Solving for the Rubinstein Equilibrium

Introduction

Welcome back. In our previous lesson, we established the logical foundation for solving the Rubinstein bargaining model. We saw that while the game has an infinite horizon, its repeating, self-similar structure allows us to define the equilibrium payoffs recursively. This led us to a system of two linear equations representing the payoffs $v_{1}$ and $v_{2}$ that each player expects to get when it is their turn to propose:

{v_{1} = 1 - δ_{2} v_{2} v_{2} = 1 - δ_{1} v_{1}

Today, our objective is to solve this system of equations to find the unique subgame perfect equilibrium (SPE) of the Rubinstein model. By finding the explicit values of $v_{1}$ and $v_{2}$ , we will uncover the precise division of the "pie" that rational, patient players will agree upon. This moves us from the abstract structure of the game to its concrete, calculated outcome.

Solving the System of Equations

We have a standard system of two linear equations with two unknowns, $v_{1}$ and $v_{2}$ . Given your background in mathematics, solving this should be straightforward. We will use the method of substitution.

Substitute Equation (2) into Equation (1):
We take the expression for $v_{2}$ from the second equation and plug it into the first equation.
$v_{1} = 1 - δ_{2} (1 - δ_{1} v_{1})$
Solve for $v_{1}$ :
Now we have a single equation with only one unknown, $v_{1}$ . We can solve for it algebraically.
$v_{1} = 1 - δ_{2} + δ_{1} δ_{2} v_{1}$
Next, we gather the terms involving $v_{1}$ on one side:
$v_{1} - δ_{1} δ_{2} v_{1} = 1 - δ_{2}$
Factor out $v_{1}$ :
$v_{1} (1 - δ_{1} δ_{2}) = 1 - δ_{2}$
Finally, isolate $v_{1}$ to find its equilibrium value:
$v_{1}^{*} = \frac{1 - δ _{2}}{1 - δ _{1} δ _{2}}$
Solve for $v_{2}$ :
We can follow the exact same procedure by substituting equation (1) into equation (2). By symmetry, we can infer the result by swapping the player indices (1 and 2).
$v_{2}^{*} = \frac{1 - δ _{1}}{1 - δ _{1} δ _{2}}$

So, we have found the unique solutions for $v_{1}^{*}$ and $v_{2}^{*}$ . These represent the total share of the pie a player receives in any subgame where they are the first to make an offer.

The Equilibrium Outcome

Now, let's connect these values back to the actual game, which begins at period $t = 0$ with Player 1 making the first offer.

Player 1 wants to offer Player 2 the smallest possible share that she will accept.
Player 2 knows that if she rejects the offer, the game moves to period $t = 1$ , where she will be the proposer. The value of that subgame to her is $v_{2}^{*}$ .
Because of discounting, the value of that future opportunity at period t=0 is $δ_{2} v_{2}^{*}$ . This is her reservation value. She will reject any offer less than this.
Therefore, Player 1's optimal strategy is to offer Player 2 exactly $δ_{2} v_{2}^{*}$ .

The equilibrium agreement is reached immediately in period $t = 0$ . Let's calculate the shares:

Player 2's share is $δ_{2} v_{2}^{*} = δ_{2} (\frac{1 - δ _{1}}{1 - δ _{1} δ _{2}}) = \frac{δ _{2} ( 1 - δ _{1} )}{1 - δ _{1} δ _{2}}$ .
Player 1's share is the remainder: $1 - (Player 2’s share)$ .
$x_{1}^{*} = 1 - \frac{δ _{2} ( 1 - δ _{1} )}{1 - δ _{1} δ _{2}} = \frac{( 1 - δ _{1} δ _{2} ) - ( δ _{2} - δ _{1} δ _{2} )}{1 - δ _{1} δ _{2}} = \frac{1 - δ _{1} δ _{2} - δ _{2} + δ _{1} δ _{2}}{1 - δ _{1} δ _{2}} = \frac{1 - δ _{2}}{1 - δ _{1} δ _{2}}$

Notice that Player 1's share, $x_{1}^{*}$ , is precisely equal to $v_{1}^{*}$ . This confirms the consistency of our recursive logic. The value Player 1 gets from starting the game is exactly the share he secures for himself in the first round.

