常用概率分布的矩母函数、特征函数以及期望、方差的推导

一、定义与性质

$$设X为随机变量，I是一个包含0的(有限或无限的)开区间，对任意t\in I，期望E{e}^{tx}存在$$
$$则称函数M_X(t)=E(e^{tX})={\int }_{-\infty }^{+\infty }{e}^{tx}dF(x),t\in I为X的矩母函数$$
$$设X为任意随机变量，称函数{\phi }_X(t)=E(e^{itX})={\int }_{-\infty }^{+\infty }{e}^{itx}dF(x)为X的特征函数$$
$$一个随机变量的矩母函数不一定存在，但是特征函数一定存在。$$
$$随机变量与特征函数存在一一对应的关系$$

二、离散型随机变量的分布

0、退化分布(Degenerate distribution)

$$若X服从参数为a的退化分布，那么f(k;a)=\{\begin{array}{c}1,k=a\\ 0,k̸=a\end{array}$$
$$M(t)=e^{ta}$$
$$\phi (t)=e^{ita}$$
$$M^{\prime }(t)=a{e}^{ta}$$
$$EX=M^{\prime }(0)=a$$
$$M^{\prime \prime }(t)=a^2{e}^{ta}$$
$$E{X}^2=M^{\prime \prime }(0)=a^2$$
$$DX=E{X}^2-(EX{)}^2=0$$

1、离散型均匀分布(Discrete uniform distribution)

