A Mellin transform
In this appendix, we use the Mellin transform method to obtain an analytic solution to the Cauchy problem for the Black-Scholes type partial differential operator. The definition and basic properties of Mellin transformation are summarized in Appendix A in [9]. Also, the reader can refer to [6, 22] and [2].
Theorem 4
(Non-homogeneous Black-Scholes type PDE) We consider the following Cauchy problem:
$$\begin{aligned} \begin{aligned}&\left( \dfrac{\partial }{\partial t} + {\mathcal {L}} \right) \phi (t,y) = f(t,y), \\&\phi (T,y) = g(y), \end{aligned} \end{aligned}$$
(5.1)
on the domain \(\{(t,y) \mid 0<t<T,\;0<y<\infty \}\), where the partial differential operator \({\mathcal {L}}\) is given in (3.7). Then, \(\phi (t,y)\) can be represented by
$$\begin{aligned}\begin{aligned} \phi (t,y)=\int _{0}^{\infty }g(v){\mathcal {G}}(T-t,\dfrac{y}{v})\dfrac{1}{v}dv -\int _{t}^{T}\int _{0}^{\infty }f(\eta ,v){\mathcal {G}}(\eta -t,\dfrac{y}{v})\dfrac{1}{v}dvd\eta , \end{aligned} \end{aligned}$$
where the kernel function \({\mathcal {G}}(\cdot ,\cdot )\) is given by
$$\begin{aligned} \begin{aligned} {\mathcal {G}}(t,y) = e^{-\frac{1}{2}\left\{ \left( \frac{1-k_2}{2}\right) ^2+k_1\right\} \theta ^2t}\cdot \dfrac{y^{\frac{1-k_2}{2}}}{\theta \sqrt{2\pi t}} \exp {\left\{ -\dfrac{1}{2}\dfrac{(\log {y})^2}{\theta ^2 t}\right\} }, \end{aligned} \end{aligned}$$
(5.2)
and \(k_1 = {2\beta }/{\theta ^2}\), \(k_2 = {2(\beta -r)}/{\theta ^2}\).
Proof
We denote \(\hat{\phi }(t,y^{*})\) as the Mellin transform of \(\phi (t,y)\). Then, by the inverse Mellin transform, we derive
$$\begin{aligned}\begin{aligned} \phi (t,y)=\dfrac{1}{2\pi i}\int _{c-i\infty }^{c+i\infty }\hat{\phi }(t,y^{*})y^{-y^{*}}dy^{*}. \end{aligned} \end{aligned}$$
The PDE in (5.1) is changed by the following ODE:
$$\begin{aligned} \begin{aligned} \dfrac{d\hat{\phi }}{dt}(t,y^{*})+\dfrac{1}{2}\theta ^2 Q(y^{*})\hat{\phi }(t,y^{*})&={\hat{f}}(t,y^{*})\\ Q(y^{*})&=\left( y^{*}\right) ^2+y^{*}(1-k_2)-k_1, \end{aligned} \end{aligned}$$
(5.3)
where \({\hat{f}}(t,y^{*})\) is the Mellin transform of f(t, y), and \(k_1={2\beta }/{\theta ^2}\), \(k_2={2(\beta -r)}/{\theta ^2}\).
The non-homogeneous ODE (5.3) yields
$$\begin{aligned}\begin{aligned} \hat{\phi }(t,y^{*})=&e^{\frac{1}{2}\theta ^2 Q(y^{*})(T-t)}{\hat{g}}(y^{*})\\&-\int _{t}^{T}e^{\frac{1}{2}\theta ^2 Q(y^{*})(\eta -t)}{\hat{f}}(\eta ,y^{*})d\eta . \end{aligned} \end{aligned}$$
and consequently we obtain
$$\begin{aligned} \begin{aligned} {\phi }(t,y)=&\dfrac{1}{2\pi i}\int _{c-i\infty }^{c+i\infty }e^{\frac{1}{2}\theta ^2 Q(y^{*})(T-t)}{\hat{g}}(y^{*})y^{-y^{*}}dy^{*} \\&-\dfrac{1}{2\pi i}\int _{c-i\infty }^{c+i\infty }\int _{t}^{T}e^{\frac{1}{2}\theta ^2 Q(y^{*})(\eta -t)}{\hat{f}}(\eta ,y^{*})y^{-y^{*}}d\eta dy^{*}, \end{aligned} \end{aligned}$$
(5.4)
where \({\hat{g}}(y^{*})\) is the Mellin transform of g(y).
