x)

One should note that the previous equality holds only for p ∈ [x, y]. Obviously for p ≤ x we have that P{v ≥ p} = 0 and for p ≥ y we have that P{v ≥ p} = 1. Taking into account the previous observation, we can diﬀerentiate with respect to p and get the F.O.C., since the objective function is strictly concave in p we know that the maximum will be achieved at a corner (that is, p ∈ {x, y}), if the solution to the F.O.C. yields a p 6∈ [x, y]. Taking the derivative, we get:

∂ 3 1 {v ≥ p}· [ v − p|v ≥ p] = (4x − 2p) ∂pP E 2 4(y − x) ∂ 3 ⇒ {v ≥ p}· [ v − p|v ≥ p] = 0 ⇐⇒ p = 2x ∂pP E 2 Thus, we have that the buyer proposes the following price depending on the distribution of v. ( 2x 2x ≤ y p∗ = y 2x > y

(b) If the buyer knows the value v, then the unique subgame perfect Nash equilibrium is that the buyer offers v and the seller accepts. There are two things to note. First, this is a complete information game, thus the right solution concept is subgame perfect Nash equilibrium. Note that there are other Nash equilibrium which are not sub-game perfect. For example, the buyer offers 0 and the seller rejects all offers. If neither agent know the valuation v, then the game can also be treated as a complete information game. In this case, the buyer offers E[v] = (y + x)/2, and the sellet accepts. Note that without asymmetric information, the game can be treated as complete information game, in which the value of buyer and seller are 3E[v]/2 and E[v] respectively. (c) From part (a) it is clear that the seller accepts if and only if v ≥ p. On the other hand, the buyer accepts if and only if he gets non-negative

2 proﬁts by accepting. Thus, the buyer accepts if and only if 3 p − x {v ≥ p}· [ v − p|v ≥ p] = (3x − p) ≥ 0. P E 2 4(y − x)

To maximize the probability of trade, we just need to ﬁnd the maximum p such that the buyer gets non-negative proﬁts. Clearly, we can just equate the previous inequality to 0 to get the price that maximizes trade. That is,

p∗ − x (3x − p∗) = 0 4(y − x) The quadratic equation has two solutions, as the buyer also gets 0 proﬁts when there is 0 probability of trade. Thus, we consider the other solution. Thus, the price that maximizes the probability of trade is: ( 3x 3x ≤ y p∗ = , [y, 3x] 3x ≥ y where we note that whenever 3x ≥ y we have a continuum of prices that ensure trade with probability 1.

2. (Akerlof’s Lemon in a General Equilibrium Model) Suppose there is a continuum of buyers with a characteristic vb ∈ [0, 1] distributed according to a distribution function G(vb). There is also a continuum of sellers with mass one, all with the same characteristic vs ∈ (0, 1). Each seller owns an object of value θ ∈ [0, 1] which is distributed according to a distribution function F (θ) on the unit interval, independent across sellers. The value θ is known to the seller and unknown to the buyer. The valuations for the object are

us = vsθ

and ub = vbθ. (a) Determine the efficient volume of trade and the efficient mix of sellers (in terms of types of θ) and buyer (in terms of types of vb). (b) Suppose for the moment that the characteristic of each object is observable and hence that information is symmetric across sellers and buyers. What would be the equilibrium price for the object with value θ be and would trade be efficient. (c) Under asymmetric information, compute the demand and supply functions. Explain what the demand is not necessarily downward sloping (despite the absence of income effects.)

3 (d) Solve for the competitive equilibrium for f and g uniform on [0, 1]. Verify that trade is suboptimal. (e) Argue that, in general, there can exist multiple equilibria. Then assume that it is the case. Show that a higher price equilibrium Pareto dominates a lower price one. Compare with the ﬁrst fundamental theorem of welfare economics.

