Dynamic contracts and learning by doing

Abstract

This paper studies the design of optimal contracts in dynamic environments where agents learn by doing. We derive a condition under which contracts are fully incentive compatible. A closed-form solution is obtained when agents have CARA utility. It shows that human capital accumulation strengthens the power of incentives and allows the principal to provide the agent with better insurance against transitory risks.

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Notes

  1. 1.

    See the seminal work of [1].

  2. 2.

    Fernandes and Phelan [4] propose a framework where the state space remains manageable. They assume that output is only correlated from one period to the next, and that agents can take one of only two actions. Then the relevant deviation is unique, and it is sufficient to keep track of a single “threat keeping constraint”.

  3. 3.

    He et al. [7] extends Jovanovic and Prat (2013) by introducing hidden savings and effort costs which are convex instead of linear.

  4. 4.

    The companion paper [13] studies a dynamic moral hazard problem where the agent has access to states which the principal cannot observe. As an illustration of his general results, [13] solves a model where the hidden states corresponds to the amount of savings accumulated by the agent.

  5. 5.

    By contrast, our closed form example uses a utility function that is not separable in consumption and leisure.

  6. 6.

    It is straightforward to generalize the production function by adding a multiplicative term \(\alpha \in \left[ 0,1\right] \) in front of effort, i.e., \(Y_{t}=\int _{0}^{t}(\alpha a_{s}+h_{s})ds+\int _{0}^{t}\sigma dZ_{s}.\ \) All the derivations presented in this paper go through under this more general specification. Notice that [10] focuses on cases where \( \alpha =0\) so that effort affects output solely through its impact on the stock variable \(h\).

  7. 7.

    Our incentive constraint is also closely related to the one in [10] which reads \(-u_{a}\left( \cdot \right) =E_{t}^{a}\left[ \int _{t}^{T}e^{-\rho \left( s-t\right) }f\left( s-t\right) \gamma _{s}ds \right] \!.\,\)The function \(f\left( s-t\right) \) captures the effect of past action \(a_{s}\) on the value at time \(t\) of the stock variable. It is therefore equal to \(\exp (-\delta \left( s-t\right) )\) in our model. More importantly, the volatility coefficient \(\gamma _{t}\) does not anymore appear on the right hand side of the incentive constraint. This is because, as discussed in footnote 6, [10] assumes that effort has no direct effect on output, i.e., \(Y_{t}=\int _{0}^{t}h_{s}ds+\int _{0}^{t}\sigma dZ_{s}.\)

  8. 8.

    The argument is mostly heuristic because both wages and effort are endogenous. Yet, we will see that its intuition is valid when we solve for the optimal contract under CARA utility.

  9. 9.

    This constructive approach shows that the solution in Proposition is not invalidated by the lack of mutually absolutely continuous measure in the infinite horizon limit.

  10. 10.

    In Williams’ [12] model, the variable \(p\) corresponds to the co-state associated to the variable \(z\) capturing the covariation between reports and the stock of lies.

  11. 11.

    The sensitivity coefficient \(\gamma \) is unambiguously positive because \(v\) is negative and, as shown in the proof of Corollary 1, \(\rho >k.\)

  12. 12.

    We show in Corollary 1 that \(\rho >k\), whereas \(v<0\) follows from the specification (11) of the utility function.

  13. 13.

    This will not necessarily be true if we allowed effort to have a cumulative effect, i.e., if we let \(\delta \) be negative. Then persistence may even lead to backloaded transfers.

  14. 14.

    To see that this conclusion holds true in our set-up, set \(\delta \) equal to infinity. It follows that \(\lim _{\delta \rightarrow \infty }p_{t}=\lim _{\delta \rightarrow \infty }E_{t}\left[ \int _{t}^{T}e^{-\left( \rho +\delta \right) \left( s-t\right) }\gamma _{s}ds\right] =0.\) Thus the volatility coefficient \(\vartheta _{t}=0\), which implies in turn that the sufficient condition is satisfied since \(2u_{aa}\left( w_{t},a_{t}\right) \le 0=\vartheta _{t}.\)

  15. 15.

    We cannot directly apply the standard Martingale Representation theorem because we are considering weak solutions, so that \(\left\{ Z_{t}^{a}\right\} \) does not necessarily generate \(\left\{ \mathcal {F} _{t}^{Y}\right\} \).

  16. 16.

    The additional expectation term vanishes because both\(\ \nabla A_{s}\) and \( \Delta a_{s}\) are bounded and so

    $$\begin{aligned} \left( \int _{0}^{t}U\left( \tau ,Y_{\cdot },a_{\tau }\right) d\tau \right) E_{t}^{a}\left[ \int _{t}^{T}\left( \nabla A_{s}+\Delta a_{s}\right) dZ_{s}^{a}\right] =0. \end{aligned}$$
  17. 17.

    Square integrability of \(\Gamma _{s}^{2}\) can be established for any \( \varepsilon \in \left[ 0,\varepsilon _{0}\right) \) following the same steps as in Lemma 7.3 of [2].

  18. 18.

    \(\chi ^{*}\) is predictable since both \( \xi ^{*}\) and \(h^{*}\) are \(\mathbb {F}^{Y}-\)predictable.

  19. 19.

    In our particular problem, switching from a weak formulation of the agent’s problem, to a strong formulation of the principal’s problem, does not raise measurability issues because the agent’s action is constant over time, and so does not directly depends on the Brownian motion.

  20. 20.

    To establish this claim analytically, one can differentiate \(L\left( k\right) \) and \(R\left( k\right) \) to obtain: \(L^{\prime \prime }\left( k\right) =\left( \theta \lambda \sigma \right) ^{2}\) and \(R^{\prime \prime }\left( k\right) =2\left[ \theta \lambda \sigma \left( \delta +1\right) \right] ^{2}\left( k+\delta +1\right) ^{-3}\). Hence there exists a unique \( \hat{k}\) such that \(L^{\prime \prime }\left( k\right) \gtrless R^{\prime \prime }\left( k\right) \) for all \(k\gtrless \hat{k},\) which rules out the possibility that \(R\left( k\right) \) intersects twice \(L\left( k\right) \) from above.

  21. 21.

    The second equality below follows from the definition of \(k\) as

    $$\begin{aligned} k\left( \frac{k+\delta }{k+\delta +1}\right) =\frac{\rho -k}{\left( \theta \lambda \sigma \right) ^{2}k}. \end{aligned}$$

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Acknowledgments

I thank Boyan Jovanovic for his invaluable support and advice. I am grateful to an anonymous referee and to seminar participants at the University of Zurich, CERGE-EI and Federal Reserve of Minneapolis for comments and suggestions. I acknowledge the support of the Labex Ecodec (ANR-11-LABX-0047) and of the Barcelona GSE.

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Correspondence to Julien Prat.

Appendices

Appendix A

Proof of Proposition 1

Consider the Brownian motion \(Z^{0}\) under some probability space with probability measure \(Q\), and let \(\mathbb {F }^{Z^{0}}\triangleq \left\{ \mathcal {F}_{t}^{Z^{0}}\right\} _{0\le t\le T}\) denote the suitably augmented filtration generated by \(Z^{0}\). Let

$$\begin{aligned} Y_{t}=\int _{0}^{t}\sigma dZ_{s}^{0}, \end{aligned}$$

so that \(Y_{t}\) is also a Brownian motion under \(Q\). Since expected output is linear in cumulative output, the exponential local martingale

$$\begin{aligned} \Lambda _{t,\tau }^{a}\triangleq \exp \left( \int _{t}^{\tau }\left( \frac{ h_{s}+a_{s}}{\sigma }\right) dZ_{s}^{0}-\frac{1}{2}\int _{t}^{\tau }\left| \frac{h_{s}+a_{s}}{\sigma }\right| ^{2}ds\right) , \end{aligned}$$

for\(\ t\le \tau \le T,\ \)is a martingale, i.e., \(E_{t}\left[ \Lambda _{t,T}^{a}\right] =1\). Hence Girsanov theorem holds and ensures that

$$\begin{aligned} Z_{t}^{a}\triangleq Z_{t}^{0}-\int _{0}^{t}\frac{h_{s}+a_{s}}{\sigma }ds \end{aligned}$$

is a Brownian motion under the new probability measure \(dQ^{a}/dQ\triangleq \Lambda _{0,T}^{a}\). Given that both measures are equivalent, the triple \( \left( Y,Z^{a},Q^{a}\right) \) is a weak solution of the SDE

$$\begin{aligned} Y_{t}=\int _{0}^{t}\left( h_{s}+a_{s}\right) ds+\int _{0}^{t}\sigma dZ_{s}^{a}. \end{aligned}$$

