常见概率分布及python实现

总结

1.两点分布——离散型概率分布

概念:一次试验,若成功随机变量取值为1,成功概率为p; 若失败随机变量取0,失败概率为1-p
期望\(E(X)=1*p+0*(1-p)=p\)
方差

\[\begin{aligned} D(X)&=p*(1-p)^2+(1-p)*(0-p)^2\\ &=p(1-p) \end{aligned} \]

2.二项分布——离散型概率分布

概念:进行n次伯努利试验。(n>=1),当n=1,二项分布就是伯努利分布
n次试验中总共成功的次数为k的概率 \(P(X=k;n,p)=C_n^k*p^k*(1-p)^{n-k}\)
期望 $ E(X)=np $
期望的推导

\[\begin{aligned} E(X) &= \sum_{k=0}^{n}k*P(X=k)\\ &=\sum_{k=0}^{n}k*\frac{n!}{k!(n-k)!}p^k (1-p)^{n-k}\\ &=np\sum_{k=1}^{n}\frac{(n-1)!}{(k-1)!(n-k)!}p^{k-1} (1-p)^{(n-k)}\\ &=np \end{aligned} \]

方差$ D(X)=np(1-p) $
方差的推导

\[\begin{aligned} D(X) &= E(X^2)-E^2(X)\\ &=E[X(X-1)+X]-n^2p^2=E[X(X-1)]+np-n^2p^2 \end{aligned} \]

\[\begin{aligned} E[X(X-1)] &= \sum_{k=0}^{n}k(k-1)*P(X=k)\\ &= \sum_{k=0}^{n}k(k-1)*\frac{n!}{k!(n-k)!}p^k(1-p)^{n-k}\\ &=n(n-1)p^2\sum_{k=2}^{n}\frac{(n-2)!}{(k-2)!(n-k)!}p^{k-2}(1-p)^{n-k}\\ &=n(n-1)p^2 \end{aligned} \]

\[\begin{aligned} D(X)&=n(n-1)p^2+np-n^2p^2\\ &=np(1-p) \end{aligned} \]

3.泊松分布——离散型概率分布

泰勒展开式

\[\begin{aligned} e^x&=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots+ \frac{x^n}{n!}+R_n\\ 1&=e^{-x}+xe^{-x}+\frac{x^2}{2!}e^{-x}+\frac{x^3}{3!}e^{-x}+\cdots+\frac{x^n}{n!}e^{-x}+R_ne^{-n} \end{aligned} \]

通项 $ \frac{xk}{k!}e $ ---> \(\frac{\lambda^k}{k!}e^{-\lambda}\)
概率分布\(P(X=k)=\frac{\lambda^k}{k!}e^{-\lambda},\lambda>0,k=0,1,2,\cdots\)
期望

\[\begin{aligned} E(X)&=\sum_{k=0}^{\infty}k*f(x)\\ &=\sum_{k=0}^{\infty}k*\frac{\lambda^k}{k!}e^{-\lambda}\\ &=\lambda\sum_{k=1}^{\infty}\frac{\lambda^{k-1}}{(k-1)!}e^{-\lambda}\\ &=\lambda \end{aligned} \]

方差

\[\begin{aligned} D(X) &= E(X^2)-E^2(X)\\ &=E[X(X-1)+X]-E^2(X)\\ &=E[X(X-1)]+E(X)-E^2(X)\\ &=E[X(X-1)+\lambda-\lambda^2 \end{aligned} \]

\[\begin{aligned} E[X(X-1)&=\sum_{k=0}^{\infty}k(k-1)*\frac{\lambda^k}{k!}e^{-\lambda}\\ &=\lambda^2\sum_{k=2}^{\infty}\frac{\lambda^{k-2}}{(k-2)!}e^{-\lambda}\\ &=\lambda^2 \end{aligned} \]

\[\begin{aligned} D(X)&=E[X(X-1)]+E(X)-E^2(X)=\lambda^2+\lambda-\lambda^2=\lambda \end{aligned} \]

泊松分布的期望和方差都是参数\(\lambda\)!

import numpy as np
a = np.random.poisson(55,size=(4,))
print(a)
print(type(a))
>>> [46 50 39 57]
<class 'numpy.ndarray'>

4.均匀分布——连续型概率分布

概率密度函数为

\[f(x)=\left\{ \begin{aligned} &\frac{1}{b-a},&a<x<b\\ &0,&others \end{aligned} \right. \]

期望\(E(X)=\int_{-\infty}^{\infty}x*f(x)dx=\frac{a+b}{2}\)
方差\(D(X)=E(X^2)-E^2(X)=\int_{a}^{b}x^2*\frac{1}{b-a}dx-\frac{(a+b)^2}{4}=\frac{(b-a)^2}{12}\)

#np.random.uniform(low=0.0, high=1.0, size=None)

a = np.random.uniform(20,50,size=(2,6))
print(a)
print(type(a))
>>> [[ 45.20217569  43.75312926  26.52703807  41.91200572  42.85374841
   29.24479553]
 [ 45.12516381  30.12544796  35.53555014  32.28527649  21.76682194
   46.33104556]]
<class 'numpy.ndarray'>

5.指数分布——连续型概率分布

概率密度函数为

\[f(x)=\left\{ \begin{aligned} &\frac{1}{\theta}e^{-\frac{x}{\theta}},&x>0,\\ &0,&x\leq0 \end{aligned} \right. \]

其中\(\theta>0\)
期望

\[\begin{aligned} E(X)&=\int_0^{+\infty}x*f(x)dx\\ &=\int_0^{\infty}x\frac{1}{\theta}e^{-\frac{x}{\theta}}dx\\ &=-\int_0^{\infty}xd(e^{-\frac{x}{\theta}})\\ &=-[xe^{-\frac{x}{\theta}}|_0^{\infty}-\int_0^{\infty}e^{-\frac{x}{\theta}}dx]\\ &=\theta \end{aligned} \]

方差

\[\begin{aligned} D(X)&=E(X^2)-E^2(X)\\ &=\int_0^{+\infty}x^2\frac{1}{\theta}e^{-\frac{x}{\theta}}-\theta^2\\ &=2\theta^2-\theta^2\\ &=\theta^2 \end{aligned} \]

6.正态分布/高斯分布

设随机变量X服从正态分布,即X~\(N(\mu,\sigma^2)\)
概率密度函数为

\[f(x)=\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}} \]

期望\(E(X)=\mu\)
方差\(D(X)=\sigma^2\)

a = np.random.normal(40,3,size=(5,2))
print(a)
print(type(a))
>>>[[ 42.75053239  36.92362467]
 [ 42.90588338  38.58249427]
 [ 42.91278062  39.05507689]
 [ 39.69794259  40.26237062]
 [ 38.90643225  42.94278753]]
<class 'numpy.ndarray'>
posted @ 2017-12-13 09:35  米小粥  阅读(3341)  评论(0编辑  收藏  举报