(f) Show that a minimum quality standard s0 > 0, if enforceable, may improve welfare. [SOLUTION] (a) If you are the benevolent social planner, you will want trade to occur only when ub ≥ us, that is, when vb ≥ vs. Thus, the probability of trade is Pr (Trade) = Pr (vb > vs) = 1−G (vs) . In other words, you are only going to match high-value buyers to some sellers. To which sellers? To the ones that allow you (the planner) to maximize gains from trade θ (vb − vs) . Therefore, you will choose the highest-valuation sellers (the best objects) to sell. The set of matched sellers must have the same measure of the set of buyers. Let θ∗ be the lowest seller that gets matched. Then,

∗ 1 − F (θ ) = 1 − G (vs) ∗ −1 θ = F G (vs) .

So the eﬃcient mix of sellers and buyers (in terms of maximizing social ∗ ∗ welfare) involves buyers with vb ≥ vs and sellers with θ ≥ θ , with θ as deﬁned above. (b) Let p(θ) be the price of an object of value θ (since θ is observable we can condition on it). In equilibrium, prices p(θ) must be such that market clears. We expect p(θ) to be strictly increasing in θ. Suppose not, say there are two sellers with v2 > v1 and p(v2) < p(v1), then clearly good 2 will “dominate” good 1 (2 has a higher quality which is valued by consumers and a lower price).

Also note that, the higher vb buyers have higher valuation for quality and hence are willing to pay more for quality. Therefore, we expect high θ goods, which have higher prices, to be matched with high vb buyers. By the market clearing condition we require that for all items sold in equilibrium G(vb) = F (θ) which in turn implies the matching between sellers and buyers

−1 vb(θ) = G (F (θ)) (1)

−1 That is, a good with quality θ is matched with a buyer with vb = G (F (θ)). From the participation conditions (i.e., matched buyer and seller both willing to trade), we can immediately see that, if θ is sold in equilibrium, p(θ)

4 should satisfy −1 vsθ ≤ p(θ) ≤ G (F (θ))θ (2) But not all functions satisfying (2) will be equilibrium prices1 because, to support the matching, buyers have to be willing to buy the θ corresponding to them and not other. Type-vb buyers, if in equilibrium buying, are going to choose a good so that max vbθ − p(θ) θ From the FOC we have, 0 p (θ) = vb (3) Combining conditions (1) and (3) we get that, in equilibrium, p(θ) should satisfy the following condition, p0(θ) = G−1(F (θ)). (4)

Can we further characterize p(θ)? Yes, we can put additional restrictions on the last unit sold. Let θ be the last unit sold, that is, for θ < θ @p(θ) such that (2) and (4) hold. −1 Proposition: Let θ be the last unit sold, and define vb ≡ G (F (θ)). Then, −1 in equilibrium, θ = F (G(vs)) and p(θ) = vsθ. −1 Proof. Let’s first show that θ = F (G(vs)) by contradiction. Say θ < −1 F (G(vs)). This is equivalent to vb < vs. Then, from (2) −1 vsθ ≤ p(θ) ≤ G (F (θ))θ = vbθ < vsθ −1 a contradiction. Now, say θ > F (G(vs)). Then, it is enough to show that we can match a set of buyers and sellers with positive mass, willing to trade goods with quality ˜θ < θ at prices satisfying (2) and (4). −1 −1 Consider the sets Θ˜ = [vs, vb], and V˜ = [F (G(vs)),F (G(vb))]. Since −1 θ > F (G(vs)), we have that vb > vs, which in turn implies that P r(θ ∈ V˜ ) = P r(vb ∈ Θ)˜ > 0. Now, for θ ∈ V˜ consider the following prices: R θ −1 −1 p(θ) = −1 G (F (x))dx + vsF (G(vs)). It is easy to check that F (G(vs)) p(θ) satisfies (2) and (4), contradicting the fact that θ was the last item sold.

Now, let’s show that p(θ) = vsθ. From the previous argument we know −1 that θ = F (G(vs)). Then, from (2) we have −1 −1 −1 vsθ ≤ p(θ) ≤ G (F (θ))θ = G (F (F (G(vs))))θ = vsθ

which implies p(θ) = vsθ. Note that this price by construction satisﬁes (2), and also (4), since

0 p (θ) = vs = G−1(F (θ)).

