The Power of Patience: Discount Factors in Bargaining

Introduction

In our last lesson, we successfully solved the Rubinstein bargaining model and derived the explicit formulas for the unique subgame perfect equilibrium division. We found that when Player 1 makes the initial offer, the game ends immediately with the 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}})

Where $δ_{1}$ and $δ_{2}$ are the discount factors representing the patience of Player 1 and Player 2, respectively.

Today, we will move from derivation to interpretation. Our goal is to analyze how patience, modeled by the discount factor, affects bargaining outcomes and power. By dissecting these formulas, we can uncover one of the most fundamental insights of strategic bargaining: patience is power.

The Intuition: Why Patience is Power

Before we dive into the mathematics, let's build our intuition. A discount factor $δ$ close to 1 signifies a patient player, who doesn't mind waiting for a future payoff. A discount factor close to 0 signifies an impatient player, for whom future payoffs are heavily devalued.

In bargaining, the power to reject an offer credibly is crucial.

An impatient player is desperate to reach a deal quickly. The prospect of delay is costly, making them more likely to accept a mediocre offer today rather than wait for a potentially better one tomorrow. Their threat to reject is weak.
A patient player, on the other hand, is less affected by delay. They can afford to reject an unfavorable offer, confident that the value of their counteroffer in the next period will not have diminished much. Their threat to reject is strong and credible.

This implies that a more patient player should be able to secure a larger share of the pie.

Rubinstein Bargaining Model: Payoffs and Discount Factors — This diagram illustrates how payoffs are discounted over time. An agreement reached in period t is worth only $\delta^{t-1}$ of its nominal value, highlighting the cost of delay that impatient players seek to avoid.

To formalize this intuition, let's start with a clear, concise summary from a set of lecture notes.

Bargaining and Repeated Games

The following reading from Jonathan Levin's notes on sequential bargaining elegantly summarizes the relationship between patience and payoffs.

Please read 'Comment 1' on page 3 of the PDF. This section begins with 'A few comments on the Rubinstein model...'. Focus on how player one's payoff is explicitly described as 'increasing in δ₁ and decreasing in δ₂'.

The Mathematical Proof

The reading confirmed our intuition. Now, as you appreciate mathematical rigor, let's prove this relationship formally. We will use calculus to analyze how Player 1's share, $x_{1}^{*}$ , changes with respect to each player's discount factor.

1. How does Player 1's share change with their own patience?

We want to determine the sign of the partial derivative $\frac{\partial x _{1}^{*}}{\partial δ _{1}}$ .

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

Using the chain rule for differentiation:

\frac{\partial x _{1}^{*}}{\partial δ _{1}} = (1 - δ_{2}) \cdot [- 1 (1 - δ_{1} δ_{2})^{- 2}] \cdot (- δ_{2})

\frac{\partial x _{1}^{*}}{\partial δ _{1}} = \frac{δ _{2} ( 1 - δ _{2} )}{( 1 - δ _{1} δ _{2} ) ^{2}}

Since both discount factors $δ_{1}, δ_{2}$ are in the interval $(0, 1)$ , the term $(1 - δ_{2})$ is positive, and so is $δ_{2}$ . The denominator is a squared term, so it is also positive. Therefore:

\frac{\partial x _{1}^{*}}{\partial δ _{1}} > 0

This confirms that Player 1's share of the pie is strictly increasing with their own patience. The more patient Player 1 is, the larger their share.

2. How does Player 1's share change with their opponent's patience?

Next, we evaluate the partial derivative $\frac{\partial x _{1}^{*}}{\partial δ _{2}}$ using the quotient rule. Let $u = 1 - δ_{2}$ and $v = 1 - δ_{1} δ_{2}$ . Then $u^{'} = - 1$ and $v^{'} = - δ_{1}$ .

