常用概率分佈的矩母函數、特徵函數以及期望、方差的推導

一、定義與性質

$設X爲隨機變量，I是一個包含0的(有限或無限的)開區間，對任意t∈I，期望Ee^{tx}存在$
$則稱函數M_{X}(t)=E(e^{tX})=\int_{-\infin}^{+\infin}e^{tx}dF(x),t∈I爲X的矩母函數$
$設X爲任意隨機變量，稱函數\varphi_{X}(t)=E(e^{itX})=\int_{-\infin}^{+\infin}e^{itx}dF(x)爲X的特徵函數$
$一個隨機變量的矩母函數不一定存在，但是特徵函數一定存在。$
$隨機變量與特徵函數存在一一對應的關係$

二、離散型隨機變量的分佈

0、退化分佈(Degenerate distribution)

$若X服從參數爲a的退化分佈，那麼f(k;a)=\left\{\begin{matrix} 1,k=a \\ 0,k\neq a \end{matrix}\right.$
$M(t)=e^{ta}$
$\varphi(t)=e^{ita}$
$M'(t)=ae^{ta}$
$EX=M'(0)=a$
$M''(t)=a^2e^{ta}$
$EX^2=M''(0)=a^2$
$DX=EX^2-(EX)^2=0$

1、離散型均勻分佈(Discrete uniform distribution)

