Subjective evaluation versus public information

Abstract

This paper studies a principal–agent relation in which the principal’s private information about the agent’s effort choice is more accurate than a noisy public performance measure. For some contingencies the optimal contract has to specify ex post inefficiencies in the form of inefficient termination (firing the agent) or wasteful activities that are formally equivalent to third-party payments (money burning). Under the optimal contract, the use of these instruments depends not only on the precision of public information but also on job characteristics. Money burning is used at most in addition to firing and only if the loss from termination is small. The agent’s wage may depend only on the principal’s report and not on the public signal. Nonetheless, public information is valuable as it facilitates truthful subjective evaluation by the principal.

Appendix

Proof of Proposition 1

Suppose that $\theta =b=0$, $w_{HH}=w_{HL}$, and $w_{LH}=w_{LL}$. Then the principal’s incentive constraints (8) are obviously satisfied. Let the difference of the wages satisfy

$$\begin{aligned} w_{HH}-w_{LL}=\frac{c^{\prime }\left( \tilde{e}\right) }{2\sigma -1}. \end{aligned}$$

(32)

Then by (13) the agent will choose the first-best effort $\tilde{e}$. In addition, by unlimited liability one can choose the wage $w_{LL}$ such that the agent’s individual rationality constraint holds with equality:

$$\begin{aligned} w_{LL}=c\left( \tilde{e}\right) -\left[ \frac{1-\sigma }{2\sigma -1}+\tilde{e}\right] c^{\prime }\left( \tilde{e}\right) . \end{aligned}$$

(33)

This contract implements the first-best effort $\tilde{e}$. Moreover, the principal’s payoff is equal to the first-best surplus $S\left( \tilde{e}\right)$ because the agent receives his outside option payoff. Obviously, the payoff of the principal cannot be higher; thus the contract considered here is optimal. Moreover, any optimal contract must implement the first-best effort $\tilde{e}$, for otherwise the principal’s payoff must be lower than the first-best surplus $S\left( \tilde{e}\right)$.

It remains to show that $\theta =b=0$ in any optimal contract. By assumption (1), $\tilde{e}\in \left( 0,1\right)$. Since $\sigma <1$, this implies that all four possible combinations of output and the public signal occur with positive possibility. Therefore, whenever $\theta \not =0$ or $b\not =0$, total surplus is below the first-best surplus $S\left( \tilde{e}\right)$, and hence the principal’s payoff is below $S\left( \tilde{e} \right)$ as well. $\square$

Proof of Proposition 2

The agent’s utility is

$$\begin{aligned} U\left( \gamma ,e\right) =\max _{e^{\prime }}U\left( \gamma ,e^{\prime }\right) \ge U\left( \gamma ,0\right) . \end{aligned}$$

(34)

Since $w\ge 0$ and $c\left( 0\right) =0$, $U\left( \gamma ,0\right) \ge 0$. Thus, in what follows we can ignore constraint (14) because it is automatically satisfied.

If $\left( \gamma ,e\right)$ solves problem (16), then obviously $\gamma$ must maximize $V(\gamma ,e)$ subject to the constraints in (16) when e is treated as a fixed parameter. The latter is a linear optimization problem since $V\left( \gamma ,e\right)$ and all constraints are linear in $\gamma$, and the Kuhn–Tucker conditions are both necessary and sufficient for a maximum. Following a standard method, we temporarily ignore that $\gamma$ has to satisfy the inequality $V_{L}\left( \gamma , \hat{x}_{L}\right) \ge V_{L}\left( \gamma ,\hat{x} _{H}\right)$ in (8), and show below that this constraint is automatically satisfied. Consider the Lagrangian

$$\begin{aligned} L \equiv V(\gamma ,e)+\lambda \left( V_{H}(\gamma ,\hat{x}_{H})-V_{H}(\gamma ,\hat{x}_{L})\right) +\mu \left( U_{H}(\gamma )-U_{L}(\gamma )-c^{\prime }(e)\right) \end{aligned}$$

(35)

with $\lambda \ge 0$. Note that $\mu >0$ as the agent’s incentive constraint must be binding.

Straightforward differentiation shows that $w_{HL}= w_{LL} = 0$ because

$$\begin{aligned} \frac{\partial L}{\partial w_{HL}}= & {} -\left( 1-e\right) \left( 1-\sigma \right) +\lambda \sigma -\mu \left( 1-\sigma \right) < \frac{\partial L}{\partial b_{HL}}=-\left( 1-e\right) \left( 1-\sigma \right) +\lambda \sigma , \nonumber \\\end{aligned}$$

(36)

$$\begin{aligned} \frac{\partial L}{\partial w_{LL}}= & {} -\left( 1-e\right) \sigma +\lambda \left( 1-\sigma \right) -\mu \sigma < \frac{\partial L}{\partial b_{LL}}=-\left( 1-e\right) \sigma +\lambda \left( 1-\sigma \right) . \end{aligned}$$

(37)

Moreover, $\theta _{HH}=b_{HH}=\theta _{LH}=b_{LH}=0$ because

$$\begin{aligned} \frac{\partial L}{\partial \theta _{HH}}= & {} \alpha x_{H} \, \frac{\partial L}{\partial b_{HH}} =- \alpha \sigma x_{H}\left( e + \lambda \right) <0, \end{aligned}$$

(38)

$$\begin{aligned} \frac{\partial L}{\partial \theta _{LH}}= & {} \alpha x_{H} \, \frac{\partial L}{\partial b_{LH}} =- \alpha \left( 1-\sigma \right) x_{H}\left( e +\lambda \right) <0. \end{aligned}$$

(39)

Finally, $b_{LL}=0$ because $\sigma > 1/2$ implies that

$$\begin{aligned} \frac{\partial L}{\partial b_{LL}} =-\left( 1-e\right) \sigma +\lambda \left( 1-\sigma \right) < \frac{\partial L}{\partial b_{HL}} =-\left( 1-e\right) \left( 1-\sigma \right) +\lambda \sigma . \end{aligned}$$

(40)

The above arguments prove part (b) of the proposition. To complete the proof of part (a), note that the agent’s incentive constraint (13) reduces to (17), because $w_{HL}= w_{LL} = 0$. Changing $w_{HH}$ and $w_{LH}$ such that Eq. (17) continues to hold leaves the principal’s payoff constant and does not interfere with any of the constraints. Therefore, the optimal wages $w_{HH}$ and $w_{LH}$ are not unique.

