A More General Framework to Analyze Whether Voluntary Disclosure is Insufficient or Excessive

Abstract

I analyze if the excessive quality disclosure finding of the “classical literature” extends to environments in which consumers have a downward-sloping demand. While the answer is affirmative, there are at least two situations under which disclosure is socially insufficient: (1) when there are quality levels that are too low to generate any positive demand; and (2) when the prior beliefs place sufficiently higher weight on lower qualities. In both cases, non-disclosure by the seller leads to a severe reduction in the perceived quality, thereby significantly lowering the demand and the quantity consumed.

Appendix

1.1 Maximization of $W\left( \theta ^{s}\right)$ for $\alpha >\delta$ and $\mu \in \left[ 0,1\right]$

In this section of the “Appendix”, I derive and simplify the first-order condition that is associated with the maximization of the ex-ante expected welfare function given in Eq. (8) for $\alpha >\delta$ and any $\mu \in \left[ 0,1\right]$. For convenience, rewrite Eq. (8):

$$\begin{aligned} W\left( \theta ^{s}\right)&= \mu \left[ \left( \alpha -\delta \right) q(\tilde{\theta }^{s})-\frac{\beta }{2}\left( q(\tilde{\theta }^{s})\right) ^{2}\right] \\&+\left( 1-\mu \right) \int \limits _{0}^{\theta ^{s}}\left[ \left( \alpha -\left( 1-\theta \right) \delta \right) q(\tilde{\theta }^{s})-\frac{\beta }{2}\left( q(\tilde{\theta }^{s})\right) ^{2}\right] d\theta \\&+\left( 1-\mu \right) \int \limits _{\theta ^{s}}^{1}\left[ \left( \alpha -\left( 1-\theta \right) \delta \right) q\left( \theta \right) -\frac{\beta }{2}\left( q\left( \theta \right) \right) ^{2}-D\right] d\theta \text {.} \end{aligned}$$

The second term above can be rewritten as

$$\begin{aligned} \left( 1-\mu \right) \theta ^{s}\left[ \left( \alpha -\left( 1-\frac{\theta ^{s}}{2}\right) \delta \right) q(\tilde{\theta }^{s})-\frac{\beta }{2}\left( q(\tilde{\theta }^{s})\right) ^{2}\right] \text {.} \end{aligned}$$

Adding this term to the first term and noting that $\tilde{\theta } ^{s}=\left( \frac{\left( 1-\mu \right) \theta ^{s}}{\mu +\left( 1-\mu \right) \theta ^{s}}\right) \frac{\theta ^{s}}{2}$ leads to

$$\begin{aligned} \left( \mu +\left( 1-\mu \right) \theta ^{s}\right) \left[ \left( \alpha -(1-\tilde{\theta }^{s})\delta \right) q(\tilde{\theta }^{s})-\frac{\beta }{2}\left( q(\tilde{\theta }^{s})\right) ^{2}\right] \text {.} \end{aligned}$$

Given that $q(\tilde{\theta }^{s})=\frac{\alpha -\left( 1-\tilde{\theta }^{s}\right) \delta }{2\beta }$, this term can further be simplified as

$$\begin{aligned} \frac{3}{8\beta }\left( \mu +\left( 1-\mu \right) \theta ^{s}\right) \left( \alpha -(1-\tilde{\theta }^{s})\delta \right) ^{2}\text {.} \end{aligned}$$

Similarly, given that $q\left( \theta \right) =\frac{\alpha -\left( 1-\theta \right) \delta }{2\beta }$, the third term in (8) can be rewritten as

$$\begin{aligned} \left( 1-\mu \right) \int \limits _{\theta ^{s}}^{1}\left[ \frac{3}{8\beta }\left( \alpha -\left( 1-\theta \right) \delta \right) ^{2}-D\right] d\theta \text {.} \end{aligned}$$

Hence,

$$\begin{aligned} W\left( \theta ^{s}\right)&= \frac{3}{8\beta }\left( \mu +\left( 1-\mu \right) \theta ^{s}\right) \left( \alpha -(1-\tilde{\theta }^{s})\delta \right) ^{2}\\&+\left( 1-\mu \right) \int \limits _{\theta ^{s}}^{1}\left[ \frac{3 }{8\beta }\left( \alpha -\left( 1-\theta \right) \delta \right) ^{2}-D\right] d\theta \text {.} \end{aligned}$$

Maximizing this expression with respect to $\theta ^{s}$ yields the following first-order condition:

$$\begin{aligned}&\frac{3}{8\beta }\left( 1-\mu \right) \left( \alpha -(1-\tilde{\theta }^{s})\delta \right) ^{2}+\frac{6\delta }{8\beta }\left( \mu +\left( 1-\mu \right) \theta ^{s}\right) \left( \alpha -(1-\tilde{\theta }^{s})\delta \right) \frac{d\tilde{\theta }^{s}}{d\theta ^{s}} \\&= \left( 1-\mu \right) \left[ \frac{3}{8\beta }\left( \alpha -\left( 1-\theta ^{s}\right) \delta \right) ^{2}-D\right] \end{aligned}$$

