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Optimal finite horizon contract with limited commitment

Abstract

We study a finite horizon optimal contracting problem with limited commitment. A risk-neutral principal enters into an insurance contract with a risk-averse agent who receives a stochastic income stream and cannot commit to keeping the contract. We consider a general concave utility function and a general process. We use the dual approach and the Lagrangian method to solve our optimization problem by transforming the dual problem into an infinite series of optimal stopping problems. We derive the optimal contract by representing the optimal intermediate and terminal payments from the principal to the agent in a closed-form. We show that the contract begins with a low level of payment to the agent and ratchets up the payment if the stochastic income of the agent rises above a pre-specified threshold level. In particular, if the agent’s income follows a geometric Brownian motion, the threshold level is a deterministic decreasing function of time. We also show that the final payment depends on the history of the agent’s income and the sale value of the production facility at the final time .

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Notes

  1. Ai and Li [2], and Bolton et al. [3] use dynamic programming, while Miao and Zhang [26] use the dual approach to study the problem.

  2. We can extend the model to accommodate termination of the production facility at a random time \(\tau >T\).

  3. In Appendix A, we provide a heuristic derivation of (16).

  4. We use \(D_{T-}\) in the Lagrangian, which is equivalent to assuming that D is continuous at T. We show in Appendix B that the assumption is without loss of generality.

  5. Theorem 5.5.29 in Karatzas and Shreve [20] provides appropriate necessary and sufficient conditions for the non-explosive.

  6. The CRRA utility function has been commonly employed by economists, since it allows for the examination of risk sharing between a risk-neutral principal and a risk-averse agent in a parsimonious model (e.g., Grochulski and Zhang [13]).

  7. In Appendix K, following the ideas and proofs in Yang and Koo [35], we fully characterize the analytic properties of the variational inequality in (51) and provide the integral representation of M(tz) and the corresponding free boundary z(t) .

  8. For the accuracy and efficiency of RIM, see Huang et al. [15], Jeon et al. [17], Jeon and Kwak [19], and Jeon et al. [18].

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Correspondence to Kyunghyun Park.

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Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

We thank an anonymous referee, and the Associate Editor for their comments and discussions. Junkee Jeon is supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government [Grant No. NRF-2020R1C1C1A01007313]. Hyeng Keun Koo is supported by NRF grant [Grant No. NRF-2020R1A2C1A01006134]

Appendices

Appendix

Heuristic Derivation of (16)

In this section, we provide the informal heuristic derivation of the Lagrangian \({\mathfrak {L}}(D)\) in (16) by ignoring some technical conditions.

The key is to write the part of the Lagrangian corresponding to the participation constraint (7), which is comprised of infinite individual constraints. By utilizing a method similar to He and Pagés [14] and Miao and Zhang [26], we write an integral which incorporates the infinite number of constraints implied by (7). The Lagrangian takes the following form:

$$\begin{aligned} {\mathfrak {L}}(D)\,\equiv \,&{\mathbb {E}}\left[ \int _0^T e^{-r t} (Y_t-C_t)dt+e^{-rT}({\varPi }_T-X_T) \right] \nonumber \\&+ \lambda \left( {\mathbb {E}}\left[ \int _0^T e^{-\rho t} u(C_t)dt +e^{-\rho T}{{\mathfrak {B}}}(X_T)\right] -w_0\right) \nonumber \\&+{\mathbb {E}}\left[ \int _0^T e^{(\rho -r)t} \eta _t \left( \int _t^T e^{-\rho s}(u(C_s)-u(Y_s))ds + e^{-\rho T}({{\mathfrak {B}}}(X_T)-U_d(T,Y))\right) dt \right] , \end{aligned}$$
(66)

where \(\lambda >0\) is the Lagrange multiplier associated with the initial participation constraint (8), and \(e^{(\rho -r)t}\eta _t \ge 0 \) is the Lagrange multiplier associated with the participation constraint (7) at each time \(t \in [0,T)\).

Using integration by parts, the third term of right-hand side in (66) can be written as follows:

$$\begin{aligned}&{\mathbb {E}}\left[ \int _0^T e^{(\rho -r)t} \eta _t \left( \int _t^T e^{-\rho s}(u(C_s)-u(Y_s))ds + e^{-\rho T}({{\mathfrak {B}}}(X_T)-U_d(T,Y)) \right) dt \right] \nonumber \\&\,=\,{\mathbb {E}}\left[ \int _0^T d\left( \int _0^t e^{(\rho -r)s} \eta _s ds\right) \left( \int _t^T e^{-\rho s}(u(C_s)-u(Y_s))ds \right) dt \right] \nonumber \\&+{\mathbb {E}}\left[ \left( \int _0^Te^{(\rho -r)t}\eta _t dt \right) e^{-\rho T}({{\mathfrak {B}}}(X_T)-U_d(T,Y))\right] \nonumber \\&\,=\,{\mathbb {E}}\left[ \int _0^T\left( \int _0^te^{(\rho -r)s}\eta _s ds\right) e^{-\rho t}(u(C_t)-u(Y_t))dt\right] \nonumber \\&+{\mathbb {E}}\left[ \left( \int _0^Te^{(\rho -r)t} \eta _t dt \right) e^{-\rho T}({{\mathfrak {B}}}(X_T)-U_d(T,Y))\right] . \end{aligned}$$
(67)

Plugging equation (67) into the Lagrangian in (66), we obtain

$$\begin{aligned} {\mathfrak {L}}\,=\,&{\mathbb {E}}\left[ \int _0^T e^{-r t} (Y_t-C_t)dt +e^{-rT}({\varPi }_T-X_T)\right] \nonumber \\&+{\mathbb {E}}\left[ \int _0^T\left( \int _0^te^{(\rho -r)s}\eta _s ds+\lambda \right) e^{-\rho t}(u(C_t)-u(Y_t))dt\right] \nonumber \\&+{\mathbb {E}}\left[ \left( \int _0^Te^{(\rho -r)t}\eta _t dt +\lambda \right) e^{-\rho T}({{\mathfrak {B}}}(X_T)-U_d(T,Y))\right] -\lambda \left( w_0-{\mathbb {E}}\left[ \int _0^\infty e^{-\rho t}u(Y_t)dt\right] \right) . \end{aligned}$$
(68)

We define a shadow price process \(\{D_t\}_{t=0}^T\) as the cumulative Lagrange multiplier:

$$\begin{aligned} D_t \equiv \int _0^t e^{(\rho -r) s}\eta _s ds + \lambda , \,\,\,\,D_0=\lambda . \end{aligned}$$
(69)

It follows that the Lagrangian can be rewritten as

$$\begin{aligned} {\mathfrak {L}} =&{\mathbb {E}}\left[ \int _0^T e^{-r t} (Y_t-C_t)dt+\int _T^\infty e^{-rt}\alpha Y_t dt-e^{-rT}X_T \right] \nonumber \\+&{\mathbb {E}}\left[ \int _0^T D_t e^{-\rho t}(u(C_t)-u(Y_t))dt+D_{T-} \left( {{\mathfrak {B}}}(X_T)-U_d(T,Y)\right) \right] -\lambda \left( w_0-U_d(0,Y)\right) . \end{aligned}$$
(70)

Proof of Proposition 1

By the definition of \({\tilde{u}}(\cdot )\) and \(\widetilde{{{\mathfrak {B}}}}(\cdot )\), for any \(x>0\)

$$\begin{aligned} {\tilde{u}}\left( \frac{e^{(\rho -r)t}}{x}\right) \ge u(C_t)-C_t\left( \frac{e^{(\rho -r)t}}{x}\right) \,\,\text{ and }\, \,\widetilde{{{\mathfrak {B}}}}\left( \frac{e^{(\rho -r)T}}{x}\right) \ge {{\mathfrak {B}}}(X_T)-X_T\left( \frac{e^{(\rho -r)T}}{x}\right) . \end{aligned}$$

For any enforceable consumption and severance plan \((C,X)\in \varGamma (w_0;Y)\) and shadow price process \(D\in {\mathcal {I}}(\lambda )\), we deduce that

$$\begin{aligned}&{\mathbb {E}}\left[ \int _0^T e^{-r t}(Y_t-C_t)dt+e^{-rT}({\varPi }_T-X_T) \right] \nonumber \\&\quad \le \,{\mathbb {E}}\left[ \int _{0}^{T} e^{-r t}(Y_t-C_t)dt +e^{-rT}({\varPi }_T-X_T)\right] +\lambda (U_0^a(C,X)-w_0)\nonumber \\&\qquad +{\mathbb {E}} \left[ \int _{0}^{T}e^{-\rho t}\left( U_t^a(C,X)-U_d(t,Y)\right) dD_t\right] \nonumber \\&\quad =\,{\mathbb {E}}\left[ \int _{0}^{T} e^{-r t}(Y_t-C_t)dt+e^{-rT}({\varPi }_T-X_T)\right] \nonumber \\&\qquad +{\mathbb {E}}\left[ \int _{0}^{T}\left( \int _{t}^{T}e^{-\rho s}\left( u(C_{s})-u(Y_{s})\right) ds +e^{-\rho T}({{\mathfrak {B}}}(X_T)-U_d(T,Y)\right) dD_t\right] \nonumber \\&\qquad +\lambda \left( {\mathbb {E}}\left[ \int _0^T e^{-\rho t}u(C_t)dt +e^{-\rho T}{{\mathfrak {B}}}(X_T)\right] -w_0\right) \nonumber \\&\quad =\,{\mathbb {E}}\left[ \int _{0}^{T} e^{-r t}(Y_t-C_t)dt+e^{-rT}({\varPi }_T-X_T) \right] \nonumber \\&\qquad +{\mathbb {E}}\left[ \int _0^T e^{-\rho t}D_t (u(C_t)-u(Y_t))dt +e^{-\rho T}D_{T}({{\mathfrak {B}}}(X_T)-U_d(T,Y))\right] \nonumber \\&\qquad +\lambda \left( {\mathbb {E}}\left[ \int _0^\infty e^{-\rho t}u(Y_t)dt \right] -w_0\right) . \end{aligned}$$
(71)

where the first inequality follows from the fact that C is enforceable and \(D\in {\mathcal {I}}(\lambda )\) and the second equality follows from the integration by parts.

Since \({\mathfrak {B}}(X_T)-U_d(T,Y)\ge 0\), \(D_T= (D_{T}-D_{T-})+D_{T-}\), and \(D_{T}\ge D_{T-}\), we deduce that

$$\begin{aligned}&{\mathbb {E}}\left[ \int _0^T e^{-r t}(Y_t-C_t)dt+e^{-rT}({\varPi }_T-X_T) \right] \nonumber \\&\quad \le \inf _{D\in {{{\mathcal {I}}}}(\lambda )}\left\{ {\mathbb {E}} \left[ \int _{0}^{T} e^{-r t}(Y_t-C_t)dt+e^{-rT}({\varPi }_T-X_T)\right] \right. \nonumber \\&\qquad \left. +{\mathbb {E}}\left[ \int _0^T e^{-\rho t}D_t (u(C_t)-u(Y_t))dt +e^{-\rho T}D_{T}({{\mathfrak {B}}}(X_T)-U_d(T,Y))\right] \right. \nonumber \\&\qquad +\left. \lambda \left( {\mathbb {E}}\left[ \int _0^\infty e^{-\rho t}u(Y_t)dt \right] -w_0\right) \right\} \end{aligned}$$
(72)
$$\begin{aligned}&\quad = \inf _{D\in {{{\mathcal {I}}}}(\lambda )}\left\{ {\mathbb {E}} \left[ \int _{0}^{T} e^{-r t}(Y_t-C_t)dt+e^{-rT}({\varPi }_T-X_T)\right] \right. \nonumber \\&\qquad \left. +{\mathbb {E}}\left[ \int _0^T e^{-\rho t}D_t (u(C_t)-u(Y_t))dt +e^{-\rho T}D_{T-}({{\mathfrak {B}}}(X_T)-U_d(T,Y))\right] \right. \nonumber \\&\qquad +\left. \lambda \left( {\mathbb {E}}\left[ \int _0^\infty e^{-\rho t}u(Y_t)dt \right] -w_0\right) \right\} \nonumber \\&\quad =\,\inf _{D\in {{{\mathcal {I}}}}(\lambda )}\left\{ {\mathbb {E}} \left[ \int _0^T e^{-\rho t }D_t\Big (u(C_t)-C_t\dfrac{e^{(\rho -r)t}}{D_t} -u(Y_t)\Big )dt\right. \right. \nonumber \\&\qquad \left. \left. +e^{-\rho T}D_{T-}\left( {{\mathfrak {B}}}(X_T)-X_T \left( \frac{e^{(\rho -r)T}}{D_{T-}}\right) -U_d(T,Y)\right) \right] \right. \nonumber \\&\qquad +\left. {\mathbb {E}}\left[ \int _0^Te^{-r t}Y_t\right] +{\mathbb {E}} \left[ \int _T^\infty e^{-r t}\alpha Y_t\right] +\lambda U_d(0,Y)-\lambda w_0 \right\} \nonumber \\&\quad \le \, \inf _{D\in {{{\mathcal {I}}}}(\lambda )}{{{\mathcal {J}}}}(\lambda ; Y,D)-\lambda w_0. \end{aligned}$$
(73)

Notice that (72) is valid if we assume D is continuous at T. Thus, replacing \(D_T\) by \(D_{T-}\) is without loss of generality.

