1. No category

MIDTERM 1
Your name:
Question 1. Let s ∈ S be a Nash Equilibrium strategy profile. Show that any
a ∈ Ai s.t. si (ai ) > 0 is rationalizable. (In words, show that any action used with a
positive probability in a mixed strategy Nash Equilibrium is rationalizable.)
i
Hint: You may find it useful to use the fact that a Nash Equilibrium mixed
strategy is concentrated on the set of pure best responses.
Hint 2: An action ai ∈ Ai is rationalizable if there exists a vector (R1 , ..., RN )
such that:
1. ai ∈ Ri
2. For all j, Rj ⊂ Aj
3. ∀j, bj ∈ Rj ,
Aj .
∃µ(bj ) ∈ ∆(A−j ) (with support R−j ) s.t. uj (bj , µ) ≥ uj (aj , µ) ∀aj ∈
You need to show that every part of the definition can be satisfied. That is, for
any ai ∈ Ai s.t. si (ai ) > 0, you need to find (R1 , ..., RN ) such that all parts of the
definition hold.
Solution: Let s be a NE strategy. Consider any i, ai ∈ Ai s.t. si (ai ) > 0.
Let R = (R1 , R2 , ..., RN ) with Rj = P BRj (s) for all j.
1. ai ∈ P BRi (s) since any si ∈ BRi (s) and BRi (s) is concentrated on P BRi (s).
2. ∀j, P BRj (s) ⊂ Aj .
3. ∀j, bj ∈ P BRj (s), let µ(bj ) = P rs−j (·) ∈ ∆(A−j ). I.e., µ(bj ) ∈ ∆(A−j ) is a
distribution on the other players’ actions implied by the Nash Equilibrium s. Note
that µ(bj ) (which is a product on NE mixed strategies) is concentrated on R−j , the
product of the PBR sets of players other than j.
It remains to show that
uj (bj , µ(bj )) ≥ uj (aj , µ(bj )) ∀aj ∈ Aj .
1
By definition of P BRj (s),
uj (bj , s−j ) ≥ uj (aj , s−j ) ∀aj ∈ Aj .
Since µ(bj ) =
Q
k6=j
sk , it follows that:
uj (bj , µ(bj )) ≥ uj (aj , µ(bj )) ∀aj ∈ Aj .
This is what we needed to show.
2
Question 2. Consider the following game:
a1
a2
a3
b1
(1, −1)
(−1, 1)
(0, 3)
b2
(−1, 1)
(1, −1)
(−2, 4)
b3
(7, −2)
(2, 0)
(5, 1)
Question: What is the set of all (pure and mixed) Nash Equilibria? Show your
work.
Answer: By Question 1, any action played in a Nash Equilibrium is rationalizable. Rationalizable actions survive IESDA. Hence, we can focus on the game that
survives IESDA:
a1
a2
b1
b2
(1, −1) (−1, 1)
(−1, 1) (1, −1)
Clearly, this game has no pure strategy NE. To find the mixed strategy equilibria,
let (p, 1 − p) be the mixed strategy of the row player and (q, 1 − q) the mixed strategy
of the column player. NE solves
q − (1 − q) = (1 − q) = q
and
(1 − p) − p = p − (1 − p).
Hence, p = q = 12 . I.e., the mixed strategy Nash is (( 12 , 12 ), ( 12 , 12 )).
Question: What is the set of all correlated equilibria? Show your work.
The first thing to notice is that a strictly dominated action will not be played in
any correlated equilibrium.
To see this, recall that the definition of correlated equilibrium implies that:
ui (ai , µ(a−i |ai )) ≥ ui (bi , µ(a−i |ai )) ∀bi ∈ Ai .
Hence,
ui (ai , µ(a−i |ai )) ≥ ui (si , µ(a−i |ai )) ∀si ∈ S i .
If ai is strictly dominated, then ∃ˆ
si s.t. ui (ˆ
si , µ(a−i |ai )) > ui (ai , µ(a−i |ai )), which
contradicts the statement above.
3
It follows that to find all correlated equilibria, we can concentrate on the following
game:
a1
a2
b1
b2
(1, −1) (−1, 1)
(−1, 1) (1, −1)
A correlated equilibrium (p11 , p12 , p21 , p22 ) satisfies the following inequalities:
p11 − p12 ≥ p12 − p11
p22 − p21 ≥ p21 − p22
p21 − p11 ≥ p11 − p21
p12 − p22 ≥ p22 − p12
Hence,
p11 ≥ p12
p22 ≥ p21
p21 ≥ p11
p12 ≥ p22
Hence,
p11 ≥ p12 ≥ p22 ≥ p21 ≥ p11 .
Hence,
p11 = p12 = p21 = p22 .
4