Incentive contracting under ambiguity aversion

Abstract

This paper studies a principal–agent model in which the information on future firm performance is ambiguous and the agent is averse to ambiguity. We show that if firm risk is ambiguous, while stocks always induce the agent to perceive a high risk, options can induce him to perceive a low risk. As a result, options can be less costly in incentivizing the agent than stocks in the presence of ambiguity. In addition, we show that providing the agent with more incentives would induce the agent to perceive a higher risk, and there is a discontinuous jump in the compensation cost as incentives increase, which makes the principal reluctant to reset contracts frequently when underlying fundamentals change. Thus, compensation contracts exhibit an inertia property. Lastly, the model sheds some light on the use of relative performance evaluation and provides a rationale for the puzzle of pay-for-luck in the presence of ambiguity.

Appendix

Proof of Proposition 1

For stock compensation, the agent’s utility is given by $E[u]=\alpha _s+ \beta _{s}(a+l)-\frac{1}{2}\lambda \beta _{s}^{2}\sigma ^{2}-\frac{1}{2}a^{2}$. So to induce the effort a, we must have $\beta _{s}=a$. For a fixed target effort a, the principal wants to minimize the cost. So $\alpha _{s}$ must be set so that the IR constraint is binding. Thus, it is easy to derive that the cost of the contract (dropping the constant $\frac{1}{2}a^{2}$) is: $C_{s}=\frac{1}{2}\lambda a^{2}\sigma ^{2}$.

We derive the agent’s utility for an option compensation as follows. $E[u]=\alpha _{k}+\beta _{k}m_{k}-\frac{1}{2}\lambda \beta _{k}^{2}\sigma _{k}^{2}-\frac{1}{2}a^{2}.$ Taking partial derivative of $m_{k}$ and $\sigma _{k}^{2}$ w.r.t. a yields that

$$\begin{aligned} \frac{\partial m_{k}}{\partial a}= & {} 1-\Phi \\ \frac{\partial \sigma _{k}^{2}}{\partial a}= & {} 2\Phi m_{k} \end{aligned}$$

To induce the effort a, we must have that $\frac{\partial E[u]}{\partial a}\ge 0$, which yields that

$$\begin{aligned} \frac{(1-\Phi )-\sqrt{\Delta }}{2\lambda \Phi m_{k}}\le \beta _{k}\le \frac{(1-\Phi )+\sqrt{\Delta }}{2\lambda \Phi m_{k}}, \end{aligned}$$

where $\Delta =(1-\Phi )^{2}-4a\lambda \Phi m_{k}$. Similarly, we have the binding IR constraint. So the cost is

$$\begin{aligned} C_{k}\ge \frac{1}{2}\lambda \left( \frac{(1-\Phi )-\sqrt{\Delta }}{2\lambda \Phi m_{k}}\right) ^{2}\sigma _{k}^{2}=\frac{2\lambda a^{2}\sigma ^{2}y_{1}}{(1+\sqrt{1-4a\lambda \sigma y_{2}})^{2}}, \end{aligned}$$

(9)

where

$$\begin{aligned} y_{1}= & {} \frac{(1-\Phi -\phi ^{2})-\phi (2\Phi -1)\eta +\Phi (1-\Phi )\eta ^{2}}{(1-\Phi )^{2}},\\ y_{2}= & {} \frac{\Phi [\phi -(1-\Phi )\eta ]}{(1-\Phi )^{2}}. \end{aligned}$$

It can be shown that $y_{1}^{\prime }(\eta )\ge 0$ and $y_{2}^{\prime }(\eta )\ge 0$ (see Feltham and Wu 2001). Thus, $C_k$ is increasing in k. As k goes to $-\infty$, $\Phi$ goes to zero. So $\sigma ^2y_1=\frac{\sigma _k^2}{(1-\Phi )^{2}}\rightarrow \sigma ^2$, and $\sigma y_2=\frac{\Phi m_k}{(1-\Phi )^{2}}\rightarrow 0$. Hence, the right-hand side of (9) goes to $\frac{1}{2}\lambda a^2\sigma ^2$ when k goes to $-\infty$. Thus, we obtain that $C_{s}<C_{k}$.$\square$

Proof of Lemma 2

If the agent is granted with stocks, his utility $E[u]=\beta _{s}(a+l)-\frac{1}{2}\lambda \beta _{s}^{2}\sigma ^{2}-\frac{1}{2}a^{2}$ is minimized at $\sigma =\sigma _{2}$. So he will always perceive the highest risk $\sigma _{2}$. Following the proof of Proposition 1, we can show that $C_{s}=\frac{1}{2}\lambda a^{2}\sigma _{2}^{2}$.

