A theory of voluntary disclosure and cost of capital

Abstract

This paper explores the links between firms’ voluntary disclosures and their cost of capital. Existing studies investigate the relation between mandatory disclosures and cost of capital and find no cross-sectional effect but a negative association in time-series. In this paper, I find that when disclosure is voluntary firms that disclose their information have a lower cost of capital than firms that do not disclose, but the association between voluntary disclosure and cost of capital for disclosing and nondisclosing firms is positive in aggregate. I further examine whether reductions in cost of capital indicate improved risk-sharing or investment efficiency. I also find that high (low) disclosure frictions lead to overinvestment (underinvestment) relative to first-best. As average cost of capital proxies for risk-sharing but not investment efficiency, the relation between cost of capital and ex ante efficiency may be ambiguous and often irrelevant.

Appendix: Technical supplements

Proof of Proposition 1:

The social planner solves the following maximization problem:

$$ \max_{\tilde{\epsilon}} \int f(y) U\left(\int\limits_{\tilde{\epsilon}}^{+\infty}(\epsilon+y)h(\epsilon)d\epsilon\right) dy $$

(5)

The first order condition (FOC) of the maximization problem (5) yields:

$$ -h(\epsilon^{FB}) \int\limits f(y) (\epsilon^{FB}+y) U'\left(\int\limits_{\epsilon^{FB}}^{+\infty}(\epsilon+y)h(\epsilon)d\epsilon\right) dy=0 $$

As $h(\epsilon^{FB})>0$, one can rewrite this FOC as $\Upphi(\epsilon^{FB})=0$ where:

$$ \begin{aligned} \Upphi(\epsilon^{FB})&=-\int\limits f(y) (\epsilon^{FB}+y) U'\left(\int\limits_{\epsilon^{FB}}^{+\infty}(\epsilon+y)h(\epsilon)d\epsilon\right) dy \\ &= -\epsilon^{FB}\int\limits f(y) U'\left(\int\limits_{\epsilon^{FB}}^{+\infty}(\epsilon+y)h(\epsilon)d\epsilon\right) dy +\int\limits y f(y) U'\left(\int\limits_{\epsilon^{FB}}^{+\infty}(\epsilon+y)h(\epsilon)d\epsilon\right) dy \end{aligned} $$

Noting that U′ > 0, the fact that $\Upphi(\epsilon^{FB})=0$ implies that $\epsilon^{FB}=-{\frac{\int y f(y) U'\left(\int_{\epsilon^{FB}}^{+\infty}( \epsilon+y)h(\epsilon)d\epsilon\right)}{\int f(y) U'\left(\int_{\epsilon^{FB}}^{+\infty}(\epsilon+y)h(\epsilon)d\epsilon\right)}} dy$. To prove that $\epsilon^{FB}$ is unique and in (0, θ), it is sufficient to show that: (i) $\Upphi'<0$ and, (ii) $\Upphi(0)>0$ and $\Upphi(\theta)<0$.

$$ \Upphi'(\tilde{\epsilon})=\int\limits f(y) (\tilde{\epsilon}+y)^{2} h(\tilde{\epsilon}) U''\left(\int\limits_{\tilde{\epsilon}}^{+\infty}(\epsilon+y)h(\epsilon)d\epsilon\right) dy-\int\limits f(y) U'\left(\int\limits_{\tilde{\epsilon}}^{+\infty}(\epsilon+y)h(\epsilon)d\epsilon\right) dy $$

Since U′ > 0 and $U''<0, \Upphi'<0$.

$$ \begin{aligned} \Upphi(0)&=-\int\limits y f(y) U'\left(\int\limits_{0}^{+\infty}(\epsilon+y)h(\epsilon)d\epsilon\right) dy\\ & =- \int\limits_{0}^{+\infty} y f(y)U'\left(\int\limits_{0}^{+\infty}(\epsilon+y)h(\epsilon)d\epsilon\right)dy-\int\limits_{\underline {y}}^{0} y f(y)U'\left(\int\limits_{0}^{+\infty}(\epsilon+y)h(\epsilon)d\epsilon\right)dy\\ &> - \int\limits_{0}^{+\infty} y f(y)U'\left(\int\limits_{0}^{+\infty}\epsilon h(\epsilon)d\epsilon\right)dy-\int\limits_{\underline {y}}^{0} y f(y)U'\left(\int\limits_{0}^{+\infty}\epsilon h(\epsilon)d\epsilon\right)dy\\ &> -U'\left(\int\limits_{0}^{+\infty}\epsilon h(\epsilon)d\epsilon\right)\int\limits_{\underline {y}}^{+\infty} y f(y) dy = 0 \end{aligned} $$

where the inequality follows from the concavity of U. Moreover, θ + y > 0 implies that:

$$ \Upphi(\theta)=-\int\limits f(y) (\theta+y) U'\left(\int\limits_{\theta}^{+\infty}(\epsilon+y)h(\epsilon)d\epsilon\right) dy < 0 $$

Proof of Lemma 1

The disclosure threshold satisfies $P_{\epsilon^{over}}-P_{\emptyset}=0$. This implies, by Eqs. (2) and (3) that ${\epsilon^{over}-\mathbb{E}(\epsilon|ND)=0}$. The non disclosure conditional expectation ${\mathbb{E}(\epsilon|ND)}$ can be expanded by Bayesian updating to:

$$ {\mathbb{E}}(\epsilon|ND)={\frac{\eta \theta +(1-\eta)H(\epsilon^{over}){\mathbb{E}}(\epsilon|\epsilon\leq \epsilon^{over})}{\eta+(1-\eta)H(\epsilon^{over})}} $$

