A two-dimensional control problem arising from dynamic contracting theory

Abstract

We study a dynamic corporate finance contracting model in which the firm’s profitability fluctuates and is impacted by the unobservable managerial effort. Thereby, we introduce in an agency framework the issue of strategic liquidation. We show that the principal’s problem takes the form of a two-dimensional fully degenerate Markov control problem. We prove regularity properties of the value function and derive explicitly the optimal contract that implements full effort. Our regularity results appear in some recent studies, but with heuristic proofs that do not clarify the importance of the regularity of the value function at the boundaries.

Appendix

Proof of Lemma 4.4

First observe that for every pair $(x,w) \in (x^{*},+\infty )\times (0,+\infty )$, there is some $C>0$ such that $u(x,w)\le C(1+x)$. Indeed,

$$\begin{aligned} u(x,w) &\le \mathbb{E}^{0}\left [\int_{0}^{\infty }e^{-rs}\vert x+ \sigma Z_{s}\vert \,ds\right ] \\ &\le \frac{x}{r} + \sigma \mathbb{E}^{0}\left [\int_{0}^{\infty }e ^{-rs}\vert Z_{s}\vert \,ds\right ] \\ &= \frac{x}{r} + \sigma \sqrt{\frac{2}{\pi }}\mathbb{E}^{0}\left [\int _{0}^{\infty }e^{-rs}\sqrt{s} \,ds \right ] \\ &\le C(1+x). \end{aligned}$$

Therefore, Lemma 4.4 holds for $w \ge 1$. Let us now consider $w \in (0,1)$. We decompose $u(x,w)$ as $u(x,w)=u_{1}(x,w)+u _{2}(x,w)$, with

$$\begin{array}{rl}{u}_1(x,w)& ={\mathbb{E}}^0[{\mathbb{1}}_{\{{\tau }_0^w<{\tau }_1^w\}}{\int }_0^{{\tau }^{\ast }\wedge {\tau }_0^w}{e}^{-rs}(x+\sigma Z_s)ds],\\ u_2(x,w)& ={\mathbb{E}}^0[{\mathbb{1}}_{\{{\tau }_0^w>{\tau }_1^w\}}{\int }_0^{{\tau }^{\ast }\wedge {\tau }_0^w}{e}^{-rs}(x+\sigma Z_s)ds],\end{array}$$

where $\tau_{1}^{w}=\inf \{ t \ge 0: W_{t}^{w}=1 \}$. On the event $\{\tau_{0}^{w} < \tau_{1}^{w} \}$, we have for every $t \le \tau^{*} \wedge \tau_{0}^{w} < \tau_{1}^{w}$ the inequality $X_{t} \le \frac{1}{ \lambda } + x$. Therefore,

$$u_1(x,w)\le (\frac{1}{\lambda }+x){\mathbb{E}}^0\left[{\mathbb{1}}_{\{{\tau }_0^w<{\tau }_1^w\}}{\int }_0^{{\tau }_0^w}{e}^{-rs}ds\right].$$

Conditioning the process $(W_{t})_{t \in [0,\tau_{1}^{w}]}$ on the event $\{\tau_{0}^{w} < \tau_{1}^{w} \}$ and using Doob’s $h$-transform (see Rogers and Williams [19, Chap. IV, Sect. 39] for a definition) makes $(W_{t})_{t \in [0,\tau_{1}^{w}]}$ a diffusion absorbed at 0 with generator

$$ \tilde{\mathcal{L}}=\frac{\lambda^{2} \sigma^{2}}{2}\frac{\partial^{2}}{ \partial w^{2}}+\left (rw+\lambda \sigma \frac{h^{\prime }(w)}{h(w)}\right )\frac{ \partial }{\partial w}, $$

where

$$ h(w)=\mathbb{P}^{0}[\tau_{0}^{w} < \tau_{1}^{w}]=\frac{\int_{w}^{1}e ^{-rs^{2}} \, ds}{\int_{0}^{1}e^{-rs^{2}} \, ds}. $$

Let us denote $\tilde{\tau }_{0}^{w}=\inf \{t \ge 0:\tilde{W}_{t}^{w}=0 \}$, where

$$ d\tilde{W}_{t}=\bigg( r\tilde{W}_{t} +\lambda \sigma \frac{h^{\prime }(\tilde{W}_{t})}{h(\tilde{W}_{t})} \bigg)\,dt +\lambda \sigma\, dZ_{t}. $$

