题型 1. 协方差与相关系数的计算

3.1

设随机变量 $X$ 和 $Y$ 的联合概率分布为

Y \ X	-1	0	1
0	0.07	0.18	0.15
1	0.08	0.32	0.20
则 $X$ 和 $Y$ 的相关系数 $\rho =$.
$X^2$ 和 $Y^2$ 的协方差 $\mathrm{Cov}\left(X^2,Y^2\right)=$.

解

1

$X$ 的分布律

X	0	1
P	0.4	0.6

$Y$ 的分布律

Y	-1	0	1
P	0.15	0.5	0.35

$$E\left(X\right)=0.6,$$ $$E\left(Y\right)=0.35-0.15=0.2,$$ $$E\left(XY\right)=\sum _i\sum _j{x}_i{y}_j{p}_{ij}=-0.08+0.20=0.12,$$

所以

$$\mathrm{Cov}\left(X,Y\right)=E\left(XY\right)-E\left(X\right)E\left(Y\right)=0.$$

2

$$E\left(Y^2\right)=0.5,$$ $$E\left(X^2{Y}^2\right)=\sum _i\sum _j{x}_i^2{y}_j^2{p}_{ij}=0.28.$$

所以 $\mathrm{Cov}\left(X^2,Y^2\right)=E\left(X^2{Y}^2\right)-E\left(X^2\right)E\left(Y^2\right)=-0.02.$

3.2

已知 $X\sim \left[\begin{array}{cc}-1& 1\\ \frac{1}{2}& \frac{1}{2}\end{array}\right]$, $Y\sim \left[\begin{array}{cc}0& 1\\ \frac{1}{4}& \frac{3}{4}\end{array}\right]$, $P\{X=Y\}=\frac{1}{4}$, 则 $r(X,Y)=$ ___.

解

由 $P\{X=Y\}=P\{X=1,Y=1\}=\frac{1}{4}$,可求得 $\left(X,Y\right)$ 联合分布律

Y	0	1
-1	0	$\frac{1}{2}$
1	$\frac{1}{4}$	$\frac{1}{4}$

故 $E\left(XY\right)=-\frac{1}{4}$, 又

$$EX=0,DX=1,EY=\frac{3}{4},DY=\frac{3}{16},$$

则

$$\begin{array}{rl}{\rho }_{XY}& =\frac{\mathrm{Cov}\left(X,Y\right)}{\sqrt{DX}\cdot \sqrt{DY}}\\ & =\frac{E\left(XY\right)-EX\cdot EY}{\sqrt{DX}\cdot \sqrt{DY}}\\ & =-\frac{\sqrt{3}}{3}\end{array}$$

3.3

设随机变量 $\left(X,Y\right)$ 具有概率密度函数

$$f\left(x,y\right)=\{\begin{array}{ll}\frac{1}{8}\left(x+y\right),& 0\le x\le 2,0\le y\le 2\\ 0,& \text{ 其他 }\end{array}$$

求 $E\left(X\right),E\left(Y\right),\mathrm{Cov}\left(X,Y\right),{\rho }_{XY},D\left(X+Y\right)$.

解

由于

$$\begin{array}{rl}E\left(X\right)& ={\int }_{-\infty }^{+\infty }{\int }_{-\infty }^{+\infty }xf\left(x,y\right)\mathrm{d}x\mathrm{d}y\\ & ={\int }_0^2\mathrm{d}x{\int }_0^2\frac{1}{8}x\left(x+y\right)\mathrm{d}y\\ & =\frac{7}{6}\end{array}$$ $$\begin{array}{rl}E\left(Y\right)& ={\int }_{-\infty }^{+\infty }{\int }_{-\infty }^{+\infty }yf\left(x,y\right)\mathrm{d}x\mathrm{d}y\\ & ={\int }_0^2\mathrm{d}x{\int }_0^2\frac{1}{8}y\left(x+y\right)\mathrm{d}y\\ & =\frac{7}{6}\end{array}$$ $$\begin{array}{rl}E\left(X^2\right)& ={\int }_{-\infty }^{+\infty }{\int }_{-\infty }^{+\infty }{x}^{2}f\left(x,y\right)\mathrm{d}x\mathrm{d}y\\ & ={\int }_0^2\mathrm{d}x{\int }_0^2\frac{1}{8}{x}^2\left(x+y\right)\mathrm{d}y\\ & =\frac{10}{6}\end{array}$$

