Biased voluntary disclosure

Abstract

We provide a bridge between the voluntary disclosure and the earnings management literature. Voluntary disclosure models focus on managers’ discretion in deciding whether or not to provide truthful voluntary disclosure to the capital market. Earnings management models, on the other hand, concentrate on managers’ discretion in deciding how to bias their mandatory disclosure. By analyzing managers’ disclosure strategy when disclosure is voluntary and not necessarily truthful, we show the robustness of voluntary disclosure theory to the relaxation of the standard assumption of truthful reporting. We also demonstrate the sensitivity of earnings management theory to the commonly made mandatory disclosure assumption.

Appendix: Proofs

The appendix presents the proofs of Observation 1, Observation 2, Proposition 3, Corollary 4, and Corollary 5. The proofs are based on two lemmata that are stated and proved below.

Lemma A

Let $F(\pi ,x) = E[\tilde{v}\left| {\tilde{s} = x} \right.] - E \, [\tilde{v}\left| { \, \tilde{s} \le x} \right.] + \frac{1 - \pi }{{1 - \pi \,{\text{prob}}(\tilde{s} \ge x)}}(E \, [\tilde{v}\left| { \, \tilde{s} \le x} \right.] - \mu )$ for any $\pi \in (0,1]$ and $x \in \Re$. Then, $F(\pi ,x)$ is increasing in x and in π, where $\mathop {\lim }\limits_{x \to - \infty } F(\pi ,x) = - \infty$ for any $\pi \in (0,1)$ , $\mathop {\lim }\limits_{x \to - \infty } F(1,x) = 0$ and $\mathop {\lim }\limits_{x \to + \infty } F(\pi ,x) = + \infty$ for any $\pi \in (0,1]$.

