Impact of Social Connectivity on Herding Behavior

Abstract

Information cascades have been studied in the literature where myopic selfish users sequentially appear and make a decision to buy a product based on their private observation about the value of the product and actions of their predecessors. Bikhchandani et. al (1992) and Banerjee (1992) introduced such a model and showed that after a finite time, almost surely, users discard their private information and herd on an action asymptotically. In this paper, we study a generalization of that model where we assume users are connected through a random tree, which locally acts as an approximation for Erdös–Rényi random graph when the degree distribution of each vertex of the tree is binomial and as the number of nodes grows large. We show that informational cascades on such tree-structured networks may be analyzed by studying the extinction probability of a certain branching process. We use the theory of multi-type Galton–Watson branching process and calculate the probability of the tree network falling into a cascade. More specifically, we find conditions when this probability is strictly smaller than 1 that are in terms of the degree distributions of the vertices in the tree. Our results indicate that groups that are less tightly knit, i.e., have lesser connection probability (and as a result have lesser diversity of thought), tend to herd more than the groups that have more social connections.

Supported in part by Department of Defense grant W911NF1510225 and Simon’s Foundation grant 26-7523-99.

Appendices

Appendix A

Proof

$$\begin{aligned}&F^v(q) = P(Q_t^k <q|V=v)\end{aligned}$$

(26a)

$$\begin{aligned}&= P (P(V=1|X_t^k) < q | V=v)\end{aligned}$$

(26b)

$$\begin{aligned}&= P \left( \frac{P(V=1)Q(X_t^k|V=1)}{P(V=1)Q(X_t^k|V=1) + P(V=0)Q(X_t^k|V=0)} < q | V=v\right) \end{aligned}$$

(26c)

$$\begin{aligned}&= P(\{x_t^k : \frac{Q(X_t^k = x_t^k|V=1)}{Q(X_t^k = x_t^k|V=0)} <\frac{q}{1-q} \}|V=v)\end{aligned}$$

(26d)

$$\begin{aligned}&= F_X^v \left( \left( \frac{Q(\cdot |V=1)}{Q(\cdot |V=0)}\right) ^{-1} \left( \frac{q}{1-q}\right) \right) \end{aligned}$$

(26e)

where $\left( \frac{Q(\cdot |V=1)}{Q(\cdot |V=0)}\right)$ inverse exists since it is non-decreasing as implied the monotone likelihood assumption, and $F_X^v(a) := P(X\le a|V=v)$.

Appendix B

Proof

Under the equilibrium policy g defined in (8), belief $\pi _t^k$ can be updated using Bayes’ rule as follows.

$$\begin{aligned} \pi _{t+1}^k&= P^g(V=1|a^{\mathcal {P}(p_{t+1}^k)})\end{aligned}$$

(27a)

$$\begin{aligned}&= \frac{P^g(V=1,a_{t}^k|a^{\mathcal {P}(p_t^k)})}{\sum _v P^g(V=v,a_{t}^k|a^{\mathcal {P}(p_t^k)})}\end{aligned}$$

(27b)

$$\begin{aligned}&=\frac{\pi _t^k P^g(a_{t}^k|V=1,a^{\mathcal {P}(p_t^k)})}{(1-\pi _t^k)P^g(a_{t}^k|V=0,a^{\mathcal {P}(p_t^k)}) +\pi _t^k P^g(a_{t}^k|V=1,a^{\mathcal {P}(p_t^k)})}\end{aligned}$$

(27c)

$$\begin{aligned}&=: \psi _{a_{t}^k }(\pi _t^k) \end{aligned}$$

(27d)

Appendix C

Proof

$$\begin{aligned}&P(V=1|a^{\mathcal {P}(p_t^k)},x_t^k) \nonumber \\&= \frac{P(V=1,x_t^k|a^{\mathcal {P}(p_t^k)})}{P(V=0,x_t^k|a^{\mathcal {P}(p_t^k)}) + P(V=1,x_t^k|a^{\mathcal {P}(p_t^k)})}\end{aligned}$$

(28a)

$$\begin{aligned}&= \frac{\pi _t^k P(x_t^k|V=1)}{(1-\pi _t^k) P(x_t^k|V=0) + \pi _t^k P(x_t^k|V=1)}\end{aligned}$$

(28b)

$$\begin{aligned}&= \frac{\pi _t^k P(V=1|x_t^k)}{(1-\pi _t^k) P(V=0|x_t^k) + \pi _t^k P(V=1|x_t^k)}\end{aligned}$$

