﻿ 跳扩散模型下亚式期权定价的柳树法研究
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 同济大学学报(自然科学版)  2018, Vol. 46 Issue (12): 1761-1769.  DOI: 10.11908/j.issn.0253-374x.2018.12.020 0

### 引用本文

YAO Yi, LI Shuaifang, XU Wei. Efficient Willow Tree Method for Asian Option Pricing Under Merton Jump-Diffusion Model[J]. Journal of Tongji University (Natural Science), 2018, 46(12): 1761-1769. DOI: 10.11908/j.issn.0253-374x.2018.12.020

### 文章历史

Efficient Willow Tree Method for Asian Option Pricing Under Merton Jump-Diffusion Model
YAO Yi , LI Shuaifang , XU Wei
School of Mathematical Sciences, Tongji University, Shanghai 200092, China
Abstract: The logarithm of the stock price is described as a combination of a Brownian motion and a compound Poisson process in the jump diffusion model proposed by Merton, which can capture the negative skewness and high kurtosis of stock returns observed from the financial market. However, existing methods for the Asian option pricing under the jump diffusion model is quite expensive. Thus, an efficient and accurate willow tree method is proposed in this paper and its theoretical convergence is analyzed. Besides, some numerical experiments are conducted to demonstrate the efficiency and accuracy of the proposed method.
Key words: Asian option    jump-diffusion model    willow tree method    Monte Carlo method

 图 1 包含5个时刻、5个资产价格节点的柳树结构图 Fig.1 Graphical depiction of the willow tree lattice with 5 space nodes and 5 time nodes

1 跳扩散模型下亚式期权的柳树法定价

 $\frac{{{\rm{d}}{S_t}}}{{{S_t}}} = \left( {r - \rho - \lambda k} \right){\rm{d}}t + \sigma {\rm{d}}{W_t} + \left( {y - 1} \right){\rm{d}}{N_t}$ (1)

1.1 资产价格估计

 $\begin{array}{*{20}{c}} {X = \left( {r - \rho - \lambda \kappa - \frac{1}{2}{\sigma ^2}} \right)t + \sigma \sqrt t Z + \sum\limits_{i = 1}^{{N_t}} {\ln {y_i}} ,}\\ {Z \sim N\left( {0,1} \right),\;\;\;\ln y \sim N\left( {{\alpha _{\rm{J}}},\sigma _{\rm{J}}^2} \right)} \end{array}$

 $\left\{ \begin{array}{l} \mu = \left[ {r - {\sigma ^2}/2 - \lambda \kappa - \rho + \lambda {\alpha _{\rm{J}}}} \right]t\\ \upsilon = \left( {{\sigma ^2} + \lambda \alpha _{\rm{J}}^2 - \lambda \sigma _{\rm{J}}^2} \right)t\\ {\kappa _3} = \frac{1}{{\sqrt t }}\left( {\frac{{\lambda \left( {\alpha _{\rm{J}}^3 + 3{\alpha _{\rm{J}}}\sigma _{\rm{J}}^2} \right)}}{{{{\left( {{\sigma ^2} + \lambda \alpha _{\rm{J}}^2 - \lambda \sigma _{\rm{J}}^2} \right)}^{3/2}}}}} \right)\\ {\kappa _4} = 3 + \frac{1}{t}\left( {\frac{{\lambda \left( {\alpha _{\rm{J}}^4 + 6\alpha _{\rm{J}}^2\sigma _{\rm{J}}^2 + 3\sigma _{\rm{J}}^4} \right)}}{{{{\left( {{\sigma ^2} + \lambda \alpha _{\rm{J}}^2 + \lambda \sigma _{\rm{J}}^2} \right)}^2}}}} \right) \end{array} \right.$ (2)

Johnson提出的方法[13]可以将任意连续随机变量X转换成正态分布的随机变量Z.其主要原理是通过计算已知变量的四阶矩，然后代入统一的公式中估计该变量的离散值.该模型可以灵活地匹配任意变量的期望、方差、偏度和峰度，并且根据偏度和峰度便可唯一确定模型中所需函数的具体类型.Johnson曲线公式如下：

 $Z = a + bg\left( {\frac{{X - c}}{d}} \right)$

 $X = c + d \cdot {g^{ - 1}}\left( {\frac{{Z - a}}{b}} \right)$ (3)

1.2 转移概率计算

 ${p_{\rm{J}}}\left( {\ln {S_{t + \Delta t}}\left| {\ln {S_t}} \right.} \right) = \sum\limits_{d = 0}^\infty {\frac{{{{\rm{e}}^{ - \lambda \Delta t}}{{\left( {\lambda \Delta t} \right)}^d}}}{{d!}}{\varphi _N}\left( {{\mu _d},\sigma _d^2} \right)}$

