Multi-task agency with unawareness

Abstract

The paper introduces the problem of unawareness into multi-dimensional Principal–Agent theory. We introduce two key parameters to describe the problem, the extent and the effect of unawareness, show under what conditions it is optimal for the Principal to propose an incomplete or a complete contract, and characterize the incentive power of optimal linear contracts. If Agents differ in their unawareness, optimal incentive schemes can be distorted for both aware and unaware Agents, because, different from standard contract theory, the single-crossing property fails to hold. In this case, even aware Agents can be subject to inefficiently high or low incentives.

Appendix

1.1 Proof of Proposition 2

Proposition 2 (with explicit expressions for the optimal contracts)

Let

$$\begin{aligned} \underline{\tau }&= \underline{\tau }(\lambda )=\frac{3(1-\lambda )(\sigma ^{2}+1)+1+\lambda }{(\sigma ^{2}+2)(\lambda +2(1-\lambda )(\sigma ^{2}+1))}\\ \overline{\tau }&= \overline{\tau }(\lambda )=\frac{(1-\lambda )(\sigma ^{2}+1)+3\lambda -1}{\lambda (\sigma ^{2}+2)} \end{aligned}$$

We have $\frac{\mathrm{d}}{\mathrm{d}\lambda }\underline{\tau }>0, \frac{\mathrm{d}}{\mathrm{d}\lambda } \overline{\tau }<0$ for all $\lambda \in [0,1], \underline{\tau } (0)<\tau _{\min }<\tau _{\max }<\overline{\tau }(0)$ and $\underline{\tau }(1)=\overline{\tau }(1)=t_{2\mathrm{F}}^{\mathrm{A}}.$ The solution of the problem (22)-(ICU) is unique and given as follows:

• 1. If $\tau <\underline{\tau }$ or $\tau >\overline{\tau }$ the incentive constraint (ICU) is slack and the solution is separating, with

  $$\begin{aligned} \alpha ^{\mathrm{A}}=\frac{2}{2+\sigma ^{2}},\qquad \text { } \beta ^{\mathrm{A}}&= \frac{4\left( \sigma ^{2}-2\right) }{2(2+\sigma ^{2})^{2}}+\frac{1}{2}\left( 1-\lambda \right) ^{2}\frac{\left( 1-(\sigma ^{2}+1)\tau -1\right) ^{2}}{\left( 1+\sigma ^{2}(1-\lambda )\right) ^{2}},\\ \alpha ^{\mathrm{U}}\!=\!\frac{1+\lambda (\tau -1)}{1+\sigma ^{2}(1-\lambda )},\text { } \beta ^{\mathrm{U}}&= \frac{1}{2}\tau ^{2}\!-\!\tau \frac{1+\lambda (\tau -1)}{1+\sigma ^{2}(1-\lambda )}-\frac{(1-\sigma ^{2})\left( 1\!+\!\lambda (\tau -1)\right) ^{2} }{2\left( 1+\sigma ^{2}(1-\lambda )\right) ^{2}}. \end{aligned}$$

• 2. If $\underline{\tau }\le \tau \le \frac{2}{2+\sigma ^{2}}$, the incentive constraint (ICU) is binding, with $\alpha ^{\mathrm{A}}-\tau =\tau -\alpha ^{\mathrm{U}}$ and

