Portfolio Optimization for Assets with Stochastic Yields and Stochastic Volatility

Abstract

In this paper, we consider a stochastic portfolio optimization model for investment on a risky asset with stochastic yields and stochastic volatility. The problem is formulated as a stochastic control problem, and the goal is to choose the optimal investment and consumption controls to maximize the investor’s expected total discounted utility. The Hamilton–Jacobi–Bellman equation is derived by virtue of the dynamic programming principle, which is a second-order nonlinear equation. Using the subsolution–supersolution method, we establish the existence result of the classical solution of the equation. Finally, we verify that the solution is equal to the value function and derive and verify the optimal investment and consumption controls.

Appendices

Appendices

Appendix A: Proof of Lemma 2.1

Proof

First consider $\rho \in [0,1]$. It is easy to get that $q(y,z)> \sigma _2^2(z) \ge \tilde{\sigma }_2^2>q_0> 0$. Now consider $\rho _0 \le \rho <0$. The minimum of q with respect to y is given by the function $\sigma _2^2(z)(1-\rho ^2)$. So we have

$$\begin{aligned} q(y,z) \ge \sigma _2^2(z)(1-\rho ^2) \ge \tilde{\sigma }_2^2(1-\rho _0^2) =q_0> 0. \end{aligned}$$

So (24) holds for all $\rho \in [\rho _0,1].$$\square$

Appendix B: Proof of Lemma 2.2

Proof

To prove this, we consider two cases. First, let $0 \le \rho \le 1$. Then, by virtue of (26) and (15), we can get

$$\begin{aligned} \varPsi (y,z)= & {} \displaystyle \frac{(be^{-y} + \mu - r)^2}{2(1-\gamma )(\sigma _1^2e^{-2y} + 2\rho \sigma _1\sigma _2(z) e^{-y} + \sigma _2^2(z))}\\\le & {} \displaystyle \frac{(be^{-y}+\mu - r)^2}{2(1-\gamma )(\sigma _1^2e^{-2y} + \tilde{\sigma }_2^2)} \le \displaystyle \frac{2b^2e^{-2y} + 2(\mu - r)^2}{2(1-\gamma )(\sigma _1^2e^{-2y} + \tilde{\sigma }_2^2)} \\= & {} \frac{b^2}{\sigma _1^2(1-\gamma )} + \frac{(\mu - r)^2 - \frac{b^2\tilde{\sigma }_2^2}{\sigma _1^2}}{(1-\gamma )(\sigma _1^2e^{-2y} + \tilde{\sigma }_2^2)}. \end{aligned}$$

If $(\mu - r)^2 - \frac{b^2\tilde{\sigma }_2^2}{\sigma _1^2} \le 0$, then, we have $\varPsi (y,z) \le \frac{b^2}{\sigma _1^2(1-\gamma )}$. Otherwise, we have

$$\begin{aligned} \varPsi (y,z) \le \frac{b^2}{\sigma _1^2(1-\gamma )} + \frac{(\mu - r)^2 - \frac{b^2\tilde{\sigma }_2^2}{\sigma _1^2}}{(1-\gamma )\tilde{\sigma }_2^2}= \frac{(\mu - r)^2}{\tilde{\sigma }_2^2(1-\gamma )}. \end{aligned}$$

Thus, if $0\le \rho \le 1$, we have

$$\begin{aligned} \varPsi (y,z) \le \max \left\{ \frac{b^2}{\sigma _1^2(1-\gamma )}, \frac{(\mu - r)^2}{\tilde{\sigma }_2^2(1-\gamma )} \right\} \equiv {\tilde{\varPsi }}. \end{aligned}$$

We can get a similar bound if $\rho$ is negative. If $\rho _0 \le \rho < 0,$ then $\rho (\sigma _1 e^{-y} - \sigma _2(z))^2 \le 0$ implies

$$\begin{aligned} 2\rho \sigma _1\sigma _2(z) e^{-y} \ge \rho \sigma _1^2 e^{-2y} + \rho \sigma _2^2(z), \end{aligned}$$

and so we can get

$$\begin{aligned} \varPsi (y,z)= & {} \displaystyle \frac{(be^{-y} + \mu - r)^2}{2(1-\gamma )(\sigma _1^2e^{-2y} + 2\rho \sigma _1\sigma _2(z) e^{-y} + {\tilde{\sigma }}_1^2(z))}\\\le & {} \displaystyle \frac{(be^{-y}+\mu - r)^2}{2(1-\gamma )(\sigma _1^2e^{-2y} +\rho \sigma _1^2e^{-2y} + \rho \sigma _2^2(z) + \sigma _2^2(z))}\\= & {} \frac{(be^{-y}+\mu - r)^2}{2(1-\gamma )(1+\rho )(\sigma _1^2e^{-2y} + \sigma _2^2(z))}\\\le & {} \frac{2b^2e^{-2y} + 2(\mu - r)^2}{2(1-\gamma )(1+\rho )(\sigma _1^2e^{-2y} + \tilde{\sigma }_2^2)}\\\le & {} \frac{1}{1+\rho }\max \left\{ \frac{b^2}{\sigma _1^2(1-\gamma )}, \frac{(\mu - r)^2}{\tilde{\sigma }_2^2(1-\gamma )} \right\} \le \frac{1}{1+\rho _0} {\tilde{\varPsi }}. \end{aligned}$$

