Debt contracts in the presence of performance manipulation

Abstract

Empirical evidence suggests that firms often manipulate reported numbers to avoid debt covenant violations. We study how a firm’s ability to manipulate reports affects the terms of its debt contracts and the resulting investment and manipulation decisions that the firm implements. Our model generates novel empirical predictions regarding the use and the level of debt covenant, the interest rate, the efficiency of investment decisions, and the likelihood of covenant violations. For example, the model predicts that the optimal debt contract for firms with relatively strong (weak) corporate governance (i.e., cost of manipulation) induces overinvestment (underinvestment). Moreover, for firms with strong (weak) corporate governance, an increase in corporate governance quality leads to tighter (looser) covenant, more (less) frequent covenant violations and lower (higher) interest rate. Our model highlights that the interest rate, which is a common proxy for the cost of debt, neither accounts for the distortion of investment efficiency nor the expected manipulation costs arising under debt financing. We propose a measure of cost of debt capital that accounts for these effects.

Appendix

Proof of Lemma 1

Observe that

$$E\left( \min \left( x,K\right) |s\right) =\left\{ \begin{array}{ccc} \rho s+\left( 1-\rho \right) \left( {{\int}_{0}^{K}}\frac{x}{h}dx+{{\int}_{K}^{h}} \frac{K}{h}dx\right) & \text{if} & s<K \\ \rho K+\left( 1-\rho \right) \left( {{\int}_{0}^{K}}\frac{x}{h}dx+{{\int}_{K}^{h}} \frac{K}{h}dx\right) & \text{if} & s>K \end{array} \right. . $$

Using the lender’s participation constraint, while assuming $\tau <K,$ yields

$$\frac{\tau }{h}L+{\int}_{\tau }^{h}E\left[ \min \left( x,\mathcal{K}\right) |s \right] f\left( s\right) ds=I. $$

Solving for $\mathcal {\mathcal {K}}$ yields

$$\mathcal{K}\left( \tau \right) =h-\sqrt{\frac{2h\left( \tau L-Ih\right) +\left( h-\tau \right) \left( \rho \tau h+h^{2}\right) }{\rho \tau +h-\tau }} . $$

The function $\mathcal {K}\left (\cdot \right )$ is defined over $[0,\hat {\tau }]$ where $\hat {\tau }\in \left (0,h\right )$ is given by the solution to

$$\mathcal{K}\left( \hat{\tau}\right) =h. $$

Furthermore, observe that $\frac {d\mathcal {K}\left (\tau \right ) }{d\tau } |_{\tau = 0}=h-\sqrt {h^{2}-2Ih}<0$. Also, it’s easy to verify that $\frac {d \mathcal {K}\left (\tau \right ) }{d\tau }|_{\tau =\hat {\tau }}=\infty$ . The equation $\frac {d\mathcal {K}\left (\tau \right ) }{d\tau }= 0$ has two solutions. We argue by contradiction that only one of the two solutions is in $[0,\hat {\tau }]$. By the Intermediate Value Theorem, at least one solution must lie in this interval, given that $\frac {d\mathcal {K}}{d\tau }|_{\tau = 0}<0$ and $\frac {d\mathcal {K}}{d\tau } |_{\tau =\hat {\tau }}>0$. If both solutions lay in this interval, then the equation

$$\frac{d\mathcal{K}\left( \tau \right) }{d\tau }= 0 $$

would have at least three solutions, which is a contradiction. Hence, the function $\mathcal {K}\left (\tau \right )$ has a unique minimum over the interval $\left [ 0,\hat {\tau }\right ]$, denoted $\tau ^{+}$, given by

$$\tau^{+}\equiv \arg\underset{\tau \in \left[ 0,\hat{\tau}\right]}{\min}\mathcal{K} \left( \tau \right) . $$

□

Remark 1

In equilibrium $\tau ^{\ast }\leq K^{\ast }$.

