Uncertain random portfolio optimization with non-dominated sorting genetic algorithm-II and optimal solution criterion

Abstract

The complexity of the financial systems inevitably leads to the uncertain information and random information simultaneously. Because asset returns frequently show excessive kurtosis and tend to be skewed, we consider an uncertain random higher moments portfolio optimization problem in this paper, in which uncertain and random return rates exist simultaneously. First, the concept of kurtosis for uncertain random variable is defined and the deterministic expressions of kurtosis under three kinds of distributions are derived. Then, an uncertain random mean-variance-skewness-kurtosis-entropy model is formulated with two auxiliary models for portfolio optimization problem. After solving the equivalent deterministic model with NSGA-II algorithm, we propose a new optimal solution criterion for finding a single optimal solution in Pareto optimal solution set. Finally, we present a numerical simulation and obtain the following results: (i) the practicability and the validity of the proposed model, the NSGA-II algorithm and the optimal selection criterion have verified; (ii) the size of population has an obvious influence on the single optimal solution; (iii) the parameter adjustment has a significant impact on the results, and the results are in perfect agreement with the actual situation.

Appendix

The proof of Theorem 2

Proof

Based on Definition 5 in Liu (2013a) and Definition 3, we have

$$\begin{aligned} K[\eta ]=\displaystyle \int _{0}^{+\infty }Ch\left\{ \left( \eta -E\left[ \eta \right] \right) ^{4}\ge r\right\} \textrm{d}r-\displaystyle \int _{-\infty }^{0}Ch\left\{ \left( \eta -E\left[ \eta \right] \right) ^{4}\le r\right\} \textrm{d}r, \end{aligned}$$

where Ch is the chance measure of uncertain random event. Because $\left( \eta -E[\eta ]\right) ^{4}$ is non-negative, it holds that

$$\begin{aligned} K[\eta ]&=\displaystyle \int _{0}^{+\infty }Ch\left\{ \left( \eta -E\left[ \eta \right] \right) ^{4}\ge r\right\} \textrm{d}r\\[4pt]&=\displaystyle \int _{0}^{+\infty }Ch\left\{ \left( \eta \ge E\left[ \eta \right] +\root 4 \of {r}\right) \cup \left( \eta \le E\left[ \eta \right] -\root 4 \of {r}\right) \right\} \textrm{d}r\\[4pt]&\le \displaystyle \int _{0}^{+\infty }Ch\left\{ \eta \ge E\left[ \eta \right] +\root 4 \of {r}\right\} \textrm{d}r+\displaystyle \int _{0}^{+\infty }Ch\left\{ \eta \le E\left[ \eta \right] -\root 4 \of {r}\right\} \textrm{d}r. \end{aligned}$$

Here, it follows from Stipulation 2.3 in Liu (2007) that

$$\begin{aligned} K[\eta ]&=\displaystyle \int _{0}^{+\infty }Ch\left\{ \eta \ge E\left[ \eta \right] +\root 4 \of {r}\right\} \textrm{d}r+\displaystyle \int _{0}^{+\infty }Ch\left\{ \eta \le E\left[ \eta \right] -\root 4 \of {r}\right\} \textrm{d}r\\[4pt]&=\displaystyle \int _{0}^{+\infty }\left[ 1-\Psi \left( E\left[ \eta \right] +\root 4 \of {r}\right) \right] \textrm{d}r+\displaystyle \int _{0}^{+\infty }\Psi \left( E\left[ \eta \right] -\root 4 \of {r}\right) \textrm{d}r. \end{aligned}$$

Letting $x=E[\eta ]+\root 4 \of {r}$ ($r=\left( x-E[\eta ]\right) ^{4}$) and taking integration by parts, we have