To prove part (c), note that $b_{HL}>0$ implies

$$\begin{aligned} \frac{\partial L}{\partial b_{HL}}=-\left( 1-e\right) \left( 1-\sigma \right) +\lambda \sigma =0, \end{aligned}$$

(41)

and thus $\lambda = {\left( 1-e\right) \left( 1-\sigma \right) }/{\sigma }$. This implies

$$\begin{aligned} \frac{\partial L}{\partial \theta _{HL}}= & {} -\left( 1-e\right) \left( 1-\sigma \right) \alpha x_{L}+\lambda \sigma \alpha x_{H} \end{aligned}$$

(42)

$$\begin{aligned}= & {} \left( 1-e\right) \left( 1-\sigma \right) \alpha \left( x_{H} - x_{L}\right) >0. \end{aligned}$$

(43)

Therefore $b_{HL}>0$ implies $\theta _{HL}=1$.

We now show that, as claimed above, only the second inequality in the ICP constraints (8) is binding. By parts (a) and (b), the principal’s incentive constraint $V_{H}(\gamma ,\hat{x} _{H})\ge V_{H}(\gamma ,\hat{x}_{L})$ in (8) simplifies to

$$\begin{aligned} \sigma w_{HH}+\left( 1-\sigma \right) w_{LH}\le \sigma \left( \theta _{HL}\alpha x_{H}+b_{HL}\right) +\left( 1-\sigma \right) \theta _{LL}\alpha x_{H}. \end{aligned}$$

(44)

Suppose this constraint is not binding. Then $\theta _{HL}$ must be strictly positive, since by limited liability the left hand side of (44) is non-negative, and by part (c) the right hand side can be positive only if $\theta _{HL}>0$. Thus one must have ${\partial L}/{\partial \theta _{HL}} \ge 0$, and by (42) this implies $\lambda > 0$. This proves that the constraint (44) must be binding and $V_{H}(\gamma ,\hat{x}_{H}) \ge V_{H}(\gamma ,\hat{x}_{L})$ must hold with equality.

Using part (a) and (b) the inequality $V_{L}(\gamma ,\hat{x}_{L})\ge V_{L}(\gamma ,\hat{x}_{H})$ in (8) reduces to

$$\begin{aligned} \left( 1-\sigma \right) \left( \theta _{HL}\alpha x_{L}+b_{HL}\right) +\sigma \theta _{LL}\alpha x_{L}\le \sigma w_{LH}+\left( 1-\sigma \right) w_{HH}. \end{aligned}$$

(45)

Since (44) is binding,

$$\begin{aligned}&\sigma w_{LH}+\left( 1-\sigma \right) w_{HH} =\frac{2\sigma -1}{\sigma } w_{LH}+\frac{1-\sigma }{\sigma }\left[ \sigma \left( \alpha \theta _{HL}x_{H}+b_{HL}\right) \right. \nonumber \\&\left. \qquad +\left( 1-\sigma \right) \alpha \theta _{LL}x_{H}\right] \nonumber \\&\quad \ge \left( 1-\sigma \right) \left( \alpha \theta _{HL}x_{H}+b_{HL}\right) +\frac{\left( 1-\sigma \right) ^{2}}{\sigma }\alpha \theta _{LL}x_{H} \end{aligned}$$

(46)

where the inequality follows from $w_{LH}\ge 0$. Subtracting the left hand side of (45) shows that

$$\begin{aligned}&\sigma w_{LH}+\left( 1-\sigma \right) w_{HH}-\left( \left( 1-\sigma \right) \left( \theta _{HL}\alpha x_{L}+b_{HL}\right) +\sigma \theta _{LL}\alpha x_{L}\right) \nonumber \\&\quad \ge \left( 1-\sigma \right) \alpha \theta _{HL}\left( x_{H}-x_{L}\right) +\left[ {\left( 1-\sigma \right) ^{2}}x_{H}/{\sigma } -\sigma x_{L}\right] \alpha \theta _{LL}. \end{aligned}$$

(47)

If $\theta _{LL}=0$, this implies that (45) is satisfied. Similarly, if $\left( 1-\sigma \right) ^{2}x_{H}/\sigma \ge \sigma x_{L}$, (45) is satisfied. To complete the argument, we show that one cannot have that $\left( 1-\sigma \right) ^{2}x_{H}/\sigma <\sigma x_{L}$ and $\theta _{LL}>0$. Indeed, $\theta _{LL}>0$ implies

$$\begin{aligned} \frac{\partial L}{\partial \theta _{LL}}=-\left( 1-e\right) \sigma \alpha x_{L}+\lambda \left( 1-\sigma \right) \alpha x_{H}\ge 0, \end{aligned}$$

(48)

and so

$$\begin{aligned} \lambda \ge \frac{\left( 1-e\right) \sigma x_{L}}{\left( 1-\sigma \right) x_{H}}. \end{aligned}$$

(49)

By the first equality in (41) this implies

$$\begin{aligned} \frac{\partial L}{\partial b_{HL}}\ge \left( 1-e\right) \left[ -\left( 1-\sigma \right) +\frac{\sigma x_{L}}{\left( 1-\sigma \right) x_{H}}\sigma \right] > 0, \end{aligned}$$

(50)

where the second inequality holds if $\left( 1-\sigma \right) ^{2}x_{H}/\sigma <\sigma x_{L}$. Since this would imply $b_{HL} = \infty$, we have shown that $\theta _{LL} = 0$ if $\left( 1-\sigma \right) ^{2}x_{H}/\sigma <\sigma x_{L}$.

