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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.

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Notes

  1. 1.

    To see the intuition, suppose there are \(n>1\) possible quality levels. If the product has the highest possible quality, the seller would surely reveal it because staying silent would simply induce a lower perceived quality and a lower willingness-to-pay. Thus, when consumers see non-disclosure, they will place zero probability on the highest quality and as a result the expected quality in case of non-disclosure will be strictly below the second highest quality. The seller, therefore, would as well choose full disclosure when the product has the second highest quality. Continuing in a similar fashion, one can eliminate all qualities but the lowest one. As a result, if quality information is ever withheld, it must be that the product has the lowest possible quality.

  2. 2.

    By “classical literature,” I refer to Grossman (1981), Grossman and Hart (1980), Milgrom (1981) and Jovanovic (1982).

  3. 3.

    Note that when disclosure is costless, all disclosure policies are welfare-equivalent.

  4. 4.

    Given that the seller charges a higher price under full disclosure, consumer surplus would increase only if the quantity consumed goes up sufficiently.

  5. 5.

    See Dranove and Jin (2010) for an excellent survey on verifiable information disclosure.

  6. 6.

    See Proposition 4 (p. 985) in Daughety and Reinganum (2008).

  7. 7.

    See the following website for their subsequent recognition that their claim is too sweeping: http://www.vanderbilt.edu/econ/faculty/Daughety/Daughetypreviouspubs.html.

  8. 8.

    If, for instance, marginal cost increases with quality, then price may carry a signaling role in situations where quality is not revealed through direct disclosure. Daughety and Reinganum (2008) focus on fully separating prices in this context and characterize an equilibrium in which both price signaling and direct disclosure are used. There is, however, a plethora of other equilibria that involve semi-separating or pooling pricing strategies in the non-disclosure subgame. When marginal cost does not depend on quality, these problems do not arise; all seller types that choose non-disclosure charge the same price.

  9. 9.

    One can interpret this as the consumer’s learning through a process of trial and error in observing when the seller discloses and when he does not.

  10. 10.

    Alternatively, one can simply evaluate \(W\left( \theta ^{s}\right) \) at \(\mu =0\) and \(\tilde{\theta }^{s}=\frac{\theta ^{s}}{2}\), and then maximize the resulting expression with respect to \(\theta ^{s}\). I adopt the current approach because I use it later in scenario 3 as well.

  11. 11.

    Recall that a lower \(\alpha \) or \(\tilde{\theta }\), or a higher \(\delta \) are all associated with a more elastic demand.

  12. 12.

    The only difference is that the term \(\left( \frac{4\left( \alpha -\delta \right) }{3\delta }\right) ^{2}\) in Eq. (11) will come out of the square root as \(\frac{4\left( \delta -\alpha \right) }{3\delta }\).

  13. 13.

    See the proof for further details.

  14. 14.

    This observation also implies that disclosure is always insufficient when \(\alpha \le \frac{\delta }{2}\) unless \(D\) is zero or sufficiently large.

  15. 15.

    For a formal derivation of this condition, see the section “Maximization of \(W\left( \theta ^{s}\right) \) for \(\alpha >\delta \) and \(\mu \in \left[ 0,1\right] \)” of the “Appendix”.

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Acknowledgments

I would like to thank the editor Lawrence J. White for providing many helpful comments that greatly improved the article. I also thank Andrew Daughety, Jennifer Reinganum, and two anonymous referees for helpful comments. All errors are my own.

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Correspondence to Levent Celik.

Appendix

Appendix

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

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

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 \(\frac{6\left( \alpha -\delta \right) ^{2}}{\beta }<D\le \frac{3\delta ^{2}}{8\beta }\). \(\square \)

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Celik, L. A More General Framework to Analyze Whether Voluntary Disclosure is Insufficient or Excessive. Rev Ind Organ 44, 161–178 (2014). https://doi.org/10.1007/s11151-013-9409-5

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Keywords

  • Monopoly
  • Quality uncertainty
  • Verifiable information disclosure

JEL Classification

  • D82
  • D83
  • L12
  • L15