Let

$$\begin{aligned} {\bar{\varPsi }} \equiv \max \left\{ {\tilde{\varPsi }}, \displaystyle \frac{{\tilde{\varPsi }}}{1+\rho _0}\right\} . \end{aligned}$$

(88)

Then, we can get that $0 \le \varPsi (y,z) \le {\bar{\varPsi }} < \infty$. Thus, $\varPsi (y,z)$ is bounded. $\square$

Appendix C: Proof of Lemma 3.2

Proof

Recall from (25) that for $k^*> 0$, G is defined by

$$\begin{aligned} G(y,z,p) = \displaystyle \frac{[be^{-y} + \mu - r + (\sigma _2^2(z) +\rho \sigma _1\sigma _2(z)e^{-y}) p ]^2}{2(1-\gamma )q(y,z)} \ge 0, \end{aligned}$$

(89)

or in the trivial case when $k^*=0$, $G\equiv 0$ otherwise. Consider the expression for G given above. We can expand G to a quadratic form in p:

$$\begin{aligned} G(y,z,p) = g_2(y,z) p^2 + g_1(y,z) p + g_0(y,z), \end{aligned}$$

(90)

where

$$\begin{aligned} g_2(y,z)= & {} \displaystyle \frac{(\sigma _2^2(z) + \rho \sigma _1\sigma _2(z)e^{-y})^2}{2(1-\gamma )q(y,z)}, \quad \nonumber \\ g_1(y,z)= & {} \displaystyle \frac{(be^{-y}+\mu - r)( \sigma _2^2(z) + \rho \sigma _1\sigma _2(z)e^{-y})}{(1-\gamma )q(y,z)}, \quad \text {and} \nonumber \\ g_0(y,z)= & {} \displaystyle \frac{(be^{-y} + \mu - r)^2}{2(1-\gamma )q(y,z)}. \end{aligned}$$

(91)

By the definition of q(y, z) [see (15)], it is not hard to show that $|g_2|, |g_1|,$ and $|g_0|$ are bounded for all $(y,z) \in \mathbb {R}^2$. Then, we can get (36) very easily. $\square$

Appendix D: Proof of Lemma 3.3

Proof

Suppose $\displaystyle \sup _{(y,z) \in \bar{B}_R} (\tilde{Q} - \hat{Q})(y,z) > 0$. Since $\tilde{Q} \le \hat{Q}$ on $\partial B_R,$ this implies that $\tilde{Q} - \hat{Q}$ reaches its maximum at $(y_0, z_0) \in B_R$. So we have that

$$\begin{aligned}&(\tilde{Q} - \hat{Q})_{yy}(y_0,z_0)< 0, \quad (\tilde{Q} - \hat{Q})_{zz}(y_0,z_0) < 0, \end{aligned}$$

(92)

$$\begin{aligned}&\tilde{Q}_y(y_0,z_0) = \hat{Q}_y(y_0,z_0), \quad \text {and} \quad \tilde{Q}_z(y_0,z_0) = \hat{Q}_z(y_0,z_0). \end{aligned}$$

(93)

By (37) we have that, for $(y,z) \in B_R,$

$$\begin{aligned}&\displaystyle \frac{\sigma _2^2(z)}{2}(\tilde{Q}_{yy} - \hat{Q}_{yy}) + \displaystyle \frac{\sigma _3^2}{2}(\tilde{Q}_{zz} - \hat{Q}_{zz}) \\&\quad - H(y,z,\tilde{Q}, \tilde{Q}_y, \tilde{Q}_z, \tau ) + H(y,z,\hat{Q}, \hat{Q}_y, \hat{Q}_z, \tau ) \ge 0. \end{aligned}$$

Evaluating this inequality at $(y_0,z_0)$ and applying (92), (93) and (32), we can get

$$\begin{aligned} -(1-\tau \gamma )e^{\frac{\hat{Q}(y_0,z_0)}{\tau \gamma - 1}} > -(1-\tau \gamma )e^{\frac{{{\underline{Q}}}(y_0,z_0)}{\tau \gamma - 1}}, \end{aligned}$$

or equivalently $\hat{Q}(y_0,z_0) > \tilde{Q}(y_0,z_0),$ which is a contradiction. Therefore, $\tilde{Q} \le \hat{Q}$ on ${\bar{B}}_R$. $\square$