1 For example, p(θ) = vsθ satisﬁes (2), but it is not an equilibrium price. It is enough to see that all buyers with vb > vs prefer buying the good with θ = 1 to all other goods, and we will observe excess demand.

5 Let’s check now an example. Let θ and vb be distributed uniformly between 0 and 1, that is F (θ) = θ and G(vb) = vb. From (4), p0(θ) = G−1(F (θ)) = θ hence 1 p(θ) = θ2 + κ 2 where κ is a constant of integration. From the Proposition we also know 2 1 that p(vs) = vs , hence κ = 2 vs. Therefore, 1 1 p(θ) = θ2 + v 2 2 s

p (c) At a price p, sellers are willing to trade iﬀ p ≥ vsθ, or θ ≤ . Therefore, vs the supply function is p F vs

Analogously, a buyer wants to trade iﬀ vbE (θ | p) ≥ p. The expected quality of the good (conditional on the seller being willing to trade at price p) is p R vs ¯ p 0 θf (θ) dθ θ = E θ | θ ≤ = vs F p vs Therefore, the demand function is   pF p p p vs Pr vb ≥ = 1 − G = 1 − G  p  ¯θ ¯θ R vs 0 θf (θ) dθ

Note that, although G(.) is an increasing function, without further knowl- edge of F (.), we cannot say that the demand is downward sloping since pF p ( vs ) both the numerator and denominator of p are increasing func- R vs 0 θf(θ)dθ tions of p. (d) For the uniform case p ¯θ (p) = 2vs D (p) = 1 − 2vs p S (p) = vs

1 Therefore, for vs ≤ 2 , there exists an equilibrium in which

2 p = vs − 2vs

6 The volume of trade in this case is 1 − 2vs which is less than the eﬃcient volume of trade (1 − vs). (e) Notice that without imposing restrictions on F and G, there can exist multiple equilibria. This is the case since, although the supply is an increasing function of the price, the demand is not necessarily decreasing in the price, hence multiple intersections of supply and demand can occur.

Now let there be two equilibria at prices p1 and p2 with p2 > p1. Note that sellers obviously prefer the high price. Moreover, at p2 more sellers are going to be willing to trade. In equilibrium, this means demand will have to be higher also. Therefore, the threshold buyer will have a lower taste, i.e.: p1 p2 ¯ > ¯ θ1 θ2

This implies, ∀vb ¯ ¯ θ2 θ1 vb > vb p2 p1 ¯ ¯ θ2 θ1 vb − 1 > vb − 1 p2 p1 ¯ ¯ θ2 ¯ vbθ2 − p2 > p1 vb − 1 > vbθ1 − p1 p2 which shows that the expected utility of all trading buyers is higher when the price is higher. (f) It is enough to show an example. Consider the uniform case in part (d), but introduce a minimum quality standard s0 ∈ [0, 1]. We now have p S(p, s0) = − s0 vs pvs D(p, s0) = 1 − 2 . p + s0vs We can solve for the equilibrium price by equating demand and supply,

pvs p 1 − 2 = − s0. p + s0vs vs We ﬁnd two roots, 1 h i p∗(s ) = (v − 2v 2) ± ((v − 2v 2)2 + 4s v (s v + v ))1/2 , 0 2 s s s s 0 s 0 s s ∗ but since we require p (s0) ≥ 0 we can discard the negative root. More- ∗ over, there is also an upper bound on the price, since p ≤ vs. Therefore, the equilibrium price will be

1 h i p∗(s ) = min (v − 2v 2) + ((v − 2v 2)2 + 4s v (s v + v ))1/2 , v . 0 2 s s s s 0 s 0 s s s

7 Since in equilibrium only the lowest quality items are traded, in order to show that welfare increases it is enough to show that the volume of trade ∗ increases. Let V (s0) denote the (equilibrium) volume of trade at quality standard s0, then ∗ ∗ p (s0) V (s0) = − s0. vs

2 For small s0, we will show that a minimum quality standard s0 > 0 increases the volume of trade, that is,

∗ ∗ V (s0) − V (0) > 0.

