如何计算具有对立变量的估计量的标准误差

如何解决如何计算具有对立变量的估计量的标准误差

我使用 Longstaff 和 Schwartz 最小二乘法为美式期权定价。

当使用以下 Python 代码时，我获得的价格和标准误差与 通过模拟评估美式期权几乎相同： Longstaff 和 Schwartz 的简单最小二乘法（2001 年）：

import numpy as np
import numpy.random as npr
import warnings
warnings.simplefilter('ignore')
from numpy.polynomial.laguerre import lagfit,lagval

# Same parameters as in the original paper
class par: pass
par.S0 = 36
par.K = 40
par.r = 0.06
par.sigma = 0.2
par.T = 1.0
par.I = 100000
par.M = 50

def gen_sn(par,anti_path):
    ''' Function to generate random numbers for simulation.
    Parameters
    ==========
    M : int
    number of time intervals for discretization
    I : int
    number of paths to be simulated
    '''
    np.random.seed(1)
    if anti_path is True:
        sn = npr.standard_normal((par.M + 1,par.I//2))
    else:
        sn = npr.standard_normal((par.M + 1,par.I))
    return sn



def gbm_mcs_amer(par):
    ''' Valuation of American option in Black-Scholes-Merton
    by Monte Carlo simulation by LS algorithm. 
    Parameters
    ==========
    S0 : spot price
    K : strike float
    r : riskless interest rate
    I : int,number of paths to be simulated
    T : time to maturity,in years
    M : int,number of time intervals for discretization
    sigma : vol
    Returns
    =======
    C0 : float
    estimated present value of American call option
    '''
    dt = par.T / par.M 
    df = np.exp(-par.r * dt) # discount function

    # Generation of underlying asset process
    # Stock Price Paths
    S = par.S0 * np.exp(np.cumsum((par.r - 0.5 * par.sigma ** 2) * dt
    + par.sigma * np.sqrt(dt) * sn,axis=0)) # by exponentiating the Brownian motion
    S[0] = par.S0 # Initiliazing underlying path
    
    # put option pay-off
    h = np.maximum(par.K - S,0)
    # LS algorithm
    V = np.copy(h)
    for t in range(par.M - 1,-1):
        reg = lagfit(S[t],V[t + 1] * df,10)
        C = lagval(S[t],reg)
        V[t] = np.where(C > h[t],h[t])
        
        # MCS estimator
        y_i = df * V[1]
        C0 = np.mean(y_i)
        SE = np.std(y_i) / np.sqrt(par.I)
    
    return C0,SE
# Regular Estimate loop
sn = gen_sn(par,False)


print("Reg","T:",par.T,"sigma:",par.sigma)
for par.S0 in range(36,44+1,2):
    print("S0:",par.S0,"Price,SE:",gbm_mcs_amer_reg(par)[0],gbm_mcs_amer_reg(par)[1])

然后，我尝试实施 Glasserman 提出的对立变量价格 Antithetic paths estimator 和 Boyle 和 Glasserman 提出的标准误差 Antithetic paths standard error：

def gbm_mcs_amer_AP(par):
dt = par.T / par.M 
df = np.exp(-par.r * dt) # discount function

# Generation of underlying asset process
# Stock Price Paths
S = par.S0 * np.exp(np.cumsum((par.r - 0.5 * par.sigma ** 2) * dt
+ par.sigma * np.sqrt(dt) * sn,axis=0)) # by exponentiating the Brownian motion
S[0] = par.S0
S1 = par.S0 * np.exp(np.cumsum((par.r - 0.5 * par.sigma ** 2) * dt
+ par.sigma * np.sqrt(dt) * -sn,axis=0)) # Antithetic paths
S1[0] = par.S0

# put option pay-off
h = np.maximum(par.K - S,0)
h1 = np.maximum(par.K - S1,0)
# LS algorithm
V = np.copy(h)
V1 = np.copy(h1)
for t in range(par.M - 1,-1):
    reg = lagfit(S[t],10)
    C = lagval(S[t],reg)
    V[t] = np.where(C > h[t],h[t])
    reg1 = lagfit(S1[t],V1[t + 1] * df,10)
    C1 = lagval(S1[t],reg1)
    V1[t] = np.where(C1 > h1[t],h1[t])
    
    # MCS estimator
    y_i = df * (V[1]+V1[1])/2 # avg. pairs
    C0 = np.mean(y_i)
    SE = np.std(y_i) / np.sqrt(par.I) # Sample std. dev. of avg. pairs

return C0,SE
  
# AP Estimate loop
sn = gen_sn(par,True)

print("AP",gbm_mcs_amer_AP(par)[0],gbm_mcs_amer_AP(par)[1])

对立变量的价格与 Longstaff 和 Schwartz 论文中的价格接近，但标准误差似乎太小了，因为它们比控制变量减少了更多的方差，并且在随机数生成中匹配了两个矩。

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