Proof of Remark 1

In equilibrium, the lender’s participation constraint is binding (otherwise, slightly decreasing the contract’s face value K, while holding constant the termination threshold $\tau$, would satisfy the lender’s participation constraint and increase the manager’s expected payoff, despite increasing the expected manipulation cost). Hence,

$$I=\Pr \left( s<\tau^{\ast }\right) L+\Pr \left( s>\tau^{\ast }\right) E\left( \min \left\{ x,K\right\} |s>\tau^{\ast }\right) . $$

For any $s>K$ the continuation value to the lender is independent of s since

$$E\left( \min \left\{ x,K\right\} |s\right) =\rho K+\left( 1-\rho \right) {{\int}_{0}^{h}}\min \left\{ x,K\right\} f\left( x\right) dx. $$

Suppose $\tau ^{\ast }>K$. This implies that the lender’s expected payoff from continuation, given $\tau ^{\ast }$, is strictly greater than his payoff upon termination, which is L. That is

$$E\left( \min \left\{ x,K\right\} |s=\tau^{\ast }\right) >I>L. $$

Therefore, the lender strictly prefers to continue the project when $s=\tau ^{\ast }$. The manager always prefers to continue the project. This implies there is a feasible contract that offers the lender the same face value and a lower threshold, and does not induce higher expected manipulation cost. Such a contract Pareto dominates the assumed contract; hence, a contract with $\tau ^{\ast }>K^{\ast }$ is suboptimal. □

Proof of Lemma 2

We need to show that in any equilibrium $\mathcal {K}^{\prime }\left (\tau \right ) \leq 0$. Let $\left \{ z^{\ast },K^{\ast }\right \}$ be the equilibrium threshold and face value, then the manager’s expected payoff can be written as

$$ {\int}_{\tau^{\ast }}^{h}\left( \mathbb{E}\left[ \max \left( x-K^{\ast },0\right) |s\right] f\left( s\right) ds-c\max \left( z^{\ast }-s,0\right) \right) ds, $$

(10)

where $z^{\ast }$ is the solution to the manager’s misreporting constraint,

$$ c\left( z^{\ast }-\tau^{\ast }\right) =\mathbb{E}\left( \left( x-K^{\ast }\right)^{+} |s=\tau^{\ast }\right) . $$

(11)

Let us consider an alternative contract $\left \{ z^{o},K^{\ast }\right \}$ such that $z^{o}=z^{\ast }-\varepsilon$ for small ε > 0. Define $s^{o}$ as the solution to Eq. (11) such that $c\left (z^{o}-s^{o}\right ) =\mathbb {E}\left (\max \left (x-K^{\ast }\right ) |s^{o}\right )$. From Eq. (11), we see that $z^{o}<z^{\ast }$ implies $s^{o}<\tau ^{\ast }$. If $\mathcal {K}^{\prime }\left (\tau ^{\ast }\right ) >0$ then the contract $\left \{ z^{o},K^{\ast }\right \}$ is feasible. Indeed, recall that the set of feasible contracts is defined by

$$\left\{ \left\{ \tau ,K\right\} :K\geq \mathcal{K}\left( \tau \right) ,\tau \in \left[ 0,\hat{\tau}\right] \right\} , $$

where, as shown before, $\mathcal {K}\left (\cdot \right )$ is a U -shaped function over $\left [ 0,\hat {\tau }\right ]$. We will consider two cases: $\tau ^{\ast }\leq K$ and τ^∗ > K.

When $\tau ^{\ast }\leq K$ a manager with a signal $s=\tau ^{\ast }$ obtains positive expected payoff only if the signal is wrong, in which case his payoff from continuation is independent of $\tau ^{\ast }$. This implies that $z^{\ast }-\tau ^{\ast }$ is independent of $\tau ^{\ast }$. In such a case, offering the contract $\left \{ z^{o},K^{\ast }\right \}$ is preferable to the lender, increases the likelihood of continuation (which is beneficial to the manager), and will have no effect on the manager’s expected manipulation cost. As such, the contract $\left \{ z^{o},K^{\ast }\right \}$ is feasible and strictly dominates the contract $\left \{ z^{\ast },K^{\ast }\right \}$ for which $\mathcal {K}^{\prime }\left (\tau ^{\ast }\right ) >0$.

