﻿ 随机游走行情期权复制的研究并推导出若干公式 Research on the Replication of the Option of Random Walk Market and Deducing Some Formulas

Vol. 08  No. 04 ( 2019 ), Article ID: 29994 , 12 pages
10.12677/AAM.2019.84093

Research on the Replication of the Option of Random Walk Market and Deducing Some Formulas

Dawei Tang, Zhizhuo Cai

Zhejiang Zhuosheng Xianghong Asset Management Co., Ltd., Jiaxing Zhejiang

Received: Apr. 6th, 2019; accepted: Apr. 21st, 2019; published: Apr. 28th, 2019

ABSTRACT

First, this paper deduces the option pricing formula of random walks, and then proposes three propositions and proves them separately. Proposition 1 explains the principle of option replication from a new perspective. Proposition 2 finds two inequalities and gives the upper and lower bounds of the cumulative function of the standard normal distribution. Combining Proposition 2 and Proposition 3, four elementary functions that approximate the cumulative function of the standard normal distribution are found. The second function has the highest precision: the maximum absolute error is 0.0000406.

Keywords:Option, Random Walk, Inequality, Approximation, Error

1. 引言

Black-Scholes期权定价模型不太适用零和博弈市场如期货市场。且BS公式在计算到期时间短的期权价格时偏差较大。金融市场长线行情或可预测但短线行情往往非常近似随机游走行情。所以本文从随机游走行情推导期权定价公式并提出三个命题，给出证明并推导出若干公式。

2. 正文

$\phi$ 为标准正态分布概率密度函数：

$\phi \left(x\right)=\frac{{e}^{\frac{-{x}^{2}}{2}}}{\sqrt{2\text{π}}}$ [1]

N为标准正态分布变量的累积概率分布函数：

$N\left(x\right)={\int }_{-\infty }^{x}\frac{1}{\sqrt{2\text{π}}}{e}^{-\frac{{u}^{2}}{2}}du=P\left\{\xi \le x\right\}$ [1]

$C={\int }_{K}^{+\infty }\left(x-K\right)\cdot \left(\frac{1}{\omega \sqrt{T}}\cdot \phi \left(\frac{x-S-0}{\omega \sqrt{T}}\right)\right)\text{d}x={\int }_{K}^{+\infty }\frac{x-S}{\omega \sqrt{T}}\phi \left(\frac{x-S}{\omega \sqrt{T}}\right)\text{d}x+\left(S-K\right){\int }_{K}^{+\infty }\frac{1}{\omega \sqrt{T}}\cdot \phi \left(\frac{x-S}{\omega \sqrt{T}}\right)\text{d}x$

$\phi \left(x\right)=\phi \left(-x\right)$ ，所以C亦可写成：

$C=\omega \sqrt{T}\cdot \phi \left(\frac{S-K}{\omega \sqrt{T}}\right)+\left(S-K\right)\cdot N\left(\frac{S-K}{\omega \sqrt{T}}\right)$

$\frac{dC}{dS}$

$\frac{dC}{dS}=N\left(\frac{S-K}{\omega \sqrt{T}}\right)$

$\omega =1,T=1,S=0时$ ，此时该标的资产单位时间涨跌幅度均值为0，标准差为1，当前价格为0，

$C=\phi \left(K\right)-K\cdot N\left(-K\right)$

$\frac{dC}{dS}=N\left(-K\right)$

$C+Mx-0$

$\text{C}+Mx-\left(x-K\right)$

$\alpha \left(M\right)={\int }_{-\infty }^{K}{\left(\left(C+Mx-0\right)-0\right)}^{2}\phi \left(x\right)\text{d}x+{\int }_{K}^{+\infty }{\left(\left(C+Mx-\left(x-K\right)\right)-0\right)}^{2}\phi \left(x\right)\text{d}x$

$\alpha \left(M\right)$ 的导数(M是自变量) ${\alpha }^{\prime }\left(M\right)=0$ 时， $\alpha \left(M\right)$ 最小。

