On the consumer problem under an informational externality

Abstract

We use the Hendricks and Kovenock (RAND J Econ 20(2):164–182, 1989) framework to study the consumer problem under an informational externality. The informational externality arises when each consumer of a social network is endowed with private information regarding the quality of a good. In such situations, the past purchasing decisions of the consumers are informative and, thus, are used as partially revealing signals of private information. Asymmetric information and the observability of actions render the consumer problem dynamic and strategic because the purchasing decision of a consumer affects the other consumers’ future payoffs through the learning process. We show that there exists a unique symmetric Bayesian–Nash equilibrium. The informational externality increases the likelihood for a consumer to refrain from purchasing the good immediately to make a more informed decision in the future.

Proofs

The proofs are in the spirit of Hendricks and Kovenock (1989).

Proof of Proposition 3.4

From (7),

$$\begin{aligned}&\Delta ({\hat{s}}_{A},{\mathcal {X}}_{B}) = v_{1}({\hat{s}}_{A}) - \beta W({\hat{s}}_{A},{\mathcal {X}}_{B}), \end{aligned}$$

(28)

$$\begin{aligned}&\quad = \int _{\mu \in {\mathbb {R}}} \mu \xi (\mu \vert {\hat{s}}_{A}) \text {d} \mu - \beta \max \left\{ 0,\int _{\mu \in {\mathbb {R}}} \mu \Phi (S_{B} \in {\mathcal {X}}_{B} \vert \mu ) \xi (\mu \vert {\hat{s}}_{A}) \text {d} \mu \right\} \nonumber \\&\qquad - \beta \max \left\{ 0,\int _{\mu \in {\mathbb {R}}} \mu \Phi (S_{B} \notin {\mathcal {X}}_{B} \vert \mu ) \xi (\mu \vert {\hat{s}}_{A}) \text {d} \mu \right\} , \end{aligned}$$

(29)

$$\begin{aligned}&\quad = \int _{\mu \in {\mathbb {R}}} \mu \left[ 1 - 1_{[{\hat{s}}_{A} \ge t_{1}]} \beta \Phi (S_{B} \in {\mathcal {X}}_{B} \vert \mu ) - 1_{[ {\hat{s}}_{A} \ge t_{2} ]} \beta \Phi ( S_{B} \notin {\mathcal {X}}_{B} \vert \mu ) \right] \xi (\mu \vert {\hat{s}}_{A}) \text {d} \mu , \end{aligned}$$

(30)

$$\begin{aligned}&\quad = \int _{\mu \in {\mathbb {R}}} \mu \left[ 1 - 1_{[{\hat{s}}_{A}> t_{1}]} \beta \Phi (S_{B} \in {\mathcal {X}}_{B} \vert \mu ) - 1_{[ {\hat{s}}_{A} > t_{2} ]} \beta \Phi (S_{B} \notin {\mathcal {X}}_{B} \vert \mu ) \right] \xi (\mu \vert {\hat{s}}_{A}) \text {d} \mu \end{aligned}$$

(31)

be the difference between the value of purchasing the good in period 1, and the value of exercising the outside option in period 1. Note that $\Delta ({\hat{s}}_{A},{\mathcal {X}}_{B}) < 0$ for ${\hat{s}}_{A} \le t_{0}$, and $\Delta ({\hat{s}}_{A},{\mathcal {X}}_{B}) > 0$ for ${\hat{s}}_{A} \ge \max \{t_{1},t_{2}\}$. Since $\Delta ({\hat{s}}_{A},{\mathcal {X}}_{B})$ is continuous in ${\hat{s}}_{A}$, there is at least one $c_{A} \in (t_{0},\max \{t_{1},t_{2}\})$, such that $\Delta (c_{A},{\mathcal {X}}_{B}) = 0$. It remains to show the uniqueness of $c_{A} \in (t_{0},\max \{t_{1},t_{2}\})$, such that $\Delta (c_{A},{\mathcal {X}}_{B}) = 0$. $\square$

