Voluntary versus mandatory disclosure

Abstract

We develop a theory of asymmetries between voluntary and mandatory disclosure. Efficiently designed mandatory disclosure policies are substitutes for excessive voluntary disclosures. The efficient policy takes the form of a lower threshold below which firms must disclose bad news and an upper threshold above which firms voluntarily disclose good news. Hence mandatory disclosures are asymmetric and feature conservative reporting of bad news. The threshold to recognize bad news increases when information is more precise. We also characterize interactions between disclosures and real decisions in environments where information has social value: investment decisions, optimal liquidations, and adverse selection in a lemons market.

Appendix

Proof of Lemma 1

Let $D_{m}^{*}$ be an efficient policy with $\tau \equiv \tau (D_{m}^{*})$ and λ ≡{x : x > τ}. Suppose that there exists $(a_{1},a_{2})\subset D_{m}^{*}$ and 𝜖 > 0, where $-\infty <a_{1}<a_{2}\leq \tau$ and $[a_{1}-\epsilon , a_{1}]\cap D_{m}^{*}=\emptyset$. Below, we construct an alternative disclosure set with a lower probability of disclosure.

Define 𝜖₁,𝜖₂ > 0 such that (i) 𝜖₁ < 𝜖, (ii) a₁ + 𝜖₂ < a₂, and (iii) F(a₁ + 𝜖₂) − F(a₁) = F(a₁) − F(a₁ − 𝜖₁) and an alternative mandatory disclosure set $D_{m}^{\prime }$ by $D_{m}^{\prime }=D_{m}^{*}\cup (a_{1}-\epsilon _{1},a_{1}) \setminus [a_{1},a_{1}+\epsilon _{2}]$. By construction, $\tau -c-\mathbb {E}(\tilde {x}|\tilde {x}\leq \tau ,\tilde {x}\notin D_{m}^{\prime })<\tau -c-\mathbb {E}(\tilde {x}|\tilde {x}\leq \tau ,\tilde {x}\notin D_{m}^{*})=0$. Moreover, $t-c-\mathbb {E}(\tilde {x}|\tilde {x}\leq t,\tilde {x}\notin D_{m}^{\prime })$ is positive for $t>\mathbb {E}(\tilde {x}|\tilde {x}\notin D_{m}^{\prime })+c$. Hence $\tau ^{\prime }\equiv \tau (D_{m}^{\prime })$ exists by intermediate value theorem, and it is true that $\tau ^{\prime }>\tau$. Denote $\lambda ^{\prime }\equiv \{x:x>\tau ^{\prime }\}$. It then follows that

$$ {\int}_{x\in D_{m}^{\prime}\cup \lambda^{\prime}} f(x)\mathrm{d}x={\int}_{x\in D_{m}^{*}\cup \lambda^{\prime}} f(x)\mathrm{d}x<{\int}_{x\in D_{m}^{*}\cup \lambda } f(x)\mathrm{d}x. $$

Therefore the set $D_{m}^{\prime }$ would feature lower expected disclosure costs than $D_{m}^{*}$, a contradiction. □

Proof of Proposition 1

Since $t-\mathbb E(\tilde x|\theta \leq \tilde x\leq t)$ is positive for sufficiently large t, it must be that

$$ \frac{\partial (t-\mathbb E(\tilde x|\theta\leq \tilde x\leq t))}{\partial t}|_{t=\tau_{\theta}}\geq 0. $$

For a particular $\theta \in \mathbb R$, there are two cases we need to consider.

Case 1

$\frac {\partial (t-\mathbb E(\tilde x|\theta \leq \tilde x\leq t))}{\partial t}|_{t=\tau _{\theta }}> 0$.

$$ \begin{array}{@{}rcl@{}} \frac{\partial (t-\mathbb E(\tilde x|\theta\leq \tilde x\leq t))}{\partial t}|_{t=\tau_{\theta}}&=&1-\frac{f(\tau_{\theta})\tau_{\theta}{\int}^{\tau_{\theta}}_{\theta} f(x)\mathrm{d}x-f(\tau_{\theta}){\int}^{\tau_{\theta}}_{\theta} xf(x)\mathrm{d}x }{({\int}_{\theta}^{\tau_{\theta}} f(x)\mathrm{d}x)^{2}}\\ &=&1-\frac{f(\tau_{\theta})}{{\int}_{\theta}^{\tau_{\theta}} f(x)\mathrm{d}x}\underbrace{(\tau_{\theta}-\frac{{\int}_{\theta}^{\tau_{\theta}} xf(x)\mathrm{d}x}{{\int}_{\theta}^{\tau_{\theta}} f(x)\mathrm{d}x})}_{=c \text{ (from (3.2))}} >0, \end{array} $$

which can be rewritten as F(τ_𝜃) − F(𝜃) − cf(τ_𝜃) > 0.

Applying the implicit function theorem on Γ₁(τ_𝜃,𝜃) = 0, it then follows that

$$ \begin{array}{@{}rcl@{}} \frac{\partial \tau_{\theta}}{\partial \theta}&=&-\frac{\partial {\Gamma}_{1}(t,\theta)}{\partial \theta}/\frac{\partial {\Gamma}_{1}(t,\theta)}{\partial t}|_{t=\tau_{\theta}}\\ &=&\frac{(\tau_{\theta}-\theta-c)f(\theta)}{F(\tau_{\theta})-F(\theta)-cf(\tau_{\theta})}> 0, \end{array} $$

(7.1)

because τ_𝜃 − c = P_𝜃(nd) > 𝜃.

Let $\mathcal C(\theta )$ denote the expected disclosure cost as a function of 𝜃:

$$ \begin{array}{@{}rcl@{}} \mathcal C(\theta)&=&F(\theta)c+(1-F(\tau_{\theta}))c;\\ \mathcal C^{\prime}(\theta)&=&\left( f(\theta)-\left( \frac{\partial \tau_{\theta}}{\partial \theta}\right)f(\tau_{\theta})\right)c \\ &=&\left( f(\theta)-\frac{(\tau_{\theta}-\theta-c)f(\theta)}{F(\tau_{\theta})-F(\theta)-cf(\tau_{\theta})}f(\tau_{\theta})\right)c \\ &=&f(\theta)c\left( \frac{F(\tau_{\theta})-F(\theta)-cf(\tau_{\theta})-(\tau_{\theta}-\theta-c)f(\tau_{\theta})}{F(\tau_{\theta})-F(\theta)-cf(\tau_{\theta})}\right) \\ &=&f(\theta)c\left( \frac{F(\tau_{\theta})-F(\theta)-(\tau_{\theta}-\theta)f(\tau_{\theta})}{F(\tau_{\theta})-F(\theta)-c f(\tau_{\theta})}\right). \end{array} $$

(7.2)

Because the denominator in the right-hand side of Eq. 7.2 is positive, the sign of $\mathcal C^{\prime }(\theta )$ is equal to the sign of F(τ_𝜃) − F(𝜃) − (τ_𝜃 − 𝜃)f(τ_𝜃). If τ_𝜃 is bounded below by $\underline \tau >-\infty$, it is obvious that F(τ_𝜃) − F(𝜃) − (τ_𝜃 − 𝜃)f(τ_𝜃) is negative for 𝜃 sufficiently small. If τ_𝜃 is not bounded from below, it is true that τ_𝜃 < m when 𝜃 is sufficiently small, which suggests that $F(\tau _{\theta })-F(\theta )\equiv {\int \limits }_{\theta }^{\tau _{\theta }}f(x)\mathrm d x<{\int \limits }_{\theta }^{\tau _{\theta }}f(\tau _{\theta })\mathrm d x\equiv (\tau _{\theta }-\theta )f(\tau _{\theta })$ for 𝜃 is sufficiently small.

