矩量母函式介紹

1 矩量母函式

?矩量母函式又稱矩母函式（Moment Generating Function）又稱動差生成函式，是一種建構式，其定義為：

隨機變數 X X X是連續型隨機變數時，其矩量母函式為： M X ( t ) = E ( e t X ) = ∫ ? ∞ + ∞ e t x f ( x ) d x M_X(t)=\mathrm{E}(e^{tX})=\int_{-\infty}^{+\infty}e^{tx}f(x)dx MX?(t)=E(etX)=∫?∞+∞?etxf(x)dx隨機變數 X X X是離散型隨機變數時，其矩量母函式為： M X ( t ) = E ( e t X ) = ∑ x i e t x i p ( x i ) M_X(t)=\mathrm{E}(e^{tX})=\sum\limits_{x_i}e^{tx_i}p(x_i) MX?(t)=E(etX)=xi?∑?etxi?p(xi?)

?由泰勒級數可知 e x = 1 + x + x 2 2 ! + x 3 3 ! + x 4 4 ! + ? + x n n ! + ? e^{x}=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots+\frac{x^n}{n!}+\cdots ex=1+x+2!x2?+3!x3?+4!x4?+?+n!xn?+?得到： M X ( t ) = ∫ ? ∞ + ∞ ( 1 + x t + x 2 t 2 2 ! + x 3 t 3 3 ! + x 4 t 4 4 ! + ? + x n t n n ! + ? ? ) f ( x ) d x = ∫ ? ∞ + ∞ f ( x ) d x + t ∫ ? ∞ + ∞ x f ( x ) d x + t 2 2 ! ∫ ? ∞ + ∞ x 2 f ( x ) d x + ? t n n ! ∫ ? ∞ + ∞ x n f ( x ) d x + ? = t 0 M 0 + t 1 M 1 + t 2 2 ! M 2 + ? + t n n ! M n + ? \begin{aligned}M_X(t)&=\int_{-\infty}^{+\infty}(1+xt+\frac{x^2t^2}{2!}+\frac{x^3t^3}{3!}+\frac{x^4t^4}{4!}+\cdots+\frac{x^nt^n}{n!}+\cdots)f(x)dx\\&=\int_{-\infty}^{+\infty}f(x)dx+t\int_{-\infty}^{+\infty}xf(x)dx+\frac{t^2}{2!}\int_{-\infty}^{+\infty}x^2f(x)dx+\cdots \frac{t^n}{n!}\int_{-\infty}^{+\infty}x^nf(x)dx+\cdots\\&=t^0 M_0 + t^1 M_1 +\frac{t^2}{2!}M_2+\cdots+\frac{t^n}{n!}M_n+\cdots\end{aligned} MX?(t)?=∫?∞+∞?(1+xt+2!x2t2?+3!x3t3?+4!x4t4?+?+n!xntn?+?)f(x)dx=∫?∞+∞?f(x)dx+t∫?∞+∞?xf(x)dx+2!t2?∫?∞+∞?x2f(x)dx+?n!tn?∫?∞+∞?xnf(x)dx+?=t0M0?+t1M1?+2!t2?M2?+?+n!tn?Mn?+??其中， M n M_n Mn?即為 X X X的 n n n階中心距，
?矩量母函式對 t t t求 n n n階導可得 M X ( n ) ( t ) = ( t 0 M 0 ) ( n ) + ( t 1 M 1 ) ( n ) + ( t 2 2 ! M 2 ) ( n ) + ? + ( t n n ! M n ) ( n ) + ( t ( n + 1 ) ( n + 1 ) ! M n + 1 ) ( n ) + ? = 0 + 0 + 0 + ? + M n + ( n + 1 ) ? n ? ( n ? 1 ) ? 2 ? t ( n + 1 ) ! M n + 1 + ? \begin{aligned}M^{(n)}_X(t)&=(t^0M_0)^{(n)}+(t^1M_1)^{(n)}+\left(\frac{t^2}{2!}M_2\right)^{(n)}+\cdots+\left(\frac{t^n}{n!}M_n\right)^{(n)}+\left(\frac{t^{(n+1)}}{(n+1)!}M_{n+1}\right)^{(n)}+\cdots\\&=0+0+0+\cdots+M_n+\frac{(n+1)\cdot n \cdot (n-1) \cdots 2 \cdot t}{(n+1)!}M_{n+1}+\cdots\end{aligned} MX(n)?(t)?=(t0M0?)(n)+(t1M1?)(n)+(2!t2?M2?)(n)+?+(n!tn?Mn?)(n)+((n+1)!t(n+1)?Mn+1?)(n)+?=0+0+0+?+Mn?+(n+1)!(n+1)?n?(n?1)?2?t?Mn+1?+??當 t = 0 t=0 t=0時，則有
M X ( n ) ( 0 ) = 0 + 0 + 0 + ? + M n + ( n + 1 ) ? n ? ( n ? 1 ) ? 2 ? 0 ( n + 1 ) ! M n + 1 + ? 0 + ? = M n \begin{aligned}M^{(n)}_X(0)&=0+0+0+\cdots+M_n+\frac{(n+1)\cdot n \cdot (n-1) \cdots 2 \cdot 0}{(n+1)!}M_{n+1}+\cdots 0 +\cdots\\&=M_n\end{aligned} MX(n)?(0)?=0+0+0+?+Mn?+(n+1)!(n+1)?n?(n?1)?2?0?Mn+1?+?0+?=Mn??由此可知隨機變數 X X X的均值和方差分別為： E ( X ) = M X ( 1 ) ( 0 ) = ∫ ? ∞ + ∞ x f ( x ) d x = M 1 \mathrm{E}(X)=M_X^{(1)}(0)=\int_{-\infty}^{+\infty}xf(x)dx=M_1 E(X)=MX(1)?(0)=∫?∞+∞?xf(x)dx=M1? V a r ( X ) = E ( X 2 ) ? [ E ( X ) ] 2 = ∫ ? ∞ + ∞ x 2 p ( x ) d x ? ( ∫ ? ∞ + ∞ x p ( x ) d x ) 2 = M 2 ? ( M 1 ) 2 \mathrm{Var}(X)=\mathrm{E}(X^2)-[\mathrm{E}(X)]^2=\int_{-\infty}^{+\infty}x^2p(x)dx-\left(\int_{-\infty}^{+\infty}xp(x)dx\right)^2=M_2-(M_1)^2 Var(X)=E(X2)?[E(X)]2=∫?∞+∞?x2p(x)dx?(∫?∞+∞?xp(x)dx)2=M2??(M1?)2

