西南师范大学学报(自然科学版)   2019, Vol. 44 Issue (7): 110-116.  DOI: 10.13718/j.cnki.xsxb.2019.07.017
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  • 非线性Black-Scholes模型下利差期权定价    [PDF全文]
    韩婵 1, 陈东立 2     
    1. 西安建筑科技大学 华清学院, 西安 710043;
    2. 西安建筑科技大学 理学院, 西安 710043
    摘要:研究了原生资产价格遵循非线性Black-Scholes模型时的利差期权定价问题.利用扰动理论中单参数摄动展开方法,给出了利差期权的近似定价公式.最后,结合Feyman-Kac公式分析了近似定价公式的误差估计问题,结果表明近似解一致收敛于相应期权价格的精确解.
    关键词非线性Black-Scholes模型    障碍期权    近似定价公式    误差分析    

    利差期权是一种欧式多资产期权[1-10].为了更好地描述“波动率微笑现象”, 文献[11-14]在波动率及其平方满足Lipschitz条件和线性增长条件下研究了非线性Black-Scholes模型下的期权定价问题.

    相比之下, 本文采用摄动方法研究了利差期权定价问题, 并采用Feymann-Kac公式分析了近似结论的误差估计, 结果表明, 当波动率及其平方不满足Lipschitz条件和线性增长条件时, 文献[11-14]中的误差估计结论依然成立.

    1 利差期权定价模型

    由文献[1, 15]可知, 利差期权的价格C(t, S1, S2)适定下面的抛物初值问题

    $ \left\{ \begin{gathered} {\partial _t}C + \Delta C + r{S_1}{\partial _{{S_1}}}C + r{S_2}{\partial _{{S_2}}}C - rC = 0,\left( {t,S} \right) \in \left[ {0,T} \right) \times {\mathbb{R}_ + } \hfill \\ C(T,S) = \max \left( {{S_1} - {S_2},0} \right),{S_i} \in {\mathbb{R}_ + },i = 1,2 \hfill \\ \end{gathered} \right. $ (1)

    这里0 < ε < 1,

    $ \Delta C = \frac{1}{2}{\sigma _{11}}\left( {t,{S_1},{S_2};\varepsilon } \right)S_1^2{\partial _{{S_1}{S_1}}}C + {\sigma _{12}}\left( {t,{S_1},{S_2};\varepsilon } \right){S_1}{S_2}{\partial _{{S_1}{S_2}}}C +\\ \frac{1}{2}{\sigma _{22}}\left( {t,{S_1},{S_2};\varepsilon } \right)S_1^2{\partial _{{S_2}{S_2}}}C $

    为了便于阐述问题, 定义

    $ \xi = \frac{{{S_1}}}{{{S_2}}},u\left( {t,\xi } \right) = \frac{{C\left( {t,{S_1},{S_2}} \right)}}{{{S_2}}} $ (2)
    $ {\sigma _{11}}\left( {t,{S_1},{S_2};\varepsilon } \right) = {\sigma _{11}}\left( {t,\xi ;\varepsilon } \right),{\sigma _{22}}\left( {t,{S_1},{S_2};\varepsilon } \right) =\\ {\sigma _{22}}\left( {t,\xi ;\varepsilon } \right),{\sigma _{12}}\left( {t,{S_1},{S_2};\varepsilon } \right) = {\sigma _{12}}\left( {t,\xi ;\varepsilon } \right) $

    在上述变换之下, 可以得到

    $ {\partial _{{S_1}}}C = {S_2}\frac{{\partial u}}{{\partial \xi }}\frac{{\partial \xi }}{{\partial {S_1}}} = \frac{{\partial u}}{{\partial \xi }},{\partial _{{S_2}}}C = u + {S_2}\frac{{\partial u}}{{\partial \xi }}\frac{{\partial \xi }}{{\partial {S_2}}} = u - \frac{{\partial u}}{{\partial \xi }} $
    $ {\partial _{{S_1}{S_1}}}C = \frac{1}{{{S_2}}}\frac{{{\partial ^2}u}}{{\partial {\xi ^2}}},{\partial _{{S_1}{S_2}}}C = \frac{{{\partial ^2}u}}{{\partial {\xi ^2}}}\frac{{\partial \xi }}{{\partial {S_2}}} = - \frac{\xi }{{{S_2}}}\frac{{{\partial ^2}u}}{{\partial {\xi ^2}}} $
    $ {\partial _{{S_2}{S_2}}}C = \frac{{\partial u}}{{\partial \xi }}\frac{{\partial \xi }}{{\partial {S_2}}} - \frac{\xi }{{{S_2}}}\frac{{\partial u}}{{\partial \xi }} - \xi \frac{{{\partial ^2}u}}{{\partial {\xi ^2}}}\frac{{\partial \xi }}{{\partial {S_2}}} = \frac{{{\xi ^2}}}{{{S_2}}}\frac{{{\partial ^2}u}}{{\partial {\xi ^2}}} $