$若X服從離散型均勻分佈DU(a,b) ,則X分佈函數爲F(k;a,b)=\frac{\lfloor k\rfloor -a+1}{b-a+1}$
$則矩母函數M(t)=\sum_{k=a}^{b} e^{tk}P(x=k)$
$=(\sum_{k=a}^{b} e^{tk})\frac{1}{b-a+1}$
$=\frac{e^{at}-e^{(b+1)t}}{(1-e^{t})(b-a+1)}$
$特徵函數\varphi(t)=\sum_{k=a}^{b} e^{itk}P(x=k)$
$=(\sum_{k=a}^{b} e^{itk})\frac{1}{b-a+1}$
$=\frac{e^{ait}-e^{(b+1)it}}{(1-e^{it})(b-a+1)}$
$M'(t)=\frac{1}{b-a+1}\frac{(ae^{at}-(b+1)e^{(b+1)t})(1-e^t)+(e^{at}-e^{(b+1)t})e^t}{(e^{t}-1)^{2}}$
$t=0爲M'(t)的可去間斷點，補充定義M'(0)=\lim_{t\rightarrow0}M'(t)$
$EX=M'(0)=\lim_{t\rightarrow0}\frac{1}{b-a+1}\frac{(a^2e^{at}-(b+1)^2e^{(b+1)t})(1-e^t)+(e^{at}-e^{(b+1)t})e^t}{2(e^{t}-1)e^t}$
$=\lim_{t\rightarrow0}\frac{1}{b-a+1}\frac{(a^2e^{at}-(b+1)^2e^{(b+1)t})(e^{-t}-1)+(e^{at}-e^{(b+1)t})}{2(e^{t}-1)}$
$=\lim_{t\rightarrow0}\frac{1}{b-a+1}\frac{(a^3e^{at}-(b+1)^3e^{(b+1)t})(e^{-t}-1)-(a^2e^{at}-(b+1)^2e^{(b+1)t})e^{-t}+(ae^{at}-(b+1)e^{(b+1)t})}{2e^{t}}$
$=\frac{-a^2+(b+1)^2+a-(b+1)}{2(b-a+1)}$
$=\frac{-a^2+(b+1)^2}{2(b-a+1)}-\frac{1}{2}$
$=\frac{(b+1-a)(b+1+a)}{2(b-a+1)}-\frac{1}{2}$
$=\frac{b+1+a}{2}-\frac{1}{2}$
$=\frac{b+a}{2}$
$由於對M'(t)求導得到M''(t)，再求M''(0)的方法比較繁瑣，而我們只需要t=0時M的二階導數值，$
$因此可以考慮使用Taylor公式計算M''(0)$
$令1-e^t=u,t=0時,u=0$
$M(t)=\frac{e^{at}-e^{(b+1)t}}{(1-e^{t})(b-a+1)}$
$=\frac{1}{b-a+1}\frac{u^a-u^{b+1}}{u}$
$=\frac{1}{b-a+1}\frac{1+\frac{a}{1!}(-u)+\frac{a(a-1)}{2!}u^2+\frac{a(a-1)(a-2)}{3!}(-u^3)+o(u^3)-1-\frac{b+1}{1!}(-u)-\frac{(b+1)b}{2!}u^2-\frac{(b+1)b(b-1)}{3!}(-u^3)-o(u^3)}{u}$
$=\frac{1}{b-a+1}\frac{\frac{a}{1!}(-u)+\frac{a(a-1)}{2!}u^2+\frac{a(a-1)(a-2)}{3!}(-u^3)+o(u^3)-\frac{b+1}{1!}(-u)-\frac{(b+1)b}{2!}u^2-\frac{(b+1)b(b-1)}{3!}(-u^3)}{u}$
$=\frac{1}{b-a+1}((b+1-a)+\frac{a(a-1)}{2!}u+\frac{a(a-1)(a-2)}{3!}(-u^2)+o(u^2)-\frac{(b+1)b}{2!}u-\frac{(b+1)b(b-1)}{3!}(-u^2))$
$=1+\frac{a(a-1)-(b+1)b}{2!(b-a+1)}u+\frac{(b+1)b(b-1)-a(a-1)(a-2)}{3!(b-a+1)}u^2+o(u^2)$
$而u=1-e^t=-t-\frac{t^2}{2!}+o(t^2)$
$因此M(t)=1-\frac{a(a-1)-(b+1)b}{2!(b-a+1)}t-\frac{a(a-1)-(b+1)b}{2!(b-a+1)}\frac{t^2}{2!}+\frac{(b+1)b(b-1)-a(a-1)(a-2)}{3!(b-a+1)}t^2+o(t^2)$
$又因爲M(t)=M(0)+M'(0)t+\frac{M''(0)}{2!}t^2+o(t^2)$
$因此M'(0)=-\frac{a(a-1)-(b+1)b}{2!(b-a+1)}=\frac{a+b}{2}$
$EX=M'(0)=\frac{a+b}{2}$
$而M''(0)=2!*(-\frac{a(a-1)-(b+1)b}{4(b-a+1)}+\frac{(b+1)b(b-1)-a(a-1)(a-2)}{3!(b-a+1)})$
$=\frac{a+b}{2}+\frac{(b+1-a)(b^2+ab-b+a^2-2a)}{3(b-a+1)}$
$=\frac{a+b}{2}+\frac{b^2+ab-b+a^2-2a}{3}$
$=\frac{2a^2+2b^2+2ab+b-a}{6}$
$DX=EX^2-(EX)^2=M''(0)-(EX)^2$
$=\frac{2a^2+2b^2+2ab+b-a}{6}-\frac{a^2+2ab+b^2}{4}$
$=\frac{(b-a+1)^2-1}{12}$

2、伯努利分佈/兩點分佈(Bernoulli distribution)

$若X服從伯努利分佈B(1,p) ,則X滿足P(x=1)=p, P(x=0)=1-p=q$
$M(t)=pe^{t}+1-p$
$\varphi(t)=pe^{it}+1-p$
$M'(t)=pe^{t}$
$EX=M'(0)=p$
$M''(t)=pe^{t}$
$EX^{2}=M''(0)=p$
$DX=EX^{2}-(EX)^{2}=p(1-p)$

3、二項分佈(Binomial distribution)