Appendix E: Proof of Corollary 3.1

Proof

Suppose $Q^{(1)}, Q^{(2)} \in C^2({\bar{B}}_R)$ are both solutions of (34). Then, $Q^{(1)} = Q^{(2)}$ on $\partial B_R$. Further, for $(y,z)\in B_R$, we have

$$\begin{aligned}&\displaystyle \frac{\sigma _2^2(z)}{2}(Q^{(2)}_{yy} - Q^{(1)}_{yy}) + \displaystyle \frac{\sigma _3^2}{2}(Q^{(2)}_{zz} - Q^{(1)}_{zz}) \nonumber \\&\quad - H(y,z, Q^{(2)}, Q^{(2)}_y, Q^{(2)}_z, \tau ) + H(y,z, Q^{(1)}, Q^{(1)}_y, Q^{(1)}_z, \tau ) = 0. \end{aligned}$$

(94)

Now we assume that $\displaystyle \sup _{(y,z)\in B_R} (Q^{(2)}-Q^{(1)}) > 0$. Then, $Q^{(2)} - Q^{(1)}$ attains its maximum at some $(y_0,z_0) \in B_R$. Then,

$$\begin{aligned}&(Q^{(2)}_{yy} - Q^{(1)}_{yy})(y_0,z_0)< 0, \quad (Q^{(2)}_{zz} - Q^{(1)}_{zz})(y_0,z_0) < 0, \end{aligned}$$

(95)

$$\begin{aligned}&Q^{(1)}_y(y_0,z_0) = Q^{(2)}_y(y_0,z_0), \quad \text {and} \quad Q^{(1)}_z(y_0,z_0) = Q^{(2)}_z(y_0,z_0). \end{aligned}$$

(96)

Evaluating (94) at $(y_0,z_0)$ and applying (96) and (95), we arrive at

$$\begin{aligned} (1-\tau \gamma )e^{\frac{Q^{(2)}(y_0,z_0)}{\tau \gamma - 1}} > (1-\tau \gamma )e^{\frac{Q^{(1)}(y_0,z_0)}{\tau \gamma - 1}}, \end{aligned}$$

or $Q^{(2)}(y_0,z_0) < Q^{(1)}(y_0,z_0),$ which is a contradiction. Therefore, we must have $Q^{(2)} \le Q^{(1)}$ in $B_R$. Similarly, we can show that $Q^{(2)} \ge Q^{(1)}$ in $B_R$ by assuming that

$$\begin{aligned} \displaystyle \sup _{(y,z)\in B_R} (Q^{(1)}-Q^{(2)}) > 0. \end{aligned}$$

Therefore, $Q^{(1)} \equiv Q^{(2)}$ on ${\bar{B}}_R$. $\square$

Appendix F: Proof of Lemma 3.4

Proof

By the Cauchy–Schwarz Inequality, we have

$$\begin{aligned}&\displaystyle \sum _{i,j,k} D_k a^{ij} D_k Q D_{ij} Q \\&\quad = \sum _{i,j} \left( \sum _k D_k a^{ij} D_k Q \right) D_{ij} Q \le \sum _{i,j} |Da^{ij}| |DQ| \cdot D_{ij} Q\\&\quad = \sum _{i,j}\frac{1}{\sqrt{\epsilon }} |Da^{ij}| |DQ| \cdot \sqrt{\epsilon } D_{ij} Q \le \sum _{i,j} \left( \frac{1}{2\epsilon } |Da^{ij}|^2 |DQ|^2 + \frac{\epsilon }{2} (D_{ij}Q)^2 \right) \\&\quad = \frac{1}{2\epsilon } \left( \sum _{i,j} |Da^{ij}|^2\right) |DQ|^2 + \frac{\epsilon }{2} |D^2 Q|^2. \end{aligned}$$

This completes the proof. $\square$

Appendix G: Proof of Theorem 3.3

Proof

We first prove inequality (41). Let

$$\begin{aligned} \tilde{Q}^\tau = - \log \left( \max \left\{ \frac{\beta }{1-\gamma }, 1 \right\} \right) - \sup _{B_R} \{|\psi (y,z)|\}. \end{aligned}$$

(97)

Then, it is easy to show that $\tilde{Q}^\tau$ is a subsolution of (34). We also have that $\tilde{Q}^\tau \le \tau \psi = Q^\tau$ on $\partial B_R$. Therefore, $\tilde{Q}^\tau$ and $Q^\tau$ satisfy

$$\begin{aligned} \begin{array}{ll} \frac{\sigma _2^2(z)}{2} \tilde{Q}^\tau _{yy} + \frac{\sigma _3^2}{2} \tilde{Q}^\tau _{zz} - H(y,z, \tilde{Q}^\tau , \tilde{Q}^\tau _y, \tilde{Q}^\tau _z, \tau ) \ge 0, \quad &{}{\mathrm{on}} \quad B_R,\\ \frac{\sigma _2^2(z)}{2} Q^\tau _{yy} + \frac{\sigma _3^2}{2} Q^\tau _{zz} - H(y,z, Q^\tau , Q^\tau _y, Q^\tau _z, \tau ) = 0, \quad &{}{\mathrm{on}} \quad B_R,\\ \tilde{Q}^\tau \le Q^\tau . \quad &{}{\mathrm{on}} \quad \partial B_R. \end{array} \end{aligned}$$

