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Optimal Stopping Time of a Portfolio Selection Problem with Multi-assets

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Abstract

In this work, we study a right time for an investor to stop the investment among multi-assets over a given investment horizon so as to obtain maximum profit. We formulate it to a two-stage problem. The main problem is not a standard optimal stopping problem due to the non-adapted term in the objective function, and we turn it to a standard one by stochastic analysis. The subproblem with control variable in the drift and volatility terms is solved first via stochastic control method. A numerical example is presented to illustrate the efficiency of the theoretical results.

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Acknowledgements

We acknowledge the contribution of Professor Xun Li in the Hong Kong Polytechnic University for his valuable comments and suggestions to this paper.

Author information

Authors and Affiliations

  1. School of Applied Mathematics, Guangdong University of Technology, Guangzhou, 510006, China

    Xian-Ping Wu

  2. Department of Mathematics, University of Macau, Macau, China

    Seakweng Vong

  3. Department of Applied Mathematics, The Hong Kong Polytechnic University, Hong Kong, China

    Wen-Xin Zhou

Authors
  1. Xian-Ping Wu
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  2. Seakweng Vong
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  3. Wen-Xin Zhou
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Corresponding author

Correspondence to Xian-Ping Wu.

Additional information

This work is supported by the National Natural Science Foundation of China (Nos. 11571124 and 11671158), the doctoral start-up Grant of Natural Science Foundation of Guangdong Province, China (No. 2017A030310167), the Opening Project of Guangdong Province Key Laboratory of Computational Science at the Sun Yat-sen University (No. 201808) and Unversity of Macau (No. MYGR2018-00047-FST).

Appendices

Appendix

Expression of Function \(G_1\)

We now derive the explicit expression of the function \(G_1\), defined by

$$\begin{aligned} \begin{aligned} G_1(t,y)&= \mathbb {E}\Big [\exp \Big (\displaystyle \max \Big \{y,\max _{0\leqslant u\leqslant T-t}(\nu u + B(u))\Big \}\Big )\Big ] \\&= \displaystyle \int _y^\infty \mathrm{e}^z {\mathrm{d}}{P}\Big (\max _{0\leqslant u\leqslant T-t}(\nu u + B(u)) \leqslant z\Big ) + \mathrm{e}^y {P}\Big (\max _{0\leqslant u\leqslant T-t}(\nu u + B(u)) \leqslant y\Big ). \end{aligned} \end{aligned}$$

Note that

$$\begin{aligned} {P}\Big (\displaystyle \max _{0\leqslant u\leqslant T-t}(\nu u + B(u)) \leqslant z\Big ) = \Phi \Big (\frac{z - \nu (T-t)}{\sqrt{T-t}}\Big ) - \mathrm{e}^{2\nu z}\Phi \Big (\frac{-z - \nu (T-t)}{\sqrt{T-t}}\Big ). \end{aligned}$$

According to the standard normal distribution, we have

$$\begin{aligned} \begin{aligned} \displaystyle \int _y^\infty \mathrm{e}^z {\mathrm{d}}\Phi \Big (\frac{z - \nu (T-t)}{\sqrt{T-t}}\Big )&= \displaystyle \int _y^\infty \mathrm{e}^z \frac{1}{\sqrt{2\pi (T-t)}}\mathrm{e}^{-\frac{(z - \nu (T-t))^2}{2(T-t)}}{\mathrm{d}}z \\&= \displaystyle \mathrm{e}^{(\nu + \frac{1}{2})(T-t)}\Big [1 - \Phi \Big (\frac{y - (\nu +1)(T-t)}{\sqrt{T-t}}\Big )\Big ]. \end{aligned} \end{aligned}$$

Assume that \(\nu \ne -\frac{1}{2}\). Then

$$\begin{aligned} \begin{aligned}&\ \ \qquad \displaystyle \int _y^\infty \mathrm{e}^z {\mathrm{d}}\Big [\mathrm{e}^{2\nu z}\Phi \Big (\frac{-z - \nu (T-t)}{\sqrt{T-t}}\Big )\Big ] \\&\quad = \displaystyle \int _y^\infty 2\nu \mathrm{e}^{(1+2\nu )z} \Phi \Big (\frac{-z - \nu (T-t)}{\sqrt{T-t}}\Big ){\mathrm{d}}z + \int _y^\infty \mathrm{e}^{(1+2\nu )z} {\mathrm{d}}\Phi \Big (\frac{-z - \nu (T-t)}{\sqrt{T-t}}\Big ) \\&\quad = - \displaystyle \frac{2\nu }{1 + 2\nu }\mathrm{e}^{(1+2\nu )y} \Phi \Big (\frac{-y - \nu (T-t)}{\sqrt{T-t}}\Big ) \\&\qquad - \frac{1}{1 + 2\nu }\mathrm{e}^{(\nu + \frac{1}{2})(T-t)}\Big [1 - \Phi \Big (\frac{y - (\nu +1)(T-t)}{\sqrt{T-t}}\Big )\Big ]. \end{aligned} \end{aligned}$$

