Backward Induction in the Rubinstein Model

Introduction

In our last lesson, you learned to describe the structure and rules of the Rubinstein alternating-offers bargaining model. We established the key components: two players, a "pie" to divide, an infinite sequence of alternating offers, and discount factors ( $δ_{1}, δ_{2}$ ) that make waiting costly.

Today, we move from description to analysis. Our goal is to solve this game. The learning outcome for this lesson is to apply backward induction to identify the recursive relationship between offers in the Rubinstein model.

You've used backward induction before in Module 4 to solve finite sequential games. The challenge here is the infinite horizon—there is no "last move" to start from. The key insight, which we will uncover, is that the game has a self-referential or recursive structure. The subgame that starts two periods from now looks just like the game we start with today. This self-similarity is what allows us to find a solution.

Building Intuition with a Finite Horizon

To make the logic of backward induction in a bargaining context concrete, let's first analyze a simpler, finite version of the game. Imagine the negotiation must end after just two periods.

Period 0: Player 1 makes an offer. Player 2 accepts or rejects.
Period 1: If Player 2 rejected, she makes a final, take-it-or-leave-it counter-offer. If Player 1 rejects this, both get zero.

This finite structure provides a clear endpoint from which to work backward.

(AGT3E7) [Game Theory] Solving Rubinstein's Alternating Offer Bargaining Game: Two-Period Version

The following video provides an excellent step-by-step solution for this two-period bargaining game. It demonstrates precisely how a player's decision in the first period is dictated by the anticipated outcome of the second.

Watch the segment from 06:28 to 20:00. As you watch, focus on these two steps of the backward induction process: Analysis of Period 1 (starting at 08:32): What is the optimal offer for Player 2 to make in the final period, and why will Player 1 accept it? Analysis of Period 0 (starting at 12:13): Knowing the outcome of Period 1, what is the minimum share Player 1 must offer Player 2 to prevent her from rejecting and moving to the next period? How does the discount factor \delta_2 play a crucial role here?

Let's recap the logic you just saw.

In the final period (t=1): Player 2 makes a take-it-or-leave-it offer. Since Player 1 gets 0 if he rejects, he will accept any non-negative offer. Player 2's optimal move is to offer Player 1 a share of $y = 0$ and keep the entire pie (1) for herself. The outcome is (0, 1).
Rolling back to the first period (t=0): Player 1 is deciding what offer $(x, 1 - x)$ to make. He knows that if Player 2 rejects, the game proceeds to t=1, where she will secure a payoff of 1. However, that payoff is one period in the future, so its present value to Player 2 is $δ_{2}^{1} \cdot 1 = δ_{2}$ .
- This value, $δ_{2}$ , becomes Player 2's reservation value. She will reject any offer that gives her less than $δ_{2}$ .
- To ensure his offer is accepted, Player 1 must offer Player 2 a share of at least $δ_{2}$ .
- To maximize his own share, Player 1 offers Player 2 exactly $δ_{2}$ .

The unique subgame perfect equilibrium (SPE) outcome is an immediate agreement in Period 0 on the division $(1 - δ_{2}, δ_{2})$ . Notice the core relationship: the offer in Period 0 is determined by the discounted value of the outcome in Period 1.

The Recursive Structure of the Infinite Game

Now, how do we apply this logic to the infinite-horizon game? There's no final period. The breakthrough insight from Rubinstein is that the structure of the game repeats itself.

The subgame beginning in period 2, where it is Player 1's turn to make an offer, is identical to the entire game starting in period 0.
The subgame beginning in period 1, where it is Player 2's turn to make an offer, is identical to the entire game if Player 2 had been the first mover.

This allows us to define a player's expected equilibrium payoff not as a fixed number, but in terms of the other player's expected payoff. This creates a recursive relationship.

Let's define two key variables:

Let $v_{1}$ be the equilibrium payoff Player 1 receives in any subgame that starts with Player 1 making the offer.
Let $v_{2}$ be the equilibrium payoff Player 2 receives in any subgame that starts with Player 2 making the offer.

Now, let's use backward induction-style reasoning.

Consider a subgame where Player 1 offers (e.g., at t=0):
Player 1 proposes a division $(x, 1 - x)$ . Player 2 considers her options:

Accept: She receives a payoff of $1 - x$ immediately.
Reject: The game moves to the next period (t=1), where it will be her turn to make an offer. By our definition, the value she can expect to get from that subgame is $v_{2}$ . However, this occurs one period in the future, so its present value is $δ_{2} v_{2}$ .

A rational Player 2 will accept the offer if and only if $1 - x \geq δ_{2} v_{2}$ .

