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
0 < x < y. The good is worth 32 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 backward 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
P{v ≥ p} · E[ v − p|v ≥ p].
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{v ≥ p} · E[ v − p|v ≥ p]
2
=
=
=
=
(p − x)
3
(E[ v|v ≤ p] − p)
y−x
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
∂
P{v ≥ p} · E[ v − p|v ≥ p] =
(4x − 2p)
∂p
2
4(y − x)
⇒
∂
3
P{v ≥ p} · E[ v − p|v ≥ p] = 0 ⇐⇒ p = 2x
∂p
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
P{v ≥ p} · E[ v − p|v ≥ p] =
(3x − p) ≥ 0.
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:
(
∗
p =
3x
3x ≤ y
,
[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 = vs θ
and
ub = vb θ.
(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 )
∗
=
F −1 G (vs ) .
θ
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
vb (θ) = G−1 (F (θ))
(1)
That is, a good with quality θ is matched with a buyer with vb = G−1 (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
vs θ ≤ p(θ) ≤ G−1 (F (θ))θ
(2)
1
But not all functions satisfying (2) will be equilibrium prices 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 vb θ − p(θ)
θ
From the FOC we have,
p0 (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.
Proposition: Let θ be the last unit sold, and define vb ≡ G−1 (F (θ)). Then,
in equilibrium, θ = F −1 (G(vs )) and p(θ) = vs θ.
Proof. Let’s first show that θ = F −1 (G(vs )) by contradiction. Say θ <
F −1 (G(vs )). This is equivalent to vb < vs . Then, from (2)
vs θ ≤ p(θ) ≤ G−1 (F (θ))θ = vb θ < vs θ
a contradiction. Now, say θ > F −1 (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 θ̃ < θ at prices satisfying (2) and (4).
Consider the sets Θ̃ = [vs , vb ], and Ṽ = [F −1 (G(vs )), F −1 (G(vb ))]. Since
θ > F −1 (G(vs )), we have that vb > vs , which in turn implies that P r(θ ∈
Ṽ ) = P r(vb ∈ Θ̃) > 0. Now, for θ ∈ Ṽ consider the following prices:
Rθ
p(θ) = F −1 (G(vs )) G−1 (F (x))dx + vs F −1 (G(vs )). It is easy to check that
p(θ) satisfies (2) and (4), contradicting the fact that θ was the last item
sold.
Now, let’s show that p(θ) = vs θ. From the previous argument we know
that θ = F −1 (G(vs )). Then, from (2) we have
vs θ ≤ p(θ) ≤ G−1 (F (θ))θ = G−1 (F (F −1 (G(vs ))))θ = vs θ
which implies p(θ) = vs θ. Note that this price by construction satisfies
(2), and also (4), since
p0 (θ)
= vs
= G−1 (F (θ)).
1 For example, p(θ) = v θ satisfies (2), but it is not an equilibrium price. It is enough to
s
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 2
θ +κ
2
where κ is a constant of integration. From the Proposition we also know
that p(vs ) = vs 2 , hence κ = 21 vs . Therefore,
p(θ) =
p(θ) =
1 2 1
θ + vs
2
2
(c) At a price p, sellers are willing to trade iff p ≥ vs θ, or θ ≤
the supply function is
p
F
vs
p
vs .
Therefore,
Analogously, a buyer wants to trade iff vb E (θ | p) ≥ p. The expected
quality of the good (conditional on the seller being willing to trade at
price p) is
R vps
θf (θ) dθ
p
= 0 θ̄ = E θ | θ ≤
vs
F vps
Therefore, the demand function is

Pr vb ≥
p
θ̄
=1−G
p
θ̄
= 1 − G R
pF
p
vs
0
p
vs


θf (θ) dθ
Note that, although G(.) is an increasing function, without further knowledge of F (.), we cannot say that the demand is downward sloping since
pF ( p )
are increasing funcboth the numerator and denominator of R p vs
vs
0
θf (θ)dθ
tions of p.