The equation can be rearranged as follows,

$$ \eta(\theta-\epsilon^{over})=(1-\eta)(H(\epsilon^{over})-\int\limits_{-\infty}^{\epsilon^{over}}\epsilon dH(\epsilon))= (1-\eta)\int\limits_{-\infty}^{\epsilon^{over}} H(\epsilon)d\epsilon $$

This corresponds to Eq. (7), p. 149, in Jung and Kwon (1988) and

$$ \frac{\partial{\epsilon^{over}}}{{\partial \eta}}=\frac{\int_{\epsilon^{over}}^{+\infty}(1-H(\epsilon^{over}))d\epsilon }{\eta+(1-\eta)H(\epsilon^{over})} $$

(6)

Proof of Proposition 2

I solve the maximization problem of an investor:

$$ (\Upgamma_\delta)\quad \max\limits_{\gamma} \int f(y) (P_{\delta}\gamma+{\frac{(1-\gamma)P_{\delta}}{P_{m}}}CF_m(y))^{1-\alpha}dy $$

The dependence on δ is due to the price P _δ . But from the program $\Upgamma_{\delta}, P_{\delta}$ is only a constant multiplicative term of the objective function, and the maximizer γ does not depend on P _δ , and so does not depend on δ. Thus $\gamma^{over}_{\emptyset}=\gamma^{over}_{ \epsilon}=\gamma^{over}$.

The aggregate demand of all investors in the risk-free asset AD is equal to:

$$ \begin{aligned} AD&=(\eta+(1-\eta)H(\epsilon^{over}))P_{\emptyset}\gamma^{over}_{\emptyset }+(1-\eta)\int\limits_{\epsilon^{over}}^{+\infty} P(\epsilon)\gamma^{over}_{\epsilon }h(\epsilon)d\epsilon \\ &=\gamma^{over}\left((\eta+(1-\eta)H(\epsilon^{over})) \left(\epsilon^{over}-\theta+{\frac{\theta}{{\mathbb{E}}(R_{m})}}\right)\right)\\ & \quad +\gamma^{over}\left((1-\eta)\int\limits_{\epsilon^{over}}^{+\infty} (\epsilon-\theta+{\frac{\theta}{{\mathbb{E}}(R_{m})}})h(\epsilon)d\epsilon \right)\\ &=\gamma^{over}\left({\frac{\theta}{{\mathbb{E}}(R_{m})}}-\theta+(\eta+(1-\eta)H(\epsilon^{over}))\epsilon^{over}\right)+\gamma^{over} (1-\eta)\int\limits_{\epsilon^{over}}^{+\infty} \epsilon h(\epsilon)d\epsilon \end{aligned} $$

(7)

I rewrite expression $\int_{\epsilon^{over}}^{+\infty} \epsilon h(\epsilon)d\epsilon$ by integrating by parts as follows:

$$ \int\limits_{\epsilon^{over}}^{+\infty} \epsilon h(\epsilon)d\epsilon=\int\limits_{\epsilon^{over}}^{+\infty} (1-H(\epsilon))d\epsilon+\epsilon^{over} (1-H(\epsilon^{over})) $$

Substituting this expression into Eq. (7)

$$ \begin{aligned} AD&=\gamma^{over}\left({\frac{\theta}{{\mathbb{E}}(R_{m})}}-\theta+(\eta+(1-\eta)H(\epsilon^{over}))\epsilon^{over}\right)\\ & \quad + \gamma^{over}(1-\eta)\left(\int\limits_{\epsilon^{over}}^{+\infty} (1-H(\epsilon))d\epsilon+\epsilon^{over} (1-H(\epsilon^{over}))\right) \\ &=\gamma^{over}\left({\frac{\theta}{{\mathbb{E}}(R_{m})}}-\theta+\epsilon^{over}+\int\limits_{\epsilon^{over}}^{+\infty} (1-H(\epsilon))d\epsilon-\int\limits_{-\infty}^{\epsilon^{over}} H(\epsilon)d\epsilon\right) \end{aligned} $$

(8)

The mean θ can be rewritten as:

$$ \begin{aligned} \theta&=\int\limits_{-\infty}^{+\infty} \epsilon h(\epsilon) d\epsilon=\int\limits_{\epsilon^{over}}^{+\infty} \epsilon h(\epsilon) d\epsilon+\int\limits_{-\infty}^{\epsilon^{over}} \epsilon h(\epsilon) d\epsilon\\ &=\int\limits_{\epsilon^{over}}^{+\infty} (1-H(\epsilon))d\epsilon+\epsilon^{over}(1-H(\epsilon^{over}))-\int\limits^{\epsilon^{over}}_{-\infty} H(\epsilon)d\epsilon+\epsilon^{over}H(\epsilon^{over}) \end{aligned} $$

Finally expression (8) is equal to ${\gamma^{over}(\theta+{\frac{\theta}{\mathbb{E}(R_{m})}}-\theta)=\gamma^{over} {\frac{\theta}{\mathbb{E}(R_{m})}}}$. As the net supply is equal to zero, it yields γ ^over = 0. I determine next the expression of the expected market portfolio return in this economy. Simplifying the maximization problem $(\Upgamma_{\delta})$, it yields:

$$ \max_{\gamma} \int f(y) \left\{\gamma +(1-\gamma){\frac{(\theta+y)}{P_{m}}}\right\}^{1-\alpha}dy $$

The FOC from the investor’s maximization problem with respect to γ is equal to:

$$ {\mathbb{E}}\left\{\left\{1-{\frac{(\theta+y)}{P_{m}}}\right\}\left\{\gamma+(1-\gamma){\frac{(\theta+y)}{P_{m}}}\right\}^{-\alpha}\right\}=0 $$

