Nash Bargaining

EC487_Advanced_Microeconomics

Nash Bargaining involves Threat Values, Cooperative Surplus and Bargaining Power

Unlike the initial discussion about the Noncooperative Game Theory, the Nash Bargaining is a Cooperative Game Theory problem.

The famous example: Prisoner’s Dilemma

In the previous study in Nash Equilibrium, we know that the equilibrium happens in $(D,D)$ , which is not efficient $(C,C)$ ,that means people act non-cooperatively. But we know that in the Prisoner’s Dilemma, people do so is because they do not have Bargaining Power, if they could, they may reach the better outcome (efficiency).

In this case, the Threat Values would be $(0,0)$ , the Cooperative Surplus is $2$. They could reach the efficient outcome by binding contract which is $(C,C)$

Binding Contract means no cheating, everyone do as they say.

But this is the case happens when they have equal bargaining power , suppose player 1 has full Bargaining Power but player 2 has none. Then player 1 could ask player 2 to pay $1or$0.999999 to him, since 0.000001 > -1, player 2 would agree the contract, in this case, player 1 get the full Cooperative Surplus.

The same is in the pollution problem.

Usually in such problems, we follow those steps:

1. What happens in the efficient outcome? (when no rights allocation)
2. What happens without bargaining?
3. What is the Cooperative Surplus
4. How about the results after bargaining?
5. Provide the results.

The More formal explanation from EC487 Advanced Microeconomics

Definition: A bargaining problem is a tuple $(X,(u^j{)}_{j=1}^2,d)$ where:

• $X\subseteq R^n$ is a compact set of outcomes. It could be understood as the possible contracts that two companies could reach.
• $u^1,u^2:X\to \mathbb{R}$ are two continuous utility functions.
• $d\in X$ is a disagreement outcome so that $u^1(x)\ge u^1(d)$ and $u^2(x)\ge u^2(d)$ for all $x\in X$.

The definition of $X$ to make it compact is saying that the outcome is bounded and closed, this allows us to find the maximum utility for both players. So this is an important assumption.

Definition The Nash Bargaining Solution to a bargaining problem $(X,(u^j{)}_{j=1}^2,x_0)$ is an outcome $x^{\ast }\in X$ so that for all $j\in \{1,2\}$ and $(p,x)\in [0,1]\times X$

$$p{u}^j(x)+(1-p)u^j(d)>u^j(x^{\ast })⟹p{u}^{-j}(x^{\ast })+(1-p)u^{-j}(d)<u^{-j}(x)$$

$(p,x)$ here means a threat point, suppose at the current state $x^{\ast }$, player $j$ is not that satisfy with the current status, then he could propose a new $x$, this deal could happen with probability $p$, and with probability $1-p$ the deal fails and both players get the disagreement outcome $d$. The definition says that if player $j$ could get a better outcome by proposing $(p,x)$, then the other player $-j$ would try to avoid this deal by making sure that his expected utility from the current state $x^{\ast }$ is higher than the expected utility from the new proposed deal $(p,x)$.

Proposition $x^{\ast }\in X$ is a Nash Bargaining Solution for $(X,(u^j{)}_{j=1}^2,x_0)$ if an only if

$$x^{\ast }\in argma{x}_{x\in X}(u^1(x)-u^1(d))(u^2(x)-u^2(d))$$

Proof

From definition, we could easily have $u^1(x)>u^1(d)$ and $u^2(x)>u^2(d)$ for any $x\in X$. Thus it is wlog to assume that $u(d)=0$.

($\Rightarrow$) Suppose $x^{\ast }$ is a Nash Bargaining Solution, which is $x^{\ast }$. Suppose that there exists one $x$ such that $u^1(x)u^2(x)>u^1(x^{\ast })u^2(x^{\ast })$.

It’s obvious at least one player is better off. Also wlog that we assume it is player $1$, which means $u^1(x)>u^1(x^{\ast })$ . Then we have

$$u^2(x)>u^2(x^{\ast })\frac{u^1(x^{\ast })}{u^1(x)}=u^2(x^{\ast })p$$

where $p\in (0,1)$.

By the definition of Nash Bargaining, we have

$$p{u}^1(x)>u^1(x^{\ast })⟹p{u}^2(x^{\ast })<u^2(x)$$

Thus by definition, it is not a Nash Bargaining Solution, which contradicts our assumption.

($\Leftarrow$) Suppose we have $x^{\ast }\in argma{x}_{x\in X}{u}^1(x)u^2(x)$, we want to show it is the Nash Bargaining solution.

Since $x^{\ast }\in argma{x}_{x\in X}{u}^1(x)u^2(x)$, we have

$$u^1(x^{\ast })u^2(x^{\ast })\ge u^1(x)u^2(x)$$

Consider a $p\in [0,1]$ such that $p{u}^j(x)>u^j(x^{\ast })$

Thus we have:

$$u^2(x)\le \frac{u^1(x^{\ast })}{u^1(x)}{u}^2(x^{\ast })$$

Since $p{u}^j(x)>u^j(x^{\ast })$ we have

$$u^2(x)\le \frac{u^1(x^{\ast })}{u^1(x)}{u}^2(x^{\ast })<p{u}^2(x^{\ast })$$