Case 2

$\frac {\partial (t-\mathbb E(\tilde x|\theta \leq \tilde x\leq t))}{\partial t}|_{t=\tau _{\theta }}=0$.

In this case, $\frac {\partial {\Gamma }_{1}(t,\theta )}{\partial t}|_{t=\tau _{\theta }}= 0$, and the implicit function theorem fails. But we can consider the right-derivative:

$$ \begin{array}{@{}rcl@{}} \lim\limits_{\varepsilon \to 0^{+}} \frac{{\Gamma}_{1}(\tau_{\theta+\varepsilon},\theta+\varepsilon)-{\Gamma}_{1}(\tau_{\theta},\theta)}{\varepsilon}=0, \end{array} $$

which implies that

$$ \begin{array}{@{}rcl@{}} &&\lim\limits_{\varepsilon \to 0^{+}}\frac{{\Gamma}_{1}(\tau_{\theta+\varepsilon},\theta+\varepsilon)-{\Gamma}_{1}(\tau_{\theta},\theta)}{\varepsilon}\\ &=&\lim\limits_{\varepsilon \to 0^{+}}\frac{{\Gamma}_{1}(\tau_{\theta+\varepsilon},\theta+\varepsilon)-{\Gamma}_{1}(\tau_{\theta+\varepsilon},\theta)}{\varepsilon}+\lim\limits_{\varepsilon \to 0^{+}}\frac{{\Gamma}_{1}(\tau_{\theta+\varepsilon},\theta)-{\Gamma}_{1}(\tau_{\theta},\theta)}{\varepsilon}\\ &=&\underbrace{\frac{\partial {\Gamma}_{1}(t,\theta)}{\partial \theta}|_{t=\tau_{\theta}}}_{\in \mathbb R^{++}}+\left( \lim\limits_{\varepsilon \to 0^{+}}\frac{\tau_{\theta+\varepsilon}-\tau_{\theta}}{\varepsilon}\right)\underbrace{\left( \lim\limits_{\varepsilon \to 0^{+}}\frac{{\Gamma}_{1}(\tau_{\theta}+\epsilon,\theta)-{\Gamma}_{1}(\tau_{\theta},\theta)}{\varepsilon} \right)}_{= 0^{-}}=0. \end{array} $$

Hence it must be true that $\lim _{\varepsilon \to 0^{+}}\frac {\tau _{\theta +\varepsilon }-\tau _{\theta }}{\varepsilon }=\infty$, and

$$ \lim\limits_{\varepsilon \to 0^{+}} \frac{\mathcal C(\theta+\varepsilon)- \mathcal C(\theta)}{\varepsilon} = \left( f(\theta)-\left( \lim\limits_{\varepsilon \to 0^{+}}\frac{\tau_{\theta+\varepsilon}-\tau_{\theta}}{\varepsilon}\right)f(\tau_{\theta})\right)c=-\infty. $$

As a result, $\theta ^{*}=-\infty$ is not optimal.

With the c.d.f. strictly log-concave, $\frac {\partial (t-\mathbb E(\tilde x|\theta \leq x\leq t))}{\partial t}> 0$ for any $\theta \in \mathbb R,$ and τ_𝜃 is the unique solution to Eq. 3.2 (Bagnoli 2005). By Eq. 7.2, the mandatory disclosure threshold 𝜃^∗ must satisfy the following first-order condition:

$$ F(\tau_{\theta^{*}})-F(\theta^{*})-(\tau_{\theta^{*}}-\theta^{*})f(\tau_{\theta^{*}})=0. $$

(7.3)

From the mean value theorem, there exists $z^{*}\in (\theta ^{*},\tau _{\theta ^{*}})$ such that

$$ f(z^{*})=\frac{F(\tau_{\theta^{*}})-F(\theta^{*})}{\tau_{\theta^{*}}-\theta^{*}}=f(\tau_{\theta^{*}}). $$

(7.4)

Since the p.d.f. is single-peaked, z^∗ and $\tau _{\theta ^{*}}$ must be on the opposite sides of the mode m, i.e., $z^{*}<m<\tau _{\theta ^{*}}$, which implies: $\theta ^{*}<m<\tau _{\theta ^{*}}$ and

$$ f(\theta^{*})<f(\tau_{\theta^{*}}). $$

(7.5)

We prove next that the solution to Eq. 7.3 is unique. For any 𝜃 < m, define τ₂(𝜃) as a solution to Γ₂(t,𝜃) = 0 with 𝜃 < m < t and f(𝜃) < f(t), where

$${\Gamma}_{2}(t,\theta)\equiv F(t)-F(\theta)-(t-\theta)f(t).$$

Note that (i) F(m) − F(𝜃) − (m − 𝜃)f(m) < 0, because $F(m)-F(\theta )= {\int \limits }_{\theta }^{m}f(x)\mathrm d x<{\int \limits }_{\theta }^{m} f(m)\mathrm dx= (m-\theta )f(m)$, (ii) $\lim _{t\to \infty }{\Gamma }_{2}(t,\theta )=\lim _{t\to \infty }F(t)-F(\theta )-\lim _{t\to \infty }tf(t)+\lim _{t\to \infty }\theta f(t)=1-F(\theta )>0$, because $\lim _{t\to \infty }tf(t)= 0$, and (iii) $\partial {\Gamma }_{2}/\partial t=-(t-\theta )f^{\prime }(t)$ is strictly positive for t > m. Hence τ₂(𝜃) is unique and is a function from $(-\infty ,m)$ to $(m,\infty )$. Further, the voluntary threshold $\tau _{\theta ^{*}}$ must satisfy the cost-minimizing optimality condition τ₂(𝜃^∗), which implies that $\tau _{2}(\theta ^{*})=\tau _{\theta ^{*}}$. We know from Eq. 7.1 that τ_𝜃 is increasing in 𝜃 and, applying the implicit function theorem on Γ₂(τ₂(𝜃),𝜃) = 0,

$$\tau_{2}^{\prime}(\theta)=\frac{(\tau_{2}(\theta)-\theta)f^{\prime}(\tau_{2}(\theta))}{f(\tau_{2}(\theta))-f(\theta)}<0,\text{ for $\tau_{2}(\theta)>m>\theta$ and $f(\tau_{2}(\theta))>f(\theta)$.}$$

This implies that τ_𝜃 and τ₂(𝜃) cross once and hence the solution to $\tau _{\theta ^{*}}=\tau _{2}(\theta ^{*})$ is unique. □

Proof of Corollary 1

The fact that $\theta ^{*}<m<\tau _{\theta ^{*}}$ is established in the proof of Proposition 1. For the special case in which the distribution of $\tilde {x}$ is symmetric, $\tau _{\theta ^{*}}-m=m-z^{*}$, where z^∗ is given by Eq. 7.4. Thus $Prob(\tilde {x}< z^{*})=Prob(\tilde {x}> \tau _{\theta ^{*}})$, implying that the probability of mandatory disclosure $Prob(\tilde {x}< \theta ^{*})<Prob(\tilde {x}< z^{*})$ is lower than the probability of voluntary disclosure. □