Adopting a weak formulation allows us to view the choice of control \(a\) as determining the probability measure \(Q^{a}\). In order to define the agent’s optimization problem, let \(R^{a}\left( t\right) \) denote the reward from time \(t\) onwards so that

$$\begin{aligned} R^{a}\left( t\right) \triangleq e^{\rho t}\left[ \int _{t}^{T}u\left( s,Y_{\cdot },a_{s}\right) ds+U\left( T,Y_{\cdot }\right) \right] \!, \end{aligned}$$

where the output path is denoted by \(Y_{\cdot }\) and, with a slight abuse of notation, \(u\left( s,Y_{\cdot },a_{s}\right) \triangleq e^{-\rho s}u\left( w\left( Y_{\cdot }\right) ,a_{s}\right) \) and \(U\left( T,Y_{\cdot }\right) \triangleq e^{-\rho T}U\left( Y_{\cdot }\right) \) are utilities at time \(t\) discounted from time \(0\). The agent’s objective is to find an admissible control process that maximizes the expected reward \(E^{a}\left[ R^{a}\left( 0\right) \right] \) over all admissible controls \(a\in \mathcal {A}\). In other words, the agent solves the following problem

$$\begin{aligned} v_{t}=\sup _{a_{t}\in \mathcal {A}}V^{a}(t)\triangleq \sup _{a_{t}\in \mathcal {A }}E_{t}^{a}\left[ R^{a}\left( t\right) \right] \!,\quad \text {for all }\,0\le t\le T. \end{aligned}$$

The objective function can be recast as

$$\begin{aligned} V^{a}(t)=E_{t}^{a}\left[ R^{a}\left( t\right) \right] =E_{t}\left[ \Lambda _{t,T}^{a}R^{a}\left( t\right) \right] \!, \end{aligned}$$
(16)

where the operators \(E^{a}\left[ \cdot \right] \) and \(E\left[ \cdot \right] \) are expectations under the probability measure \(Q^{a}\) and \(Q\), respectively. One can see from (16) that varying \( a \) is indeed equivalent to changing the probability measure. The key advantage of the weak formulation is that, under the reference measure \(Q\), the output process does not depend on \(a\). Hence, we can treat it as fixed which enables us to solve our problem in spite of its non-Markovian structure.

Our derivation of the necessary conditions builds on the variational argument in [2]. Define the control perturbation

$$\begin{aligned} a^{\varepsilon }\triangleq a+\varepsilon \Delta a, \end{aligned}$$

We assume that there exists an \(\varepsilon _{0}>0\) for which any \( \varepsilon \in \left[ 0,\varepsilon _{0}\right) \) satisfy \(\left| a^{\varepsilon }\right| ^{4},\ \left| u^{a^{\varepsilon }}\right| ^{4},\ \left| u_{a}^{a^{\varepsilon }}\right| ^{4},\left| \Lambda _{t,\tau }^{a^{\varepsilon }}\right| ^{4}\), \( \left( \mathcal {U}_{t,\tau }^{a^{\varepsilon }}\right) ^{2}\) and \(\left( \partial _{a}\mathcal {U}_{t,\tau }^{a^{\varepsilon }}\right) ^{2}\) being uniformly integrable in \(L^{1}\left( Q\right) \) where

$$\begin{aligned} \mathcal {U}_{t,\tau }^{a}\triangleq \int _{t}^{\tau }u\left( s,Y_{\cdot },a_{s}\right) ds. \end{aligned}$$

We introduce the following shorthand notations for “variations”

$$\begin{aligned}&\displaystyle \nabla \mathcal {U}_{t,\tau }^{a} \triangleq \int _{t}^{\tau }u_{a}\left( s,Y_{\cdot },a_{s}\right) \Delta a_{s}ds, \end{aligned}$$
(17)
$$\begin{aligned}&\displaystyle \nabla h_{t} \triangleq \int _{0}^{t}e^{-\delta \left( t-s\right) }\Delta a_{s}ds, \end{aligned}$$
(18)
$$\begin{aligned} \nabla \Lambda _{t,\tau }^{a}&\triangleq \frac{\Lambda _{t,\tau }^{a}}{ \sigma }\left[ \int _{t}^{\tau }\left( \nabla h_{s}+\Delta a_{s}\right) dZ_{s}^{0}-\int _{t}^{\tau }\left( h_{t}+a_{s}\right) \left( \nabla h_{s}+\Delta a_{s}\right) ds\right] \nonumber \\&=\frac{\Lambda _{t,\tau }^{a}}{\sigma }\int _{t}^{\tau }\left( \nabla h_{s}+\Delta a_{s}\right) dZ_{s}^{a}. \end{aligned}$$
(19)

Step 1 We first characterize the variations of the agent’s objective with respect to \(\varepsilon \)

$$\begin{aligned} \frac{V^{a^{\varepsilon }}(t)-V^{a}(t)}{\varepsilon }&= E\left[ \Lambda _{t,T}^{a^{\varepsilon }}R^{a^{\varepsilon }}\left( t\right) -\Lambda _{t,T}^{a}R^{a}\left( t\right) \right] \\&= E\left[ \left( \frac{\Lambda _{t,T}^{a^{\varepsilon }}-\Lambda _{t,T}^{a} }{\varepsilon }\right) R^{a^{\varepsilon }}\left( t\right) +\Lambda _{t,T}^{a}\left( \frac{R^{a^{\varepsilon }}\left( t\right) -R^{a}\left( t\right) }{\varepsilon }\right) \right] \\&= E\left[ \nabla \Lambda _{t,T}^{a^{\varepsilon }}R^{a^{\varepsilon }}\left( t\right) +\Lambda _{t,T}^{a}\left( \frac{R^{a^{\varepsilon }}\left( t\right) -R^{a}\left( t\right) }{\varepsilon }\right) \right] . \end{aligned}$$

To obtain the limit of the first term as \(\varepsilon \) goes to zero, observe that

$$\begin{aligned} \nabla \Lambda _{t,T}^{a^{\varepsilon }}R^{a^{\varepsilon }}\left( t\right) -\nabla \Lambda _{t,T}^{a}R^{a}\left( t\right) =\left[ \nabla \Lambda _{t,T}^{a^{\varepsilon }}-\Lambda _{t,T}^{a}\right] R^{a}\left( t\right) +\nabla \Lambda _{t,T}^{a^{\varepsilon }}\left[ R^{a^{\varepsilon }}\left( t\right) -R^{a}\left( t\right) \right] . \end{aligned}$$

As shown in [2], for any \(\varepsilon \in \left[ 0,\varepsilon _{0}\right) \), this expression is integrable uniformly with respect to \(\varepsilon \) and so

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0}E\left[ \nabla \Lambda _{t,T}^{a^{\varepsilon }}R^{a^{\varepsilon }}\left( t\right) \right] =E\left[ \nabla \Lambda _{t,T}^{a}R^{a}\left( t\right) \right] . \end{aligned}$$

The limit of the second term reads

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0}\frac{R^{a^{\varepsilon }}\left( t\right) -R^{a}\left( t\right) }{\varepsilon }=e^{\rho t}\nabla \mathcal {U} _{t,T}^{a}. \end{aligned}$$

Due to the uniform integrability of \(\Lambda _{t,T}^{a}\left( R^{a^{\varepsilon }}\left( t\right) -R^{a}\left( t\right) \right) /\varepsilon \), the expectation is also well defined. Combining the two expressions above, we finally obtain

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0}\frac{V^{a^{\varepsilon }}(t)-V^{a}(t)}{ \varepsilon }=E\left[ \nabla \Lambda _{t,T}^{a}R^{a}\left( t\right) +\Lambda _{t,T}^{a}e^{\rho t}\nabla \mathcal {U}_{t,T}^{a}\right] \triangleq \nabla V^{a}(t). \end{aligned}$$
(20)

Step 2 We are now in a position to derive the necessary condition. Consider total earnings as of date \(0\)

$$\begin{aligned} I^{a}(t)\triangleq E_{t}^{a}\left[ \int _{0}^{T}u\left( s,Y_{\cdot },a_{s}\right) ds+U\left( T,Y_{\cdot }\right) \right] =\int _{0}^{t}u\left( s,Y_{\cdot },a_{s}\right) ds+e^{-\rho t}V^{a}(t). \end{aligned}$$
(21)

By definition, it is a \(Q^{a}-\)martingale. According to the extended Martingale Representation TheoremFootnote 15 of [5], all square integrable \(Q^{a}-\)martingales are stochastic integrals of \(\left\{ Z_{t}^{a}\right\} \) and there exists a unique process \(\zeta \) in \( L^{2}\left( Q^{a}\right) \) such that

$$\begin{aligned} I^{a}(T)=I^{a}(t)+\int _{t}^{T}\zeta _{s}\sigma dZ_{s}^{a}. \end{aligned}$$
(22)