$$若X服从离散型均匀分布DU(a,b),则X分布函数为F(k;a,b)=\frac{\lfloor k\rfloor -a+1}{b-a+1}$$
$$则矩母函数M(t)=\sum _{k=a}^b{e}^{tk}P(x=k)$$
$$=(\sum _{k=a}^b{e}^{tk})\frac{1}{b-a+1}$$
$$=\frac{e^{at}-e^{(b+1)t}}{(1-e^t)(b-a+1)}$$
$$特征函数\phi (t)=\sum _{k=a}^b{e}^{itk}P(x=k)$$
$$=(\sum _{k=a}^b{e}^{itk})\frac{1}{b-a+1}$$
$$=\frac{e^{ait}-e^{(b+1)it}}{(1-e^{it})(b-a+1)}$$
$$M^{\prime }(t)=\frac{1}{b-a+1}\frac{(a{e}^{at}-(b+1)e^{(b+1)t})(1-e^t)+(e^{at}-e^{(b+1)t})e^t}{(e^t-1{)}^2}$$
$$t=0为M^{\prime }(t)的可去间断点，补充定义M^{\prime }(0)=\underset{t\to 0}{\lim }{M}^{\prime }(t)$$
$$EX=M^{\prime }(0)=\underset{t\to 0}{\lim }\frac{1}{b-a+1}\frac{(a^2{e}^{at}-(b+1{)}^2{e}^{(b+1)t})(1-e^t)+(e^{at}-e^{(b+1)t})e^t}{2(e^t-1)e^t}$$
$$=\underset{t\to 0}{\lim }\frac{1}{b-a+1}\frac{(a^2{e}^{at}-(b+1{)}^2{e}^{(b+1)t})(e^{-t}-1)+(e^{at}-e^{(b+1)t})}{2(e^t-1)}$$
$$=\underset{t\to 0}{\lim }\frac{1}{b-a+1}\frac{(a^3{e}^{at}-(b+1{)}^3{e}^{(b+1)t})(e^{-t}-1)-(a^2{e}^{at}-(b+1{)}^2{e}^{(b+1)t})e^{-t}+(a{e}^{at}-(b+1)e^{(b+1)t})}{2e^t}$$
$$=\frac{-a^2+(b+1{)}^2+a-(b+1)}{2(b-a+1)}$$
$$=\frac{-a^2+(b+1{)}^2}{2(b-a+1)}-\frac{1}{2}$$
$$=\frac{(b+1-a)(b+1+a)}{2(b-a+1)}-\frac{1}{2}$$
$$=\frac{b+1+a}{2}-\frac{1}{2}$$
$$=\frac{b+a}{2}$$
$$由于对M^{\prime }(t)求导得到M^{\prime \prime }(t)，再求M^{\prime \prime }(0)的方法比较繁琐，而我们只需要t=0时M的二阶导数值，$$
$$因此可以考虑使用Taylor公式计算M^{\prime \prime }(0)$$
$$令1-e^t=u,t=0时,u=0$$
$$M(t)=\frac{e^{at}-e^{(b+1)t}}{(1-e^t)(b-a+1)}$$
$$=\frac{1}{b-a+1}\frac{u^a-u^{b+1}}{u}$$
$$=\frac{1}{b-a+1}\frac{1+\frac{a}{1!}(-u)+\frac{a(a-1)}{2!}{u}^2+\frac{a(a-1)(a-2)}{3!}(-u^3)+o(u^3)-1-\frac{b+1}{1!}(-u)-\frac{(b+1)b}{2!}{u}^2-\frac{(b+1)b(b-1)}{3!}(-u^3)-o(u^3)}{u}$$
$$=\frac{1}{b-a+1}\frac{\frac{a}{1!}(-u)+\frac{a(a-1)}{2!}{u}^2+\frac{a(a-1)(a-2)}{3!}(-u^3)+o(u^3)-\frac{b+1}{1!}(-u)-\frac{(b+1)b}{2!}{u}^2-\frac{(b+1)b(b-1)}{3!}(-u^3)}{u}$$
$$=\frac{1}{b-a+1}((b+1-a)+\frac{a(a-1)}{2!}u+\frac{a(a-1)(a-2)}{3!}(-u^2)+o(u^2)-\frac{(b+1)b}{2!}u-\frac{(b+1)b(b-1)}{3!}(-u^2))$$
$$=1+\frac{a(a-1)-(b+1)b}{2!(b-a+1)}u+\frac{(b+1)b(b-1)-a(a-1)(a-2)}{3!(b-a+1)}{u}^2+o(u^2)$$
$$而u=1-e^t=-t-\frac{t^2}{2!}+o(t^2)$$
$$因此M(t)=1-\frac{a(a-1)-(b+1)b}{2!(b-a+1)}t-\frac{a(a-1)-(b+1)b}{2!(b-a+1)}\frac{t^2}{2!}+\frac{(b+1)b(b-1)-a(a-1)(a-2)}{3!(b-a+1)}{t}^2+o(t^2)$$
$$又因为M(t)=M(0)+M^{\prime }(0)t+\frac{M^{\prime \prime }(0)}{2!}{t}^2+o(t^2)$$
$$因此M^{\prime }(0)=-\frac{a(a-1)-(b+1)b}{2!(b-a+1)}=\frac{a+b}{2}$$
$$EX=M^{\prime }(0)=\frac{a+b}{2}$$
$$而M^{\prime \prime }(0)=2!\ast (-\frac{a(a-1)-(b+1)b}{4(b-a+1)}+\frac{(b+1)b(b-1)-a(a-1)(a-2)}{3!(b-a+1)})$$
$$=\frac{a+b}{2}+\frac{(b+1-a)(b^2+ab-b+a^2-2a)}{3(b-a+1)}$$
$$=\frac{a+b}{2}+\frac{b^2+ab-b+a^2-2a}{3}$$
$$=\frac{2a^2+2b^2+2ab+b-a}{6}$$
$$DX=E{X}^2-(EX{)}^2=M^{\prime \prime }(0)-(EX{)}^2$$
$$=\frac{2a^2+2b^2+2ab+b-a}{6}-\frac{a^2+2ab+b^2}{4}$$
$$=\frac{(b-a+1{)}^2-1}{12}$$

2、伯努利分布/两点分布(Bernoulli distribution)

$$若X服从伯努利分布B(1,p),则X满足P(x=1)=p,P(x=0)=1-p=q$$
$$M(t)=p{e}^t+1-p$$
$$\phi (t)=p{e}^{it}+1-p$$
$$M^{\prime }(t)=p{e}^t$$
$$EX=M^{\prime }(0)=p$$
$$M^{\prime \prime }(t)=p{e}^t$$
$$E{X}^2=M^{\prime \prime }(0)=p$$
$$DX=E{X}^2-(EX{)}^2=p(1-p)$$

3、二项分布(Binomial distribution)

$$若X服从二项分布B(n,p),则X满足f(k;n,p)=P(x=k)=C_n^k{p}^k(1-p{)}^{n-k}(n为整数)$$
$$因为服从二项分布的变量可以看作n个独立相同的服从伯努利分布的变量之和$$
$$因此M(t)=(p{e}^t+1-p{)}^n$$
$$\phi (t)=(p{e}^{it}+1-p{)}^n$$
$$M^{\prime }(t)=np(p{e}^t+1-p{)}^{n-1}{e}^t$$
$$EX=M^{\prime }(0)=np$$
$$M^{\prime \prime }(t)=n(n-1)p^2(p{e}^t+1-p{)}^{n-2}{e}^{2t}+np(p{e}^t+1-p{)}^{n-1}{e}^t$$
$$E{X}^2=M^{\prime \prime }(0)=n(n-1)p^2+np$$
$$DX=E{X}^2-(EX{)}^2=np(1-p)$$

4、几何分布(Geometric distribution)

$$若X服从几何分布Ge(p),则X满足f(k;p)=P(x=k)=(1-p{)}^{k-1}p(k=1,2,3......)$$
$$M(t)=\sum _{k=1}^{\infty }(1-p{)}^{k-1}p{e}^{tk}$$
$$=p{e}^t\sum _{k=1}^{\infty }((1-p)e^t{)}^{k-1}$$
$$=\frac{p{e}^t}{1-(1-p)e^t}$$
$$\phi (t)=\sum _{k=1}^{\infty }(1-p{)}^{k-1}p{e}^{itk}$$
$$=p{e}^{it}\sum _{k=1}^{\infty }((1-p)e^{it}{)}^{k-1}$$
$$=\frac{p{e}^{it}}{1-(1-p)e^{it}}$$
$$M^{\prime }(t)=\frac{p{e}^t}{(1-(1-p)e^t{)}^2}$$
$$EX=M^{\prime }(0)=\frac{1}{p}$$
$$M^{\prime \prime }(t)=\frac{p{e}^t(e^t-p{e}^t+1)}{(1-(1-p)e^t{)}^3}$$
$$E{X}^2=M^{\prime \prime }(0)=\frac{2-p}{p^2}$$
$$DX=E{X}^2-(EX{)}^2=\frac{1-p}{p^2}$$

5、负二项分布(Negative binomial distribution)

$$若X服从负二项分布NB(r,p),则X满足f(k;r,p)=\left(\genfrac{}{}{0ex}{}{k+r-1}{k}\right)p^k(1-p{)}^r,k=0,1,2,3......$$
$$(r可以为实数，此时的分布称为波利亚分布)$$
$$M(t)=\sum _{k=0}^{\infty }\left(\genfrac{}{}{0ex}{}{k+r-1}{k}\right)p^k(1-p{)}^r{e}^{tk}$$
$$=\sum _{k=0}^{\infty }(-1{)}^k(\genfrac{}{}{0ex}{}{-r}{k})p^k(1-p{)}^r{e}^{tk}$$
$$=\sum _{k=0}^{\infty }(-p{e}^t{)}^k(\genfrac{}{}{0ex}{}{-r}{k})(1-p{)}^r$$
$$=(1-p{)}^r\sum _{k=0}^{\infty }(-p{e}^t{)}^k\left(\genfrac{}{}{0ex}{}{-r}{k}\right)1^{-r-k}$$
$$=(1-p{)}^r(1-p{e}^t{)}^{-r}$$
$$\phi (t)=\sum _{k=0}^{\infty }\left(\genfrac{}{}{0ex}{}{k+r-1}{k}\right)p^k(1-p{)}^r{e}^{itk}$$
$$=\sum _{k=0}^{\infty }(-1{)}^k(\genfrac{}{}{0ex}{}{-r}{k})p^k(1-p{)}^r{e}^{itk}$$
$$=\sum _{k=0}^{\infty }(-p{e}^{it}{)}^k(\genfrac{}{}{0ex}{}{-r}{k})(1-p{)}^r$$
$$=(1-p{)}^r\sum _{k=0}^{\infty }(-p{e}^{it}{)}^k\left(\genfrac{}{}{0ex}{}{-r}{k}\right)1^{-r-k}$$
$$=(1-p{)}^r(1-p{e}^{it}{)}^{-r}$$
$$M^{\prime }(t)=(1-p{)}^r(-r)(1-p{e}^t{)}^{-r-1}(-p{e}^t)$$
$$=rp(1-p{)}^r{e}^t(1-p{e}^t{)}^{-r-1}$$
$$EX=M^{\prime }(0)=\frac{rp}{1-p}$$
$$M^{\prime \prime }(t)=rp(1-p{)}^r{e}^t(1-p{e}^t{)}^{-r-1}+rp(1-p{)}^r{e}^t(-r-1)(1-p{e}^t{)}^{-r-2}(-p{e}^t)$$
$$E{X}^2=rp(1-p{)}^{-1}+r(r+1)p^2(1-p{)}^{-2}$$
$$=\frac{rp(1-p)+r(r+1)p^2}{(1-p{)}^2}$$
$$=\frac{rp+r^2{p}^2}{(1-p{)}^2}$$
$$DX=E{X}^2-(EX{)}^2=\frac{pr}{(1-p{)}^2}$$