When $\tau ^{\ast }>K$ a manager with a signal $s=\tau ^{\ast }$ obtains positive expected payoff both when the signal is informative and when it is uninformative. In this case, the manager’s expected payoff from continuation is increasing in $\tau ^{\ast }$. This implies that $z^{\ast }-\tau ^{\ast }$ is also increasing in $\tau ^{\ast }$. To show that the contract $\left \{ z^{o},K^{\ast }\right \}$ dominates $\left \{ z^{\ast },K^{\ast }\right \}$ for which $\mathcal {K}^{\prime }\left (\tau ^{\ast }\right ) >0$ note that decreasing τ^∗ to $s^{0}$ decreases the magnitude of the manipulation for each $s\in \left [ \tau ^{\ast },h\right ]$; hence, the expected payoff of these types is higher under the contract $\left \{ z^{o},K^{\ast }\right \}$ . All types $s\in \left (s^{0},\tau ^{\ast }\right )$ (who didn’t manipulate and terminated the project under $\left \{ z^{\ast },K^{\ast }\right \}$) prefer to manipulate and continue the project and hence prefer $\left \{ z^{o},K^{\ast }\right \}$ over $\left \{ z^{\ast },K^{\ast }\right \}$. Since the lender also prefers the contract $\left \{ z^{o},K^{\ast }\right \}$, this contract is feasible and strictly dominates the contract $\left \{ z^{\ast },K^{\ast }\right \}$ for which $\mathcal {K}^{\prime }\left (\tau ^{\ast }\right ) >0$. □

Lemma 3

$z^{\ast }>h\rightarrow \frac {dz^{\ast }}{dc}<0$ .

Proof of Lemma 3

Let $\zeta \left (\tau \right )$ be the covenant that induces a cutoff τ when the face value is $\mathcal {K}\left (\tau \right )$, or:

$$\zeta \left( \tau \right) =\frac{\left( 1-\rho \right) }{2c}\frac{\left( \mathcal{K}\left( \tau \right) -h\right)^{2}}{h}+\tau . $$

Now, $\frac {\partial \zeta \left (\tau \right ) }{\partial c}<0$. Also we know that $\mathcal {K}^{\prime }\left (\tau ^{\ast }\right ) \leq 0,$ which implies that $\left . \frac {d\zeta \left (\tau \right ) }{d\tau }\right \vert _{\tau =\tau ^{\ast }}>0$ (given Lemma 2). Finally, by Lemma 4, we know that $z^{\ast }>h\Rightarrow \frac { d\tau ^{\ast }}{dc}$ $<0$. Taken together these results imply that when $z^{\ast }=h,$

$$\frac{dz^{\ast }}{dc}=\left. \frac{\partial \zeta \left( \tau \right) }{ \partial c}\right\vert_{\tau =\tau^{\ast }}+\left. \frac{\partial \zeta \left( \tau \right) }{\partial \tau }\right\vert_{\tau =\tau^{\ast }}\frac{ d\tau^{\ast }}{dc}<0. $$

□

Lemma 4

Suppose $\rho \geq \hat {\rho }$ , then there is a unique $\tilde {c}>\hat {c}$ such that $z^{\ast }>h$ if and only if $c<\tilde {c}$ .

Proof of Lemma 4

Clearly $\lim _{c\rightarrow \infty }z^{\ast }=\tau ^{FB}<h$. Also, when ρ is large and c is small we have z^∗ > h and

$$\tau^{\ast }=\arg \underset{\tau \in \left[ 0,\hat{\tau}\right]}{\max}\left\{ V\left( \tau \right) -{\int}_{\tau }^{h}c\left( \zeta \left( \tau \right) -s\right) f\left( s\right) ds\right\} . $$

Taking limits $\lim _{c\rightarrow 0}\tau ^{\ast }=\frac {h-\sqrt {\frac {h\left (h\rho -4\rho I + 2\rho L-2L + 2\rho ^{2}I + 2I\right )}{\rho }}}{1-\rho }>\tau ^{FB},$ hence $\lim _{c\rightarrow 0}z^{\ast }=\infty$.