This decomposition allows us to solve for \(\nabla V^{a}(t)\). Reinserting (17), (18) and (19) into (20) yieldsFootnote 16

$$\begin{aligned} \nabla V^{a}(t)&= E_{t}\left[ R^{a}\left( t\right) \frac{\Lambda _{t,T}^{a} }{\sigma }\int _{t}^{T}\left( \nabla h_{s}+\Delta a_{s}\right) dZ_{s}^{a}+\Lambda _{t,T}^{a}e^{\rho t}\int _{t}^{T}u_{a}\left( \cdot \right) \Delta a_{s}ds\right] \\&= e^{\rho t}E_{t}^{a}\left[ \frac{I^{a}(T)}{\sigma }\int _{t}^{T}\left( \nabla h_{s}+\Delta a_{s}\right) dZ_{s}^{a}{\small +}\int _{t}^{T}u_{a}\left( \cdot \right) \Delta a_{s}ds\right] . \end{aligned}$$

where subscripts denote derivatives and arguments are omitted for brevity. Given the law of motion (22), applying Ito’s rule to the first term yields

$$\begin{aligned}&d\left( \frac{I^{a}(\tau )}{\sigma }\int _{t}^{\tau }\left( \nabla h_{s}+\Delta a_{s}\right) dZ_{s}^{a}\right) =\left[ \zeta _{\tau }\left( \nabla h_{\tau }+\Delta a_{\tau }\right) \right] d\tau \\&\quad +\left[ \zeta _{\tau }\int _{t}^{\tau }\left( \nabla h_{s}+\Delta a_{s}\right) dZ_{s}^{a}+I_{t}^{a}(\tau )\left( \nabla h_{\tau }+\Delta a_{\tau }\right) \right] dZ_{\tau }^{a}. \end{aligned}$$

Hence \(\nabla V^{a}(t)\) can be represented as

$$\begin{aligned} e^{-\rho t}\nabla V^{a}(t)=E_{t}^{a}\left[ \int _{t}^{T}\Gamma _{s}^{1}ds+\int _{t}^{T}\Gamma _{s}^{2}dZ_{s}^{a}\right] \!, \end{aligned}$$

where

$$\begin{aligned} \Gamma _{s}^{1}&\triangleq \zeta _{s}\left[ \int _{0}^{s}e^{-\delta \left( s-\tau \right) }\Delta a_{\tau }d\tau +\Delta a_{s}\right] +u_{a}\left( s,Y_{\cdot },a_{s}\right) \Delta a_{s}, \\ \Gamma _{s}^{2}&\triangleq \zeta _{s}\left[ \int _{t}^{s}\left( \int _{0}^{\tau }e^{-\delta \left( \tau -r\right) }\Delta a_{r}dr+\Delta a_{\tau }\right) dZ_{\tau }^{a}\right] +I_{t}^{a}(s)\left( \int _{0}^{s}e^{-\delta \left( s-\tau \right) }\Delta a_{\tau }d\tau +\Delta a_{s}\right) . \end{aligned}$$

Given that \(\Gamma _{s}^{2}\) is square integrable,Footnote 17 we have \(E_{t}^{a} \left[ \int _{t}^{T}\Gamma _{s}^{2}dZ_{s}^{a}\right] =0\). As for the deterministic term, collecting the effect of each perturbation \(\Delta a_{s}\) yields

$$\begin{aligned} e^{-\rho t}\nabla V^{a}(t)=E_{t}^{a}\left[ \int _{t}^{T}\left( \int _{s}^{T}e^{-\delta \left( \tau -s\right) }\zeta _{\tau }d\tau +\zeta _{s}+u_{a}\left( s,Y_{\cdot },a_{s}\right) \right) \Delta a_{s}ds\right] \!. \end{aligned}$$

Finally, noticing that \(\Delta a_{s}\) was arbitrary, optimality of \(a_{t}\) requires that

$$\begin{aligned} \left( E_{t}^{a}\left[ \int _{t}^{T}e^{-\delta \left( s-t\right) }\zeta _{s}ds \right] +\zeta _{t}+u_{a}\left( t,Y_{\cdot },a_{t}\right) \right) \left( a-a_{t}\right) \le 0. \end{aligned}$$
(23)

We now rewrite the necessary condition as a function of the volatility \(\gamma \) of the promised value \(v_{t}\). Differentiating (21) with respect to time yields

$$\begin{aligned} dI^{a}(t)=e^{-\rho t}\left[ dv_{t}-\rho v_{t}+u\left( Y_{\cdot },a_{t}\right) \right] =\zeta _{t}\sigma dZ_{t}^{a}, \end{aligned}$$

so that

$$\begin{aligned} dv_{t}=\left( \rho v_{t}-u\left( Y_{\cdot },a_{t}\right) \right) dt+e^{\rho t}\zeta _{t}\sigma dZ_{t}^{a}. \end{aligned}$$

Replacing \(\gamma _{t}=e^{\rho t}\zeta _{t}\) into (23) and collecting the exponential terms, one obtains the necessary condition (6).\(\square \)

Proof Proposition 2

We wish to compare rewards along the equilibrium path \(\left\{ a_{t}^{*}\right\} _{t=0}^{T}\) with those derived following an arbitrary strategy \(\left\{ a_{t}\right\} _{t=0}^{T}\). As in the proof of Proposition 1 let \(\Delta a_{t}\triangleq a_{t}-a_{t}^{*}\) denote the local effort deviation. Similarly, let

$$\begin{aligned} \nabla h_{t}\triangleq h_{t}-h_{t}^{*}=\int _{0}^{t}e^{-\delta (t-s)}\Delta a_{s}ds, \end{aligned}$$

denote the resulting differences in human capital at each date. The Brownian motions generated by the two effort policies satisfy

$$\begin{aligned} \sigma dZ_{t}^{^{a^{*}}}=\sigma dZ_{t}^{a}+\left[ a_{t}+h_{t}-\left( a_{t}^{*}+h_{t}^{*}\right) \right] dt=\sigma dZ_{t}^{a}+\left( \Delta a_{t}+\nabla h_{t}\right) dt. \end{aligned}$$

The total rewards \(I^{a^{*}}(t)\) from strategy \(a\) as of date \(0\) can therefore be decomposed as follows

$$\begin{aligned} I^{a^{*}}\left( T\right)&= \int _{0}^{T}e^{-\rho t}u\left( Y_{\cdot },a_{t}\right) dt+e^{-\rho T}U\left( Y_{\cdot }\right) =V^{a^{*}}\left( 0\right) +\int _{0}^{T}e^{-\rho t}\gamma _{t}^{*}\sigma dZ_{t}^{a^{*}}\\&= V^{a^{*}}\left( 0\right) +\int _{0}^{T}e^{-\rho t}\gamma _{t}^{*}\left( \Delta a_{t}+\nabla h_{t}\right) dt+\int _{0}^{T}e^{-\rho t}\gamma _{t}^{*}\sigma dZ_{t}^{a}. \end{aligned}$$

Hence, the total reward from the arbitrary policy is given by

$$\begin{aligned} I^{a}\left( T\right)&= \int _{0}^{T}e^{-\rho t}\left[ u\left( Y_{\cdot },a_{t}\right) -u\left( Y_{\cdot },a_{t}^{*}\right) \right] dt+I^{a^{*}}\left( T\right) \\&= \int _{0}^{T}e^{-\rho t}\left[ u\left( Y_{\cdot },a_{t}\right) -u\left( Y_{\cdot },a_{t}^{*}\right) \right] dt+V^{a^{*}}\left( 0\right) \\&+\int _{0}^{T}e^{-\rho t}\gamma _{t}^{*}\left( \Delta a_{t}+\nabla h_{t}\right) dt+\int _{0}^{T}e^{-\rho t}\gamma _{t}^{*}\sigma dZ_{t}^{a}. \end{aligned}$$

