Internal financing, managerial compensation and multiple tasks

Abstract

We study the optimal capital budgeting policy of a firm taking into account the choice between internal and external financing. The manager can dedicate effort either to increase short-term profitability, thus generating greater immediate cash-flow, or to improve long-term perspectives. When both types of effort are observable, low productivity firms end up using internal funds, while high productivity firms use external capital markets. When effort to boost short-term cash flow is observable, while effort to boost long-term profitability is not, non-monotonic policies may be optimal. In such cases financing switches back and forth between internal and external funds as the quality of the project increases.

Appendices

Appendix I

Proof of Proposition 1

Consider problem (2). Since $\alpha \in \left( 0,1\right)$ and $\gamma \in \left( 0,1\right)$ the marginal return of $e_p$ and k at zero goes to infinity, so the non-negativity constraints for those two variables can be ignored. Since the function is differentiable, the solution must satisfy the first order conditions of the Lagrangian

$$\begin{aligned} L\left( e_c,e_p,k,\lambda \right) =\left( \theta +e_{p}^{\alpha }\right) k^{\gamma }-\left( k - e_{c}\right) -\frac{1}{2}\left( e_{c}+e_{p}\right) ^{2} + \lambda e_c \end{aligned}$$

(14)

and the complementary slackness conditions $\lambda e_c=0$, $e_c \ge 0$ and $\lambda \ge 0$.

The first order conditions are

$$\begin{aligned} \gamma \left( \theta +e_{p}^{\alpha }\right)= & {} k^{1-\gamma } \end{aligned}$$

(15)

$$\begin{aligned} \alpha e_{p}^{\alpha -1}k^{\gamma }= & {} e_{c}+e_{p} \end{aligned}$$

(16)

$$\begin{aligned} e_{c}+e_{p}= & {} 1+\lambda \end{aligned}$$

(17)

$$\begin{aligned} \lambda \ge 0,\qquad e_{c}\ge 0,\qquad \lambda e_{c}= & {} 0. \end{aligned}$$

(18)

If the solution is such that $\lambda = 0$ then $e_c + e_p =1$ and (16) implies $k = \alpha ^{-\frac{1}{\gamma }} e_p^{\frac{1-\alpha }{\gamma }}$. Substituting k into (15) we obtain the equation:

$$\begin{aligned} \theta e_{p}^{\frac{\left( 1-\alpha \right) \left( \gamma -1\right) }{\gamma }}+e_{p}^{\frac{\alpha +\gamma -1}{\gamma }}= \gamma ^{-1} \alpha ^{ \frac{\gamma - 1}{\gamma }}. \end{aligned}$$

(19)

Since $\alpha +\gamma <1$, the left hand side is strictly decreasing in $e_{p}$ and it goes from $+\infty$ to 0 as $e_{p}$ moves from 0 to $+\infty$. The right hand side is strictly positive so there is a unique solution. Call $\hat{e}_p \left( \theta \right)$ the solution and observe, by inspection, that it is strictly increasing in $\theta$. It is feasible if $\hat{e}_{c}\left( \theta \right) =1-\hat{e}_{p}\left( \theta \right) \ge 0$ or $\hat{e}_{p} \left( \theta \right) \le 1$. There is a unique value $\theta ^{+}$ such that $e_{p}^{*}\left( \theta ^{+} \right) = 1$, which is given by

$$\begin{aligned} \theta ^+ = \gamma ^{-1} \alpha ^{-\frac{1-\gamma }{\gamma }} -1 \end{aligned}$$

(20)

Thus, the solution obtained for $\lambda = 0$ is acceptable if $\theta \le \theta ^{+}$.

Suppose now that $\lambda >0$, so that $e_c = 0$ and $e_p > 1$. In this case (16) implies $k = \alpha ^{-\frac{1}{\gamma }} e_p^{\frac{2-\alpha }{\gamma }}$. Substituting k into (15) we obtain

$$\begin{aligned} \theta e_{p}^{\frac{\left( 2-\alpha \right) \left( \gamma -1\right) }{\gamma }}+e_{p}^{\frac{\alpha +2\gamma -2}{\gamma }}=\gamma ^{-1} \alpha ^{\frac{\gamma - 1}{\gamma }}. \end{aligned}$$

(21)

Equation (21) has a unique solution, call it $\overline{e}_{p}\left( \theta \right)$, which is strictly increasing in $\theta$ and it satisfies $\overline{e}_{p} \left( \theta ^+\right) = 1$. Thus, the solution is feasible only if $\theta >\theta ^{+}$. We conclude that for each $\theta$ there is only one stationary point. Let $e^{*}_p \left( \theta \right)$ be such solution, i.e. $e^{*}_p \left( \theta \right) = \hat{e}_p \left( \theta \right)$ if $\theta \le \theta ^{+}$ and $e^{*}_p \left( \theta \right) = \overline{e}_p \left( \theta \right)$ if $\theta > \theta ^{+}$.

