题型 4. 统计量求数字特征

方法与技巧

统计量求期望, 方差等数字特征是数理统计中的基本题型. 另外在后面的有关估计量的无偏性及有效性内容当中也会用到此类计算. 统计量求数字特征时经常用到以下公式, 需熟记:

$$E\left(\bar{X}\right)=E\left(X\right)\left(\text{ 或 }\mu \right),$$ $$D\left(\bar{X}\right)=\frac{D\left(X\right)}{n}\left(\text{ 或 }\frac{{\sigma }^2}{n}\right),$$ $$E\left(S^2\right)=D\left(X\right)\text{(或}{\sigma }^2\text{),}$$

1.17

设总体 $X$ 服从正态分布 $N\left(\mu ,{\sigma }^2\right),X_1,X_2,\cdots ,X_n$ 为其样本, $\bar{X}$ 为均值, $S^2$ 为样本方差. 试求:

(1) $\bar{X}$ 的数学期望与方差.

(2) $S^2$ 的数学期望.

分析

本题既可以利用公式 $E\left(\bar{X}\right)=E\left(X\right)\text{、}D\left(\bar{X}\right)=\frac{D\left(X\right)}{n}\text{、}E\left(S^2\right)=D\left(X\right)$ 直接计算,也可利用期望与方差的性质推导,其中在求 $S^2$ 的期望时,采用 $S^2$ 的另一种表达式,即

$$S^2=\frac{1}{n-1}\left(\sum _{i=1}^n{X}_i^2-n{\bar{X}}^2\right),$$

问题就变得简单了.

解

(1)

$X_1,X_2,\cdots ,X_n$ 相互独立且与 $X$ 有相同的分布,所以

$$E\left(X_i\right)=E\left(X\right),$$ $$\begin{array}{rl}E\left(\bar{X}\right)& =E\left(\frac{1}{n}\sum _{i=1}^n{X}_i\right)\\ & =\frac{1}{n}\sum _{i=1}^{n}E\left(X_i\right)\\ & =\frac{1}{n}nE\left(X\right)\\ & =E\left(X\right).\end{array}$$

故 $E\left(\bar{X}\right)=E\left(X\right)=\mu$.

又因为 $X_1,X_2,\cdots ,X_n$ 相互独立,而且与 $X$ 分布相同,所以

$$D\left(X_i\right)=D\left(X\right)\left(i=1,2,\cdots ,n\right),$$ $$\begin{array}{rl}D\left(\bar{X}\right)& =D\left(\frac{1}{n}\sum _{i=1}^n{X}_i\right)\\ & =\frac{1}{n^2}\sum _{i=1}^{n}D\left(X_i\right)\\ & =\frac{1}{n^2}nD\left(X\right)\\ & =\frac{1}{n}D\left(X\right).\end{array}$$

故 $D\left(\bar{X}\right)=\frac{{\sigma }^2}{n}$.

(2)

$$\begin{array}{rl}E\left(S^2\right)& =E\left[\frac{1}{n-1}\left(\sum _{i=1}^n{X}_i^2-n{\bar{X}}^2\right)\right]\\ & =\frac{1}{n-1}\left[\sum _{i=1}^{n}E\left(X_i^2\right)-nE\left({\bar{X}}^2\right)\right].\end{array}$$

因为

$$E\left(X_i^2\right)=D\left(X_i\right)+{\left[E\left(X_i\right)\right]}^2\left(i=1,2,\cdots ,n\right),$$ $$E\left({\bar{X}}^2\right)=D\left(\bar{X}\right)+{\left[E\left(\bar{X}\right)\right]}^2,$$

因此

$$E\left(X_i^2\right)=D\left(X\right)+{\left[E\left(X\right)\right]}^2\left(i=1,2,\cdots ,n\right),$$ $$E\left({\bar{X}}^2\right)=\frac{1}{n}D\left(X\right)+{\left[E\left(X\right)\right]}^2,$$

故

$$\begin{array}{rl}E\left(S^2\right)& =\frac{1}{n-1}\left\{nD\left(X\right)+n{\left[E\left(X\right)\right]}^2-D\left(X\right)-n{\left[E\left(X\right)\right]}^2\right\}\\ & =D\left(X\right).\end{array}$$

所以 $E\left(S^2\right)={\sigma }^2$.

1.18

设 $X_1,X_2,\cdots ,X_m$ 为来自二项分布总体 $B\left(n,p\right)$ 的简单随机样本, $\bar{X}$ 和 $S^2$ 分别为样本均值和样本方差. 记统计量 $T=\bar{X}-S^2$, 则 $E\left(T\right)=$ ___ .

解

因为 $X\sim B\left(n,p\right)$,则 $EX=np,DX=np\left(1-p\right)$.

