(一)连续随机量的生成-基于分布函数

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)](#1. 概率积分变换方法(分布函数))
  • [2. Python编程实现指数分布的采样](#2. Python编程实现指数分布的采样)

1. 概率积分变换方法(分布函数)

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

Example : Exponential distribution Exp ⁡ ( 1 ) \operatorname{Exp} (1) Exp(1).

Exp (1) has a probability density function: f ( z ) = { e − z , z ⩾ 0 , 0 , z < 0. f(z)= \begin{cases}e^{-z}, & z \geqslant 0, \\ 0, & z<0 .\end{cases} f(z)={e−z,0,z⩾0,z<0.

Its distribution function is F ( z ) = { 1 − e − z , z ⩾ 0 , 0 , z < 0. F(z)= \begin{cases}1-e^{-z}, & z \geqslant 0, \\ 0, & z<0 .\end{cases} F(z)={1−e−z,0,z⩾0,z<0.

We only need to concentrate on F ( z ) F(z) F(z) on [ 0 , ∞ ) [0, \infty) [0,∞), and have
F − 1 ( z ) = − log ⁡ ( 1 − z ) . F^{-1}(z)=-\log (1-z). F−1(z)=−log(1−z).

So F − 1 ( U ) = − log ⁡ ( 1 − U ) F^{-1}(U)=-\log (1-U) F−1(U)=−log(1−U) has a probability distribution Exp ( 1 ) (1) (1). Because 1 − U ∼ U ( [ 0 , 1 ] ) 1-U \sim U([0,1]) 1−U∼U([0,1]), we have − log ⁡ U ∼ Exp ⁡ ( 1 ) -\log U \sim \operatorname{Exp}(1) −logU∼Exp(1).

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

2. Python编程实现指数分布的采样

Assignment: Sample a random quantity Z ∼ Exp ⁡ ( λ ) Z \sim \operatorname{Exp}(\lambda) Z∼Exp(λ) for some λ > 0 \lambda>0 λ>0.

python 复制代码
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()
相关推荐
软件开发技术深度爱好者22 分钟前
使用Python实现播放“.gif”文件增强版
开发语言·python
感哥43 分钟前
Django Model高级特性
python·django
李辉20031 小时前
Python简介及Pycharm
开发语言·python·pycharm
赵谨言1 小时前
基于python大数据的城市扬尘数宇化监控系统的设计与开发
大数据·开发语言·经验分享·python
云和数据.ChenGuang1 小时前
parser_error UnicodeDecodeError: ‘utf-8‘ codec can‘t decode bytes
python
zhangfeng11331 小时前
R和python 哪个更适合生物信息分析,或者更擅长做什么工作
开发语言·python·r语言·生物信息
liliangcsdn2 小时前
如何结合langchain、neo4j实现关联检索问答
开发语言·python·langchain·neo4j
j七七2 小时前
5分钟搭微信自动回复机器人5分钟搭微信自动回复机器人
运维·服务器·开发语言·前端·python·微信
lgbisha2 小时前
Dify异步接口调用优化实践:解决长时任务处理与网络超时问题
人工智能·python·ai·语言模型
Hs_QY_FX2 小时前
幸福指数数据分析与预测:从数据预处理到模型构建完整案例
开发语言·python·机器学习