In addition, in order to calculate \(\phi (t,y)\) in (5.4), we consider
$$\begin{aligned} {\mathcal {G}}(t,y)=\dfrac{1}{2\pi i}\int _{c-i\infty }^{c+i\infty }e^{\frac{1}{2}\theta ^2 Q(y^{*})(T-t)}y^{-y^{*}}dy^{*}. \end{aligned}$$
Since \(e^{\frac{1}{2}\theta ^2 Q(y^{*})(T-t)}\), \({\hat{g}}(y^{*})\), and \({\hat{f}}(\eta ,y^{*})\) are the Mellin transforms of \({\mathcal {G}}(t,y)\), g(y), and f(t, y), respectively, by using the Mellin convolution theorem in [2] we obtain
$$\begin{aligned}\begin{aligned} \phi (t,y)=\int _{0}^{\infty }g(v){\mathcal {G}}(T-t,\dfrac{y}{v})\dfrac{1}{v}dv-\int _{t}^{T}\int _{0}^{\infty }f(\eta ,v){\mathcal {G}}(\eta -t,\dfrac{y}{v})\dfrac{1}{v}dvd\eta , \end{aligned} \end{aligned}$$
where
$$\begin{aligned}\begin{aligned} {\mathcal {G}}(t,y)=e^{-\frac{1}{2}\left\{ \left( \frac{1-k_2}{2}\right) ^2 +k_1\right\} \theta ^2 t}\cdot \dfrac{y^{\frac{1-k_2}{2}}}{\theta \sqrt{2\pi t}}\exp {\left\{ -\frac{1}{2}\dfrac{(\log {y})^2}{\theta ^2 {t}}\right\} }. \end{aligned} \end{aligned}$$
\(\square\)
The following lemmas are also useful.
Lemma 1
For any real number \(\alpha\) and the kernel function \({\mathcal {G}}(t,y)\) in (5.2), we have
$$\begin{aligned}\begin{aligned}&\int _{0}^{b}v^{-\alpha }{\mathcal {G}}(t,\dfrac{y}{v})\dfrac{1}{v}dv = y^{-\alpha }e^{-\frac{1}{2}\{k_1-(1-k_2)\alpha -\alpha ^2\}\theta ^2 t}{\mathcal {N}}\left( \frac{-\log {\dfrac{y}{b}}+\left( \frac{1-k_2}{2}+\alpha \right) \theta ^2 t}{\theta \sqrt{t}}\right) ,\\&\quad \int _{b}^{\infty }v^{-\alpha }{\mathcal {G}}(t,\dfrac{y}{v})\dfrac{1}{v}dv = y^{-\alpha }e^{-\frac{1}{2}\{k_1-(1-k_2)\alpha -\alpha ^2\}\theta ^2 t}{\mathcal {N}}\left( \frac{\log {\dfrac{y}{b}}-\left( \frac{1-k_2}{2}+\alpha \right) \theta ^2 t}{\theta \sqrt{t}}\right) , \end{aligned} \end{aligned}$$
where \({\mathcal {N}}(\cdot )\) is a cumulative distribution function of the standard normal distribution.
Proof
First we consider
$$\begin{aligned}\begin{aligned}&\int _{0}^{b}v^{-\alpha }{\mathcal {G}}(t,\frac{y}{v})\frac{1}{v}dv\\&=\int _{0}^{b}v^{-\alpha }e^{-\frac{1 }{2}\left\{ \left( \frac{1-k_2}{2}\right) ^2 +k_1\right\} \theta ^2t}\cdot \dfrac{\left( \frac{y}{v}\right) ^{\frac{1-k_2}{2}}}{\theta \sqrt{2\pi t}}e^{-\frac{1}{2}\left( \frac{{\log {(y/v)}}}{{\theta \sqrt{t}}}\right) ^2} \dfrac{1}{v} dv \\&=-y^{-\alpha }e^{-\frac{1 }{2}\left\{ \left( \frac{1-k_2}{2}\right) ^2 +k_1\right\} \theta ^2t}\int _{\infty }^{\log {\frac{y}{b}}} e^{\alpha w} \dfrac{e^{\left( \frac{1-k_2}{2}\right) w}}{\theta \sqrt{2\pi t}}e^{-\frac{1}{2}\frac{w^2}{\theta ^2 t}}dw \\&=-y^{-\alpha }e^{-\frac{1}{2} \left\{ \left( \frac{1-k_2}{2}\right) ^2+k_1-\left( \frac{1-k_2}{2}+\alpha \right) ^2\right\} \theta ^2t}\\&\quad \int _{\infty }^{\log {\frac{y}{b}}}\dfrac{1}{\theta \sqrt{2\pi t}}\exp {\left\{ -\frac{1}{2}\left( \frac{w-\theta ^2t\left( \frac{1-k_2}{2}+\alpha \right) }{\theta \sqrt{t}} \right) ^2 \right\} }\; dw\\&=y^{-\alpha }e^{-\frac{1}{2}\left\{ k_1-(1-k_2)\alpha -\alpha ^2\right\} \theta ^2 t}{\mathcal {N}}\left( \dfrac{-\log {\frac{y}{b}}+\theta ^2 t \left( \frac{1-k_2}{2}+\alpha \right) }{\theta \sqrt{t}} \right) , \end{aligned} \end{aligned}$$
where the second equality is obtained from the transformation \(w=\log (y/v)\). Similarly we obtain
$$\begin{aligned}\begin{aligned}&\int _{b}^{\infty }v^{-\alpha }{\mathcal {G}}(t,\dfrac{y}{v})\dfrac{1}{v}dv\\&\quad = y^{-\alpha }e^{-\frac{1}{2}\{k_1-(1-k_2)\alpha -\alpha ^2\}\theta ^2 t}\\&\qquad {\mathcal {N}}\left( \frac{\log {\dfrac{y}{b}}-\left( \frac{1-k_2}{2}+\alpha \right) \theta ^2 t}{\theta \sqrt{t}}\right) . \end{aligned} \end{aligned}$$
\(\square\)
Lemma 2
For any real number \(\alpha\) and the kernel function \({\mathcal {G}}(t,y)\) in (5.2), we have
$$\begin{aligned}\begin{aligned}&\int _{0}^{b}v^{-\alpha }\log {v}\cdot {\mathcal {G}}(t,\dfrac{y}{v})\dfrac{1}{v}dv\\&\quad =y^{-\alpha }\left( \log {y}-\left( \dfrac{1-k_2}{2}+\alpha \right) \theta ^2 t\right) e^{-\frac{1}{2}\{k_1-(1-k_2)\alpha -\alpha ^2\}\theta ^2 t}\\&\qquad {\mathcal {N}}\left( \frac{-\log {\dfrac{y}{b}}+\left( \frac{1-k_2}{2}+\alpha \right) \theta ^2 t}{\theta \sqrt{t}}\right) \\&\qquad -y^{-\alpha }e^{-\frac{1}{2}\{k_1-(1-k_2)\alpha -\alpha ^2\}\theta ^2 t}\theta \sqrt{t}\cdot \mathbf{n}\left( \frac{-\log {\dfrac{y}{b}}+\left( \frac{1-k_2}{2}+\alpha \right) \theta ^2 t}{\theta \sqrt{t}}\right) , \end{aligned}\end{aligned}$$
where \(\mathbf{n}(\cdot )\) is a probability density function of the standard normal distribution.
Proof
We see that
$$\begin{aligned} \begin{aligned}&\int _{0}^{b}v^{-\alpha }\log {v}\cdot {\mathcal {G}}(t,\dfrac{y}{v})\dfrac{1}{v}dv\\&\quad =\int _{0}^{b}v^{-\alpha }(\log {y}-\log {\dfrac{y}{v}})\cdot {\mathcal {G}}(t,\dfrac{y}{v})\dfrac{1}{v}dv\\&\quad =\log {y}\int _{0}^{b}v^{-\alpha }{\mathcal {G}}(t,\dfrac{y}{v})\dfrac{1}{v}dv - \int _{0}^{b}v^{-\alpha }\log {\dfrac{y}{v}}\cdot {\mathcal {G}}(t,\dfrac{y}{v})\dfrac{1}{v}dv\\&\quad =y^{-\alpha }(\log {y})e^{-\frac{1}{2}\{k_1-(1-k_2)\alpha -\alpha ^2\}\theta ^2 t}{\mathcal {N}}\left( \frac{-\log {\dfrac{y}{b}}+\left( \frac{1-k_2}{2}+\alpha \right) \theta ^2 t}{\theta \sqrt{t}}\right) \\&\qquad - \int _{0}^{b}v^{-\alpha }\log {\dfrac{y}{v}}\cdot {\mathcal {G}}(t,\dfrac{y}{v})\dfrac{1}{v}dv, \end{aligned} \end{aligned}$$
(5.5)
where the last equality in (5.5) is due to Lemma 1. The integral of the last equality in (5.5) can be written as