则

$$\begin{array}{rl}E\left(T\right)& =E\left(\bar{X}-S^2\right)\\ & =E\left(\bar{X}\right)-E\left(S^2\right)\\ & =EX-DX\\ & =np-np(1-p)\\ & =n{p}^2.\end{array}$$

故应填 $n{p}^2$.

1.19

设 $X_1,X_2,\cdots ,X_n$ 为来自总体 $N\left(\mu ,{\sigma }^2\right)$ 的简单随机样本,记统计量 $T=\frac{1}{n}\sum _{i=1}^n{X}_i^2$, 则 $E\left(T\right)=$ ___ .

解

因为 $X_i\sim N\left(\mu ,{\sigma }^2\right)$,所以 $E{X}_i=\mu ,D{X}_i={\sigma }^2$. 则

$$\begin{array}{rl}E\left(T\right)& =E\left(\frac{1}{n}\sum _{i=1}^n{X}_i^2\right)\\ & =\frac{1}{n}\sum _{i=1}^{n}E\left(X_i^2\right)\\ & =\frac{1}{n}\sum _{i=1}^n\left[D{X}_i+{\left(E{X}_i\right)}^2\right]\\ & =\frac{1}{n}\sum _{i=1}^n\left({\sigma }^2+{\mu }^2\right)\\ & ={\sigma }^2+{\mu }^2.\end{array}$$

1. 20

设总体 $X$ 的概率密度为 $f\left(x\right)=\frac{1}{2}{\mathrm{e}}^{-|x|}\left(-\infty <x<+\infty \right),X_1,X_2,\cdots ,X_n$ 为总体 $X$ 的简单随机样本,其样本方差为 $S^2$,则 $E{S}^2=$ ___ .

解

因为 $E\left(S^2\right)=DX$,而

$$EX={\int }_{-\infty }^{+\infty }xf\left(x\right)\mathrm{d}x=0,$$

$E\left(X^2\right)={\int }_{-\infty }^{+\infty }{x}^{2}f\left(x\right)\mathrm{d}x={\int }_{-\infty }^{+\infty }{x}^2\cdot \frac{1}{2}{\mathrm{e}}^{-|x|}\mathrm{d}x$

$$={\int }_0^{+\infty }{x}^2{\mathrm{e}}^{-x}\mathrm{d}x=-{x^2{\mathrm{e}}^{-x}|}_0^{+\infty }+{\int }_0^{+\infty }2x{\mathrm{e}}^{-x}\mathrm{d}x=2,$$ $$DX=E\left(X^2\right)-{\left(EX\right)}^2=2,$$

则 $E\left(S^2\right)=2$.

1.21

设总体 $X$ 服从正态分布 $N\left(\mu ,{\sigma }^2\right)\left(\sigma >0\right)$. 从该总体中抽取简单随机样本 $X_1,X_2$, $\cdots ,X_{2n}\left(n\ge 2\right)$. 其样本均值为 $\bar{X}=\frac{1}{2n}\sum _{i=1}^{2n}{X}_i$,试求统计量

$$Y=\sum _{i=1}^n{\left(X_i+X_{n+i}-2\bar{X}\right)}^2$$

的数学期望 $E\left(Y\right)$.

解法一

由已知条件 $X_1,X_2,\cdots ,X_{2n}$ 均服从 $N\left(\mu ,{\sigma }^2\right)$ 且相互独立,

所以 $\left(X_1+X_{n+1}\right),\left(X_2+X_{n+2}\right),\cdots ,\left(X_n+X_{2n}\right)$ 相互独立且服从 $N\left(2\mu ,2{\sigma }^2\right)$,故

$$\left(X_1+X_{n+1}\right),\left(X_2+X_{n+2}\right),\cdots ,\left(X_n+X_{2n}\right)$$

可作为来自总体 $N\left(2\mu ,2{\sigma }^2\right)$ 的样本.

其样本均值为

$$\frac{1}{n}\sum _{i=1}^n\left(X_i+X_{n+i}\right)=\frac{1}{n}\sum _{i=1}^{2n}{X}_i=2\bar{X},$$

其样本方差为

$$\frac{1}{n-1}\sum _{i=1}^n{\left(X_i+X_{n+i}-2\bar{X}\right)}^2=\frac{1}{n-1}Y,$$

因为 $E\left(S^2\right)={\sigma }^2$,故

$$E\left(\frac{1}{n-1}Y\right)=2{\sigma }^2,\text{ 得 }E\left(Y\right)=2\left(n-1\right){\sigma }^2.$$

解法二

记 ${\bar{X}}^{\prime }=\frac{1}{n}\sum _{i=1}^n{X}_i,{\bar{X}}^{\prime \prime }=\frac{1}{n}\sum _{i=1}^n{X}_{n+i}$,

显然有 $2\bar{X}={\bar{X}}^{\prime }+{\bar{X}}^{\prime \prime }$. 因此

$$\begin{array}{rl}E\left(Y\right)& =E\left[\sum _{i=1}^n{\left(X_i+X_{n+i}-2\bar{X}\right)}^2\right]\\ & =E\left\{\sum _{i=1}^n{\left[\left(X_i-{\bar{X}}^{\prime }\right)+\left(X_{n+i}-{\bar{X}}^{\prime \prime }\right)\right]}^2\right\}\\ & =E\left\{\sum _{i=1}^n\left[{\left(X_i-{\bar{X}}^{\prime }\right)}^2+2\left(X_i-{\bar{X}}^{\prime }\right)\left(X_{n+i}-{\bar{X}}^{\prime \prime }\right)+{\left(X_{n+i}-{\bar{X}}^{\prime \prime }\right)}^2\right]\right\}\\ & =E\left[\sum _{i=1}^n{\left(X_i-{\bar{X}}^{\prime }\right)}^2\right]+0+E\left[\sum _{i=1}^n{\left(X_{n+i}-{\bar{X}}^{\prime \prime }\right)}^2\right]\\ & =(n-1){\sigma }^2+(n-1){\sigma }^2\\ & =2(n-1){\sigma }^2.\end{array}$$

解法三

$$\begin{array}{rl}Y& =\sum _{i=1}^n{\left(X_i+X_{n+i}-2\bar{X}\right)}^2\\ & =\sum _{i=1}^n\left(X_i^2+X_{n+i}^2+2X_i{X}_{n+i}-4\bar{X}{X}_i-4\bar{X}{X}_{n+i}+4{\bar{X}}^2\right)\\ & =\sum _{i=1}^{2n}{X}_i^2+2\sum _{i=1}^n{X}_i{X}_{n+i}-4\bar{X}\sum _{i=1}^{2n}{X}_i+4n{\bar{X}}^2\\ & =\sum _{i=1}^{2n}{X}_i^2+2\sum _{i=1}^n{X}_i{X}_{n+i}-4n{\bar{X}}^2.\end{array}$$

又由 $D\bar{X}=E\left({\bar{X}}^2\right)-{\left(E\bar{X}\right)}^2$ 可得, $E\left({\bar{X}}^2\right)=\frac{{\sigma }^2}{2n}+{\mu }^2$.

同理 $E{X}_i^2={\mu }^2+{\sigma }^2$.

因此

$$\begin{array}{rl}E\left(Y\right)& =\sum _{i=1}^{2n}E{X}_i^2+2\sum _{i=1}^{n}E{X}_{i}E{X}_{n+i}-4nE\left({\bar{X}}^2\right)\\ & =2n\left({\sigma }^2+{\mu }^2\right)+2n{\mu }^2-4n\left(\frac{{\sigma }^2}{2n}+{\mu }^2\right)\\ & =2(n-1){\sigma }^2.\end{array}$$

点评

本题方法一和方法二都用到了 $E\left(S^2\right)={\sigma }^2$ 这个结论,非常方便.

1.22

设总体 $X$ 服从参数 $\lambda \left(\lambda >0\right)$ 的泊松分布, $X_1,X_2,\cdots ,X_n\left(n\ge 2\right)$ 为来自总体的简单随机样本,则对于统计量 $T_1=\frac{1}{n}\sum _{i=1}^n{X}_i,T_2=\frac{1}{n-1}\sum _{i=1}^{n-1}{X}_i+\frac{1}{n}{X}_n$ 有_____.

(A) $E{T}_1>E{T}_2,D{T}_1>D{T}_2$ (B) $E{T}_1>E{T}_2,D{T}_1<D{T}_2$

(C) $E{T}_1<E{T}_2,D{T}_1>D{T}_2$ (D) $E{T}_1<E{T}_2,D{T}_1<D{T}_2$

解

由 $X_1,\cdots ,X_n\sim P\left(\lambda \right)$ 知

$$E{X}_i=\lambda ,D{X}_i=\lambda ,i=1,2,\cdots ,n.$$

从而 $E{T}_1=E\left(\frac{1}{n}\sum _{i=1}^n{X}_i\right)=\lambda$,

$$E{T}_2=\frac{1}{n-1}\sum _{i=1}^{n-1}E{X}_i+\frac{1}{n}E{X}_n=\lambda +\frac{\lambda }{n},$$

故 $E{T}_1<E{T}_2$,

$$D{T}_1=\frac{1}{n^2}D\left(\sum _{i=1}^n{X}_i\right)=\frac{n\lambda }{n^2}=\frac{\lambda }{n},$$ $$D{T}_2=\frac{1}{{\left(n-1\right)}^2}\sum _{i=1}^{n-1}D{X}_i+\frac{1}{n^2}D{X}_n$$ $$=\frac{\lambda }{n-1}+\frac{\lambda }{n^2}>\frac{\lambda }{n}=D{T}_1.$$

故应选(D).

