Appendix A
1.1 A.1 Proof of Lemma 3.4
Note that for any \(\nu >0\),
$$\begin{aligned} u\big(I(\nu )\big)-\nu I(\nu )=\tilde{u}(\nu )\qquad \text{and}\qquad \tilde{u}'(\nu )=-I(\nu ), \end{aligned}$$
and thus we have
$$\begin{aligned} u\big(I(\xi )\big)=u\big(I(\nu )\big)-\nu I(\nu )+\xi I(\xi )+\int _{ \xi}^{\nu }I(\zeta )d\zeta . \end{aligned}$$
It follows that
$$\begin{aligned} &\int _{0}^{\nu }\xi ^{-n_{2}-1}\big| u\big(I(\xi )\big)\big| d\xi + \int _{\nu}^{\infty }\xi ^{-n_{1}-1}\big|u\big(I(\xi )\big)\big|d\xi \\ &\le -\frac{1}{n_{2}}\nu ^{-n_{2}}\big|u\big(I(\nu )\big)-\nu I(\nu ) \big|+\int _{0}^{\nu }\xi ^{-n_{2}}I(\xi )d\xi +\int _{0}^{\nu }\int _{ \xi}^{\nu }\xi ^{-n_{2}-1}I(\zeta )d\zeta d\xi \\ & \phantom{=:} +\frac{1}{n_{1}}\nu ^{-n_{1}}\big|u\big(I(\nu )\big)-\nu I(\nu )\big|+ \int _{\nu}^{\infty }\xi ^{-n_{1}}I(\xi )d\xi +\int _{\nu}^{\infty } \int _{\nu}^{\xi }\xi ^{-n_{1}-1}I(\zeta )d\zeta d\xi \\ &=-\frac{1}{n_{2}}\nu ^{-n_{2}}\big|u\big(I(\nu )\big)-\nu I(\nu ) \big|+\frac{1}{n_{1}}\nu ^{-n_{1}}\big|u\big(I(\nu )\big)-\nu I(\nu ) \big| \\ & \phantom{=:} +\bigg(1+\frac{1}{n_{1}}\bigg)\int _{\nu}^{\infty }\xi ^{-n_{1}}I( \xi )d\xi +\bigg(1-\frac{1}{n_{2}}\bigg)\int _{0}^{\nu }\xi ^{-n_{2}}I( \xi )d\xi < \infty , \end{aligned}$$
(A.1)
where the inequality in (A.1) comes from (2.6) and Assumption 2.3 and we have used Fubini’s theorem in the last equality. Moreover, from (2.6) and Assumption 2.3, it is easy to check that
$$ \int _{0}^{\nu }\xi ^{-n_{2}}I(\xi )d\xi +\int _{\nu}^{\infty }\xi ^{-n_{1}}I( \xi )d\xi < \infty . $$
Therefore, it follows from \(|\tilde{u}(\xi )|\le |u (I(\xi ) )|+I(\xi )\) that
$$ \int _{0}^{\nu }\xi ^{-n_{2}-1}|\tilde{u}(\xi )|d\xi +\int _{\nu}^{ \infty }\xi ^{-n_{1}-1}|\tilde{u}(\xi )|d\xi < \infty . $$
□
1.2 A.2 Proof of Lemma 3.5
Note that
$$\begin{aligned} \frac{d}{d\nu}\bigg((1+\nu )\tilde{u}\Big(\frac{y}{1+\nu}\Big)\bigg)= \tilde{u}\bigg(\frac{y}{1+\nu}\bigg)+\frac{y}{1+\nu}I\bigg( \frac{y}{1+\nu}\bigg)=u\bigg(I\Big(\frac{y}{1+\nu}\Big)\bigg). \end{aligned}$$
By integrating both sides and rearranging terms, we have
$$\begin{aligned} (1+{\mathcal{Z}}_{t})\tilde{u}\bigg(\frac{y}{1+{\mathcal{Z}}_{t}}\bigg)=\int _{z}^{{ \mathcal{Z}}_{t}}u\bigg(I\Big(\frac{y}{1+\nu}\Big)\bigg)d\nu +(1+z) \tilde{u}\bigg(\frac{y}{1+z}\bigg). \end{aligned}$$
It follows that
$$\begin{aligned} &\mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t} \Big|(1+{\mathcal{Z}}_{t}) \tilde{u}\Big(\frac{{\mathcal{Y}}_{t}^{y}}{1+{\mathcal{Z}}_{t}}\Big)\Big|dt \bigg| {\mathcal{Z}}_{0}=z\bigg] \\ &= \mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}\bigg|\int _{z}^{{ \mathcal{Z}}_{t}}u\bigg(I\Big(\frac{{\mathcal{Y}}_{t}^{y}}{1+\nu}\Big)\bigg)d \nu +(1+z)\tilde{u}\bigg(\frac{{\mathcal{Y}}_{t}^{y}}{1+z}\bigg)\bigg|dt \bigg] \\ &\le \mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}\bigg|\int _{z}^{{ \mathcal{Z}}_{t}}u\bigg(I\Big(\frac{{\mathcal{Y}}_{t}^{y}}{1+\nu}\Big)\bigg)d \nu \bigg|dt \bigg] \\ & \phantom{=:} +\mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t} \bigg|(1+z)\tilde{u} \bigg(\frac{{\mathcal{Y}}_{t}^{y}}{1+z}\bigg)\bigg|dt \bigg] \\ &\le \mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}\int _{z}^{{\mathcal{Z}}_{t}} \bigg|u\bigg(I\Big(\frac{{\mathcal{Y}}_{t}^{y}}{1+\nu}\Big)\bigg)\bigg| d \nu dt \bigg]+ (1+z)\Gamma _{|\tilde{u}|}\bigg(\frac{y}{1+z}\bigg) \\ &< \infty , \end{aligned}$$
where the last inequality comes from the integrability condition (3.6), Lemma 3.4 and Proposition 3.1. Consequently, for any \({\mathcal{Z}}\in \Pi (0)\),
$$\begin{aligned} |J(y)|&\le \mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}\bigg|(1+{ \mathcal{Z}}_{t})\tilde{u}\bigg(\frac{{\mathcal{Y}}_{t}^{y}}{1+{\mathcal{Z}}_{t}} \bigg)\bigg|dt\bigg] + \frac{\epsilon }{r} y + |P| \mathbb{E}\bigg[ \int _{0}^{\infty }e^{-\beta t}d{\mathcal{Z}}_{t}\bigg] \\ &< \infty \end{aligned}$$
under the integrability condition (3.7). □
1.3 A.3 Proof of Proposition 3.6
For given \(y>0\), any admissible strategy \((c,\pi )\in{\mathcal{A}}(x)\) and any shadow price process \({\mathcal{Z}}\in \Pi (0)\), we have
$$\begin{aligned} (1+{\mathcal{Z}}_{t})u(c_{t})-{\mathcal{Y}}_{t}^{y} c_{t} \le (1+{\mathcal{Z}}_{t}) \tilde{u}\bigg(\frac{{\mathcal{Y}}_{t}^{y}}{1+{\mathcal{Z}}_{t}}\bigg) \end{aligned}$$
by the definition of \(\tilde{u}(\, \cdot \,)\). This implies that
$$\begin{aligned} (1+{\mathcal{Z}}_{t})\big(u(c_{t})\big)^{\pm}\le (1+{\mathcal{Z}}_{t} )\big| \big(u(c_{t})\big)\big| \le {\mathcal{Y}}_{t}^{y} c_{t} + (1+{\mathcal{Z}}_{t}) \bigg|\tilde{u}\bigg(\frac{{\mathcal{Y}}_{t}^{y}}{1+{\mathcal{Z}}_{t}}\bigg) \bigg|, \end{aligned}$$
where \((u(c_{t}))^{+}\) and \((u(c_{t}))^{-}\) are the positive and negative parts of \(u(c_{t})\), respectively. It follows from Lemma 3.5 that
$$\begin{aligned} &\mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}(1+{\mathcal{Z}}_{t})|u(c_{t})|dt \bigg] \\ &\le \mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}{\mathcal{Y}}_{t}^{y} c_{t}dt \bigg]+\mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t} (1+{\mathcal{Z}}_{t}) \bigg|\tilde{u}\bigg(\frac{{\mathcal{Y}}_{t}^{y}}{1+{\mathcal{Z}}_{t}}\bigg) \Big|dt \bigg] \\ &\le y\mathbb{E}\bigg[\int _{0}^{\infty }{\mathcal{H}}_{t} c_{t}dt\bigg]+ \mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t} (1+{\mathcal{Z}}_{t}) \bigg|\tilde{u}\bigg(\frac{{\mathcal{Y}}_{t}^{y}}{1+{\mathcal{Z}}_{t}}\bigg) \bigg|dt \bigg]< \infty . \end{aligned}$$
Hence
$$\begin{aligned} \mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t} |u(c_{t})|dt\bigg] \leq \mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}(1+{\mathcal{Z}}_{t})|u(c_{t})|dt \bigg]< \infty \end{aligned}$$
(A.2)
and
$$\begin{aligned} \mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}(1+{\mathcal{Z}}_{t})\big(u(c_{t}) \big)^{\pm}dt\bigg]< \infty . \end{aligned}$$
Then integration by parts yields that for a fixed \(T>0\),
$$\begin{aligned} &\mathbb{E}\bigg[\int _{0}^{T} e^{-\beta t}(1+{\mathcal{Z}}_{t})\big(u(c_{t}) \big)^{+}dt\bigg] \\ &=-\mathbb{E}\bigg[(1+{\mathcal{Z}}_{t})\int _{t}^{T}e^{-\beta s}\big(u(c_{s}) \big)^{+}ds\bigg]\bigg|_{t=0}^{T} \\ & \phantom{=:} +\mathbb{E}\bigg[\int _{0}^{T} e^{-\beta t}\mathbb{E}_{t}\Big[\int _{t}^{T} e^{-\beta (s-t)}\big(u(c_{s})\big)^{+}ds\Big]d{\mathcal{Z}}_{t}\bigg] \\ &=\mathbb{E}\bigg[\int _{0}^{T}e^{-\beta s}\big(u(c_{s})\big)^{+}ds \bigg]+\mathbb{E}\bigg[\int _{0}^{T} e^{-\beta t}\mathbb{E}_{t}\Big[ \int _{t}^{T} e^{-\beta (s-t)}\big(u(c_{s})\big)^{+}ds\Big]d{\mathcal{Z}}_{t} \bigg]. \end{aligned}$$
Letting \(T\to \infty \), the monotone convergence theorem implies that
$$\begin{aligned} &\mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}\big(u(c_{t})\big)^{+}dt \bigg]+\mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}\mathbb{E}_{t} \Big[\int _{t}^{\infty }e^{-\beta (s-t)}\big(u(c_{s})\big)^{+}ds\Big]d{ \mathcal{Z}}_{t}\bigg] \\ &=\mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}(1+{\mathcal{Z}}_{t}) \big(u(c_{t})\big)^{+}dt\bigg]< \infty . \end{aligned}$$
Thus we obtain
$$\begin{aligned} \mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}\mathbb{E}_{t}\Big[ \int _{t}^{\infty }e^{-\beta (s-t)}\big(u(c_{s})\big)^{+}ds\Big]d{ \mathcal{Z}}_{t}\bigg]< \infty . \end{aligned}$$
(A.3)
Note that for any \(t\ge 0\),
$$\begin{aligned} P&\le \mathbb{E}_{t}\bigg[\int _{t}^{\infty }e^{-\beta (s-t)}u(c_{s})ds \bigg] \\ &=\mathbb{E}_{t}\bigg[\int _{t}^{\infty }e^{-\beta (s-t)}\big(u(c_{s}) \big)^{+}ds\bigg]-\mathbb{E}_{t}\bigg[\int _{t}^{\infty }e^{-\beta (s-t)} \big(u(c_{s})\big)^{-}ds\bigg]. \end{aligned}$$
It follows from (A.3) that
$$\begin{aligned} &\mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}\mathbb{E}_{t}\Big[ \int _{t}^{\infty }e^{-\beta (s-t)}\big(u(c_{s})\big)^{-}ds\Big]d{ \mathcal{Z}}_{t}\bigg]+\mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}Pd{ \mathcal{Z}}_{t}\bigg] \\ &\le \mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}\mathbb{E}_{t} \Big[\int _{t}^{\infty }e^{-\beta (s-t)}\big(u(c_{s})\big)^{+}ds\Big]d{ \mathcal{Z}}_{t}\bigg]< \infty . \end{aligned}$$
Since \(\mathbb{E} [\int _{0}^{\infty }e^{-\beta t}Pd{\mathcal{Z}}_{t} ]<\infty \) by (3.7), we have
$$\begin{aligned} \mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}\mathbb{E}_{t}\Big[ \int _{t}^{\infty }e^{-\beta (s-t)}\big(u(c_{s})\big)^{-}ds\Big]d{ \mathcal{Z}}_{t}\bigg]< \infty . \end{aligned}$$
(A.4)
From (A.3) and (A.4), we get
$$\begin{aligned} \mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}\mathbb{E}_{t}\Big[ \int _{t}^{\infty }e^{-\beta (s-t)}|u(c_{s})|ds\Big]d{\mathcal{Z}}_{t} \bigg]< \infty . \end{aligned}$$
(A.5)
Thus we deduce that for a fixed \(T>0\),
$$\begin{aligned} &\mathbb{E}\bigg[\int _{0}^{T} e^{-\beta t}(1+{\mathcal{Z}}_{t})u(c_{t})dt \bigg] \\ &=-\mathbb{E}\bigg[(1+{\mathcal{Z}}_{t})\int _{t}^{T} e^{-\beta s}u(c_{s})ds \bigg]\bigg|_{t=0}^{T} \\ & \phantom{=:} +\mathbb{E}\bigg[\int _{0}^{T} e^{-\beta t}\mathbb{E}_{t}\Big[\int _{t}^{T} e^{-\beta (s-t)}u(c_{s})ds\Big]d{\mathcal{Z}}_{t}\bigg] \\ &=\mathbb{E}\bigg[ \int _{0}^{T} e^{-\beta t}u(c_{t})dt\bigg]+ \mathbb{E}\bigg[\int _{0}^{T} e^{-\beta t}\mathbb{E}_{t}\Big[\int _{t}^{T} e^{-\beta (s-t)}u(c_{s})ds\Big]d{\mathcal{Z}}_{t}\bigg]. \end{aligned}$$
From (A.2) and (A.5), the dominated convergence theorem implies that
$$\begin{aligned} &\mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t} (1+{\mathcal{Z}}_{t})u(c_{t})dt \bigg] \\ &=\mathbb{E}\bigg[ \int _{0}^{\infty }e^{-\beta t}u(c_{t})dt\bigg]+ \mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}\mathbb{E}_{t}\Big[ \int _{t}^{\infty }e^{-\beta (s-t)}u(c_{s})ds\Big]d{\mathcal{Z}}_{t}\bigg]. \end{aligned}$$
(A.6)
Using the budget constraint (3.2), the welfare constraints (2.3), the definition of \(\tilde{u}(\, \cdot \,)\) and (A.6), we derive that
$$\begin{aligned} &\mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t} u(c_{t})dt\bigg] \\ &\le \mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t} u(c_{t})dt\bigg]+y \bigg(x-\mathbb{E}\bigg[\int _{0}^{\infty }{\mathcal{H}}_{t}(c_{t}- \epsilon )dt\bigg]\bigg) \\ & \phantom{=:} +\mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}\bigg(\mathbb{E}_{t} \Big[\int _{t}^{\infty }e^{-\beta (s-t)}u(c_{s})ds\Big]-P\bigg)d{ \mathcal{Z}}_{t}\bigg] \\ &= \mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t} (1+{\mathcal{Z}}_{t})u(c_{t})dt \bigg] \\ & \phantom{=:} +y\bigg(x-\mathbb{E}\bigg[\int _{0}^{\infty }{\mathcal{H}}_{t}(c_{t}- \epsilon )dt\bigg]\bigg) -\mathbb{E}\bigg[P\int _{0}^{\infty }e^{- \beta t}d{\mathcal{Z}}_{t}\bigg] \\ &= \mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}\bigg((1+{\mathcal{Z}}_{t}) \Big(u(c_{t})-c_{t}\frac{{\mathcal{Y}}_{t}^{y}}{1+{\mathcal{Z}}_{t}}\Big)+ { \mathcal{Y}}_{t}^{y} \epsilon \bigg)dt- P \int _{0}^{\infty }e^{-\beta t}d{ \mathcal{Z}}_{t}\bigg] \\ & \phantom{=:} +yx \\ &\le \mathfrak{L}(y,{\mathcal{Z}}) \\ &= \mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}\bigg((1+{\mathcal{Z}}_{t}) \tilde{u}\Big(\frac{{\mathcal{Y}}_{t}^{y}}{1+{\mathcal{Z}}_{t}}\Big)+{\mathcal{Y}}_{t}^{y} \epsilon \bigg)dt -P\int _{0}^{\infty }e^{-\beta t}d{\mathcal{Z}}_{t} \bigg]+yx. \end{aligned}$$
□
1.4 A.4 Proof of Lemma 3.7
Since
$$ \frac{d}{dz}\bigg((1+z)\tilde{u}\Big(\frac{y}{1+z}\Big)\bigg)= u \bigg(I\Big(\frac{y}{1+z}\Big) \bigg), $$
it is clear that \((1+z)\tilde{u}(y/(1+z))\) is strictly convex in \(z\ge 0\).
Let \(\lambda \in (0,1)\). For any \(z_{1}, z_{2} \ge 0\) and any processes \({\mathcal{Z}}^{i} \in \Pi (z_{i})\) for \(i =1,2\), define \(z_{3}:= \lambda z_{1} + (1-\lambda )z_{2}\) and \({\mathcal{Z}}^{3}:= \lambda {\mathcal{Z}}^{1} + (1-\lambda ) {\mathcal{Z}}^{2}\). Since \({\mathcal{Z}}^{1} \in \Pi (z_{1})\) and \({\mathcal{Z}}^{2}\in \Pi (z_{2})\), it is easy to check that \({\mathcal{Z}}^{3}\in \Pi (z_{3})\).
Since \((1+z)\tilde{u}(y/(1+z))\) is strictly convex in \(z\), we deduce that
$$\begin{aligned} &\mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}\bigg((1+{\mathcal{Z}}_{t}^{3}) \tilde{u}\Big(\frac{{\mathcal{Y}}_{t}^{y}}{1+{\mathcal{Z}}_{t}^{3}}\Big)+{\mathcal{Y}}_{t}^{y} \epsilon \bigg)dt -P\int _{0}^{\infty }e^{-\beta t}d{\mathcal{Z}}_{t}^{3} \bigg] \\ &\le \lambda \mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}\bigg((1+{ \mathcal{Z}}_{t}^{1})\tilde{u}\Big( \frac{{\mathcal{Y}}_{t}^{y}}{1+{\mathcal{Z}}_{t}^{1}}\Big)+{\mathcal{Y}}_{t}^{y} \epsilon \bigg)dt -P\int _{0}^{\infty }e^{-\beta t}d{\mathcal{Z}}_{t}^{1} \bigg] \\ & \phantom{=:} +(1-\lambda ) \mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}\bigg((1+{ \mathcal{Z}}_{t}^{2})\tilde{u}\Big( \frac{{\mathcal{Y}}_{t}^{y}}{1+{\mathcal{Z}}_{t}^{2}}\Big)+{\mathcal{Y}}_{t}^{y} \epsilon \bigg)dt -P\int _{0}^{\infty }e^{-\beta t}d{\mathcal{Z}}_{t}^{2} \bigg]. \end{aligned}$$
Taking the infimum on both sides yields
$$\begin{aligned} \mathfrak{J}(y,z_{3}) \le \lambda \mathfrak{J}(y,z_{1}) + (1-\lambda ) \mathfrak{J}(y,z_{2}), \end{aligned}$$
which means that \(\mathfrak{J}(y,z)\) is convex in \(z\ge 0\). □
1.5 A.5 Proof of Proposition 4.1
Because \({\mathcal{M}}(\nu )-\nu{\mathcal{M}}'(\nu )=P\) for all \(\nu \ge \bar{\nu}\) and \(({\mathcal{M}}(\nu )-\nu{\mathcal{M}}'(\nu ))'=-\nu {\mathcal{M}}''(\nu )\), it follows that
$$ {\mathcal{M}}''(\nu )=0\qquad \text{for } \nu \ge \bar{\nu}. $$
This implies that \({\mathcal{M}}'(\nu )\) is constant on \(\{\nu \ge \bar{\nu}\}\), that is,
$$ {\mathcal{M}}'(\nu )=\chi \qquad \text{for } \nu \ge \bar{\nu} $$
for some constant \(\chi \). Since we are constructing \({\mathcal{M}}(\nu )\) which is \(C^{2}\) in \(\nu \), we have
$$\begin{aligned} {\mathcal{L}}{\mathcal{M}}(\bar{\nu})+\tilde{u}(\bar{\nu})+{\bar{\nu}}\epsilon =0. \end{aligned}$$
Since \({\mathcal{M}}''(\bar{\nu})=0\) and \({\mathcal{M}}(\bar{\nu})-\bar{\nu}{\mathcal{M}}'(\bar{\nu})=P\), we have
$$\begin{aligned} \chi = \frac{1}{r\bar{\nu}}\big(\tilde{u}(\bar{\nu})-\beta P + \bar{\nu} \epsilon \big). \end{aligned}$$
This leads to
$$\begin{aligned} {\mathcal{M}}(\nu )= \nu {\mathcal{M}}'(\nu )+P=\big(\tilde{u}(\bar{\nu})- \beta P +\bar{\nu} \epsilon \big)\frac{\nu}{r\bar{\nu}}+P\qquad \text{for } \nu \ge \bar{\nu}. \end{aligned}$$
In the region \(\{0<\nu <\bar{\nu}\}\), a solution to the FBP (4.6) can be expressed as the sum of a general solution to the homogeneous equation and a particular solution, that is,
$$\begin{aligned} {\mathcal{M}}(\nu )=D_{1} \nu ^{n_{1}} + D_{2} \nu ^{n_{2}} + \frac{2}{\theta ^{2}(n_{1}-n_{2})}\bigg(&\nu ^{n_{2}}\int _{0}^{\nu} \xi ^{-n_{2}-1}\big(\tilde{u}(\xi )+\xi \epsilon \big)d\xi \\ &+\nu ^{n_{1}}\int _{\nu}^{\infty}\xi ^{-n_{1}-1}\big(\tilde{u}(\xi )+ \xi \epsilon \big)d\xi \bigg). \end{aligned}$$
From Proposition 3.1 and Lemma 3.4, we deduce that
$$\begin{aligned} &\frac{2}{\theta ^{2}(n_{1}-n_{2})}\bigg(\nu ^{n_{2}}\int _{0}^{\nu} \xi ^{-n_{2}-1}\big(\tilde{u}(\xi )+\xi \epsilon \big)d\xi +\nu ^{n_{1}} \int _{\nu}^{\infty}\xi ^{-n_{1}-1}\big(\tilde{u}(\xi )+\xi \epsilon \big)d\xi \bigg) \\ &=\frac{2}{\theta ^{2}(n_{1}-n_{2})}\bigg(\nu ^{n_{2}}\int _{0}^{\nu} \xi ^{-n_{2}-1}\tilde{u}(\xi )d\xi +\nu ^{n_{1}}\int _{\nu}^{\infty} \xi ^{-n_{1}-1}\tilde{u}(\xi )d\xi \bigg)+\frac{\epsilon }{r}\nu \end{aligned}$$
is well defined, where the above equation holds because \((n_{1}-1)(1-n_{2}) = 2 r /\theta ^{2}\). As \({\mathcal{Q}}(y,z)\) should satisfy the transversality condition \(\lim _{t\to \infty}\mathbb{E}[e^{-\beta t}{\mathcal{Q}}({\mathcal{Y}}_{t}^{y}, { \mathcal{Z}}_{t})]=0\), we set \(D_{2}=0\). Thus we can rewrite \({\mathcal{M}}(\nu )\) as
$$\begin{aligned} {\mathcal{M}}(\nu )=D_{1} \nu ^{n_{1}} +\frac{2}{\theta ^{2}(n_{1}-n_{2})} \bigg(&\nu ^{n_{2}}\int _{0}^{\nu}\xi ^{-n_{2}-1}\tilde{u}(\xi )d\xi \\ &+\nu ^{n_{1}}\int _{\nu}^{\infty}\xi ^{-n_{1}-1}\tilde{u}(\xi )d\xi \bigg)+\frac{\epsilon }{r}\nu \end{aligned}$$
for \(0<\nu <\bar{\nu}\).
From Lemma 3.4 and part (a) of Proposition 3.1, we have
$$\begin{aligned} \liminf _{y\downarrow 0} y^{-n_{2}+1}I(y) = 0. \end{aligned}$$
It follows that
$$\begin{aligned} \nu{\mathcal{M}}'(\nu )= n_{1} D_{1} \nu ^{n_{1}} + \frac{2}{\theta ^{2}(n_{1}-n_{2})}\bigg(&n_{2}\nu ^{n_{2}}\int _{0}^{ \nu}\xi ^{-n_{2}-1}\big(\tilde{u}(\xi )+\xi \epsilon \big)d\xi \\ &+n_{1}\nu ^{n_{1}}\int _{\nu}^{\infty}\xi ^{-n_{1}-1}\big(\tilde{u}( \xi )+\xi \epsilon \big)d\xi \bigg) \\ = n_{1} {D_{1}}\nu ^{n_{1}}+\frac{2}{\theta ^{2}(n_{1}-n_{2})}\bigg(& \nu ^{n_{2}}\int _{0}^{\nu}\xi ^{-n_{2}-1}\big(-\xi I(\xi )+\xi \epsilon \big)d\xi \\ &+\nu ^{n_{1}}\int _{\nu}^{\infty}\xi ^{-n_{1}-1}\big(-\xi I(\xi )+ \xi \epsilon \big)d\xi \bigg) , \end{aligned}$$
where we used integration by parts for the Riemann–Stieltjes integrals in the second equality. Because \(\tilde{u}(\xi )=u (I(\xi ) )-\xi I(\xi )\), we obtain for \(0<\nu \le \bar{\nu}\) that
$$\begin{aligned} &{\mathcal{M}}(\nu )-\nu{\mathcal{M}}'(\nu ) \\ &=(1-n_{1}) {D_{1}} \nu ^{n_{1}}+\frac{2}{\theta ^{2}(n_{1}-n_{2})} \bigg(\nu ^{n_{2}}\int _{0}^{\nu}\xi ^{-n_{2}-1}u\big(I(\xi )\big)d \xi \\ & \hphantom{:=(1-n_{1}) {D_{1}} \nu ^{n_{1}}+\frac{2}{\theta ^{2}(n_{1}-n_{2})}\bigg(} +\nu ^{n_{1}}\int _{\nu}^{\infty}\xi ^{-n_{1}-1}u\big(I(\xi )\big)d \xi \bigg). \end{aligned}$$
(A.7)
The conditions in (4.6) imply that \({\mathcal{M}}(\bar{\nu})-\bar{\nu}{\mathcal{M}}'(\bar{\nu})=P\) and \(({\mathcal{M}}-\nu{\mathcal{M}}')'(\bar{\nu})=0\), and we can deduce that
$$\begin{aligned} D_{1}&=\frac{2}{\theta ^{2}(n_{1}-1)(n_{1}-n_{2})}\int _{\bar{\nu}}^{ \infty }\xi ^{-n_{1}-1}u\big(I(\xi )\big)d\xi - \frac{-n_{2} P}{(n_{1}-1)(n_{1}-n_{2})}\bar{\nu}^{-n_{1}} \\ &={\frac{2}{\theta ^{2}(n_{1}-1)(n_{1}-n_{2})}\int _{\bar{\nu}}^{ \infty }\xi ^{-n_{1}-1}\Big(u\big(I(\xi )\big)-\beta P\Big)d\xi} \end{aligned}$$
and
$$\begin{aligned} 0&=\int _{0}^{\bar{\nu}} \xi ^{-n_{2}-1}u\big(I(\xi )\big)d\xi - \dfrac{\theta ^{2}}{2}n_{1} \bar{\nu}^{-n_{2}}P \\ &=\int _{0}^{\bar{\nu}} \xi ^{-n_{2}-1}\Big(u\big(I(\xi )\big)-\beta P \Big)d\xi =\Psi (\bar{\nu}). \end{aligned}$$
It remains to show that there is a unique free boundary \(\bar{\nu}>0\) satisfying (4.9). Lemma 3.4 implies that \(\Psi (\nu )\) is well defined for any \(\nu >0\). By Assumption 2.5, \(\lim _{\nu \to \infty}u (I(\nu ) )-\beta P < 0\) and \(u (I(0) )-\beta P >0\). Since \(u (I(\nu ) )\) is decreasing in \(\nu \), there exists a unique \(\hat{\nu}>0\) such that
$$\begin{aligned} u\big(I(\hat{\nu})\big) - \beta P &= 0, \\ u\big(I(\nu )\big)-\beta P &> 0\qquad \text{for}~\nu < {\hat{\nu}}, \\ u\big(I(\nu )\big)-\beta P &< 0\qquad \text{for}~\nu > {\hat{\nu}}. \end{aligned}$$
Then it is easy to confirm that \(\Psi (\nu )\) increases in \(\nu \) on \((0,\hat{\nu})\) and decreases on \((\hat{\nu},\infty )\). Moreover, we have that
$$ \Psi (\nu ) >0\qquad \text{for } \nu \in (0,\hat{\nu}]. $$
For a sufficiently large \(M>0\), there exists a constant \(\delta >0\) such that
$$\begin{aligned} u\big(I(\nu )\big) - \beta P < -\delta \qquad \text{for } \nu \ge M. \end{aligned}$$
Since
$$\begin{aligned} \int _{M}^{\infty }\xi ^{-n_{2}-1}\Big(u\big(I(\xi )\big)-\beta P \Big)d\xi &\le -\delta \int _{M}^{\infty }\xi ^{-n_{2}-1}d\xi \\ & =-\delta \bigg(-\frac{\xi ^{-n_{2}}}{n_{2}}\bigg) \bigg|_{\xi =M}^{ \infty }= -\infty , \end{aligned}$$
we have
$$\begin{aligned} \lim _{\nu \to \infty} \Psi (\nu ) &= \lim _{\nu \to \infty}\int _{0}^{ \nu }\xi ^{-n_{2}-1}\Big(u\big(I(\xi )\big)-\beta P\Big)d\xi \\ &=\int _{0}^{M} \xi ^{-n_{2}-1}\Big(u\big(I(\xi )\big)-\beta P\Big) d \xi \\ & \phantom{=:} +\lim _{\nu \to \infty}\int _{M}^{\nu }\xi ^{-n_{2}-1}\Big(u\big(I( \xi )\big)-\beta P\Big)d\xi =-\infty . \end{aligned}$$
Therefore we can conclude that there exists a unique \(\bar{\nu} >\hat{\nu}\) such that \(\Psi (\bar{\nu})=0\). Clearly, \(u (I(\bar{\nu}) )-\beta P <0\). □
1.6 A.6 Proof of Proposition 4.2