\frac{\partial x _{1}^{*}}{\partial δ _{2}} = \frac{u ^{'} v - u v ^{'}}{v ^{2}} = \frac{( - 1 ) ( 1 - δ _{1} δ _{2} ) - ( 1 - δ _{2} ) ( - δ _{1} )}{( 1 - δ _{1} δ _{2} ) ^{2}}

Simplifying the numerator:

(- 1 + δ_{1} δ_{2}) - (- δ_{1} + δ_{1} δ_{2}) = - 1 + δ_{1} δ_{2} + δ_{1} - δ_{1} δ_{2} = δ_{1} - 1

So, the derivative is:

\frac{\partial x _{1}^{*}}{\partial δ _{2}} = \frac{δ _{1} - 1}{( 1 - δ _{1} δ _{2} ) ^{2}}

Since $δ_{1} \in (0, 1)$ , the numerator $(δ_{1} - 1)$ is negative. The denominator is positive. Therefore:

\frac{\partial x _{1}^{*}}{\partial δ _{2}} < 0

This confirms that Player 1's share is strictly decreasing with Player 2's patience. The more patient your opponent is, the smaller your share.

These two results provide the mathematical foundation for the strategic importance of patience in negotiations.

Special Cases and First-Mover Advantage

Let's explore what the model predicts in two important scenarios.

Scenario 1: Equally Patient Players

Suppose both players have the same discount factor, $δ_{1} = δ_{2} = δ$ . The payoff for Player 1 (the first-mover) becomes:

x_{1}^{*} = \frac{1 - δ}{1 - δ ^{2}} = \frac{1 - δ}{( 1 - δ ) ( 1 + δ )} = \frac{1}{1 + δ}

The payoff for Player 2 is the remainder:

x_{2}^{*} = 1 - x_{1}^{*} = 1 - \frac{1}{1 + δ} = \frac{1 + δ - 1}{1 + δ} = \frac{δ}{1 + δ}

Since $δ < 1$ , we have $1 > δ$ , which means $\frac{1}{1 + δ} > \frac{δ}{1 + δ}$ . Player 1 gets a larger share. This is the first-mover advantage. All else being equal, the player who gets to make the first proposal has an edge.

Scenario 2: Infinitely Patient Players

What happens in the limit as players become perfectly patient, i.e., $δ \to 1$ ?

δ \to 1 lim x_{1}^{*} = δ \to 1 lim \frac{1}{1 + δ} = \frac{1}{2}

δ \to 1 lim x_{2}^{*} = δ \to 1 lim \frac{δ}{1 + δ} = \frac{1}{2}

As the cost of delay approaches zero, the first-mover advantage vanishes and the players agree to an even split.

The following video provides an alternative derivation for the symmetric case and closes by framing the exact question we've just answered about the role of patience.

Bargaining 101 (#12): Rubinstein Bargaining

This video by William Spaniel uses a 'continuation value' approach to solve the symmetric case. It's a great way to see the same result derived from a slightly different logical path.

Please watch from 05:09 to 13:08 to see the derivation of the symmetric (\delta_1 = \delta_2) equilibrium. Then, watch from 14:42 to 15:28, where the presenter discusses the importance of analyzing the case with different discount factors, reinforcing the main theme of our lesson.

Conclusion

In this lesson, we analyzed the equilibrium of the Rubinstein bargaining model to understand the strategic role of patience. By combining intuition with a formal mathematical proof, we've solidified one of game theory's most important insights about negotiation.

Key Takeaways:

Patience is Bargaining Power: A player's equilibrium share increases as they become more patient (their $δ$ increases) and decreases as their opponent becomes more patient.
Mathematical Verification: We proved this relationship by showing that $\frac{\partial x _{1}^{*}}{\partial δ _{1}} > 0$ and $\frac{\partial x _{1}^{*}}{\partial δ _{2}} < 0$ .
First-Mover Advantage: When players are equally patient, the player who makes the first offer secures a larger share.
Vanishing Advantage: This first-mover advantage disappears as players become infinitely patient ( $δ \to 1$ ), leading to an equal split.

So far, our model has assumed that the only alternative to a negotiated agreement is to continue bargaining indefinitely. But what if one player could simply walk away to take another deal? In our next lesson, we will analyze how a player's outside options influence their bargaining power and the final agreement, adding another layer of realism to our model.

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