Answers to Problem Set 5
Economics 703
Spring 2016
1. Let’s write down what a common prior, if it exists, must satisfy. So suppose p is a
common prior. Then it must be true that the conditional probabilities it generates match
the beliefs of the two types of the players. In particular, it must satisfy
1
p(a, c)
= p1 (c | a) =
p(a, c) + p(a, d)
4
p(b, c)
3
= p1 (c | b) =
p(b, c) + p(b, d)
4
p(a, c)
1
= p2 (a | c) =
p(a, c) + p(b, c)
4
3
p(a, d)
= p2 (a | d) =
p(a, d) + p(b, d)
4
We can rearrange these equations as
3p(a, c) = p(a, d)
p(b, c) = 3p(b, d)
3p(a, c) = p(b, c)
p(a, d) = 3p(b, d)
The first three equations together imply 3p(a, c) = p(a, d) = p(b, c) = 3p(b, d). This plus
the condition that the probabilities sum to 1 is enough to pin down the prior. Specifically,
let x = p(a, d). Then we have p(b, c) = x and p(a, c) = p(b, d) = x/3. Hence the condition
that the probabilities sum to 1 implies 2x + (2x/3) = 1 or 8x = 3, so x = 3/8. Hence
p(a, d) = p(b, c) = 3/8 and p(a, c) = p(b, d) = 1/8. Note that these probabilities satisfy
the fourth equation, so this gives a common prior.
b) After seeing (a), you probably have a good idea why we’re not going to have
a common prior in (b). In (a), the common prior was pinned down by the first three
equations. Hence it was sheer luck (or clever construction of the problem) that it satisfied
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the fourth. Here we won’t be so fortunate. To see this, suppose that there is a common
prior p. Then it must satisfy
p(a, c)
1
= p1 (c | a) =
p(a, c) + p(a, d)
2
p(b, c)
1
= p1 (c | b) =
p(b, c) + p(b, d)
2
p(a, c)
1
= p2 (a | c) =
p(a, c) + p(b, c)
3
p(a, d)
2
= p2 (a | d) =
p(a, d) + p(b, d)
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We can rearrange these equations as
p(a, c) = p(a, d)
p(b, c) = p(b, d)
2p(a, c) = p(b, c)
p(a, d) = 2p(b, d)
The first three equations together imply 2p(a, c) = 2p(a, d) = p(b, d) = p(b, c). This plus
the condition that the probabilities sum to 1 is enough to pin down the prior. Specifically,
let x = p(b, d). Then we have p(b, c) = x and p(a, c) = p(a, d) = x/2. Hence the condition
that the probabilities sum to 1 implies 3x = 1 or x = 1/3. Hence p(b, d) = p(b, c) = 1/3
and p(a, c) = p(a, d) = 1/6. However, plugging these numbers into the last equation
gives 1/6 = 2(1/3) which is obviously false. Hence no common prior exists.
2. a) The simplest way to describe to do this one is to suppose that player 2 has two
types, t1 and t2 , both of which have probability 1/2. He is type t1 when he knows the
game is G1 and type t2 when he knows it is G2 . In other words, the utility function for
player j given that player 2 is type ti is the one defined by the matrix Gi .
Clearly, if when player 2 is type t1 , he has a strictly dominant strategy of b2 . Hence
σ2 (t1 ) = b2 . Suppose that σ2 (t2 ) also equals b2 . Then player 1’s best strategy is a2 since
he knows this will yield a payoff of 3. Since b2 is not the best reply to a2 for player 2 when
his type is t2 , this is not an equilibrium. Hence if there is a pure strategy equilibrium,
we must have σ2 (t2 ) = b1 . Since the two types are equally likely, it is not hard to show
player 1’s best reply to this strategy is a3 . Since player 2’s best reply when his type is t2
is b1 , this is an equilibrium. Hence the unique pure strategy Bayes–Nash equilibrium is
σ1 = a3 , σ2 (t1 ) = b2 , and σ2 (t2 ) = b1 .
b) Now we have several types for player 1. The easiest way to describe the game is to
i2
suppose that there are three possible types for player 1, say tu1 , ti1
1 , and t1 , and two for
2
1
i2 2
player 2, say t12 and t22 . The common prior is p(tu1 , t12 ) = p(tu1 , t12 ) = p(ti1
1 , t2 ) = p(t1 , t2 ) =
i1
u
1/4. More intuitively, think of t1 as player 1’s type when he is uninformed, t1 as his
type when he is informed that player 2 is his first type, and ti2
1 as his type when he is
informed that player 2 is his second type. Just as in part (a), player 1 is of type t12 when
he is informed that the game is G1 and type t22 when he is informed that the game is
G2 . Again, we can define payoffs as a function of types and actions analogously to what
we did in (a). You can use Bayes’ Rule to verify that if player 1 is told his type, he
updates to 1/2 probability of t12 if his type is tu1 , updates to probability 1 on t12 if his type
i2
1
is ti1
1 , and updates to probability 0 on t2 if his type is t1 . Similarly, player 2 always puts
probability 1/2 on player 1 being informed and knows that if player 1 is uninformed,
player 1 believes each game has probability 1/2.