Proof of the Lemma A

Utilizing the properties of the normal distribution, and using the notation $\hat{x} = \frac{x - \mu }{{\sqrt {\sigma^{2} + \sigma_{\varepsilon }^{2} } }}$, we can substitute $E[\tilde{v}\left| {\tilde{s} = x} \right.] = \mu + \frac{{\sigma^{2} }}{{\sqrt {\sigma^{2} + \sigma_{\varepsilon }^{2} } }} \cdot \hat{x}$, $E \, [\tilde{v}\left| { \, \tilde{s} \le x} \right.] = \mu - \frac{{\sigma^{2} }}{{\sqrt {\sigma^{2} + \sigma_{\varepsilon }^{2} } }} \cdot \frac{{z(\hat{x})}}{{\Upphi (\hat{x})}}$ and get $F(\pi ,x) = \frac{{\sigma^{2} }}{{\sqrt {\sigma^{2} + \sigma_{\varepsilon }^{2} } }}\left( {\hat{x} + \frac{{\pi \, z(\hat{x})}}{{1 - \pi (1 - \, \Upphi (\hat{\tilde{x}}))}}} \right)$, where z and Φ are, respectively, the probability density function and the cumulative distribution function for a standard normal variable. Using $z^{\prime} (\hat{x}) = - \hat{x}z(\hat{x})$ and $\Upphi^{\prime} (\hat{x}) = z(\hat{x})$, we get $\frac{\partial F}{\partial x} = \frac{{\sigma^{2} }}{{\sqrt {\sigma^{2} + \sigma_{\varepsilon }^{2} } }}\left( { \, 1 - \frac{{\pi \, z(\hat{x})}}{{1 - \pi + \pi \, \Upphi (\hat{x})}}\left( {\hat{x} + \frac{{\pi \, z(\hat{x})}}{{1 - \pi + \pi \, \Upphi (\hat{x})}}} \right)} \right)$. It thus follows that $\frac{\partial F}{\partial x} > 0$ for any $x \in \Re$ such that $x + \frac{{\pi \, z(\hat{x})}}{{1 - \pi + \pi \, \Upphi (\hat{x})}} \le 0$. It also follows that $\frac{\partial F}{\partial x} > \frac{{\sigma^{2} }}{{\sqrt {\sigma^{2} + \sigma_{\varepsilon }^{2} } }}\,\left( {1 - \frac{{z(\hat{x})}}{{\Upphi (\hat{x})}}\left( {\hat{x} + \frac{{z(\hat{x})}}{{\Upphi (\hat{x})}}} \right)} \right)$ for $x + \frac{{\pi \, z(\hat{x})}}{{1 - \pi + \pi \, \Upphi (\hat{x})}} > 0$, so using Sampford’s (1953) inequality, $0 < \frac{{z(\hat{x})}}{{\Upphi (\hat{x})}}\left( {\hat{x} + \frac{{z(\hat{x})}}{{\Upphi (\hat{x})}}} \right) < 1$, we get $\frac{\partial F}{\partial x} > 0$. Also, $\frac{\partial F}{\partial \pi } = \, \frac{{\sigma^{2} }}{{\sqrt {\sigma^{2} + \sigma_{\varepsilon }^{2} } }} \cdot \frac{{z(\hat{x})}}{{(1 - \pi + \pi \, \Upphi (\hat{x}))^{2} }} > 0$. Using $\mathop {\lim }\limits_{{\hat{x} \to - \infty }} z(\hat{x}) = \mathop {\lim }\limits_{{\hat{x} \to + \infty }} z(\hat{x}) = \mathop {\lim }\limits_{{\hat{x} \to - \infty }} \Upphi (\hat{x}) = 0$ and $\mathop {\lim }\limits_{{\hat{x} \to + \infty }} \Upphi (\hat{x}) = 1$, we get $\mathop {\lim }\limits_{x \to - \infty } F(\pi ,x) = - \infty$ for any $\pi \in (0,1)$ and $\mathop {\lim }\limits_{x \to + \infty } F(\pi ,x) = \infty$ for any $\pi \in (0,1]$. Lastly, using l’Hopital’s rule repeatedly, we get $\mathop {\lim }\limits_{x \to - \infty } x\Upphi (x) = \mathop {\lim }\limits_{x \to - \infty } \frac{\Upphi (x)}{{{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle x$}}}} = \mathop {\lim }\limits_{x \to - \infty } \frac{z(x)}{{ - {\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle {x^{2} }$}}}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{x^{2} }}{{ - {\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle {z(x)}$}}}} = \mathop {\lim }\limits_{x \to - \infty } \frac{2x}{{{\raise0.5ex\hbox{$\scriptstyle {z^{\prime}(x)}$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle {z^{2} (x)}$}}}} = \mathop {\lim }\limits_{x \to - \infty } - 2z(x) = 0$ and $\mathop {\lim }\limits_{x \to - \infty } \frac{x\Upphi (x) + z(x)}{\Upphi (x)} = \mathop {\lim }\limits_{x \to - \infty } \frac{\Upphi (x) + xz(x) + z^{\prime} (x)}{z(x)} = \mathop {\lim }\limits_{x \to - \infty } \frac{\Upphi (x)}{z(x)} = \mathop {\lim }\limits_{x \to - \infty } \frac{z(x)}{{z^{\prime}(x)}} = \mathop {\lim }\limits_{x \to - \infty } - \frac{1}{x} = 0$, so $\mathop {\lim }\limits_{x \to - \infty } F(1,x) = \frac{{\sigma^{2} }}{{\sqrt {\sigma^{2} + \sigma_{\varepsilon }^{2} } }}\mathop {\lim }\limits_{x \to - \infty } \frac{x\Upphi (x) + z(x)}{\Upphi (x)} = 0$. □

Lemma B

Let $G(x) = - x - \left( {2c\frac{{\sigma^{2} + \sigma_{\varepsilon }^{2} }}{{\sigma^{2} }}} \right)^{ - 1} \ln (1 - 2c\frac{{\sigma^{2} + \sigma_{\varepsilon }^{2} }}{{\sigma^{2} }}x)$ for any $x < \frac{{\sigma^{2} }}{{2c(\sigma^{2} + \sigma_{\varepsilon }^{2} )}}$. Then, G(x) is decreasing in x for negative values of x, reaching a minimum of zero at x = 0, and then increasing in x for positive values of x, where $\mathop {\lim }\limits_{{x \to \frac{{\sigma^{2} }}{{2c(\sigma^{2} + \sigma_{\varepsilon }^{2} )}}}} G(x) = + \infty$.

