Borrowing constraints, effective flexibility in labor supply, and portfolio selection

Abstract

We study optimal job switching and consumption/investment policies of an economic agent under the borrowing constraints against future labor income in a continuous and infinite time horizon. The agent’s preference is given by the Cobb–Douglas utility function whose arguments are consumption and leisure, and the jobs are characterized by the trade-off between labor income and leisure. We obtain a closed-form solution to the optimization problem by using the martingale and duality method, and investigate theoretical implications of it. The most interesting finding for the optimal job switching policy is that the borrowing constraints in the financial market can decrease the effective flexibility in labor supply of the agent in the labor market. Thus, an environment of the financial market can affect an agent’s decision making in the labor market. We also show how the effects of the borrowing constraints on the optimal consumption/investment policy reinforce or compete with those of the job switching opportunities.

Appendices

A Proof of Lemma 1

The proof is similar to that of the Eq. (5) in Lim and Shin [12] and that of the Eq. (17) in He and Pagès [8]:

For any fixed $T > 0$, we can define an equivalent martingale probability measure $\widetilde{{\mathbb {P}}}^T$ as

$$\begin{aligned} \widetilde{\mathbb {P}}^T(A) = {\mathbb {E}}[e^{-\frac{1}{2}\theta ^2T-\theta B_T}{\mathbf {1}}_A], ~~\text{ for } \text{ any }~~ A \in {\mathcal {F}}_T. \end{aligned}$$

(45)

Then, by Girsanov theorem, the process ${\widetilde{B}}^T_t \triangleq B_t + \theta t$ is a standard Brownian motion under $\widetilde{\mathbb {P}}^T$ for $t \in [0, T]$. By Itô’s formula, we have

$$\begin{aligned} d(e^{-rt}X_t) = e^{-rt}\left( -c_t+Y_0 {\mathbf {1}}_{\{\varTheta _t=A_0\}}+Y_1 {\mathbf {1}}_{\{\varTheta _t=A_1\}}\right) dt+e^{-rt}\pi _t\sigma d{\widetilde{B}}^T_t. \end{aligned}$$

(46)

Since $\varvec{\delta }$ is of bounded variation and $e^{-rt}X_t$ evolving according to (46) has continuous sample paths, we can apply integration by parts to get

$$\begin{aligned}&e^{-rt}X_t\delta _t +\int _0^te^{-rs}\left( c_s-Y_0 {\mathbf {1}}_{\{\varTheta _s=A_0\}}-Y_1 {\mathbf {1}}_{\{\varTheta _s=A_1\}}\right) \delta _sds - \int _0^te^{-rs}X_s\,d\delta _s\\&\quad = x+\int _0^te^{-rs}\pi _s\sigma \delta _s d{\widetilde{B}}^T_s. \end{aligned}$$

Since $X_t \ge 0$ a.s. for $t \ge 0$ and $\varvec{\delta }$ is a decreasing process, $\int _0^te^{-rs}X_s\,d\delta _s \le 0$. Therefore, the left-hand side of the above equation is a continuous local martingale bounded from below and hence a supermartingale under $\widetilde{\mathbb {P}}^T$ for $t \in [0, T]$ so that

$$\begin{aligned} \widetilde{\mathbb {E}}^T\left[ e^{-rT}X_T\delta _T+\int _0^Te^{-rt}\left( c_t-Y_0 {\mathbf {1}}_{\{\varTheta _t=A_0\}}-Y_1 {\mathbf {1}}_{\{\varTheta _t=A_1\}}\right) \delta _t dt - \int _0^Te^{-rt}X_t\,d\delta _t\right] \le x, \end{aligned}$$

where $\widetilde{{\mathbb {E}}}^T$ denotes the expectation operator under the probability measure $\widetilde{{\mathbb {P}}}^T$. Since $\int _0^te^{-rs}X_s\,d\delta _s \le 0$, we have

$$\begin{aligned} \widetilde{{\mathbb {E}}}^T\left[ e^{-rT}X_T\delta _T+\int _0^Te^{-rt}\left( c_t-Y_0 {\mathbf {1}}_{\{\varTheta _t=A_0\}}-Y_1 {\mathbf {1}}_{\{\varTheta _t=A_1\}}\right) \delta _t dt \right] \le x. \end{aligned}$$

(47)

Note that

$$\begin{aligned}&\widetilde{{\mathbb {E}}}^T\left[ e^{-rT}X_T\delta _T+\int _0^Te^{-rt}\left( c_t-Y_0 {\mathbf {1}}_{\{\varTheta _t=A_0\}}-Y_1 {\mathbf {1}}_{\{\varTheta _t=A_1\}}\right) \delta _t dt \right] \nonumber \\&\quad ={\mathbb {E}}\left[ e^{-\frac{1}{2}\theta ^2T-\theta B_T}e^{-rT}X_T\delta _T+e^{-\frac{1}{2}\theta ^2T-\theta B_T}\int _0^Te^{-rt}\left( c_t-Y_0 {\mathbf {1}}_{\{\varTheta _t=A_0\}}-Y_1 {\mathbf {1}}_{\{\varTheta _t=A_1\}}\right) \delta _tdt\right] \nonumber \\&\quad ={\mathbb {E}}\left[ H_TX_T\delta _T+\int _0^Te^{-rt}\left( c_t-Y_0 {\mathbf {1}}_{\{\varTheta _t=A_0\}}-Y_1 {\mathbf {1}}_{\{\varTheta _t=A_1\}}\right) \delta _te^{-\frac{1}{2}\theta ^2T-\theta B_T}dt\right] \nonumber \\&\quad ={\mathbb {E}}[H_TX_T\delta _T]+\int _0^T{\mathbb {E}}\left[ e^{-rt}\left( c_t-Y_0 {\mathbf {1}}_{\{\varTheta _t=A_0\}}-Y_1 {\mathbf {1}}_{\{\varTheta _t=A_1\}}\right) \delta _te^{-\frac{1}{2}\theta ^2T-\theta B_T}\right] dt\nonumber \\&\quad ={\mathbb {E}}[H_TX_T\delta _T]+\int _0^T{\mathbb {E}}\left[ {\mathbb {E}}\left[ \left. e^{-rt}\left( c_t-Y_0 {\mathbf {1}}_{\{\varTheta _t=A_0\}}-Y_1 {\mathbf {1}}_{\{\varTheta _t=A_1\}}\right) \delta _te^{-\frac{1}{2}\theta ^2T-\theta B_T}\right| {\mathcal {F}}_t\right] \right] dt\nonumber \\&\quad ={\mathbb {E}}[H_TX_T\delta _T]+\int _0^T{\mathbb {E}}\left[ \left( c_t-Y_0 {\mathbf {1}}_{\{\varTheta _t=A_0\}}-Y_1 {\mathbf {1}}_{\{\varTheta _t=A_1\}}\right) \delta _tH_t{\mathbb {E}}\left[ \left. e^{-\frac{1}{2}\theta ^2(T-t)-\theta (B_T-B_t)}\right| {\mathcal {F}}_t\right] \right] dt\nonumber \\&\quad ={\mathbb {E}}[H_TX_T\delta _T]+\int _0^T{\mathbb {E}}\left[ \left( c_t-Y_0 {\mathbf {1}}_{\{\varTheta _t=A_0\}}-Y_1 {\mathbf {1}}_{\{\varTheta _t=A_1\}}\right) \delta _tH_t{\mathbb {E}}\left[ e^{-\frac{1}{2}\theta ^2(T-t)-\theta (B_T-B_t)}\right] \right] dt\nonumber \\&\quad ={\mathbb {E}}[H_TX_T\delta _T]+\int _0^T{\mathbb {E}}\left[ \left( c_t-Y_0 {\mathbf {1}}_{\{\varTheta _t=A_0\}}-Y_1 {\mathbf {1}}_{\{\varTheta _t=A_1\}}\right) \delta _tH_t\right] dt\nonumber \\&\quad ={\mathbb {E}}\left[ H_TX_T\delta _T+\int _0^T\left( c_t-Y_0 {\mathbf {1}}_{\{\varTheta _t=A_0\}}-Y_1 {\mathbf {1}}_{\{\varTheta _t=A_1\}}\right) \delta _tH_tdt\right] , \end{aligned}$$

(48)

where the first equality comes from (45), the second from (8), the third from Fubini theorem, the forth from the tower property, the fifth from the fact that $\left( c_t-Y_0{\mathbf {1}}_{\{\varTheta _t=A_0\}}-Y_1 {\mathbf {1}}_{\{\varTheta _t=A_1\}}\right) \delta _tH_t$ is ${\mathcal {F}}_t$-measurable, the sixth from the independent increment of a Brownian motion, the seventh from the fact that ${\mathbb {E}}\left[ e^{-\frac{1}{2}\theta ^2(T-t)-\theta (B_T-B_t)}\right] = 1$, the last again from Fubini theorem. Combining (47) and (48), we have

$$\begin{aligned} {\mathbb {E}}\left[ H_TX_T\delta _T+\int _0^T\left( c_t-Y_0 {\mathbf {1}}_{\{\varTheta _t=A_0\}}-Y_1 {\mathbf {1}}_{\{\varTheta _t=A_1\}}\right) \delta _tH_tdt\right] \le x. \end{aligned}$$

Since $H_TX_T\delta _T \ge 0$ by (2) and (8), we have

$$\begin{aligned} {\mathbb {E}}\left[ \int _0^T\left( c_t-Y_0 {\mathbf {1}}_{\{\varTheta _t=A_0\}}-Y_1 {\mathbf {1}}_{\{\varTheta _t=A_1\}}\right) \delta _tH_tdt\right] \le x. \end{aligned}$$

(49)

Since (49) holds for any $T \ge 0$, we get (11). $\square$

B Proof of Lemma 2

The proof is similar to that of Lemma 1 in He and Pagès [8]. Let

$$\begin{aligned} \varGamma _t \triangleq \sup _{T \in [t, \infty )}{\mathbb {E}}\left[ \left. \int _0^T\left( c_s-Y_0 {\mathbf {1}}_{\{\varTheta _s=A_0\}}-Y_1 {\mathbf {1}}_{\{\varTheta _s=A_1\}}\right) H_sds\right| {\mathcal {F}}_t\right] \quad \text{ for }~~t\ge 0. \end{aligned}$$

(50)

Then,

$$\begin{aligned} -\left( Y_0+Y_1\right) {\mathbb {E}}\left[ \left. \int _0^\infty H_sds\right| {\mathcal {F}}_t\right] \le \varGamma _t \le {\mathbb {E}}\left[ \left. \int _0^{\infty }c_sH_sds\right| {\mathcal {F}}_t\right] . \end{aligned}$$

Taking the expectation on each term of the above inequalities, we have, by the tower property,

$$\begin{aligned} -\left( Y_0+Y_1\right) {\mathbb {E}}\left[ \int _0^\infty H_sds\right] \le {\mathbb {E}}\left[ \varGamma _t\right] \le {\mathbb {E}}\left[ \int _0^{\infty }c_sH_sds\right] . \end{aligned}$$

(51)

However,

$$\begin{aligned} -\left( Y_0+Y_1\right) {\mathbb {E}}\left[ \int _0^\infty H_sds\right] = -\left( Y_0+Y_1\right) \int _0^\infty {\mathbb {E}}\left[ H_s\right] ds = -\frac{Y_0+Y_1}{r}, \end{aligned}$$

(52)

and

$$\begin{aligned} {\mathbb {E}}\left[ \int _0^{\infty }c_sH_sds\right]= & {} {\mathbb {E}}\left[ \int _0^\infty \left( c_s-Y_0 {\mathbf {1}}_{\{\varTheta _s=A_0\}}-Y_1 {\mathbf {1}}_{\{\varTheta _s=A_1\}}\right) H_sds\right] \nonumber \\&+ \,{\mathbb {E}}\left[ \int _0^\infty \left( Y_0 {\mathbf {1}}_{\{\varTheta _s=A_0\}}+Y_1 {\mathbf {1}}_{\{\varTheta _s=A_1\}}\right) H_sds\right] \nonumber \\\le & {} {\mathbb {E}}\left[ \int _0^\infty \left( c_s-Y_0 {\mathbf {1}}_{\{\varTheta _s=A_0\}}-Y_1 {\mathbf {1}}_{\{\varTheta _s=A_1\}}\right) H_sds\right] \nonumber \\&+\left( Y_0+Y_1\right) {\mathbb {E}}\left[ \int _0^\infty H_sds\right] \nonumber \\\le & {} x + \frac{Y_0+Y_1}{r}, \end{aligned}$$

(53)

where the last inequality (53) comes from (13) and (52). By (51), (52), and (53), $\varGamma _t$ is integrable for every $t \ge 0$, and

$$\begin{aligned} \varGamma _0 = \sup _{T \in [0, \infty )}{\mathbb {E}}\left[ \int _0^T\left( c_s-Y_0 {\mathbf {1}}_{\{\varTheta _s=A_0\}}-Y_1 {\mathbf {1}}_{\{\varTheta _s=A_1\}}\right) H_sds\right] \le x, \end{aligned}$$

(54)

where the inequality holds since $(\varvec{\varTheta }, {\mathbf {c}})$ satisfies the budget constraint (12). Note that

$$\begin{aligned} \varGamma _t \ge \int _{0}^{t}\left( c_s-Y_0 {\mathbf {1}}_{\{\varTheta _s=A_0\}}-Y_1 {\mathbf {1}}_{\{\varTheta _s=A_1\}}\right) H_sds. \end{aligned}$$

(55)