Clearly, the all equalities in (72) hold if and only if the following equality holds: for every \(\lambda > 0\) and every \(t\in [0,T]\),

$$\begin{aligned}&U_0^a (C,X)=w_0,\,\, D_t e^{-(\rho -r) t }u'(C_t)-1=0,\,\,D_{T-} e^{-(\rho -r) T }{{\mathfrak {B}}}'(X_T)-1=0\\&\,\,\text{ and }\,\,\int _{0}^{t}e^{-\rho (s-t)} (U_s^a(C,X)-U_d(I_s))dD_{s}=0. \end{aligned}$$

Since the inequality (71) holds for all enforceable consumption plan, we deduce that

$$\begin{aligned} V(w_0)\le \inf _{\lambda >0} \left( J(\lambda )-\lambda w_0 \right) . \end{aligned}$$
(74)

Proof of Proposition 2

Let us consider the left-continuous inverse \(\tau (\cdot )\) of the non-decreasing and RCLL process \(D_t\). Then, \(\tau (\cdot )\) is a \({\mathcal {F}}\)-stopping time and given by \(\tau (\nu ) = \inf \{t\ge 0 \mid D_t \ge \nu \}\). Thus, the following relation holds:

$$\begin{aligned} \{ \nu \le D_t \} = \{ \tau (\nu )\le t\}\,\,\text{ and }\,\,\{ \nu \le D_{T-} \} = \{ \tau (\nu )< T\}. \end{aligned}$$
(75)

Using the first equivanlent relations in (75), we obtain the following transformation:

$$\begin{aligned}&D_t \left( {\tilde{u}}\left( \frac{e^{(\rho -r)t}}{D_t}\right) -u(Y_t)\right) \\ =&\int _{\lambda }^{D_t} \left\{ u\left( I \left( \frac{e^{(\rho -r)t}}{\nu } \right) \right) -u(Y_t)\right\} d\nu +\lambda \left( {\tilde{u}} \left( \frac{e^{(\rho -r)t}}{\lambda }\right) -u(Y_t)\right) \\ =&\int _{\lambda }^{\infty } \left\{ u\left( I \left( \frac{e^{(\rho -r)t}}{\nu } \right) \right) -u(Y_t)\right\} \mathbf{1}_{\{ \nu \le D_t \}}dx+\lambda \left( {\tilde{u}}\left( \frac{e^{(\rho -r)t}}{\lambda }\right) -u(Y_t)\right) \\ =&\int _{\lambda }^{\infty } \left\{ u\left( I \left( \frac{e^{(\rho -r)t}}{\nu }\right) \right) -u(Y_t)\right\} \mathbf{1}_{\{ \tau (\nu ) \le t \}}d\nu +\lambda \left( {\tilde{u}}\left( \frac{e^{(\rho -r)t}}{\lambda }\right) -u(Y_t)\right) . \end{aligned}$$

Similarly, the second relations in (75) implies

$$\begin{aligned} D_{T-}\left( \widetilde{{{\mathfrak {B}}}}\left( \dfrac{e^{(\rho -r)T}}{D_{T-}}\right) -U_d(T,Y)\right) =&\int _{\lambda }^\infty \left\{ {{\mathfrak {B}}}\left( I_{{\mathfrak {B}}} \left( \frac{e^{(\rho -r)T}}{\nu }\right) \right) -U_d(T,Y)\right\} \mathbf{1}_{\{ \tau (\nu ) < T \}}d\nu \\ +&\lambda \left( \widetilde{{\mathfrak {B}}}\left( \frac{e^{(\rho -r)T}}{\lambda }\right) -U_d(T,Y)\right) . \end{aligned}$$

From the above transformations, we obtain

$$\begin{aligned}&{\mathbb {E}}\left[ \int _0^T e^{-\rho t} D_t \Big ( {\tilde{u}} \left( \frac{e^{(\rho -r)t}}{D_t}\right) - u(I_t)\Big )dt +e^{-\rho T}D_{T-} \left( \widetilde{{{\mathfrak {B}}}}\left( \dfrac{e^{(\rho -r)T}}{D_{T-}} \right) -U_d(I_T)\right) \right] \\ =&\int _{\lambda }^{\infty }{\mathbb {E}}\left[ \int _{\tau (\nu )}^T {e^{-\rho t}} \left\{ u\left( I \left( \frac{e^{(\rho -r)t}}{\nu }\right) \right) -u(Y_t) \right\} dt+{e^{-\rho T}}\left\{ {{\mathfrak {B}}}\left( I_{{\mathfrak {B}}} \left( \frac{e^{(\rho -r)T}}{\nu }\right) \right) -U_d(T,Y)\right\} \right. \\&\quad \left. \mathbf{1}_{\{ \tau (\nu ) < T \}}\right] d\nu \\ +&{\mathbb {E}}\left[ \int _0^Te^{-\rho t}\lambda \left( {\tilde{u}}\left( \frac{e^{(\rho -r)t}}{\lambda }\right) -u(Y_t)\right) dt+e^{-\rho T}\lambda \left( \widetilde{{\mathfrak {B}}}\left( \frac{e^{(\rho -r)T}}{\lambda }\right) -U_d(T, Y)\right) \right] . \end{aligned}$$

Since

$$\begin{aligned}&{\mathbb {E}}\left[ \int _{\tau (\nu )}^T {e^{-\rho t}}\left\{ u\left( I \left( \frac{e^{(\rho -r)t}}{\nu }\right) \right) -u(Y_t)\right\} dt+{e^{-\rho T}}\left\{ {{\mathfrak {B}}}\left( I_{{\mathfrak {B}}} \left( \frac{e^{(\rho -r)T}}{\nu }\right) \right) -U_d(T,Y)\right\} \mathbf{1}_{\{ \tau (\nu )< T \}}\right] \\&\quad \ge \inf _{\tau \in {\mathcal {S}}}{\mathbb {E}}\left[ \int _{\tau }^T {e^{-\rho t}}\left\{ u\left( I \left( \frac{e^{(\rho -r)t}}{\nu }\right) \right) -u(Y_t)\right\} dt+{e^{-\rho T}}\left\{ {{\mathfrak {B}}}\left( I_{{\mathfrak {B}}} \left( \frac{e^{(\rho -r)T}}{\nu }\right) \right) -U_d(T,Y)\right\} \mathbf{1}_{\{ \tau < T \}}\right] , \end{aligned}$$

we deduce that

$$\begin{aligned} \int _{\lambda }^\infty {\mathcal {G}}_0(\nu )d\nu + J^{FB}_0(\lambda )\le&{\mathbb {E}}\left[ \int _0^T e^{-\rho t} D_t \Big ( {\tilde{u}}\left( \frac{e^{(\rho -r)t}}{D_t}\right) - u(Y_t)\Big )dt\right. \nonumber \\&\left. +e^{-\rho T}D_{T-}\left( \widetilde{{{\mathfrak {B}}}}\left( \dfrac{e^{(\rho -r)T}}{D_{T-}}\right) -U_d(T,Y)\right) \right] \nonumber \\ +&{\mathbb {E}}\left[ \int _0^T e^{-rt}Y_tdt\right] +{\mathbb {E}}\left[ \int _T^\infty e^{-rt}\alpha Y_tdt\right] +\lambda U_d(0,Y) \end{aligned}$$
(76)

where

$$\begin{aligned} {\mathcal {G}}_0(\nu )= & {} \inf _{\tau \in {\mathcal {S}}} {\mathbb {E}}\left[ \int _{\tau }^T {e^{-\rho t}}\left\{ u\left( I \left( \frac{e^{(\rho -r)t}}{\nu }\right) \right) -u(Y_t)\right\} dt+{e^{-\rho T}}\right. \nonumber \\&\left. \left\{ {{\mathfrak {B}}}\left( I_{{\mathfrak {B}}} \left( \frac{e^{(\rho -r)T}}{\nu }\right) \right) -U_d(T,Y)\right\} \mathbf{1}_{\{ \tau < T \}}\right] \end{aligned}$$
(77)

By minimizing the right-side of (76) over \(D\in {\mathcal {I}}(\lambda )\), we obtain

$$\begin{aligned} \begin{aligned} J(\lambda ) \ge \int _{\lambda }^\infty {\mathcal {G}}_0(\nu )d\nu + J_0^{FB}(\lambda ). \end{aligned} \end{aligned}$$
(78)

For given \(\nu \ge \lambda \), let \(\zeta (\nu )\) be the solution of the optimal stopping problem (77), i.e.,

$$\begin{aligned} \begin{aligned} {\mathcal {G}}_0(\nu )&={\mathbb {E}}\left[ \int _{\zeta (\nu )}^T {e^{-\rho t}}\left\{ u\left( I \left( \frac{e^{(\rho -r)t}}{\nu }\right) \right) -u(Y_t)\right\} dt\right. \\&\quad \left. +{e^{-\rho T}}\left\{ {{\mathfrak {B}}}\left( I_{{\mathfrak {B}}} \left( \frac{e^{(\rho -r)T}}{\nu }\right) \right) -U_d(T,Y)\right\} \mathbf{1}_{\{ \zeta (\nu ) < T \}}\right] . \end{aligned} \end{aligned}$$

Note that \(\zeta (\nu )\) is non-decreasing and left-continuous (see Lemma 1). Let \(\varUpsilon \) be the right continuous inverse of \(\zeta (\cdot )\), i.e.,

$$\begin{aligned} \varUpsilon (t) = \inf \{\nu \ge 0 \mid \zeta (\nu ) > t \} \end{aligned}$$

which satisfies \(\{\zeta (\nu )< t\}=\{\nu <\varUpsilon (t)\}\).

From this, we can derive a non-decreasing and RCLL process \({D}^*(\lambda )\) which is the right continuous inverse of the map \(\nu \in [\lambda ,\infty )\,\rightarrow \,\zeta (\nu )\), given by

$$\begin{aligned} D_t^*(\lambda ) = \max \left\{ \lambda , \varUpsilon (t) \right\} \quad \text{ for }\quad t\in (0,T]. \end{aligned}$$
(79)

By the definition of the dual value function in (17), we can deduce that

$$\begin{aligned} J(\lambda ) \le&{\mathbb {E}}\left[ \int _0^T e^{-\rho t} D_t^*(\lambda ) \Big ( {\tilde{u}}\left( \frac{e^{(\rho -r)t}}{D_t^*(\lambda )}\right) - u(Y_t)\Big )dt +e^{-\rho T}D_{T-}^*(\lambda )\left( \widetilde{{{\mathfrak {B}}}} \left( \dfrac{e^{(\rho -r)T}}{D_{T-}^*(\lambda )}\right) -U_d(T,Y)\right) \right] \nonumber \\ +&{\mathbb {E}}\left[ \int _0^T e^{-rt}Y_tdt\right] +{\mathbb {E}}\left[ \int _T^\infty e^{-rt}\alpha Y_tdt\right] +\lambda {\mathbb {E}}\left[ \int _0^\infty e^{-\rho t}u(Y_t)dt \right] \nonumber \\ =&\int _{\lambda }^\infty {\mathcal {G}}_0(\nu )d\nu + J^{FB}(\lambda ). \end{aligned}$$
(80)

From (78) and (80), \((D_t^*(\lambda ))_{t=0}^T\in {\mathcal {I}}(\lambda )\) is the optimal solution to Problem 3 and

$$\begin{aligned} J(\lambda ) = \int _{\lambda }^\infty {\mathcal {G}}_0(\nu )d\nu + J_0^{FB}(\lambda ). \end{aligned}$$
(81)

Proof of Lemma 1

Proof of (a)

Since \(u(I(e^{(\rho -r)s}/\nu ))\) is increasing in \(\nu \), \(Q(t,\nu e^{-(\rho -r)t})\) is also increasing in \(\nu \). It follows that the function \(\nu \,\rightarrow \,\zeta (\nu )\) is increasing. To show the left-continuity of \(\zeta \), let us define a sequence \((\nu _n)\) such that \(\nu _n <\nu \) and \(\nu _n\,\rightarrow \,\nu \). Then,

$$\begin{aligned} \zeta ^* \triangleq \lim _{n\rightarrow \infty } \zeta (\nu _n) \le \zeta (\nu ). \end{aligned}$$

Moreover, for any \(\tau \in {{{\mathcal {S}}}}\),

$$\begin{aligned}&{\mathbb {E}}\left[ \int _0^{\tau } e^{-\rho t}\left( u\left( I\left( \frac{e^{(\rho -r)t}}{\nu _n}\right) \right) -u(Y_t)\right) dt+e^{-\rho T}\left( {{\mathfrak {B}}}\left( I_{{\mathfrak {B}}}\left( \frac{e^{(\rho -r)T}}{\nu _n}\right) \right) -U_d(T,Y)\right) \mathbf{1}_{\{\tau = T\}}\right] \\&\quad \le {\mathbb {E}}\left[ \int _0^{\zeta (\nu _n)} e^{-\rho t} \left( u\left( I\left( \frac{e^{(\rho -r)t}}{\nu _n}\right) \right) -u(Y_t) \right) dt+e^{-\rho T}\left( {{\mathfrak {B}}}\left( I_{{\mathfrak {B}}} \left( \frac{e^{(\rho -r)T}}{\nu _n}\right) \right) -U_d(T,Y)\right) \mathbf{1}_{\{\zeta (\nu _n) = T\}}\right] \\&\quad \le {\mathbb {E}}\left[ \int _0^{\zeta (\nu _n)} e^{-\rho t}\left( u\left( I\left( \frac{e^{(\rho -r)t}}{\nu }\right) \right) -u(Y_t) \right) dt+e^{-\rho T}\left( {{\mathfrak {B}}}\left( I_{{\mathfrak {B}}}\left( \frac{e^{(\rho -r)T}}{\nu }\right) \right) -U_d(T,Y)\right) \mathbf{1}_{\{\zeta (\nu _n) = T\}}\right] . \end{aligned}$$

Letting \(n\rightarrow \infty \), the Monotone Convergence Theorem implies that

$$\begin{aligned}&{\mathbb {E}}\left[ \int _0^{\tau } e^{-\rho t}\left( u\left( I \left( \frac{e^{(\rho -r)t}}{\nu _n}\right) \right) -u(Y_t)\right) dt +e^{-\rho T}\left( {{\mathfrak {B}}}\left( I_{{\mathfrak {B}}} \left( \frac{e^{(\rho -r)T}}{\nu _n}\right) \right) -U_d(T,Y)\right) \mathbf{1}_{\{\tau = T\}}\right] \\&\quad \le {\mathbb {E}}\left[ \int _0^{\zeta ^*} e^{-\rho t}\left( u\left( I\left( \frac{e^{(\rho -r)t}}{\nu }\right) \right) -u(Y_t) \right) dt+e^{-\rho T}\left( {{\mathfrak {B}}}\left( I_{{\mathfrak {B}}}\left( \frac{e^{(\rho -r)T}}{\nu }\right) \right) -U_d(T,Y)\right) \mathbf{1}_{\{\zeta ^* = T\}}\right] . \end{aligned}$$