To show that the stationary point is a constrained maximum, observe that the objective function is strictly concave in $\left( k,e_p\right)$ for each $e_c$ (the objective function is not concave in the three variables $\left( k,e_p,e_c\right)$). This can be checked looking at the Hessian matrix

$$\begin{aligned} \begin{bmatrix} \gamma \left( \gamma - 1\right) \left( \theta + e_p^{\alpha }\right) k^{\gamma - 2} &{} \quad \gamma \alpha e_{p}^{\alpha - 1}k^{\gamma - 1} \\ \gamma \alpha e_{p}^{\alpha - 1}k^{\gamma - 1} &{} \quad \alpha \left( \alpha - 1\right) e_p^{\alpha - 2} k^{\gamma } - 1 \end{bmatrix} \end{aligned}$$

The values on the main diagonal are negative and the determinant is positive under the assumption $\gamma +\alpha <1$. Thus, for any fixed $e_c$ the unique stationary point is a maximum. Let $k\left( e_c,\theta \right)$ and $e_p \left( e_c,\theta \right)$ be the points at which the maximum is attained and define

$$\begin{aligned} F \left( e_c,\theta \right) = \left( \theta + e^{\alpha }_p \left( e_c,\theta \right) \right) k^{\gamma } \left( e_c, \theta \right) - \left( k \left( e_c, \theta \right) - e_c \right) - \frac{1}{2} \left( e_c + e_p\left( e_c,\theta \right) \right) ^2 \end{aligned}$$

(22)

By inspection, $\lim _{e_c \rightarrow + \infty } F\left( e_c,\theta \right) = - \infty$. Consider now the problem

$$\begin{aligned} \max _{e_c \ge 0} \; \; F \left( e_c, \theta \right) \end{aligned}$$

(23)

with Lagrangian

$$\begin{aligned} L \left( e_c,\theta ,\lambda \right) = F \left( e_c, \theta ,\right) + \lambda e_c \end{aligned}$$

(24)

The envelope theorem implies

$$\begin{aligned} \frac{dL}{de_c} = 1-\left( e_c+e_p\left( e_c, \theta \right) \right) + \lambda . \end{aligned}$$

so the first order and complementary slackness conditions are

$$\begin{aligned} 1-\left( e_c+e_p\left( e_c, \theta \right) \right) + \lambda= & {} 0 \\ \lambda e_c= & {} 0, \; \; \; \lambda \ge 0, \; \; \; e_c \ge 0 \end{aligned}$$

Suppose first $\theta > \theta ^{+}$. In that case we established that $\left. \frac{dF }{d e_c} \right| _{e_c=0} <0$ and there is no point $\theta > \theta ^+$ at which $\left( e_c+e_p\left( e_c, \theta \right) \right) =1$. We conclude that F is always decreasing and the maximum is achieved at $e_c =0$.

If instead $\theta < \theta ^+$ then we established $\left. \frac{dF }{d e_c} \right| _{e_c=0} > 0$, so that the solution must involve $e_c >0$. Since there is a unique stationary point at which $e_c>0$ and the objective function goes to $- \infty$ as $e_c$ gets large, the point must be a maximum. $\square$

Proof of Proposition 2

Existence of a solution follows from the results in Kadan et al. (2017). Given an optimal mechanism $w \left( V,\hat{\theta }\right) , k \left( \hat{\theta }\right) , e_c \left( \hat{\theta }\right) , e_p \left( \hat{\theta }\right)$, define $w^e \left( \theta , \hat{\theta }, e_p\right)$, $U\left( \theta , \hat{\theta }, e_p \right)$, $U\left( \theta \right)$ as in main text and $w^e \left( \theta \right) \equiv w^e \left( \theta , \theta , e_p \left( \theta \right) \right)$.