$若X服從二項分佈B(n,p) ,則X滿足f(k;n,p)=P(x=k)=C_{n}^{k}p^k(1-p)^{n-k} (n爲整數)$
$因爲服從二項分佈的變量可以看作n個獨立相同的服從伯努利分佈的變量之和$
$因此M(t)=(pe^{t}+1-p)^{n}$
$\varphi(t)=(pe^{it}+1-p)^{n}$
$M'(t)=np(pe^{t}+1-p)^{n-1}e^{t}$
$EX=M'(0)=np$
$M''(t)=n(n-1)p^{2}(pe^{t}+1-p)^{n-2}e^{2t}+np(pe^{t}+1-p)^{n-1}e^{t}$
$EX^{2}=M''(0)=n(n-1)p^{2}+np$
$DX=EX^{2}-(EX)^{2}=np(1-p)$

4、幾何分佈(Geometric distribution)

$若X服從幾何分佈Ge(p), 則X滿足f(k;p)=P(x=k)=(1-p)^{k-1}p (k=1,2,3......)$
$M(t)=\sum_{k=1}^{\infin}(1-p)^{k-1}pe^{tk}$
$=pe^{t}\sum_{k=1}^{\infin}((1-p)e^t)^{k-1}$
$=\frac{pe^{t}}{1-(1-p)e^{t}}$
$\varphi(t)=\sum_{k=1}^{\infin}(1-p)^{k-1}pe^{itk}$
$=pe^{it}\sum_{k=1}^{\infin}((1-p)e^{it})^{k-1}$
$=\frac{pe^{it}}{1-(1-p)e^{it}}$
$M'(t)=\frac{pe^t}{(1-(1-p)e^t)^2}$
$EX=M'(0)=\frac{1}{p}$
$M''(t)=\frac{pe^t(e^t-pe^t+1)}{(1-(1-p)e^t)^3}$
$EX^{2}=M''(0)=\frac{2-p}{p^2}$
$DX=EX^{2}-(EX)^{2}=\frac{1-p}{p^2}$

5、負二項分佈(Negative binomial distribution)

$若X服從負二項分佈NB(r,p), 則X滿足f(k;r,p)=\binom{k+r-1}{k}p^{k}(1-p)^{r} , k=0,1,2,3......$
$(r可以爲實數，此時的分佈稱爲波利亞分佈)$
$M(t)=\sum_{k=0}^{\infin}\binom{k+r-1}{k}p^k(1-p)^re^{tk}$
$=\sum_{k=0}^{\infin}(-1)^k\binom{-r}{k}p^k(1-p)^re^{tk}$
$=\sum_{k=0}^{\infin}(-pe^t)^k\binom{-r}{k}(1-p)^r$
$=(1-p)^r\sum_{k=0}^{\infin}(-pe^t)^k\binom{-r}{k}1^{-r-k}$
$=(1-p)^r(1-pe^t)^{-r}$
$\varphi(t)=\sum_{k=0}^{\infin}\binom{k+r-1}{k}p^k(1-p)^re^{itk}$
$=\sum_{k=0}^{\infin}(-1)^k\binom{-r}{k}p^k(1-p)^re^{itk}$
$=\sum_{k=0}^{\infin}(-pe^{it})^k\binom{-r}{k}(1-p)^r$
$=(1-p)^r\sum_{k=0}^{\infin}(-pe^{it})^k\binom{-r}{k}1^{-r-k}$
$=(1-p)^r(1-pe^{it})^{-r}$
$M'(t)=(1-p)^r(-r)(1-pe^{t})^{-r-1}(-pe^t)$
$=rp(1-p)^re^t(1-pe^t)^{-r-1}$
$EX=M'(0)=\frac{rp}{1-p}$
$M''(t)=rp(1-p)^re^t(1-pe^t)^{-r-1}+rp(1-p)^re^t(-r-1)(1-pe^t)^{-r-2}(-pe^t)$
$EX^2=rp(1-p)^{-1}+r(r+1)p^2(1-p)^{-2}$
$=\frac{rp(1-p)+r(r+1)p^2}{(1-p)^2}$
$=\frac{rp+r^2p^2}{(1-p)^2}$
$DX=EX^2-(EX)^2=\frac{pr}{(1-p)^2}$

