Robust Optimal Investment Problem with Delay under Heston’s Model

Abstract

This paper considers a robust optimal portfolio problem under Heston model in which the risky asset price is related to the historical performance. The finance market includes a riskless asset and a risky asset whose price is controlled by a stochastic delay equation. The objective is to choose the investment strategy to maximize the minimal expected utility of terminal wealth. By employing dynamic programming principle and Hamilton-Jacobin-Bellman (HJB) equation, we obtain the specific expression of the optimal control and the explicit solution of the corresponding HJB equation. Besides, a verification theorem is provided to ensure the value function is indeed the solution of the HJB equation. Finally, we use numerical examples to illustrate the relationship between the optimal strategy and parameters.

Appendices

Solve the Problem 36

For solving the problem 36, we consider the following three cases: $\rho =1$, $\rho =-1$ and $\rho ^2\ne 1$.

Case (1): $\rho =1$. In this case, problem 36 has the following form:

$$\begin{aligned} {\left\{ \begin{array}{ll} \dfrac{d\phi _2(t)}{dt}-\phi _2(t)k+\dfrac{\lambda ^2}{2\left( \gamma +\zeta _1\right) }-\lambda \phi _2(t)\sigma =0,\\ \phi _2(T)=0. \end{array}\right. } \end{aligned}$$

(72)

By variation of constant approach, the solutions of problem 72 is

$$\begin{aligned} \phi _2(s)=\dfrac{\lambda ^2}{2(\gamma +\zeta _1)}\dfrac{1-e^{(k+\lambda \sigma )(s-T)}}{k+\lambda \sigma }. \end{aligned}$$

Case (2): $\rho =-1$. If $k\ne \lambda \sigma$, problem 36 reduces to

$$\begin{aligned} {\left\{ \begin{array}{ll} \dfrac{d\phi _2(t)}{dt}-\phi _2(t)k+\dfrac{\lambda ^2}{2\left( \gamma +\zeta _1\right) }+\lambda \phi _2(t)\sigma =0,\\ \phi _2(T)=0. \end{array}\right. } \end{aligned}$$

(73)

Then, we have

$$\begin{aligned} \phi _2(s)=\dfrac{\lambda ^2}{2(\gamma +\zeta _1)}\dfrac{1-e^{(k-\lambda \sigma )(s-T)}}{k-\lambda \sigma }. \end{aligned}$$

If $k=\lambda \sigma$, problem 36 is simplified as

$$\begin{aligned} {\left\{ \begin{array}{ll} \dfrac{d\phi _2(t)}{dt}+\dfrac{\lambda ^2}{2\left( \gamma +\zeta _1\right) }=0,\\ \phi _2(T)=0. \end{array}\right. } \end{aligned}$$

(74)

We have the solution for problem 74

$$\begin{aligned} \phi _2(s)=\dfrac{\lambda ^2}{2(\gamma +\zeta _1)}(T-s). \end{aligned}$$

Case (3): $\rho ^2\ne 1$. Consider the following equation

$$\begin{aligned} {} \dfrac{\sigma ^2}{2}\left[ (1-\rho ^2)(\zeta _2+\gamma )\right] \phi _2^2(t)+(k+\lambda \sigma \rho )\phi _2(t)-\dfrac{\lambda ^2}{2(\gamma +\zeta _1)}=0. \end{aligned}$$

(75)

Let $\Delta =(k+\lambda \sigma \rho )^2+\dfrac{\sigma ^2\lambda ^2}{(\gamma +\zeta _1)(1-\rho ^2)(\gamma +\zeta _2)}$ is the discriminant of the equation 75. For $k,\lambda ,\sigma ,\zeta _1,\zeta _2$ are all positive and $-1<\rho < 1$, $\Delta >0$. Therefore, the equation 75 has two different real roots, which are

$$\begin{aligned} n_1=\dfrac{-(k+\lambda \sigma \rho )+\sqrt{\Delta }}{(1-\rho ^2)(\gamma +\zeta _2)\sigma ^2},\quad \quad \quad n_2=\dfrac{-(k+\lambda \sigma \rho )-\sqrt{\Delta }}{(1-\rho ^2)(\gamma +\zeta _2)\sigma ^2}. \end{aligned}$$

The problem 36 can be written as

$$\begin{aligned} {} \left\{ \begin{aligned} \dfrac{d\phi _2(t)}{dt}=\dfrac{\sigma ^2}{2}(1-\rho ^2)(\gamma +\zeta _2)(\phi _2(t)-n_1)(\phi _2(t)-n_2),\\ \phi _2(T)=0. \end{aligned}\right. \end{aligned}$$

(76)

Using the method of variable separation, 76 can be transformed into

$$\begin{aligned} {} \dfrac{d\phi _2(t)}{(\phi _2(t)-n_1)(\phi _2(t)-n_2)}=\dfrac{\sigma ^2}{2}(1-\rho ^2)(\zeta _2+\gamma )dt. \end{aligned}$$