Observe that

$$\begin{aligned} w^{e} \left( \theta , \hat{\theta }, e_p\right) =\int _{0}^{+\infty }w\left( \left( \theta +e_{p}^{\alpha }\right) k^{\gamma }\left( \widehat{ \theta }\right) u,\widehat{\theta }\right) f\left( u\right) du \end{aligned}$$

can be written, using the change of variable $v=\left( \theta +e_{p}^{\alpha }\right) k^{\gamma }\left( \widehat{\theta }\right) u$, as

$$\begin{aligned} w^e \left( \theta , \hat{\theta }, e_p\right) =\int _{0}^{+\infty }w\left( v,\widehat{\theta }\right) \frac{f\left( \frac{v}{\left( \theta +e_{p}^{\alpha }\right) k^{\gamma }\left( \widehat{\theta }\right) }\right) }{\left( \theta +e_{p}^{\alpha }\right) k^{\gamma }\left( \widehat{\theta } \right) }dv, \end{aligned}$$

(25)

so $w^e \left( \theta , \hat{\theta }, e_p\right)$ is differentiable with respect to $\theta$ and $e_p$. The derivative with respect to $\theta$ is

$$\begin{aligned} \frac{\partial w^e \left( \theta , \hat{\theta }, e_p\right) }{\partial \theta } = - \frac{\int _{0}^{+\infty }w\left( \left( \theta + e_p^{\alpha }\right) k^{\gamma } \left( \hat{\theta }\right) u,\widehat{\theta }\right) \left[ f'\left( u\right) u + f\left( u\right) \right] du}{\left( \theta + e_p^{\alpha } \right) ^2 k^{\gamma }\left( \widehat{\theta }\right) }. \end{aligned}$$

(26)

The integrals in (25) and (26) must be well defined in an optimal mechanism in order to ensure that the maximization problem for the agent is solvable.

$\underline{\mathbf{Step 1: } U\left( \theta \right) \hbox {is continuous}}$. Suppose not. Then there is a sequence $\left\{ \theta _n \right\}$ converging to a value $\theta ^*$ and $\varepsilon >0$ such that $\left| U \left( \theta ^*\right) - U \left( \theta _n\right) \right| > \varepsilon$ each n. This in turn implies

$$\begin{aligned} \left| U \left( \theta ^*\right) - U \left( \theta _n,\theta ^*, e_p \left( \theta ^*\right) \right) \right| + \left| U \left( \theta _n,\theta ^*, e_p \left( \theta ^*\right) \right) - U \left( \theta _n\right) \right| > \varepsilon \end{aligned}$$

Continuity of $w^e$ in its first argument implies that there is N such that, for $n>N$ we have

$$\begin{aligned} \left| U \left( \theta ^*\right) - U\left( \theta _n, \theta ^*, e_p \left( \theta ^*\right) \right) \right| < \frac{\varepsilon }{2}. \end{aligned}$$

In turn, incentive compatibility implies

$$\begin{aligned} U\left( \theta _n\right) > U\left( \theta _n, \theta ^*, e_p \left( \theta ^*\right) \right) + \frac{\varepsilon }{2} \end{aligned}$$

and

$$\begin{aligned} U\left( \theta ^*\right) > U\left( \theta ^*, \theta _n, e_p \left( \theta _n\right) \right) . \end{aligned}$$

Summing the two inequalities side by side we obtain

$$\begin{aligned} \left[ U\left( \theta _n\right) - U\left( \theta ^*, \theta _n, e_p \left( \theta _n\right) \right) \right] + \left[ U\left( \theta ^*\right) - U\left( \theta _n, \theta ^*, e_p \left( \theta ^*\right) \right) \right] > \frac{\varepsilon }{2} \end{aligned}$$

which can be written as

$$\begin{aligned} \left[ w^e \left( \theta _n \right) - w^e \left( \theta ^*, \theta _n, e_p\left( \theta _n\right) \right) \right] + \left[ w^e \left( \theta ^* \right) - w^e \left( \theta _n, \theta ^*, e_p\left( \theta ^*\right) \right) \right] > \frac{\varepsilon }{2} \end{aligned}$$

Since $w^{e} \left( \theta ,\hat{\theta },e_p \right)$ is continuous in $\theta$, both terms on the LHS go to zero as n goes to infinity, a contradiction. This establishes the continuity of $U \left( \theta \right)$.

$\underline{\mathbf{Step 2: } U \left( \theta \right) \hbox {is non-decreasing}.}$ Since the mechanism maximizes expected profit, it should not be possible to implement the same allocation at a lower expected cost. We will show that if U is strictly decreasing on an interval $\left( \theta _a, \theta _b\right)$ then it is possible to offer a strictly lower expected salary and implement the same allocation, thus contradicting the optimality of the mechanism.

Individual rationality implies $U\left( \theta _a \right) > U\left( \theta _b \right) \ge 0$. Since U is continuous and strictly decreasing on $\left( \theta _a, \theta _b\right)$, it is almost everywhere differentiable on that interval. Let

$$\begin{aligned} \delta = \sup _{\theta \in \left( \theta _a, \theta _b\right) } \max \left\{ U'_{-}\left( \theta \right) ,U'_{+}\left( \theta \right) \right\} . \end{aligned}$$

where $U'_{-}\left( \theta \right)$ and $U'_{+}\left( \theta \right)$ are the left and right derivative of U. The interval $\left( \theta _a, \theta _b\right)$ can be chosen so that $\delta < 0$.