Finally, Lemma 3 proves that $z^{\ast }=h\rightarrow \frac { dz^{\ast }}{dc}<0,$ so there is only one value $\tilde {c}$ such that $z^{\ast }=h,$ and $z^{\ast }<h$ if and only if $c>\tilde {c}$. Now, when $z^{\ast }=h,$ the optimal threshold satisfies

$$\begin{array}{@{}rcl@{}} V^{\prime }\left( \tau^{\ast }\right) &=&\chi^{\prime }\left( \tau^{\ast }\right) \\ &=&-2c\frac{\left( \frac{1-\rho }{2c}\right)^{2}\left( h-\mathcal{K}\left( \tau^{\ast }\right) \right)^{3}}{h^{3}}\mathcal{K}^{\prime }\left( \tau^{\ast }\right) >0, \end{array} $$

so at $c=\tilde {c}$ there is over-continuation, i.e., $\tau ^{\ast }\left (\tilde {c}\right ) <\tau ^{FB}$. □

Lemma 5

If $c\geq \tilde {c}$ the equilibrium entails over-continuation. Formally $\tau ^{\ast }\leq \tau ^{FB}$ .

Proof of Lemma 5

See proof of Proposition 1. □

Corollary 2

There is $\bar \rho$ such that for all $\rho \geq \bar \rho$ and all $c>0$ the optimal debt contract includes a covenant.

Proof of Corollary 2

First, we show that using a covenant is optimal when $\rho \in \left (1-\varepsilon ,1\right )$ , even as $c\rightarrow 0$. The optimal threshold, in the absence of manipulation, is defined as:

$$\tau^{FB}\equiv \arg \underset{\tau}{\max}\left\{ F\left( \tau \right) L+{\int}_{\tau }^{h}E\left( x|s\right) f\left( s\right) ds\right\} . $$

By assumption $V^{FB}>\mathbb {E}\left (x\right )$. We argue that for $\rho$ high enough, the debt contract includes a covenant, even as $c\rightarrow 0$. Suppose we implement τ^FB as the contract’s threshold (perhaps sub-optimally) and set the face value accordingly at $\mathcal {K}\left (\tau ^{FB}\right )$ to satisfy the lender’s participation constraint. Assuming the contract leads to $z\left (\tau ^{FB}\right ) >h$ (which we can always guarantee by making c small enough), then the expected payoff of the manager is

$$\begin{array}{@{}rcl@{}} {\Pi} \left( \tau^{FB}|\rho \right) +I &=&V^{FB}-\chi \left( \tau^{FB}\right) \\ &>&V^{FB}-\left( h-\tau \right) \left( 1-\rho \right) \frac{\left( 2h\left( \tau L-Ih\right) +\left( h-\tau \right) \left( \rho \tau h+h^{2}\right) \right) }{2h^{2}\left( \rho \tau +h-\tau \right) } \end{array} $$

Now, since

$$\begin{array}{@{}rcl@{}} &&\underset{\rho \rightarrow 1}{\lim}\left\{ V^{FB}-\left( h-\tau \right) \left( 1-\rho \right) \frac{\left( 2h\left( \tau L-Ih\right) +\left( h-\tau \right) \left( \rho \tau h+h^{2}\right) \right) }{2h^{2}\left( \rho \tau +h-\tau \right) }\right\} \\ &=&V^{FB}>E\left( x\right) . \end{array} $$

This means we can select $\rho \$close to 1, denoted $\rho ^{+}$, to ensure

$${\Pi} \left( \tau^{FB}|\rho^{+}\right) +I>E(x)\text{.} $$

Of course, for a fixed $c,$ a large $\rho$ may lead to $\zeta \left (\tau ^{FB}\right ) <h$. So to ensure $\zeta \left (\tau ^{FB}\right ) >h,$ we pick $c$ small enough, say $c_{0}$, such that

$$\begin{array}{@{}rcl@{}} {\Pi} \left( \tau^{FB}|\rho^{+}\right) \,+\,I \!&=&\!\underset{>E\left( x\right) }{ \underbrace{V^{FB}\,-\,\left( h\,-\,\tau^{FB}\right) \left( 1\,-\,\rho^{+}\right) \frac{\left( 2h\left( \tau^{FB}L\,-\,Ih\right) \,+\,\left( h\,-\,\tau^{FB}\right) h\left( \rho^{+}\tau^{FB}+h\right) \right) }{2h^{2}\left( \rho^{+}\tau \,+\,h\,-\,\tau^{FB}\right) }}} \\ +\,\frac{1}{2}\frac{c_{0}\left( h-\tau^{FB}\right)^{2}}{h} &>&\mathbb{E} \left( x\right) . \end{array} $$

