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 npimport mathimport seaborn as snsfrom matplotlib import pylabfrom 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 spfrom 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:2    def __init__(self, model, payoff, T, M, N):3        self.model = model4        self.T = T5        self.M = M6        self.N = N7        self.dt = self.T / self.M8        self.payoff = payoff9        self.x_min = model.x_min(self.T)10        self.x_max = model.x_max(self.T)11        self.dx = (self.x_max - self.x_min) / self.N12        self.C = np.zeros((self.N+1, self.M+1)) # 全部网格13        self.xArray = np.linspace(self.x_min, self.x_max, self.N+1)14        self.C[:,0] = map(self.payoff.ic, np.exp(self.xArray))15        16        sigma_square = self.model.sigma*self.model.sigma17        r = self.model.r18        19        self.l_j = -(0.5*sigma_square*self.dt/self.dx/self.dx - 0.5 * (r - 0.5 * sigma_square)*self.dt/self.dx)20        self.c_j = 1.0 + sigma_square*self.dt/self.dx/self.dx + r*self.dt21        self.u_j = -(0.5*sigma_square*self.dt/self.dx/self.dx + 0.5 * (r - 0.5 * sigma_square)*self.dt/self.dx)22        23    def roll_back(self):24        25        for k in range(0, self.M):26            udiag = np.ones(self.N-1) * self.u_j27            ldiag =  np.ones(self.N-1) * self.l_j28            cdiag =  np.ones(self.N-1) * self.c_j29            30            mat = np.zeros((3,self.N-1))31            mat[0,:] = udiag32            mat[1,:] = cdiag33            mat[2,:] = ldiag34            rhs = np.copy(self.C[1:self.N,k])35            36            # 应用左端边值条件37            v1 = self.payoff.bcl(math.exp(self.x_min), (k+1)*self.dt, self.model)38            rhs[0] -= self.l_j * v139            40            # 应用右端边值条件41            v2 = self.payoff.bcr(math.exp(self.x_max), (k+1)*self.dt, self.model)42            rhs[-1] -= self.u_j * v243            44            x = solve_banded((1,1), mat, rhs)45            self.C[1:self.N, k+1] = x46            self.C[0][k+1] = v147            self.C[self.N][k+1] = v248            49    def mesh_grids(self):50        tArray = np.linspace(0, self.T, self.M+1)51        tGrids, xGrids = np.meshgrid(tArray, self.xArray)52        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 pylabpylab.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)