It follows from the definition of \(\zeta (\nu )\) that

$$\begin{aligned} \zeta ^*=\zeta (\nu )=\lim _{n\rightarrow \infty }\zeta (\nu _n). \end{aligned}$$

That is, the map \(\nu \,\rightarrow \,\zeta (\nu )\) is left-continuous function of \(\nu \). \(\square \)

Proof of (b)

In Remark 4, it follows from the similar arguments in Lemma 2 that \(Q_t(\nu )\) is non-negative, non-decreasing and continuous with respect to \(\nu \). This implies that there exists an adapted process \(\chi (t)\) defined by

$$\begin{aligned} \chi (t)=\sup \{\nu >0\mid Q_t(\nu )=0\}\quad \text{ for }\quad t\in [0,T) , \end{aligned}$$

such that

$$\begin{aligned} \{\nu \mid Q_t(\nu )=0\}=\{\nu \mid 0\le \nu \le \chi (t) \}. \end{aligned}$$

It follows that

$$\begin{aligned} \inf \big \{ t\in [0,T) \mid Q_t(\nu e^{-(\rho -r)t})=0 \big \} = \inf \big \{t\in [0,T) \mid \nu e^{-(\rho -r)t} \le \chi (t) \big \}. \end{aligned}$$

Moreover, when \(\nu \) goes to infinity, the term \(I(e^{(\rho -r)s}/\nu )\) also goes to infinity. In other words, for sufficiently large \(\nu >0\), it follows from Assumption 2 that \(Q_t(\nu e^{-(\rho -r)t})\) is strictly greater than 0. Thus, \(\chi (t)<+\infty \) for any \(t<T\). \(\square \)

Proof of (c)

Let us temporarily denote \({{{\mathcal {R}}}}_t\) by

$$\begin{aligned} {{{\mathcal {R}}}}_t = \sup _{0\le s \le t}{e^{(\rho -r)s}}\chi (s). \end{aligned}$$

Since \(\nu \ge {{{\mathcal {R}}}}_t\,\text{ on }\,\{\zeta (\nu )>t\}\) for any \(\nu >0\), it follows from the definition of \(\varUpsilon _t\) in Appendix C that \(\varUpsilon _t \ge {{{\mathcal {R}}}}_t\). If \(\nu >{{{\mathcal {R}}}}_t\), it follows from \(\nu >\sup _{0\le s\le t}e^{(\rho -r)s}\chi (s)\) that \(\zeta (\nu )>t\) and \(\nu >\varUpsilon _t\). From this, we deduce that \({{{\mathcal {R}}}}_t\ge \varUpsilon _t\). This completes the proof. \(\square \)

Proof of Proposition 3

It follows from Lemma 1 and (79) that for \(t\in [0,T)\)

$$\begin{aligned} D_t^*(\lambda ) = \max \left\{ \lambda , \sup _{0\le s \le t}e^{(\rho -r)s} \chi (s) \right\} . \end{aligned}$$

And, (78) and (80) imply that \(D_t^*(\lambda )\) attains the minimization of \(J(\lambda )\).

Proof of Proposition 4

Given the optimal process \(D^*(\lambda )\in {{{\mathcal {I}}}}(\lambda )\) of the minimization problem in (17), define

$$\begin{aligned} D^{\pm \varkappa } \equiv D^*(\lambda )\pm \varkappa \,\,\,\text{ and }\,\,\,D^{1\pm \varkappa }\equiv D^*(\lambda )(1\pm \varkappa ). \end{aligned}$$

For sufficiently small \(\delta >0\), it follows from Remark 1 that

$$\begin{aligned}&{\mathbb {E}}\left[ \int _0^T e^{-\rho t}D^{\pm \delta }_t\left| {\tilde{u}} \left( \dfrac{e^{(\rho -r)t}}{D^{\pm \delta }_t}\right) \right| dt\right]<\infty ,\,\,{\mathbb {E}}\left[ \int _0^T e^{-\rho t}D^{\pm \delta }_t|u(Y_t)|dt \right]<\infty \\&{\mathbb {E}}\left[ e^{-\rho T}D^{\pm \delta }_T\left| \widetilde{{\mathfrak {B}}}\left( \dfrac{e^{(\rho -r)T}}{D^{\pm \delta }_T}\right) \right| \right]<\infty ,\,\,\,{\mathbb {E}}[D^{\pm \delta }_T |U_d(T,Y)|]<\infty , \end{aligned}$$

and

$$\begin{aligned}&{\mathbb {E}}\left[ \int _0^T e^{-\rho t}D^{1\pm \delta }_t\left| {\tilde{u}} \left( \dfrac{e^{(\rho -r)t}}{D^{1\pm \delta }_t}\right) \right| dt\right]<\infty ,\,\,{\mathbb {E}}\left[ \int _0^T e^{-\rho t}D^{1\pm \delta }_t|u(Y_t)|dt \right]<\infty \\&{\mathbb {E}}\left[ e^{-\rho T}D^{1\pm \delta }_T\left| \widetilde{{\mathfrak {B}}}\left( \dfrac{e^{(\rho -r)T}}{D^{1\pm \delta }_T}\right) \right| \right]<\infty ,\,\,\,{\mathbb {E}}[D^{1\pm \delta }_T |U_d(T,Y)|]<\infty . \end{aligned}$$

Since \(x{\tilde{u}}(1/x)\) and \(x\widetilde{{\mathfrak {B}}}(1/x)\) are convex for \(x>0\), we deduce that for any \(\epsilon \in (0, \delta )\)

$$\begin{aligned}&\dfrac{D^{-\delta }_t{\tilde{u}}\left( \dfrac{e^{(\rho -r)t}}{D^{-\delta }_t}\right) -D^{*}_t{\tilde{u}}\left( \dfrac{e^{(\rho -r)t}}{D^{*}_t}\right) }{-\delta }\le \dfrac{D^{\pm \epsilon }_t{\tilde{u}}\left( \dfrac{e^{(\rho -r)t}}{D^{\pm \epsilon }_t}\right) -D^{*}_t{\tilde{u}}\left( \dfrac{e^{(\rho -r)t}}{D^{*}_t}\right) }{\pm \epsilon }\\&\quad \le \dfrac{D^{+\delta }_t{\tilde{u}}\left( \dfrac{e^{(\rho -r)t}}{D^{+\delta }_t}\right) -D^{*}_t{\tilde{u}}\left( \dfrac{e^{(\rho -r)t}}{D^{*}_t}\right) }{\delta } \end{aligned}$$

and

$$\begin{aligned}&\dfrac{D^{-\delta }_T\widetilde{{\mathfrak {B}}}\left( \dfrac{e^{(\rho -r)T}}{D^{-\delta }_T}\right) -D^{*}_T\widetilde{{\mathfrak {B}}} \left( \dfrac{e^{(\rho -r)T}}{D^{*}_T}\right) }{-\delta }\le \dfrac{D^{\pm \epsilon }_{T}\widetilde{{\mathfrak {B}}}\left( \dfrac{e^{(\rho -r)T}}{D^{\pm \epsilon }_T}\right) -D^{*}_T\widetilde{{\mathfrak {B}}} \left( \dfrac{e^{(\rho -r)T}}{D^{*}_T}\right) }{\pm \epsilon }\\&\quad \le \dfrac{D^{+\delta }_T\widetilde{{\mathfrak {B}}}\left( \dfrac{e^{(\rho -r)T}}{D^{+\delta }_T}\right) -D^{*}_T\widetilde{{\mathfrak {B}}} \left( \dfrac{e^{(\rho -r)T}}{D^{*}_T}\right) }{\delta }. \end{aligned}$$

Moreover, it follows from \(D^*(\lambda +\epsilon )\le D^\epsilon \) and \(D^*(\lambda -\epsilon )\ge D^{-\epsilon }\) that

$$\begin{aligned} \dfrac{J(\lambda +\epsilon )-J(\lambda )}{\epsilon }= & {} \dfrac{{{{\mathcal {J}}}}(\lambda +\epsilon ;D^*(\lambda +\epsilon ),Y) -{{{\mathcal {J}}}}(\lambda ;D^*(\lambda ),Y)}{\epsilon }\\\le & {} \dfrac{{{{\mathcal {J}}}} (\lambda +\epsilon ;D^\epsilon ,Y)-{{{\mathcal {J}}}}(\lambda ;D^*(\lambda ),Y)}{\epsilon }, \end{aligned}$$

and

$$\begin{aligned} \dfrac{J(\lambda -\epsilon )-J(\lambda )}{-\epsilon }= & {} \dfrac{{{{\mathcal {J}}}}(\lambda -\epsilon ;D^*(\lambda -\epsilon ),Y) -{{{\mathcal {J}}}}(\lambda ;D^*(\lambda ),Y)}{-\epsilon }\\\ge & {} \dfrac{{{{\mathcal {J}}}} (\lambda -\epsilon ;D^{-\epsilon },Y)-{{{\mathcal {J}}}}(\lambda ;D^*(\lambda ),Y)}{-\epsilon }. \end{aligned}$$

Note that \(J(\lambda )\) is differentiable with respect to \(\lambda \) and

$$\begin{aligned} J'(\lambda ) = -{{{\mathcal {G}}}}_0(\lambda ) +(J_0^{FB}(\lambda ))'= Q_0(\lambda ) + U_d(0,Y). \end{aligned}$$

where we use that \({{{\mathcal {G}}}}_0(\lambda )=-Q_0(\lambda )+{{{\mathcal {G}}}}_0^{FB}(\lambda )\) and \((J_0^{FB}(\lambda ))'={{{\mathcal {G}}}}_0^{FB}(\lambda )+U_d(0,Y)\).

By the Dominated Convergence Theorem, we have

$$\begin{aligned} \lim _{\epsilon \downarrow 0} \dfrac{{{{\mathcal {J}}}}(\lambda -\epsilon ;D^{-\epsilon },Y)-{{{\mathcal {J}}}} (\lambda ;D^*(\lambda ),Y)}{-\epsilon }\le J'(\lambda ) \le \lim _{\epsilon \downarrow 0}\dfrac{{{{\mathcal {J}}}}(\lambda +\epsilon ;D^\epsilon ,Y)-{{{\mathcal {J}}}} (\lambda ;D^*(\lambda ),Y)}{\epsilon }, \end{aligned}$$

Furthermore, since

$$\begin{aligned}&\lim _{\epsilon \downarrow 0}\dfrac{{{{\mathcal {J}}}}(\lambda +\epsilon ;D^\epsilon ,Y) -{{{\mathcal {J}}}}(\lambda ;D^*(\lambda ),Y)}{\epsilon }=\lim _{\epsilon \downarrow 0} \dfrac{{{{\mathcal {J}}}}(\lambda -\epsilon ;D^{-\epsilon },Y)-{{{\mathcal {J}}}}(\lambda ;D^*(\lambda ),Y)}{-\epsilon }\\&\quad ={\mathbb {E}}\left[ \int _{0}^T e^{-\rho t}\left( u\left( I\left( \frac{e^{(\rho -r)t}}{D_t^*(\lambda )}\right) \right) -u(Y_t)\right) dt+e^{-\rho T}\left( {{\mathfrak {B}}}\left( I_{{\mathfrak {B}}}\left( \frac{e^{(\rho -r)T}}{D_{T-}^{*}(\lambda )}\right) \right) -U_d(T,Y)\right) \right] \\&\qquad +U_d(0,Y), \end{aligned}$$

we obtain

$$\begin{aligned} Q_0(\lambda )&={\mathbb {E}}\left[ \int _{0}^T e^{-\rho t}\left( u\left( I\left( \frac{e^{(\rho -r)t}}{D_t^*(\lambda )}\right) \right) -u(Y_t)\right) dt+e^{-\rho T}\right. \\&\left. \left( {{\mathfrak {B}}}\left( I_{{\mathfrak {B}}} \left( \frac{e^{(\rho -r)T}}{D_{T-}^{*}(\lambda )}\right) \right) -U_d(T,Y)\right) \right] . \end{aligned}$$

Proof of Lemma 2

Proof of (a)

Recall that

$$\begin{aligned} Q_0(\lambda )=&{\mathbb {E}}\left[ \int _{0}^T e^{-\rho t}\left( u \left( I\left( \frac{e^{(\rho -r)t}}{D_t^*(\lambda )}\right) \right) -u(Y_t)\right) dt +e^{-\rho T}\left( {{\mathfrak {B}}}\left( I_{{\mathfrak {B}}} \left( \frac{e^{(\rho -r)T}}{D_{T-}^{*}(\lambda )}\right) \right) -U_d(T,Y)\right) \right] \\ =&{\mathbb {E}}\left[ \int _{0}^T e^{-\rho t}u\left( I\left( \frac{e^{(\rho -r)t}}{D_t^*(\lambda )}\right) \right) dt+e^{-\rho T}{{\mathfrak {B}}}\left( I_{{\mathfrak {B}}}\left( \frac{e^{(\rho -r)T}}{D_{T-}^{*}(\lambda )}\right) \right) \right] -U_d(0,Y). \end{aligned}$$

Since \(D_t^*(\lambda )=\max \left\{ \lambda , \varUpsilon _t \right\} \), \(D_t^*(\lambda )\) is continuous and increasing with respect to \(\lambda >0\). If \(\lambda >\chi (0)\), then the right continuity of \(\varUpsilon _t\) with respect to t implies that there exists a sufficiently small \(t_\epsilon >0\) such that \(\lambda >\varUpsilon _t\) in \(t\in (0, t_\epsilon )\).