Let us focus on the third term on the right hand side

$$\begin{aligned} \int _{0}^{T}e^{-\rho t}\gamma _{t}^{*}\nabla h_{t}dt&= \int _{0}^{T}e^{-\rho t}\gamma _{t}^{*}\left( \int _{0}^{t}e^{-\delta (t-s)}\Delta a_{s}ds\right) dt\\&= \int _{0}^{T}e^{\delta t}\Delta a_{t}\left( \int _{t}^{T}e^{-\left( \delta +\rho \right) s}\gamma _{s}^{*}ds\right) dt\\&= \int _{0}^{T}e^{\delta t}\Delta a_{t}\left[ e^{-\left( \delta +\rho \right) t}p_{t}^{*}+\int _{t}^{T}e^{-\left( \delta +\rho \right) s}\vartheta _{s}^{*}\sigma dZ_{s}^{a^{*}}\right] dt, \end{aligned}$$

where the last equality follows from the definition of \(p\) and \(\vartheta \). Changing the Brownian motion in the last term on the RHS and taking expectation yields

$$\begin{aligned}&E_{0}^{a}\left[ \int _{0}^{T}e^{\delta t}\Delta a_{t}\left( \int _{t}^{T}e^{-\left( \delta +\rho \right) s}\vartheta _{s}^{*}\sigma dZ_{s}^{a^{*}}\right) dt\right] \\&\quad =E_{0}^{a}\left[ \int _{0}^{T}e^{\delta t}\Delta a_{t}\left( \int _{t}^{T}e^{-\left( \delta +\rho \right) s}\vartheta _{s}^{*}\left( \Delta a_{s}+\nabla h_{s}\right) ds\right) dt\right] \\&\quad =E_{0}^{a}\left[ \int _{0}^{T}e^{-\rho t}\vartheta _{t}^{*}\left( \Delta a_{t}+\nabla h_{t}\right) \left( \int _{0}^{t}e^{-\delta \left( t-s\right) }\Delta a_{s}\right) dt\right] \\&\quad =E_{0}^{a}\left[ \int _{0}^{T}e^{-\rho t}\vartheta _{t}^{*}\left( \Delta a_{t}+\nabla h_{t}\right) \nabla h_{t}dt\right] \!. \end{aligned}$$

Hence we have

$$\begin{aligned} V^{a}\left( 0\right) -V^{a^{*}}\left( 0\right)&= E_{0}^{a}\bigg [ \int _{0}^{T}e^{-\rho t}\left( u\left( w_{t},a_{t}\right) -u\left( w_{t},a_{t}^{*}\right) +\left( \gamma _{t}^{*}+p_{t}^{*}\right) \Delta a_{t}\right. \\&\left. +\vartheta _{t}^{*}\left( \Delta a_{t}+\nabla h_{t}\right) \nabla h_{t}\right) dt\bigg ]. \end{aligned}$$

We know from the optimization property of \(a_{t}^{*}\) that the first expectation term is at most equal to zero. On the other hand, the sign of the second expectation term is ambiguous. In order to bound it, we introduce the predictable processFootnote 18 \(\chi _{t}^{*}\triangleq \gamma _{t}^{*}-\vartheta _{t}^{*}h_{t}^{*}\) and define the function

$$\begin{aligned} H\left( a,h;\chi ^{*},\vartheta ^{*},p^{*}\right) \triangleq u\left( w,a\right) +\left( \chi ^{*}+\vartheta ^{*}h\right) a+\vartheta ^{*}h^{2}+p^{*}a. \end{aligned}$$

Taking a linear approximation of \(H\left( \cdot \right) \) around \(A^{*}\) yields

$$\begin{aligned}&H\left( a_{t},h_{t}\right) -H\left( a_{t}^{*},h_{t}^{*}\right) - \frac{\partial H\left( a_{t}^{*},h_{t}^{*}\right) }{\partial h} \nabla h_{t} \\&= u\left( w_{t},a_{t}\right) -u\left( w_{t},a_{t}^{*}\right) +\left( \chi _{t}^{*}+\vartheta _{t}^{*}h_{t}^{*}\right) \Delta a_{t}+a_{t}\vartheta _{t}^{*}\left( h_{t}-h_{t}^{*}\right) \\&+\vartheta _{t}^{*}\left( h_{t}^{2}-h_{t}^{*2}\right) +p_{t}^{*}\Delta a_{t}-\left[ \vartheta _{t}^{*}a_{t}^{*}+2\vartheta _{t}^{*}h_{t}^{*}\right] \nabla h_{t} \\&= u\left( w_{t},a_{t}\right) -u\left( w_{t},a_{t}^{*}\right) +\left( \gamma _{t}^{*}+p_{t}^{*}\right) \Delta a_{t}+\vartheta _{t}^{*}\nabla h_{t}\left( a_{t}-a_{t}^{*}\right) +\vartheta _{t}^{*}\left( h_{t}^{2}-h_{t}^{*2}-2\Delta _{t}h_{t}^{*}\right) \\&= u\left( w_{t},a_{t}\right) -u\left( w_{t},a_{t}^{*}\right) +\left( \gamma _{t}^{*}+p_{t}^{*}\right) \Delta a_{t}+\vartheta _{t}^{*}\nabla h_{t}\left( \Delta a_{t}+\nabla h_{t}\right) . \end{aligned}$$

The expected benefits of following an alternative strategy can therefore be written as

$$\begin{aligned} V^{a}\left( 0\right) -V^{a^{*}}\left( 0\right) =E_{0}^{a}\left[ \int _{0}^{T}e^{-\rho t}\left( H\left( a_{t},h_{t}\right) -H\left( a_{t}^{*},h_{t}^{*}\right) -\frac{\partial H\left( a_{t}^{*},h_{t}^{*}\right) }{\partial h}\nabla h_{t}\right) dt\right] \!, \end{aligned}$$

which is negative when \(H\left( \cdot \right) \) is jointly concave. Given that the agent seeks to maximize expected returns, imposing concavity ensures that \(a^{*}\) dominates any alternative effort path. Concavity is established considering the Hessian matrix of \(H\left( \cdot \right) \)

$$\begin{aligned} \mathcal {H}\left( t,a,h\right) =\left( \begin{array}{cc} u_{aa}\left( w_{t},a_{t}\right) &{} \vartheta _{t} \\ \vartheta _{t} &{} 2\vartheta _{t} \end{array} \right) \!, \end{aligned}$$

which is negative semi-definite when \(2u_{aa}\left( w_{t},a_{t}\right) \le \vartheta _{t}\), as stated in (10).\(\square \)

Proof of Proposition 3

In order to simplify the algebra, we start by focusing on contracts which extract full effort. Then we can omit actions \(a\) from the list of control and recast the principal’s optimization problem as

$$\begin{aligned} B_{t}=\underset{\left\{ w,\gamma ,\vartheta \right\} }{\min }E_{t}^{a}\left[ \int _{t}^{\infty }e^{-\rho (s-t)}w_{s}ds\right] \!, \end{aligned}$$

subject to

$$\begin{aligned} dv_{t}&= \left[ \rho v_{t}-u\left( w_{t},1\right) \right] dt+\gamma _{t}\sigma dZ_{t}, \end{aligned}$$
(24)
$$\begin{aligned} dp_{t}&= \left[ \left( \rho +\delta \right) p_{t}-\gamma _{t}\right] dt+\vartheta _{t}\sigma dZ_{t}, \end{aligned}$$
(25)
$$\begin{aligned} \gamma _{t}&= -u_{a}\left( w_{t},1\right) -p_{t}, \end{aligned}$$
(26)

i.e., the two promise-keeping constraints and the necessary condition under which income volatility is minimized.

The Hamilton-Jacobi-Bellman (HJB hereafter) equation associated to the principal’s optimal control problem readsFootnote 19

$$\begin{aligned} \rho B\left( v,p\right) =\underset{\left\{ w,\vartheta \right\} }{\min } \left\{ \begin{array}{c} w+\frac{\partial B}{\partial v}\left( \rho v-u\left( w,1\right) \right) + \frac{\partial B}{\partial p}\left( \left( \rho +\delta \right) p-\gamma \right) \\ +\frac{\sigma ^{2}}{2}\left[ \frac{\partial ^{2}B}{\partial v^{2}}\gamma \left( p,w\right) ^{2}+\frac{\partial ^{2}B}{\partial p^{2}}\vartheta ^{2}+2 \frac{\partial ^{2}B}{\partial v\partial p}\gamma \left( p,w\right) \vartheta \right] \end{array} \right\} , \end{aligned}$$
(27)

where we have used (26) to express \(\gamma \) as a function of \(p\) and \(w.\) We seek a solution to the HJB equation of the following form

$$\begin{aligned} \rho B\left( v\right) =\lambda -\frac{\ln \left( -\rho v\right) }{\theta } -C, \end{aligned}$$
(28)

and guess that the optimal wage schedule is equal to

$$\begin{aligned} w(v)=-\frac{\ln (-kv)}{\theta }+\lambda \Rightarrow u(w(v),1)=kv, \end{aligned}$$
(29)

while the value of private information is given by

$$\begin{aligned} p\left( v\right) =-\theta \lambda \frac{k}{k+\delta +1}v. \end{aligned}$$
(30)

