Solution
Bargaining with Unequal Patience
Show the problem again
Players 1 and 2 bargain over one dollar. In round 1 player 1 proposes a split $(x,1-x)$ and player 2 accepts or rejects. If player 2 rejects, then in round 2 player 2 proposes a split and player 1 accepts or rejects, and so on without any deadline, alternating forever. An agreement reached in round $t$ gives player $i$ a payoff of $\delta_i^{\,t-1}$ times the money received, with $\delta_1 = 4/5$ and $\delta_2 = 3/5$. If no agreement is ever reached, both get 0. A player who is indifferent between accepting and rejecting accepts. In the unique subgame-perfect equilibrium, agreement is reached immediately. What share of the dollar does player 1 receive? Give a reduced fraction.
Worked solution
The answer is 10/13. By stationarity, let $a$ be the equilibrium share of whoever proposes for player 1, and $b$ the share of whoever proposes for player 2 (the same in every round where that player proposes). When player 1 proposes, player 2 can reject and become the proposer one round later, worth $\delta_2 b$ to him, so player 1 offers exactly that: $a = 1-\delta_2 b$. Symmetrically $b = 1-\delta_1 a$. Solving, $a = 1-\delta_2(1-\delta_1 a)$, so $a = \dfrac{1-\delta_2}{1-\delta_1\delta_2} = \dfrac{2/5}{1-12/25} = \dfrac{2/5}{13/25} = \dfrac{10}{13}$. Note that the opponent's patience, not the proposer's own, determines the proposer's share: the impatient player 2 is exploited. A backward-induction computation on a truncated game with 60 rounds already agrees to nine decimals.
Source: Rubinstein (1982), 'Perfect equilibrium in a bargaining model', Econometrica, with different discount factors for the two players. Statement written for AxiomIQ.