Finally, statement (d) directly follows by combining (17) with (44), which holds with equality as shown above. $\square$

Proof of Lemma 1

(a) As shown in the last part of the proof Proposition 2, $\theta _{LL} = 0$ if $\left( 1-\sigma \right) ^{2}x_{H}/\sigma <\sigma x_{L}$. As this inequality is equivalent to $\sigma > \bar{\sigma }$, this proves part (a).

(b) Let $b_{HL}>0$ and $\sigma <\bar{\sigma }$. Then ${\partial L}/{\partial b_{HL}} = 0$ and so by the second equality in (36)

$$\begin{aligned} \lambda =\frac{\left( 1-e\right) \left( 1-\sigma \right) }{\sigma }. \end{aligned}$$

(51)

By the equality in (48) this implies

$$\begin{aligned} \frac{\partial L}{\partial \theta _{LL}}= & {} -\left( 1-e\right) \sigma \alpha x_{L}+\frac{\left( 1-e\right) \left( 1-\sigma \right) }{\sigma }\left( 1-\sigma \right) \alpha x_{H} \end{aligned}$$

(52)

$$\begin{aligned}= & {} \left( 1-e\right) \alpha \left[ -\sigma x_{L}+\frac{\left( 1-\sigma \right) }{\sigma }\left( 1-\sigma \right) x_{H}\right] > 0, \end{aligned}$$

(53)

where the last inequality holds because $\sigma <\bar{\sigma }$. Therefore, $\theta _{LL}=1$.

As shown in the proof of Proposition 2, $\theta _{LL}>0$ implies (49). Therefore, by (42)

$$\begin{aligned} \frac{\partial L}{\partial \theta _{HL}}\ge & {} -\left( 1-e\right) \left( 1-\sigma \right) \alpha x_{L}+\frac{\left( 1-e\right) \sigma x_{L}}{\left( 1-\sigma \right) x_{H}}\sigma \alpha x_{H} \end{aligned}$$

(54)

$$\begin{aligned}= & {} \left( 1-e\right) \alpha x_{L}\frac{2\sigma -1}{1-\sigma }>0. \end{aligned}$$

(55)

Therefore $\theta _{LL}>0$ implies $\theta _{HL}=1$. $\square$

Proof of Proposition 3

We substitute out all choice variable except e from the principal’s problem, and then optimize with respect to e. By parts (a) and (b) of Proposition 2, the principal’s profit $V\left( e,\gamma \right)$ equals

$$\begin{aligned} e x_{H}+\left( 1-e\right) x_{L}-e c^{\prime }\left( e\right) -\left( 1-e\right) \left[ \left( 1-\sigma \right) \left( \alpha \theta _{HL}x_{L}+b_{HL}\right) +\sigma \theta _{LL}\alpha x_{L}\right] . \end{aligned}$$

(56)

By Lemma 1, $\theta _{LL}=0$ and hence by (19),

$$\begin{aligned} \sigma \left( \alpha \, \theta _{HL}\,x_{H}+b_{HL}\right) =c^{\prime }(e). \end{aligned}$$

(57)

There are two possible cases. First, suppose that $\sigma \alpha x_{H}\ge c^{\prime }\left( e\right)$. This is equivalent to $e \le \hat{e}$, where $\hat{e} \equiv c^{\prime -1}\left( \sigma \alpha x_{H}\right)$. In this case, (57) and Proposition 2 (c) imply that $b_{HL}=0$ and $\theta _{HL}=c^{\prime }\left( e\right) /\left( \sigma \alpha x_{H}\right)$. Profit equals

$$\begin{aligned} \phi _{1}\left( e\right) \equiv e x_{H}+\left( 1-e\right) x_{L}-\left[ ec^{\prime }\left( e\right) +\left( 1-e\right) \left( 1-\sigma \right) \frac{c^{\prime }(e)x_{L}}{\sigma x_{H}}\right] . \end{aligned}$$

(58)

Second, suppose that $\sigma \alpha x_{H} < c^{\prime }\left( e\right) ,\$or, equivalently, $e>\hat{e}$. Then $\theta _{HL}=1$ and so by (57) $b_{HL} = {c^{\prime }(e)}/{\sigma }-\alpha x_{H}>0$. In this case, the principal’s payoff is

$$\begin{aligned} \phi _{2}\left( e\right) \equiv e x_{H}+\left( 1-e\right) x_{L}- ec^{\prime }\left( e\right) - \left( 1-e\right) \left( 1-\sigma \right) \left[ \frac{c^{\prime }(e)}{\sigma }-\alpha \left( x_{H}-x_{L}\right) \right] . \end{aligned}$$

(59)

Note that $\phi _{1}\left( e\right) \le \phi _{2}\left( e\right)$ if and only if $c^{\prime }\left( e\right) \le \sigma \alpha x_{H}$. Therefore, the principal’s payoff as a function of e can be written as

$$\begin{aligned} \tilde{V}\left( e\right) \equiv \min \left\{ \phi _{1}\left( e\right) ,\phi _{2}\left( e\right) \right\} . \end{aligned}$$

(60)

The functions $\phi _{1}$ and $\phi _{2}$ are strictly concave in e.²⁶ The minimum of two strictly concave functions is strictly concave, hence $\tilde{V}$ is strictly concave.