(9)

As from the market clearing condition, I showed that γ ^over = 0., the FOC is reduced to:

$$ {\mathbb{E}}\left\{\left\{{\mathbb{E}}(R_{m})(\theta+y)-\theta\right\}(\theta+y)^{-\alpha}\right\}=0 $$

(10)

Simplifying,

$$ {\mathbb{E}}(R^{over}_{m})=\frac{\theta {\mathbb{E}}((\theta+y)^{-\alpha})}{{\mathbb{E}}((\theta+y)(\theta+y)^{-\alpha})}= \frac{\theta}{\theta+\underbrace{ {\mathbb{E}}(y(\theta+y)^{-\alpha})/{\mathbb{E}}((\theta+y)^{-\alpha})}_{Q^{over}}} $$

I prove next that Q ^over is negative where $Q^{over}=\int {\frac{y f(y)U'(\theta+y)}{\int f(y)U'(\theta+y)dy}} dy$. By additivity of the integral,

$$ \int\limits_{\underline {y}}^{+\infty} y f(y)U'(\theta+y)dy=\int\limits_{0}^{+\infty} y f(y)U'(\theta+y)dy+\int\limits_{\underline {y}}^{0} y f(y)U'(\theta+y)dy $$

Moreover $\int_{0}^{+\infty} y f(y)U'(\theta+y)dy<\int_{0}^{+\infty} y f(y)U'(\theta)dy$ as U is strictly concave. Likewise $\int_{\underline {y}}^{0} y f(y)U'(\theta+y)dy<\int_{\underline {y}}^{0} y f(y)U'(\theta)dy$. By assumption $\int_{\underline {y}}^{+\infty} y f(y)dy=E(\tilde{y})=0$, and it yields

$$ \int\limits_{\underline {y}}^{+\infty} y f(y)U'(\theta+y)dy< U'(\theta)\int\limits_{\underline {y}}^{+\infty} y f(y)dy=0 $$

In the overinvestment equilibrium, the nondisclosing price $P_{\emptyset}(\epsilon^{over})$ is equal to $P(\epsilon^{over})=\epsilon^{over}+Q^{over}$. As Q ^over is independent of η, the derivative of $P(\epsilon^{over})$ w.r.t to η is equal to $\frac{\partial{\epsilon^{over}}}{{\partial \eta}}\geq 0$. Thus the nondisclosing price is increasing in the disclosure friction η. Further it implies that there exists a unique η ^over such that $P_\emptyset(\epsilon^{over}(\eta^{over}))=0$ and $\epsilon^{over}$ is always positive.

Proof of Proposition 3:

If the investor has a nondisclosing firm, its wealth is equal to zero, and it cannot invest in the risk-free asset nor in the market portfolio. However if an investor has a disclosing firm then he solves the maximization problem $(\Upgamma_{\epsilon})$.

$$ (\Upgamma_\epsilon) \quad \max_{\gamma} \int f(y) (P_{\delta}\gamma+{\frac{(1-\gamma)P_{\delta}}{P_{m}}}CF(y))^{1-\alpha}dy $$

Simplifying,

$$ (\Upgamma_{\epsilon}) \quad \max\limits_{\gamma} \int\limits f(y) (\gamma+{\frac{(1-\gamma)}{P_{m}}}(1-\eta)\int\limits_{\epsilon^{under}}^{+\infty}(\epsilon+y)h(\epsilon)d\epsilon)^{1-\alpha}dy $$

The maximizers of an individual investor having a disclosing firm do not depend on $P_{\epsilon}$ and thus $\epsilon$. Therefore $\forall \epsilon, \gamma^{under}_{\epsilon }=\gamma^{under}$. The aggregate demand of all investors for the risk free asset is equal to:

$$ (\eta+(1-\eta)H(\epsilon^{under}))\gamma^{under}_{\emptyset } P_{\emptyset}+(1-\eta)\int\limits_{\epsilon^{under}}^{+\infty} \gamma^{under}_{\epsilon }P(\epsilon) d\epsilon=\gamma^{under} (1- \eta)\int\limits_{\epsilon^{under}}^{+\infty} P(\epsilon) d\epsilon $$

As the net supply is equal to zero, it yields γ ^under = 0. Taking the FOC of the investor problem, it yields

$$ {\mathbb{E}}\left\{\left({\frac{(1-\eta)\int\limits_{\epsilon^{under}}^{+\infty}(\epsilon+y)h(\epsilon)d\epsilon}{P_{m}}}-1\right)\right. \left.\left(\gamma+{\frac{(1-\gamma)}{P_{m}}}(1-\eta)\int\limits_{\epsilon^{under}}^{+\infty}(\epsilon+y)h(\epsilon)d\epsilon)^{1-\alpha}\right)^{-\alpha}\right\}=0 $$