Proof of Corollary 2

Applying the implicit function theorem on ${\Gamma }_{1}(\tau _{\theta ^{*}},\theta ^{*})=0$ and ${\Gamma }_{2}(\tau _{\theta ^{*}},\theta ^{*})=0$:

$$ \begin{array}{@{}rcl@{}} \left( \begin{array}{ccc} \frac{\partial\theta^{*}}{\partial c} \\ \frac{\partial\tau_{\theta^{*}}}{\partial c} \end{array} \right) &=&-\left( \begin{array}{ccc} \frac{\partial {\Gamma}_{1}(t_,\theta)}{\partial \theta}|_{\theta=\theta^{*},t=\tau_{\theta^{*}}} & \frac{\partial {\Gamma}_{1}(t_,\theta)}{\partial t}|_{\theta=\theta^{*},t=\tau_{\theta^{*}}} \\ \frac{\partial {\Gamma}_{2}(t_,\theta)}{\partial \theta}|_{\theta=\theta^{*},t=\tau_{\theta^{*}}} & \frac{\partial {\Gamma}_{2}(t,\theta)}{\partial t}|_{\theta=\theta^{*},t=\tau_{\theta^{*}}} \end{array} \right)^{-1}\left( \begin{array}{ccc} \frac{\partial{\Gamma}_{1}(t,\theta)}{\partial c}|_{\theta=\theta^{*},t=\tau_{\theta^{*}}} \\ \frac{\partial{\Gamma}_{2}(t,\theta)}{\partial c}|_{\theta=\theta^{*},t=\tau_{\theta^{*}}} \end{array} \right)\\ &=&-\left( \begin{array}{ccc} (\tau_{\theta^{*}}-\theta^{*}-c)f(\theta^{*}) & cf(\tau_{\theta^{*}})-(F(\tau_{\theta^{*}})-F(\theta^{*})) \\ -f(\theta^{*})+f(\tau_{\theta^{*}}) & -(\tau_{\theta^{*}}-\theta^{*})f^{\prime}(\tau_{\theta^{*}}) \end{array} \right)^{-1}\left( \begin{array}{ccc} {\Delta} \\ 0 \end{array} \right)\\ &=&\left( \begin{array}{ccc} \frac{F(\tau_{\theta^{*}})-F(\theta^{*})}{Z}(\tau_{\theta^{*}}-\theta^{*})f^{\prime}(\tau_{\theta^{*}}) \\ \frac{F(\tau_{\theta^{*}})-F(\theta^{*})}{Z}(f(\tau_{\theta^{*}})-f(\theta^{*})) \end{array} \right), \end{array} $$

where ${\Delta }=F(\tau _{\theta }^{*})-F(\theta ^{*})$ and

$$ Z=(f(\tau_{\theta^{*}})-f(\theta^{*}))(F(\tau_{\theta^{*}})-F(\theta^{*})-cf(\tau_{\theta^{*}}))+(\tau_{\theta^{*}}-\theta^{*})(c-\tau_{\theta^{*}}+\theta^{*})f(\theta^{*})f^{\prime}(\tau_{\theta^{*}}). $$

(7.6)

We know that (i) $f^{\prime }(\tau _{\theta ^{*}})<0$ from Corollary 1 and (ii) $f(\tau _{\theta ^{*}})-f(\theta ^{*})>0$ from Eq. 7.5 so that $Sign(\frac {\partial \tau _{\theta ^{*}}}{\partial c})=-Sign(\frac {\partial \theta ^{*}}{\partial c})=Sign(Z)$, and Z > 0 is implied by (iii) $\tau _{\theta ^{*}}-c>\theta ^{*}$ from Eq. 3.1, and (iv) $F(\tau _{\theta ^{*}})-F(\theta ^{*})-cf(\tau _{\theta ^{*}})>0$ from (iii) and Eq. 3.3. □

Proof of Corollary 3

Without loss of generality, we set μ = 0. Denote the p.d.f. and c.d.f. of the standard normal distribution as ϕ(⋅) and Φ(⋅) respectively. The probability of mandatory and voluntary disclosure can be written respectively as:

$$ \begin{array}{@{}rcl@{}} Prob(x<\theta^{*})&\equiv&{\Phi}\left( \frac{\theta^{*}}{\sigma}\right);\\ Prob(x>\tau_{\theta^{*}})&\equiv&1-{\Phi}\left( \frac{\tau_{\theta^{*}}}{\sigma}\right). \end{array} $$

We use the following change of variable on ${\Gamma }_{1}(\tau _{\theta ^{*}},\theta ^{*})=0$ and ${\Gamma }_{2}(\tau _{\theta ^{*}},\theta ^{*})=0$:

$$ \begin{array}{@{}rcl@{}} {\int}_{\theta^{*}/\sigma}^{\tau_{\theta^{*}}/\sigma} \! \phi\left( u\right)u \mathrm{d}u-\left( \frac{\tau_{\theta^{*}}}{\sigma}-\frac{c}{\sigma}\right){\int}_{\theta^{*}/\sigma}^{\tau_{\theta^{*}}/\sigma} \! \phi\left( u\right)\mathrm{d}u=0;\\ \\ {\Phi}\left( \frac{\tau_{\theta^{*}}}{\sigma}\right)-{\Phi}\left( \frac{\theta^{*}}{\sigma}\right)-\left( \frac{\tau_{\theta^{*}}}{\sigma}-\frac{\theta^{*}}{\sigma}\right)\phi\left( \frac{\tau_{\theta^{*}}}{\sigma}\right)=0. \end{array} $$

From the proof of Corollary 2, we have that $\frac {\theta ^{*}}{\sigma }$ decreases in $\frac {c}{\sigma }$ and $\frac {\tau _{\theta ^{*}}}{\sigma }$ increases in $\frac {c}{\sigma }$. Hence it proves that $\frac {\theta ^{*}}{\sigma }$ increases in σ and $\frac {\tau _{\theta ^{*}}}{\sigma }$ decreases in σ. □

Proof of Proposition 2

We prove this result in several steps.

Step 1::

We argue that $\tau ^{*}={\max \limits } \{\tau : \phi (\tau )-c-\phi (\mathbb {E}(\tilde {x}|\tilde {x}\leq \tau ,{\Theta }^{*}(\tilde {x})=1))=0\}$, i.e., τ^∗ is the maximal threshold that can be sustained in equilibrium. To see this, note that, for any two solutions $\tau ^{\prime \prime }>\tau ^{\prime }$ to $\phi (\tau )-c-\phi (\mathbb {E}(\tilde {x}|\tilde {x}\leq \tau ,{\Theta }^{*}(\tilde {x})=1))=0$, it holds that (i) $\phi (\mathbb {E}(\tilde {x}|\tilde {x}\leq \tau ^{\prime \prime },{\Theta }^{*}(\tilde {x})=1))>\phi (\mathbb {E}(\tilde {x}|\tilde {x}\leq \tau ^{\prime },{\Theta }^{*}(\tilde {x})=1))$, and (ii) $\phi (\mathbb {E}(\tilde {x}|\tilde {x}\leq \tau ^{\prime \prime },{\Theta }^{*}(\tilde {x})=1))\geq \phi (x)-c$ for any $x\in (\tau ^{\prime },\tau ^{\prime \prime }]$. This implies that the objective function is greater under $\tau ^{\prime \prime }$ than under $\tau ^{\prime }$.