To do so, note that for small s0

∗ ∗ V (s0) − V (0) > 0

∗ ∗ p (s0) − p (0) ⇔ − s0 > 0 vs 2 2 2 1/2 ∗ ⇔ (vs − 2vs ) + ((vs − 2vs ) + 4s0vs(s0vs + vs)) − 2p (0) − 2s0vs > 0 2 2 1/2 2 ⇔ ((vs − 2vs ) + 4s0vs(s0vs + vs)) > (vs − 2vs ) + 2s0vs 2 2 2 2 2 2 2 ⇔ (vs −2vs ) +4s0vs(s0vs +vs) > (vs −2vs ) +4(vs −2vs )s0vs +4s0vs 3 ⇔ 8vs s0 > 0

which is always true since vs, s0 > 0. 3. Consider the following game of public good provision with private costs ci ≥ 0:

CD Contribute 1 − c1,1 − c2 1 − c1,1 Don’t Contribute 1,1 − c2 0,0

where the cost ci is i.i.d. distributed with a uniform density on [0, 2]. (a) Deﬁne the notion of a mixed strategy in this Bayesian game and deﬁne the notion of a Bayesian equilibrium for this game. (b) Compute a Bayesian Nash equilibrium of this game in pure strategies. (c) Consider the same public good provision game, but now analyze it with the solution concept of iteratively eliminating strictly dominated strategies rather than the Bayesian equilibrium. Where does the process of iterative elimination of strictly dominated strategies lead to?

2 1 2 2 2 1/2 s0 ∈ [0, 1] such that such 2 (vs − 2vs ) + ((vs − 2vs ) + 4s0vs(s0vs + vs)) ≤ vs.

8 [SOLUTION]

(a) Let type ci of player i contributing be denoted by si(ci) = 1, and not contributing by si(ci) = 0. Then we can write the (ex post) net utility as

ui(s1(c1), s2(c2), c1, c2) = max(s1(c1), s2(c2)) − cisi(ci).

A mixed strategy σi for player i in this game is given by

σi : Ti → ∆(Ai) where Ti = [0, 2] and Ai = {0, 1}. Hence it is enough to characterize a mixed strategy to specify for each player and for each type the probability of contributing, which we can denote by qi(ci) ∈ [0, 1]. Abusing notation, we can deﬁne a mixed strategy BNE. A strategy proﬁle ∗ ∗ σ is a (mixed strategy) BNE if σi (ci) maximizes

0 ∗ Ec−i ui(σi, σ−i(c−i), c) for all ci and all i. ∗ ∗ (b) A strategy proﬁle s is a (pure strategy) BNE if si (ci) maximizes

0 ∗ 0 Ec−i max(si, s−i(c−i)) − cisi for all ci and all i.

Now note that the payoff from choosing si(ci) = 1 is 1 − ci, and the payoff from choosing si(ci) = 0 is P r(s−i(c−i) = 1). Thus, the payoff from si = 1 is decreasing in ci, and the payoff of si = 0 is independent of ci. Hence, look at monotonic cutoff strategies of the form

∗ 1 if ci ≤ c si(ci) = ∗ 0 if ci > c

In particular, let’s look at symmetric c∗. Type c∗ of player i must be indiﬀerent between contributing and not, so

∗ ∗ ∗ 1 − c = P r(sj(cj) = 1) = P r(cj ≤ c ) = c /2 or 2 c∗ = 3 Therefore, all players with private cost below 2/3 contribute while players with ci > 2/3 do not.