 ${\mu _d} = \ln {S_t} + \left( {r - \rho - \lambda k - 0.5{\sigma ^2}} \right)\Delta t + d{\alpha _{\rm{J}}}$
 $\sigma _d^2 = {\sigma ^2}\Delta t + d\sigma _{\rm{J}}^2$

 $\begin{array}{l} {P_{\rm{J}}}\left( {\ln {S_{t + \Delta t}} \le A\left| {\ln {S_t}} \right.} \right) = \\ \;\;\;\;\;\;\;\sum\limits_{d = 0}^\infty {\frac{{{{\rm{e}}^{ - \lambda \Delta t}}{{\left( {\lambda \Delta t} \right)}^d}}}{{d!}}N\left( {\frac{{A - {\mu _d}}}{{{\sigma _d}}}} \right)} \end{array}$

 $\begin{array}{l} p_{ij}^t = {P_{\rm{J}}}\left( {\ln S_j^{t + \Delta t} \le B\left| {\ln S_i^t} \right.} \right) - \\ \;\;\;\;\;\;\;{P_{\rm{J}}}\left( {\ln S_j^{t + \Delta t} \le A\left| {\ln S_i^t} \right.} \right), \end{array}$
 $i = 1,2, \cdots ,m,\;\;\;j = 1,2, \cdots ,m$

(1) $p_{ij}^t \ge 0$

(2) $\sum\limits_{j = 1}^m {p_{ij}^t = 1}$

(3) ${{\rm{e}}^{ - \left( {r - p} \right)\Delta t}}\;\sum\limits_{j = 1}^m {p_{ij}^tS_j^{t + \Delta t} = S_i^t}$.

 $\left\{ \begin{array}{l} E\left[ \eta \right] = \int_{ - \infty }^\infty {\eta f\left( \eta \right){\rm{d}}\eta } = {\alpha _{\rm{J}}}\\ E\left[ {{\eta ^2}} \right] = \int_{ - \infty }^\infty {{\eta ^2}f\left( \eta \right){\rm{d}}\eta } = \alpha _{\rm{J}}^2 + \sigma _{\rm{J}}^2\\ E\left[ {{\eta ^3}} \right] = \int_{ - \infty }^\infty {{\eta ^3}f\left( \eta \right){\rm{d}}\eta } = 3{\alpha _{\rm{J}}}\sigma _{\rm{J}}^2 + \alpha _{\rm{J}}^3\\ E\left[ {{\eta ^4}} \right] = \int_{ - \infty }^\infty {{\eta ^4}f\left( \eta \right){\rm{d}}\eta } = \alpha _{\rm{J}}^4 + 6\alpha _{\rm{J}}^2\sigma _{\rm{J}}^2 + 3\alpha _{\rm{J}}^4\\ E\left[ {{\eta ^5}} \right] = \int_{ - \infty }^\infty {{\eta ^5}f\left( \eta \right){\rm{d}}\eta } = \alpha _{\rm{J}}^5 + 10\alpha _{\rm{J}}^3\sigma _{\rm{J}}^2 + 15{\alpha _{\rm{J}}}\sigma _{\rm{J}}^4 \end{array} \right.$

${\tilde V}$(Xi+η, tn)在(Xi, tn)处四阶泰勒展开得到

 $\begin{array}{*{20}{c}} {\tilde V\left( {{X_i} + \eta ,{t_n}} \right) = \tilde V\left( {{X_i},{t_n}} \right) + {{\tilde V}_X}\eta + \frac{1}{2}{{\tilde V}_{XX}}{\eta ^2} + }\\ {\frac{1}{6}{{\tilde V}_{XXX}}{\eta ^3} + \frac{1}{{24}}{{\tilde V}_{XXXX}}{\eta ^4} + {R_4}\left( \xi \right)} \end{array}$

