By Martin J. Osborne and Ariel Rubinstein
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Extra info for A Course in Game Theory. SOLUTIONS
Let h \ h(h ) be the part of h subsequent to Ii . Then, Pr(β , µ|Ii )(Ii ) = µ(Ii )(h(h )) · Pβ (h \ h(h )). h ∈Ii Since (β, µ) is consistent there is a sequence of completely mixed assessments (β n , µn ) with µn → µ and β n → β as n → ∞ and for all n the belief µn is derived from β n using Bayes’ rule. For each n we have ˆ · Pβ (a ) µn (Ii )(h) n h ∈Ii µ (Ii )(h(h )) · Pβ (h \ h(h )) ˆ a)= µn (Ii )(h, since Pr(β , µ|Ii )(Ii ) > 0. Taking the limit as n → ∞ and using Pβ (a , a ) = Pβ (a )·Pβ (a ) we conclude that O(β , µ|Ii )(h) = O(β , µ|Ii )(h) · Pr(β , µ|Ii )(Ii ).
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. Comment This is not an exercise in game theory. 3). Now, the concavity of f (1, k) implies that k(f (1, k) − f (1, k − 1)) ≤ 2(f (1, k) − f (1, k/2)) (since k (f (1, j) − f (1, j − 1)) f (1, k) − f (1, k/2) = j=k/2+1 k (f (1, k) − f (1, k − 1)) ≥ j=k/2+1 ≥ (k/2)[f (1, k) − k(1, k − 1)]). Since f is bounded we have f (1, k) − f (1, k/2) → 0, establishing the result.
Suppose that a game form G with outcome function g Nash-implements f . Then (G, ) has a Nash equilibrium, say (s1 , s2 ), for which g(s1 , s2 ) = a. Since (s1 , s2 ) is a Nash equilibrium, g(s1 , s2 ) 2 a for all actions s2 of player 2, so that g(s1 , s2 ) = a for all actions s2 of player 2. That is, by choosing s1 , player 1 guarantees that the outcome is a. Since a 1 b, it follows that (G, ) has no Nash equilibrium (t1 , t2 ) for which g(t1 , t2 ) = b. We conclude that f is not Nash-implementable.
A Course in Game Theory. SOLUTIONS by Martin J. Osborne and Ariel Rubinstein