intermediate 9 min read
Social sciences · Topic
Bargaining
optimization · convex optimization · probability theory · dynamical systems
Bargaining theory models how two parties divide a jointly created surplus. Nash's axiomatic approach characterises the unique split satisfying four fairness axioms, while Rubinstein's alternating-offers model provides strategic foundations through subgame perfect equilibrium, showing that patience and outside options determine bargaining power.

Nash Bargaining Solution

Let $U \subseteq \mathbb{R}^2$ be the feasible utility set and $(d_1, d_2)$ the disagreement point. The Nash bargaining solution is the unique outcome satisfying efficiency, symmetry, invariance to affine transformations, and independence of irrelevant alternatives:

\[(\phi_1^*, \phi_2^*) = \arg\max_{(u_1, u_2) \in U,\; u_i \geq d_i} (u_1 - d_1)(u_2 - d_2)\]

The maximiser is found where the Pareto frontier is tangent to a hyperbola. With $U$ defined by $u_1 + u_2 \leq S$ and $d = (0,0)$, the solution is the equal split $\phi^* = (S/2, S/2)$.

Rubinstein Alternating-Offers Model

In Rubinstein (1982), two players alternate making offers $(x, 1-x)$ over a pie of size 1. Both discount at rate $\delta \in (0,1)$ per period. The unique subgame perfect equilibrium involves immediate agreement in the first period. The proposer (player 1) receives:

\[x_1^* = \frac{1 - \delta}{1 - \delta^2} = \frac{1}{1 + \delta}\]

As $\delta \to 1$ (infinite patience), $x_1^* \to 1/2$ and the outcome converges to the Nash bargaining solution. First-mover advantage vanishes with patience.

Outside Options and Asymmetric Power

With outside options $(d_1, d_2)$, the Nash bargaining solution generalises to the asymmetric Nash bargaining solution:

\[\max_{u \in U} (u_1 - d_1)^\alpha (u_2 - d_2)^{1-\alpha}\]

where $\alpha \in (0,1)$ reflects player 1’s relative bargaining power. Player 1 receives $d_1 + \alpha(S - d_1 - d_2)$, so better outside options and higher $\alpha$ both raise one’s share. This links axiomatic and strategic approaches when $\alpha$ is identified with $1/(1+\delta)$ in the Rubinstein model.