By the market clearing condition, γ ^under = 0, and the FOC is then reduced to:

$$ {\mathbb{E}} \left\{\left({\mathbb{E}}(R_{m})\int\limits_{\epsilon^{under}}^{+\infty}(\epsilon+y)h(\epsilon)d\epsilon-\int\limits_{\epsilon^{under}}^{+\infty}\epsilon h(\epsilon)d\epsilon\right) \right. \left. \left(\int\limits_{\epsilon^{under}}^{+\infty}(\epsilon+y)h(\epsilon)d\epsilon\right)^{-\alpha}\right\}=0\\ $$

$$ \begin{aligned} \hbox{Simplifying}\,\,{\mathbb{E}}(R^{under}_{m})&={\frac{{\mathbb{E}}\left((\int_{\epsilon^{under}}^{+\infty}\epsilon h(\epsilon)d\epsilon)(\int_{\epsilon^{under}}^{+\infty}(\epsilon+y)h(\epsilon)d\epsilon)^{-\alpha}\right)}{{\mathbb{E}}\left(\int_{\epsilon^{under}}^{+\infty}(\epsilon+y)h(\epsilon)d\epsilon (\int_{\epsilon^{under}}^{+\infty}(\epsilon+y)h(\epsilon)d\epsilon)^{-\alpha}\right)}}\\ \hbox{Rearranging,}\,\, {\mathbb{E}}(R^{under}_{m})&=\frac{\int_{\epsilon^{under}}^{+\infty}\epsilon h(\epsilon)d\epsilon}{\int_{\epsilon^{under}}^{+\infty}\epsilon h(\epsilon)d\epsilon+(1-H(\epsilon^{under}))\frac{{\mathbb{E}} \left(y(\int_{\epsilon^{under}}^{+\infty}(\epsilon+y) h(\epsilon)d\epsilon)^{-\alpha}\right)}{{\mathbb{E}}\left ((\int_{\epsilon^{under}}^{+\infty}(\epsilon+y)h(\epsilon)d\epsilon)^{-\alpha}\right)}} \end{aligned} $$

(11)

I now turn to the determination of the disclosure threshold $\epsilon^{under}$. The firm observing $\epsilon^{under}$ has a cash flow:

$$ \epsilon^{under}+y= \underbrace{\epsilon^{under}-{\frac{\int_{\epsilon^{under}}^{+\infty} \epsilon h(\epsilon)d\epsilon}{ 1-H(\epsilon^{under})}}}_{\rm units\,of\,the\,risk\,free\,asset}+ \underbrace{{\frac{1}{(1-\eta)(1-H(\epsilon^{under}))}}}_{\rm units\,of\,the\,market\,portfolio}CF_{m} $$

Its price is then $\epsilon^{under}-{\frac{\int_{\epsilon^{under}}^{+\infty} \epsilon h(\epsilon)d\epsilon}{ 1-H(\epsilon^{under})}} +{\frac{1}{(1-\eta)(1-H(\epsilon^{under}))}}P_{m}^{under}=0$ as by definition this firm has a market price of zero, where

$$ \begin{aligned} P_{m}^{under}&=(1-\eta) \int\limits_{\epsilon^{under}}^{+\infty} \epsilon h(\epsilon)d\epsilon+(1-\eta) (1-H(\epsilon^{under}))Q^{under}\\ \hbox{with}\quad Q^{under}&=\int\limits_{\underline {y}}^{+\infty} {\frac{y f(y)\left(\int_{\epsilon^{under}}^{+\infty}(\epsilon+y)h(\epsilon)d\epsilon\right)^{-\alpha}}{\int_{\underline {y}}^{+\infty} f(y)\left(\int_{\epsilon^{under}}^{+\infty}(\epsilon+y)h(\epsilon)d\epsilon\right)^{-\alpha} dy}} dy \end{aligned} $$

Replacing P ^under _m by its expression and simplifying it yields $\epsilon^{under}=-{\frac{\int y f(y)(\int_{\epsilon^{under}}^{+\infty}(\epsilon+y)h(\epsilon)d\epsilon)^{-\alpha}dy}{\int f(\tilde{y})(\int_{\epsilon^{under}}^{+\infty}(\epsilon+\tilde{y})h(\epsilon)d\epsilon)^{-\alpha}d\tilde{y}}}$. The disclosure threshold $\epsilon^{under}$ is independent of the disclosure friction η.

I now prove that Q ^under is negative. This is similar to the proof of Q ^over < 0.

$$ Q^{under}=\int\limits_{\underline {y}}^{+\infty} {\frac{y f(y)\left(\int_{\epsilon^{under}}^{+\infty}(\epsilon+y)h(\epsilon)d\epsilon\right)^{-\alpha}}{\int_{\underline {y}}^{+\infty} f(y)\left(\int_{\epsilon^{under}}^{+\infty}(\epsilon+y)h(\epsilon)d\epsilon\right)^{-\alpha} dy}} dy $$

As the utility function is strictly concave,

$$ \int\limits_{0}^{+\infty} y f(y)\left(\int\limits_{\epsilon^{under}}^{+\infty}(\epsilon+y)h(\epsilon)d\epsilon\right)^{-\alpha}dy< \int\limits_{0}^{+\infty} y f(y)\left(\int\limits_{\epsilon^{under}}^{+\infty}\epsilon h(\epsilon)d\epsilon\right)^{-\alpha}dy $$

(12)

and,

$$\int\limits_{\underline {y}}^{0} y f(y)\left(\int\limits_{\epsilon^{under}}^{+\infty}(\epsilon+y)h(\epsilon)d\epsilon\right)^{-\alpha}dy< \int\limits_{\underline {y}}^{0} y f(y)\left(\int\limits_{\epsilon^{under}}^{+\infty}\epsilon h(\epsilon)d\epsilon\right)^{-\alpha}dy $$

(13)

By assumption ${\int_{\underline {y}}^{+\infty} y f(y)dy=\mathbb{E}(\tilde{y})=0}$, which yields

$$ Q^{under}<\left(\int\limits_{\epsilon^{under}}^{+\infty}\epsilon h(\epsilon)d\epsilon\right)^{-\alpha} \int\limits_{\underline {y}}^{+\infty} y f(y)dy=0 $$

Further in the underinvestment region, the turning point η ^under is determined by Eq. (4) such that the price for nondisclosing firms is equal to zero. Given that Q ^under is not equal to Q ^over, there exists η ^under≠ η ^over such that $\epsilon^{under}=-Q^{under}$.