Under our premise that \(p\) is equal to \(v\) times a constant, we can eliminate \(p\) from the HJB equation and rewrite it as follows

$$\begin{aligned} \rho B\left( v\right) =\underset{w}{\min }\left\{ w+B^{\prime }\left( v\right) \left( \rho v-u\left( w,1\right) \right) +\frac{\sigma ^{2}}{2} B^{\prime \prime }\left( v\right) \gamma \left( v,w\right) ^{2}\right\} . \end{aligned}$$
(31)

The FOC with respect to wages reads

$$\begin{aligned} 1-B^{\prime }\left( v\right) u_{w}\left( w,1\right) +\sigma ^{2}B^{\prime \prime }\left( v\right) \gamma \left( v,w\right) \frac{\partial \gamma \left( v,w\right) }{\partial w}=0. \end{aligned}$$
(32)

In order to obtain the expression of \(\partial \gamma /\partial w\), we replace our guess into (26) to obtain

$$\begin{aligned} \gamma =-u_{a}\left( w(v),1\right) -p=-\theta \lambda \left( k-\frac{k}{ k+\delta +1}\right) v=-\theta \lambda k\left( \frac{k+\delta }{k+\delta +1} \right) v. \end{aligned}$$

Differentiating this expression with respect to wages yields

$$\begin{aligned} \frac{\partial \gamma \left( v,w\right) }{\partial w}=-u_{aw}\left( w(v),1\right) =-\theta \gamma +\theta ^{2}\lambda \frac{k}{k+\delta +1}v. \end{aligned}$$

The FOC (32) for wages is therefore equivalent to

$$\begin{aligned} 1+B^{\prime }\left( v\right) \theta vk-\sigma ^{2}B^{\prime \prime }\left( v\right) \theta ^{3}\left( \lambda v\right) ^{2}\left[ \left( k-\frac{k}{ k+\delta +1}\right) ^{2}\!+\!\left( k\!-\!\frac{k}{k+\delta +1}\right) \frac{k}{ k\!+\delta +\!1}\!\right] =0, \end{aligned}$$

or

$$\begin{aligned} \rho -k-\left( \theta \lambda \sigma \right) ^{2}\left[ k\left( k-\frac{k}{ k+\delta +1}\right) \right] =\rho -k-\left( \theta \lambda \sigma \right) ^{2}k^{2}\left( \frac{k+\delta }{k+\delta +1}\right) =0. \end{aligned}$$

Thus \(k\) solves the following cubic equation

$$\begin{aligned} k^{3}\left( \theta \lambda \sigma \right) ^{2}+k^{2}\left[ \left( \theta \lambda \sigma \right) ^{2}\delta +1\right] +k\left( \delta +1-\rho \right) -\rho \left( \delta +1\right) =0. \end{aligned}$$
(33)

The relevant solution is given by the positive root because wages are not defined when \(k\) is negative. The existence and uniqueness of \(k\) are established below in the proof of Corollary 1.

We now verify that the dynamic programming equation is indeed satisfied

$$\begin{aligned} \rho B\left( v\right)&=w+B^{\prime }\left( v\right) \left( \rho v-u\left( w\left( v\right) ,1\right) \right) +B^{\prime \prime }\left( v\right) \frac{ \sigma ^{2}}{2}\gamma ^{2} \\&=\lambda -\frac{\ln (-kv)}{\theta }-\frac{\rho -k}{\rho \theta }+\frac{ \left( \sigma \lambda \right) ^{2}\theta }{2\rho }\left[ k\left( \frac{ k+\delta }{k+\delta +1}\right) \right] ^{2}. \end{aligned}$$

Replacing our guess for \(B\left( v\right) \) on the left hand side, one finds that the HJB holds true for all promised value \(v\) as long as

$$\begin{aligned} C=\frac{\rho -k}{\rho \theta }-\frac{\left( \sigma \lambda \right) ^{2}\theta }{2\rho }\left[ k\left( \frac{k+\delta }{k+\delta +1}\right) \right] ^{2}. \end{aligned}$$
(34)

We still have to verify our guess (30) for \(p\). Reinserting the Incentive Constraint (26) into the law of motion of \(p\), we find that

$$\begin{aligned} dp_{t}=\left[ \left( \rho +\delta +1\right) p_{t}+u_{a}\left( w_{t},1\right) \right] dt+\vartheta _{t}\sigma dZ_{t}, \end{aligned}$$

Integrating this expression with respect to time yields

$$\begin{aligned} p_{t}=-E\left[ \int _{t}^{\infty }e^{-\left( \rho +\delta +1\right) (s-t)}u_{a}\left( w_{s},1\right) ds\right] \!, \end{aligned}$$

which, using our specification (11) of the utility function and wage function (29), yields

$$\begin{aligned} p_{t}=-\theta \lambda E_{t}\left[ \int _{t}^{\infty }e^{-\left( \rho +\delta +1\right) \left( s-t\right) }u\left( w_{s},1\right) ds\right] =-\theta \lambda \int _{t}^{\infty }e^{-\left( \rho +\delta +1\right) \left( s-t\right) }kE_{t}\left[ v_{s}\right] ds. \end{aligned}$$
(35)

The expression of \(E_{t}\left[ v_{s}\right] \) follows from the law of motion of \(v\)

$$\begin{aligned} dv_{t}=\left[ \rho v_{t}-u\left( w_{t},1\right) \right] dt+\gamma _{t}\sigma dZ_{t}=v_{t}\left( \rho -k\right) dt+\gamma _{t}\sigma dZ_{t}. \end{aligned}$$

Since \(\gamma _{t}\) is square integrable, we have \(E_{t}\left[ v_{s}\right] =v_{t}\exp \left( \left( \rho -k\right) \left( s-t\right) \right) .\) Replacing this expression into (35), we finally obtain

$$\begin{aligned} p_{t}=-\theta \lambda kv_{t}\int _{t}^{\infty }e^{-\left( k+\delta +1\right) \left( s-t\right) }ds=-\theta \lambda \frac{k}{k+\delta +1}v_{t}, \end{aligned}$$

as conjectured in (30).

The last step consists in establishing that the contract maximizes the principal’s profit. First notice that the profit function \(\pi \) can be rewritten as follows

$$\begin{aligned} \pi \left( h_{_{t}},v_{t}\right)&= \max _{\left\{ a,w,\gamma ,\vartheta \right\} }E_{t}\left[ \int _{t}^{\infty }e^{-\rho (s-t)}\left( a_{s}+h_{s}-w_{s}\right) ds\right] \\&\!=\!&h_{t}\int _{t}^{\infty }e^{\!-\!\left( \rho +\delta \right) (s-t)}ds+\max _{\left\{ a,w,\gamma ,\vartheta \right\} }E_{t} \left[ \int _{t}^{\infty }e^{-\rho (s-t)}\left( a_{s}\!+\!\int _{t}^{s}e^{-\delta (s-\tau )}a_{\tau }d\tau \!-\!w_{s}\right) ds\right] \\&= \frac{h_{t}}{\rho +\delta }+\pi \left( 0,v_{t}\right) . \end{aligned}$$

Since the agent’s incentive constraint is independent of the level of human capital, neither the principal’s objective function nor its constraint depend on \(h\), and Bellman’s principle of optimality implies that it is equivalent to maximize \(\pi \left( 0,v_{t}\right) \) or \(\pi \left( h_{t},v_{t}\right) \). We can therefore rewrite the principal problem as

$$\begin{aligned} \pi \left( v_{t}\right) =\max _{\left\{ a,w,\gamma ,\vartheta \right\} }E_{t} \left[ \int _{t}^{\infty }e^{-\rho (s-t)}\left( a_{s}\left[ 1+\frac{1}{\rho +\delta }\right] -w_{s}\right) ds\right] . \end{aligned}$$

subject to (24), (25) and (26). According to the steps above, our guess for the value function

$$\begin{aligned} \rho \pi \left( v\right) =1+\frac{1}{\rho +\delta }-\rho B\left( v\right) \!, \end{aligned}$$

is a solution of the dynamic programming equation

$$\begin{aligned} \rho \pi \left( v\right) =\sup _{\left\{ a,w\right\} }\left\{ a\left( 1+\frac{ 1}{\rho +\delta }\right) -w+\pi ^{\prime }\left( v\right) \left( \rho v-u\left( w,a\right) \right) +\pi ^{\prime \prime }\left( v\right) \frac{ \sigma ^{2}}{2}\gamma \left( v,w,a\right) ^{2}\right\} \!. \end{aligned}$$
(36)

