Voluntary ambiguity in incentive contracts

Abstract

Because accounting standards allow for some discretion in their application, accounting numbers often result from the choice of one possible interpretation among several. This paper investigates the optimal choice of a principal when this information is used for stewardship: whether committing ex-ante to using one specific interpretation, the more informative one, or keeping the ambiguity and opportunistically choosing ex-post the performance measure the least favorable to the agent. The choice between ambiguity and commitment involves a trade-off between the potential windfall for the agent (higher with commitment) and the risk of undeserved punishment (higher with ambiguity). The optimal policy depends on the risk aversion of the agent and of the extent of the ambiguity: high risk aversion and a large number of interpretations being in favor of ex-ante commitment.

Appendix

1.1 Proof of Lemma 1

First, the proof of (6). Note that:

$$\begin{aligned}&\frac{s_1(e)}{s_1(0)} < \frac{m(e)}{m(0)} \\&\quad \quad \Leftrightarrow (p(e)\pi _1+(1-p(e))(1-\pi _1)) \left( p(0)\prod _{i=1}^{N}\pi _i +(1-p(0))\prod _{i=1}^{N}(1-\pi _i) \right) \\&\quad < (p(0)\pi _1+(1-p(0))(1-\pi _1)) \left( p(e) \prod _{i=1}^{N}\pi _i +(1-p(e))\prod _{i=1}^{N}(1-\pi _i) \right) \\&\quad \Leftrightarrow p(e)\pi _1(1-\pi _1) \prod _{i=2}^{N}(1-\pi _i) + p(0)\pi _1(1-\pi _1) \prod _{i=2}^{N}\pi _i \\&\quad < p(0)\pi _1(1-\pi _1) \prod _{i=2}^{N}(1-\pi _i) + p(e)\pi _1(1-\pi _1) \prod _{i=2}^{N}\pi _i \end{aligned}$$

Thus, as $1-\pi _1>0$,

$$\begin{aligned} \frac{s_1(e)}{s_1(0)} < \frac{m(e)}{m(0)} \Leftrightarrow \prod _{i=2}^{N}(1-\pi _i) <\prod _{i=2}^{N}\pi _i \end{aligned}$$

which is clearly the case because $\pi _2>1/2$ and the others $\pi \ge 1/2$. Thus (6) holds.

Now, the proof of the inequality (7):

$$\begin{aligned}&\frac{1-s_1(e)}{1-s_1(0)} < \frac{1-m(e)}{1-m(0)}\\&\quad \Leftrightarrow ((1-p(e))\pi _1+p(e)(1-\pi _1)) \left( 1-p(0)\prod _{i=1}^{N}\pi _i -(1-p(0))\prod _{i=1}^{N}(1-\pi _i)\right) \\&\quad < ((1-p(0))\pi _1+p(0)(1-\pi _1)) \left( 1-p(e) \prod _{i=1}^{N}\pi _i-(1-p(e))\prod _{i=1}^{N}(1-\pi _i)\right) \\&\quad \Leftrightarrow p(e)(1-2\pi _1)-p(0)\pi _1^2 \prod _{i=2}^{N}\pi _i-p(e)(1-\pi _1)^2\prod _{i=2}^{N}(1-\pi _i)\\&\quad < p(0)(1-2\pi _1)-p(e)\pi _1^2 \prod _{i=2}^{N}\pi _i -p(0)(1-\pi _1)^2\prod _{i=2}^{N}(1-\pi _i)\\&\quad \Leftrightarrow \pi _1^2 \prod _{i=2}^{N}\pi _i < (1-\pi _1)^2\prod _{i=2}^{N}(1-\pi _i) + (2\pi _1-1) \end{aligned}$$

As $\pi _1 < 1$, the inequality (7) is equivalent to

$$\begin{aligned} \pi _1^2 \left( \prod _{i=2}^{N}\pi _i -1\right)&< (1-\pi _1)^2\left( \prod _{i=2}^{N}(1-\pi _i) -1\right) \nonumber \\&\Leftrightarrow \frac{1 -\prod _{i=2}^{N}(1-\pi _i ) }{1-\prod _{i=2}^{N}\pi _i } < \frac{\pi _1^2}{(1-\pi _1)^2} \end{aligned}$$

(10)

Thus, if (10) holds, inequality (7) will follow. I am going to show it by iteration (more precisely, that for all $1\ge \pi _1 \ge \pi _2 \ge \cdots \ge \pi _N \ge 1/2$ with $\pi _1>1/2$, (10) holds).

