1

几何布朗运动 (gBM)是一个随机过程,可以被认为是标准布朗运动的扩展。

我正在尝试编写一个函数来模拟ntrajgBM 的不同路径(路径),然后在 list 中指定的某些点绘制直方图tcheck。一旦绘制了这些图,该函数就意味着每次在图上叠加一个对数正态分布。

输出看起来像这样 在此处输入图像描述

除了 gBM 而不是标准的布朗运动过程。到目前为止,我有一个生成多个 gBM 路径的功能,

def oneDGeometricBM(nTraj=100,n=100,T=1.0,sigma=1,mu=0):
    '''
    DOCSTRING:
    1D geomwtric brownian motion
    INPUTS:
    ntraj = "number of trajectories"
    n = "length of a trajectory"
    T = "last time point, i.e final tradjectory t = {0,1,...,T}"
    sigma= volatility
    mu= percentage drift

    '''
    np.random.seed(52323)
    S_0 = 0

    # Discretize, dt =  time step = $t_{j+1}- t_{j}$
    dt = T/(n)  
    sqrtdt = np.sqrt(dt)

    # Container for different colors for each trajectory
    colors = plt.cm.jet(np.linspace(0,1,nTraj))
    # Container for trajectories
    xtraj=np.zeros(n+1,float)
    ztraj=np.zeros(n+1,float)
    trange=np.linspace(start = 0,stop = T ,num = n+1)

    # Simulation
    # Random Variable $X_{n}$ is distributed np.sqrt(dt)* N(mu=0,sigma=1) 
    for j in range(nTraj):
        # Loop over time
        for i in range(n):
            xtraj[i+1]=xtraj[i]+ sqrtdt * np.random.randn() + dt*mu
        # Loop again over time in order to make geometric drift
        ztraj = S_0 * np.exp(xtraj) # ztraj[z+1]=  ztraj[0]+ np.exp(xtraj[z])

        plt.plot(trange , xtraj,'b-',alpha=0.2, color=colors[j], lw=3.0,label="$\sigma$={}, $\mu$={:.5f}".format(sigma,mu))
    plt.title("1D Geometric Brownian Motion:\n nTraj={}, T={},\n $\Delta t$={:.3f}, $\sigma$={}, $\mu$={:.3f}".format(nTraj,T,dt,sigma,mu))    
    plt.xlabel(r'$t$')
    plt.ylabel(r'$Z_t$');

oneDGeometricBM(nTraj=5,n=10**3,T=10.0,sigma=0.8,mu=1.1)

在此处输入图像描述

我已经看到很多关于如何绘制 gBM 多条路径的问题的答案,但我对如何在特定时间查看直方图然后查看分布感兴趣。以下是我到目前为止的功能。它不起作用,但我无法弄清楚我做错了什么。我还添加了我得到的输出。

import yfinance as yf
import pandas as pd
import matplotlib.pyplot as plt
import numpy as np
import math
from scipy.stats import norm, lognorm
ntraj = 10000
S_0 =0
sigma=1
mu=1
tfinal = 4.0
tcheck = [0.5, 1.0, 4.0]
dt = 0.01
xv = 1.0
'''
ntraj = 10**4
tfinal = 4.0
tcheck = [0.5, 1.0, 4.0]
dt = 0.01
xv = 5.0 # limits
'''
n=int(tfinal/dt)
sqrtdt = np.sqrt(dt)

x=np.zeros(shape=[ntraj,n+1], dtype=float)
z=np.zeros(shape=[ntraj,n+1], dtype=float)
zrange=np.arange(start=-xv, stop=xv, step=dt)

# Calculate the number of the bins 
binval = math.ceil(np.sqrt(ntraj))
# Nested for loop to create Drifted BM
for i in range(n):
    for j in range(ntraj):
        x[j,i+1]=x[j,i]+ sqrtdt*np.random.randn()


 #Nested loop to create gBM
for j0 in range(ntraj):
    for i0 in range(n+1):
        z[j0,i0] = 0 + np.exp(x[j0,i0])