6、泊松分佈(Poisson distribution)

$若X服從泊松分佈P(\lambda),則P(X=k)=\frac{e^{-\lambda}\lambda^k}{k!},k=0,1,2......$
$M(t)=\sum_{k=0}^{\infin}\frac{e^{-\lambda}\lambda^k}{k!}e^{tk}$
$=e^{-\lambda}\sum_{k=0}^{\infin}\frac{(\lambda e^t)^k}{k!}$
$=e^{-\lambda}e^{\lambda e^t}$
$=e^{\lambda (e^t-1)}$
$\varphi(t)=\sum_{k=0}^{\infin}\frac{e^{-\lambda}\lambda^k}{k!}e^{itk}$
$=e^{-\lambda}\sum_{k=0}^{\infin}\frac{(\lambda e^{it})^k}{k!}$
$=e^{-\lambda}e^{\lambda e^{it}}$
$=e^{\lambda (e^{it}-1)}$
$M'(t)=e^{\lambda (e^t-1)}\lambda e^t$
$EX=M'(0)=\lambda$
$M''(t)=e^{\lambda (e^t-1)}\lambda e^t+e^{\lambda (e^t-1)}\lambda e^t\lambda e^t$
$EX^2=M''(0)=\lambda+\lambda^2$
$DX=EX^2-(EX)^2=\lambda$

三、連續型隨機變量的分佈

1、連續型均勻分佈(Uniform distribution (continuous))

$若X服從連續型均勻分佈U(a,b),則f(x)=\frac{1}{b-a}I_{[a,b]}(x)$
$M(t)=\int_{a}^{b}\frac{1}{b-a}e^{tx}dx$
$=\frac{1}{b-a}\int_{a}^{b}e^{tx}dx$
$=\frac{1}{b-a}(\frac{1}{t}e^{tx}\mid_{a}^{b})$
$=\frac{e^{tb}-e^{ta}}{t(b-a)}$
$\varphi(t)=\int_{a}^{b}\frac{1}{b-a}e^{itx}dx$
$=\frac{1}{b-a}\int_{a}^{b}e^{itx}dx$
$=\frac{1}{b-a}(\frac{1}{it}e^{itx}\mid_{a}^{b})$
$=\frac{e^{itb}-e^{ita}}{it(b-a)}$
$M'(t)=\frac{1}{b-a}\frac{(be^{tb}-ae^{ta})t-(e^{tb}-e^{ta})}{t^2}$
$t=0爲M'(t)的可去間斷點，補充定義M'(0)=\lim_{t\rightarrow0}M'(t)$
$EX=M'(0)=\lim_{t\rightarrow0}\frac{(be^{tb}-ae^{ta})+(b^2e^{tb}-a^2e^{ta})t-(be^{tb}-ae^{ta})}{2t(b-a)}$
$=\lim_{t\rightarrow0}\frac{(b^2e^{tb}-a^2e^{ta})}{2(b-a)}$
$=\frac{b^2-a^2}{2(b-a)}$
$=\frac{a+b}{2}$
$M''(t)=\frac{1}{b-a}\frac{((b^2e^{tb}-a^2e^{ta})t+(be^{tb}-ae^{ta})-(be^{tb}-ae^{ta}))t-2((be^{tb}-ae^{ta})t-(e^{tb}-e^{ta}))}{t^3}$
$=\frac{1}{b-a}\frac{t^2(b^2e^{tb}-a^2e^{ta})-2t(be^{tb}-ae^{ta})+2(e^{tb}-e^{ta})}{t^3}$
$t=0爲M''(t)的可去間斷點，補充定義M''(0)=\lim_{t\rightarrow0}M''(t)$
$EX^2=M''(0)=\lim_{t\rightarrow0}\frac{1}{b-a}\frac{t^2(b^3e^{tb}-a^3e^{ta})+2t(b^2e^{tb}-a^2e^{ta})-2t(b^2e^{tb}-a^2e^{ta})-2(be^{tb}-ae^{ta})+2(be^{tb}-ae^{ta})}{3t^2}$
$=\frac{1}{b-a}\lim_{t\rightarrow0}\frac{t^2(b^3e^{tb}-a^3e^{ta})}{3t^2}$
$=\frac{1}{b-a}\lim_{t\rightarrow0}\frac{(b^3e^{tb}-a^3e^{ta})}{3}$
$=\frac{1}{b-a}\frac{(b^3-a^3)}{3}$
$=\frac{b^2+ab+a^2}{3}$
$DX=EX^2-(EX)^2=\frac{(b-a)^2}{12}$