Furthermore, for each fixed \(v\in \mathbb {R} \), the supremum with respect to wages is attained when \(w(v)=-\ln (-kv)/\theta +\lambda .\ \)For optimal effort to be equal to one, we must have

$$\begin{aligned} 1+\frac{1}{\rho +\delta }+B^{\prime }(v)u_{a}\left( w,1\right) -B^{\prime \prime }(v)\sigma ^{2}\gamma \left( v,w,1\right) \frac{\partial \gamma \left( v,w,1\right) }{\partial a}\ge 0. \end{aligned}$$
(37)

To see that the inequality is indeed satisfied, differentiate the incentive constraint (26) to obtain \(\partial \gamma /\partial a=-\lambda \partial \gamma /\partial w\). Reinserting this equality along with \(u_{a}\left( \cdot \right) =-\lambda u_{w}\left( \cdot \right) \) into the FOC (32) for wages, we find that (37) is strictly positive. Hence the supremum with respect to effort is attained when \(a=1\), which completes the verification argument.\(\square \)

Proof of Corollary 1

The first step consists in characterizing the cubic equation defining \(k\)

$$\begin{aligned} k^{3}\left( \theta \lambda \sigma \right) ^{2}+k^{2}\left[ \left( \theta \lambda \sigma \right) ^{2}\delta +1\right] +k\left( \delta +1-\rho \right) -\rho \left( \delta +1\right) =0. \end{aligned}$$
(38)

Observe that it can be rewritten as follows

$$\begin{aligned} \underset{\triangleq L\left( k\right) }{\underbrace{\left( \theta \lambda \sigma \right) ^{2}k^{2}-\rho }}\ =\ \underset{\triangleq R(k)}{\underbrace{ \frac{k^{2}\left[ \left( \theta \lambda \sigma \right) ^{2}-1\right] -\left( \delta +1\right) k}{k+\delta +1}}} \end{aligned}$$
(39)

We distinguish two cases:

  1. 1.

    \(\theta \lambda \sigma \le 1:\) The function \(R(k)\) on the RHS of (39) is decreasing for all \(k\in \mathbb {R} ^{+}\) and \(R(0)=0\). By contrast, the function \(L\left( k\right) \) on the LHS of (39) is always increasing for all \(k\in \mathbb {R} ^{+}\). Furthermore, since \(L\left( 0\right) =-\rho \) and \(lim_{k\rightarrow \infty }L\left( k\right) =\infty ,\) (39) always admits a unique positive real solution.

  2. 2.

    \(\theta \lambda \sigma >1:\) The function \(R(k)\) on the RHS of (39) is not anymore decreasing for all \(k\in \mathbb {R} ^{+}.\) However, \(R(0)>L\left( 0\right) \) still holds true and since \( lim_{k\rightarrow \infty }R\left( k\right) <lim_{k\rightarrow \infty }L\left( k\right) \), (39) always admits a positive real solution. Furthermore, these two inequalities imply that the number of positive solutions of (39) cannot be equal to two. This means that either one root or all three of them are positive real numbers. But the last possibility being impossible too,Footnote 20 the only remaining option is that (39) has a unique positive real solution.

Having established the uniqueness of \(k,\) we proceed by showing that \(\partial k/\partial \delta <0\). Again, this is most easily established considering (39). Given that \(\partial L\left( k\right) /\partial \delta =0\) while \(\partial R\left( k\right) /\partial \delta =-k^{2}\left( \theta \lambda \sigma \right) ^{2}\left( k+\delta +1\right) ^{-2}<0\), the solution to (39) must be decreasing in \(\delta .\) Figure 3 illustrates the comparative statics exercise.

Fig. 3
figure3

Positive root of the cubic equation (39) for different values of \(\delta \)

Finally it will prove useful to show that \(k<\rho .\) Let \(\bar{k}\) denote the positive root of (38) when \(\delta =0\), i.e.,

$$\begin{aligned} \bar{k}^{3}\left( \theta \lambda \sigma \right) ^{2}+\bar{k}^{2}+\bar{k} \left( 1-\rho \right) -\rho =0. \end{aligned}$$
(40)

Since \(\partial k/\partial \delta <0\) and \(\delta \ge 0\), we have \(k<\bar{k} \). Assume for the sake of contradiction that \(\bar{k}>\rho ,\) then \(\bar{k} ^{2}-\bar{k}\rho >0\) and \(\bar{k}^{3}\left( \theta \lambda \sigma \right) ^{2}+\bar{k}^{2}+\bar{k}\left( 1-\rho \right) >\bar{k}^{3}\left( \theta \lambda \sigma \right) ^{2}+\bar{k}>\rho ,\) which contradicts the definition in (40) of \(\bar{k}\). Hence it holds true that\(\ \rho >\bar{k}>k\) for all \(\delta \ge 0.\)

We can finally prove Corollary 1. Differentiating the incentive costs \(C\) with respect to \(k\), we getFootnote 21

$$\begin{aligned} \frac{\partial C}{\partial k}&= \frac{1}{\theta k}-\frac{1}{\theta \rho }- \frac{\left( \sigma \lambda \right) ^{2}\theta }{2\rho }2\left[ k\left( \frac{k+\delta }{k+\delta +1}\right) \right] \frac{\partial }{\partial k} \left[ k\left( \frac{k+\delta }{k+\delta +1}\right) \right] \\&= \frac{1}{\theta }\left[ \frac{1}{k}-\frac{1}{\rho }\right] +\frac{\left( \sigma \lambda \right) ^{2}\theta }{2\rho }2\frac{1}{\left( \theta \lambda \sigma \right) ^{2}}\left[ \frac{\rho }{k}-1\right] \frac{1}{\left( \theta \lambda \sigma \right) ^{2}}\frac{\rho }{k^{2}}>0. \end{aligned}$$

The positive sign follows from \(k<\rho \), and the corollary holds true since

$$\begin{aligned} \frac{\partial C}{\partial \delta }=\underset{+}{\underbrace{\frac{\partial C }{\partial k}}}\underset{\mathbf {-}}{\underbrace{\frac{\partial k}{\partial \delta }}}-\frac{\left( \sigma \lambda \right) ^{2}\theta }{2\rho } k^{2}2\left( \frac{k+\delta }{k+\delta +1}\right) \frac{1}{\left( k+\delta +1\right) ^{2}}<0. \end{aligned}$$

\(\square \)

Proof of Corollary 2

The sufficiency condition derived in Proposition 2 is satisfied when \(2u_{aa}\left( w_{t},1\right) \le \vartheta _{t}.\ \)The volatility coefficient \(\vartheta \) of \(p\) follows from our finding in the proof of Proposition 3 that \(p_{t}=-v_{t}\theta \lambda k/\left( k+\delta +1\right) \). Applying Ito’s lemma, we get

$$\begin{aligned} \vartheta _{t}=-\theta \lambda \frac{k}{k+\delta +1}\gamma _{t}=\left( \frac{ \theta \lambda k}{k+\delta +1}\right) ^{2}\left( k+\delta \right) v_{t}. \end{aligned}$$

We have also shown that \(u\left( w_{t},1\right) =kv_{t}\). Given that \( u_{aa}\left( w_{t},1\right) =\left( \theta \lambda \right) ^{2}u\left( w_{t},1\right) \), the sufficient condition is equivalent to

$$\begin{aligned} 2u_{aa}\left( w_{t},1\right) \le \vartheta _{t}\Longleftrightarrow 2\ge \frac{k\left( k+\delta \right) }{\left( k+\delta +1\right) ^{2}}, \end{aligned}$$

which is true for all \(\delta \ge 0.\) \(\square \)

Proof of Corollary 3

We first show that it is optimal to retire the agent when \(1+1/\left( \rho +\delta \right) <\lambda -C\). Let \( R\left( v\right) \) denote the discounted cost of retiring an agent with promised value \(v\). Since there is no need to motivate the agent, the principal perfectly insures her; in other words \(\gamma _{t}=0\) and \( dv_{t}/dt=\rho \left( v_{t}\right) -u(w_{t},0)=0\) for all \(t\ge 0\). Evaluating the principal’s value function \(R\) associated to retirement yields

$$\begin{aligned} \rho R\left( v_{t}\right) =-\rho \int _{t}^{\infty }e^{-\rho (s-t)}w\left( v_{t}\right) ds=-w\left( v_{t}\right) =\frac{\ln \left( -\rho v\right) }{ \theta }. \end{aligned}$$

Retirement at date \(0\ \)dominates incentive provision when

$$\begin{aligned} R\left( v\right) -\pi \left( 0,v\right) =-1-\frac{1}{\rho +\delta } -C+\lambda >0, \end{aligned}$$
(41)

as stated in Corollary 3. We still have to prove that there is no optimal stopping rule such that retirement follows a period with effort exertion. We have shown in the proof of Proposition that

$$\begin{aligned} \pi _{t}\left( v\right) =\frac{h_{t}}{\delta \left( \rho +\delta \right) } +\pi _{0}\left( v\right) =\frac{1-e^{-\delta t}}{\delta \left( \rho +\delta \right) }+\pi _{0}\left( v\right) , \end{aligned}$$

where we have used the fact that \(h\) is deterministic to substitute it with \( t\). Thus, if the agent is not immediately retired at date \(0\), we must have \(0>R\left( v\right) -\pi _{0}\left( v\right) >R\left( v\right) -\pi _{t}\left( v\right) \), showing that it can never be optimal to retire the agent at a future date. \(\square \)

Appendix B

The closed-from solution described in Proposition 3 holds when the contracting horizon is infinite. We show in this Appendix that the same expression can be obtained by solving for fully incentive contracts with a finite horizon and then letting the retirement date go to infinity.