(98)

Then, by Lemma 3.3, $\tilde{Q}^\tau \le Q^\tau$ holds in ${\bar{B}}_R,$ which gives us (41). Next we prove (38). It is equivalent to show that

$$\begin{aligned} \frac{1}{\tau \gamma }(e^{Q^\tau } - f) \le \frac{1}{\gamma }(e^{\hat{Q}} - f). \end{aligned}$$

(99)

Define $V(x,y,z) \equiv \frac{1}{\gamma }x^\gamma e^{\hat{Q}(y,z)} \quad \text {and} \quad V^\tau (x,y,z) \equiv \frac{1}{\tau \gamma }x^{\tau \gamma } e^{Q^\tau (y,z)}$. Then, we have

$$\begin{aligned}&\max _{c\ge 0}\left[ \frac{1}{\gamma }(cx)^{\gamma } - cxV_x \right] = (1-\gamma )e^\frac{\hat{Q}}{\gamma - 1} V, \text{ and } \quad \\&\quad \max _{c\ge 0}\left[ \frac{1}{\tau \gamma }(cx)^{\tau \gamma } - cxV^\tau _x \right] = (1-\tau \gamma )e^\frac{Q^\tau }{\tau \gamma - 1}V^\tau . \end{aligned}$$

Now, we define

$$\begin{aligned} V_0(x,y,z)\equiv & {} \frac{1}{\gamma }(x^\gamma e^{\hat{Q}(y,z)} - f(y,z)),\quad \text{ and }\quad \\ V_0^\tau (x,y,z)\equiv & {} \frac{1}{\tau \gamma }(x^{\tau \gamma } e^{Q^\tau (y,z)} - f(y,z)). \end{aligned}$$

Since f does not depend on x,  we have that $(V_0)_x = V_x$ and $(V_0^\tau )_x = V^\tau _x$. Thus, we can get

$$\begin{aligned}&\max _{c \ge 0} \left[ \frac{1}{\gamma }(cx)^\gamma - cx(V_0)_x \right] = (1-\gamma )e^\frac{\hat{Q}}{\gamma - 1} V, \\&\quad \text{ and }\quad \max _{c \ge 0} \left[ \frac{1}{\tau \gamma }(cx)^{\tau \gamma } - cx(V_0)_x \right] = (1-\tau \gamma )e^\frac{Q^\tau }{\tau \gamma - 1} V^\tau . \end{aligned}$$

Define the operator $\mathcal {L}^{k, c}$ as

$$\begin{aligned} \mathcal {L}^{k,c} V(x,y,z)\equiv & {} \frac{k^2x^2}{2}q(y,z)V_{xx} +\frac{\sigma _2^2(z)}{2} V_{yy} + \frac{\sigma _3^2}{2} V_{zz}\\&+\, kx(\sigma _2^2(z) + \rho \sigma _1\sigma _2(z)e^{-y})V_{xy} \\&+\, (be^{-y} + \mu - r)kxV_x + (r-c)xV_x + {\tilde{\mu }}(z)V_y + a(\bar{z} - z)V_z. \end{aligned}$$

Noting that f satisfies (39) and $\hat{Q}$ is a supersolution of (34) for $\tau =1$, we can get that, for $(y, z) \in B_R$,

$$\begin{aligned}&-\beta V_0 + \max _{k, c\ge 0} \left[ \mathcal {L}^{k, c} V_0 +\frac{1}{\gamma } (cx)^\gamma \right] -\frac{1}{\gamma }\nonumber \\&\quad = V\left[ \frac{\sigma _2^2(z)}{2} \hat{Q}_{yy} + \frac{\sigma _3^2}{2} \hat{Q}_{zz} -H(y,z,\hat{Q}, \hat{Q}_y, \hat{Q}_z, 1)\right] \nonumber \\&\qquad -\, \frac{1}{\gamma }\left[ \frac{\sigma _2^2(z)}{2} f_{yy} + \frac{\sigma _3^2}{2} f_{zz} + rxf_x + {\tilde{\mu }}(z)f_y + a(\bar{z} - z)f_z - \beta f + 1 \right] \nonumber \\&\quad = V\left[ \frac{\sigma _2^2(z)}{2} \hat{Q}_{yy} + \frac{\sigma _3^2}{2} \hat{Q}_{zz} -H(y,z,\hat{Q}, \hat{Q}_y, \hat{Q}_z, 1)\right] - \frac{1}{\gamma }[0] \nonumber \\&\quad \le 0. \end{aligned}$$

(100)