It follows that \(Q_0(\lambda )\) is continuous and increasing in \(\lambda >0\). In particular, \(Q_0(\lambda )\) is strictly increasing in \(\lambda \in (\chi (0),\infty )\). \(\square \)

Proof of (b)

Since \(Q_0(\lambda )\) is strictly increasing in \(\lambda \in (\chi (0),\infty )\), it follows from the definition of \(\chi (t)\), i.e.,

$$\begin{aligned} \chi (t)= \sup \{\nu \mid Q_t(\nu )=0\} \quad \text{ for }\quad t\in [0,T) \end{aligned}$$

that \(\lim _{\lambda \rightarrow \chi (0)}Q_0(\lambda )=0.\)

Since

$$\begin{aligned} \lim _{\lambda \rightarrow \infty }I\left( \frac{e^{(\rho -r)t}}{D_t^*(\lambda )}\right) =\lim _{\lambda \rightarrow \infty }I_{{\mathfrak {B}}}\left( \frac{e^{(\rho -r)T}}{D_{T-}^{*}(\lambda )} \right) =\infty , \end{aligned}$$

we can easily obtain that

$$\begin{aligned} \lim _{\lambda \rightarrow \infty }\left( Q_0(\lambda )+U_d(0,Y)\right) = \dfrac{1-e^{-\rho T}}{\rho }\lim _{c\rightarrow \infty }u(c)+e^{-\rho T} \lim _{x\rightarrow \infty }{{\mathfrak {B}}}(x). \end{aligned}$$

\(\square \)

Proof of Theorem 1

It follows from Lemma 2 that for given \(w_0\) satisfying the condition (9), there exists a unique \(\lambda ^*\) such that

$$\begin{aligned}w_0 = J'(\lambda ^*) = Q_0(\lambda ^*)+U_d(0,Y).\end{aligned}$$

For \(\lambda ^*>0\), let us consider the contract strategy \(((C_t^{D^*(\lambda ^*)})_{t=0}^T, X_T^{D^*(\lambda ^*)})\) defined in Corollary 1 by

$$\begin{aligned} C_t^{D^*(\lambda ^*)}=I\left( \dfrac{e^{(\rho -r)t}}{D_t^*(\lambda ^*)}\right) \,\,\text{ and }\,\, X_T^{D^*(\lambda ^*)}=I_{{\mathfrak {B}}}\left( \dfrac{e^{(\rho -r)T}}{D_{T-}^{*} (\lambda ^*)}\right) . \end{aligned}$$
(82)

By Proposition 4, we deduce that

$$\begin{aligned}&w_0 = Q_0(\lambda ^*)+U_d(0,Y)\nonumber \\&\quad ={\mathbb {E}}\left[ \int _{0}^T e^{-\rho t}u\left( I\left( \frac{e^{(\rho -r)t}}{D_t^*(\lambda ^*)}\right) \right) dt+e^{-\rho T}{{\mathfrak {B}}}\left( I_{{\mathfrak {B}}}\left( \frac{e^{(\rho -r)T}}{D_{T-}^{*}(\lambda ^*)}\right) \right) \right] . \end{aligned}$$
(83)

From Remark 4, we deduce that

$$\begin{aligned} Q_t(e^{-(\rho -r)t}D_t^*(\lambda ^*))=&{\mathbb {E}}\left[ \int _{t}^T e^{-\rho (s-t)}\left( u(C_s^{D^*(\lambda ^*)})-u(Y_s)\right) ds+ e^{-\rho (T-t)}\right. \nonumber \\&\quad \left. \left( {{\mathfrak {B}}}(X_T^{D^*(\lambda ^*)})-U_d(T,Y)\right) \right] \nonumber \\ =&U_t^a(C^{D^*(\lambda ^*)},X^{D^*(\lambda ^*)})-U_d(t,Y). \end{aligned}$$
(84)

Since \(Q_t(e^{-(\rho -r)t}D_t^*(\lambda ^*))\ge 0\) for any \(t\in [0,T)\), we have

$$\begin{aligned} U_t^a(C^{D^*(\lambda ^*)},X^{D^*(\lambda ^*)})\ge U_d(t,Y). \end{aligned}$$
(85)

Clearly,

$$\begin{aligned} {\mathbb {E}}\left[ \int _0^T e^{-r t } C_t^{D^*(\lambda ^*)}dt\right]< \infty \,\,\,\text{ and }\,\,\,{\mathbb {E}}\left[ X_T^{D^*(\lambda ^*)}\right] < \infty . \end{aligned}$$
(86)

From (83), (85) and (86), \(((C_t^{D^*(\lambda ^*)})_{t=0}^T, X_T^{D^*(\lambda ^*)})\) is enforceable, i.e.

$$\begin{aligned} ((C_t^{D^*(\lambda ^*)})_{t=0}^T, X_T^{D^*(\lambda ^*)})\in \varGamma (w_0;Y). \end{aligned}$$

It follows from Corollary 1 that

$$\begin{aligned} {\mathbb {E}}\left[ \int _{0}^T\left( U_t^a(C^{D^*(\lambda ^*)},X^{D^*(\lambda ^*)}) -U_d(t,Y)\right) dD_t^*(\lambda ^*)\right] =0. \end{aligned}$$
(87)

From Proposition 1, we conclude that the following duality holds:

$$\begin{aligned} V(w_0) = \inf _{\lambda>0} \big (J(\lambda )-\lambda w_0\big )=\min _{\lambda >0}\big (J(\lambda )-\lambda w_0\big )=J(\lambda ^*)-\lambda ^*w_0. \end{aligned}$$
(88)

Moreover, \(((C_t^{D^*(\lambda ^*)})_{t=0}^T, X_T^{D^*(\lambda ^*)})\) is optimal to Problem 2.

Proof of Proposition 5

Proof of (a)

Let us denote the process \((D_s^{t,*})_{s=t}^T\) by

$$\begin{aligned} D_s^{t,*} = e^{-(\rho -r)t}D_s^*(\lambda ^*)\,\,\text{ with }\,\,D_t^{t,*}=\lambda _t^*. \end{aligned}$$

It follows from Remark 4 that

$$\begin{aligned} {{{\mathcal {W}}}}_t={\mathbb {E}}_t\left[ \int _t^T e^{-\rho (s-t)}u(C_t^*)ds+e^{-\rho (T-t)}{{\mathfrak {B}}}(X_T^*) \right] =Q_t(\lambda _t^*)+U_d(t,Y) \end{aligned}$$
(89)

and

$$\begin{aligned} {\mathbb {E}}\left[ \int _{t}^T\left( U_s^a({C^*},{X^*})-U_d(s,Y) \right) dD_s^{t,*}\right] =0. \end{aligned}$$
(90)

Thus, we deduce that

$$\begin{aligned} {{{\mathcal {V}}}}_t=&{\mathbb {E}}_t\left[ \int _t^Te^{-r(s-t)}(Y_s-C_s^*)ds +e^{-r(T-t)}({\varPi }_T-X_T^*)\right] \nonumber \\ =&{\mathbb {E}}_t\left[ \int _t^Te^{-r(s-t)}(Y_s-C_s^*)ds +e^{-r(T-t)} ({\varPi }_T-X_T^*)\right] \nonumber \\&+\lambda _t^*\left( {\mathbb {E}}_t \left[ \int _t^T e^{-\rho (s-t)}u(C_s^*)ds+e^{-\rho (T-t)} {{\mathfrak {B}}}(X_T^*)\right] -{{{\mathcal {W}}}}_t\right) \nonumber \\ +&{\mathbb {E}}_t\left[ \int _{t}^T\left( U_s^a({C^*},{X^*})-U_d(s,Y) \right) dD_s^{t,*}\right] \nonumber \\ =&{\mathbb {E}}_t\left[ \int _t^T e^{-\rho (s-t)}D_s^{t,*}\left( u(C_s^*) -\frac{e^{(\rho -r)(s-t)}}{D_s^{t,*}}C_s^*-u(Y_s)\right) ds +e^{-\rho (T-t)}D_{T-}^{t,*}\right. \nonumber \\&\quad \left. \left( {{\mathfrak {B}}}(X_T^*) -\frac{e^{(\rho -r)(T-t)}}{D_{T-}^{t,*}}X_T^*-U_d(T,Y)\right) \right] \nonumber \\ +&{\mathbb {E}}_t\left[ \int _t^T e^{-r(s-t)}Y_sds\right] +{\mathbb {E}}_t \left[ \int _T^\infty e^{-r(s-t)}\alpha Y_sds\right] +\lambda _t^* U_d(t,Y) -{{{\mathcal {W}}}}_t\lambda _t^*\nonumber \\ =&{\mathbb {E}}\left[ \int _t^T e^{-\rho (s-t)} D_s^{t,*} \Big ( {\tilde{u}}\left( \frac{e^{(\rho -r)(s-t)}}{D_s^{t,*}}\right) - u(Y_s)\Big )ds +e^{-\rho (T-t)}D_{T-}^{t,*}\right. \nonumber \\&\quad \left. \left( \widetilde{{\mathfrak {B}}} \left( \dfrac{e^{(\rho -r)(T-t)}}{D_{T-}^{t,*}}\right) -U_d(T,Y)\right) \right] \nonumber \\ +&{\mathbb {E}}_t\left[ \int _t^T e^{-r(s-t)}Y_sds\right] +{\mathbb {E}}_t\left[ \int _T^\infty e^{-r(s-t)}\alpha Y_sds\right] +\lambda _t^* U_d(t,Y)-{{{\mathcal {W}}}}_t\lambda _t^*. \end{aligned}$$
(91)

By using the similar arguments in Subsection 3.3, we deduce that

$$\begin{aligned}&{\mathbb {E}}\left[ \int _t^T e^{-\rho (s-t)} D_s^{t,*} \Big ( {\tilde{u}}\left( \frac{e^{(\rho -r)(s-t)}}{D_s^{t,*}}\right) - u(Y_s)\Big )ds +e^{-\rho (T-t)}D_{T-}^{t,*}\right. \nonumber \\&\quad \left. \left( \widetilde{{\mathfrak {B}}}\left( \dfrac{e^{(\rho -r)(T-t)}}{D_{T-}^{t,*}}\right) -U_d(T,Y)\right) \right] \nonumber \\ =&\int _{\lambda _t^*}^\infty \inf _{\tau \in {\mathcal {S}}(t,T)}{\mathbb {E}} \left[ \int _{\tau }^T e^{-\rho (s-t)}\left( u\left( I\left( \frac{e^{(\rho -r) (s-t)}}{\nu }\right) \right) -u(Y_s)\right) ds+e^{-\rho (T-t)}\right. \nonumber \\&\quad \left. \left( {{\mathfrak {B}}}\left( I_{{\mathfrak {B}}}\left( \frac{e^{(\rho -r) (T-t)}}{\nu }\right) \right) -U_d(T,Y)\right) \mathbf{1}_{\{\tau < T\}}\right] d\nu \nonumber \\ +&\int _{t}^{T}{e^{-\rho (s-t)}}\lambda _t^*{\tilde{u}} \left( \dfrac{e^{(\rho -r)(s-t)}}{\lambda _t^*}\right) ds +e^{-\rho (T-t)}\lambda _t^*\widetilde{{\mathfrak {B}}} \left( \dfrac{e^{(\rho -r)(T-t)}}{\lambda _t^*}\right) -\lambda _t^* U_d(t,Y) \nonumber \\ =&\int _{\lambda ^*_t}^\infty \left( {{{\mathcal {G}}}}_t^{FB}(\nu )-Q_t(\nu )\right) d\nu +\int _{t}^{T}{e^{-\rho (s-t)}}\lambda _t^*{\tilde{u}}\left( \dfrac{e^{(\rho -r)(s-t)}}{\lambda _t^*} \right) ds+e^{-\rho (T-t)}\lambda _t^*\nonumber \\&\quad \widetilde{{\mathfrak {B}}}\left( \dfrac{e^{(\rho -r)(T-t)}}{\lambda _t^*}\right) -\lambda _t^* U_d(t,Y). \end{aligned}$$
(92)

It follows from (91) and (92) that

$$\begin{aligned} {{{\mathcal {V}}}}_t = \int _{\lambda ^*}^\infty \left( {{{\mathcal {G}}}}_t^{FB} (\nu )-Q_t(\nu )\right) d\nu + J_t^{FB}(\lambda ^*). \end{aligned}$$

\(\square \)

Proof of (b)

It follows from Theorem 23.5 in Rockafeller [32] that the map \(\lambda \rightarrow J_t(\lambda )-\lambda {{{\mathcal {W}}}}_t\) is minimized at \(\lambda =x\) if and only if \(-x\in \partial \widetilde{{{{\mathcal {V}}}}}_t ({{{\mathcal {W}}}}_t)\), where \(\partial \widetilde{{{{\mathcal {V}}}}}_t ({{{\mathcal {W}}}}_t)\) is the set of all sub-gradient of \(\widetilde{{{{\mathcal {V}}}}}_t({{{\mathcal {W}}}}_t)\) and \(\widetilde{{{{\mathcal {V}}}}}_t({{{\mathcal {W}}}}_t)\) is given by

$$\begin{aligned} \widetilde{{{{\mathcal {V}}}}}_t({{{\mathcal {W}}}}_t) = \inf _{\lambda >0}\left( J_t(\lambda ) - \lambda {{{\mathcal {W}}}}_t\right) . \end{aligned}$$

Note that \({J}_t(\lambda )\) is continuously differentiable and convex in \(\lambda >0\).

However, the map \(\lambda \rightarrow J_t(\lambda )-\lambda {{{\mathcal {W}}}}_t\) is uniquely minimized by \(\lambda =(J_t^\prime )^{-1}({{{\mathcal {W}}}}_t)\). By Theorem 25.1 in Rockafeller [32], we deduce that \(\widetilde{{{{\mathcal {V}}}}}_t({{{\mathcal {W}}}}_t)\) is differentiable with respect to \({{{\mathcal {W}}}}_t\) and

$$\begin{aligned} (\widetilde{{{{\mathcal {V}}}}}_t({{{\mathcal {W}}}}_t))^\prime = -\lambda . \end{aligned}$$

It follows from \({{{\mathcal {V}}}}_t({{{\mathcal {W}}}}_t^*) = J_t(\lambda _t^*)-\lambda _t^* {{{\mathcal {W}}}}_t\) with \(\lambda _t^*({{{\mathcal {W}}}}_t)=(J_t^\prime )^{-1}({{{\mathcal {W}}}}_t)\) that

$$\begin{aligned} {{{\mathcal {V}}}}_t'({{{\mathcal {W}}}}_t^*)= - \lambda _t^* <0. \end{aligned}$$

Since \({{{\mathcal {W}}}}_t\) is strictly increasing in \(\lambda _t^*\), we conclude that the principal’s profit \({{{\mathcal {V}}}}_t\) is strictly decreasing in the agent’s continuation value \({{{\mathcal {W}}}}_t\). \(\square \)

Proof of (c)

Since \({{{\mathcal {W}}}}_t = Q_t(\lambda _t^*)+U_d(t,Y)\ge U_d(t,Y)\), it follows from Proposition 5.(a) that \({{{\mathcal {V}}}}_t \le {\bar{p}}(t)\).