$$\begin{aligned} \begin{aligned}&\int _{0}^{b}v^{-\alpha }\log {\dfrac{y}{v}}\cdot {\mathcal {G}}(t,\dfrac{y}{v})\dfrac{1}{v}dv \\&\quad =y^{-\alpha }e^{-\frac{1}{2}\left\{ \left( \frac{1-k_2}{2}\right) ^2 +k_1\right\} \theta ^2 t}\int _{0}^{b}\left( \dfrac{y}{v}\right) ^{\alpha }\log {\dfrac{y}{v}}\cdot \dfrac{\left( \frac{y}{v}\right) ^{\frac{1-k_2}{2}}}{\theta \sqrt{2\pi t}}e^{-\frac{1}{2}\left( \frac{{\log {(y/v)}}}{{\theta \sqrt{t}}}\right) ^2} \dfrac{1}{v} dv \\&\quad =-y^{-\alpha }e^{-\frac{1}{2}\left\{ \left( \frac{1-k_2}{2}\right) ^2 +k_1\right\} \theta ^2 t}\int _{\infty }^{\log {\frac{y}{b}}}\dfrac{1}{\theta \sqrt{2\pi t}}we^{\left( \alpha +\frac{1-k_2}{2}\right) w}e^{-\frac{w^2}{2\theta ^2 t}}dw\\&\quad =-y^{-\alpha }e^{-\frac{1}{2} \{k_1-(1-k_2)\alpha -\alpha ^2\}\theta ^2 t}\int _{\infty }^{\log {\frac{y}{b}}}\dfrac{1}{\theta \sqrt{2\pi t}}w\exp {\left\{ -\frac{1}{2}\left( \dfrac{w-\left( \frac{1-k_2}{2}+\alpha \right) \theta ^2 t}{\theta \sqrt{t}}\right) ^2\right\} }dw, \end{aligned} \end{aligned}$$
(5.6)
where the second equality is obtained from the transformation \(w=\log (y/v)\). The integral without the coefficient \(-y^{-\alpha }e^{-\frac{1}{2}\{k_1-(1-k_2)\alpha -\alpha ^2\}\theta ^2 t}\) of the last equality in the equation(5.6) can be given by
$$\begin{aligned} \begin{aligned}&\int _{\infty }^{\log {\frac{y}{b}}}\dfrac{1}{\theta \sqrt{2\pi t}}w\exp {\left\{ -\frac{1}{2}\left( \dfrac{w-\left( \frac{1-k_2}{2}+\alpha \right) \theta ^2 t}{\theta \sqrt{t}}\right) ^2\right\} }dw\\&\quad =\int _{\infty }^{\log {\frac{y}{b}}}\dfrac{1}{\theta \sqrt{2\pi t}}\left( w-\left( \dfrac{1-k_2}{2}+\alpha \right) \theta ^2 t\right) \exp {\left\{ -\frac{1}{2}\left( \dfrac{w-\left( \frac{1-k_2}{2}+\alpha \right) \theta ^2 t}{\theta \sqrt{t}}\right) ^2\right\} }dw\\&\qquad +\left( \dfrac{1-k_2}{2}+\alpha \right) \theta ^2 t\int _{\infty }^{\log {\frac{y}{b}}}\dfrac{1}{\theta \sqrt{2\pi t}}\exp {\left\{ -\frac{1}{2}\left( \dfrac{w-\left( \frac{1-k_2}{2}+\alpha \right) \theta ^2 t}{\theta \sqrt{t}}\right) ^2\right\} }dw\\&\quad =\left[ -\dfrac{\theta \sqrt{t}}{\sqrt{2\pi }} \exp {\left\{ -\frac{1}{2}\left( \dfrac{w-(\frac{1-k_2}{2}+\alpha )\theta ^2 t}{\theta \sqrt{t}}\right) ^2 \right\} }\right] _{\infty }^{\log {\frac{y}{b}}} -\left( \dfrac{1-k_2}{2}+\alpha \right) \theta ^2 t \\&\qquad \times {\mathcal {N}}\left( \frac{-\log {\dfrac{y}{b}}+\left( \frac{1-k_2}{2}+\alpha \right) \theta ^2 t}{\theta \sqrt{t}}\right) \\&\quad =-\theta \sqrt{t}\cdot \mathbf{n}\left( \frac{-\log {\dfrac{y}{b}}+\left( \frac{1-k_2}{2}+\alpha \right) \theta ^2 t}{\theta \sqrt{t}}\right) \\&\qquad -\left( \dfrac{1-k_2}{2}+\alpha \right) \theta ^2 t \cdot {\mathcal {N}}\left( \frac{-\log {\dfrac{y}{b}}+\left( \frac{1-k_2}{2}+\alpha \right) \theta ^2 t}{\theta \sqrt{t}}\right) . \end{aligned}\end{aligned}$$
(5.7)
Therefore, by (5.5), (5.6), and (5.7), we obtain the desired result. \(\square\)
B Proof of Theorem 1
We will prove Theorem 1 in the following steps.
