题目
设_(1),(X)_(2),... 是独立同分布随机变量序列,且_(1),(X)_(2),... ,对任给_(1),(X)_(2),... ,均有_(1),(X)_(2),... _________.
设
是独立同分布随机变量序列,且
,对任给
,均有
_________.
题目解答
答案
是独立同分布随机变量序列,且
,则
,
,则
,则对任给
,均有
.
解析
步骤 1:计算$\sum _{i=1}^{n}{X}_{i}$的期望和方差
由于$X_1, X_2, \cdots$是独立同分布的随机变量序列,且$E(X_1) = \mu$,$Var(X_1) = \sigma^2$,则$\sum _{i=1}^{n}{X}_{i}$的期望和方差分别为:
$$E(\sum _{i=1}^{n}{X}_{i}) = \sum _{i=1}^{n}E({X}_{i}) = \sum _{i=1}^{n}\mu = n\mu$$
$$Var(\sum _{i=1}^{n}{X}_{i}) = \sum _{i=1}^{n}Var({X}_{i}) = \sum _{i=1}^{n}{\sigma}^{2} = n{\sigma}^{2}$$
步骤 2:确定$\sum _{i=1}^{n}{X}_{i}$的分布
由于$X_1, X_2, \cdots$是独立同分布的随机变量序列,且$E(X_1) = \mu$,$Var(X_1) = \sigma^2$,则$\sum _{i=1}^{n}{X}_{i}$的分布为:
$$\sum _{i=1}^{n}{X}_{i} \sim N(n\mu, n{\sigma}^{2})$$
步骤 3:确定$\dfrac {\sum _{i=1}^{n}{X}_{i}-n\mu }{\sigma \sqrt {n}}$的分布
由于$\sum _{i=1}^{n}{X}_{i} \sim N(n\mu, n{\sigma}^{2})$,则$\dfrac {\sum _{i=1}^{n}{X}_{i}-n\mu }{\sigma \sqrt {n}}$的分布为:
$$\dfrac {\sum _{i=1}^{n}{X}_{i}-n\mu }{\sigma \sqrt {n}} \sim N(0,1)$$
步骤 4:计算$\lim _{n\rightarrow \infty }P\{ \dfrac {\sum _{i=1}^{n}{X}_{i}-n\mu }{\sigma \sqrt {n}}\leqslant x\}$
由于$\dfrac {\sum _{i=1}^{n}{X}_{i}-n\mu }{\sigma \sqrt {n}} \sim N(0,1)$,则对任给$x\in (-\infty ,\infty )$,均有:
$$\lim _{n\rightarrow \infty }P\{ \dfrac {\sum _{i=1}^{n}{X}_{i}-n\mu }{\sigma \sqrt {n}}\leqslant x\} = \phi(x)$$
其中$\phi(x)$是标准正态分布的累积分布函数。
由于$X_1, X_2, \cdots$是独立同分布的随机变量序列,且$E(X_1) = \mu$,$Var(X_1) = \sigma^2$,则$\sum _{i=1}^{n}{X}_{i}$的期望和方差分别为:
$$E(\sum _{i=1}^{n}{X}_{i}) = \sum _{i=1}^{n}E({X}_{i}) = \sum _{i=1}^{n}\mu = n\mu$$
$$Var(\sum _{i=1}^{n}{X}_{i}) = \sum _{i=1}^{n}Var({X}_{i}) = \sum _{i=1}^{n}{\sigma}^{2} = n{\sigma}^{2}$$
步骤 2:确定$\sum _{i=1}^{n}{X}_{i}$的分布
由于$X_1, X_2, \cdots$是独立同分布的随机变量序列,且$E(X_1) = \mu$,$Var(X_1) = \sigma^2$,则$\sum _{i=1}^{n}{X}_{i}$的分布为:
$$\sum _{i=1}^{n}{X}_{i} \sim N(n\mu, n{\sigma}^{2})$$
步骤 3:确定$\dfrac {\sum _{i=1}^{n}{X}_{i}-n\mu }{\sigma \sqrt {n}}$的分布
由于$\sum _{i=1}^{n}{X}_{i} \sim N(n\mu, n{\sigma}^{2})$,则$\dfrac {\sum _{i=1}^{n}{X}_{i}-n\mu }{\sigma \sqrt {n}}$的分布为:
$$\dfrac {\sum _{i=1}^{n}{X}_{i}-n\mu }{\sigma \sqrt {n}} \sim N(0,1)$$
步骤 4:计算$\lim _{n\rightarrow \infty }P\{ \dfrac {\sum _{i=1}^{n}{X}_{i}-n\mu }{\sigma \sqrt {n}}\leqslant x\}$
由于$\dfrac {\sum _{i=1}^{n}{X}_{i}-n\mu }{\sigma \sqrt {n}} \sim N(0,1)$,则对任给$x\in (-\infty ,\infty )$,均有:
$$\lim _{n\rightarrow \infty }P\{ \dfrac {\sum _{i=1}^{n}{X}_{i}-n\mu }{\sigma \sqrt {n}}\leqslant x\} = \phi(x)$$
其中$\phi(x)$是标准正态分布的累积分布函数。