是两个随机变量$X$和$Y$应该是独立的？如果是这样，很容易证明的分布函数$Z=X-Y$由卷积给出

$$F_Z(z) = P(X-Y\leq z) = \int F_X(z+y) \, dF_Y(y) \, .$$ 因此，一种想法是计算经验分布函数$\hat{F}_m$的$(x_1,\dots,x_m)$， 和$\hat{G}_n$的$(y_1,\dots,y_n)$, 并使用 $$\hat{H}(z) = \int \hat{F}_m(z+y)\,d\hat{G}_n(y) = \frac{1}{m\,n}\sum_{i=1}^n \sum_{j=1}^m I_{[x_j,\infty)}(z + y_i)$$ 作为估计$F_Z(z)$. 请注意，相应的估计量对于每个$z$.

我认为您不需要特殊的软件包来执行此操作；普通的numpy就足够了。我在下面附加了示例代码及其输出。请注意，(AB) 的 cdf 看起来与 A 和 B 的 cdf 非常相似，但实际上并非如此。您可以看到大约 +/- 2 或 3 sigma 的细微差别。(AB) 的 cdf 比 A 和 B 的单独 cdf 宽一点。

import numpy as np
import matplotlib.pyplot as plt

#!/usr/bin/env python

# Number of random draws to use
ndraws = 1000
# Set this distance (in sigmas) large enough to capture all of the outliers
plotrange = 5
# Number of bins to use for pdf/cdf
nbin = 100
# Get random draws from a Gaussian
A = np.random.randn(1,ndraws)
B = np.random.randn(1,ndraws)
dfAB = A - B

# Calculate cdfs of A and B
Apdf, edges = np.histogram(A, bins=nbin, range=(-plotrange, plotrange))
Bpdf, edges = np.histogram(B, bins=nbin, range=(-plotrange, plotrange))
dfABpdf, edges = np.histogram(dfAB, bins=nbin, range=(-plotrange, plotrange))
xrng = (edges[0:-1] + edges[1:]) / 2

Acdf = np.cumsum(map(float, Apdf)) / ndraws
Bcdf = np.cumsum(map(float, Bpdf)) / ndraws
dfABcdf = np.cumsum(map(float,dfABpdf)) / ndraws

# Plot cdfs and differences of cdfs
fig = plt.figure()
ax1 = fig.add_subplot(2,2,1)
ax1.plot(xrng, Acdf)
ax1.set_title("A cdf")

ax2 = fig.add_subplot(2,2,2)
ax2.plot(xrng, Bcdf)
ax2.set_title("B cdf")

ax3 = fig.add_subplot(2,2,3)
ax3.plot(xrng, dfABcdf)
ax3.set_title("(A-B) cdf")

ax4 = fig.add_subplot(2,2,4)
ax4.plot(xrng, Acdf - Bcdf)
ax4.set_title("(A cdf) - (B cdf)")

plt.show()