We have

$$\begin{array}{rl}{\mathbb{E}}^0\left[{\mathbb{1}}_{\{{\tau }_0^w<{\tau }_1^w\}}{\int }_0^{{\tau }_0^w}{e}^{-rs}ds\right]& ={\mathbb{P}}^0[{\tau }_0^w<{\tau }_1^w]{\mathbb{E}}^0\left[{\int }_0^{{\tilde{\tau }}_0^w}{e}^{-rs}ds\right]\\ & \le {\mathbb{E}}^0\left[{\int }_0^{{\tilde{\tau }}_0^w}{e}^{-rs}ds\right]\\ & =\tilde{\varphi }(w).\end{array}$$

The function $\tilde{\phi }$ satisfies

$$ \tilde{\mathcal{L}}\tilde{\phi }- r\tilde{\phi }= 0 $$

with $\tilde{\phi }(0)=0$. A computation shows that $rw+\lambda \sigma \frac{h^{\prime }(w)}{h(w)}<0$, and because $\tilde{\phi }$ is nondecreasing, we get that $\tilde{\phi }$ is convex. We deduce that $\tilde{\phi }(w)\le C w$ for some positive constant $C$ which implies $u_{1}(x,w)\le C(1+x)w$. Now, we decompose $u_{2}$ as

$$\begin{array}{rl}{u}_2(x,w)& ={\mathbb{E}}^0[{\mathbb{1}}_{\{{\tau }_0^w>{\tau }_1^w\}}{\int }_0^{{\tau }^{\ast }\wedge {\tau }_1^w}{e}^{-rs}(x+\sigma Z_s)ds]\\ & \phantom{=:}+{\mathbb{E}}^0[{\mathbb{1}}_{\{{\tau }_0^w>{\tau }_1^w\}}{\int }_{{\tau }^{\ast }\wedge {\tau }_1^w}^{{\tau }^{\ast }\wedge {\tau }_0^w}{e}^{-rs}(x+\sigma Z_s)ds]\\ & =u_3(x,w)+u_4(x,w).\end{array}$$

Because for every $s \le \tau_{1}^{w}$, we have $X_{s}\le \frac{1}{ \lambda } +x$, it follows that

$$\begin{aligned} u_{3}(x,w) &\le \left ( \frac{1}{\lambda } +x \right )\int_{0}^{\infty }e^{-rs}\,ds \;\mathbb{P}^{0}[\tau_{0}^{w} > \tau_{1}^{w}] \\ &=\frac{\frac{1}{\lambda } +x}{r}\mathbb{P}^{0}[\tau_{0}^{w} > \tau _{1}^{w}] \\ &=\frac{\frac{1}{\lambda } +x}{r}\big(1-h(w)\big) \\ &\le C(1+x)w. \end{aligned}$$

Finally, the strong Markov property implies

$$\begin{array}{rl}{u}_4(x,w)& ={\mathbb{E}}^0[{\mathbb{1}}_{\{{\tau }_0^w>{\tau }_1^w\}}{e}^{-r({\tau }^{\ast }\wedge {\tau }_1^w)}u(X_{{\tau }^{\ast }\wedge {\tau }_1^w},W_{{\tau }^{\ast }\wedge {\tau }_1^w})]\\ & \le {\mathbb{E}}^0[{\mathbb{1}}_{\{{\tau }_0^w>{\tau }_1^w\}}{e}^{-r{\tau }_1^w}u(X_{{\tau }_1^w},1)]\\ & \le {\mathbb{E}}^0[{\mathbb{1}}_{\{{\tau }_0^w>{\tau }_1^w\}}{e}^{-r{\tau }_1^w}C(1+X_{{\tau }_1^w})]\\ & \le C(1+\frac{1}{\lambda }+x){\mathbb{P}}^0[{\tau }_0^w>{\tau }_1^w]\\ & \le C(1+x)w,\end{array}$$

where the first inequality holds because $u(x^{*},W_{\tau^{*}})=0$. This ends the proof of Lemma 4.4. □

Proof of Proposition 4.5

We show that $u(x,w) = c (x) \, w + o(w)$ for $w$ sufficiently small, where $c (x)$ is a real constant. This will prove that $\frac{\partial u}{\partial w} (x,0)< +\infty$.