Proof of the Lemma B

The function G(x) is defined only for $x < \frac{{\sigma^{2} }}{{2c(\sigma^{2} + \sigma_{\varepsilon }^{2} )}}$. Also, $\frac{dG}{dx} = 2c\frac{{\sigma^{2} + \sigma_{\varepsilon }^{2} }}{{\sigma^{2} }}x\left( {1 - 2c\frac{{\sigma^{2} + \sigma_{\varepsilon }^{2} }}{{\sigma^{2} }}x} \right)^{ - 1}$. Hence, $\frac{dG}{dx}$ is negative when x < 0, it is zero when x = 0, and it is positive when $0 < x < \frac{{\sigma^{2} }}{{2c(\sigma^{2} + \sigma_{\varepsilon }^{2} )}}$. Lastly, G(0) = 0 and $\mathop {\lim }\limits_{{x \to \frac{{\sigma^{2} }}{{2c(\sigma^{2} + \sigma_{\varepsilon }^{2} )}}}} G(x) = - \frac{{\sigma^{2} }}{{2c(\sigma^{2} + \sigma_{\varepsilon }^{2} )}} - \left( {2c\frac{{\sigma^{2} + \sigma_{\varepsilon }^{2} }}{{\sigma^{2} }}} \right)^{ - 1} \mathop {\lim }\limits_{y \to 0} \ln (y) = + \infty$. □

Proof of Observation 1

Due to the truthful reporting assumption, the manager’s report upon disclosure equals the true realization of her private signal $\tilde{s}$, and thus the firm price upon disclosure with a report $r \in \Re$ equals $P^{B1} (r) = E[\tilde{v}\left| {\tilde{s} = r]} \right.$. The positive correlation between the firm’s value $\tilde{v}$ and the signal $\tilde{s}$ implies that the price $P^{B1} (r) = E[\tilde{v}\left| {\tilde{s} = r]} \right.$ upon disclosure is an increasing function of the report r, which equals the true realization s of the signal $\tilde{s}$. Also, the price $P^{B1} (\phi ) = E[\tilde{v}\left| {\hat{D}^{B1} (\tilde{t},\tilde{s}) = \phi ]} \right.$ in the absence of disclosure is independent of the actual realization s of the signal $\tilde{s}$. As this shape of the market price rule is rationally inferred by the manager in equilibrium ($\hat{P}^{B1} = P^{B1}$), the disclosure strategy of an informed manager must be an upper-tailed strategy with a minimal threshold s ^B1.

As the investors also rationally infer the disclosure strategy of the manager in equilibrium ($\hat{D}^{B1} = D^{B1}$), ${\text{prob}}(\tilde{s} \ge s^{B1} )$ is the investors’ estimate of the probability that an informed manager will provide disclosure. Their estimate of the probability that the manager is informed is τ. Hence, they attribute a probability $\tau \,{\text{prob}}(\tilde{s} \ge s^{B1} )$ to disclosure occurrence and a probability $1 - \tau \,{\text{prob}}(\tilde{s} \ge s^{B1} )$ to its absence. Conditioned on the absence of disclosure, $\frac{1 - \tau }{{1 - \tau \,{\text{prob}}(\tilde{s} \ge s^{B1} )}}$ is the probability that the investors attribute to the manager being uninformed, while $\frac{{\tau \, (1 - {\text{prob}}(\tilde{s} \ge s^{B1} ))}}{{1 - \tau \,{\text{prob}}(\tilde{s} \ge s^{B1} )}}$ is the probability that they attribute to the manager being informed and choosing to withhold her information. Thus, the firm price in the absence of disclosure is $P^{B1} (\phi ) = E[\tilde{v}\left| {\hat{D}^{B1} (\tilde{t},\tilde{s}) = \phi ]} \right. = \frac{1 - \tau }{{1 - \tau \,{\text{prob}}(\tilde{s} \ge s^{B1} )}}\mu + \frac{{\tau \, (1 - {\text{prob}}(\tilde{s} \ge s^{B1} ))}}{{1 - \tau \,{\text{prob}}(\tilde{s} \ge s^{B1} )}}E \, [\tilde{v}\left| { \, \tilde{s} \le s^{B1} } \right.].$