Consider any fixed t and $t^{\prime }$ with $t < t^{\prime }$. For any given $\epsilon > 0$, there exists $T^{\prime } \ge t^{\prime }$ such that

$$\begin{aligned} \varGamma _{t^{\prime }} - {\mathbb {E}}\left[ \left. \int _{0}^{T^{\prime }}\left( c_s-Y_0 {\mathbf {1}}_{\{\varTheta _s=A_0\}}-Y_1 {\mathbf {1}}_{\{\varTheta _s=A_1\}}\right) H_sds\right| {\mathcal {F}}_{t^{\prime }}\right] < \epsilon . \end{aligned}$$

(56)

Thus, we have

$$\begin{aligned} {\mathbb {E}}[\varGamma _{t^{\prime }} | {\mathcal {F}}_t]= & {} {\mathbb {E}}\left[ \left. \varGamma _{t^{\prime }} - {\mathbb {E}}\left[ \left. \int _{0}^{T^{\prime }}\left( c_s-Y_0 {\mathbf {1}}_{\{\varTheta _s=A_0\}}-Y_1 {\mathbf {1}}_{\{\varTheta _s=A_1\}}\right) H_sds\right| {\mathcal {F}}_{t^{\prime }}\right] \right| {\mathcal {F}}_t\right] \nonumber \\&+\, {\mathbb {E}}\left[ \left. {\mathbb {E}}\left[ \left. \int _{0}^{T^{\prime }}\left( c_s-Y_0 {\mathbf {1}}_{\{\varTheta _s=A_0\}}-Y_1 {\mathbf {1}}_{\{\varTheta _s=A_1\}}\right) H_sds\right| {\mathcal {F}}_{t^{\prime }}\right] \right| {\mathcal {F}}_t\right] \end{aligned}$$

(57)

$$\begin{aligned}\le & {} \epsilon + {\mathbb {E}}\left[ \left. \int _{0}^{T^{\prime }}\left( c_s-Y_0 {\mathbf {1}}_{\{\varTheta _s=A_0\}}-Y_1 {\mathbf {1}}_{\{\varTheta _s=A_1\}}\right) H_sds\right| {\mathcal {F}}_t\right] \nonumber \\\le & {} \epsilon + \varGamma _t, \end{aligned}$$

(58)

where the first inequality holds by (56) and the tower property, the second since $T^{\prime } \ge t^{\prime } > t$. Since the inequality (58) holds holds for arbitrary $\epsilon > 0$, we have ${\mathbb {E}}[\varGamma _{t^{\prime }} | {\mathcal {F}}_t] \le \varGamma _t$. That is, $(\varGamma _t)_{t\ge 0}$ is a supermartingale under ${\mathbb {P}}$ so that the Doob-Meyer Decomposition Theorem allows us to write

$$\begin{aligned} \varGamma _t = \varGamma _0 + M_t - K_t, \end{aligned}$$

(59)

where $(M_t)_{t\ge 0}$ is a martingale under ${\mathbb {P}}$ with $M_0 = 0$ and $(K_t)_{t\ge 0}$ is an increasing process with $K_0 = 0$. By the martingale representation theorem, we can find an ${\mathcal {F}}_t$-adapted process $(\psi _t)_{t\ge 0}$ with $\int _0^t\psi ^2_tds<\infty ,~\text{ a.s. }$, such that

$$\begin{aligned} M_t = \int _0^t\psi _sdB_s,~\quad \text{ for }~ t\ge 0. \end{aligned}$$

(60)

Consider the portfolio process $\varvec{\pi }$ given by

$$\begin{aligned} \pi _t = \frac{\psi _t}{\sigma H_t} + \frac{\theta }{\sigma }X_t,\quad t \ge 0, \end{aligned}$$

where $X_t$ is the agent’s wealth at time t. With the policy $(\varvec{\varTheta }, {\mathbf {c}}, \varvec{\pi })$, the stochastic differential equation (10) becomes

$$\begin{aligned} d(H_tX_t) = \left[ -c_t+Y_0 {\mathbf {1}}_{\{\varTheta _t=A_0\}}+Y_1 {\mathbf {1}}_{\{\varTheta _t=A_1\}}\right] H_tdt+\psi _tdB_t. \end{aligned}$$

Thus,

$$\begin{aligned} H_tX_t= & {} x - \int _{0}^{t}\left( c_s-Y_0 {\mathbf {1}}_{\{\varTheta _s=A_0\}}-Y_1 {\mathbf {1}}_{\{\varTheta _s=A_1\}}\right) H_sds + \int _0^t\psi _sdB_s\\= & {} x - \int _{0}^{t}\left( c_s-Y_0 {\mathbf {1}}_{\{\varTheta _s=A_0\}}-Y_1 {\mathbf {1}}_{\{\varTheta _s=A_1\}}\right) H_sds + M_t\\= & {} x -\varGamma _0 + \varGamma _t - \int _{0}^{t}\left( c_s-Y_0 {\mathbf {1}}_{\{\varTheta _s=A_0\}}-Y_1 {\mathbf {1}}_{\{\varTheta _s=A_1\}}\right) H_sds + K_t\\\ge & {} 0, \end{aligned}$$

where the second equality comes from (60), the third from (59), and the inequality from (54), (55), and the fact that $K_t \ge 0$. Thus, $X_t$ satisfies (2). $\Box$

C Proof of Lemma 3

For any triple $(\varvec{\varTheta }, {\mathbf {c}}, \varvec{\pi })\in {\mathcal {A}}(x)$, $\varvec{\delta }\in {\mathcal {D}}$, and $\lambda > 0$, we have

$$\begin{aligned} J(x;\varvec{\varTheta }, \mathbf {c}, \varvec{\pi })&=\mathbb {E}\left[ \int _0^\infty e^{-\rho t}\left( L_0^{\gamma _1-\gamma }\frac{c_t^{1-\gamma _1}}{1-\gamma _1}\mathbf {1}_{ \{\varTheta _t=A_0\}}+L_1^{\gamma _1-\gamma }\frac{c_t^{1-\gamma _1}}{1-\gamma _1}\mathbf {1}_{\{ \varTheta _t=A_1\}}\right) dt\right] \nonumber \\&\le \mathbb {E}\left[ \int _0^\infty e^{-\rho t}\left\{ \left( \widetilde{u}_0(z_t)+Y_0z_t\right) \mathbf {1}_{\{\varTheta _t=A_0\}}+ \left( \widetilde{u}_1(z_t)+Y_1z_t\right) \mathbf {1}_{\{ \varTheta _t=A_1\}}\right\} dt\right] \nonumber \\&\quad + \lambda \mathbb {E}\left[ \int _0^\infty \left( c_t-Y_0 \mathbf {1}_{\{\varTheta _t=A_0\}}-Y_1 \mathbf {1}_{\{\varTheta _t=A_1\}}\right) \delta _tH_tdt\right] \end{aligned}$$

(61)

$$\begin{aligned}&\le \mathbb {E}\left[ \int _0^\infty e^{-\rho t}\left\{ \left( \widetilde{u}_0(z_t)+Y_0z_t\right) \mathbf {1}_{\{ \varTheta _t=A_0\}}+\left( \widetilde{u}_1(z_t)+Y_1z_t\right) \mathbf {1}_{\{ \varTheta _t=A_1\}}\right\} dt\right] \nonumber \\&\quad + \lambda x \end{aligned}$$

(62)

$$\begin{aligned}&\le \mathbb {E}\left[ \int _0^\infty e^{-\rho t}\left\{ \left( \widetilde{u}_0(z_t)+Y_0z_t\right) \mathbf {1}_{\{0<z_t \le \bar{z}\}}+\left( \widetilde{u}_1(z_t)+Y_1z_t\right) \mathbf {1}_{\{z_t> \bar{z}\}}\right\} dt\right] \nonumber \\&\quad +\lambda x \nonumber \\&= \varPhi (\varvec{\delta }, \lambda ), \end{aligned}$$

(63)

where the first inequality comes from (5), the second inequality from Remark 3, and the last inequality from Remark 2. Thus, the inequality (16) holds for any triple $(\varvec{\varTheta }, {\mathbf {c}}, \varvec{\pi })\in {\mathcal {A}}(x)$, $\varvec{\delta }\in {\mathcal {D}}$, and $\lambda > 0$. By (6), the inequality (61) holds with equality if and only if

$$\begin{aligned} c_t=\left\{ \begin{array}{ll} f_0(z_t) = L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}z_t^{-\frac{1}{\gamma _1}}, &{}\quad \text{ for } \varTheta _t=A_0,\\ f_1(z_t) = L_1^{\frac{\gamma _1-\gamma }{\gamma _1}}z_t^{-\frac{1}{\gamma _1}}, &{}\quad \text{ for } \varTheta _t=A_1. \end{array}\right. \end{aligned}$$

The inequality (62) holds with equality if and only if

$$\begin{aligned} {\mathbb {E}}\left[ \int _0^\infty \left( c_t-Y_0 {\mathbf {1}}_{\{\varTheta _t=A_0\}}-Y_1 {\mathbf {1}}_{\{\varTheta _t=A_1\}}\right) \delta _tH_tdt\right] = x. \end{aligned}$$

The inequality (63) holds with equality if and only if

$$\begin{aligned} \varTheta _t=\left\{ \begin{array}{ll} A_0, &{}\quad \text{ if } 0<z_t\le {\bar{z}},\\ A_1, &{}\quad \text{ if } z_t>{\bar{z}}. \end{array}\right. \end{aligned}$$

Thus, we have shown the second statement of the lemma. For the proof of the last statement, note that, for any triple $(\varvec{\varTheta }, {\mathbf {c}}, \varvec{\pi })\in {\mathcal {A}}(x)$, we have

$$\begin{aligned} J(x;\varvec{\varTheta }, {\mathbf {c}}, \varvec{\pi }) \le \varPhi (\varvec{\delta }^*, \lambda ^*) = J(x;\varvec{\varTheta }^*, {\mathbf {c}}^*, \varvec{\pi }^*), \end{aligned}$$

where the inequality holds since $(\varvec{\varTheta }, {\mathbf {c}}, \varvec{\pi })\in {\mathcal {A}}(x)$, $\varvec{\delta }\in {\mathcal {D}}$, and $\lambda > 0$ are arbitrary in the inequality(16). Thus, the last statement of the lemma holds. $\Box$

D Proof of Theorem 1

The proof is similar to that of Lemma 1 in He and Pagès [8]: Fix any $T > 0$ and let $\varvec{\delta }^{\epsilon } \triangleq (\delta ^{\epsilon }_t)_{t\ge 0}$ with $\delta ^{\epsilon }_t = (\lambda ^*\delta ^*_t + \epsilon {\mathbf {1}}_{[0,T)}(t))/(\lambda ^* + \epsilon )$ for $t\ge 0$, where $\epsilon > 0$. Then, $\varvec{\delta }^{\epsilon } \in {\mathcal {D}}$. Let $\lambda ^{\epsilon } \triangleq \lambda ^* + \epsilon$ and let $z^{\epsilon }_t \triangleq z_t^{\varvec{\delta }^{\epsilon }, \lambda ^{\epsilon }}$ for $t\ge 0$. Then, $z^{\epsilon }_t = z^*_t + \epsilon e^{\rho t}H_t$ for $0 \le t < T$ and $z^{\epsilon }_t = z^*_t$ for $t \ge T$. Define

$$\begin{aligned} {\mathcal {N}}^{\epsilon } \triangleq {\mathbb {E}}\left[ \int _0^\infty e^{-\rho t}\left\{ \left( {\widetilde{u}}_0(z^{*}_t)+Y_0z^{*}_t\right) { \mathbf {1}}_{\{0<z^{\epsilon }_t\le {\bar{z}}\}}+\left( { \widetilde{u}}_1(z^{*}_t)+Y_1z^{*}_t\right) {\mathbf {1}}_{\{z^{\epsilon }_t>{ \bar{z}}\}}\right\} dt\right] +\lambda ^{*}x. \end{aligned}$$

Since ${\mathcal {N}}^{\epsilon } \le \varPhi (\varvec{\delta }^{*}, \lambda ^*)$ by Remark 2 and $\varPhi (\varvec{\delta }^{*}, \lambda ^*) \le \varPhi (\varvec{\delta }^{\epsilon }, \lambda ^{\epsilon })$, we have

$$\begin{aligned} \frac{\varPhi (\varvec{\delta }^{\epsilon }, \lambda ^{\epsilon }) - {\mathcal {N}}^{\epsilon }}{\epsilon } \ge 0. \end{aligned}$$

That is,

$$\begin{aligned} 0\le & {} {\mathbb {E}}\Bigg [\int _0^T e^{-\rho t}\Bigg \{\Big (\frac{{\widetilde{u}}_0(z^*_t+\epsilon e^{\rho t}H_t)-{\widetilde{u}}_0(z^*_t)}{\epsilon }+Y_0e^{\rho t}H_t\Big ){\mathbf {1}}_{\{0<z^{*}_t+\epsilon e^{\rho t}H_t\le {\bar{z}}\}}\nonumber \\&+\,\Big (\frac{{\widetilde{u}}_1(z^*_t+\epsilon e^{\rho t}H_t)-{\widetilde{u}}_1(z^*_t)}{\epsilon }+Y_1e^{\rho t}H_t\Big ){\mathbf {1}}_{\{z^{*}_t+\epsilon e^{\rho t}H_t>{\bar{z}}\}}\Bigg \}dt\Bigg ]+ x. \end{aligned}$$

(64)

Note that

$$\begin{aligned} 0 \le \int _0^T Y_0H_t{\mathbf {1}}_{\{0<z^{*}_t+\epsilon e^{\rho t}H_t\le {\bar{z}}\}}\,dt \le Y_0\int _0^T H_t\,dt. \end{aligned}$$