  $$\begin{aligned} \alpha ^{\mathrm{A}}&= \frac{1}{1+2\lambda +\sigma ^{2}}\left( 2\tau (1+\sigma ^{2})(1-\lambda )-1+3\lambda +\tau \lambda \right) ,\\ \beta ^{\mathrm{A}}&= \frac{1}{2}\left[ \frac{1}{1+2\lambda +\sigma ^{2}}\left( 1-3\lambda +(5+2\sigma ^{2})\tau \lambda \right) \right] ^{2} \\&-\frac{\tau }{1+2\lambda +\sigma ^{2}}\left( 1-3\lambda +(5+2\sigma ^{2})\tau \lambda \right) \\&+\frac{\tau ^{2}}{2}-\frac{1}{2}(2-\sigma ^{2})\left[ \frac{1}{1+2\lambda +\sigma ^{2}}\left( 2\tau (1\!+\!\sigma ^{2})(1-\lambda )-1\!+\!3\lambda +\tau \lambda \right) \right] ^{2},\\ \alpha ^{\mathrm{U}}&= 2\tau -\alpha ^{\mathrm{A}}=\frac{1}{1+2\lambda +\sigma ^{2}}\left( 1-3\lambda +(5+2\sigma ^{2})\tau \lambda \right) ,\\ \beta ^{\mathrm{U}}&= \frac{1}{2}\tau ^{2}-\frac{1}{2}(1-\sigma ^{2})\left[ \frac{1}{1+2\lambda +\sigma ^{2}}\left( 1-3\lambda +(5+2\sigma ^{2})\tau \lambda \right) \right] ^{2}\\&-\frac{\tau }{1+2\lambda +\sigma ^{2}}\left( 1-3\lambda +(5+2\sigma ^{2} )\tau \lambda \right) . \end{aligned}$$

• 3. If $\frac{2}{2+\sigma ^{2}}\le \tau \le \overline{\tau }$ the solution is pooling, with

  $$\begin{aligned} \alpha ^{\mathrm{A}}&= \alpha ^{\mathrm{U}}=\frac{1+\lambda +\tau \lambda }{1+2\lambda +\sigma ^{2} },\\ \beta ^{\mathrm{A}}&= \beta ^{\mathrm{U}}=\frac{1}{2}\tau ^{2}-\tau \frac{1+\lambda +\tau \lambda }{1+2\lambda +\sigma ^{2}}-\frac{1}{2}\frac{\left( 1+\lambda +\tau \lambda \right) ^{2}}{\left( 1+2\lambda +\sigma ^{2}\right) ^{2}}\left( 1-\sigma ^{2}\right) . \end{aligned}$$

Proof

Using Lemmas 2 and 3, one can eliminate the fixed payment $\beta$ from the problem and express the contracting problem solely in terms of the incentive component $\alpha$:

$$\begin{aligned}&\max _{\alpha ^{\mathrm{A}},\alpha ^{\mathrm{U}}}\lambda \left[ 4\alpha ^{\mathrm{A}}\!-\!(\sigma ^{2} +2)(\alpha ^{\mathrm{A}})^{2}\!+\!2\tau \alpha ^{\mathrm{U}}-(\alpha ^{\mathrm{U}})^{2}\right] +(1-\lambda )\left[ 2\alpha ^{\mathrm{U}}\!-\!(\sigma ^{2}\!+\!1)(\alpha ^{\mathrm{U}})^{2}\right] \nonumber \\\end{aligned}$$

(23)

$$\begin{aligned}&\quad \text {s.t. }\,\,(\alpha ^{\mathrm{A}}-\tau )^{2}\ge (\alpha ^{\mathrm{U}}-\tau )^{2} \end{aligned}$$

(24)

By straightforward differentiation, the unconstrained solution to the maximization problem (23) and (24) is

$$\begin{aligned} \alpha ^{\mathrm{A}}=\frac{2}{2+\sigma ^{2}},\quad \alpha ^{\mathrm{U}}=\frac{1+\lambda (\tau -1)}{1+\sigma ^{2}(1-\lambda )} \end{aligned}$$

(25)

This solution satisfies the constraint (24) strictly if and only if

$$\begin{aligned} (\tau (\sigma ^{2}+2)-2)^{2}(\lambda +(1-\lambda )(\sigma ^{2}+1))^{2} >(1-\lambda )^{2}(\tau (\sigma ^{2}+1)-1)^{2}(\sigma ^{2}+2)^{2} \end{aligned}$$

Viewed as a quadratic inequality in $\tau$, this is equivalent to $\tau <\underline{\tau }$ or $\tau >\overline{\tau }$. Hence, (25) yields the separating solution of the proposition.

