常用随机变量的数学期望和方差

0-1分布

X	0	1
P	1-P	P

$$\begin{gathered} E(X)=0\times (1-p)+1\times p=p \\ E(X^2)=0^2\times (1-p)+1^2\times p=p \\ D(X)=E(X^2)-E^2(X)=p-p^2=p(1-p) \end{gathered}$$

二项分布，$X\sim B(n,p)$

可看作n次0-1分布，设$X=X_1+X_2+\dots +X_n$
$$\begin{gathered} E(X_i)=p,D(X_i)=p(1-p) \\ E(X)=nE(X_i)=np,D(X)=nD(X_i)=np(1-p) \end{gathered}$$

泊松分布，$X\sim P(\lambda )$

$$\begin{gathered} P(X=k)=\frac{{\lambda }^k{e}^{-\lambda }}{k!},k=0,1,\dots \\ E(X)=\sum _{k=0}^{+\infty }kP(X=k)=\sum _{k=0}^{+\infty }k\frac{{\lambda }^k{e}^{-\lambda }}{k!}=\sum _{k=1}^{+\infty }k\frac{{\lambda }^k{e}^{-\lambda }}{k!}=\sum _{k=1}^{+\infty }\frac{{\lambda }^k{e}^{-\lambda }}{(k-1)!}=\sum _{k=0}^{+\infty }\frac{{\lambda }^{k+1}{e}^{-\lambda }}{k!}=\lambda e^{-\lambda }\sum _{k=0}^{+\infty }\frac{{\lambda }^k}{k!}=\lambda e^{-\lambda }{e}^{\lambda }=\lambda \\ E(X^2)=\sum _{k=0}^{+\infty }{k}^{2}P(X=k)=\sum _{k=0}^{+\infty }{k}^2\frac{{\lambda }^k{e}^{-\lambda }}{k!}=\sum _{k=1}^{+\infty }{k}^2\frac{{\lambda }^k{e}^{-\lambda }}{k!}=\sum _{k=1}^{+\infty }k\frac{{\lambda }^k{e}^{-\lambda }}{(k-1)!}=\sum _{k=0}^{+\infty }(k+1)\frac{{\lambda }^{k+1}{e}^{-\lambda }}{k!}=\sum _{k=0}^{+\infty }k\frac{{\lambda }^{k+1}{e}^{-\lambda }}{k!}+\sum _{k=0}^{+\infty }\frac{{\lambda }^{k+1}{e}^{-\lambda }}{k!}=\sum _{k=1}^{+\infty }\frac{{\lambda }^{k+1}{e}^{-\lambda }}{(k-1)!}+\sum _{k=0}^{+\infty }\frac{{\lambda }^{k+1}{e}^{-\lambda }}{k!}=\sum _{k=0}^{+\infty }\frac{{\lambda }^{k+2}{e}^{-\lambda }}{k!}+\sum _{k=0}^{+\infty }\frac{{\lambda }^{k+1}{e}^{-\lambda }}{k!}={\lambda }^2{e}^{-\lambda }\sum _{k=0}^{+\infty }\frac{{\lambda }^k}{k!}+\lambda e^{-\lambda }\sum _{k=0}^{+\infty }\frac{{\lambda }^k}{k!}={\lambda }^2{e}^{-\lambda }{e}^{\lambda }+\lambda e^{-\lambda }{e}^{\lambda }={\lambda }^2+\lambda \\ D(X)=E(X^2)-E^2(X)={\lambda }^2+\lambda -{\lambda }^2=\lambda \end{gathered}$$

