14.12: Problem Set 5 Solution
Ruitian Lang
December 3, 2010
1
(a) There are two players, 1 and 2, with
A
1
=
{
U,D
}
and
A
2
=
{
L,R
}
. The type spaces are
T
1
=
{
1
,
1
}
,T
2
=
{
1
,
1
}
, and the beliefs are
p
1
(
t
2

t
1
) =
p
2
(
t
1

t
2
) = 1
/
2 for all
t
1
∈
T
1
and
t
2
∈
T
2
. The
payoﬀ function is as follows:
u
i
(
D,L
;
t
1
,t
2
) =
u
i
(
U,R
;
t
1
,t
2
) = 0 for
i
= 1
,
2 and all
t
1
,t
2
;
u
1
(
U,L
;
t
1
,t
2
) =
2
t
1
,u
1
(
D,R
;
t
1
,t
2
) =
t
1
,u
2
(
U,L
;
t
1
,t
2
) =
t
2
and
u
2
(
D,R
;
t
1
,t
2
) = 2
t
2
. In the notation of the problem,
t
1
is
θ
, and
t
2
is
λ
.
(b) Suppose that in a BNE Player 2 always plays
L
. Then Player 1’s best response is
U
when
θ
= 1 and
D
when
θ
=

1, which must coincide with Player 1’s strategy in the BNE. Now if Player 2 sees that
λ
= 1,
her payoﬀ by playing
L
is
∑
θ
∈
T
1
u
2
(
s
1
(
θ
)
,L
;
θ,λ
= 1)
p
2
(
θ

λ
= 1) = 1
/
2 while her payoﬀ by playing
R
is
∑
θ
∈
T
1
u
2
(
s
1
(
θ
)
,R
;
θ,λ
= 1)
p
2
(
θ

λ
= 1) = 1. Therefore, her best response is
R
, a contradiction.
Similarly, we can see that Player 2 cannot always play
R
in a BNE, since then Player 1 plays
U
when
θ
=

1 and
D
when
θ
= 1 and Player 2’s best response is to play
L
when
λ
=

1.
Therefore, in a pure strategy Nash equilibrium, Player 2 must play diﬀerent strategies for diﬀerent types.
Since
p
1
(
λ

θ
) = 1
/
2 for all
λ
and
θ
,
∑
λ
∈
T
2
u
1
(
a
1
,s
2
(
λ
);
θ,λ
)
p
1
(
λ

θ
) =
1
2
u
1
(
a
1
,L
;
θ,λ
) +
1
2
u
1
(
a
1
,R
;
θ,λ
). In
other words, Player 1 always believes that Player 2 plays
L
with probability 1
/
2. Therefore, Player 1’s best
response is to play
U
when
θ
= 1 and
D
when
θ
=

1. Given this, Player 2’s best response is to play
L
when
λ
=

1 and
R
when
λ
= 1. Therefore, the only possible BNE is that
s
1
(
θ
= 1) =
U,s
1
(
θ
=

1) =
D,s
2
(
λ
= 1) =
R
and
s
2
(
λ
=

1) =
L
. We have also seen that this is indeed a BNE.