 $\begin{array}{l} \int_{ - \infty }^\infty {\tilde V\left( {{X_i} + \eta ,{t_n}} \right)f\left( \eta \right){\rm{d}}\eta } = \int_{ - \infty }^\infty {\left[ {\tilde V\left( {x,t} \right) + } \right.} \\ \;\;\;\;\;\;{{\tilde V}_X}\eta + \frac{1}{2}{{\tilde V}_{XX}}{\eta ^2} + \frac{1}{6}{{\tilde V}_{XXX}}{\eta ^3} + \\ \;\;\;\;\;\;\left. {\frac{1}{{24}}{{\tilde V}_{XXXX}}{\eta ^4}} \right]f\left( \eta \right){\rm{d}}\eta = \tilde V\left( {x,t} \right) + {\alpha _{\rm{J}}}{{\tilde V}_X} + \\ \;\;\;\;\;\;\frac{1}{2}\left( {\alpha _{\rm{J}}^2 + \sigma _{\rm{J}}^2} \right){{\tilde V}_{XX}} + \frac{1}{6}\left( {\alpha _{\rm{J}}^3 + 3{\alpha _{\rm{J}}}\sigma _{\rm{J}}^2} \right){{\tilde V}_{XXX}} + \\ \;\;\;\;\;\;\frac{1}{{24}}\left( {\alpha _{\rm{J}}^4 + 6\alpha _{\rm{J}}^2\sigma _{\rm{J}}^2 + 3\sigma _{\rm{J}}^4} \right){{\tilde V}_{XXXX}} + \\ \;\;\;\;\;\;\int_{ - \infty }^\infty {{R_4}\left( \xi \right)f\left( \eta \right){\rm{d}}\eta } \end{array}$ (15)

 $\begin{array}{*{20}{c}} {\int_{ - \infty }^\infty {{R_4}\left( \xi \right)f\left( \eta \right){\rm{d}}\eta } = \int_{ - \infty }^\infty {{{\tilde V}^{\left( 5 \right)}}\left( \xi \right)\frac{{{\eta ^5}}}{{5!}}f\left( \eta \right){\rm{d}}\eta } = }\\ {\frac{1}{{120}}{{\tilde V}^{\left( 5 \right)}}\left( \xi \right)\left( {\alpha _{\rm{J}}^5 + 10\alpha _{\rm{J}}^3\sigma _{\rm{J}}^2 + 15{\alpha _{\rm{J}}}\sigma _{\rm{J}}^4} \right)} \end{array}$

ξ∈(Xi, Xi+η).假设| ${\tilde V}$(5)(ξ)|≤M1M1为常数.则

 $\begin{array}{*{20}{c}} {\lambda \int_{ - \infty }^\infty {{R_4}\left( \xi \right)f\left( \eta \right){\rm{d}}\eta } = O\left( {\alpha _{\rm{J}}^5 + 10\alpha _{\rm{J}}^3\sigma _{\rm{J}}^2 + } \right.}\\ {\left. {15{\alpha _{\rm{J}}}\sigma _{\rm{J}}^4} \right) = O\left( {E\left[ {{\eta ^5}} \right]} \right)} \end{array}$ (16)

 $\begin{array}{l} \tilde V_i^n = \tilde V_i^n + \left\{ {{{\tilde V}_t} + \frac{{{\sigma ^2}}}{2}{{\tilde V}_{XX}} + \left( {r - \rho - \lambda \kappa } \right){{\tilde V}_X} - } \right.\\ \;\;\;\;\;\;\;\;\left. {\left( {r + \lambda } \right)\tilde V + \lambda \int_{ - \infty }^\infty {\tilde V\left( {{X_i} + \eta ,{t_n}} \right)f\left( \eta \right){\rm{d}}\eta } } \right\}\Delta t - \\ \;\;\;\;\;\;\;\;\lambda \int_{ - \infty }^\infty {{R_4}\left( \xi \right)f\left( \eta \right){\rm{d}}\eta } \cdot \Delta t + \varepsilon \Delta t + O\left( {\Delta {t^2}} \right) \end{array}$

 $\begin{array}{*{20}{c}} {{{\tilde V}_t} + \frac{{{\sigma ^2}}}{2}{{\tilde V}_{XX}} + \left( {r - \rho - \lambda \kappa } \right){{\tilde V}_X} - \left( {r + \lambda } \right)\tilde V + }\\ {\lambda \int_{ - \infty }^\infty {\tilde V\left( {{X_i} + \eta ,{t_n}} \right)f\left( \eta \right){\rm{d}}\eta } + \varepsilon + O\left( {E\left[ {{\eta ^5}} \right]} \right) = 0} \end{array}$

2.2 柳树法的插值误差

 $\bar V_{jk}^{n + 1} = \alpha _{jk}^{n + 1}V_{j{k^ + }}^{n + 1} + \left( {1 - \alpha _{jk}^{n + 1}} \right)V_{j{k^ - }}^{n + 1}$

Vjkn+1的插值误差βjkn+1

 $\begin{array}{l} \beta _{jk}^{n + 1} = \left| {V_{j\bar k}^{n + 1} - \bar V_{jk}^{n + 1}} \right| = \\ \;\;\;\;\;\;\;\;\;\;\frac{{\left( {\bar A_{jk}^{n + 1} - A_{{k^ - }}^{n + 1}} \right)\left( {\bar A_{jk}^{n + 1} - A_{{k^ + }}^{n + 1}} \right)}}{2}\frac{{{\partial ^2}V_{j\bar k}^{n + 1}\left( \eta \right)}}{{\partial {A^2}}} \end{array}$ (17)

