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{x^k}{k!}e^{-x} $ ---> \(\frac{\lambda^k}{k!}e^{-\lambda}\) 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 npa = np.random.poisson(55,size=(4,))print(a)print(type(a))>>> [46 50 39 57]
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]]
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]]