6、泊松分布(Poisson distribution)

$$若X服从泊松分布P(\lambda ),则P(X=k)=\frac{e^{-\lambda }{\lambda }^k}{k!},k=0,1,2......$$
$$M(t)=\sum _{k=0}^{\infty }\frac{e^{-\lambda }{\lambda }^k}{k!}{e}^{tk}$$
$$=e^{-\lambda }\sum _{k=0}^{\infty }\frac{(\lambda e^t{)}^k}{k!}$$
$$=e^{-\lambda }{e}^{\lambda e^t}$$
$$=e^{\lambda (e^t-1)}$$
$$\phi (t)=\sum _{k=0}^{\infty }\frac{e^{-\lambda }{\lambda }^k}{k!}{e}^{itk}$$
$$=e^{-\lambda }\sum _{k=0}^{\infty }\frac{(\lambda e^{it}{)}^k}{k!}$$
$$=e^{-\lambda }{e}^{\lambda e^{it}}$$
$$=e^{\lambda (e^{it}-1)}$$
$$M^{\prime }(t)=e^{\lambda (e^t-1)}\lambda e^t$$
$$EX=M^{\prime }(0)=\lambda$$
$$M^{\prime \prime }(t)=e^{\lambda (e^t-1)}\lambda e^t+e^{\lambda (e^t-1)}\lambda e^t\lambda e^t$$
$$E{X}^2=M^{\prime \prime }(0)=\lambda +{\lambda }^2$$
$$DX=E{X}^2-(EX{)}^2=\lambda$$

三、连续型随机变量的分布

1、连续型均匀分布(Uniform distribution (continuous))