We have

$$\begin{aligned} \begin{aligned}[b] u(x,w) &= \mathbb{E}^{0} \bigg[\int_{0}^{\tau_{0}^{w}} e^{-rs} (x + \sigma Z_{s}) \, ds\bigg]- \mathbb{E}^{0} \bigg[\int_{\tau_{R}}^{\tau _{0}^{w} }e^{-rs} (x + \sigma Z_{s}) \, ds\bigg] \\ &=A-B. \end{aligned} \end{aligned}$$

(A.1)

First, observe that

$$ A= \frac{x}{r} \big(1 - \ell (w)\big) + \mathbb{E}^{0} \bigg[\int_{0} ^{\tau_{0}^{w}} e^{-rs} \sigma Z_{s} \, ds \bigg], $$

(A.2)

where we set $\ell (w) =\mathbb{E}^{0} [e^{-r\tau_{0}^{w}}]$. Following [3, Chap. 2, Sect. 1.10], we introduce the function $\phi$ defined as the nonincreasing fundamental solution on ℝ of

$$\begin{aligned} \frac{\sigma^{2}\lambda^{2}}{2} \phi^{\prime \prime } + r w \phi^{ \prime } - r \phi =0, \\ \hbox{ with }\phi (0) =1, \; \; \lim_{w\to +\infty }\phi (w) = 0. \end{aligned}$$

The function $\ell$ coincides with $\phi$ on $(0,+\infty )$, and in particular $\ell$ is twice continuously differentiable over $(0, \infty )$ with $\ell^{\prime }(0+)<+\infty$. It follows that for $w$ small enough, $1 - \ell (w) = - \ell^{\prime }(0+) \, w = o(w)$.

We now study the second term on the right-hand side of (A.2). Let us define the $\mathbb{P}^{0}$-uniformly integrable martingale

$$ N_{t}=\int_{0}^{t} e^{-rs}dZ_{s}=e^{-r t}Z_{t}-r\int_{0}^{t} e^{-rs}Z _{s}\,ds, \qquad t \geq 0. $$

The optional sampling theorem gives $\mathbb{E}^{0}[N_{\tau_{0}^{w} \wedge T}]=0$. Then, letting $T \to +\infty$,

$$ \mathbb{E}^{0} \bigg[\int_{0}^{\tau_{0}^{w}} e^{-rs} Z_{s} \, ds \bigg] = -\frac{1}{r}\mathbb{E}^{0}[e^{-r\tau_{0}^{w} } Z_{\tau_{0} ^{w}}]. $$

(A.3)

Observe that $e^{-rt} W_{t} = w+\lambda \sigma N_{t}$, $t \geq 0$, is thus under $\mathbb{P}^{0}$ a uniformly integrable martingale. Using the dynamics of $(W_{t})$, we deduce that

$$ w e^{-rt} + r e^{-rt} \int_{0}^{t} W_{s} \, ds + \lambda \sigma e^{-rt} Z_{t}, \qquad t \geq 0, $$

is a $\mathbb{P}^{0}$-uniformly integrable martingale. Thus the optional sampling theorem yields

$$\begin{aligned} w &=\mathbb{E}^{0}[e^{-r(T\wedge \tau_{0}^{w})} W_{T\wedge \tau_{0} ^{w}}] \\ &= w \mathbb{E}^{0} [e^{-r(T\wedge \tau_{0}^{w})}] + r \mathbb{E} ^{0} \bigg[e^{-r(T\wedge \tau_{0}^{w})} \int_{0}^{T\wedge \tau_{0} ^{w}} W_{s} \, ds\bigg] \\ & \phantom{=:}+ \lambda \sigma \mathbb{E}^{0}[ e^{-r(T\wedge \tau_{0} ^{w})} Z_{T\wedge \tau_{0}^{w}}] . \end{aligned}$$

(A.4)

Letting $T$ tend to $\infty$, we obtain

$$ w=w \mathbb{E}^{0} (e^{-r\tau_{0}^{w}}) + r \mathbb{E}^{0} \bigg[e ^{-r \tau_{0}^{w}} \int_{0}^{ \tau_{0}^{w}} W_{s} \, ds\bigg] + \lambda \sigma \mathbb{E}^{0}[ e^{-r \tau_{0}^{w}} Z_{\tau_{0}^{w}}]. $$