As in the main text, we restrict our attention to the exponential utility functions

$$\begin{aligned} u(w,a)=-\exp (-\theta \left( w-\lambda a)\right) ,\quad \text {with}\,\lambda \in \left( 0,1\right) ,\ \theta >0, \end{aligned}$$
(42)

with \(a\in \left[ 0,1\right] .\) Since the agent continuation value is given by

$$\begin{aligned} v_{t}\triangleq \max _{a_{t}\in \mathcal {A}}E_{t}^{a}\left[ \int _{t}^{T}e^{-\rho \left( s-t\right) }u\left( w_{s},a_{s}\right) ds+e^{-\rho \left( T-t\right) }U\left( W_{T}\right) \right] \!, \end{aligned}$$

we also have to specify the termination utility \(U\left( W\right) \). We assume that it lies on the Pareto frontier under zero effort, i.e.,

$$\begin{aligned} U(W)=-\frac{\exp \left( -\theta \rho W\right) }{\rho }. \end{aligned}$$
(43)

This is equivalent to assuming that agents are infinitely lived and that they retire at date \(T\). While retired, they consume the perpetual annuity derived from \(W\) and provide zero effort. Aside from its natural interpretation, this specification has been chosen for its tractability since the particular form of \(U(W)\) will become immaterial to our results as we shall let the contracting horizon diverge to infinity.

Proposition 4

Assume that: (i) \(u\left( w,a\right) \ \)and \(U\left( W\right) \) are as specified in (42) and (43); (ii) Recommended effort is set equal to its first-best level \(a_{t}^{*}=1 \). The expected wage bill as a function of the current date \(t\) and of the retirement date \(T\) reads

$$\begin{aligned} \rho B^{T}\left( t,v\right) =f^{T}\left( t\right) -\frac{\ln \left( -\rho v\right) }{\theta }. \end{aligned}$$
(44)

The function\(\ f^{T}\left( t\right) \) is given by

$$\begin{aligned} f^{T}\left( t\right) =-\int _{t}^{T}e^{-\rho \left( s-t\right) }\left[ \rho \left( \frac{\ln (k_{t}^{T}/\rho )}{\theta }-\lambda \right) -\frac{\left( \sigma \lambda \right) ^{2}\theta }{2}\left( \left( \varphi _{s}^{T}\right) ^{2}-\left( k_{s}^{T}\right) ^{2}\right) \right] ds, \end{aligned}$$
(45)

where \(k_{t}^{T}\) is the positive solution of

$$\begin{aligned} \rho -k_{t}^{T}-\left( \theta \lambda \sigma \right) ^{2}\left[ k_{t}^{T}\left( k_{t}^{T}-\varphi _{t}^{T}\right) \right] =0,\ \text {with\ }\varphi _{t}^{T}=\int _{t}^{T}e^{-\int _{t}^{s}\left( k_{\tau }^{T}+\delta +1\right) d\tau }k_{s}^{T}ds. \end{aligned}$$
(46)

Furthermore, when \(T\) goes to infinity, \(B^{T}\left( \cdot \right) \) converges pointwise to

$$\begin{aligned} \lim _{T\rightarrow \infty }B^{T}\left( t,v\right) =B\left( v\right) =\lambda -\frac{\ln \left( -\rho v\right) }{\theta }-C, \end{aligned}$$

where \(C\) is defined in Proposition 3.

Proof of Proposition 4

Step 1: Initial guess—We seek a solution to the HJB equation of the following form

$$\begin{aligned} \rho b^{T}\left( t,v\right) =f_{t}^{T}-\frac{\ln \left( -\rho v\right) }{ \theta }, \end{aligned}$$
(47)

and guess that the associated wage schedule and value of private information are given by

$$\begin{aligned} w_{t}(v)&= -\frac{\ln (-k_{t}^{T}v)}{\theta }+\lambda \Rightarrow u(w_{t}(v),1)=k_{t}^{T}v, \\ p_{t}\left( v\right)&= -\theta \lambda \varphi _{t}^{T}v, \end{aligned}$$

where \(f_{t}^{T}\), \(k_{t}^{T}\) and \(\varphi _{t}^{T}\) are continuously differentiable functions that depends solely on time \(t\). We impose the boundary condition \(\rho b^{T}\left( T,v\right) =-\ln \left( -\rho v\right) /\theta .\)

Step 2: Incentive Constraint—According to our guess, the Incentive Constraint reads

$$\begin{aligned} \gamma _{t}=-u_{a}\left( w_{t}(v),a_{t}\right) -p_{t}=-\theta \lambda \left( k_{t}^{T}-\varphi _{t}^{T}\right) v_{t}. \end{aligned}$$

Differentiating it with respect to wages yields

$$\begin{aligned} \frac{\partial \gamma \left( t,v,w,a\right) }{\partial w}=-u_{aw}\left( w_{t}(v),a_{t}\right) =-\theta \left( \gamma _{t}-\theta \lambda \varphi _{t}^{T}v_{t}\right) . \end{aligned}$$

The FOC (32) for wages is therefore equivalent to

$$\begin{aligned} 1+\frac{\partial b^{T}\left( t,v\right) }{\partial v}\theta k_{t}^{T}v-\sigma ^{2}\frac{\partial ^{2}b^{T}\left( t,v\right) }{\partial v^{2}}\theta ^{3}\left( \lambda v\right) ^{2}\left[ k_{t}^{T}\left( k_{t}^{T}-\varphi _{t}^{T}\right) \right] =0, \end{aligned}$$

Replacing the derivatives of \(b^{T}\left( \cdot \right) \) with respect to \( v,\ \)we obtain the characteristic equation

$$\begin{aligned} \rho -k_{t}^{T}-\left( \theta \lambda \sigma \right) ^{2}\left[ k_{t}^{T}\left( k_{t}^{T}-\varphi _{t}^{T}\right) \right] =0. \end{aligned}$$
(48)

The relevant solution \(k_{t}^{T}\) is given by the positive root because wages are not defined when \(k_{t}^{T}\) is negative.

Step 3: HJB equation—The dynamic programming equation reads

$$\begin{aligned} \rho b^{T}\left( t,v\right)&=w_{t}(v)+\frac{\partial b^{T}\left( t,v\right) }{\partial t}+\frac{\partial b^{T}\left( t,v\right) }{\partial v} \left( \rho v-u\left( w_{t}(v),1\right) \right) +\frac{\partial ^{2}b^{T}\left( t,v\right) }{\partial v^{2}}\frac{\sigma ^{2}}{2}\gamma ^{2} \\&=\lambda -\frac{\ln (-k_{t}^{T}v)}{\theta }+\frac{1}{\rho }\frac{ df^{T}\left( t\right) }{dt}-\frac{\rho -k_{t}^{T}}{\rho \theta }+\frac{ \left( \sigma \lambda \right) ^{2}\theta }{2\rho }\left( k_{t}^{T}-\varphi _{t}^{T}\right) ^{2}. \end{aligned}$$

The HJB is satisfied for all promised value \(v\) when

$$\begin{aligned} \frac{df^{T}\left( t\right) }{dt}-\rho f^{T}\left( t\right) =\rho \left( \frac{\ln (k_{t}^{T}/\rho )}{\theta }-\lambda \right) -\frac{\left( \sigma \lambda \right) ^{2}\theta }{2}\left( \left( \varphi _{t}^{T}\right) ^{2}-\left( k_{t}^{T}\right) ^{2}\right) , \end{aligned}$$
(49)

or

$$\begin{aligned} f^{T}\left( t\right) =-\int _{t}^{T}e^{-\rho \left( s-t\right) }\underset{ \triangleq \psi _{s}^{T}}{\underbrace{\left[ \rho \left( \frac{\ln (k_{t}^{T}/\rho )}{\theta }-\lambda \right) -\frac{\left( \sigma \lambda \right) ^{2}\theta }{2}\left( \left( \varphi _{s}^{T}\right) ^{2}-\left( k_{s}^{T}\right) ^{2}\right) \right] }}ds. \end{aligned}$$
(50)