Similarly, for $V_0^\tau$, we have

$$\begin{aligned}&-\beta V_0^\tau + \max _{k, c\ge 0} \left[ \mathcal {L}^{k, c} V_0^\tau +\frac{1}{\tau \gamma } (cx)^{\tau \gamma }\right] -\frac{1}{\tau \gamma }\nonumber \\&\quad =V^\tau \left[ \frac{\sigma _2^2(z)}{2} Q_{yy}^\tau + \frac{\sigma _3^2}{2} Q_{zz}^\tau - H(y,z,Q^\tau , Q^\tau _y, Q^\tau _z, \tau )\right] \nonumber \\&\qquad -\, \frac{1}{\tau \gamma }\left[ \frac{\sigma _2^2(z)}{2} f_{yy} + \frac{\sigma _3^2}{2} f_{zz} + rxf_x + {\tilde{\mu }}(z)f_y + a(\bar{z} - z)f_z - \beta f + 1 \right] \nonumber \\&\quad = 0. \end{aligned}$$

(101)

Suppose that the optimal controls for the above equation are given by ${\tilde{k}}$ and ${\tilde{c}}$. Then, we can rewrite (101) as

$$\begin{aligned} -\beta V_0^\tau + \left[ \mathcal {L}^{\tilde{k}, \tilde{c}} V_0^\tau +\frac{1}{\tau \gamma } (cx)^{\tau \gamma }\right] -\frac{1}{\tau \gamma } =0. \end{aligned}$$

(102)

At the same time, from (100) we obtain

$$\begin{aligned} -\beta V_0 + \left[ \mathcal {L}^{\tilde{k}, \tilde{c}} V_0 +\frac{1}{\gamma } (cx)^{\gamma }\right] -\frac{1}{\gamma } \le 0. \end{aligned}$$

(103)

Define $g(\tau ;\theta ) = \frac{1}{\tau \gamma }(\theta ^\tau - 1), \quad \forall 0<\tau \le 1.$ Then, the difference between Eqs. (102) and (103) is the function $g (\tau ; \theta )$, with $\theta = (cx)^\gamma > 0$ for $\tau =1$ in (102) and $\tau <1$ in (103). It is not hard to verify that $g'(\tau )>0$, so $g(\tau )$ is a nonincreasing function with respect to $\tau$. Therefore,

$$\begin{aligned} g(\tau ) \le g(1), \end{aligned}$$

(104)

or equivalently,

$$\begin{aligned} \frac{1}{\tau \gamma }((cx)^{\tau \gamma } - 1) \le \frac{1}{\gamma }((cx)^{\gamma } - 1). \end{aligned}$$

(105)

Therefore, from (103) we can get

$$\begin{aligned} -\beta V_0 + \left[ \mathcal {L}^{\tilde{k}, \tilde{c}} V_0 +\frac{1}{\tau \gamma } (cx)^{\tau \gamma }\right] -\frac{1}{\tau \gamma } \le 0. \end{aligned}$$

Subtracting the above inequality from (102), we have

$$\begin{aligned} \mathcal {L}^{\tilde{k}, \tilde{c}} (V_0^\tau -V_0) \ge 0. \end{aligned}$$

(106)

Note that (106) holds for $x > 0, (y,z) \in \bar{B}_R$. This equation is used later in this proof to show a contradiction.

To prove the estimate in (99), we wish to show that for $x >0, (y,z) \in {\bar{B}}_R,$

$$\begin{aligned}&V_0^\tau (x,y,z) \le V_0(x,y,z), \quad \text{ or } \text{ equivalently, } \quad \frac{1}{\tau \gamma }(x^{\tau \gamma }e^{Q^\tau } - f) \nonumber \\&\quad \le \frac{1}{\gamma }(x^{\gamma }e^{\hat{Q}} - f). \end{aligned}$$

(107)

We then take $x = 1$ to get the desired result.

From the definition of f [see (40)], we can get

$$\begin{aligned} f(y,z) = \frac{1}{\beta } + \left( 1 - \frac{1}{\beta }\right) \mathbf {E}_{y,z}[ e^{-\beta t_R}] =\frac{1}{\beta }\Big (1 - \mathbf {E}_{y,z}[ e^{-\beta t_R}]\Big ) + \mathbf {E}_{y,z}[ e^{-\beta t_R}]. \end{aligned}$$

Hence $f > 0$. Therefore,

$$\begin{aligned} - \frac{1}{\tau \gamma } f < -\frac{1}{\gamma }f. \end{aligned}$$

(108)

On the boundary $(y,z) \in \partial B_R$, we have that $\hat{Q} \ge \psi$. Using (104) with $\theta = x^\gamma e^{\psi }$, we can get that

$$\begin{aligned} V_0^\tau= & {} \frac{1}{\tau \gamma }(x^{\tau \gamma }e^{\tau \psi } - 1) \le \frac{1}{\gamma }(x^\gamma e^{\psi } - 1) \nonumber \\\le & {} \frac{1}{\gamma }(x^\gamma e^{\hat{Q}} - 1) = V_0, \quad \forall x > 0, (y,z) \in \partial B_R. \end{aligned}$$

(109)