As disclosure is truthful, the only cost associated with disclosure is λ. So, the manager’s utility upon disclosure of r is $P^{B1} (r) - \lambda = E[\tilde{v}\left| {\tilde{s} = r} \right.] - \lambda$ for any report $r \in \Re$. Her utility in the absence of disclosure is $P^{B1} (\phi ) = \frac{1 - \tau }{{1 - \tau \,{\text{prob}}(\tilde{s} \ge s^{B1} )}}\mu + \frac{{\tau \, (1 - {\text{prob}}(\tilde{s} \ge s^{B1} ))}}{{1 - \tau \,{\text{prob}}(\tilde{s} \ge s^{B1} )}}E \, [\tilde{v}\left| { \, \tilde{s} \le s^{B1} } \right.]$. When the realization of her private signal $\tilde{s}$ equals the threshold s ^B1, the informed manager is indifferent between truthfully disclosing and withholding information. This yields the equation $E[\tilde{v}\left| {\tilde{s} = s^{B1} } \right.] - \lambda = \frac{1 - \tau }{{1 - \tau \,{\text{prob}}(\tilde{s} \ge s^{B1} )}}\mu + \frac{{\tau (1 - {\text{prob}}(\tilde{s} \ge s^{B1} ))}}{{1 - \tau \,{\text{prob}}(\tilde{s} \ge s^{B1} )}}E \, [\tilde{v}\left| { \, \tilde{s} \le s^{B1} } \right.]$, which can be rewritten as $E[\tilde{v}\left| {\tilde{s} = s^{B1} } \right.] - E \, [\tilde{v}\left| { \, \tilde{s} \le s^{B1} } \right.] + \frac{1 - \tau }{{1 - \tau \,{\text{prob}}(\tilde{s} \ge s^{B1} )}}(E \, [\tilde{v}\left| { \, \tilde{s} \le s^{B1} } \right.] - \mu ) = \lambda$ or

$$ F(\tau ,s^{B1} ) = \lambda . $$

(1)

By Lemma A, the left side of Eq. 1 is monotonically increasing in s ^B1, converging to −∞ when s ^B1 converges to −∞ and converging to +∞ when s ^B1 converges to +∞. The right side of the equation is a positive constant, which is independent of s ^B1. Therefore, there is only one value of s ^B1 that solves Eq. 1. The unique solution s ^B1 of Eq. 1 implies the following equilibrium: $D^{B1} (0,s) = \hat{D}^{B1} (0,s) = \phi$, $D^{B1} (1,s) = \hat{D}^{B1} (1,s) = \left\{ {\begin{array}{*{20}c} \phi & {{\text{if}}\;s < s^{B1} } \\ s & {\text{otherwise}} \\ \end{array} } \right.$, $P^{B1} (\phi ) = \hat{P}^{B1} (\phi ) = \frac{1 - \tau }{{1 - \tau \,{\text{prob}}(\tilde{s} \ge s^{B1} )}}\mu + \frac{{\tau \,{\text{prob}}(\tilde{s} \le s^{B1} )}}{{1 - \tau \,{\text{prob}}(\tilde{s} \ge s^{B1} )}}E \, [\tilde{v}\left| { \, \tilde{s} \le s^{B1} } \right.]$ and $P^{B1} (r) = \hat{P}^{B1} (r) = E[\tilde{v}\left| {\tilde{s} = r} \right.]$ for any $s \in \Re$ and r ≥ s ^B1. Since the left side of Eq. 1 is increasing in τ by Lemma A, whereas the right side of the equation is increasing in λ, the disclosure threshold s ^B1 is increasing in λ and decreasing in τ.

When λ converges to 0 and τ converges to 1, Eq. 1 is reduced to

$$ F(1,s^{B1} ) = 0. $$

(2)

Again using Lemma A, the left side of Eq. 2 is increasing in s ^B1, converging to 0 when s ^B1 converges to −∞ and converging to +∞ when t converges to +∞. Hence, the one and only solution of Eq. 2 is $s^{B1} = - \infty$, implying full disclosure. □