Step 2::

In this step, we show that it is not optimal to disclose an interval adjacent to the left of τ^∗. By contradiction, suppose that Θ^∗(x) = 0 for x ∈ (τ^∗− ε,τ^∗) for some ε > 0. Set Θ^∗∗(x) ≡Θ^∗(x) except that Θ^∗∗(x) = 1 for x ∈ (τ^∗− ε,τ^∗∗), where τ^∗∗ solves $\phi (\tau ^{**})-c-\phi (\mathbb {E}(\tilde {x}|\tilde {x}\leq \tau ^{**},{\Theta }^{**}(\tilde {x})=1))=0$. Θ^∗∗(x) is more efficient than Θ^∗(x) because:

• (i) $\phi (\tau ^{*})-c-\phi (\mathbb {E}(\tilde {x}|\tilde {x}\leq \tau ^{*},{\Theta }^{**}(\tilde {x})=1))<0$ implies τ^∗∗ > τ^∗;

• (ii) $\phi (\mathbb {E}(\tilde {x}|\tilde {x}\leq \tau ^{**},{\Theta }^{**}(\tilde {x})=1))>\phi (\mathbb {E}(\tilde {x}|\tilde {x}\leq \tau ^{*},{\Theta }^{*}(\tilde {x})=1))$; and

• (iii) for any x ∈ (τ^∗− ε,τ^∗∗), $\phi (\mathbb {E}(\tilde {x}|\tilde {x}\leq \tau ^{**},{\Theta }^{**}(\tilde {x})=1))\geq \phi (x)-c$.

Step 3::

We show a necessary property of τ^∗.

From step 1, τ^∗ is the maximal solution to the constraint (4.1) at optimality. Also

$$ \lim\limits_{\tau\to+\infty} \phi(\tau)-c-\phi\left( \frac{{\int}_{-\infty}^{\tau} {\Theta}^{*}(x)xf(x)\mathrm{d}x}{{\int}_{-\infty}^{\tau} {\Theta}^{*}(x)f(x)\mathrm d x}\right)=+\infty. $$

Thus the following inequality holds:

$$ \frac{\partial \left( \phi(\tau)-c-\phi\left( \frac{{\int}_{-\infty}^{\tau} {\Theta}^{*}(x)xf(x)\mathrm{d}x}{{\int}_{-\infty}^{\tau} {\Theta}^{*}(x)f(x)\mathrm d x}\right)\right)}{\partial \tau}|_{\tau=\tau^{*}}>0. $$

Expanding the above inequality, we get

$$ \begin{array}{@{}rcl@{}} &&\frac{\partial \left( \phi(\tau)-c-\phi\left( \frac{{\int}_{-\infty}^{\tau} {\Theta}^{*}(x)xf(x)\mathrm{d}x}{{\int}_{-\infty}^{\tau} {\Theta}^{*}(x)f(x)\mathrm d x}\right)\right)}{\partial \tau}|_{\tau=\tau^{*}}\\ &=& \phi^{\prime}(\tau^{*})-\phi^{\prime}(x^{*}_{nd})\left( \frac{{\Theta}^{*}(\tau^{*})f(\tau^{*})}{{\int}_{-\infty}^{\tau^{*}}{\Theta}^{*}(x)f(x)\mathrm d x}\right)\left( \tau^{*}-\frac{{\int}_{-\infty}^{\tau^{*}}{\Theta}^{*}(x)xf(x)\mathrm d x}{{\int}_{-\infty}^{\tau^{*}}{\Theta}^{*}(x)f(x)\mathrm d x}\right)\\ &=&\phi^{\prime}(\tau^{*})-\phi^{\prime}(x^{*}_{nd})\left( \frac{{\Theta}^{*}(\tau^{*})f(\tau^{*})}{{\int}_{-\infty}^{\tau^{*}}{\Theta}^{*}(x)f(x)\mathrm d x}\right)(\tau^{*}-x^{*}_{nd})\\ &=&\frac{\phi^{\prime}(\tau^{*}){\int}_{-\infty}^{\tau^{*}}{\Theta}^{*}(x)f(x)\mathrm d x-\phi^{\prime}(x^{*}_{nd})f(\tau^{*})(\tau^{*}-x^{*}_{nd})}{{\int}_{-\infty}^{\tau^{*}}{\Theta}^{*}(x)f(x)\mathrm d x}>0. \end{array} $$

The above implies that τ^∗ satisfies

$$ \begin{array}{@{}rcl@{}} \phi^{\prime}(\tau^{*}){\int}_{-\infty}^{\tau^{*}}{\Theta}^{*}(x)f(x)\mathrm d x-\phi^{\prime}(x^{*}_{nd})f(\tau^{*})(\tau^{*}-x^{*}_{nd})>0. \end{array} $$

(7.7)

Step 4::

Taking the F.O.C.s of the Lagrangian, we get:

$$ \begin{array}{@{}rcl@{}} &&\frac{\partial \mathcal L}{\partial {\Theta}(x)}=f(x)\left( \phi(x^{*}_{nd})-\phi(x)+\gamma_{2} (x^{*}_{nd}-x)+c\right),\text{ for each \textit{x}}\leq \tau^{*};\\ &&\frac{\partial \mathcal L}{\partial \tau}|_{\tau=\tau^{*}}= \gamma_{1}\phi^{\prime}(\tau^{*})+\gamma_{2}f(\tau^{*})\left( x^{*}_{nd}-\tau^{*}\right)=0;\\ &&\frac{\partial \mathcal L}{\partial x_{nd}}|_{x_{nd}=x_{nd}^{*}}=\phi^{\prime}(x^{*}_{nd}){\int}_{-\infty}^{\tau^{*}} {\Theta}^{*}(x) f(x)\mathrm{d}x-\gamma_{1}\phi^{\prime}(x^{*}_{nd})+\gamma_{2} {\int}_{-\infty}^{\tau^{*}} {\Theta}^{*}(x)f(x)\mathrm{d}x=0. \end{array} $$

Define

$$ S\equiv\frac{\phi^{\prime}(\tau^{*}){\int}_{-\infty}^{\tau^{*}}{\Theta}^{*}(x)f(x)\mathrm d x \ }{\phi^{\prime}(\tau^{*}){\int}_{-\infty}^{\tau^{*}}{\Theta}^{*}(x)f(x)\mathrm d x-\phi^{\prime}(x^{*}_{nd})f(\tau^{*})(\tau^{*}-x^{*}_{nd})}>1, \text{ from inequality (7.7).} $$

Solving for γ₂, we get

$$ \begin{array}{@{}rcl@{}} \gamma_{2}=-S\phi^{\prime}(x^{*}_{nd}). \end{array} $$

Hence the sign of $\frac {\partial L}{\partial {\Theta }(x)}$ depends on the sign of

$$ \begin{array}{@{}rcl@{}} {\Sigma}(x)\equiv\phi(x^{*}_{nd})-\phi(x)-S \phi^{\prime}(x^{*}_{nd})(x^{*}_{nd}-x)+c. \end{array} $$

(7.8)

ϕ(⋅) is convex, i.e., $\phi ^{\prime \prime }(x)\geq 0$, which implies that:

$$ \begin{array}{@{}rcl@{}} \frac{\partial{\Sigma}(x)}{\partial x}&=&S\phi^{\prime}(x^{*}_{nd})-\phi^{\prime}(x)>0 \text{ for } x\leq x^{*}_{nd}, (\text{ from } S>1 \text{ and } \phi^{\prime\prime}(\cdot)\geq0);\\ \frac{\partial^{2}{\Sigma}(x)}{\partial x^{2}}&=&-\phi^{\prime\prime}(x)\leq 0. \end{array} $$

Hence, in this case, Σ(x) is concave and increasing for $x<x^{*}_{nd}\in \mathbb R$, with ${\Sigma }(x^{*}_{nd})=c>0$, which implies that there exists a $\theta ^{*}\in \mathbb (-\infty ,x^{*}_{nd})$ such that Σ(x) < 0 for x < 𝜃^∗. In step 2, we have shown that it is not optimal to disclose an interval adjacent to the left of $(\tau ^{*},+\infty )$.