Player 1, wanting to maximize his own share $x$ , will make Player 2's share $1 - x$ as small as possible while still ensuring acceptance. Thus, he will set her share to be exactly equal to her reservation value:

1 - x = δ_{2} v_{2}

This means Player 1's share will be $x = 1 - δ_{2} v_{2}$ . Since $v_{1}$ is defined as Player 1's payoff in this situation, we have our first recursive equation:

v_{1} = 1 - δ_{2} v_{2} (1)

Consider a subgame where Player 2 offers (e.g., at t=1):
By symmetric logic, Player 2 proposes a division $(y, 1 - y)$ . Player 1 considers his options:

Accept: He receives a payoff of $y$ immediately.
Reject: The game moves to the next period (t=2), where it's his turn again. The value he can expect from that subgame is $v_{1}$ , and its present value is $δ_{1} v_{1}$ .

Player 1 will accept if and only if $y \geq δ_{1} v_{1}$ .

Player 2, wanting to maximize her share $1 - y$ , will offer Player 1 exactly his reservation value:

y = δ_{1} v_{1}

This means Player 2's share will be $1 - y = 1 - δ_{1} v_{1}$ . Since $v_{2}$ is Player 2's payoff in this situation, we have our second recursive equation:

v_{2} = 1 - δ_{1} v_{1} (2)

We have now successfully identified the recursive relationship at the heart of the Rubinstein model. It's captured by this system of two linear equations:

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

Each player's equilibrium payoff is precisely one minus the discounted equilibrium payoff of their opponent.

A More Formal Derivation

Given your preference for mathematical rigor, let's walk through a more formal argument that not only identifies this relationship but also proves that the equilibrium is unique. This approach uses bounds on the potential equilibrium payoffs.

[PDF] MA300.2 Game Theory II

This document by Debraj Ray provides a concise and elegant proof of the unique SPE. We will focus on the argument that establishes the recursive relationship between payoffs.

Please read Section 3.3, 'Perfect Equilibrium in the Rubinstein Model' (pages 5-7). The argument is dense but powerful. Focus on understanding the logic behind inequalities (3) and (4), as they form the core of the recursive argument.

Let's break down that proof's logic, as it's a beautiful piece of game-theoretic reasoning. Let $M_{i}$ be the maximum possible SPE payoff for player $i$ , and $m_{i}$ be the minimum possible SPE payoff for player $i$ , across all possible subgames.

Finding a lower bound for $m_{i}$ :
When it's player $i$ 's turn to offer, she can always guarantee herself a certain amount. Her opponent, $j$ , will get at most $M_{j}$ in any SPE. If player $j$ rejects the current offer, the best he can hope for is a payoff of $M_{j}$ in the next period, which has a present value of $β_{j} M_{j}$ . Therefore, player $j$ must accept any offer that gives him slightly more than $β_{j} M_{j}$ . This means player $i$ can secure for herself (and thus her minimum payoff $m_{i}$ must be at least) what's left over:
$m_{i} \geq 1 - β_{j} M_{j}$
This is inequality (3) from the text.
Finding an upper bound for $M_{j}$ :
When it's player $i$ 's turn to offer, what's the most player $j$ could possibly get? Player $j$ can either accept the offer or reject. If he rejects, he becomes the proposer. But he must make an offer that player $i$ will accept. Player $i$ will not accept anything less than her discounted minimum payoff, $β_{i} m_{i}$ . So, the most player $j$ can get after rejecting is $1 - β_{i} m_{i}$ . Therefore, player $j$ 's maximum payoff $M_{j}$ can't be more than this value.
$M_{j} \leq 1 - β_{i} m_{i}$
This is inequality (4) from the text.

These two inequalities establish the recursive relationship between the bounds on players' payoffs. The proof then brilliantly shows that these inequalities can only hold if the upper and lower bounds are the same ( $m_{i} = M_{i}$ ), which forces a unique solution. Substituting $v_{i}$ for $m_{i} = M_{i}$ and $δ$ for $β$ , these inequalities become the very equations we derived earlier:

$v_{i} = 1 - δ_{j} v_{j}$
$v_{j} = 1 - δ_{i} v_{i}$

Conclusion

In this lesson, we made the crucial leap from describing the bargaining game to setting up its solution. We used the principle of backward induction, adapting it from a finite context to the infinite-horizon Rubinstein model.

Key Takeaways:

In a finite bargaining game, backward induction works by starting at the final offer and rolling back. The offer in each period is determined by the discounted value of the rejection outcome.
In the infinite-horizon game, there is no final period. However, the game's self-similar structure allows for a solution. The subgame starting at period $t + 2$ is identical to the one starting at period $t$ .
This self-similarity allows us to define a player's expected payoff in their proposing turn ( $v_{i}$ ) in terms of their opponent's expected payoff in their proposing turn ( $v_{j}$ ).
This leads to a system of two recursive equations: $v_{1} = 1 - δ_{2} v_{2}$ and $v_{2} = 1 - δ_{1} v_{1}$ . This system is the mathematical expression of the recursive relationship between offers.

We have successfully identified the core logic that underpins the model's solution. The next step is straightforward. In our next lesson, we will solve this system of equations to find the unique subgame perfect equilibrium of the Rubinstein model, revealing the precise shares each player will receive.

Recommended resources

Introduction

Building Intuition with a Finite Horizon

The Recursive Structure of the Infinite Game

A More Formal Derivation

Conclusion