Proof of Observation 2

As in monotonic signaling games, the D1 criterion filters out all pooling and partially separating equilibria. Under the D1 criterion, a pooling region, which contains different information realizations that result in the same report, cannot survive because a manager with the highest information realization in the pooling region is better off deviating and providing a slightly higher report. We thus look for fully separating equilibria, where the market can adjust for the manager’s bias, so the market price of the firm equals $P^{B1} (s + b(s)) = E[\tilde{v}\left| {\tilde{s} = s]} \right. = \frac{{\sigma_{\varepsilon }^{2} }}{{\sigma^{2} + \sigma_{\varepsilon }^{2} }} \cdot \mu + \frac{{\sigma^{2} }}{{\sigma^{2} + \sigma_{\varepsilon }^{2} }} \cdot s$, where b(s) is the reporting bias for a realization s of the signal $\tilde{s}$. The biasing function $b:\Re \to \Re$ satisfies two conditions for any $s \in \Re$ and $\Updelta s > 0$. Denoting ∆b = b(s + ∆s) − b(s), the first condition is $\frac{{\sigma_{\varepsilon }^{2} }}{{\sigma^{2} + \sigma_{\varepsilon }^{2} }} \cdot \mu + \frac{{\sigma^{2} }}{{\sigma^{2} + \sigma_{\varepsilon }^{2} }} \cdot s - cb(s)^{2} \ge \frac{{\sigma_{\varepsilon }^{2} }}{{\sigma^{2} + \sigma_{\varepsilon }^{2} }} \cdot \mu + \frac{{\sigma^{2} }}{{\sigma^{2} + \sigma_{\varepsilon }^{2} }} \cdot (s + \Updelta s) - c(b(s) + \Updelta s + \Updelta b)^{2}$, implying that it is not beneficial for a manager with an information realization s to mimic the report of a manager with an information realization $s + \Updelta s$ by choosing a bias of $b(s) + \Updelta s + \Updelta b$. The second condition is $\frac{{\sigma_{\varepsilon }^{2} }}{{\sigma^{2} + \sigma_{\varepsilon }^{2} }} \cdot \mu + \frac{{\sigma^{2} }}{{\sigma^{2} + \sigma_{\varepsilon }^{2} }} \cdot (s + \Updelta s) - c(b(s) + \Updelta b)^{2} \ge \frac{{\sigma_{\varepsilon }^{2} }}{{\sigma^{2} + \sigma_{\varepsilon }^{2} }} \cdot \mu + \frac{{\sigma^{2} }}{{\sigma^{2} + \sigma_{\varepsilon }^{2} }} \cdot s - c(b(s) - \Updelta s)^{2}$, implying that it is not beneficial for a manager with an information realization $s + \Updelta s$ to mimic the report of a manager with an information realization s by choosing a bias of $b(s) - \Updelta s$.

After rearranging that first and second conditions, we get $\frac{\Updelta b}{\Updelta s} \ge \frac{{\sqrt {\frac{{\sigma^{2} }}{{\sigma^{2} + \sigma_{\varepsilon }^{2} }} \cdot \Updelta s/c + b(s)^{2} } - b(s)}}{\Updelta s} - 1$ and $\frac{\Updelta b}{\Updelta s} \le \frac{{\sqrt {\frac{{\sigma^{2} }}{{\sigma^{2} + \sigma_{\varepsilon }^{2} }} \cdot \Updelta s/c + (b(s) - \Updelta s)^{2} } - b(s)}}{\Updelta s}$, respectively. Using l’Hopital’s rule, $\mathop {\lim }\limits_{\Updelta s \to 0} \frac{{\sqrt {\frac{{\sigma^{2} }}{{\sigma^{2} + \sigma_{\varepsilon }^{2} }} \cdot \Updelta s/c + b(s)^{2} } - b(s)}}{\Updelta s} - 1$ equals $\mathop {\lim }\limits_{\Updelta s \to 0} \frac{{\frac{{\sigma^{2} }}{{2c(\sigma^{2} + \sigma_{\varepsilon }^{2} )}}\left( {\frac{{\sigma^{2} }}{{\sigma^{2} + \sigma_{\varepsilon }^{2} }} \cdot \Updelta s/c + b(s)^{2} } \right)^{{ - {\raise0.7ex\hbox{$1$} \!\mathord{\left/ {\vphantom {1 2}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{$2$}}}} }}{1} - 1$ or $\frac{{\sigma^{2} }}{{2c(\sigma^{2} + \sigma_{\varepsilon }^{2} )b(s)}} - 1$. Again using l’Hopital’s rule, $\mathop {\lim }\limits_{\Updelta s \to 0} \frac{{\sqrt {\frac{{\sigma^{2} }}{{\sigma^{2} + \sigma_{\varepsilon }^{2} }} \cdot \Updelta s/c + (b(s) - \Updelta s)^{2} } - b(s)}}{\Updelta s}$ equals $\mathop {\lim }\limits_{\Updelta s \to 0} \frac{{\frac{1}{2}\left( {\frac{{\sigma^{2} }}{{c(\sigma^{2} + \sigma_{\varepsilon }^{2} )}} - 2(b(s) - \Updelta s)} \right)\left( {\frac{{\sigma^{2} }}{{\sigma^{2} + \sigma_{\varepsilon }^{2} }} \cdot \Updelta s/c + (b(s) - \Updelta s)^{2} } \right)^{{ - {\raise0.7ex\hbox{$1$} \!\mathord{\left/ {\vphantom {1 2}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{$2$}}}} }}{1}$, which is reduced to $\frac{1}{2}{{\left( {\frac{{\sigma^{2} }}{{c(\sigma^{2} + \sigma_{\varepsilon }^{2} )}} - 2b(s)} \right)} \mathord{\left/ {\vphantom {{\left( {\frac{{\sigma^{2} }}{{c(\sigma^{2} + \sigma_{\varepsilon }^{2} )}} - 2b(s)} \right)} {b(s)}}} \right. \kern-\nulldelimiterspace} {b(s)}}$ or $\frac{{\sigma^{2} }}{{2c(\sigma^{2} + \sigma_{\varepsilon }^{2} )b(s)}} - 1$. We thus conclude that $\frac{{\sigma^{2} }}{{2c(\sigma^{2} + \sigma_{\varepsilon }^{2} )b(s)}} - 1 \le \frac{db}{ds} \le \frac{{\sigma^{2} }}{{2c(\sigma^{2} + \sigma_{\varepsilon }^{2} )b(s)}} - 1$, implying the differential equation