For each positive integer n define the function

$$\begin{aligned} q_n \left( V,\hat{\theta }\right) = \frac{1}{ {\mathbb {E}}\left[ u^n\right] } \left( \frac{V}{ k^{\gamma }\left( \hat{\theta }\right) }\right) ^n \end{aligned}$$

where $V=\left( \theta + e^{\alpha }_p\right) k \left( \hat{\theta }\right) u$. Consider now polynomial functions of the form

$$\begin{aligned} q^*\left( V,\hat{\theta }\right) = \sum _{n=0}^{m} \gamma _n q_n \left( V,\hat{\theta }\right) . \end{aligned}$$

When $e_p = 0$ we have ${\mathbb {E}} \left[ q_n \left( V,\hat{\theta }\right) \right] = \theta ^n$. This implies that we can build any polynomial function

$$\begin{aligned} q\left( \theta \right) = \sum _{n=0}^{m} \gamma _n \theta ^n ={\mathbb {E}} \left[ q^*\left( V,\hat{\theta }\right) \right] . \end{aligned}$$

By the Stone–Weierstrass theorem any continuous function on the interval $\left[ \underline{\theta },\overline{\theta }\right]$ can be approximated to any degree desired by a polynomial function. Thus, we can define a function

$$\begin{aligned} \eta \left( V,\hat{\theta } \right) =\sum _{n=0}^{\infty } \gamma _n q_n\left( V,\hat{\theta } \right) \end{aligned}$$

such that

$$\begin{aligned} \eta \left( \theta \right) \equiv {\mathbb {E}}\left[ \eta \left( V,\hat{\theta } \right) \right] = \max \left\{ 0, \left( \theta _a-\theta \right) \left( \theta -\theta _b\right) \right\} . \end{aligned}$$

The derivative of $\eta$ on the interval $\left( \theta _a, \theta _b \right)$ is given by $\eta ' \left( \theta \right) = \theta _a + \theta _b - 2 \theta$. Now consider a new mechanism which is identical to the initial one except that the wage function is replaced by

$$\begin{aligned} \hat{w} \left( V,\hat{\theta }\right) = w\left( V,\hat{\theta }\right) - \varepsilon \eta \left( V,\hat{\theta } \right) \end{aligned}$$

where $\varepsilon >0$ is chosen to be sufficiently small so that $\varepsilon \eta ' \left( \theta \right) > \delta$ for each $\theta \in \left( \theta _a, \theta _b\right)$. This implies that $U \left( \theta \right)$ remains strictly decreasing on $\left( \theta _a, \theta _b \right)$.

The new mechanism gives the same incentives as the previous one. In particular notice that, since ${\mathbb {E}} \left[ \eta \left( V,\hat{\theta } \right) \right]$ does not depend on $\hat{\theta }$, truth-telling must remain optimal. Also, the choice of $e_p$ remains optimal and in particular it remains optimal to select $e_p=0$ in the interval $\left( \theta _a,\theta _b\right)$.

We conclude that the new mechanism satisfies both incentive compatibility and individual rationality and it implements the same allocation at a strictly lower cost, contradicting the optimality of the original scheme. Thus, $U \left( \theta \right)$ is non-decreasing.

$\underline{\mathbf{Step 3: } \hbox {A formula for } U' \left( \theta \right) \hbox {when } e_p \left( \theta \right) >0}$. Since $U \left( \theta \right)$ is continuous and non-decreasing on the compact interval $\left[ \underline{\theta },\overline{\theta }\right]$ it is differentiable almost everywhere. By the envelope theorem the derivative must be

$$\begin{aligned} U' \left( \theta \right) = \left. \frac{\partial w^e \left( s,\hat{\theta }, e_p \right) }{\partial s } \right| _{\left( s,\hat{\theta },e_p\right) = \left( \theta ,\theta ,e_p \left( \theta \right) \right) } \end{aligned}$$

Using the expression of $w^{e}\left( \theta ,\widehat{\theta } ,e_{p} \right)$ given in (25) and the assumption that the density $f \left( u\right)$ is differentiable it is immediate to see that whenever $e_p > 0$ we have

$$\begin{aligned} \frac{\partial w^{e}}{\partial e_{p}}=\frac{\partial w^{e}}{\partial \theta } \alpha e_{p}^{\alpha -1}. \end{aligned}$$

(27)

Suppose now that at $\theta$ the project improving effort is $e_{p}\left( \theta \right) >0$. A necessary condition for implementability is that $e_{p}\left( \theta \right)$ maximizes the expected utility of the manager when the true $\theta$ is reported, i.e.