By using Fubini Theorem, we get

$$\begin{aligned} {\mathbb {E}}\Bigg [Y_0\int _0^T H_t\,dt\Bigg ] = Y_0\int _0^T {\mathbb {E}}[H_t]\,dt = Y_0\int _0^T e^{-rt}\,dt = \frac{Y_0}{r}(1-e^{-rT}) < \infty . \end{aligned}$$

By the dominated convergence theorem, we have

$$\begin{aligned} \lim _{\epsilon \downarrow 0}{\mathbb {E}}\Bigg [\int _0^T Y_0H_t{\mathbf {1}}_{\{0<z^{*}_t+\epsilon e^{\rho t}H_t\le {\bar{z}}\}}\,dt\Bigg ] = {\mathbb {E}}\Bigg [\lim _{\epsilon \downarrow 0}\int _0^T Y_0H_t{\mathbf {1}}_{\{0<z^{*}_t+\epsilon e^{\rho t}H_t\le {\bar{z}}\}}\,dt\Bigg ]. \end{aligned}$$

Since $Y_0H_t{\mathbf {1}}_{\{0<z^{*}_t+\epsilon e^{\rho t}H_t\le {\bar{z}}\}} \le Y_0H_t$ and $\int _0^T Y_0H_t\,dt < \infty$ a. s., the limit and the integration in the second term of the above equation can be interchanged by the dominated convergence theorem so that we have

$$\begin{aligned} \lim _{\epsilon \downarrow 0}{\mathbb {E}}\Bigg [\int _0^T Y_0H_t{\mathbf {1}}_{\{0<z^{*}_t+\epsilon e^{\rho t}H_t\le {\bar{z}}\}}\,dt\Bigg ] = {\mathbb {E}}\Bigg [\int _0^T Y_0H_t{\mathbf {1}}_{\{0<z^{*}_t\le {\bar{z}}\}}\,dt\Bigg ]. \end{aligned}$$

(65)

Similarly, we have

$$\begin{aligned} \lim _{\epsilon \downarrow 0}{\mathbb {E}}\Bigg [\int _0^T Y_1H_t{\mathbf {1}}_{\{z^{*}_t+\epsilon e^{\rho t}H_t> {\bar{z}}\}}\,dt\Bigg ] = {\mathbb {E}}\Bigg [\int _0^T Y_1H_t{\mathbf {1}}_{\{z^{*}_t > {\bar{z}}\}}\,dt\Bigg ]. \end{aligned}$$

(66)

By using (65) and (66), and by applying Fatou’s Lemma which is possible since ${\widetilde{u}}_i(z^*_t+\epsilon e^{\rho t}H_t)-{\widetilde{u}}_i(z^*_t) \le 0$ for $i = 0, 1$, in (64), we get

$$\begin{aligned} 0\le & {} {\mathbb {E}}\Bigg [\int _0^T e^{-\rho t}\Bigg \{\Big (e^{\rho t}H_t{\widetilde{u}}_0^{\prime }(z^*_t)+Y_0e^{\rho t}H_t\Big ){\mathbf {1}}_{\{0<z^*_t\le {\bar{z}}\}}\\&+\Big (e^{\rho t}H_t{\widetilde{u}}_1^{\prime }(z^*_t)+Y_1e^{\rho t}H_t\Big ){\mathbf {1}}_{\{z^*_t>{\bar{z}}\}}\Bigg \}dt\Bigg ]+ x\\= & {} -{\mathbb {E}}\left[ \int _0^T\left( c^*_t-Y_0 {\mathbf {1}}_{\{\varTheta ^*_t=A_0\}}-Y_1 {\mathbf {1}}_{\{\varTheta ^*_t=A_1\}}\right) H_tdt\right] + x, \end{aligned}$$

where the equality comes from (7) and (19). Therefore, we have

$$\begin{aligned} {\mathbb {E}}\left[ \int _0^T\left( c^*_t-Y_0 {\mathbf {1}}_{\{\varTheta ^*_t=A_0\}}-Y_1 {\mathbf {1}}_{\{\varTheta ^*_t=A_1\}}\right) H_tdt\right] \le x. \end{aligned}$$

(67)

Since this holds for any $T \ge 0$, the job and consumption policies $(\varvec{\varTheta }^*, {\mathbf {c}}^*)$ satisfy the budget constraint (12). Hence, by Lemma 2, there exists a portfolio process $\varvec{\pi }^*$ such that

$$\begin{aligned} (\varvec{\varTheta }^*, {\mathbf {c}}^*, \varvec{\pi }^*)\in {\mathcal {A}}(x). \end{aligned}$$

(68)

Let $\lambda ^{\epsilon } \triangleq (1+\epsilon )\lambda ^*$ where $-1 < \epsilon \ne 0$. Let ${\bar{z}}^{\epsilon }_t \triangleq z_t^{\varvec{\delta }^*, \lambda ^{\epsilon }} = (1+\epsilon )z_t^*$ for $t\ge 0$, and let

$$\begin{aligned} \bar{{\mathcal {N}}}^{\epsilon } \triangleq {\mathbb {E}}\left[ \int _0^\infty e^{-\rho t}\left\{ \left( {\widetilde{u}}_0(z^{*}_t)+Y_0z^{*}_t\right) {\mathbf {1}}_{ \{0<{\bar{z}}^{\epsilon }_t\le {\bar{z}}\}}+\left( {\widetilde{u}}_1(z^{*}_t)+Y_1z^{*}_t \right) {\mathbf {1}}_{\{{\bar{z}}^{\epsilon }_t>{\bar{z}}\}}\right\} dt\right] + \lambda ^{*}x. \end{aligned}$$

Since $\bar{{\mathcal {N}}}^{\epsilon } \le \varPhi (\varvec{\delta }^{*}, \lambda ^*)$ by Remark 2 and $\varPhi (\varvec{\delta }^{*}, \lambda ^*) \le \varPhi (\varvec{\delta }^{*}, \lambda ^{\epsilon })$, we have

$$\begin{aligned} \limsup _{\epsilon \downarrow 0}\frac{\varPhi (\varvec{\delta }^{*}, \lambda ^{\epsilon }) - \bar{{\mathcal {N}}}^{\epsilon }}{\epsilon } \ge 0 ~~\text{ and }~~ \liminf _{\epsilon \uparrow 0}\frac{\varPhi (\varvec{\delta }^{*}, \lambda ^{\epsilon }) - \bar{{\mathcal {N}}}^{\epsilon }}{\epsilon } \le 0, \end{aligned}$$

(69)

where

$$\begin{aligned}&\frac{\varPhi (\varvec{\delta }^{*}, \lambda ^{\epsilon }) - \bar{{\mathcal {N}}}^{\epsilon }}{\epsilon }\\&\quad ={\mathbb {E}}\Bigg [\int _0^{\infty } e^{-\rho t}\Bigg \{\Big (\frac{{\widetilde{u}}_0(z^*_t+\epsilon \lambda ^*\delta ^*_te^{\rho t}H_t)-{\widetilde{u}}_0(z^*_t)}{\epsilon }+Y_0\lambda ^*\delta ^*_te^{\rho t}H_t\Big ){\mathbf {1}}_{\{0<{\bar{z}}^{\epsilon }_t\le {\bar{z}}\}}\\&\qquad +\,\Big (\frac{{\widetilde{u}}_1(z^*_t+\epsilon \lambda ^*\delta ^*_te^{\rho t}H_t)-{\widetilde{u}}_1(z^*_t)}{\epsilon }+Y_1\lambda ^*\delta ^*_te^{\rho t}H_t\Big ){\mathbf {1}}_{\{{\bar{z}}^{\epsilon }_t>{\bar{z}}\}}\Bigg \}dt\Bigg ]+ \lambda ^*x. \end{aligned}$$

Note that $0 < \lambda ^*\delta ^*_t \le \lambda ^*$ for all $t \ge 0$. Similarly to the way of deriving (67), by applying Fatou’s Lemma to (69), we get

$$\begin{aligned} {\mathbb {E}}\left[ \int _0^{\infty }\left( c^*_t-Y_0 {\mathbf {1}}_{\{\varTheta ^*_t=A_0\}}-Y_1 {\mathbf {1}}_{\{\varTheta ^*_t=A_1\}}\right) \delta ^*_tH_tdt\right] = x. \end{aligned}$$

(70)

By (68), (19), (70), and Lemma 3, the theorem holds. $\Box$

E Proof of Lemma 4

The inequality (21) implies $D_2>0.$ Define a function F(z) for $z > 0$ as

$$\begin{aligned} F(z)\triangleq (n_+-n_-)n_-D_2z^{n_--1}-\frac{n_+\gamma _1+1-\gamma _1}{ \gamma _1K_1}L_1^{\frac{\gamma _1-\gamma }{\gamma _1}}z^{-\frac{1}{\gamma _1}}+(n_+-1)\frac{Y_1}{r}. \end{aligned}$$

(71)

Then, $F^{\prime }(z)>0$ and $\displaystyle {\lim _{z\uparrow \infty }F(z)=(n_+-1){Y_1}/{r}>0}.$ We have

$$\begin{aligned} F({\bar{z}})&=(n_+-n_-)n_-D_2{\bar{z}}^{n_--1}-\frac{n_+\gamma _1+1-\gamma _1}{ \gamma _1K_1}L_1^{\frac{\gamma _1-\gamma }{\gamma _1}}{\bar{z}}^{-\frac{1}{\gamma _1}}+(n_+-1) \frac{Y_1}{r}\\&=\frac{n_-(n_+\gamma _1+1-\gamma _1)}{\gamma _1K_1}\frac{L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}}{L_0^{ \frac{\gamma _1-\gamma }{\gamma _1}}-L_1^{\frac{\gamma _1-\gamma }{\gamma _1}}}(Y_1-Y_0)\\&\quad -\frac{(n_+\gamma _1+1- \gamma _1)(n_-\gamma _1+1-\gamma _1)}{\gamma _1^2K_1}\frac{L_1^{\frac{\gamma _1-\gamma }{\gamma _1}}}{L_0^{ \frac{\gamma _1-\gamma }{\gamma _1}}-L_1^{\frac{\gamma _1-\gamma }{\gamma _1}}}(Y_1-Y_0)\\&\quad -\frac{(1-n_-)(1-n_+)}{r}(Y_1-Y_0)+\frac{(n_+-1)Y_0}{r}\\&=-\frac{(1-\gamma _1)(n_+\gamma _1+1-\gamma _1)}{\gamma _1^2K_1}\frac{L_0^{\frac{\gamma _1- \gamma }{\gamma _1}}}{L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}-L_1^{\frac{\gamma _1-\gamma }{\gamma _1}}}(Y_1-Y_0)+ \frac{(n_+-1)Y_0}{r}\\&<0, \end{aligned}$$

where the second equality is obtained from plugging $D_2$ in (25) and ${\bar{z}}$ in Remark 2 into the given equation, the third equality from the equality (22) in Remark 6, and the inequality from the condition (24). Thus $F(\cdot )=0$ has a unique solution ${\tilde{z}} > {\bar{z}}$. The function v is continuous at $z = {\tilde{z}}$ since it is defined to be $v(z) = v({\tilde{z}})$ for $z\ge {\tilde{z}}$.

The second order ordinary differential equation of v, $-\rho v(z) + {\mathcal {L}}v(z)+\left( {\widetilde{u}}_0(z)+Y_0z\right) =0$ for $0<z<{\bar{z}}$, has two homogeneous solutions $z^{n_+}$ and $z^{n_-}$, and a particular solution $L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}\frac{\gamma _1}{(1-\gamma _1)K_1}z^{-\frac{1-\gamma _1}{\gamma _1}}+\frac{Y_0}{r}z$, so that the general solution of it is

$$\begin{aligned} const_1\cdot z^{n_+} + const_2z^{n_-} + L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}\frac{\gamma _1}{(1-\gamma _1)K_1}z^{-\frac{1-\gamma _1}{\gamma _1}}+ \frac{Y_0}{r}z, \end{aligned}$$

where $const_1$ and $const_2$ are arbitrary constants. Therefore, v satisfies the first one in the Bellman equations (30). Here, we are conjecturing that the term $const_2z^{n_-}$ is discarded due to the rapid growth of $z^{n_-}$ as $z\downarrow 0$. Similarly, v satisfies also the second one in the Bellman equations. Thus, the function v satisfies the Bellman equations (30).

By calculation, it can be shown that v is continuously differentiable at $z = {\bar{z}}$ and twice continuously differentiable at $z = {\tilde{z}}$. Since $n_+ > 1$, we have

$$\begin{aligned} \lim _{z\downarrow 0}v^{\prime }(z) = \lim _{z\downarrow 0}\left[ C_1n_+z^{n_+-1}-L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}\frac{1}{K_1}z^{-\frac{1}{\gamma _1}}+ \frac{Y_0}{r}\right] = -\infty . \end{aligned}$$

Since v(z) is twice continuously differentiable at $z={\tilde{z}}$ and constant for $z\ge {\tilde{z}}$, we have

$$\begin{aligned} v^{\prime }(z) = v^{\prime \prime }(z) = 0 ~~\text{ for } z\ge {\tilde{z}}. \end{aligned}$$

(72)

Now we show that v is strictly decreasing and strictly convex on $(0, {\tilde{z}}]$. To do it, it suffice to show that $v^{\prime \prime }(z) > 0$ for $z \in (0, {\bar{z}})\cup ({\bar{z}}, {\tilde{z}})$, since v is smoothly pasted at ${\bar{z}}$ and $v^{\prime }({\tilde{z}}) = 0$.