    代入公式(1), 有

    $ {\partial _t}u + \frac{1}{2}\sigma \left( {t,\xi ;\varepsilon } \right){\xi ^2}{\partial _{\xi \xi }}u = 0 $ (3)

    其中σ(t, ξ; ε)=σ11(t, ξ; ε)+σ12(t, ξ; ε)+σ22(t, ξ; ε).此外在变换(2)之下, 终值条件转化为

    $ u\left( {T,\xi } \right) = \frac{1}{{{S_2}}}C\left( {T,{S_1},{S_2}} \right) = \frac{1}{{{S_2}}}\max \left( {{S_1} - {S_2},0} \right) = {\left( {\xi - 1} \right)^ + } $ (4)

    为了方便证明, 同时定义

    $ {L_0}V = {\partial _t}V + \frac{1}{2}{\sigma _0}{\partial _{xx}}V - \frac{1}{2}{\sigma _0}{\partial _x}V $ (5)
    $ {L_\varepsilon }V = {\partial _t}V + \frac{1}{2}f\left( {t,x;\varepsilon } \right)\left( {{\partial _{xx}}V - {\partial _x}V} \right) $ (6)

    其中

    $ f(t, x ; \varepsilon)=\sigma(t, \exp (x) ; \varepsilon) $

    进一步作变换

    $ x=\ln \xi, u(t, \xi)=V(t, x) $ (7)

    则抛物方程问题(1)变为

    $ \left\{\begin{array}{l}{L_{\varepsilon} V=0,(x, t) \in(0, T) \times \mathbb{R}} \\ {V(T, x)=(\exp (x)-1)+, x \in \mathbb{R}}\end{array}\right. $ (8)
    2 摄动方法

    下面考察抛物问题(8)的单参数摄动展开, 假设V(t, x)在ε=0附近可以幂级数展开

    $ V\left( {t,x} \right) = {V_0} + \varepsilon {V_1} + {\varepsilon ^2}{V_2} + \cdots + {\varepsilon ^n}{V_n} + \cdots $ (9)

    假设当ε=0时, f(t, x; ε)为常数, 即f(t, x; 0)=σ0, f(t, x; ε)可以根据泰勒公式在ε=0邻域内具备如下形式的幂级数

    $ f(t,x;\varepsilon ) = {f_0}(0,0) + \varepsilon {f_1}(t,x) + \cdots + \frac{{{\varepsilon ^n}}}{{n!}}{f_n}(t,x) + \cdots $ (10)

    其中${f_n}\left( {t, x} \right) = \frac{{{\partial ^n}f\left( {t, x;\varepsilon } \right)}}{{\partial {\varepsilon ^n}}}\left| {_{\varepsilon = 0}} \right.$.注意

    $ L_{\varepsilon} V=L_{0} V+\frac{1}{2}\left(f(t, x ; \varepsilon)-\sigma_{0}\right)\left(\partial_{x x} V-\partial_{x} V\right) $ (11)

    从而将公式(9)代入抛物问题(8), 整理之后可以得到

    $ \begin{array}{l} 0 = {\varepsilon ^0}{L_0}{V_0} + \varepsilon \left( {{L_0}{V_1} + \frac{1}{2}{f_1}\left( {t,x} \right)\left( {{\partial _{xx}}{V_0} - {\partial _x}{V_0}} \right)} \right) + \\ \;\;\;\;\;{\varepsilon ^2}\left( {{L_0}{V_2} + \frac{1}{2}{f_2}\left( {t,x} \right)\left( {{\partial _{xx}}{V_0} - {\partial _x}{V_0}} \right) + \frac{1}{2}{f_0}\left( {t,x} \right)\left( {{\partial _{xx}}{V_1} - {\partial _x}{V_1}} \right)} \right) + \cdots + \\ \;\;\;\;\;{\varepsilon ^n}\left( {{L_0}{V_n} + \frac{1}{2}\sum\limits_{k = 1}^n {\frac{1}{{k!}}{f_k}\left( {t,x} \right)\left( {{\partial _{xx}}{V_{n - k}} - {\partial _x}{V_{n - k}}} \right)} } \right) + \cdots \end{array} $