If (ICU) is binding, there are two possibilities: $\alpha ^{\mathrm{U}}=\alpha ^{\mathrm{A}}$ (pooling) or $\alpha ^{\mathrm{U}}+\alpha ^{\mathrm{A}}=2\tau$ (constrained separating). Direct comparison shows that when $\frac{2}{2+\sigma ^{2}}\le \tau \le \overline{\tau }(\lambda )$ we have the pooling solution, and when $\underline{\tau } (\lambda )\le \tau \le \frac{2}{2+\sigma ^{2}}$ we have the constrained separating solution.

The monotonicity of $\underline{\tau }(\lambda )$ and of $\overline{\tau }(\lambda )$ follows by differentiation, and the statements about $\underline{\tau }(0),\overline{\tau }(0),\underline{\tau }(1)$, and $\overline{\tau }(1)$ by direct computation. $\square$

1.2 Proof of Proposition 3

Proof

We must compare the value $\pi _{\mathrm{S}}(\lambda ,\tau )$ of the screening problem solved in Proposition 2 to the profit from making all Agents aware, $\pi ^{\mathrm{A}}=\frac{2}{2+\sigma ^{2}}$. We do this by discussing the three possible cases derived in Proposition 2 in turn.

• (1) The separating case: Straightforward computation shows that $\pi ^{\mathrm{A}} >\pi _{\mathrm{S}}(\lambda ,\tau )$ if and only if

  $$\begin{aligned} \tau >R_{1}=\frac{X_{1}+Y_{1}}{Z_{1}} \end{aligned}$$

  or

  $$\begin{aligned} \tau <L_{1}=\frac{X_{1}-Y_{1}}{Z_{1}} \end{aligned}$$

  where

  $$\begin{aligned}&\displaystyle X_{1}=\left( 2+\sigma ^{2}\right) \left( 1+\lambda +\sigma ^{2}\left( 1-\lambda \right) \right) ,\\&\displaystyle Y_{1}=\sigma ^{2}\sqrt{\left( 2+\sigma ^{2}\right) \left( 1-\lambda \right) \left( 1+\sigma ^{2}\left( 1-\lambda \right) \right) }, \end{aligned}$$

  and

  $$\begin{aligned} Z_{1}=\left( 2+\sigma ^{2}\right) \left( 1+\lambda +\sigma ^{2}\right) . \end{aligned}$$

  Comparing these boundaries to those of Proposition 2, it is straightforward to show that $R_{1}>\overline{\tau }$ if and only if $\lambda >\frac{1}{2}$, and that $L_{1}<\underline{\tau }$ if and only if $\lambda >\frac{1}{2}$. Thus, in the separating case, when $\lambda >\frac{1}{2},$ the Principal makes all Agents aware if and only if $\tau >R_{1}$ or $\tau <L_{1}$. When $\lambda <\frac{1}{2},$ the Principal makes all Agents aware.

• (2) The pooling case: We have $\pi ^{A}>\pi _{S}(\lambda ,\tau )$ if and only if

  $$\begin{aligned} \tau >R_{2}=\frac{X_{2}+Y_{2}}{Z_{2}} \end{aligned}$$

  or

  $$\begin{aligned} \tau <L_{2}=\frac{X_{2}-Y_{2}}{Z_{2}} \end{aligned}$$

  where

  $$\begin{aligned}&\displaystyle X_{2}=\left( 2+\sigma ^{2}\right) \left( 1+\lambda \left( 2-\lambda \right) +\sigma ^{2}\left( 1-\lambda \right) \right) ,\\&\displaystyle Y_{2}=\left( 1-\lambda \right) \sigma ^{2}\sqrt{\left( 2+\sigma ^{2}\right) \left( 1+2\lambda +\sigma ^{2}\right) }, \end{aligned}$$

  and

  $$\begin{aligned} Z_{2}=\left( 2+\sigma ^{2}\right) \left( 1+\lambda \left( 2-\lambda \right) +\sigma ^{2}\right) . \end{aligned}$$

