寻找一种方法来简化以下内容。计算一些涉及正常 cdf 的导数。
当尝试简化时,我希望它们以范数 pdf [0,1] 和范数 cdf [0,1] 表示。如何在没有 erf 的情况下获得规范 cdf 规范 pdf 的符号答案。
例如,Delta 计算应简化为 Norm.cdf(d1) 见下图;
对于另一个例子,theta 函数应该简化为范数 pdf 和 cdf 的函数,请参见下图。
下面附有代码,请告知如何获得所需的简化。
import sympy as sp
from sympy.stats import Normal, cdf
#spot, strike, volatility, time, interest rate, time at expiry, d1, and d2
S, K, sigma, t, r, T, d1, d2 = sp.symbols('S_t,K,sigma,t,r,T,d_1,d_2')
#define a symbol to represent the normal CDF
N = sp.Function('N')
#Black and Scholes price
C = S * N(d1) - N(d2) * K * sp.exp(-r * (T-t))
#expanded d1 and d2 for substitution:
d1_sub = (sp.ln(S / K) + (r + sp.Rational(1,2) * sigma ** 2) * (T-t)) / (sigma * sp.sqrt(T-t))
d2_sub = d1 - sigma * sp.sqrt(T-t)
#instance a standard normal distribution:
Norm = Normal('N',0.0, 1.0)
#define the long form b-s equation with all substitutions:
bs = C.subs(N, cdf(Norm)).subs(d2, d2_sub).subs(d1, d1_sub)
#Callable function for black and scholes price:
#example usage: bs_calc(100, 98, 0.15, 0, 0.03, 0.5)
bs_calc = sp.lambdify((S, K, sigma, t, r, T), bs)
print("Delta -> Mess")
print(sp.diff(bs,S))
print('Delta -> mess unresolved')
print(sp.simplify(sp.diff(bs,S)))
print("Theta -> Even worse Mess")
print(sp.diff(bs,T))
print('Theta -> Even Worse unresolved')
print(sp.simplify(sp.diff(bs,T)))
解决方案集将其打印到控制台:
运行文件('我的文件')
Delta -> 混乱 -0.5sqrt(2)Kexp(-r(T - t))exp(-0.5(-sigmasqrt(T - t) + ((T - t)(r + sigma) 2/2) + log(S_t/K))/(sigmasqrt(T - t)))**2)/(sqrt(pi)S_tsigmasqrt(T - t)) + erf(0.5 sqrt(2)((T - t)*(r + sigma2/2) + log(S_t/K))/(sigmasqrt(T - t)))/2 + 1/2 + 0.5 sqrt(2)exp(-0.5((T - t)(r + sigma**2/2) + log(S_t/K))2/(sigma2(T - t)) )/(sqrt(pi)sigmasqrt(T - t))
Delta -> 混乱尚未解决 -0.5sqrt(2)Kexp(-r(T - t))exp(-0.5(sigmasqrt(T - t)) - ((T - t)(r + sigma2) /2) + log(S_t/K))/(sigmasqrt(T - t)))**2)/(sqrt(pi)S_tsigmasqrt(T - t)) + erf(0.5 sqrt(2)((T - t)*(r + sigma2/2) + log(S_t/K))/(sigmasqrt(T - t)))/2 + 1/2 + 0.5 sqrt(2)exp(-0.5((T - t)(r + sigma**2/2) + log(S_t/K))2/(sigma2(T - t))) /(sqrt(pi)sigmasqrt(T - t))
Theta -> 更糟糕的混乱 Kr(erf(0.5sqrt(2)(-sigmasqrt(T - t) + ((T - t)(r + sigma2/2) + log(S_t/K))/ (sigmasqrt(T - t))))/2 + 1/2)exp(-r(T - t)) - 0.5sqrt(2)K(-sigma/(2*sqrt( T - t)) + (r + sigma2/2)/(sigmasqrt(T - t)) - ((T - t)(r + sigma2/2) + log(S_t/K) )/(2sigma(T - t)(3/2)))exp(-r(T - t))exp(-0.5(-sigmasqrt(T - t) + ( (T - t)(r + sigma2/2) + log(S_t/K))/(sigmasqrt(T - t)))2)/sqrt(pi) + S_t(0.5) sqrt(2)(r + sigma2/2)/(sigmasqrt(T - t)) - 0.25sqrt(2)((T - t)(r + sigma2/2) + log(S_t/K))/(sigma*(T - t)(3/2)))exp(-0.5((T - t)(r + sigma**2/2) + log( S_t/K))2/(sigma2(T - t)))/sqrt(pi)