First, for the rank $N=2$, this is equivalent to

$$\begin{aligned} \frac{\pi _2}{1-\pi _2} < \frac{\pi _1^2}{(1-\pi _1)^2} \end{aligned}$$

As $1 \ge \pi _1\ge \pi _2 \ge 0,\,\frac{\pi _2}{(1-\pi _2)} \le \frac{\pi _1}{(1-\pi _1)}$. And as $\pi _1>1/2$, $\frac{\pi _1}{(1-\pi _1)}<\frac{\pi _1^2}{(1-\pi _1)^2}$. The result holds for the rank $N=2$.

Assume the assumption true at the rank $N-1$. If $\pi _2=1/2$ then $\pi _N = \cdots =\pi _2=1/2$. Moreover $\pi _1>1/2$. Thus

$$\begin{aligned} \frac{1 -\prod _{i=2}^{N}(1-\pi _i ) }{1-\prod _{i=2}^{N}\pi _i } =\frac{1 -\prod _{i=2}^{N}1/2 }{1-\prod _{i=2}^{N} 1/2 } = 1 < \frac{\pi _1^2}{(1-\pi _1)^2} \end{aligned}$$

and (10) holds at the rank $N$. If $\pi _2>1/2$, it is possible to apply (10) at the rank $N-1$ to the collection $\pi _2 \ge \cdots \ge \pi _N$:

$$\begin{aligned} \frac{1 -\prod _{i=3}^{N}(1-\pi _i ) }{1-\prod _{i=3}^{N}\pi _i } < \frac{\pi _2^2}{(1-\pi _2)^2} \end{aligned}$$

(11)

Moreover,

$$\begin{aligned}&\frac{1 -\prod _{i=2}^{N}(1-\pi _i ) }{1-\prod _{i=2}^{N}\pi _i }< \frac{\pi _2}{1-\pi _2} \nonumber \\&\quad \Leftrightarrow \frac{1 -\prod _{i=2}^{N}(1-\pi _i ) }{\pi _2}< \frac{1-\prod _{i=2}^{N}\pi _i }{1-\pi _2}\nonumber \\&\quad \Leftrightarrow \frac{1 -(1-\pi _2)+(1-\pi _2)-\prod _{i=2}^{N}(1-\pi _i) }{\pi _2}< \frac{1-\pi _2+\pi _2-\prod _{i=2}^{N}\pi _i }{1-\pi _2}\nonumber \\&\quad \Leftrightarrow \frac{1-\pi _2}{\pi _2} \left( 1 -\prod _{i=3}^{N}(1-\pi _i )\right) < \frac{\pi _2}{1-\pi _2} \left( 1-\prod _{i=3}^{N}\pi _i \right) \end{aligned}$$

(12)

which holds, given (11). Thus (12) holds. As $\frac{\pi _2}{(1-\pi _2)} \le \frac{\pi _1}{(1-\pi _1)} < \frac{\pi _1^2}{(1-\pi _1)^2}$, (10) holds at the rank $N$.

1.2 Proof of Lemma 2

$\frac{q(e)}{q(0)} < \frac{h(e)}{h(0)} \Leftrightarrow \frac{h(e)-h(0)}{h(0)}> \frac{q(e)-q(0)}{q(0)} \Leftrightarrow -\frac{h(0)}{h(e)-h(0)} > -\frac{q(0)}{q(e)-q(0)} \Leftrightarrow u^-_{q} < u^-_h$

$\frac{1-q(e)}{1-q(0)} < \frac{1-h(e)}{1-h(0)} \Leftrightarrow \frac{q(e)-q(0)}{1-q(0)} > \frac{h(e)-h(0)}{1-h(0)} \Leftrightarrow \frac{1-q(0)}{q(e)-q(0)} < \frac{1-h(0)}{h(e)-h(0)} \Leftrightarrow u^+_{q} < u^+_h$

1.3 Proof of Proposition 1

I am going to provide two sets of parameters, one for which ex-ante commitment is optimal and the other for which ambiguity is optimal.