Note that \(\lambda _t^* = e^{-(\rho -r)t}D_t^* =e^{-(\rho -r)t}\max \{\lambda ^*, \sup _{0\le s\le t}e^{(\rho -r)s}\chi (s)\}\ge \chi (t)\) for all \(t\in [0,T)\). From the definition of the stochastic boundary \(\chi (t)\), we deduce that

$$\begin{aligned} {{{\mathcal {W}}}}_t= U_d(t,Y)\,\,&\Longleftrightarrow \,\,Q_t(\lambda _t^*)=0\\&\Longleftrightarrow \,\,\lambda _t^*=\chi (t)\\&\Longleftrightarrow \,\,\lambda _t^*\,\,\text{ attains } \text{ a } \text{ new } \text{ maximum } \text{ at } \text{ time } \,\,t\\&\Longleftrightarrow \,\,C_t^*=I(1/\lambda _t^*)\,\,\text{ increases } \text{ at } \text{ time }\,\,t. \end{aligned}$$

This completes the proof. \(\square \)

Proof of Theorem 2

Note that

$$\begin{aligned} \varPsi _0 = {\mathbb {E}}\left[ \int _0^T e^{-rt}(C_t^*-Y_t)dt +e^{-rT}(X_T^*-\varPi _T)\right] . \end{aligned}$$

As similar to Theorem 3.5 in Karatzas and Shreve [20], we deduce that there exists a hedging position process \(\pi \) such that \((C^*,\pi ^*)\) is admissible and

$$\begin{aligned} \varPsi _t = {\mathbb {E}}\left[ \int _t^T e^{-r(s-t)}(C_s^*-Y_s)ds +e^{-r(T-t)}(X_T^*-\varPi _T)\right] =-{{{\mathcal {V}}}}_t({{{\mathcal {W}}}}_t). \end{aligned}$$

with \(\varPsi _T=X_T^*-\varPi _T\).

Suppose that there exists an optimal strategy \(\{{\widetilde{C}}_t, {\tilde{\pi }}_t\}\in {{\mathfrak {A}}(\varPsi _0)}\) with bank account balance process \({\widetilde{\varPsi }}_t\) such that

  1. (i)

    \(\displaystyle {\widetilde{w}}_0={\mathbb {E}}\left[ \int _0^T e^{-\rho t}u({\widetilde{C}}_t)dt+e^{-\rho T}{\mathfrak {B}}(X_T^*)\right] >w_0, \)

  2. (ii)

    \( d{\widetilde{\varPsi }}_t = (r{\widetilde{\varPsi }}_t - {\widetilde{C}}_t+Y_t)dt +{\tilde{\pi }}_t dB_t\,\,\text{ with }\,\,{\widetilde{\varPsi }}_T = {\widetilde{X}}_T-\varPi _T. \) with \({\widetilde{X}}_T\ge 0\),

  3. (iii)

    for any \(t\in [0,T]\)

    $$\begin{aligned} {\widetilde{\varPsi }}_t \ge -{\text{ CL }}_t. \end{aligned}$$

Note that

$$\begin{aligned} (({\widetilde{C}}_t)_{t=0}^T,{\widetilde{X}}_T)\in \varGamma ({\widetilde{w}}_0;Y). \end{aligned}$$

Proposition 5 implies that

$$\begin{aligned} -\varPsi _0=V({w}_0)>V({\tilde{w}}_0)=&\sup _{(C,X)\in \varGamma ({\tilde{w}}_0;Y)} {\mathbb {E}}\left[ \int _0^T e^{-r t} (Y_t-C_t)dt+e^{-rT}({\varPi }_T-X_T) \right] \nonumber \\\ge&{\mathbb {E}}\left[ \int _0^Te^{-rt}(Y_t-{\widetilde{C}}_t)dt+e^{-rT} (\varPi _T-{\widetilde{X}}_T)\right] . \end{aligned}$$
(93)

Since

$$\begin{aligned} d(e^{-rt}{\widetilde{\varPsi }}_t)+e^{-rt}({\widetilde{C}}_t-Y_t)dt = e^{-rt}{\tilde{\pi }}_tdB_t, \end{aligned}$$

we have

$$\begin{aligned} {\varPsi }_0+\int _0^Te^{-rt}{\tilde{\pi }}_tdB_t= & {} {\varPsi }_0+\int _0^T d(e^{-rt} {\widetilde{\varPsi }}_t) +\int _0^Te^{-rt}({\widetilde{C}}_t-Y_t)dt\\= & {} e^{-rT}({\widetilde{X}}_T-\varPi _T)+\int _0^Te^{-rt} ({\widetilde{C}}_t-Y_t)dt\\\ge & {} -{\mathbb {E}}\left[ \int _0^\infty e^{-rt}Y_tdt\right] . \end{aligned}$$

It follows from Assumption 2 that the left-hand side of the above equation is a continuous local martingale bounded from below and hence a supermartingale under \({\mathbb {P}}\).

Thus, we deduce that

$$\begin{aligned} {\mathbb {E}}\left[ \int _0^Te^{-rt}({\widetilde{C}}_t-Y_t)dt +e^{-rT}({\widetilde{X}}_T-\varPi _T)\right] \le {\varPsi }_0. \end{aligned}$$

It is a contradiction with the inequality (93). Hence, we conclude that \((C_t^*, \pi _t^*)\) is optimal solution to Problem 5.

Properties of the variational inequality (51)

We provide a complete self-contained derivation of the solution to VI (51) by borrowing the ideas and proofs in Yang and Koo [35].

First, we show the existence and uniqueness of \(W^{1,2}_{p,loc}\)-solution \((p\ge 1)\) to VI (51) and describe properties of the solution.

Lemma 3

  VI (51) has a unique strong solution \({{{\mathcal {M}}}}\) satisfying the following properties:

  1. (a)

    \({{{\mathcal {M}}}} \in W^{1,2}_{p,loc}({\mathcal {R}}_T) \cap C(\widetilde{{\mathcal {R}}}_T)\) for any \(p \ge 1 \) and \(\partial _z {{{\mathcal {M}}}} \in C(\widetilde{{\mathcal {R}}}_T)\), where \(\widetilde{{\mathcal {R}}}_T=[0,T]\times (0,+\infty )\).

  2. (b)

    \(\partial _z {{{\mathcal {M}}}} \le 0\) and \(\partial _z({{{\mathcal {M}}}}-{{{\mathcal {M}}}}_0)\ge 0\) in \(\widetilde{{\mathcal {R}}}_T\). Here, \({\mathcal {M}}_0(t,z)\) is defined by

    $$\begin{aligned} {\mathcal {M}}_0(t,z) \equiv&{\mathbb {E}}_t^{{\mathbb {Q}}}\left[ \int _{t}^T e^{-\rho _Y (s-t)}h({\mathcal {Z}}_s)ds+e^{-\rho _Y(T-t)}{{{\mathcal {H}}}} ({{{\mathcal {Z}}}}_T)\bigm | {\mathcal {Z}}_t=z\right] \nonumber \\ =&\dfrac{1}{1-\gamma }\left( \dfrac{1-e^{-K(T-t)}}{K}z^{-\frac{1-\gamma }{\gamma }} -\dfrac{1-e^{-\rho _Y (T-t)}}{\rho _Y}\right) \nonumber \\&\quad +\dfrac{1}{1-\gamma }\left( e^{-K(T-t)} {\kappa _{{\mathfrak {B}}}^{\frac{1-\gamma }{\gamma }}}z^{-\frac{1-\gamma }{\gamma }} -e^{-\rho _Y(T-t)}\dfrac{1}{\rho _Y}\right) , \end{aligned}$$
    (94)

    where \( K\equiv r+\frac{\rho -r}{\gamma }\).

Proof

(a) Since the differential operator \({\mathcal {L}}\) in VI (42) is degenerate, we need the following transformation to eliminate the degeneration:

$$\begin{aligned} \widehat{{{{\mathcal {M}}}}}(t,x)={{{\mathcal {M}}}}(t,z)\quad \text{ with }\quad x=\log z. \end{aligned}$$

Then \(\widehat{{{{\mathcal {M}}}}}(t,x )\) satisfies

$$\begin{aligned} \left\{ \begin{aligned}&-\partial _t \widehat{{{{\mathcal {M}}}}}-\widehat{{{{\mathcal {L}}}}} \widehat{{{{\mathcal {M}}}}}=\dfrac{1}{1-\gamma } \left( e^{x(-\frac{1-\gamma }{\gamma })}-1\right) ,~&\text{ if }~~\widehat{{{{\mathcal {M}}}}}(t,x)>0~\text {and}~{(t,x)\in [0,T)\times {\mathbb {R}}}, \\&-\partial _t \widehat{{{{\mathcal {M}}}}}-\widehat{{{{\mathcal {L}}}}} \widehat{{{{\mathcal {M}}}}}\ge \dfrac{1}{1-\gamma }\left( e^{x(-\frac{1-\gamma }{\gamma })}-1\right) ,~&\text{ if }~~\widehat{{{{\mathcal {M}}}}}(t,x)=0 ~\text {and}~{(t,x)\in [0,T)\times {\mathbb {R}}}, \\&\widehat{{{{\mathcal {M}}}}}(T,x)={\dfrac{1}{1-\gamma } \left( {\kappa _{{\mathfrak {B}}}^{\frac{1-\gamma }{\gamma }}}e^{x(-\frac{1-\gamma }{\gamma })}-\dfrac{1}{\rho _Y}\right) },\quad&\forall \,x\in {\mathbb {R}}, \end{aligned} \right. \nonumber \\ \end{aligned}$$
(95)

where \( \widehat{{{{\mathcal {L}}}}}= \frac{\gamma ^2\sigma ^2}{2} \partial _{xx}+(\rho _Y-r_Y-\frac{\sigma ^2\gamma ^2}{2})\partial _x-\rho _Y.\) Since the inhomogeneous term , the lower obstacle and the terminal value are all smooth functions, we can easily show that VI (95) has a unique solution satisfying \(\widehat{{{{\mathcal {M}}}}} \in W^{1,2}_{p,loc}([0,T)\times (-\infty ,\infty )) \cap C([0,T]\times (-\infty ,\infty ))\) for any \(p \ge 1 \) and \(\partial _x \widehat{{{{\mathcal {M}}}}} \in C([0,T]\times (-\infty ,\infty ))\) (See Friedman [12]).

(b) For any \(\eta >1\), let us denote

$$\begin{aligned} {{{\mathcal {M}}}}_1(t,z)={{{\mathcal {M}}}}(t,\eta z). \end{aligned}$$

Then \({{{\mathcal {M}}}}_1\) satisfies following VI:

$$\begin{aligned} \left\{ \begin{aligned}&-\partial _t {{{\mathcal {M}}}}_1-{{{\mathcal {L}}}} {{{\mathcal {M}}}}_1=\dfrac{1}{1-\gamma }\left( (\eta z)^{-\frac{1-\gamma }{\gamma }}-1\right) ,~~&\text{ if }~~{{{\mathcal {M}}}}_1(t,z)>0~~\text {and}~~(t,z) \in {\mathcal {M}}_T,\\&-\partial _t {{{\mathcal {M}}}}_1-{{{\mathcal {L}}}} {{{\mathcal {M}}}}_1\ge \dfrac{1}{1-\gamma }\left( (\eta z)^{-\frac{1-\gamma }{\gamma }}-1\right) ,~~&\text{ if }~~{{{\mathcal {M}}}}_1(t,z)=0 ~~\text {and}~~(t,z) \in {\mathcal {M}}_T,\\&{{{\mathcal {M}}}}_1(T,z)={\dfrac{1}{1-\gamma } \left( {\kappa _{{\mathfrak {B}}}^{\frac{1-\gamma }{\gamma }}}(\eta z)^{-\frac{1-\gamma }{\gamma }}-\dfrac{1}{\rho _Y}\right) }\quad&\forall \,z\in (0,+\infty ). \end{aligned} \right. \nonumber \\ \end{aligned}$$
(96)

For any \(\eta >1\), we can easily check that the inhomogeneous term of \({{{\mathcal {M}}}}_1\) is lower than that of \({{{\mathcal {M}}}}\), i.e.,

$$\begin{aligned} \begin{aligned}&\frac{1}{1-\gamma }\left( (\eta z)^{-\frac{1-\gamma }{\gamma }}-1\right)< \frac{1}{1-\gamma }\left( z^{-\frac{1-\gamma }{\gamma }}-1\right) \,\, \text{ and }\,\,\frac{1}{1-\gamma } \left( {\kappa _{{\mathfrak {B}}}^{\frac{1-\gamma }{\gamma }}}(\eta z)^{-\frac{1-\gamma }{\gamma }}-\dfrac{1}{\rho _Y }\right) \\&\quad < \frac{1}{1-\gamma }\left( {\kappa _{{\mathfrak {B}}}^{\frac{1-\gamma }{\gamma }}}z^{-\frac{1-\gamma }{\gamma }}-\dfrac{1}{\rho _Y}\right) . \end{aligned} \end{aligned}$$

The comparison theory for VI implies that \({{{\mathcal {M}}}}_1(t,z)={{{\mathcal {M}}}}(t,\eta z) \le {{{\mathcal {M}}}}(t,z)\) for any \(\eta >1\) and \((t,z) \in {\mathcal {R}}_T\). So we obtain \(\partial _z {{{\mathcal {M}}}} \le 0\) in \({\mathcal {R}}_T\).