It proves that an efficient policy features $D_{m}^{*}=(-\infty ,\theta ^{*})$, where $-\infty <\theta ^{*}<\tau ^{*}<\infty$. □

Proof of Lemma 2

By contradiction, suppose that there exists $\underline l>\overline l$ such that $(\underline l, \overline l)\subseteq (-\infty ,0)$ and $(\underline l, \overline l)\notin D_{m}^{*}$. Denote the set of of types liquidated as L. For any x < c, disclosure by mandate would induce liquidation. Also, nondisclosure price has to be positive if there is nondisclosure in equilibrium.

Case 1

The preferred equilibrium is such that (i) the voluntary disclosure set is $(c,\infty )$, and (ii) $L=(-\infty ,c)$. In this equilibrium, there is no nondisclosure on equilibrium path. Define $D_{m}^{\prime }\equiv (-\infty ,0)\cup D_{m}^{*}$. Then, given this $D^{\prime }_{m}$, those with x < 0 liquidate. In the preferred equilibrium, there is a positive nondisclosure price $P_{D_{m}^{\prime }}(nd)=\tau (D_{m}^{\prime })-c>0$ such that those with $x>\tau (D^{\prime }_{m})$ voluntarily disclose, those with $x\in [0,c]\cap D_{m}^{\prime }$ liquidate, and those with $x\in [0,\tau (D^{\prime }_{m})]\setminus D_{m}^{\prime }$ sell without disclosure at price $P_{D_{m}^{\prime }}(nd)=\mathbb E(\tilde {x}|\tilde {x}\notin D^{\prime }_{m},\tilde x\leq \tau (D^{\prime }_{m}))$. This equilibrium features weakly more trade for positive types (i.e., those in $[0,c]\setminus D_{m}^{*}$ now trade) and less disclosure (i.e., $\tau (D_{m}^{\prime })>c$). Hence $D_{m}^{\prime }$ is more efficient than $D^{*}_{m}$, a contradiction.

Case 2

The preferred equilibrium is such that there is a positive nondisclosure price $P_{D_{m}^{*}}(nd)=\tau (D_{m}^{*})-c>0$ such that (i) the voluntary disclosure set is $(\tau (D_{m}^{*}),\infty )$, where $\tau (D_{m}^{*})>c$, (ii) the liquidation set is $L\equiv (-\infty ,c)\cap D_{m}^{*}$, and (iii) others sell at $P_{D^{*}_{m}}(nd)$ without disclosure with the price being determined by $P_{D^{*}_{m}}(nd)=\mathbb E(\tilde {x}|\tilde {x}\notin D^{*}_{m},\tilde x\leq \tau (D^{*}_{m}))>0$. In this case, no sellers would voluntarily liquidate unless mandated to disclose for the reason that a positive nondisclosure price is always a better alternative to liquidation. Define $D_{m}^{\prime }\equiv (-\infty ,0)\cup D_{m}^{*}$. Given $D_{m}^{\prime }$, $x\in (\underline l, \overline l)$ would be liquidated, so are others with $x\in (-\infty ,0)$. Moreover, the preferred equilibrium, given this $D_{m}^{\prime }$, must feature a positive nondisclosure price $P_{D_{m}^{\prime }}(nd)>0$ such that $P_{D_{m}^{\prime }}(nd)>P_{D_{m}^{*}}(nd)$. Hence the policy $D_{m}^{\prime }$ features less trade for x < 0 and less disclosure, which makes it more efficient than $D_{m}^{*}$, a contradiction.

We have proved that $(-\infty ,0)\subseteq D_{m}^{*}$. For x₂ > x₁ ≥ 0, liquidating x₁ implies lower loss in trade efficiency than liquidating x₁. Hence the argument in the proof for lemma 1 holds here as well; that is, one can always construct a more efficient policy if the $D_{m}^{*}$ is not a threshold type.

□

Proof of Proposition 3

For 𝜃 > c, 𝜃 maximizes

$$ \begin{array}{@{}rcl@{}} &&\mathcal P_{1}(\theta)\equiv{\int}_{-\infty}^{c} 0f(x)\mathrm{d}x+{\int}_{c}^{\theta} (x-c) f(x)\mathrm{d}x+{\int}_{\theta}^{\tau_{\theta}} P_{\theta}(nd) f(x)\mathrm{d}x+{\int}_{\tau_{\theta}}^{+\infty} (x-c) g(x)\mathrm{d}x\\ &=& {\int}_{c}^{+\infty} x f(x)\mathrm{d}x-{\int}_{c}^{\theta} c f(x)\mathrm{d}x-{\int}_{\tau_{\theta}}^{+\infty} c f(x)\mathrm{d}x\\ &=& {\int}_{c}^{+\infty} x f(x)\mathrm{d}x + {\int}_{-\infty}^{c} c f(x)\mathrm{d}x - \underbrace{\left( {\int}_{-\infty}^{\theta} c f(x)\mathrm{d}x + {\int}_{\tau_{\theta}}^{+\infty} c f(x)\mathrm{d}x\right)}_{\mathcal C(\theta) \text{ in the proof of Proposition 1}}, \end{array} $$

where τ_𝜃 uniquely solves (3.2) and $P_{\theta }(nd)=\frac {{\int \limits }_{\theta }^{\tau _{\theta }} xf(x) \mathrm {d}x}{{\int \limits }_{\theta }^{\tau _{\theta }} f(x) \mathrm {d}x}$.

Hence, for 𝜃 > c, the objective function is equivalent to the objective function in the baseline, i.e., ¹⁷

$$ \mathcal P_{1}(\theta)\equiv Constant -\mathcal C(\theta). $$

From Eq. 7.2,

$$ \begin{array}{@{}rcl@{}} \mathcal P_{1}^{\prime}(\theta)&=&-\mathcal C^{\prime}(\theta)\\ &=&cf(\tau_{\theta})\left( \frac{\partial \tau_{\theta}}{\partial \theta}\right)-cf(\theta)\\ &=& cf(\tau_{\theta})\left( \frac{(\tau_{\theta}-\theta-c)f(\theta)}{F(\tau_{\theta})-F(\theta)-cf(\tau_{\theta})}\right)-cf(\theta)\\ & = &\frac{f(\theta)c((\tau_{\theta}-\theta)f(\tau_{\theta})-(F(\tau_{\theta})-F(\theta)))}{F(\tau_{\theta})-F(\theta)-c f(\tau_{\theta})}. \end{array} $$

For 0 ≤ 𝜃 ≤ c, 𝜃 should maximize

$$ \begin{array}{@{}rcl@{}} \mathcal P_{2}(\theta) &\equiv& {\int}_{-\infty}^{\theta} 0f(x)\mathrm{d}x+{\int}_{\theta}^{\tau_{\theta}} P_{\theta}(nd) f(x)\mathrm{d}x+{\int}_{\tau_{\theta}}^{+\infty} (x-c)f(x)\mathrm{d}x\\ &=& {\int}_{\theta}^{\tau_{\theta}} xf(x) \mathrm{d}x+{\int}_{\tau_{\theta}}^{+\infty} (x-c)f(x)\mathrm{d}x\\ &=& {\int}_{\theta}^{+\infty} xf(x) \mathrm{d}x - {\int}_{\tau_{\theta}}^{+\infty} cf(x)\mathrm{d}x, \end{array} $$

where τ_𝜃 uniquely solves (3.2) and $P_{\theta }(nd)=\frac {{\int \limits }_{\theta }^{\tau _{\theta }} xf(x) \mathrm {d}x}{{\int \limits }_{\theta }^{\tau _{\theta }} f(x) \mathrm {d}x}$.