$$若X服从连续型均匀分布U(a,b),则f(x)=\frac{1}{b-a}{I}_{[a,b]}(x)$$
$$M(t)={\int }_a^b\frac{1}{b-a}{e}^{tx}dx$$
$$=\frac{1}{b-a}{\int }_a^b{e}^{tx}dx$$
$$=\frac{1}{b-a}(\frac{1}{t}{e}^{tx}{\mid }_a^b)$$
$$=\frac{e^{tb}-e^{ta}}{t(b-a)}$$
$$\phi (t)={\int }_a^b\frac{1}{b-a}{e}^{itx}dx$$
$$=\frac{1}{b-a}{\int }_a^b{e}^{itx}dx$$
$$=\frac{1}{b-a}(\frac{1}{it}{e}^{itx}{\mid }_a^b)$$
$$=\frac{e^{itb}-e^{ita}}{it(b-a)}$$
$$M^{\prime }(t)=\frac{1}{b-a}\frac{(b{e}^{tb}-a{e}^{ta})t-(e^{tb}-e^{ta})}{t^2}$$
$$t=0为M^{\prime }(t)的可去间断点，补充定义M^{\prime }(0)=\underset{t\to 0}{\lim }{M}^{\prime }(t)$$
$$EX=M^{\prime }(0)=\underset{t\to 0}{\lim }\frac{(b{e}^{tb}-a{e}^{ta})+(b^2{e}^{tb}-a^2{e}^{ta})t-(b{e}^{tb}-a{e}^{ta})}{2t(b-a)}$$
$$=\underset{t\to 0}{\lim }\frac{(b^2{e}^{tb}-a^2{e}^{ta})}{2(b-a)}$$
$$=\frac{b^2-a^2}{2(b-a)}$$
$$=\frac{a+b}{2}$$
$$M^{\prime \prime }(t)=\frac{1}{b-a}\frac{((b^2{e}^{tb}-a^2{e}^{ta})t+(b{e}^{tb}-a{e}^{ta})-(b{e}^{tb}-a{e}^{ta}))t-2((b{e}^{tb}-a{e}^{ta})t-(e^{tb}-e^{ta}))}{t^3}$$
$$=\frac{1}{b-a}\frac{t^2(b^2{e}^{tb}-a^2{e}^{ta})-2t(b{e}^{tb}-a{e}^{ta})+2(e^{tb}-e^{ta})}{t^3}$$
$$t=0为M^{\prime \prime }(t)的可去间断点，补充定义M^{\prime \prime }(0)=\underset{t\to 0}{\lim }{M}^{\prime \prime }(t)$$
$$E{X}^2=M^{\prime \prime }(0)=\underset{t\to 0}{\lim }\frac{1}{b-a}\frac{t^2(b^3{e}^{tb}-a^3{e}^{ta})+2t(b^2{e}^{tb}-a^2{e}^{ta})-2t(b^2{e}^{tb}-a^2{e}^{ta})-2(b{e}^{tb}-a{e}^{ta})+2(b{e}^{tb}-a{e}^{ta})}{3t^2}$$
$$=\frac{1}{b-a}\underset{t\to 0}{\lim }\frac{t^2(b^3{e}^{tb}-a^3{e}^{ta})}{3t^2}$$
$$=\frac{1}{b-a}\underset{t\to 0}{\lim }\frac{(b^3{e}^{tb}-a^3{e}^{ta})}{3}$$
$$=\frac{1}{b-a}\frac{(b^3-a^3)}{3}$$
$$=\frac{b^2+ab+a^2}{3}$$
$$DX=E{X}^2-(EX{)}^2=\frac{(b-a{)}^2}{12}$$

2、指数分布(Exponential distribution)

$$若X服从指数分布E(\lambda )，则f(x)=\lambda e^{-\lambda x}{I}_{[0,+\infty )}(x)$$
$$M(t)={\int }_0^{+\infty }\lambda e^{-\lambda x}{e}^{tx}dx$$
$$=\lambda {\int }_0^{+\infty }{e}^{(t-\lambda )x}dx$$
$$=\frac{\lambda }{t-\lambda }(e^{(t-\lambda )x}{\mid }_0^{+\infty })$$
$$t<\lambda 时，M(t)=\frac{\lambda }{t-\lambda }(0-1)$$
$$=\frac{\lambda }{\lambda -t}$$
$$\phi (t)=\frac{\lambda }{\lambda -it}$$
$$M^{\prime }(t)=\frac{\lambda }{(\lambda -t{)}^2}$$
$$EX=M^{\prime }(0)=\frac{1}{\lambda }$$
$$M^{\prime \prime }(t)=\frac{2\lambda }{(\lambda -t{)}^3}$$
$$E{X}^2=M^{\prime \prime }(0)=\frac{2}{{\lambda }^2}$$
$$DX=E{X}^2-(EX{)}^2=\frac{1}{{\lambda }^2}$$

3、正态分布(Normal distribution)

