（一）连续随机量的生成-基于分布函数

1. 概率积分变换方法（分布函数）

Consider drawing a random quantity $X$ from a continuous probability distribution with the distribution function $F$. We know $F$ is a continues nondecreasing function if $F$ has an inverse $F^{-1}$, then $Z=F^{-1}(U)$, where $U$ is a random quantity drawn from $U([0,1])$, is a random quantity as desired. Indeed,
$$P(X⩽z)=P\left(F^{-1}(U)⩽z\right)=P(U⩽F(z))=F(z),\forall z\in \mathbb{R}$$

Example: Exponential distribution $\mathrm{Exp}(1)$.
Exp (1) has a probability density function: $f(z)=\{\begin{array}{ll}{e}^{-z},& z⩾0,\\ 0,& z<0.\end{array}$
Its distribution function is $F(z)=\{\begin{array}{ll}1-e^{-z},& z⩾0,\\ 0,& z<0.\end{array}$
We only need to concentrate on $F(z)$ on $[0,\infty )$, and have
$$F^{-1}(z)=-\log (1-z).$$
So $F^{-1}(U)=-\log (1-U)$ has a probability distribution Exp $(1)$. Because $1-U\sim U([0,1])$, we have $-\log U\sim \mathrm{Exp}(1)$.

For a distribution function which does not have an inverse, we define a generalized inverse as the following:
$$F^{-}(z)=\inf \{x\in \mathbb{R}:F(x)⩾z\}.$$

2. Python编程实现指数分布的采样

Assignment: Sample a random quantity $Z\sim \mathrm{Exp}(\lambda )$ for some $\lambda >0$.

import numpy as np
import matplotlib.pyplot as plt

# Parameter for the exponential distribution
lambda_value = 0.5

# Generate random quantity using CDF method
u = np.random.rand(1000)  # Uniform random numbers between 0 and 1
Z = -np.log(1 - u) / lambda_value

# Plot histogram
plt.hist(Z, bins=30, density=True, alpha=0.6, color='b', label='Sampled Data')
plt.xlabel('Value')
plt.ylabel('Density')
plt.title('Histogram of Exponential Distribution (Generated using CDF)')
plt.legend()
plt.grid(True)
plt.show()