The unique SPE outcome of the Rubinstein bargaining model is that the game ends in period 0, with Player 1 proposing the division $(x_{1}^{*}, x_{2}^{*})$ and Player 2 immediately accepting, where:

Payoffs = (x_{1}^{*}, x_{2}^{*}) = (\frac{1 - δ _{2}}{1 - δ _{1} δ _{2}}, \frac{δ _{2} ( 1 - δ _{1} )}{1 - δ _{1} δ _{2}})

This immediate agreement is often called the no-delay property and is a key feature of the model's solution.

Proving Uniqueness

We have found a solution to the system of equations. But how do we know it's the only one? The power of the Rubinstein model lies in the fact that subgame perfection pins down a single, unique outcome. In the last lesson, you read a proof in Debraj Ray's notes that used the supremum ( $M_{i}$ ) and infimum ( $m_{i}$ ) of possible payoffs to show they must be equal.

Let's now watch a video that walks through a very similar proof. This approach reinforces the core argument for uniqueness by demonstrating that the highest possible payoff a player can get in any SPE is identical to the lowest possible payoff they can get.

(AGT3E8) [Game Theory] Solving Rubinstein's Infinite Horizon Alternating Offer Bargaining Game

This video by Selcuk Ozyurt provides a rigorous, step-by-step derivation that proves the uniqueness of the SPE payoff. It uses a method of bounding the best and worst possible equilibrium payoffs for a player and then showing that these bounds must converge to a single value.

Please watch from 04:34 to 23:14. As you watch, focus on how the logic unfolds: Setup (04:34 - 08:40): The presenter defines the lowest possible SPE payoff for Player 1 (v1 lower bar) and the highest (v1 upper bar) in subgames where Player 1 makes the offer. Deriving Bounds (08:40 - 20:33): Follow the reasoning that uses the opponent's rational acceptance/rejection decisions to create inequalities that constrain v1 upper bar and v1 lower bar. The Squeeze (20:33 - 23:14): Observe how the two inequalities are solved, leading to the conclusion that the upper and lower bounds are the same, forcing a unique solution. This is the crucial step that proves uniqueness.

The argument you just witnessed is a cornerstone of modern game theory. Unlike the Folk Theorem in repeated games, which allows for a vast number of equilibria, the structure of alternating-offers bargaining combined with the discipline of subgame perfection leads to a remarkably precise and unique prediction. The "threats" to reject offers are only credible if they are optimal in the future, and this constraint is so tight that it leaves no room for any other outcome.

Conclusion

In this lesson, we successfully solved the Rubinstein bargaining model and derived its unique subgame perfect equilibrium.

Key Takeaways:

The recursive system of equations $v_{1} = 1 - δ_{2} v_{2}$ and $v_{2} = 1 - δ_{1} v_{1}$ can be solved using simple algebra.
The unique SPE payoffs for a player initiating a subgame are $v_{1}^{*} = \frac{1 - δ _{2}}{1 - δ _{1} δ _{2}}$ for Player 1 and $v_{2}^{*} = \frac{1 - δ _{1}}{1 - δ _{1} δ _{2}}$ for Player 2.
The equilibrium involves an immediate agreement in the first period (the "no-delay" property). Player 1 offers Player 2 a share equal to the discounted value of her making a counteroffer ( $δ_{2} v_{2}^{*}$ ), and Player 2 accepts.
The argument that the maximum and minimum possible SPE payoffs must be equal proves that this solution is unique.

We now have the explicit formulas for the equilibrium division. In our next lesson, we will analyze how patience, modeled by the discount factors $δ_{1}$ and $δ_{2}$ , affects bargaining outcomes and power. By examining these formulas, we can make precise predictions about who gets more of the pie and why.

Recommended resources

Introduction

Solving the System of Equations

The Equilibrium Outcome

Proving Uniqueness

Conclusion