Let us temporarily denote \({{{\mathcal {M}}}}_2(t,z)\) by

$$\begin{aligned} {{{\mathcal {M}}}}_2(t,z) = {{{\mathcal {M}}}}(t,z) - {{{\mathcal {M}}}}_0(t,z). \end{aligned}$$

where \({{{\mathcal {M}}}}_0(t,z)\) in (94).

Then, \({{{\mathcal {M}}}}_2(t,z)\) satisfies the following VI:

$$\begin{aligned} \left\{ \begin{aligned}&-\partial _t {{{\mathcal {M}}}}_2-{{{\mathcal {L}}}} {{{\mathcal {M}}}}_2=0,~~&\text{ if }~~{{{\mathcal {M}}}}_2(t,z)>-{{{\mathcal {M}}}}_0(t,z)~ ~\text {and}~~(t,z)\in {\mathcal {R}}_T, \\&-\partial _t {{{\mathcal {M}}}}_2-{{{\mathcal {L}}}} {{{\mathcal {M}}}}_2\ge 0,~~&\text{ if }~~{{{\mathcal {M}}}}_2(t,z)=-{{{\mathcal {M}}}}_0(t,z) ~~\text {and}~~(t,z)\in {\mathcal {R}}_T, \\&{{{\mathcal {M}}}}_2(T,z)=0\quad&\forall \,z\in (0,+\infty ). \end{aligned} \right. \end{aligned}$$
(97)

Note that

$$\begin{aligned} -\dfrac{\partial {{{\mathcal {M}}}}_0}{\partial z}=\dfrac{1}{\gamma }\dfrac{1-e^{-K(T-t)}}{K}z^{-\frac{1}{\gamma }} +\dfrac{1}{\gamma }e^{-K(T-t)}{\kappa _{{\mathfrak {B}}}^{\frac{1-\gamma }{\gamma }}}z^{-\frac{1}{\gamma }}>0, \end{aligned}$$

with recalling K defined in Lemma 3.

Similarly, we can easily obtain that \(\partial _z{{{\mathcal {M}}}}_2\ge 0\).

This completes the proof. \(\square \)

Now, we can define the increasing region \(\mathbf{IR}_z\) and the continuation region \({\text{ CL }}_z\) as follow:

$$\begin{aligned} \mathbf{IR}_z\equiv \left\{ (t,z)~ |~ {{{\mathcal {M}}}}(t,z)={{{{\mathcal {H}}}}(z)} \right\} ,~~~{\text{ CL }}_z\equiv \left\{ (t,z)~ |~ {{{\mathcal {M}}}}(t,z)>{{{{\mathcal {H}}}}(z)} \right\} . \end{aligned}$$
(98)

It follows from \(\partial _z {{{\mathcal {M}}}} \le 0\) that we can define the free boundary z(t) as follows:

$$\begin{aligned} z(t)\equiv \inf \left\{ z >0 ~|~ {{{\mathcal {M}}}}(t,z) = {{{{\mathcal {H}}}}(z)} \right\} ,~~~~\forall t\in [0,T). \end{aligned}$$
(99)

In terms of the free boundary, the two regions can be rewritten as follows:

$$\begin{aligned} \begin{aligned} \mathbf{IR}_z&=\left\{ (t,z)~ |~ z\ge z(t),~ t \in [0,T) \right\} ,\\ \mathbf{CR}_z&=\left\{ (t,z)~ |~ 0< z < z(t),~ t \in [0,T) \right\} . \end{aligned} \end{aligned}$$
(100)

By using the estimation (b) in Lemma 3, we can provide the integral equation representation of the solution M(tz) to VI (51) in the following lemma.

Lemma 4

  The value function \({{{\mathcal {M}}}}(t,z)\) has the integral equation representation in (52) and the free boundary z(t) satisfies the integral equation in (53).

Proof

From Lemma 3, we know that \({{{\mathcal {M}}}} \in W^{1,2}_{p,loc}({\mathcal {R}}_T) \cap C(\widetilde{{\mathcal {R}}}_T)\) for any \(p \ge 1\). By applying Itô lemma to \(e^{-\rho _Y s}{{{\mathcal {M}}}}(s,{\mathcal {Z}}_s)\) (see Krylov [23]),

$$\begin{aligned} \int _{t}^{T}d\left( e^{-{\rho _Y}s}{{{\mathcal {M}}}}(s,{\mathcal {Z}}_s)\right) =\int _{t}^{T}e^{-{\rho _Y}s}\left( \dfrac{\partial {{{\mathcal {M}}}}}{\partial s} + {\mathcal {L}}{{{\mathcal {M}}}}\right) ds + {\gamma \sigma }\int _{t}^{T} e^{-{\rho _Y}s}{\mathcal {Z}}_s \dfrac{\partial {{{\mathcal {M}}}}}{\partial z}dB^{{\mathbb {Q}}}_{s}.\nonumber \\ \end{aligned}$$
(101)

From Lemma 3(b), we deduce that

$$\begin{aligned} -\dfrac{1}{\gamma }\left( \dfrac{1-e^{-K(T-t)}}{K}+e^{-K(T-t)} {\kappa _{{\mathfrak {B}}}^{\frac{1-\gamma }{\gamma }}}\right) z^{-\frac{1-\gamma }{\gamma }}\le z\partial _z {{{\mathcal {M}}}}(t,z) \le 0. \end{aligned}$$

It follows

$$\begin{aligned} {\mathbb {E}}^{{\mathbb {Q}}}\left[ \int _{0}^{T}\left( \gamma \sigma e^{-\rho _Y t}{\mathcal {Z}}_t\dfrac{\partial {{{\mathcal {M}}}}}{\partial z}\right) ^2dt\right] \le C \int _{0}^{T}\left( e^{-\rho _Y t}{{\mathcal {Z}}_t}^{\frac{\gamma -1}{\gamma }}\right) ^2 dt <\infty \end{aligned}$$

for some positive constant C.

This implies that the diffusion term \({\gamma \sigma }\int _{t}^{T} e^{-{\rho _Y}s}{\mathcal {Z}}_s \frac{\partial {{{\mathcal {M}}}}}{\partial z}dB^{{\mathbb {Q}}}_{s}\) is a martingale under \({\mathbb {Q}}\)-measure, i.e.,

$$\begin{aligned} {\mathbb {E}}_t^{{\mathbb {Q}}}\left[ {\gamma \sigma }\int _{t}^{T} e^{-{\rho _Y}s}{\mathcal {Z}}_s \dfrac{\partial {{{\mathcal {M}}}}}{\partial z}dB^{{\mathbb {Q}}}_{s}\right] =0. \end{aligned}$$

(see Chapter 3 in Oksendal [27]).

By taking expectation to both-side of the equation (101),

$$\begin{aligned} {{{\mathcal {M}}}}(t,z)=&{\mathbb {E}}_t^{{\mathbb {Q}}} \left[ e^{-\rho _Y(T-t)}{{{\mathcal {M}}}}(T,{\mathcal {Z}}_T)\right] -{\mathbb {E}}_t^{{\mathbb {Q}}}\left[ \int _{t}^{T}e^{-{\rho _Y}(s-t)} \left( \dfrac{\partial {{{\mathcal {M}}}}}{\partial s} + {\mathcal {L}}{{{\mathcal {M}}}}\right) ds\right] \\ =&{\dfrac{1}{1-\gamma }\left( e^{-K(T-t)}{\kappa _{{\mathfrak {B}}}^{\frac{1-\gamma }{\gamma }}}z^{-\frac{1-\gamma }{\gamma }}-e^{-\rho _Y(T-t)}\dfrac{1}{\rho _Y}\right) }\\&\quad -{\mathbb {E}}^{{\mathbb {Q}}}_t\left[ \int _{t}^{T}e^{-{\rho _Y}(s-t)} \left( \dfrac{\partial {{{\mathcal {M}}}}}{\partial s} + {\mathcal {L}} {{{\mathcal {M}}}}\right) \mathbf{1}_{\{(s,{\mathcal {Z}}_s)\in {\mathbf{CR}_z}\}}ds \right] \\ =&{\dfrac{1}{1-\gamma }\left( e^{-K(T-t)}{\kappa _{{\mathfrak {B}}}^{\frac{1-\gamma }{\gamma }}}z^{-\frac{1-\gamma }{\gamma }}-e^{-\rho _Y(T-t)}\dfrac{1}{\rho _Y}\right) }\\&\quad -{\mathbb {E}}^{{\mathbb {Q}}}_t\left[ \int _{t}^{T}e^{-{\rho _YY}(s-t)} \left( \dfrac{\partial {{{\mathcal {M}}}}}{\partial s} + {\mathcal {L}}{{{\mathcal {M}}}}\right) \mathbf{1}_{\{{\mathcal {Z}}_s\le z(s)\}}ds\right] \\ =&{\dfrac{1}{1-\gamma }\left( e^{-K(T-t)}{\kappa _{{\mathfrak {B}}}^{\frac{1-\gamma }{\gamma }}}z^{-\frac{1-\gamma }{\gamma }}-e^{-\rho _Y(T-t)}\dfrac{1}{\rho _Y}\right) }\\&\quad +\dfrac{1}{1-\gamma }{\mathbb {E}}_t^{{\mathbb {Q}}}\left[ \int _{t}^{T} e^{-{\rho _Y}(s-t)}\left( {{\mathcal {Z}}_s}^{-\frac{1-\gamma }{\gamma }}-1\right) \mathbf{1}_{\{{\mathcal {Z}}_s\le z(s)\}}ds\right] \\ =&{\dfrac{1}{1-\gamma }\left( e^{-K(T-t)}{\kappa _{{\mathfrak {B}}}^{\frac{1-\gamma }{\gamma }}}z^{-\frac{1-\gamma }{\gamma }}-e^{-\rho _Y(T-t)}\dfrac{1}{\rho _Y}\right) }\\&\quad +\dfrac{1}{1-\gamma }{\mathbb {E}}_t^{{\mathbb {Q}}}\left[ \int _{t}^{T}e^{-{\rho _Y}(s-t)} {{\mathcal {Z}}_s}^{-\frac{1-\gamma }{\gamma }}{} \mathbf{1}_{\{{\mathcal {Z}}_s\le z(s)\}}ds \right] \\ -&\dfrac{1}{1-\gamma }{\mathbb {E}}^{{\mathbb {Q}}}_t\left[ \int _{t}^{T} e^{-{\rho _Y}(s-t)}\mathbf{1}_{\{{\mathcal {Z}}_s\le z(s)\}}ds\right] . \end{aligned}$$

Since \(\dfrac{d{\mathbb {P}}}{d{\mathbb {Q}}}=\exp {\left\{ -\frac{1}{2}(1-\gamma )^2 \sigma ^2 T-(1-\gamma )\sigma B_T^{{\mathbb {Q}}}\right\} }\) and \(B_t=B_t^{{\mathbb {Q}}}+(1-\gamma )\sigma t\) for \(t\in [0,T]\),

$$\begin{aligned}&{\mathbb {E}}^{{\mathbb {Q}}}_t\left[ \int _{t}^{T}e^{-{\rho _Y}(s-t)} {{\mathcal {Z}}_s}^{-\frac{1-\gamma }{\gamma }}{} \mathbf{1}_{\{{\mathcal {Z}}_s \le z(s)\}}ds\right] \\=&z^{-\frac{1-\gamma }{\gamma }}{\mathbb {E}}_t \left[ \int _{t}^{T}e^{-K(s-t)}{} \mathbf{1}_{\{{\mathcal {Z}}_s\le z(s)\}}ds\right] \\ =&z^{-\frac{1}{\gamma }-1}\int _{t}^{T}e^{-K(s-t)}{\mathbb {P}}({\mathcal {Z}}_s \le z(s))ds\\ =&z^{-\frac{1}{\gamma }-1}\int _{t}^{T}e^{-K(s-t)}{\mathcal {N}} \left( -\dfrac{\log {\frac{z}{z(s)}}+(\rho _Y-r_Y+\frac{1}{2}(\gamma \sigma )^2 -\frac{1}{\gamma }(\gamma \sigma )^2)(s-t)}{\gamma \sigma \sqrt{s-t}}\right) ds. \end{aligned}$$

Similarly,

$$\begin{aligned}&{\mathbb {E}}_t^{{\mathbb {Q}}}\left[ \int _{t}^{T}e^{-{\rho _Y}(s-t)} \mathbf{1}_{\{{\mathcal {Z}}_s\le z(s)\}}ds\right] \left( =\int _{t}^{T}e^{-\rho _Y(s-t)}{\mathbb {Q}}({\mathcal {Z}}_s\le z(s))ds \right) \\ =&\int _{t}^{T}e^{-\rho _Y(s-t)}{\mathcal {N}} \left( -\dfrac{\log {\frac{z}{z(s)}}+(\rho _Y-r_Y-\frac{1}{2}(\gamma \sigma )^2) (s-t)}{\gamma \sigma \sqrt{s-t}}\right) ds. \end{aligned}$$

Thus,

$$\begin{aligned} {{{\mathcal {M}}}}(t,z)=&{\dfrac{1}{1-\gamma }\left( e^{-K(T-t)} {\kappa _{{\mathfrak {B}}}^{\frac{1-\gamma }{\gamma }}}y^{-\frac{1-\gamma }{\gamma }} -e^{-\rho _Y(T-t)}\dfrac{1}{\rho _Y}\right) }\\ +&\dfrac{z^{-\frac{1-\gamma }{\gamma }}}{1-\gamma }\int _{t}^{T}e^{-K(s-t)}{\mathcal {N}} \left( -\dfrac{\log {\frac{z}{z(s)}}+(\rho _Y-r_Y+\frac{1}{2}(\gamma \sigma )^2 -\frac{1}{\gamma }(\gamma \sigma )^2)(s-t)}{\gamma \sigma \sqrt{s-t}}\right) ds\\ -&\frac{1}{1-\gamma }\int _{t}^{T}e^{-\rho _Y(s-t)}{\mathcal {N}} \left( -\dfrac{\log {\frac{z}{z(s)}}+(\rho _Y-r_Y-\frac{1}{2}(\gamma \sigma )^2) (s-t)}{\gamma \sigma \sqrt{s-t}}\right) ds. \end{aligned}$$