Taking the first-order derivative of the above objective function, we get

$$ \begin{array}{@{}rcl@{}} \mathcal P_{2}^{\prime}(\theta) &=& cf(\tau_{\theta})\left( \frac{\partial \tau_{\theta}}{\partial \theta}\right)-\theta f(\theta)\\ &=& c f(\tau_{\theta}) \left( \frac{(\tau_{\theta}-\theta-c)f(\theta)}{F(\tau_{\theta})-F(\theta)-cf(\tau_{\theta})}\right)-\theta f(\theta) \text{ (from (7.1))}\\ &=& \frac{f(\theta)(c(\tau_{\theta}-c)f(\tau_{\theta})-\theta (F(\tau_{\theta})-F(\theta))}{F(\tau_{\theta})-F(\theta)-cf(\tau_{\theta})}. \end{array} $$

From the proof in Proposition 1, we have shown that F(τ_𝜃) − F(𝜃) − cf(τ_𝜃) > 0. Hence

$$ \mathcal P_{2}^{\prime}(0)= \frac{f(0)c(\tau_{0}-c)f(\tau_{0})}{F(\tau_{0})-F(0)-cf(\tau_{0})}>0 \ \text{(because $\tau_{0}-c>0$)}, \text{which proves that $0<k^{*}\leq \theta^{*}$.} $$

An immediate observation is that $\mathcal P_{1}(c)=\mathcal P_{2}(c)$ and $\mathcal P_{1}^{\prime }(c)=\mathcal P_{2}^{\prime }(c)$. Further

$$ P_{1}^{\prime}(c)=\mathcal P_{2}^{\prime}(c)=f(c)c\left( \frac{\chi(c)}{F(\tau_{c})-F(c)-cf(\tau_{c})}\right), $$

where

$$\chi(c)\equiv (\tau_{c}-c)f(\tau_{c})-(F(\tau_{c})-F(c)).$$

Case 1

χ(c) > 0: We know from the proof of uniqueness in Proposition 1 that

$$ \begin{array}{@{}rcl@{}} \mathcal P_{1}^{\prime}(\theta)=cf(\tau_{\theta})\left( \frac{\partial \tau_{\theta}}{\partial \theta}\right)-cf(\theta)>0 \text{, for $\theta \leq c$.} \end{array} $$

This implies that

$$ \begin{array}{@{}rcl@{}} \mathcal P_{2}^{\prime}(\theta)=cf(\tau_{\theta})\left( \frac{\partial \tau_{\theta}}{\partial \theta}\right)-\theta f(\theta)\geq cf(\tau_{\theta})\left( \frac{\partial \tau_{\theta}}{\partial \theta}\right)-c f(\theta)>0\text{, for $\theta \leq c$.} \end{array} $$

In this case, the solution to 𝜃^∗ > c is given by Eq. 3.3.

Case 2

χ(c) < 0: We know from the proof of uniqueness in Proposition 1 that

$$ \begin{array}{@{}rcl@{}} \mathcal P_{1}^{\prime}(\theta)=cf(\tau_{\theta})\left( \frac{\partial \tau_{\theta}}{\partial \theta}\right)-cf(\theta)<0 \text{, for $\theta \geq c$.} \end{array} $$

This implies that

$$ \begin{array}{@{}rcl@{}} P_{2}^{\prime}(\theta)=cf(\tau_{\theta})\left( \frac{\partial \tau_{\theta}}{\partial \theta}\right)-\theta f(\theta)\leq cf(\tau_{\theta})\left( \frac{\partial \tau_{\theta}}{\partial \theta}\right)-c f(\theta)<0\text{, for $\theta \geq c$.} \end{array} $$

In this case, the solution to 𝜃^∗ < c is given by Eq. 4.3. We have shown that $0=\mathcal P_{2}^{\prime }(\theta ^{*})>\mathcal P_{1}^{\prime }(\theta ^{*})$ because 𝜃^∗ < c. Hence 𝜃^∗ in this case is greater than the baseline mandatory disclosure threshold.

From Corollary 2, we have that χ(c) ≥ (≤)0 when c ≤ (≥)c₀ where c₀ uniquely solves χ(c₀) = 0.

□

Proof of Proposition 4

The proof that the efficient policy is a threshold type of rule follows along the lines of Lemma 1.

Λ(k,𝜃) is positive for sufficiently large k:

$$ \begin{array}{@{}rcl@{}} &&\frac{\partial {\Lambda}(k,\theta)}{\partial k}|_{k=\kappa(\theta)}\\ &=&\delta-\frac{\kappa(\theta)f(\kappa(\theta)){\int}_{\theta}^{\kappa(\theta)}f(x)\mathrm dx-{\int}_{\theta}^{\kappa(\theta)}xf(x)\mathrm dx f(\kappa(\theta))}{\left( {\int}_{\theta}^{_{\kappa(\theta)}}f(x)\mathrm dx \right)^{2}}\\ &=&\delta-\frac{\kappa(\theta)f(\kappa(\theta))-\left( \frac{{\int}_{\theta}^{\kappa(\theta)}xf(x)\mathrm dx}{{\int}_{\theta}^{\kappa(\theta)}f(x)\mathrm dx}\right)f(\kappa(\theta))}{{\int}_{\theta}^{_{\kappa(\theta)}}f(x)\mathrm dx}\\ &=&\delta - \frac{\kappa(\theta)f(\kappa(\theta))-\delta \kappa(\theta)f(\kappa(\theta))}{{\int}_{\theta}^{_{\kappa(\theta)}}f(x)\mathrm dx}\\ &=&\frac{\delta(F(\kappa (\theta))-F(\theta))-f(\kappa(\theta))\kappa(\theta)(1-\delta)}{F(\kappa (\theta))-F(\theta)}>0. \end{array} $$

Hence we can conclude that δ(F(κ(𝜃)) − F(𝜃)) − f(κ(𝜃))κ(𝜃)(1 − δ) > 0. Applying the implicit function theorem on Λ(κ(𝜃),𝜃) = 0, we get

$$ \frac{\partial \kappa (\theta)}{\partial \theta}=-\frac{\partial {\Lambda}/\partial \theta}{\partial {\Lambda}/\partial k}|_{k=\kappa(\theta)}=\frac{f(\theta)(\delta \kappa(\theta)-\theta )}{\delta(F(\kappa (\theta))-F(\theta))-f(\kappa(\theta))\kappa(\theta)(1-\delta)}> 0, $$

because $\delta \kappa (\theta )=\mathbb {E}(\tilde x| \tilde x\in [\theta ,\kappa (\theta )])>\theta$.

First, if κ(𝜃) < c/(1 − δ), those with x < 𝜃 would retain the asset to avoid disclosure because 𝜃 < κ(𝜃) < c/(1 − δ). Those with x ∈ [𝜃,κ(𝜃)] would sell without disclosure, those with x ∈ (κ(𝜃),c/(1 − δ)] would retain the asset, and those with x > c/(1 − δ) would sell with voluntary disclosure.