    由于xt是任意常数, 于是有

    $ L_{0} V_{0}=0 $ (12)
    $ L_{0} V_{1}+\frac{1}{2} f_{1}(t, x)\left(\partial_{x x} V_{0}-\partial_{x} V_{0}\right)=0 $ (13)
    $ L_{0} V_{n}+g_{n}(t, x)=0 $ (14)

    其中

    $ \begin{array}{l} {g_n}\left( {t,x} \right) = \frac{1}{2}\sum\limits_{k = 1}^n {\frac{1}{{k!}}{f_k}\left( {t,x} \right)\left( {{\partial _{xx}}{V_{n - k}} - {\partial _x}{V_{n - k}}} \right)} = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{1}{2}\sum\limits_{k = 1}^{n - 1} {\left( {{\partial _{xx}}{V_k} - {\partial _x}{V_k}} \right)\frac{1}{{\left( {n - k} \right)!}}{f_{n - k}}\left( {t,x} \right)} \end{array} $ (15)

    进一步, 根据公式(9), 将抛物方程(8)中初边值条件进行划分

    $ V_{0}(T, x)=(\exp (x)-1)^{+}, V_{n}(T, x)=0, n=1,2, \cdots $

    如此, 抛物型方程(8)归结为下面一系列常系数抛物方程初边值问题的线性叠加, 其中V0(t, x)适合抛物初边值问题

    $ \left\{ {\begin{array}{*{20}{c}} \begin{gathered} {L_0}{V_0} = 0 \hfill \\ {V_0}\left( {T,x} \right) = {\left( {\exp \left( x \right) - 1} \right)^ + } \hfill \\ \end{gathered} &\begin{gathered} \left( {t,x} \right) \in \left( {0,T} \right) \times \mathbb{R} \hfill \\ x \in \mathbb{R} \hfill \\ \end{gathered} \end{array}} \right. $ (16)

    Vn(t, x)为如下偏微分方程初边值的解

    $ \left\{ {\begin{array}{*{20}{c}} \begin{gathered} {L_0}{V_0} + {g_n} = 0 \hfill \\ {V_n}\left( {T,x} \right) = 0 \hfill \\ \end{gathered} &\begin{gathered} \left( {t,x} \right) \in \left( {0,T} \right) \times \mathbb{R} \hfill \\ x \in \mathbb{R} \hfill \\ \end{gathered} \end{array}} \right. $ (17)

    先处理抛物方程问题(17), 令

    $ y=x-\frac{1}{2} \sigma_{0}(T-t), s=T-t, V(t, x)=U(s, y) $ (18)

    $ \frac{\partial V(t, x)}{\partial t}=\frac{\partial U(s, y)}{\partial s} \frac{\partial s}{\partial t}+\frac{\partial U(s, y)}{\partial y} \frac{\partial y}{\partial t}=-\partial_{s} U(s, y)+\frac{1}{2} \sigma_{0} \partial_{y} U(s, y) $
    $ \frac{\partial V(t, x)}{\partial x}=\frac{\partial U(s, y)}{\partial y} \frac{\partial y}{\partial x}=\partial_{y} U(s, y), \frac{\partial^{2} V(t, x)}{\partial x^{2}}=\frac{\partial^{2} U(s, y)}{\partial y^{2}} \frac{\partial^{2} y}{\partial x^{2}}=\partial_{y y} U(s, y) $

    从而抛物方程问题(17)简化为

    $ \left\{ \begin{gathered} {\partial _s}U = \frac{1}{2}{\sigma _0}{\partial _{yy}}U\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {s,x} \right) \in \left( {0,T} \right) \times \mathbb{R} \hfill \\ U\left( {0,y} \right) = {\left( {\exp \left( y \right) - 1} \right)^ + }\;\;\;\;\;\;x \in \mathbb{R} \hfill \\ \end{gathered} \right. $