1.23

设 $X_1,X_2,\cdots ,X_n$ 是取自 $N\left(0,{\sigma }^2\right)$ 的简单样本, ${\bar{X}}_k=\frac{1}{k}\sum _{i=1}^k{X}_i,1\le k\le n$,则 $\mathrm{Cov}\left({\bar{X}}_k,{\bar{X}}_{k+1}\right)=$ ( ).

(A) ${\sigma }^2$ (B) $\frac{{\sigma }^2}{k}$ (C) $\frac{{\sigma }^2}{k+1}$ (D) $\frac{{\sigma }^2}{k\left(k+1\right)}$

解

$\mathrm{Cov}\left({\bar{X}}_k,{\bar{X}}_{k+1}\right)=\frac{1}{k\left(k+1\right)}\mathrm{Cov}\left(\sum _{i=1}^k{X}_i,\sum _{i=1}^k{X}_i+X_{k+1}\right)$

$=\frac{1}{k\left(k+1\right)}\mathrm{Cov}\left(\sum _{i=1}^k{X}_i,\sum _{i=1}^k{X}_i\right)=\frac{1}{k\left(k+1\right)}D\left(\sum _{i=1}^k{X}_i\right)$

$=\frac{1}{k\left(k+1\right)}\cdot k{\sigma }^2=\frac{{\sigma }^2}{k+1}.$

故应选(C).

1.24

设 $X_1,X_2,\cdots ,X_n$ 是来自总体 $N\left(\mu ,{\sigma }^2\right)$ 的样本,记 $Y=\frac{1}{n}\sum _{i=1}^n\left|X_i-\mu \right|$,试证:

$$E\left(Y\right)=\sqrt{\frac{2}{\pi }}\sigma ,D\left(Y\right)=\left(1-\frac{2}{\pi }\right)\frac{{\sigma }^2}{n}.$$

分析

$Y_i=X_i-\mu \sim N\left(0,{\sigma }^2\right),E\left(\left|Y_i\right|\right)=\frac{2}{\sqrt{2\pi }\sigma }{\int }_0^{+\infty }y{\mathrm{e}}^{\frac{y^2}{2{\sigma }^2}}\mathrm{d}y$.

证 记 $Y_i=X_i-\mu$,得 $Y_i\sim N\left(0,{\sigma }^2\right),i=1,2,\cdots ,n$.

$$E\left(\left|X_i-\mu \right|\right)=E\left(\left|Y_i\right|\right)=\frac{1}{\sqrt{2\pi }\sigma }{\int }_{-\infty }^{+\infty }\left|y\right|{\mathrm{e}}^{-\frac{y^2}{2{\sigma }^2}}\mathrm{d}y=\frac{2}{\sqrt{2\pi }\sigma }{\int }_0^{+\infty }y{\mathrm{e}}^{-\frac{y^2}{2{\sigma }^2}}\mathrm{d}y$$ $$=-{\frac{2\sigma }{\sqrt{2\pi }}{\mathrm{e}}^{-\frac{y^2}{2{\sigma }^2}}|}_0^{+\infty }=\sqrt{\frac{2}{\pi }}\sigma .$$ $$D\left(\left|X_i-\mu \right|\right)=D\left(\left|Y_i\right|\right)=E\left(Y_i^2\right)-{\left[E\left(\left|Y_i\right|\right)\right]}^2$$ $$=D\left(Y_i\right)+{\left(E{Y}_i\right)}^2-{\left(\sqrt{\frac{2}{\pi }}\sigma \right)}^2$$ $$={\sigma }^2+0-\frac{2}{\pi }{\sigma }^2=\left(1-\frac{2}{\pi }\right){\sigma }^2.$$

所以 $E\left(Y\right)=E\left(\frac{1}{n}\sum _{i=1}^n\left|X_i-\mu \right|\right)=\frac{1}{n}\sum _{i=1}^{n}E\left(\left|X_i-\mu \right|\right)$

$$=\frac{1}{n}\cdot n\sqrt{\frac{2}{\pi }}\sigma =\sqrt{\frac{2}{\pi }}\sigma ,$$ $$D\left(Y\right)=D\left(\frac{1}{n}\sum _{i=1}^n\left|X_i-\mu \right|\right)=\frac{1}{n^2}\sum _{i=1}^{n}D\left(\left|X_i-\mu \right|\right)$$ $$=\left(1-\frac{2}{\pi }\right)\frac{{\sigma }^2}{n}.$$