Moral Hazard Under Ambiguity

Abstract

In this paper, we extend the classical Holmström and Milgrom contracting problem, by adding uncertainty on the volatility of the output for both the Agent and the Principal. We study more precisely the impact of the “Nature” playing against the Agent and the Principal, by choosing the worst possible volatility of the output. We solve the first-best and the second-best problems in this framework, and we show that optimal contracts are in a class of contracts linear with respect to the output and its quadratic variation. We also present a general modus operandi to apply our method.

Appendix

Proof of Proposition 4.1

Our fist step is to look at the dynamic version of the value function of the Agent. Fix some $a\in {\mathcal {A}}$. We refer to the papers [44, 63] for the proofs that, for any ${{\mathcal {F}}}_T$-measurable contract $\xi \in {\mathcal {C}}$, one can define a process, which we denote by $u_t^A(\xi ,a)$ (denoted by $Y_t$ in [63]), which is càdlàg, ${\mathbb {G}}^{{{\mathcal {P}}}_A,+}$-adapted (recall that for any $a\in {\mathcal {A}}$, ${\mathbb {G}}^{{{\mathcal {P}}}^a_A}={\mathbb {G}}^{{{\mathcal {P}}}_A}$, since the polar sets of ${{\mathcal {P}}}^a_A$ are the same as the polar sets of ${{\mathcal {P}}}_A$) and such that

$$\begin{aligned} u_t^A(\xi ,a)=\underset{{\mathbb {P}}^{'}\in {\mathcal {P}}^a_A({\mathbb {P}},t^+)}{\mathrm{essinf}^{\mathbb {P}}}\ {\mathbb {E}}^{{\mathbb {P}}^{'}}\bigg [{\mathcal {U}}_A\bigg (\xi -\int _t^Tk(a_s)\mathrm{d}s\bigg )\bigg |{\mathcal {F}}_t\bigg ],\ {\mathbb {P}}-a.s., \text { for all }{\mathbb {P}}\in {\mathcal {P}}^a_A.\nonumber \\ \end{aligned}$$

(43)

Notice that since $\xi \in {\mathcal {C}}$, it has exponential moments of any order, so that since in addition the effort process a is bounded, we have that $u^A(\xi ,a)$ has moments of any order, in the sense that

$$\begin{aligned} \underset{a\in {{\mathcal {A}}}}{\sup }\ \underset{{\mathbb {P}}\in {{\mathcal {P}}}^a_{{\mathcal {A}}}}{\sup }{\mathbb {E}}^{\mathbb {P}}\bigg [\underset{0\le t\le T}{\sup }\big |u_t^A(\xi ,a)\big |^p\bigg ]<+\infty ,\ \text {for all } p\ge 0, \end{aligned}$$

(44)

where we have used the generalised Doob inequality for sub-linear expectations given in Proposition A.1 in [64].

Moreover, by [63, step 2 in the proof of Theorem 2.3], $e^{R_A\int _0^tk(a_s)\mathrm{d}s}u_t^A(\xi ,a)$ is a $({\mathbb {P}}, {\mathbb {G}}^{{{\mathcal {P}}}_A,+})$-sub-martingale for every ${\mathbb {P}}\in {\mathcal {P}}^a_A$, and by [63, step 3 in the proof of Theorem 2.3], there is a ${\mathbb {G}}^{{{\mathcal {P}}}_A}$-predictable process ${\widetilde{Z}}$, and a family of non-decreasing and ${\mathbb {F}}^{\mathbb {P}}$-predictable processes $({\widetilde{K}}^{\mathbb {P}})_{{\mathbb {P}}\in {\mathcal {P}}^a_A}$, such that, $\text {for all }{\mathbb {P}}\in {\mathcal {P}}^a_A$

$$\begin{aligned} e^{R_A\int _0^tk(a_s)\mathrm{d}s}u_t^A(\xi ,a)=e^{R_A\int _0^Tk(a_s)\mathrm{d}s}{\mathcal {U}}_A(\xi )-\int _t^T{\widetilde{Z}}_s{{\widehat{\alpha }}}_s^{\frac{1}{2}}\mathrm{d}W^{a}_s-{\widetilde{K}}_T^{\mathbb {P}}+{\widetilde{K}}_t^{\mathbb {P}},\ {\mathbb {P}}-a.s. \end{aligned}$$