If the agent is granted with options, his utility is given by

$$\begin{aligned} E[u]=\alpha _{k}+\beta _{k}(\sigma _{1})m_{k}(\sigma _{a})-\frac{1}{2}\lambda \beta _{k}(\sigma _{1})^{2}\sigma _{k}(\sigma _{a})^2-\frac{1}{2}a^{2}. \end{aligned}$$

Let

$$\begin{aligned} A_{1}=\{\hat{a}: E[u(\hat{a})| \sigma _{a}=\sigma _{1}]\le \min _{\sigma _{a}\in (\sigma _{1},\sigma _{2}]}E[u(\hat{a})]\} \end{aligned}$$

be the set of effort for which the agent’s utility is minimized at the lowest risk $\sigma _{1}$. By condition (5), for any $\sigma \in (\sigma _{1},\sigma _{2}]$, we have $\lambda <\frac{2}{\beta _{k}(\sigma _{1})} \frac{m_{k}(\sigma )-m_{k}(\sigma _{1})}{\sigma _{k}^{2}(\sigma )-\sigma _{k}^{2}(\sigma _{1})}$. This implies that $\alpha _{k}+\beta _{k}(\sigma _{1})m_{k}(\sigma _{1})-\frac{1}{2}\lambda \beta _{k}(\sigma _{1})^{2}\sigma _{k}(\sigma _{1})^2-\frac{1}{2}a^{2}<\alpha _{k}+\beta _{k}(\sigma _{1})m_{k}(\sigma )-\frac{1}{2}\lambda \beta _{k}(\sigma _{1})^{2}\sigma _{k}(\sigma )^2-\frac{1}{2}a^{2}$, i.e., $E[u(a) | \sigma _{a}=\sigma _{1}] < E[u(a) | \sigma _{a}=\sigma ]$. So the target effort a belongs to $A_{1}$, i.e., $a\in A_{1}$. Note that $\beta _{k}(\sigma _{1})$ is exactly the number of options to induce the effort a given $\sigma _{a}=\sigma _{1}$. If the agent perceives another risk $\sigma \ne \sigma _{1}$ and exerts an effort $\hat{a}$, we must have

$$\begin{aligned} E[u(\hat{a}) | \sigma _{a}=\sigma ]\le E[u(\hat{a}) | \sigma _{a}=\sigma _{1}]\le E[u(a) | \sigma _{a}=\sigma _{1}], \end{aligned}$$

where $E[u(\hat{a}) | \sigma _{a}=\sigma ]\le E[u(\hat{a}) | \sigma _{a}=\sigma _{1}]$ follows from the fact that the agent has a max-min preference, which means that under the same level of effort, his perception of risk should minimize his expected utility. Therefore at the optimum, the agent will perceive the lowest risk $\sigma _{1}$ and exert the target effort a. The binding IR constraint yields that $\alpha _{k}+\beta _{k}(\sigma _{1})m_{k}(\sigma _{1}) =\frac{1}{2}\lambda \beta _{k}(\sigma _{1})^{2}\sigma _{k}(\sigma _{1})^2+\frac{1}{2}a^{2}$. From the principal’s perspective, the cost (dropping the constant $\frac{1}{2}a^{2}$) is $C_{k}=\alpha _{k}+\beta _{k}(\sigma _{1})m_{k}(\sigma _{p})=\frac{1}{2}\lambda \beta _{k}(\sigma _{1})^{2}\sigma _{k}(\sigma _{1})^2 +\beta _{k}(\sigma _{1})[m_{k}(\sigma _{p})-m_{k}(\sigma _{1})].$ $\square$

Proof of Proposition 2

By Lemma 2, the right-hand side of (6) ensures that an option of exercise price k induces the agent to perceive $\sigma _{1}$ and exert the effort a. The cost of the contract is $C_{k}=\frac{1}{2}\lambda \beta _{k}(\sigma _{1})^{2}\sigma _{k}(\sigma _{1})^2 +\beta _{k}(\sigma _{1})[m_{k}(\sigma _{p})-m_{k}(\sigma _{1})].$ Since the cost of stock compensation is $C_{s}=\frac{1}{2}\lambda a^{2}\sigma _{2}^{2}$, the left-hand side of (6) ensures that the option compensation is less costly.