(Step 1) For given \(x>\int _t^T \gamma _s^t R(s)ds\), there exists a unique \(y^*\) such that
$$\begin{aligned} x= -\dfrac{\partial {\widetilde{V}}}{\partial y}(t,y^*). \end{aligned}$$
Proof of (Step 1). By Feynman-Kac formula, \({\widetilde{V}}(t,y)\) satisfies the following PDE:
$$\begin{aligned} \begin{aligned} {\left\{ \begin{array}{ll} &{}\left( \dfrac{\partial }{\partial t} + {\mathcal {L}} \right) {\widetilde{V}}(t,y) + {\widetilde{u}}(t,y) = 0, \\ &{}{\widetilde{V}}(T,y) = 0. \end{array}\right. } \end{aligned} \end{aligned}$$
(6.1)
Let us denote
$$\begin{aligned} h(t,y) \equiv -\dfrac{\partial {\widetilde{V}}}{\partial y}(t,y). \end{aligned}$$
By differentiating the PDE (6.1) with respect to y, it is easy to confirm that h(t, y) is the solution to the following PDE:
$$\begin{aligned} \begin{aligned} {\left\{ \begin{array}{ll} &{}\left( \dfrac{\partial }{\partial t} + {{\mathcal {L}}_1}\right) h(t,y) + I(y - (y - {\widetilde{y}}(t))^{+}) = 0 \\ &{}h(T,y) = 0, \end{array}\right. } \end{aligned} \end{aligned}$$
(6.2)
where the partial differential operator \({{\mathcal {L}}_1}\) is given by
$$\begin{aligned} \begin{aligned} {{\mathcal {L}}_1} \triangleq \dfrac{1}{2}\theta ^2 y^2 \dfrac{\partial ^2}{\partial y^2}+\left( \beta - r+\theta ^2\right) y\dfrac{\partial }{\partial y} - r {\mathcal {I}}. \end{aligned} \end{aligned}$$
(6.3)
For the operator \({{\mathcal {L}}_1}\), applying Theorem 4 in Appendix A, h(t, y) is represented by
$$\begin{aligned}\begin{aligned} h(t,y)&=\int _{t}^{T}\int _{0}^{{\widetilde{y}}(\eta )}I(v)\cdot \widetilde{{\mathcal {G}}}(\eta -t,\dfrac{y}{v})\dfrac{1}{v}dvd\eta +\int _{t}^{T}\int _{{\widetilde{y}}(\eta )}^{\infty }R(\eta )\cdot \widetilde{{\mathcal {G}}}(\eta -t,\dfrac{y}{v})\dfrac{1}{v}dvd\eta \\ \end{aligned} \end{aligned}$$
where
$$\begin{aligned}\begin{aligned} \widetilde{{\mathcal {G}}}(t,y) = e^{-\frac{1}{2}\left\{ \left( \frac{1-{\widetilde{k}}_2}{2}\right) ^2+{\widetilde{k}}_1\right\} \theta ^2t} \dfrac{y^{\frac{1-{\widetilde{k}}_2}{2}}}{\theta \sqrt{2\pi t}} \exp {\left\{ -\dfrac{1}{2}\dfrac{(\log {y})^2}{\theta ^2 t}\right\} } \end{aligned} \end{aligned}$$
and \({\widetilde{k}}_1 = {2r}/{\theta ^2}\), \({\widetilde{k}}_2 = {2(\beta -r+\theta ^2)}/{\theta ^2}\). Indeed we see that
$$\begin{aligned}\begin{aligned} \widetilde{{\mathcal {G}}}(t,y) = \dfrac{1}{y}{\mathcal {G}}(t,y), \end{aligned} \end{aligned}$$
where \({\mathcal {G}}(t,y)\) is defined in (5.2). Therefore, by Lemma 1 in Appendix A, h(t, y) is given by
$$\begin{aligned} \begin{aligned} h(t,y)=&\dfrac{1}{y}\int _{0}^{\tau }\int _{0}^{{\widetilde{y}}(\xi +t)}vI(v)\cdot {{\mathcal {G}}}(\xi ,\dfrac{y}{v})\dfrac{1}{v}dvd\xi \\&+\dfrac{1}{y} \int _{t}^{T}\int _{{\widetilde{y}}(\eta )}^{\infty }R(\eta )v\cdot {{\mathcal {G}}}(\eta -t,\dfrac{y}{v})\dfrac{1}{v}dvd\eta \\ =&\dfrac{1}{y}\int _{0}^{\tau }\int _{0}^{{\widetilde{y}}(\xi +t)}vI(v)\cdot {{\mathcal {G}}}(\xi ,\dfrac{y}{v})\dfrac{1}{v}dvd\xi \\&+ \int _{0}^{\tau }R(\xi +t)\cdot e^{-r\xi } {\mathcal {N}}\left( d^{+}(\xi ,\dfrac{y}{{\widetilde{y}}(\xi +t)})\right) d\xi , \end{aligned} \end{aligned}$$
(6.4)
where \(\xi =\eta -t\) and \(\tau =T-t\).
Since \({\mathcal {L}}_1\left( y\dfrac{\partial }{\partial y}\right) =y\dfrac{\partial }{\partial y} {\mathcal {L}}_1\), we deduce that \(y\dfrac{\partial h}{\partial y}\) satisfies the following PDE:
$$\begin{aligned} \begin{aligned} {\left\{ \begin{array}{ll} &{}\left( \dfrac{\partial }{\partial t} + {{\mathcal {L}}_1}\right) \left( y\dfrac{\partial h}{\partial y}\right) + \dfrac{y}{u''(I(y))}{} \mathbf{1}_{\{y\le {\tilde{y}}(t)\}} = 0, \\ &{}\left( y\dfrac{\partial h}{\partial y}\right) (T,y) = 0. \end{array}\right. } \end{aligned} \end{aligned}$$
(6.5)
Similarly, we can obtain
$$\begin{aligned} \begin{aligned} y\dfrac{\partial h}{\partial y}(t,y) = \int _{0}^{\tau }\int _{0}^{{\widetilde{y}}(\xi +t)}\underbrace{\dfrac{v}{u''(I(v))}}_{<0}\cdot \underbrace{{{\mathcal {G}}}(\xi ,\dfrac{y}{v})\dfrac{1}{v}}_{>0}dvd\xi <0. \end{aligned} \end{aligned}$$
(6.6)
This implies that the dual value function \({\widetilde{V}}(t,y)\) is strictly convex in \(y>0\).