Notice also that since every probability measure in ${\mathcal {P}}_A$ is equivalent, by definition, to a probability measure in ${{\mathcal {P}}}^a_A$ (and conversely), the above also holds ${\mathbb {P}}-a.s.$, for any ${\mathbb {P}}\in {\mathcal {P}}_A$, with the convention that we will still denote by ${\widetilde{K}}^{\mathbb {P}}$ the non-decreasing process associated with ${\mathbb {P}}\in {{\mathcal {P}}}^a_A$ or ${{\mathcal {P}}}_A$. Moreover, using the aggregation result of [40], we can actually aggregate the family ${\widetilde{K}}^{\mathbb {P}}$ into a universal process, which is ${\mathbb {G}}^{{{\mathcal {P}}}_A}$-predictable, and which we denote by ${\widetilde{K}}$.

Define

$$\begin{aligned} Y_t^a:= & {} -\frac{\ln \left( -u_t^A(\xi ,a)\right) }{R_A},\; Z^a_t:=-\frac{e^{-R_A\int _0^tk(a_s)\mathrm{d}s}}{R_Au_t(\xi ,a)}{\widetilde{Z}}_t,\; K_t^{a}\\:= & {} -\int _0^t\frac{e^{-R_A\int _0^sk(a_r)dr}}{R_Au^A_t(\xi ,a)}d{\widetilde{K}}_r. \end{aligned}$$

We have, after some computations, for all ${\mathbb {P}}\in {\mathcal {P}}_A$

$$\begin{aligned} Y_t^a= & {} \xi -\int _t^T\left( \frac{R_A}{2}\left| Z_s^a\right| ^2{{\widehat{\alpha }}}_s+k(a_s)-a_sZ^a_s\right) \mathrm{d}s-\int _t^TZ_s^a{{\widehat{\alpha }}}_s^{1/2}\mathrm{d}W_s\\&-\int _t^T\mathrm{d}K_s^{a},\ {\mathbb {P}}-a.s. \end{aligned}$$

Now notice that by (44), we immediately have

$$\begin{aligned} \underset{a\in {{\mathcal {A}}}}{\sup }\ \underset{{\mathbb {P}}\in {{\mathcal {P}}}_A}{\sup }{\mathbb {E}}^{\mathbb {P}}\bigg [\exp \bigg (p\underset{0\le t\le T}{\sup }\left| Y_t^a\right| \bigg )\bigg ]<+\infty ,\ \text {for every } p\ge 0. \end{aligned}$$

Moreover, remember that by (43), we also have for every ${\mathbb {P}}\in {\mathcal {P}}^a_A$, by the exact same arguments as above applied under any fixed measure ${\mathbb {P}}\in {{\mathcal {P}}}_A$, that

$$\begin{aligned} Y_t^{a}=\underset{{\mathbb {P}}^{'}\in {\mathcal {P}}^a_A({\mathbb {P}},t^+)}{\mathrm{essinf}^{\mathbb {P}}}\ {{\mathcal {Y}}}^{{\mathbb {P}}^{'},a}_t,\ {\mathbb {P}}-a.s., \end{aligned}$$