Theta -> 更糟糕的是尚未解决 (-piKrsigma(T - t)(9/2)(erf(sqrt(2)(0.5*sigma2*(T - t) - 0.25*(T - t)) (2r + sigma2) - 0.5log(S_t/K))/(sigmasqrt(T - t))) - 1)exp((rsigma2*(T - t)) 2 + 0.125((T - t)(2r + 西格玛**2) + 2log(S_t/K))2 + 0.125(2sigma2*(T - t) ) - (T - t)(2r + sigma2) - 2log(S_t/K))2)/(sigma2(T - t))) + 0.25sqrt(2) sqrt(pi)K(T - t)3(2sigma**2(T - t) - (T - t)(2r + sigma**2) + 2 log(S_t/K))exp((rsigma2*(T - t)2 + 0.125((T - t)(2*r + sigma2) + 2log(S_t) /K))2)/(sigma2(T - t))) + 2sqrt(2)sqrt(pi)S_t(T - t)3(0.125(T - t)(2r + sigma2) - 0.25log(S_t/K))exp((2rsigma2*(T - t)2 + 0.125(2sigma) 2*(T - t) - (T - t)(2r + sigma2) - 2log(S_t/K))2)/(sigma2(T - t)))) exp(-(2rsigma**2(T - t)2 + 0.125((T - t)(2r + sigma**2) + 2log(S_t/K ))2 + 0.125(2sigma2*(T - t) - (T - t)(2r + sigma2) - 2log(S_t/K))2)/ (西格玛2(T - t)))/(2pi西格玛*(T - t)**(9/2))
尽可能避免使用浮点数,因此最好使用
Normal('N', 0, 1)
,就像我对下面的输出所做的那样。
sympy 中没有用于正态分布 cdf 的 Phi 函数,因为它是用 erf 表示的。不过,您可以定义自己的 Phi 函数:
In [108]: phi = Function('phi')
In [109]: class Phi(Function):
...: def _eval_derivative(self, x):
...: return phi(self.args[0]) * self.args[0].diff(x)
...:
In [110]: bs.subs(erf, Lambda(x, 2*Phi(x*sqrt(2))-1))
Out[110]:
⎛ ⎛ 2⎞ ⎞ ⎛ ⎛ 2⎞ ⎞
⎜ ⎜ σ ⎟ ⎛Sₜ⎞⎟ ⎜ ⎜ σ ⎟ ⎛Sₜ⎞⎟
⎜ (T - t)⋅⎜r + ──⎟ + log⎜──⎟⎟ ⎜(T - t)⋅⎜r + ──⎟ + log⎜──⎟⎟
⎜ _______ ⎝ 2 ⎠ ⎝K ⎠⎟ -r⋅(T - t) ⎜ ⎝ 2 ⎠ ⎝K ⎠⎟
- K⋅Φ⎜- σ⋅╲╱ T - t + ──────────────────────────⎟⋅ℯ + Sₜ⋅Φ⎜──────────────────────────⎟
⎜ _______ ⎟ ⎜ _______ ⎟
⎝ σ⋅╲╱ T - t ⎠ ⎝ σ⋅╲╱ T - t ⎠
In [111]: bs.subs(erf, Lambda(x, 2*Phi(x*sqrt(2))-1)).diff(S)
Out[111]:
⎛ ⎛ 2⎞ ⎞ ⎛ ⎛ 2⎞ ⎞
⎜ ⎜ σ ⎟ ⎛Sₜ⎞⎟ ⎜ ⎜ σ ⎟ ⎛Sₜ⎞⎟
⎜ (T - t)⋅⎜r + ──⎟ + log⎜──⎟⎟ ⎜(T - t)⋅⎜r + ──⎟ + log⎜──⎟⎟
⎜ _______ ⎝ 2 ⎠ ⎝K ⎠⎟ -r⋅(T - t) ⎛ ⎛ 2⎞ ⎞ ⎜ ⎝ 2 ⎠ ⎝K ⎠⎟
K⋅φ⎜- σ⋅╲╱ T - t + ──────────────────────────⎟⋅ℯ ⎜ ⎜ σ ⎟ ⎛Sₜ⎞⎟ φ⎜──────────────────────────⎟
⎜ _______ ⎟ ⎜(T - t)⋅⎜r + ──⎟ + log⎜──⎟⎟ ⎜ _______ ⎟
⎝ σ⋅╲╱ T - t ⎠ ⎜ ⎝ 2 ⎠ ⎝K ⎠⎟ ⎝ σ⋅╲╱ T - t ⎠
- ─────────────────────────────────────────────────────────── + Φ⎜──────────────────────────⎟ + ─────────────────────────────
_______ ⎜ _______ ⎟ _______
Sₜ⋅σ⋅╲╱ T - t ⎝ σ⋅╲╱ T - t ⎠ σ⋅╲╱ T - t
扩展上述解决方案,对于那些希望进行进一步导数或数值评估的人:
from sympy import diff, Function, sqrt, exp, pi
from sympy.stats import P, Normal, density
class Phi(Function):
def _eval_derivative(self, x):
return phi(self.args[0]) * self.args[0].diff(x)
@classmethod
def eval(cls, x):
Z = Normal('Z', 0, 1)
return P(Z<x)
class phi(Function):
def _eval_derivative(self, x):
return diff(
density(Normal("x", 0, 1))(x),
x
)
# for printing out derivatives without 'erf':
# comment the one above and uncomment this one
# return -0.5*sqrt(2)*x*exp(-0.5*x**2)/sqrt(pi)
@classmethod
def eval(cls, x):
return density(Normal("x", 0, 1))(x)