  Direct computation shows that $\overline{t}_{2}^{\mathrm{A}}>L_{2}$. Because necessarily $\tau \ge \overline{t}_{2}^{\mathrm{A}}$ under pooling, $\tau <L_{2}$ is impossible. Furthermore, $R_{2}<\overline{\tau }$ if and only if $\lambda <\frac{1}{2}$. Hence, if pooling is optimal in the screening problem, when $\lambda <\frac{1}{2}$ the Principal makes all Agents aware if and only if$\ \tau >R_{2}$. When $\lambda >\frac{1}{2},$ the Principal only uses the pooling solution.

• (3) The constrained separating case: We have $\pi ^{\mathrm{A}}>\pi _{\mathrm{S}}(\lambda ,\tau )$ if and only if

  $$\begin{aligned} \tau >R_{3}=\frac{X_{3}+Y_{3}}{Z_{3}}\end{aligned}$$

  or

  $$\begin{aligned} \tau <L_{3}=\frac{X_{3}-Y_{3}}{Z_{3}} \end{aligned}$$

where

$$\begin{aligned}&\displaystyle X_{3}=\left( 2+\sigma ^{2}\right) \left( \sigma ^{2}\left( 6\lambda +1\right) \left( \lambda -1\right) +9\lambda ^{2}-10\lambda -1\right) ,\\&\displaystyle Y_{3}=\left( 1-\lambda \right) \sigma ^{2}\sqrt{\left( 2+\sigma ^{2}\right) \left( 1+2\lambda +\sigma ^{2}\right) }, \end{aligned}$$

and

$$\begin{aligned} Z_{3}=\left( 2+\sigma ^{2}\right) \left( \sigma ^{2}\left( 4\lambda \sigma ^{2}\left( \lambda -1\right) +12\lambda ^{2}-12\lambda -1\right) +9\lambda ^{2}-10\lambda -1\right) . \end{aligned}$$

We have $\overline{t}_{2}^{\mathrm{A}}<R_{3}$ Since necessarily $\tau \le \overline{t}_{2}^{\mathrm{A}}, \tau >R_{3}$ is impossible. Furthermore, $L_{3}>\underline{\tau }$ if and only if$\ \lambda <\frac{1}{2}$. Hence, in the constrained-separating case, when $\lambda <\frac{1}{2}$ the Principal makes all Agents aware if and only if$\ \tau <L_{3}$. When $\lambda >\frac{1}{2},$ the Principal only uses the constrained separating solution.

The above three case discussions establish the awareness-thresholds for each $\lambda$. Formally, we have $\tau _{R}(\lambda )=R_{1}$ and $\tau _{L} (\lambda )=L_{1}$ for $\lambda \ge \frac{1}{2}$, and $\tau _{R}(\lambda )=R_{2}$ and $\tau _{L}(\lambda )=L_{3}$ for $\lambda \le \frac{1}{2}$. It is straightforward to show that these $\tau _{R}$ and $\tau _{L}$ are continuous at $\frac{1}{2}$, that $R_{1}$ and $R_{2}$ are decreasing in $\lambda$, and $L_{1}$ and $L_{3}$ are increasing in $\lambda$.

To complete the picture given in Fig. 9, for $\lambda =1$ one calculates $R_{1}=L_{1}=\frac{2}{2+\sigma ^{2}}=\overline{t}_{2}^{A}$, hence $\tau _{R}(1)=\tau _{L}(1)=\overline{t}_{2}^{A}$. Similarly, for $\lambda =0, R_{2}=\tau _{\max }$ and $L_{3}=\tau _{\min }$. $\square$