(45)

where for any ${\mathbb {P}}\in {{\mathcal {P}}}_A^a$, $({{\mathcal {Y}}}^{{\mathbb {P}},a}, {{\mathcal {Z}}}^{{\mathbb {P}},a})$ is the unique¹⁰ solution to the following BSDE defined under ${\mathbb {P}}$

$$\begin{aligned} {{\mathcal {Y}}}^{{\mathbb {P}},a}_t= & {} \xi -\int _t^T\left( \frac{R_A}{2}\left| {{\mathcal {Z}}}_s^{{\mathbb {P}},a}\right| ^2{{\widehat{\alpha }}}_s+k(a_s)-a_s{{\mathcal {Z}}}^{{\mathbb {P}},a}_s\right) \mathrm{d}s\\&-\int _t^T{{\mathcal {Z}}}_s^{{\mathbb {P}},a}{{\widehat{\alpha }}}_s^{1/2}\mathrm{d}W_s,\ {\mathbb {P}}-a.s. \end{aligned}$$

Then, using (44), we can follow the proof of Lemma 3.1 in [65]¹¹ to obtain that $Z^a$ actually belongs to the BMO space defined in [65] (see Section 2.3.2). Then, we can follow exactly the proof of Theorem 6.1 in [65] to obtain with (45) that for any ${\mathbb {P}}\in {{\mathcal {P}}}_A^a$

$$\begin{aligned} K_t^a=\underset{{\mathbb {P}}^{'}\in {\mathcal {P}}^a_A({\mathbb {P}},t^+)}{\mathrm{essinf}^{\mathbb {P}}}{\mathbb {E}}^{{\mathbb {P}}^{'}}\big [K_T^a\big |{\mathcal {F}}_t\big ],\ {\mathbb {P}}-a.s. \end{aligned}$$

Therefore, $(Y_t^a,Z^a_t)$ is the unique solution to the (quadratic linear) 2BSDE with terminal condition $\xi$ and generator $R_A/2z^2{{\widehat{\alpha }}}_s+k(a_s)-a_sz$ (see, for instance, Definition 2.3 of [65]).

The final step of the proof is now to relate the family $(Y^a)_{a\in {{\mathcal {A}}}}$ to the solution of the 2BSDE (18). Before proceeding, let us explain why the 2BSDE (18) does indeed admit a maximal solution. First of all, the corresponding quadratic BSDEs admit a maximal solution, because, since the infimum in the generator is over a compact set, the generator of the BSDE is bounded from above by a function with linear growth in z. The existence of a maximal solution is then direct from Proposition 4 of [66]. Furthermore, since this maximal solution is obtained as a monotone approximation of Lipschitz BSDEs, it satisfies a comparison theorem. Hence, we can apply first Proposition 2.1 of [36] to obtain the existence of a maximal solution of the 2BSDE, in the sense of Definition 4.1 of [36], and then use Remark 4.1 of [36] to aggregate the family of non-decreasing processes into K. (We remind the reader that all the measures in ${{\mathcal {P}}}_A$ satisfy the predictable martingale representation property.)

In particular, we have the following representation for any ${\mathbb {P}}\in {{\mathcal {P}}}_A$,

$$\begin{aligned} Y_t=\underset{{\mathbb {P}}^{'}\in {\mathcal {P}}^a_A({\mathbb {P}},t^+)}{\mathrm{essinf}^{\mathbb {P}}}\ {{\mathcal {Y}}}^{{\mathbb {P}}^{'}}_t,\ {\mathbb {P}}-a.s., \end{aligned}$$

(46)

where for any ${\mathbb {P}}\in {{\mathcal {P}}}_A^a$, $({{\mathcal {Y}}}^{{\mathbb {P}}}, {{\mathcal {Z}}}^{{\mathbb {P}}})$ is the maximal solution of the quadratic BSDE

$$\begin{aligned} {{\mathcal {Y}}}_t^{\mathbb {P}}= & {} \xi -\int _t^T\left( \frac{R_A}{2}\left| {{\mathcal {Z}}}^{\mathbb {P}}_s\right| ^2{{\widehat{\alpha }}}_s+\underset{a\in [0,a_{\max }]}{\inf }\left\{ k(a)-a{{\mathcal {Z}}}^{\mathbb {P}}_s\right\} \right) \mathrm{d}s\\&-\int _t^T{{\mathcal {Z}}}^{\mathbb {P}}_s{{\widehat{\alpha }}}_s^{1/2}\mathrm{d}W_s,\ {\mathbb {P}}-a.s. \end{aligned}$$

