前言

简介

       本文是对概率论中常见分布包括二项分布、0-1分布、泊松分布、均匀分布、正态分布、指数分布的期望和方差的证明整合，附加自己的推导或理解。

导览

      二项分布 (Binomial Distribution)
      泊松分布 (Poisson’s Distribution)
      均匀分布 (Uniform Distribution)
      正态分布 (Normal Distribution)
      指数分布 (Exponential Distribution)

总结

      二项分布 (Binomial Distribution) X~B(n,p)：E(X)=np，D(X)=np(1-p)=npq 。
      0-1分布 X~B(1,p)：E(X)=p，D(X)=p(1-p)=pq 。
      泊松分布 (Poisson’s Distribution) X~P($\lambda$λ)：E(X)=λ，D(X)=λ 。
      均匀分布 (Uniform Distribution) X~U(a,b) ：E(X)=$(a+b)/2$(a+b)/2，D(X)=$(b-a)^2/12$(b−a)2/12 。
      正态分布 (Normal Distribution) X~N(μ,σ)：E(X)=μ，D(X)=$σ^2$σ2。
      指数分布 (Exponential Distribution)X~Γ(1,β)：E(X)=$1/λ$1/λ，D(X)=$1/λ^2$1/λ2 。

正文

二项分布 (Binomial Distribution) From QUETAL and chs007chs

Part Ⅰ From QUETAL

X~B(n,p)

分布律：$P(X=k) = {n\choose k}p^kq^{n-k}，k = 0,1,2,..,n，q = 1-p$P(X=k)=(kn​)pkqn−k，k=0,1,2,..,n，q=1−p
期望：
$EX = \sum_{k=0}^n k {n\choose k}p^kq^{n-k} \\ = \sum_{k=1}^n k {n\choose k}p^kq^{n-k} \\ = \sum_{k=1}^n k {\frac{n!}{k!(n-k)!}}p^kq^{n-k} \\ = np\sum_{k=1}^n {\frac{(n-1)!}{(k-1)!(n-k)!}}p^{k-1}q^{(n-1)-(k-1)} \\ = np\sum_{k=1}^n{n-1\choose k-1}p^{k-1}q^{(n-1)-(k-1)}\\ = np[{n-1\choose 0}p^0q^{n-1}+{n-1\choose 1}p^1q^{n-2}+...+{n-1\choose n-1}p^{n-1}q^0] \\ = np$EX=k=0∑n​k(kn​)pkqn−k=k=1∑n​k(kn​)pkqn−k=k=1∑n​kk!(n−k)!n!​pkqn−k=npk=1∑n​(k−1)!(n−k)!(n−1)!​pk−1q(n−1)−(k−1)=npk=1∑n​(k−1n−1​)pk−1q(n−1)−(k−1)=np[(0n−1​)p0qn−1+(1n−1​)p1qn−2+...+(n−1n−1​)pn−1q0]=np
方差：$DX = EX^2-(EX)^2$DX=EX2−(EX)2
计算EX^2：
$EX^2 = \sum_{k=1}^nk^2{n\choose k}p^kq^{n-k}, k = 0,1,2,..,n,q = 1-p\\ = \sum_{k=1}^n[k(k-1)+k]{n\choose k}p^kq^{n-k}\\ = \sum_{k=1}^nk(k-1){n\choose k}p^kq^{n-k} + \sum_{k=1}^nk{n\choose k}p^kq^{n-k}\\ 其中， \sum_{k=1}^nk{n\choose k}p^kq^{n-k} = EX = np\\ \sum_{k=1}^nk(k-1){n\choose k}p^kq^{n-k} \\ = \sum_{k=1}^nk(k-1){\frac{n!}{k!(n-k)!}}p^2p^{k-2}q^{n-k} \\ = \sum_{k=2}^nk(k-1){\frac{n!}{k!(n-k)!}}p^2p^{k-2}q^{n-k} \\$EX2=k=1∑n​k2(kn​)pkqn−k,k=0,1,2,..,n,q=1−p=k=1∑n​[k(k−1)+k](kn​)pkqn−k=k=1∑n​k(k−1)(kn​)pkqn−k+k=1∑n​k(kn​)pkqn−k其中，k=1∑n​k(kn​)pkqn−k=EX=npk=1∑n​k(k−1)(kn​)pkqn−k=k=1∑n​k(k−1)k!(n−k)!n!​p2pk−2qn−k=k=2∑n​k(k−1)k!(n−k)!n!​p2pk−2qn−k注：特别注意这里k=1时项为0，所以可以从k=2开始计算。$\\ = \sum_{k=1}^n{\frac{n(n-1)(n-2)!}{(k-2)!(n-k)!}}p^2p^{k-2}q^{[(n-2)-(k-2)]} \\ = n(n-1)p^2\sum_{k=2}^n{\frac{(n-2)!}{(k-2)!(n-k)!}}p^{k-2}q^{[(n-2)-(k-2)]}\\ = n(n-1)p^2\sum_{k=2}^n{n-2\choose k-2}p^{k-2}q^{[(n-2)-(k-2)]}\\ = n(n-1)p^2 \\ \rightarrow EX^2 = n(n-1)p^2+np \\$=k=1∑n​(k−2)!(n−k)!n(n−1)(n−2)!​p2pk−2q[(n−2)−(k−2)]=n(n−1)p2k=2∑n​(k−2)!(n−k)!(n−2)!​pk−2q[(n−2)−(k−2)]=n(n−1)p2k=2∑n​(k−2n−2​)pk−2q[(n−2)−(k−2)]=n(n−1)p2→EX2=n(n−1)p2+np$\rightarrow DX = EX^2-(EX)^2 = np-np^2 = np(1-p)$→DX=EX2−(EX)2=np−np2=np(1−p)

