矩生成函數

（重定向自動差生成函數

定義

${\displaystyle M_{X}(t)=\mathbb {E} \left(e^{tX}\right),\quad t\in \mathbb {R} }$

计算

${\displaystyle M_{X}(t)=\int _{-\infty }^{\infty }e^{tx}f(x)\,\mathrm {d} x}$
${\displaystyle =\int _{-\infty }^{\infty }\left(1+tx+{\frac {t^{2}x^{2}}{2!}}+\cdots \right)f(x)\,\mathrm {d} x}$
${\displaystyle =1+tm_{1}+{\frac {t^{2}m_{2}}{2!}}+\cdots }$

${\displaystyle M_{X}(t)=\int _{0}^{1}e^{tx}\,dF(x)}$

${\displaystyle S_{n}=\sum _{i=1}^{n}a_{i}X_{i}}$

${\displaystyle M_{S_{n}}(t)=M_{X_{1}}(a_{1}t)M_{X_{2}}(a_{2}t)\cdots M_{X_{n}}(a_{n}t)}$ 　。

${\displaystyle M_{X}(\mathbf {t} )=\operatorname {E} \left(e^{\langle \mathbf {t} ,\mathbf {X} \rangle }\right)}$

意义

${\displaystyle \operatorname {\mathbb {E} } \left(X^{n}\right)=M_{X}^{(n)}(0)=\left.{\frac {\mathrm {d} ^{n}M_{X}(t)}{\mathrm {d} t^{n}}}\right|_{t=0}}$ 　。

例子

${\displaystyle \forall t<-\ln(1-p)}$
${\displaystyle {\frac {pe^{it}}{1-(1-p)\,e^{it}}}}$

Noncentral chi-squared ${\displaystyle \chi _{k}^{2}(\lambda )}$  ${\displaystyle e^{\lambda t/(1-2t)}(1-2t)^{-{\frac {k}{2}}}}$  ${\displaystyle e^{i\lambda t/(1-2it)}(1-2it)^{-{\frac {k}{2}}}}$

Multivariate Cauchy

${\displaystyle \operatorname {MultiCauchy} (\mu ,\Sigma )}$ [1]

参考文献

1. ^ Kotz et al.[需要完整来源] p. 37 using 1 as the number of degree of freedom to recover the Cauchy distribution