$$若X服从正态分布N(\mu ,{\sigma }^2),则f(x)=\frac{1}{\sqrt{2\pi }\sigma }{e}^{-\frac{(x-\mu {)}^2}{2{\sigma }^2}}$$
$$引理1：{\int }_{-\infty }^{+\infty }{e}^{-\frac{t^2}{2}}dt=\sqrt{2\pi }$$
$$证明：({\int }_{-\infty }^{+\infty }{e}^{-\frac{t^2}{2}}dt{)}^2={\int }_{-\infty }^{+\infty }{\int }_{-\infty }^{+\infty }{e}^{-\frac{x^2+y^2}{2}}dxdy$$
$$={\int }_0^{2\pi }d\theta {\int }_0^{+\infty }{e}^{-\frac{r^2}{2}}rdr$$
$$=2\pi {\int }_0^{+\infty }{e}^{-\frac{r^2}{2}}rdr$$
$$=2\pi (-e^{-\frac{r^2}{2}}{\mid }_0^{+\infty })$$
$$=2\pi$$
$$因此{\int }_{-\infty }^{+\infty }{e}^{-\frac{t^2}{2}}dt=\sqrt{2\pi }$$
$$M(t)={\int }_{-\infty }^{+\infty }\frac{1}{\sqrt{2\pi }\sigma }{e}^{-\frac{(x-\mu {)}^2}{2{\sigma }^2}}{e}^{tx}dx$$
$$=\frac{1}{\sqrt{2\pi }\sigma }{\int }_{-\infty }^{+\infty }{e}^{-\frac{(x-\mu {)}^2}{2{\sigma }^2}+tx}dx$$
$$令w=\frac{x-\mu }{\sigma }$$
$$原式=\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{+\infty }{e}^{-\frac{w^2}{2}+t(w\sigma +\mu )}dw$$
$$=e^{\mu t}\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{+\infty }{e}^{-\frac{w^2}{2}+t\sigma w}dw$$
$$=e^{\mu t}\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{+\infty }{e}^{-\frac{(w-t\sigma {)}^2-t^2{\sigma }^2}{2}}dw$$
$$=e^{\mu t+\frac{t^2{\sigma }^2}{2}}\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{+\infty }{e}^{-\frac{(w-t\sigma {)}^2}{2}}dw$$
$$=e^{\mu t+\frac{t^2{\sigma }^2}{2}}\frac{1}{\sqrt{2\pi }}\sqrt{2\pi }$$
$$=e^{\mu t+\frac{t^2{\sigma }^2}{2}}$$
$$\phi (t)={\int }_{-\infty }^{+\infty }\frac{1}{\sqrt{2\pi }\sigma }{e}^{-\frac{(x-\mu {)}^2}{2{\sigma }^2}}{e}^{itx}dx$$
$$=\frac{1}{\sqrt{2\pi }\sigma }{\int }_{-\infty }^{+\infty }{e}^{-\frac{(x-\mu {)}^2}{2{\sigma }^2}+itx}dx$$
$$令w=\frac{x-\mu }{\sigma }$$
$$原式=\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{+\infty }{e}^{-\frac{w^2}{2}+it(w\sigma +\mu )}dw$$
$$=e^{i\mu t}\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{+\infty }{e}^{-\frac{w^2}{2}+it\sigma w}dw$$
$$=e^{i\mu t}\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{+\infty }{e}^{-\frac{(w-it\sigma {)}^2+t^2{\sigma }^2}{2}}dw$$
$$=e^{i\mu t-\frac{t^2{\sigma }^2}{2}}\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{+\infty }{e}^{-\frac{(w-it\sigma {)}^2}{2}}dw$$
$$=e^{i\mu t-\frac{t^2{\sigma }^2}{2}}\frac{1}{\sqrt{2\pi }}\sqrt{2\pi }$$
$$=e^{i\mu t-\frac{t^2{\sigma }^2}{2}}$$
$$M^{\prime }(t)=e^{\mu t+\frac{t^2{\sigma }^2}{2}}(\mu +{\sigma }^{2}t)$$
$$EX=M^{\prime }(0)=\mu$$
$$M^{\prime \prime }(t)=e^{\mu t+\frac{t^2{\sigma }^2}{2}}(\mu +{\sigma }^{2}t{)}^2+e^{\mu t+\frac{t^2{\sigma }^2}{2}}{\sigma }^2$$
$$E{X}^2=M^{\prime \prime }(0)={\mu }^2+{\sigma }^2$$
$$DX=E{X}^2-(EX{)}^2={\sigma }^2$$
$$特别地,X服从标准正态分布N(0,1)时$$
$$M(t)=e^{\frac{t^2}{2}}$$
$$\phi (t)=e^{-\frac{t^2}{2}}$$
$$EX=0,DX=1$$