(1) Firstly we will show that $v^{\prime \prime }(z) > 0$ for $0< z < {\bar{z}}$. We see that, for $0<z<{\bar{z}}$,

$$\begin{aligned} v^{\prime }(z)&=n_+C_1z^{n_+-1}-L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}\frac{1}{K_1}z^{-\frac{1}{\gamma _1}}+\frac{Y_0}{r},\\ v^{\prime \prime }(z)&=z^{-\frac{1}{\gamma _1}-1}\left[ n_+(n_+-1)C_1z^{n_++\frac{1}{\gamma _1}-1}+L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}\frac{1}{\gamma _1 K_1}\right] . \end{aligned}$$

Let

$$\begin{aligned} h_1(z):=n_+(n_+-1)C_1z^{n_++\frac{1}{\gamma _1}-1}+L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}\frac{1}{\gamma _1 K_1}. \end{aligned}$$

If $C_1\ge 0$, then $h_1(z)>0$ and hence $v^{\prime \prime }(z) > 0$ for $0< z < {\bar{z}}$. If $C_1<0$, then $h_1(z)$ is decreasing, so it is enough to show that

$$\begin{aligned} h_1({\bar{z}})>0. \end{aligned}$$

From the coefficients $D_2$ in (25), $D_1$ in (27), and $C_1$ in (28), we see that

$$\begin{aligned}&\displaystyle D_1=-\frac{n_-}{n_+}\frac{\frac{n_+\gamma _1+1-\gamma _1}{\gamma _1K_1}+\frac{1-n_+}{r}}{(n_+-n_-){ \bar{z}}^{n_--1}}(Y_1-Y_0){\tilde{z}}^{n_--n_+}+\frac{L_1^{\frac{\gamma _1- \gamma }{\gamma _1}}}{n_+K_1}{\tilde{z}}^{1-n_+-\frac{1}{\gamma _1}}- \frac{Y_1}{n_+r}{\tilde{z}}^{1-n_+},\\&\displaystyle C_1=D_1+\frac{\frac{n_-\gamma _1+1-\gamma _1}{\gamma _1K_1}+\frac{1-n_-}{r}}{(n_+-n_-){ \bar{z}}^{n_+-1}}(Y_1-Y_0). \end{aligned}$$

Since ${\bar{z}}<{\tilde{z}}$, we have the following inequalities:

$$\begin{aligned} \left( \frac{{\bar{z}}}{{\tilde{z}}}\right) ^{n_++\frac{1}{\gamma _1}-n_-}< \left( \frac{{\bar{z}}}{{\tilde{z}}}\right) ^{n_+-n_-}<\left( \frac{{\bar{z}}}{{ \tilde{z}}}\right) ^{n_++\frac{1}{\gamma _1}-1}<\left( \frac{{\bar{z}}}{{ \tilde{z}}}\right) ^{n_+-1}<1. \end{aligned}$$

(73)

Then $h_1({\bar{z}})$ is given by

$$\begin{aligned} h_1({\bar{z}})&=-n_-(n_+-1)\frac{\frac{n_+\gamma _1+1-\gamma _1}{\gamma _1K_1}+ \frac{1-n_+}{r}}{(n_+-n_-)}(Y_1-Y_0){{\bar{z}}^{\frac{1}{\gamma _1}}}{ \left( \frac{{\bar{z}}}{\tilde{z}}\right) ^{n_+-n_-}}\\&\quad +(n_+-1)\frac{L_1^{\frac{\gamma _1-\gamma }{\gamma _1}}}{K_1}{\left( \frac{{ \bar{z}}}{\tilde{z}}\right) ^{n_++\frac{1}{\gamma _1}-1}}-(n_+-1)\frac{Y_1}{r}{{ \bar{z}}^{\frac{1}{\gamma _1}}}{\left( \frac{{\bar{z}}}{\tilde{z}}\right) ^{n_+-1}}\\&\quad +n_+(n_+-1)\frac{\frac{n_-\gamma _1+1-\gamma _1}{\gamma _1K_1}+\frac{1-n_-}{r}}{(n_+-n_-)}( Y_1-Y_0){{\bar{z}}^{\frac{1}{\gamma _1}}}+L_0^{\frac{\gamma _1-\gamma }{\gamma _1}} \frac{1}{\gamma _1 K_1}\\&=\frac{n_-(1-n_-)}{n_+-n_-}\left[ \frac{n_+\gamma _1+1-\gamma _1}{\gamma _1K_1}+\frac{1-n_+}{r} \right] (Y_1-Y_0) {{\tilde{z}}^{\frac{1}{\gamma _1}}}{\left( \frac{{\bar{z}}}{ \tilde{z}}\right) ^{n_++\frac{1}{\gamma _1}-n_-}}\\&\quad -\frac{1}{\gamma _1 K_1}L_1^{\frac{\gamma _1-\gamma }{\gamma _1}}{\left( \frac{{\bar{z}}}{\tilde{z}} \right) ^{n_++\frac{1}{\gamma _1}-1}}+n_+(n_+-1)\frac{\frac{n_-\gamma _1+1-\gamma _1}{ \gamma _1K_1}+\frac{1-n_-}{r}}{(n_+-n_-)}(Y_1-Y_0){{\bar{z}}^{\frac{1}{ \gamma _1}}}\\&\quad +L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}\frac{1}{\gamma _1 K_1}, \end{aligned}$$

where the second equality is from $F({\tilde{z}})=0$ in (71).

If $\gamma >1$, then $\gamma _1>1$ and

$$\begin{aligned} n_+(n_+-1)\frac{\frac{n_-\gamma _1+1-\gamma _1}{\gamma _1K_1}+\frac{1-n_-}{r}}{(n_+-n_-)}\frac{\gamma _1}{\gamma _1-1}>\frac{1}{\gamma _1 K_1}, \end{aligned}$$

(74)

which follows from (22), (20), and the fact that $n_+>1$. If $\gamma >1$, then $\gamma>\gamma _1>1$ and $L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}<L_1^{\frac{\gamma _1-\gamma }{\gamma _1}}$, and hence $h_1({\bar{z}}) > 0$ by

$$\begin{aligned} h_1({\bar{z}})&=\frac{n_-(1-n_-)}{n_+-n_-}\left[ \frac{n_+\gamma _1+1-\gamma _1}{\gamma _1K_1}+\frac{1-n_+}{r}\right] (Y_1-Y_0) {{\tilde{z}}^{\frac{1}{\gamma _1}}}{\left( \frac{{\bar{z}}}{{\tilde{z}}}\right) ^{n_++\frac{1}{\gamma _1}-n_-}}\\&\quad -\frac{1}{\gamma _1 K_1}L_1^{\frac{\gamma _1-\gamma }{\gamma _1}}{\left( \frac{{\bar{z}}}{{\tilde{z}}}\right) ^{n_++\frac{1}{\gamma _1}-1}}+n_+(n_+-1)\frac{\frac{n_-\gamma _1+1-\gamma _1}{\gamma _1K_1}+\frac{1-n_-}{r}}{(n_+-n_-)}(Y_1-Y_0){{\bar{z}}^{\frac{1}{\gamma _1}}}+L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}\frac{1}{\gamma _1 K_1}\\&=-\frac{1-\gamma _1}{\gamma _1^2K_1}(Y_1-Y_0){{\bar{z}}^{\frac{1}{\gamma _1}}}{\left( \frac{{\bar{z}}}{{\tilde{z}}}\right) ^{n_+-n_-}}-\frac{1}{\gamma _1 K_1}L_1^{\frac{\gamma _1-\gamma }{\gamma _1}}{\left( \frac{{\bar{z}}}{{\tilde{z}}}\right) ^{n_++\frac{1}{\gamma _1}-1}}\\&\quad +n_+(n_+-1)\frac{\frac{n_-\gamma _1+1-\gamma _1}{\gamma _1K_1}+\frac{1-n_-}{r}}{(n_+-n_-)}\frac{\gamma _1}{\gamma _1-1}\left( L_1^{\frac{\gamma _1-\gamma }{\gamma _1}}-L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}\right) {\left( 1-\left( \frac{{\bar{z}}}{{\tilde{z}}}\right) ^{n_+-n_-}\right) }+L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}\frac{1}{\gamma _1 K_1}\\&{>}-\frac{1}{\gamma _1K_1}\left( L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}-L_1^{\frac{\gamma _1-\gamma }{\gamma _1}}\right) {\left( \frac{{\bar{z}}}{{\tilde{z}}}\right) ^{n_+-n_-}}-\frac{1}{\gamma _1 K_1}L_1^{\frac{\gamma _1-\gamma }{\gamma _1}}{\left( \frac{{\bar{z}}}{{\tilde{z}}}\right) ^{n_++\frac{1}{\gamma _1}-1}}\\&\quad +\frac{1}{\gamma _1K_1}\left( L_1^{\frac{\gamma _1-\gamma }{\gamma _1}}-L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}\right) {\left( 1-\left( \frac{{\bar{z}}}{{\tilde{z}}}\right) ^{n_+-n_-}\right) }+L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}\frac{1}{\gamma _1 K_1}\\&=\frac{1}{\gamma _1K_1}\left[ L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}\left( 1-{\left( \frac{{\bar{z}}}{{\tilde{z}}}\right) ^{n_+-n_-}}-\left( 1-\left( \frac{{\bar{z}}}{{\tilde{z}}}\right) ^{n_+-n_-}\right) \right) \right. \\&\quad \left. +L_1^{\frac{\gamma _1-\gamma }{\gamma _1}}\left( {\left( \frac{{\bar{z}}}{{\tilde{z}}}\right) ^{n_+-n_-}}-{\left( \frac{{\bar{z}}}{{\tilde{z}}}\right) ^{n_++\frac{1}{\gamma _1}-1}}+\left( 1-\left( \frac{{\bar{z}}}{{\tilde{z}}}\right) ^{n_+-n_-}\right) \right) \right] \\&=\frac{1}{\gamma _1K_1}L_1^{\frac{\gamma _1-\gamma }{\gamma _1}}\left( 1-\left( \frac{{\bar{z}}}{{\tilde{z}}}\right) ^{n_++\frac{1}{\gamma _1}-1}\right) \\&>0, \end{aligned}$$

where the second equality follows from the equality (22), the first inequality from (74), and the last inequality from (73).

If $0<\gamma <1$, then $0<\gamma<\gamma _1<1$ and $L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}>L_1^{\frac{\gamma _1-\gamma }{\gamma _1}}$, and hence $h_1({\bar{z}}) > 0$ by

$$\begin{aligned} h_1({\bar{z}})&=\frac{n_-(1-n_-)}{n_+-n_-}\left[ \frac{n_+\gamma _1+1-\gamma _1}{\gamma _1K_1}+\frac{1-n_+}{r}\right] (Y_1-Y_0) {{\tilde{z}}^{\frac{1}{\gamma _1}}}{\left( \frac{{\bar{z}}}{{\tilde{z}}}\right) ^{n_++\frac{1}{\gamma _1}-n_-}}\\&\quad -\frac{1}{\gamma _1 K_1}L_1^{\frac{\gamma _1-\gamma }{\gamma _1}}{\left( \frac{{\bar{z}}}{{\tilde{z}}}\right) ^{n_++\frac{1}{\gamma _1}-1}}+n_+(n_+-1)\frac{\frac{n_-\gamma _1+1-\gamma _1}{\gamma _1K_1}+\frac{1-n_-}{r}}{(n_+-n_-)}(Y_1-Y_0){{\bar{z}}^{\frac{1}{\gamma _1}}}+L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}\frac{1}{\gamma _1 K_1}\\&>\frac{n_-(1-n_-)}{n_+-n_-}\left[ \frac{n_+\gamma _1+1-\gamma _1}{\gamma _1K_1}+\frac{1-n_+}{r}\right] (Y_1-Y_0) {{\tilde{z}}^{\frac{1}{\gamma _1}}}{\left( \frac{{\bar{z}}}{{\tilde{z}}}\right) ^{n_++\frac{1}{\gamma _1}-n_-}}\\&\quad -\frac{1}{\gamma _1 K_1}L_1^{\frac{\gamma _1-\gamma }{\gamma _1}}{\left( \frac{{\bar{z}}}{{\tilde{z}}}\right) ^{n_++\frac{1}{\gamma _1}-1}}\\&\quad +n_+(n_+-1)\frac{\frac{n_-\gamma _1+1-\gamma _1}{\gamma _1K_1}+\frac{1-n_-}{r}}{(n_+-n_-)}(Y_1-Y_0){{\tilde{z}}^{\frac{1}{\gamma _1}}}{\left( \frac{{\bar{z}}}{{\tilde{z}}}\right) ^{n_++\frac{1}{\gamma _1}-n_-}}+L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}\frac{1}{\gamma _1 K_1}\\&=-\frac{1-\gamma _1}{\gamma _1^2K_1}(Y_1-Y_0){{\bar{z}}^{\frac{1}{\gamma _1}}}{\left( \frac{{\bar{z}}}{{\tilde{z}}}\right) ^{n_+-n_-}}-\frac{1}{\gamma _1 K_1}L_1^{\frac{\gamma _1-\gamma }{\gamma _1}}{\left( \frac{{\bar{z}}}{{\tilde{z}}}\right) ^{n_++\frac{1}{\gamma _1}-1}}+L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}\frac{1}{\gamma _1 K_1}\\&=-\frac{1}{\gamma _1K_1}\left( L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}-L_1^{\frac{\gamma _1-\gamma }{\gamma _1}}\right) {\left( \frac{{\bar{z}}}{{\tilde{z}}}\right) ^{n_+-n_-}}-\frac{1}{\gamma _1 K_1}L_1^{\frac{\gamma _1-\gamma }{\gamma _1}}{\left( \frac{{\bar{z}}}{{\tilde{z}}}\right) ^{n_++\frac{1}{\gamma _1}-1}}+L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}\frac{1}{\gamma _1 K_1}\\&=\frac{1}{\gamma _1K_1}\left[ L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}\left( 1-{\left( \frac{{\bar{z}}}{{\tilde{z}}}\right) ^{n_+-n_-}}\right) +L_1^{\frac{\gamma _1-\gamma }{\gamma _1}}\left( {\left( \frac{{\bar{z}}}{{\tilde{z}}}\right) ^{n_+-n_-}}-{\left( \frac{{\bar{z}}}{{\tilde{z}}}\right) ^{n_++\frac{1}{\gamma _1}-1}}\right) \right] \\&>\frac{1}{\gamma _1K_1}\left[ L_1^{\frac{\gamma _1-\gamma }{\gamma _1}}\left( 1-{\left( \frac{{\bar{z}}}{{\tilde{z}}}\right) ^{n_++\frac{1}{\gamma _1}-1}}\right) \right] \\&>0, \end{aligned}$$

where the first inequality is from (73) for the fourth term in left-hand side, the second equality from (22), and the second and the last inequalities from (73).