Next we prove that $V_0^\tau \le V_0$ for $(y,z) \in {\bar{B}}_R$. Suppose on the contrary that

$$\begin{aligned} \sup _{x>0, |(y,z)| \le R} \{ V_0^\tau (x,y,z) - V_0(x,y,z) \} > 0. \end{aligned}$$

(110)

Then, the maximum is attained at some $(x_0,y_0,z_0)$, where $x_0 > 0$ and $|(y_0,z_0)| < R$. So at $(x_0,y_0,z_0)$ we have the following:

$$\begin{aligned}&V_0(x_0,y_0,z_0) < V_0^\tau (x_0,y_0,z_0), \end{aligned}$$

(111)

$$\begin{aligned}&(V_0^\tau - V_0)_x(x_0,y_0,z_0) = (V_0^\tau - V_0)_y(x_0,y_0,z_0) = (V_0^\tau - V_0)_z(x_0,y_0,z_0) = 0,\nonumber \\ \end{aligned}$$

(112)

and $D^2(V_0^\tau - V_0)$ is negative semi-definite at $(x_0,y_0,z_0)$, where $D^2$ is the $3 \times 3$ matrix operator of second derivatives (the Hessian). That is, for any $\eta \in \mathbb {R}^3,$

$$\begin{aligned} \eta ^T D^2(V_0^\tau - V_0)(x_0,y_0,z_0) \eta \le 0, \end{aligned}$$

(113)

or in expanded form, at $(x_0, y_0, z_0)$ for any $\eta _1, \eta _2, \eta _3 \in \mathbb {R}$,

$$\begin{aligned}&(V_0^\tau - V_0)_{xx} \eta _1^2 + (V_0^\tau - V_0)_{yy}\eta _2^2 + (V_0^\tau - V_0)_{zz} \eta _3^2 \nonumber \\&\quad +\, 2(V_0^\tau - V_0)_{xy} \eta _1\eta _2 + 2(V_0^\tau - V_0)_{xz} \eta _1\eta _3 + 2(V_0^\tau - V_0)_{yz} \eta _2\eta _3 \le 0. \end{aligned}$$

(114)

Note that this implies

$$\begin{aligned} (V_0^\tau - V_0)_{xx}, \,\, (V_0^\tau - V_0)_{yy}, \,\, (V_0^\tau - V_0)_{zz} \le 0 \end{aligned}$$

(115)

at the point $(x_0,y_0,z_0)$. We also note that

$$\begin{aligned} q(y,z)= & {} \sigma _2^2(z) + 2\rho \sigma _1\sigma _2 e^{-y} + \sigma _1^2 e^{-2y} \nonumber \\\ge & {} \sigma _2^2(z) + 2\rho \sigma _1\sigma _2 e^{-y} + \rho ^2\sigma _1^2 e^{-2y} =(\sigma _2(z) + \rho \sigma _1 e^{-y})^2. \end{aligned}$$

(116)

Now we evaluate (106) at $(x_0,y_0,z_0)$ and apply conditions (111) and (112). Then, at the point $(x_0,y_0,z_0)$, we have

$$\begin{aligned}&\frac{\sigma _2^2(z)}{2} (V_0^\tau - V_0)_{yy} + \frac{\sigma _3^2}{2} (V_0^\tau - V_0)_{zz} + \frac{{\tilde{k}}^2x^2}{2}q(y,z)(V_0^\tau - V_0)_{xx} \nonumber \\&\quad +\, {\tilde{k}}x\sigma _2(z)(\sigma _2(z) + \rho \sigma _1 e^{-y})(V_0^\tau - V_0)_{xy} > 0. \end{aligned}$$

(117)

Applying (115)–(117), we have

$$\begin{aligned}&\frac{\sigma _2^2(z)}{2} (V_0^\tau - V_0)_{yy} + \frac{{\tilde{k}}^2x^2}{2}(\sigma _2(z) + \rho \sigma _1 e^{-y})^2(V_0^\tau - V_0)_{xx} \\&\quad +\, {\tilde{k}}x\sigma _2(z)(\sigma _2(z) + \rho \sigma _1 e^{-y})(V_0^\tau - V_0)_{xy} > 0. \end{aligned}$$

at $(x_0,y_0,z_0)$. Taking $\eta _1 = \frac{{\tilde{k}} x}{\sqrt{2}}(\sigma _2 + \rho \sigma _1 e^{-y}), \eta _2 = \frac{\sigma _2(z)}{\sqrt{2}}$, and $\eta _3 = 0$, we get a contradiction to (114). Therefore, $V_0^\tau \le V_0$ for $(y,z) \in {\bar{B}}_R$. Thus, (107) holds. We take $x=1$ in (107) to arrive at (38). $\square$