$$\begin{aligned} e_{p}\left( \theta \right) \in \arg \max _{e_{p} \ge 0}\quad w^{e}\left( \theta ,\theta , e_{p} \right) -\frac{\left( e_{c}\left( \theta \right) +e_{p}\right) ^{2}}{2} \end{aligned}$$

Since $w^{e}$ is differentiable, a necessary condition for optimality is

$$\begin{aligned} \frac{\partial w^{e}\left( \theta ,\theta ,e_{p} \right) }{\partial \theta } \alpha e_{p}^{\alpha -1}=e_{c}\left( \theta \right) +e_{p} \end{aligned}$$

where we made use of (27). This implies

$$\begin{aligned} \frac{\partial w^{e}\left( \theta ,\theta , e_{p} \right) }{\partial \theta }= \frac{e_{c}\left( \theta \right) +e_{p}\left( \theta \right) }{\alpha } e_{p}^{1-\alpha }\left( \theta \right) \end{aligned}$$

(28)

whenever $e_{p}\left( \theta \right) >0$.

$\underline{\mathbf{Step 4: } \frac{\partial w^e}{\partial \theta } =0 \hbox { when } e_p \left( \theta \right) =0}$. Suppose the optimal policy prescribes $e_p \left( \theta \right) = 0$ on an interval $\left( \theta _a,\theta _b\right)$. A necessary condition for $e_{p}=0$ to be an optimal choice is

$$\begin{aligned} \lim _{e_p \downarrow 0} \left[ \frac{\partial w^{e}\left( \theta ,\theta ,e_{p} \right) }{\partial \theta } \alpha e_{p}^{\alpha -1} - \left( e_{c}\left( \theta \right) +e_{p} \right) \right] \le 0. \end{aligned}$$

(29)

Since $e_{p}^{\alpha -1}$ goes to $+\infty$ as $e_{p}$ goes to zero, inequality (29) implies $\frac{\partial w^{e}\left( \theta ,\theta ,e_{p} \right) }{\partial \theta }\le 0$. Suppose the inequality is strict on $\left( \theta _a,\theta _b\right)$. Since $U \left( \theta \right)$ is non-decreasing, it must be the case that $e_c \left( \theta \right)$ is strictly decreasing.

Furthermore, it must be $e_c \left( \theta \right) \le 1$. In fact, since $e_c \left( \theta \right)$ is strictly decreasing on the interval, it must be $e_c \left( \theta \right) <1$ almost everywhere. If not, reducing $e_c$ by $\varepsilon < 2\left( e_c-1\right)$ and decreasing the wage by $\varepsilon ' = e_c \varepsilon - 0.5 \varepsilon ^2$ would leave the manager with the same utility, would not affect incentive compatibility and it would yield a higher profit. This is sub-optimal. Increasing the effort $e_c$ by $\varepsilon$ and increasing the compensation by $\varepsilon ' = 0.5 \varepsilon ^2 +e_c \varepsilon$ in order to keep utility constant results in a higher profit for $\varepsilon$ sufficiently small. The change does not violate incentive compatibility.

$\underline{\mathbf{Step 5: } \frac{\partial w^e \left( \theta , \hat{\theta }, e_p \right) }{\partial \theta } \hbox { is absolutely continuous wrt } \theta }$. A sufficient condition for absolute continuity of a function over a compact interval is that the derivative is bounded. Equation (28) and the fact that $\frac{\partial w^e}{\partial \theta } =0$ when $e_p \left( \theta \right) =0$ imply that the expression (28) holds both when $e_p \left( \theta \right) >0$ and when $e_p \left( \theta \right) =0$. Thus, $\frac{\partial w^e}{\partial \theta }$ is bounded below by zero and above by

$$\begin{aligned} \sup _{\theta \in \left[ \underline{\theta }, \overline{\theta }\right] } \; \; \frac{e_{c}\left( \theta \right) +e_{p}\left( \theta \right) }{\alpha } e_{p}^{1-\alpha }\left( \theta \right) \end{aligned}$$

which must be bounded in any optimal mechanism.