$\begin{array}{c}{\alpha }^{\prime }\left(M\right)={\int }_{-\infty }^{K}2x\left(C+Mx\right)\phi \left(x\right)\text{d}x+{\int }_{K}^{+\infty }2x\left(C+Mx-\left(x-K\right)\right)\phi \left(x\right)\text{d}x\\ =C\cdot {\int }_{-\infty }^{+\infty }2x\phi \left(x\right)\text{d}x+{\int }_{-\infty }^{+\infty }2M{x}^{2}\phi \left(x\right)\text{d}x-{\int }_{K}^{+\infty }2x\left(x-K\right)\phi \left(x\right)\text{d}x\\ ={\int }_{-\infty }^{+\infty }2M{x}^{2}\phi \left(x\right)\text{d}x-{\int }_{K}^{+\infty }2x\left(x-K\right)\phi \left(x\right)\text{d}x\\ =0\end{array}$

${\int }_{-\infty }^{+\infty }{x}^{2}\phi \left(x\right)\text{d}x$ 为方差的定义，因方差为1，所以 ${\int }_{-\infty }^{+\infty }{x}^{2}\phi \left(x\right)\text{d}x=1$ ，上式整理为：

${\alpha }^{\prime }\left(M\right)=2M-{\int }_{K}^{+\infty }2x\left(x-K\right)\phi \left(x\right)\text{d}x=0$

$-\phi \left(-K\right)+\phi \left(K\right)-{K}^{2}\phi \left(K\right)$

$-{K}^{2}\phi \left(K\right)$

$N\left(-K\right)={\int }_{K}^{+\infty }x\left(x-K\right)\phi \left(x\right)\text{d}x,\forall K\ge 0等式成立$

${M}_{1}$ 满足如下式子：

$\underset{0}{\overset{K}{\int }}\left(C+{M}_{1}x-0\right)\phi \left(x\right)\text{d}x+{\int }_{K}^{+\infty }\left(C+{M}_{1}x-\left(x-K\right)\right)\phi \left(x\right)\text{d}x=0$

${M}_{1}\approx \frac{\phi \left(K\right)}{2\phi \left(0\right)+K}$

${M}_{2}$ 满足如下式子：

${\int }_{K}^{+\infty }\left(C+{M}_{2}x-\left(x-K\right)\right)\phi \left(x\right)\text{d}x=0$

${M}_{2}\approx \frac{1}{2}-\frac{\sqrt{{K}^{2}+4{\phi }^{2}\left(K\right)}-2\phi \left(K\right)}{2K}$

$\frac{\phi \left(K\right)}{2\phi \left(0\right)+K}\le N\left(-K\right)\le \frac{1}{2}-\frac{\sqrt{{K}^{2}+4{\phi }^{2}\left(K\right)}-2\phi \left(K\right)}{2K},对于\forall K\ge 0成立,当K=0时,等号成立.$

$N\left(x\right)\ge \frac{1}{2}+\frac{\sqrt{{x}^{2}+4{\phi }^{2}\left(x\right)}-2\phi \left(x\right)}{2x},x\ge 0时不等式成立$

$N\left(x\right)\le 1-\frac{\phi \left(x\right)}{\frac{2}{\sqrt{2\text{π}}}+x},x\ge 0时不等式成立$

Figure 1. The lower bound function minus $N\left(x\right)$

$L\left(x\right)=\frac{1}{2}+\frac{\sqrt{{x}^{2}+4{\phi }^{2}\left(x\right)}-2\phi \left(x\right)}{2x}-N\left(x\right),x\ge 0$

$L\left(x\right)$ 的函数图像即是图1

Figure 2. The upper bound function minus $N\left(x\right)$

$U\left(x\right)=1-\frac{\phi \left(x\right)}{\frac{2}{\sqrt{2\text{π}}}+x}-N\left(x\right)$$U\left(x\right)$ 的函数图像即是图2

