In: Economics
Question on Rubinstein Bargaining in Game Theory
Suppose players alternate in making offers until offer T ∈ N (N are Natural Numbers) is rejected, at which point the game ends with payoffs (0,0). Find the unique Subgame Perfect Equilibrium payoffs. Do they converge as T → ∞?
T is replaced with K here. In the final assignment, please replace time period K with T.
Consider the following finite horizon bargaining game.
Two players i = 1, 2 are trying to allocate $1 between them.
The game lasts for K < ∞ periods. Periods are counted backward, so the game starts in period K and ends in period 1. In an odd period k, player 1 makes an offer x1 ∈ [0, 1], player 2 rejects the offer. The game continues to the period k −1. In any even period k, player 2 makes an offer x2 ∈ [0, 1], player 1 rejects the offer. Otherwise the game continues to period k − 2. If the offer is rejected in period 1, then both players receive 0 payoff. Player i’s discount factor is δi ∈ (0, 1).
Suppose that K = 2.
Applying backward induction. In period 1, every strictly positive offer is accepted by player 2. Hence the equilibrium offer must be 0, which must be accepted by player 2. In period 2, player 1 accepts any offer strictly larger than δ1 and rejects any offer strictly smaller than δ1. Hence the equilibrium offer must be exactly δ1, which must be accepted by player 1. Hence the SPE is unique. Player 1: offer 0 in period 1, accept x2 if and only if x2 ≥ δ1 in period 2. Player 2: offer δ1 in period 2, accept any offer in period 1. There are many other NE.
Suppose that K = 3.
Again applying BI.
In period 1, every strictly positive offer is accepted by player 2. Hence the equilibrium offer must be 0, which must be accepted by player 2.
In period 2, player 1 accepts an offer if and only if the offer is larger than or equal to δ1. Player 2 always offers δ1.
In period 3, player 2 accepts an offer if and only if the offer is larger than or equal to δ2(1 − δ1). Player 1 offers δ2(1 − δ1).
This is the Subgame Perfect Equilibrium payoffs which converges as K→ ∞