(2) Secondly we show that $v^{\prime \prime }(z) > 0$ for ${\bar{z}}< z < {\tilde{z}}$. We see that

$$\begin{aligned} v'(z)&=n_+D_1z^{n_+-1}+n_-D_2z^{n_--1}-L_1^{\frac{\gamma _1-\gamma }{\gamma _1}}\frac{1}{K_1}z^{-\frac{1}{\gamma _1}}+\frac{Y_1}{r},\\ v''(z)&=z^{-\frac{1}{\gamma _1}-1}\left[ n_+(n_+-1)D_1z^{n_++\frac{1}{\gamma _1}-1}+n_-(n_--1)D_2z^{n_-+\frac{1}{\gamma _1}-1}+L_1^{\frac{\gamma _1-\gamma }{\gamma _1}}\frac{1}{\gamma _1 K_1}\right] \end{aligned}$$

Let

$$\begin{aligned} h_2(z)&:=n_+(n_+-1)D_1z^{n_++\frac{1}{\gamma _1}-1}+n_-(n_--1)D_2z^{n_-+\frac{1}{\gamma _1}-1}+L_1^{\frac{\gamma _1-\gamma }{\gamma _1}}\frac{1}{\gamma _1 K_1}\\&=z^{n_-+\frac{1}{\gamma _1}-1}\left[ n_+(n_+-1)D_1z^{n_+-n_-}+n_-(n_--1)D_2\right] +L_1^{\frac{\gamma _1-\gamma }{\gamma _1}}\frac{1}{\gamma _1 K_1}. \end{aligned}$$

Since $D_1<0,~D_2>0$, and $n_-+\frac{1}{\gamma _1}-1<0$, $h_2(z)$ is strictly decreasing, so that

$$\begin{aligned} h_2(z) > h_2({\tilde{z}}) = 0, ~~\text{ for }~~{\bar{z}}< z < {\tilde{z}}, \end{aligned}$$

where the equality holds since $v^{\prime \prime }({\tilde{z}}) = 0$. Thus $v''(z)> 0$ for ${\bar{z}}<z<{\tilde{z}}$. $\Box$

F Proof of Theorem 2

Let

$$\begin{aligned} z^*_t \triangleq {\mathcal {Z}}(X_t), \quad t \ge 0, \end{aligned}$$

(75)

and let

$$\begin{aligned} \lambda ^* \triangleq z^*_0 = {\mathcal {Z}}(x). \end{aligned}$$

(76)

Since $x = X_0 \ge 0$, $\lambda ^*$ in (76) is well defined and $0 < \lambda ^* \le {\tilde{z}}$. By (33) and (75), we have

$$\begin{aligned} X_t\ge {\bar{x}} \Leftrightarrow 0<z^*_t\le {\bar{z}} ~~\text{ and }~~ 0\le X_t<{\bar{x}} \Leftrightarrow z^*_t > {\bar{z}}. \end{aligned}$$

With the strategy $(\varvec{\varTheta }^*,\varvec{c}^*,\varvec{\pi }^*)$ in the theorem, the wealth process (1) satisfies

$$\begin{aligned} dX_t = \left[ rX_t+\theta ^2z^*_tv^{\prime \prime }(z^*_t)-c^*_t+Y_0 {\mathbf {1}}_{\{0<z^*_t\le {\bar{z}}\}}+Y_1 {\mathbf {1}}_{\{z^*_t>{\bar{z}}\}}\right] dt+\theta z^*_tv^{\prime \prime }(z^*_t)dB_t. \end{aligned}$$

If $X_t = 0$ for $\exists t \ge 0$, then $z^*_t = {\mathcal {Z}}(0) = {\tilde{z}}$ so that

$$\begin{aligned} dX_t = [-f_1({\tilde{z}}) + Y_1]dt. \end{aligned}$$

(77)

However, by using (21) and the fact that ${\tilde{z}}$ a solution to (26) and $D_2 > 0$ as stated in Remark 7, we can show

$$\begin{aligned} -f_1({\tilde{z}}) + Y_1 > 0. \end{aligned}$$

(78)

Therefore, $X_t = 0$ is an upward reflection boundary so that $X_t \ge 0$ for all $t\ge 0$ with the strategy $(\varvec{\varTheta }^*, {\mathbf {c}}^*, \varvec{\pi }^*)$. Thus, the relation (75) is well defined, $0 < z^*_t \le {\tilde{z}}$, and

$$\begin{aligned} (\varvec{\varTheta }^*, {\mathbf {c}}^*, \varvec{\pi }^*)\in {\mathcal {A}}(x). \end{aligned}$$

(79)

Since the Bellman equations (30) hold for all $z \in (0, {\bar{z}}) \cup ({\bar{z}}, {\tilde{z}})$, we have, by differentiation with respect to z,

$$\begin{aligned} \left\{ \begin{array}{ll} -rv^{\prime }(z) + (\rho -r+\theta ^2)zv^{\prime \prime }(z)+ \frac{1}{2}\theta ^2z^2v^{\prime \prime \prime }(z) -f_0(z)+Y_0 =0,&{}\quad \text{ for } 0<z<{\bar{z}},\\ -rv^{\prime }(z) + (\rho -r+\theta ^2)zv^{\prime \prime }(z)+ \frac{1}{2}\theta ^2z^2v^{\prime \prime \prime }(z) -f_1(z)+Y_1=0,&{}\quad \text{ for } {\bar{z}}<z<{\tilde{z}}, \end{array}\right. \end{aligned}$$

which, by using (31), can be rewritten as

$$\begin{aligned} \left\{ \begin{array}{ll} r{\mathcal {X}}(z) -(\rho -r+\theta ^2)z{\mathcal {X}}^{\prime }(z)- \frac{1}{2}\theta ^2z^2{\mathcal {X}}^{\prime \prime }(z) -f_0(z)+Y_0 =0,&{}\quad \text{ for } 0<z<{\bar{z}},\\ r{\mathcal {X}}(z) -(\rho -r+\theta ^2)z{\mathcal {X}}^{\prime }(z)- \frac{1}{2}\theta ^2z^2{\mathcal {X}}^{\prime \prime }(z) -f_1(z)+Y_1=0,&{}\quad \text{ for } {\bar{z}}<z<{\tilde{z}}. \end{array}\right. \nonumber \\ \end{aligned}$$

(80)

Let $\delta ^*_t \triangleq e^{-\rho t}H_t^{-1}z^*_t/\lambda ^*$ for $t\ge 0$. Thus, $\delta ^*_t > 0$ for $t\ge 0$ with $\delta ^*_0 = 1$, and

$$\begin{aligned} z^*_t = \lambda ^*\delta ^*_t e^{\rho t}H_t. \end{aligned}$$

(81)

Suppose $0< z^*_{t_0} = \lambda ^*\delta ^*_{t_0} e^{\rho t_0}H_{t_0} < {\tilde{z}}$ (equivalently $X_{t_0} > 0$) at $\exists t_0 \ge 0$. Then, the process $(z^*_t)_{t\ge t_0}$, before $z^*_{t}$ touching ${\tilde{z}}$, evolves according to

$$\begin{aligned} dz^*_t= & {} d{\mathcal {Z}}(X_t)\\= & {} {\mathcal {Z}}^{\prime }(X_t)\left\{ \left[ rX_t+\theta ^2z^*_tv^{\prime \prime }(z^*_t)-c^*_t+Y_0 {\mathbf {1}}_{\{0<z^*_t\le {\bar{z}}\}}+Y_1 {\mathbf {1}}_{\{z^*_t>{\bar{z}}\}}\right] dt+\theta z^*_tv^{\prime \prime }(z^*_t)dB_t\right\} \\&+\, \frac{1}{2}{\mathcal {Z}}^{\prime \prime }(X_t)(\theta z^*_tv^{\prime \prime }(z^*_t))^2 dt\\= & {} {\mathcal {Z}}^{\prime }(X_t)\left\{ \left[ r{\mathcal {X}}(z^*_t)-\theta ^2z^*_t{\mathcal {X}}^{\prime }(z^*_t)-c^*_t+Y_0 {\mathbf {1}}_{\{0<z^*_t\le {\bar{z}}\}}+Y_1 {\mathbf {1}}_{\{z^*_t>{\bar{z}}\}}\right] dt-\theta z^*_t{\mathcal {X}}^{\prime }(z^*_t)dB_t\right\} \\&+\, \frac{1}{2}{\mathcal {Z}}^{\prime \prime }(X_t)(\theta z^*_t{\mathcal {X}}^{\prime }(z^*_t))^2dt\\= & {} {\mathcal {Z}}^{\prime }(X_t)\left[ r{\mathcal {X}}(z^*_t)-\theta ^2z^*_t{\mathcal {X}}^{\prime }(z^*_t)+ \frac{1}{2}{\mathcal {Z}}^{\prime \prime }(X_t)(\theta z^*_t)^2({\mathcal {X}}^{\prime }(z^*_t))^3-c^*_t+Y_0 {\mathbf {1}}_{\{0<z^*_t\le {\bar{z}}\}}+Y_1 {\mathbf {1}}_{\{z^*_t>{\bar{z}}\}}\right] dt\\&-\,{\mathcal {Z}}^{\prime }(X_t)\theta z^*_t{\mathcal {X}}^{\prime }(z^*_t)dB_t\\= & {} {\mathcal {Z}}^{\prime }(X_t)\left[ (\rho -r)z^*_t{\mathcal {X}}^{\prime }(z^*_t)+ \frac{1}{2}(\theta z^*_t)^2{\mathcal {X}}^{\prime \prime }(z^*_t) + \frac{1}{2}{\mathcal {Z}}^{\prime \prime }(X_t)(\theta z^*_t)^2({\mathcal {X}}^{\prime }(z^*_t))^3\right] dt\\&-\,{\mathcal {Z}}^{\prime }(X_t)\theta z^*_t{\mathcal {X}}^{\prime }(z^*_t)dB_t\\= & {} \left[ (\rho -r)z^*_t + \frac{1}{2}(\theta z^*_t)^2\left\{ {\mathcal {X}}^{\prime \prime }(z^*_t) {\mathcal {Z}}^{\prime }(X_t) + {\mathcal {Z}}^{\prime \prime }(X_t)({\mathcal {X}}^{\prime }(z^*_t))^2\right\} \right] dt\\&-\,\theta z^*_tdB_t\\= & {} z^*_t[(\rho -r) dt -\theta dB_t], \end{aligned}$$

where the second equality comes from the generalized Itô’s rule, the third from (31), (32), and (75), the fifth from the definition of $c^*_t$ and (80), the sixth from the fact that ${\mathcal {X}}^{\prime }({\mathcal {Z}}(X_t)){\mathcal {Z}}^{\prime }(X_t) = 1$ by the inverse relationship of ${\mathcal {X}}$ and ${\mathcal {Z}}$, and the last holds since ${\mathcal {X}}^{\prime \prime }({\mathcal {Z}}(X_t)) {\mathcal {Z}}^{\prime }(X_t) + {\mathcal {Z}}^{\prime \prime }(X_t)({\mathcal {X}}^{\prime }({\mathcal {Z}}(X_t)))^2 = {\mathcal {X}}^{\prime }({\mathcal {Z}}(X_t))[{\mathcal {X}}^{\prime \prime }({\mathcal {Z}}(X_t)) ({\mathcal {Z}}^{\prime }(X_t))^2 + {\mathcal {X}}^{\prime }({\mathcal {Z}}(X_t)){\mathcal {Z}}^{\prime \prime }(X_t)]= 0$ by the inverse relationship of ${\mathcal {X}}$ and ${\mathcal {Z}}$. Therefore, we have

$$\begin{aligned} z^*_t= & {} z^*_{t_0}e^{(\rho -r - \frac{\theta ^2}{2})(t-t_0) - \theta (B_t-B_{t_0})}\\= & {} \lambda ^*\delta ^*_{t_0} e^{\rho t_0}H_{t_0}e^{(\rho -r - \frac{\theta ^2}{2})(t-t_0) - \theta (B_t-B_{t_0})}\\= & {} \lambda ^*\delta ^*_{t_0} e^{(\rho -r - \frac{\theta ^2}{2})t - \theta B_t}\\= & {} \lambda ^*\delta ^*_{t_0}e^{\rho t}H_{t}, \end{aligned}$$

which means $\delta ^*_{t}$ in (81) stays constant before $z^*_{t}$ touching ${\tilde{z}}$. That is,

$$\begin{aligned} d\delta ^*_{t} = 0 ~~ \text{ when }~~ 0< z^*_t < {\tilde{z}}. \end{aligned}$$

However, once $z^*_t$ hits ${\tilde{z}}$, that is, once $X_t$ hits 0, $X_t$ moves in the upward direction by (77) and (78) so that $z^*_t$ moves in the downward direction. That is, $z^*_t = {\tilde{z}}$ is a downward reflection boundary since $X_t = 0$ is an upward reflection boundary. Actually, $z^*_t$ moves in the downward direction in an infinite instantaneous speed, since ${\mathcal {Z}}^{\prime }(0) \triangleq {\mathcal {Z}}^{\prime }(0+) = 1/{\mathcal {X}}^{\prime }({\tilde{z}}-) = -1/v^{\prime \prime }({\tilde{z}}-) = -\infty$. More concretely, suppose that $z^*_{t_0} = \lambda ^*\delta ^*_{t_0} e^{\rho t_0}H_{t_0} = {\tilde{z}}$ for $\exists t_0\ge 0$, that is, $X_{t_0} = 0$, then, by (77) and (78),

$$\begin{aligned} dz^*_{t_0} = d{\mathcal {Z}}(X_{t_0}) = {\mathcal {Z}}^{\prime }(X_{t_0})\left[ -f_1({\tilde{z}}) + Y_1\right] dt = {\mathcal {Z}}^{\prime }(X_{t_0})\left[ -f_1({\tilde{z}}) + Y_1\right] dt = -\infty dt, \end{aligned}$$

which implies $\delta ^*_t$ also decreases in an infinite instantaneous speed at $t_0$. Thus, $\varvec{\delta }^* \triangleq (\delta ^*_t)_{t\ge 0} \in {\mathcal {D}}$ and it decreases only when $z^*_t$ hits ${\tilde{z}}$. As in He and Pagès [8], the decreases of $\delta ^*_t$ when $z^*_t$ hitting ${\tilde{z}}$, regulate $(z^*_t)_{t\ge 0}$ to stay in $(0, {\tilde{z}})$, thus $\varvec{\delta }^*$ can be characterized by the local times of $(z^*_t)_{t\ge 0}$ at ${\tilde{z}}$.

