Stochastic Value Formation

Abstract

We propose a simple model of value formation in two societies (communities). When forming values, individuals face conformity pressure within their own society and they are intolerant toward the other society. When such a value formation process is noisy, the interaction between conformity, intolerance and noise can give rise to interesting dynamic outcomes including two polarized societies and polarization across the two societies.

Appendix

Proof of Proposition 1

For any $(\mu _a, \mu _b)$ that satisfies $\mu _a(\{x\})=\mu _b(\{x\})=1$ for some $x\in [0, 1]$, we have $\frac{\partial u_i(x_i)}{\partial x_i}|_{x_i=x}=0$, for $i \in \{a, b\}$. Therefore, it is a steady state. This gives us Proposition 1 (1).

If $(\mu _a, \mu _b)$ satisfies $\mu _a(\{0\})=1$ and $\mu _b(\{1\})=1$, we have $\frac{\partial u_a(x_a)}{\partial x_a}|_{x_a=0}=-\beta <0$ and $\frac{\partial u_i(x_b)}{\partial x_b}|_{x_b=1}=\beta >0$. Hence, it is a steady state. Similarly, If $(\mu _a, \mu _b)$ satisfies $\mu _a(\{1\})=1$ and $\mu _b(\{0\})=1$, we have $\frac{\partial u_a(x_a)}{\partial x_a}|_{x_a=1}=\beta >0$ and $\frac{\partial u_i(x_b)}{\partial x_b}|_{x_b=0}=-\beta <0$. Hence, it is a steady state. This gives us Proposition 1 (2).

When $\alpha =\beta$, we have $\frac{\partial u_i(x_i)}{\partial x_i}=E_i-E_j$, for $i ,j \in \{a, b\}$ and $i \ne j$. Hence, any $(\mu _a, \mu _b)$ that satisfies $E_a=E_b$ is a steady state. This gives us Proposition 1 (3).

Finally, we prove Proposition 1 (4). Suppose that $(\mu _a, \mu _b)$ such that $\mu _a(\{0\})+\mu _a(\{1\})=1$, $\mu _b(\{0\})+\mu _b(\{1\})=1$, and $\mu _a(\{1\}), \mu _b(\{1\}) \in (0, 1)$ is a steady state. Then, it must satisfy that, for $i \in \{a, b\}$

$$\begin{aligned}&\frac{\partial u_i(x_i)}{\partial x_i}|_{x_i=0}=\alpha \mu _i(\{1\})-\beta \mu _j(\{1\})\le 0, \end{aligned}$$

(6)

$$\begin{aligned}&\frac{\partial u_i(x_i)}{\partial x_i}|_{x_i=1}=(\beta -\alpha )+\alpha \mu _i(\{1\})-\beta \mu _j(\{1\})\ge 0, \ \text {where} \ j \in \{a, b\}, \ j \ne i. \end{aligned}$$

(7)

These give us

$$\begin{aligned}&\beta -\alpha \ge \beta \mu _b(\{1\}) -\alpha \mu _a(\{1\}) \ge 0, \end{aligned}$$

(8)

$$\begin{aligned}&\beta -\alpha \ge \beta \mu _a(\{1\}) -\alpha \mu _b(\{1\}) \ge 0. \end{aligned}$$

(9)

When $\beta >\alpha$, both can hold at the same time. Q.E.D. $\square$

Proof of Proposition 2

Proving the following four statements would give us Proposition 2:

• (1) $(\mu _a, \mu _b)$ that satisfies either $\mu _a(\{0\})=\mu _b(\{0\})=1$, or $\mu _a(\{1\})=\mu _b(\{1\})=1$, is not stable.

• (2) There is no stable state in which there exists $x \in (0, 1)$, such that for one of the society $i\in \{a, b\}$, $\frac{\partial u_i(x_i)}{\partial x_i}|_{x_i=x}=0$ and x is in the support of $\mu _i$.

• (3) $(\mu _a, \mu _b)$ that satisfies for $i \in \{a, b\}$, such that $\mu _i(\{0\})+\mu _i(\{1\})=1$ and $\mu _i(\{0\}), \mu _i(\{1\})>0$, is not stable.