几何分布 $P(X=k)=p(1-p{)}^{k-1},k=1,2,\dots ,0<p<1$

$$\begin{gathered} E(X)=\sum _{k=1}^{+\infty }kP(X=k)=\sum _{k=1}^{+\infty }kp(1-p{)}^{k-1}=p\sum _{k=1}^{+\infty }k(1-p{)}^{k-1}=p[\sum _{k=1}^{+\infty }(1-p{)}^k{]}^{\prime }=p[\frac{1-p}{1-(1-p)}{]}^{\prime }=p(\frac{x}{1-x}{)}^{\prime }=p\frac{1}{(1-x{)}^2}=\frac{1}{p} \\ E(X^2)=\sum _{k=1}^{+\infty }{k}^{2}P(X=k)=\sum _{k=1}^{+\infty }{k}^{2}p(1-p{)}^{k-1}=p\sum _{k=1}^{+\infty }{k}^2(1-p{)}^{k-1}=p\sum _{k=1}^{+\infty }(k^2+k-k)(1-p{)}^{k-1}=p\sum _{k=1}^{+\infty }[k(k+1)-k](1-p{)}^{k-1}=p[\sum _{k=1}^{+\infty }(1-p{)}^{k+1}{]}^{\prime \prime }-p[\sum _{k=1}^{+\infty }(1-p{)}^k{]}^{\prime }=p[\frac{(1-p{)}^2}{1-(1-p)}{]}^{\prime \prime }-E(X)=\frac{1-p}{p^2} \end{gathered}$$

均匀分布 $X\sim U(a,b)$

$$\begin{gathered} f_X(x)=\frac{1}{b-a},a<x<b \\ E(X)={\int }_a^b\frac{x}{b-a}dx=\frac{a+b}{2} \\ E(X^2)={\int }_a^b\frac{x^2}{b-a}dx=\frac{a^2+ab+b^2}{3} \\ D(X)=\frac{a^2+ab+b^2}{3}-(\frac{a+b}{2}{)}^2=\frac{(b-a{)}^2}{12} \end{gathered}$$

指数分布，$X\sim E(\lambda )$,利用伽马函数的性质

$$\begin{gathered} f_X(x)=\lambda e^{-\lambda x},x\ge 0 \\ E(X)={\int }_0^{+\infty }xf(x)dx={\int }_0^{+\infty }\lambda x{e}^{-\lambda x}dx=\frac{1}{\lambda }{\int }_0^{+\infty }(\lambda x)e^{-(\lambda x)}d(\lambda x)=\frac{T(2)}{\lambda }=\frac{1}{\lambda } \\ E(X^2)={\int }_0^{+\infty }{x}^{2}f(x)dx={\int }_0^{+\infty }\lambda x^2{e}^{-\lambda x}dx=\frac{1}{{\lambda }^2}{\int }_0^{+\infty }(\lambda x{)}^2{e}^{-(\lambda x)}d(\lambda x)=\frac{T(3)}{{\lambda }^2}=\frac{2}{{\lambda }^2} \\ D(X)=\frac{2}{{\lambda }^2}-\frac{1}{{\lambda }^2}=\frac{1}{{\lambda }^2} \end{gathered}$$

正态分布，$X\sim N(\mu ,{\sigma }^{)}$

$$\begin{gathered} f_X(x)=\frac{1}{\sqrt{2\pi }\sigma }{e}^{-\frac{(x-\mu {)}^2}{2{\sigma }^2}},x\in R \\ E(X)={\int }_{-\infty }^{+\infty }x\frac{1}{\sqrt{2\pi }\sigma }{e}^{-\frac{(x-\mu {)}^2}{2{\sigma }^2}}dx={\int }_{-\infty }^{+\infty }(x-\mu )\frac{1}{\sqrt{2\pi }\sigma }{e}^{-\frac{(x-\mu {)}^2}{2{\sigma }^2}}d(x-\mu )+{\int }_{-\infty }^{+\infty }\mu \frac{1}{\sqrt{2\pi }\sigma }{e}^{-\frac{(x-\mu {)}^2}{2{\sigma }^2}}dx=0+\mu =\mu \\ D(X)=E(X-EX{)}^2=E(X-\mu {)}^2={\int }_{-\infty }^{+\infty }(x-\mu {)}^2\frac{1}{\sqrt{2\pi }\sigma }{e}^{-\frac{(x-\mu {)}^2}{2{\sigma }^2}}dx \\ 令u=\frac{x-\mu }{\sqrt{2}\sigma } \\ D(X)=\sqrt{2}\sigma {\int }_{-\infty }^{+\infty }\frac{2{\sigma }^2}{\sqrt{2\pi }\sigma }{u}^2{e}^{-u^2}du=\frac{2{\sigma }^2}{\sqrt{\pi }}{\int }_{-\infty }^{+\infty }{u}^2{e}^{-u^2}du=\frac{2{\sigma }^2}{\sqrt{\pi }}\cdot \frac{\sqrt{\pi }}{2}={\sigma }^2 \end{gathered}$$