Part Ⅱ 0-1分布 From chs007chs

X~B(1,p)

也可以从上式直接推导得到

泊松分布 (Poisson’s Distribution) From saltriver

X~P($\lambda$λ)

分布律：$P(X=k)=\frac{\lambda ^{k}e^{-\lambda }}{k!}$P(X=k)=k!λke−λ​
期望：
$E(X)=\sum_{k=0}^{\infty }k\cdot \frac{\lambda ^{k}e^{-\lambda }}{k!} \\$E(X)=k=0∑∞​k⋅k!λke−λ​$因为k=0时， k⋅λke−λk!=0 \\$因为k=0时，k⋅λke−λk!=0$E(X)=\sum_{k=1}^{\infty }k\cdot \frac{\lambda ^{k}e^{-\lambda }}{k!} \\$E(X)=k=1∑∞​k⋅k!λke−λ​$E(X)=\sum_{k=1}^{\infty }k\cdot \frac{\lambda ^{k}e^{-\lambda }}{k!}=\sum_{k=1}^{\infty } \frac{\lambda ^{k}e^{-\lambda }}{(k-1)!}=\sum_{k=1}^{\infty } \frac{\lambda ^{k-1}\lambda e^{-\lambda }}{(k-1)!}=\lambda e^{-\lambda }\sum_{k=1}^{\infty } \frac{\lambda ^{k-1}}{(k-1)!} \\$E(X)=k=1∑∞​k⋅k!λke−λ​=k=1∑∞​(k−1)!λke−λ​=k=1∑∞​(k−1)!λk−1λe−λ​=λe−λk=1∑∞​(k−1)!λk−1​$用到泰勒展开式：e^{x}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+...+\frac{x^{n}}{n!}+...=\sum_{k=1}^{\infty } \frac{x ^{k-1}}{(k-1)!} \\$用到泰勒展开式：ex=1+x+2!x2​+3!x3​+...+n!xn​+...=k=1∑∞​(k−1)!xk−1​$E(X)=\lambda e^{-\lambda }\sum_{k=1}^{\infty } \frac{\lambda ^{k-1}}{(k-1)!}=\lambda e^{-\lambda }e^{\lambda }=\lambda$E(X)=λe−λk=1∑∞​(k−1)!λk−1​=λe−λeλ=λ
方差：$DX = EX^2-(EX)^2$DX=EX2−(EX)2
计算EX^2：
$E(X^2)=\sum_{k=0}^{\infty }k^2 \cdot \frac{\lambda ^{k}e^{-\lambda }}{k!}=\lambda e^{-\lambda} \sum_{k=1}^{\infty } \frac{k \lambda ^{k-1}}{(k-1)!}=\lambda e^{-\lambda} \sum_{k=1}^{\infty } \frac{(k-1+1) \lambda ^{k-1}}{(k-1)!} \\$E(X2)=k=0∑∞​k2⋅k!λke−λ​=λe−λk=1∑∞​(k−1)!kλk−1​=λe−λk=1∑∞​(k−1)!(k−1+1)λk−1​$=\lambda e^{-\lambda} (\sum_{m=0}^{\infty } \frac{m \cdot \lambda ^{m}}{m!}+\sum_{m=0}^{\infty } \frac{ \lambda ^{m}}{m!}) (m=k-1) \\$=λe−λ(m=0∑∞​m!m⋅λm​+m=0∑∞​m!λm​)(m=k−1)$=\lambda e^{-\lambda} ( \lambda \cdot \sum_{m=1}^{\infty } \frac{\lambda ^{m-1}}{(m-1)!}+\sum_{m=0}^{\infty } \frac{ \lambda ^{m}}{m!}) \\$=λe−λ(λ⋅m=1∑∞​(m−1)!λm−1​+m=0∑∞​m!λm​)$=\lambda e^{-\lambda}(\lambda e^{\lambda}+e^\lambda)=\lambda(\lambda+1) \\$=λe−λ(λeλ+eλ)=λ(λ+1)$\rightarrow D(X)=E(X^2)-(E(X))^2=\lambda(\lambda+1)-\lambda^2=\lambda$→D(X)=E(X2)−(E(X))2=λ(λ+1)−λ2=λ