Differentiating $\phi _{2}$ yields

$$\begin{aligned} \phi _{2}^{\prime }\left( e\right) = x_{H}-x_{L}-\frac{2\sigma -1}{\sigma } c^{\prime }\left( e\right) -\left[ e + \left( 1-e\right) \frac{\left( 1-\sigma \right) }{\sigma }\right] c^{\prime \prime }\left( e\right) -\alpha \left( 1-\sigma \right) \left( x_{H}-x_{L}\right) . \end{aligned}$$

(61)

Define $e_{2}^{*}$ implicitly by $\phi _{2}^{\prime }\left( e_{2}^{*}\right) =0$. Since $\phi _{2}$ is strictly concave, $e_{2}^{*}$ is unique. Since $c^{\prime \prime \prime }(e) \ge 0$ and $c^{\prime }(0) = 0$, $e c^{\prime \prime }(e) \ge c^{\prime }(e)$ for all e. By (1) therefore $c^{\prime \prime }(1) \ge c^{\prime }(1) > x_H - x_L$. This implies $\phi _{2}^{\prime }\left( 1\right) <0$ and so $e_{2}^{*}<1$.

If $e_{2}^{*}>\hat{e}$, then $e_{2}^{*}$ maximizes the principal’s payoff $\tilde{V}\left( e\right)$. Moreover, if the optimal contract involves $b_{HL}>0$, then $e_{2}^{*}> \hat{e}$. We use the intermediate value theorem to show that $e_{2}^{*}> \hat{e}$ if and only if $\alpha$ is strictly smaller than a critical value $\bar{\alpha }\in \left( 0,1\right)$. The argument proceeds in three steps:

• 1. At $\alpha =0$, $\phi _{2}^{\prime }\left( 0\right) >0$ by assumptions (1) and (2). Moreover, if $\alpha =0$, the critical value $\hat{e}$ equals zero. Therefore, if $\alpha =0$, then $e_{2}^{*} > \hat{e}$.

• 2. The critical value $\hat{e}$ is continuous and strictly increasing in $\alpha$. Moreover, $e_{2}^{*}$ is continuous and strictly decreasing in $\alpha$:

  $$\begin{aligned} \frac{de_{2}^{*}}{\mathrm{d}\alpha }=-\frac{{\partial \phi _{2}^{\prime } \left( e_{2}^{*}\right) }/{\partial \alpha }}{\phi _{2}^{\prime \prime } \left( e_{2}^{*}\right) }=\frac{\left( 1-\sigma \right) \left( x_{H}-x_{L}\right) }{\phi _{2}^{\prime \prime }\left( e_{2}^{*}\right) } <0. \end{aligned}$$

  (62)

• 3. If $\alpha =1$, $e_{2}^{*}$ solves

  $$\begin{aligned} \frac{2\sigma -1}{\sigma }c^{\prime }\left( e_{2}^{*}\right) +\left[ e_{2}^{*}+\left( 1-e_{2}^{*}\right) \frac{\left( 1-\sigma \right) }{ \sigma }\right] c^{\prime \prime }\left( e_{2}^{*}\right) =\sigma \left( x_{H}-x_{L}\right) . \end{aligned}$$

  (63)

  Since $c^{\prime \prime \prime }\left( e\right) \ge 0 = c^{\prime }\left( 0\right)$, we have $ec^{\prime \prime }\left( e\right) \ge c^{\prime }\left( e\right)$ and thus

  $$\begin{aligned} \left[ \frac{2\sigma -1}{\sigma }+1\right] c^{\prime }\left( e_{2}^{*}\right) <\sigma \left( x_{H}-x_{L}\right) . \end{aligned}$$

  (64)

  By $\sigma >1/2$, it follows that $c^{\prime }\left( e_{2}^{*}\right) <\sigma x_{H}$. We conclude that, if $\alpha =1$, $c^{\prime }\left( e_{2}^{*}\right) <\sigma \alpha x_{H}$ and thus $e_{2}^{*}<\hat{e}$.

From steps 1–3, it follows that there exists a critical value $\bar{\alpha } \in \left( 0,1\right)$ such that $e^{*}>\hat{e}$ holds if and only if $\alpha <\bar{\alpha }$. As argued above, this implies that $b_{HL} > 0$ if and only if $\alpha <\bar{\alpha }$. $\square$

Proof of Proposition 4

There are three cases corresponding to statements (a), (b) and (c). First, suppose that $c^{\prime }\left( e\right) \le \sigma \alpha x_{H}$, or equivalently, $e\le \hat{e}=c^{\prime -1}\left( \alpha \sigma x_{H}\right)$. Then, by Proposition 2 (c) and (d), $\theta _{HL}=c^{\prime }\left( e\right) /\left( \sigma \alpha x_{H}\right)$ and $\theta _{LL}=b_{HL}=0$. Profit equals $\phi _{1}\left( e\right)$, as defined in the proof of Proposition 3. Second, suppose that $\sigma \alpha x_{H}<c^{\prime }\left( e\right) \le \alpha x_{H}$. This is equivalent to $e\in \left( \hat{e},\bar{e}\right] ,\$where $\bar{e}:=c^{\prime -1}\left( \alpha x_{H}\right)$. In this case $\theta _{HL}=1$, and $b_{HL}=0$, and $\theta _{LL}=\left[ c^{\prime }(e)-\sigma \alpha x_{H}\right] /\left[ \left( 1-\sigma \right) \alpha x_{H}\right]$. Using (56), the principal’s profit is

$$\begin{aligned} \phi _{3}\left( e\right) \equiv ex_{H}+\left( 1-e\right) x_{L}-ec^{\prime }\left( e\right) -\left( 1-e\right) \left[ \frac{\sigma x_{L}}{\left( 1-\sigma \right) x_{H}}c^{\prime }(e)-\alpha x_{L}\frac{2\sigma -1}{1-\sigma }\right] . \end{aligned}$$

