Optimal Portfolio and Consumption Rule with a CIR Model Under HARA Utility

Abstract

In the real-world environments, different individuals have different risk preferences. This paper investigates the optimal portfolio and consumption rule with a Cox–Ingersoll–Ross (CIR) model in a more general utility framework. After consumption, an individual invests his wealth into the financial market with one risk-free asset and multiple risky assets, where the short-term rate is driven by the CIR model and stock price dynamics are simultaneously influenced by random sources from both stochastic interest rate and stock market itself. The individual hopes to optimize their portfolios and consumption rules to maximize expected utility of terminal wealth and intermediate consumption. Risk preference of individual is assumed to satisfy hyperbolic absolute risk aversion (HARA) utility, which contains power utility, logarithm utility, and exponential utility as special cases. By using the principle of stochastic optimality and Legendre transform-dual theory, the explicit expressions of the optimal portfolio and consumption rule are obtained. The sensitivity of the optimal strategies to main parameters is analysed by a numerical example. In addition, economic implications are also presented. Our research results show that Legendre transform-dual theory is an effective methodology in dealing with the portfolio selection problems with HARA utility and interest rate risk can be completely hedged by constructing specific portfolios.

Appendix

(A1) Proof of Lemma 4.1 Introducing the following variational operator on any function f(t, r):

$$\begin{aligned} \begin{aligned} \nabla f(t,r)&=\left( {\frac{1}{p-1}\beta -\frac{p}{p-1}r+\frac{p}{2(p-1)^2}\left( \left\| {\lambda _s } \right\| ^2+\lambda _r^2 r\right) } \right) f\\&\quad +\left( {a+\frac{p}{p-1}k\lambda _r r-br} \right) f_r +\frac{1}{2}k^2rf_{rr}, \end{aligned} \end{aligned}$$

we can rewrite Eq. (4.5) as

$$\begin{aligned} \frac{\partial f(t,r)}{\partial t}+\nabla f(t,r)+\left( {\frac{\alpha }{1-\alpha }} \right) ^{-\frac{1}{p-1}}=0, \quad f(T,r)=1. \end{aligned}$$

(a1)

In fact, according to $f(t,r)=\left( {\frac{\alpha }{1-\alpha }} \right) ^{-\frac{1}{p-1}}\int _t^{T} {{\hat{f}}(s,r)\hbox {d}s} +{\hat{f}}(t,r)$, we have

$$\begin{aligned} \frac{\partial f(t,r)}{\partial t}= & {} -\left( {\frac{\alpha }{1-\alpha }} \right) ^{-\frac{1}{p-1}}{\hat{f}}(t,r)+\frac{\partial \hat{f}(t,r)}{\partial t}\nonumber \\= & {} \left( {\frac{\alpha }{1-\alpha }} \right) ^{-\frac{1}{p-1}}\left( {\int _t^{T} {\frac{\partial {\hat{f}}(s,r)}{\partial s}\hbox {d}s} -{\hat{f}}(T,r)} \right) +\frac{\partial {\hat{f}}(t,r)}{\partial t}, \end{aligned}$$

(a2)

$$\begin{aligned} \nabla f(t,r)= & {} \left( {\frac{\alpha }{1-\alpha }} \right) ^{-\frac{1}{p-1}}\int _t^{T} {\nabla {\hat{f}}(s,r)\hbox {d}s} +\nabla \hat{f}(t,r). \end{aligned}$$

(a3)

Putting (a2) and (a3) into (a1), we obtain

$$\begin{aligned}&\left( {\frac{\alpha }{1-\alpha }} \right) ^{-\frac{1}{p-1}}\left( {\int _t^{T} {\left( {\frac{\partial {\hat{f}}(s,r)}{\partial s}+\nabla {\hat{f}}(s,r)} \right) \hbox {d}s} -{\hat{f}}(T,r)+1} \right) \nonumber \\&\qquad +\frac{\partial {\hat{f}}(t,r)}{\partial t}+\nabla {\hat{f}}(t,r)=0. \end{aligned}$$

(a4)

Comparing the coefficients on both sides of (a4), we get

$$\begin{aligned} \frac{\partial {\hat{f}}(t,r)}{\partial t}+\nabla {\hat{f}}(t,r)=0, \quad {\hat{f}}(T,r)=1. \end{aligned}$$

(a5)

Therefore, Eq. (4.7) holds.

(A2) Proof of Lemma 4.2 Substituting ${\hat{f}}(t,r)=\hbox {e}^{D_1 (t)+D_2 (t)r}$ back into (4.7) and separating the variables, we arrive at the following two ODEs:

$$\begin{aligned}&{\dot{D}}_2 (t)+\frac{p}{2(p-1)^2}\lambda _r^2 -\frac{p}{p-1}+\left( {\frac{p}{p-1}k\lambda _r -b} \right) D_2 (t)\nonumber \\&\qquad +\frac{1}{2}k^2D_2^2 (t)=0, \quad D_2 (T)=0; \end{aligned}$$

(a6)

$$\begin{aligned}&{\dot{D}}_1 (t)+\frac{1}{p-1}\beta +\frac{p}{2(p-1)^2}\left\| {\lambda _s } \right\| ^2+aD_2 (t)=0, \quad D_1 (T)=0. \end{aligned}$$

(a7)

By investigating the following quadratic equation:

$$\begin{aligned} -\frac{1}{2}k^2D_2^2 (t)-\left( {\frac{p}{p-1}k\lambda _r -b} \right) D_2 (t)-\left( {\frac{p}{2(p-1)^2}\lambda _r^2 -\frac{p}{p-1}} \right) =0, \end{aligned}$$

(a8)

we get its discriminant:

$$\begin{aligned} \Delta _1 =\frac{p}{p-1}\left( {(k\lambda _r -b)^2+2k^2-\frac{1}{p}b^2} \right) . \end{aligned}$$

(a9)

• (i) If $0<p<1$ and $p<\frac{b^2}{(k\lambda _r -b)^2+2k^2}$, i.e. $0<p<\min \left\{ {\frac{b^2}{(k\lambda _r -b)^2+2k^2},\;1} \right\}$, we have $\Delta _1 >0$; if $p<0$, we have $\Delta _1 >0$. To sum up, we obtain $p<\min \left\{ {\frac{b^2}{(k\lambda _r -b)^2+2k^2},\;1} \right\}$ and $p\ne 0$, we thus have $\Delta _1 >0$.

