Moral hazard and compensation packages: does reshuffling matter?

Abstract

We study a moral hazard model in which the agent receives a compensation package made up of multiple commodities. We allow for the possibility that commodities are traded on the market and consider two scenarios. When trade in commodities is verifiable, the agent cannot reshuffle the compensation package prescribed by the principal and simply selects the hidden action which is optimal given that package. When trade in commodities is, instead, not verifiable, the agent can reshuffle the prescribed package by trading it for another one and can select a different action. We prove that an optimal contract (i.e., a contract which maximizes the principal’s expected payoff) when trade is verifiable remains optimal when trade is not verifiable if agent’s preferences for commodities are independent of the action performed. When, instead, preference independence fails, we show it is always possible to find prices of commodities such that an optimal contract under trade verifiability cannot be optimal under nonverifiability.

Appendix

Proof of Proposition 1

As explained in Sect. 3, we prove the result in three steps. In these steps, we will use the fact that Assumption I is equivalent to assuming that there exist functions $f\,:\,X\,\rightarrow \,{\mathbb {R}}$ and $g\,:\,{\mathcal {R}}(f)\times \mathcal {A }\,\rightarrow \,{\mathbb {R}}$ (where ${\mathcal {R}}(f)$ is the range of f) such that $u(x_{s},a)=g(f(x_{s}),a)$. Therefore, under Assumption I, a bundle $x_{s}\in X$ can be assigned a sub-utility level $f(x_{s})$ independent of a. Given the assumptions on the utility function u, it follows that g is strictly increasing in f.¹⁶

Step 1: If $\left( x^{*},a^{*}\right) \in X^{S}\times {\mathcal {A}}$ is a solution to (P1), i.e., $\left( x^{*},a^{*}\right) \in \beta (p,u_{\circ })$, then $x_{s}^{*}$ is a solution to (A4) given $a^{*}$ and ${\bar{u}}_{s}=u(x_{s}^{*},a^{*})$; that is it must be true that $x_{s}^{*}\in h_{s}(p,u(x_{s}^{*},a^{*}),a^{*})$ for $s=1,\ldots ,S$.