Proof of Proposition 4

Recall that Q ^under is as follows

$$ \begin{aligned} Q^{under}&={\frac{\int y f(y)(\int_{\epsilon^{under}}^{+\infty} h(\epsilon)(\epsilon+y)d\epsilon)^{-\gamma}dy}{\int f(y) (\int_{\epsilon^{under}}^{+\infty} h(\epsilon)(\epsilon+y)d\epsilon)^{-\gamma}dy}} \\ &= {\frac{\int y f(y)({\frac{\int_{\epsilon^{under}}^{+\infty} h(\epsilon)(\epsilon+y)d\epsilon}{\int_{\epsilon^{under}}^{+\infty} h(\epsilon) d\epsilon}})^{-\gamma}dy}{\int f(y) ({\frac{\int_{\epsilon^{under}}^{+\infty} h(\epsilon)(\epsilon+y)d\epsilon}{\int_{\epsilon^{under}}^{+\infty} h(\epsilon) d\epsilon}})^{-\gamma}dy}} \\ &= {\frac{\int y f(y)(y+k(\epsilon^{under}))^{-\gamma}dy}{\int f(y) \int_a^{+\infty} h(\epsilon)(y+k(\epsilon^{under}))^{-\gamma}dy}}\\ &= \int y z(y;k(\epsilon^{under}))dy \end{aligned} $$

$$ where\quad k(x)={\mathbb{E}}(\epsilon|\epsilon\geq x)\quad \hbox{and} \quad z(y;a)= \frac{f(y)(y+a)^{-\gamma}}{\int f(y')(y'+a)^{-\gamma}dy'} $$

Q ^under can also be interpreted as Q ^over in an economy with an endowment process $y+k(\epsilon^{under})$. The function z(.;a) is a well-defined probability density function. It is commonly referred to as the risk-neutral probability measure. I shall argue next that the distribution decreases in the sense of the first-order stochastic dominance when a increases. To see this, define ψ(X;a) as the cumulative distribution function:

$$ \psi(X;a)={\frac{\int\nolimits_{\underline {y}}^X f(y)(y+a)^{-\gamma}dy}{\int f(y)(y+a)^{-\gamma}dy}}\quad and \quad \psi_a(X;A)= \gamma {\frac{K_1}{(\int f(y)(y+a)^{-\gamma}dy)^2}} $$

with $K_1=\int f(y)(y+a)^{-\gamma-1}dy\int_{\underline {y}}^X f(y)(y+a)^{-\gamma}dy -\int_{\underline {y}}^X f(y)(y+a)^{-\gamma-1}dy \int f(y)(y+a)^{-\gamma}dy.$

To prove first-order stochastic dominance, I need to show that ψ _a  < 0, i.e., K ₁ < 0.

$$ \begin{aligned} {\frac{\partial K_1}{\partial X}}&= \int\limits f(y)(y+a)^{-\gamma-1}dy f(X)(X+a)^{-\gamma}\\ &- f(X)(X+a)^{-\gamma-1} \int\limits f(y)(y+a)^{-\gamma}dy \\ &= f(X)(X+k(a))^{-\gamma-1}\underbrace{\int\limits f(y)(y+k(a))^{-\gamma-1} (X-y)dy}_{K_2} \end{aligned} $$

The term K ₂ is strictly increasing in X and is strictly negative (positive) at $X=\underline {y} (X=\overline{y}).$ Therefore, K ₁ must be first decreasing and then increasing, which in turn implies that K ₁ < 0 for any X. It follows that ψ _a (.) < 0 and thus, for two a′ > a, the risk-neutral probability measure induced by the density z(.;a′) first-order stochastically dominates that induced by the density z(.;a′). Using this and the fact that $\theta<k(\epsilon^{under})$,

$$ Q^{under}=\int\limits y z(y;k(\epsilon^{under}))dy>\int\limits y z(y;\theta )dy = Q^{over} $$

This also implies that η ^under < η ^over. Finally, comparing the risk premium in each equilibrium:

$$ \begin{aligned} {\mathbb{E}}(R^{under}_{m})&={\frac{\int_{\epsilon^{under}}^{+\infty} \epsilon h(\epsilon)d\epsilon}{\int_{\epsilon^{under}}^{+\infty} \epsilon h(\epsilon)d\epsilon+(1-H(\epsilon^{under}))Q^{under}}}\\ &={\frac{{\mathbb{E}}(\epsilon|\epsilon\geq \epsilon^{under})}{{\mathbb{E}}(\epsilon|\epsilon\geq \epsilon^{under})+Q^{under}}}< {\frac{\theta}{\theta+Q^{over}}}={\mathbb{E}}(R_m^{over}) \end{aligned} $$

Proof of Lemma 2

The costs of capital of nondisclosing firms and disclosing firms are respectively:

$$ R_{\emptyset}={\frac{{\mathbb{E}}(\epsilon|ND)}{P_{\emptyset}}}={\frac{\epsilon^{over}}{P(\epsilon^{over})}} \quad \hbox{and}\quad R_{D}=\int\limits_{\epsilon^{over}}^{+\infty} {\frac{\epsilon }{P(\epsilon)}}\,{\frac{h(\epsilon)}{(1-H(\epsilon^{over}))}}d\epsilon $$