(77)

77 can be further rewritten as

$$\begin{aligned} {} \dfrac{d\phi _2(t)}{\phi _2(t)-n_2}-\dfrac{d\phi _2(t)}{\phi _2(t)-n_1}=-\sqrt{\Delta }dt. \end{aligned}$$

(78)

Integrating 78 on both sides from s to T, and according to the terminal conditions $\phi _2(T)=0$, the solution of 36 is

$$\begin{aligned} \phi _2(s)=\dfrac{n_1n_2\left( 1-e^{\sqrt{\Delta }(T-s)}\right) }{n_1-n_2e^{\sqrt{\Delta }(T-s)}}. \end{aligned}$$

(79)

The proof is completed.

Verify $K^*(s)>0$

Based on $K(t)=\pi (t)X(t)$, we prove that $K^*(s)>0$ from the following three cases: $\rho =1,\ \rho =-1\ and\ \rho ^2\ne 1$.

Firstly, consider $\rho =1$. In this case, the optimal investment strategy can be written as

$$\begin{aligned} {\begin{matrix} {K}^*(s)&{}=\left( \dfrac{\lambda }{\gamma +\zeta _1}-\dfrac{\sigma \lambda ^2(1-e^{(k+\lambda \sigma )(s-T)})}{2(\gamma +\zeta _1)(k+\lambda \sigma )}\right) e^{(r+\mu _{2}e^{\delta h})(s-T)}\\ &{}=\dfrac{2k\lambda +\lambda ^2\sigma +\lambda ^2\sigma e^{(k+\lambda \sigma )(s-T)}}{2(\gamma +\zeta _1)(k+\lambda \sigma )}e^{(r+\mu _{2}e^{\delta h})(s-T)}. \end{matrix}} \end{aligned}$$

(80)

Obviously, ${K}^*(s)>0$ for posivitive constants $k,\ \lambda ,\ \sigma ,\ \gamma \ and\ \zeta _1$ as well as $s\in [0,T]$.

Secondly, investigate the case that $\rho =-1$. If $k\ne \lambda \sigma$, the optimal investment strategy can be expressed as

$$\begin{aligned} {K}^*(s)=\left( \dfrac{\lambda }{\gamma +\zeta _1}+\dfrac{\sigma \lambda ^2(1-e^{(k-\lambda \sigma )(s-T)})}{2(\gamma +\zeta _1)(k-\lambda \sigma )}\right) e^{(r+\mu _{2}e^{\delta h})(s-T)}. \end{aligned}$$

(81)

It is easy to see that ${K}^*(s)>0$. If $k=\lambda \sigma$, then ${K}^*(s)=\left( \dfrac{\lambda }{\gamma +\zeta _1}+\dfrac{\sigma \lambda ^2(T-s)}{2(\gamma +\zeta _1)}\right) e^{(r+\mu _{2}e^{\delta h})(s-T)}>0, \forall s\in [0,T]$ with posivitive constants $k,\ \lambda ,\ \sigma ,\ \gamma \ and\ \zeta _1$.

Finally, discuss the case that $\rho ^2\ne 1$. The optimal investment strategy can be represented as

$$\begin{aligned} {\begin{matrix}& {K}^*(s){}=\left( \dfrac{\lambda }{\gamma +\zeta _1}-\dfrac{\sigma \rho n_1n_2(1-e^{\sqrt{\Delta }(T-s)})}{n_1-n_2e^{\sqrt{\Delta }(T-s)}}\right) e^{(r+\mu _{2}e^{\delta h})(s-T)}\\ &{}=\left( \dfrac{\lambda }{\gamma +\zeta _1}-\dfrac{\lambda ^2\sigma \rho \left( e^{\sqrt{\Delta }(T-s)}-1\right) }{(\gamma +\zeta _1)\left( (k+\lambda \sigma \rho +\sqrt{\Delta })(e^{\sqrt{\Delta }(T-s)}-1)+2\sqrt{\Delta }\right) }\right) e^{(r+\mu _{2}e^{\delta h})(s-T)}\\ &{}=\dfrac{\lambda (e^{\sqrt{\Delta }(T-s)}-1)(k+\sqrt{\Delta })+2\lambda \sqrt{\Delta }}{(\gamma +\zeta _1)\left( (k+\lambda \sigma \rho +\sqrt{\Delta })(e^{\sqrt{\Delta }(T-s)}-1)+2\sqrt{\Delta }\right) }e^{(r+\mu _{2}e^{\delta h})(s-T)}. \end{matrix}} \end{aligned}$$

(82)

Since $k,\ \lambda ,\ \sigma ,\ \gamma$ and $\zeta _1$ are positive, and by the definition of $\Delta$, we deduce that ${K}^*(s)>0,\ \forall s\in [0,T]$.

To summarize, ${K}^*(s)>0$, for any $s\in [0,T]$.