We first consider the case where $\gamma _1 > 1$. Fix $T > 0$. We have

$$\begin{aligned}&\int _0^{T} e^{-\rho t}\Big \{\left( {\widetilde{u}}_0(z^*_t)+Y_0z^*_t\right) {\mathbf {1}}_{\{0<z^*_t\le {\bar{z}}\}} +\left( {\widetilde{u}}_1(z^*_t)+Y_1z^*_t\right) {\mathbf {1}}_{\{z^*_t>{\bar{z}}\}}\Big \}\,dt\nonumber \\&\qquad + \,e^{-\rho T}v(z^*_{T}) \end{aligned}$$

(82)

$$\begin{aligned}= & {} \int _0^{T} e^{-\rho t}\Big \{\left( {\widetilde{u}}_0(z^*_t)+Y_0z^*_t\right) {\mathbf {1}}_{\{0<z^*_t\le {\bar{z}}\}} +\left( {\widetilde{u}}_1(z^*_t)+Y_1z^*_t\right) {\mathbf {1}}_{\{z^*_t>{\bar{z}}\}}\Big \}\,dt + v(\lambda ^*)\nonumber \\&\,+\int _0^{T}e^{-\rho t}\Big \{-\rho v(z^*_t) + {\mathcal {L}}v(z^*_t)\Big \}\,dt+\int _0^{T}v^{\prime }(z^*_t)\lambda ^*H_t\,d\delta ^*_t - \int _0^{T}e^{-\rho t}v^{\prime }(z^*_t)\theta z^*_t\,dB_t\nonumber \\= & {} v(\lambda ^*) - \int _0^{T}e^{-\rho t}v^{\prime }(z^*_t)\theta z^*_t\,dB_t, \end{aligned}$$

(83)

where the first equality comes from the generalized Itô’s rule using (81) and (15), the second from the fact that v satisfies the Bellman equations (30) and that $\int _0^Tv^{\prime }(z^*_t)\lambda ^*H_t\,d\delta ^*_t = 0$ since $(\delta ^*_t)_{t\ge 0} \in {\mathcal {D}}$ decreases only when $z^*_t$ hits ${\tilde{z}}$ with $v^{\prime }({\tilde{z}}) = 0.$ By using (29), we have

$$\begin{aligned} zv^{\prime }(z)=\left\{ \begin{array}{ll} n_+C_1z^{n_+}-L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}\frac{1}{K_1}z^{1-\frac{1}{\gamma _1}}+ \frac{Y_0}{r}z, \quad 0<z\le {\bar{z}},\\ n_+D_1z^{n_+}+n_-D_2z^{n_-}-L_1^{\frac{\gamma _1-\gamma }{\gamma _1}}\frac{1}{K_1}z^{1- \frac{1}{\gamma _1}}+\frac{Y_1}{r}z, \quad {\bar{z}}<z\le {\tilde{z}}, \end{array}\right. \end{aligned}$$

(84)

which is bounded for $z \in (0, {\tilde{z}}]$ since $\lim _{z\downarrow 0}zv^{\prime }(z) = 0$ with $\gamma _1 > 1$. Therefore, with $\gamma _1 > 1$,

$$\begin{aligned} {\mathbb {E}}\Big [\int _0^{T}\big (e^{-\rho t}v^{\prime }(z^*_t)\theta z^*_t\big )^2\,dt\Big ] < \infty . \end{aligned}$$

Thus, the process $(\int _0^{t}e^{-\rho s}v^{\prime }(z^*_s)\theta z^*_s\,dB_s)_{t\ge 0}$ is a martingale under ${\mathbb {P}}$, with $\gamma _1 > 1$, so that

$$\begin{aligned} {\mathbb {E}}\Big [\int _0^{T}e^{-\rho t}v^{\prime }(z^*_t)\theta z^*_t\,dB_t\Big ] = 0. \end{aligned}$$

Hence, by taking the expectation operator on (82) and (83), we have, with $\gamma _1 > 1$,

$$\begin{aligned} {\mathbb {E}}\Big [\int _0^{T} e^{-\rho t}\Big \{\left( {\widetilde{u}}_0(z^*_t)+Y_0z^*_t\right) {\mathbf {1}}_{\{0<z^*_t \le {\bar{z}}\}} +\left( {\widetilde{u}}_1(z^*_t)+Y_1z^*_t\right) {\mathbf {1}}_{\{z^*_t>{ \bar{z}}\}}\Big \}\,dt\Big ] + {\mathbb {E}}\Big [e^{-\rho T}v(z^*_{T})\Big ] = v(\lambda ^*). \end{aligned}$$

(85)

When $\gamma _1 > 1$, the function v defined as (29) is bounded for $z \in (0, {\tilde{z}}]$ since $\lim _{z\downarrow 0}v(z) = 0$, which implies

$$\begin{aligned} \lim _{T\uparrow \infty }{\mathbb {E}}\Big [e^{-\rho T}v(z^*_{T})\Big ] = 0. \end{aligned}$$

(86)

Letting $T\uparrow \infty$ on both sides of the equality (85), by applying the monotone convergence theorem and by (86), we get, when $\gamma _1 > 1$,

$$\begin{aligned} v(\lambda ^*) = {\mathbb {E}}\Big [\int _0^{\infty } e^{-\rho t}\Big \{\left( {\widetilde{u}}_0(z^*_t)+Y_0z^*_t\right) {\mathbf {1}}_{\{0<z^*_t \le {\bar{z}}\}} +\left( {\widetilde{u}}_1(z^*_t)+Y_1z^*_t\right) {\mathbf {1}}_{\{z^*_t>{ \bar{z}}\}}\Big \}\,dt\Big ]. \end{aligned}$$

(87)

By the generalized Itô’s rule using (10), we have, with $T > 0$,

$$\begin{aligned}&\int _0^T\left( c^*_t-Y_0 {\mathbf {1}}_{\{0<z^*_t\le {\bar{z}}\}}-Y_1 { \mathbf {1}}_{\{z^*_t>{\bar{z}}\}}\right) \delta ^*_tH_t\,dt\\&\quad = x - H_TX_T\delta ^*_T + \int _0^TH_tX_t\,d\delta ^*_t + \int _0^T(\sigma \pi ^*_t-\theta X_t)\delta ^*_tH_tdB_t\\&\quad = x + \frac{1}{\lambda ^*}e^{-\rho T}v^{\prime }(z^*_T)z^*_T - \int _0^TH_tv^{ \prime }(z^*_t)\,d\delta ^*_t + \frac{1}{\lambda ^*}\int _0^Te^{-\rho t}\big (\sigma z^*_tv^{ \prime \prime }(z^*_t)\frac{\theta }{\sigma }+\theta v^{\prime }(z^*_t)\big )z^*_t\,dB_t\\&\quad = x + \frac{1}{\lambda ^*}e^{-\rho T}v^{\prime }(z^*_T)z^*_T + \frac{1}{\lambda ^*}\int _0^Te^{-\rho t}\big (\sigma z^*_tv^{\prime \prime }(z^*_t)\frac{\theta }{\sigma }+\theta v^{\prime }(z^*_t)\big )z^*_t\,dB_t, \end{aligned}$$

where the second equality holds by (75), (81), and the formula of $\pi ^*_t$ in the theorem, the third by the fact that $\int _0^TH_tv^{\prime }(z^*_t)\,d\delta ^*_t = 0$. Similarly to the derivation of (87), we can derive

$$\begin{aligned} {\mathbb {E}}\left[ \int _0^\infty \left( c^*_t-Y_0 {\mathbf {1}}_{\{0<z^*_t\le {\bar{z}}\}}-Y_1 {\mathbf {1}}_{\{z^*_t>{\bar{z}}\}}\right) \delta ^*_tH_tdt\right] = x. \end{aligned}$$

(88)

By (79), (34), (88), and Lemma 3, the triple $(\varvec{\varTheta }^*, {\mathbf {c}}^*, \varvec{\pi }^*)$ of the policies given in the theorem is optimal and

$$\begin{aligned} V(x) = \varPhi (\varvec{\delta }^*, \lambda ^*) = v(\lambda ^*) + \lambda ^*x = v({\mathcal {Z}}(x)) + x{\mathcal {Z}}(x), \end{aligned}$$

where the second equality comes from (87) and the third from (76).

Now we consider the case where $0< \gamma _1 < 1$. Note that $z^*_t = {\mathcal {Z}}(X_t)$ has continuous sample paths and $z^*_0 = \lambda ^*$. Let $\xi _{\epsilon } \triangleq \inf \{t\ge 0: z^*_t = \epsilon \}$ where $0< \epsilon < \lambda ^*$ is fixed. Let $\tau = T \wedge \xi _{\epsilon } \triangleq \min [T, \xi _{\epsilon }]$ where $T > 0$ is fixed. Since $zv^{\prime }(z)$ is bounded for $z \in [\epsilon , {\tilde{z}}]$, similarly to (85), we get

$$\begin{aligned} {\mathbb {E}}\Big [\int _0^{\tau } e^{-\rho t}\Big \{\left( {\widetilde{u}}_0(z^*_t)+Y_0z^*_t\right) {\mathbf {1}}_{\{0<z^*_t\le {\bar{z}}\}} +\left( {\widetilde{u}}_1(z^*_t)+Y_1z^*_t\right) {\mathbf {1}}_{\{z^*_t>{\bar{z}}\}}\Big \}\,dt\Big ] + {\mathbb {E}}\Big [e^{-\rho \tau }v(z^*_{\tau })\Big ] = v(\lambda ^*). \end{aligned}$$

(89)

Letting $\epsilon \downarrow 0$ and $T\uparrow \infty$, then $\tau \uparrow \infty$. When $0< \gamma _1 < 1$, ${\widetilde{u}}_i(z) \ge 0$ by (5), and we can show that $v(z) \ge v({\tilde{z}}) \ge 0$ by using (27), (29), $n_+ > 1$, and the fact that $D_2>0$ as mentioned in Remark 7. Thus, by the monotone convergence theorem, the first term on the left-hand side of (89) has a limit as $\epsilon \downarrow 0$ and $T\uparrow \infty$, and the limit is finite since it is between $[0, v(\lambda ^*)]$. Therefore, ${\mathbb {E}}[e^{-\rho \tau }v(z^*_{\tau })]$ also has a finite limit as $\epsilon \downarrow 0$ and $T\uparrow \infty$, which is nonnegative. However, we have, by (29),

$$\begin{aligned}&\lim _{T\uparrow \infty , \epsilon \downarrow 0}{\mathbb {E}}\Big [e^{-\rho \tau }v(z^*_{\tau })\Big ]\nonumber \\&\quad = \lim _{T\uparrow \infty , \epsilon \downarrow 0}{\mathbb {E}}\Big [e^{-\rho \tau }v(z^*_{\tau }){\mathbf {1}}_{\{0<z^*_t\le {\bar{z}}\}} + e^{-\rho \tau }v(z^*_{\tau }){\mathbf {1}}_{\{z^*_t>{\bar{z}}\}}\Big ]\nonumber \\&\quad = \lim _{T\uparrow \infty , \epsilon \downarrow 0}{\mathbb {E}}\Big [e^{-\rho \tau }v(z^*_{\tau }){\mathbf {1}}_{\{0<z^*_t\le {\bar{z}}\}}\Big ]\nonumber \\&\quad =\lim _{T\uparrow \infty , \epsilon \downarrow 0}{\mathbb {E}}\Big [e^{-\rho \tau }L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}\frac{\gamma _1}{(1- \gamma _1)K_1}(z^*_{\tau })^{-\frac{1-\gamma _1}{\gamma _1}}{\mathbf {1}}_{\{0<z^*_t\le {\bar{z}}\}}\Big ]\nonumber \\&\quad =\frac{\gamma _1}{(1-\gamma _1)K_1}\lim _{T\uparrow \infty , \epsilon \downarrow 0}{\mathbb {E}}\Big [e^{-\rho \tau }L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}(z^*_{\tau })^{-\frac{1-\gamma _1}{ \gamma _1}}{\mathbf {1}}_{\{0<z^*_t\le {\bar{z}}\}}\Big ], \end{aligned}$$