Proof of Step 1: Suppose there exists $(x^{*},a^{*})\in \beta (p,u_{\circ })$ such that $x_{s}^{*}\notin h_{s}(p,u(x_{s}^{*},a^{*}),a^{*})$ for some s. Observe that it must be $x_{s}^{*}\ne 0_{s}$,¹⁷ since $x_{s}^{*}=0_{s}$ would imply $p\cdot x_{s}^{*}\leqslant p\cdot x_{s}$ for every $x_{s}\in X$ and, a fortiori, for all $x_{s}\in X$ such that $u(x_{s},a^{*})\geqslant u(x_{s}^{*},a^{*})$.¹⁸ Pick $\hat{x}_{s}\in h_{s}(p,u(x_{s}^{*},a^{*}),a^{*})$, so that $p\cdot \hat{x}_{s}<p\cdot x_{s}^{*}$. Suppose that $u(\hat{x}_{s},a^{*})>u(x_{s}^{*},a^{*})$. Because of monotonicity, $\hat{x}_{s}\ne 0_{s}$.¹⁹ Consider the bundle $\alpha \hat{x}_{s}$ for $\alpha \in (0,1)$. Clearly, $\alpha \hat{x}_{s}\in X$ and, since $\hat{x}_{s}\ne 0_{s}$, $p\cdot (\alpha \hat{x}_{s})<p\cdot \hat{x}_{s}$. Furthermore, when $\alpha$ is close enough to 1, $u(\alpha \hat{x}_{s},a^{*})\geqslant u(x_{s}^{*},a^{*})$ follows from the continuity of the utility function. These conditions, however, would contradict the fact that $\hat{x} _{s}\in h_{s}(p,u(x_{s}^{*},a^{*}),a^{*})$. Therefore, it must be that $u(\hat{x}_{s},a^{*})=u(x_{s}^{*},a^{*})$. This equality implies that $f(\hat{x}_{s})=f(x_{s}^{*})$. Therefore²⁰$\pi _{s}(a)g(f(\hat{x}_{s}),a)+\sum _{s^{ \prime }\ne s}\pi _{s^{\prime }}(a)g(f(x_{s^{\prime }}^{*}),a)=v(\hat{x} _{s},x_{-s}^{*},a)$ is equal to $\pi _{s}(a)g(f(x_{s}^{*}),a)+\sum _{s^{\prime }\ne s}\pi _{s^{\prime }}(a)g(f(x_{s^{\prime }}^{*}),a)=v(x_{s}^{*},x_{-s}^{*},a)$ for every $a\in {\mathcal {A}}$. Hence, it follows that $a(\hat{x}_{s},x_{-s}^{*})=a(x_{s}^{*},x_{-s}^{*})$. Since $a^{*}\in a(x_{s}^{*},x_{-s}^{*})$, this last equality implies $a^{*}\in a(\hat{x}_{s},x_{-s}^{*})$. It follows that $(\hat{x}_{s},x_{-s}^{*},a^{*})$ satisfies (P1.a)-(P1.c), and therefore, it must be true that $\sum _{s}\pi _{s}(a^{*})(\varrho _{s}-p\cdot x_{s}^{*})\geqslant \pi _{s}(a^{*})(\varrho _{s}-p\cdot \hat{x}_{s})+\sum _{s^{\prime }\ne s}\pi _{s}(a^{*})(\varrho _{s^{\prime }}-p\cdot x_{s^{\prime }}^{*})$. However, this inequality implies that $\varrho _{s}-p\cdot x_{s}^{*}\geqslant \varrho _{s}-p\cdot \hat{x}_{s}$, hence that $p\cdot x_{s}^{*}\leqslant p\cdot \hat{x}_{s}$, which is a contradiction.

Step 2: If $\left( x^{*},a^{*}\right) \in X^{S}\times {\mathcal {A}}$ is a solution to (P1), i.e., $\left( x^{*},a^{*}\right) \in \beta (p,u_{\circ })$, then $x_{s}^{*}$ is a solution to (A3) with $\omega _{s}=p\cdot x_{s}^{*}$ given $a^{*}$; that is, it must be true that $x_{s}^{*}\in x_{s}(p,p\cdot x_{s}^{*},a^{*})$ for $s=1,\ldots ,S$.

Proof of Step 2: Suppose there exists $(x^{*},a^{*})\in \beta (p,u_{\circ })$ such that $x_{s}^{*}\notin x_{s}(p,p\cdot x_{s}^{*},a^{*})$ for some s. Observe that it must be $x_{s}^{*}\ne 0_{s}$, since $x_{s}^{*}=0_{s}$ would imply $p\cdot x_{s}^{*}=0$ and, therefore, $x_{s}(p,p\cdot x_{s}^{*},a^{*})=\{x_{s}^{*}\}$. Pick $\hat{x}_{s}\in x_{s}(p,p\cdot x_{s}^{*},a^{*})$, so that $u(\hat{x}_{s},a^{*})>u(x_{s}^{*},a^{*})$ and, because of monotonicity, $\hat{x}_{s}\ne 0_{s}$. Consider the bundle $\alpha \hat{x}_{s}$ for $\alpha \in (0,1)$. Clearly, $\alpha \hat{x}_{s}\in X$ and, since $\hat{x}_{s}\ne 0_{s}$, $p\cdot (\alpha \hat{x}_{s})<p\cdot \hat{x}_{s}=p\cdot x_{s}^{*}$. Furthermore, when $\alpha$ is close enough to 1, $u(\alpha \hat{x}_{s},a^{*})\geqslant u(x_{s}^{*},a^{*})$ follows from the continuity of the utility function. These conditions, however, would contradict the fact that $x_{s}^{*}\in h_{s}(p,u(x_{s}^{*},a^{*}),a^{*})$, as established in the previous step.