Using (A.3) and (A.4), one gets

$$\begin{array}{rcl}{\mathbb{E}}^0\left[{\int }_0^{{\tau }_0^w}{e}^{-rs}{Z}_{s}ds\right]& =& \frac{1}{\lambda \sigma r}w({\mathbb{E}}^0[e^{-r{\tau }_0^w}]-1)+\frac{1}{\lambda \sigma }{\mathbb{E}}^0\left[e^{-r{\tau }_0^w}{\int }_0^{{\tau }_0^w}{W}_{s}ds\right]\\ & =& \frac{1}{\lambda \sigma }\left(\frac{1}{r}w\right(\ell (w)-1)+{\mathbb{E}}^0\left[{\int }_0^{\mathrm{\infty }}{e}^{-r{\tau }_0^w}{\mathbb{1}}_{\{s\le {\tau }_0^w\}}{W}_{s}ds\right])\\ & =& \frac{1}{\lambda \sigma }\left(\frac{1}{r}w\right(\ell (w)-1)\\ & & \phantom{=:(}+{\mathbb{E}}^0[{\int }_0^{\mathrm{\infty }}{\mathbb{E}}^0[e^{-r{\tau }_0^w}|{\mathcal{F}}_s]{\mathbb{1}}_{\{s\le {\tau }_0^w\}}{W}_{s}ds\left]\right)\\ & =& \frac{1}{\lambda \sigma }\left(\frac{1}{r}w\right(\ell (w)-1)+{\mathbb{E}}^0[{\int }_0^{{\tau }_0^w}{e}^{-rs}\ell (W_s)W_{s}ds\left]\right),\end{array}$$

where the last equality follows from the strong Markov property. Now, it remains to prove that

$$ g(w) := \mathbb{E}^{0}\bigg[\int_{0}^{\tau_{0}^{w}} e^{-rs} \ell (W _{s}) W_{s} \, ds \bigg] = cw + o(w) $$

for $w$ small enough. First, we show that the function $w\mapsto w \ell (w)$ is bounded. Let us consider the real function $k(w) = (\int _{w}^{\infty }e^{-\frac{r}{\sigma^{2}\lambda^{2}} t^{2}} \, dt) / ( \int_{0}^{\infty }e^{-\frac{r}{\sigma^{2}\lambda^{2}} t^{2}} \, dt)$ which is the smooth solution to

$$\begin{aligned} \frac{\sigma^{2}\lambda^{2}}{2} k{''} + r w k' &=0, \\ k(0) =1, \; \lim_{w\to +\infty }k(w) &=0. \end{aligned}$$

Note that the function $\theta =k-\ell$ is twice continuously differentiable, bounded over $(0, \infty )$ and satisfies $\theta (0) = \lim_{w\to +\infty }\theta (w) = 0$ together with

$$ \frac{\sigma^{2}\lambda^{2}}{2} \theta {''} + r w \theta ' = -r\ell \leq 0. $$

Then the process $(\theta (W_{t}))_{t \geq 0}$ is a bounded $\mathbb{P}^{0}$-supermartingale and thus for every $T>0$,

$$ \mathbb{E}^{0} [\theta (W_{\tau_{0}^{w}\wedge T})] \le \theta (w). $$

Letting $T \to +\infty$, we conclude that $\theta (w) \ge 0$ because $\theta$ is bounded with $\theta (0)=0$. It follows that $\ell (w) w \leq k(w) w$ over $[0, \infty )$. We observe that $\lim_{w \to \infty } w k(w) =0$ and thus $\lim_{w \to \infty } w \ell (w) =0$. Therefore, the function $w \mapsto w \ell (w)$ is bounded on $[0, \infty )$ and the function $g$ is well defined.

Assume for a while the existence of a bounded $C^{2}$ solution $f$ on ℝ to the differential equation

$$ \frac{\sigma^{2} \lambda^{2}}{2} f{''} + r w f{'} - r f + w \ell (w) =0, \qquad f(0) =0. $$

(A.5)

Itô’s formula yields for every $T>0$ that

$$ \mathbb{E}^{0}\big[ e^{-r(T\wedge \tau_{0}^{w})}f(W_{T\wedge \tau_{0} ^{w}}) \big]=f(w)-\mathbb{E}^{0} \bigg[\int_{0}^{T\wedge \tau_{0}^{w}} e^{-rs} \ell (W_{s}) W_{s} \, ds\bigg]. $$

(A.6)

Observe that because $f(0)=0$,

$${\mathbb{E}}^0[e^{-r(T\wedge {\tau }_0^w)}f(W_{T\wedge {\tau }_0^w})]={\mathbb{E}}^0[e^{-rT}f(W_T){\mathbb{1}}_{\{T\le {\tau }_0^w\}}]\le {\parallel f\parallel }_{\mathrm{\infty }}{e}^{-rT}.$$

The monotone convergence theorem gives

$$ \lim_{T\to +\infty }\mathbb{E}^{0} \bigg[\int_{0}^{T\wedge \tau_{0} ^{w}} e^{-rs} \ell (W_{s}) W_{s} \, ds\bigg]=g(w). $$

Letting $T$ tend to $+\infty$ in (A.6), we have that $g$ coincides with $f$ on $(0,+\infty )$. Therefore, $g$ is a bounded, twice continuously differentiable function over $(0, \infty )$ and thus $g(w) = g' (0+) w + o(w)$.