• 1. For ${\hat{s}}_{A} > c_{A}$. Let

  $$\begin{aligned} f(\mu ,{\hat{s}}_{A}) = \mu \left[ 1 - 1_{[{\hat{s}}_{A} \ge t_{1}]} \beta \Phi (S_{B} \in {\mathcal {X}}_{B} \vert \mu ) - 1_{[{\hat{s}}_{A} \ge t_{2}]} \beta \Phi (S_{B} \notin {\mathcal {X}}_{B} \vert \mu ) \right] \nonumber \\ \end{aligned}$$

  (32)

  and $\rho = \xi (0 \vert c_{A}) / \xi (0 \vert {\hat{s}}_{A})$ when $\mu = 0$. Then,

  $$\begin{aligned} \rho \Delta ({\hat{s}}_{A},{\mathcal {X}}_{B})&= \rho \int _{\mu \in {\mathbb {R}}} f(\mu ,{\hat{s}}_{A}) \xi (\mu \vert {\hat{s}}_{A}) \text {d} \mu , \end{aligned}$$

  (33)
  $$\begin{aligned}&= \rho \int _{\mu \in {\mathbb {R}}} f(\mu ,{\hat{s}}_{A}) \xi (\mu \vert {\hat{s}}_{A}) \text {d} \mu - \int _{\mu \in {\mathbb {R}}} f(\mu ,c_{A}) \xi (\mu \vert c_{A}) \text {d} \mu , \end{aligned}$$

  (34)
  $$\begin{aligned}&= \int _{\mu \in {\mathbb {R}}} f(\mu ,{\hat{s}}_{A}) \left( \rho - \frac{f(\mu ,c_{A})}{f(\mu ,{\hat{s}}_{A})} \frac{\xi (\mu \vert c_{A})}{\xi (\mu \vert {\hat{s}}_{A})} \right) \xi (\mu \vert {\hat{s}}_{A}) \text {d} \mu . \end{aligned}$$

  (35)

  Given the strict monotone likelihood ratio property, $\rho > \xi (\mu \vert c_{A}) / \xi (\mu \vert {\hat{s}}_{A})$ if and only if $\mu > 0$. Since $f(\mu ,{\hat{s}}_{A})$ is right continuous in ${\hat{s}}_{A}$, there is $\varepsilon > 0$ such that

  $$\begin{aligned} \rho - \frac{f(\mu ,c_{A})}{f(\mu ,{\hat{s}}_{A})} \frac{\xi (\mu \vert c_{A})}{\xi ( \mu \vert {\hat{s}}_{A})} > 0 \end{aligned}$$

  (36)

  for ${\hat{s}}_{A} \in (c_{A},c_{A} + \varepsilon )$ if and only if $\mu > 0$. This, combined with the fact that $f(\mu ,{\hat{s}}_{A}) > 0$ if and only if $\mu > 0$, implies that $\rho \Delta ({\hat{s}}_{A},{\mathcal {X}}_{B}) > 0$ for ${\hat{s}}_{A} \in (c_{A},c_{A} + \varepsilon )$. Since $\rho > 0$, $\Delta ({\hat{s}}_{A},{\mathcal {X}}_{B}) > 0$ for ${\hat{s}}_{A} \in (c_{A},c_{A} + \varepsilon )$.

• 2. For ${\hat{s}}_{A} < c_{A}$. Use

  $$\begin{aligned} g(\mu ,{\hat{s}}_{A}) = \mu \left[ 1 - 1_{[{\hat{s}}_{A}> t_{1}]} \beta \Phi (S_{B} \in {\mathcal {X}}_{B} \vert \mu ) - 1_{[{\hat{s}}_{A} > t_{2}]} \beta \Phi (S_{B} \notin {\mathcal {X}}_{B} \vert \mu ) \right] \nonumber \\ \end{aligned}$$

  (37)

  instead of $f(\mu ,{\hat{s}}_{A})$, because g is left continuous in ${\hat{s}}_{A}$. The steps are identical to 1. Since $g(\mu ,{\hat{s}}_{A})$ is left continuous in ${\hat{s}}_{A}$, there is $\varepsilon > 0$ such that $\Delta ({\hat{s}}_{A},{\mathcal {X}}_{B}) < 0$ for ${\hat{s}}_{A} \in (c_{A},c_{A} - \varepsilon )$.