2、指數分佈(Exponential distribution)

$若X服從指數分佈E(\lambda)，則f(x)=\lambda e^{-\lambda x}I_{[0,+\infin)}(x)$
$M(t)=\int_{0}^{+\infin} \lambda e^{-\lambda x}e^{tx}dx$
$=\lambda \int_{0}^{+\infin} e^{(t-\lambda)x}dx$
$=\frac{\lambda}{t-\lambda}(e^{(t-\lambda)x}\mid_{0}^{+\infin})$
$t<\lambda時，M(t)=\frac{\lambda}{t-\lambda}(0-1)$
$=\frac{\lambda}{\lambda-t}$
$\varphi(t)=\frac{\lambda}{\lambda-it}$
$M'(t)=\frac{\lambda}{(\lambda-t)^2}$
$EX=M'(0)=\frac{1}{\lambda}$
$M''(t)=\frac{2\lambda}{(\lambda-t)^3}$
$EX^2=M''(0)=\frac{2}{\lambda^2}$
$DX=EX^2-(EX)^2=\frac{1}{\lambda^2}$

3、正態分佈(Normal distribution)

$若X服從正態分佈N(\mu,\sigma^2),則f(x)=\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$
$引理1：\int_{-\infin}^{+\infin}e^{-\frac{t^2}{2}}dt=\sqrt{2\pi}$
$證明：(\int_{-\infin}^{+\infin}e^{-\frac{t^2}{2}}dt)^2=\int_{-\infin}^{+\infin}\int_{-\infin}^{+\infin}e^{-\frac{x^2+y^2}{2}}dxdy$
$=\int_{0}^{2\pi}d\theta \int_{0}^{+\infin}e^{-\frac{r^2}{2}}rdr$
$=2\pi \int_{0}^{+\infin}e^{-\frac{r^2}{2}}rdr$
$=2\pi (-e^{-\frac{r^2}{2}}\mid_{0}^{+\infin})$
$=2\pi$
$因此\int_{-\infin}^{+\infin}e^{-\frac{t^2}{2}}dt=\sqrt{2\pi}$
$M(t)=\int_{-\infin}^{+\infin}\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}}e^{tx}dx$
$=\frac{1}{\sqrt{2\pi}\sigma}\int_{-\infin}^{+\infin}e^{-\frac{(x-\mu)^2}{2\sigma^2}+tx}dx$
$令w=\frac{x-\mu}{\sigma}$
$原式=\frac{1}{\sqrt{2\pi}}\int_{-\infin}^{+\infin}e^{-\frac{w^2}{2}+t(w\sigma+\mu)}dw$
$=e^{\mu t}\frac{1}{\sqrt{2\pi}}\int_{-\infin}^{+\infin}e^{-\frac{w^2}{2}+t\sigma w}dw$
$=e^{\mu t}\frac{1}{\sqrt{2\pi}}\int_{-\infin}^{+\infin}e^{-\frac{(w-t\sigma)^2-t^2\sigma^2}{2}}dw$
$=e^{\mu t+\frac{t^2\sigma^2}{2}}\frac{1}{\sqrt{2\pi}}\int_{-\infin}^{+\infin}e^{-\frac{(w-t\sigma)^2}{2}}dw$
$=e^{\mu t+\frac{t^2\sigma^2}{2}}\frac{1}{\sqrt{2\pi}}\sqrt{2\pi}$
$=e^{\mu t+\frac{t^2\sigma^2}{2}}$
$\varphi(t)=\int_{-\infin}^{+\infin}\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}}e^{itx}dx$
$=\frac{1}{\sqrt{2\pi}\sigma}\int_{-\infin}^{+\infin}e^{-\frac{(x-\mu)^2}{2\sigma^2}+itx}dx$
$令w=\frac{x-\mu}{\sigma}$