Thus

$$\begin{aligned} \begin{aligned} G_1(t,y)&= \displaystyle e^y\Phi \Big (\frac{y - \nu (T-t)}{\sqrt{T-t}}\Big ) - \frac{1}{1 + 2\nu }\mathrm{e}^{(1+2\nu )y}\Phi \Big (\frac{-y - \nu (T-t)}{\sqrt{T-t}}\Big ) \\&\quad + \displaystyle \frac{2(1+\nu )}{1 + 2\nu }\mathrm{e}^{(\nu + \frac{1}{2})(T-t)}\Big [1 - \Phi \Big (\frac{y - (\nu +1)(T-t)}{\sqrt{T-t}}\Big )\Big ]. \end{aligned} \end{aligned}$$

In addition, note that when \(\nu = -\frac{1}{2}\),

$$\begin{aligned} \begin{aligned}&\ \ \qquad \displaystyle \int _y^\infty \mathrm{e}^z {\mathrm{d}}\Big [\mathrm{e}^{2\nu z}\Phi \Big (\frac{-z - \nu (T-t)}{\sqrt{T-t}}\Big )\Big ] \\&\quad = \displaystyle \int _y^\infty \mathrm{e}^z {\mathrm{d}}\Big [\mathrm{e}^{-z}\Phi \Big (\frac{-z - \nu (T-t)}{\sqrt{T-t}}\Big )\Big ] \\&\quad = - \displaystyle \int _y^\infty \Phi \Big (\frac{-z - \nu (T-t)}{\sqrt{T-t}}\Big ){\mathrm{d}}z + \int _y^\infty {\mathrm{d}}\Phi \Big (\frac{-z - \nu (T-t)}{\sqrt{T-t}}\Big ) \\&\quad = \displaystyle y\Phi \Big (\frac{-y - \nu (T-t)}{\sqrt{T-t}}\Big ) -\frac{\sqrt{T-t}}{\sqrt{2\pi }}\mathrm{e}^{-\frac{(x + \nu (T-t))^2}{2(T-t)}} \\&\qquad + \nu (T-t)\Big [1 - \Phi \Big (\frac{y + \nu (T-t)}{\sqrt{T-t}}\Big )\Big ] \\&\qquad -\Phi \Big (\frac{-y - \nu (T-t)}{\sqrt{T-t}}\Big ). \end{aligned} \end{aligned}$$

Thus

$$\begin{aligned} \begin{aligned} G_1(t,y)&= 1 - \displaystyle \Phi \Big (\frac{y - (\nu +1)(T-t)}{\sqrt{T-t}}\Big ) - y\Phi \Big (\frac{-y - \nu (T-t)}{\sqrt{T-t}}\Big ) \\&\quad +\frac{\sqrt{T-t}}{\sqrt{2\pi }}\mathrm{e}^{-\frac{(y + \nu (T-t))^2}{2(T-t)}} \\&\quad - \displaystyle \nu (T-t)\Big [1 - \Phi \Big (\frac{y + \nu (T-t)}{\sqrt{T-t}}\Big )\Big ] + \mathrm{e}^y\Phi \Big (\frac{y - \nu (T-t)}{\sqrt{T-t}}\Big ). \end{aligned} \end{aligned}$$

Expression of Function \(G_2\)

We now derive the explicit expression of the function \(G_2\), defined by

$$\begin{aligned} \begin{aligned} G_2(t,y)&= \mathbb {E}\Big [\exp \Big (\displaystyle \max \Big \{y,\max _{0\leqslant u\leqslant T-t}(\nu u + B(u))\Big \}\Big )^2\Big ] \\&= \displaystyle \int _y^\infty \mathrm{e}^{2z} {\mathrm{d}}{P}\Big (\max _{0\leqslant u\leqslant T-t}(\nu u + B(u)) \leqslant z\Big ) + \mathrm{e}^{2y}{P}\Big (\max _{0\leqslant u\leqslant T-t}(\nu u + B(u)) \leqslant y\Big ). \end{aligned} \end{aligned}$$

Note that

$$\begin{aligned} {P}\Big (\displaystyle \max _{0\leqslant u\leqslant T-t}(\nu u + B(u)) \leqslant z\Big ) = \Phi \Big (\frac{z - \nu (T-t)}{\sqrt{T-t}}\Big ) - \mathrm{e}^{2\nu z}\Phi \Big (\frac{-z - \nu (T-t)}{\sqrt{T-t}}\Big ). \end{aligned}$$

According to the standard normal distribution, we have

$$\begin{aligned} \begin{aligned} \displaystyle \int _y^\infty \mathrm{e}^{2z} {\mathrm{d}}\Phi \Big (\frac{z - \nu (T-t)}{\sqrt{T-t}}\Big )&= \displaystyle \int _y^\infty \mathrm{e}^{2z} \frac{1}{\sqrt{2\pi (T-t)}}\mathrm{e}^{-\frac{(z - \nu (T-t))^2}{2(T-t)}}{\mathrm{d}}z \\&= \displaystyle \mathrm{e}^{2(\nu + 1)(T-t)}\Big [1 - \Phi \Big (\frac{y - (\nu +2)(T-t)}{\sqrt{T-t}}\Big )\Big ]. \end{aligned} \end{aligned}$$

