使用Python计算累积密度函数的导数

Question

累积分布函数的精确导数是否就是概率密度函数（PDF）？我正在使用numpy.diff()计算导数，这样做正确吗？请参见下面的代码：

import scipy.stats as s
import matplotlib.pyplot as plt
import numpy as np

wei = s.weibull_min(2, 0, 2) # shape, loc, scale - creates weibull object
sample = wei.rvs(1000)
shape, loc, scale = s.weibull_min.fit(sample, floc=0) 

x = np.linspace(np.min(sample), np.max(sample))

plt.hist(sample, normed=True, fc="none", ec="grey", label="frequency")
plt.plot(x, wei.cdf(x), label="cdf")
plt.plot(x, wei.pdf(x), label="pdf")
plt.plot(x[1:], np.diff(wei.cdf(x)), label="derivative")
plt.legend(loc=1)
plt.show()

那么，我如何将导数缩放到与PDF等效呢？

- kungphil

Answer 1

CDF的导数是PDF。

这里是CDF导数的近似值：

dx = x[1]-x[0]
deriv = np.diff(wei.cdf(x))/dx

---

import scipy.stats as s
import matplotlib.pyplot as plt
import numpy as np

wei = s.weibull_min(2, 0, 2) # shape, loc, scale - creates weibull object
sample = wei.rvs(1000)
shape, loc, scale = s.weibull_min.fit(sample, floc=0) 

x = np.linspace(np.min(sample), np.max(sample))
dx = x[1]-x[0]
deriv = np.diff(wei.cdf(x))/dx
plt.hist(sample, normed=True, fc="none", ec="grey", label="frequency")
plt.plot(x, wei.cdf(x), label="cdf")
plt.plot(x, wei.pdf(x), label="pdf")
plt.plot(x[1:]-dx/2, deriv, label="derivative")
plt.legend(loc=1)
plt.show()

收益率

请注意，与deriv相关的x-locations已经被dx/2移动，因此近似值位于用于计算它的值之间。

- unutbu