常见分布期望方差推导
0-1分布
$p_k=p^k(1-p)^{1-k},k=0,1$
$E(X)=\sum\limits_{k=0}^1 kp^k(1-p)^{1-k}=p$
$V(X)=E(X^2)-E(X)^2=p-p^2=p(1-p)$
二项分布$b(n,p)$
$p_k=\binom{n}{k}p^k(1-p)^{n-k},k=0,1,…,n$
$E(X)=\sum\limits_{k=0}^n k\binom{n}{k}p^k(1-p)^{n-k}=np\sum\limits_{k=1}^n\binom{n-1}{k-1}p^{k-1}(1-p)^{(n-1)-(k-1)}=np(p+(1-p))^n=np $
$V(X)=E(X^2)-E(X)^2=\sum\limits_{k=0}^nk^2\binom{n}{k}p^k(1-p)^{n-k}-E(X)^2\\=\sum\limits_{k=2}^nk(k-1)\binom{n}{k}p^k(1-p)^{n-k}+\sum\limits_{k=0}^nk\binom{n}{k}p^k(1-p)^{n-k}-E(X)^2=n(n-1)p^2+np-n^2p^2=np(1-p)$
泊松分布$P(\lambda)$
$p_k=\frac{\lambda^k}{k!}e^{-\lambda k},k=0,1,…$
$E(X)=\sum\limits_{k=0}^\infty k\frac{\lambda^k}{k!}e^{-\lambda}=\lambda e^{-\lambda}\sum\limits_{k=0}^\infty\frac{\lambda^k}{k!}=\lambda e^{-\lambda}e^\lambda=\lambda$
$E(X^2)=\sum\limits_{k=0}^\infty k^2\frac{\lambda^k}{k!}e^{-\lambda}=\sum\limits_{k=0}^\infty [k(k-1)+k]\frac{\lambda^k}{k!}e^{-\lambda}=e^{-\lambda}\lambda^2\sum\limits_{k=0}^\infty \frac{\lambda^k}{k!}+\lambda=\lambda^2+\lambda$
$V(X)=E(X^2)-E(X)^2=\lambda$
几何分布$Ge(p)$
$p_k=(1-p)^{k-1}p,k=1,2,…$
$E(X)=\sum\limits_{k=0}^\infty k(1-p)^{k-1}p=p\sum\limits_{k=0}^\infty [-(1-p)^k]’=p[\sum\limits_{k=0}^\infty -(1-p)^k]’=p(-\frac{1}{p})’=\frac{1}{p}$
$E(X^2)=\sum\limits_{k=0}^\infty k^2(1-p)^{k-1}p=p\sum\limits_{k=0}^\infty k(k-1)+k^{k-1}$
其中$\sum\limits_{k=0}^\infty k(k-1)(1-p)^{k-1}=(1-p)\sum\limits_{k=0}^\infty [(1-p)^k]’’=(1-p)[\sum\limits_{k=0}^\infty (1-p)^k]’’=2p^{-3}(1-p)$
所以$E(X^2)=2p^{-2}(1-p)+\frac{1}{p}$
$V(X)=E(X^2)-E(X)^2=\frac{2(1-p)}{p^2}+\frac{1}{p}-\frac{1}{p^2}=\frac{1-p}{p^2}$
负二项分布$Nb(r,p)$
$p_k=\binom{k-1}{r-1}(1-p)^{k-r}p^r,k=r,r+1,…$
$E(X)=\sum\limits_{k=r}^\infty k\binom{k-1}{r-1}p^r(1-p)^{k-r}=\frac{r}{p}\sum\limits_{k=r}^\infty \frac{k!}{(k-r)!r!}p^{r+1}(1-p)^{k-r}=\frac{r}{p}\sum\limits_{k=r+1}^\infty \binom{k-1}{r}p^{r+1}(1-p)^{k-1-r}=\frac{r}{p}$
$E(X^2)=\sum\limits_{k=r}^\infty k^2\binom{k-1}{r-1}p^r(1-p)^{k-r}=\frac{r}{p}\sum\limits_{k=r+1}^\infty (k-1)\binom{k-1}{r}p^{r+1}(1-p)^{k-1-r}=\frac{r}{p}(\frac{r+1}{p}-1)=\frac{r^2+r-rp}{p^2}$
$V(X)=E(X^2)-E(X)^2=\frac{r(1-p)}{p^2}$
