By Martin J. Osborne and Ariel Rubinstein

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If it chooses F then challenger k + 1 believes that the probability that the chain-store is tough is max{bK−k , µCS (h)(T )} and stays out. Thus if the chain-store chooses F then it obtains −1 against challenger k and a against challenger k + 1. Thus it is optimal to choose F . • Suppose that t(h) = k ≤ K − 1 and µCS (h)(T ) < bK−k . Then if the chain-store chooses C it obtains 0 in the future. If it chooses F then challenger k + 1 believes that the probability that the chain-store is tough is max{bK−k , µCS (h)(T )} = bK−k and chooses Out with probability 1/a.

For each i ∈ N let Si be a winning coalition that does not contain i; let x be a payoff profile in the core. Then x(N \ {i}) ≥ x(Si ) ≥ v(Si ) = 1, so that i∈N x(N \ {i}) ≥ |N |. On the other hand x(N \ {i}) = (|N | − 1) i∈N xi = |N | − 1, i∈N 46 Chapter 13. The Core a contradiction. b. Let V be the set of veto players. Let x be a nonnegative feasible payoff profile for which xi = 0 for all i ∈ N \ V . If S is not a winning coalition then v(S) = 0 so that certainly x(S) ≥ v(S); if S is a winning coalition then x(S) = 1 = v(S).

I∈N Since ( i∈N zi∗ + 1j , i∈N fi (zi∗ )) ∈ X, where 1j is the jth unit vector, we have αj ≤ 0 for all j. We now show that β > 0. Since i∈N ωi > 0 there exists ∈ R++ and δ > 0 such that ( i∈N zi∗ − , i∈N fi (zi∗ ) − δ) ∈ X, so that −α · − βδ ≤ 0 or βδ ≥ −α · . If α = 0 then we conclude that β ≥ 0, and since (α, β) = 0 it follows that β > 0. If αj < 0 for some j then we conclude directly that β > 0. Now let p = −α/β ≥ 0. Since ( i∈N zi∗ − zk∗ + zk , i∈N fi (zi∗ ) − fk (zk∗ ) + fk (zk )) ∈ X for any zk ∈ R+ we have fk (zk∗ ) − pzk∗ ≥ fk (zk ) − pzk for all zk ∈ R+ , so that (p, (zi∗ )i∈N ) is a competitive equilibrium.