Combining the facts that there is a $c_{A} \in (t_{0},\max \{t_{1},t_{2}\})$, such that $\Delta (c_{A},{\mathcal {X}}_{B}) = 0$, $\Delta ({\hat{s}}_{A},{\mathcal {X}}_{B})$ is continuous in ${\hat{s}}_{A}$, $\Delta ({\hat{s}}_{A},{\mathcal {X}}_{B}) < 0$ for ${\hat{s}}_{A} \le t_{0}$, $\Delta ({\hat{s}}_{A},{\mathcal {X}}_{B}) > 0$ for ${\hat{s}}_{A} \ge \max \{t_{1},t_{2}\}$, $\Delta ({\hat{s}}_{A},{\mathcal {X}}_{B}) > 0$ for ${\hat{s}}_{A} \in (c_{A},c_{A} + \varepsilon )$, and $\Delta ({\hat{s}}_{A},{\mathcal {X}}_{B}) < 0$ for ${\hat{s}}_{A} \in (c_{A},c_{A} - \varepsilon )$ implies that $c_{A} \in (t_{0},\max \{t_{1},t_{2}\})$ is unique. Thus, a unique response function $R_{A}({\hat{s}}_{A},{\mathcal {X}}_{B}) = 1_{[{\hat{s}}_{A} \ge c_{A}]}$ to consumer B’s strategy ${\mathcal {X}}_{B}$ exists.

Proof of Proposition 3.5

Given Proposition 3.4, consumers A and B’s strategies in a symmetric Bayesian–Nash equilibrium are ${\mathcal {X}}^{*} = \left\{ {\hat{s}} : \chi ({\hat{s}}) = 1_{[{\hat{s}} \ge c^{*}] }\right\}$, where $c^{*}$ is defined by $v_{1}(c^{*}) = \beta W(c^{*},{\mathcal {X}}^{*})$. From (7), let

$$\begin{aligned} \Gamma (c)&= v_{1}(c) - \beta W(c,{\mathcal {X}}), \end{aligned}$$

(38)

$$\begin{aligned}&= \int _{\mu \in {\mathbb {R}}} \mu \xi (\mu \vert c) \text {d} \mu - \beta \max \left\{ 0,\int _{\mu \in {\mathbb {R}}} \mu \Phi (S_{B} \ge c \vert \mu ) \xi (\mu \vert c) \text {d} \mu \right\} \nonumber \\&\quad - \beta \max \left\{ 0,\int _{\mu \in {\mathbb {R}}} \mu \Phi (S_{B} < c \vert \mu ) \xi (\mu \vert c) \text {d} \mu \right\} , \end{aligned}$$

(39)

$$\begin{aligned}&= \int _{\mu \in {\mathbb {R}}} \mu \left[ 1 - 1_{[c \ge t_{1}]} \beta \Phi (S_{B} \ge c \vert \mu ) - 1_{[c \ge t_{2}]} \beta \Phi (S_{B} < c \vert \mu ) \right] \xi (\mu \vert c) \text {d} \mu , \end{aligned}$$

(40)

$$\begin{aligned}&= \int _{\mu \in {\mathbb {R}}} \mu \left[ 1 - 1_{[ c> t_{1}] } \beta \Phi (S_{B} \ge c \vert \mu ) - 1_{[ c > t_{2}]} \beta \Phi (S_{B} < c \vert \mu ) \right] \xi (\mu \vert c) \text {d} \mu , \end{aligned}$$