Example 1—ex-ante commitment is optimal. Assume $p(e)=0.75,\,p(0)=0.25,\,C=8,\,N=2,\,\pi _1=0.8,\,\pi _2=0.6$, $u_\mathrm{res}=10$ and $\alpha =0.5$. Then $\mathrm{IC}_1=121$ and $\mathrm{IC}_m=175$: ex-ante commitment is optimal.

Example 2—ambiguity is optimal. Assume $p(e)=0.75$, $p(0)=0.6,\,C=10,\,N=2,\,\pi _1=0.6,\,\pi _2=0.59,\,u_\mathrm{res}=300$ and $\alpha =0.99$. Then $\mathrm{IC}_1=325.70$ and $\mathrm{IC}_m=325.59$: ambiguity is optimal.

1.4 Proof of Proposition 2

First the proof that, when risk aversion is large enough, ex-ante commitment is optimal. Ex-ante commitment is optimal if and only if $\mathrm{IC}_m>\mathrm{IC}_1$ that is

$$\begin{aligned} m(e)(u^+_m)^{1/\alpha } + (1-m(e)) (u^-_m)^{1/\alpha } - s_1(e)(u^+_1)^{1/\alpha } - (1-s_1(e)) (u^-_1)^{1/\alpha }>0\nonumber \\ \end{aligned}$$

(13)

where the four $u^i_j$ do not depend on $\alpha$. Let $D(\alpha )$ denote the LHS of the above inequality. Lemmas 1 and 2 entail that $u^+_m$ is strictly higher than the other three $u^i_j$. Thus $D(\alpha )/(u^+_m)^{1/\alpha }$ goes to $m(e)$ when $\alpha$ goes to zero. As $m(e)>0$ there is an $\alpha _{\lim }$ such that $\forall \alpha \le \alpha _{\lim },\, D(\alpha )/(u^+_m)^{1/\alpha }>0$. Thus $\forall \alpha \le \alpha _{\lim }$, that is if as soon as the agent is sufficiently risk averse, ex-ante commitment is optimal.

Now the proof of the second part of the proposition: if $N$ is large enough, then, again, ex-ante commitment is optimal.

The only property of the power utility function, $u(y)=y^\alpha /\alpha$ that is needed is $u'(y) \rightarrow _{y \rightarrow \infty } 0$. Denote $x_\mathrm{max}= \lim _{y \rightarrow \infty }u(y)$. With the power utility function $x_\mathrm{max}$ is infinite but it might be finite with other utility functions.

As $f'(x)=1/u'(f(x)),\,u'(y) \rightarrow _{y \rightarrow \infty } 0 \Leftrightarrow f'(x) \rightarrow _{x \rightarrow x_\mathrm{max}} + \infty$. I am going to show, first, that

$$\begin{aligned} f(x)/x \rightarrow _{x \rightarrow x_\mathrm{max}} + \infty \end{aligned}$$

(14)

Let $M>0$. As $f'(x) \rightarrow _{x \rightarrow x_\mathrm{max}} + \infty$, $\exists x_M>0$ such that $f'(x) \ge 2M$ for all $x$ above $x_M$. Thus $\forall x \ge x_M,\,f(x)-f(x_M) \ge 2M(x-x_M)$. Moreover, $f(x_M) \ge 0$. Thus $\forall x \ge 2x_M,\,f(x)/x \ge 2M(x-x_M)/x \ge M$.