(28c)

$$\begin{aligned}&= \frac{q_t^k \pi _t^k}{(1-q_t^k) (1-\pi _t^k) + q_t^k \pi _t^k} \end{aligned}$$

(28d)

Thus,

$$\begin{aligned} \frac{q_t^k \pi _t^k}{(1-q_t^k) (1-\pi _t^k) + q_t^k \pi _t^k}&\ge \frac{1}{2}\end{aligned}$$

(28e)

$$\begin{aligned} q_t^k \pi _t^k&\ge (1-q_t^k) (1-\pi _t^k)\end{aligned}$$

(28f)

$$\begin{aligned} \pi _t^k + q_t^k&\ge 1 \end{aligned}$$

(28g)

Appendix D

Let $f(y) = p\phi _D(p+ \bar{p}\phi _D(y))+ \bar{p}\phi _D(p\phi _D(y)+ \bar{p}).$ It is shown in Appendix E that f(y) is monotonically increasing and strictly convex for $y>0$.

Since $f(0) = p (\phi _D(p))+ \bar{p}\phi _D(\bar{p} ) > 0$ and f(y) is monotonically increasing and strictly convex, there exists another fixed point of $f(y) = y$ in the range (0, 1) if and only if $\frac{d}{dy} f(y) \mid _{y=1}>1$. The smallest fixed point is stable since the derivative of f(y) is less than 1 at that point, and thus by Theorem 1, it represents the probability of cascading. The derivative of f(y) at $y=1$ is given by,

$$\begin{aligned} \frac{d}{dy} f(y)&= \frac{d}{dy} p (\phi _D(p + \bar{p}\phi _D(y)))+ \bar{p}\phi _D(p \phi _D(y)+ \bar{p} )\end{aligned}$$

(29)

$$\begin{aligned} \frac{d}{dy} f(y)&= p (\phi _D'(p + \bar{p}\phi _D(y)))\bar{p}\phi _D'(y)+ \bar{p}\phi _D'(p \phi _D(y)+ \bar{p} )p\phi _D'(y)\end{aligned}$$

(30)

$$\begin{aligned} \frac{d}{dy} f(y) \mid _{y=1}&= 2p\bar{p} \phi _D'(1) \phi _D'(1 ) \end{aligned}$$

(31)

Thus, the tree cascades with probability 1 if and only if $2p\bar{p}(\phi _D^{'}(1))^2\le 1$, i.e., $\phi _D^{'}(1)=\mathbb {E}[D]\le \frac{1}{\sqrt{2p\bar{p}}}$. Otherwise the tree cascades with a nonzero probability which is the smallest fixed point of $f(y)=y$.

Appendix E

Lemma 4

$f(y)=p\phi _D(p+ \bar{p}\phi _D(y))+ \bar{p}\phi _D(p\phi _D(y)+ \bar{p})$ is monotonically increasing and strictly convex for $y>0$.

Proof

We will show that $p\phi _D(p+ \bar{p}\phi _D(y))$ is strictly increasing and convex, and the proof of the other part is identical.

$$\begin{aligned}&\frac{d}{dy} p\phi _D(p+ \bar{p}\phi _D(y)) \end{aligned}$$

(32a)

$$\begin{aligned}&= p\phi _D^{'}(p+ \bar{p}\phi _D(y))\bar{p}\phi _D^{'}(y) \end{aligned}$$

(32b)

$$\begin{aligned}&>0 \end{aligned}$$

(32c)

where the last inequality is true since $\phi _D(y) = \mathbb {E}[y^D]>0$ and $\frac{d}{dy} \phi _D(y) = \mathbb {E}[Dy^{D-1}]>0$ since $y>0$ and $D\ge 1$ a.s..

$$\begin{aligned}&\frac{d^2}{dy^2} p\phi _D(p+ \bar{p}\phi _D(y)) \end{aligned}$$

(33)

$$\begin{aligned}&=\frac{d}{dy} p\phi _D^{'}(p+ \bar{p}\phi _D(y))\bar{p}\phi _D^{'}(y)\end{aligned}$$

(34)

$$\begin{aligned}&=p\bar{p}\left( \phi _D^{'}(p+ \bar{p}\phi _D(y)) \phi _D^{''}(y) + \phi _D^{''}(p+ \bar{p}\phi _D(y))\bar{p}(\phi _D^{'}(y))^2 \right) \end{aligned}$$

(35)

$$\begin{aligned}&>0 \end{aligned}$$

(36)

where the last inequality is true by similar arguments as before.