By the smooth-pasting condition (or \(C^1\)-condition) at \(z=z(t)\) (according to Lemma 3),

$$\begin{aligned} 0=&{\dfrac{1}{1-\gamma }\left( e^{-K(T-t)}{\kappa _{{\mathfrak {B}}}^{\frac{1-\gamma }{\gamma }}}(z(t))^{-\frac{1-\gamma }{\gamma }}-e^{-\rho _Y(T-t)}\dfrac{1}{\rho _Y} \right) }\\+&\frac{(z(t))^{-\frac{1-\gamma }{\gamma }}}{1-\gamma } \int _{t}^{T}e^{-K(s-t)}{\mathcal {N}}\left( -\dfrac{\log {\frac{z(t)}{z(s)}} +(\rho _Y-r_Y+\frac{1}{2}(\gamma \sigma )^2-\frac{1}{\gamma }(\gamma \sigma )^2) (s-t)}{\gamma \sigma \sqrt{s-t}}\right) ds\\ -&\frac{1}{1-\gamma }\int _{t}^{T}e^{-\rho _Y(s-t)}{\mathcal {N}} \left( -\dfrac{\log {\frac{z(t)}{z(s)}}+(\rho _Y-r_Y-\frac{1}{2}(\gamma \sigma )^2)(s-t)}{\gamma \sigma \sqrt{s-t}}\right) ds. \end{aligned}$$

\(\square \)

Lemma 5

When time goes to maturity, the free boundary approaches \(\kappa _{{\mathfrak {B}}} (\rho _Y)^{\gamma /(1-\gamma )}\), i.e.,

$$\begin{aligned} \lim _{t\rightarrow T-}z(t)= \kappa _{{\mathfrak {B}}}( \rho _Y)^{\frac{\gamma }{1-\gamma }}. \end{aligned}$$

Proof

Since

$$\begin{aligned}&\lim _{t\rightarrow T-}\int _{t}^{T}e^{-K(s-t)}{\mathcal {N}}\left( -\dfrac{\log {\frac{z(t)}{z(s)}}+(\rho _Y-r_Y+\frac{1}{2}(\gamma \sigma )^2-\frac{1}{\gamma } (\gamma \sigma )^2)(s-t)}{\gamma \sigma \sqrt{s-t}}\right) ds\\ =&\lim _{t\rightarrow T-}\int _{t}^{T}e^{-\rho _Y(s-t)}{\mathcal {N}}\left( -\dfrac{\log {\frac{z(t)}{z(s)}}+(\rho _Y-r_Y-\frac{1}{2}(\gamma \sigma )^2)(s-t)}{\gamma \sigma \sqrt{s-t}}\right) ds=0, \end{aligned}$$

we have

$$\begin{aligned} 0= & {} \lim _{t\rightarrow T-} {\dfrac{1}{1-\gamma }\left( e^{-K(T-t)}{\kappa _{{\mathfrak {B}}}^{\frac{1-\gamma }{\gamma }}}(z(t))^{-\frac{1-\gamma }{\gamma }}-e^{-\rho _Y(T-t)}\dfrac{1}{\rho _Y}\right) }\\= & {} \dfrac{1}{1-\gamma }\left( \kappa _{{\mathfrak {B}}}^{\frac{1-\gamma }{\gamma }} \lim _{t\rightarrow T-}(z(t))^{-\frac{1-\gamma }{\gamma }} - \dfrac{1}{\rho _Y}\right) . \end{aligned}$$

Hence, we have

$$\begin{aligned} \lim _{t\rightarrow T-}z(t)=\kappa _{{\mathfrak {B}}} (\rho _Y)^{\frac{\gamma }{1-\gamma }}. \end{aligned}$$

\(\square \)

Proposition 7

The optimal stopping time \(\tau ^*\) of Problem (50) is characterized as

$$\begin{aligned} \tau ^* = \inf \{s\ge t\mid {\mathcal {Z}}_s \ge z(s)\}\wedge T. \end{aligned}$$
(102)

Proof

It follows from Lemma 3 that

$$\begin{aligned} {{{\mathcal {M}}}}\in W^{1,2}_{p,{loc}}({\mathcal {R}}_T) \cap C(\widetilde{{\mathcal {R}}}_T) ~~\text{ for } \text{ any }\quad p \ge 1. \end{aligned}$$

For any stopping time \(\tau \in {\mathcal {S}}(t,T)\), the generalized Itô’s lemma for Sobolev space (e.g., Krylov [23]) implies that

$$\begin{aligned} \begin{aligned}&e^{-\rho _Y(\tau -t)}{{{\mathcal {M}}}}(\tau ,{\mathcal {Z}}_{\tau }) +\int _t^\tau e^{-\rho _Y (s-t)}\dfrac{1}{1-\gamma } \left( ({\mathcal {Z}}_s)^{-\frac{1-\gamma }{\gamma }}-1\right) ds\\ =&{{{\mathcal {M}}}}(t,z) + \int _{t}^{\tau }\left( \partial _s {{{\mathcal {M}}}}(s,{\mathcal {Z}}_s) + {\mathcal {L}}{{{\mathcal {M}}}}(s,{\mathcal {Z}}_s)+\dfrac{1}{1-\gamma } \left( ({\mathcal {Z}}_s)^{-\frac{1-\gamma }{\gamma }}-1\right) \right) ds\\ +&(\gamma \sigma )\int _{t}^\tau e^{-\rho _Y(s-t)}{\mathcal {Z}}_s\partial _z {{{\mathcal {M}}}}(s,{\mathcal {Z}}_s)dB_s^{{\mathbb {Q}}}. \end{aligned} \end{aligned}$$
(103)

Note that

$$\begin{aligned} {\mathbb {E}}_t^{{\mathbb {Q}}}\left[ \int _t^\tau e^{-\rho _Y(s-t)}{\mathcal {Z}}_s\partial _z {{{\mathcal {M}}}}(s,{\mathcal {Z}}_s)dB_s^{{\mathbb {Q}}}\right] =0. \end{aligned}$$

By taking the expectation on the both sides of the equation (103), we have

$$\begin{aligned}&{{{\mathcal {M}}}}(t,z) + {\mathbb {E}}_t^{{\mathbb {Q}}} \left[ \int _t^\tau e^{-\rho _Y(s-t)}\left( \partial _s {{{\mathcal {M}}}} (s,{\mathcal {Z}}_s) + {\mathcal {L}}{{{\mathcal {M}}}}(s,{\mathcal {Z}}_s) +\dfrac{1}{1-\gamma }\left( ({\mathcal {Z}}_s)^{-\frac{1-\gamma }{\gamma }} -1\right) \right) ds\right] \\&={\mathbb {E}}_t^{{\mathbb {Q}}}\left[ e^{-\rho _Y(\tau -t)}{{{\mathcal {M}}}} (\tau ,{\mathcal {Z}}_{\tau })+\int _t^\tau e^{-\rho _Y(s-t)}\dfrac{1}{1-\gamma }\left( {\mathcal {Z}}_s^{-\frac{1-\gamma }{\gamma }} -1\right) ds\right] . \end{aligned}$$

It follows from the VI (51) that

$$\begin{aligned} {{{\mathcal {M}}}}(t,z) \ge {\mathbb {E}}_t^{{\mathbb {Q}}} \left[ \int _t^\tau e^{-\rho _Y(s-t)}\dfrac{1}{1-\gamma } \left( {\mathcal {Z}}_s^{-\frac{1-\gamma }{\gamma }}-1\right) ds + e^{-\rho _Y(T-t)} {{{\mathcal {H}}}}({{{\mathcal {Z}}}}_T)\mathbf{1}_{\{\tau \ge T\}}\right] , \end{aligned}$$

for any stopping time \(\tau \in {\mathcal {S}}(t,T)\). That is,

$$\begin{aligned} {{{\mathcal {M}}}}(t,z) \ge \sup _{\tau \in {\mathcal {S}}(t,T)}{\mathbb {E}}_t^{{\mathbb {Q}}} \left[ \int _t^\tau e^{-\rho _Y(s-t)}\dfrac{1}{1-\gamma }\left( {\mathcal {Z}}_s^{-\frac{1-\gamma }{\gamma }} -1\right) ds+e^{-\rho _Y(T-t)}{{{\mathcal {H}}}}({{{\mathcal {Z}}}}_T)\mathbf{1}_{\{\tau \ge T\}}\right] .\nonumber \\ \end{aligned}$$
(104)

On the other hand, let us define

$$\begin{aligned} \tau ^*= \inf \{s\ge t\mid {\mathcal {Z}}_s \ge z(s)\} \wedge T.\end{aligned}$$

If \(z={\mathcal {Z}}_t \ge z(t)\), then \(\tau ^*=t\), and

$$\begin{aligned}&{{{\mathcal {M}}}}(t,z)={{{\mathcal {H}}}}({{{\mathcal {Z}}}}_T)\mathbf{1}_{\{t\ge T\}}\\&\quad ={\mathbb {E}}_t^{{\mathbb {Q}}}\left[ \int _t^{\tau ^*} e^{-\rho _Y(s-t)}\dfrac{1}{1-\gamma }\left( {\mathcal {Z}}_s^{-\frac{1-\gamma }{\gamma }} -1\right) ds+e^{-\rho _Y(T-\tau ^*)}{{{\mathcal {H}}}}({{{\mathcal {Z}}}}_T)\mathbf{1}_{\{\tau ^*\ge T\}}\right] . \end{aligned}$$

If \(0<z< z(t)\), it follows from the definition of \(\tau ^*\) and VI (51) that

$$\begin{aligned}\begin{aligned} {\left\{ \begin{array}{ll} \partial _s {{{\mathcal {M}}}}(s, {\mathcal {Z}}_s) + {\mathcal {L}} {{{\mathcal {M}}}}(s,{\mathcal {Z}}_s)+ \dfrac{1}{1-\gamma } \left( {\mathcal {Z}}_s^{-\frac{1-\gamma }{\gamma }}-1\right) =0\, \,\text{ for }\,\,s\in (t,\tau ^*),\\ {{{\mathcal {M}}}}(\tau ^*, {\mathcal {Z}}_{\tau ^*}) = {{{\mathcal {H}}}}({{{\mathcal {Z}}}}_T)\mathbf{1}_{\{\tau ^*\ge T\}}. \end{array}\right. } \end{aligned} \end{aligned}$$

Inserting \(\tau =\tau ^*\) in (103), we deduce

$$\begin{aligned} \begin{aligned} {{{\mathcal {M}}}}(t,z) ={\mathbb {E}}_t^{{\mathbb {Q}}}\left[ \int _t^{\tau ^*} e^{-\rho _Y(s-t)}h({{{\mathcal {Z}}}}_s)ds+e^{-\rho _Y(T-\tau ^*)} {{{\mathcal {H}}}}({{{\mathcal {Z}}}}_T)\mathbf{1}_{\{\tau ^*\ge T\}}\right] . \end{aligned} \end{aligned}$$
(105)

It follows from (104) and (105) that

$$\begin{aligned} {{{\mathcal {M}}}}(t,z) = \sup _{\tau \in {\mathcal {S}}(t,T)}{\mathbb {E}}_t^{{\mathbb {Q}}} \left[ \int _t^\tau e^{-\rho _Y(s-t)}h({{{\mathcal {Z}}}}_s)ds+e^{-\rho _Y(T-\tau )} {{{\mathcal {H}}}}({{{\mathcal {Z}}}}_T)\mathbf{1}_{\{\tau ^*\ge T\}}\right] . \end{aligned}$$
(106)

\(\square \)

Proof of Theorem 3

Since the proof of parts (a), (b), and (d) is straightforward, it suffice to show the part (c) of Theorem 3.

It follows from (a) in Proposition 5 and the integral equation representation of \(Q_t(e^{-(\rho -r)t}\lambda )\) in (54) that

$$\begin{aligned} J_t=&Y_t^{1-\gamma }\int _{\lambda _t^*}^\infty \left[ \dfrac{1}{1-\gamma } \left( \frac{Y_t^\gamma }{x}\right) ^{-\frac{1-\gamma }{\gamma }}\int _{t}^{T}e^{-K (s-t)}{\mathcal {N}}\left( d^{\gamma }\left( s-t,\frac{Y_t^\gamma }{z(s)x}\right) \right) ds\right. \nonumber \&\left. \dfrac{1}{1-\gamma }\int _{t}^{T}e^{-\rho _Y (s-t)}{\mathcal {N}}\left( d^{-}\left( s-t,\frac{Y_t^\gamma }{z(s)x}\right) \right) ds\right] dx + J_t^{FB}(\lambda _t^*). \end{aligned}$$
(107)

By the integration by parts, we obtain

$$\begin{aligned}&\int _{\lambda _t^*}^\infty x^{\frac{1-\gamma }{\gamma }}e^{-K(s-t)}{\mathcal {N}} \left( d^{\gamma }\left( s-t,\frac{Y_t^\gamma }{z(s)x}\right) \right) dx\nonumber \\ =&\left[ e^{-K(s-t)}\gamma x^{\frac{1}{\gamma }}{\mathcal {N}}\left( d^{\gamma } \left( s-t,\frac{Y_t^\gamma }{z(s)x}\right) \right) \right] _{x=\lambda _t^*}^\infty \nonumber \\&\quad +\gamma e^{-K(s-t)}\int _{\lambda _t^*}^\infty x^{\frac{1}{\gamma }}{} \mathbf{n} \left( d^\gamma \left( s-t,\frac{Y_t^\gamma }{z(s) x}\right) \right) \dfrac{1}{x(\gamma \sigma )\sqrt{s-t}}dx\nonumber \\ =&-e^{-K(s-t)}\gamma (\lambda _t^*)^{\frac{1}{\gamma }}{\mathcal {N}}\left( d^\gamma \left( s-t,\frac{Y_t^\gamma }{z(s)\lambda _t^*)}\right) \right) \nonumber \\&\quad +\gamma e^{-r_Y (s-t)} {(z(s))^{-\frac{1}{\gamma }}}Y_t \int _{\lambda _t^*}^\infty \mathbf{n} \left( d^+\left( s-t,\frac{Y_t^\gamma }{z(s)x}\right) \right) \dfrac{1}{x(\gamma \sigma ) \sqrt{s-t}}dx\nonumber \\ =&-e^{-K(s-t)}\gamma (\lambda _t^*)^{\frac{1}{\gamma }}{\mathcal {N}}\left( d^\gamma \left( s-t,\frac{Y_t^\gamma }{z(s)\lambda _t^*}\right) \right) +\gamma e^{-r_Y (s-t)}{(z(s))^{-\frac{1}{\gamma }}}Y_t{\mathcal {N}}\left( d^+\left( s-t, \frac{Y_t^\gamma }{z(s)\lambda }\right) \right) , \end{aligned}$$
(108)

where \(\mathbf{n}(\cdot )\) is a probability density function of the standard normal random variable and we have used the fact (see Lemma 6 in Section N):