• (4) $(\mu _a, \mu _b)$ that satisfies either $\mu _a(\{0\})=1$ and $\mu _b(\{1\})=1$, or $\mu _a(\{1\})=1$ and $\mu _b(\{0\})=1$, is stable.

For $(\mu _a, \mu _b)$ that satisfies $\mu _a(\{0\})=\mu _b(\{0\})=1$, consider an alternative state $(\nu _a, \nu _b)$ such that $\nu _a(\{\varepsilon \})=1$ and $\nu _b(\{0\})=1$, for some $\varepsilon \in (0, 1)$. Then, under $(\nu _a, \nu _b)$

$$\begin{aligned}&\frac{\partial u_a(x_a)}{\partial x_a}|_{x_i=\varepsilon }=\beta \varepsilon > 0, \ \frac{\partial u_b(x_b)}{\partial x_b}|_{x_b=0}=-\beta \varepsilon <0, \end{aligned}$$

(10)

implying that individuals in society a moves toward higher values, and individuals in society b remain holding value 0. Since this is true for any $\varepsilon \in (0, 1)$, the dynamic keeps evolving until everyone in society a holds value 1 while society b stays at 0. Hence, $(\mu _a, \mu _b)$ that satisfies $\mu _a(\{0\})=\mu _b(\{0\})=1$ is not stable. Similarly, $(\mu _a, \mu _b)$ that satisfies $\mu _a(\{1\})=\mu _b(\{1\})=1$ is not stable either. This gives us statement (1).

Suppose $(\mu _a, \mu _b)$ is a steady state such that there exists $x \in (0, 1)$, such that for one of the society $i\in \{a, b\}$, $\frac{\partial u_i(x_i)}{\partial x_i}|_{x_i=x}=0$ and x is in the support of $\mu _i$. First, assume that $\alpha =\beta$. In this case, we must have $E_a=E_b=\bar{x}$, for some $\bar{x} \in (0, 1)$. Consider a shift from $(\mu _a, \mu _b)$ to $(\nu _a, \nu _b)$ such that $E_a=\bar{x}-\varepsilon$ for some arbitrarily small positive $\varepsilon$ and $\nu _b=\mu _b$. Then, we have

$$\begin{aligned}&\frac{\partial u_a(x_a)}{\partial x_a}=-\varepsilon < 0, \ \frac{\partial u_b(x_b)}{\partial x_b}=\varepsilon >0, \end{aligned}$$

(11)

implying that the entire distribution in society a shifts to the left, and the entire distribution in society b shifts to the right, which further ensures that $E_a<E_b$. Hence, the dynamic keeps evolving until everyone in society a holds value 0 and everyone in society b holds value 1. Therefore, when $\alpha =\beta$, there is no stable state in which there exists $x \in (0, 1)$, such that for one of the society $i\in \{a, b\}$, $\frac{\partial u_i(x_i)}{\partial x_i}|_{x_i=x}=0$ and x is in the support of $\mu _i$.

Next, assume that $\alpha \ne \beta$. Without loss of generality, assume that there exists a steady state $(\mu _a, \mu _b)$ such that a $x\in (0, 1)$ is in the support of $\mu _a$. It must satisfy:

$$\begin{aligned} \frac{\partial u_a(x_a)}{\partial x_a}|_{x_a=x}=(\beta -\alpha )x+\alpha E_a-\beta E_b=0. \end{aligned}$$

(12)

We obtain that $x=\frac{\beta E_b-\alpha E_a}{\beta -\alpha }$, which is unique. Hence, there exists no other value in (0, 1) in the support of $\mu _a$. This implies that $\mu _a$ must satisfy $\mu _a(\{x\})>0$ and $\mu _a(\{0\})+\mu _a(\{x\})+\mu _a(\{1\})=1$.