$\underline{\mathbf{Step 6: } \hbox {The partial derivative of} w^{e} \hbox {is non-decreasing}}$. Since $w^e \left( \theta , \hat{\theta }, e_p \right)$ is differentiable with respect to $\theta$, for each pair $\left( \hat{\theta },e_p\right)$ we can write

$$\begin{aligned} w^e \left( \theta , \hat{\theta }, e_p \right) = w^e \left( \underline{\theta }, \hat{\theta }, e_p \right) + \int _{\underline{\theta }}^{\theta } \frac{\partial w^e \left( s,\hat{\theta },e_p\right) }{\partial s} ds. \end{aligned}$$

(30)

Incentive compatibility implies

$$\begin{aligned} w^{e} \left( \theta ,\theta ,e_p\left( \theta \right) \right) -\frac{1}{2}\left( e_c\left( \theta \right) +e_p \left( \theta \right) \right) ^2 \ge w^{e} \left( \theta ,\hat{\theta },e_p\left( \hat{\theta }\right) \right) -\frac{1}{2}\left( e_c\left( \hat{\theta }\right) +e_p \left( \hat{\theta }\right) \right) ^2 \end{aligned}$$

(31)

and

$$\begin{aligned}&w^{e} \left( \hat{\theta },\hat{\theta },e_p\left( \hat{\theta }\right) \right) -\frac{1}{2}\left( e_c\left( \hat{\theta }\right) +e_p \left( \hat{\theta }\right) \right) ^2 \nonumber \\&\quad \ge w^{e} \left( \hat{\theta },\theta ,e_p\left( \theta \right) \right) -\frac{1}{2}\left( e_c\left( \theta \right) +e_p \left( \theta \right) \right) ^2 \end{aligned}$$

(32)

Now set $\hat{\theta }= \theta +\varepsilon$. Summing (31) and (32) side by side and using expression (30) for $w^e \left( \theta ,\hat{\theta },e_p\right)$ we obtain

$$\begin{aligned} \int _{\theta }^{\theta +\varepsilon } \left[ \frac{\partial w^e \left( s,\theta +\varepsilon ,e_p \left( \theta +\varepsilon \right) \right) }{\partial s} - \frac{\partial w^e \left( s,\theta ,e_p \left( \theta \right) \right) }{\partial s}\right] ds \ge 0 \end{aligned}$$

(33)

for each $\varepsilon > 0$.

Define the functions

$$\begin{aligned} h \left( \theta , \hat{\theta }\right) \equiv \frac{\partial w^e \left( \theta , \hat{\theta }, e_p \left( \hat{\theta }\right) \right) }{\partial \theta } \;\;\;\; \;\;\;\; \;\;\;\; \;\;\;\; \;\;\;\; h\left( \theta \right) \equiv h\left( \theta , \theta \right) \end{aligned}$$

and suppose that there is an interval $\left( \theta _a,\theta _b\right)$ such that $h \left( \theta _b \right) < h\left( \theta _a\right)$. Define

$$\begin{aligned} \kappa = \frac{h\left( \theta _b\right) - h\left( \theta _a \right) }{\theta _b - \theta _a} \end{aligned}$$

and notice that $\kappa < 0$. Notice also that for each value $\delta _h < \theta _b - \theta _a$ we can find an interval $\left( \theta '_a, \theta '_b \right) \subset \left( \theta _a, \theta _b \right)$ with the properties that $\theta '_b - \theta '_a < \delta _h$ and $\frac{h\left( \theta '_b\right) - h\left( \theta '_a \right) }{\theta '_b - \theta '_a} \le k$.

For each $\theta$ the function $h \left( s,\theta \right)$ is absolutely continuous in s. This means that for each $\varepsilon$ there is $\delta$ such that $\left| s - \theta \right| < \delta$ implies $\left| h(s,\theta ) - h\left( \theta \right) \right| < \varepsilon$.

Now pick an interval $\left( \theta '_a, \theta _b' \right)$ such that $\frac{h\left( \theta '_b\right) - h\left( \theta '_a \right) }{\theta '_b - \theta '_a} \le \kappa$ and for each $s \in \left( \theta _a', \theta _b'\right)$, $\left| s- \theta \right| < \theta _b' - \theta _a'$ implies $\left| h(s,\theta _a) - h \left( \theta _a\right) \right| < \frac{\left| \kappa \right| }{4}$ and $\left| h(s,\theta _b) - h \left( \theta _b\right) \right| < \frac{\left| \kappa \right| }{4}$.