# Loop to plot the distribution of gBM tradjectories at different times       
for i1 in range(n):
    # Compute histogram at every tsample , sample at time t
    t=(i1+1)*dt
    if t in tcheck:
       # Plot histogram on sample
       plt.hist(z[:,i1],bins=30,density=False,alpha=0.6,label=['t ={}'.format(t)] )
       # Superimpose each samples mean
       xbar = np.average(z[:,i1])
       plt.axvline(xbar, color='RED', linestyle='dashed', linewidth=2) 
       # Plot theoretic distribution { N(0, sqrt[t] ) }
       #plt.plot(xrange,norm.pdf(xrange,0.0,np.sqrt(t)),'k--')

在此处输入图像描述

所以总结一下我的问题。我正在尝试模拟 gBM 的多个轨迹,将我的结果存储在一个数组中,然后循环遍历这个数组并使用 matplotlib 在特定点上绘制一个直方图,然后最后在我的直方图上叠加一个对数正态分布。

编辑1:

如果可能,我需要在 GBM 和 Cauchy 上叠加对数正态分布。我的问题是,当我编辑@Paul Harris 的更正时,我得到了, 在此处输入图像描述

def oneDGeometricBM(nTraj=100,n=100,T=1.0,sigma=1,mu=0):
    '''
    DOCSTRING:
    INPUTS:
    ntraj = "number of trajectories"
    n = "length of a trajectory"
    T = "last time point, i.e final tradjectory t = {0,1,...,T}"
    sigma= volatility
    mu= percentage drift

    '''
    np.random.seed(52323)
    S0 = 10

    # Discretize, dt =  time step = $t_{j+1}- t_{j}$
    dt = T/(n)  
    sqrtdt = np.sqrt(dt)

    # Container for different colors for each trajectory
    colors = plt.cm.jet(np.linspace(0,1,nTraj))
    # Container for trajectories
    xtraj=np.zeros(n+1,float)
    ztraj=np.zeros(n+1,float)
    trange=np.linspace(start = 0,stop = T ,num = n+1)

    out = []
    # Simulation
    # Random Variable $X_{n}$ is distributed np.sqrt(dt)* N(mu=0,sigma=1) 
    for j in range(nTraj):
        # Loop over time
        for i in range(n):
            xtraj[i+1]=xtraj[i]+ sqrtdt * np.random.randn() + dt*mu

        # Loop again over time in order to make geometric drift
    ztraj = S0 * np.exp(xtraj)
    # Return gBM 
    return ztraj

# Plotting
fig, ax = plt.subplots(ncols=2, figsize=plt.figaspect(1./2))

colors = ['k', 'r', 'b']
T =  [1.0, 2.0, 5.0]
sigma=0.8
mu=1.1 

for c, T in zip(colors, T):
    ztraj = oneDGeometricBM(nTraj=5,n=10**4,T=T,sigma=0.8,mu=1.1)
    # Plot Emperical Values
    xrange = range(0,80,1)
    ax[0].hist(ztraj, bins=100, alpha=0.5, label=f'T={T}', density=True, color=c, range=(0, 95))

    # Plot the theoretical values
    theoretic_mean = math.exp(mu * T + 0.5 * sigma**2 * T)
    theoretic_var = math.exp(2* mu * T + sigma**2 * T)* (math.exp(sigma**2 * T) - 1)
    ax[0].plot(xrange,lognorm.pdf(xrange, theoretic_mean , theoretic_var ),'k--')

    # Plot the differences between consecutive elements of gBM (an array)
    diff = np.ediff1d(ztraj)
    ax[1].hist(diff, bins=100, alpha=0.5, label=f'T={T}', density=True, color=c, range=(-5, 5))


ax[0].set_xlabel('z')
ax[0].set_ylabel('$p(z,T)$')
ax[0].set_title('Histogram of ztraj positions')

ax[1].set_xlabel('dz')
ax[1].set_ylabel('$p(dz,T)$')
ax[1].set_title('Histogram of d(ztraj) positions\nbetween time steps')

ax[0].legend()
fig.tight_layout()

所以总结一下,我需要叠加每个时间点的分布,即 gBM 的理论分布,即对数正态分布。

4

1 回答 1

0

所以我看了你的问题。我已经编辑了您的函数以停止绘图并返回xtraj,我认为这是您的布朗运动:

def oneDGeometricBM(nTraj=100,n=100,T=1.0,sigma=1,mu=0):
    '''
    DOCSTRING:
    1D geomwtric brownian motion
    INPUTS:
    ntraj = "number of trajectories"
    n = "length of a trajectory"
    T = "last time point, i.e final tradjectory t = {0,1,...,T}"
    sigma= volatility
    mu= percentage drift