$原式=\frac{1}{\sqrt{2\pi}}\int_{-\infin}^{+\infin}e^{-\frac{w^2}{2}+it(w\sigma+\mu)}dw$
$=e^{i\mu t}\frac{1}{\sqrt{2\pi}}\int_{-\infin}^{+\infin}e^{-\frac{w^2}{2}+it\sigma w}dw$
$=e^{i\mu t}\frac{1}{\sqrt{2\pi}}\int_{-\infin}^{+\infin}e^{-\frac{(w-it\sigma)^2+t^2\sigma^2}{2}}dw$
$=e^{i\mu t-\frac{t^2\sigma^2}{2}}\frac{1}{\sqrt{2\pi}}\int_{-\infin}^{+\infin}e^{-\frac{(w-it\sigma)^2}{2}}dw$
$=e^{i\mu t-\frac{t^2\sigma^2}{2}}\frac{1}{\sqrt{2\pi}}\sqrt{2\pi}$
$=e^{i\mu t-\frac{t^2\sigma^2}{2}}$
$M'(t)=e^{\mu t+\frac{t^2\sigma^2}{2}}(\mu+\sigma^2t)$
$EX=M'(0)=\mu$
$M''(t)=e^{\mu t+\frac{t^2\sigma^2}{2}}(\mu+\sigma^2t)^2+e^{\mu t+\frac{t^2\sigma^2}{2}}\sigma^2$
$EX^2=M''(0)=\mu^2+\sigma^2$
$DX=EX^2-(EX)^2=\sigma^2$
$特別地,X服從標準正態分佈N(0,1)時$
$M(t)=e^{\frac{t^2}{2}}$
$\varphi(t)=e^{-\frac{t^2}{2}}$
$EX=0,DX=1$

4、伽馬分佈(Gamma distribution)

$若X服從伽馬分佈\Gamma(\alpha,\beta)(\alpha,\beta>0),則f(x)=\frac{\beta^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\beta x}I_{(0,+\infin)}(x)$
$其中，\Gamma(\alpha)=\int_{0}^{+\infin}t^{\alpha-1}e^{-t}dt,\alpha>0$
$指數分佈E(\lambda)是伽馬分佈\Gamma(1,\lambda),\chi^2分佈\chi^2_n是伽馬分佈\Gamma(\frac{n}{2},\frac{1}{2})$
$M(t)=\int_{0}^{+\infin}\frac{\beta^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\beta x}e^{tx}dx$
$=\int_{0}^{+\infin}\frac{\beta^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{(t-\beta) x}dx$
$=\beta^\alpha\int_{0}^{+\infin}\frac{1}{\Gamma(\alpha)}x^{\alpha-1}e^{(t-\beta) x}dx$
$t<\beta時，令v=(\beta-t)x，原式=\frac{\beta^\alpha}{\beta-t}\int_{0}^{+\infin}\frac{1}{\Gamma(\alpha)}(\frac{v}{\beta-t})^{\alpha-1}e^{-v}dv$
$=(\frac{\beta}{\beta-t})^\alpha\frac{1}{\Gamma(\alpha)}\int_{0}^{+\infin}v^{\alpha-1}e^{-v}dv$
$=(\frac{\beta}{\beta-t})^\alpha\frac{1}{\Gamma(\alpha)}\Gamma(\alpha)$
$=(\frac{\beta}{\beta-t})^\alpha$
$\varphi(t)=(\frac{\beta}{\beta-it})^\alpha$
$M'(t)=\beta^\alpha(\beta-t)^{-\alpha-1}\alpha$
$EX=M'(0)=\frac{\alpha}{\beta}$
$M''(t)=\beta^\alpha(\beta-t)^{-\alpha-2}\alpha(\alpha+1)$
$EX^2=\frac{\alpha(\alpha+1)}{\beta^2}$
$DX=EX^2-(EX)^2=\frac{\alpha}{\beta^2}$