The objective function in this case is

$$ \mathcal{W}_{3}(\theta) = {\int}_{0}^{\theta}f(x)\delta x \mathrm dx + {\int}_{\theta}^{\kappa(\theta)}xf(x)\mathrm dx + {\int}_{\kappa(\theta)}^{c/(1-\delta)}f(x)\delta x \mathrm dx+ {\int}_{c/(1-\delta)}^{+\infty}f(x)(x-c) \mathrm dx. $$

Taking the first-order condition yields

$$ \begin{array}{@{}rcl@{}} \mathcal{W}_{3}^{\prime}(\theta) &=& f(\theta)\delta \theta +\kappa(\theta)f(\kappa(\theta))\frac{\partial \kappa (\theta)}{\partial \theta} - \theta f(\theta)-f(\kappa(\theta))\delta \kappa(\theta)\frac{\partial \kappa (\theta)}{\partial \theta},\\ &=& (1-\delta)\kappa(\theta)f(\kappa(\theta))\frac{\partial \kappa (\theta)}{\partial \theta}-(1-\delta)f(\theta) \theta ,\\ &=& \left( \frac{(1-\delta)\kappa(\theta)f(\kappa(\theta))f(\theta)(\delta \kappa(\theta)-\theta )}{\delta(F(\kappa (\theta))-F(\theta))-f(\kappa(\theta))\kappa(\theta)(1-\delta)}\right)-(1-\delta)f(\theta) \theta,\\ &=&(1-\delta)\left( \left( \frac{\kappa(\theta)f(\kappa(\theta))f(\theta)(\delta \kappa(\theta)-\theta )}{\delta(F(\kappa (\theta))-F(\theta))-f(\kappa(\theta))\kappa(\theta)(1-\delta)}\right)-f(\theta) \theta \right), \\ &=&\left( \frac{\kappa(\theta)f(\kappa(\theta))(\kappa(\theta)-\theta )-\theta(F(\kappa (\theta))-F(\theta))}{\delta(F(\kappa (\theta))-F(\theta))-f(\kappa(\theta))\kappa(\theta)(1-\delta)}\right)f(\theta)(1-\delta)\delta. \end{array} $$

Evaluating the numerator at 𝜃 = 0 yields

$$ \begin{array}{@{}rcl@{}} &&\kappa(0)f(\kappa(0))(\kappa(0)-0)-0(F(\kappa (0))-F(0))\\ &=&\kappa(0)f(\kappa(0)) \kappa(0)>0. \end{array} $$

We denote the 𝜃 that solves the first order condition as 𝜃₃; that is,

$$ \kappa(\theta_{3})f(\kappa(\theta_{3}))(\kappa(\theta_{3})-\theta_{3} )-\theta_{3}(F(\kappa (\theta_{3}))-F(\theta_{3}))=0. $$

Second, if 𝜃 < c/(1 − δ) but κ(𝜃) > c/(1 − δ), the firm in the middle sell without disclosure. The objective function in this case is

$$ \begin{array}{@{}rcl@{}} \mathcal{W}_{2}(\theta)&=& {\int}_{0}^{\theta}f(x)\delta x \mathrm dx + {\int}_{\theta}^{\tau_{\theta}}xf(x)\mathrm dx + {\int}_{\tau_{\theta}}^{+\infty}f(x)(x-c) \mathrm dx, \end{array} $$

where τ_𝜃 is given by Eq. 3.2.

Taking the first-order condition yields

$$ \begin{array}{@{}rcl@{}} \mathcal{W}^{\prime}_{2}(\theta)&=&f(\theta)\delta \theta +\tau_{\theta} f(\tau_{\theta})\frac{\partial\tau_{\theta}}{\partial \theta}-\theta f(\theta)-f(\tau_{\theta})(\tau_{\theta}-c)\frac{\partial\tau_{\theta}}{\partial \theta},\\ &=& f(\theta)\delta \theta+cf(\tau_{\theta})\left( \frac{(\tau_{\theta}-\theta-c)f(\theta)}{F(\tau_{\theta})-F(\theta)-cf(\tau_{\theta})}\right)-\theta f(\theta),\\ &=&f(\theta)\frac{cf(\tau_{\theta})(\tau_{\theta}-c)-\theta(F(\tau_{\theta})-F(\theta))+\delta\theta(F(\tau_{\theta})-F(\theta)-cf(\tau_{\theta})) }{F(\tau_{\theta})-F(\theta)-cf(\tau_{\theta})},\\ &=&f(\theta)\frac{cf(\tau_{\theta})(\tau_{\theta}-c-\delta\theta)-(1-\delta)\theta(F(\tau_{\theta})-F(\theta)) }{F(\tau_{\theta})-F(\theta)-cf(\tau_{\theta})}. \end{array} $$

Evaluating the numerator at 𝜃 = 0:

$$ \begin{array}{@{}rcl@{}} &&cf(\tau_{0})(\tau_{0}-c-\delta0)-(1-\delta)0(F(\tau_{0})-F(0)),\\ &=& cf(\tau_{0})(\tau_{0}-c)>0. \end{array} $$

We denote 𝜃 that solves the first-order condition as 𝜃₂; that is,

$$ cf(\tau_{\theta_{2}})(\tau_{\theta_{2}}-c-\delta\theta_{2})-(1-\delta)\theta_{2}(F(\tau_{\theta_{2}})-F(\theta_{2}))=0. $$

The third case to consider is when 𝜃 > c/(1 − δ). The objective function is

$$ \mathcal W_{1}(\theta)={\int}_{0}^{c/(1-\delta)}f(x)\delta x \mathrm dx +{\int}_{c/(1-\delta)}^{\theta}f(x)(x-c) \mathrm dx +{\int}_{\theta}^{\tau_{\theta}}xf(x)\mathrm dx + {\int}_{\tau_{\theta}}^{+\infty}f(x)(x-c) \mathrm dx. $$

The first-order condition is the same as the baseline model:

$$ \begin{array}{@{}rcl@{}} \mathcal W_{1}^{\prime}(\theta) & = &f(\theta)\frac{c((\tau_{\theta}-\theta)f(\tau_{\theta})-(F(\tau_{\theta})-F(\theta)))}{F(\tau_{\theta})-F(\theta)-c f(\tau_{\theta})}. \end{array} $$

Denote the solution to the first-order condition as 𝜃₁.