To find an equilibrium, again, note that σ2 (t12 ) must be b2 . As a result, we must have
2
σ1 (ti1
1 ) = a2 . Suppose σ2 (t2 ) = b2 . Then every type of player 1 chooses a2 . Since b2 is not
then a best reply by t22 , this is not an equilibrium. So suppose σ2 (t22 ) = b1 . As in part (a),
the best response for player 1 if his type is tu1 is a3 . If his type is ti2
2 , it is a1 . Hence this
strategy for type t22 is optimal iff (1/2)(0) + (1/2)(2) ≥ (1/2)(1) + (1/2)(0) which is true.
(Recall that if player 2 is type t22 , he puts probability 1/2 on player 1 being type tu1 and
probability 1/2 on player 1 being type ti2
2 .) Therefore, this is the unique pure strategy
Bayes–Nash equilibrium.
3. First, let’s look for separating equilibria. Try m(t1 ) = m1 and m(t2 ) = m2 . If this is
1’s strategy, 2’s strategy must be a(m1 ) = a2 and a(m2 ) = a1 . Given this strategy for 2,
is 1’s strategy optimal? First consider type t1 . If he sends m1 , his payoff is −1, while if
he sends m2 , his payoff is 0. Hence he prefers message m2 , so this is not an equilibrium.
Now let’s try m(t1 ) = m2 and m(t2 ) = m1 . If this is 1’s strategy, 2’s best strategy
is a(m1 ) = a1 and a(m2 ) = a2 . If this is 2’s strategy, what is optimal for 1? Type t1
gets 5 if he uses m1 and gets −1 if he uses m2 . Hence he prefers m1 so this is not an
equilibrium. Hence there are no separating equilibria.
Now let’s turn to pooling equilibria. First, suppose m(t1 ) = m(t2 ) = m1 . Then 2’s
belief in response to m1 must be the same as the prior which is to put probability p < 1/2
on t1 . Hence his optimal action in response to m1 must be a1 . Suppose his response to
m2 is also a1 . Then both types prefer m1 to m2 since m1 gives each type 5, while m2
would give each type 0. So if 2’s beliefs are that he puts probability p on t1 in response
to either message, he’ll play a1 in response to either and 1’s optimal strategy will be to
always send message m1 . Hence this is a pooling equilibrium.
Could we have a pooling equilibrium with m(t1 ) = m(t2 ) = m1 , a(m1 ) = a1 , and
a(m2 ) = a2 — that is, can we change 2’s reaction off the equilibrium path? Yes. This
change lowers the incentives of player 1 to play m2 since this reaction gets him a payoff
of −1. Hence this is another pooling equilibrium.
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Can we have a pooling equilibrium where m(t1 ) = m(t2 ) = m2 ? The beliefs in
response to this message would have to be the prior, so 2 would be putting probability p
on t1 and would play a1 . If he also plays a1 in response to m1 , this is not an equilibrium
since both types of player 1 would deviate to m1 . If he plays a2 in response to m1 ,
though, it is an equilibrium. Neither type would deviate to m1 since this gives a payoff of
−1, while m2 gives a payoff of 0. Again, we can pick 2’s beliefs to make this sequentially
rational for him.
Summarizing, then, there are three pooling equilibria: one with m(t1 ) = m(t2 ) = m1
and a(m1 ) = a(m2 ) = a1 , one with m(t1 ) = m(t2 ) = m1 , a(m1 ) = a1 , and a(m2 ) = a2 ,
and one with m(t1 ) = m(t2 ) = m2 , a(m1 ) = a2 , and a(m2 ) = a1 .
Some people would argue that this last equilibrium is not terribly intuitive. Since
both types really hate message m2 , why are they pooling at that message?
4. Note first that there is no way to get type t1 to pick m2 — his payoff is −4, while at
worst he gets 0 from m1 .
Let’s start with separating. As noted above, we can’t have m(t1 ) = m2 , so the only
possibility for separating is m(t1 ) = m1 and m(t2 ) = m2 . If this is 1’s strategy, 2’s best
response to m1 is a1 since this gives 1 instead of 0. His best response to m2 is also a1 .
So we’d have to have a(m1 ) = a(m2 ) = a1 . If this is 2’s strategy, then 1’s strategy is
optimal: type t1 certainly wants to play m1 , while type t2 is indifferent and so is willing
to play m2 .
Now let’s try pooling. Again, we know we can’t have m(t1 ) = m2 , so the only
possibility for pooling is m(t1 ) = m(t2 ) = m1 . If this is 1’s strategy, 2’s beliefs in response
to m1 must be the same as the prior which puts probability p < 1/2 on t1 . Hence 2’s
best action would be a2 . What should we make the reply to m2 be? If we make it a1 ,
type t2 will certainly deviate: he gets 1 from playing m1 and getting a response of a2 ,
while the deviation would then give him 3. So we must have a(m1 ) = a(m2 ) = a2 . If this
is 2’s strategy, then it is optimal for both types of player 1 to choose m1 . Notice that
the only beliefs that will get player 2 to respond to m2 with a2 are probability 1 on type
t1 . Weak perfect Bayesian will certainly let us use these beliefs and even sequential will
allow them (which you can verify). However, it seems very weird: type t1 is the type who
would never use message m2 , but 2’s beliefs in this equilibrium have to put probability
1 on t1 when he sees m2 .
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