(41)

where ${\mathcal {X}} = \{{\hat{s}} : \chi ({\hat{s}}) = 1_{[{\hat{s}} \ge c]}\}$. Note that $\Gamma (c) < 0$ for $c \le t_{0}$, because $v_{1}(t_{0}) = 0 < v_{2}(t_{0},s_{B} \ge t_{0})$, and $\Gamma (c) > 0$ for $c \ge \max \{t_{1},t_{2}\}$. Since $\Gamma (c)$ is continuous in c, there is at least one $c^{*} \in (t_{0},\max \{t_{1},t_{2}\})$, such that $\Gamma (c^{*}) = 0$. It remains to show the uniqueness of $c^{*} \in (t_{0},\max \{t_{1},t_{2}\})$, such that $\Gamma (c^{*}) = 0$. $\square$

• 1. For $c > c^{*}$. Let $f(\mu ,c) = \mu \left[ 1 - 1_{[c \ge t_{1}]} \beta \Phi (S_{B} \ge c \vert \mu ) - 1_{[c \ge t_{2}]} \beta \Phi (S_{B} < c \vert \mu ) \right]$ and $\rho = \xi (0 \vert c^{*}) / \xi (0 \vert c)$ when $\mu = 0$. Then

  $$\begin{aligned} \rho \Gamma (c) =&\rho \int _{\mu \in {\mathbb {R}}} f(\mu ,c) \xi (\mu \vert c) \text {d} \mu = \rho \int _{\mu \in {\mathbb {R}}} f(\mu ,c) \xi (\mu \vert c) \text {d} \mu \nonumber \\&- \int _{\mu \in {\mathbb {R}}} f(\mu ,c^{*}) \xi (\mu \vert c^{*}) \text {d} \mu , \end{aligned}$$

  (42)
  $$\begin{aligned} =&\rho \int _{\mu \in {\mathbb {R}}} f(\mu ,c) \left( \rho - \frac{f(\mu ,c^{*})}{f(\mu ,c)} \frac{\xi (\mu \vert c^{*})}{\xi (\mu \vert c)}\right) \xi ( \mu \vert c) \text {d} \mu . \end{aligned}$$

  (43)

  Given the strict monotone likelihood ratio property, $\rho > \xi ( \mu \vert c^{*}) / \xi ( \mu \vert c)$ if and only if $\mu > 0$. Since $f(\mu ,c)$ is right continuous in c, there is $\varepsilon > 0$ such that

  $$\begin{aligned} \rho - \frac{f(\mu ,c^{*})}{f(\mu ,c)} \frac{\xi (\mu \vert c^{*})}{\xi (\mu \vert c)} > 0 \end{aligned}$$

  (44)

  for $c \in (c^{*},c^{*} + \varepsilon )$ if and only if $\mu > 0$. This, combined with the fact that $f(\mu ,c) > 0$ if and only if $\mu > 0$, implies that $\rho \Gamma (c) > 0$ for $c \in (c^{*},c^{*} + \varepsilon )$. Since $\rho > 0$, $\Gamma (c) > 0$ for $c \in (c^{*},c^{*} + \varepsilon )$.

• 2. For $c < c^{*}$. Let $g(\mu ,c) = \mu \left[ 1 - 1_{[c> t_{1}]} \beta \Phi (S_{B} \ge c \vert \mu ) - 1_{[c > t_{2}]} \beta \Phi (S_{B} < c \vert \mu ) \right]$ instead of $f(\mu ,c)$, because g is left continuous in c. The steps are identical to 1. Since $g(\mu ,c)$ is left continuous in c, there is $\varepsilon > 0$ such that $\Gamma (c) > 0$ for $c \in (c^{*},c^{*} - \varepsilon )$.

Combining the facts that there is a $c^{*} \in (t_{0},\max \{t_{1},t_{2}\})$, such that $\Gamma (c^{*}) = 0$, $\Gamma (c)$ is continuous in c, $\Gamma (c) < 0$ for $c \le t_{0}$, $\Gamma (c) > 0$ for $c \ge \max \{t_{1},t_{2}\}$, $\Gamma (c) > 0$ for $c \in (c^{*},c^{*} + \varepsilon )$, and $\Gamma (c) < 0$ for $c \in (c^{*},c^{*} - \varepsilon )$ implies that $c^{*} \in (t_{0},\max \{t_{1},t_{2}\})$ is unique.