It can be easily shown that $\mathcal W_{2}(c/(1- \delta ))=\mathcal W_{1}(c/(1- \delta ))$. Further,

$$ \begin{array}{@{}rcl@{}} && \mathcal W_{2}^{\prime}(c/(1- \delta))=\mathcal W_{1}^{\prime}(c/(1- \delta))\\ &=& f(c/(1-\delta))\frac{\psi(c/(1-\delta))}{F(\tau_{c/(1-\delta)})-F(c/(1-\delta))-c f(\tau_{c/(1-\delta)})}, \end{array} $$

where

$$ \psi (c/(1-\delta))\equiv c((\tau_{c/(1-\delta)}-c/(1-\delta))f(\tau_{c/(1-\delta)})-(F(\tau_{c/(1-\delta)})-F(c/(1-\delta)))). $$

Case 1

ψ(c/(1 − δ)) > 0. We know from the proof of uniqueness in Proposition 1 that $\mathcal W_{1}^{\prime }(\theta ) = f(\theta )\frac {c((\tau _{\theta }-\theta )f(\tau _{\theta })-(F(\tau _{\theta })-F(\theta )))}{F(\tau _{\theta })-F(\theta )-c f(\tau _{\theta })}>0$ for 𝜃 ≤ c/(1 − δ). This implies that

$$ \begin{array}{@{}rcl@{}} \mathcal W_{2}^{\prime}(\theta)\geq W_{1}^{\prime}(\theta)>0 \text{ for $\theta \leq c/(1-\delta)$}. \end{array} $$

Consider the numerator of $\mathcal W_{3}^{\prime }(\theta )$ for 𝜃 < κ(𝜃) ≤ c/(1 − δ):

$$ \begin{array}{@{}rcl@{}} &&\kappa(\theta)f(\kappa(\theta))(\kappa(\theta)-\theta )-\theta(F(\kappa (\theta))-F(\theta))\\ &&>\theta \left( f(\kappa(\theta))(\kappa(\theta)-\theta )-(F(\kappa (\theta))-F(\theta))\right). \end{array} $$

From corollary 1, c/(1 − δ) < 𝜃₁ < m, which implies that 𝜃 < κ(𝜃) < m. Hence f(κ(𝜃))(κ(𝜃) − 𝜃) − (F(κ(𝜃)) − F(𝜃)) > 0; that is, $\mathcal W_{3}^{\prime }(\theta )>0$ for 𝜃 < κ(𝜃) ≤ c/(1 − δ).

This proves that the optimal threshold is the baseline one; that is, 𝜃^∗ = 𝜃₁.

Case 2

ψ(c/(1 − δ)) < 0. We know from the proof of uniqueness in Proposition 1 that $\mathcal W_{1}^{\prime }(\theta ) = f(\theta )\frac {c((\tau _{\theta }-\theta )f(\tau _{\theta })-(F(\tau _{\theta })-F(\theta )))}{F(\tau _{\theta })-F(\theta )-c f(\tau _{\theta })}<0$ for 𝜃 ≥ c/(1 − δ). This implies that

$$ \begin{array}{@{}rcl@{}} \mathcal W_{2}^{\prime}(\theta)\leq W_{1}^{\prime}(\theta)<0 \text{ for $\theta\geq c/(1-\delta)$}. \end{array} $$

In this case, the optimal threshold is 𝜃^∗ < c/(1 − δ).

When m > 0, from the proof of corollary 2, ψ(c/(1 − δ)) > (<)0 when c < (>)c₁, where c₁ solves ψ(c₁/(1 − δ)) = 0. When m = 0, it must be true that ψ(c/(1 − δ)) < 0.

□

Proof of Proposition 5

As in the benchmark, there is a unique equilibrium for any 𝜃, because one can think of this problem as the original problem with disclosure cost equal to [c − h(𝜃)] > 0. It then follows from the implicit function theorem that

$$ \begin{array}{@{}rcl@{}} \frac{\partial \tau_{\theta}}{\partial \theta} &=&\frac{(\tau_{\theta}-\theta-(c-h(\theta)))f(\theta)-h^{\prime}(\theta)(F(\tau_{\theta})-F(\theta))}{F(\tau_{\theta})-F(\theta)-(c-h(\theta))f(\tau_{\theta})}. \end{array} $$

(7.9)

The denominator is positive from log-concavity. The sign of the numerator is undetermined. As 𝜃 increases, the net cost of disclosure decreases, which may yield more voluntary disclosure.

Then the first-order condition is

$$ \begin{array}{@{}rcl@{}} \frac{\partial \mathcal C(\theta)}{\partial \theta}&=&c\left( f(\theta)-\frac{\partial \tau_{\theta}}{\partial \theta}f(\tau_{\theta}) \right) + h^{\prime}(\theta)(F(\tau_{\theta})-F(\theta)) +h(\theta) \left( \frac{\partial \tau_{\theta}}{\partial \theta}f(\tau_{\theta}) -f(\theta) \right),\\ &=&(c-h(\theta))\left( f(\theta)-\frac{\partial \tau_{\theta}}{\partial \theta}f(\tau_{\theta}) \right) + h^{\prime}(\theta)(F(\tau_{\theta})-F(\theta)). \end{array} $$

Since $\lim _{\theta \to -\infty } h(\theta )= 0$ and $\lim _{\theta \to -\infty } h^{\prime }(\theta )= 0$, the first-order condition converges to that in the benchmark as $\theta \to -\infty$. Hence $\theta =-\infty$ is never optimal. □

Proof of Proposition 6

It can be easily checked that Lemma 1 applies. We prove the if and only if parts separately.

If As shown in the proof of Proposition 1, the sign of the F.O.C. for minimizing the expected disclosure cost when evaluated at $\theta =\underline x$ is determined by the sign of $[F(\tau _{\underline {x}})-F(\underline {x})-(\tau _{\underline {x}}-\underline {x})f(\tau _{\underline {x}})]$. The set $\{(F(\cdot ),c)|F(\tau _{\underline {x}})-F(\underline {x})-(\tau _{\underline {x}}-\underline {x})f(\tau _{\underline {x}})<0\}$ is not empty, as shown below.

Assuming $\underline {x}<m<+\infty$, $F(\tau _{\underline {x}})-F(\underline {x})-(\tau _{\underline {x}}-\underline {x})f(\tau _{\underline {x}})<0$ if $\underline {x}<\tau _{\underline {x}}\leq m$. If $\tau _{\underline {x}}>m$, $F(\tau _{\underline {x}})-F(\underline {x})-(\tau _{\underline {x}}-\underline {x})f(\tau _{\underline {x}})$ is increasing in $\tau _{\underline {x}}$, i.e., $\frac {\partial \left (F(\tau _{\underline {x}})-F(\underline {x})-(\tau _{\underline {x}}-\underline {x})f(\tau _{\underline {x}})\right ) }{\partial \tau _{\underline {x}}}=-(\tau _{\underline x}-\underline {x})f^{\prime }(\tau _{\underline {x}})>0$. Also $\lim _{\tau _{\underline {x}\to +\infty }}F(\tau _{\underline {x}})-F(\underline {x})+(\underline {x}-\tau _{\underline {x}})f(\tau _{\underline {x}})=F(+\infty )-F(\underline {x})+(\underline {x}-\infty )f(+\infty )=1-F(\underline {x})>0$. By intermediate value theorem, the exists a unique $\hat \tau _{\underline {x}}> m$ such that $F(\tau _{\underline {x}})-F(\underline {x})-(\tau _{\underline {x}}-\underline {x})f(\tau _{\underline {x}})\geq (\leq )0$ if $\tau _{\underline x}\geq (\leq ) \hat \tau _{\underline x}$. Since $\tau _{\underline {x}}$ increases in c, there exists a unique $\overline c>0$ such that $F(\tau _{\underline {x}})-F(\underline {x})-(\tau _{\underline {x}}-\underline {x})f(\tau _{\underline {x}})<0$ if (and only if) $c<\overline c$.

$F(\tau _{\underline {x}})-F(\underline {x})-(\tau _{\underline {x}}-\underline {x})f(\tau _{\underline {x}})<0$ implies that $\underline x<m<+\infty$. The proof of uniqueness is identical to the uniqueness proof in Proposition 1, with the the support replaced by $[\underline x,+\infty )$.

Only if $F(\tau _{\underline {x}})-F(\underline {x})-(\tau _{\underline {x}}-\underline {x})f(\tau _{\underline {x}})\geq 0$ contradicts the fact that there exists a unique solution to Eq. 3.3 in $[\underline x,+\infty )$. □