Consider a shift from $(\mu _a, \mu _b)$ to $(\nu _a, \nu _b)$ such that $\nu _a(\{0\})=\mu _a(\{0\})$, $\nu _a(\{1\})=\mu _a(\{1\})$, $\nu _a(\{x-\varepsilon \})=\mu _a(\{x\})$ for some arbitrarily small positive $\varepsilon$ and $\nu _b=\mu _b$. Then, we have

$$\begin{aligned}&\frac{\partial u_a(x_a)}{\partial x_a}|_{x_a=x-\varepsilon }=(\alpha (1-\mu _a(\{x\}))-\beta )\varepsilon . \end{aligned}$$

(13)

When $\alpha >\beta$, we have

$$\begin{aligned}&\frac{\partial u_a(x_a)}{\partial x_a}|_{x_a=0}=\alpha E_a-\beta E_b>\frac{\partial u_a(x_a)}{\partial x_a}|_{x_a=x}=0, \end{aligned}$$

(14)

$$\begin{aligned}&\frac{\partial u_a(x_a)}{\partial x_a}|_{x_a=1}=(\beta -\alpha )+\alpha E_a-\beta E_b<\frac{\partial u_a(x_a)}{\partial x_a}|_{x_a=x}=0. \end{aligned}$$

(15)

Therefore, we must have $\mu _a(\{0\})=\mu _a(\{1\})=0$ and $\mu _a(\{x\})=1$. Hence, $\frac{\partial u_a(x_a)}{\partial x_a}|_{x_a=x-\varepsilon }<0$. When $\alpha <\beta$, $\frac{\partial u_a(x_a)}{\partial x_a}|_{x_a=x-\varepsilon }=(\alpha (1-\mu _a(\{x\}))-\beta )\varepsilon<(\alpha -\beta )\varepsilon <0$. Therefore, the mass on value $x-\varepsilon$ in society a is moving to the left.

If $\mu _a(\{0\})>0$, we must have $\frac{\partial u_a(x_a)}{\partial x_a}|_{x_a=0}=\alpha E_a-\beta E_b=\alpha (\mu _a(\{x\})x+\mu _a(\{1\}))-\beta E_b\le 0$ under $(\mu _a, \mu _b)$. Then, under $(\nu _a, \nu _b)$, $\frac{\partial u_a(x_a)}{\partial x_a}|_{x_a=0}=\alpha (\mu _a(\{x\})(x-\varepsilon )+\mu _a(\{1\}))-\beta E_b<0$, implying that the mass on value 0 in society a if there is any remains the same. If $\mu _a(\{1\})>0$, we must have $\frac{\partial u_a(x_a)}{\partial x_a}|_{x_a=1}=(\beta -\alpha )+\alpha E_a-\beta E_b=(\beta -\alpha )+\alpha (\mu _a(\{x\})x+\mu _a(\{1\}))-\beta E_b\ge 0$ under $(\mu _a, \mu _b)$. Then, under $(\nu _a, \nu _b)$, $\frac{\partial u_a(x_a)}{\partial x_a}|_{x_a=0}=(\beta -\alpha )+\alpha (\mu _a(\{x\})(x-\varepsilon )+\mu _a(\{1\}))-\beta E_b$, which can be negative or positive. Hence, the mass on value 1 in society a if there is any either remains the same or is moving to the left. Hence, $E_a$ is decreasing.

In society b, if there exists $y \in (0, 1)$, such that $\frac{\partial u_b(x_b)}{\partial x_b}|_{x_b=y}=0$ and y is in the support of $\mu _b$, then because $E_a$ is decreasing, the mass on value y in society b is moving to the right, the mass on value 0 in society b if there is any is moving to the right or remains the same, and the mass on value 1 on society b if there is any remains the same. Hence, the dynamic keeps evolving until the two societies reach a state with polarization across the two societies ($\mu _a(\{0\})=1, \mu _b(\{1\})=1$), or with at least one of the society is polarized ( there exists a $i \in \{a, b\}$, such that $\mu _i(\{0\})+\mu _i(\{1\})=1$ and $\mu _i(\{0\}), \mu _i(\{1\})>0$). Therefore, when $\alpha \ne \beta$, there is no stable state in which there exists $x \in (0, 1)$, such that for one of the society $i\in \{a, b\}$, $\frac{\partial u_i(x_i)}{\partial x_i}|_{x_i=x}=0$ and x is in the support of $\mu _i$. This concludes the proof of statement (2).