Analysis of Condition (6)

We first solve for the right-hand side of (6) to obtain an upper bound of $\lambda$. Let $M=\min _{\sigma \in [\sigma _{1},\sigma _{2}]}\frac{m_{k}(\sigma )-m_{k}(\sigma _{1})}{\sigma _{k}^{2}(\sigma )-\sigma _{k}^{2}(\sigma _{1})}$, which does not depend on $\lambda$. From the proof of Proposition 1, we know that $\beta _{k}=\frac{(1-\Phi )-\sqrt{\Delta }}{2\lambda \Phi m_{k}}$, where $\Delta =(1-\Phi )^{2}-4a\lambda \Phi m_{k}$. Plugging the formula of $\beta _{k}$ into the right-hand side of (6), we can solve that the right-hand side of (6) is equivalent to $\lambda <\frac{K(2-2\Phi (\sigma _1)-K)}{4a\Phi (\sigma _1)m_{k}(\sigma _1)}$, where $K=4M\Phi (\sigma _1)m_{k}(\sigma _1)$.

For the left-hand side of (6), it can be simplified to $\lambda \beta _{k}^{2}(\sigma _{1})\sigma _{k}^2(\sigma _{1})+2[m_{k}(\sigma _{p})-m_{k}(\sigma _{1})]\beta _{k}(\sigma _{1})-\lambda a^{*2}\sigma _{2}^{2}<0$. Note that $\lambda \beta _k^2=\frac{1-\Phi }{\Phi m_k}\beta _k-\frac{a}{\Phi m_k}$, so the left-hand side of (6) is equivalent to

$$\begin{aligned} \left[ \frac{(1-\Phi (\sigma _{1}))\sigma _{k}^2(\sigma _{1})}{\Phi (\sigma _{1}) m_k(\sigma _{1})}+2[m_{k}(\sigma _{p})-m_{k}(\sigma _{1})] \right] \beta _{k}(\sigma _{1})-\frac{a\sigma _{k}^2(\sigma _{1})}{\Phi (\sigma _{1}) m_k(\sigma _{1})}-\lambda a^{2}\sigma _{2}^{2}<0. \end{aligned}$$

(10)

It is easy to check that the left-hand side of (10) is convex in $\lambda$ and is nonnegative when $\lambda =0$. So there could be three cases: Case 1: the left-hand side of (10) never crosses with the x-axis, which means that (10) will never be satisfied. In this case, the condition (6) cannot be satisfied. Case 2: the left-hand side of (10) crosses with the x-axis once. Let $\lambda _1$ denote the crossed point, then the condition (6) is equivalent to $\lambda _1<\lambda <\frac{K(2-2\Phi (\sigma _1)-K)}{4a\Phi (\sigma _1)m_{k}(\sigma _1)}$. Case 3: the left-hand side of (10) crosses with the x-axis twice. Let $\lambda _1<\lambda _2$ denote the two crossed points, then the condition (6) is equivalent to $\lambda _1<\lambda <\min \left( \frac{K(2-2\Phi (\sigma _1)-K)}{4a\Phi (\sigma _1)m_{k}(\sigma _1)}, \lambda _2 \right)$. $\square$

Proof of Lemma3

Suppose the principal uses a contract $w=\alpha +\beta _{s}x+\beta _{k}\max (x-k,0)$ to induce the agent to perceive $\sigma _{1}$ and exert the target effort $a_{2}$. Then we must have that for any $\sigma _{a}\in (\sigma _{1},\sigma _{2}]$,

$$\begin{aligned}&\alpha +\beta _{s}(a_2+l)+\beta _{k}m_{k}(\sigma _{1})-\frac{1}{2}\lambda \left[ \beta _{s}^{2}\sigma _{1}^{2}+\beta _{k}^{2}\sigma _{k}(\sigma _{1})^{2} +2\beta _{s}\beta _{k}\sigma _{0,k}(\sigma _{1})\right] \\&\quad \le \alpha +\beta _{s}(a_2+l)+\beta _{k}m_{k}(\sigma _{a})-\frac{1}{2}\lambda \left[ \beta _{s}^{2}\sigma _{a}^{2}+\beta _{k}^{2}\sigma _{k}(\sigma _{a})^{2} +2\beta _{s}\beta _{k}\sigma _{0,k}(\sigma _{a})\right] , \end{aligned}$$

where $\sigma _{0,k}$ denotes the covariance between stocks and options. The inequality can be simplified to

$$\begin{aligned}&\frac{1}{2}\lambda \left[ \frac{\beta _{s}^2}{\beta _k}(\sigma _{a}^{2}-\sigma _{1}^{2})+\beta _{k}(\sigma _{k}(\sigma _{a})^{2}-\sigma _{k}(\sigma _{1})^{2})+2\beta _{s}(\sigma _{0,k}(\sigma _{a})-\sigma _{0,k}(\sigma _{1}))\right] \\&\quad \le m_{k}(\sigma _{a})-m_{k}(\sigma _{1}). \end{aligned}$$

Note that $\sigma _{k}(\sigma )^{2}$, $\sigma _{0,k}(\sigma )$ and $m_{k}(\sigma )$ are increasing in $\sigma$. As we gradually reduce $\beta _{s}$ and $\beta _{k}$, the inequality above can still hold, so the agent will continue to perceive the lowest risk $\sigma _{1}$. We keep reducing $\beta _{s}$ and $\beta _{k}$ until $a_{1}$ is induced. Then we are done.$\square$

Proof of Lemma 4

When the effort a is induced, the agent’s utility is

$$\begin{aligned} E[u|\sigma _{a}]= & {} \alpha +\beta _{s}(a+l)+\beta _{k}m_{k}(\sigma _{a})\\&\quad -\frac{1}{2}\lambda \left[ \beta _{s}^{2}\sigma _{a}^{2}+\beta _{k}^{2}\sigma _{k}(\sigma _{a})^{2} +2\beta _{s}\beta _{k}\sigma _{0,k}(\sigma _{a})\right] -\frac{1}{2}a^2. \end{aligned}$$

We can calculate that $\frac{\partial ^{2}m_{k}(\sigma )}{\partial \sigma ^{2}}=\frac{\phi \eta ^{2}}{\sigma }$; $\frac{\partial ^{2}\sigma _{k}(\sigma )^{2}}{\partial \sigma ^{2}}=2(1-\Phi -\phi ^{2}+\phi \eta -\phi ^{2}\eta ^{2}+\phi \eta ^{3}-\phi \Phi \eta ^{3})$; $\frac{\partial ^{2}\sigma _{0,k}(\sigma )}{\partial \sigma ^{2}}=2(1-\Phi )+2\phi \eta +\phi \eta ^{3}$.

Since a is induced, we must have

$$\begin{aligned} \max _{\sigma \in [\sigma _{1},\sigma _{2}]}\frac{\partial E[u|\sigma ]}{\partial a}\ge 0. \end{aligned}$$

Let $\sigma _{0}\in \arg \max _{\sigma \in [\sigma _{1},\sigma _{2}]}\frac{\partial E[u|\sigma ]}{\partial a}$. Then, we obtain that

$$\begin{aligned} \beta _{s}+\beta _{k}(1-\Phi (\sigma _{0}))-\frac{1}{2}\lambda \left[ \beta _{k}^{2}\frac{\partial \sigma _{k}(\sigma _{0})^{2}}{\partial a}+ 2\beta _{s}\beta _{k}\frac{\partial \sigma _{0,k}(\sigma _{0})}{\partial a}\right] -a\ge 0. \end{aligned}$$

Since $\frac{\partial \sigma _{k}(\sigma _{0})^{2}}{\partial a}>0$ and $\frac{\partial \sigma _{0,k}(\sigma _{0})}{\partial a}>0$, $\beta _{s}+\beta _{k}>a$.

