Nash's solution and Rubinstein's model

From my doctoral course

This note comes from the doctoral course “Game Theory and Bargaining”. I rewrote the explanations in my own words and recomputed the numerical examples.

Two views of bargaining

There are two traditions: axiomatisation (Nash, 1950): “what properties does a fair agreement have?” and the strategic model (Rubinstein, 1982): “what happens if players make offers in turn?” The important point is that the results of both are close.

Nash’s bargaining solution

If $d=(d_1,d_2)$ is the disagreement point (the utility when negotiation breaks down), Nash’s solution is the outcome that maximises the product of the gains above disagreement:

${\max }_{(u_1,u_2)}(u_1-d_1)(u_2-d_2)$

This is the only solution satisfying four reasonable axioms:

1. Individual rationality: nobody accepts less than disagreement.
2. Efficiency (Pareto): no surplus is wasted.
3. Symmetry: if the two players are exactly alike, equal shares.
4. Independence of irrelevant alternatives: removing unacceptable agreements does not change the answer. (Together with invariance to linear transformation of utility.)

Dividing a cake of size 1 with linear utility, the answer is: each gets their disagreement point and the rest of the surplus is split equally. That is, raising the outside option raises the share.

Rubinstein’s model: alternating offers

A cake of size 1 and two players. In period 1 player 1 makes an offer; if player 2 accepts, it is over. If they reject, a period passes and player 2 makes an offer, and so on until agreement. The passing of time has a cost: each player $i$ discounts the next period’s cake by a factor ${\delta }_i\in (0,1)$ (impatience).

Result (unique subgame-perfect equilibrium): player 1 makes an offer in the very first period and it is accepted; no delay occurs. Their share:

$x_1^{\ast }=\frac{1-{\delta }_2}{1-{\delta }_1{\delta }_2},x_2^{\ast }=1-x_1^{\ast }$

If ${\delta }_1={\delta }_2=\delta$, then $x_1^{\ast }=1/(1+\delta )$; the first proposer gets more than half, because next period’s cake is smaller. When periods become very short ($\delta \to 1$), the shares approach 50-50 and reach Nash’s solution.

The proof is by backward induction: suppose time is unlimited and see what minimum each player accepts; so we lay out expectations from the end to the start.

Numerical example

${\delta }_1$	${\delta }_2$	Player 1’s share	Player 2’s share
0.9	0.9	0.526	0.474
0.8	0.9	0.357	0.643
0.9	0.8	0.714	0.286

Second row: player 1 is more impatient (${\delta }_1=0.8$) and although they propose first, they get only 36%. Patience beats turn order. This turns exactly the “patience” factor in Bargaining: power factors and the cost of waiting into a number.

Critique

Rubinstein’s model assumes a world free of lies: complete information, alternating offers. In practice negotiations have delays (strikes) and that does not fit the model; the explanation is asymmetric information. Still, I think its key result should stay with me: in bargaining, the one in less of a hurry wins.

Related: Bayesian games and signaling.

rubinsteinnash_solutionbargaining