Game Theory

Download A Course in Game Theory. SOLUTIONS by Martin J. Osborne and Ariel Rubinstein PDF

By Martin J. Osborne and Ariel Rubinstein

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Extra info for A Course in Game Theory. SOLUTIONS

Example text

Step 3. m1 ≥ 1 − max{b, δM2 } and M2 ≤ 1 − δm1 . The proof is analogous to those for Steps 1 and 2. Step 4. If δ/(1 + δ) ≥ b then mi ≤ 1/(1 + δ) ≤ Mi for i = 1, 2. Proof. These inequalities follow from the fact that in the subgame perfect equilibrium described in the text player 1 obtains the payoff 1/(1 + δ) in any subgame in which she makes the first offer, and player 2 obtains the same utility in any subgame in which he makes the first offer. Step 5. If δ/(1 + δ) ≥ b then M1 = m1 = 1/(1 + δ) and M2 = m2 = 1/(1 + δ).

Again using Step 4 we have δM2 ≥ δ/(1 + δ) ≥ b, and hence by Step 3 we have m1 ≥ 1 − δM2 ≥ 1 − δ(1 − δm1 ). Thus m1 ≥ 1/(1 + δ). Hence m1 = 1/(1 + δ) by Step 4. Finally, by Step 3 we have M2 ≤ 1 − δm1 = 1/(1 + δ), so that M2 = 1/(1 + δ) by Step 4. Step 6. If b ≥ δ/(1 + δ) then m1 ≤ 1 − b ≤ M1 and m2 ≤ 1 − δ(1 − b) ≤ M2 . 24 Chapter 7. A Model of Bargaining Proof. These inequalities follow from the subgame perfect equilibrium described in the text (as in Step 4). Step 7. If b ≥ δ/(1 + δ) then M1 = m1 = 1 − b and M2 = m2 = 1 − δ(1 − b).

Thus it is optimal for the tough chain-store to choose F . Bayesian updating of beliefs: • If k ≤ K −1 and µCS (h)(T ) ≥ bK−k then both types of chain-store fight challenger k if it enters. Thus the probability µCS (h, hk )(T ) assigned by challenger k+1 is µCS (h)(T ) when hk = (In, F ). • If k ≤ K − 1 and µCS (h)(T ) < bK−k then the tough chain-store fights challenger k if it enters and the regular chain-store accommodates with positive probability pk = (1 − bK−k )µCS (h)(T )/((1 − µCS (h)(T ))bK−k ).

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