  The two different roots of Eq. (a8) are given by

  $$\begin{aligned} m_{1,2} =\frac{\frac{p}{p-1}k\lambda _r -b}{-k^2}\pm \frac{\sqrt{\Delta _1 } }{-k^2}. \end{aligned}$$

  (a10)

  The solution to Eq. (a6) is derived as follows.

  $$\begin{aligned} \begin{aligned}&{\dot{D}}_2 (t)=-\frac{1}{2}k^2D_2^2 (t)-\left( {\frac{p}{p-1}k\lambda _r -b} \right) D_2 (t)-\left( {\frac{p}{2(p-1)^2}\lambda _r^2 -\frac{p}{p-1}} \right) \\&\quad \Rightarrow {\dot{D}}_2 (t)=-\frac{1}{2}k^2(D_2 (t)-m_1 )(D_2 (t)-m_2 )\\&\quad \Rightarrow \frac{1}{m_1 -m_2 }\int _t^{T} {\left( {\frac{1}{D_2 (t)-m_1 }-\frac{1}{D_2 (t)-m_2 }} \right) } \hbox {d}D_2 (t)=-\frac{1}{2}k^2(T-t). \end{aligned} \end{aligned}$$

  Solving the above integration, we obtain (4.8). Further, plugging (4.8) into Eq. (a7), we have (4.9).

• (ii) If $p=\frac{b^2}{(k\lambda _r -b)^2+2k^2}$ and $p<1$, we have $\Delta _1 =0$. Then, the unique root of (a8) is given by

  $$\begin{aligned} m_3 =\frac{\frac{p}{p-1}k\lambda _r -b}{-k^2}. \end{aligned}$$

  (a11)

  We derive the following process to solve Eq. (a6):

  $$\begin{aligned} \begin{aligned}&{\dot{D}}_2 (t)=-\frac{1}{2}k^2(D_2 (t)-m_3 )^2\\&\quad \Rightarrow \int _t^{T} {\frac{1}{(D_2 (t)-m_3 )^2}\mathrm{d}D_2 (t)} =-\frac{1}{2}k^2(T-t). \end{aligned} \end{aligned}$$

  Solving the above integration yields (4.10). Putting (4.10) in Eq. (a7), we get (4.11).

• (iii) If $\frac{b^2}{(k\lambda _r -b)^2+2k^2}<p<1$, we have $\Delta _1 <0$. Then, for Eq. (a6), we adopt the following solving technique:

  $$\begin{aligned} \begin{aligned}&{\dot{D}}_2 (t)=-\frac{1}{2}k^2D_2^2 (t)-\left( {\frac{p}{p-1}k\lambda _r -b} \right) D_2 (t)-\left( {\frac{p}{2(p-1)^2}\lambda _r^2 -\frac{p}{p-1}} \right) \\&\quad \Rightarrow {\dot{D}}_2 (t)=-\frac{1}{2}k^2\left( {\left( {D_2 (t)-m_3 } \right) ^2+\frac{-\Delta _1 }{k^4}} \right) \\&\quad \Rightarrow \int _t^{T} {\frac{1}{\frac{\sqrt{-\Delta _1 } }{k^2}\left( {\left( {\frac{k^2}{\sqrt{-\Delta _1 } }\left( {D_2 (t)-m_3 } \right) } \right) ^2+1} \right) }\mathrm{d}\left( {\frac{k^2}{\sqrt{-\Delta _1 } }\left( {D_2 (t)-m_3 } \right) } \right) }\\&\quad =-\frac{1}{2}k^2\left( {T-t} \right) . \end{aligned} \end{aligned}$$

  Therefore, we obtain (4.12). Further, we have (4.13). Thus, the proof is completed.

(A3) Proof of Lemma 4.4 Plugging ${\hat{h}}(t,r)=\hbox {e}^{D_3 (t)+D_4 (t)r}$ into Eq. (4.14) and separating the variables, we get

$$\begin{aligned}&{\dot{D}}_4 (t)-1+(k\lambda _r -b)D_4 (t)+\frac{1}{2}k^2D_4^2 (t)=0, \quad D_4 (T)=0; \end{aligned}$$

(a12)

$$\begin{aligned}&{\dot{D}}_3 (t)+aD_4 (t)=0, \quad D_3 (T)=0. \end{aligned}$$

(a13)

The discriminant for the quadratic equation $-\frac{1}{2}k^2D_4^2 (t)-(k\lambda _r -b)D_4 (t)+1=0$ is given by $\Delta _2 =(k\lambda _r -b)^2+2k^2>0$. So, the two different roots are represented as

$$\begin{aligned} m_{4,5} =\frac{k\lambda _r -b}{-k^2}\pm \frac{\sqrt{\Delta _2 } }{-k^2}. \end{aligned}$$

(a14)

By using the same technique as Lemma 4.2, we derive (4.15) and (4.16).

Therefore, we easily complete the proof.