The existence of a bounded solution of (A.5) comes from the general form of solutions given by the method of variation of constants,

$$ f(x)=\alpha x+ \beta \ell (x)+\ell (x)\int_{0}^{x} \frac{u^{2}\ell (u)}{J(u)}\,du-x\int_{x}^{\infty } \frac{u\ell^{2}(u)}{J(u)}\,du, $$

where $J(x)=\frac{2}{\sigma^{2}}(\ell (x)-x\ell^{\prime }(x))$ is the Wronskian. Because we have $\ell \le k$ and $J(x)=Ce^{-\frac{r}{ \sigma^{2}\lambda^{2}}x^{2}}$ (see [3, Chap. 2, Sect. 1.11]), the two integrals $\int_{0}^{x} \frac{u^{2}\ell (u)}{J(u)}\,du$ and $\int_{0}^{x} \frac{u\ell^{2}(u)}{J(u)}\,du$ converge for $x \to \infty$. Then, it suffices to choose $\beta =-\int_{0}^{\infty }\frac{u ^{2}\ell (u)}{J(u)}\,du$ and $\alpha =0$ to conclude.

Summing up our results, we have obtained that

$$ \tilde{u}(x,w) := \mathbb{E}^{0}\bigg[ \int_{0}^{\tau_{0}^{w}} e^{-rs} (x + \sigma Z_{s}) \,ds\bigg] = c(x) w + w\eta (w), \qquad \lim_{w \to 0} \eta (w)=0. $$

Moreover, observe that $\eta$ is a continuous bounded function on $(0,+\infty )$.

We now turn to the second term of (A.1). We show that for $w$ sufficiently small,

$$ B= \mathbb{E}^{0}\bigg[ \int_{\tau_{R}}^{\tau_{0}^{w}} e^{-rs} (x + \sigma Z_{s}) \,ds\bigg] = Cw+o(w), $$

where $C$ is a constant that may depend on $x$. The strong Markov property yields

$$\begin{array}{rl}B& ={\mathbb{E}}^0[{\mathbb{1}}_{\{{\tau }^{\ast }<{\tau }_0^w\}}{\int }_{{\tau }^{\ast }}^{{\tau }_0^w}{e}^{-rs}(x+\sigma Z_s)ds]\\ & ={\mathbb{E}}^0[{\mathbb{1}}_{\{{\tau }^{\ast }<{\tau }_0^w\}}{e}^{-r{\tau }^{\ast }}\tilde{u}(x^{\ast },W_{{\tau }^{\ast }}^w)]\\ & ={\mathbb{E}}^0\left[{\mathbb{1}}_{\{{\tau }^{\ast }<{\tau }_0^w\}}{e}^{-r{\tau }^{\ast }}\right(C{W}_{{\tau }^{\ast }}+W_{{\tau }^{\ast }}^w\eta (W_{{\tau }^{\ast }}^w)\left)\right]\\ & =C{\mathbb{E}}^0\left[{\mathbb{1}}_{\{{\tau }^{\ast }<{\tau }_0^w\}}{e}^{-r{\tau }^{\ast }}{W}_{{\tau }^{\ast }}^w\right]+{\mathbb{E}}^0[{\mathbb{1}}_{\{{\tau }^{\ast }<{\tau }_0^w\}}{e}^{-r{\tau }^{\ast }}{W}_{{\tau }^{\ast }}^w\eta (W_{{\tau }^{\ast }}^w)]\\ & =Cw+{\mathbb{E}}^0[{\mathbb{1}}_{\{{\tau }^{\ast }<{\tau }_0^w\}}{e}^{-r{\tau }^{\ast }}{W}_{{\tau }^{\ast }}^w\eta (W_{{\tau }^{\ast }}^w)],\end{array}$$

where the last equality used again the optional sampling theorem with the martingale $(e^{-rt} W_{t})_{t \geq 0}$ because $\tau^{*}$ is almost surely finite. We end the proof by showing that

$${\mathbb{E}}^0[{\mathbb{1}}_{\{{\tau }^{\ast }<{\tau }_0^w\}}{e}^{-r{\tau }^{\ast }}{W}_{{\tau }^{\ast }}^w\eta (W_{{\tau }^{\ast }}^w)]=o(w)$$

for sufficiently small $w$. For every $\varepsilon >0$, there is some $\delta >0$ such that $\vert \eta (w) \vert \le \varepsilon$ for all $w < \delta$. Now, fix $\varepsilon >0$, $w<\delta$ and introduce the stopping time

