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Game Theory with Applications to Finance and Marketing

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e ∈ (0, r), x i (σ i )+(e, e, · · · , e) ∈ X i , <strong>and</strong> it strictly dominates σ i . Next,<br />

define<br />

Y i ≡ {y − x i (σ i ) : y ∈ X i }.<br />

It follows that zero is a boundary point of the nonempty, convex, compact<br />

set Y i . Consider the nonempty Z ⊂ R n j<br />

defined by<br />

Z ≡ {z ∈ R n j<br />

: z ≫ 0 nj ×1},<br />

where z ≫ 0 means that for all k = 1, 2, · · · , n j , the k-th element of z,<br />

denoted z k , is strictly positive. Note that Z ⋂ Y i = ∅. Moreover, Z is<br />

convex. One version of the separating hyperplane theorem implies the<br />

presence of some non-zero vec<strong>to</strong>r p ∈ R n j<br />

such that p ′ y ≤ 0 ≤ p ′ z for<br />

all y ∈ Y i <strong>and</strong> z ∈ Z. Now we claim that for all k = 1, 2, · · · , n j , the<br />

k-th element of p, denoted p k , is non-negative. To see this, suppose<br />

that p k < 0 for some k. This implies that for some l, p l > 0 (so that<br />

n j ≥ 2). Let m be the largest l <strong>with</strong> p l > 0. Pick z ∗ ∈ Z such that<br />

zk ∗ > (n j − 1)p m <strong>and</strong> zq ∗ = 1 for all q ≠ k. It follows that, for this z∗ ,<br />

p ′ z ∗ < 0, which is a contradiction.<br />

Thus we have shown the existence of a positive vec<strong>to</strong>r p ∈ R n j<br />

, of<br />

which not all elements are zero, such that p defines a hyperplane (or<br />

a functional) separating the sets Z <strong>and</strong> Y i . We can normalize this<br />

functional by letting p be such that ∑ n j<br />

k=1 p k = 1, so that p is a legitimate<br />

mixed strategy for player j. Given p, since p ′ y ≤ 0 for all y ∈ Y i , we<br />

have shown that σ i is a best response of player i <strong>to</strong> player j’s mixed<br />

strategy p. As in the first proof for proposition 4, this argument can<br />

be iterated <strong>to</strong> show that the set of profiles surviving iterated strict<br />

dominance is included in the set of rationalizable profiles, so that the<br />

two solution concepts coincide in two-player finite strategic games.<br />

14. The above proof for proposition 4 fails if I > 2 because not all prob.<br />

distributions over S −i are products of independent prob. distributions<br />

over S j , for all j ≠ i. (Recall that an NE in mixed strategy assumes<br />

independent r<strong>and</strong>omization among players.) However, the equivalence<br />

between the two concepts stated in Proposition 4 is res<strong>to</strong>red if players’<br />

r<strong>and</strong>omization can be correlated.<br />

Definition 12. Given a game in normal form,<br />

G = (I ⊂ R; {S i ; i ∈ I}; {u i : Π i∈I S i → R; i ∈ I}),<br />

13

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