2 引數為 n n n和 p p p的二項分布

?離散隨機變數 X X X服從引數為 n n n和 p p p的二項分布，則其矩母函式為 ? ( t ) = E [ e t X ] = ∑ k = 0 n e t k ( n k ) p k ( 1 ? p ) n ? k = ∑ k = 0 n ( n k ) ( p e t ) k ( 1 ? p ) n ? k = ( p e t + 1 ? p ) n \begin{aligned}\phi(t)&=\mathrm{E}[e^{tX}]=\sum\limits_{k=0}^ne^{tk}\left(\begin{array}{c}n\\k\end{array}\right)p^k (1-p)^{n-k}\\&=\sum\limits_{k=0}^n\left(\begin{array}{c}n\\k\end{array}\right)(pe^t)^k(1-p)^{n-k}\\&=(pe^t+1-p)^n\end{aligned} ?(t)?=E[etX]=k=0∑n?etk(nk?)pk(1?p)n?k=k=0∑n?(nk?)(pet)k(1?p)n?k=(pet+1?p)n?因此 ? ′ ( t ) = n ( p e t + 1 ? p ) n ? 1 p e t \phi^{\prime}(t)=n(pe^t+1-p)^{n-1}pe^t ?′(t)=n(pet+1?p)n?1pet所以則有 E [ X ] = ? ′ ( 0 ) = n p \mathrm{E}[X]=\phi^{\prime}(0)=np E[X]=?′(0)=np求二階導則有 ? ′ ′ ( t ) = n ( n ? 1 ) ( p e t + 1 ? p ) n ? 2 ( p e t ) 2 + n ( p e t + 1 ? p ) n ? 1 p e t \phi^{\prime\prime}(t)=n(n-1)(pe^t+1-p)^{n-2}(pe^t)^2+n(pe^t+1-p)^{n-1}pe^t ?′′(t)=n(n?1)(pet+1?p)n?2(pet)2+n(pet+1?p)n?1pet所以 E [ X 2 ] = ? ′ ′ ( 0 ) = n ( n ? 1 ) p 2 + n p \mathrm{E}[X^2]=\phi^{\prime\prime}(0)=n(n-1)p^2+np E[X2]=?′′(0)=n(n?1)p2+np因此， X X X的方差為 V a r ( X ) = E [ X 2 ] ? ( E [ X ] ) 2 = n ( n ? 1 ) p 2 + n p ? n 2 p 2 = n p ( 1 ? p ) \mathrm{Var}(X)=\mathrm{E}[X^2]-(\mathrm{E}[X])^2=n(n-1)p^2+np-n^2p^2=np(1-p) Var(X)=E[X2]?(E[X])2=n(n?1)p2+np?n2p2=np(1?p)