4、伽马分布(Gamma distribution)

$$若X服从伽马分布\Gamma (\alpha ,\beta )(\alpha ,\beta >0),则f(x)=\frac{{\beta }^{\alpha }}{\Gamma (\alpha )}{x}^{\alpha -1}{e}^{-\beta x}{I}_{(0,+\infty )}(x)$$
$$其中，\Gamma (\alpha )={\int }_0^{+\infty }{t}^{\alpha -1}{e}^{-t}dt,\alpha >0$$
$$指数分布E(\lambda )是伽马分布\Gamma (1,\lambda ),{\chi }^2分布{\chi }_n^2是伽马分布\Gamma (\frac{n}{2},\frac{1}{2})$$
$$M(t)={\int }_0^{+\infty }\frac{{\beta }^{\alpha }}{\Gamma (\alpha )}{x}^{\alpha -1}{e}^{-\beta x}{e}^{tx}dx$$
$$={\int }_0^{+\infty }\frac{{\beta }^{\alpha }}{\Gamma (\alpha )}{x}^{\alpha -1}{e}^{(t-\beta )x}dx$$
$$={\beta }^{\alpha }{\int }_0^{+\infty }\frac{1}{\Gamma (\alpha )}{x}^{\alpha -1}{e}^{(t-\beta )x}dx$$
$$t<\beta 时，令v=(\beta -t)x，原式=\frac{{\beta }^{\alpha }}{\beta -t}{\int }_0^{+\infty }\frac{1}{\Gamma (\alpha )}(\frac{v}{\beta -t}{)}^{\alpha -1}{e}^{-v}dv$$
$$=(\frac{\beta }{\beta -t}{)}^{\alpha }\frac{1}{\Gamma (\alpha )}{\int }_0^{+\infty }{v}^{\alpha -1}{e}^{-v}dv$$
$$=(\frac{\beta }{\beta -t}{)}^{\alpha }\frac{1}{\Gamma (\alpha )}\Gamma (\alpha )$$
$$=(\frac{\beta }{\beta -t}{)}^{\alpha }$$
$$\phi (t)=(\frac{\beta }{\beta -it}{)}^{\alpha }$$
$$M^{\prime }(t)={\beta }^{\alpha }(\beta -t{)}^{-\alpha -1}\alpha$$
$$EX=M^{\prime }(0)=\frac{\alpha }{\beta }$$
$$M^{\prime \prime }(t)={\beta }^{\alpha }(\beta -t{)}^{-\alpha -2}\alpha (\alpha +1)$$
$$E{X}^2=\frac{\alpha (\alpha +1)}{{\beta }^2}$$
$$DX=E{X}^2-(EX{)}^2=\frac{\alpha }{{\beta }^2}$$

5、贝塔分布(Beta distribution)

$$若X服从贝塔分布\mathrm{B}e(\alpha ,\beta )(\alpha ,\beta >0)，则f(x)=\frac{x^{\alpha -1}(1-x{)}^{\beta -1}}{{\int }_0^1{u}^{\alpha -1}(1-u{)}^{\beta -1}du}{I}_{(0,1)}(x)$$
$$=\frac{\Gamma (\alpha +\beta )}{\Gamma (\alpha )\Gamma (\beta )}{x}^{\alpha -1}(1-x{)}^{\beta -1}{I}_{(0,1)}(x)=\frac{1}{\mathrm{B}(\alpha ,\beta )}{x}^{\alpha -1}(1-x{)}^{\beta -1}{I}_{(0,1)}(x)$$
$$其中,\mathrm{B}(\alpha ,\beta )={\int }_0^1{u}^{\alpha -1}(1-u{)}^{\beta -1}du,(\alpha ,\beta >0)$$

$$M(t)={\int }_0^1\frac{1}{\mathrm{B}(\alpha ,\beta )}{x}^{\alpha -1}(1-x{)}^{\beta -1}{e}^{tx}dx$$
$$E(X)={\int }_0^1\frac{1}{\mathrm{B}(\alpha ,\beta )}{x}^{\alpha -1}(1-x{)}^{\beta -1}xdx$$
$$={\int }_0^1\frac{1}{\mathrm{B}(\alpha ,\beta )}{x}^{\alpha }(1-x{)}^{\beta -1}xdx$$
$$=\frac{{\int }_0^1{x}^{\alpha }(1-x{)}^{\beta -1}xdx}{\mathrm{B}(\alpha ,\beta )}$$
$$=\frac{\mathrm{B}(\alpha +1,\beta )}{\mathrm{B}(\alpha ,\beta )}$$
$$=\frac{\frac{\Gamma (\alpha +1)\Gamma (\beta )}{\Gamma (\alpha +\beta +1)}}{\frac{\Gamma (\alpha )\Gamma (\beta )}{\Gamma (\alpha +\beta )}}$$
=