# 中心极限定理

（重定向自中心極限定理

## 历史

Tijms (2004, p.169) 写到：

## 棣莫佛－拉普拉斯定理

### 内容

${\displaystyle X\sim B(n,p)}$ ${\displaystyle n}$ 伯努利实验中事件 A 出现的次数，每次試驗成功的機率為 ${\displaystyle p}$ ，且 ${\displaystyle q=1-p}$ ，则对任意有限区间 ${\displaystyle [a,b]}$

${\displaystyle x_{k}\equiv {\frac {k-np}{\sqrt {npq}}}}$ ，當${\displaystyle n\to {\infty }}$

(i) ${\displaystyle P(X=k)\to {\frac {1}{\sqrt {npq}}}\cdot {\frac {1}{\sqrt {2\pi }}}e^{-{\frac {1}{2}}x_{\mu _{n}}^{2}}}$

(ii) ${\displaystyle P(a\leq {\frac {X-np}{\sqrt {npq}}}\leq {b})\to \int _{a}^{b}\varphi (x)dx}$ ，其中${\displaystyle \varphi (x)={\frac {1}{\sqrt {2\pi }}}e^{-{\frac {x^{2}}{2}}}(-\infty

## 林德伯格－列维定理

### 内容

${\displaystyle {\bar {X}}={\frac {1}{n}}\sum _{i=1}^{n}X_{i}}$ ${\displaystyle \zeta _{n}={\frac {{\bar {X}}-\mu }{\sigma /{\sqrt {n}}}}}$ ，则 ${\displaystyle \lim _{n\rightarrow \infty }P\left(\zeta _{n}\leq z\right)=\Phi \left(z\right)}$

### 证明

${\displaystyle X_{k}-\mu }$ 特征函数${\displaystyle \varphi (t)}$ ，根据傅里叶变换，样本空间中的褶积在特征函数空间变为乘积，因此${\displaystyle \zeta _{n}}$ 的特征函数为${\displaystyle {\left[\varphi {\left({\frac {t}{\sigma {\sqrt {n}}}}\right)}\right]}^{n}}$ .由于${\displaystyle E(X_{k})=\mu ,D(X_{k})=\sigma ^{2}}$ ${\displaystyle \varphi '(0)=0,\varphi ''(0)=-\sigma ^{2}.}$ 因此

${\displaystyle \varphi (t)=1-{\frac {1}{2}}\sigma ^{2}t^{2}+o(t^{2})}$

${\displaystyle {\left[\varphi {\left({\frac {t}{\sigma {\sqrt {n}}}}\right)}\right]}^{n}=\left[1-{\frac {1}{2n}}t^{2}+o\left({\frac {t^{2}}{n}}\right)\right]^{n}\to {e^{-t^{2}/2}}}$

${\displaystyle \lim _{n\rightarrow \infty }P\left(\zeta _{n}\leq z\right)\to \Phi \left(z\right)}$

## 林德伯格-费勒定理

### 内容

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

${\displaystyle s_{i}^{2}={\rm {Var}}(X_{i})}$

${\displaystyle \sigma _{n}^{2}=\sum _{i=1}^{n}s_{i}^{2}={\rm {Var}}(S_{n})}$ .

${\displaystyle \lim _{n\rightarrow \infty }{1 \over \sigma _{n}^{2}}\sum _{i=1}^{n}E[X_{i}^{2};\{|X_{i}|>\epsilon \sigma _{n}\}]=0}$

${\displaystyle S_{n}/\sigma _{n}{\stackrel {d}{\rightarrow }}N(0,1)}$

${\displaystyle E[|X_{i}|^{3}]<\infty ,\,\lim _{n\rightarrow \infty }{1 \over \sigma _{n}^{3}}\sum _{i=1}^{n}E[|X_{i}|^{3}]=0}$

### 证明

${\displaystyle \left|\varphi _{S_{n}/\sigma _{n}}(t)-e^{-t^{2}/2}\right|=\left|\prod _{k=1}^{n}\varphi _{X_{k}}(t/\sigma _{n})-\prod _{k=1}^{n}e^{-t^{2}s_{k}^{2}/2\sigma _{n}^{2}}\right|\leq \sum _{k=1}^{n}\left|\varphi _{X_{k}}(t/\sigma _{n})-e^{-t^{2}s_{k}^{2}/2\sigma _{n}^{2}}\right|}$

${\displaystyle \sum _{k=1}^{n}\left|{\frac {i^{3}t^{3}E[X_{k}^{3}]}{6\sigma _{n}^{3}}}+{\frac {t^{4}s_{k}^{4}}{8\sigma _{n}^{4}}}\right|\leq {|t|^{3} \over 6\sigma _{n}^{3}}\sum _{k=1}^{n}E[|X_{k}|^{3}]+{\frac {t^{4}}{8\sigma _{n}^{4}}}\sum _{k=1}^{n}s_{k}^{4}\leq {|t|^{3} \over 6\sigma _{n}^{3}}\sum _{k=1}^{n}E[|X_{k}|^{3}]+{\frac {t^{4}}{8}}\max _{1\leq k\leq n}{s_{k}^{2} \over \sigma _{n}^{2}}}$

${\displaystyle k_{n}={\rm {arg}}\max _{1\leq k\leq n}s_{k}^{2}/\sigma _{n}^{2}}$ ，则由李雅普诺夫不等式

${\displaystyle (s_{k_{n}}/\sigma _{n})^{3/2}\leq E[|X_{k_{n}}/\sigma _{n}|^{3}]\leq {\frac {1}{\sigma _{n}^{3}}}\sum _{k=1}^{n}E[|X_{k}|^{3}]}$

## 参考文献

• 李贤平，概率论基础（第二版），高等教育出版社
• Olav Kallenberg，现代概率论基础（第二版），Springer（2002）。