同理 $E\left(Y^2\right)=\frac{10}{6}$ 故

$$\begin{array}{rl}D\left(X\right)& =E\left(X^2\right)-{\left(EX\right)}^2\\ & =\frac{10}{6}-\frac{49}{36}\\ & =\frac{11}{36}.\end{array}$$

同理 $DY=\frac{11}{36}$.

$$\begin{array}{rl}E\left(XY\right)& ={\int }_{-\infty }^{+\infty }{\int }_{-\infty }^{+\infty }xyf\left(x,y\right)\mathrm{d}x\mathrm{d}y\\ & ={\int }_0^2\mathrm{d}x{\int }_0^2\frac{1}{8}xy\left(x+y\right)\mathrm{d}y\\ & =\frac{8}{6}\end{array}$$

故

$$\begin{array}{rl}\mathrm{Cov}\left(X,Y\right)& =E\left(XY\right)-EX\cdot EY\\ & =\frac{8}{6}-\frac{49}{36}\\ & =-\frac{1}{36}\end{array}$$ $$\begin{array}{rl}{\rho }_{XY}& =\frac{\mathrm{Cov}\left(X,Y\right)}{\sqrt{DX}\cdot \sqrt{DY}}\\ & =-\frac{\frac{1}{36}}{\sqrt{\frac{11}{36}\cdot \frac{11}{36}}}\\ & =-\frac{1}{11}\end{array}$$ $$\begin{array}{rl}D\left(X+Y\right)& =DX+DY+2\mathrm{Cov}\left(X,Y\right)\\ & =\frac{11}{36}+\frac{11}{36}-\frac{2}{36}\\ & =\frac{5}{9}.\end{array}$$

3.4

某箱装有 100 件产品,其中一、二和三等品分别为 80、10 和 10 件,现在从中随机抽取一件, 记

$$X_i=\{\begin{array}{ll}1,& \text{ 若抽到 }i\text{ 等品 }\\ 0,& \text{ 其他 }\end{array}\left(i=1,2,3\right)$$

试求 (1) 随机变量 $X_1$ 和 $X_2$ 的联合分布;

(2)随机变量 $X_1$ 与 $X_2$ 的相关系数 $\rho$.

解

(1) 设事件 $A_i=$ “抽到 $i$ 等品” $\left(i=1,2,3\right)$. 由题意知 $A_1,A_2,A_3$ 两两互不相容.

$P\left(A_1\right)=0.8,P\left(A_2\right)=P\left(A_3\right)=0.1.$

易见,

$P\left\{X_1=0,X_2=0\right\}=P\left(A_3\right)=0.1,$

$P\left\{X_1=0,X_2=1\right\}=P\left(A_2\right)=0.1;$

$P\left\{X_1=1,X_2=0\right\}=P\left(A_1\right)=0.8,$

$P\left\{X_1=1,X_2=1\right\}=P\left(\varnothing \right)=0.$

故 $X_1$ 和 $X_2$ 的联合分布为

$X_1$ $X_2$	0	1
0	0.1	0.8
1	0.1	0
(2) $E{X}_1=0.8,E{X}_2=0.1$,

$D{X}_1=0.8\times 0.2=0.16,D{X}_2=0.1\times 0.9=0.09,$

$E\left(X_1{X}_2\right)=0\times 0\times 0.1+0\times 1\times 0.1+1\times 0\times 0.8+1\times 1\times 0=0,$

$\mathrm{Cov}\left(X_1,X_2\right)=E\left(X_1{X}_2\right)-E{X}_1\cdot E{X}_2=0-0.8\times 0.1=-0.08,$

$\rho =\frac{\mathrm{Cov}\left(X_1,X_2\right)}{\sqrt{D{X}_1\cdot D{X}_2}}=\frac{-0.08}{\sqrt{0.16\times 0.09}}=-\frac{2}{3}.$