(d) For the uniform case
θ̄ (p)
=
D (p)
=
S (p)
=
p
2vs
1 − 2vs
p
vs
Therefore, for vs ≤ 12 , there exists an equilibrium in which
p = vs − 2vs 2
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.:
p2
p1
>
θ̄1
θ̄2
This implies, ∀vb
vb
θ̄2
θ̄1
> vb
p2
p1
θ̄1
θ̄2
− 1 > vb − 1
p2
p1
θ̄2
vb θ̄2 − p2 > p1 vb − 1 > vb θ̄1 − p1
p2
vb
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
− s0
vs
pvs
D(p, s0 ) = 1 − 2
.
p + s0 vs
S(p, s0 ) =
We can solve for the equilibrium price by equating demand and supply,
1−2
pvs
p
=
− s0 .
p + s0 vs
vs
We find two roots,
i
1h
p∗ (s0 ) =
(vs − 2vs 2 ) ± ((vs − 2vs 2 )2 + 4s0 vs (s0 vs + vs ))1/2 ,
2
but since we require p∗ (s0 ) ≥ 0 we can discard the negative root. Moreover, there is also an upper bound on the price, since p∗ ≤ vs . Therefore,
the equilibrium price will be
h
i
1
∗
2
2 2
1/2
p (s0 ) = min
(vs − 2vs ) + ((vs − 2vs ) + 4s0 vs (s0 vs + vs ))
, vs .
2
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
For small s0 ,2 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
⇔ (vs − 2vs 2 ) + ((vs − 2vs 2 )2 + 4s0 vs (s0 vs + vs ))1/2 − 2p∗ (0) − 2s0 vs > 0
⇔ ((vs − 2vs 2 )2 + 4s0 vs (s0 vs + vs ))1/2 > (vs − 2vs 2 ) + 2s0 vs
⇔ (vs − 2vs 2 )2 + 4s0 vs (s0 vs + vs ) > (vs − 2vs 2 )2 + 4(vs − 2vs 2 )s0 vs + 4s20 vs 2
⇔ 8vs 3 s0 > 0
which is always true since vs , s0 > 0.
3. Consider the following game of public good provision with private costs
ci ≥ 0:
C
1 − c1 ,1 − c2
1,1 − c2
Contribute
Don’t Contribute
D
1 − c1 ,1
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?
2s
0
∈ [0, 1] such that such
1
2
(vs − 2vs 2 ) + ((vs − 2vs 2 )2 + 4s0 vs (s0 vs + vs ))1/2 ≤ 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 )) − ci si (ci ).
A mixed strategy σ i for player i in this game is given by
σ i : 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 σ ∗i (ci ) maximizes
Ec−i ui (σ 0i , σ ∗−i (c−i ), c)
for all ci and all i.
(b) A strategy profile s∗ is a (pure strategy) BNE if s∗i (ci ) maximizes
Ec−i max(s0i , s∗−i (c−i )) − ci s0i
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
c∗ =
2
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
above 1 − 21 (1 − x2 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 definitions 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
For each player i, let ui (s) ≡
payoff of strategy profile s.
R
p (t) gi (s (t) , t) represent the ex-ante
t∈[0,1]I
Assume first that (5) holds for all i = 1, ..., I, almost all ti ∈ [0, 1] , and all
ai ∈ Ai . Now note that, since Ai is finite,
s−1
i (·) 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
X Z
=
p (t) gi ai , s∗−i (t−i ) , t
ai ∈Ai
=
ai ∈Ai
=
s−1
i (ai )×T−i
X Z
X
Z
s−1
i (ai )
T−i
p s−1
i (ai )
p s−1
i (ai )
Z
≤
=
p (tj |ti ) gi
s∗i (ti ) , s∗−i (t−i ) , t
T−i
However, we have
Z
∗ ∗
ui si , s−i
=
p (t) gi
T
X Z
=
X
ai , s∗−i (t−i ) , t
Z
ai ∈Ai
ai ∈Ai
p (tj |ti ) gi
T−i
ai ∈Ai
X
ai , s∗−i (t−i ) , t
p (tj |ti ) gi
p (ti )
s∗i (ti ) , s∗−i (t−i ) , t
Z
p (t)
p (tj |ti ) gi
s−1
i (ai )×T−i
p s−1
i (ai )
s∗i (ti ) , s∗−i (t−i ) , t
T−i
Z
p (tj |ti ) gi
s∗i (ti ) , s∗−i (t−i ) , t
T−i
ai ∈Ai
Therefore, ui s∗i , 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}
and some subset Ti0 ⊆ [0, 1] such that p (Ti0 ) > 0, there exists an a0i ∈ Ai
such that, for all ti ∈ Ti0 ,
Z
Z
p (t−i |ti ) gi s∗i (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
