# Chapter 12. Bargaining and Power in Networks

[TOC]

## 12.1 Power in Social Networks
In a social network like
We may say $B$ is more powerful
> how to define power?
- *Dependence*
- $A$ and $C$ are completely dependent on $B$
- *Exclusion*
- $B$ has the ability to isolate $A$ or $C$
- *Satiation*
- implicitly psychological principle
- *Betweenness*

## 12.2 Experimental Studies of Power and Exchange
### The Network Exchange Experiment
- Fixed a small graph
- nodes
- test subjects
- edges
- exchange messages
- *resource pool*
- $1 on each edge
- divided by two endpoints by negotiation
- *one-exchange rule*
- one can only make a deal with one neighbor
- in other word, those edges form a matching
- time limit
- stop negotiation after an agreement
- known information
- neighbor
- the whole graph?
- others' deal?
- others' communication?
- multiple rounds
## 12.3 Result of Network Exchange Experiments
- 
- Half half
- 
- theoretically, $B$ gets 1, $A,C$ gets 0
- actually, $B$ gets roughly $\frac{5}{6}$
- 
- if $B$ exclude $A$, $C$ has the power to pair with $D$
- $B$ is less powerful than 3-node path
- $B$ gets roughly $\frac{7}{12}$
- 
- $A$ tries to pair with $B$, $E$ tries to pair with $D$
- $C$ is weak as well
- 
- $C,D$ tend to trade with each other
- $B$ is not that powerful
- 
- unstable
## 12.4 A connection to Buyer-Seller Networks
:::info
In a bipartite graph, **Network Exchange** is equivalent to **Buyer-Seller Network** that buyer valuate 0 for the item, and seller valuate 1 for all items
:::

## 12.5 Modeling Two-Person Interaction: The Nash Bargaining Solution

### The Nash Bargaining Solution
- Assume $x+y\leq1$, otherwise at least one of $A$ and $B$ don't need to negotiate
- Let $s=1-x-y$
- *The Nash Bargaining Solution* says that $A$ and $B$ divided $s$ equally
- $x+\frac{1}{s}=\frac{x-y+1}{2}$ to $A$ and $y+\frac{1}{s}=\frac{y-x+1}{2}$ to $B$
- in network exchange theory, it is called *equidependent*
## 12.6 Modeling Two-Person Intersection: The Ultimatum Game
### The Ultimatum Game
- person $A$ and $B$
- $A$ has $1 in hand
- $A$ proposes a division to $B$
- $x$ for $A$, $1-x$ for $B$
- if $B$ agrees, then $A$ gets $x$, $B$ gets $1-x$
- if $B$ disagrees, both get nothing

### Theoretical
- $B$ always agree
- $A$ gives $B$ as least as possible

### Reality
- $x=\frac{1}{3}$
- negative emotional payoff

## 12.7 Modeling Network Exchange: Stable Outcomes
### Outcome
- **matching**
- ***value*** of node
### Stable Outcomes
**Instability**: An edge $(X,Y)$ not in the matching, $X.value+Y.value \lt 1$.
**Stability**: An outcome is *stable* $\Leftrightarrow$ no *instabilities*.

### Applications of Stable Outcomes
- Extreme power imbalances
- Only stable outcomes on the 3-node path: $B$ gets 1.
- Identifying situations that have no stable outcome.
- 
- Suppose $C.value=0$

## 12.8 Modeling Network Exchange: Balanced Outcomes
***Balanced Outcome***: An outcome is balanced if, for each edge in the matching, the split of the money represents the Nash bargaining outcome, given the best outside options for each node provided by the values in the rest of the network.
## 12.9 Advanced Material: A Game-Theoretic Approach to Bargain
### Formulating Bargaining as a Dynamic Game
(Period 1) A: (.70, .30)? B: Reject.
(Period 2) B: (.60, .40)? A: Reject.
(Period 3) A: (.66, .34)? B: Reject.
(Period 4) B: (.64, .36)? A: Accept.
- probability p > 0 that negotiations abruptly break down
- payoff: accepted split of the dollar or outside options
### Analyzing the Game: An Overview
- The set of available strategies is infinite
- infinite-horizon game
### A First Step: Analyzing a Two-Period Version of Bargaining
- Game will end in period two at the latest.
- Solve it backward
- Period 2:
- $B$: $(a_2, b_2)$ $A$: accepts if $a_2\space \ge$ $x$ ($A$’s outside option).
- $(a_2, b_2)\space = \space(x,1-x)$ And $x+y<1\Rightarrow 1−x>y$
- Period 1:
- If $B$ rejects, expected payoff $z$ = $py + (1 − p)(1 − x)$.
- $A:(a_1,b_1)$ $B$: accepts if $b_1\ge z$
- $A:(a_1,b_1)=(1-z,z)$ and $z = py + (1 − p)(1 − x)\lt p(1-x)+(1-p)(1-x)=1-x$ $\Rightarrow 1-z\gt x$

### Back to the Infinite-Horizon Bargaining Game

- stationary strategies: those in which each of A and B plans to propose the same split in every period in which they are scheduled to propose, and each of A and B also has a fixed amount that they require in order to accept a proposal.
- Stationary equilibrium:
An equilibrium that uses
stationary strategies is called a stationary equilibrium.
### Analyzing the Game: A Stationary Equilibrium
- Any pair of stationary strategies for $A$ and $B$ can be represented by following:
- $A$ : $(a_1, b_1)$ when $A$'s turn
- $B$ : $(a_2, b_2)$ when $B$'s turn
- Reservation amounts $\bar{a}$ and $\bar{b}$, the minimum offers that A and B will accept.
- $a_1+b_1=1$

Show p converges to 0, the payoffs to A and B converge to the Nash bargaining outcome.
- $A$ will set $b_1=\bar{b}$.
- Similarly, $B$ will set $a_2=\bar{a}$.
- $b1 = py + (1 − p)b2$
- $a2 = px + (1 − p)a1$
- In this equilibrium, $A$’s initial offer is accepted, so $A$ gets a payoff $a_1$, and $B$ gets a payoff $b_1 = 1 − a_1 = \frac{y + (1 − p)(1 − x)}{2-p}$
- When $p\approx 1$, $a_1 \approx 1 − y$(almost all the surplus) and $b_1\approx y$(close to outside option)
- As p converges to 0, the players can expect the negotiations to continue for a long time, the opening offer is still accepted in this stationary equilibrium, payoffs converge to$(\frac{x + 1 − y}{2},\frac{y + 1 − x}{2})$