$$ \frac{db}{ds} = \frac{{\sigma^{2} }}{{2c(\sigma^{2} + \sigma_{\varepsilon }^{2} )b(s)}} - 1. $$

(3)

To derive the biasing function $b:\Re \to \Re$, we now need to solve Eq. 3. One solution of the differential equation is the constant biasing function $b(s) = \frac{{\sigma^{2} }}{{2c(\sigma^{2} + \sigma_{\varepsilon }^{2} )}}$ for any $s \in \Re$.

Other solutions to the differential Eq. 3, where $\frac{db}{ds} \ne 0$, satisfy the equation $\left( {\frac{{\sigma^{2} }}{{2c(\sigma^{2} + \sigma_{\varepsilon }^{2} )b}} - 1} \right)^{ - 1} db = ds$ or $2c\frac{{\sigma^{2} + \sigma_{\varepsilon }^{2} }}{{\sigma^{2} }}b\left( {1 - 2c\frac{{\sigma^{2} + \sigma_{\varepsilon }^{2} }}{{\sigma^{2} }}b} \right)^{ - 1} db = ds$. This implies $\int {2c\frac{{\sigma^{2} + \sigma_{\varepsilon }^{2} }}{{\sigma^{2} }}b\left( {1 - 2c\frac{{\sigma^{2} + \sigma_{\varepsilon }^{2} }}{{\sigma^{2} }}b} \right)^{ - 1} db} = \int {ds}$. So, the biasing function b(s) is defined for any $s \in \Re$ by the implicit equation

$$ G(b(s)) = s + \omega , $$

(4)

where ω could be any scalar. By Lemma B, the function G(x), which is defined only for $x < \frac{{\sigma^{2} }}{{2c(\sigma^{2} + \sigma_{\varepsilon }^{2} )}}$, gets only nonnegative values. So, there is no solution b(s) to Eq. 4 for any $s < - \omega$. It follows that the constant biasing function, where $b(s) = \frac{{\sigma^{2} }}{{2c(\sigma^{2} + \sigma_{\varepsilon }^{2} )}}$ for any $s \in \Re$, is the only biasing function that exists in equilibrium. The constant reporting bias $\frac{{\sigma^{2} }}{{2c(\sigma^{2} + \sigma_{\varepsilon }^{2} )}}$, denoted b ^B2, implies the following fully revealing equilibrium $D^{B2} (0,s) = \hat{D}^{B2} (0,s) = \phi$ , $D^{B2} (1,s) = \hat{D}^{B2} (1,s) = s + b^{B2}$ , $P^{B2} (\phi ) = \hat{P}^{B2} (\phi ) = \mu$ and $P^{B2} (r) = \hat{P}^{B2} (r) = E[\tilde{v}\left| {\tilde{s} = r} \right. - b^{B2} ]$ for any $s,r \in \Re$. □

