$$$$

0-1 分布

若 $X\sim B(1,p)$ 则

$$\begin{array}{c}P(X=1)=p,P(X=0)=1-p\\ \mathbb{E}(X)=p\\ \mathbb{V}(X)=p(1-p)\end{array}$$

证

$$\begin{array}{c}\mathbb{E}(X)=0\cdot P(X=0)+1\cdot P(X=1)=p\\ \mathbb{E}\left(X^2\right)=0^2\cdot P(X=0)+1^2\cdot P(X=1)=p\\ \mathbb{V}(X)=\mathbb{E}\left(X^2\right)-[\mathbb{E}(X){]}^2=p^2-p=p(1-p)\end{array}$$

二项分布

若 $X\sim B(n,p)$ 则

$$\begin{array}{c}P(X=k)=(\genfrac{}{}{0ex}{}{n}{k})p^k(1-p{)}^{n-k}(k=0,1,2,\cdots ,n)\\ \mathbb{E}(X)=np\\ \mathbb{V}(X)=np(1-p)\end{array}$$

证

根据二项分布的可加性，设 $\{X_i{\}}_{i=1}^n$ 相互独立，均服从 0-1 分布，且 $X=\sum _{i=1}^n{X}_i$，则

$$\begin{array}{c}\mathbb{E}(X)=\sum _{k=0}^n\mathbb{E}(X_k)=np\\ \mathbb{V}(X)=\sum _{k=0}^n\mathbb{V}(X_k)=np(1-p)\end{array}$$

泊松分布

若 $X\sim P(\lambda )$ 则

$$\begin{array}{c}P(X=k)={\mathrm{e}}^{-\lambda }\frac{{\lambda }^k}{k!}(k=0,1,2,\cdots )\\ \mathbb{E}(X)=\lambda \\ \mathbb{V}(X)=\lambda \end{array}$$

证

$$\begin{array}{c}\begin{array}{rl}\mathbb{E}(X)& =\sum _{k=0}^{\mathrm{\infty }}k\cdot P(X=k)=\sum _{k=0}^{\mathrm{\infty }}k\cdot {\mathrm{e}}^{-\lambda }\frac{{\lambda }^k}{k!}\\ & =\lambda {\mathrm{e}}^{-\lambda }\sum _{k=1}^{\mathrm{\infty }}\frac{{\lambda }^{k-1}}{(k-1)!}=\lambda {\mathrm{e}}^{-\lambda }{\mathrm{e}}^{\lambda }=\lambda \end{array}\\ \begin{array}{rl}\mathbb{E}\left(X^2\right)& =\sum _{k=0}^{\mathrm{\infty }}{k}^2\cdot P(X=k)=\sum _{k=0}^{\mathrm{\infty }}{k}^2\cdot {\mathrm{e}}^{-\lambda }\frac{{\lambda }^k}{k!}\\ & =\sum _{k=0}^{\mathrm{\infty }}k(k-1)\cdot {\mathrm{e}}^{-\lambda }\frac{{\lambda }^k}{k!}+\sum _{k=0}^{\mathrm{\infty }}k\cdot {\mathrm{e}}^{-\lambda }\frac{{\lambda }^k}{k!}\\ & ={\lambda }^2{\mathrm{e}}^{-\lambda }\sum _{k=2}^{\mathrm{\infty }}\frac{{\lambda }^{k-2}}{(k-2)!}+\mathbb{E}(X)\\ & ={\lambda }^2{\mathrm{e}}^{-\lambda }{e}^{\lambda }+\lambda ={\lambda }^2+\lambda \end{array}\\ \mathbb{V}(X)=\mathbb{E}\left(X^2\right)-[\mathbb{E}(X){]}^2={\lambda }^2+\lambda -\lambda =\lambda \end{array}$$