Appendix H: Proof of Theorem 3.4

Proof

Since $Q_\tau ^0$ is a solution of (42), by virtue of the definition of H, we have that

$$\begin{aligned}&\frac{\sigma _2^2(z)}{2}(Q^0_\tau )_{yy} + \frac{\sigma _3^2}{2}(Q^0_\tau )_{zz} + \tau \bigg [\frac{\sigma _2^2(z)}{2}(Q^0_\tau )^2_{y} + \frac{\sigma _3^2}{2}(Q^0_\tau )^2_{z} \\&\quad +\, {\tilde{\mu }}(z)(Q^0_\tau )_{y} + a(\bar{z} - z)(Q^0_\tau )_{z} - \beta + e^{-Q^0_\tau } \bigg ] = 0. \end{aligned}$$

Then, we can get

$$\begin{aligned} \frac{\sigma _2^2(z)}{2}(Q^0_\tau )_{yy} + \frac{\sigma _3^2}{2}(Q^0_\tau )_{zz} + \tau {\tilde{\mu }}(z)(Q^0_\tau )_{y} + \tau a(\bar{z} - z)(Q^0_\tau )_{z} - \tau \beta \le 0. \end{aligned}$$

(118)

Applying Ito’s rule to $Q_\tau ^0(\hat{Y}_t,\hat{Z}_t)$ and using (118), we have that, for $0 \le t \le {\bar{t}}_R,$

$$\begin{aligned} dQ_\tau ^0(\hat{Y}_t,\hat{Z}_t)= & {} (Q_\tau ^0)_{y}\Big [\tau {\tilde{\mu }}(\hat{Z}_t)\mathrm {d}t + \sigma _2(\hat{Z}_t)\mathrm {d}B_{2,t}\Big ] \\&+ \,(Q_\tau ^0)_{z} \Big [\tau a(\bar{z} - \hat{Z}_t)\mathrm {d}t + \sigma _3 \mathrm {d}B_{3,t}\Big ] \\&+ \,\frac{1}{2}(Q_\tau ^0)_{yy} \sigma _2^2(\hat{Z}_t)\mathrm {d}t + \frac{1}{2}(Q_\tau ^0)_{zz} \sigma _3^2 \mathrm {d}t \\= & {} \left[ \frac{\sigma _2^2(\hat{Z}_t)}{2}(Q^0_\tau )_{yy} + \frac{\sigma _3^2}{2}(Q^0_\tau )_{zz}\right. \\&\left. + \,\tau {\tilde{\mu }}(\hat{Z}_t)(Q^0_\tau )_{y} + \tau a(\bar{z} - \hat{Z}_t)(Q^0_\tau )_{z} \right] \mathrm {d}t \\&+ \,\sigma _2(\hat{Z}_t) (Q_\tau ^0)_y \mathrm {d} B_{2,t} + \sigma _3(Q_\tau ^0)_z \mathrm {d}B_{3,t} \\\le & {} \, \tau \beta \mathrm {d}t + \sigma _2(\hat{Z}_t) (Q_\tau ^0)_y \mathrm {d} B_{2,t} + \sigma _3(Q_\tau ^0)_z \mathrm {d}B_{3,t}. \end{aligned}$$

Integrate it from 0 to $\bar{t}_R$, and we can get

$$\begin{aligned} Q_\tau ^0(\hat{Y}_{{\bar{t}}_R},\hat{Z}_{{\bar{t}}_R}) - Q_\tau ^0(y,z)\le & {} \int _0^{{\bar{t}}_R} \tau \beta \mathrm{d}t + \int _0^{{\bar{t}}_R} \sigma _2(\hat{Z}_t) (Q_\tau ^0)_y \mathrm{d} B_{2,t} \\&+ \int _0^{{\bar{t}}_R} \sigma _3(Q_\tau ^0)_z \mathrm{d}B_{3,t}. \end{aligned}$$

Since $Q^0_\tau = 0$ on $\partial B_R$, we can take expectations to the above inequality and get

$$\begin{aligned} -Q_\tau ^0(y,z) \le \tau \beta \mathbf {E}[{\bar{t}}_R] \le \beta \mathbf {E}[{\bar{t}}_R], \end{aligned}$$

which proves the first inequality in (43). Define $\phi (y,z) = e^{Q_\tau ^0(y,z)}$. Then,

$$\begin{aligned} \phi _y= & {} (Q_\tau ^0)_y \phi , \quad \phi _{z} = (Q_\tau ^0)_z \phi ,\quad \phi _{yy} = ( (Q_\tau ^0)_{yy} + (Q_\tau ^0)_y^2) \phi , \quad \\ \phi _{zz}= & {} ((Q_\tau ^0)_{zz} + (Q_\tau ^0)_z^2)\phi . \end{aligned}$$

Then, it is easy to check that $\phi$ satisfies

$$\begin{aligned} \begin{array}{ll} \frac{\sigma _2^2(z)}{2}\phi _{yy} + \frac{\sigma _3^2}{2}\phi _{zz} - \frac{1-\tau }{2\phi } (\sigma _2^2(z) \phi ^2_y + \sigma _3^2 \phi ^2_z) &{}\\ \quad \quad + \, \tau \big ({\tilde{\mu }}(z)\phi _y + a(\bar{z} - z)\phi _z\big ) + \tau (1-\beta \phi ) = 0, \quad &{}\text {on} \quad B_R,\\ \phi = 1. &{}\text {on} \quad \partial B_R. \end{array} \end{aligned}$$