 $\left| {\beta _{jk}^{n + 1}} \right| \le \left| {\left( {\bar A_{jk}^{n + 1} - A_{{k^ - }}^{n + 1}} \right)\left( {\bar A_{jk}^{n + 1} - A_{{k^ + }}^{n + 1}} \right)} \right| \cdot {M_2}$

 $\left\{ \begin{array}{l} A_{\min }^{n + 1} = \left( {{S^0} + \sum\limits_{j = 1}^{n + 1} {S_1^j} } \right)/\left( {n + 2} \right)\\ A_{\max }^{n + 1} = \left( {{S^0} + \sum\limits_{j = 1}^{n + 1} {S_m^j} } \right)/\left( {n + 2} \right) \end{array} \right.$

 $\begin{array}{l} \bar h = \frac{{A_{\max }^{n + 1} - A_{\min }^{n + 1}}}{{{k_{\rm{a}}} - 1}} = \frac{{\sum\limits_{j = 1}^{n + 1} {\left( {S_m^j - S_1^j} \right)} }}{{\left( {n + 2} \right)\left( {{k_{\rm{a}}} - 1} \right)}} < \\ \;\;\;\;\;\;\frac{{n + 1}}{{\left( {n + 2} \right)\left( {{k_{\rm{a}}} - 1} \right)}}H \end{array}$

 $\left| {\beta _{jk}^{n + 1}} \right| \le M \cdot {{\bar h}^2} < M{H^2}\frac{{{{\left( {n + 1} \right)}^2}}}{{{{\left( {n + 2} \right)}^2}{{\left( {{k_{\rm{a}}} - 1} \right)}^2}}}$ (18)

 $\begin{array}{l} V_{ik}^n = {{\rm{e}}^{ - r\Delta t}}\sum\limits_{j = 1}^m {p_{ij}^n\left[ {\alpha _{jk}^{n + 1}V_{j{k^ + }}^{n + 1} + } \right.} \\ \;\;\;\;\;\;\;\left. {\left( {1 - \alpha _{jk}^{n + 1}} \right)V_{j{k^ - }}^{n + 1}} \right] + {i_{{\rm{err}}}} + {t_{{\rm{err}}}} \end{array}$

 $\begin{array}{l} E_{ik}^n = {{\rm{e}}^{ - r\Delta t}}\sum\limits_{j = 1}^m {p_{ij}^n\left[ {\alpha _{jk}^{n + 1}E_{j{k^ + }}^{n + 1} + } \right.} \\ \;\;\;\;\;\;\;\left. {\left( {1 - \alpha _{jk}^{n + 1}} \right)E_{j{k^ - }}^{n + 1}} \right] + {i_{{\rm{err}}}} + {t_{{\rm{err}}}} \end{array}$

 $\begin{array}{l} \left\| {{E^n}} \right\| \le {{\rm{e}}^{ - r\Delta t}}\sum\limits_{j = 1}^m {p_{ij}^n\left[ {\alpha _{jk}^{n + 1} + \left( {1 - \alpha _{jk}^{n + 1}} \right)} \right]} \cdot \\ \left\| {{E^{n + 1}}} \right\| + {i_{{\rm{err}}}} + {t_{{\rm{err}}}} \le \left\| {{E^{n + 1}}} \right\| + {i_{{\rm{err}}}} + {t_{{\rm{err}}}} \end{array}$

 $\begin{array}{l} {\left\| {{E^0}} \right\|^I} \le {{\rm{e}}^{ - r\Delta t}}\sum\limits_{n = 0}^{N - 1} {\sum\limits_{j = 1}^m {p_{ij}^n \cdot \mathop {\max }\limits_k \left| {\beta _{jk}^{n + 1}} \right|} } \le \\ \;\;\;\;\;\;\;\;\;\;\;{{\rm{e}}^{ - r\Delta t}}{M_2}{H^2}\sum\limits_{n = 0}^{N - 1} {\frac{{{{\left( {n + 1} \right)}^2}}}{{{{\left( {n + 2} \right)}^2}{{\left( {{k_{\rm{a}}} - 1} \right)}^2}}}} \le \\ \;\;\;\;\;\;\;\;\;\;\;{M_2}{H^2}{T^2}\frac{1}{{{{\left( {{k_{\rm{a}}} - 1} \right)}^2} \cdot \Delta t}} = O\left( {\frac{1}{{k_{\rm{a}}^2 \cdot \Delta t}}} \right) \end{array}$

3 Merton跳扩散模型下亚式期权定价的数值实验

4 总结

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