均匀分布 (Uniform Distribution) From chs007chs

X~U(a,b)

推导较为简单。

正态分布 (Normal Distribution) From 一只驽马

X~N(μ,σ)

概率密度函数：$f_X(x) =\frac{1}{\sigma\sqrt{2\pi}}\exp\left\{-\frac{(x-\mu)^2}{2\sigma^2}\right\}$fX​(x)=σ2π​1​exp{−2σ2(x−μ)2​}

期望：$E(x) = \mu\int_{-\infty}^{+\infty}\mathcal{N}(x|\mu' = 0, \sigma^2)dx = \mu$E(x)=μ∫−∞+∞​N(x∣μ′=0,σ2)dx=μ

方差：$\Rightarrow V(X) = \sigma^2\frac{4}{\sqrt{\pi}}\frac 12 \frac {\sqrt \pi}{2} = \sigma^2$⇒V(X)=σ2π​4​21​2π​​=σ2

指数分布 (Exponential Distribution) From saltriver

X~Γ(1,β)

概率密度函数：
期望：
$E(X)=\int_{-\infty }^{\infty }|x|f(x)dx=\int_{0}^{\infty }xf(x)dx=\int_{0}^{\infty }x\cdot\lambda e^{-\lambda x}dx=\frac {1} {\lambda}\int_{0}^{\infty }\lambda xe^{-\lambda x}d\lambda x \\ 令u=λx，E(X)=\frac {1} {\lambda}\int_{0}^{\infty }ue^{-u}du=\frac {1} {\lambda}[(-e^{-u}-ue^{-u})|(\infty,0)]=\frac {1} {\lambda} \\$E(X)=∫−∞∞​∣x∣f(x)dx=∫0∞​xf(x)dx=∫0∞​x⋅λe−λxdx=λ1​∫0∞​λxe−λxdλx令u=λx，E(X)=λ1​∫0∞​ue−udu=λ1​[(−e−u−ue−u)∣(∞,0)]=λ1​
方差：$DX = EX^2-(EX)^2$DX=EX2−(EX)2
计算EX^2：
$E(X^2)=\int_{-\infty }^{\infty }|x^2|f(x)dx=\int_{0}^{\infty }x^2f(x)dx=\int_{0}^{\infty }x^2\cdot\lambda e^{-\lambda x}dx \\ E(X^2)=\frac {1} {\lambda^2}\int_{0}^{\infty }\lambda x \lambda xe^{-\lambda x}d\lambda x \\ 令u=λx，E(X^2)=\frac {1} {\lambda^2}\int_{0}^{\infty }u^2e^{-u}du=\frac {1} {\lambda^2}[(-2e^{-u}-2ue^{-u}-u^2e^{-u})|(\infty,0)]=\frac {1} {\lambda^2}\cdot 2=\frac {2} {\lambda^2} \\ \rightarrow D(X)=E(X^2)-(E(X))^2=\frac {2} {\lambda^2}-(\frac {1} {\lambda})^2=\frac {1} {\lambda^2}$E(X2)=∫−∞∞​∣x2∣f(x)dx=∫0∞​x2f(x)dx=∫0∞​x2⋅λe−λxdxE(X2)=λ21​∫0∞​λxλxe−λxdλx令u=λx，E(X2)=λ21​∫0∞​u2e−udu=λ21​[(−2e−u−2ue−u−u2e−u)∣(∞,0)]=λ21​⋅2=λ22​→D(X)=E(X2)−(E(X))2=λ22​−(λ1​)2=λ21​