Now it is a classical result dating back to [67,68,69] (see also [70] for a similar result using 2BSDEs) that, using the comparison theorem satisfied by the maximal solution of the 2BSDEs (which is automatically inherited from the one satisfied by the BSDEs)

$$\begin{aligned} Y_0=\underset{a\in {{\mathcal {A}}}}{\sup }\ Y_0^{a}= & {} \underset{a\in {{\mathcal {A}}}}{\sup }\ \underset{{\mathbb {P}}\in {\mathcal {P}}_A^a}{\inf }\ {{\mathcal {Y}}}^{{\mathbb {P}},a}_0\\= & {} \underset{a\in {{\mathcal {A}}}}{\sup }\ \underset{{\mathbb {P}}\in {\mathcal {P}}_A^a}{\inf }\left\{ -\frac{1}{R_A}\log \left( -{\mathbb {E}}^{{\mathbb {P}}}\left[ {\mathcal {U}}_A\left( \xi -\int _0^Tk(a_s)\mathrm{d}s\right) \right] \right) \right\} , \end{aligned}$$

so that $U_0^A(\xi )=-\exp (-R_AY_0).$ Furthermore, it is then clear, since the function k is strictly convex that there is some $a^\star (Z_\cdot )\in {{\mathcal {A}}}$ such that

$$\begin{aligned} \underset{a\in [0,a_{\max }]}{\inf }\left\{ k(a)-aZ_s\right\} = k(a^\star (Z_s))-a^\star (Z_s)Z_s, \; s\in [0,T]. \end{aligned}$$

This implies that $Y_0=\underset{{\mathbb {P}}\in {\mathcal {P}}_A^{a^\star (Z_\cdot )}}{\inf }\ {{\mathcal {Y}}}^{{\mathbb {P}},a^\star (Z_\cdot )}_0.$ $\square$

Proof of Theorem 4.1

We recall Definition (42) for any $\alpha \ge 0$

$$\begin{aligned} z^\star (\alpha ):=\frac{1+k\alpha R_P}{1+\alpha k(R_A+R_P)}. \end{aligned}$$

We begin with the proof of (i). Assume that ${\underline{\alpha }}^A\le {{\overline{\alpha }}}^P\le {\overline{\alpha }}^A$, then

$$\begin{aligned} {\underline{U}}^P_0\ge & {} \underset{{\mathbb {P}}\in {{\mathcal {P}}}_P^{a^\star (z^\star ({{\widehat{\alpha }}}_\cdot ))}}{\inf }{\mathbb {E}}^{\mathbb {P}}\left[ -{\mathcal {E}}\left( -R_P\int _0^T{{\widehat{\alpha }}}_s^{\frac{1}{2}}(1\right. \right. \\&\left. \left. -z^\star ({{\widehat{\alpha }}}_s))\mathrm{d}W_s^{a^\star (z^\star )}\right) e^{R_P\left( R_0-\int _0^TH({{\widehat{\alpha }}}_s, z^\star ({{\widehat{\alpha }}}_s), 0)\mathrm{d}s\right) }\right] . \end{aligned}$$

Then we have for any $\alpha \ge 0$

$$\begin{aligned} H(\alpha ,z^\star (\alpha ),0)&=-\frac{\alpha }{2} R_P+ \frac{(1+\alpha k R_P)^2}{2k(1+\alpha k(R_A+R_P))}. \end{aligned}$$