(65)

Third, suppose that $c^{\prime }\left( e\right) >\alpha x_{H}$, or, equivalently, $e>\bar{e}$. Then $\theta _{LL}=\theta _{HL}=1$ and $b_{HL}=\left( c^{\prime }(e)-\alpha x_{H}\right) /\sigma$, and profit equals

$$\begin{aligned} \phi _{4}\left( e\right) \equiv ex_{H}+\left( 1-e\right) x_{L}-ec^{\prime }\left( e\right) -\left( 1-e\right) \left[ \alpha x_{L}+\frac{\left( 1-\sigma \right) \left( c^{\prime }(e)-\alpha x_{H}\right) }{\sigma }\right] . \end{aligned}$$

(66)

Note that $\phi _{1}\left( e\right) \le \phi _{3}\left( e\right)$ if and only if $c^{\prime }(e) \le \sigma \alpha x_H$. Moreover $\phi _{3}\left( e\right) \le \phi _{4}\left( e\right)$ if and only if $c^{\prime }(e) \le \alpha x_H$. Thus $c^{\prime }\left( e\right) \le \alpha \sigma x_{H}$ if and only if $\phi _{1}\left( e\right) =\min \left\{ \phi _{1}\left( e\right) ,\phi _{3}\left( e\right) ,\phi _{4}\left( e\right) \right\}$. Moreover, $\sigma \alpha x_{H}<c^{\prime }\left( e\right) \le \sigma x_{H}$ if and only if $\phi _{3}\left( e\right) <\phi _{1}\left( e\right)$ and $\phi _{3}\left( e\right) \le \phi _{4}\left( e\right)$. Therefore, the principal’s payoff can be written

$$\begin{aligned} \tilde{V}\left( e\right) =\min \left\{ \phi _{1}\left( e\right) ,\phi _{3}\left( e\right) ,\phi _{4}\left( e\right) \right\} . \end{aligned}$$

(67)

The functions $\phi _{1}$, $\phi _{3}$, and $\phi _{4}$ are strictly concave in e.²⁷ Therefore, $\tilde{V}$ is strictly concave in e.

Define $e_{4}^{*}$ implicitly by $\phi _{4}^{\prime }\left( e_{4}^{*}\right) =0$. Since $\phi _{4}$ is strictly concave, $e_{4}^{*}$ is unique. By assumption (1), $e_{4}^{*}<1$. If $e_{4}^{*} > \bar{e}$, then $e_{4}^{*}$ maximizes $\tilde{V}\left( e\right)$. Moreover, if the optimal contract involves $b_{HL}>0$, then $e_{4}^{*}>\bar{e}$.

We use the intermediate value theorem to show that $e_{4}^{*}>\bar{e}$ if and only if $\alpha$ is strictly smaller than a critical value $\bar{\alpha } _{1}\in \left( 0, 1\right)$. The argument proceeds in three steps:

• 1. If $\alpha =0$, $\phi _{4}^{\prime }\left( 0\right) >0 = \bar{e}$. Thus if $\alpha =0$, $e_{4}^{*}>\bar{e}$.

• 2. $e_{4}^{*}$ is continuous and strictly decreasing in $\alpha$. Moreover, $\bar{e}=c^{\prime -1}\left( \alpha x_{H}\right)$ is continuous and strictly increasing in $\alpha$.

• 3. At $\alpha =1$, $e_{4}^{*}$ solves

  $$\begin{aligned} x_{H}=c^{\prime }\left( e_{4}^{*}\right) + \left[ e_{4}^{*}\sigma +\left( 1-e_{4}^{*}\right) \left( 1-\sigma \right) \right] c^{\prime \prime }\left( e_{4}^{*}\right) \frac{\sigma }{2 \sigma -1}. \end{aligned}$$

  (68)

  Therefore, at $\alpha =1$, $\alpha x_{H}>c^{\prime }\left( e_{4}^{*}\right)$ and so $e_{4}^{*} < \bar{e}$.

Thus there exists an $\bar{\alpha }_{1}\in \left( 0,1\right)$ such that if $\alpha =\bar{\alpha }_{1}$, $e_{4}^{*}$ $=\bar{e}$. For all $\alpha <\bar{\alpha }_{1}$, the optimal effort is $e_{4}^{*}>\bar{e};$ moreover $b_{HL}>0$ and $\theta _{HL}=\theta _{LL}=1$. On the other hand, for all $\alpha \ge \bar{\alpha }_{1}$, $b_{HL}=0$.

It remains to show that there exists $\bar{\alpha }_{2}\in \left( \bar{\alpha } _{1},1\right)$ such that $\theta _{LL}>0$ if and only if $\alpha <\bar{\alpha }_{2}$. Note that $\theta _{LL}>0$ if and only $\phi _{3}^{\prime }\left( \hat{e}\right) >0$. To see this, first suppose that $\phi _{3}^{\prime }\left( \hat{e}\right) \le 0$. Then $\tilde{V}\left( e\right) < \tilde{V}\left( \hat{e}\right)$ for all $e>\hat{e}$ since $\tilde{V}(\cdot )$ is strictly concave. Hence the optimal effort is no larger than $\hat{e}$ and $\theta _{LL}=0$. Second, if $\phi _{3}^{\prime }\left( \hat{e}\right) >0$, then the optimal effort is strictly bigger than $\hat{e}$ and $\theta _{LL}>0$.

