Alternative equilibria in two-period ultimatum bargaining with envy

Abstract

A two-period ultimatum bargaining game is developed in which parties experience an envy-type externality coming from the surplus captured by their counterparts. Our assumptions on envy levels and outside opportunities allow us to characterize a richer set of bargaining outcomes than that identified by the prior literature, which includes a novel agreement equilibrium which we label Type I agreement. As this novel agreement solution is delivered by a negotiation resembling a one-shot ultimatum game, only characteristics of the second-moving player shape the sources of bargaining power. This property contrasts with that of Type II agreement—an agreement solution previously reported by related literature—in which characteristics of both players influence negotiating strengths. Numerical simulations are performed to illustrate the interplay between envy, impatience rates and outside opportunities as well as the degree of inequity generated by each agreement type.

Appendix

Proof of Proposition 1

By the backward induction principle, we proceed as follows: Period $t=2$ At this stage player, A accepts player B’s counteroffer if

and rejects it otherwise. In turn, player B chooses $x_{2}$ such that

$$\begin{aligned} \underset{x_{2}}{Max}\, \pi -(1+\theta _{B})x_{2} \\ s.t. \\ 0\le x_{2}\le \pi , \\ \pi -(1+\theta _{B})x_{2}\ge U_{B}, \\ x_{2}\ge \frac{U_{A}+\theta _{A}\pi }{1+\theta _{A}}. \end{aligned}$$

This problem is equivalent to the following program:

$$\begin{aligned} \underset{x_{2}}{Min}\,(1+\theta _{B})x_{2} \end{aligned}$$

(16)

$$\begin{aligned} s.t. \nonumber \\ \frac{U_{A}+\theta _{A}\pi }{1+\theta _{A}}\le x_{2}\le \frac{\pi -U_{B}}{ (1+\theta _{B})}. \end{aligned}$$

(17)

The interval of $x_{2}$ defined by constraint (17) turns out to be empty because when (1) is not satisfied, it follows that

$$\begin{aligned} \frac{U_{A}+\theta _{A}\pi }{1+\theta _{A}}>\frac{\pi -U_{B}}{(1+\theta _{B})}. \end{aligned}$$

Hence, a negotiation breakdown would take place at $t=2$. The optimal strategy for player B is therefore:

$$\begin{aligned} x_{2}^{*}=\frac{U_{A}+\theta _{A}\pi }{1+\theta _{A}}-\varepsilon _{B}, \end{aligned}$$

for some $\varepsilon _{B}>0$.

Period $t=1$ At this stage, player B accepts player A’s offer if

and rejects it otherwise. In turn, player A chooses $x_{1}^{*}$ such that

$$\begin{aligned} \underset{x_{1}}{M}ax u_{1}^{A}=\left\{ \begin{array}{ll} (1+\theta _{A})x_{1}-\theta _{A}\pi &{} \quad \text { if } \; x_{1}\le \overline{x_{1}} \\ \delta _{A}U_{A} &{} \quad \text { otherwise} \end{array} \right. . \end{aligned}$$

If we make the substitution $x_{1}=\overline{x_{1}}$, the previous problem implies that player A finally chooses $x_{1}^{*}=\overline{x_{1}}$ if

(18)

Note that condition (18) holds if it is true that

which is in fact condition (2) identified in the main text. Thus, the optimal strategy of player A is described by

$$\begin{aligned} x_{1}^{*}=\left\{ \begin{array}{ll} \overline{x_{1}} &{} \quad \mathrm{if\,(2)\,is\,verified} \\ \overline{x_{1}}+\varepsilon _{A} &{} \quad \text { otherwise} \end{array} \right. \end{aligned}$$

for some $\varepsilon _{A}>0$. $\square$

Proof of Corollary 1

(i) Since condition (1) is not satisfied and condition (2) is, it follows from Proposition 1 that the equilibrium path is such that at $t=1$, player A offers $x_{1}^{*}= \overline{x_{1}}$ and player B accepts it. (ii) Since conditions (1) and (2) are not satisfied, it follows from Proposition 1 that the equilibrium path is such that:

• At $t=1$, player A offers $x_{1}^{*}=\overline{x_{1}}+\varepsilon _{A}$ and player B rejects it.

• At $t=2$, player B counteroffers

  $$\begin{aligned} x_{2}^{*}=\frac{U_{A}+\theta _{A}\pi }{1+\theta _{A}}-\varepsilon _{B}, \end{aligned}$$

and player A rejects it. $\square$

Proof of Corollary 2

If $\Theta >1$ then $\pi (1-\theta _{A}\theta _{B})$ is negative. Thus, since $\delta _{A}>0$, $U_{A}\ge 0$ and $\theta _{B}\ge 0$ for all $i=A,B$, it is the case that

$$\begin{aligned} \pi (1-\theta _{A}\theta _{B})<\delta _{A}U_{A}(1+\theta _{B})+\delta _{B}U_{B}(1+\theta _{A}), \end{aligned}$$

(19)

which is a sufficient condition for both conditions (1) and (2) not to hold. Applying Corollary 1, Part (ii), the statement is finally demonstrated. $\square$

Proof of Proposition 2

By backward induction, we proceed as follows:Period $t=2$

We follow a similar line of reasoning to that used in proof of Proposition 1. Thus, whereas the optimal decision rule for player A remains the same, player B solves the program described by Eqs. (16) and (17). Note, however, that now the interval for $x_{2}$ defined by constraint (17) is nonempty, which follows from assuming that condition (1) is satisfied.

Thus, the optimal strategy for player B is given by

$$\begin{aligned} x_{2}^{*}=\frac{U_{A}+\theta _{A}\pi }{1+\theta _{A}}. \end{aligned}$$

Period $t=1$

At this stage, player B accepts player A’s offer if

and rejects it otherwise. Player A chooses $x_{1}^{*}$ such that

$$\begin{aligned} \underset{x_{1}}{Max} u_{1}^{A}=\left\{ \begin{array}{ll} (1+\theta _{A})x_{1}-\theta _{A}\pi &{} \quad \text {if } \; x_{1}\le \overline{ \overline{x_{1}}} \\ \delta _{A}\left[ (1+\theta _{A})x_{2}^{*}-\theta _{A}\pi \right] &{} \quad \text { otherwise} \end{array}. \right. \end{aligned}$$

If we make substitution $x_{1}=\overline{\overline{x_{1}}}$, the above program implies that player A finally chooses $x_{1}^{*}=\overline{ \overline{x_{1}}}$ as long as

$$\begin{aligned} (1+\theta _{A})\overline{\overline{x_{1}}}-\theta _{A}\pi \ge \delta _{A} \left[ (1+\theta _{A})x_{2}^{*}-\theta _{A}\pi \right] . \end{aligned}$$

which after substituting $x_{2}^{*}$ is equivalent to

$$\begin{aligned} \overline{\overline{x_{1}}}\ge \frac{\delta _{A}U_{A}+\theta _{A}\pi }{ 1+\theta _{A}}. \end{aligned}$$

(20)

Note, however, that condition (20) holds if

Since condition (1) suffices for the above inequality to hold, checking that it is always satisfied is straightforward.¹⁰ Hence, the optimal strategy of player A becomes

$$\begin{aligned} x_{1}^{*}=\overline{\overline{x_{1}}}\text {,} \end{aligned}$$

which completes the proof. $\square$

Proof of Corollary 3

Since condition (1) holds, the equilibrium path according to Proposition 2 is such that at $t=1$, player A offers $x_{1}^{*}=\overline{\overline{x_{1}}}$ and player B accepts it. $\square$