3 均值為 λ \lambda λ的泊松分布

?離散隨機變數 X X X服從均值為 λ \lambda λ的泊松分布，則其矩母函式為 ? ( t ) = E [ e t X ] = ∑ k = 0 ∞ e t n e ? λ λ n n ! = e ? λ ∑ n = 0 ∞ ( λ e t ) n n ! = e ? λ e λ e t = exp ? { λ ( e t ? 1 ) } \phi(t)=\mathrm{E}[e^{tX}]=\sum\limits_{k=0}^\infty\frac{e^{tn}e^{-\lambda}\lambda^n}{n!}=e^{-\lambda}\sum\limits_{n=0}^{\infty}\frac{(\lambda e^t)^n}{n!}=e^{-\lambda}e^{\lambda e^t}=\exp\{\lambda(e^t-1)\} ?(t)=E[etX]=k=0∑∞?n!etne?λλn?=e?λn=0∑∞?n!(λet)n?=e?λeλet=exp{λ(et?1)}求微分可得 ? ′ ( t ) = λ e t exp ? { λ ( e t ? 1 ) } ? ′ ′ ( t ) = ( λ e t ) 2 exp ? { λ ( e t ? 1 ) } + λ e t { λ ( e t ? 1 ) } \begin{aligned}\phi^{\prime}(t)&=\lambda e^t\exp\{\lambda(e^t-1)\}\\\phi^{\prime\prime}(t)&=(\lambda e^t)^2 \exp\{\lambda(e^t-1)\}+\lambda e^t\{\lambda(e^t-1)\}\end{aligned} ?′(t)?′′(t)?=λetexp{λ(et?1)}=(λet)2exp{λ(et?1)}+λet{λ(et?1)}?所以則有 E [ X ] = ? ′ ( 0 ) = λ E [ X ] = ? ′ ′ ( 0 ) = λ 2 + λ V a r ( X ) = E [ X 2 ] ? ( E [ X ] ) 2 = λ \begin{aligned}\mathrm{E}[X]&=\phi^{\prime}(0)=\lambda\\\mathrm{E}[X]&=\phi^{\prime\prime}(0)=\lambda^2+\lambda\\ \mathrm{Var}(X)&=\mathrm{E}[X^2]-(\mathrm{E}[X])^2=\lambda\end{aligned} E[X]E[X]Var(X)?=?′(0)=λ=?′′(0)=λ2+λ=E[X2]?(E[X])2=λ?因此，泊松分布的均值和方差都是 λ \lambda λ，

4 引數為 λ \lambda λ的指數分布

?離散隨機變數 X X X服從引數為 λ \lambda λ的指數分布，則其矩母函式為 ? ( t ) = E [ e t X ] = ∫ 0 ∞ e t x λ e ? λ x d x = λ ∫ 0 ∞ e ? ( λ ? t ) x d x = λ λ ? t , t < λ \phi(t)=\mathrm{E}[e^{tX}]=\int_0^{\infty}e^{tx}\lambda e^{-\lambda x}dx=\lambda \int_{0}^{\infty}e^{-(\lambda -t)x}dx =\frac{\lambda}{\lambda -t},\quad t < \lambda ?(t)=E[etX]=∫0∞?etxλe?λxdx=λ∫0∞?e?(λ?t)xdx=λ?tλ?,t<λ從上面的推導可以發現，對于指數分布， ? ( t ) \phi(t) ?(t)只對小于 λ \lambda λ的 t t t值定義，對 ? ( t ) \phi(t) ?(t)微分可以得到 ? ′ ( t ) = λ ( λ ? t ) 2 , ? ′ ′ ( t ) = 2 λ ( λ ? t ) 3 \phi^{\prime}(t)=\frac{\lambda}{(\lambda - t)^2},\quad \phi^{\prime\prime}(t)=\frac{2\lambda}{(\lambda - t)^3} ?′(t)=(λ?t)2λ?,?′′(t)=(λ?t)32λ?因此 E [ X ] = ? ′ ( 0 ) = 1 λ , E [ X 2 ] = ? ′ ′ ( 0 ) = 2 λ 2 \mathrm{E}[X]=\phi^{\prime}(0)=\frac{1}{\lambda},\quad \mathrm{E}[X^2]=\phi^{\prime\prime}(0)=\frac{2}{\lambda^2} E[X]=?′(0)=λ1?,E[X2]=?′′(0)=λ22?于是 X X X的方差為 V a r ( X ) = E [ X 2 ] ? ( E [ X ] ) 2 = 1 λ 2 \mathrm{Var}(X)=\mathrm{E}[X^2]-(\mathrm{E}[X])^2=\frac{1}{\lambda^2} Var(X)=E[X2]?(E[X])2=λ21?

