Question

我将Blacks Scholes方程转换为Heat方程。我尝试使用显式有限差分方法来解决此PDE并获得看涨期权的价格。我也通过使用黑色学习方程式来解决这个问题＆＃34;分析＆＃34;。

问题在于我无法在数值结果中得到更准确的结果。这是我的Python代码。

以下是我的算法说明： https://drive.google.com/file/d/0B5h3oewtgjFgdVFpNFJRNTB5LXM/view?usp=sharing



import math
import numpy as np
from scipy.stats import norm

s0 = 15
sigma = 0.2
r = 0.01
t = 1
Xmax = 10

'''B-S price''' 

def C(s,k,t):
    d1 = (math.log(s/k)+(r+sigma*sigma/2)*t)/(sigma*math.sqrt(t))
    d2 = (math.log(s/k)+(r-sigma*sigma/2)*t)/(sigma*math.sqrt(t))
    return s*norm.cdf(d1)-math.exp(-r*t)*k*norm.cdf(d2)  

print('B-S',  C(s0,10,t))

'''Explicit_finite_difference'''

EFD_n_x = 500
EFD_n_t = 100
EFD_k = Xmax/EFD_n_x
EFD_h = t/EFD_n_t
EFD_xx = np.linspace(Xmax,-Xmax, 2 * EFD_n_x + 1)
EFD_xx = EFD_xx[1:2 * EFD_n_x]

def EFD_T0_Bound(x):
    return max(math.exp(x)-10*math.exp(-r*t),0) 
def EFD_U_Bound(tao):
    return math.exp(Xmax)-10*math.exp(-r*(t-tao))
def EFD_L_Bound(tao):
    return 0
EFD_T0bound = np.vectorize(EFD_T0_Bound)
EFD_lambda = EFD_h*sigma*sigma/2/EFD_k/EFD_k

EFD_A = (np.eye(2 * EFD_n_x - 1) * (1-2*EFD_lambda)
        + np.eye(2 * EFD_n_x - 1, k=1)*EFD_lambda
        + np.eye(2 * EFD_n_x - 1, k=-1)*EFD_lambda)

EFD_Y = np.zeros(2 * EFD_n_x - 1)
EFD_U = EFD_T0bound(EFD_xx)

for i in range(EFD_n_t):
    EFD_Y[0] = EFD_lambda*EFD_U_Bound(EFD_h*i)
    EFD_Y[2 * EFD_n_x - 2] = EFD_lambda*EFD_L_Bound(EFD_h*i)
    EFD_U = np.dot(EFD_A,EFD_U) + EFD_Y       #U_t_i+1 = A * U_t_i + Y

print('Explicit_finite_difference',EFD_U[EFD_n_x - 1 - round(math.log(s0)/EFD_k)])

import math import numpy as np from scipy.stats import norm s0 = 15 sigma = 0.2 r = 0.01 t = 1 Xmax = 10 '''B-S price''' def C(s,k,t): d1 = (math.log(s/k)+(r+sigma*sigma/2)*t)/(sigma*math.sqrt(t)) d2 = (math.log(s/k)+(r-sigma*sigma/2)*t)/(sigma*math.sqrt(t)) return s*norm.cdf(d1)-math.exp(-r*t)*k*norm.cdf(d2) print('B-S', C(s0,10,t)) '''Explicit_finite_difference''' EFD_n_x = 500 EFD_n_t = 100 EFD_k = Xmax/EFD_n_x EFD_h = t/EFD_n_t EFD_xx = np.linspace(Xmax,-Xmax, 2 * EFD_n_x + 1) EFD_xx = EFD_xx[1:2 * EFD_n_x] def EFD_T0_Bound(x): return max(math.exp(x)-10*math.exp(-r*t),0) def EFD_U_Bound(tao): return math.exp(Xmax)-10*math.exp(-r*(t-tao)) def EFD_L_Bound(tao): return 0 EFD_T0bound = np.vectorize(EFD_T0_Bound) EFD_lambda = EFD_h*sigma*sigma/2/EFD_k/EFD_k EFD_A = (np.eye(2 * EFD_n_x - 1) * (1-2*EFD_lambda) + np.eye(2 * EFD_n_x - 1, k=1)*EFD_lambda + np.eye(2 * EFD_n_x - 1, k=-1)*EFD_lambda) EFD_Y = np.zeros(2 * EFD_n_x - 1) EFD_U = EFD_T0bound(EFD_xx) for i in range(EFD_n_t): EFD_Y[0] = EFD_lambda*EFD_U_Bound(EFD_h*i) EFD_Y[2 * EFD_n_x - 2] = EFD_lambda*EFD_L_Bound(EFD_h*i) EFD_U = np.dot(EFD_A,EFD_U) + EFD_Y #U_t_i+1 = A * U_t_i + Y print('Explicit_finite_difference',EFD_U[EFD_n_x - 1 - round(math.log(s0)/EFD_k)])

Answer 1

在我看来，既然你正在使用一个明确的方案（这不是无条件的稳定），你就不能独立设置你的nb资产步骤和你的nb时间步骤。通常人们将它们与波动性联系在一起以保持方案稳定

Black Scholes方程的有限差分不准确

1 个答案: