Game Development Reference

In-Depth Information

this context we talk about a
pure Nash equilibrium
, when each of the

constituent strategies is pure, and refer to an arbitrary Nash equilibrium of

the mixed extension as a
Nash equilibrium in mixed strategies
of the

initial finite game. In what follows, when we use the letter
m
we implicitly

refer to the latter Nash equilibrium.

Lemma 1.13
(Characterisation)
Consider a finite strategic game

(
S
1
,...,S
n
,p
1
,...,p
n
)
. The following statements are equivalent:

(i) m is a Nash equilibrium in mixed strategies, i.e.,

p
i
(
m
)
≥ p
i
(
m
i
,m
−i
)

and all m
i
∈

for all i

∈{

1
,...,n

}

Δ
S
i
,

(ii) for all i

∈{

1
,...,n

}

and all s
i
∈

S
i

p
i
(
m
)

≥

p
i
(
s
i
,m
−i
)
,

(iii) for all i

∈{

1
,...,n

}

and all s
i

∈

support
(
m
i
)

p
i
(
m
)=
p
i
(
s
i
,m
−i
)

and for all i ∈{
1
,...,n} and all s
i
∈ support
(
m
i
)

p
i
(
m
)

≥

p
i
(
s
i
,m
−i
)
.

Exercise 1.5

Provide the proof.

Note that the equivalence between (
i
) and (
ii
) implies that each Nash

equilibrium of the initial game is a pure Nash equilibrium of the mixed

extension. In turn, the equivalence between (
i
) and (
iii
) provides us with

a straightforward way of testing whether a joint mixed strategy is a Nash

equilibrium.

We now illustrate the use of the above theorem by finding in the Battle of

the Sexes game a Nash equilibrium in mixed strategies, in addition to the

two pure ones exhibited in Section 1.3. Take

m
1
:=
r
1

·

F
+(1

−

r
1
)

·

B,

m
2
:=
r
2

·

F
+(1

−

r
2
)

·

B,

where 0
<r
1
,r
2
<
1. By definition

p
1
(
m
1
,m
2
)=2

·

r
1
·

r
2
+(1

−

r
1
)

·

(1

−

r
2
)
,

p
2
(
m
1
,m
2
)=
r
1
·

r
2
+2

·

(1

−

r
1
)

·

(1

−

r
2
)
.

Suppose now that (
m
1
,m
2
) is a Nash equilibrium in mixed strategies. By

the equivalence between (
i
) and (
iii
) of the Characterisation Lemma 1.13

p
1
(
F, m
2
)=
p
1
(
B, m
2
), i.e., (using
r
1
= 1 and
r
1
= 0 in the above formula

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