This proves that using a covenant is optimal for sufficiently high $\rho ,$ even as $c\rightarrow 0$. □

Lemma 6

$\lim _{c\rightarrow 0}\tau ^{\ast }=\frac {h-\sqrt {\frac { h\left (h\rho -4\rho I + 2\rho L-2L + 2\rho ^{2}I + 2I\right ) }{\rho }}}{1-\rho }>\tau ^{FB}$ and when $z^{\ast }>h,$ $\lim _{\rho \rightarrow 1}\tau ^{\ast }=\frac {L-ch}{1-c}<\tau ^{FB}$ .

Proof of Lemma Lemma 6

To obtain the $\lim _{c\rightarrow 0}\tau ^{\ast }$ we take the first order condition of the manager’s optimization program with respect to $\tau$, assuming $\tau ^{\ast }>h$:

$$V^{\prime }\left( \tau \right) -\chi^{\prime }\left( \tau \right) = 0, $$

which leads to the first order condition

$$\begin{array}{@{}rcl@{}} &&\frac{\left( 2c\rho^{2}-4c\rho + 2c\right) \tau^{3}+\left( h\rho^{2}-h\rho^{3}-6ch+ 8ch\rho -2ch\rho^{2}\right) \allowbreak \tau^{2}}{ 2h\left( \rho \tau +h-\tau \right)^{2}} \\ &&+\,\frac{\left( 6ch^{2}-4ch^{2}\rho -2h^{2}\rho^{2}\right) \tau +\left( 2h^{2}\rho^{2}I-2ch^{3}+ 2Lh^{2}\rho -2h^{2}\rho I\right) \allowbreak }{ 2h\left( \rho \tau +h-\tau \right)^{2}} \\ &=&0 \end{array} $$

Now, taking the limit as $c\rightarrow 0$ and solving for $\tau$ yields

$$\underset{c\rightarrow 0}{\lim}\tau^{\ast }=\frac{h-\sqrt{\frac{h\left( h\rho -4\rho I + 2\rho L-2L + 2\rho^{2}I + 2I\right) }{\rho }}}{1-\rho }. $$

We argue that $\lim _{c\rightarrow 0}\tau ^{\ast }>\tau ^{FB}$. In effect,

$$\underset{\rho \rightarrow 1}{\lim}\left[\underset{c\rightarrow 0}{\lim}\tau^{\ast }-\tau^{FB} \right] = 0. $$

and

$$\underset{\rho \rightarrow 1}{\lim}\left[ \frac{d\left( \lim_{c\rightarrow 0}\tau^{\ast }-\tau^{FB}\right) }{d\rho }\right] =-\frac{1}{2}\frac{ h^{2}+L^{2}-2Ih}{h}<0. $$

When $\rho >\hat {\rho },$ and $c\leq \tilde {c}$, the optimal debt contract includes a covenant. Taking the limit of the FOC as $\rho \rightarrow 1$ and solving for $\tau ^{\ast }$ yields

$$\underset{\rho \rightarrow 1}{\lim}\tau^{\ast }=\frac{L-ch}{1-c}<\underset{\rho \rightarrow 1}{\lim}\tau^{FB}=L\text{.} $$

□

Corollary 3

When $z^{\ast }\leq h,\ \frac {\partial \tau ^{\ast }}{\partial c}>0$ and $\frac {\partial K^{\ast }}{\partial c}<0$ .