Step 3: For any $(x^{*},a^{*})\in X^{S}\times {\mathcal {A}}$ and for every $s=1,\ldots ,S$, if $a^{*}$ is a solution to (A1) given $x^{*}$, that is $a^{*}\in a(x^{*})$, and $x_{s}^{*}$ is a solution to (A3) with $\omega _{s}=p\cdot x_{s}^{*}$ given $a^{*}$, that is $x_{s}^{*}\in x_{s}(p,p\cdot x_{s}^{*},a^{*})$, then $(x^{*},a^{*})$ is a solution to (A2) with $\omega _{s}=p\cdot x_{s}^{*}$ for $s=1,\ldots ,S$; that is $(x^{*},a^{*})\in \varphi (p,p\otimes x^{*})$.²¹

Proof of Step 3: We prove the result by contraposition, i.e., we show that if $(x^{*},a^{*})\notin \varphi (p,p\otimes x^{*})$ then either $a^{*}\notin a(x^{*})$ or $x_{s}^{*}\notin x_{s}(p,p\cdot x_{s}^{*},a^{*})$ for at least one s. Clearly, if $a^{*}\notin {\mathcal {A}}$, then a fortiori $a^{*}\notin a(x^{*})$. Therefore, assume $a^{*}\in {\mathcal {A}}$. Since $(x^{*},a^{*})\notin \varphi (p,p\otimes x^{*})$, it follows that there exist $(\hat{x},\hat{a})$ such that $\hat{a}\in {\mathcal {A}}$, $p\cdot \hat{x}_{s}\leqslant p\cdot x_{s}^{*}$ for every s, and $v(\hat{x}, \hat{a})>v(x^{*},a^{*})$. Recall that Assumption I holds. Suppose $f(x_{s}^{*})\geqslant f(\hat{x}_{s})$ for every s. In this case, $u(x_{s}^{*},\hat{a})\geqslant u(\hat{x}_{s},\hat{a})$ for every s, hence $v(x^{*},\hat{a})\geqslant v(\hat{x},\hat{a})$. This implies $v(x^{*},\hat{a})>v(x^{*},a^{*})$ and therefore $a^{*}\notin a(x^{*})$. Suppose instead that $f(x_{s}^{*})<f(\hat{x}_{s})$ for some s. In this case, $u(x_{s}^{*},a^{*})<u(\hat{x}_{s},a^{*})$ and, therefore, $x_{s}^{*}\notin x_{s}(p,p\cdot x_{s}^{*},a^{*})$. $\square$

Proof of Proposition 2

Fix a price vector $\bar{ p}\in {\mathbb {R}}_{++}^{L}$. Select a pair of actions $\hat{a}$ and ${\tilde{a}}$ and pick $\hat{x}\in H(\hat{a},{\bar{p}},u_{\circ })$ and ${\tilde{x}}\in H( {\tilde{a}},{\bar{p}},u_{\circ })$. Choose a state of nature s such that $\bar{ p}^{\hat{a}}\cdot \hat{x}_{s}^{\hat{a}}>0$, which is always possible because of property (1) in Remark 3. Take a price vector p which is related to ${\bar{p}}$ as follows: $p_{l}=\delta {\bar{p}}_{l}$ for every $l\in {\mathcal {L}}( {\tilde{a}})$ and $p_{l}={\bar{p}}_{l}$ for every $l\in {\mathcal {L}}\backslash {\mathcal {L}}({\tilde{a}})$. Choose $\delta \in (0,1]$ such that $p^{\hat{a} }\cdot \hat{x}_{s}^{\hat{a}}>\delta \left( {\bar{p}}^{{\tilde{a}}}\cdot {\tilde{x}} _{s}^{{\tilde{a}}}\right) =p^{{\tilde{a}}}\cdot {\tilde{x}}_{s}^{{\tilde{a}}}$, which is possible since $p^{\hat{a}}\cdot \hat{x}_{s}^{\hat{a}}={\bar{p}}^{ \hat{a}}\cdot \hat{x}_{s}^{\hat{a}}>0$. Note that $H({\tilde{a}},p,u_{\circ })=H({\tilde{a}},{\bar{p}},u_{\circ })$ and $H(\hat{a},p,u_{\circ })=H(\hat{a}, {\bar{p}},u_{\circ })$. Therefore, $\hat{x}\in H(\hat{a},p,u_{\circ })$ and ${\tilde{x}}\in H({\tilde{a}},p,u_{\circ })$. From properties (2) and (3) in Remark 3, it follows $u_{\circ }=v(\hat{x},\hat{a})$ and also

