중요하다(i.i.d 조건)
If \(X_n,n\ge 1\) are i.i.d random variables with \(E(X_1)=c\) and \(\text{Var}(X_1)=\sigma^2\), where \(0<\sigma^2<\infty\), then \[ \frac{S_n-nc}{\sigma\sqrt{n}}\leadsto Z\sim N(0,1). \]
우리가 가장 많이 봐왔던 CLT이다 \(\left(\frac{\bar X_n-E(X_1)}{\sigma/\sqrt{n}}=\frac{S_n-nc}{\sigma\sqrt{n}}\leadsto Z\sim N(0,1)\right)\).
i.i.d 조건이 있다는 것에 주목해야 한다.
매우 중요하다(independent 조건)
Suppose that for \(n\ge 1\), \(X_{n,1},\ldots, X_{n,r_n}\) are independent random variables with \[ E(X_{n,k})=0\mbox{ }\mbox{ }\mbox{ }\mbox{ and }\mbox{ }\mbox{ }\mbox{ } \sigma_{n,k}^2=E(X_{n,k}^2)<\infty,\mbox{ }\mbox{ }k=1,\ldots,r_n. \] Let \[ S_n=X_{n,1}+\cdots+X_{n,r_n}\mbox{ }\mbox{ }\mbox{ and }\mbox{ }\mbox{ }\mbox{ }s_n^2=\sigma_{n,1}^2+\cdots+\sigma_{n,r_n}^2, \] and assume that \(s_n^2>0\) for \(n\) large. Then
\(\frac{S_n}{s_n}\leadsto Z\sim (0,1)\) as \(n\rightarrow \infty\);
\(\frac{\max_{1\le k\le r_n}\sigma^2_{n,k}}{s_n^2}\rightarrow 0\) as \(n\rightarrow \infty\),
are necessary and sufficient that the Linderberg condition hold: \[ \lim_{n\rightarrow \infty}\frac{1}{s_n^2}\sum_{k=1}^{r_n}\int_{\{|X_{n,k}|\ge \epsilon s_n \}}X_{n,k}^2\mbox{ }dP=0\mbox{ }\mbox{ }\mbox{ for all }\mbox{ }\mbox{ }\epsilon>0. \]
증명은 복잡하므로 생략한다.
우리는 항상 Linderberg Feller CLT를 사용할때 변수를 \(r_n=n\)에 대한 \(X_{n,k}=X_k\), \(1\le k\le n\)로 생각하여 적용할 것이다.
주로 주어진 독립변수들에 대해 Linderberg 조건이 만족하는지 check하고 만족한다면 \(\frac{S_n}{s_n}\leadsto Z\sim(0,1)\)로 문제를 해결할 것이다.
If \(X_n,n\ge 1\) are independent random variables with \[ E(X_n)=0\mbox{ }\mbox{ }\mbox{ }\mbox{ and }\mbox{ }\mbox{ }\mbox{ }\text{Var}(X_n)=\sigma_n^2, \mbox{ }\mbox{ }\mbox{ }n\ge 1, \] and \(s_n^2=\sum_{k=1}^n\sigma_k^2\), then the Linderberg condition \[ \lim_{n\rightarrow \infty}\frac{1}{s_n^2}\sum_{k=1}^{r_n}\int_{\{|X_{n,k}|\ge \epsilon s_n \}}X_{n,k}^2\mbox{ }dP=0\mbox{ }\mbox{ }\mbox{ for all }\mbox{ }\mbox{ }\epsilon>0, \] is equivalent to \[ \lim_{n\rightarrow \infty}\frac{1}{s_n^2}\sum_{k=1}^{r_n}\int_{\{|X_{n,k}|\ge \epsilon s_k \}}X_{n,k}\mbox{ }dP=0\mbox{ }\mbox{ }\mbox{ for all }\mbox{ }\mbox{ }\epsilon>0. \]
For each \(n\ge 1\), let \(X_{n,1},\ldots X_{n,{r_n}}\) be independent random variables with \[ E[X_{n,k}]=0\mbox{ }\mbox{ }\mbox{ }\mbox{ and }\mbox{ }\mbox{ }\mbox{ }\text{Var}[X_{n,k}]=\sigma_{n,k}^2, \mbox{ }\mbox{ }\mbox{ }k=1,\ldots,r_n., \] and let \(s_n^2=\sum_{k=1}^{r_n}\sigma_{n,k}^2\). If there exists a constant \(\delta>0\) such that Lyapounov’s condition \[ \lim_{n\rightarrow \infty}\frac{1}{s_n^{2+\delta}}\sum_{k=1}^{r_n}E\left[|X_{n,k}|^{2+\delta}\right]=0 \] holds, then \(S_n/s_n\leadsto Z\sim N(0,1)\).
Suppose that \(X_n,n\ge 1\) are independent, uniformly bounded random variables with mean 0. If \[ s_n^2=\text{Var}(S_n)\rightarrow \infty \mbox{ }\mbox{ }\mbox{ as }\mbox{ }n\rightarrow \infty, \] then, \(S_n/s_n\leadsto Z\sim N(0,1)\)
\[ \frac{1}{s_n^3}\sum_{k=1}^nE\left[|X_k|^3\right]\le \frac{B}{s_n^3}\sum_{k=1}^nE\left[X_k^2\right]=\frac{B}{s_n}\rightarrow 0. \] 때문에 Lyapounov condition이 만족한다.
Let \(1\le \alpha<\infty\) and let \(X_n,n\ge1\) be independent random variables with \[ P(X_n=n^\alpha)=P(X_n=-n^\alpha)=\frac{1}{6}n^{-2(\alpha-1)}, \] and \[ P(X_n=0)=1-\frac{1}{3}n^{-2(\alpha-1)}. \] Then, \[ \frac{S_n}{s_n}\leadsto Z\sim N(0,1)\mbox{ }\mbox{ }\iff \mbox{ }\mbox{ } 1\le \alpha <\frac{3}{2}. \]