$$ \tau^{w}_{\delta }=\inf \{ t \ge 0: W^{w}_{t}=\delta \}. $$

We have

$$\begin{array}{rl}\left|{\mathbb{E}}^0\right[{\mathbb{1}}_{\{{\tau }^{\ast }<{\tau }_0^w\}}{e}^{-r{\tau }^{\ast }}{W}_{{\tau }^{\ast }}^w\eta (W_{{\tau }^{\ast }}^w)\left]\right|& =\left|{\mathbb{E}}^0\right[e^{-r{\tau }^{\ast }}{W}_{{\tau }^{\ast }}^w\eta (W_{{\tau }^{\ast }}^w){\mathbb{1}}_{\{{\tau }^{\ast }<{\tau }_0^w\}}{\mathbb{1}}_{\{{\tau }^{\ast }\le {\tau }_{\delta }^w\}}]\\ & \phantom{=|}+{\mathbb{E}}^0[e^{-r{\tau }^{\ast }}{W}_{{\tau }^{\ast }}^w\eta (W_{{\tau }^{\ast }}^w){\mathbb{1}}_{\{{\tau }^{\ast }<{\tau }_0^w\}}{\mathbb{1}}_{\{{\tau }^{\ast }>{\tau }_{\delta }^w\}}]|\\ & \le \epsilon {\mathbb{E}}^0\left[e^{-r{\tau }^{\ast }}{W}_{{\tau }^{\ast }}^w{\mathbb{1}}_{\{{\tau }^{\ast }<{\tau }_0^w\}}\right]\\ & \phantom{=:}+{\parallel \eta \parallel }_{\mathrm{\infty }}{\mathbb{E}}^0\left[e^{-r{\tau }^{\ast }}{W}_{{\tau }^{\ast }}^w{\mathbb{1}}_{\{{\tau }^{\ast }<{\tau }_0^w\}}{\mathbb{1}}_{\{{\tau }^{\ast }>{\tau }_{\delta }^w\}}\right]\\ & \le \epsilon w+{\parallel \eta \parallel }_{\mathrm{\infty }}{\mathbb{E}}^0\left[e^{-r{\tau }_{\delta }^w}{W}_{{\tau }_{\delta }^w}^w{\mathbb{1}}_{\{{\tau }_{\delta }^w<{\tau }_0^w\}}\right].\end{array}$$

Proceeding analogously as in the proof of Lemma 4.4, we define on the set $\{\tau_{\delta }^{w} < \tau_{0}^{w}\}$ the diffusion $(\hat{W}_{t})_{t \le \tau_{0}}$ absorbed at $\delta$ by using Doob’s $h$-transform and obtain

$${\mathbb{E}}^0\left[e^{-r{\tau }_{\delta }^w}{W}_{{\tau }_{\delta }^w}{\mathbb{1}}_{\{{\tau }_{\delta }^w<{\tau }_0^w\}}\right]={\mathbb{P}}^0[{\tau }_{\delta }^w<{\tau }_0^w]{\mathbb{E}}^0\left[e^{-r{\hat{\tau }}_{\delta }^w}{\hat{W}}_{{\hat{\tau }}_{\delta }^w}\right].$$

Now, both expressions $\mathbb{P}^{0}[ \tau_{\delta }^{w} < \tau_{0} ^{w}]$ and $\mathbb{E}^{0} [ e^{-r\hat{\tau }_{\delta }^{w}} \hat{W} _{\hat{\tau }_{\delta }^{w}}]= \delta \mathbb{E}^{0} [ e^{-r \hat{\tau }_{\delta }^{w}} ]$ converge to zero when $w \to 0$. Moreover, as in Lemma 4.4, $\mathbb{P}^{0}[ \tau_{\delta }^{w} < \tau_{0}^{w}]=Cw+o(w)$. This ends the proof that $\frac{\partial u }{\partial w}(x,0)$ exists and is finite. Furthermore, we observe that the function $x \mapsto \frac{\partial u }{\partial w}(x,0)$ is nondecreasing because $x \mapsto u(x,w)$ is nondecreasing.

With these preparations, we are now ready to prove (i)–(iii).