Consider $(\mu _a, \mu _b)$ that satisfies for $i \in \{a, b\}$, such that $\mu _i(\{0\})+\mu _i(\{1\})=1$ and $\mu _i(\{0\}), \mu _i(\{1\})>0$. Suppose $(\mu _a, \mu _b)$ shifts to $(\nu _a, \nu _b)$ such that $\nu _a(\{0\})=\mu _a(\{0\})+\varepsilon$ and $\nu _a(\{1\})=\mu _a(\{1\})-\varepsilon$, for some arbitrarily small positive $\varepsilon$ and $\nu _b=\mu _b$. Under $(\nu _a, \nu _b)$, $\frac{\partial u_a(x_a)}{\partial x_a}|_{x_a=0}=\alpha (\mu _a(\{1\})-\varepsilon )-\beta \mu _b(\{1\})<\alpha \mu _a(\{1\})-\beta \mu _b(\{1\}) \le 0$. Hence, the mass on 0 in society a remains the same. $\frac{\partial u_a(x_a)}{\partial x_a}|_{x_a=1}=(\beta -\alpha )+\alpha (\mu _a(\{1\})-\varepsilon )-\beta \mu _b(\{1\})$, which can be either positive or negative. Hence, the mass on value 1 in society a either remains the same or is moving to the left. Since $E_a$ is smaller under $(\nu _a, \nu _b)$ than that under $(\mu _a, \mu _b)$, the mass on value 0 in society b remains the same or is moving to the right, the mass on value 1 in society b remains the same, and $E_b$ under $(\nu _a, \nu _b)$ is either equal to or bigger than that under $(\mu _a, \mu _b)$. Therefore, either $(\nu _a, \nu _b)$ is a new steady state, or it sets the motion for the dynamic to move away from $(\mu _a, \mu _b)$. This gives us statement (3).

Finally, consider $(\mu _a, \mu _b)$ that satisfies $\mu _a(\{0\})=1$ and $\mu _b(\{1\})=1$. For any $(\nu _a, \nu _b) \in U^\varepsilon$, given the definition of the Prohorov metric, we must have $\kappa (\mu _a, \nu _a)<\varepsilon$ and $\kappa (\mu _b, \nu _b)<\varepsilon$. Hence, under $(\nu _a, \nu _b)$, for any $x \in [0, 1]$, and for sufficiently small positive $\varepsilon$,

$$\begin{aligned}&\frac{\partial u_a(x_a)}{\partial x_a}|_{x_a=x}= (\beta -\alpha )x+\alpha E_a-\beta E_b \nonumber \\&\quad \le {\left\{ \begin{array}{ll} \beta -\alpha +\alpha \varepsilon -\beta (1-\varepsilon )=(\alpha +\beta )\varepsilon -\alpha<0 &{} \text {if } \beta -\alpha \ge 0, \\ \alpha \varepsilon -\beta (1-\varepsilon )=(\alpha +\beta )\varepsilon -\beta<0 &{} \text {if } \beta -\alpha <0. \end{array}\right. } \end{aligned}$$

(16)

$$\begin{aligned}&\frac{\partial u_b(x_b)}{\partial x_b}|_{x_a=x}= (\beta -\alpha )x+\alpha E_b-\beta E_a \nonumber \\&\quad \ge {\left\{ \begin{array}{ll} \alpha (1-\varepsilon )-\beta \varepsilon =\alpha -(\alpha +\beta )\varepsilon>0 &{} \text {if } \beta -\alpha \ge 0, \\ (\beta -\alpha )+\alpha (1-\varepsilon )-\beta \varepsilon =\beta -(\alpha +\beta )\varepsilon >0 &{} \text {if } \beta -\alpha <0. \end{array}\right. } \end{aligned}$$

(17)

Hence, the distribution in society a moves back to $\mu _a$ and the distribution in society b moves back to $\mu _b$, implying that $(\mu _a, \mu _b)$ that satisfies $\mu _a(\{0\})=1$ and $\mu _b(\{1\})=1$, is stable. Similarly, $(\mu _a, \mu _b)$ that satisfies $\mu _a(\{1\})=1$ and $\mu _b(\{0\})=1$, is stable as well. This concludes the proof of statement (4). Q.E.D. $\square$