Similarly, we have

$$\begin{aligned}&\int _{\lambda _t^*}^\infty e^{-\rho _Y (s-t)}{\mathcal {N}}\left( d^{-} \left( s-t,\frac{Y_t^\gamma }{z(s)x}\right) \right) dx\nonumber \\ =&\left[ e^{-\rho _Y (s-t)} x{\mathcal {N}}\left( d^{-} \left( s-t,\frac{Y_t^\gamma }{z(s)x}\right) \right) \right] _{x=\lambda _t^*}^\infty + e^{-\rho _Y (s-t)}\int _{\lambda _t^*}^\infty x\mathbf{n}\left( d^{-} \left( s-t,\frac{Y_t^\gamma }{z(s) x}\right) \right) \nonumber \\&\quad \dfrac{1}{x(\gamma \sigma ) \sqrt{s-t}}dx\nonumber \\ =&-e^{-\rho _Y(s-t)} \lambda _t^* {\mathcal {N}}\left( d^{-}\left( s-t, \frac{Y_t^\gamma }{z(s)\lambda _t^*}\right) \right) + e^{-r_Y (s-t)} \frac{Y_t^\gamma }{z(s)} \int _{\lambda _t^*}^\infty \mathbf{n} \left( d^+\left( s-t,\frac{Y_t^\gamma }{z(s)x}\right) \right) \nonumber \\&\quad \dfrac{1}{x(\gamma \sigma ) \sqrt{s-t}}dx\nonumber \\ =&-e^{-\rho _Y(s-t)} \lambda _t^*{\mathcal {N}}\left( d^{-}\left( s-t,\frac{Y_t^\gamma }{z(s) \lambda _t^*}\right) \right) + e^{-r_Y (s-t)}\frac{Y_t^\gamma }{z(s)}{\mathcal {N}}\left( d^+\left( s-t, \frac{Y_t^\gamma }{z(s)\lambda _t^*}\right) \right) . \end{aligned}$$
(109)

Since \({{{\mathcal {V}}}}_t=J_t (\lambda _t^*)-\lambda _t^* {{{\mathcal {W}}}}_t\) and

$$\begin{aligned} {\mathcal {W}}_t=&Y_t^{1-\gamma }\cdot \left\{ \dfrac{1}{1-\gamma }e^{-K(T-t)}{\kappa _{{\mathfrak {B}}}^{\frac{1-\gamma }{\gamma }}}\left( \dfrac{Y_t^\gamma }{\lambda _t^*}\right) ^{-\frac{1-\gamma }{\gamma }}+\dfrac{1}{1-\gamma }\int _t^T e^{-\rho _Y (s-t)}{\mathcal {N}}\left( d^{-}\left( s-t,\frac{Y_t^\gamma }{z(s)\lambda _t^*} \right) \right) ds\right. \\ +&\left. \dfrac{1}{1-\gamma }\left( \frac{Y_t^\gamma }{\lambda _t^*}\right) ^{-\frac{1-\gamma }{\gamma }}\int _t^T e^{-K (s-t)}{\mathcal {N}}\left( -d^{\gamma }\left( s-t,\frac{Y_t^\gamma }{z(s)\lambda _t^*}\right) \right) ds\right\} , \end{aligned}$$

it follows from (107), (108) and (109) that

$$\begin{aligned} {\mathcal {V}}_t=&Y_t\left[ -\left( \frac{Y_t^\gamma }{\lambda _t^*} \right) ^{-\frac{1}{\gamma }}\int _t^Te^{-K (s-t)}{\mathcal {N}}\left( -d^\gamma \left( s-t,\frac{Y_t^\gamma }{ z(s)\lambda ^*_t}\right) \right) ds\right. \\&\left. +\dfrac{\gamma }{1-\gamma }\int _t^T e^{-r_Y (s-t)}(z(s))^{-\frac{1}{\gamma }} {\mathcal {N}}\left( d^+\left( s-t,\frac{Y_t^\gamma }{z(s)\lambda _t^*}\right) \right) ds\right. \\&\left. -\dfrac{1}{1-\gamma }\int _t^Te^{-r_Y (s-t)}(z(s))^{-1}{\mathcal {N}}\left( d^{+}\left( s-t,\frac{Y_t^\gamma }{z(s) \lambda _t^*}\right) \right) ds+\frac{1-e^{-r_Y (T-t)}}{r_Y}+\alpha \dfrac{e^{-r_Y (T-t)}}{r_Y}\right] . \end{aligned}$$

Note that \(U_d(Y_t) = {{{\mathcal {W}}}}_t(z(t),Y_t)\). This implies that

$$\begin{aligned} {\bar{p}}(t)\equiv {{{\mathcal {V}}}}_t(U_d(t,Y))=J_t(z(t))-z(t)U_d(Y_t) \end{aligned}$$

and thus we have obtained the desired results.

Derivation of the explicit form \({{{\mathcal {M}}}}(z)\) in (60)

M(z) should satisfy

$$\begin{aligned} \begin{aligned} {\left\{ \begin{array}{ll} -{{{\mathcal {L}}}} {{{\mathcal {M}}}}(z) \ge \dfrac{1}{1-\gamma } \left( z^{-\frac{1-\gamma }{\gamma }}-1\right) \,\,\,&{}\text{ if }\, \,{{{\mathcal {M}}}}(z)=0,\\ -{{{\mathcal {L}}}} {{{\mathcal {M}}}}(z) = \dfrac{1}{1-\gamma } \left( z^{-\frac{1-\gamma }{\gamma }}-1\right) \,\,\,&{}\text{ if }\, \,{{{\mathcal {M}}}}(z)>0\\ \end{array}\right. } \end{aligned} \end{aligned}$$
(110)

with

$$\begin{aligned} \lim _{t\rightarrow \infty } e^{-\rho _Y t}{\mathbb {E}}^{\mathbb {Q}}\left[ {{{\mathcal {M}}}}({{{\mathcal {Z}}}}_t) \right] =0\,\,\,(\text{ transversality } \text{ condition}). \end{aligned}$$

To solve the variational inequality (110), we consider the following free boundary problem for \({{{\mathcal {M}}}}(z)\):

$$\begin{aligned} \begin{aligned} {\left\{ \begin{array}{ll} {{{\mathcal {L}}}}{{{\mathcal {M}}}}(z) + \dfrac{1}{1-\gamma } \left( z^{-\frac{1-\gamma }{\gamma }}-1\right) =0\,\,\,&{}\text{ for }\, \,0<z<{\bar{z}},\\ {{{\mathcal {M}}}}(y) = 0 \,\,\,\text{ for }\,\, {\bar{z}}\le z,\\ {{{\mathcal {M}}}}({\bar{z}}) = {{{\mathcal {M}}}}'({\bar{z}})=0 \end{array}\right. } \end{aligned} \end{aligned}$$
(111)

For \(0<z<{\bar{z}}\), a general solution to the FBP (111) can be represented as the sum of a general solution to the homogeneous equation and a particular solution as

$$\begin{aligned} {{{\mathcal {M}}}}(z) = E_1z^{\alpha _1}+E_2z^{\alpha _2} +\frac{1}{K}\frac{1}{1-\gamma }z^{-\frac{1-\gamma }{\gamma }}-\frac{1}{\rho _Y} \frac{1}{1-\gamma }, \end{aligned}$$
(112)

where \(\alpha _1>1\) and \(\alpha _2<0\) are the roots of the following quadratic equation:

$$\begin{aligned}\frac{(\gamma \sigma )^2}{2}\alpha ^2+\left( \rho _Y-r_Y -\frac{(\gamma \sigma )^2}{2}\right) \alpha -\rho _Y=0. \end{aligned}$$

To satisfy the transversality condition in (110) for \({{{\mathcal {M}}}}(Z)\), the coefficient \(E_2\) of \(z^{\alpha _2}\) should be zero, i.e,

$$\begin{aligned} E_2= 0. \end{aligned}$$

Thus, we can write down \({{{\mathcal {M}}}}(z)\) as follows:

$$\begin{aligned} {{{\mathcal {M}}}}(z) = E y^{\alpha _1} +\frac{1}{K}\frac{1}{1-\gamma }z^{-\frac{1-\gamma }{\gamma }}-\frac{1}{\rho _Y} \frac{1}{1-\gamma }. \end{aligned}$$
(113)

By using the smooth pasting condition (\({{{\mathcal {M}}}}({\bar{z}})={{{\mathcal {M}}}}'({\bar{z}})=0\)), we can easily derive

$$\begin{aligned} {\bar{z}}= \left( \frac{1-\gamma +\gamma \alpha _2}{\gamma \alpha _2}\right) ^{-\frac{\gamma }{1-\gamma }} \end{aligned}$$

and

$$\begin{aligned} M(z)=\frac{1}{\rho _Y}\frac{1}{1-\gamma +\gamma \alpha _1}\left( \frac{z}{{\bar{z}}} \right) ^{\alpha _1}+\frac{1}{K}\frac{1}{1-\gamma }z^{-\frac{1-\gamma }{\gamma }} -\frac{1}{\rho _Y}\frac{1}{1-\gamma }\,\,\text{ for }\,\,0<z<{\bar{z}}. \end{aligned}$$

Miscellaneous

Lemma 6

For any \(\kappa \ne 0\), \(\delta \ne 0\) and \(\eta \in {\mathbb {R}}\),

$$\begin{aligned} \begin{aligned} \lim _{\xi \rightarrow 0+} \xi ^{\kappa }{} \mathbf{n}(\delta \log {\xi } + \eta ) =0,\quad \lim _{\xi \rightarrow \infty } \xi ^{\kappa }{} \mathbf{n}(\delta \log {\xi } + \eta ) =0 \end{aligned} \end{aligned}$$
(114)

and

$$\begin{aligned}&\lim _{\xi \rightarrow 0+} \xi ^{\kappa }{\mathcal {N}}(\delta \log {\xi } +\eta )= {\left\{ \begin{array}{ll} 0\,\,\,&{}\kappa>0,\\ 0\,\,\,&{}\kappa<0,\,\delta >0,\\ \infty \,\,\,&{}\kappa<0,\,\delta <0, \end{array}\right. } \end{aligned}$$
(115)
$$\begin{aligned}&\lim _{\xi \rightarrow \infty } \xi ^{\kappa }{\mathcal {N}}(\delta \log {\xi } +\eta )= {\left\{ \begin{array}{ll} 0\,\,\,&{}\kappa<0,\\ \infty \,\,\,&{}\kappa>0,\,\delta>0,\\ 0\,\,\,&{}\kappa >0,\,\delta <0, \end{array}\right. } \end{aligned}$$
(116)

where \(\mathbf{n}(\cdot )\) is a probability density function of the standard normal random variable.

Proof

Since

$$\begin{aligned} \begin{aligned}&\xi ^{\kappa }{} \mathbf{n} (\delta \log {\xi } + \eta ) \\&=\dfrac{1}{\sqrt{2\pi }}e^{-\frac{1}{2}(\delta \log {\xi } + \eta )^2}e^{\kappa \log {\xi }}=\dfrac{1}{\sqrt{2\pi }}e^{-\frac{1}{2}(\delta \log {\xi } + (\eta -\frac{\kappa }{\delta }))^2}e^{-2\frac{\eta \kappa }{\delta } -\frac{\kappa ^2}{\delta ^2}}, \end{aligned} \end{aligned}$$
(117)

we have

$$\begin{aligned} \lim _{\xi \rightarrow 0+} \xi ^{\kappa }{} \mathbf{n}(\delta \log {\xi } + \eta ) =0,\quad \lim _{\xi \rightarrow \infty } \xi ^{\kappa }{} \mathbf{n}(\delta \log {\xi } + \eta ) =0. \end{aligned}$$

If \(\kappa >0\), it is clear that

$$\begin{aligned} \lim _{\xi \rightarrow 0+} \xi ^{\kappa } {\mathcal {N}}(\delta \log {\xi }+\eta ) = 0. \end{aligned}$$

If \(\kappa <0\) and \(\delta >0\), L’Hôpital’s theorem implies that

$$\begin{aligned} \begin{aligned} \lim _{\xi \rightarrow 0+} \xi ^{\kappa } {\mathcal {N}}(\delta \log {\xi }+\eta ) = \lim _{\xi \rightarrow 0+} \dfrac{{\mathcal {N}}(\delta \log {\xi }+\eta )}{\xi ^{-\kappa }}=\lim _{\xi \rightarrow 0+}\dfrac{\delta \mathbf{n}(\delta \log {\xi }+\eta )}{-\kappa \xi ^{-\kappa }}=0. \end{aligned} \end{aligned}$$
(118)

Hence, we can obtain the results. \(\square \)

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Jeon, J., Koo, H.K. & Park, K. Optimal finite horizon contract with limited commitment. Math Finan Econ 16, 267–315 (2022). https://doi.org/10.1007/s11579-021-00309-x

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Keywords

  • Optimal contract
  • Limited commitment
  • Principal-Agent problem
  • Optimal stopping problem
  • Singular control problem

JEL Classification

  • C61
  • D86
  • E21
  • D15