Since \({\mathcal{M}}(\nu ) = (\tilde{u}(\bar{\nu})-\beta P +\bar{\nu} \epsilon ) \frac{\nu}{r\bar{\nu}}+P\) for \(\nu \ge \bar{\nu}\), it is clear that \({\mathcal{M}}(\nu )-\nu {\mathcal{M}}'(\nu ) =P\) for \(\nu \ge \bar{\nu}\). Moreover, for \(\nu \ge \bar{\nu}\),
$$\begin{aligned} {\mathcal{L}}{\mathcal{M}}(\nu ) + \tilde{u}(\nu )+\nu \epsilon &=-\big( \tilde{u}(\bar{\nu})-\beta P +\bar{\nu} \epsilon \big) \frac{\nu}{\bar{\nu}}-\beta P + \tilde{u}(\nu ) + \epsilon \nu \\ &=-\big(\tilde{u}(\bar{\nu})-\beta P \big)\frac{\nu}{\bar{\nu}}- \beta P + \tilde{u}(\nu ) . \end{aligned}$$
Let us temporarily define \(\ell (\nu )\) as
$$\begin{aligned} \ell (\nu ):=-\big(\tilde{u}(\bar{\nu})-\beta P \big) \frac{\nu}{\bar{\nu}}-\beta P + \tilde{u}(\nu ). \end{aligned}$$
Then we deduce that
$$\begin{aligned} \ell '(\nu )=-\big(\tilde{u}(\bar{\nu})-\beta P \big) \frac{1}{\bar{\nu}}-I(\nu )=-\Big(u\big(I(\bar{\nu})\big)-\beta P \Big)\frac{1}{\bar{\nu}}+\big(I(\bar{\nu})-I(\nu )\big)>0 \end{aligned}$$
for \(\nu \ge \bar{\nu}\), where we have used the fact provided in (4.10) that \(u (I(\bar{\nu}) )-\beta P <0\). That is, \(\ell (\nu )\) is increasing for \(\nu \ge \bar{\nu}\) and thus
$$\begin{aligned} {\mathcal{L}}{\mathcal{M}}(\nu ) + \tilde{u}(\nu )+\nu \epsilon =\ell (\nu ) \ge \ell (\bar{\nu}) =0\qquad \text{for } \nu \ge \bar{\nu}. \end{aligned}$$
Since \({\mathcal{M}}(\nu )\) in (4.7) is the solution of the FBP (4.6), it is clear that
$$\begin{aligned} {\mathcal{L}}{\mathcal{M}}(\nu )+ \tilde{u}(\nu )+\epsilon \nu =0 \qquad \text{for } 0< \nu < \bar{\nu}. \end{aligned}$$
Thus it remains to show that \({\mathcal{M}}(\nu )-\nu {\mathcal{M}}'(\nu ) - P>0\) for \(0<\nu <\bar{\nu}\) to prove that \(M(\nu )\) in (4.7) satisfies the HJB equation (4.4). For \(0<\nu <\bar{\nu}\), it follows from (A.7), (4.7) and (4.8) that
$$\begin{aligned} &{\mathcal{M}}(\nu )-\nu {\mathcal{M}}'(\nu ) \\ &=-\frac{2}{\theta ^{2}(n_{1}-n_{2})}\int _{\bar{\nu}}^{\infty }\xi ^{-n_{1}-1} \Big(u\big(I(\xi )\big)-\beta P\Big)d\xi \\ & \phantom{=:} +\frac{2}{\theta ^{2}(n_{1}-n_{2})}\bigg(\nu ^{n_{2}}\int _{0}^{\nu} \xi ^{-n_{2}-1}u\big(I(\xi )\big)d\xi +\nu ^{n_{1}}\int _{\nu}^{ \infty}\xi ^{-n_{1}-1}u\big(I(\xi )\big)d\xi \bigg) \\ &=\frac{2}{\theta ^{2}(n_{1}-n_{2})}\bigg(\nu ^{n_{2}}\int _{0}^{\nu} \xi ^{-n_{2}-1}u\big(I(\xi )\big)d\xi +\nu ^{n_{1}}\int _{\nu}^{ \bar{\nu}}\xi ^{-n_{1}-1}u\big(I(\xi )\big)d\xi \bigg) \\ & \phantom{=:} + \frac{2}{\theta ^{2}(n_{1}-n_{2})} \int _{\bar{\nu}}^{\infty} \xi ^{-n_{1}-1} \beta P d\xi . \end{aligned}$$
(A.8)
Note that
$$\begin{aligned} P= \frac{2}{\theta ^{2}(n_{1}-n_{2})}\bigg(\nu ^{n_{2}}\int _{0}^{\nu} \xi ^{-n_{2}-1}\beta Pd\xi +\nu ^{n_{1}}\int _{\nu}^{\infty}\xi ^{-n_{1}-1} \beta Pd\xi \bigg). \end{aligned}$$
(A.9)
From (A.8) and (A.9), we have that for \(0<\nu <\bar{\nu}\),
$$\begin{aligned} {\mathcal{M}}(\nu )-\nu {\mathcal{M}}'(\nu ) - P&= \frac{2}{\theta ^{2}(n_{1}-n_{2})} \\ & \phantom{=:} \times \bigg(\nu ^{n_{2}}\int _{0}^{\nu}\xi ^{-n_{2}-1}\Big(u\big(I( \xi )\big)-\beta P\Big)d\xi \\ & \hphantom{=:\times \bigg(} +\nu ^{n_{1}}\int _{\nu}^{\bar{\nu}}\xi ^{-n_{1}-1}\Big(u\big(I(\xi ) \big)-\beta P\Big)d\xi \bigg). \end{aligned}$$
(A.10)
Note that since \({\mathcal{M}}(\bar{\nu})-\bar{\nu} {\mathcal{M}}'(\bar{\nu}) - P=0\), it is enough to show that
$$ \big({\mathcal{M}}(\nu )-\nu {\mathcal{M}}'(\nu ) - P\big)'< 0 $$
for \(0<\nu <\bar{\nu}\). To do so, let us temporarily define \(g(\nu )\) as
$$\begin{aligned} g(\nu )&:=\frac{\theta ^{2}}{2}\nu ^{-n_{2}+1}\big({\mathcal{M}}(\nu )- \nu {\mathcal{M}}'(\nu ) - P\big)' \\ & \phantom{:} =\frac{1}{(n_{1}-n_{2})}\bigg(n_{2}\int _{0}^{\nu}\xi ^{-n_{2}-1} \Big(u\big(I(\xi )\big)-\beta P\Big)d\xi \\ & \hphantom{=:\frac{1}{(n_{1}-n_{2})}\bigg[} +n_{1}\nu ^{n_{1}-n_{2}}\int _{\nu}^{\bar{\nu}}\xi ^{-n_{1}-1}\Big(u \big(I(\xi )\big)-\beta P\Big)d\xi \bigg). \end{aligned}$$
Since \(g(\nu )\) and \(({\mathcal{M}}(\nu )-\nu {\mathcal{M}}'(\nu ) - P )'\) have the same sign, we show that \(g(\nu )<0\) for \(0<\nu <\bar{\nu}\).
We can compute that
$$\begin{aligned} &g'(\nu ) \\ &= -\nu ^{-n_{2}-1}\Big(u\big(I(\nu )\big)- \beta P\Big) +n_{1} \nu ^{n_{1}-n_{2}-1}\int _{\nu}^{\bar{\nu}}\xi ^{-n_{1}-1} \Big(u\big(I(\xi )\big)-\beta P\Big)d\xi \\ &=-\bar{\nu}^{-n_{2}-1}\Big(u\big(I(\bar{\nu})\big)-\beta P\Big)+\nu ^{n_{1}-n_{2}-1} \int _{\nu}^{\bar{\nu}}\xi ^{-n_{1}}d\Big(u\big(I(\xi )\big)-\beta P \Big), \end{aligned}$$
(A.11)
where the last equality follows from integration by parts. Since \(n_{1}>1\), \(n_{2}<0\) and \(u (I(\xi ) )\) is decreasing in \(\xi \), we can deduce from (A.11) that \(g'(\nu )\) is strictly increasing for \(\nu \in (0,\bar{\nu})\). By (4.10), we have
$$ g'(\bar{\nu})=-\bar{\nu}^{-n_{2}-1}\Big(u\big(I(\bar{\nu})\big)- \beta P\Big)>0. $$
On the other hand, since we have shown in the proof of Proposition 4.1 that there exists a unique \(\hat{\nu}\in (0,\bar{\nu})\) such that \(u (I(\hat{\nu}) )-\beta P =0\) and \(u (I(\nu ) )-\beta P <0\) for \(\nu >\hat{\nu}\), we have
$$\begin{aligned} g'(\hat{\nu})= n_{1}\hat{\nu}^{n_{1}-n_{2}}\int _{\hat{\nu}}^{ \bar{\nu}}\xi ^{-n_{1}-1}\Big(u\big(I(\xi )\big)-\beta P\Big)d\xi < 0. \end{aligned}$$
Thus there exists a unique \(\nu _{1}\in (\hat{\nu},\bar{\nu})\) such that \(g'(\nu _{1})=0\). Moreover,
$$\begin{aligned} g'(\nu ) \textstyle\begin{cases} < 0 &\quad \text{if } 0< \nu < \nu _{1}, \\ =0 &\quad \text{if } \nu =\nu _{1}, \\ >0 &\quad \text{if } \nu >\nu _{1}. \end{cases}\displaystyle \end{aligned}$$
(A.12)
This implies that \(g(\nu )\) is strictly decreasing for \(\nu \in (0,\nu _{1})\) and strictly increasing for \(\nu \in (\nu _{1},\bar{\nu})\).
Since \(u (I(\xi ) )-\beta P \) is decreasing in \(\xi \) and \(u (I(\xi ) )-\beta P >0\) for \(0<\xi <\hat{\nu}\), there exists a sufficiently small \(m\in (0,\hat{\nu})\) such that for \(\xi < m\), we have \(u (I(\xi ) )-\beta P > \delta \) for some \(\delta >0\). Then for \(\nu < m\), we have
$$\begin{aligned} \int _{\nu}^{m} \xi ^{-n_{1}-1} \Big(u\big(I(\xi )\big)-\beta P\Big) d \xi > \int _{\nu}^{m} \xi ^{-n_{1}-1} \delta d\xi = \frac{\delta }{n_{1}}(\nu ^{-n_{1}}-m^{-n_{1}}), \end{aligned}$$
and it follows that
$$\begin{aligned} \lim _{\nu \to 0+}\int _{\nu}^{m}\xi ^{-n_{1}-1}\Big(u\big(I(\xi ) \big)-\beta P\Big)d\xi = \infty . \end{aligned}$$
Thus we have
$$\begin{aligned} \lim _{\nu \to 0+}\int _{\nu}^{\bar{\nu}}\xi ^{-n_{1}-1}\Big(u\big(I( \xi )\big)-\beta P\Big)d\xi = \infty , \end{aligned}$$
and it follows from L’Hôpital’s rule that
$$\begin{aligned} g(0+)&=\lim _{\nu \to 0+}\frac{n_{1}}{n_{1}-n_{2}}\bigg( \frac{1}{\nu ^{n_{2}-n_{1}}}\int _{\nu}^{\bar{\nu}}\xi ^{-n_{1}-1} \Big(u\big(I(\xi )\big)-\beta P\Big)d\xi \bigg) \\ &=\frac{n_{1}}{n_{1}-n_{2}}\lim _{\nu \to 0+} \frac{-\nu ^{-n_{1}-1} (u (I(\nu ) )-\beta P )}{(n_{2}-n_{1})\nu ^{n_{2}-n_{1}-1}} \\ &=\frac{n_{1}}{(n_{1}-n_{2})^{2}}\lim _{\nu \to 0+}\nu ^{-n_{2}}\Big(u \big(I(\nu )\big)-\beta P\Big)=0, \end{aligned}$$
(A.13)
where we used the fact that by part (a) of Proposition 3.1 and Lemma 3.4,
$$ \lim _{\nu \to 0+}\nu ^{-n_{2}}\Big(u\big(I(\nu )\big)-\beta P\Big)=0. $$
It is clear from (4.9) that
$$\begin{aligned} g(\bar{\nu})=\frac{n_{2}}{n_{1}-n_{2}}\int _{0}^{\bar{\nu}}\xi ^{-n_{2}-1} \Big(u\big(I(\xi )\big)-\beta P\Big)d\xi =0. \end{aligned}$$
(A.14)
Combining (A.12)–(A.14), we deduce that
$$\begin{aligned} g(\nu )< 0\qquad \text{for } 0< \nu < \bar{\nu}, \end{aligned}$$
or equivalently,
$$\begin{aligned} \big({\mathcal{M}}(\nu )-\nu {\mathcal{M}}'(\nu ) - P\big)'< 0\qquad \text{for } 0< \nu < \bar{\nu}, \end{aligned}$$
(A.15)
and this shows that
$$ {\mathcal{M}}(\nu )-\nu {\mathcal{M}}'(\nu ) - P>0\qquad \text{for } \nu \in (0, \bar{\nu}). $$
In conclusion, \({\mathcal{M}}(z)\) satisfies the HJB equation with mixed boundary condition
$$\begin{aligned} \min \{\mathcal{L}{\mathcal{M}}(\nu )+\tilde{u}(\nu )+\nu \epsilon , { \mathcal{M}}(\nu )-\nu{\mathcal{M}}'(\nu )-P \}=0\qquad \text{for } \nu \in (0, \infty ). \end{aligned}$$
Note that \(({\mathcal{M}}(\nu )-\nu {\mathcal{M}}'(\nu ) - P )' = -\nu {\mathcal{M}}''(\nu )\) and (A.15) mean that \({\mathcal{M}}''(\nu )>0\) for \(0<\nu <\bar{\nu}\). Moreover, we already know from \({\mathcal{M}}(\nu )-\nu {\mathcal{M}}'(\nu ) - P = 0\) for \(\nu \geq \bar{\nu}\) that \({\mathcal{M}}''=0\) for \(\nu \geq \bar{\nu}\). Overall, we can conclude that \({\mathcal{M}}(\nu )\) in (4.7) is convex, and this completes the proof. □
1.7 A.7 Proof of Lemma 4.4
We prove this result in the following steps.
