1 矩量母函数
矩量母函数又称矩母函数（Moment Generating Function）又称动差生成函数，是一种构造函数，其定义为：
随机变量
X X X是连续型随机变量时，其矩量母函数为： 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随机变量 X X X是离散型随机变量时，其矩量母函数为： 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∑etxip(xi)
由泰勒级数可知
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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：
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\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!t2M2+⋯+n!tnMn+⋯其中，
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M_n
Mn即为
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X的
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n阶中心距。
矩量母函数对
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n阶导可得
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\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!t2M2)(n)+⋯+(n!tnMn)(n)+((n+1)!t(n+1)Mn+1)(n)+⋯=0+0+0+⋯+Mn+(n+1)!(n+1)⋅n⋅(n−1)⋯2⋅tMn+1+⋯当
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t=0时，则有
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\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⋅0Mn+1+⋯0+⋯=Mn由此可知随机变量
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X的均值和方差分别为：
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\mathrm{E}(X)=M_X^{(1)}(0)=\int_{-\infty}^{+\infty}xf(x)dx=M_1
E(X)=MX(1)(0)=∫−∞+∞xf(x)dx=M1
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\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 参数为
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n和
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p的二项分布
离散随机变量
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X服从参数为
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p的二项分布，则其矩母函数为
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\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}
ϕ(t)=E[etX]=k=0∑netk(nk)pk(1−p)n−k=k=0∑n(nk)(pet)k(1−p)n−k=(pet+1−p)n因此
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\phi^{\prime}(t)=n(pe^t+1-p)^{n-1}pe^t
ϕ′(t)=n(pet+1−p)n−1pet所以则有
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\mathrm{E}[X]=\phi^{\prime}(0)=np
E[X]=ϕ′(0)=np求二阶导则有
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\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
ϕ′′(t)=n(n−1)(pet+1−p)n−2(pet)2+n(pet+1−p)n−1pet所以
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\mathrm{E}[X^2]=\phi^{\prime\prime}(0)=n(n-1)p^2+np
E[X2]=ϕ′′(0)=n(n−1)p2+np因此，
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\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 均值为
λ
\lambda
λ的泊松分布
离散随机变量
X
X
X服从均值为
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\lambda
λ的泊松分布，则其矩母函数为
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\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)\}
ϕ(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
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\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}
ϕ′(t)ϕ′′(t)=λetexp{λ(et−1)}=(λet)2exp{λ(et−1)}+λet{λ(et−1)}所以则有
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\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)=ϕ′(0)=λ=ϕ′′(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 参数为
λ
\lambda
λ的指数分布
离散随机变量
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X服从参数为
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\lambda
λ的指数分布，则其矩母函数为
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\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
ϕ(t)=E[etX]=∫0∞etxλe−λxdx=λ∫0∞e−(λ−t)xdx=λ−tλ,t<λ从上面的推导可以发现，对于指数分布，
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ϕ(t)只对小于
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ϕ(t)微分可以得到
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\phi^{\prime}(t)=\frac{\lambda}{(\lambda - t)^2},\quad \phi^{\prime\prime}(t)=\frac{2\lambda}{(\lambda - t)^3}
ϕ′(t)=(λ−t)2λ,ϕ′′(t)=(λ−t)32λ因此
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\mathrm{E}[X]=\phi^{\prime}(0)=\frac{1}{\lambda},\quad \mathrm{E}[X^2]=\phi^{\prime\prime}(0)=\frac{2}{\lambda^2}
E[X]=ϕ′(0)=λ1,E[X2]=ϕ′′(0)=λ22于是
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\mathrm{Var}(X)=\mathrm{E}[X^2]-(\mathrm{E}[X])^2=\frac{1}{\lambda^2}
Var(X)=E[X2]−(E[X])2=λ21
5 参数为
μ
\mu
μ和
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2
\delta^2
δ2的正态分布
标准正态随机变量
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X
X的矩母函数如下所示
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\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]=2π1∫−∞+∞etxe−x2/2dx=2π1∫−∞+∞e−(x2−2tx)dx=et2/22π1∫−∞+∞e−(x−t)2/2dx=et2/2如果
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Z=\sigma X +\mu
Z=σX+μ就是参数为
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\mu
μ和
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\sigma^2
σ2的正态分布，则有
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\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\}
ϕ(t)=E[etZ]=E[et(σX+μ)]=etuE[etσX]=exp{2σ2t2+μt}经过微分可以得到
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ϕ
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exp
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exp
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\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}
ϕ′(t)ϕ′′(t)=(μ+tσ2)exp{2σ2t2+μt}=(μ+tσ2)2exp{2σ2t2+μt}+σ2exp{2σ2t2+μt}所以则有
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\mathrm{E}[X]=\phi^{\prime}(0)=\mu,\quad \mathrm{E}[X^2]=\phi^{\prime\prime}(0)=\mu^2+\sigma^2
E[X]=ϕ′(0)=μ,E[X2]=ϕ′′(0)=μ2+σ2方差为
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\mathrm{Var(X)}=\mathrm{E}[X^2]-(\mathrm{E}[X])^2=\sigma^2
Var(X)=E[X2]−(E[X])2=σ2
本文转载自: https://blog.csdn.net/qq_38406029/article/details/122659445
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版权归原作者 鬼道2022 所有， 如有侵权，请联系我们删除。