Proof of Proposition 3

We first show that the equilibrium disclosure strategy takes the upper-tailed shape, so that the manager provides disclosure only when the realization of her private signal is sufficiently high. Suppose by contradiction that there exist $s_{1} ,s_{2} \in \Re$, such that s ₁ < s ₂, D(1, s ₁) = s ₁ + b ₁ and D(1, s ₂) = φ where b ₁ ≥ 0 and s ₂ − s ₁ ≤ b ₁. This implies that $P(\phi ) \le P(s_{1} + b_{1} ) - \lambda - cb_{1}^{2}$ and $P(\phi ) \ge \max_{{b_{2} \in \Re }} \{ P(s_{2} + b_{2} ) - \lambda - cb_{2}^{2} \}$. Consequently, $\max_{{b_{2} \in \Re }} \{ P(s_{2} + b_{2} ) - \lambda - cb_{2}^{2} \} \le P(\phi ) \le P(s_{1} + b_{1} ) - \lambda - cb_{1}^{2}$. However, since s ₁ < s ₂, we get for b ₂ = b ₁ − (s ₂ − s ₁) that P(s ₂ + b ₂) = P(s ₁ + b ₁) and $cb_{2}^{2} < cb_{1}^{2}$, which implies $P(s_{2} + b_{2} ) - \lambda - cb_{2}^{2} > P(s_{1} + b_{1} ) - \lambda - cb_{1}^{2}$—a contradiction. The disclosure strategy is thus characterized by a threshold $s_{0} \in \Re$, such that $D(1,s) \ne \phi$ iff s ≥ s ₀. Using the same arguments as in Observation 2, the D1 criterion allows us to focus on equilibria that are separating upon disclosure, where any two different information realizations that belong to the disclosure region are associated with two different reports.

We next show that the equilibrium reporting bias at the threshold s ₀ is zero. That is, b(s ₀) = 0. Suppose by contradiction that b(s ₀) > 0. At the threshold s ₀, the informed manager is indifferent between providing disclosure and keeping quiet, so the manager’s utility P(s ₀ + b(s ₀)) − λ − cb(s ₀)² upon disclosure equals her utility P(φ) in the absence of disclosure. Hence, $E[\tilde{v}\left| {\tilde{s} = s_{0} } \right.] - \lambda - cb(s_{0} )^{2} = P(\phi )$. For any 0 < ε < b(s ₀), there exists δ(ε) > 0, such that $E[\tilde{v}\left| {\tilde{s} = s_{0} - \delta (\varepsilon )} \right.] - \lambda - c(b(s_{0} ) - \varepsilon )^{2} = P(\phi )$, where δ(ε) is strictly increasing in ε. Therefore, a manager with information realization of s ₀ is better off deviating to the out-of-equilibrium report s ₀ + b(s ₀) − ε (where 0 < ε < b(s ₀)) for any market response P(s ₀ + b(s ₀) − ε) that exceeds $E[\tilde{v}\left| {\tilde{s} = s_{0} - \delta (\varepsilon )} \right.]$. Managers with a slightly lower information realization $s_{0} - \Updelta$, such that $0 < \Updelta < \varepsilon$, are better off or at least no worse off deviating to the out-of-equilibrium report s ₀ + b(s ₀) − ε only for higher market responses that equal or exceed $E[\tilde{v}\left| {\tilde{s} = s_{0} - \delta (\varepsilon - \Updelta )} \right.]$, where $\delta (\varepsilon - \Updelta ) < \delta (\varepsilon )$. Managers with an even lower information realization $s_{0} - \Updelta$, such that $\Updelta \ge \varepsilon$, are better off or at least no worse off providing the out-of-equilibrium report s ₀ + b(s ₀) − ε only for much higher market reactions (at the level of $E[\tilde{v}\left| {\tilde{s} = s_{0} } \right.]$ or even higher). Hence, a manager with information realization s ₀ is more likely to make a deviation to the out-of-equilibrium report s ₀ + b(s ₀) − ε then managers with lower information realizations. Utilizing the D1 criterion of Cho and Kreps (1987), it follows that the out-of-equilibrium report s ₀ + b(s ₀) − ε must be ascribed by the market to managers with information realization that is at least s ₀ for any 0 < ε < b(s ₀). This implies $P(s_{0} + b(s_{0} ) - \varepsilon ) \ge E[\tilde{v}\left| {\tilde{s} = s_{0} } \right.]$. A contradiction now arises as a manager with an information realization of s ₀ can decrease her bias by an amount 0 < ε < b(s ₀) and thereby improve her utility from P(φ) to at least $E[\tilde{v}\left| {\tilde{s} = s_{0} } \right.] - \lambda - c(b(s_{0} ) - \varepsilon )^{2}$, which is higher then her equilibrium utility $E[\tilde{v}\left| {\tilde{s} = s_{0} } \right.] - \lambda - cb(s_{0} )^{2} = P(\phi )$.