Step 1: The function \(\varphi (y)\) given in (4.16) is continuously differentiable for \(y>0\) and satisfies the HJB equation
$$\begin{aligned} \min \{{\mathcal{L}}\varphi (y),\Gamma _{(u\circ I -\beta P)}(y)-\varphi (y) \}=0. \end{aligned}$$
(A.16)
Moreover, in terms of \(\varphi (y)\), the two regions WR and IR defined in (4.5) can be written as
$$\begin{aligned} {\mathbf{WR}}&=\{y>0: \Gamma _{(u\circ I -\beta P)}(y)>\varphi (y)\} , \\ {\mathbf{IR}}&=\{y>0: \Gamma _{(u\circ I -\beta P)}(y)=\varphi (y)\}. \end{aligned}$$
(A.17)
Proof of Step 1: It is obvious from (4.16) that \(\varphi (y)\) is continuous at \(y=\bar{\nu}\). We can compute that
$$\begin{aligned} \varphi '(\bar{\nu})&=\frac{2}{\theta ^{2}(n_{1}-n_{2})} n_{1} \bar{\nu}^{n_{1}-1}\int _{\bar{\nu}}^{\infty}\xi ^{-n_{1}-1}\Big(u \big(I(\xi )\big) -\beta P\Big)d\xi , \\ \varphi '(\bar{\nu}-)&=\Gamma _{(u\circ I - \beta P)}(\bar{\nu}) \frac{n_{1}}{\bar{\nu}}. \end{aligned}$$
From (4.9), we know that
$$\begin{aligned} \Gamma _{(u\circ I -\beta P)}(\bar{\nu})= \frac{2}{\theta ^{2}(n_{1}-n_{2})}\bar{\nu}^{n_{1}}\int _{\bar{\nu} }^{ \infty }\xi ^{-n_{1}-1}\Big(u\big(I(\xi )\big) -\beta P\Big)d\xi . \end{aligned}$$
Thus it follows that \(\varphi '(y)\) is continuous at \(y=\bar{\nu}\), that is, \(\varphi (y)\) is continuously differentiable for \(y>0\).
Recall that \(u (I(y) ) -\beta P\) is decreasing in \(y\) and \(u (I(\bar{\nu}) )-\beta P<0\) in (4.10). By using part (c) of Proposition 3.1, we have
$$\begin{aligned} {\mathcal{L}}\Gamma _{(u\circ I -\beta P)}(y) = -\Big(u\big(I(y)\big) - \beta P\Big)\ge -\Big(u\big(I(\bar{\nu})\big) -\beta P\Big)>0. \end{aligned}$$
Since \(\varphi (y)=\Gamma _{(u\circ I -\beta P)}(y)\) for \(y\geq \bar{\nu}\), we have \({\mathcal{L}}\varphi (y) >0\) for \(y\geq \bar{\nu}\) and (A.16) holds for \(y\geq \bar{\nu}\).
On the other hand, it follows from (A.10) that for \(0< y<\bar{\nu}\),
$$\begin{aligned} \Gamma _{(u\circ I -\beta P)}(y)-\varphi (y)&= \frac{2}{\theta ^{2}(n_{1}-n_{2})}\bigg(y^{n_{2}}\int _{0}^{y}\xi ^{-n_{2}-1} \Big(u\big(I(\xi )\big) -\beta P\Big)d\xi \\ & \hphantom{=\frac{2}{\theta ^{2}(n_{1}-n_{2})}\bigg[} +y^{n_{1}}\int _{y}^{\bar{\nu}} \xi ^{-n_{1}-1}\Big(u\big(I(\xi ) \big) -\beta P\Big)d\xi \bigg) \\ &={\mathcal{M}}(y)-y{\mathcal{M}}'(y)-P>0. \end{aligned}$$
Clearly, by the definition of \(n_{1}\), we have \({\mathcal{L}}\varphi (y) =0\) for \(0< y<\bar{\nu}\). Thus (A.16) also holds for \(0< y<\bar{\nu}\). Based on the above discussion, it is clear that WR and IR can be written as (A.17), and this completes the proof of Step 1.
Step 2: We have the inequality
$$\begin{aligned} \inf _{\tau \in{\mathcal{S}}}\mathbb{E} [e^{-\beta \tau}\Gamma _{(u\circ I - \beta P)}({\mathcal{Y}}_{\tau}^{y}) ]\ge \varphi (y). \end{aligned}$$
Proof of Step 2: Even though \(\varphi (y)\) is \(C^{1}\) on \((0,\infty )\) and \(C^{2}\) on \((0,\infty )\backslash \{\bar{\nu}\}\), one can still apply Itô’s lemma (see Karatzas and Shreve [18, Exercise 3.6.24]), and it follows that
$$\begin{aligned} d\big(e^{-\beta t}\varphi ({\mathcal{Y}}_{t}^{y})\big)=e^{-\beta t}\bigg(& \Big((\beta -r){\mathcal{Y}}_{t}^{y}\varphi '({\mathcal{Y}}_{t}^{y})+ \frac{\theta ^{2}}{2}({\mathcal{Y}}_{t}^{y})^{2}\varphi ''({\mathcal{Y}}_{t}^{y})- \beta \varphi ({\mathcal{Y}}_{t}^{y})\Big)dt \\ &-\theta {\mathcal{Y}}_{t}^{y}\varphi '({\mathcal{Y}}_{t}^{y})dB_{t}\bigg). \end{aligned}$$
Then for any \(\tau \in {\mathcal{S}}\), we have
$$\begin{aligned} e^{-\beta (\tau \wedge t)}\varphi ({\mathcal{Y}}_{\tau \wedge t}^{y})- \varphi (y)&=\int _{0}^{\tau \wedge t}e^{-\beta s}{\mathcal{L}}\varphi ({ \mathcal{Y}}_{s}^{y})ds \\ & \phantom{=:} +\int _{0}^{\tau \wedge t}e^{-\beta s}(-\theta ){\mathcal{Y}}_{s}^{y} \varphi '({\mathcal{Y}}_{s}^{y})dB_{s}. \end{aligned}$$
(A.18)
By part (d) of Proposition 3.1, there exists a constant \(C_{1}>0\) such that
$$ |\Gamma _{(u\circ I -\beta P)}'(y)| \le C_{1}(y^{n_{1}-1}+y^{n_{2}-1}). $$
It follows from (4.16) that for some constant \(C_{2}>0\), we have
$$\begin{aligned} |\varphi '(y)| \le C_{2}(y^{n_{1}-1}+y^{n_{2}-1}). \end{aligned}$$
Thus we can easily show that for every constant \(T>0\), the process
$$ \int _{0}^{T\wedge t} e^{-\beta s}(-\theta ){\mathcal{Y}}_{s}^{y} \varphi '({ \mathcal{Y}}_{s}^{y})dB_{s}, \qquad t \geq 0, $$
is a martingale (see Knudsen et al. [20, Lemma 3.4] for the details). Therefore, by taking expectations on both sides of (A.18), we deduce that
$$\begin{aligned} \varphi (y)+\mathbb{E}\bigg[\int _{0}^{\tau \wedge t}e^{-\beta s}{ \mathcal{L}}\varphi ({\mathcal{Y}}_{s}^{y})ds\bigg] =\mathbb{E} [e^{-\beta ( \tau \wedge t)}\varphi ({\mathcal{Y}}_{\tau \wedge t}^{y}) ]. \end{aligned}$$
(A.19)
Since \(\varphi (y)\) satisfies the HJB equation (A.16), it follows that
$$\begin{aligned} \varphi (y)\le \mathbb{E} [e^{-\beta (\tau \wedge t)}\Gamma _{(u \circ I -\beta P)}({\mathcal{Y}}_{\tau \wedge t}^{y}) ]. \end{aligned}$$
The dominated convergence theorem implies that
$$\begin{aligned} \varphi (y)\le \lim _{t\to \infty} \mathbb{E} [e^{-\beta (\tau \wedge t)}\Gamma _{(u\circ I -\beta P)}({\mathcal{Y}}_{\tau \wedge t}^{y}) ]= \mathbb{E} [e^{-\beta \tau }\Gamma _{(u\circ I -\beta P)}({\mathcal{Y}}_{ \tau}^{y}) ] \end{aligned}$$
for any \(\tau \in{\mathcal{S}}\). Thus we have
$$\begin{aligned} \inf _{\tau \in{\mathcal{S}}}\mathbb{E} [e^{-\beta \tau}\Gamma _{(u\circ I - \beta P)}({\mathcal{Y}}_{\tau}^{y}) ]\ge \varphi (y). \end{aligned}$$
(A.20)
Step 3: The stopping time \(\hat{\tau}(y)\) defined in (4.14) is the solution to the optimal stopping problem in (4.15).
Proof of Step 3: Replacing \(\tau \) in (A.19) by \(\hat{\tau}(y)\) defined in (4.14), we can deduce that
$$\begin{aligned} \varphi (y)+\mathbb{E}\bigg[\int _{0}^{\hat{\tau}(y)\wedge t}e^{- \beta s}{\mathcal{L}}\varphi ({\mathcal{Y}}_{s}^{y})ds\bigg] =\mathbb{E} [e^{- \beta (\hat{\tau}(y)\wedge t)}\varphi ({\mathcal{Y}}_{\hat{\tau}(y)\wedge t}^{y}) ]. \end{aligned}$$
By Step 1,
$$\begin{aligned} {\mathcal{L}}\varphi ({\mathcal{Y}}_{s}^{y}) =0 \qquad \text{for } 0\le s < \hat{\tau}(y). \end{aligned}$$
Thus we can get
$$\begin{aligned} \varphi (y) =\mathbb{E} [e^{-\beta (\hat{\tau}(y)\wedge t)}\varphi ({ \mathcal{Y}}_{\hat{\tau}(y)\wedge t}^{y}) ]. \end{aligned}$$
When \(0< y<\bar{\nu}\), we have \({\mathcal{Y}}_{\hat{\tau}(y)\wedge t}^{y} \le \bar{\nu}\) for all \(t\ge 0\). Thus
$$ \mathbb{E} [e^{-\beta (\hat{\tau}(y)\wedge t)}|\varphi ({\mathcal{Y}}_{ \hat{\tau}(y)\wedge t}^{y})| ]\le |\Gamma _{(u\circ I -\beta P)}( \bar{\nu})| < \infty \qquad \text{for } t\ge 0. $$
By the dominated convergence theorem, we have
$$\begin{aligned} \varphi (y) &=\lim _{t\to \infty}\mathbb{E} [e^{-\beta (\hat{\tau}(y) \wedge t)}\varphi ({\mathcal{Y}}_{\hat{\tau}(y)\wedge t}^{y}) ] \\ &=\mathbb{E} [e^{-\beta \hat{\tau}(y)}\varphi ({\mathcal{Y}}_{\hat{\tau}(y)}^{y}) ]=\mathbb{E} [e^{-\beta \hat{\tau}(y)}\Gamma _{(u\circ I -\beta P)}({ \mathcal{Y}}_{\hat{\tau}(y)}^{y}) ]. \end{aligned}$$
When \(y\geq \bar{\nu}\), it is obvious that \({\mathcal{Y}}_{\hat{\tau}(y)\wedge t}^{y} =y\) by the definition of \(\hat{\tau}(y)\). Because we have \(\varphi (y)=\Gamma _{(u\circ I - \beta P)}(y)\) for \(y\geq \bar{\nu}\) in (4.16), we can deduce that
$$\begin{aligned} \varphi (y) = \Gamma _{(u\circ I -\beta P)}(y) = \mathbb{E} [e^{- \beta \hat{\tau}(y)}\Gamma _{(u\circ I -\beta P)}({\mathcal{Y}}_{ \hat{\tau}(y)}^{y}) ]. \end{aligned}$$
Overall, for any \(y\), we have \(\varphi (y)= \mathbb{E} [e^{-\beta \hat{\tau}(y)}\Gamma _{(u\circ I - \beta P)}({\mathcal{Y}}_{\hat{\tau}(y)}^{y}) ]\). Since \(\hat{\tau}(y) \in {\mathcal{S}}\),
$$\begin{aligned} \varphi (y)=\mathbb{E} [e^{-\beta \hat{\tau}(y)}\Gamma _{(u\circ I - \beta P)}({\mathcal{Y}}_{\hat{\tau}(y)}^{y}) ]\ge \inf _{\tau \in{\mathcal{S}}} \mathbb{E} [e^{-\beta \tau}\Gamma _{(u\circ I -\beta P)}({\mathcal{Y}}_{ \tau}^{y}) ]. \end{aligned}$$
(A.21)
From (A.20) and (A.21), we can conclude that
$$ \varphi (y)=\mathbb{E} [e^{-\beta \hat{\tau}(y)}\Gamma _{(u\circ I - \beta P)}({\mathcal{Y}}_{\hat{\tau}(y)}^{y}) ]= \inf _{\tau \in{\mathcal{S}}} \mathbb{E} [e^{-\beta \tau}\Gamma _{(u\circ I -\beta P)}({\mathcal{Y}}_{ \tau}^{y}) ], $$
and this completes the proof. □
1.8 A.8 Proof of Theorem 4.5
Before proceeding to the proof of Theorem 4.5, we introduce the following lemma.