Proof of Corollary 3

From Lemma 5, we know that $z^{\ast }\leq h\Rightarrow$ $\chi ^{\prime }\left (\tau ^{\ast }\right ) >0$. Hence $\frac {\partial \chi ^{\prime }\left (\tau ^{\ast }\right ) }{\partial c}=-\frac {1}{c}\chi ^{\prime }\left (\tau ^{\ast }\right ) <0$. Now the first order condition is

$$\begin{array}{@{}rcl@{}} {\Pi}_{\tau }^{\ast } &=&0\Rightarrow \frac{\partial \tau^{\ast }}{\partial c }=-\frac{{\Pi}_{\tau c}^{\ast }}{{\Pi}_{\tau \tau }^{\ast }}=-\frac{-\frac{ \partial \chi^{\prime }\left( \tau^{\ast }\right) }{\partial c}}{{\Pi} _{\tau \tau }^{\ast }} \\ &\Rightarrow &\text{sign}\left( \frac{\partial \tau^{\ast }}{\partial c} \right) =\text{sign}\left( \frac{-\partial \chi^{\prime }\left( \tau^{\ast }\right) }{\partial c}\right) >0 \end{array} $$

This in turn implies that the face value decreases in c. □

Corollary 4

When $z^{\ast }\geq h$ , $\frac {\partial \tau ^{\ast }}{\partial c}<0$ and $\frac {\partial K^{\ast }}{\partial c}>0$ .

Proof of Corollary 4

For a given $\tau$ (such that $\zeta \left (\tau \right ) \geq h$) we can write the manager’s expected payoff as

$${\Pi} \left( \tau |c\right) +I\equiv V\left( \tau \right) -\frac{1}{2}c\left( h-\tau \right) \frac{\frac{\left( 1-\rho \right) }{c}\frac{\frac{2h\left( \tau L-Ih\right) +\left( h-\tau \right) \left( \rho \tau h+h^{2}\right) }{ \rho \tau +h-\tau }}{h}+\tau -h}{h}. $$

The cross partial derivative is

$${\Pi}_{\tau c}=-\frac{h-\tau }{h}<0, $$

which means that in equilibrium

$$\frac{\partial \tau^{\ast }}{\partial c}=-\frac{{\Pi}_{\tau c}^{\ast }}{{\Pi} _{\tau \tau }^{\ast }}<0. $$

□

Proof of Proposition 1

First we show that the contract includes a covenant if and only if precision ρ is high enough. Recall that $\hat {\rho }$ is defined by

$$\underset{\tau \in \left[ 0,\tau^{+}\right]}{\max}V\left( \tau |\hat{\rho}\right) -\chi \left( \tau |\hat{\rho}\right) =E\left( x\right) . $$

$\hat {\rho }$ is the precision level such that the firm is indifferent between using a covenant and not using one. Next we argue that the firm will use a covenant if and only if $\rho \geq \hat {\rho }$. Suppose, by contradiction, that for some $\rho \in \left (\hat {\rho },1\right )$ the firm does not use a covenant. Then we have that

$$\underset{\tau}{\max}\left\{ V\left( \tau |\rho \right) -\chi \left( \tau |\rho \right) \right\} <E\left( x\right) . $$

To prove that this leads to a contradiction, we construct a feasible contract that yields payoffs above $E\left (x\right ) -I$. Indeed, consider a contract implementing the same threshold $\tau ^{\ast }\left (\hat {\rho } \right )$ and face value $K^{\ast }\left (\hat {\rho }\right )$ as under precision $\hat {\rho }$. This new contract strictly satisfies the lender’s participation constraint. Furthermore,

$$\begin{array}{@{}rcl@{}} {\int}_{\tau^{\ast }\left( \hat{\rho}\right) }^{h}E^{\rho }\left[ \left( x\,-\,K\right)^{+}|s\right] f\left( s\right) ds-{\int}_{\tau^{\ast }\left( \hat{ \rho}\right) }^{z\left( \hat{\rho}|\rho \right) }c\left( z\left( \hat{\rho} |\rho \right) -s\right) f\left( s\right) ds &>&\underset{\tau \in \left[ 0,\tau^{+}\right]}{\max}V\left( \tau |\hat{\rho}\right) -\chi \left( \tau |\hat{\rho} \right) \\ &\geq &E\left( x\right) , \end{array} $$