We use the intermediate value theorem to show that there exists $\bar{\alpha }_{2} \in \left( \bar{\alpha }_{1}, 1\right)$ such that $\phi _{3}^{\prime }\left( \hat{e}\right) >0$ if and only if $\alpha <\bar{\alpha }_{2}$. The argument proceeds in three steps:

• 1. By definition of $\bar{\alpha }_{1}$, if $\alpha =\bar{\alpha }_{1}$, then $\phi _{4}^{\prime }\left( e_{4}^{*}\right) =\phi _{4}^{\prime }\left( \bar{e}\right) =0$. Since $\phi _{3}$ and $\tilde{V}$ are strictly concave, and $\bar{e}>\hat{e}$, it follows that $\phi _{4}^{\prime }\left( \bar{e} \right) \le \phi _{3}^{\prime }\left( \bar{e}\right) <\phi _{3}^{\prime }\left( \hat{e}\right)$. Therefore, if $\alpha =\bar{\alpha }_{1}$, then $\phi _{3}^{\prime }\left( \hat{e}\right) >0$.

• 2. $\phi _{3}^{\prime }\left( \hat{e}\right)$ is continuous and strictly decreasing in $\alpha :$

  $$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}\alpha }\phi _{3}^{\prime }\left( \hat{e}\right) =\left[ \frac{ \partial }{\partial \alpha }\phi _{3}^{\prime }\left( e\right) \right] _{e= \hat{e}}+\phi _{3}^{\prime \prime }\left( \hat{e}\right) \frac{\mathrm{d}\hat{e}}{ \mathrm{d}\alpha }<0. \end{aligned}$$

  (69)

• 3. Suppose $\alpha =1$. Since $e c^{\prime \prime }\left( e\right) \ge c^{\prime }\left( e\right)$,

  $$\begin{aligned} \phi _{3}^{\prime }\left( \hat{e}\right)= & {} x_{H}-x_{L}-x_{L}\frac{2\sigma -1}{1-\sigma }-\left[ 1-\frac{\sigma x_{L}}{\left( 1-\sigma \right) x_{H}} \right] c^{\prime }\left( \hat{e}\right) \end{aligned}$$

  (70)
  $$\begin{aligned}&-\left[ \hat{e}+\left( 1-\hat{e}\right) \frac{\sigma x_{L}}{\left( 1-\sigma \right) x_{H}}\right] c^{\prime \prime }\left( \hat{e}\right) \end{aligned}$$

  (71)
  $$\begin{aligned}< & {} x_{H}-x_{L}-x_{L}\frac{2\sigma -1}{1-\sigma }-\left[ 2-\frac{\sigma x_{L} }{\left( 1-\sigma \right) x_{H}}\right] \sigma x_{H} \end{aligned}$$

  (72)
  $$\begin{aligned}= & {} x_{H}-2\sigma x_{H}-\sigma x_{L}<0. \end{aligned}$$

  (73)

  Hence if $\alpha =1$, then $\phi _{3}^{\prime }\left( \hat{e}\right) <0$.

$\square$

Proof of Proposition 5

The principal’s expected payoff $\tilde{V}\left( e\right)$ defined in (60) and (67) is strictly increasing in $\sigma$ because the functions $\phi _1, \phi _2, \phi _3$ and $\phi _4$ are strictly increasing in $\sigma$. If $\theta _{HL} < 1$, then $\tilde{V}\left( e\right) = \phi _1(e)$, as defined in (58), and therefore $\partial \tilde{V}\left( e\right) /\partial \alpha = 0$. If $\theta _{HL} = 1$, then $\tilde{V}\left( e\right) = \phi _2(e)$ in the case of Proposition 3 and $\partial \tilde{V}\left( e\right) /\partial \alpha > 0$ because $\phi _2$ is strictly increasing in $\alpha$. In the case of Proposition 4, $\tilde{V}\left( e\right) = \min \left\{ \phi _3(e), \phi _4(e)\right\}$. As $\sigma < \bar{\sigma }$, $\phi _3$ and $\phi _4$ are strictly increasing in $\alpha$ and so $\partial \tilde{V}\left( e\right) /\partial \alpha > 0$. $\square$

Proof of Proposition 6

Suppose $\sigma >\bar{\sigma }$. By Proposition 3, $b_{HL}$ is strictly positive if and only if $\alpha <\bar{\alpha }$. Moreover, the proof of Proposition 3 establishes that, when $\alpha <\bar{\alpha }$, the profit maximizing effort is equal to $e_{2}^{*}$ which is strictly decreasing in $\alpha$ [see Eq. (62)], and $\theta _{HL}=1$ and $\theta _{LL}=0$. By Eq. (19) it follows that

$$\begin{aligned} b_{HL}=\frac{c^{\prime }\left( e_{2}^{*}\right) }{\sigma }-\alpha x_{H}, \end{aligned}$$

(74)

which is strictly decreasing in $\alpha$.

By the proof of Proposition 3, the principal’s profit from implementing effort e is $\tilde{V}\left( e\right) =\phi _{1}\left( e\right)$ if $e\le \hat{e}\equiv c^{\prime -1}\left( \sigma \alpha x_{H}\right)$, and $\tilde{V}\left( e\right) =\phi _{2}\left( e\right)$ if $e>\hat{e}$. We show that there exists an $\varepsilon ^{\prime }>0$ such that, if $\alpha \in \left( \bar{\alpha },\bar{\alpha }+\varepsilon ^{\prime }\right)$, the profit maximizing effort is equal to $\hat{e}$ and thus strictly increasing in $\alpha$.