Lemma 8.1
For any \(({\mathcal{Z}}_{t})_{t \geq 0} \in \Pi (z)\),
$$\begin{aligned} \lim _{t\to \infty} e^{-\beta t} \mathbb{E}[{\mathcal{Z}}_{t}]=0\qquad \textit{and}\qquad z+\mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}d{ \mathcal{Z}}_{t}\bigg]=\beta \mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}{ \mathcal{Z}}_{t}dt\bigg]. \end{aligned}$$
Proof
For a given \(T>0\), integration by parts implies
$$\begin{aligned} \mathbb{E}\bigg[\int _{0}^{T} e^{-\beta t} d{\mathcal{Z}}_{t}\bigg]&= \mathbb{E}[e^{-\beta T}{\mathcal{Z}}_{T}]-z +\beta \mathbb{E}\bigg[\int _{0}^{T} e^{-\beta t} {\mathcal{Z}}_{t}dt\bigg]. \end{aligned}$$
Since (3.7) holds for \(({\mathcal{Z}}_{t})_{t \geq 0} \in \Pi (z)\), the monotone convergence theorem gives
$$\begin{aligned} \infty > z+\mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t} d{\mathcal{Z}}_{t} \bigg]=\lim _{T\to \infty}\mathbb{E}[e^{-\beta T}{\mathcal{Z}}_{T}] + \beta \mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t} {\mathcal{Z}}_{t}dt \bigg]. \end{aligned}$$
(A.22)
Thus we have
$$\begin{aligned} \lim _{T\to \infty}\mathbb{E}[e^{-\beta T}{\mathcal{Z}}_{T}]< \infty \qquad \text{and}\qquad \mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t} {\mathcal{Z}}_{t}dt\bigg]< \infty . \end{aligned}$$
(A.23)
Suppose that \(\lim _{T\to \infty}e^{-\beta T}\mathbb{E}[{\mathcal{Z}}_{T}]>0\). Then there exist positive constants \(\widehat{C}\) and \(\widehat{T}\) such that
$$ e^{-\beta t}\mathbb{E}[{\mathcal{Z}}_{t}] \ge \widehat{C}\qquad \text{for all } t\ge \widehat{T}. $$
It follows that
$$\begin{aligned} \mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t} {\mathcal{Z}}_{t}dt\bigg] \ge \int _{\widehat{T}}^{\infty }\widehat{C} dt =\infty , \end{aligned}$$
which contradicts (A.23). Hence we deduce that \(\lim _{t\to \infty} e^{-\beta t} \mathbb{E}[{\mathcal{Z}}_{t}]=0\) and it follows from (A.22) that
$$\begin{aligned} z+\mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}d{\mathcal{Z}}_{t}\bigg]= \beta \mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}{\mathcal{Z}}_{t}dt \bigg]. \end{aligned}$$
□
Proof
of Theorem 4.5 (a) Since it is clear that \(\widehat{\mathcal{Z}}_{t}^{z}(y)\) is nonnegative, nondecreasing and continuous in \(t\), it is sufficient to show that
$$\begin{aligned} \mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t} d\widehat{\mathcal{Z}}_{t}^{z}(y) \bigg] < \infty \end{aligned}$$
and
$$\begin{aligned} \mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}\bigg|\bigl(1+ \widehat{{\mathcal{Z}}}_{t}^{z}(y)\bigr)\tilde{u}\bigg( \frac{{\mathcal{Y}}_{t}^{y}}{1+\widehat{{\mathcal{Z}}}_{t}^{z}(y)}\bigg)\bigg|dt \bigg]< \infty . \end{aligned}$$
For a given \(T>0\), it follows from integration by parts that
$$\begin{aligned} \mathbb{E}\bigg[\int _{0}^{T} e^{-\beta t} d\widehat{\mathcal{Z}}_{t}^{z}(y) \bigg]&=\mathbb{E} [e^{-\beta T}(1+\widehat{\mathcal{Z}}_{T}^{z}) ]-(1+z) \\ & \phantom{=:} +\beta \mathbb{E}\bigg[\int _{0}^{T} e^{-\beta t}\big(1+ \widehat{\mathcal{Z}}_{t}^{z}(y)\big)dt\bigg]. \end{aligned}$$
Note that for any \(t \ge 0\),
$$\begin{aligned} 1+\widehat{\mathcal{Z}}_{t}^{z}(y) \le 1+z + \sup _{0\le s \le t} \frac{{\mathcal{Y}}_{s}^{y}}{\bar{\nu}} = 1+z + \frac{1}{\bar{\nu}} \overline{\mathcal{Y}}_{t}^{y}, \end{aligned}$$
(A.24)
where \(\overline{\mathcal{Y}}_{t}^{y} = \sup _{0\le s \le t} {\mathcal{Y}}_{s}^{y}\). For any given \(\lambda \in (0,n_{1})\), Merhi and Zervos [22, Lemma 1] implies that there exist \(\epsilon _{1}\in (0,\beta )\) and \(\epsilon _{2}>0\) such that
$$\begin{aligned} \mathbb{E} [e^{-\beta t} (\overline{\mathcal{Y}}_{t}^{y})^{\lambda }]&\le \frac{\theta ^{2} \lambda ^{2}/2+\epsilon _{2}}{\epsilon _{2}}y^{ \lambda }e^{-\epsilon _{1} t}, \\ \mathbb{E}\Big[\sup _{t\ge 0} e^{-\beta t}(\overline{\mathcal{Y}}_{t}^{y})^{ \lambda }\Big]&\le \frac{\theta ^{2}\lambda ^{2}/2 +\epsilon _{2}}{\epsilon _{2}}y^{ \lambda }. \end{aligned}$$
(A.25)
Since \(0<1<n_{1}\), using (A.24) and putting \(\lambda =1\) for (A.25) gives
$$\begin{aligned} \mathbb{E}\bigg[\int _{0}^{T} e^{-\beta t} d\widehat{\mathcal{Z}}_{t}^{z}(y) \bigg]&\le \frac{1}{\bar{\nu}}\bigg(\mathbb{E}[e^{-\beta T} \overline{\mathcal{Y}}_{T}^{y}]+\beta \int _{0}^{T}\mathbb{E}[e^{-\beta t} \overline{\mathcal{Y}}_{t}^{y}]dt\bigg) \\ &\le \frac{\beta }{\bar{\nu}} \frac{\theta ^{2}/2+\epsilon _{2}}{\epsilon _{1}\epsilon _{2}}y < \infty . \end{aligned}$$
Using the same arguments as in the proof of Lemma 3.5, we have
$$\begin{aligned} &\mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t} \bigg|\big(1+ \widehat{\mathcal{Z}}_{t}^{z}(y)\big)\tilde{u}\bigg( \frac{{\mathcal{Y}}_{t}^{y}}{1+\widehat{\mathcal{Z}}_{t}^{z}(y)}\bigg)\bigg|dt \bigg] \\ &\le \mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}\int _{z}^{ \widehat{\mathcal{Z}}_{t}^{z}(y)}\bigg|u\bigg(I\Big( \frac{{\mathcal{Y}}_{t}^{y}}{1+\nu}\Big)\bigg)\bigg| d\nu dt \bigg]+ (1+z) \Gamma _{|\tilde{u}|}\bigg(\frac{y}{1+z}\bigg). \end{aligned}$$
Since \((1+z)\Gamma _{|\tilde{u}|} (\frac{y}{1+z} )<\infty \) by Lemma 3.4 and Proposition 3.1, it is enough to show that
$$ \Phi (y,z):=\mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}\int _{z}^{ \widehat{\mathcal{Z}}_{t}^{z}(y)}\bigg|u\bigg(I\Big( \frac{{\mathcal{Y}}_{t}^{y}}{1+\nu}\Big)\bigg)\bigg| d\nu dt \bigg]< \infty . $$
The Fubini–Tonelli theorem implies that
$$\begin{aligned} \Phi (y,z)&=\mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}\int _{z}^{ \widehat{\mathcal{Z}}_{t}^{z}(y)}\big|u\big(I({\mathcal{Y}}_{t}^{ \frac{y}{1+\nu}})\big)\big| d\nu dt \bigg] \\ &=\int _{z}^{\infty}\mathbb{E}\bigg[\int _{\hat{\tau}(\frac{y}{1+\nu})}^{ \infty} e^{-\beta t}\big|u\big(I({\mathcal{Y}}_{t}^{\frac{y}{1+\nu}})\big) \big|dt \bigg] d\nu . \end{aligned}$$
(A.26)
Let us define \(\Lambda (y)\) by
$$\begin{aligned} \Lambda (y)&:= \mathbb{E}\bigg[\int _{\hat{\tau}(y)}^{\infty} e^{- \beta t}\big|u\big(I({\mathcal{Y}}_{t}^{y})\big)\big|dt \bigg] \\ & \phantom{:} =\mathbb{E}\bigg[e^{-\beta \hat{\tau}(y)} \mathbb{E}_{\hat{\tau}(y)} \Big[\int _{\hat{\tau}(y)}^{\infty} e^{-\beta (t-\hat{\tau}(y))}\big|u \big(I({\mathcal{Y}}_{t}^{y})\big)\big|dt\Big] \bigg] \\ & \phantom{:} =\mathbb{E} [e^{-\beta \hat{\tau}(y)} \Gamma _{|u\circ I|}({\mathcal{Y}}_{ \hat{\tau}(y)}^{y}) ], \end{aligned}$$
where the last equality comes from the strong Markov property. Note that \(\Gamma _{|u\circ I|}(y)\) is twice differentiable by Lemma 3.4 and part (c) of Proposition 3.1. For \(y\ge \bar{\nu}\), it is clear from the definition of \(\hat{z}(y)\) in (4.14) that
$$\begin{aligned} \Lambda (y) = \mathbb{E}\bigg[\int _{0}^{\infty} e^{-\beta t}\big|u \big(I({\mathcal{Y}}_{t}^{y})\big)\big|dt \bigg]=\Gamma _{|u\circ I|}(y). \end{aligned}$$
(A.27)
For \(0< y<\bar{\nu}\), \(\Lambda (y)\) satisfies the ordinary differential equation (ODE)
$$ {\mathcal{L}}\Lambda (y)=0 \quad \text{for } 0< y< \bar{\nu}, \qquad \Lambda ( \bar{\nu}) = \Gamma _{|u\circ I|}(\bar{\nu}). $$
(A.28)
Moreover, we have by part (e) of Proposition 3.1 the transversality condition
$$\begin{aligned} \lim _{t\to \infty}e^{-\beta t} \mathbb{E} [\Lambda ({\mathcal{Y}}_{t}^{y}) ]=0. \end{aligned}$$
(A.29)
By solving the ODE (A.28) with condition (A.29), we can easily confirm that
$$\begin{aligned} \Lambda (y) = \Gamma _{|u\circ I|}(\bar{\nu})\bigg( \frac{y}{\bar{\nu}}\bigg)^{n_{1}}\qquad \text{for } 0< y< \bar{\nu}. \end{aligned}$$
(A.30)
From (A.27) and (A.30), we have
$$\begin{aligned} \Lambda (y)= \textstyle\begin{cases} \Gamma _{|u\circ I|}(y)\quad &\text{for } y\ge \bar{\nu}, \\ \Gamma _{|u\circ I|}(\bar{\nu}) (\frac{y}{\bar{\nu}} )^{n_{1}}\quad & \text{for } 0< y< \bar{\nu}. \end{cases}\displaystyle \end{aligned}$$
From (A.26), we obtain
$$\begin{aligned} \Phi (y,z)=\int _{z}^{\infty }\Lambda \bigg(\frac{y}{1+\nu}\bigg) d \nu , \end{aligned}$$
where
$$\begin{aligned} \Lambda \bigg(\frac{y}{1+\nu}\bigg)= \textstyle\begin{cases} \Gamma _{|u\circ I|} (\frac{y}{1+\nu} )\quad &\text{for } 0\leq \nu \leq \hat{z}(y), \\ \Gamma _{|u\circ I|}(\bar{\nu}) (\frac{y}{(1+\nu )\bar{\nu}} )^{n_{1}} \quad &\text{for } \nu > \hat{z}(y), \end{cases}\displaystyle \end{aligned}$$
with \(\hat{z}(y)\) given in (4.11). Consequently, we have
$$\begin{aligned} \Phi (y,z)&=\int _{z}^{z\vee \hat{z}(y)} \Gamma _{|u\circ I|}\bigg( \frac{y}{1+\nu}\bigg)d\nu +\int _{z\vee \hat{z}(y)}^{\infty }\Gamma _{|u \circ I|}(\bar{\nu})\bigg(\frac{y}{(1+\nu )\bar{\nu}}\bigg)^{n_{1}}d \nu \\ &=\int _{z}^{z\vee \hat{z}(y)} \Gamma _{|u\circ I|}\bigg( \frac{y}{1+\nu}\bigg)d\nu \\ & \phantom{=:} +\frac{1}{n_{1}-1}\Big(\big(z\vee \hat{z}(y)\big)+1\Big)\Gamma _{|u \circ I|}(\bar{\nu})\bigg(\frac{\hat{z}(y)+1}{ (z\vee \hat{z}(y) )+1} \bigg)^{n_{1}} \\ &< \infty . \end{aligned}$$
(b) Let us define \(\widehat{\mathcal{Q}}(y,z)\) by
$$\begin{aligned} \widehat{\mathcal{Q}}(y,z) &:= \mathbb{E}\bigg[\int _{0}^{\infty }e^{- \beta t}\bigg(\big(1+\widehat{{\mathcal{Z}}}_{t}^{z}(y)\big)\tilde{u}\Big( \frac{{\mathcal{Y}}_{t}^{y}}{1+\widehat{{\mathcal{Z}}}_{t}^{z}(y)}\Big)+{\mathcal{Y}}_{t}^{y} \epsilon \bigg)dt \\ & \hphantom{=:\mathbb{E}\bigg[} -P\int _{0}^{\infty }e^{-\beta t}d\widehat{{\mathcal{Z}}}_{t}^{z}(y)\bigg]. \end{aligned}$$
We show that \({\mathcal{Q}}(y,z)=\widehat{\mathcal{Q}}(y,z)\). Using Lemma 8.1 and rearranging terms, we have
$$\begin{aligned} \widehat{\mathcal{Q}}(y,z) &=\mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t} \bigg(\big(1+\widehat{{\mathcal{Z}}}_{t}^{z}(y)\big)\tilde{u}\Big( \frac{{\mathcal{Y}}_{t}^{y}}{1+\widehat{{\mathcal{Z}}}_{t}^{z}(y)}\Big)+{\mathcal{Y}}_{t}^{y} \epsilon -\beta P \widehat{\mathcal{Z}}_{t}^{z}(y) \bigg)dt\bigg] \\ & \hphantom{=:} +Pz \\ &=\mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}\bigg(\big(1+ \widehat{{\mathcal{Z}}}_{t}^{z}(y)\big)\tilde{u}\Big( \frac{{\mathcal{Y}}_{t}^{y}}{1+\widehat{{\mathcal{Z}}}_{t}^{z}(y)}\Big) -\beta P \big(1+\widehat{\mathcal{Z}}_{t}^{z}(y)\big) \bigg)dt\bigg] \\ & \phantom{=:} +\mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}{\mathcal{Y}}_{t}^{y} \epsilon dt\bigg]+P(1+z). \end{aligned}$$
(A.31)
As in the proof of Lemma 3.5, we have
$$\begin{aligned} &\mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}\bigg(\big(1+ \widehat{{\mathcal{Z}}}_{t}^{z}(y)\big)\tilde{u}\Big( \frac{{\mathcal{Y}}_{t}^{y}}{1+\widehat{{\mathcal{Z}}}_{t}^{z}(y)}\Big) -\beta P \big(1+\widehat{\mathcal{Z}}_{t}^{z}(y)\big) \bigg)dt\bigg] \\ &=\mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}\int _{z}^{ \widehat{{\mathcal{Z}}}_{t}^{z}(y)} \bigg(u\bigg(I\Big( \frac{{\mathcal{Y}}_{t}^{y}}{1+\nu}\Big)\bigg)-\beta P\bigg)d\nu dt\bigg] \\ & \phantom{=:} +\mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t} \bigg((1+z)\tilde{u} \Big(\frac{{\mathcal{Y}}_{t}^{y}}{1+z}\Big) -\beta P (1+z) \bigg)dt\bigg]. \end{aligned}$$