    '''
    np.random.seed(52323)
    S_0 = 10

    # Discretize, dt =  time step = $t_{j+1}- t_{j}$
    dt = T/(n)  
    sqrtdt = np.sqrt(dt)

    # Container for different colors for each trajectory
    colors = plt.cm.jet(np.linspace(0,1,nTraj))
    # Container for trajectories
    xtraj=np.zeros(n+1,float)
    ztraj=np.zeros(n+1,float)
    trange=np.linspace(start = 0,stop = T ,num = n+1)

    out = []
    # Simulation
    # Random Variable $X_{n}$ is distributed np.sqrt(dt)* N(mu=0,sigma=1) 
    for j in range(nTraj):
        # Loop over time
        for i in range(n):
            xtraj[i+1]=xtraj[i]+ sqrtdt * np.random.randn() + dt*mu

        # Loop again over time in order to make geometric drift
    ztraj = S_0 * np.exp(xtraj) # ztraj[z+1]=  ztraj[0]+ np.exp(xtraj[z])

    return ztraj

每个时间步长的位移就是数组 xtraj: 内的差异dx = np.ediff1d(oneDGeometricBM(...)),因此我们计算这些值的直方图:

fig, ax = plt.subplots()

ax.hist(np.ediff1d(oneDGeometricBM(nTraj=5,n=10**3,T=10.0,sigma=0.8,mu=1.1)), bins=50, alpha=0.5, label='T=10', density=True)
ax.hist(np.ediff1d(oneDGeometricBM(nTraj=5,n=10**3,T=1.0,sigma=0.8,mu=1.1)), bins=50, alpha=0.5, color='k', label='T=1', density=True)
ax.hist(np.ediff1d(oneDGeometricBM(nTraj=5,n=10**3,T=5.0,sigma=0.8,mu=1.1)), bins=50, alpha=0.5, color='r', label='T=5', density=True)

ax.set_xlabel('x')
ax.set_ylabel('$p(x,T)$')

ax.legend()

输出

如示例所示,我使用T了 3 个不同的值。标准化直方图,使 y 轴现在代表概率p(x,T),即。总之p*x = 1,我们使用density=True论点。

编辑

我已经编辑了oneDGeometricBM要返回的函数ztraj = S0*np.exp(xtraj)。您的初始S0值为0,所以我将其设为非零。

您可以将差异绘制ztraj为:

fig, ax = plt.subplots()

colors = ['k', 'r', 'b']
T = [1.0, 2.0, 5.0]

for c, T in zip(colors, T):
    ztraj = oneDGeometricBM(nTraj=5,n=10**3,T=T,sigma=0.8,mu=1.1)
    diff = np.ediff1d(ztraj)
    ax.hist(diff, bins=100, alpha=0.5, label=f'T={T}', density=True, color=c, range=(-10, 10))

ax.set_xlabel('x')
ax.set_ylabel('$p(x,T)$')

ax.legend()

兹特拉伊

编辑2

通过更仔细地查看您生成的直方图,我认为您的建模是正确的,只需将绘图的 xrange 调整为ztrajlarge T,您可以使用参数限制直方图range。所以我已经为三个单独ztraj的. 确实似乎大致遵循对数正态分布,并且差异似乎大致遵循洛伦兹分布(必须检查那个理论,可能是高斯分布)。重现代码:d(ztraj)Tztrajztraj

fig, ax = plt.subplots(ncols=2, figsize=plt.figaspect(1./2))

colors = ['k', 'r', 'b']
T = [1.0, 2.0, 5.0]

for c, T in zip(colors, T):
    ztraj = oneDGeometricBM(nTraj=5,n=10**4,T=T,sigma=0.8,mu=1.1)

    ax[0].hist(ztraj, bins=100, alpha=0.5, label=f'T={T}', density=True, color=c, range=(0, 95))