$$\begin{aligned} K[\eta ]&=\displaystyle \int _{0}^{+\infty }\left[ 1-\Psi \left( E\left[ \eta \right] +\root 4 \of {r}\right) \right] \textrm{d}r+\displaystyle \int _{0}^{+\infty }\Psi \left( E\left[ \eta \right] -\root 4 \of {r}\right) \textrm{d}r\\[4pt]&=\displaystyle \int _{E\left[ \eta \right] }^{+\infty }\left( 1-\Psi \left( x\right) \right) \textrm{d}\left( x-E[\eta ]\right) ^{4}+\displaystyle \int _{E[\eta ]}^{-\infty }\Psi \left( x\right) \textrm{d}\left( x-E[\eta ]\right) ^{4}\\[4pt]&=\left( 1-\Psi \left( x\right) \right) \left( x-E[\eta ]\right) ^{4}\mid ^{+\infty }_{E[\eta ]}-\displaystyle \int _{E[\eta ]}^{+\infty }\left( x-E[\eta ]\right) ^{4}\textrm{d}\left( 1-\Psi (x)\right) \\[4pt]&\quad +\Psi \left( x\right) \left( x-E[\eta ]\right) ^{4}\mid ^{-\infty }_{E[\eta ]}-\displaystyle \int _{E[\eta ]}^{-\infty }\left( x-E[\eta ]\right) ^{4}\textrm{d}\Psi \left( x\right) \\[4pt]&=\displaystyle \int _{E[\eta ]}^{+\infty }\left( x-E[\eta ]\right) ^{4}\textrm{d}\Psi \left( x\right) +\displaystyle \int _{-\infty }^{E[\eta ]}\left( x-E[\eta ]\right) ^{4}\textrm{d}\Psi \left( x\right) \\[4pt]&=\displaystyle \int _{-\infty }^{+\infty }\left( x-E[\eta ]\right) ^{4}\textrm{d}\Psi (x). \end{aligned}$$

The proof is completed. $\square$

The proof of Theorem 3

Proof

According to Definition 3, we have

$$\begin{aligned} K[a\eta +b]&=E\displaystyle \left[ \displaystyle \left[ \left( a\eta +b\right) -E\left[ a\eta +b\right] \displaystyle \right] ^{4}\displaystyle \right] \\[6pt]&=E\displaystyle \left[ \displaystyle \left[ (a\eta +b)-\displaystyle \left( aE[\eta ]+b\displaystyle \right) \displaystyle \right] ^{4}\displaystyle \right] \\[6pt]&=E\displaystyle \left[ \displaystyle \left( a\eta -aE[\eta ]+b-b\displaystyle \right) ^{4}\displaystyle \right] \\[6pt]&=E\displaystyle \left[ a^{4}\displaystyle \left( \eta -E[\eta ]\displaystyle \right) ^{4}\displaystyle \right] \\[6pt]&=a^{4}E\displaystyle \left[ (\eta -E[\eta ])^{4}\displaystyle \right] \\[6pt]&=a^{4}K[\eta ]. \end{aligned}$$

The proof is completed. $\square$

The proof of Theorem 7

Proof

We construct a Lagrange function to calculate the maximum value of investment diversification in (5.8)

$$\begin{aligned} L(u_{1},u_{2},\cdots ,u_{m},u_{m+1},u_{m+2},\cdots ,u_{m+n},\lambda )&=L(x_{1},x_{2},\cdots ,x_{m},y_{1},x_{2},\cdots ,y_{n},\lambda )\\[6pt]&=-\sum \limits _{i=1}^{m+n}u_{i}\ln (u_{i}+\varepsilon )+\lambda (\sum \limits _{i=1}^{m+n}u_{i}-1). \end{aligned}$$

Taking the partial derivatives of $L(u_{1},u_{2},\cdots ,u_{m},u_{m+1},u_{m+2},\cdots ,u_{m+n},\lambda )$ with respect to $u_{1},u_{2},$ $\cdots ,u_{m+n}$ and $\lambda$, we can get

$$\begin{aligned}&\displaystyle \frac{\partial L(u_{1},u_{2},\cdots ,u_{m},u_{m+1},u_{m+2},\cdots ,u_{m+n},\lambda )}{\partial u_{i}}=-\ln (u_{i}+\varepsilon )-\displaystyle \frac{u_{i}}{u_{i}+\varepsilon }+\lambda , i=1,2,\cdots ,m+n,\\&\displaystyle \frac{\partial L(u_{1},u_{2},\cdots ,u_{m},u_{m+1},u_{m+2},\cdots ,u_{m+n},\lambda )}{\partial \lambda }=\sum \limits _{i=1}^{m+n}u_{i}-1. \end{aligned}$$

Then, set all the partial derivatives equal to zero, it follows that

$$\begin{aligned}&\ln \left( \displaystyle \frac{u_{i}+\varepsilon }{u_{j}+\varepsilon } \right) +\displaystyle \frac{u_{i}}{u_{i}+\varepsilon }-\displaystyle \frac{u_{j}}{u_{j}+\varepsilon }=0, i,j=1,2,\cdots ,m+n,\\&\sum \limits _{i=1}^{m+n}u_{i}-1=0. \end{aligned}$$

We can get the result that $u_{1}=u_{2}=\cdots =u_{m+n}=\displaystyle \frac{1}{m+n}$, and the maximum value of the corresponding investment diversification is $-\ln \left( \displaystyle \frac{1}{m+n}+\varepsilon \right)$. Conversely, when investors invest all their assets in a single asset, the value of investment diversification will reach a minimum of zero. The proof is completed. $\square$