(A.32)
It follows from (A.31) and (A.32) that
$$\begin{aligned} \widehat{\mathcal{Q}}(y,z) &=\mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t} \int _{z}^{\widehat{\mathcal{Z}}_{t}^{z}(y)} \bigg(u\bigg(I\Big( \frac{{\mathcal{Y}}_{t}^{y}}{1+\nu}\Big)\bigg)-\beta P\bigg)d\nu dt\bigg] \\ & \phantom{=:} +(1+z)\mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t} \big(\tilde{u}({ \mathcal{Y}}_{t}^{\frac{y}{1+z}})+ {\mathcal{Y}}_{t}^{\frac{y}{1+z}}\epsilon \big)dt\bigg] \\ &=\widehat{\Phi}(y,z) +(1+z)\bigg(\Gamma _{\tilde{u}}\Big( \frac{y}{1+z}\Big)+\frac{\epsilon }{r}\Big(\frac{y}{1+z}\Big)\bigg), \end{aligned}$$
(A.33)
where
$$\begin{aligned} \widehat{\Phi}(y,z):=\mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t} \int _{z}^{\widehat{\mathcal{Z}}_{t}^{z}(y)} \Big(u\big(I({\mathcal{Y}}_{t}^{ \frac{y}{1+\nu}})\big)-\beta P\Big)d\nu dt\bigg]. \end{aligned}$$
Since \((\widehat{\mathcal{Z}}_{t}^{z}(y))_{t \geq 0} \in \Pi (z)\), Fubini’s theorem implies that
$$\begin{aligned} &\widehat{\Phi}(y,z) \\ &=\int _{z}^{\infty}\mathbb{E}\bigg[\int _{ \hat{\tau}(\frac{y}{1+\nu})}^{\infty }e^{-\beta t} \Big(u\big(I({{ \mathcal{Y}}_{t}^{\frac{y}{{1+\nu}}}})\big)-\beta P\Big)dt\bigg]d\nu \\ &=\int _{z}^{\infty}\mathbb{E}\bigg[ e^{-\beta \hat{\tau}( \frac{y}{1+\nu})}\mathbb{E}_{\hat{\tau}(\frac{y}{1+\nu})}\Big[\int _{ \hat{\tau}(\frac{y}{1+\nu})}^{\infty }e^{-\beta (t-\hat{\tau}( \frac{y}{1+\nu}) )} \\ & \hphantom{=:\int _{z}^{\infty}\mathbb{E}\bigg[ e^{-\beta \hat{\tau}(\frac{y}{1+\nu})}\mathbb{E}_{\hat{\tau}(\frac{y}{1+\nu})}\Big[\int _{\hat{\tau}(\frac{y}{1+\nu})}^{\infty}} \times \Big(u\big(I({{\mathcal{Y}}_{t}^{\frac{y}{{1+\nu}}}})\big)-\beta P \Big)dt\Big]\bigg]d\nu , \end{aligned}$$
(A.34)
where \(\hat{\tau}(y)\) is given in (4.14). By Lemma 4.4 and the strong Markov property,
$$\begin{aligned} \varphi \bigg(\frac{y}{1+\nu}\bigg) &=\mathbb{E} \big[ e^{-\beta \hat{\tau}(\frac{y}{1+\nu})}\Gamma _{(u\circ I -\beta P)}({\mathcal{Y}}_{ \hat{\tau}(\frac{y}{1+\nu})}^{\frac{y}{1+\nu}}) \big] \\ &=\mathbb{E}\bigg[ e^{-\beta \hat{\tau}(\frac{y}{1+\nu})}\mathbb{E}_{ \hat{\tau}(\frac{y}{1+\nu})}\Big[\int _{\hat{\tau}(\frac{y}{1+\nu})}^{ \infty }e^{-\beta (t-\hat{\tau}(\frac{y}{1+\nu}) )} \\ & \hphantom{=:\mathbb{E}\bigg[ e^{-\beta \hat{\tau}(\frac{y}{1+\nu})}\mathbb{E}_{\hat{\tau}(\frac{y}{1+\nu})}\Big[\int _{\hat{\tau}(\frac{y}{1+\nu})}^{\infty}} \times \Big(u\big(I({{\mathcal{Y}}_{t}^{\frac{y}{{1+\nu}}}})\big)-\beta P \Big)dt\Big]\bigg]. \end{aligned}$$
(A.35)
It follows from (A.33)–(A.35) that
$$\begin{aligned} \widehat{Q}(y,z)=\int _{z}^{\infty }\varphi \bigg(\frac{y}{1+\nu} \bigg)d\nu + (1+z)\Gamma _{\tilde{u}}\bigg(\frac{y}{1+z}\bigg)+ \frac{\epsilon }{r}y. \end{aligned}$$
Note that
$$\begin{aligned} \varphi \bigg(\frac{y}{1+\nu}\bigg)= \textstyle\begin{cases} \Gamma _{(u\circ I -\beta P)} (\frac{y}{{1+\nu}} )\quad &\text{for } 0 \leq \nu \leq \hat{z}(y), \\ \Gamma _{(u\circ I -\beta P)}(\bar{\nu}) (\frac{y}{\bar{\nu}(1+\nu )} )^{n_{1}} \quad &\text{for } \nu > \hat{z}(y). \end{cases}\displaystyle \end{aligned}$$
(A.36)
Moreover, using (4.9) and (4.8), we deduce that
$$\begin{aligned} \Gamma _{(u\circ I -\beta P)}(\bar{\nu})&= \frac{2}{\theta ^{2}(n_{1}-n_{2})}\bigg(\bar{\nu}^{n_{2}}\int _{0}^{ \bar{\nu} }\xi ^{-n_{2}-1}\Big(u\big(I(\xi )\big) -\beta P\Big)d\xi \\ & \hphantom{=\frac{2}{\theta ^{2}(n_{1}-n_{2})}\bigg[} +\bar{\nu}^{n_{1}}\int _{\bar{\nu}}^{\infty }\xi ^{-n_{1}-1}\Big(u \big(I(\xi )\big) -\beta P\Big)d\xi \bigg) \\ &=\frac{2}{\theta ^{2}(n_{1}-n_{2})}\bar{\nu}^{n_{1}}\int _{\bar{\nu} }^{ \infty }\xi ^{-n_{1}-1}\Big(u\big(I(\xi )\big) -\beta P\Big)d\xi \\ &=(n_{1}-1)\bar{\nu}^{n_{1}}D_{1}. \end{aligned}$$
(A.37)
When \(\hat{z}(y)\le z\), using (A.36), (A.37) and part (b) of Proposition 3.1 gives
$$\begin{aligned} \widehat{Q}(y,z) &=\int _{z}^{\infty }\Gamma _{(u\circ I -\beta P)}( \bar{\nu})\bigg(\frac{y}{\bar{\nu}(1+\nu )}\bigg)^{n_{1}} d\nu + (1+z) \Gamma _{\tilde{u}}\bigg(\frac{y}{1+z}\bigg)+\frac{\epsilon }{r}y \\ &=D_{1} y^{n_{1}}(1+z)^{1-n_{1}} + (1+z)\Gamma _{\tilde{u}}\bigg( \frac{y}{1+z}\bigg)+\frac{\epsilon }{r}y \\ &= (1+z)\bigg({D_{1}}\Big(\frac{y}{1+z}\Big)^{n_{1}} \\ & \hphantom{=:(1+z)\bigg(} +\frac{2}{\theta ^{2}(n_{1}-n_{2})}\bigg(\Big(\frac{y}{1+z}\Big)^{n_{2}} \int _{0}^{\frac{y}{1+z}}\xi ^{-n_{2}-1}\tilde{u}(\xi )d\xi \\ & \hphantom{:(1+z)\bigg(+\frac{2}{\theta ^{2}(n_{1}-n_{2})}\bigg(\Big(} +\Big(\frac{y}{1+z}\Big)^{n_{1}}\int _{\frac{y}{1+z}}^{\infty}\xi ^{-n_{1}-1} \tilde{u}(\xi )d\xi \bigg) \\ & \hphantom{=:(1+z)\bigg(} +\frac{\epsilon }{r}\Big(\frac{y}{1+z}\Big)\bigg). \end{aligned}$$
(A.38)
When \(0\le z <\hat{z}(y)\), from (A.36), we have
$$\begin{aligned} \widehat{Q}(y,z)&=\int _{z}^{\hat{z}(y)} \Gamma _{(u\circ I -\beta P)} \bigg(\frac{y}{{1+\nu}}\bigg)d\nu +\int _{\hat{z}(y)}^{\infty } \Gamma _{(u\circ I -\beta P)}(\bar{\nu})\bigg( \frac{y}{\bar{\nu}(1+\nu )}\bigg)^{n_{1}} d\nu \\ & \phantom{=:} + (1+z)\Gamma _{\tilde{u}}\bigg(\frac{y}{1+z}\bigg)+ \frac{\epsilon }{r}y. \end{aligned}$$
(A.39)
Note that
$$\begin{aligned} \frac{d}{dz}\bigg((1+z)\Gamma _{\tilde{u}}\Big(\frac{y}{1+z}\Big) \bigg)&=\Gamma _{\tilde{u}}\bigg(\frac{y}{1+z}\bigg)-\frac{y}{1+z} \Gamma _{\tilde{u}}'\bigg(\frac{y}{1+z}\bigg) \\ &=\Gamma _{u\circ I}\bigg(\frac{y}{1+z}\bigg) \\ &=\Gamma _{(u\circ I-\beta P)}\bigg(\frac{y}{1+z}\bigg) +P, \end{aligned}$$
where we have used the fact that
$$\begin{aligned} \nu \Gamma _{\tilde{u}}'(\nu )&=\frac{2}{\theta ^{2}(n_{1}-n_{2})} \bigg(n_{2}\nu ^{n_{2}}\int _{0}^{\nu}\xi ^{-n_{2}-1}\tilde{u}(\xi )d \xi +\nu ^{n_{1}}\int _{\nu}^{\infty}\xi ^{-n_{1}-1}\tilde{u}(\xi )d \xi \bigg) \\ &=-\frac{2}{\theta ^{2}(n_{1}-n_{2})}\bigg(\nu ^{n_{2}}\int _{0}^{\nu} \xi ^{-n_{2}} I(\xi )d\xi +\nu ^{n_{1}}\int _{\nu}^{\infty}\xi ^{-n_{1}}I( \xi )d\xi \bigg) \\ &=-\Gamma _{u\circ I}(\nu ) + \Gamma _{\tilde{u}}(\nu ), \end{aligned}$$
(A.40)
which comes from integration by parts, part (b) of Proposition 3.1 and the relation
$$\begin{aligned} \Gamma _{(u\circ I -\beta P)}(\nu )=\Gamma _{u\circ I}(\nu )-P. \end{aligned}$$
(A.41)
This implies that
$$\begin{aligned} &\int _{z}^{\hat{z}(y)} \Gamma _{(u\circ I -\beta P)}\bigg( \frac{y}{{1+\nu}}\bigg)d\nu \\ &=\big(1+\hat{z}(y)\big)\Gamma _{\tilde{u}}\bigg( \frac{y}{1+\hat{z}(y)}\bigg)-(1+z)\Gamma _{\tilde{u}}\bigg( \frac{y}{1+z}\bigg)+P\big(z-\hat{z}(y)\big). \end{aligned}$$
(A.42)
It follows from (A.39) and (A.42) that
$$\begin{aligned} \widehat{\mathcal{Q}}(y,z) &=\int _{\hat{z}(y)}^{\infty }\Gamma _{(u\circ I -\beta P)}(\bar{\nu})\bigg(\frac{y}{\bar{\nu}(1+\nu )}\bigg)^{n_{1}} d \nu \\ & \phantom{=:} + \big(1+\hat{z}(y)\big)\Gamma _{\tilde{u}}\bigg( \frac{y}{1+\hat{z}(y)}\bigg)+\frac{\epsilon }{r}y+P\big(z-\hat{z}(y) \big) \\ &=\widehat{\mathcal{Q}}\big(y,\hat{z}(y)\big)+P\big(z-\hat{z}(y)\big) \qquad \text{for } 0\le z < \hat{z}(y). \end{aligned}$$
(A.43)
Comparing \({\mathcal{Q}}(y,z)\) in (4.12), (4.13) with \(\widehat{\mathcal{Q}}(y,z)\) in (A.38), (A.43) yields
$$ {\mathcal{Q}}(y,z)=\widehat{\mathcal{Q}}(y,z). $$
(c) For any \({\mathcal{Z}} \in \Pi (z)\), using the same arguments applied to \(\widehat{\mathcal{Q}}(y,z)\) to obtain (A.33) in the proof of part (b), we have
$$\begin{aligned} &\mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}\bigg((1+{\mathcal{Z}}_{t}) \tilde{u}\Big(\frac{{\mathcal{Y}}_{t}^{y}}{1+{\mathcal{Z}}_{t}}\Big)+{\mathcal{Y}}_{t}^{y} \epsilon \bigg)dt -P\int _{0}^{\infty }e^{-\beta t}d{\mathcal{Z}}_{t} \bigg] \\ &=\mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}\int _{z}^{{{\mathcal{Z}}}_{t}} \Big(u\big(I({\mathcal{Y}}_{t}^{\frac{y}{1+\nu}})\big)-\beta P\Big)d\nu dt \bigg] \\ & \phantom{=:} +(1+z)\Gamma _{\tilde{u}}\bigg(\frac{y}{1+z}\bigg)+ \frac{\epsilon }{r}y. \end{aligned}$$
(A.44)
Since (3.6) and (3.7) hold, we can easily deduce that
$$ \mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}\int _{z}^{{{\mathcal{Z}}}_{t}} \Big|\Big(u\big(I({\mathcal{Y}}_{t}^{\frac{y}{1+\nu}})\big)-\beta P\Big) \Big|d\nu dt\bigg]< \infty . $$
Fubini’s theorem implies that
$$\begin{aligned} &\mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}\int _{z}^{{{\mathcal{Z}}}_{t}} \Big(u\big(I({{\mathcal{Y}}_{t}^{\frac{y}{{1+\nu}}}})\big)-\beta P\Big)d \nu dt\bigg] \\ &=\int _{z}^{\infty}\mathbb{E}\bigg[\int _{\kappa (\nu )}^{\infty }e^{- \beta t} \Big(u\big(I({{\mathcal{Y}}_{t}^{\frac{y}{{1+\nu}}}})\big)-\beta P \Big)dt\bigg] d\nu , \end{aligned}$$
(A.45)
where the stopping time \(\kappa (\nu )\) for \(\nu \ge z\) is defined as
$$\begin{aligned} \kappa (\nu )=\inf \{t\ge 0: {\mathcal{Z}}_{t} \ge \nu \}. \end{aligned}$$
Moreover, by Lemma 4.4, we deduce that for any \(\nu \ge z\),
$$\begin{aligned} \mathbb{E}\bigg[\int _{\kappa (\nu )}^{\infty }e^{-\beta t} \Big(u \big(I({{\mathcal{Y}}_{t}^{\frac{y}{{1+\nu}}}})\big)-\beta P\Big)dt\bigg]& \ge \inf _{\tau \in{\mathcal{S}}}\bigg[\int _{\tau}^{\infty }e^{-\beta t} \Big(u\big(I({{\mathcal{Y}}_{t}^{\frac{y}{{1+\nu}}}})\big)-\beta P\Big)dt \bigg] \\ &=\varphi \bigg(\frac{y}{1+\nu}\bigg) \end{aligned}$$
(A.46)
It follows from (A.44)–(A.46) and part (b) of Theorem 4.5 that for any \({\mathcal{Z}}\in \Pi (z)\),
$$\begin{aligned} &\mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}\bigg((1+{\mathcal{Z}}_{t}) \tilde{u}\Big(\frac{{\mathcal{Y}}_{t}^{y}}{1+{\mathcal{Z}}_{t}}\Big)+{\mathcal{Y}}_{t}^{y} \epsilon \bigg)dt -P\int _{0}^{\infty }e^{-\beta t}d{\mathcal{Z}}_{t} \bigg] \\ &\ge \int _{z}^{\infty }\varphi \bigg(\frac{y}{1+\nu}\bigg)d\nu +(1+z) \Gamma _{\tilde{u}}\bigg(\frac{y}{1+z}\bigg)+\frac{\epsilon }{r}y={ \mathcal{Q}}(y,z). \end{aligned}$$
By the definition of \(\mathfrak{J}(y,z)\) in (3.8), we have
$$ \mathfrak{J}(y,z) \geq {\mathcal{Q}}(y,z). $$
At the same time, it follows from parts (a) and (b) of Theorem 4.5 that
$$\begin{aligned} &{\mathcal{Q}}(y,z) \\ &=\mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}\bigg(\big(1+ \widehat{{\mathcal{Z}}}_{t}^{z}(y)\big)\tilde{u}\Big( \frac{{\mathcal{Y}}_{t}^{y}}{1+\widehat{{\mathcal{Z}}}_{t}^{z}(y)}\Big)+{\mathcal{Y}}_{t}^{y} \epsilon \bigg)dt -P\int _{0}^{\infty }e^{-\beta t}d \widehat{{\mathcal{Z}}}_{t}^{z}(y)\bigg] \\ &\ge \inf _{{\mathcal{Z}}_{t}\in \Pi (z)}\mathbb{E}\bigg[\int _{0}^{ \infty }e^{-\beta t}\bigg((1+{\mathcal{Z}}_{t})\tilde{u}\Big( \frac{{\mathcal{Y}}_{t}^{y}}{1+{\mathcal{Z}}_{t}}\Big)+{\mathcal{Y}}_{t}^{y} \epsilon \bigg)dt -P\int _{0}^{\infty }e^{-\beta t}d{\mathcal{Z}}_{t} \bigg] \\ &= \mathfrak{J}(y,z). \end{aligned}$$
Thus we conclude that \(\mathfrak{J}(y,z)={\mathcal{Q}}(y,z) \).