Also, from the definition of \({\mathcal {G}}(t,y)\) in (3.8), we see that, for any \(v>0\),
$$\begin{aligned} \lim _{y\rightarrow 0+}\dfrac{{\mathcal {G}}(t,\frac{y}{v})}{y}&=e^{-\frac{1}{2}\left\{ \left( \frac{1-k_2}{2}\right) ^2+k_1\right\} \theta ^2 t}\cdot \dfrac{1}{v \theta \sqrt{2\pi t}}\lim _{y\rightarrow 0+}\dfrac{\left( \frac{y}{v}\right) ^{\frac{-1-k_2}{2}}}{\exp {\left\{ \dfrac{1}{2}\dfrac{(\log {\frac{y}{v}})^2}{\theta ^2 t}\right\} }}=+\infty ,\\ \lim _{y\rightarrow +\infty }\dfrac{{\mathcal {G}}(t,\frac{y}{v})}{y}&=e^{-\frac{1}{2}\left\{ \left( \frac{1-k_2}{2}\right) ^2+k_1\right\} \theta ^2 t}\cdot \dfrac{1}{v \theta \sqrt{2\pi t}}\lim _{y\rightarrow +\infty }\dfrac{\left( \frac{y}{v}\right) ^{\frac{-1-k_2}{2}}}{\exp {\left\{ \dfrac{1}{2}\dfrac{(\log {\frac{y}{v}})^2}{\theta ^2 t}\right\} }}=0 \end{aligned}$$
under Assumption 2. Thus we obtain
$$\begin{aligned} \lim _{y\rightarrow 0+}h(t,y)&=+\infty ,\\ \lim _{y\rightarrow +\infty }h(t,y)&=\int _{0}^{\tau }R(t+\xi )\cdot e^{-r\xi } d\xi >0. \end{aligned}$$
Hence, we can conclude that there exists a unique \(y^*\) such that
$$\begin{aligned} x= -\dfrac{\partial {\widetilde{V}}}{\partial y}(t,y^*). \end{aligned}$$
(Step 2) The following duality relationship is established:
$$\begin{aligned} V(t,x) = \inf _{y >0} \left( {\widetilde{V}}(t,y)+ y x \right) . \end{aligned}$$
(6.7)
Moreover, for \(s\in [t,T]\), the optimal wealth \(X_s^*\), the optimal consumption \(c_s^*\) and the optimal portfolio \(\pi _s^*\) at time s are given by
$$\begin{aligned} X_s^*=-\dfrac{\partial {\widetilde{V}}}{\partial }(s,Y_s^*),\;\;c_s^* = I(Y_s^* - (Y_s^* - {\tilde{y}}(s))^+)\;\;\;\text{ and }\;\;\;\pi _s^* = \dfrac{\theta }{\sigma }Y_s^* \dfrac{\partial ^2 {\widetilde{V}}}{\partial y^2}(s, Y_s^*) \end{aligned}$$
(6.8)
with \(Y_s^* = y^* e^{\beta (s-t)}H_s^t\).
Proof of (Step 2). For \(s\in [t,T]\), let us denote \(\mathcal {X}_s \equiv h(s,Y_s^*)\) with \(Y_s^* = y^* e^{\beta (s-t)}H_s^t\). By applying Itô’s formula to \(\mathcal {X}_s\), we obtain that
$$\begin{aligned} \begin{aligned} d\mathcal {X}_s =&\dfrac{\partial h}{\partial s}(s,Y_s^*)ds + \dfrac{\partial h}{\partial y}(s,Y_s^*) dY_s^* +\dfrac{1}{2}\dfrac{\partial ^2 h}{\partial y^2}(dY_s^*)^2\\ =&\left( \dfrac{\partial h}{\partial s}+\dfrac{\theta ^2}{2}(Y_s^*)^2\dfrac{\partial ^2 h}{\partial y^2}\right. \\&\left. +(\beta -r+\theta ^2)Y_s^*\dfrac{\partial h}{\partial y}\right) ds - \theta Y_s^*\dfrac{\partial h}{\partial y}(s,Y_s^*)(dB_s + \theta ds)\\ =&\underbrace{\left( \dfrac{\partial h}{\partial s}+\dfrac{\theta ^2}{2}(Y_s^*)^2\dfrac{\partial ^2 h}{\partial y^2} +(\beta -r+\theta ^2)Y_s^*\dfrac{\partial h}{\partial y}-r h(s,Y_s) +I(Y_s^*-(Y_s^*-{\tilde{y}}(s))^+)\right) }_{=0}ds\\&+\left( r h(s,Y_s)-I(Y_s^*-(Y_s^*-{\tilde{y}}(s))^+)+(\mu -r)\left( -\dfrac{\theta }{\sigma }\dfrac{\partial ^2 h}{\partial y^2}\right) \right) ds\\&- \theta Y_s^*\dfrac{\partial h}{\partial y}(s,Y_s^*)dB_s\\ =&(r\mathcal {X}_s -c_s^* +(\mu -r)\pi _s^*)ds + \sigma \pi _s^* dB_s, \end{aligned} \end{aligned}$$
where
