量化分析师的Python日记【Q Quant兵器谱之偏微分方程3的具体金融学运用】

欢迎来到
Black - Scholes — Merton
的世界！本篇中我们将把第11天学习到的知识应用到这个金融学的具体方程之上！

import numpy as np
import math
import seaborn as sns
from matplotlib import pylab
from CAL.PyCAL import *
font.set_size(15)

1. 问题的提出

BSM模型可以设置为如下的偏微分方差初值问题：

{∂C(S,t)∂t+rS∂C(S,t)∂S+12σ2S2∂2C(S,t)∂S2−rC(S,t)=0,0≤t≤T\[5pt]C(S,T)=max(S−K,0)

做变量替换τ=T−t，并且设置上下边界条件:

⎧⎩⎨⎪⎪⎪⎪∂C(x,τ)∂τ\[5pt]C(x,0)\[5pt]C(xmax,τ)\[5pt]C(xmin,τ)=(r−12σ2)∂C(x,τ)∂x+12σ2∂2C(x,τ)∂x2−rC(x,τ)=0,0≤τ≤T=max(ex−K,0),=exmax−Ke−τ,=0

2. 算法

按照之前介绍的隐式差分格式的方法，用离散差分格式代替连续微分：

\begin{align}&\frac{C_{j,k+1} - C_{j,k}}{\Delta \tau} = (r - \frac{1}{2}\sigma^2)\frac{C_{j+1,k+1} - C_{j-1,k+1}}{2\Delta x} + \frac{1}{2}\sigma^2 \frac{C_{j+1,k+1} - 2C_{j,k+1} + C_{j-1,k+1}}{\Delta x^2} - rC_{j,k+1}, \\\[5pt]\Rightarrow& \quad
C_{j,k+1} - C_{j,k} - (r - \frac{1}{2}\sigma^2)\frac{\Delta \tau}{2\Delta x}(C_{j+1,k+1} - C_{j-1,k+1}) - \frac{1}{2}\sigma^2 \frac{\Delta \tau}{\Delta x^2}(C_{j+1,k+1} - 2C_{j,k+1} + C_{j-1,k+1}) + r\Delta \tau C_{j,k+1} = 0, \\\[5pt]\Rightarrow& \quad -
(\frac{1}{2}(r - \frac{1}{2}\sigma^2)\frac{\Delta \tau}{\Delta x} + \frac{1}{2}\sigma^2\frac{\Delta \tau}{\Delta x^2})C_{j+1,k+1} + (1 + \sigma^2\frac{\Delta \tau}{\Delta x^2} + r\Delta \tau)C_{j,k+1} - (\frac{1}{2}\sigma^2\frac{\Delta \tau}{\Delta x^2} -
\frac{1}{2}(r - \frac{1}{2}\sigma^2 )\frac{\Delta \tau}{\Delta x})C_{j-1,k+1} = C_{j,k}, \\\[5pt]\Rightarrow& \quad l_j C_{j-1,k+1} + c_j C_{j,k+1} + u_j C_{j+1,k+1} = C_{j,k}\end{align}

其中\begin{align}l_j &= - (\frac{1}{2}\sigma^2\frac{\Delta \tau}{\Delta x^2} - \frac{1}{2}(r - \frac{1}{2}\sigma^2 )\frac{\Delta \tau}{\Delta x}), \\\[5pt]c_j &= 1 + \sigma^2\frac{\Delta \tau}{\Delta x^2} + r\Delta \tau, \\\[5pt]u_j &= - (\frac{1}{2}(r
- \frac{1}{2}\sigma^2)\frac{\Delta \tau}{\Delta x} + \frac{1}{2}\sigma^2\frac{\Delta \tau}{\Delta x^2})\end{align}

以上即为差分方程组。

这里有些细节需要处理，就是左右边界条件，我们这里使用Dirichlet边界条件，则：

\begin{align*}C_{0,k} = C(x_{min},\tau), \\\[5pt]C_{N,k} = C(x_{max},\tau)

\end{align*}

3.实现

import scipy as sp
from scipy.linalg import solve_banded

描述BSM方程结构的类：BSModel

class BSMModel:

def __init__(self, s0, r, sigma):
self.s0 = s0
self.x0 = math.log(s0)
self.r = r
self.sigma = sigma

def log_expectation(self, T):
return self.x0 + (self.r - 0.5 * self.sigma * self.sigma) * T

def expectation(self, T):
return math.exp(self.log_expectation(T))

def x_max(self, T):
return self.log_expectation(T) + 4.0 * self.sigma * math.sqrt(T)

def x_min(self, T):
return self.log_expectation(T) - 4.0 * self.sigma * math.sqrt(T)