Fig. 4

Two functions

Taking the second-order derivative of $E[u|\sigma _{a}]$ w.r.t. $\sigma _{a}$ yields that

$$\begin{aligned} \frac{\partial ^{2}E[u]}{\partial \sigma _{a}^{2}}= & {} \beta _k\frac{\phi \eta ^2}{\sigma _a} -\frac{1}{2}\lambda \left[ 2\beta _s^2 \right. \\&+\, 2\beta _k^2(1-\Phi -\phi ^{2}+\phi \eta -\phi ^{2}\eta ^{2}+\phi \eta ^{3}-\phi \Phi \eta ^{3}) \\&\left. +\, 2\beta _s \beta _k(2(1-\Phi )+2\phi \eta +\phi \eta ^{3})\right] \\< & {} \beta _k\left[ \frac{\phi \eta ^2}{\sigma _a} -\frac{1}{2}\lambda \left[ 2\beta _k(1-\Phi -\phi ^{2}+\phi \eta -\phi ^{2}\eta ^{2}+\phi \eta ^{3}-\phi \Phi \eta ^{3}) \right. \right. \\&\left. \left. +\, 2\beta _s (2(1-\Phi )+2\phi \eta +\phi \eta ^{3})\right] \right] \end{aligned}$$

If $|k-a-l|<\sigma _{1}\cdot \min \left( 1, \sqrt{\frac{1}{2}\lambda a\sigma _{1}}\right)$, then $|\eta (\sigma _a)| < \min \left( 1, \sqrt{\frac{1}{2}\lambda a\sigma _{1}}\right)$. Thus, $\frac{\phi \eta ^2}{\sigma _a}<\frac{\phi \cdot \frac{1}{2}\lambda a\sigma _1}{\sigma _1}=\frac{1}{2}\lambda a \phi$. Using MATLAB, we can draw the graph for the functions $1-\Phi -\phi ^{2}+\phi \eta -\phi ^{2}\eta ^{2}+\phi \eta ^{3}-\phi \Phi \eta ^{3}$ and $2(1-\Phi )+2\phi \eta +\phi \eta ^{3}$ with $|\eta |<1$. From Fig. 4, we can see that both functions are greater than 0.5 when $|\eta |<1$. Note that $\phi <\frac{1}{\sqrt{2\pi } }$ for any $\eta$. Thus,

$$\begin{aligned} \frac{\partial ^{2}E[u]}{\partial \sigma _{a}^{2}}<\frac{1}{2}\lambda \beta _k \left[ \frac{1}{\sqrt{2\pi } }a -(\beta _s +\beta _k) \right] <0. \end{aligned}$$

Thus, we prove that if $|k-a-l|<\sigma _{1}\cdot \min \left( 1, \sqrt{\frac{1}{2}\lambda a\sigma _{1}}\right) ,$ $\frac{\partial ^{2}E[u]}{\partial \sigma _{a}^{2}}<0.$ Since $E[u|\sigma _{a}]$ is concave in $\sigma _{a}$, $E[u|\sigma _{a}]$ must be minimized at $\sigma _{a}=\sigma _{1}$ or $\sigma _{a}=\sigma _{2}$. So the agent will perceive either the lowest risk $\sigma _{1}$ or the highest risk $\sigma _{2}$.$\square$

Proof of Proposition 4

It is straightforward from the proofs of Lemma 2 and Proposition 2.$\square$

Proof of Lemma 6

We can calculate that

$$\begin{aligned} E[u(\alpha _s+\beta _s x)]= & {} \int _{-\infty }^{\infty }u(\alpha _s+\beta _s (a+l+y))\frac{1}{\sigma \sqrt{2\pi }}e^{-\frac{y^2}{2\sigma ^2}}\mathrm{d}y, \text { let }z=\frac{y}{\sigma }\\= & {} \int _{-\infty }^{\infty }u(\alpha _s+\beta _s (a+l+z\sigma ))\frac{1}{\sqrt{2\pi }}e^{-\frac{z^2}{2}}\mathrm{d}z. \end{aligned}$$

Thus, taking the derivative w.r.t. $\sigma$ yields that

$$\begin{aligned}&\frac{\partial }{\partial \sigma }E[u(\alpha _s+\beta _s x)] \\&\quad =\int _{-\infty }^{\infty }u'(\alpha _s+\beta _s (a+l+z\sigma ))\beta _s z\frac{1}{\sqrt{2\pi }}e^{-\frac{z^2}{2}}\mathrm{d}z,\\&\quad =\int _{0}^{\infty }u'(\alpha _s+\beta _s (a+l+z\sigma ))\beta _s z\frac{1}{\sqrt{2\pi }}e^{-\frac{z^2}{2}}\mathrm{d}z \\&\qquad -\int _{0}^{\infty }u'(\alpha _s+\beta _s (a+l-z\sigma ))\beta _s z\frac{1}{\sqrt{2\pi }}e^{-\frac{z^2}{2}}\mathrm{d}z. \end{aligned}$$