Since $\tilde{V}\left( e\right)$ is strictly concave in e, it is sufficient to prove that $\phi _{1}^{\prime }\left( \hat{e}\right)>0>\phi _{2}^{\prime }\left( \hat{e}\right)$ if $\alpha \in \left( \bar{\alpha } ,\bar{\alpha }+\varepsilon ^{\prime }\right)$ for some $\varepsilon ^{\prime }>0$. Straightforward calculations show that

$$\begin{aligned} \phi _{1}\left( e\right) -\phi _{2}\left( e\right) =-\left( 1-e\right) \left( 1-\sigma \right) \left[ \frac{c^{\prime }(e)}{\sigma }\left( \frac{x_{L}}{x_{H}}-1\right) +\alpha \left( x_{H}-x_{L}\right) \right] . \end{aligned}$$

(75)

Since $\phi _{1}\left( \hat{e}\right) =\phi _{2}\left( \hat{e}\right)$,

$$\begin{aligned} \frac{c^{\prime }(\hat{e})}{\sigma }\left( \frac{x_{L}}{x_{H}}-1\right) +\alpha \left( x_{H}-x_{L}\right) =0. \end{aligned}$$

(76)

Moreover, if $\alpha =\bar{\alpha }$, then $\phi _{2}^{\prime }\left( \hat{e}\right) =0$ by construction of $\bar{\alpha }$. Thus $\alpha =\bar{\alpha }$ implies

$$\begin{aligned} \phi _{1}^{\prime }\left( \hat{e}\right) =\phi _{1}^{\prime }\left( \hat{e}\right) -\phi _{2}^{\prime }\left( \hat{e}\right) =-\left( 1-\hat{e}\right) \left( 1-\sigma \right) \frac{c^{\prime \prime }(\hat{e})}{\sigma }\left( \frac{x_{L}}{x_{H}}-1\right) >0. \end{aligned}$$

(77)

By continuity, there exists $\varepsilon ^{\prime }>0$ such that $\phi _{1}^{\prime }\left( \hat{e}\right) >0$ when $\alpha \in \left( \bar{\alpha },\bar{\alpha }+\varepsilon ^{\prime }\right)$. It remains to show that $\phi _{2}^{\prime }\left( \hat{e}\right) <0$ when $\alpha \in \left( \bar{\alpha },\bar{\alpha }+\varepsilon ^{\prime }\right)$. Equation (61) implies that $\phi _{2}^{\prime }\left( e\right)$ is strictly decreasing in $\alpha$ (holding e constant). Moreover, $\hat{e}$ is strictly increasing in $\alpha$, and $\phi _{2}^{\prime \prime }\left( e\right) <0$. Since $\alpha =\bar{\alpha }$ implies $\phi _{2}^{\prime }\left( \hat{e}\right) =0$, it follows that $\phi _{2}^{\prime }\left( \hat{e}\right) <0$ when $\alpha \in \left( \bar{\alpha },\bar{\alpha }+\varepsilon ^{\prime }\right)$.

Suppose $\sigma <\bar{\sigma }$. By Proposition 4, $b_{HL}$ is strictly positive if and only if $\alpha <\bar{\alpha }_{1}$. Moreover, the proof of Proposition 4 establishes that, when $\alpha <\bar{\alpha }_{1}$, the profit maximizing effort is equal to $e_{4}^{*}$ which is strictly decreasing in $\alpha$, and $\theta _{LL}=\theta _{HL}=1$. By Eq. (19), it follows that

$$\begin{aligned} b_{HL}=\frac{c^{\prime }\left( e_{4}^{*}\right) -\alpha x_{H}}{\sigma }, \end{aligned}$$

(78)

which is strictly decreasing in $\alpha$.

By the proof of Proposition 4, the principal’s profit from implementing effort e is $\tilde{V}\left( e\right) =\phi _{1}\left( e\right)$ if $e\le \hat{e}$, $\tilde{V}\left( e\right) =\phi _{3}\left( e\right)$ if $\hat{e}<e\le \bar{e}\equiv c^{\prime -1}\left( \alpha x_{H}\right)$, and $\tilde{V}\left( e\right) =\phi _{4}\left( e\right)$ if $e>\bar{e}$. We show that there exists an $\varepsilon ^{\prime \prime }>0$ such that, if $\alpha \in \left( \bar{\alpha }_{1},\bar{\alpha }_{1} +\varepsilon ^{\prime \prime }\right)$, the profit maximizing effort is equal to $\bar{e}$ and thus strictly increasing in $\alpha$.

Since $\tilde{V}\left( e\right)$ is strictly concave in e, it is sufficient to prove that $\phi _{3}^{\prime }\left( \bar{e}\right)>0>\phi _{4}^{\prime }\left( \bar{e}\right)$ if $\alpha \in \left( \bar{\alpha } _{1},\bar{\alpha }_{1}+\varepsilon ^{\prime \prime }\right)$ for some $\varepsilon ^{\prime \prime }>0$. Straightforward calculations show that

$$\begin{aligned} \phi _{3}\left( e\right) -\phi _{4}\left( e\right) =-\left( 1-e\right) \left[ \frac{\sigma x_{L}}{\left( 1-\sigma \right) x_{H}}c^{\prime }(e)-\alpha x_{L}\frac{\sigma }{1-\sigma } -\frac{\left( 1-\sigma \right) \left( c^{\prime }(e)-\alpha x_{H}\right) }{\sigma }\right] .\nonumber \\ \end{aligned}$$

(79)