As $P_{\emptyset}=P(\epsilon^{over})=\epsilon^{over}+Q^{over}$ and $P(\epsilon)=\epsilon+Q^{over}$, a simple rewriting of the costs of capital yields:

$$ R_{\emptyset}=1-{\frac{Q^{over}}{P_{\emptyset}}}\quad and \quad R_{D}=1-Q^{over}\int\limits_{\epsilon^{over}}^{+\infty} {\frac{h(\epsilon)}{P(\epsilon) (1-H(\epsilon^{over}))}}d\epsilon $$

Looking closely at the prices, I can express them as a CAPM formulation:

$$ \begin{aligned} R_{\emptyset}&=\underbrace{1}_{\rm Riskfree}+\underbrace{{\frac{P^{over}_{m}}{P_{\emptyset}}}}_{\beta_{\emptyset }}\underbrace{({\mathbb{E}}(R^{over}_{m})-1)}_{\rm Risk\,Premium}\\ R_{D}&=\underbrace{1}_{\rm Riskfree}+\underbrace{\int\limits_{\epsilon^{over}}^{+\infty} {\frac{P^{over}_{m}}{P(\epsilon)}}\,{\frac{h(\epsilon)}{ (1-H(\epsilon^{over}))}}d\epsilon}_{\beta_{D}}\underbrace{({\mathbb{E}}(R^{over}_{m})-1)}_{\rm Risk\,Premium} \end{aligned} $$

I derive the beta β as an expression of the covariance between the firm’s return and the market portfolio return over the variance of the market portfolio return.

$$ \begin{aligned} \beta_{\emptyset}&={\frac{P^{over}_{m}}{P_{\emptyset}}}=\underbrace{{\frac{V(y)}{P_{\emptyset}P^{over}_{m}}}}_{\rm{covariance}}\left/\underbrace{{\frac{V(y)}{P^{over 2 }_{m}}}}_{\rm market\,variance}\right.\\ \beta_{D}&= \int\limits_{\epsilon^{over}}^{+\infty} {\frac{P^{over}_{m}}{P(\epsilon)}}\,{\frac{h(\epsilon)}{ (1-H(\epsilon^{over}))}}d\epsilon=\underbrace{\int\limits_{\epsilon^{over}}^{+\infty}{\frac{V(y)}{ P(\epsilon)P^{over}_{m}}}\,{\frac{h(\epsilon)}{ (1-H(\epsilon^{over}))}}d\epsilon }_{\rm covariance}\left/\underbrace{{\frac{V(y)}{P^{over 2}_{m}}}}_{\rm market\,variance}\right. \end{aligned} $$

Proof of Proposition 5

Further $P_{\emptyset}(\epsilon^{over})=P(\epsilon^{over})$. As $\forall \epsilon\geq\epsilon^{over}, P(\epsilon^{over})\leq P(\epsilon)$, R _∅ ≥ R _D .

Proof of Lemma 3

• (i) Let us prove that if η ≥ η ^over, for $\eta<\eta^{'}, \Updelta (.;\eta')$ second-order stochastically dominates: $\Updelta(.;\eta)$. For a price $p<P_{\emptyset}(\epsilon^{over}(\eta))$, $\Updelta (p;\eta)=0$ otherwise for a price $P_{\emptyset}(\epsilon^{over}(\eta))\leq p\leq \overline{p}, \Updelta (p;\eta)=\left\{\eta+(1-\eta) H(\epsilon^{over}(\eta))\right\}+(1-\eta)\left\{H(p-Q^{over})-H(\epsilon^{over}(\eta))\right\}$. Let us define η < η′ and $T(\overline{p})$ as the area between the two curves $\Updelta (;\eta)$ and $\Updelta (;\eta')$ in $[P_{\emptyset}(\epsilon^{over}(\eta)), \overline{p}]$, specifically $T(\overline{p}) = \int_{P_{\emptyset}(\epsilon^{over}(\eta))}^{\overline{p}}(\Updelta (p;\eta) - \Updelta (p;\eta'))dp$. By additivity of the integral I rewrite $T(\overline{p})$:

  $$ T(\overline{p})=\int\limits_{P_{\emptyset}(\epsilon^{over}(\eta))}^{P_{\emptyset}(\epsilon^{over}(\eta'))}(\Updelta (p;\eta) - 0)dp+\int\limits_{P_{\emptyset}(\epsilon^{over}(\eta'))}^{\overline{p}}(\Updelta (p;\eta) - \Updelta (p;\eta'))dp $$

  Consider η′ = η + μ, then $T(\overline{p})$ becomes:

  $$ T(\overline{p})=\int\limits_{P_{\emptyset}(\epsilon^{over}(\eta))}^{P_{\emptyset}(\epsilon^{over}(\eta+\mu))} \Updelta (p;\eta) dp+\int\limits_{P_{\emptyset}(\epsilon^{over}(\eta+\mu))}^{\overline{p}}(\Updelta (p;\eta) - \Updelta (p;\eta+\mu))dp $$