(d) Since \(J(y)=\mathfrak{J}(y,0)={\mathcal{Q}}(y,0)\), it is clear that
$$\begin{aligned} J(y)&={\mathcal{Q}}(y,0) \\ &= \textstyle\begin{cases} D_{1} y^{n_{1}} +\frac{2}{\theta ^{2}(n_{1}-n_{2})} (y^{n_{2}}\int _{0}^{y} \xi ^{-n_{2}-1} (\tilde{u}(\xi )+\xi \epsilon )d\xi \\ \phantom{D_{1} y^{n_{1}} +\frac{2}{\theta ^{2}(n_{1}-n_{2})} } +y^{n_{1}}\int _{y}^{\infty}\xi ^{-n_{1}-1} (\tilde{u}(\xi )+\xi \epsilon )d\xi )&\quad \text{for } 0< y< \bar{\nu}, \\ (\tilde{u}(\bar{\nu})-\beta P +\bar{\nu} \epsilon ) \frac{y}{r\bar{\nu}}+P&\quad \text{for } y\ge \bar{\nu}. \end{cases}\displaystyle \end{aligned}$$
□
1.9 A.9 Proof of Lemma 4.6
For sufficiently small \(\delta >0\), it follows from part (a) of Theorem 4.5 that
$$\begin{aligned} \mathfrak{J}(y\pm \delta ,0) &=\inf _{{\mathcal{Z}}\in \Pi (0)}\mathbb{E} \bigg[\int _{0}^{\infty }e^{-\beta t}\bigg((1+{\mathcal{Z}}_{t})\tilde{u} \Big(\frac{{\mathcal{Y}}_{t}^{y\pm \delta }}{1+{\mathcal{Z}}_{t}}\Big)+{\mathcal{Y}}_{t}^{y \pm \delta } \epsilon \bigg)dt \\ & \hphantom{=:\inf _{{\mathcal{Z}}\in \Pi (0)}\mathbb{E}\bigg[} -P\int _{0}^{\infty }e^{-\beta t}d{\mathcal{Z}}_{t}\bigg] \\ &\le \mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}\bigg(\big(1+ \widehat{\mathcal{Z}}_{t}^{0}(y)\big)\tilde{u}\Big( \frac{{\mathcal{Y}}_{t}^{y\pm \delta }}{1+\widehat{\mathcal{Z}}_{t}^{0}(y)} \Big)+{\mathcal{Y}}_{t}^{y\pm \delta } \epsilon \bigg)dt \\ & \hphantom{=:\mathbb{E}\bigg[} -P\int _{0}^{\infty }e^{-\beta t}d\widehat{\mathcal{Z}}_{t}^{0}(y)\bigg]. \end{aligned}$$
It follows that
$$\begin{aligned} &\mathfrak{J}(y\pm \delta ,0)-\mathfrak{J}(y,0) \\ &\le \mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}\bigg((1+ \widehat{\mathcal{Z}}_{t}^{0})\tilde{u}\Big( \frac{{\mathcal{Y}}_{t}^{y\pm \delta }}{1+\widehat{\mathcal{Z}}_{t}^{0}}\Big)-(1+ \widehat{\mathcal{Z}}_{t}^{0})\tilde{u}\Big( \frac{{\mathcal{Y}}_{t}^{y}}{1+\widehat{\mathcal{Z}}_{t}^{0}}\Big)\pm \delta e^{ \beta t}{\mathcal{H}}_{t} \epsilon \bigg)dt\bigg]. \end{aligned}$$
Note that
$$\begin{aligned} \lim _{\delta \to 0+} \frac{\tilde{u} (\frac{{\mathcal{Y}}_{t}^{y\pm \delta }}{1+\widehat{\mathcal{Z}}_{t}^{0}} )-\tilde{u} (\frac{{\mathcal{Y}}_{t}^{y}}{1+\widehat{\mathcal{Z}}_{t}^{0}} )}{\pm \delta } = \tilde{u}'\bigg(\frac{{\mathcal{Y}}_{t}^{y}}{1+\widehat{\mathcal{Z}}_{t}^{0}} \bigg)\frac{{\mathcal{Y}}_{t}^{1}}{1+\widehat{\mathcal{Z}}_{t}^{0}} = -I \bigg( \frac{{\mathcal{Y}}_{t}^{y}}{1+\widehat{\mathcal{Z}}_{t}^{0}}\bigg) \frac{{\mathcal{Y}}_{t}^{1}}{1+\widehat{\mathcal{Z}}_{t}^{0}} \end{aligned}$$
and \({\mathcal{H}}_{t} = e^{-\beta t} {\mathcal{Y}}_{t}^{1}\). By the dominated convergence theorem, we have
$$\begin{aligned} J'(y)&=\lim _{\delta \to 0+} \frac{\mathfrak{J}(y+\delta ,0)-\mathfrak{J}(y,0)}{\delta }\le \mathbb{E}\bigg[\int _{0}^{\infty}{\mathcal{H}}_{t}\bigg(-I\Big( \frac{{\mathcal{Y}}_{t}^{y}}{1+\widehat{\mathcal{Z}}_{t}^{0}(y)}\Big)+\epsilon \bigg)dt\bigg] \end{aligned}$$
and
$$\begin{aligned} J'(y)&=\lim _{\delta \to 0+} \frac{\mathfrak{J}(y,0)-\mathfrak{J}(y-\delta ,0)}{-\delta }\ge \mathbb{E}\bigg[\int _{0}^{\infty}{\mathcal{H}}_{t}\bigg(-I\Big( \frac{{\mathcal{Y}}_{t}^{y}}{1+\widehat{\mathcal{Z}}_{t}^{0}(y)}\Big)+\epsilon \bigg)dt\bigg]. \end{aligned}$$
Hence we get
$$ \mathbb{E}\bigg[\int _{0}^{\infty}{\mathcal{H}}_{t}\bigg(I\Big( \frac{{\mathcal{Y}}_{t}^{y}}{1+\widehat{\mathcal{Z}}_{t}^{0}(y)}\Big)-\epsilon \bigg)dt\bigg]=-J'(y). $$
Assume that \(z>0\). For sufficiently small \(\delta >0\), define \(\widehat{\mathcal{Z}}^{z,\pm \delta }(y)\) by
$$\begin{aligned} \widehat{\mathcal{Z}}_{t}^{z,\pm \delta }(y) = \widehat{\mathcal{Z}}_{t}^{z}(y) \pm \delta . \end{aligned}$$
As in the proof of part (a) of Theorem 4.5, we can deduce that
$$\begin{aligned} \widehat{\mathcal{Z}}_{t}^{z, \pm \delta }(y)\in \Pi (z\pm \delta ). \end{aligned}$$
Since \((1+z)\tilde{u}(y/(1+z))\) is convex in \(z\ge 0\) for given \(y>0\), we deduce that for any \(\rho \in (0,\delta )\),
$$\begin{aligned} & \frac{(1+\widehat{\mathcal{Z}}_{t}^{z, -\delta }(y))\tilde{u} (\frac{{\mathcal{Y}}_{t}^{y}}{1+\widehat{\mathcal{Z}}_{t}^{z, -\delta }(y)} )-(1+\widehat{\mathcal{Z}}_{t}^{z}(y))\tilde{u} (\frac{{\mathcal{Y}}_{t}^{y}}{1+\widehat{\mathcal{Z}}_{t}^{z}(y)} )}{-\delta } \\ &\le \frac{(1+\widehat{\mathcal{Z}}_{t}^{z, \pm \rho})\tilde{u} (\frac{{\mathcal{Y}}_{t}^{y}}{1+\widehat{\mathcal{Z}}_{t}^{z, \pm \rho}(y)} )-(1+\widehat{\mathcal{Z}}_{t}^{z}(y))\tilde{u} (\frac{{\mathcal{Y}}_{t}^{y}}{1+\widehat{\mathcal{Z}}_{t}^{z}(y)} )}{\pm \rho} \\ &\le \frac{(1+\widehat{\mathcal{Z}}_{t}^{z, +\delta }(y))\tilde{u} (\frac{{\mathcal{Y}}_{t}^{y}}{1+\widehat{\mathcal{Z}}_{t}^{z, +\delta }(y)} )-(1+\widehat{\mathcal{Z}}_{t}^{z}(y))\tilde{u} (\frac{{\mathcal{Y}}_{t}^{y}}{1+\widehat{\mathcal{Z}}_{t}^{z}(y)} )}{\delta }. \end{aligned}$$
(A.47)
By Theorem 4.5,
$$\begin{aligned} &\mathfrak{J}(y,z\pm \rho ) \\ &=\inf _{{\mathcal{Z}}\in \Pi (z\pm \rho )}\mathbb{E}\bigg[\int _{0}^{ \infty }e^{-\beta t}\bigg((1+{\mathcal{Z}}_{t})\tilde{u}\Big( \frac{{\mathcal{Y}}_{t}^{y}}{1+{\mathcal{Z}}_{t}}\Big)+{\mathcal{Y}}_{t}^{y} \epsilon \bigg)dt -P\int _{0}^{\infty }e^{-\beta t}d{\mathcal{Z}}_{t} \bigg] \\ &\le \mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}\bigg((1+ \widehat{\mathcal{Z}}_{t}^{z, \pm \rho })\tilde{u}\Big( \frac{{\mathcal{Y}}_{t}^{y}}{1+\widehat{\mathcal{Z}}_{t}^{z, \pm \rho }}\Big)+{ \mathcal{Y}}_{t}^{y} \epsilon \bigg)dt -P\int _{0}^{\infty }e^{-\beta t}d \widehat{\mathcal{Z}}_{t}^{z, \pm \rho }\bigg]. \end{aligned}$$
It follows that
$$\begin{aligned} &\frac{\mathfrak{J}(y,z+\rho )-\mathfrak{J}(y,z)}{\rho} \\ &\le \mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t} \frac{(1+\widehat{\mathcal{Z}}_{t}^{z, +\rho}(y))\tilde{u} (\frac{{\mathcal{Y}}_{t}^{y}}{1+\widehat{\mathcal{Z}}_{t}^{z, +\rho}(y)} )-(1+\widehat{\mathcal{Z}}_{t}^{z}(y))\tilde{u} (\frac{{\mathcal{Y}}_{t}^{y}}{1+\widehat{\mathcal{Z}}_{t}^{z}(y)} )}{+\rho}dt \bigg] \end{aligned}$$
and
$$\begin{aligned} &\frac{\mathfrak{J}(y,z-\rho )-\mathfrak{J}(y,z)}{-\rho} \\ &\ge \mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t} \frac{(1+\widehat{\mathcal{Z}}_{t}^{z, -\rho}(y))\tilde{u} (\frac{{\mathcal{Y}}_{t}^{y}}{1+\widehat{\mathcal{Z}}_{t}^{z, -\rho}(y)} )-(1+\widehat{\mathcal{Z}}_{t}^{z}(y))\tilde{u} (\frac{{\mathcal{Y}}_{t}^{y}}{1+\widehat{\mathcal{Z}}_{t}^{z}(y)} )}{-\rho}dt \bigg]. \end{aligned}$$
Note that
$$\begin{aligned} &\lim _{\rho \to 0+} \frac{(1+\widehat{\mathcal{Z}}_{t}^{z, \pm \rho }(y))\tilde{u} (\frac{{\mathcal{Y}}_{t}^{y}}{1+\widehat{\mathcal{Z}}_{t}^{z, \pm \rho }(y)} )-(1+\widehat{\mathcal{Z}}_{t}^{z}(y))\tilde{u} (\frac{{\mathcal{Y}}_{t}^{y}}{1+\widehat{\mathcal{Z}}_{t}^{z}(y)} )}{\pm \rho } \\ &= u\bigg(I\Big( \frac{{\mathcal{Y}}_{t}^{y}}{1+\widehat{\mathcal{Z}}_{t}^{z}(y)}\Big) \bigg). \end{aligned}$$
(A.48)
Moreover, also note that by Lemma 3.5, \(\widehat{\mathcal{Z}}^{z, \pm \delta }(y)\in \Pi (z\pm \delta )\) implies
$$\begin{aligned} \mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t} \big(1+ \widehat{\mathcal{Z}}_{t}^{z, \pm \delta }(y)\big)\bigg|\tilde{u}\bigg( \frac{{\mathcal{Y}}_{t}^{y}}{1+\widehat{\mathcal{Z}}_{t}^{z, \pm \delta }(y)} \bigg)\bigg|dt\bigg]< \infty . \end{aligned}$$
(A.49)
From (A.47)–(A.49), the dominated convergence theorem yields
$$\begin{aligned} \mathfrak{J}_{z}(y,z)=\mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}u \bigg(I\Big(\frac{{\mathcal{Y}}_{t}^{y}}{1+\widehat{\mathcal{Z}}_{t}^{z}(y)} \Big)\bigg)dt\bigg]. \end{aligned}$$
The monotone convergence theorem implies that
$$\begin{aligned} J(y)-yJ'(y)=\mathfrak{J}_{z}(y,0)&=\lim _{z\to 0+}\mathbb{E}\bigg[ \int _{0}^{\infty }e^{-\beta t}u\bigg(I\Big( \frac{{\mathcal{Y}}_{t}^{y}}{1+\widehat{\mathcal{Z}}_{t}^{z}(y)}\Big)\bigg)dt \bigg] \\ &=\mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}u\bigg(I\Big( \frac{{\mathcal{Y}}_{t}^{y}}{1+\widehat{\mathcal{Z}}_{t}^{0}(y)}\Big)\bigg)dt \bigg]. \end{aligned}$$
□
1.10 A.10 Proof of Proposition 5.4
(a) From (4.9), we have
$$\begin{aligned} 0=\int _{0}^{\bar{\nu}} \xi ^{-n_{2}-1}\Big(u\big(I(\xi )\big)-\beta P \Big)d\xi =\int _{0}^{\bar{\nu}} \xi ^{-n_{2}-1}u\big(I(\xi )\big)d \xi +\beta P\frac{\bar{\nu}^{-n_{2}}}{n_{2}}. \end{aligned}$$
By differentiating the above equation with respect to \(P\), we obtain
$$\begin{aligned} 0&=\bar{\nu}^{-n_{2}-1}u\big(I(\bar{\nu})\big)\frac{d\bar{\nu}}{dP}- \beta P\bar{\nu}^{-n_{2}-1}\frac{d\bar{\nu}}{dP}+\beta \frac{\bar{\nu}^{-n_{2}}}{n_{2}} \\ &=\bar{\nu}^{-n_{2}-1}\Big(u\big(I(\bar{\nu})\big)-\beta P\Big) \frac{d\bar{\nu}}{dP}+\beta \frac{\bar{\nu}^{-n_{2}}}{n_{2}}. \end{aligned}$$
It follows that
$$\begin{aligned} \frac{d\bar{\nu}}{dP}=-\beta \frac{\bar{\nu}^{-n_{2}}}{n_{2}} \frac{1}{\bar{\nu}^{-n_{2}-1} (u (I(\bar{\nu}) )-\beta P )}< 0, \end{aligned}$$
(A.50)
where we have used the inequality (4.10). Then by differentiating (5.1) with respect to \(P\), we deduce that