(i) We have for $\varepsilon >0$ that

$$ u(x,w +\varepsilon )-u(x,w)=\mathbb{E}^{0}\bigg[\int_{0}^{\tau_{R} ^{\varepsilon }}e^{-rs}X_{s}\,ds\bigg]-\mathbb{E}^{0}\bigg[\int_{0} ^{\tau_{R}}e^{-rs}X_{s}\,ds\bigg], $$

where $\tau_{R}^{\varepsilon }=\inf \{t \ge 0:(x + \sigma Z_{t}, w + \varepsilon + \int_{0}^{t} r W_{s} \, ds + \lambda \sigma Z_{t}) \notin R\} \ge \tau_{R}$. The strong Markov property gives for the first term

$$ \mathbb{E}^{0}\bigg[\int_{0}^{\tau_{R}^{\varepsilon }}e^{-rs}X_{s}\,ds \bigg]=\mathbb{E}^{0}\bigg[\int_{0}^{\tau_{R}}e^{-rs}X_{s}\,ds\bigg]+ \mathbb{E}^{0} \big[e^{-r(\tau^{*}\wedge \tau^{w}_{0})}u\big(X_{\tau ^{*}\wedge \tau^{w}_{0}},W^{w+\varepsilon }_{\tau^{*}\wedge \tau^{w} _{0}}\big)\big]. $$

Using $u(x^{*},w)=0$ for all $w>0$, we get

$$\frac{1}{\epsilon }(u(x,w+\epsilon )-u(x,w))=\frac{1}{\epsilon }{\mathbb{E}}^0\left[e^{-r{\tau }_0^w}u\right(X_{{\tau }_0^w},W_{{\tau }_0^w}^{w+\epsilon }\left){\mathbb{1}}_{\{{\tau }^{\ast }\ge {\tau }_0^w\}}\right].$$

Now, observe that $W^{w+\varepsilon }_{\tau_{0}}=\varepsilon e^{r \tau _{0}^{w}}$ and thus

$$\frac{1}{\epsilon }(u(x,w+\epsilon )-u(x,w))={\mathbb{E}}^0\left[\frac{u(X_{{\tau }_0^w},\epsilon e^{r{\tau }_0^w})}{\epsilon e^{r{\tau }_0^w}}{\mathbb{1}}_{\{{\tau }^{\ast }\ge {\tau }_0^w\}}\right]\ge 0.$$

(A.7)

Because we have just proved that $\frac{\partial u}{\partial w}(x,0)$ exists for every $x>0$, we know that the random variables $\frac{u(X_{{\tau }_0^w},\epsilon e^{r{\tau }_0^w})}{\epsilon e^{r{\tau }_0^w}}{\mathbb{1}}_{\{{\tau }^{\ast }\ge {\tau }_0^w\}}$ converge to $\frac{\partial u}{\partial w}(X_{{\tau }_0^w},0){\mathbb{1}}_{\{{\tau }^{\ast }\ge {\tau }_0^w\}}$ almost surely when $\varepsilon$ tends to zero. Moreover, up to a constant, they are bounded above by $1+X_{\tau^{w} _{0}}$ by Lemma 4.4. Now observe that

$$ \lambda X_{\tau^{w}_{0}} \le \lambda x -w $$

on the set $\{\tau^{*}\ge \tau^{w}_{0} \}$ and thus $1+X_{\tau^{w} _{0}}$ is bounded on this set. Then the dominated convergence theorem yields assertion (i) by letting $\varepsilon$ tend to zero in (A.7).

(ii) First, we prove that $X^{x}_{\tau_{0}^{w_{0}}} \ge X^{x}_{\tau _{0}^{w_{1}}}$ for any $w_{0} \le w_{1}$ on the set $\{ \tau_{0}^{w _{1}}<+\infty \}$. Note that $\tau_{0}^{w_{0}} \le \tau_{0}^{w_{1}}$ almost surely for $w_{0} \le w_{1}$. We integrate $\lambda dX_{t} = dW _{t}-rW_{t}\,dt$ on the interval $(\tau_{0}^{w_{0}},\tau_{0}^{w_{1}})$ on the set $\{ \tau_{0}^{w_{1}}<+\infty \}$. We obtain

$$ \lambda ( X^{x}_{\tau_{0}^{w_{1}}}-X^{x}_{\tau_{0}^{w_{0}}})= -W^{w _{1}}_{\tau_{0}^{w_{0}}}-r\int_{\tau_{0}^{w_{0}}}^{\tau_{0}^{w_{1}}}W _{s}^{w_{1}}\,ds \le 0. $$

