Game Development Reference
In-Depth Information
To conclude this section we clarify the connections between the notions of
dominance and of best response.
Clearly, if a strategy is strictly dominated, then it is a never best response.
However, the converse fails. Further, there is no relation between the notions
of weak dominance and never best response. Indeed, in the game considered
in Subsection 1.3.3 strategy C is a never best response, yet it is neither
strictly nor weakly dominated. Further, in the game given in Example 1.9
strategy M is weakly dominated and is also a best response to B .
The situation changes in the case of mixed extensions of two-player finite
games. Below, by a totally mixed strategy we mean a mixed strategy with
full support, i.e., one in which each strategy is used with a strictly positive
probability. The following results were established by Pearce [1984].
Theorem 1.24
Consider a finite two-player strategic game.
(i) A pure strategy is strictly dominated by a mixed strategy iff it is not a
best response to a mixed strategy.
(ii) A pure strategy is weakly dominated by a mixed strategy iff it is not a
best response to a totally mixed strategy.
We only prove here part ( i ). Pearce [1984] provides a short, but a bit
tricky proof based on Nash's Theorem 1.14. The proof we provide, due to
Fudenberg and Tirole [1991], is a bit more intuitive.
We shall use the following result, see, e.g., Rockafellar [1996].
Theorem 1.25 (Separating Hyperplane)
Let A and B be disjoint convex
k . Then there exists a non-zero c
k and d
subsets of
R
R
R
such that
c
·
x
d for all x
A,
c
·
y
d for all y
B.
Proof of Theorem 1.24 ( i ).
Clearly, if a pure strategy is strictly dominated by a mixed strategy, then
it is not a best response to a mixed strategy. To prove the converse, fix a
two-player strategic game ( S 1 ,S 2 ,p 1 ,p 2 ). Also fix i ∈{ 1 , 2 } and abbreviate
3 − i to −i .
Suppose that a strategy s i ∈ S i is not strictly dominated by a mixed
strategy. Let
R |S −i | |∀
A :=
{
x
s −i
S −i x s −i > 0
}
and
B :=
{
( p i ( m i ,s −i )
p i ( s i ,s −i )) s −i ∈S −i |
m i
Δ S i }
.