Step 4: Verification of the guess for \(p_{t}\)—Following the steps leading to equation (35), we find that

$$\begin{aligned} p_{t}=-E_{t}\left[ \int _{t}^{T}e^{-\left( \rho +\delta +1\right) (s-t)}u_{a}\left( w_{s},a_{s}\right) ds\right] =-\theta \lambda \int _{t}^{T}e^{-\left( \rho +\delta +1\right) (s-t)}k_{s}^{T}E_{t}\left[ v_{s}\right] ds. \end{aligned}$$

The SDE for \(v_{t}\) reads

$$\begin{aligned} dv_{t}=v_{t}\left[ \left( \rho -k_{t}^{T}\right) dt-\Gamma _{t}^{T}\sigma dZ_{t}\right] \!, \end{aligned}$$

where \(\Gamma _{t}^{T}\triangleq \theta \lambda \left( k_{t}^{T}-\varphi _{t}^{T}\right) \). Whenever \(\Gamma _{t}^{T}\) is bounded, a conjecture that will be justified below, we have \(E_{t}\left[ v_{s}\right] =v_{t}\exp \left( \int _{t}^{s}\left( \rho -k_{\tau }^{T}\right) d\tau \right) \ \)and

$$\begin{aligned} p_{t}=-\theta \lambda \underset{=\varphi _{t}^{T}}{\underbrace{\left( \int _{t}^{T}e^{-\int _{t}^{s}\left( k_{\tau }^{T}+\delta +1\right) d\tau }k_{s}^{T}ds\right) }}v_{t}. \end{aligned}$$

Given the definition of \(k_{t}^{T}\), the coefficient \(\varphi _{t}^{T}\) is positive and bounded which implies in turn that, as conjectured above, \( \Gamma _{t}^{T}\) is bounded. Hence, we have verified our guess for the functional form of \(p_{t}\). Note that the value of private information \( p_{T}=\varphi _{T}^{T}=0\) because earnings are constant after date \(T\).

Step 5: Verification theorem—Our guess for the value function \( b^{T}\in \mathcal {C}^{1,2}\left( [ 0,T[ \times \mathbb {R} \right) \cap \mathcal {C}\left( \left[ 0,T\right] \times \mathbb {R} \right) \) is a solution of the dynamic programming equation

$$\begin{aligned} \rho b^{T}\left( t,v\right)&= \frac{\partial b^{T}\left( t,v\right) }{\partial t}+\inf _{w}\Bigg \{ w+\frac{\partial b^{T}\left( t,v\right) }{ \partial v}\left( \rho v-u\left( w,1\right) \right) \nonumber \\&+\frac{\partial ^{2}b^{T}\left( t,v\right) }{\partial v^{2}}\frac{\sigma ^{2}}{2}\gamma \left( t,v,w,1\right) ^{2}\Bigg \}, \end{aligned}$$
(51)

with boundary condition \(\rho b^{T}\left( T,v\right) =-\ln \left( -\rho v\right) /\theta .\ \)Furthermore, for each fixed \((t,v)\in [ 0,T[ \times \mathbb {R} \), the infimum in the expression

$$\begin{aligned} \inf _{w}\left\{ w+\frac{\partial b^{T}\left( t,v\right) }{\partial v}\left( \rho v-u\left( w,1\right) \right) +\frac{\partial ^{2}b^{T}\left( t,v\right) }{\partial v^{2}}\frac{\sigma ^{2}}{2}\gamma \left( t,v,w,1\right) ^{2}\right\} \end{aligned}$$

is attained by \(w^{*}(t,v)=-\ln (-k_{t}^{T}v)/\theta +\lambda .\ \)Thus the verification theorem holds and the value function for the control problem \(B^{T}\left( t,v\right) =b^{T}\left( t,v\right) \) while \(w^{*}(t,v)\) is the optimal markovian control.

Step 6: Incentive compatibility—The sufficient condition \( 2u_{aa}\left( w_{t},1\right) \le \vartheta _{t}\) is satisfied when \(2\ge \varphi _{t}^{T}\left( 1-\varphi _{t}^{T}/k_{t}^{T}\right) \!.\) Since both \( \varphi _{t}^{T}\) and \(k_{t}^{T}\) are positive, \(2u_{aa}\left( w_{t},1\right) \le \vartheta _{t}\) whenever \(\varphi _{t}^{T}\le 1\). To show that this restriction holds true, note that \(\partial \varphi _{t}^{T}/\partial T=k_{t}^{T}-\varphi _{t}^{T}\left( k_{t}^{T}+\delta +1\right) \!.\) This differential equation implies that \(\partial \varphi _{t}^{T}/\partial T=0\) whenever \(\varphi _{t}^{T}=k_{t}^{T}/\left( k_{t}^{T}+\delta +1\right) \). Given that the definition (48) of \(k_{t}^{T}\) is such that \(\partial k_{t}^{T}/\partial T=0\) when \(\partial \varphi _{t}^{T}/\partial T=0,\) we have \(\partial ^{2}\varphi _{t}^{T}/\partial T^{2}=\partial \varphi _{t}^{T}/\partial T=0\) whenever \( \varphi _{t}^{T}=k_{t}^{T}/\left( k_{t}^{T}+\delta +1\right) \!.\) This shows that \(\varphi _{t}^{T}\) is bounded, either from above or below, by \( k_{t}^{T}/\left( k_{t}^{T}+\delta +1\right) \). But we know that \(\varphi _{t}^{t}=0\). Hence it must be the case that \(\varphi _{t}^{T}\le k_{t}^{T}/\left( k_{t}^{T}+\delta +1\right) <1\) and, as explained before, the sufficient condition (10) is fulfilled.

Step 7: Convergence as \(T \ \) goes to infinity—The solutions for \(k_{t}^{T}\) and \(\varphi _{t}^{T}\) are found by backward induction using the terminal condition \(\varphi _{T}^{T}=0\) and reinserting the expression for \(\varphi _{t}^{T}\) into the quadratic equation (48) defining \(k_{t}^{T}\). Let \( k\triangleq \lim _{T\rightarrow \infty }k_{t}^{T}\) so that \(\varphi \triangleq \lim _{T\rightarrow \infty }\varphi _{t}^{T}=k/\left( k+\delta +1\right) \!.\) Reinserting the expression of \(\varphi _{t}^{T}\) into (48), we find that \(k\) is indeed given by the positive solution of the cubic equation (14). The pointwise convergence of \(k_{t}^{T}\) and \(\varphi _{t}^{T}\) ensure that \(\psi _{t}^{T}\) converges pointwise to

$$\begin{aligned} \psi \triangleq \lim _{T\rightarrow \infty }\psi _{t}^{T}=\rho \left( \frac{ \ln (k/\rho )}{\theta }-\lambda \right) -\frac{\left( \sigma \lambda \right) ^{2}\theta }{2}\left( \left( \frac{k}{k+\delta +1}\right) ^{2}-k^{2}\right) \!. \end{aligned}$$
(52)

Applying the dominated convergence theorem to evaluate the limit of the function \(f_{t}^{T}\) defined in (50), we obtain

$$\begin{aligned} \lim _{T\rightarrow \infty }f_{t}^{T}&= -\int _{t}^{T}e^{-\rho \left( s-t\right) }\psi ds=\lambda -\frac{\ln (k/\rho )}{\theta }+\frac{\left( \sigma \lambda \right) ^{2}\theta }{2\rho }\left( \left( \frac{k}{k+\delta +1 }\right) ^{2}-k^{2}\right) \\&= \lambda -\frac{\ln (k/\rho )}{\theta }-\frac{\rho -k_{t}^{T}}{\rho \theta }+\frac{\left( \sigma \lambda \right) ^{2}\theta }{2\rho }\left[ k\left( \frac{k}{k+\delta +1}\right) \right] ^{2}. \end{aligned}$$

Given that\(\ \lim _{T\rightarrow \infty }f_{t}^{T}=\lambda -C,\) we have \( \lim _{T\rightarrow \infty }B^{T}\left( t,v\right) =B\left( v\right) \) and the value function described in Proposition 3 is indeed the limit of a sequence of finite horizon problems. \(\square \)

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Prat, J. Dynamic contracts and learning by doing. Math Finan Econ 9, 169–193 (2015). https://doi.org/10.1007/s11579-014-0120-6

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Keywords

  • Principal agent model
  • Moral hazard
  • Human capital
  • Dynamic incentives