    由文献[9]可知, u(s, y)可表示为

    $ U\left( {s,y} \right) = \frac{1}{{\sqrt {\pi {\sigma _0}s} }}\int_\mathbb{R} {{{\left( {\exp \left( \rho \right) - 1} \right)}^ + }\exp \left\{ { - \frac{{{{\left( {y - \rho } \right)}^2}}}{{2{\sigma _0}s}}} \right\}{\text{d}}\rho } $

    进行多次积分换元, 可以得到

    $ U(s,y) = \exp \left( {y + \frac{1}{2}{\sigma _0}s} \right)\mathit{\Phi }\left( {\frac{{y + {\sigma _0}s}}{{\sqrt {{\sigma _0}s} }}} \right) - \mathit{\Phi }\left( {\frac{y}{{\sqrt {{\sigma _0}s} }}} \right) $ (19)

    将变换(18), (17)和(2)依次回代, 有

    $ {C_0}\left( {t,{S_1},{S_2}} \right) = {S_1}\mathit{\Phi }\left( {{a_1}} \right) - {S_2}\mathit{\Phi }\left( {{a_2}} \right) $ (20)

    其中

    $ {a_1} = \frac{{\ln {S_1} - \ln {S_2} + \frac{1}{2}{\sigma _0}(T - t)}}{{\sqrt {{\sigma _0}(T - t)} }},{a_2} = \frac{{\ln {S_1} - \ln {S_2} - \frac{1}{2}{\sigma _0}(T - t)}}{{\sqrt {{\sigma _0}(T - t)} }} $

    接下来考虑Vn(t, x).采用递推法, 有如下结论.

    引理1 依据摄动递推方法抛物初边值问题(14)的解Vn(t, x)满足

    $ {V_n}\left( {t,x} \right) = \exp \left( {\frac{1}{2}x - \frac{1}{8}{\sigma _0}\left( {T - t} \right)} \right)\int_0^{ + \infty } {\int_t^T {{{\tilde g}_n}\left( {T - s,x} \right){G_1}\left( {x,\xi ,s} \right){\text{d}}s{\text{d}}\xi } } $ (21)

    其中

    $ {\tilde g_n}\left( {\tau ,x} \right) = \exp \left( { - \frac{1}{2}x + \frac{1}{8}{\sigma _0}\tau } \right){g_n}\left( {T - \tau ,x} \right),{G_1}\left( {x,\xi ,t} \right) \\ = \frac{1}{{\sqrt {2\pi {\sigma _0}\left( {T - t} \right)} }}\exp \left( { - \frac{{{{\left( {x - \xi } \right)}^2}}}{{2{\sigma _0}\left( {T - t} \right)}}} \right) $

     令

    $ V_{n}(t, x)=\exp \left(\frac{1}{2} x-\frac{1}{8} \sigma_{0}(T-t)\right) u_{n}(t, x) $ (22)

    再令τ=Tt, 则初值问题(14)化为

    $ \left\{ \begin{gathered} {\partial _\tau }{u_n} - \frac{1}{2}{\sigma _0}{\partial _{xx}}{u_n} = {{\tilde g}_n}\left( {\tau ,x} \right)\;\;\;\;\;\;\;\;\;\left( {\tau ,x} \right) \in \left( {0,T} \right) \times \mathbb{R} \hfill \\ {u_n}\left( {\tau ,0} \right) = 0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x \in \mathbb{R} \hfill \\ \end{gathered} \right. $ (23)

    注意当完成n-1次递推之后, gn(t, x)和${\tilde g_n}\left( {\tau, x} \right)$是明确已知的, 根据文献[9], 抛物问题(23)的解可表示为

    $ {u_n}\left( {t,x} \right) = \int_0^{ + \infty } {\int_t^T {{{\tilde g}_n}\left( {T - s,x} \right){G_1}\left( {x,\xi ,s} \right){\text{d}}s{\text{d}}\xi } } $

    考虑(22)式的逆变换, 即可得结论成立.证毕.

    综合引理1和式(20), 有如下结论成立.