$$\begin{aligned} 0\le & {} \int _{\theta _a'}^{\theta _b'} \left[ h\left( s, \theta _b'\right) - h\left( s, \theta _a' \right) \right] ds = \int _{\theta _a'}^{\theta _b'} \left[ h\left( s, \theta _b'\right) - h\left( \theta _b' \right) \right] ds \\&\quad + \int _{\theta _a'}^{\theta _b'} \left[ h\left( \theta _a' \right) - h\left( s, \theta _a'\right) \right] ds + \left( h\left( \theta _b'\right) - h\left( \theta _a'\right) \right) \left( \theta _b' - \theta _a'\right) \\\le & {} \left( \theta '_b-\theta '_a\right) \left( \frac{\left| \kappa \right| }{4} + \frac{\left| \kappa \right| }{4} +\kappa \right) = \left( \theta '_b-\theta '_a\right) \frac{\kappa }{2} < 0 \end{aligned}$$

a contradiction. We conclude that we must have

$$\begin{aligned} \left. \frac{\partial w^{e} \left( s,\hat{\theta }, e_p \right) }{\partial s } \right| _{\left( s,\hat{\theta },e_p\right) = \left( \theta ',\theta ',e_p \left( \theta '\right) \right) } \ge \left. \frac{\partial w^{e} \left( s,\hat{\theta }, e_p \right) }{\partial s } \right| _{\left( s,\hat{\theta },e_p\right) = \left( \theta ,\theta ,e_p \left( \theta \right) \right) } \end{aligned}$$

whenever $\theta ' > \theta$.

At last, notice that this implies that $U' \left( \theta \right)$ is non-decreasing. $\square$

Proof of Proposition 3

In any optimal mechanism the expression $U^{\prime }\left( \theta \right) =\frac{1}{\alpha } \Big (e_{c}\left( \theta \right) +e_{p}\left( \theta \right) \Big ) e_{p}^{1-\alpha }\left( \theta \right)$ must be non-decreasing. Thus, if $U^{\prime }\left( \theta \right)$ is strictly positive at $\theta$ it cannot be zero at any $\theta ' > \theta$. It immediately follows that $e_p \left( \theta \right) > 0$ implies $e_p \left( \theta ' \right) > 0$ for each $\theta ' > \theta$. Thus, there is a $\theta ^*$ such that $e_p=0$ on $\left[ \underline{\theta }, \theta ^*\right)$ and $e_p > 0$ for $\theta \ge \theta ^*$.

If the interval $e_p=0$ on $\left[ \underline{\theta }, \theta ^*\right)$ is non-empty then no incentive rent is paid on that interval. Thus, the solution must be the same as in the complete information case under the constraint $e_p = 0$.

At last, to see that $\theta ^* < \overline{\theta }$ observe that at $\overline{\theta }$ the solution must coincide with the complete information solution and that implies $e_p \left( \overline{\theta }\right) > 0$. By standard arguments, this has to be true also on a left-neighborhood of $\overline{\theta }$ of strictly positive measure. $\square$

Proof of Proposition 4

Assume that when $e_c \ne e_c \left( \hat{\theta }\right)$ wage is zero, so that it is always optimal to pick $e_c \left( \hat{\theta }\right)$. Consider the maximization problem of the manager when the wage function is given by (11). The expected utility when the true value of the parameter is $\theta ,$ the announced value is $\hat{\theta }$ and $e_p$ is chosen is

$$\begin{aligned} U \left( \theta , \hat{\theta },e_p\right) = a \left( \hat{\theta } \right) + U'\left( \hat{\theta }\right) \left( \theta +e_p^{\alpha }\right) -\frac{1}{2} \left( e_c \left( \hat{\theta }\right) + e_p\right) ^2 \end{aligned}$$

(34)

where $a\left( \widehat{\theta }\right)$ is given by (12). For each $\hat{\theta }$ such that $U'\left( \hat{\theta }\right) =0$, the value of $e_p$ that maximizes expected utility is $e_p=0$. Since $U'\left( \hat{\theta }\right) =0$ is possible only if $e_p\left( \hat{\theta }\right) =0$, this implies that expected utility is maximized at the value of $e_p$ prescribed by the mechanism.

When $U'\left( \hat{\theta }\right) >0$ the optimal $e_p$ must be obtained solving the first order condition

$$\begin{aligned} U'\left( \hat{\theta }\right) \alpha e_p^{\alpha - 1} = e_c \left( \hat{\theta }\right) + e_p. \end{aligned}$$

(35)

Using the definition of $U'\left( \hat{\theta }\right)$, the condition can be written as

$$\begin{aligned} \left( e_c\left( \hat{\theta }\right) +e_p\left( \hat{\theta }\right) \right) e^{1-\alpha }_p \left( \hat{\theta }\right) = \left( e_c\left( \hat{\theta }\right) +e_p\right) e^{1-\alpha }_p \end{aligned}$$

(36)

The LHS is strictly positive and constant in $e_p$, while the RHS is strictly increasing in $e_p$ and it goes from 0 to $+\infty$. Thus there is a unique solution and, by inspection, the solution is $e_p = e_p\left( \hat{\theta }\right)$.