According to assertion (i),

$$\begin{array}{rcl}\frac{\partial u}{\partial w}(x,w_0)& =& {\mathbb{E}}^0\left[{\mathbb{1}}_{\{{\tau }_0^{w_0}\le {\tau }^{\ast }\}}\frac{\partial u}{\partial w}\right(X_{{\tau }_0^{w_0}}^x,0\left)\right]\\ & \ge & {\mathbb{E}}^0\left[{\mathbb{1}}_{\{{\tau }_0^{w_1}\le {\tau }^{\ast }\}}\frac{\partial u}{\partial w}\right(X_{{\tau }_0^{w_0}}^x,0\left)\right]\\ & \ge & {\mathbb{E}}^0\left[{\mathbb{1}}_{\{{\tau }_0^{w_1}\le {\tau }^{\ast }\}}\frac{\partial u}{\partial w}\right(X_{{\tau }_0^{w_1}}^x,0\left)\right]\\ & =& \frac{\partial u}{\partial w}(x,w_1),\end{array}$$

where the last inequality comes from the fact that $x \mapsto \frac{ \partial u}{\partial w}(x,0)$ is nondecreasing. Thus, the function $w \mapsto \frac{\partial u}{\partial w}$ is a decreasing function. Because we know that $u$ is twice continuously differentiable over $R$, we get assertion (ii).

(iii) Let us consider $f$ defined as

$$ f(x)=\frac{\partial u}{\partial w}\big(x,\lambda (x-c)\big) \qquad \hbox{for } x\ge x^{*}. $$

To prove assertion (iii), we show that $f$ is nonincreasing for any $c$ such that $(x,\lambda (x-c))$ is in $R$. Take $x_{0} \le x_{1}$ and $w_{i}=\lambda (x_{i}-c)$ for $i=0,1$. From assertion (ii), we have

$$f(x_0)={\mathbb{E}}^0\left[{\mathbb{1}}_{\{{\tau }_0^{w_0}\le {\tau }^{\ast ,x_0}\}}\frac{\partial u}{\partial w}\right(X_{{\tau }_0^{w_0}}^{x_0},0\left)\right].$$

We show that ${\mathbb{1}}_{\{{\tau }_0^{w_0}\le {\tau }^{\ast ,x_0}\}}\ge {\mathbb{1}}_{\{{\tau }_0^{w_1}\le {\tau }^{\ast ,x_1}\}}$ or, equivalently, that

$$ \{ \tau_{0}^{w_{0}} > \tau^{*, x_{0}} \} \subseteq \{ \tau_{0}^{w_{1}} > \tau^{*, x_{1}} \}. $$

On the set $\{ \tau_{0}^{w_{0}} > \tau^{*,x_{0}} \}$, we have

$$\begin{aligned} X_{\tau^{*, x_{0}} }^{x_{1}} &=x^{*}+x_{1}-x_{0}, \\ W_{\tau^{*, x_{0}} }^{w_{1}} &=W_{\tau^{*, x_{0}}}^{w_{0}}+(w_{1}-w _{0})e^{r\tau^{*, x_{0}}}. \end{aligned}$$

Therefore,

$$\begin{aligned} W_{\tau^{*, x_{0}} }^{w_{1}}-\lambda X_{\tau^{*, x_{0}} }^{x_{1}} =&W _{\tau^{*, x_{0}} }^{w_{0}}+(w_{1}-w_{0})e^{r\tau^{*, x_{0}} }-\lambda (x^{*}+x_{1}-x_{0}) \\ \ge &\lambda (x_{1}-x_{0})(e^{r\tau^{*, x_{0}} }-1)-\lambda x^{*} \\ \ge &-\lambda x^{*}, \end{aligned}$$

and thus for all $t \ge \tau^{*, x_{0}}$, we have $W_{t}^{w_{1}} \ge \lambda (X_{t}^{x_{1}}-x^{*})$. This implies $\tau_{0}^{w_{1}} > \tau^{*, x_{1}}$. Consequently,

$$f(x_0)\ge {\mathbb{E}}^0\left[{\mathbb{1}}_{\{{\tau }_0^{w_1}\le {\tau }^{\ast ,x_1}\}}\frac{\partial u}{\partial w}\right(X_{{\tau }_0^{w_0}}^{x_0},0\left)\right].$$

Proceeding as previously, on the set $\{\tau_{0}^{w_{1}} < +\infty \}$, we have

$$ \lambda \big(X^{x_{0}}_{\tau_{0}^{w_{0}}} - X^{x_{1}}_{\tau_{0}^{w _{1}}}\big)= r\bigg(\int_{0}^{\tau_{0}^{w_{1}}}W^{w_{1}}_{s}\,ds-\int _{0}^{\tau_{0}^{w_{0}}}W^{w_{0}}_{s}\,ds \bigg)\ge 0. $$

Thus,

$$f(x_0)\ge {\mathbb{E}}^0\left[{\mathbb{1}}_{\{{\tau }_0^{w_1}\le {\tau }^{\ast ,x_1}\}}\frac{\partial u}{\partial w}\right(X_{{\tau }_0^{w_1}}^{x_1},0\left)\right]=f(x_1).$$

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