    定理1 利差期权的定价公式P(t, S)有如下半解析形式

    $ C\left( {t,{S_1},{S_2}} \right) = {C_0}\left( {t,{S_1},{S_2}} \right) + \varepsilon {C_1}\left( {t,{S_1},{S_2}} \right) + \cdots + {\varepsilon ^n}{C_n}\left( {t,{S_1},{S_2}} \right) + \cdots $

    其中C0(t, S1, S2)见式(19), ${C_n}\left( {t, {S_1}, {S_2}} \right) = {S_2}{V_n}\left( {t, \frac{{{S_1}}}{{{S_2}}}} \right)$, Vn(t, x)见式(21).

    3 误差估计

    f(t, x; ε)在ε=0处进行幂级数展开, 从而对任意的$\left( {t, x} \right) \in \left[{0, T} \right] \times {\mathbb{R}_ + }$

    $ \left| {f\left( {t,x;\varepsilon } \right) - \sum\limits_{i = 0}^n {\frac{{{\varepsilon ^i}}}{{i!}}{f_i}\left( {t,x} \right)} } \right| \leqslant \frac{{{M_0}}}{{\left( {n + 1} \right)!}}{\varepsilon ^{n + 1}} $ (24)

    本节假设M0是非负常数, 并在此假设下, 考察近似结论(定理1)的误差估计.

    引理2[14]  设ε为常数, ε∈(0, 1), 则对任意的$x \in {\mathbb{R}_ + }$, 有

    $ \left| {\exp \left( { - \varepsilon \ln x} \right) - 1} \right| \leqslant \varepsilon \max \left\{ {\ln x,{x^{ - 1}}} \right\} $

    引理3 设ε为常数, ε∈(0, 1), 则对任意的$x \in {\mathbb{R}_ + }$, 有

    $ {\partial _{xx}}{V_0}\left( {t,x} \right) - {\partial _x}{V_0}\left( {t,x} \right) = \frac{1}{{\sqrt {{\sigma _0}\left( {T - t} \right)} }}\exp \left( x \right)n\left( {{b_1}} \right) $

     由式(19), 可以得到V0(t, x)=exp(x)Φ(b1)-Φ(b2), 其中

    $ {b_1} = \frac{{x + \frac{1}{2}{\sigma _0}\left( {T - t} \right)}}{{\sqrt {{\sigma _0}\left( {T - t} \right)} }},{b_2} = \frac{{x - \frac{1}{2}{\sigma _0}\left( {T - t} \right)}}{{\sqrt {{\sigma _0}\left( {T - t} \right)} }} $

    从而

    $ {\partial _x}{V_0}\left( {t,x} \right) = \frac{1}{{\sqrt {{\sigma _0}\left( {T - t} \right)} }}n\left( {{b_1}} \right) + \exp \left( x \right)\mathit{\Phi }\left( {{b_1}} \right) - \frac{1}{{\sqrt {{\sigma _0}\left( {T - t} \right)} }}n\left( {{b_2}} \right) $

    显然标准正态分布的密度函数n(·)满足$n\left( x \right) = \frac{1}{{\sqrt {2\pi } }}\exp \left( {-\frac{{{x^2}}}{2}} \right)$.容易验证n(a2)=exp(x)n(a1), 从而

    $ \partial_{x} V_{0}(t, x)=\exp (x) \Phi\left(b_{1}\right) $ (25)

    进一步, 有

    $ {\partial _{xx}}{V_0}(t,x) = \exp (x)\mathit{\Phi }\left( {{a_2}} \right) + \frac{1}{{\sqrt {{\sigma _0}(T - t)} }}\exp (x)n\left( {{b_1}} \right) $ (26)

    联立式(25)和式(26), 可以得到

    $ \partial_{x x} V_{0}(t, x)-\partial_{x} V_{0}(t, x)=\frac{1}{\sqrt{\sigma_{0}(T-t)}} \exp (x) n\left(b_{1}\right) $

    证毕

    定理2 设ε为足够小的正常数, 则存在不依赖时间t的正常数M, 使得

    $ \left|C\left(t, S_{1}, S_{2}\right)-C_{0}\left(t, S_{1}, S_{2}\right)\right| \leqslant M \varepsilon $