Hence,

$$\begin{aligned} \frac{\partial H}{\partial \alpha }(\alpha ,z^\star (\alpha ),0)=\frac{-R_A\left( 1+2k\alpha R_P +k^2\alpha ^2 R_P(R_A+R_P)\right) }{2(1+\alpha k(R_A+R_P))^2}\le 0, \ \forall \alpha \in [{{\underline{\alpha }}}^P, {{\overline{\alpha }}}^P]. \end{aligned}$$

Therefore, ${\underline{U}}^P_0\ge -e^{R_P R_0}e^{-R_P\int _0^TH({{\overline{\alpha }}}^P, z^\star ({{\overline{\alpha }}}^P), 0)\mathrm{d}s}.$ Indeed, $z^\star ({{\widehat{\alpha }}}_s)$ is bounded so that the stochastic exponential is trivially a true martingale. We now turn to the converse inequality; we have

$$\begin{aligned} {\underline{U}}_0^P\le & {} \underset{(Z,{\varGamma })\in {\mathfrak {K}}}{\sup }\ {\mathbb {E}}^{{\mathbb {P}}_{a^\star (Z_\cdot )}^{{{\overline{\alpha }}}^P}}\left[ -{\mathcal {E}}\left( -R_P\int _0^T({\overline{\alpha }}^P)^{1/2}(1\right. \right. \\&\left. \left. -Z_s)\mathrm{d}W_s^{a^\star (Z_\cdot )}\right) e^{R_P\left( R_0-\int _0^TH({{\overline{\alpha }}}^P,Z_s,{\varGamma }_s)\mathrm{d}s\right) }\right] . \end{aligned}$$

According to Lemma 4.2(i), we obtain ${\underline{U}}_0^P\le -e^{R_P R_0} e^{-R_PTH({{\overline{\alpha }}}^P, z^\star ({{\overline{\alpha }}}^P), 0)}.$ Hence, if ${{\underline{\alpha }}}^A\le {{\overline{\alpha }}}^P\le {{\overline{\alpha }}}^A$, then ${\underline{U}}_0^P= -e^{R_PR_0}e^{-R_P\int _0^TH({{\overline{\alpha }}}^P, z^\star ({{\overline{\alpha }}}^P), 0)\mathrm{d}s}.$ We now prove that the contract $\xi ^{R_0,z^\star ({{{\overline{\alpha }}}^P}), 0}\in {\mathfrak {C}}^{\mathrm{SB}}$ is indeed optimal. We have

$$\begin{aligned}&\underset{{\mathbb {P}}\in {{\mathcal {P}}}_P^{a^\star (z^\star ({{\overline{\alpha }}}^P))}}{\inf }{\mathbb {E}}^{\mathbb {P}}\left[ -{\mathcal {E}}\left( -R_P\int _0^T{{\widehat{\alpha }}}_s^{\frac{1}{2}}(1\right. \right. \\&\quad \left. \left. -z^\star ({{\overline{\alpha }}}^P))\mathrm{d}W_s^{a^\star (z^\star ({{\overline{\alpha }}}^P))}\right) e^{R_P\left( R_0-\int _0^TH({{\widehat{\alpha }}}_s, z^\star ({{\overline{\alpha }}}^P), 0)\mathrm{d}s\right) }\right] ={\underline{U}}_0^P, \end{aligned}$$

since by definition (40) of H, $\alpha \longmapsto H(\alpha , z^\star ({{\overline{\alpha }}}^P),0)$ is decreasing, so that the above infimum is attained for the measure ${\mathbb {P}}^{{{\overline{\alpha }}}^P}_{a^\star (z^\star ({{\overline{\alpha }}}^P))}$.