Note that

$$\begin{aligned} \frac{d\tilde{\theta }^{s}}{d\theta ^{s}}&= \frac{\left( 1-\mu \right) \theta ^{s}}{2\left( \mu +\left( 1-\mu \right) \theta ^{s}\right) }+\frac{ \mu \left( 1-\mu \right) \theta ^{s}}{2\left( \mu +\left( 1-\mu \right) \theta ^{s}\right) ^{2}} \\&= \frac{\left( 1-\mu \right) }{\mu +\left( 1-\mu \right) \theta ^{s}}\left[ \frac{\theta ^{s}}{2}+\frac{\mu \theta ^{s}}{2\left( \mu +\left( 1-\mu \right) \theta ^{s}\right) }\right] \\&= \frac{\left( 1-\mu \right) }{\mu +\left( 1-\mu \right) \theta ^{s}}\left[ \theta ^{s}+\frac{\mu \theta ^{s}}{2\left( \mu +\left( 1-\mu \right) \theta ^{s}\right) }-\frac{\theta ^{s}}{2}\right] \\&= \frac{\left( 1-\mu \right) }{\mu +\left( 1-\mu \right) \theta ^{s}}\left[ \theta ^{s}-\frac{\left( 1-\mu \right) \theta ^{s}}{\mu +\left( 1-\mu \right) \theta ^{s}}\frac{\theta ^{s}}{2}\right] \\&= \frac{\left( 1-\mu \right) }{\mu +\left( 1-\mu \right) \theta ^{s}}\left( \theta ^{s}-\tilde{\theta }^{s}\right) \text {.} \end{aligned}$$

So, the first-order condition above can be rewritten as

$$\begin{aligned}&\frac{3}{8\beta }\left( 1-\mu \right) \left( \alpha -(1-\tilde{\theta } ^{s})\delta \right) ^{2}+\frac{3\delta }{4\beta }\left( 1-\mu \right) \left( \alpha -(1-\tilde{\theta }^{s})\delta \right) (\theta ^{s}-\tilde{\theta }^{s}) \\&\quad =\left( 1-\mu \right) \left[ \frac{3}{8\beta }\left( \alpha -\left( 1-\theta ^{s}\right) \delta \right) ^{2}-D\right] \\&\quad \Leftrightarrow \!\frac{3}{8\beta }\left[ \left( \alpha \!-\!\left( 1\!-\!\theta ^{s}\right) \delta \right) ^{2}\!-\!\left( \alpha \!-\!(1\!-\!\tilde{\theta }^{s})\delta \right) ^{2}\right] \!-\!\frac{3\delta }{4\beta }\left( \alpha \!-\!(1\!-\!\tilde{\theta } ^{s})\delta \right) (\theta ^{s}-\tilde{\theta }^{s})=D\\&\quad \Leftrightarrow \! \frac{3\delta }{4\beta }\left( \alpha \!-\!\left( 1-\frac{\theta ^{s}+\tilde{\theta }^{s}}{2}\right) \delta \right) (\theta ^{s}-\tilde{\theta } ^{s})-\frac{3\delta }{4\beta }\left( \alpha -(1-\tilde{\theta }^{s})\delta \right) (\theta ^{s}-\tilde{\theta }^{s})=D\text {.} \end{aligned}$$

This finally leads to

$$\begin{aligned} \frac{3\delta ^{2}}{8\beta }(\theta ^{s}-\tilde{\theta }^{s})^{2}=D\text {.} \end{aligned}$$

1.2 Proofs of the Propositions

In this section, I present the proofs of the propositions stated in the main text:

Proof of Proposition 1

As stated in the main text, a simple comparison of $\theta ^{*}$ and $\theta ^{s}$ in expressions (12) and (14) yields that $\theta ^{*}=\theta ^{s}=1$ when $D\ge \frac{(4\alpha -\delta )\delta }{16\beta }$ and that $0<\theta ^{*}<\theta ^{s}=1$ when $\frac{3\delta ^{2}}{32\beta }\le D<\frac{(4\alpha -\delta )\delta }{16\beta }$. It only remains to compare $\theta ^{*}$ and $\theta ^{s}$ for $\frac{3\delta ^{2}}{32\beta }\le D<\frac{(4\alpha -\delta )\delta }{16\beta }$ where $0<\theta ^{*}<\theta ^{s}<1$. First, rewrite $\theta ^{s}$ as $\theta ^{s}=\frac{2\left( \alpha -\delta \right) }{3\delta }\sqrt{\frac{24\beta D}{\left( \alpha -\delta \right) ^{2}}}$. Then,

