题目
设((x)_(1),(x)_(2),... ,(x)_(n))是来自总体X的样本值。记((x)_(1),(x)_(2),... ,(x)_(n))((x)_(1),(x)_(2),... ,(x)_(n))现又获得第((x)_(1),(x)_(2),... ,(x)_(n))个观测值((x)_(1),(x)_(2),... ,(x)_(n)).证明：((x)_(1),(x)_(2),... ,(x)_(n))((x)_(1),(x)_(2),... ,(x)_(n))
设
是来自总体X的样本值。记
现又获得第
个观测值
.证明：
题目解答
答案
（1）已知所以，
(2)因为
，所以，
因此，由方差的定义知，
因为
,所以，
解析
步骤 1：计算样本均值$\overline {{x}_{n+1}}$
已知${z}_{n}=\dfrac {1}{n}\sum _{i=1}^{n}{x}_{i}$,所以，$\sum _{i=1}^{n}{x}_{i}=n\overline {{x}_{n}},$${z}_{n+1}=\dfrac {\sum _{i=1}^{n-1}{x}_{i}}{n+1}=\dfrac {\sum _{i=1}^{n}{x}_{i}+{x}_{n+1}}{n+1}$$=\dfrac {n\overrightarrow {{x}_{n}+{x}_{n+1}}{n+1}=\dfrac {(n+1-1){x}_{n}+{x}_{n+1}}{n+1}$$=\dfrac {(n+1)\overline {{x}_{n}}+{x}_{n+1}-\overline {{x}_{n}}}{n+1}$$=\dfrac {1}{n+1}({x}_{n+1}-\overline {{x}_{n}})+\overline {{x}_{n}}$
步骤 2：计算样本方差${{S}_{n+1}}^{2}$
因为${{S}_{n}}^{2}=\dfrac {1}{n}\sum _{i=1}^{n}{({x}_{i}-\overline {{x}_{n}})}^{2}$，所以，>(x1-xn )^2 ^2=nSn^2 =1
因此，由方差的定义知，${{S}_{n+1}}^{2}=\dfrac {1}{n+1}\sum _{i=1}^{n+1}({x}_{i}-{\overline {x}}_{n+1})$$=\dfrac {1}{n+1}\sum _{i=1}^{n}{({x}_{i}-\overline {{x}_{n+1}})}^{2}+{({x}_{n+1}-\overline {{x}_{n+1}})}^{2}] $$=\dfrac {1}{n+1}\sum _{i=1}^{n}{({x}_{i}-\overline {{x}_{n}}+\overline {{x}_{n}}-\overline {{x}_{n+1}})}^{2}$$+{({x}_{n+1}-\overline {{x}_{n+1}})}^{2}] $$=\dfrac {1}{n+1}\sum _{i=1}^{n}{({x}_{i}-\overline {{x}_{n}})}^{2}+\sum _{i=1}^{n}{({x}_{n}-\overline {{x}_{n+1}})}^{2}$n +2 ${({x}_{i}-\overline {{x}_{n}})}_{{y}_{1}}({x}_{n}-{x}_{n}+1)$ $\hat {a}=1$$+{({x}_{n+1}-\overline {{x}_{n+1}})}^{2}] $
因为>(x1-xn )(xn-xn+1) =1=(xn-xn+1) $\sum _{i=1}^{n}{x}_{i}-n\overline {{x}_{n}})=0$, i==1 (xn-xn+1 )^2=5 [n+1 (xn+1-xn)]^2$=\dfrac {n{({x}_{n+1}-{x}_{n})}^{2}}{{(n+1)}^{2}}$,所以，${S}_{n+1}$$=\dfrac {\sum _{i=1}^{n}{({x}_{i}-\overline {{x}_{n}})}^{2}+\dfrac {n{({({n}_{n}-1)}^{2}}+{({x}_{n-1}-1-\overline {{x}_{n+1})}^{2}}{n+$$=\dfrac {n}{n+1}[ {{S}_{n}}^{2}+\dfrac {1}{n+1}{({x}_{n+1}-\overline {{x}_{n}})}^{2}] $
已知${z}_{n}=\dfrac {1}{n}\sum _{i=1}^{n}{x}_{i}$,所以，$\sum _{i=1}^{n}{x}_{i}=n\overline {{x}_{n}},$${z}_{n+1}=\dfrac {\sum _{i=1}^{n-1}{x}_{i}}{n+1}=\dfrac {\sum _{i=1}^{n}{x}_{i}+{x}_{n+1}}{n+1}$$=\dfrac {n\overrightarrow {{x}_{n}+{x}_{n+1}}{n+1}=\dfrac {(n+1-1){x}_{n}+{x}_{n+1}}{n+1}$$=\dfrac {(n+1)\overline {{x}_{n}}+{x}_{n+1}-\overline {{x}_{n}}}{n+1}$$=\dfrac {1}{n+1}({x}_{n+1}-\overline {{x}_{n}})+\overline {{x}_{n}}$
步骤 2：计算样本方差${{S}_{n+1}}^{2}$
因为${{S}_{n}}^{2}=\dfrac {1}{n}\sum _{i=1}^{n}{({x}_{i}-\overline {{x}_{n}})}^{2}$，所以，>(x1-xn )^2 ^2=nSn^2 =1
因此，由方差的定义知，${{S}_{n+1}}^{2}=\dfrac {1}{n+1}\sum _{i=1}^{n+1}({x}_{i}-{\overline {x}}_{n+1})$$=\dfrac {1}{n+1}\sum _{i=1}^{n}{({x}_{i}-\overline {{x}_{n+1}})}^{2}+{({x}_{n+1}-\overline {{x}_{n+1}})}^{2}] $$=\dfrac {1}{n+1}\sum _{i=1}^{n}{({x}_{i}-\overline {{x}_{n}}+\overline {{x}_{n}}-\overline {{x}_{n+1}})}^{2}$$+{({x}_{n+1}-\overline {{x}_{n+1}})}^{2}] $$=\dfrac {1}{n+1}\sum _{i=1}^{n}{({x}_{i}-\overline {{x}_{n}})}^{2}+\sum _{i=1}^{n}{({x}_{n}-\overline {{x}_{n+1}})}^{2}$n +2 ${({x}_{i}-\overline {{x}_{n}})}_{{y}_{1}}({x}_{n}-{x}_{n}+1)$ $\hat {a}=1$$+{({x}_{n+1}-\overline {{x}_{n+1}})}^{2}] $
因为>(x1-xn )(xn-xn+1) =1=(xn-xn+1) $\sum _{i=1}^{n}{x}_{i}-n\overline {{x}_{n}})=0$, i==1 (xn-xn+1 )^2=5 [n+1 (xn+1-xn)]^2$=\dfrac {n{({x}_{n+1}-{x}_{n})}^{2}}{{(n+1)}^{2}}$,所以，${S}_{n+1}$$=\dfrac {\sum _{i=1}^{n}{({x}_{i}-\overline {{x}_{n}})}^{2}+\dfrac {n{({({n}_{n}-1)}^{2}}+{({x}_{n-1}-1-\overline {{x}_{n+1})}^{2}}{n+$$=\dfrac {n}{n+1}[ {{S}_{n}}^{2}+\dfrac {1}{n+1}{({x}_{n+1}-\overline {{x}_{n}})}^{2}] $