5、貝塔分佈(Beta distribution)

$若X服從貝塔分佈\Beta e(\alpha,\beta)(\alpha,\beta>0)，則f(x)=\frac{x^{\alpha-1}(1-x)^{\beta-1}}{\int_{0}^{1} u^{\alpha-1}(1-u)^{\beta-1}du}I_{(0,1)}(x)$
$=\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}x^{\alpha-1}(1-x)^{\beta-1}I_{(0,1)}(x)=\frac{1}{\Beta(\alpha,\beta)}x^{\alpha-1}(1-x)^{\beta-1}I_{(0,1)}(x)$
$其中,\Beta(\alpha,\beta)=\int_{0}^{1} u^{\alpha-1}(1-u)^{\beta-1}du,(\alpha,\beta>0)$

$M(t)=\int_{0}^{1}\frac{1}{\Beta(\alpha,\beta)}x^{\alpha-1}(1-x)^{\beta-1}e^{tx}dx$
$E(X)=\int_{0}^{1}\frac{1}{\Beta(\alpha,\beta)}x^{\alpha-1}(1-x)^{\beta-1}xdx$
$=\int_{0}^{1}\frac{1}{\Beta(\alpha,\beta)}x^{\alpha}(1-x)^{\beta-1}xdx$
$=\frac{\int_{0}^{1}x^{\alpha}(1-x)^{\beta-1}xdx}{\Beta(\alpha,\beta)}$
$=\frac{\Beta(\alpha+1,\beta)}{\Beta(\alpha,\beta)}$
$=\frac{\frac{\Gamma(\alpha+1)\Gamma(\beta)}{\Gamma(\alpha+\beta+1)}}{\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}}$
$=\frac{\alpha}{\alpha+\beta}$
$E(X^2)=\int_{0}^{1}\frac{1}{\Beta(\alpha,\beta)}x^{\alpha-1}(1-x)^{\beta-1}x^2dx$
$=\frac{\int_{0}^{1}x^{\alpha+1}(1-x)^{\beta-1}dx}{\Beta(\alpha,\beta)}$
$=\frac{\Beta(\alpha+2,\beta)}{\Beta(\alpha,\beta)}$
$=\frac{\frac{\Gamma(\alpha+2)\Gamma(\beta)}{\Gamma(\alpha+\beta+2)}}{\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}}$
$=\frac{\alpha(\alpha+1)}{(\alpha+\beta+1)(\alpha+\beta)}$
$D(X)=E(X^2)-(EX)^2$
$=\frac{\alpha(\alpha+1)}{(\alpha+\beta+1)(\alpha+\beta)}-\frac{\alpha^2}{(\alpha+\beta)^2}$
$=\frac{\alpha\beta}{(\alpha+\beta+1)(\alpha+\beta)^2}$

6、t分佈(Student’s t-distribution)

$若X服從自由度爲n的t分佈t(x,n)，則f(x)=\frac{\Gamma(\frac{n+1}{2})}{\sqrt{n\pi}\Gamma(\frac{n}{2})}(1+\frac{x^2}{n})^{-\frac{n+1}{2}},x∈R$

7 、F分佈(F-distribution)