$$\begin{aligned} c_s^* = I(Y_s^* - (Y_s^* - {\tilde{y}}(s))^+)\;\;\;\text{ and }\;\;\;\pi _s^* = \dfrac{\theta }{\sigma }Y_s^* \dfrac{\partial ^2 {\widetilde{V}}}{\partial y^2}(s, Y_s^*). \end{aligned}$$
Since \(\mathcal {X}_t = h(t,y) = x\), the uniqueness of the solution to SDE implies that
$$\begin{aligned} X_s^*= \mathcal {X}_s \;\;\text{ for }\;\;s\in [t,T]. \end{aligned}$$
It is not difficult to show that for \(s\in [t,T]\),
$$\begin{aligned} \int _s^T \theta Y_u^*\dfrac{\partial h}{\partial y}(u,Y_u^*)dB_u \end{aligned}$$
is a martingale. Thus, we can derive that for all \(s\in [t,T]\),
$$\begin{aligned} \begin{aligned} X_s^* = {\mathbb {E}}_s\left[ \int _s^T H_u^s c_u^* du\right] . \end{aligned} \end{aligned}$$
(6.9)
Especially,
$$\begin{aligned} \begin{aligned} x = {\mathbb {E}}_t\left[ \int _t^T H_s^t c_s^* ds\right] . \end{aligned} \end{aligned}$$
(6.10)
This implies that
$$\begin{aligned} \begin{aligned} V(t,x)\ge&~{\mathbb {E}}_t\left[ \int _t^T e^{-\beta (s-t)}u(c_s^*)ds\right] \\ =&~{\mathbb {E}}_t\left[ \int _t^T e^{-\beta (s-t)}u(c_s^*)ds\right] +y^*\left( x-{\mathbb {E}}_t\left[ \int _t^T H_s^t c_s^* ds\right] \right) \\ =&~{\mathbb {E}}_t\left[ \int _t^T e^{-\beta (s-t)}{\tilde{u}}(s,Y_s^*)ds\right] +y^* x\\ =&~{\widetilde{V}}(t,y^*) + y^* x \\ \ge&~\inf _{y >0} [{\widetilde{V}}(t,y)+yx]. \end{aligned} \end{aligned}$$
(6.11)
From (3.3), we conclude that
$$\begin{aligned} V(t,x) = {\widetilde{V}}(t,y^*) + y^* x. \end{aligned}$$
(6.12)
Thus, the strategies \(X_s^*\), \(c_s^*\) and \(\pi _s^*\) are optimal.
C Proof of Theorem 2
Proof
By Theorem 2 and Lemma 1 in Appendix A, the value function V(t, x) is given by
$$\begin{aligned}\begin{aligned} V(t,x)&=\dfrac{(y^{*})^{-\frac{1-\gamma }{\gamma }}}{1-\gamma }\int _{0}^{\tau }e^{-K\xi }{\mathcal {N}}\left( -d^{-}_{\delta }(\xi ,\frac{y^{*}}{R(t+\xi )^{-\gamma }}) \right) d\xi \\&\;\;+ \int _{0}^{\tau }\dfrac{{R}(t+\xi )^{1-\gamma }}{1-\gamma }\cdot e^{-\beta \xi }{\mathcal {N}}\left( d^{-}(\xi ,\frac{y^{*}}{R(t+\xi )^{-\gamma }})\right) d\xi , \end{aligned}\end{aligned}$$
where \(y^{*}\) is the solution to the following algebraic equation:
$$\begin{aligned} \begin{aligned} x=&(y^{*})^{-\frac{1}{\gamma }}\int _{0}^{\tau }e^{-K\xi }{\mathcal {N}}\left( -d^{-}_{\delta }(\xi ,\frac{y^{*}}{R(t+\xi )^{-\gamma }})\right) d\xi \\&+ \int _{0}^{\tau }R(t+\xi )e^{-r\xi }{\mathcal {N}}\left( d^{+}(\xi ,\frac{y^{*}}{R(t+\xi )^{-\gamma }})\right) d\xi . \end{aligned} \end{aligned}$$
(6.13)
We can rewrite (6.13) as
$$\begin{aligned} \begin{aligned}&x-\int _{0}^{\tau }R(t+\xi )e^{-r\xi }d\xi \\&\quad =(y^{*})^{-\frac{1}{\gamma }}\int _{0}^{\tau }e^{-K\xi }{\mathcal {N}}\left( -d^{-}_{\delta }(\xi ,\frac{y^{*}}{R(t+\xi )^{-\gamma }})\right) d\xi \\&\qquad -\int _{0}^{\tau }R(t+\xi )e^{-r\xi }{\mathcal {N}}\left( -d^{+}(\xi ,\frac{y^{*}}{R(t+\xi )^{-\gamma }})\right) d\xi . \end{aligned} \end{aligned}$$
(6.14)
Here
$$\begin{aligned} \int _{0}^{\tau }R(t+\xi )e^{-r\xi }d\xi \end{aligned}$$
(6.15)
can be considered as the annuity value of consumption if the agent keeps the rate of consumption larger than R(t) for t.