     由变换(2)和变换(7)易知, 只需证明

    $ \left|V(t, x)-V_{0}(t, x)\right| \leqslant M \varepsilon $

    成立.令E0(t, x)=V0(t, x)-V(t, x), 则

    $ E_{0}(T, x)=0 $ (27)

    考虑到LεV(t, x)=0, 从而由式(5)可得

    $ {L_\varepsilon }{E_0}(t,x) = \frac{1}{2}\left[ {f(t,x;\varepsilon ) - {\sigma _0}} \right]\left( {{\partial _{xx}}{V_0} - {\partial _x}{V_0}} \right) $ (28)

    故将式(28)代入式(27), 可以得到

    $ \left\{ {\begin{array}{*{20}{c}} \begin{gathered} {L_\varepsilon }{E_0} = {h_1}\left( {t,x} \right) \hfill \\ {E_0}\left( {T,x} \right) = 0 \hfill \\ \end{gathered} &\begin{gathered} \left( {t,x} \right) \in \left( {0,T} \right) \times \mathbb{R} \hfill \\ x \in \mathbb{R} \hfill \\ \end{gathered} \end{array}} \right. $ (29)

    其中${h_1}\left( {t, x} \right) = \frac{1}{2}\left[{f\left( {t, x;\varepsilon } \right)-{\sigma _0}} \right]\left( {{\partial _{xx}}{V_0} -{\partial _x}{V_0}} \right)$.

    另一方面利用Feynman-Kac公式[16], 抛物初边值问题(29)满足

    $ \overline{E}_{0}(t, x)=E^{Q, x}\left[\int_{t}^{T} h_{1}\left(t, \ln X_{t}\right) \mathrm{d} \tau\right] $ (30)

    其中exp(-r(Tt))XtL鞅, 注意$n\left( {{d_1}} \right) \le \sqrt {2\pi } $, 从而

    $ {\bar E_0}(t,x) \leqslant \frac{1}{{\sqrt {2\pi } }}{E^{Q,x}}\left[ {\int_t^T {\frac{1}{{\sqrt {T - \tau } }}} {X_\tau }\left( {\exp \left( { - \varepsilon \ln {X_\tau }} \right) - 1} \right){\text{d}}\tau } \right] $

    利用引理1, 有

    $ \begin{array}{l} {{\bar E}_0}(t,x) \le \frac{1}{{\sqrt {2\pi } }}{E^{Q,x}}\left[ {\int_t^T {\frac{1}{{\sqrt {T - \tau } }}} {X_\tau }\varepsilon \max \left\{ {\ln {X_\tau },X_\tau ^{ - 1}} \right\}{\rm{d}}\tau } \right]\\ \;\;\;\;\;\;\;\;\;\;\;\;\; \le \frac{1}{{\sqrt {2\pi } }}{E^{Q,x}}\left[ {\int_t^T {\frac{1}{{\sqrt {T - \tau } }}} \varepsilon \max \left\{ {X_\tau ^2,1} \right\}{\rm{d}}\tau } \right] \end{array} $

    注意瑕积分$\int_0^T {{{\left( {T-\tau } \right)}^{-\frac{1}{2}}}{\rm{d}}\tau } $是收敛的, 因此定义$X_T^* = \mathop {\max }\limits_{0 \le \tau \le T} X_\tau ^2$, 上式变成

    $ \overline{E}_{0}(t, x) \leqslant \varepsilon M_{1}\left(E^{Q, x}\left[\left(X_{T}^{*}\right)^{2}\right]+1\right) $

    考虑到exp(-r(Tt))Xt是鞅, 易见|Xt|是下鞅, 从而由Doob不等式, 可得

    $ \overline{E}_{0}(t, x) \leqslant \varepsilon M_{1}\left(E^{Q, x}\left[X_{T}^{2}\right]+1\right) $

    证毕

    定理3 对任意的正整数n存在正常数M不依赖nt, 使得

    $ \left|C\left(t, S_{1}, S_{2}\right)-\sum\limits_{i=0}^{i=n} \varepsilon^{i} C_{i}\left(t, S_{1}, S_{2}\right)\right| \leqslant M \varepsilon^{n+1} $

     类推定理2的证明过程, 只需证明|En(t, x)|≤n+1, 其中

    $ E_{n}(t, x)=V^{n}(t, x)-V(t, x), V^{n}(t, x)=\sum\limits_{i=0}^{i=n} \varepsilon^{i} V_{i}(t, x) $