Let $\mathrm{IC}_N$ denote the incentive cost with ambiguity and N signals. Denote $m_N(a)$, the probability of obtaining a good result in the ambiguity case when there are $N$ signals:

$$\begin{aligned} m_N(a)&=p(a)\prod _{i=1}^{N}\pi _i+(1-p(a))\prod _{i=1}^{N}(1-\pi _i) \end{aligned}$$

As $\forall i,\,(1-\pi _i) \le 1/2$ and $\pi _i \le \pi _1,\,m_N(a) \le p(a) (\pi _1)^N+(1-p(a)) (1/2)^N$. Thus, $m_N(a)$ goes to zero when $N$ goes to $+\infty$.

Lemma 1 ensures that $\frac{s_1(e)}{s_1(0)} < \frac{m(e)}{m(0)}$ and, thus, from Lemma 2, $u_N^- \ge u_1^-$.

$$\begin{aligned} \mathrm{IC}_N&= m_N(e) f \left( u^+_N \right) +(1-m_N(e))f(u^-_N) \\&\ge m_N(e) f \left( \frac{1-m_N(0)}{m_N(e)-m_N(0)}C + u_\mathrm{res} \right) +(1-m_N(e))f(u^-_1)\\&\ge m_N(e) f \left( \frac{(1-m_N(0))C+ u_\mathrm{res}m_N(e)}{m_N(e)} \right) +\min (0,f(u^-_1)) \end{aligned}$$

As $\forall a\,m_N(a)$ goes to zero when $N$ goes to infinity, $((1-m_N(0))C+ u_\mathrm{res}m_N(e))$ goes to $C$ and thus is above $C/2$ for $N$ high enough. Then

$$\begin{aligned} \mathrm{IC}_N&\ge \frac{C}{2} \frac{2m_N(e)}{C} f \left( \frac{C}{2m_N(e)} \right) +\min (0,f(u^-_1)) \end{aligned}$$

Since $\frac{C}{2m_N(e)}$ goes to $+\infty$, (14) implies that $\mathrm{IC}_N$ goes to $+\infty$. As a result, $\mathrm{IC}_N$ is always higher than $\mathrm{IC}_1$ for $N$ high enough.

1.5 Proof of Lemma 3

When $p(e)>1/2$, the lower $\pi$, the higher the probability of obtaining a poor result. This is the case because the probability of obtaining a poor result is $1-s(a)=p(a) - (2p(a)-1) \pi$. On the other hand, the lower $\pi$, the lower the informativeness, as both likelihood ratios (poor and good outcomes) worsen.

1.6 Proof of Lemma 4

Assume that the probability of obtaining a good future value is no more $p(a)$ but $q(a)=(1-p(a))$. Denote probability q this assumption, while the initial assumption will be denoted probability p. Denote $s_1^q(a)$, the probability that the first signal is good under probability q, while $s_1^p(a)$ is the same probability under probability p.

$$\begin{aligned} s_1^q(a)&=q(a)\pi _1+(1-q(a))(1-\pi _1) \\&= (1-p(a))\pi _1 + p(a)(1-\pi _1) \\&= 1 - (p(a)\pi _1+(1-p(a))(1-\pi _1) \\&= 1-s_1^p(a) \end{aligned}$$

Thus

$$\begin{aligned} \frac{1-s_1^p(e)}{1-s_1^p(0)} = \frac{s_1^q(e)}{s_1^q(0)} \end{aligned}$$

(15)

and

$$\begin{aligned} \frac{s_1^p(e)}{s_1^p(0)}=\frac{1-s_1^q(e)}{1-s_1^q(0)} \end{aligned}$$

(16)

Denote now $m^q(a)$ the probability that the ambiguity monitoring technology leads to a good evaluation under probability q.

$$\begin{aligned} m^q(a)&=q(a) \prod _1^N \pi _i+ (1-q(a)) \prod _1^N (1-\pi _i) \\&= (1-p(a)) \prod _1^N \pi _i+ p(a) \prod _1^N (1-\pi _i) \\ \end{aligned}$$

Denote $M^p(a)$ the probability of obtaining a good evaluation when the agent chooses the performance measure under probability p.

$$\begin{aligned} M^p(a)&= 1 - \left( p(a) \prod _1^N (1-\pi _i) + (1-p(a)) \prod _1^N \pi _i \right) \\&= 1- m^q(a) \end{aligned}$$

Thus

$$\begin{aligned} \frac{1-M^p(e)}{1-M^p(0)} = \frac{M^q(e)}{M^q(0)} \end{aligned}$$

(17)

and

$$\begin{aligned} \frac{M^p(e)}{M^p(0)}=\frac{1-M^q(e)}{1-M^q(0)} \end{aligned}$$

(18)

Given (15), (16), (17) and (18), Lemma 4 is a corollary of Lemma 1 applied to probability q.