Similar to Theorem 1, the optimal consumption and portfolio strategies \((c^{*},\pi ^{*})\) are given by
$$\begin{aligned}\begin{aligned} c^{*}_{t}=I\left( y^{{*}}- (y^{{*}}- R(t)^{-\gamma })^{+}\right) \end{aligned}\end{aligned}$$
and
$$\begin{aligned}\begin{aligned} \pi ^{*}_{t} =&\frac{\theta }{\sigma }\left[ \dfrac{1}{\gamma }(y^{*})^{-\frac{1}{\gamma }} \int _{0}^{\tau }e^{-K\xi }{\mathcal {N}}\left( -d^{-}_{\delta }(\xi ,\dfrac{y^{*}}{R(t+\xi )^{-\gamma }})\right) d\xi \right. ]\\&+\frac{1}{\sqrt{2\pi }} (y^{*})^{-\frac{1}{\gamma }}\int _{0}^{\tau }\exp {\left\{ -K\xi -\dfrac{1}{2}\left( d^{-}_{\delta }(\xi ,\dfrac{y^{*}}{R(t+\xi )^{-\gamma }})\right) ^2 \right\} }\dfrac{1}{\theta \sqrt{\xi }}d\xi \\&\left. -\frac{1}{\sqrt{2\pi }}\int _{0}^{\tau }R(t+\xi )\exp {\left\{ -r\xi -\dfrac{1}{2}\left( d^{+}(\xi ,\dfrac{y^{*}}{R(t+\xi )^{-\gamma }})\right) ^2 \right\} }\dfrac{1}{\theta \sqrt{\xi }}d\xi \right] \\ =&\frac{\theta }{\sigma \gamma }(y^{*})^{-\frac{1}{\gamma }} \int _{0}^{\tau }e^{-K\xi }{\mathcal {N}}\left( -d^{-}_{\delta }(\xi ,\dfrac{y^{*}}{R(t+\xi )^{-\gamma }})\right) d\xi . \end{aligned} \end{aligned}$$
\(\square\)
D Proof of Theorem 3
Proof
Since \(u(I(y))=-\dfrac{y}{\gamma }\), by Theorem 2 and Lemma 1 in A , the value function V(t, x) is given by
$$\begin{aligned}\begin{aligned} V(t,x)=&-\dfrac{y^{*}}{\gamma }\int _{0}^{\tau }e^{-r\xi }{\mathcal {N}}\left( -d^{+}(\xi ,\dfrac{y^{*}}{{\widetilde{y}}(t+\xi )})\right) d\xi \\&- \int _{0}^{\tau }\dfrac{e^{-\gamma R(t+\xi )}}{\gamma }e^{-\beta \xi }{\mathcal {N}}\left( d^{-}(\xi ,\dfrac{y^{*}}{{\widetilde{y}}(t+\xi )})\right) d\xi \end{aligned} \end{aligned}$$
where \(y^{*}\) is the solution to the following algebraic equation:
$$\begin{aligned} \begin{aligned} x=&\dfrac{1}{\gamma }\int _{0}^{\tau }\theta \sqrt{\xi }e^{-r\xi }\mathbf{n}\left( -d^{+}(\xi ,\dfrac{y^{*}}{{\widetilde{y}}(t+\xi )})\right) d\xi \\&+\int _{0}^{\tau }R(t+\xi )e^{-r\xi }{\mathcal {N}}\left( d^{+}(\xi ,\dfrac{y^{*}}{{\widetilde{y}}(t+\xi )})\right) d\xi \\&-\dfrac{1}{\gamma }\int _{0}^{\tau }\left( \log {y^{*}}+\left( \dfrac{1+k_2}{2}\right) \theta ^2\xi \right) e^{-r\xi }{\mathcal {N}}\left( -d^{+}(\xi ,\dfrac{y^{*}}{{\widetilde{y}}(t+\xi )})\right) d\xi . \end{aligned} \end{aligned}$$
(6.16)
Since \(\pi ^{*}_t=-\dfrac{\theta }{\sigma }y^*\dfrac{\partial x }{\partial y}(t,y^*)\), the optimal portfolio is given by
$$\begin{aligned}\begin{aligned} \pi ^{*}_t = \dfrac{\theta }{\sigma \gamma }\int _{0}^{\tau }e^{-r\xi }{\mathcal {N}}\left( -d^{+}(\xi ,\dfrac{y^*}{{\widetilde{y}}(t+\xi )})\right) d\xi . \end{aligned} \end{aligned}$$
\(\square\)