With this result, we can now consider the problem of choosing the report $\hat{\theta }$ for a given observed value $\theta$. The expected utility can now be written as

$$\begin{aligned} U \left( \theta , \hat{\theta }\right) = a \left( \hat{\theta } \right) + U'\left( \hat{\theta }\right) \left( \theta +e_p^{\alpha }\left( \hat{\theta }\right) \right) -\frac{1}{2} \left( e_c \left( \hat{\theta }\right) + e_p\left( \hat{\theta }\right) \right) ^2 \end{aligned}$$

(37)

Substituting the expression for $a\left( \widehat{\theta }\right)$ from (12) we obtain

$$\begin{aligned} U \left( \theta , \hat{\theta }\right) = U'\left( \hat{\theta }\right) \left( \theta -\hat{\theta }\right) +\int _{\underline{\theta }}^{\widehat{\theta }}U' \left( s\right) ds \end{aligned}$$

(38)

Since $U'\left( \hat{\theta }\right)$ is non-decreasing, the expression is maximized at $\widehat{\theta }=\theta$. Thus, the proposed wage function induces truthtelling, it induces the correct choice of efforts and it gives the manager expected utility $\int _{\underline{\theta }}^{\theta }U' \left( s\right) ds$, i.e. the lowest incentive rent associated to a truthtelling mechanism. $\square$

Appendix II

In this appendix we describe how the optimal policy described in Sect. 6.2.1 is computed. The objective function is everywhere differentiable and for each pair $\left( \theta ,e_{p}\right)$ the level of k that maximizes it is given by

$$\begin{aligned} k^{\frac{1}{2}}=\frac{1}{2}\left( \theta +e_{p}^{\frac{1}{3}}\right) \end{aligned}$$

where we used $\gamma =\frac{1}{2}$, $\alpha =\frac{1}{3}$. Using this fact, the objective function can be written as a function of $e_{p}$ and $e_{c}$ only:

$$\begin{aligned} W\left( e_{p},e_{c},\theta \right) =\frac{1}{4}\left( \theta +e_{p}^{\frac{1 }{3}}\right) ^{2}+e_{c}-3\left( e_{c}+e_{p}\right) e_{p}^{\frac{2}{3}}\mu \left( \theta \right) -\frac{\left( e_{c}+e_{p}\right) ^{2}}{2} \end{aligned}$$

Let $x=e_{p}^{\frac{1}{3}}$ and $y=e_{c}$. Then the objective function becomes

$$\begin{aligned} \widehat{W}\left( x,y,\theta \right) =\frac{1}{4}\left( \theta +x\right) ^{2}+y-3\left( y+x^{3}\right) x^{2}\mu \left( \theta \right) -\frac{\left( y+x^{3}\right) ^{2}}{2} \end{aligned}$$

Now notice that the function is strictly concave in y for each $\left( x,\theta \right)$ and that the optimal value of y for each given pair $\left( x,\theta \right)$ is

$$\begin{aligned} \widehat{y}\left( x,\theta \right) =\max \left\{ 1-3x^{2}\mu \left( \theta \right) -x^{3},0\right\} . \end{aligned}$$

At this point we can look at the objective function

$$\begin{aligned} W^{*}\left( x,\theta \right) =\frac{1}{4}\left( \theta +x\right) ^{2}+ \widehat{y}\left( x,\theta \right) -3\left( \widehat{y}\left( x,\theta \right) +x^{3}\right) x^{2}\mu \left( \theta \right) -\frac{\left( \widehat{y }\left( x,\theta \right) +x^{3}\right) ^{2}}{2} \end{aligned}$$

and compute numerically the optimal value $x^{*}\left( \theta \right)$ for each $\theta$. The optimal policy is then given by

$$\begin{aligned} e_{p}^{*}\left( \theta \right)= & {} \left( x^{*}\left( \theta \right) \right) ^{3} \\ e_{c}^{* }\left( \theta \right)= & {} \widehat{y}\left( x^{*}\left( \theta \right) ,\theta \right) \\ k^{* }\left( \theta \right)= & {} \frac{1}{2}\left( \theta +x^{* }\left( \theta \right) \right) \end{aligned}$$

It can also be checked numerically that the expression

$$\begin{aligned} U^{\prime }\left( \theta \right) =3\left( e_{p}^{*}\left( \theta \right) +e_{c}^{*}\left( \theta \right) \right) \left( e_{p}^{*}\left( \theta \right) \right) ^{\frac{2}{3}} \end{aligned}$$

is in fact increasing in $\theta$, as it is shown in the following picture (Fig. 3).

Fig. 3

Computation of $U'\left( \theta \right)$

At last, in order to obtain the complete information solution, it is sufficient to set $\mu \left( \theta \right) =0$ for each $\theta$ in the problem stated above.