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
[egin{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
[egin{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}\
&=npsum_{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
[egin{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}
]
[egin{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^2sum_{k=2}^{n}frac{(n-2)!}{(k-2)!(n-k)!}p^{k-2}(1-p)^{n-k}\
&=n(n-1)p^2
end{aligned}
]
[egin{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
[egin{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{-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
[egin{aligned}
E(X)&=sum_{k=0}^{infty}k*f(x)\
&=sum_{k=0}^{infty}k*frac{lambda^k}{k!}e^{-lambda}\
&=lambdasum_{k=1}^{infty}frac{lambda^{k-1}}{(k-1)!}e^{-lambda}\
&=lambda
end{aligned}
]
方差
[egin{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}
]
[egin{aligned}
E[X(X-1)&=sum_{k=0}^{infty}k(k-1)*frac{lambda^k}{k!}e^{-lambda}\
&=lambda^2sum_{k=2}^{infty}frac{lambda^{k-2}}{(k-2)!}e^{-lambda}\
&=lambda^2
end{aligned}
]
[egin{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{
egin{aligned}
&frac{1}{b-a},&a<x<b\
&0,&others
end{aligned}
ight.
]
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{
egin{aligned}
&frac{1}{ heta}e^{-frac{x}{ heta}},&x>0,\
&0,&xleq0
end{aligned}
ight.
]
其中( heta>0)
期望
[egin{aligned}
E(X)&=int_0^{+infty}x*f(x)dx\
&=int_0^{infty}xfrac{1}{ heta}e^{-frac{x}{ heta}}dx\
&=-int_0^{infty}xd(e^{-frac{x}{ heta}})\
&=-[xe^{-frac{x}{ heta}}|_0^{infty}-int_0^{infty}e^{-frac{x}{ heta}}dx]\
&= heta
end{aligned}
]
方差
[egin{aligned}
D(X)&=E(X^2)-E^2(X)\
&=int_0^{+infty}x^2frac{1}{ heta}e^{-frac{x}{ heta}}- heta^2\
&=2 heta^2- heta^2\
&= heta^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{2pi}sigma}e^{-frac{(x-mu)^2}{2sigma^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'>