(90)

where the second equality comes from the fact that v(z) is bounded for $z \in ({\bar{z}}, {\tilde{z}}]$, and the third from that $z^{n_+}$ is bounded for $z \in (0, {\bar{z}}]$. By (79) and Remark 3, we have

$$\begin{aligned} {\mathbb {E}}\left[ \int _0^\infty \left( c^*_t-Y_0 {\mathbf {1}}_{\{0<z^*_t\le {\bar{z}}\}}-Y_1 {\mathbf {1}}_{\{z^*_t>{\bar{z}}\}}\right) \delta ^*_tH_tdt\right] \le x. \end{aligned}$$

Therefore,

$$\begin{aligned} 0 \le {\mathbb {E}}\left[ \int _0^{\infty }c^*_t\delta ^*_tH_tdt\right]\le & {} x + {\mathbb {E}}\left[ \int _0^\infty \left( Y_0 {\mathbf {1}}_{\{0<z^*_t\le {\bar{z}}\}}+Y_1 {\mathbf {1}}_{\{z^*_t>{\bar{z}}\}}\right) \delta ^*_tH_tdt\right] \\\le & {} x + (Y_0 + Y_1){\mathbb {E}}\left[ \int _0^\infty H_tdt\right] \\= & {} x + \frac{Y_0 + Y_1}{r} < \infty , \end{aligned}$$

where the last inequality comes from the fact that $0 < \delta ^*_t \le 1$ for $t \ge 0$, and the equality from ${\mathbb {E}}\left[ \int _0^\infty H_tdt\right] = \int _0^\infty {\mathbb {E}}\left[ H_t\right] dt = 1/r$. That is, ${\mathbb {E}}\left[ \int _0^{\infty }c^*_t\delta ^*_tH_tdt\right]$ is finite. However, by (34), we have

$$\begin{aligned} {\mathbb {E}}\left[ \int _0^{\infty }c^*_t\delta ^*_tH_tdt\right]= & {} {\mathbb {E}}\left[ \int _0^\infty \left( L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}\left( z^*_t\right) ^{-\frac{1}{\gamma _1}}{\mathbf {1}}_{\{0<z^*_t\le {\bar{z}}\}}+L_1^{\frac{\gamma _1-\gamma }{\gamma _1}}\left( z^*_t\right) ^{-\frac{1}{\gamma _1}} {\mathbf {1}}_{\{z^*_t>{\bar{z}}\}}\right) \delta ^*_tH_tdt\right] \\= & {} \frac{1}{\lambda ^*}{\mathbb {E}}\left[ \int _0^\infty e^{-\rho t}\left( L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}\left( z^*_t\right) ^{-\frac{1-\gamma _1}{\gamma _1}}{\mathbf {1}}_{\{0<z^*_t\le {\bar{z}}\}}\!+\!L_1^{\frac{\gamma _1\!-\!\gamma }{\gamma _1}}\left( z^*_t\right) ^{-\frac{1-\gamma _1}{\gamma _1}}{\mathbf {1}}_{\{z^*_t>{\bar{z}}\}}\right) dt\right] . \end{aligned}$$

Therefore, ${\mathbb {E}}\left[ \int _0^\infty e^{-\rho t}L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}\left( z^*_t\right) ^{-\frac{1-\gamma _1}{\gamma _1}}{\mathbf {1}}_{\{0<z^*_t\le {\bar{z}}\}}dt\right] = \int _0^\infty {\mathbb {E}}\left[ e^{-\rho t}L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}\left( z^*_t\right) ^{-\frac{1-\gamma _1}{\gamma _1}}\right. \left. {\mathbf {1}}_{\{0<z^*_t\le {\bar{z}}\}}\right] dt$ is finite, which implies

$$\begin{aligned} \lim _{t\rightarrow \infty }{\mathbb {E}}\left[ e^{-\rho t}L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}\left( z^*_t\right) ^{-\frac{1-\gamma _1}{\gamma _1}}{\mathbf {1}}_{\{0<z^*_t\le {\bar{z}}\}}\right] = 0. \end{aligned}$$

Combining this with (90), we get

$$\begin{aligned} \lim _{T\uparrow \infty , \epsilon \downarrow 0}{\mathbb {E}}\Big [e^{-\rho \tau }v(z^*_{\tau })\Big ] = 0. \end{aligned}$$

Therefore, letting $\epsilon \downarrow 0$ and $T\uparrow \infty$ on both sides of (89), we find that (87) holds also for the case where $0< \gamma _1 < 1$. Similarly, we can show that (88) holds also for the case where $0< \gamma _1 < 1$. Thus, the theorem is proved also for the case where $0< \gamma _1 < 1$. $\Box$

G Proof of Lemma 5

By (20) and the fact that $n_+ > 1$, (37) is well defined and ${\tilde{z}} > 0$. By using (36), we can show ${\tilde{z}} \le {\bar{z}}$. By calculation, we can show that

$$\begin{aligned} E_1 = -\frac{Y_0}{rn_+(\gamma _1n_+-\gamma _1+1)}{\tilde{z}}^{1-n_+} < 0, \end{aligned}$$

where the inequality comes from (20). The function w is continuous at $z = {\tilde{z}}$ since it is defined to be $w(z) = w({\tilde{z}})$ for $z\ge {\tilde{z}}$. The Bellman equation (39) is a second order ordinary differential equation, which has two homogeneous solutions $z^{n_+}$ and $z^{n_-}$, and a particular solution $L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}\frac{\gamma _1}{(1-\gamma _1)K_1}z^{-\frac{1-\gamma _1}{\gamma _1}}+\frac{Y_0}{r}z$, so that the general solution of it is

$$\begin{aligned} const_1\cdot z^{n_+} + const_2z^{n_-} + L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}\frac{\gamma _1}{(1-\gamma _1)K_1}z^{-\frac{1-\gamma _1}{\gamma _1}}+\frac{Y_0}{r}z, \end{aligned}$$

where $const_1$ and $const_2$ are arbitrary constants. Therefore, w satisfies the Bellman equation (39). Here, we are conjecturing that the term $const_2z^{n_-}$ is discarded due to the rapid growth of $z^{n_-}$ as $z\downarrow 0$.

By calculation, it can be shown that w is twice continuously differentiable at $z = {\tilde{z}}$. Since $n_+ > 1$, we have

$$\begin{aligned} \lim _{z\downarrow 0}w^{\prime }(z) = \lim _{z\downarrow 0}[E_1n_+z^{n_+-1}-L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}\frac{1}{K_1}z^{-\frac{1}{\gamma _1}}+\frac{Y_0}{r}] = -\infty . \end{aligned}$$

Since w(z) is twice continuously differentiable at $z={\tilde{z}}$ and constant for $z\ge {\tilde{z}}$, we have

$$\begin{aligned} w^{\prime }(z) = w^{\prime \prime }(z) = 0 ~~\text{ for } z\ge {\tilde{z}}. \end{aligned}$$

(91)

Now we show that w is strictly decreasing and strictly convex for $z \in (0, {\tilde{z}}]$. Note that

$$\begin{aligned} w'(z)&=n_+E_1z^{n_+-1}-L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}\frac{1}{K_1}z^{-\frac{1}{\gamma _1}}+\frac{Y_0}{r},\\ w''(z)&=z^{-\frac{1}{\gamma _1}-1}\left[ n_+(n_+-1)E_1z^{n_++\frac{1}{\gamma _1}-1}+L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}\frac{1}{\gamma _1 K_1}\right] . \end{aligned}$$

Let

$$\begin{aligned} g(z):=n_+(n_+-1)E_1z^{n_++\frac{1}{\gamma _1}-1}+L_0^{\frac{\gamma _1- \gamma }{\gamma _1}}\frac{1}{\gamma _1 K_1}, \end{aligned}$$

then $g(\cdot )$ is strictly decreasing since $E_1<0$, so that

$$\begin{aligned} g(z) > g({\tilde{z}}) = 0, ~~\text{ for }~~ 0< z < {\tilde{z}}, \end{aligned}$$

where the equality holds since $w''({\tilde{z}}) = 0$. Thus $w''(z) > 0$ for $0<z<{\tilde{z}}$. That is, w is strictly convex for $z \in (0, {\tilde{z}}]$. Since $w''(z) > 0$ for $0<z<{\tilde{z}}$, $w^{\prime }(z)$ is strictly increasing for $0<z\le {\tilde{z}}$. Therefore, we have

$$\begin{aligned} w'(z)< w'({\tilde{z}}) = 0. \end{aligned}$$

Thus, w is strictly decreasing for $z \in (0, {\tilde{z}}]$. $\square$

H Proof of the assertion that ${\bar{x}}^b<0$ when the IBR is not greater than the LBR in Remark 9

Let us define

$$\begin{aligned} \varLambda \triangleq \frac{\frac{1-\gamma _1}{\gamma _1}+n_+}{K_1}+\frac{1-n_+}{r}>0,~~\varGamma \triangleq \frac{L_1^{\frac{\gamma _1-\gamma }{\gamma _1}}}{L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}-L_1^{ \frac{\gamma _1-\gamma }{\gamma _1}}}\frac{1-\gamma _1}{\gamma _1K_1}>0, \end{aligned}$$

and rewrite the threshold ${\bar{x}}^b$ in Shim and Shin [16] as

$$\begin{aligned} {\bar{x}}^b=\left( -\frac{n_-}{n_+-n_-}\varLambda +\varGamma \right) (Y_1-Y_0)-\frac{Y_1}{r}. \end{aligned}$$

(92)

The inequality $Y_1/Y_0\le \varPsi$ is then equivalent to

$$\begin{aligned} (Y_1-Y_0)\le \frac{n_+-1}{-n_-\varLambda +\frac{n_+\gamma _1-\gamma _1+1}{ \gamma _1}\varGamma }\frac{Y_1}{r}. \end{aligned}$$

(93)

According to (92) and (93), we obtain

$$\begin{aligned} {\bar{x}}^b\le \left( \frac{-\frac{n_-}{n_+-n_-}\varLambda +\varGamma }{-\frac{n_-}{n_+-n_-} \varLambda +\frac{(n_+-1)\gamma _1+1}{(n_+-1)\gamma _1}\varGamma }-1\right) \frac{Y_1}{r}<0. \end{aligned}$$

$\square$

I Proof of Proposition 1

Note that $({\mathbf {c}}^{b}, \varvec{\pi }^{b})$ without the borrowing constraints is given by (see Shim and Shin [16])

$$\begin{aligned} c_t^b=\left\{ \begin{array}{ll} L_1^{\frac{\gamma _1-\gamma }{\gamma _1}}\left( z_t^{b,\lambda _1}\right) ^{-\frac{1}{\gamma _1}}, &{}\quad \text{ for } -{Y_1}/{r}<X_t<{\bar{x}}^b,\\ L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}\left( z_t^{b,\lambda _0}\right) ^{-\frac{1}{\gamma _1}}, &{}\quad \text{ for } X_t\ge {\bar{x}}^b, \end{array}\right. \end{aligned}$$

and

$$\begin{aligned} \pi _t^b=\left\{ \begin{array}{ll} \frac{\theta }{\sigma }\left\{ n_-(n_--1)D^b_2\left( z_t^{b,\lambda _1}\right) ^{n_--1}+L_1^{\frac{\gamma _1-\gamma }{\gamma _1}}\frac{1}{\gamma _1K_1}\left( z_t^{b,\lambda _1}\right) ^{-\frac{1}{\gamma _1}}\right\} , &{}\quad \text{ for } {Y_1}/{r}<X_t<{\bar{x}}^b,\\ \frac{\theta }{\sigma }\left\{ n_+(n_+-1)C^b_1\left( z_t^{b,\lambda _0}\right) ^{n_+-1}+L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}\frac{1}{\gamma _1K_1}\left( z_t^{b,\lambda _0}\right) ^{-\frac{1}{\gamma _1}}\right\} ,&{}\quad \text{ for } X_t\ge {\bar{x}}^b, \end{array}\right. \end{aligned}$$

where

$$\begin{aligned} {\bar{x}}^b&=-n_-D_2^b{\bar{z}}^{n_--1}+L_1^{\frac{\gamma _1-\gamma }{\gamma _1}}\frac{1}{K_1}{\bar{z}}^{-\frac{1}{\gamma _1}}-\frac{Y_1}{r},\\&=\left[ \frac{\frac{-n_+n_-\gamma _1-n_-+n_-\gamma _1}{\gamma _1K_1}+\frac{n_+n_--n_-}{r}}{n_+-n_-}+\frac{L_1^{\frac{\gamma _1-\gamma }{\gamma _1}}}{L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}-L_1^{\frac{\gamma _1-\gamma }{\gamma _1}}}\frac{1-\gamma _1}{\gamma _1K_1}\right] (Y_1-Y_0)-\frac{Y_1}{r},\\ C^b_1&=D_2{\bar{z}}^{n_--n_+}+\left( L_1^{\frac{\gamma _1-\gamma }{\gamma _1}}-L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}\right) \frac{\gamma _1}{(1-\gamma _1)K_1}{\bar{z}}^{-\frac{1-\gamma _1}{\gamma _1}-n_+}+\frac{Y_1-Y_0}{r}{\bar{z}}^{1-n_+}>0,\\ D^b_2&=D_2>0, \end{aligned}$$

and $z_t^{b,\lambda _0}$ and $z_t^{b,\lambda _1}$, respectively, have a one-to-one correspondence with $X_t$ by the algebraic equations

$$\begin{aligned} X_t=-n_+C^b_1\left( z_t^{b,\lambda _0}\right) ^{n_+-1}+L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}\frac{1}{K_1}\left( z_t^{b,\lambda _0}\right) ^{-\frac{1}{\gamma _1}}-\frac{Y_0}{r}, ~~\text{ for } X_t\ge {\bar{x}}^b, \end{aligned}$$

(94)

and

$$\begin{aligned} X_t=-n_-D^b_2\left( z_t^{b,\lambda _1}\right) ^{n_--1}+L_1^{\frac{\gamma _1-\gamma }{\gamma _1}}\frac{1}{K_1}\left( z_t^{b,\lambda _1}\right) ^{-\frac{1}{\gamma _1}}-\frac{Y_1}{r}, ~~\text{ for } -{Y_1}/{r}<X_t<{\bar{x}}^b, \end{aligned}$$

(95)

respectively. Since $D_1<0$, it is straightforward to see that ${\bar{x}}^b<{\bar{x}}$ and $C_1<C_1^b$.