5 引數為 μ \mu μ和 δ 2 \delta^2 δ2的正態分布

?標準正態隨機變數 X X X的矩母函式如下所示 E [ e t X ] = 1 2 π ∫ ? ∞ + ∞ e t x e ? x 2 / 2 d x = 1 2 π ∫ ? ∞ + ∞ e ? ( x 2 ? 2 t x ) d x = e t 2 / 2 1 2 π ∫ ? ∞ + ∞ e ? ( x ? t ) 2 / 2 d x = e t 2 / 2 \begin{aligned}\mathrm{E}[e^{tX}]&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}e^{tx}e^{-x^2/2}dx=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}e^{-(x^2-2tx)}dx\\&=e^{t^2/2}\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}e^{-(x-t)^2/2}dx = e^{t^2/2}\end{aligned} E[etX]?=2π ?1?∫?∞+∞?etxe?x2/2dx=2π ?1?∫?∞+∞?e?(x2?2tx)dx=et2/22π ?1?∫?∞+∞?e?(x?t)2/2dx=et2/2?如果 X X X是標準正態分布，那么 Z = σ X + μ Z=\sigma X +\mu Z=σX+μ就是引數為 μ \mu μ和 σ 2 \sigma^2 σ2的正態分布，則有 ? ( t ) = E [ e t Z ] = E [ e t ( σ X + μ ) ] = e t u E [ e t σ X ] = exp ? { σ 2 t 2 2 + μ t } \phi(t)=\mathrm{E}[e^{tZ}]=\mathrm{E}[e^{t(\sigma X + \mu)}]=e^{tu}\mathrm{E}[e^{t\sigma X}]=\exp\left\{\frac{\sigma^2t^2}{2}+\mu t\right\} ?(t)=E[etZ]=E[et(σX+μ)]=etuE[etσX]=exp{2σ2t2?+μt}經過微分可以得到 ? ′ ( t ) = ( μ + t σ 2 ) exp ? { σ 2 t 2 2 + μ t } ? ′ ′ ( t ) = ( μ + t σ 2 ) 2 exp ? { σ 2 t 2 2 + μ t } + σ 2 exp ? { σ 2 t 2 2 + μ t } \begin{aligned}\phi^{\prime}(t)&=(\mu+t\sigma^2)\exp\{\frac{\sigma^2 t^2}{2}+\mu t\}\\\phi^{\prime\prime}(t)&=(\mu+ t\sigma^2)^2\exp\{\frac{\sigma^2 t^2}{2}+\mu t\}+\sigma^2 \exp \left\{\frac{\sigma^2 t^2}{2}+\mu t\right\}\end{aligned} ?′(t)?′′(t)?=(μ+tσ2)exp{2σ2t2?+μt}=(μ+tσ2)2exp{2σ2t2?+μt}+σ2exp{2σ2t2?+μt}?所以則有 E [ X ] = ? ′ ( 0 ) = μ , E [ X 2 ] = ? ′ ′ ( 0 ) = μ 2 + σ 2 \mathrm{E}[X]=\phi^{\prime}(0)=\mu,\quad \mathrm{E}[X^2]=\phi^{\prime\prime}(0)=\mu^2+\sigma^2 E[X]=?′(0)=μ,E[X2]=?′′(0)=μ2+σ2方差為 V a r ( X ) = E [ X 2 ] ? ( E [ X ] ) 2 = σ 2 \mathrm{Var(X)}=\mathrm{E}[X^2]-(\mathrm{E}[X])^2=\sigma^2 Var(X)=E[X2]?(E[X])2=σ2