    diff = np.ediff1d(ztraj)
    ax[1].hist(diff, bins=100, alpha=0.5, label=f'T={T}', density=True, color=c, range=(-5, 5))

ax[0].set_xlabel('z')
ax[0].set_ylabel('$p(z,T)$')
ax[0].set_title('Histogram of ztraj positions')

ax[1].set_xlabel('dz')
ax[1].set_ylabel('$p(dz,T)$')
ax[1].set_title('Histogram of d(ztraj) positions\nbetween time steps')

ax[0].legend()
fig.tight_layout()

ztraj 和 dz

这是您的数据和绘图,但限制了直方图range=(0, 10)

og图

编辑3

我已经包含了适合对数正态分布的代码,并将它们显示在您的原始图上。我们将对数正态函数定义为:

from scipy.optimize import curve_fit

def lognorm(x, x0, A, sigma):
    return A * np.exp(-(np.log(x)-x0)**2 / (2*sigma**2))

然后在最终循环中使用直方图中的值和 bin 进行拟合:

# Loop to plot the distribution of gBM tradjectories at different times       
for i1 in range(n):
    # Compute histogram at every tsample , sample at time t
    t=(i1+1)*dt
    if t in tcheck:
        # Plot histogram on sample
        v, b, patches = plt.hist(z[:,i1],bins=200,density=False,alpha=0.6,label=['t ={}'.format(t)], range=(0, 10) )

        # second term is bin centre locations rather than bin edges
        popt, pcov = curve_fit(lognorm, b[:-1] + np.ediff1d(b), v, p0=(0.1, 300, 0.3))

        # make colors match their original data but no transparency
        plt.plot(b, lognorm(b, *popt), color=patches[0].get_facecolor()[:3])

        print(f'tcheck: {t} with parameters: {popt}')

输出:

tcheck: 0.5 with parameters: [ -0.42334757 358.38545736   0.6748076 ]
tcheck: 1.0 with parameters: [ -0.90719967 321.03944864   0.96137893]
tcheck: 4.0 with parameters: [ -3.66426932 721.41708932   1.86376987]

对数正态拟合

编辑4

生成上述输出的整个代码是:

import pandas as pd
import matplotlib.pyplot as plt
import numpy as np
import math
from scipy.stats import norm, lognorm
from scipy.optimize import curve_fit

def lognorm(x, x0, A, sigma):
    return A * np.exp(-(np.log(x)-x0)**2 / (2*sigma**2))

ntraj = 10000
S_0 =0
sigma=1
mu=1
tfinal = 4.0
tcheck = [0.5, 1.0, 4.0]
dt = 0.01
xv = 1.0
'''
ntraj = 10**4
tfinal = 4.0
tcheck = [0.5, 1.0, 4.0]
dt = 0.01
xv = 5.0 # limits
'''
n=int(tfinal/dt)
sqrtdt = np.sqrt(dt)

x=np.zeros(shape=[ntraj,n+1], dtype=float)
z=np.zeros(shape=[ntraj,n+1], dtype=float)
zrange=np.arange(start=-xv, stop=xv, step=dt)

# Calculate the number of the bins 
binval = math.ceil(np.sqrt(ntraj))
# Nested for loop to create Drifted BM
for i in range(n):
    for j in range(ntraj):
        x[j,i+1]=x[j,i]+ sqrtdt*np.random.randn()


 #Nested loop to create gBM
for j0 in range(ntraj):
    for i0 in range(n+1):
        z[j0,i0] = 0 + np.exp(x[j0,i0])

# Loop to plot the distribution of gBM tradjectories at different times       
for i1 in range(n):
    # Compute histogram at every tsample , sample at time t
    t=(i1+1)*dt
    if t in tcheck:
        # Plot histogram on sample
        v, b, patches = plt.hist(z[:,i1],bins=200,density=True,alpha=0.6,label=['t ={}'.format(t)], range=(0, 10))

        popt, pcov = curve_fit(lognorm, b[:-1] + np.ediff1d(b), v, p0=(0.1, 300, 0.3))
        # make colors match their original data but no transparency
        plt.plot(b, lognorm(b, *popt), color=patches[0].get_facecolor()[:3])

        print(f'tcheck: {t} for parameters: {popt}')
于 2020-05-16T10:06:55.340 回答