$$\begin{aligned} \frac{d\bar{x}}{d P}&=\frac{\beta }{r\bar{\nu}}- \frac{-I(\bar{\nu})\bar{\nu}-\tilde{u}(\bar{\nu})+\beta P}{r\bar{\nu}^{2}} \frac{d\bar{\nu}}{dP} =\frac{\beta }{r\bar{\nu}}+ \frac{u (I(\bar{\nu}) )-\beta P}{r\bar{\nu}^{2}}\frac{d\bar{\nu}}{dP}>0. \end{aligned}$$
It follows from (4.10) that
$$\begin{aligned} I(\bar{\nu}) < u^{-1}(\beta P). \end{aligned}$$
(A.51)
By the mean value theorem, there exists \(y_{\delta }\in (I(\bar{\nu}), u^{-1}(\beta P))\) such that
$$\begin{aligned} 0> u\big(I(\bar{\nu})\big)-\beta P &=u\big(I(\bar{\nu})\big)-u\big(u^{-1}( \beta P)\big) \\ &=u'(y_{\delta }) \big(I(\bar{\nu})-u^{-1}(\beta P)\big)>\bar{\nu} \big(I(\bar{\nu})-u^{-1}(\beta P)\big), \end{aligned}$$
or equivalently
$$\begin{aligned} \big(I(\bar{\nu})-u^{-1}(\beta P)\big) < \frac{u (I(\bar{\nu}) )-\beta P}{\bar{\nu}} < 0. \end{aligned}$$
(A.52)
It is obvious that
$$ \lim _{P\downarrow \frac{1}{\beta }u(0)}u^{-1}(\beta P)=0, $$
and it follows from (A.51) that
$$ \lim _{P\downarrow \frac{1}{\beta }u(0)}I({\bar{\nu}}) \leq \lim _{P \downarrow \frac{1}{\beta }u(0)}u^{-1}(\beta P) = 0. $$
Since \(I(y)\) is continuous and \(I(y)> 0\), we have
$$ \lim _{P\downarrow \frac{1}{\beta }u(0)}I({\bar{\nu}})=0. $$
Then from (A.52), we deduce that
$$\begin{aligned} \lim _{P\downarrow \frac{1}{\beta }u(0)} \frac{u (I(\bar{\nu}) )-\beta P}{\bar{\nu}}=0, \end{aligned}$$
and it follows from (5.1) that
$$\begin{aligned} \lim _{P\downarrow \frac{1}{\beta }u(0)}\bar{x} = - \frac{\epsilon }{r}. \end{aligned}$$
On other hand, we can see that
$$ \lim _{P \uparrow \frac{1}{\beta }u(\infty )}\int _{0}^{\varrho} \xi ^{-n_{2}-1} \Big(u\big(I(\xi )\big)-\beta P\Big)d\xi < 0 $$
for any \(\varrho >0\). Hence we deduce that
$$ \lim _{P\uparrow \frac{1}{\beta }u(\infty )}\bar{\nu}=0. $$
Thus it follows from \(u (I(\bar{\nu}) )-\beta P <0\) and \(\lim _{P\uparrow \frac{1}{\beta }u(\infty )}I(\bar{\nu})=\infty \) that
$$ \lim _{P\uparrow \frac{1}{\beta }u(\infty )} \bar{x} = \frac{1}{r} \lim _{P\uparrow \frac{1}{\beta }u(\infty )}\bigg(\frac{1}{\bar{\nu}} \Big(-\Big(u\big(I(\bar{\nu})\big)-\beta P\Big)\Big) +I(\bar{\nu}) \bigg)=\infty . $$
(b) and (c) With a slight abuse of notation, the optimal consumption and investment at time \(t\) for a given wealth level \(X_{t}\geq \bar{x}\) are \(c_{t}^{*}=c^{*}(X_{t})\) and \(\pi _{t}^{*}=\pi ^{*}(X_{t})\) with
$$\begin{aligned} c^{*}(x) &= I(y^{*}), \\ \pi ^{*}(x)&=\frac{\theta}{\sigma} y^{*} J''(y^{*}), \end{aligned}$$
where \(y^{*}\in (0,\bar{\nu}]\) is the unique solution to \(x=-J'(y^{*})\).
To prove part (b) of Proposition 5.4, it is enough to show that \(\frac{d I(y^{*})}{dP} \leq 0\) for a given \(x\geq \bar{x}\). Since \(\frac{d\bar{\nu}}{dP}<0\) in (A.50), it follows that
$$\begin{aligned} \frac{dD_{1}}{dP}&=-\bar{\nu}^{-n_{1}-1}\Big(u\big(I(\bar{\nu})\big)- \beta P\Big)\frac{d\bar{\nu}}{dP} - \beta \frac{\bar{\nu}^{-n_{1}}}{n_{1}}< 0. \end{aligned}$$
(A.53)
Recall that \(J'(y)\) is given in (5.7) and \(D_{1}\) depends on \(P\). For a given \(x\geq \bar{x}\), by differentiating both sides of \(x=-J'(y^{*})\) with respect to \(P\), we have
$$\begin{aligned} 0=-J''(y^{*})\frac{dy^{*}}{dP}-n_{1}(y^{*})^{n_{1}-1} \frac{dD_{1}}{dP}, \end{aligned}$$
and it follows from the convexity of \(J(y)\) and \(dD_{1}/dP<0\) in (A.53) that
$$\begin{aligned} \frac{d{y^{*}}}{dP} = - \frac{n_{1} {(y^{*})}^{n_{1}-1}}{J''({y^{*}})} \frac{dD_{1}}{dP}>0. \end{aligned}$$
(A.54)
Since \(I(y)\) is decreasing in \(y\), we can deduce that
$$\begin{aligned} \frac{dI(y^{*})}{dP} < 0. \end{aligned}$$
To prove part (c) of Proposition 5.4, we need to show that \({d\pi ^{*}(x)}/{dP}\) and \(\theta \) always have opposite signs. For a given \(x>\bar{x}\), by using the relationship \(x=-J'(y^{*})\), we can derive
$$\begin{aligned} \pi ^{*}(x)&=\frac{\theta}{\sigma}y^{*}J''(y^{*})= \frac{\theta}{\sigma}\big(y^{*}J''(y^{*})+x+J'(y^{*})\big) \\ &=\frac{\theta}{\sigma}\bigg(x+\frac{\epsilon}{r}+D_{1} n_{1}^{2} {(y^{*})}^{n_{1}-1} \\ & \hphantom{=\frac{\theta}{\sigma}\bigg[} + \frac{2}{\theta ^{2}(n_{1}-n_{2})}\Big(n_{2}^{2} {(y^{*})}^{n_{2}-1} \int _{0}^{y^{*}} d\xi + n_{1}^{2} {(y^{*})}^{n_{1}-1} \int _{y^{*}}^{ \infty }d\xi \Big)\bigg) \\ & \phantom{=:} - \frac{2}{\theta ^{2}}\frac{\tilde{u}({y^{*}})}{{y^{*}}}. \end{aligned}$$
Taking into account that \(D_{1}\) depends on \(P\) and differentiating both sides with respect to \(P\), we have
$$\begin{aligned} \frac{d\pi ^{*}(x)}{dP} &= \frac{\theta}{\sigma}\bigg( \frac{d D_{1} }{dP} n_{1}^{2} {(y^{*})}^{n_{1}-1} + \big({y^{*}}J'''({y^{*}}) + 2J''({y^{*}})\big)\frac{d{y^{*}}}{dP}\bigg) \\ &= \frac{\theta}{\sigma} \frac{-{y^{*}}J'''({y^{*}})+(n_{1}-2)J''({y^{*}})}{J''({y^{*}})}n_{1} {(y^{*})}^{n_{1}-1} \frac{dD_{1}}{dP}, \end{aligned}$$
where the second equality comes from (A.54). We can show that
$$\begin{aligned} -{y^{*}}J'''({y^{*}})+(n_{1}-2)J''({y^{*}}) = \frac{2}{\theta ^{2}} (1-n_{2}) \int _{0}^{y^{*}} \xi ^{-n_{2}} I(\xi ) d\xi > 0 \end{aligned}$$
because \(I(\xi )>0\), and we also know that \(J''(y^{*})>0\) and \(dD_{1}/dP<0\). Thus it follows that \({d\pi ^{*}(x)}/{dP}\) and \(\theta \) always have opposite signs, which completes the proof. □
1.11 A.11 Proof of Proposition 5.5
Let \(\bar{x}\) be endogenously determined as in (5.1) by the given minimum welfare level \(P\). Since \(X_{t}^{*}=X_{t}^{c^{*},\pi ^{*}} \geq \bar{x}\) for all \(t\geq 0\) by part (b) in Theorem 5.2, it follows that \((c^{*},\pi ^{*})\in {\mathcal{A}}_{B}(x)\). Thus we have
$$\begin{aligned} V(x) &=\mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}u(c_{t}^{*})dt \bigg] \\ &\le \sup _{(c,\pi )\in{\mathcal{A}}_{B}(x)}\mathbb{E}\bigg[\int _{0}^{ \infty }e^{-\beta t}u(c_{t})dt\bigg] = V_{B}(x). \end{aligned}$$
(A.55)
For any \((c^{B}, \pi ^{B})\in{\mathcal{A}}_{B}(x)\), define
$$\begin{aligned} {\mathcal{W}}_{t}^{B}:= \mathbb{E}\bigg[\int _{t}^{\infty }e^{-\beta (s-t)}u(c_{s}^{B})ds \bigg]\qquad \text{for } t\ge 0. \end{aligned}$$
Let \(t_{0}\) be the first time such that \({\mathcal{W}}_{t_{0}}^{B} < P\), i.e.,
$$ t_{0}:=\inf \{t>0 : {\mathcal{W}}_{t}^{B} < P \}. $$
Since \(X_{t_{0}}^{c^{B}, \pi ^{B}}\ge \bar{x}\), Theorem 5.2 implies that there exists \(y_{t_{0}}\in (0,\bar{\nu}(P)]\) such that
$$ X_{t_{0}}^{c^{B},\pi ^{B}} = -J'(y_{t_{0}}), $$
where \(J(y)\) is the dual value function of our problem with welfare constraints. Here, we adopt the notation \(\bar{\nu}(P)\) to indicate the dependence of \(\bar{\nu}\) on \(P\). Let us define \({\mathcal{Y}}^{t_{0}}\) and \({\mathcal{Z}}^{t_{0}}\) by
$$ {\mathcal{Y}}_{t}^{t_{0}}:= y_{t_{0}} e^{\beta (t-t_{0})} {\mathcal{H}}_{t}/{ \mathcal{H}}_{t_{0}},\qquad {\mathcal{Z}}_{t}^{t_{0}}=\max \bigg\{ 0, \sup _{t_{0} \le s \le t}\frac{{\mathcal{Y}}_{s}^{t_{0}}}{\bar{\nu}(P)}-1 \bigg\} $$
for \(t\ge t_{0}\). We also define the consumption and portfolio strategy \((c^{t_{0}},\pi ^{t_{0}})\) for \(t\ge t_{0}\) as
$$\begin{aligned} c_{t}^{t_{0}} =I\bigg( \frac{ {\mathcal{Y}}_{t}^{t_{0}}}{1+{\mathcal{Z}}_{t}^{t_{0}}} \bigg), \qquad \pi _{t}^{t_{0}}=\frac{\theta }{\sigma } \frac{ {\mathcal{Y}}_{t}^{t_{0}}}{1+{\mathcal{Z}}_{t}^{t_{0}}} J''\bigg( \frac{ {\mathcal{Y}}_{t}^{t_{0}}}{1+{\mathcal{Z}}_{t}^{t_{0}}} \bigg). \end{aligned}$$
It follows from (4.20) and the strong Markov property that for all \(t \ge t_{0}\),
$$\begin{aligned} X_{t}^{c^{t_{0}},\pi ^{t_{0}}} = -J'\bigg( \frac{ {\mathcal{Y}}_{t}^{t_{0}}}{1+{\mathcal{Z}}_{t}^{t_{0}}} \bigg) \ge \bar{x},\qquad \mathbb{E}\bigg[\int _{t}^{\infty }e^{-\beta (s-t)} u(c_{s}^{t_{0}})ds \bigg] \ge P. \end{aligned}$$
Let us define a pair \((\tilde{c}, \tilde{\pi})\) of consumption and portfolio strategy by
$$\begin{aligned} \tilde{c}_{t}= \textstyle\begin{cases} c_{t}^{B} \qquad \text{for } 0\le t< t_{0}, \\ c_{t}^{t_{0}}\qquad \text{for } t\ge t_{0}, \end{cases}\displaystyle \text{and} \qquad \tilde{\pi}_{t}= \textstyle\begin{cases} \pi _{t}^{B} \qquad \text{for } 0\le t< t_{0}, \\ \pi _{t}^{t_{0}}\qquad \text{for } t\ge t_{0}. \end{cases}\displaystyle \end{aligned}$$
By the definition of \(t_{0}\), we have \((\tilde{c}, \tilde{\pi})\in \mathcal{A}(x)\). Thus it follows that
$$\begin{aligned} V(x) \geq \mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}u(\tilde{c}_{t})dt \bigg]&=\mathbb{E}\bigg[\int _{0}^{t_{0}} e^{-\beta t}u(c_{t}^{B})dt \bigg]+\mathbb{E}\bigg[\int _{t_{0}}^{\infty }e^{-\beta t}u(c_{t}^{t_{0}})dt \bigg] \\ &\ge \mathbb{E}\bigg[\int _{0}^{t_{0}} e^{-\beta t}u(c_{t}^{B})dt \bigg] +P \\ &>\mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}u(c_{t}^{B})dt \bigg]. \end{aligned}$$
Since the above inequality holds for any \((c^{B}, \pi ^{B})\in{\mathcal{A}}_{B}(x)\), we can conclude that
$$\begin{aligned} V(x) \geq V_{B}(x). \end{aligned}$$
(A.56)
From (A.55) and (A.56), we conclude that
$$\begin{aligned} V_{B}(x) = \mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t}u(c_{t}^{*})dt \bigg] = V(x). \end{aligned}$$
That is, \((c^{*},\pi ^{*})\) is optimal for the wealth-constrained problem \(V_{B}(x)\). □