As the reporting bias at the threshold s ₀ is zero, we get that the threshold s ₀ is the unique solution of Eq. 5 using the same arguments as in the proof of Observation 1,

$$ F(\tau ,s_{0} ) = \lambda . $$

(5)

This implies that the threshold s ₀ equals the benchmark threshold s ^B1 obtained under the assumption of truthful disclosure as the unique solution of Eq. 1.

Using the same arguments as in Observation 2, the equilibrium biasing function $b_{0} :[s_{0} , + \infty ) \to \Re$ upon disclosure satisfies Eq. 3. It also satisfies b ₀(s ₀) = 0. The condition b ₀(s ₀) = 0 precludes the benchmark constant biasing function. When considering other solutions to the differential Eq. 3, where $\frac{db}{ds} \ne 0$, and adding the condition b ₀(s ₀) = 0, Eq. 4 is reduced to

$$ G(b_{0} (s)) = s - s_{0} , $$

(6)

where s ≥ s ₀. By Lemma B, the left side of Eq. 6, which is defined only for $b_{0} (s) < \frac{{\sigma^{2} }}{{2c(\sigma^{2} + \sigma_{\varepsilon }^{2} )}}$, gets a minimum of zero at b ₀(s) = 0, and then increases in positive values of b ₀(s), converging to +∞ when b ₀(s) converges to $\frac{{\sigma^{2} }}{{2c(\sigma^{2} + \sigma_{\varepsilon }^{2} )}}$. The right side of the equation is nonnegative for any s ≥ s ₀. So, there exists a unique b ₀(s) that solves Eq. 6 for any s ≥ s ₀. As the right side of the equation is increasing in s, so is b ₀(s). The unique biasing function b ₀(s) is increasing in s, starting with a zero bias at the threshold s ₀ and converging to the asymptote $b^{B2} = \frac{{\sigma^{2} }}{{2c(\sigma^{2} + \sigma_{\varepsilon }^{2} )}}$ when s converges to +∞. □

Proof of Corollary 4

By the proof of Proposition 3, the threshold s ₀ is the unique solution of Eq. 5. When λ converges to 0 and τ converges to 1, Eq. 5 is reduced to

$$ F(1,s_{0} ) = 0 $$

(7)

Using Lemma A, the left side of Eq. 7 is increasing in s ₀, converging to 0 when s ₀ converges to −∞ and converging to +∞ when s ₀ converges to +∞. Hence, the one and only solution of Eq. 7 is s ₀ = −∞, implying full disclosure. The equilibrium biasing function under full disclosure must equal to that obtained under mandatory disclosure. □

Proof of Corollary 5

When the out-of-equilibrium beliefs are unrestricted, the bias at the disclosure threshold s ₀ is not restricted to be zero. Since the biasing function must satisfy Eq. 3, and as the function G is defined only for biases below b ^B2, b(s ₀) could be any non negative bias in the range [0, b ^B2]. For any scalar k ∈ [0, b ^B2], there exists an equilibrium with a bias of k at the disclosure threshold. For k ∈ [0, b ^B2), the implicit Eqs. 1 and 4, which define the disclosure threshold s _k and the biasing function b _k (s), respectively, take the following form after substituting b _k (s _k ) = k

$$ G(b_{k} (s)) = s + G(k) - s_{k} , $$

(8)

$$ F(\tau ,s_{k} ) = \lambda + ck^{2}. $$

(9)

Using the same arguments as in the proof of Proposition 3, Eq. 8 yields a unique biasing function b _k (s) and Eq. 9 yields a unique disclosure threshold s _k (which is increasing in k). For k = b ^B2, we need to replace Eq. 8 by b(s) = b ^B2. □