几何分布

若 $X\sim G(p)$ 则

$$\begin{array}{c}P(X=k)=(1-p{)}^{k-1}p(k=1,2,\cdots )\\ \mathbb{E}(X)=\frac{1}{p}\\ \mathbb{V}(X)=\frac{1-p}{p^2}\end{array}$$

证

$$\begin{array}{c}\begin{array}{rl}\mathbb{E}(X)& =\sum _{k=1}^{\mathrm{\infty }}k\cdot (1-p{)}^{k-1}p=p\sum _{k=1}^{\mathrm{\infty }}k(1-p{)}^{k-1}\\ & \stackrel{q=1-p}{====}p\cdot \frac{1}{(1-q{)}^2}=p\cdot \frac{1}{p^2}=\frac{1}{p}\end{array}\\ \begin{array}{rl}\mathbb{E}\left(X^2\right)& =\sum _{k=1}^{\mathrm{\infty }}{k}^2\cdot (1-p{)}^{k-1}p=p\sum _{k=1}^{\mathrm{\infty }}{k}^2{q}^{k-1}\\ & \stackrel{q=1-p}{====}p\cdot \frac{1+q}{(1-q{)}^3}=p\cdot \frac{2-p}{p^3}=\frac{2-p}{p^2}\end{array}\\ \mathbb{V}(X)=\mathbb{E}\left(X^2\right)-{[\mathbb{E}(X)]}^2=\frac{2-p}{p^2}-{\left(\frac{1}{p}\right)}^2=\frac{1-p}{p^2}\end{array}$$

超几何分布

若 $X\sim H(N,M,n)$ 则

$$\begin{array}{c}P(X=k)=\frac{(\genfrac{}{}{0ex}{}{M}{k})(\genfrac{}{}{0ex}{}{N-M}{n-k})}{(\genfrac{}{}{0ex}{}{N}{n})}(k=max\{0,n-(N-M)\},\cdots ,min\{n,M\})\\ \mathbb{E}(X)=n\frac{M}{N}\\ \mathbb{V}(X)=\frac{n(N-n)(N-M)M}{N^2(N-1)}\end{array}$$

证

没有必要。

均匀分布

若 $X\sim U(a,b)$ 则

$$\begin{array}{c}{f}_X(x)=\{\begin{array}{ll}{\displaystyle \frac{1}{b-a}}& a<x<b\\ 0& \text{other}\end{array}\\ \mathbb{E}(X)=\frac{a+b}{2}\\ \mathbb{V}(X)=\frac{1}{12}(b-a{)}^2\end{array}$$

证

$$\begin{array}{c}\mathbb{E}(X)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}x{f}_X(x)\mathrm{d}x=\frac{1}{b-a}{\int }_a^{b}x\mathrm{d}x=\frac{b^2-a^2}{2(b-a)}=\frac{a+b}{2}\\ \mathbb{E}\left(X^2\right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{x}^2{f}_X(x)\mathrm{d}x=\frac{1}{b-a}{\int }_a^b{x}^2\mathrm{d}x=\frac{b^3-a^3}{3(b-a)}=\frac{a^2+ab+b^2}{3}\\ \mathbb{V}(X)=\mathbb{E}\left(X^2\right)-[\mathbb{E}(X){]}^2=\frac{a^2+ab+b^2}{3}-{\left(\frac{a+b}{2}\right)}^2=\frac{(b-a{)}^2}{12}\end{array}$$

指数分布

若 $X\sim \mathbb{E}(\lambda )$ 则

$$\begin{array}{c}{f}_X(x)=\{\begin{array}{ll}\lambda {\mathrm{e}}^{-\lambda x}& x>0\\ 0& x\le 0\end{array}\\ \mathbb{E}(X)=\frac{1}{\lambda }\\ \mathbb{V}(X)=\frac{1}{{\lambda }^2}\end{array}$$