Since $\phi _{3}\left( \bar{e}\right) =\phi _{4}\left( \bar{e}\right)$,

$$\begin{aligned} \frac{\sigma x_{L}}{\left( 1-\sigma \right) x_{H}}c^{\prime }(\bar{e})-\alpha x_{L}\frac{\sigma }{1-\sigma }-\frac{\left( 1-\sigma \right) \left( c^{\prime }(\bar{e})-\alpha x_{H}\right) }{\sigma }=0. \end{aligned}$$

(80)

Suppose that $\alpha =\bar{\alpha }_{1}$. Then $\phi _{4}^{\prime }\left( \bar{e}\right) =0$ by construction of $\bar{\alpha }_{1}$. Thus

$$\begin{aligned} \phi _{3}^{\prime }\left( \bar{e}\right) =\phi _{3}^{\prime }\left( \bar{e}\right) -\phi _{4}^{\prime }\left( \bar{e}\right) =-\left( 1-\bar{e}\right) \left( \frac{\sigma x_{L}}{\left( 1-\sigma \right) x_{H}} -\frac{\left( 1-\sigma \right) }{\sigma }\right) c^{\prime \prime }(\bar{e}) \end{aligned}$$

(81)

which is strictly positive since $\sigma <\bar{\sigma }$. By continuity, there exists some $\varepsilon ^{\prime \prime }>0$ such that $\phi _{3}^{\prime }\left( \bar{e}\right) >0$ when $\alpha \in \left( \bar{\alpha }_{1},\bar{\alpha }_{1}+\varepsilon ^{\prime \prime }\right)$. It remains to show that $\phi _{4}^{\prime }\left( \bar{e}\right) <0$ when $\alpha \in \left( \bar{\alpha }_{1},\bar{\alpha }_{1}+\varepsilon ^{\prime \prime }\right)$. Equation (66) implies that $\frac{\partial }{\partial \alpha }\phi _{4}^{\prime }\left( e\right) =x_{L}-\frac{1-\sigma }{\sigma }x_{H}<0$, where the inequality follows from $\sigma <\bar{\sigma }$. That is, $\phi _{4}^{\prime }\left( e\right)$ is strictly decreasing in $\alpha$ (holding e constant). Moreover, $\bar{e}$ is strictly increasing in $\alpha$, and $\phi _{4}^{\prime \prime }\left( e\right) <0$. Since $\alpha =\bar{\alpha }_{1}$ implies $\phi _{4}^{\prime }\left( \bar{e}\right) =0$, it follows that $\phi _{4}^{\prime }\left( \bar{e}\right) <0$ when $\alpha \in \left( \bar{\alpha }_{1},\bar{\alpha }_{1}+\varepsilon ^{\prime \prime }\right)$.

Setting $\varepsilon :=\min \left\{ \varepsilon ^{\prime },\varepsilon ^{\prime \prime }\right\}$ concludes the proof. $\square$

Proof of Proposition 7

Since $(\gamma , e)$ solves problem (16), $\gamma$ satisfies (18) and $w_{LL} = w_{HL} = 0$ by Proposition 2 (a). This reduces the constraints in (29) to

$$\begin{aligned} V_L(\gamma , \hat{x}_L\vert s_L) - V_L(\gamma , \hat{x}_H\vert s_L)= & {} w_{LH}- \alpha \theta _{LL} x_L \ge 0, \end{aligned}$$

(82)

$$\begin{aligned} V_L(\gamma , \hat{x}_L\vert s_H) - V_L(\gamma , \hat{x}_H\vert s_H)= & {} w_{HH} -\alpha \,\theta _{HL}\, x_L - b_{HL} \ge 0, \end{aligned}$$

(83)

$$\begin{aligned} V_H(\gamma , \hat{x}_H\vert s_H) - V_H(\gamma , \hat{x}_L\vert s_H)= & {} \alpha \, \theta _{HL}\,x_H + b_{HL} - w_{HH} \ge 0, \end{aligned}$$

(84)

$$\begin{aligned} V_H(\gamma , \hat{x}_H\vert s_L) - V_H(\gamma , \hat{x}_L\vert s_L)= & {} \alpha \theta _{LL} x_H- w_{LH} \ge 0. \end{aligned}$$

(85)

Further, (17) and (19) imply

$$\begin{aligned} \sigma w_{HH} + (1-\sigma ) w_{LH} = \sigma (\alpha \,\theta _{HL}\, x_H + b_{HL}) + (1-\sigma )\alpha \,\theta _{LL}\, x_H. \end{aligned}$$

(86)

By (85) $w_{LH} \le \alpha \theta _{LL} x_H$. Suppose that this inequality is strict, i.e. $w_{LH} < \alpha \theta _{LL} x_H$. Then (86) implies that $w_{HH} > \alpha \,\theta _{HL}\, x_H + b_{HL}$, a contradiction to (84). This proves that the constraints in (29) cannot be satisfied if $w_{LH} \ne \alpha \theta _{LL} x_H$.

Now let $w_{LH} = \alpha \theta _{LL} x_H$. Then (82) and (85) hold because $x_L < x_H$. As $w_{LH} = \alpha \theta _{LL} x_H$, (86) implies

$$\begin{aligned} w_{HH} = \alpha \,\theta _{HL}\, x_H + b_{HL}. \end{aligned}$$

(87)

Thus the equality holds in (84), and the strict inequality holds in (83) because $x_L < x_H$. This proves that (82)–(85) are satisfied if $(\gamma , e)$ solves problem (16) with $w_{LH} = \alpha \theta _{LL} x_H$ and $w_{HH} = \alpha \,\theta _{HL}\, x_H + b_{HL}$. $\square$