# Chapter 16. Information Cascades

[TOC]

## 16.1 Following the Crowd

### Herding/Information Cascadess
- individuals
- private information
- decision
- decide in an order
- the decision may be influenced by former decision
- it is not 羊群效應
- 理性地根據先前的選擇判斷自己選擇的效益
- it is not Direct-Benefit Effects
- 是否購買傳真機，使用人數直接影響該選擇效益

## 16.2 A Simple Herding Experiment
- some students, $s_1,\ldots,s_n$
- an urn
- A: 50% 2 red, 1 blue
- B: 50% 1 red, 2 blue
- then $P(\mbox{red})=P(\mbox{blue})=\frac{1}{2}$
- one-by-one
- draw a ball and put it back
- all others don't know
- say the color of them $c_1,\ldots,c_n$
- guess A or B
- let all later students know
- say the guess of color $g_1,\ldots,g_n$

## Remark
- first student $s_1$
- $g_1:=c_1$
- second student $s_2$
- if $g_1=c_2$
- $g_2:=c_2$
- else
- the same case as draw twice, get one red and one blue
- guess arbitrarily
- assume he guess the color he draw $g_2=c_2$
- third student $s_3$
- if $g_1\neq g_2$ or $g_1=g_2=c_3$
- $g_3:=c_3$
- if $g_1=g_2\neq c_3$
- the same case as draw three times. get one $c_3$ and two $g_1$
- $g_3:=g_1$
- fourth student $s_4$ and after
- skip some situation
- if $g_1=g_2=g_3$
- $g_3=g_1$ no matter what he draw
- no information from $s_3$
- the same situation as $s_3$
- $g_4:=g_1$
- $P(\mbox{error})=P(1r2b)p(c_1=c_2=r\mid 1r2b)+P(2r1b)P(c_1=c_2=b\mid 2r1b)=\frac{1}{9}$
## 16.3 Bayes' Rule: A Model of Decision Making under Uncertainty
- [貝氏機率](https://zh.wikipedia.org/wiki/%E8%B2%9D%E6%B0%8F%E6%A9%9F%E7%8E%87)
$$
P(A\mid B)=\frac{P(A)\times P(B\mid A)}{P(B)}
$$
## 16.4 Bayes's Rule in the Herding Experiment
Compute $P(2b1r\mid b,b,r)$
$$
P(b,b,r)=P(2r1b)\times P(b,b,r\mid 2r1b)+P(1r2b)\times P(b,b,r\mid 1r2b)=\frac{1}{9}
$$
$$
P(b,b,r\mid 1r2b)=\frac{2}{3}\cdot\frac{2}{3}\cdot\frac{1}{3}=\frac{4}{27}
$$
$$
P(2b1r\mid b,b,r)=\frac{P(2b1r)\times P(b,b,r\mid 1r2b)}{P(b,b,r)}=\frac{\frac{1}{2}\cdot\frac{4}{27}}{\frac{1}{9}}=\frac{2}{3}>\frac{1}{2}
$$

## 16.5 A Simple, General Cascade Model
### Formulating the Model
- A group of people (numbered $1, 2, 3, . . .$) who will sequentially make decisions
- States of the World
- $G$: option is a good idea
- $B$: option is a bad idea
- $Pr [G] = p$, $Pr [B] = 1-p$
- Payoffs
- If the individual chooses rejecting
- payoff $=0$
- If the individual chooses accepting
- If the option is a good idea
- payoff $=v_g > 0$
- If the option is a bad idea
- payoff $=v_b <0$
- Assume that the expected payoff from accepting in the absence of other information $=v_gp + v_b(1 − p) = 0$
- Signals
- high signal $H$
- low signal $L$
- $q\gt \frac12$
### Individual Decisions
- Suppose that a person gets a high signal.
- expected payoff = $v_gPr [G | H] + v_bPr [B | H]$
- $Pr [G | H] = \frac{Pr [G] × Pr [H | G]}{Pr [H]}= \frac{Pr [G] ×Pr [H | G]}{ Pr [G] × Pr [H | G] + Pr [B] × Pr [H | B]}= \frac{pq}{pq+(1 −p)(1 − q)}> p$
### Multiple Signals
- Reason about an individual’s decision when they get a sequence $S$ consisting of $a$ high signals and $b$ low signals
- If $a>b$, $Pr [G | S] > Pr [G]$
(individuals should accept the option)
- If $a<b$, $Pr [G | S] < Pr [G]$
(individuals should reject the option)
- If $a=b$, $Pr [G | S] = Pr [G]$
- $Pr [G | S] = \frac{Pr [G] × Pr [S | G]}{Pr [S]}=\frac{Pr [G] × Pr [S | G]}{Pr [G] × Pr [S | G] + Pr [B] × Pr [S | B]}$$=\frac{pq^a(1 − q)^b}{pq^a(1 − q)^b + (1 − p)(1 − q)^aq^b}$
- If $a>b$, $Pr [G | S]=\frac{pq^a(1 − q)^b}{pq^a(1 − q)^b + (1 − p)(1 − q)^aq^b}$ $>\frac{pq^a(1 − q)^b}{pq^a(1 − q)^b + (1 − p)q^a(1-q)^b}$ $= \frac{pq^a(1 − q)^b}{q^a(1-q)^b}=p=Pr [G]$
- If $a<b$, $Pr [G | S]=\frac{pq^a(1 − q)^b}{pq^a(1 − q)^b + (1 − p)(1 − q)^aq^b}$ $<\frac{pq^a(1 − q)^b}{pq^a(1 − q)^b + (1 − p)q^a(1-q)^b}$ $=Pr [G]$
- If $a=b$, $Pr [G | S]=\frac{pq^a(1 − q)^b}{pq^a(1 − q)^b + (1 − p)(1 − q)^aq^b}$ $=\frac{pq^a(1 − q)^b}{pq^a(1 − q)^b + (1 − p)q^a(1-q)^b}$ $=Pr [G]$

## 16.6 Sequential Decision Making and Cascades
- Suppose that person $N$ knows that everyone before her has followed their own signal.
- If #acc before $N$ $=$ #rej before $N$, $N$ will follow her own signal.
- If $|$#acc before $N$ $-$ #rej before $N|=1$, $N$ will follow her private signal.
- $N$’s private signal will make her indifferent (diff=0)
- It will reinforce the majority signal (diff=2)
- If diff $\ge2$, $N$ will follow the earlier majority. ($N + 1$, $N + 2$, ... know that $N$ ignored her own signal, a cascade has begun)
- Argue that the probability of finding three matching signals in a row converges to 1 as $N\rightarrow \infty$.
- Divide N people into blocks of 3 consecutive people each.
- The probability that none of these blocks consists of identical signals is $(1 − q^3 − (1 − q)^3)^{N/3} \rightarrow 0$ as $N\rightarrow \infty$
## 16.7 Lessons from Cascades
- some observations
- Cascades can be wrong.
- Cascades can be based on very little information.
- Cascades are fragile