题型 2. 期望与方差性质的综合使用

2.6

设两个相互独立的随机变量 $X$ 和 $Y$ 的方差分别为 4 和 2, 则随机变量 $3X-2Y$ 的方差是 ( ).

(A) 8 (B) 16 (C) 28 (D) 44

解

由方差的性质知 $D\left(3X-2Y\right)=D\left(3X\right)+D\left(-2Y\right)=3^{2}D\left(X\right)+{\left(-2\right)}^{2}D\left(Y\right)=36+8=44.$

故应选(D).

2.7

设连续型随机变量 $X_1$ 与 $X_2$ 相互独立且方差均存在, $X_1$ 与 $X_2$ 的概率密度分别为

$f_1\left(x\right)$ 与 $f_2\left(x\right)$,随机变量 $Y_1$ 的概率密度为

$$f_{Y_1}\left(y\right)=\frac{1}{2}\left[f_1\left(y\right)+f_2\left(y\right)\right],$$

随机变量 $Y_2=\frac{1}{2}\left(X_1+X_2\right)$,则_____.

(A) $E{Y}_1>E{Y}_2,D{Y}_1>D{Y}_2$ (B) $E{Y}_1=E{Y}_2,D{Y}_1=D{Y}_2$

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

解

$E{Y}_1={\int }_{-\infty }^{+\infty }y{f}_{Y_1}\left(y\right)\mathrm{d}y=\frac{1}{2}\left[{\int }_{-\infty }^{+\infty }y{f}_1\left(y\right)\mathrm{d}y+{\int }_{-\infty }^{+\infty }y{f}_2\left(y\right)\mathrm{d}y\right]$

$=\frac{1}{2}\left(E{X}_1+E{X}_2\right),$

$E{Y}_2=\frac{1}{2}E\left(X_1+X_2\right)=\frac{1}{2}\left(E{X}_1+E{X}_2\right)$,故 $E{Y}_1=E{Y}_2$.

$D{Y}_1=E\left(Y_1^2\right)-{\left(E{Y}_1\right)}^2,D{Y}_2=E\left(Y_2^2\right)-{\left(E{Y}_2\right)}^2,$

则 $D{Y}_1-D{Y}_2=E\left(Y_1^2\right)-E\left(Y_2^2\right)$

$=\frac{1}{2}\left[{\int }_{-\infty }^{+\infty }{y}^2{f}_1\left(y\right)\mathrm{d}y+{\int }_{-\infty }^{+\infty }{y}^2{f}_2\left(y\right)\mathrm{d}y\right]-E\left[\frac{1}{4}{\left(X_1+X_2\right)}^2\right]$

$=\frac{1}{2}E\left(X_1^2\right)+\frac{1}{2}E\left(X_2^2\right)-\frac{1}{4}E\left[{\left(X_1+X_2\right)}^2\right]$

$=\frac{1}{4}E\left(X_1^2+X_2^2-2X_1{X}_2\right)$

$=\frac{1}{4}E\left[{\left(X_1-X_2\right)}^2\right]>0,$

即 $D{Y}_1>D{Y}_2$.

故应选(D).

2.8

设 $X$ 的均值、方差都存在,且 $D\left(X\right)\ne 0$,令

$$Y=\frac{X-E\left(X\right)}{\sqrt{D\left(X\right)}}$$

则 $E\left(Y\right)=$ _____, $D\left(Y\right)=$ _____.

解

$E\left(Y\right)=E\left(\frac{X-E\left(X\right)}{\sqrt{D\left(X\right)}}\right)=\frac{1}{\sqrt{D\left(X\right)}}E\left(X-E\left(X\right)\right)=\frac{1}{\sqrt{D\left(X\right)}}\left[E\left(X\right)-E\left(X\right)\right]=0$,

$D\left(Y\right)=D\left(\frac{X-E\left(X\right)}{\sqrt{D\left(X\right)}}\right)=\frac{1}{D\left(X\right)}D\left(X-E\left(X\right)\right)=\frac{1}{D\left(X\right)}\left[D\left(X\right)+D\left(-E\left(X\right)\right)\right]$

$=\frac{D\left(X\right)}{D\left(X\right)}=1.$

2.9

设 $X$ 为随机变量, $C$ 是常数,证明

$$D\left(X\right)<E\left\{{\left(X-C\right)}^2\right\},\text{ 对于 }C\ne E\left(X\right).$$

(由于 $D\left(X\right)=E{\left[X-E\left(X\right)\right]}^2$,上式表明 $E\left\{{\left(X-C\right)}^2\right\}$ 当 $C=E\left(X\right)$ 时取到最小值.

证法一

$$E\left\{{\left(X-C\right)}^2\right\}=E\left(X^2-2CX+C^2\right)=E\left(X^2\right)-2CE\left(X\right)+C^2$$ $$=E\left(X^2\right)-{\left[E\left(X\right)\right]}^2+{\left[E\left(X\right)\right]}^2-2CE\left(X\right)+C^2$$ $$=D\left(X\right)+{\left[E\left(X\right)-C\right]}^2>D\left(X\right),\text{ 当 }E\left(X\right)\ne C.$$

故当 $C=E\left(X\right)$ 时, $E{\left(X-C\right)}^2$ 取到最小值 $DX$.

证法二

$DX=E{\left(X-EX\right)}^2=E{\left[\left(X-C\right)+\left(C-EX\right)\right]}^2$

$$=E{\left(X-C\right)}^2+E{\left(C-EX\right)}^2+2E\left[\left(X-C\right)\left(C-EX\right)\right]$$ $$=E{\left(X-C\right)}^2-{\left(C-EX\right)}^2<E{\left(X-C\right)}^2.$$

2.10

设 $X_1,X_2,\cdots ,X_n$ 是 $n$ 个独立同分布的随机变量, $E\left(X_i\right)=\mu ,D\left(X_i\right)={\sigma }^2,i=1$, $2,\cdots ,n$. 设 $\bar{X}=\frac{1}{n}\sum _{i=1}^n{X}_i$,求 $E\left(\bar{X}\right)$ 和 $D\left(\bar{X}\right)$.

解

$$\begin{array}{rl}E\left(\bar{X}\right)& =E\left[\frac{1}{n}\left(X_1+X_2+\cdots +X_n\right)\right]\\ & =\frac{1}{n}\left(E{X}_1+E{X}_2+\cdots +E{X}_n\right)\\ & =\frac{1}{n}\cdot n\mu \\ & =\mu .\end{array}$$ $$\begin{array}{rlr}D\left(\bar{X}\right)& =D\left[\frac{1}{n}\left(X_1+X_2+\cdots +X_n\right)\right]& \\ & =\frac{1}{n^2}\left(D{X}_1+D{X}_2+\cdots +D{X}_n\right)& \text{(by independence)}\\ & =\frac{1}{n^2}n{\sigma }^2& \\ & =\frac{{\sigma }^2}{n}.& \end{array}$$

2.11

一台设备由三大部件构成,在设备运转中各部件需要调整的概率相应为 0.10,0.20 和 0.30,假设各部件的状态相互独立,以 $X$ 表示同时需要调整的部件数,试求 $X$ 的数学期望 $EX$ 和方差 $DX$.

解法一

先求 $X$ 的分布律,根据分布律再求期望.

根据 $X$ 的意义,显然有 $X=0,1,2,3$,事件 $A_i$ 表示第 $i$ 件需要调整, $i=1,2,3$,并注意到事件之间的独立性.

$$P\{X=0\}=P\left({\bar{A}}_1{\bar{A}}_2{\bar{A}}_3\right)=0.9\times 0.8\times 0.7=0.504\text{,}$$ $$P\{X=1\}=P\left(A_1{\bar{A}}_2{\bar{A}}_3\right)+P\left({\bar{A}}_1{A}_2{\bar{A}}_3\right)+P\left({\bar{A}}_1{\bar{A}}_2{A}_3\right)$$ $$=P\left(A_1\right)P\left({\bar{A}}_2\right)P\left({\bar{A}}_3\right)+P\left({\bar{A}}_1\right)P\left(A_2\right)P\left({\bar{A}}_3\right)+P\left({\bar{A}}_1\right)P\left({\bar{A}}_2\right)P\left(A_3\right)$$ $$=0.1\times 0.8\times 0.7+0.9\times 0.2\times 0.7+0.9\times 0.8\times 0.3=0.398\text{,}$$ $$P\{X=2\}=P\left(A_1{A}_2{\bar{A}}_3\right)+P\left(A_1{\bar{A}}_2{A}_3\right)+P\left({\bar{A}}_1{A}_2{A}_3\right)$$ $$=P\left(A_1\right)P\left(A_2\right)P\left({\bar{A}}_3\right)+P\left(A_1\right)P\left({\bar{A}}_2\right)P\left(A_3\right)+P\left({\bar{A}}_1\right)P\left(A_2\right)P\left(A_3\right)$$ $$=0.1\times 0.2\times 0.7+0.1\times 0.8\times 0.3+0.9\times 0.2\times 0.3=0.092\text{,}$$ $$P\{X=3\}=P\left(A_1{A}_2{A}_3\right)=P\left(A_1\right)P\left(A_2\right)P\left(A_3\right)=0.1\times 0.2\times 0.3=0.006\text{,}$$

所以

X	0	1	2	3
P	$\frac{3}{4}$	$\frac{9}{44}$	$\frac{9}{220}$	$\frac{1}{220}$
$E\left(X\right)=0\times 0.504+1\times 0.398+2\times 0.092+3\times 0.006=0.6$,

$D\left(X\right)=E\left(X^2\right)-{\left[E\left(X\right)\right]}^2=1^2\times 0.398+2^2\times 0.092+3^2\times 0.006-{\left(0.6\right)}^2=0.46$.

解法二

不求 $X$ 的分布律,引进新的随机变量,利用期望、方差的运算性质求出 $X$ 的期望 $E\left(X\right)$,方差 $D\left(X\right)$.

现引进新随机变量 $X_i$,定义如下:

$X_i=\{\begin{array}{ll}1,& \text{ 第 }i\text{ 个部件要调整,即 }\\ 0,& \text{ 第 }i\text{ 个部件不要调整 }\end{array}$

由此就有

$$X=\sum _{i=1}^3{X}_i$$

而

$$X_i\sim \left(0-1\right)\text{分布}E\left(X_i\right)=P\left\{X_i=1\right\}=P\left(A_i\right)$$

所以 $E\left(X\right)=\sum _{i=1}^{3}P\left(A_i\right)=P\left(A_1\right)+P\left(A_2\right)+P\left(A_3\right)=0.1+0.2+0.3=0.6$

$D\left(X_i\right)=P\left\{X_i=1\right\}P\left\{X_i=0\right\}=P\left(A_i\right)P\left({\bar{A}}_i\right),X_i$ 之间相互独立

所以

$D\left(X\right)=\sum _{i=1}^{3}D\left(X_i\right)=\sum _{i=1}^{3}P\left(A_i\right)P\left({\bar{A}}_i\right)=0.1\times 0.9+0.2\times 0.8+0.3\times 0.7=0.46.$

点评

本题中解法二比解法一简单得多,这就是利用性质求 $EX$ 和 $DX$ 的好处,但如何引进新随机变量是问题的一个难点. 一般地,总是引入 $X_i\sim \left(0-1\right)$ 分布,用 $\sum X_i$ 来解决问题.