Assume that \(\nu \ne -1\). Then

$$\begin{aligned} \begin{aligned}&\displaystyle \int _y^\infty \mathrm{e}^{2z} {\mathrm{d}}\Big [\mathrm{e}^{2\nu z}\Phi \Big (\frac{-z - \nu (T-t)}{\sqrt{T-t}}\Big )\Big ] \\&\quad = \displaystyle \int _y^\infty 2\nu \mathrm{e}^{2(1+\nu )z} \Phi \Big (\frac{-z - \nu (T-t)}{\sqrt{T-t}}\Big ){\mathrm{d}}z \\&\qquad + \int _x^\infty \mathrm{e}^{2(1+\nu )z} {\mathrm{d}}\Phi \Big (\frac{-z - \nu (T-t)}{\sqrt{T-t}}\Big ) \\&\quad = - \displaystyle \frac{\nu }{1 + \nu }\mathrm{e}^{2(1+\nu )y} \Phi \Big (\frac{-y - \nu (T-t)}{\sqrt{T-t}}\Big ) \\&\qquad - \frac{1}{1 + \nu }\mathrm{e}^{2(\nu + 1)(T-t)}\Big [1 - \Phi \Big (\frac{y - (\nu +2)(T-t)}{\sqrt{T-t}}\Big )\Big ]. \end{aligned} \end{aligned}$$

Thus

$$\begin{aligned} \begin{aligned} G_2(t,y)&= \displaystyle \mathrm{e}^{2y}\Phi \Big (\frac{y - \nu (T-t)}{\sqrt{T-t}}\Big ) - \frac{1}{1 + 2\nu }\mathrm{e}^{2(1+\nu )y}\Phi \Big (\frac{-y - \nu (T-t)}{\sqrt{T-t}}\Big ) \\&\quad + \displaystyle \frac{2+\nu }{1 + \nu }\mathrm{e}^{2(\nu + 1)(T-t)}\Big [1 - \Phi \Big (\frac{y - (\nu +2)(T-t)}{\sqrt{T-t}}\Big )\Big ]. \end{aligned} \end{aligned}$$

Also, note that when \(\nu = -1\),

$$\begin{aligned} \begin{aligned}&\displaystyle \int _y^\infty \mathrm{e}^{2z} {\mathrm{d}}\Big [\mathrm{e}^{2\nu z}\Phi \Big (\frac{-z - \nu (T-t)}{\sqrt{T-t}}\Big )\Big ] \\&\quad = \displaystyle \int _y^\infty \mathrm{e}^{2z} {\mathrm{d}}\Big [\mathrm{e}^{-2z}\Phi \Big (\frac{-z - \nu (T-t)}{\sqrt{T-t}}\Big )\Big ] \\&\quad = - 2\displaystyle \int _y^\infty \Phi \Big (\frac{-z - \nu (T-t)}{\sqrt{T-t}}\Big ){\mathrm{d}}z + {\displaystyle \int _y^\infty } {\mathrm{d}}\Phi \Big (\frac{-z - \nu (T-t)}{\sqrt{T-t}}\Big ) \\&\quad = \displaystyle 2y\Phi \Big (\frac{-y - \nu (T-t)}{\sqrt{T-t}}\Big ) - \frac{2\sqrt{T-t}}{\sqrt{2\pi }}\mathrm{e}^{-\frac{(y + \nu (T-t))^2}{2(T-t)}} \\&\qquad + 2\nu (T-t)\Big [1 - \Phi \Big (\frac{y + \nu (T-t)}{\sqrt{T-t}}\Big )\Big ] \\&\qquad - \displaystyle \Phi \Big (\frac{-y - \nu (T-t)}{\sqrt{T-t}}\Big ). \end{aligned} \end{aligned}$$

Thus

$$\begin{aligned} \begin{aligned} G_2(t,y)&= 1 - \displaystyle \Phi \Big (\frac{y - (\nu +2)(T-t)}{\sqrt{T-t}}\Big ) - 2y\Phi \Big (\frac{-y - \nu (T-t)}{\sqrt{T-t}}\Big ) \\&\quad +\frac{2\sqrt{T-t}}{\sqrt{2\pi }}\mathrm{e}^{-\frac{(y + \nu (T-t))^2}{2(T-t)}} \\&\quad - \displaystyle 2\nu (T-t)\Big [1 - \Phi \Big (\frac{y + \nu (T-t)}{\sqrt{T-t}}\Big )\Big ] + \mathrm{e}^{2y}\Phi \Big (\frac{y - \nu (T-t)}{\sqrt{T-t}}\Big ). \end{aligned} \end{aligned}$$

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Wu, XP., Vong, S. & Zhou, WX. Optimal Stopping Time of a Portfolio Selection Problem with Multi-assets. J. Oper. Res. Soc. China 9, 163–179 (2021). https://doi.org/10.1007/s40305-018-0223-5

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