${U}^{\prime }\left(x\right)=\frac{\phi \left(x\right)}{{\left(\frac{2}{\sqrt{\text{π}}}+x\right)}^{2}}\left(1-\frac{2x}{\sqrt{2\text{π}}}-\frac{2}{\text{π}}\right)$

$当x>\frac{\sqrt{2\text{π}}}{2}-\frac{2}{\sqrt{2\text{π}}}时,{U}^{\prime }\left(x\right)<0$

$\underset{x\to +\infty }{\mathrm{lim}}U\left(x\right)=0$

$N\left(x\right)\approx \frac{3}{4}+\frac{\sqrt{{x}^{2}+4{\phi }^{2}\left(x\right)}-2\phi \left(x\right)}{4x}-\frac{\phi \left(x\right)}{\frac{4}{\sqrt{2\text{π}}}+2x},x\ge 0$

x = 0时，右侧式子中有一个分母是4x，此时用极限求解，当x趋近于0得右侧式子为0.5，左侧N(0)也是0.5，此时是严格相等。

Figure 3. The error of the approximate formula one

$V$ 满足如下式子：

$N\left(-K\right)=\frac{\phi \left(K\right)}{\frac{\phi \left(V\right)}{N\left(V\right)}+K}$

$\underset{K\to 0}{\mathrm{lim}}{V}^{\prime }=\frac{\text{π}}{2}-1$

Figure 4. $\frac{N\left(-K\right)}{C}$ in a straight line approximately

Figure 5. ${\left(\mathrm{ln}\left(\frac{V}{K}\right)-\left(\frac{\text{π}}{2}-1\right)\right)}^{2}$ in a straight line approximately

Figure 6. $\frac{K}{V}$ in a straight line approximately

${N}^{2}\left(V\right)-\left(K\phi \left(V\right)+V{V}^{\prime }\phi \left(V\right)\right)\cdot N\left(V\right)-{\phi }^{2}\left(V\right)\cdot \left({V}^{\prime }+1\right)=0$

$K$ 趋向于0时，因 $0\le V\le K$ ，所以 $V$ 也趋向于0。

$K=0,V=0,N\left(0\right)=\frac{1}{2}$ 代入上式得：

${V}^{\prime }=\frac{\text{π}}{2}-1$ ，即 $\underset{K\to 0}{\mathrm{lim}}{V}^{\prime }=\frac{\text{π}}{2}-1$

$N\left(x\right)\approx 1-\frac{\phi \left(x\right)}{x+\frac{\phi \left(g\left(x\right)\right)}{0.5+g\left(x\right)\cdot \phi \left(0.5g\left(x\right)\right)}},x\ge 0$

$g\left(x\right)=x{\text{e}}^{\frac{\text{π}}{2}-1-\sqrt{\frac{x-1}{2}+\sqrt{\text{π}}}}$

Figure 7. The error of the approximate formula two

Figure 8. The comparison between the value obtained by the approximate formula two and the exact value

$N\left(x\right)\approx 1-\frac{\phi \left(x\right)}{x+\frac{\phi \left(y\left(x\right)\right)}{0.5+y\left(x\right)\cdot \phi \left(0.5y\left(x\right)\right)}},x\ge 0$

$y\left(x\right)=\frac{x}{0.385x+\frac{2}{\text{π}-2}}$

Figure 9. The error of the approximate formula three

$N\left(x\right)\approx 1-\frac{\phi \left(x\right)}{x+\frac{1}{1.2533+0.61x}},x\ge 0$

Figure 10. The error of the approximate formula four

3. 结论

Research on the Replication of the Option of Random Walk Market and Deducing Some Formulas[J]. 应用数学进展, 2019, 08(04): 826-837. https://doi.org/10.12677/AAM.2019.84093

1. 1. 盛骤, 谢式千, 潘承毅. 概率论与数理统计[M]. 北京: 高等教育出版社, 2008.