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Unformatted text preview: 0: A small increase in the price of provider 1 will generate 1
2
positive proﬁts, thus provider 1 has an incentive to deviate. 2 p = p > 0: Let x b e the corresponding ﬂow allocation. If x = 1, 1
2
1
then provider 2 has an incentive to decrease its price. If x1 < 1, then provider 1 has an incentive to decrease its price. 3 0 ≤ p < p : Player 1 has an incentive to increase its price since its 1
2
ﬂow allocation remains the same. 4 0 ≤ p < p : For � suﬃciently small, the proﬁt function of player 2, 2
1
given p1 , is strictly increasing as a function of p2 , showing that provider 2 has an incentive to increase its price. 9 Game Theory: Lecture 5 Existence Results Existence Results
We start by analyzing existence of a Nash equilibrium in ﬁnite
(strategic form) games, i.e., games with ﬁnite strategy sets.
Theorem
(Nash) Every ﬁnite game has a mixed strategy Nash equilibrium.
Implication: matching pennies game necessarily has a mixed strategy
equilibrium.
Why is this important?
Without knowing the existence of an equilibrium, it is diﬃcult (perhaps
meaningless) to try to understand its properties.
Armed with this theorem, we also know that every ﬁnite game has an
equilibrium, and thus we can simply try to locate the equilibria. 10 Game Theory: Lecture 5 Existence Results Approach
Recall that a mixed strategy proﬁle σ∗ is a NE if
ui (σi∗ , σ∗ i ) ≥ ui (σi , σ∗ i ),
−
− for all σi ∈ Σi . ∗
In other words, σ∗ is a NE if and only if σi∗ ∈ B−i (σ∗ i ) for all i ,
−
∗ ( σ ∗ ) is the b est response of player i , given that the other
where B−i −i
players’ strategies are σ∗ i .
−
We deﬁne the correspondence B : Σ � Σ such that for all σ ∈ Σ, we
have
(1)
B (σ) = [Bi (σ−i )]i ∈I The existence of a Nash equilibrium is then equivalent to the
existence of a mixed strategy σ such that σ ∈ B (σ): i.e., existence of
a ﬁxed point of the mapping B .
We will establish existence of a Nash equilibrium in ﬁnite games using
a ﬁxed point theorem.
11 Game Theory: Lecture 5 Existence Results Deﬁnitions A set in a Euclidean space is compact if and only if it is bounded and
closed.
A set S is convex...
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This document was uploaded on 03/19/2014 for the course EECS 6.254 at MIT.
 Spring '10
 AsuOzdaglar

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