We now turn to the proof of (ii). Assume that ${{\underline{\alpha }}}^P \le {{\overline{\alpha }}}^A\le {{\overline{\alpha }}}^P$. On the one hand,

$$\begin{aligned}&\underset{{\mathbb {P}}\in {{\mathcal {P}}}_P^{a^\star (z^\star ({{\overline{\alpha }}}^A))}}{\inf }{\mathbb {E}}^{\mathbb {P}}\left[ -{\mathcal {E}}\left( -R_P\int _0^T{{\widehat{\alpha }}}_s^{\frac{1}{2}}(1\right. \right. \\&\left. \left. -z^\star ({{\overline{\alpha }}}^A))\mathrm{d}W_s^{a^\star (z^\star ({{\overline{\alpha }}}^A))}\right) e^{R_P\left( R_0-\int _0^TH({{\widehat{\alpha }}}_s, z^\star ({{\overline{\alpha }}}^A), \gamma ^\star )\mathrm{d}s\right) }\right] \le {\underline{U}}_0^P, \end{aligned}$$

where $\gamma ^\star := -R_A (z^\star ({{\overline{\alpha }}}^A))^2 -R_P(1-z^\star ({{\overline{\alpha }}}^A))^2.$ Thus, using Relation (41), we have

$$\begin{aligned} {\underline{U}}_0^P\ge -e^{R_P R_0}e^{-R_PTH({{\overline{\alpha }}}^A,z^\star ({{\overline{\alpha }}}^A),0)}. \end{aligned}$$

On the other hand, since ${{\overline{\alpha }}}^A\in [{{\underline{\alpha }}}^P, {{\overline{\alpha }}}^P]$

$$\begin{aligned} {\underline{U}}_0^P\le & {} e^{R_P R_0}\underset{(Z,{\varGamma })\in {\mathfrak {U}}}{\sup }\ {\mathbb {E}}^{{\mathbb {P}}^{{{\overline{\alpha }}}^A}_{a^\star (Z_\cdot )}}\left[ -{\mathcal {E}}\left( -R_P\int _0^T{{\overline{\alpha }}^A}^{1/2}(1\right. \right. \\&\left. \left. -Z_s)\mathrm{d}W_s^{a^\star (Z_\cdot )}\right) e^{-R_P\int _0^TH({{\overline{\alpha }}}^A,Z_s,{\varGamma }_s)\mathrm{d}s}\right] . \end{aligned}$$

By using Lemma 4.2(i), we obtain ${\underline{U}}_0^P\le -e^{R_P R_0}e^{-R_PTH({{\overline{\alpha }}}^A,z^\star ({{\overline{\alpha }}}^A),0)}.$

We consider now a contract $\xi ^{R_0,z^\star ({{\overline{\alpha }}}^A), \gamma ^\star }$ and we show that ${\underline{U}}_0^P(\xi ^{R_0,z^\star ({{\overline{\alpha }}}^A), \gamma ^\star } )={\underline{U}}_0^P$. We have

$$\begin{aligned}&\underset{{\mathbb {P}}\in {{\mathcal {P}}}_P^{a^\star (z^\star ({{\overline{\alpha }}}^A)}}{\inf }{\mathbb {E}}^{\mathbb {P}}\left[ -{\mathcal {E}}\left( -R_P\int _0^T{{\widehat{\alpha }}}_s^{1/2}(1\right. \right. \\&\left. \left. \qquad -z^\star ({{\overline{\alpha }}}^A))\mathrm{d}W_s^{a^\star (z^\star ({{\overline{\alpha }}}^A))}\right) e^{R_P\left( R_0-\int _0^TH({{\widehat{\alpha }}}_s, z^\star ({{\overline{\alpha }}}^A), \gamma ^\star )\mathrm{d}s\right) }\right] \\&\quad =-e^{R_P R_0}e^{-R_PTH({{\overline{\alpha }}}^A,z^\star ({{\overline{\alpha }}}^A), 0)} ={\underline{U}}_0^P, \end{aligned}$$

since $H(\alpha , z^\star ({{\overline{\alpha }}}^A),\gamma ^\star )$ is actually independent of $\alpha$. The last two results are immediate consequences of Proposition 4.2. $\square$