Consider the case where $X_t\ge {\bar{x}}$. Recall that ${\mathcal {X}}(z) = -v^{\prime }(z)$ in (31) is a decreasing function of z with $X_t = {\mathcal {X}}({\mathcal {Z}}(X_t))$ by (32). Note that $X_t$ is a decreasing function of $z_t^{b,\lambda _0}$ in (94). That is, $X_t$ can be written as the decreasing functions of ${\mathcal {Z}}(X_t)$ and $z_t^{b,\lambda _0}$:

$$\begin{aligned} X_t&=-n_+C^b_1\left( z_t^{b,\lambda _0}\right) ^{n_+-1}+L_0^{\frac{\gamma _1-\gamma }{\gamma _1}} \frac{1}{K_1}\left( z_t^{b,\lambda _0}\right) ^{-\frac{1}{\gamma _1}}-\frac{Y_0}{r}\nonumber \\&=-n_+C_1\left( {\mathcal {Z}}(X_t)\right) ^{n_+-1}+L_0^{\frac{\gamma _1-\gamma }{\gamma _1}} \frac{1}{K_1}\left( {\mathcal {Z}}(X_t)\right) ^{-\frac{1}{\gamma _1}}-\frac{Y_0}{r}. \end{aligned}$$

(96)

Thus, since $C_1<C_1^b$, we have

$$\begin{aligned} z_t^{b,\lambda _0}<{\mathcal {Z}}(X_t) ~ \text{ for } X_t\ge {\bar{x}}, \end{aligned}$$

(97)

which implies the inequality $c_t^*<c_t^{b}$ for $X_t\ge {\bar{x}}$.

Multiplying both sides of the equality (96) by $(n_+-1)$, then we obtain

$$\begin{aligned} -n_+&(n_+-1)C^b_1\left( z_t^{b,\lambda _0}\right) ^{n_+-1}+L_0^{\frac{\gamma _1-\gamma }{ \gamma _1}}\frac{n_+-1}{K_1}\left( z_t^{b,\lambda _0}\right) ^{-\frac{1}{\gamma _1}}\\&\quad =-n_+(n_+-1)C_1\left( {\mathcal {Z}}(X_t)\right) ^{n_+-1}+L_0^{\frac{ \gamma _1-\gamma }{\gamma _1}}\frac{n_+-1}{K_1}\left( {\mathcal {Z}}(X_t)\right) ^{-\frac{1}{\gamma _1}}, \end{aligned}$$

and

$$\begin{aligned} -n_+&(n_+-1)C^b_1\left( z_t^{b,\lambda _0}\right) ^{n_+-1}-L_0^{\frac{\gamma _1-\gamma }{ \gamma _1}}\frac{1}{\gamma _1K_1}\left( z_t^{b,\lambda _0}\right) ^{-\frac{1}{\gamma _1}}+L_0^{ \frac{\gamma _1-\gamma }{\gamma _1}}\frac{\frac{1}{\gamma _1}+n_+-1}{K_1}\left( z_t^{b,\lambda _0} \right) ^{-\frac{1}{\gamma _1}}\\&\quad =-n_+(n_+\!-1\!)C_1\left( {\mathcal {Z}}(X_t)\right) ^{n_+-1}\!-\!L_0^{\frac{\gamma _1- \gamma }{\gamma _1}}\frac{1}{\gamma _1K_1}\left( {\mathcal {Z}}(X_t)\right) ^{-\frac{1}{ \gamma _1}}\!+\!L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}\frac{\frac{1}{\gamma _1}+n_+-1}{K_1} \left( {\mathcal {Z}}(X_t)\right) ^{-\frac{1}{\gamma _1}}. \end{aligned}$$

Thus, we get

$$\begin{aligned} \pi _t^b-\pi _t^*&=\frac{\theta }{\sigma }\left[ L_0^{\frac{\gamma _1-\gamma }{\gamma _1}} \frac{\frac{1}{\gamma _1}+n_+-1}{K_1}\left\{ \left( z_t^{b,\lambda _0}\right) ^{- \frac{1}{\gamma _1}}-\left( {\mathcal {Z}}(X_t)\right) ^{-\frac{1}{\gamma _1}}\right\} \right] >0, \end{aligned}$$

where the inequality holds due to (20) and (97).

Similarly, we can show that $c_t^*<c_t^{b}~ \text{ and } ~ \pi _t^*<\pi _t^b$ for $0\le X_t < {\bar{x}}^b$. $\square$

J Proof of Proposition 2

As mentioned in Remark 9, ${\bar{x}}^b < 0$ in this case. Recall that ${\mathcal {X}}(z) = -w^{\prime }(z)$ in (40) is a decreasing function of z with $X_t = {\mathcal {X}}({\mathcal {Z}}(X_t))$ by (41). Note that $X_t$ is a decreasing function of $z_t^{b,\lambda _0}$ in (94). That is, $X_t$ can be written as the decreasing functions of ${\mathcal {Z}}(X_t)$ and $z_t^{b,\lambda _0}$, for $X_t\ge 0 > {\bar{x}}^b$:

$$\begin{aligned} X_t&=-n_+E_1\left( {\mathcal {Z}}(X_t)\right) ^{n_+-1}+L_0^{\frac{\gamma _1-\gamma }{\gamma _1}} \frac{1}{K_1}\left( {\mathcal {Z}}(X_t)\right) ^{-\frac{1}{\gamma _1}}-\frac{Y_0}{r}\nonumber \\&=-n_+C^b_1\left( z_t^{b,\lambda _0}\right) ^{n_+-1}+L_0^{\frac{\gamma _1-\gamma }{\gamma _1}} \frac{1}{K_1}\left( z_t^{b,\lambda _0}\right) ^{-\frac{1}{\gamma _1}}-\frac{Y_0}{r}. \end{aligned}$$

(98)

Thus, since $E_1<0<C_1^b$, we have

$$\begin{aligned} z_t^{b,\lambda _0}<{\mathcal {Z}}(X_t)~ \text{ for } X_t\ge 0, \end{aligned}$$

(99)

which implies the inequality $c_t^*<c_t^{b}$ for $X_t\ge 0$.

Multiplying both sides of the equality (98) by $(n_+-1)$, then we obtain

$$\begin{aligned}&-n_+(n_+-1)E_1\left( {\mathcal {Z}}(X_t)\right) ^{n_+-1}+L_0^{\frac{\gamma _1-\gamma }{\gamma _1}} \frac{n_+-1}{K_1}\left( {\mathcal {Z}}(X_t)\right) ^{-\frac{1}{\gamma _1}}\\&\quad =-n_+(n_+-1)C_1^b\left( z_t^{b,\lambda _0}\right) ^{n_+-1}+L_0^{\frac{\gamma _1-\gamma }{\gamma _1}} \frac{n_+-1}{K_1}\left( z_t^{b,\lambda _0}\right) ^{-\frac{1}{\gamma _1}}, \end{aligned}$$

and

$$\begin{aligned}&-n_+(n_+-1)E_1\left( {\mathcal {Z}}(X_t)\right) ^{n_+-1}-L_0^{\frac{\gamma _1-\gamma }{\gamma _1}} \frac{1}{\gamma _1K_1}\left( {\mathcal {Z}}(X_t)\right) ^{-\frac{1}{\gamma _1}}+L_0^{\frac{\gamma _1-\gamma }{ \gamma _1}}\frac{\frac{1}{\gamma _1}+n_+-1}{K_1}\left( {\mathcal {Z}}(X_t)\right) ^{- \frac{1}{\gamma _1}}\\&\quad =-n_+(n_+-1)C_1^b\left( z_t^{b,\lambda _0}\right) ^{n_+-1}-L_0^{\frac{\gamma _1-\gamma }{\gamma _1}} \frac{1}{\gamma _1K_1}\left( z_t^{b,\lambda _0}\right) ^{-\frac{1}{\gamma _1}}+L_0^{\frac{\gamma _1-\gamma }{ \gamma _1}}\frac{\frac{1}{\gamma _1}+n_+-1}{K_1}\left( z_t^{b,\lambda _0}\right) ^{-\frac{1}{\gamma _1}}. \end{aligned}$$

Thus we get

$$\begin{aligned} \pi _t^b-\pi _t^*&=\frac{\theta }{\sigma }\left[ L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}\frac{ \frac{1}{\gamma _1}+n_+-1}{K_1}\left\{ \left( z_t^{b,\lambda _0}\right) ^{-\frac{1}{\gamma _1}}- \left( {\mathcal {Z}}(X_t)\right) ^{-\frac{1}{\gamma _1}}\right\} \right] >0, \end{aligned}$$

where the inequality holds due to (20) and (99). $\square$

K Proof of Proposition 3

Recall that $X_t = {\mathcal {X}}({\mathcal {Z}}(X_t)) = -v^{\prime }({\mathcal {Z}}(X_t))$. That is, for $0<X_t<{\bar{x}}$,

$$\begin{aligned} X_t = -n_+D_1\left( {\mathcal {Z}}(X_t)\right) ^{n_+-1}-n_-D_2\left( {\mathcal {Z}}(X_t) \right) ^{n_--1}+L_1^{\frac{\gamma _1-\gamma }{\gamma _1}}\frac{1}{K_1}\left( { \mathcal {Z}}(X_t)\right) ^{-\frac{1}{\gamma _1}}-\frac{Y_1}{r}.\nonumber \\ \end{aligned}$$

(100)

Since $D_1<0$ and $D_2>0$ we have ${\mathcal {Z}}(X_t)>z_t^{M_1}$, which yields $c_t^*<c_t^{M_1}$ for $0<X_t<{\bar{x}}$. On the other hand, we can obtain, for $0<X_t<{\bar{x}}$,

$$\begin{aligned} \frac{\sigma }{\theta }\left( \pi _t^*-\pi _t^{M_1}\right) =(n_+-1)\frac{Y_1}{r}-F \left( {\mathcal {Z}}(X_t)\right) -\left( \frac{1}{\gamma _1}+n_+-1\right) L_1^{ \frac{\gamma _1-\gamma }{\gamma _1}}\frac{1}{K_1}\left( z_t^{M_1}\right) ^{-\frac{1}{\gamma _1}}, \end{aligned}$$

if we exploit (35) along with (100), and (42) with (43), where the function F is defined as (71). Since ${\mathcal {Z}}(X_t)$ and $L_1^{\frac{\gamma _1-\gamma }{\gamma _1}}\frac{1}{K_1}\left( z_t^{M_1}\right) ^{-\frac{1}{\gamma _1}}$ go to ${\tilde{z}}$ and $Y_1/r$, respectively, as $X_t$ goes to 0, and $F\left( {\tilde{z}}\right) =0$, we have

$$\begin{aligned} \lim _{X_t\downarrow 0}\frac{\sigma }{\theta }\left( \pi _t^*-\pi _t^{M_1}\right) =\left[ (n_+-1)\frac{Y_1}{r}-F\left( {\tilde{z}}\right) -\left( \frac{1}{\gamma _1}+n_+-1\right) \frac{Y_1}{r}\right] =-\frac{1}{\gamma _1}\frac{Y_1}{r}<0, \end{aligned}$$

$\square$

L Proof of Proposition 4

If $C_1>0$, we see that ${\mathcal {Z}}(X_t)<z_t^{M_0}$ for $X_t\ge {\bar{x}}$, so we easily obtain $c_t^*>c_t^{M_0}$ for $X_t\ge {\bar{x}}$. We also have

$$\begin{aligned} \pi _t^*&=\frac{\theta }{\sigma }\left\{ n_+(n_+-1)C_1\left( {\mathcal {Z}}(X_t)\right) ^{n_+-1}+L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}\frac{1}{\gamma _1K_1}\left( {\mathcal {Z}}(X_t)\right) ^{-\frac{1}{\gamma _1}}\right\} \\&>\frac{\theta }{\sigma \gamma _1}L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}\frac{1}{K_1}\left( {\mathcal {Z}}(X_t)\right) ^{-\frac{1}{\gamma _1}} >\frac{\theta }{\sigma \gamma _1}L_0^{\frac{\gamma _1-\gamma }{\gamma _1}}\frac{1}{K_1}\left( z_t^{M_0}\right) ^{-\frac{1}{\gamma _1}}=\pi _t^{M_0}, \end{aligned}$$

for $X_t\ge {\bar{x}}$. It is straightforward to see that $c_t^*<c_t^{M_0},~\pi _t^*<\pi _t^{M_0}$ for $X_t\ge {\bar{x}}$ given $C_1<0$. $\square$