Since u is concave, it means that $u'(\cdot )$ is a decreasing function. So $\int _{0}^{\infty }u'(\alpha _s+\beta _s (a+l+z\sigma ))\beta _s z\frac{1}{\sqrt{2\pi }}e^{-\frac{z^2}{2}}\mathrm{d}z<\int _{0}^{\infty }u'(\alpha _s+\beta _s (a+l))\beta _s z\frac{1}{\sqrt{2\pi }}e^{-\frac{z^2}{2}}\mathrm{d}z$, and $\int _{0}^{\infty }u'(\alpha _s+\beta _s (a+l-z\sigma ))\beta _s z\frac{1}{\sqrt{2\pi }}e^{-\frac{z^2}{2}}\mathrm{d}z>\int _{0}^{\infty }u'(\alpha _s+\beta _s (a+l))\beta _s z\frac{1}{\sqrt{2\pi }}e^{-\frac{z^2}{2}}\mathrm{d}z$. Thus, $\frac{\partial }{\partial \sigma }E[u(\alpha _s+\beta _s x)]<0$. Hence, $E[u(\alpha _s +\beta _s x)]$ is decreasing in $\sigma$. $\square$

Proof of Lemma 7

Similar to the proof of Lemma 6, we can calculate that

$$\begin{aligned}&E[u(\alpha _k+\beta _k \max (x-k,0))] \\&\quad = \int _{-\infty }^{\infty }u(\alpha _k+\beta _k \max (a+l+\sigma z-k,0))\frac{1}{\sigma \sqrt{2\pi }}e^{-\frac{z^2}{2}}\mathrm{d}z, \\&\quad =u(\alpha _k)\Phi (\frac{k-a-l}{\sigma })+\int _{\frac{k-a-l}{\sigma }}^{\infty }u(\alpha _k+\beta _k (a+l+z\sigma -k))\frac{1}{\sqrt{2\pi }}e^{-\frac{z^2}{2}}\mathrm{d}z. \end{aligned}$$

Taking the derivative w.r.t. $\sigma$ yields that

$$\begin{aligned}&\frac{\partial }{\partial \sigma }E[u(\alpha _k+\beta _k \max (x-k,0))] \\&\quad =\int _{\frac{k-a-l}{\sigma }}^{\infty }u'(\alpha _k+\beta _k (a+l+z\sigma -k))\beta _k z\frac{1}{\sqrt{2\pi }}e^{-\frac{z^2}{2}}\mathrm{d}z \end{aligned}$$

If $k-a-l\ge 0$, then it is obvious that $\frac{\partial }{\partial \sigma }E[u(\alpha _k+\beta _k \max (x-k,0))]>0$. If $k-a-l<0$, then since $\int _{\frac{k-a-l}{\sigma }}^{\infty }u'(\alpha _k+\beta _k (a+l+z\sigma -k))\beta _k z\frac{1}{\sqrt{2\pi }}e^{-\frac{z^2}{2}}\mathrm{d}z = \int _{\frac{k-a-l}{\sigma }}^{\frac{a+l-k}{\sigma }}u'(\alpha _k+\beta _k (a+l+z\sigma -k))\beta _k z\frac{1}{\sqrt{2\pi }}e^{-\frac{z^2}{2}}\mathrm{d}z + \int _{\frac{a+l-k}{\sigma }}^{\infty }u'(\alpha _k+\beta _k (a+l+z\sigma -k))\beta _k z\frac{1}{\sqrt{2\pi }}e^{-\frac{z^2}{2}}\mathrm{d}z$. The first part equals to zero when $u''(\cdot )=0$, and the second part is always positive, so there must exist a bound B such that if $|u''(y)|<B$ for any y, then $\frac{\partial }{\partial \sigma }E[u(\alpha _k+\beta _k \max (x-k,0))]>0$, i.e., $E[u(\alpha _k +\beta _k \max (x-k,0))]$ is increasing in $\sigma$. $\square$