证

$$\begin{array}{c}\begin{array}{rl}\mathbb{E}(X)& ={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}x{f}_X(x)\mathrm{d}x={\int }_0^{+\mathrm{\infty }}x\cdot \lambda {\mathrm{e}}^{-\lambda x}\mathrm{d}x={\int }_0^{+\mathrm{\infty }}x\cdot \mathrm{d}(-{\mathrm{e}}^{-\lambda x})\\ & ={[-x{\mathrm{e}}^{-\lambda x}]}_0^{+\mathrm{\infty }}-{\int }_0^{+\mathrm{\infty }}{\mathrm{e}}^{-\lambda x}\mathrm{d}x={[-\frac{1}{\lambda }{e}^{-\lambda x}]}_0^{+\mathrm{\infty }}=\frac{1}{\lambda }\end{array}\\ \begin{array}{rl}\mathbb{E}\left(X^2\right)& ={\int }_0^{\mathrm{\infty }}{x}^2\lambda e^{-\lambda x}dx={[-x^2{e}^{-\lambda x}]}_0^{\mathrm{\infty }}+2{\int }_0^{\mathrm{\infty }}x{e}^{-\lambda x}dx\\ & =0+2\cdot \frac{1}{{\lambda }^2}=\frac{2}{{\lambda }^2}\end{array}\\ \mathbb{V}(X)=\mathbb{E}\left(X^2\right)-[\mathbb{E}(X){]}^2=\frac{2}{{\lambda }^2}-{\left(\frac{1}{\lambda }\right)}^2=\frac{1}{{\lambda }^2}\end{array}$$

正态分布

若 $X\sim N(\mu ,{\sigma }^2)$ 则

$$\begin{array}{c}{f}_X(x)=\frac{1}{\sigma \sqrt{2\pi }}{\mathrm{e}}^{-\frac{1}{2}{\left(\frac{x-\mu }{\sigma }\right)}^2}\\ \mathbb{E}(X)=\mu \\ \mathbb{V}(X)={\sigma }^2\end{array}$$

证

$$\begin{array}{c}\begin{array}{rl}\mathbb{E}(X)& ={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}x\frac{1}{\sigma \sqrt{2\pi }}{e}^{-\frac{1}{2}{\left(\frac{x-\mu }{\sigma }\right)}^2}\mathrm{d}x\\ & \stackrel{t=\frac{x-\mu }{\sigma }}{=====}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}(\sigma t+\mu )\frac{1}{\sigma \sqrt{2\pi }}{e}^{-\frac{1}{2}{t}^2}\sigma \mathrm{d}t\\ & ={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}(\sigma t+\mu )\frac{1}{\sqrt{2\pi }}{e}^{-\frac{1}{2}{t}^2}\mathrm{d}t\\ & =\mu {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{1}{\sqrt{2\pi }}{e}^{-\frac{1}{2}{t}^2}\mathrm{d}t+\sigma {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}t\frac{1}{\sqrt{2\pi }}{e}^{-\frac{1}{2}{t}^2}\mathrm{d}t\\ & =\mu \cdot 1+\sigma \cdot 0=\mu \\ \mathbb{E}[(X-\mathbb{E}(X){)}^2]& ={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}(x-\mu {)}^2\frac{1}{\sigma \sqrt{2\pi }}{e}^{-\frac{1}{2}{\left(\frac{x-\mu }{\sigma }\right)}^2}\mathrm{d}x\\ & \stackrel{t=\frac{x-\mu }{\sigma }}{=====}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{\sigma }^2{t}^2\frac{1}{\sigma \sqrt{2\pi }}{e}^{-\frac{1}{2}{t}^2}\sigma \mathrm{d}t\\ & ={\sigma }^2{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{t}^2\frac{1}{\sqrt{2\pi }}{e}^{-\frac{1}{2}{t}^2}\mathrm{d}t\\ & ={\sigma }^2\cdot 1={\sigma }^2\end{array}\\ \mathbb{V}(X)=\mathbb{E}[(X-\mathbb{E}(X){)}^2]={\sigma }^2\end{array}$$