(c) First note that if ci > 1, then 1 is a strictly dominated action and therefore we know si(ci) = 0 if ci > 1. Therefore, i knows that

P r(cj = 1) ≤ 1/2

9 So an upper bound for payoﬀ from not contributing is 1/2. Therefore, player i should contribute if ci ≤ 1/2. But then

P r(cj = 1) ≥ 1/4

Therefore i should not contribute if ci ≥ 3/4. But then

P r(cj = 1) ≤ 5/8 Continue iteratively to see that we have convergence to the BNE calculated above. To see this formally suppose that no opponent with c > x contributes. Then all types with c ≤ 1 − x/2 should contribute. But then no type 1 x above 1 − 2 (1 − 2 should contribute. The process of iterated elimination comes to a stop when 1 x x = 1 − 1 − 2 2 or 2 x = 3 For more general distribution of private costs F (c) one gets simply x = 1 − F (1 − F (x)) If this equation has a unique solution, then the game has a unique equilibrium. 4. This question asks you to show the equivalence ex ante and interim def- initions of (Bayesian) Nash equilibrium in static incomplete information games with a finite number of types. Consider a game consisting of a set of players 1, ..., I; each player has a finite set of actions Ai; each player has a type ti ∈ Ti. The profile of types t ≡ (t1, ..., tI ) is chosen according to a probability distribution p with p(t) > 0 for all t ∈ T . Payoff function for player i is a function gi : A × T → R. An incomplete information game strategy for player i is a function si : Ti → Ai. Write Si for the set of such strategies, let S = S1 × .. × SI , and write s (t) = (s1 (t1) , .., sI (tI )). (a) Show that strategy profile s∗ is an ex ante Bayesian Nash equilibrium of this game if and only if it is an interim Bayesian Nash equilibrium. (b) Does the result still hold if p(t) is only required to satisfy p(t) ≥ 0? If not, which part of the result fails to hold? [SOLUTION] (a) We need to show that the strategy profile s∗ is a (pure strategy) Nash equilibrium of this game if, and only if, for all i = 1, ..., I, for all ai ∈ Ai, and almost all ti ∈ [0, 1] , Z Z ∗ ∗ ∗ p (t−i|ti) gi si (ti) , s−i (t−i) , t ≥ p (t−i|ti) gi ai, s−i (t−i) , t . t−i∈T−i t−i∈T−i (5)

10 R For each player i, let ui (s) ≡ t∈[0,1]I p (t) gi (s (t) , t) represent the ex-ante payoﬀ of strategy proﬁle s.

Assume first that (5) holds for all i = 1, ..., I, almost all ti ∈ [0, 1] , and all −1 ai ∈ Ai. Now note that, since Ai is finite, si (·) defines a finite partition −1 of measurable sets si (ai): ai ∈ Ai of [0, 1] for any si ∈ Si. Therefore, for any i and arbitrary si ∈ Si, we have Z ∗ ∗ ui si, s−i = p (t) gi si (ti) , s−i (t−i) , t T Z X ∗ = p (t) gi ai, s−i (t−i) , t −1 s (ai)×T−i ai∈Ai i Z Z X ∗ = p (ti) p (tj|ti) gi ai, s−i (t−i) , t −1 s (ai) T−i ai∈Ai i Z X −1 ∗ = p si (ai) p (tj|ti) gi ai, s−i (t−i) , t T−i ai∈Ai Z X −1 ∗ ∗ ≤ p si (ai) p (tj|ti) gi si (ti) , s−i (t−i) , t T−i ai∈Ai However, we have Z ∗ ∗ ∗ ∗ ui si , s−i = p (t) gi si (ti) , s−i (t−i) , t T Z Z X ∗ ∗ = p (t) p (tj|ti) gi si (ti) , s−i (t−i) , t −1 s (ai)×T−i T−i ai∈Ai i Z X −1 ∗ ∗ = p si (ai) p (tj|ti) gi si (ti) , s−i (t−i) , t T−i ai∈Ai

∗ ∗ ∗ ∗ Therefore, ui si , s−i ≥ ui si, s−i . Since si and i were arbitrary, s is a (Bayesian) Nash equilibrium. Conversely, assume that (5) does not hold; that is, for some i ∈ {1, ..., I} 0 0 0 and some subset Ti ⊆ [0, 1] such that p (Ti ) > 0, there exists an ai ∈ Ai 0 such that, for all ti ∈ Ti , Z Z ∗ ∗ ∗ p (t−i|ti) gi si (ti) , s−i (t−i) , t < p (t−i|ti) gi ai, s−i (t−i) , t . t−i∈T−i t−i∈T−i