$$\begin{aligned} \theta ^{*}&< \theta ^{s}\Leftrightarrow \sqrt{1+\frac{12\beta D}{ \left( \alpha -\delta \right) ^{2}}}-1<\sqrt{\frac{24\beta D}{\left( \alpha -\delta \right) ^{2}}} \\&\Leftrightarrow 2+\frac{12\beta D}{\left( \alpha -\delta \right) ^{2}}-2 \sqrt{1+\frac{12\beta D}{\left( \alpha -\delta \right) ^{2}}}<\frac{24\beta D }{\left( \alpha -\delta \right) ^{2}} \\&\Leftrightarrow 1-\frac{6\beta D}{\left( \alpha -\delta \right) ^{2}}< \sqrt{1+\frac{12\beta D}{\left( \alpha -\delta \right) ^{2}}}\text {. } \end{aligned}$$

Since $\frac{\beta D}{\left( \alpha -\delta \right) ^{2}}>0$, the term on the left-hand side is less than $1$ while the term on the right-hand side is greater than $1$. Hence, for $\frac{3\delta ^{2}}{32\beta }\le D<\frac{ (4\alpha -\delta )\delta }{16\beta }$, it is always true that $\theta ^{*}<\theta ^{s}$, which means that the seller engages in excessive information disclosure. $\square$

Proof of Proposition 2

Comparing the values of $\theta ^{*}$ and $\theta ^{s}$ in expressions (18) and (20), it is clear that $\theta ^{*}=1-\frac{\alpha -2\sqrt{\beta D}}{\delta }>1-\frac{\alpha -2\sqrt{2\beta D/3}}{\delta }=\theta ^{s}$ when $D<\frac{\left( \delta -\alpha \right) ^{2}}{4\beta }$. For $\frac{ \left( \delta -\alpha \right) ^{2}}{4\beta }\le D<\frac{3\left( \delta -\alpha \right) ^{2}}{8\beta }$, since $\theta ^{*}$ is increasing in $D$, it follows that $\theta ^{*}>1-\frac{\alpha -2\sqrt{\beta D}}{\delta }$, so once again $\theta ^{*}>\theta ^{s}$. Next, note that

$$\begin{aligned} \frac{3\delta ^{2}}{32\beta }\ge \frac{(4\alpha -\delta )\delta }{16\beta } \Leftrightarrow \alpha \le \frac{5\delta }{8} \left( \text {with equality when }\alpha =\frac{5\delta }{8}\right) . \end{aligned}$$

When $\alpha \le \frac{5\delta }{8}$, comparing $\theta ^{*}$ and $\theta ^{s}$ for $\frac{3\left( \delta -\alpha \right) ^{2}}{8\beta }\le D< \frac{(4\alpha -\delta )\delta }{16\beta }$ yields

$$\begin{aligned} \theta ^{*}&\ge \theta ^{s}\Leftrightarrow \sqrt{1+\frac{12\beta D}{\left( \delta -\alpha \right) ^{2}}}+1\ge \sqrt{\frac{24\beta D}{\left( \delta -\alpha \right) ^{2}}} \\&\Leftrightarrow 2+\frac{12\beta D}{\left( \delta -\alpha \right) ^{2}}+2\sqrt{1+\frac{12\beta D}{\left( \delta -\alpha \right) ^{2}}}\ge \frac{24\beta D}{\left( \delta -\alpha \right) ^{2}} \\&\Leftrightarrow 1+\frac{12\beta D}{\left( \delta -\alpha \right) ^{2}}\ge \left( \frac{6\beta D}{\left( \delta -\alpha \right) ^{2}}-1\right) ^{2} \\&\Leftrightarrow \frac{24\beta D}{\left( \delta -\alpha \right) ^{2}}\ge \left( \frac{6\beta D}{\left( \delta -\alpha \right) ^{2}}\right) ^{2} \\&\Leftrightarrow D\le \frac{2\left( \delta -\alpha \right) ^{2}}{3\beta } \left( \text {with equality when }D=\frac{2\left( \delta -\alpha \right) ^{2}}{ 3\beta }\right) . \end{aligned}$$

However, it is clear by a simple comparison that

$$\begin{aligned} \frac{2\left( \delta -\alpha \right) ^{2}}{3\beta }\ge \frac{3\delta ^{2}}{32\beta }\ge \frac{(4\alpha -\delta )\delta }{16\beta } \Leftrightarrow \alpha \le \frac{5\delta }{8} \left( \text {with equality when }\alpha =\frac{5\delta }{8}\right) . \end{aligned}$$

Hence, when $\alpha \le \frac{5\delta }{8}$, it follows that $\theta ^{*}>\theta ^{s}$ for $\frac{3\left( \delta -\alpha \right) ^{2}}{8\beta }\le D<\frac{(4\alpha -\delta )\delta }{16\beta }$. For $\frac{(4\alpha -\delta )\delta }{16\beta }\le D<\frac{3\delta ^{2}}{32\beta }$, we have $\theta ^{s}<\theta ^{*}=1$. And finally, for $D>\frac{3\delta ^{2}}{32\beta }$, $\theta ^{s}=\theta ^{*}=1$.