  Further define $A(\mu)= \int_{P_{\emptyset}(\epsilon^{over}(k))}^{P_{\emptyset}(\epsilon^{over}(\eta+\mu))}\Updelta (p;\eta)dp$ and $B(\mu)= \int_{P_{\emptyset}(\epsilon^{over}(\eta+\mu))}^{\overline{p}}(\Updelta (p;\eta) - \Updelta (p;\eta+\mu))dp$. I differentiate the above expressions w.r.t μ:

  $$ \begin{aligned} A'(\mu)&={\frac{\partial{\epsilon^{over}(\eta+\mu)}}{\partial{\eta}}}\Updelta (P_{\emptyset}(\epsilon^{over}(\eta+\mu));\eta)\\ B'(\mu)&=-{\frac{\partial{\epsilon^{over}(\eta+\mu)}}{\partial{\eta}}}\left\{\Updelta (P_{\emptyset}(\epsilon^{over}(\eta+\mu));\eta)-\Updelta (P_{\emptyset}(\epsilon^{over}(\eta+\mu));\eta+\mu)\right\}\\ &\quad-\int\limits_{P_{\emptyset}(\epsilon^{over}( \eta+\mu))}^{\overline{p}}((1-H(\epsilon^{over}(\eta+\mu)))+(1-\eta-\mu) h(\epsilon^{over}(\eta+\mu)){\frac{\partial\epsilon^{over}( \eta+\mu)}{\partial{\eta}}}\\ &\quad-H(p-Q^{over})+H(\epsilon^{over}(\eta+\mu))-(1-\eta-\mu) h(\epsilon^{over}(\eta+\mu)){\frac{\partial\epsilon^{over}(\eta+\mu)}{\partial{\eta}}}) dp\\ &=-{\frac{\partial{\epsilon^{over}(\eta+\mu)}}{\partial{\eta}}}\left\{\Updelta (P_{\emptyset}(\epsilon^{over}(\eta+\mu));\eta)-\Updelta (P_{\emptyset}(\epsilon^{over}(\eta+\mu));\eta+\mu)\right\}\\ &\quad-\int\limits_{P_{\emptyset}(\epsilon^{over}(\eta+\mu))}^{\overline{p}}\left(1-H(p-Q^{over})\right) dp \end{aligned} $$

  When μ = 0, A′ can be simplified to ${\frac{\partial{\epsilon^{over}(\eta)}}{\partial{\eta}}}\Updelta (P_{\emptyset}(\epsilon^{over}(\eta));\eta)\geq 0$.

  Likewise B′ is equal to $-\int_{P_{\emptyset}(\epsilon^{over}(\eta))}^{\overline{p}}(1-H(p-Q^{over}))\leq 0$.

  Replacing ${\frac{\partial{\epsilon^{over}(\eta)}}{\partial{\eta}}}$ by expression (5) and simplifying, I compute A′(0) + B′(0):

  $$ \int\limits_{\epsilon^{over}(\eta)}^{+\infty}(1-H(\epsilon))d\epsilon -\int\limits_{P_{\emptyset}(\epsilon^{over}(\eta))}^{\overline{p}}\left(1-H(p-Q^{over})\right) dp $$

  Consider the change in variable $\epsilon=p-Q^{over}$, I further simplify the above expression by:

  $$ \int\limits_{\epsilon^{over}(\eta)}^{+\infty}(1-H(\epsilon))d\epsilon -\int\limits_{\epsilon^{over}(\eta)}^{\overline{\epsilon}}(1-H(\epsilon)) d\epsilon $$

  (14)

  If $\overline{\epsilon}$ converges to $+\infty$ then expression (14) is equal to zero. Thus $\forall \overline{\epsilon}$, expression (14) is positive. It also means that $\forall \overline{p}, T(\overline{p})\geq 0$, i.e., the probability mass of $\Updelta(;\eta)$ is more spread out than the probability mass of $\Updelta(;\eta')$, which implies that $\Updelta(;\eta)$ is "riskier" than $\Updelta(;\eta')$

• (ii) Let us prove that if η ≤ η ^under, for $\eta'<\eta, \Updelta(.;\eta')$ first-order stochastically dominates $\Updelta(.;\eta)$. It has been proven that $P(\epsilon^{under})$ does not depend on η. Thus the range of prices is the same for a given η or η′. Let us take the difference between $\Updelta (p;\eta')$ and $\Updelta (p;\eta)$ for $p\in[0, \overline{p}]$.

  $$ \begin{aligned} \Updelta (p; \eta')-\Updelta (p;\eta)&=(\eta'+(1-\eta')H(\epsilon^{under}))+(1-\eta')\left(H(p-Q^{under})-H(\epsilon^{under})\right)\cr &-\left(( \eta+(1-\eta)H(\epsilon^{under}))+(1-\eta)\left(H(p-Q^{under})-H(\epsilon^{under})\right)\right) \end{aligned} $$

  Replacing η by η′ + μ it yields

  $$ \begin{aligned} \Updelta (p;\eta')-\Updelta (p;\eta'+\mu)&= (\eta'+(1-\eta')H(\epsilon^{under}))+(1-\eta')\left(H(p-Q^{under})-H(\epsilon^{under})\right)\cr &-\left(( \eta'+\mu+(1-\eta'-\mu)H(\epsilon^{under}))+(1-\eta'-\mu)\left(H(p-Q^{under})-H(\epsilon^{under})\right)\right)\\ &=-\mu(1-H(\epsilon^{under}))+\mu\left(H(p-Q^{under})-H(\epsilon^{under})\right)\\ &=-\mu \left(1-H(p-Q^{under})\right) \leq 0 \end{aligned} $$

  Therefore $\Updelta (;\eta')$ first order stochastically dominates $\Updelta (;\eta)$.