11 0 0 0 [0,1] 0 ai if ti ∈ Ti Define si ∈ Ai by si (ti) = ∗ 0 . Then we have si (ti) if ti 6∈ Ti Z 0 ∗ 0 ∗ ui si, s−i = p (t) gi si (ti) , s−i (t−i) , t t∈T Z Z 0 ∗ = p (ti) p (t−i|ti) gi si (ti) , s−i (t−i) , t ti∈Ti t−i∈T−i Z Z 0 ∗ = p (ti) p (t−i|ti) gi si (ti) , s−i (t−i) , t 0 ti∈Ti t−i∈T−i Z Z 0 ∗ + p (ti) p (t−i|ti) gi si (ti) , s−i (t−i) , t 0 ti6∈Ti t−i∈T−i Z Z 0 ∗ = p (ti) p (t−i|ti) gi ai, s−i (t−i) , t 0 ti∈Ti t−i∈T−i Z Z ∗ ∗ + p (ti) p (t−i|ti) gi si (ti) , s−i (t−i) , t 0 ti6∈Ti t−i∈T−i Z Z ∗ ∗ > p (ti) p (t−i|ti) gi si , s−i (t−i) , t 0 ti∈Ti t−i∈T−i Z Z ∗ ∗ + p (ti) p (t−i|ti) gi si (ti) , s−i (t−i) , t 0 ti6∈Ti t−i∈T−i Z Z ∗ ∗ = p (ti) p (t−i|ti) gi si (ti) , s−i (t−i) , t ti∈Ti t−i∈T−i Z ∗ ∗ = p (t) gi (s (t) , t) = ui (s ) , t∈T where the inequality follows from by hypothesis. Therefore, s∗ cannot be a Nash equilibrium. (b) If we allow p(t) = 0 for some t, ex ante BNE no longer implies interim BNE, but interim BNE still implies ex ante BNE. 5. Ex-post version of the Bayesian-Nash equilibrium. (a) Give a natural definition of an ex-post version of the Bayesian-Nash equilibrium, where ex post refers to the fact that the equilibrium conditions are evaluated conditional on knowing the entire type profile, and not only the own type as in the interim version of the Bayesian- Nash equilibrium. (b) Given your definition of an ex-post Bayes Nash equilibrium, what is the relationship with respect to the interim version of the Bayesian- Nash equilibrium. Either state results or give examples that demon- strate how these notions are different. (c) Give a definition of dominant strategy for Bayesian games. What is the relationship between ex post Bayes Nash equilibrium and an equilibrium in dominant strategies.

12 [SOLUTION] (a) The notion of an ex-post BNE corresponds to the situation in which all players types are known by all players. Hence, there is no private information anymore. For player i, there is no need to take expectations anymore, but she simply compares the utility from the BNE to any other utility that may arise if she chooses another action ai. ∗ ∗ ∗ Definition (Ex-Post Pure-Strategy BNE): A strategy profile s = (s1, . . . , sI ) is called a pure-strategy ex-post BNE iff for all players i ∈ I and all types t = (ti, t−i) ∈ T

∗ ∗ ∗ ui((si (ti), s−i(t − i)), t) ≥ ui((ai, s−i(t − i)), t) ∀ai ∈ Ai. (6)

∗ ∗ ∗ Definition (Ex-Post Mixed-Strategy BNE): A strategy profile σ = (σ1, . . . , σI ) is called a mixed-strategy ex-post BNE iff for all players i ∈ I and all types t = (ti, t−i) ∈ T

∗ ∗ ∗ ui((σi (ti), σ−i(t − i)), t) ≥ ui((ai, σ−i(t − i)), t) ∀ai ∈ Ai. (7)

(b) Proposition: Every ex-post-BNE (in pure or mixed strategies) is also an interim-BNE (in the corresponding strategies). Proof. Consider a mixed-strategy ex-post BNE, i.e. for all players i ∈ I and all tuples of types t ∈ T , the inequality (7) holds. All of these inequalities will remain valid if both sides are multiplied by the positive number (remember assumption p(t) > 0 ∀t ∈ T ) p(t−i|ti). Hence, ∀i ∈ I, ∀t ∈ T