$$\begin{aligned} \max _{x}\ \left\{ v(x,{\tilde{a}})\ |\ p\cdot x_{s}\leqslant p\cdot {\tilde{x}} _{s}\ \text {for every}\ s\right\} =v({\tilde{x}},{\tilde{a}})=u_{\circ }\,. \end{aligned}$$

By property (4) in Remark 3, ${\hat{x}}^{-\hat{a}}=0$ and ${{\tilde{x}}}^{- {\tilde{a}}}=0$ implies that $p^{\hat{a}}\cdot {\hat{x}}_{s}^{\hat{a}}>p^{\tilde{a }}\cdot {{\tilde{x}}}_{s}^{{\tilde{a}}}$ is equivalent to $p\cdot \hat{x} _{s}>p\cdot {\tilde{x}}_{s}$. Therefore, it must be that

$$\begin{aligned} \max _{x}\ \left\{ v(x,{\tilde{a}})\ |\ p\cdot x_{s}\leqslant p\cdot \hat{x} _{s}\ \text {for every}\ s\right\} >u_{\circ } . \end{aligned}$$

(4)

The above inequality holds since the agent, conditional on choosing ${\tilde{a}}$, obtains higher utility with income equal to $p\cdot \hat{x}_{s}$ than equal to $p\cdot {\tilde{x}}_{s}$ in every s, since he can afford more in at least one state and his preferences are locally non-satiated on X. Finally, denote with $(x^{*},a^{*})$ a solution to (A2) with $\omega _{s}=p\cdot \hat{x}_{s}$ in every s. It follows that

$$\begin{aligned} v(x^{*},a^{*})\geqslant \max _{x}\ \left\{ v(x,{\tilde{a}})\ |\ p\cdot x_{s}\leqslant p\cdot \hat{x}_{s}\ \text {for every}\ s\right\} . \end{aligned}$$

(5)

Finally, (4) and (5) imply that $v(x^{*},a^{*})>u_{\circ }=v(\hat{x},\hat{a})$. This eventually means that, if the principal proposes the contract $(\hat{x},\hat{a})$ to the agent, he accepts it and, since trade in commodities is nonverifiable, is willing to reshuffle his compensation basket. In particular, he sells bundle $\hat{x}$ and buys bundle $x^{*}$ under the constraint that he can spend up to $p\cdot \hat{x}_{s}$ in every state, and chooses action $a^{*}$ rather than $\hat{a}$. To conclude the proof, we must have that $( \hat{x},\hat{a})$, with $H(\hat{a},p,u)$, is a solution to (P1) at prices p. For this, it suffices to choose $(\varrho _{1},\ldots ,\varrho _{S})$ in such a way that

$$\begin{aligned} \sum _{s}\pi _{s}(\hat{a})\varrho _{s}-c(\hat{a},p,u)>\sum _{s}\pi _{s}(a^{\prime })\varrho _{s}-c(a^{\prime },p,u)\mathbf {\ }\text {for every} \mathbf {\ }a^{\prime }\ne \hat{a}\mathbf {.} \end{aligned}$$

$\square$