1.7 Proof of Proposition 3

I am first going to use the same examples as in the proof of Proposition 1. Then I will provide an example where agent’s choice is optimal. Let $\mathrm{IC}_M$ denote the incentive cost with the agent’s choice monitoring technology.

Example 1—ex-ante commitment is optimal. Assume $p(e)=0.75,\,p(0)=0.25,\,C=8,\,N=2,\,\pi _1=0.8,\,\pi _2=0.6$, $u_\mathrm{res}=10$ and $\alpha =0.5$. Then $\mathrm{IC}_1=121,\,\mathrm{IC}_m=175$ and $\mathrm{IC}_M= 140$: ex-ante commitment is optimal.

Example 2—ambiguity is optimal. Assume $p(e)=0.75$, $p(0)=0.6,\,C=10,\,N=2,\,\pi _1=0.6,\,\pi _2=0.59,\,u_\mathrm{res}=300$ and $\alpha =0.99$. Then $\mathrm{IC}_1=325.70,\,\mathrm{IC}_m=325.59$ and $\mathrm{IC}_M=325.70$: ambiguity is optimal.

Example 3—agent’s choice is optimal. Assume $p(e)=0.75$, $p(0)=0.25,\,C=2,\,N=2,\,\pi _1=0.8,\,\pi _2=0.6,\,u_\mathrm{res}=10$ and $\alpha =0.1$. Then $\mathrm{IC}_1=24,\,\mathrm{IC}_m=151$ and $\mathrm{IC}_M= 21$: agent’s choice is optimal.

1.8 Proof of Proposition 4

In the proof of the first part of Proposition 2, the only property of the two monitoring technologies that has been used is that $u^+_m >u^+_1$, which given Lemma 2 comes from Eq. (7). Given (9), $u^+_m >u^+_M$ and $u^+_1 >u^+_M$. Thus the same proof gives that when the risk aversion of the agent is above a certain threshold, agent’s choice is optimal compared to the two other monitoring technologies.

Now, the proof of the second part of the proposition. Let $M_N(a)$ denotes the probability $M(a)$ when $N$ signals are available. As $\forall i\,1/2 \le \pi _i \le \pi _1$,

$$\begin{aligned} 1-M_N(a) \le p(a) (1/2)^N + (1-p(a)) (\pi _1)^N \end{aligned}$$

As $\pi _1<1$, when $N$ goes to $+\infty$, the RHS of the above inequality goes to zero for all $a$. Thus both $M_N(e)$ and $M_N(0)$ goes to 1 when $N$ goes to $+\infty$. As a result, $u_\mathrm{res} - M_N(0)C/(M_N(e)-M_N(0))$ goes to $-\infty$ when $N$ goes to $+\infty$. Thus for $N$ large enough: $u_\mathrm{res} < M_N(0)C/(M_N(e)-M_N(0))$. Then the optimal wages of the agent’s choice monitoring technology are no longer provided by (3) and (4). $w^-$ cannot be lower than 0 while keeping an expected utility higher than $-\infty$. The optimal rewards are then $w^+_{M_N} = f( C/ (M_N(e)-M_N(0))$ and $w^-_{M_N}=0$. As $M_N(e)$ go to 1 and $w^+_{M_N}$ goes to $+\infty$, the incentive cost goes to $+\infty$ when $N$ goes to $+\infty$, whereas the incentive cost with ex-ante commitment remains fixed. Thus, for $N$ large enough, ex-ante commitment is optimal compared to agent’s choice. Given Proposition 2, when $N$ is large enough, ex-ante commitment is also optimal compared to ambiguity.