∗ ∗ ∗ ui((σi (ti), σ−i(t − i)), t) ≥ ui((ai, σ−i(t − i)), t) ∀ai ∈ Ai

∗ ∗ ∗ ⇔ p(t−i|ti)ui((σi (ti), σ−i(t−i)), t) ≥ p(t−i|ti)ui((ai, σ−i(t−i)), t) ∀ai ∈ Ai

Then, ∀i ∈ I, ∀ti ∈ Ti

X ∗ ∗ X ∗ p(t−i|ti)ui((σi (ti), σ−i(t−i)), t) ≥ p(t−i|ti)ui((ai, σ−i(t−i)), t) ∀ai ∈ Ai t−i∈T−i t−i∈T−i which is the deﬁnition of a mixed-strategy interim-BNE, concluding the proof. Remark 1 : Once established for mixed strategies, the result follows triv- ially for pure strategies, because any pure strategy is a degenerate mixed strategy, where all probability is concentrated on one element. Remark 2 : The converse implication of the proposition is not true. Not every interim-BNE is also an ex-post BNE. This can be seen from the following simple example. Say player 1 has only one type, and player 2 1 2 has two types, t2 with probability α, and t2 with probability 1 − α. Types are private information and payoﬀs are

13 Player 2 Player 2 1 2 (type t2) (type t2) LRLR U 1,1 0,0 U 1,0 0,1 Player 1 D 0,1 1,0 D 0,0 1,1

It is easy to check that the following strategy proﬁle is an interim BNE

 U if α > 1/2  L if t = t1 s∗ = λU + (1 − λ)D for λ ∈ [0, 1] if α = 1/2 , s∗(t ) = 2 2 1 2 2 R if t = t2  D if α < 1/2 2 2

Now, let α = 1/4. In this case, player 1 chooses D. But for player 2 being 1 of type t2 (implying that player 2 will definitely choose L), player 1 will not play a best response. Thus, we have found a violation of the definition of an ex-post BNE for an equilibrium that clearly satisfies the definition of an interim BNE. (c) Definition (Weakly Domination): In a Bayesian game, we say that 0 0 for player i strategy si dominates strategy si, whith si, si ∈ Ai, if for all t ∈ T , s−i ∈ A−i 0 ui((si, s−i), t) ≥ ui((si, s−i), t) (8) and 0 ui((si, s−i), t) > ui((si, s−i), t) (9) for some s−i ∈ A−i. 0 In other words, for player i si dominates si if for all types ti and no matter 0 what the distribution of t−i is, si gives i a higher payoff than si.

Definition (Dominant Strategy): We say that strategy si is (weakly) dominant for player i if it weakly dominates every other strategy. We will now show that an equilibrium in dominant strategies is an ex post BNE. Let the strategy profile s∗ be an equilibrium in dominant strategies. ∗ Since for all i (8) and (9) hold for all s−i, in particular, they hold for s−i. Thus s∗ is an ex post BNE. The converse is not true. It is enough to show an example of an ex post BNE not in dominant strategies. Consider a simple 2 player game. Each player has one type and payoffs are

Player 2 LR U 1,1 0,0 Player 1 D 0,1 1,0

Clearly s∗ = (U, L) is an ex post BNE and player 1 in not playing a dominant strategy.

14 6. Consider the second price sealed bid auction with private values we dis- cussed in class. We showed that truthful bidding is a Bayes Nash equilibrium for all distribution, in fact it is a Bayes-Nash equilibrium in dominant strategies. Do other Bayes Nash equilbiria exist. Do other ex post Bayes- Nash equilbiria exist? [SOLUTION] Clearly there are other equilibria. For simplify assume the valuation of all players have support in [0, 1]. Then there exists a equilibrium in which one player bids 1000, and all other players bid 0. This is clearly an equilibrium, and the player bidding a 1000 pays 0 for the object in equilibrium.