描述我们这里设计到的产品的类：

CallOption

class CallOption:

def __init__(self, strike):
self.k = strike

def ic(self, spot):
return max(spot - self.k, 0.0)

def bcl(self, spot, tau, model):
return 0.0

def bcr(self, spot, tau, model):
return spot - math.exp(-model.r*tau) * self.k

完整的隐式格式：

BSMScheme

class BSMScheme:
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def __init__(self, model, payoff, T, M, N):
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self.model = model
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self.T = T
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self.M = M
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self.N = N
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self.dt = self.T / self.M
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self.payoff = payoff
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self.x_min = model.x_min(self.T)
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self.x_max = model.x_max(self.T)
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self.dx = (self.x_max - self.x_min) / self.N
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self.C = np.zeros((self.N+1, self.M+1)) # 全部网格
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self.xArray = np.linspace(self.x_min, self.x_max, self.N+1)
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self.C[:,0] = map(self.payoff.ic, np.exp(self.xArray))
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sigma_square = self.model.sigma*self.model.sigma
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r = self.model.r
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self.l_j = -(0.5*sigma_square*self.dt/self.dx/self.dx - 0.5 * (r - 0.5 * sigma_square)*self.dt/self.dx)
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self.c_j = 1.0 + sigma_square*self.dt/self.dx/self.dx + r*self.dt
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self.u_j = -(0.5*sigma_square*self.dt/self.dx/self.dx + 0.5 * (r - 0.5 * sigma_square)*self.dt/self.dx)
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def roll_back(self):
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for k in range(0, self.M):
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udiag = np.ones(self.N-1) * self.u_j
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ldiag =  np.ones(self.N-1) * self.l_j
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cdiag =  np.ones(self.N-1) * self.c_j
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mat = np.zeros((3,self.N-1))
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mat[0,:] = udiag
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mat[1,:] = cdiag
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mat[2,:] = ldiag
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rhs = np.copy(self.C[1:self.N,k])
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# 应用左端边值条件
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v1 = self.payoff.bcl(math.exp(self.x_min), (k+1)*self.dt, self.model)
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rhs[0] -= self.l_j * v1
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# 应用右端边值条件
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v2 = self.payoff.bcr(math.exp(self.x_max), (k+1)*self.dt, self.model)
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rhs[-1] -= self.u_j * v2
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x = solve_banded((1,1), mat, rhs)
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self.C[1:self.N, k+1] = x
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self.C[0][k+1] = v1
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self.C[self.N][k+1] = v2
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def mesh_grids(self):
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tArray = np.linspace(0, self.T, self.M+1)
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tGrids, xGrids = np.meshgrid(tArray, self.xArray)
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return tGrids, xGrids

应用在一起：

model = BSMModel(100.0, 0.05, 0.2)
payoff = CallOption(105.0)
scheme = BSMScheme(model, payoff, 5.0, 100, 300)

scheme.roll_back()

from matplotlib import pylab
pylab.figure(figsize=(12,8))
pylab.plot(np.exp(scheme.xArray)[50:170], scheme.C[50:170,-1])
pylab.xlabel('$S$')
pylab.ylabel('$C$')

4. 收敛性测试

首先使用BSM模型的解析解获得精确解：

analyticPrice = BSMPrice(1, 105., 100., 0.05, 0.0, 0.2, 5.)
analyticPrice



我们固定时间方向网格数为3000，分别计算不同S网格数情形下的结果：

xSteps = range(50,300,10)
finiteResult = []
for xStep in xSteps:
model = BSMModel(100.0, 0.05, 0.2)
payoff = CallOption(105.0)
scheme = BSMScheme(model, payoff, 5.0, 3000, xStep)
scheme.roll_back()

interp = CubicNaturalSpline(np.exp(scheme.xArray), scheme.C[:,-1])
price = interp(100.0)
finiteResult.append(price)
我们可以画下收敛图：

anyRes = [analyticPrice['price'][1]] * len(xSteps)

pylab.figure(figsize = (16,8))
pylab.plot(xSteps, finiteResult, '-.', marker = 'o', markersize = 10)
pylab.plot(xSteps, anyRes, '--')
pylab.legend([u'隐式差分格式', u'解析解（欧式）'], prop = font)
pylab.xlabel(u'价格方向网格步数', fontproperties = font)
pylab.title(u'Black - Scholes - Merton 有限差分法', fontproperties = font, fontsize = 20)