Hence, the contract we have constructed yields both higher continuation cash flows and lower misreporting costs under $\rho$ than the optimal debt contract under $\hat {\rho }$, hence it must dominate a no-covenant contract. To prove the other direction suppose $\rho <\hat {\rho },$ but the debt contract includes a covenant, so

$$\underset{\tau}{\max}V\left( \tau |\rho \right) -\chi \left( \tau |\rho \right) >E\left( x\right) . $$

Then consider using $\left \{ \tau ^{\ast }\left (\rho \right ) ,K^{\ast }\left (\rho \right ) \right \}$ under precision $\hat {\rho }$. If this contract was feasible under precision $\rho$ it must also be feasible under $\hat {\rho }$ given that $\hat {\rho }>\rho$. Clearly, this contract must give the manager a higher payoff under $\hat {\rho }$ than under $\rho ,$ hence

$$\begin{array}{@{}rcl@{}} {\int}_{\tau^{\ast }\left( \rho \right) }^{h}E^{\hat{\rho}}\left[ \left( x-K\right)^{+}|s\right] f\left( s\right) ds-{\int}_{\tau^{\ast }\left( \rho \right) }^{z\left( \rho |\hat{\rho}\right) }c\left( z\left( \rho |\hat{\rho} \right) -s\right) f\left( s\right) ds &>&\underset{\tau}{\max}V\left( \tau |\rho \right) -\chi \left( \tau |\rho \right) \\ &>&E\left( x\right) . \end{array} $$

which is a contradiction. Next, we show that when $\rho \geq \hat {\rho }$ there is over-continuation if and only if the cost of misreporting is higher than $\hat {c}$. Define $\tau ^{\ast }\left (c\right )$ as

$$\tau^{\ast }\left( c\right) \equiv \arg \underset{\tau}{\max}\left\{ V\left( \tau \right) -\chi \left( \tau |c\right) \right\} . $$

First, Corollary 2 demonstrates that $\lim _{c\downarrow 0}\tau ^{\ast }\left (c\right ) >\tau ^{FB}$. Also, Corollary 3 and Corollary 4 prove that $\tau ^{\ast }\left (c\right )$ decreases (resp. increases) in c for $c\geq \tilde {c}$ (resp. $c<\tilde {c}$) where $\tilde {c}$ is defined as $\zeta \left (\tau ^{\ast }\left (\tilde {c}\right ) \right ) =h$ (i.e, the value of c such that the equilibrium covenant is equal to h). Finally, $\lim _{c\rightarrow \infty }\tau ^{\ast }\left (c\right ) =\tau ^{FB}$. Taken together these observations establish that there is a unique $\hat {c}$ defined

$$\tau^{\ast }\left( \hat{c}\right) =\tau^{FB} $$

such that if $c\leq \hat {c}$ (resp. $c>\hat {c}$) there is over-termination (resp. over-continuation).

Consider uniqueness of the optimal threshold $\tau ^{\ast }$. When $z^{\ast }\geq h$ the proof follows by contradiction. In this case, the first order condition of $V^{\prime }(\tau )=\chi ^{\prime }(\tau )$ is a third order polynomial and has at most three real solutions, but the smallest solution is negative. The other two consecutive solutions cannot be both maxima, hence the maximum must be unique. When $z^{\ast }<h$ the first order condition is a fourth order polynomial, so there can be (at most) two local maxima of ${\Pi } \left (\tau \right )$ . Now, we will show that one of the maxima lies on $[\frac {h}{1-\rho },\infty ),$ being outside the relevant range. Indeed,

$$\underset{\tau \downarrow \frac{h}{1-\rho }}{\lim}{\Pi} \left( \tau \right) =-\infty $$

and

$$\underset{\tau \rightarrow \infty}{\lim}{\Pi} \left( \tau \right) =-\infty . $$

This means there is at least one local maxima in $[\frac {h}{1-\rho },\infty )$ which proves the optimal threshold is unique. □

Proof of Proposition 2

First we prove the comparative statics with respect to c. As before, we define

$$\tau^{\ast }\left( c\right) \equiv \arg \underset{\tau \in \left[ 0,h\right] }{\max}\left\{ V\left( \tau \right) -\chi \left( \tau |c\right) \right\} . $$

(Note that $\tau ^{\ast }\left (c\right )$ is not necessarily the optimal threshold since the optimal debt contract may use no covenant, in which case $z^{\ast }=\tau ^{\ast }= 0$.)

Lemma 3 proves there is a unique value of $c,$ denoted $\tilde { c},$ such that the covenant is $z^{\ast }=h$. If $c<\tilde {c}$ (resp. $c> \tilde {c}$) the covenant is larger (smaller) than h. On the other hand,

$$\frac{\partial \tau^{\ast }\left( c\right) }{\partial c}=-\frac{{\Pi}_{\tau c}}{{\Pi}_{\tau \tau }}. $$

Hence, sign$\left \{ \frac {\partial \tau ^{\ast }\left (c\right ) }{\partial c} \right \} =\textit {sign}\left (\frac {-\partial \chi ^{\prime }\left (\tau ^{\ast }\left (c\right ) \right ) }{\partial c}\right )$ . Now when $c<\tilde {c,}$ we have $\frac {-\partial \chi ^{\prime }\left (\tau ^{\ast }\left (c\right ) \right ) }{\partial c}<0,$ hence $\frac {\partial \tau ^{\ast }\left (c\right ) }{\partial c}<0$. When $c>\tilde {c},$ we have $\frac {-\partial \chi ^{\prime }\left (\tau ^{\ast }\left (c\right ) \right ) }{\partial c}=-\frac {1}{c}\chi ^{\prime }\left (\tau ^{\ast }\left (c\right ) \right ) <0,$ and $\chi ^{\prime }\left (\tau ^{\ast }\left (c\right ) \right ) >0,$ hence $\frac { \partial \tau ^{\ast }\left (c\right ) }{\partial c}>0$. Also, since $\frac { \partial K^{\ast }}{\partial c}=\mathcal {K}^{\prime }\left (\tau ^{\ast }\right ) \frac {\partial \tau ^{\ast }\left (c\right ) }{\partial c}$ the sign $\left (-\frac {\partial K^{\ast }}{\partial c}\right ) =\textit {sign}\left (\frac { \partial \tau ^{\ast }\left (c\right ) }{\partial c}\right ) ,$ since $\mathcal {K}^{\prime }\left (\tau ^{\ast }\right ) <0$. Consider the comparative statics with respect to $\rho$. Consider the $z^{\ast }>h$ case. We have:

$$\begin{array}{@{}rcl@{}} \underset{\rho \rightarrow 1}{\lim}\tau^{\ast } &=&\frac{L-hc}{1-c} \\ \underset{\rho \rightarrow 1}{\lim}\frac{\partial \tau^{\ast }}{\partial \rho } &=&\underset{\rho \rightarrow 1}{\lim}\left( -\frac{{\Pi}_{\tau \rho }}{{\Pi}_{\tau \tau }}\right) =\frac{\frac{-4\tau L + 2Ih+ 2hL+ 3\tau^{2}-4\tau h}{2h^{2}}}{-{\Pi} _{\tau \tau }}. \end{array} $$

For $c\approx 0$, we have $\tau ^{\ast }\approx L$. Plugging $\tau ^{\ast }\approx L$ above yields $\lim _{\rho \rightarrow 1}\frac {\partial \tau ^{\ast }}{\partial \rho }\approx \allowbreak \frac {-L^{2}+ 2h\left (I-L\right ) }{-{\Pi }_{\tau \tau }2h^{2}}$ which is negative as $L\rightarrow I$. This shows that $\tau ^{\ast }$ may decrease in $\rho ,$ for sufficiently high $\rho$ and small c. However, we can verify that even in this case, we have $\lim _{\rho \rightarrow 1,c\rightarrow 0}\frac {\partial K^{\ast }}{ \partial \rho }<0$. Consider the $z^{\ast }<h$ case. It’s easy to verify $\lim _{\rho \rightarrow 1}{\Pi }_{\tau \rho }^{\ast }=\frac {E\left (x\right ) -\tau ^{\ast }}{4h}$. Since $\tau ^{\ast }<E\left (x\right ) ,$ this implies that $\lim _{\rho \rightarrow 1}\frac {\partial \tau ^{\ast }}{\partial \rho } >0$. □