(119)

Since the terms $\displaystyle \frac{1-\tau }{2\phi } (\sigma _2^2(z) \phi ^2_y + \sigma _3^2 \phi ^2_z)$ and $\tau \beta \phi$ are nonnegative, we can get an inequality:

$$\begin{aligned} \frac{\sigma _2^2(z)}{2}\phi _{yy} + \frac{\sigma _3^2}{2}\phi _{zz} + \tau \big ({\tilde{\mu }}(z)\phi _y + a(\bar{z} - z)\phi _z\big ) + \tau \ge 0. \end{aligned}$$

Next, we can apply Ito’s rule to $\phi (\hat{Y}_t,\hat{Z}_t)$ and use the above inequality to obtain

$$\begin{aligned} d\phi (\hat{Y}_t,\hat{Z}_t)&= \phi _{y}\Big [\tau {\tilde{\mu }}(\hat{Z}_t)\mathrm {d}t + \sigma _2(\hat{Z}_t)\mathrm {d}B_{2,t}\Big ] + \phi _{z}\Big [\tau a(\bar{z} - \hat{Z}_t)\mathrm {d}t + \sigma _3 \mathrm {d}B_{3,t}\Big ] \\&\quad + \frac{1}{2}\phi _{yy} \sigma _2^2(\hat{Z}_t)\mathrm {d}t + \frac{1}{2}\phi _{zz} \sigma _3^2 \mathrm {d}t \\&= \left[ \frac{\sigma _2^2(\hat{Z}_t)}{2}\phi _{yy} + \frac{\sigma _3^2}{2}\phi _{zz} + \tau {\tilde{\mu }}(\hat{Z}_t)\phi _{y} + \tau a(\bar{z} - \hat{Z}_t) \phi _{z} \right] \mathrm {d}t \\&\quad + \sigma _2(\hat{Z}_t) \phi _y \mathrm {d} B_{2,t} + \sigma _3\phi _z \mathrm {d}B_{3,t} \\&\ge \, - \tau \mathrm {d}t + \sigma _2(\hat{Z}_t) \phi _y \mathrm {d} B_{2,t} + \sigma _3\phi _z \mathrm {d}B_{3,t}. \end{aligned}$$

The integral form is

$$\begin{aligned} \phi (\hat{Y}_{{\bar{t}}_R},\hat{Z}_{{\bar{t}}_R}) - \phi (y,z) \ge - \int _0^{{\bar{t}}_R} \tau \mathrm{d}t + \int _0^{{\bar{t}}_R} \sigma _2(\hat{Z}_t) \phi _y \mathrm{d} B_{2,t} + \int _0^{{\bar{t}}_R} \sigma _3\phi _z \mathrm{d}B_{3,t}.\nonumber \\ \end{aligned}$$

(120)

Using the boundary condition that $\phi =1$ on $\partial B_R$, we can get that $1 - \phi (y,z) \ge -\tau \mathbf {E}[{\bar{t}}_R]$, which implies

$$\begin{aligned} Q_\tau ^0 \le \log (1 + \tau \mathbf {E}[{\bar{t}}_R]) \le \log (1 + \mathbf {E}[{\bar{t}}_R]). \end{aligned}$$

This proves the second inequality in (43). $\square$

Appendix I: Proof of Lemma 4.1

Proof

By the definitions of $(k^*, c^*)$, Theorem 3.6, and the fact that ${\tilde{Q}}$ is bounded, it is easy to show that $k^*$ and $c^*$ are bounded. Further, by the definition of $k^*$ and q(y, z) [see (15)], we can get that $e^{-y} k^*$ is bounded. Therefore, we can assume that there is a constant $\varLambda$ such that

$$\begin{aligned} |(be^{-Y_t} + \mu - r) k^*_t + (r-c^*_t)|\le \varLambda , \quad |\sigma _1 k^*_t e^{-Y_t}| \le \varLambda , \quad |\sigma _2(Z_t) k^*_t|\le \varLambda .\nonumber \\ \end{aligned}$$

(121)

From Eq. (9), we can get that

$$\begin{aligned} X_T= & {} x\exp \bigg (\int _0^T \left[ (be^{-Y_t} + \mu - r) k^*_t + (r-c^*_t)\right. \\&\left. -\,\frac{1}{2}\left[ \sigma _1^2 (k^*_t e^{-Y_t})^2 + (\sigma _2(Z_t) k^*_t)^2\right] \right] \mathrm{d}t\\&+ \int _0^T \sigma _1 k^*_t e^{-Y_t} \mathrm{d} B_{1, t}+\int _0^T \sigma _2(Z_t) k^*_t \mathrm{d} B_{2,t}\bigg ). \end{aligned}$$

Using the above equation and (121), we can get (71). $\square$