Shaping Emotions in Negotiation: a Nash Bargaining Solution

Abstract

Modeling emotions in negotiations is an open challenge that attracted an increasing amount of attention from researchers. Bargainers look for achieving an agreement with the opposing parties and, at the same time, try to reach their own goals. This process consists of both bargaining and (game theory) problem solving. Game theory models seek to enlighten the rational negotiations between players, but these models lack the evidence of how emotional motives may influence individuals’ behavior. This paper suggests a model for shaping emotions in negotiation using Nash’s bargaining approach. We focus on the case where negotiation between players has motives of cooperating, considering eight emotions: anger, fear, joy, sadness, surprise, disgust, guilt, and disappointment. For representing the solution of the problem, we employ a homogeneous Markov game. The simplicity of the model relies on the fact that the emotions are represented by the states of the Markov chain. The relationship between the emotions is represented by a transition matrix that determines the probability of changing between the emotions (states) at any time. Because any emotion can be reached at any time with certain probability, the bargaining Markov game is ergodic. We represent naturally the emotional process of bargaining using a proximal method, which involves the bargaining Nash product for computing the equilibrium of the game. We show the convergence of the method to the emotional equilibrium point. The solution of the Nash bargaining game consists of cooperative emotional strategies, which are transformed in emotional probability distributions. Such emotional probability distributions are measured using an asymmetric distance function that determines the “emotional distance” between players in negotiations. Emotions are measured using an asymmetric distance function because they are different between players. We present a new approach for shaping emotions in negotiations employing Nash’s bargaining model. An application example shows the influence of expressing emotions in the relationship process, and those emotions are strategically selected to gain a benefit in negotiations. We show that the magnitude and direction of emotional distance matter and that feelings have an asymmetric effect on the negotiation process.

Appendix: Proof of Theorem 1

Proof

Considering that $\left (\nabla h\left (x\right ) ,\left (y-x\right ) \right ) \leq h\left (y\right ) -h\left (x\right )$ and $\left (\nabla h\left (x\right ) ,\left (x-y\right ) \right ) \geq h\left (x\right ) -h\left (y\right )$ valid for any convex function $h\left (x\right )$ and any x and y, then for any admissible points x, μ,λ, and $x_{n}^{\ast }=x^{\ast }\left (\theta _{n},\delta _{n}\right )$, $\mu _{n}^{\ast }=\mu ^{\ast }\left (\theta _{n},\delta _{n}\right ) ,$ and $\lambda _{n}^{\ast }=\lambda ^{\ast }\left (\theta _{n},\delta _{n}\right ),$ we have

$$ \begin{array}{@{}rcl@{}} &&\left( x-x_{n}^{\ast },\frac{\partial }{\partial x} \mathbb{L} _{\theta ,\delta }\left( x,\mu,\lambda|x_{n}\right) \right) \\ &&-\left( \mu-\mu_{n}^{\ast },\frac{\partial }{\partial \mu} \mathbb{L}_{\theta ,\delta }\left( x,\mu,\lambda|x_{n}\right) \right) \\ &&-\left( \lambda-\lambda_{n}^{\ast },\frac{\partial }{\partial \lambda} \mathbb{L}_{\theta ,\delta }\left( x,\mu,\lambda|x_{n}\right) \right)\\ &=&\theta_{n} \left( x-x_{n}^{\ast }\right)^{\intercal }\frac{\partial }{\partial x}(\phi(x) -\phi(x_{n}))\\ &&+\left( x-x_{n}^{\ast }\right)^{\intercal }\left[ A^{\intercal }\mu +B^{\intercal }\lambda+\delta_{n} \left( x-x_{n}\right) \right] \\ &&+\left( \mu-\mu_{n}^{\ast }\right)^{\intercal }\left( \delta_{n}\mu-Ax+a\right) \\ &&+\left( \lambda-\lambda_{n}^{\ast }\right)^{\intercal }\left( \alpha_{n}\lambda-Bx+b\right) \\ &=&\theta_{n} (\phi(x)-\phi(x_{n}))+ \mu^{\intercal }\left( Ax-a\right) \\ &&+\lambda^{\intercal }\left( Bx-b\right) + \frac{\delta_{n}}{2} \left\Vert x\text{-}x_{n}\right\Vert^{2}\text{-} \frac{\alpha_{n}}{2} \left( \left\Vert \mu\right\Vert^{2}\text{+}\left\Vert \lambda \right\Vert^{2}\right) \\ &&-\theta_{n} (\phi(x_{n}^{\ast })-\phi(x_{n}))-\left( \mu_{n}^{\ast }\right)^{\intercal }\left( Ax_{n}^{\ast }-a\right) \\ &&-\left( \lambda_{n}^{\ast }\right)^{\intercal }\left( Bx_{n}^{\ast }-b\right) - \frac{\delta_{n}}{2} \left\Vert x_{n}^{\ast }-x_{n}\right\Vert^{2}\\ &&+\frac{\alpha_{n}}{2}\left( \left\Vert \mu_{n}^{\ast }\right\Vert^{2}+ \left\Vert \lambda_{n}^{\ast }\right\Vert^{2}\right) \\ &=&\mathbb{L}_{\theta_{n},\delta_{n}}\left( x,\mu_{n},\lambda_{in}|x_{n}\right) - \mathbb{L}_{\theta ,\delta }\left( x_{n}^{\ast },\mu_{n},\lambda_{n}|x_{n}\right) \\&&+ \frac{\delta_{n}}{2} \left\Vert x-x_{n}^{\ast }\right\Vert^{2}\text{+} \frac{\alpha_{n}}{2}\left( \left\Vert \mu-\mu_{n}^{\ast }\right\Vert^{2}+\left\Vert \lambda-\lambda_{n}^{\ast }\right\Vert^{2}\right) \end{array} $$