When $\frac{5\delta }{8}<\alpha \le \delta$, on the other hand, the above calculations imply that $\theta ^{*}\gtrless \theta ^{s}$ as $D\lessgtr \frac{2\left( \alpha -\delta \right) ^{2}}{3\beta }$. Hence, disclosure is insufficient if $\alpha \le \frac{5\delta }{8}$, or otherwise if $\alpha >\frac{5\delta }{8}$ and $D<\frac{2\left( \delta -\alpha \right) ^{2}}{3\beta }$. $\square$

Proof of Proposition 3

Here, I compare the values of $\theta ^{*}$ and $\theta ^{s}$ in expressions (23) and (24). Note that

$$\begin{aligned} \frac{3\delta ^{2}}{8\beta }\ge \frac{\alpha ^{2}-\left( \alpha -\delta \right) ^{2}}{4\beta } \Leftrightarrow \alpha \le \frac{5\delta }{4} \left( \text {with equality when }\alpha =\frac{5\delta }{4}\right) . \end{aligned}$$

Rewrite $\sqrt{\frac{8\beta D}{3\delta ^{2}}}$ as $\frac{\left( \alpha -\delta \right) }{\delta }\sqrt{\frac{8\beta D}{3\left( \alpha -\delta \right) ^{2}}}$. When $\alpha \le \frac{5\delta }{4}$, comparing $\theta ^{*}$ and $\theta ^{s}$ yields

$$\begin{aligned} \theta ^{*}&\ge \theta ^{s}\Leftrightarrow \sqrt{1+\frac{4\beta D}{\left( \alpha -\delta \right) ^{2}}}-1\ge \sqrt{\frac{8\beta D}{3\left( \alpha -\delta \right) ^{2}}} \\&\Leftrightarrow 2+\frac{4\beta D}{\left( \alpha -\delta \right) ^{2}}-2\sqrt{1+\frac{4\beta D}{\left( \alpha -\delta \right) ^{2}}}\ge \frac{8\beta D}{3\left( \alpha -\delta \right) ^{2}} \\&\Leftrightarrow \left( 1+\frac{2\beta D}{3\left( \alpha -\delta \right) ^{2}}\right) ^{2}\ge 1+\frac{4\beta D}{\left( \alpha -\delta \right) ^{2}}\\&\Leftrightarrow \left( \frac{2\beta D}{3\left( \alpha -\delta \right) ^{2}}\right) ^{2}\ge \frac{8\beta D}{3\left( \alpha -\delta \right) ^{2}} \\&\Leftrightarrow D\ge \frac{6\left( \alpha -\delta \right) ^{2}}{\beta }\left( \text {with equality when }D=\frac{6\left( \alpha -\delta \right) ^{2}}{\beta }\right) . \end{aligned}$$

It is easy to see that

$$\begin{aligned} \frac{3\delta ^{2}}{8\beta }\ge \frac{\alpha ^{2}-\left( \alpha -\delta \right) ^{2}}{4\beta }\ge \frac{6\left( \alpha -\delta \right) ^{2}}{\beta } \Leftrightarrow \alpha \le \frac{5\delta }{8} \left( \text { with equality when }\alpha =\frac{5\delta }{8}\right) . \end{aligned}$$

It then follows that when $\alpha \le \frac{5\delta }{4}$, $\theta ^{*}\gtrless \theta ^{s}$ as $D\gtrless \frac{6\left( \alpha -\delta \right) ^{2}}{\beta }$. When $\alpha >\frac{5\delta }{4}$, on the other hand, the above calculations imply that $\theta ^{*}<\theta ^{s}<1$ for $D<\frac{3\delta ^{2}}{8\beta }$, $\theta ^{*}<\theta ^{s}=1$ for $\frac{3\delta ^{2}}{8\beta }\le D<\frac{\alpha ^{2}-\left( \alpha -\delta \right) ^{2}}{4\beta }$, and $\theta ^{*}=\theta ^{s}=1$ for $D\ge \frac{\alpha ^{2}-\left( \alpha -\delta \right) ^{2}}{4\beta }$. Consequently, disclosure is insufficient if $\alpha \le \frac{5\delta }{4}$ and $\frac{6\left( \alpha -\delta \right) ^{2}}{\beta }<D\le \frac{3\delta ^{2}}{8\beta }$. $\square$