Proof of Proposition 6

If η ≥ η ^over, I will prove that R _D is decreasing in η.

$$ R_{D}=1-\int\limits_{\epsilon^{over}(\eta)}^{+\infty} {\frac{Q^{over}}{P(\epsilon)}}\,{\frac{h(\epsilon)}{(1-H(\epsilon^{over}(\eta)))}}d\epsilon $$

Differentiating R _D w.r.t η, it yields

$$ -{\frac{1}{(1-H(\epsilon^{over}))^{2}}}h(\epsilon^{over})J'(\eta)\int\limits_{\epsilon^{over}}^{+\infty} {\frac{Q^{over}h(\epsilon)}{P(\epsilon)}}d\epsilon+ J'(\eta){\frac{Q^{over}h(\epsilon^{over})}{(1-H(\epsilon^{over}))P(\epsilon^{over})}} $$

Simplifying it yields

$$ -{\frac{J'(\eta)Q^{over}h(\epsilon^{over})}{(1-H(\epsilon^{over}))}}\left({\frac{1}{(1-H(\epsilon^{over}))^{}}}\int\limits_{\epsilon^{over}}^{+\infty} {\frac{h(\epsilon)}{P(\epsilon)}}d\epsilon-{\frac{1}{P(\epsilon^{over})}}\right) $$

(15)

As $\forall \epsilon>\epsilon^{over}, P(\epsilon^{over})\leq P(\epsilon)$, and ${\frac{\partial{\epsilon^{over}(\eta)}}{\partial{\eta}}}\geq 0$ and Q ^over < 0, the derivative of R _D with respect to η given by expression (15) is negative. Therefore R _D is decreasing in η.

$P_{\emptyset}(\epsilon^{over})=P(\epsilon^{over})=\epsilon^{over}+Q^{over}$. Differentiating $P(\epsilon^{over})$ with respect to η yields ${\frac{\partial{\epsilon^{over}(\eta)}}{\partial{\eta}}}\geq 0$. Thus R _∅ is decreasing in η.

If $\eta\leq\eta^{under}, \epsilon^{under}$ and Q ^under do not depend on η, so R _D is independent of η.

$$ R_{D} =1-Q^{under}\int\limits_{\epsilon^{under}}^{+\infty} {\frac{h(\epsilon)}{P(\epsilon) (1-H(\epsilon^{under}))}}d\epsilon $$

I turn to the comparative statics on the average cost of capital $\mathcal{R}$.

If η ≥ η ^over, $R_{\delta}=1-{\frac{Q^{over}}{P_{\delta}}}$. Thus ordering $\mathcal{R}(\eta)$ and $\mathcal{R}(\eta')$ boils down to ordering 1/P _δ (η) and 1/P _δ (η′). I know that the function 1/P _δ is a convex and decreasing function. I further proved that for η′ < η, $\Updelta (;\eta)$ second order stochastically dominates $\Updelta(;\eta')$. This implies that ${\mathbb{E}(1/P_{\delta}(\eta)) \leq \mathbb{E}(1/P_{\delta}(\eta'))}$. If η ≤ η ^under then $\mathcal{R}=R_{D}$ as nondisclosing firms are not financed and $R_{D}=1-Q^{under}\int\nolimits_{\epsilon^{under}}^{+\infty} {\frac{h(\epsilon)}{P(\epsilon) (1-H(\epsilon^{under}))}}d\epsilon$, independent of η. I now compare the average cost of capital to the risk premium. By applying Jensen’s inequality,

$$ {\mathbb{E}}(R_{m}^{over})\leq {\mathcal{R}}\quad \hbox{and}\quad {\mathbb{E}}(R_{m}^{under})\leq R_{D} $$

(16)

Proof of Proposition 7

For η ≥ η ^over, the investor’s indirect utility function with a firm disclosing δ is equal to

$$ {\frac{1}{1-\alpha}}P_{\delta}^{1-\alpha}\int\limits f(y)(\theta +y)^{1-\alpha}dy $$

The indirect utility is concave in P _δ . Further I know that for $\eta<\eta', \Updelta (.;\eta')$ second order stochastically dominates $\Updelta(.;\eta)$, which implies that

$$ \frac{1}{1-\alpha}\int\limits P_{\delta}^{1-\alpha}\int f(y)(\theta +y)^{1-\alpha}dy d\Updelta(P_{\delta};\eta')\geq{\frac{1}{1-\alpha}} \int P_{\delta}^{1-\alpha}\int f(y)(\theta +y)^{1-\alpha}dy d\Updelta(P_{\delta};\eta) $$

For η ≤ η ^under, the investors’ aggregate expected utility for a given η is equal to

$$ (1-\eta)\int\limits_{\epsilon^{under}}^{+\infty} {\frac{1}{1-\alpha}}P(\epsilon)^{1-\alpha}\int\limits f(y)\left(\int\limits_{\epsilon^{under}}^{+\infty} (1-\eta)(\epsilon +y)h(\epsilon)d\epsilon \right)^{1-\alpha}dy h(\epsilon)d\epsilon $$

consider η > η′

$$ (1-\eta)\int\limits_{\epsilon^{under}}^{+\infty}{\frac{1}{1-\alpha}} P(\epsilon)^{1-\alpha}\int\limits f(y)\left(\int\limits_{\epsilon^{under}}^{+\infty} (1-\eta)(\epsilon +y)h(\epsilon)d\epsilon \right)^{1-\alpha}dy h(\epsilon)d\epsilon\leq (1-\eta')\int\limits_{\epsilon^{under}}^{+\infty} {\frac{1}{1-\alpha}}P(\epsilon)^{1-\alpha}\int\limits f(y)\left(\int\limits_{\epsilon^{under}}^{+\infty}(1-\eta')(\epsilon +y)h(\epsilon)d\epsilon\right)^{1-\alpha}dy h(\epsilon)d\epsilon $$

Proof of Corollary 1

By Proposition 6, it has been proven that the average cost of capital in the economy is decreasing in the disclosure friction η, if η ≥ η ^over. Further by Proposition 7, the investors’ expected utility increases in the disclosure friction η, if η ≥ η ^over.