which by the saddle-point condition implies

$$ \begin{array}{@{}rcl@{}} &&\theta_{n} \left( x-x_{n}^{\ast }\right)^{\intercal }\frac{ \partial }{\partial x}(\phi(x)\text{-}\phi(x_{n}^{\ast }))+\left( x- x_{n}^{\ast }\right)^{\intercal } \cdot \\ &&\left[ A^{\intercal }\mu+ B^{\intercal }\lambda\text{+} \delta_{n} \left( x-x_{n}\right) \right] + \left( \mu-\mu_{n}^{\ast }\right)^{\intercal } \cdot \\ &&\left( \alpha_{n}-Ax+a\right) +\left( \lambda -\lambda_{n}^{\ast }\right)^{\intercal } \left( \alpha_{n}- Bx+b\right) \\ &\geq&\frac{\delta_{n}}{2} \left\Vert x-x^{\ast }\right\Vert^{2}+ \frac{\alpha_{n}}{2}\left( \left\Vert \mu-\mu_{n}^{\ast },\right\Vert^{2}+\left\Vert \lambda-\lambda_{n}^{\ast }\right\Vert^{2}\right) \end{array} $$

(21)

Choosing in Eq. 21x := x^∗∈ X^∗ (x^∗ is one of admissible solutions such that Ax^∗ = a and Bx^∗≤ b) and μ = μ^∗, λ = λ^∗ and considering the complementary slackness conditions we have that $\left (\lambda ^{\ast }\right )_{i}\left (Bx^{\ast }-b\right )_{i}=\left (\lambda _{n}^{\ast }\right )_{i}\left (Bx_{n}^{\ast }-b\right )_{i}=0$ obtaining

$$ \begin{array}{@{}rcl@{}} &&\theta_{n} \left( x^{\ast }-x_{n}^{\ast }\right)^{\intercal }\frac{ \partial }{\partial x}\left( \phi \left( x^{\ast }\right) -\phi(x_{n}^{\ast })\right) \\ &&+\left( x^{\ast }-x_{n}^{\ast }\right)^{\intercal }\left[ A^{\intercal }\mu^{\ast }+B^{\intercal }\lambda^{\ast }+\delta_{n} \left( x^{\ast }-x_{n}^{\ast }\right)\right] \\ &&+\left( \mu^{\ast }- \mu_{n}^{\ast }\right)^{\intercal } \left( \alpha_{n}\mu^{\ast }-Ax^{\ast }+a\right) + \left( \lambda^{\ast }-\lambda_{n}^{\ast }\right)^{\intercal } \\ &&\cdot\left( \alpha_{n}\lambda^{\ast }-Bx^{\ast }+ b\right)= \theta_{n} \left( x^{\ast }\text{-}x_{n}^{\ast }\right)^{\intercal }\frac{\partial }{\partial x}\left( \phi\left( x^{\ast }\right) \text{-}\phi(x_{n}^{\ast })\right) \\ &&+ \left( \mu^{\ast }\right)^{\intercal } \left( \left[ Ax^{\ast }\text{-}a\right] \text{-} \left[ Ax_{n}^{\ast }\text{-}a\right] \right) \text{+} \left( \lambda^{\ast }\right)^{\intercal } \left( \left[ Bx^{\ast }-b \right] \right. \\&& \left. -\left[ Bx_{n}^{\ast }-b\right] \right) + \delta_{n} \left( x^{\ast }-x_{n}^{\ast }\right)^{\intercal }\left( x^{\ast }-x_{n}^{\ast }\right) \\ &&+\alpha_{n}\left( \mu^{\ast }\text{-}\mu_{n}^{\ast }\right)^{\intercal }\mu^{\ast }+ \left( \lambda^{\ast }\text{-}\lambda_{n}^{\ast }\right)^{\intercal }\alpha_{n}\lambda^{\ast }\text{+} \left( \lambda_{n}^{\ast }\right)^{\intercal }\left( Bx^{\ast }\text{-}b\right) \\ &\geq&\frac{\delta_{n}}{2} \left\Vert x^{\ast }\text{-}x_{n}^{\ast }\right\Vert^{2}\text{+} \frac{\alpha_{n}}{2}\left( \left\Vert \mu^{\ast }-\mu_{n}^{\ast }\right\Vert^{2}+\left\Vert \lambda^{\ast }-\lambda_{n}^{\ast }\right\Vert^{2}\right) \geq 0 \end{array} $$

Simplifying the last inequality, we have

$$ \begin{array}{@{}rcl@{}} &&\theta_{n} \left( x^{\ast }-x_{n}^{\ast }\right)^{\intercal }\frac{ \partial }{\partial x}\left( f\left( x^{\ast }\right) - \phi(x_{n}^{\ast })\right) +\delta_{n} \left( x^{\ast }-x_{n}^{\ast }\right)^{\intercal }\left( x^{\ast }\right)\\ &&+\alpha_{n}\left( \mu^{\ast }-\mu_{n}^{\ast }\right)^{\intercal }\mu^{\ast }+ \left( \lambda^{\ast }-\lambda_{n}^{\ast }\right)^{\intercal }\alpha_{n}\lambda^{\ast }\geq 0 \end{array} $$

Dividing both sides of this inequality by δ_n taking α_n = δ_n, and $\frac {\theta _{n}}{\delta _{n}} \underset {n\rightarrow \infty }{\rightarrow }0$, we get

$$ \begin{array}{@{}rcl@{}} 0&\leq& \underset{n\rightarrow \infty }{\limsup }\left[ \left( x^{\ast }-x_{n}^{\ast }\right)^{\intercal }x^{\ast }+\left( \mu^{\ast }-\mu_{n}^{\ast }\right)^{\intercal }\mu^{\ast }\right.\\ &&\left.+\left( \lambda^{\ast }-\lambda_{n}^{\ast }\right)^{\intercal }\lambda^{\ast } \right] \end{array} $$

This means that there necessarily exists subsequences δ_k and 𝜃_k$\left (k\rightarrow \infty \right )$ on which there exist the limits

$$ \begin{array}{@{}rcl@{}} x_{k}^{\ast }&=&x^{\ast }\left( \theta_{k},\delta_{k}\right) \rightarrow \tilde{x}^{\ast },\text{ }\mu_{k}^{\ast }=\mu^{\ast }\left( \theta_{k},\delta_{k}\right) \rightarrow \tilde{\mu}^{\ast } \\ \lambda_{k}^{\ast }&=&\lambda^{\ast }\left( \theta_{k},\delta_{k}\right) \rightarrow \tilde{\lambda}^{\ast }\text{ { as} } k\rightarrow \infty \end{array} $$

Suppose that there exist two limit points for two different convergent subsequences, i.e., there exist the limits

$$ \begin{array}{@{}rcl@{}} x_{k^{\prime }}^{\ast }&=&x^{\ast }\left( \theta_{k^{\prime }},\delta_{k^{\prime }}\right) \rightarrow \bar{x}^{\ast },\text{ }\mu_{k^{\prime }}^{\ast }=\mu^{\ast }\left( \theta_{k^{\prime }},\delta_{k^{\prime }}\right) \rightarrow \bar{\mu}^{\ast } \\ \lambda_{k^{\prime }}^{\ast }&=&\lambda^{\ast }\left( \theta_{k^{\prime }},\delta_{k^{\prime }}\right) \rightarrow \bar{\lambda}^{\ast }\text{ { as} }k\rightarrow \infty \end{array} $$

Then, on these subsequences, one has

$$ \begin{array}{@{}rcl@{}} 0&\leq& \left( x^{\ast }-\tilde{x}^{\ast }\right)^{\intercal }x^{\ast }+\left( \mu^{\ast }-\tilde{\mu}^{\ast }\right)^{\intercal }\mu^{\ast }+\left( \lambda^{\ast }-\tilde{\lambda}^{\ast }\right)^{\intercal }\lambda^{\ast } \\ 0&\leq& \left( x^{\ast }-\bar{x}^{\ast }\right)^{\intercal }x^{\ast }+\left( \mu^{\ast }-\bar{\mu}^{\ast }\right)^{\intercal }\mu^{\ast }+\left( \lambda^{\ast }-\bar{\lambda}^{\ast }\right)^{\intercal }\lambda^{\ast } \end{array} $$

From these inequalities, it follows that points $\left (\tilde {x}^{\ast }, \tilde {\mu }^{\ast },\tilde {\lambda }^{\ast }\right )$ and $\left (\bar {x}^{\ast },\bar {\mu }^{\ast },\bar {\lambda }^{\ast }\right )$ correspond to the minimum point of the function

$$ s\left( x^{\ast },\mu^{\ast },\lambda^{\ast }\right) :=\frac{1}{2} \left( \left\Vert x^{\ast }\right\Vert^{2}+\left\Vert \mu^{\ast }\right\Vert^{2}+\left\Vert \lambda^{\ast }\right\Vert^{2}\right) $$

defined on X^∗⊗Ξ^∗ for all possible saddle-points of the Lagrange function. But the function $s\left (x^{\ast },\mu ^{\ast },\lambda ^{\ast }\right )$ is strictly convex, and, hence, its minimum is unique that gives $\tilde {x}^{\ast }$ = $\bar {x}^{\ast },$$\tilde {\mu }^{\ast }=\bar {\mu }^{\ast },$$\tilde {\lambda }^{\ast }=\bar {\lambda }^{\ast }$. □