    注意En(t, x)满足初边值条件En(t, 0)=0, En(T, x)=0, 并且

    $ L_{\varepsilon} E_{n}(t, x)=L_{0} V^{n}+\frac{1}{2}\left[f(t, x ; \varepsilon)-\sigma_{0}\right]\left(\partial_{x x} V^{n}-\partial_{x} V^{n}\right)=h_{n}(x, t) $

    这里

    $ {h_n}(x,t) = - \frac{1}{2}\sum\limits_{i = 1}^n {{\varepsilon ^i}} \sum\limits_{k = 0}^{i - 1} {\frac{1}{{(i - k)!}}} {f_{i - k}}(t,x)\left( {{\partial _{xx}}{V_k} - {\partial _x}{V_k}} \right) +\\ \frac{1}{2}\left[ {{f^{d.o}}(t,x;\varepsilon ) - {\sigma _0}} \right]\left( {{\partial _{xx}}{V^n} - {\partial _x}{V^n}} \right) $

    从而

    $ \left\{\begin{array}{ll}{L_{\varepsilon} E_{n}(t, x)=h_{n}(x, t)} & {(t, x) \in(0, T) \times \mathbb{R}} \\ {E_{n}(T, x)=0} & {x \in \mathbb{R}}\end{array}\right. $ (31)

    一方面, 类似文献[14]之推导过程, hn(x, t)满足

    $ h_{n}(x, t)=\frac{1}{2} \sum\limits_{k=0}^{n-1} \varepsilon^{k} \mathbb{R}_{k}(t, x)\left[f(t, x ; \varepsilon)-\sum\limits_{i=0}^{n-k} \frac{1}{i !} f_{i}(t, x)\right] $

    利用式(24), 对任意的$\left( {t, x} \right) \in \left[{0, T} \right] \times \mathbb{R}$, 有

    $ \left|f^{d . o}(t, x ; \varepsilon)-\sum\limits_{i=0}^{k} \frac{1}{i !} f_{i}(t, x)\right| \leqslant \frac{M_{0}}{(k+1) !} \varepsilon^{k+1} $ (32)

    另一方面, 利用Feynman-Kac公式[16]可知,

    $ E_{n}(t, x)=E^{Q, x}\left[\int_{t}^{\tau_{x}} \sum\limits_{k=0}^{n-1} h_{n}\left(\tau, \ln X_{\tau}\right) \mathrm{d} \tau\right] $ (33)

    因此令$M = \mathop {\sup }\limits_{k, t, x} {\mathbb{R}_k}\left( {t, x} \right)$, 并联立式(32)和式(33),

    $ \left|E_{n}(t, x)\right| \leqslant \frac{1}{2} M_{0} M \varepsilon^{n+1} \sum\limits_{k=0}^{n-1} \frac{1}{(n-k+1) !} \leqslant \frac{1}{2} \mathrm{e} M_{0} M \varepsilon^{n+1} $

    定理得证.

    由定理3可以看出, $\sum\limits_{i = 0}^{i = n} {{\varepsilon ^i}{C^i}\left( {t, {S_1}, {S_2}} \right)} $在定解区域$\left[{0, T} \right] \times {\mathbb{R}_ + }$上一致收敛到C(t, S1, S2).此外, 定理2和定理3的证明并不要求抛物方程(1)的初值条件是光滑的.

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    Performance Options' Pricing Under Nonlinear Black-Scholes Model
    HAN Chan 1, CHEN Dong-li 2     
    1. Huaqing College, Xi'an University of Architecture and Technology, Xi'an 710043, China;
    2. School of Science, Xi'an University of Architecture and Technologe, Xi'an 710043, China
    Abstract: In this text, the pricing problems of performance options are discussed under the condition that the price of underlying asset follows the nonlinear Black-Scholes model. The author uses the perturbation method of single-parameter to obtain asymptomatic formulae of performance options pricing problems. Finally, error estimates of these asymptotic solutions are illustrated by using the Feymann-Kac formula in which the results indicate that the asymptotic solutions uniformly converges to its exact solutions.
    Key words: nonlinear Black-Scholes model    barrier options    asymptomatic pricing formulae    error estimates    
    X