也可以把负二项分布理解成r次相互独立的几何分布，那么$E(X\sim Nb(r,p))=rE(X\sim Ge(p))=\frac{r}{p}$,$V(X\sim Nb(r,p))=rV(X\sim Ge(p))$（独立事件的期望和方差可以直接相加）
正态分布$N(\mu,\sigma^2)$
标准正态分布情况下：$p(x)=\frac{1}{\sqrt{2\pi}}\exp\{-\frac{x^2}{2}\},-\infty<x<\infty$
$E(X)=\int_{-\infty}^\infty xp(x)dx$，$xp(x)$为奇函数，积分为0，即$E(X)=0$
$E(X^2)=\int_{-\infty}^\infty x^2\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty xd(-e^{-\frac{x^2}{2}})=\frac{1}{\sqrt{2\pi}}[x(-e^{-\frac{x^2}{2}})|_{-\infty}^\infty-\int_{-\infty}^\infty -e^{-\frac{x^2}{2}}]=\frac{1}{\sqrt{2\pi}}\sqrt{2\pi}=1$
$V(X)=E(X^2)-E(X)^2=1$
若正态分布$U\sim N(\mu,\sigma^2)$，则$X=(U-\mu)/\sigma\sim N(0,1),U=\sigma X+\mu$
所以
$p(u)=\frac{1}{\sqrt{2\pi\sigma^2}}\exp\{-\frac{(x-\mu)^2}{2\sigma^2}\},-\infty<x<\infty$
$E(U)=E(\sigma X+\mu)=\mu$
$V(U)=V(\sigma X+\mu)=\sigma^2V(X)=\sigma^2$
均匀分布$U(a,b)$
$p(x)=\frac{1}{b-a},a<x<b$
$E(X)=\int_a^b x\frac{1}{b-a}dx=\frac{1}{b-a}\frac{1}{2}(b-a)(b+a)=\frac{a+b}{2}$
$E(X^2)=\int_a^bx^2\frac{1}{b-a}dx=\frac{1}{b-a}\frac{1}{3}(b^3-a^3)=\frac{1}{3(b-a)}(b-a)(b^2+ab+a^2)=\frac{a^2+ab+b^2}{3}$
$V(X)=E(X^2)-E(X)^2=\frac{(b-a)^2}{12}$
指数分布$Exp(\lambda)$
$p(x)=\lambda e^{-\lambda x},x\geqslant 0$
$E(X)=\int_0^\infty x\lambda e^{-\lambda x}dx=\lambda \int_0^\infty xd(\frac{1}{-\lambda}e^{-\lambda x})=\lambda(x\frac{1}{-\lambda}e^{-\lambda x}|_0^\infty -\int_0^\infty \frac{1}{-\lambda}e^{-\lambda x}dx)=\frac{1}{-\lambda}e^{-\lambda x}|_0^\infty=\frac{1}{\lambda}$
$E(X^2)=\int_0^\infty \lambda x^2e^{-\lambda x}dx=\frac{\Gamma(3)}{\lambda^2}\{\int_0^\infty \frac{\lambda^3}{\Gamma(3)}x^2e^{-\lambda x}dx\}=\frac{2}{\lambda^2}$
$Var(X)=E(X^2)-E(X)=\frac{1}{\lambda^2}$
伽马分布$Ga(\alpha, \lambda)$
$p(x)=\frac{\lambda^\alpha}{\Gamma(\alpha)}x^{\alpha - 1}e^{-\lambda x},x\geqslant 0$
$E(X)=\int_0^\infty \frac{\lambda^\alpha}{\Gamma(\alpha)}x^\alpha e^{-\lambda x}dx=\{\int_0^\infty \frac{\lambda^{\alpha+1}}{\Gamma(\alpha+1)}x^\alpha e^{-\lambda x}dx\}\frac{\alpha}{\lambda}=\frac{\alpha}{\lambda}$
$E(X^2)=\{\int_0^\infty \frac{\lambda^{\alpha+2}}{\Gamma(\alpha+2)}x^{\alpha+1}e^{-\lambda x}dx\}\frac{\alpha^2+\alpha}{\lambda^2}$
$Var(X)=E(X^2)-E(X)^2=\frac{\alpha}{\lambda^2}$