[0,1]
Define s0i ∈ Ai
ui s0i , s∗−i
by s0i (ti ) =
Z
=
s0i (ti ) , s∗−i (t−i ) , t
p (t) gi
Zt∈T
a0i
if ti ∈ Ti0
. Then we have
∗
si (ti ) if ti 6∈ Ti0
Z
=
p (ti )
ti ∈Ti
t
Z −i
Z
=
s0i (ti ) , s∗−i (t−i ) , t
p (t−i |ti ) gi
s0i (ti ) , s∗−i (t−i ) , t
∈T−i
p (ti )
ti ∈Ti0
p (t−i |ti ) gi
t−i ∈T−i
Z
Z
+
p (t−i |ti ) gi
p (ti )
ti 6∈Ti0
t−i ∈T−i
Z
Z
=
p (t−i |ti ) gi
p (ti )
ti ∈Ti0
Z
+
p (t−i |ti ) gi
p (ti )
ti 6∈Ti0
Z
p (t−i |ti ) gi
p (ti )
ti ∈Ti0
Z
+
p (t−i |ti ) gi
p (ti )
ti 6∈Ti0
p (t−i |ti ) gi
p (ti )
=
s∗i (ti ) , s∗−i (t−i ) , t
t−i ∈T−i
Z
Z
ti ∈Ti
s∗i , s∗−i (t−i ) , t
t−i ∈T−i
Z
Z
s∗i (ti ) , s∗−i (t−i ) , t
t−i ∈T−i
Z
=
a0i , s∗−i (t−i ) , t
t−i ∈T−i
Z
>
s0i (ti ) , s∗−i (t−i ) , t
s∗i (ti ) , s∗−i (t−i ) , t
t−i ∈T−i
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 BayesianNash equilibrium.
(b) Given your definition of an ex-post Bayes Nash equilibrium, what is
the relationship with respect to the interim version of the BayesianNash equilibrium. Either state results or give examples that demonstrate 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∗ = (s∗1 , . . . , s∗I )
is called a pure-strategy ex-post BNE iff for all players i ∈ I and all types
t = (ti , t−i ) ∈ T
ui ((s∗i (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 ∗ = (σ ∗1 , . . . , σ ∗I )
is called a mixed-strategy ex-post BNE iff for all players i ∈ I and all types
t = (ti , t−i ) ∈ T
ui ((σ ∗i (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 ((σ ∗i (ti ), σ ∗−i (t − i)), t) ≥ ui ((ai , σ ∗−i (t − i)), t) ∀ai ∈ Ai
⇔ p(t−i |ti )ui ((σ ∗i (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
p(t−i |ti )ui ((σ ∗i (ti ), σ ∗−i (t−i)), t) ≥
t−i ∈T−i
X
p(t−i |ti )ui ((ai , σ ∗−i (t−i)), t) ∀ai ∈ Ai
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 trivially 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
has two types, t12 with probability α, and t22 with probability 1 − α. Types
are private information and payoffs are
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Player 2
(type t12 )
Player 1
U
D
L
1,1
0,1
Player 2
(type t22 )
R
0,0
1,0
It is easy to check that the following

 U
λU + (1 − λ)D for λ ∈ [0, 1]
s∗1 =

D
U
D
L
1,0
0,0
R
0,1
1,1
strategy profile is an interim BNE
if α > 1/2
if α = 1/2 ,
if α < 1/2
s∗2 (t2 ) =
L
R
if t2 = t12
if t2 = t22
Now, let α = 1/4. In this case, player 1 chooses D. But for player 2 being
of type t12 (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
for player i strategy si dominates strategy s0i , whith si , s0i ∈ Ai , if for all
t ∈ T , s−i ∈ A−i
ui ((si , s−i ), t) ≥ ui ((s0i , s−i ), t)
(8)
and
ui ((si , s−i ), t) > ui ((s0i , s−i ), t)
(9)
for some s−i ∈ A−i .
In other words, for player i si dominates s0i if for all types ti and no matter
what the distribution of t−i is, si gives i a higher payoff than s0i .
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
Player 1
U
D
L
1,1
0,1
R
0,0
1,0
Clearly s∗ = (U, L) is an ex post BNE and player 1 in not playing a
dominant strategy.
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6. Consider the second price sealed bid auction with private values we discussed 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 BayesNash 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.
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