概率统计D 04.01 随机变量的数字特征

$\color{blue}{第四章 随机变量的数字特征}$

$\qquad 随机变量的概率分布反映了随机变量的统计规律性,\\ 但在实际问题中，要确定一个随机变量的分布不是件容易的\\ 事情。在许多情况下，并不需要求出随机变量的分布，只需\\ 知道从不同角度反应随机变量取值的特征的若干个数字就够\\ 了，这些数字就称为随机变量的数字特征。$
$\qquad 本章将讨论随机变量的数学期望、方差、矩以及相关\\ 系数，它们在概率论与数理统计中起着重要的作用。$

$\color{blue}{\S 4.1 数学期望}$

$\color{blue}{一、离散型随机变量的数学期望}$

$例1.一台机床加工某种零件,已知它加工出优质品、合格品和废品的概率\\ 依次为0.2,0.7和0.1.如果出售优质品和合格品,每一个零件可分别获利\\ 0.40元和0.20元;如果加工出来一件废品则要损失0.10元.问这台机床每\\ 加工出一个零件,平均可获利多少元?$
$解：$

$$\begin{array}{l | c c } X & -0.10 & 0.20 & 0.40 & \\ \hline Y & 0.1 & 0.7 & 0.2 & \end{array}$$
$现假设该机床加工n个零件,其中废品n_1件,合格品n_2件,优质品n_3件,\\ 这里n_1 + n_2 + n_3 = n.则这n个零件可获得总利润为: \\ -0.1n_1 + 0.2 n_2 + 0.4n_3, \\ 平均每个零件可获利为:\\ -0.1 \times \dfrac{n_1}{n} + 0.2 \times \dfrac{n_2}{n} + 0.4 \times \dfrac{n_3}{n} \\ 其中\dfrac{n_1}{n},\dfrac{n_2}{n}和\dfrac{n_3}{n}分别是事件\lbrace X = -0.10 \rbrace、\lbrace X = 0.20 \rbrace \\ 和 \lbrace X = 0.40 \rbrace出现的频率.当n很大时,\dfrac{n_1}{n}, \dfrac{n_2}{n}和\dfrac{n_3}{n}分别接\\ 近与0.1,0.7, 0.2,于是可以期望该机床加工出的每一个零件所获\\ 得的平均利润为:\\ -0.1 \times 0.1 + 0.2 \times 0.7 + 0.4 \times 0.2 = 0.21(元) \\ 上述结果称为随机变量X的数学期望.$

$定义1.设离散型随机变量X的分布律为$
$\begin{array}{l | c c } X & x_1 & x_2 & \cdots & x_k & \cdots & \\ \hline Y & p_1 & p_2 & \cdots & p_k & \cdots & \end{array}$
$则称$
$E(X) = \sum \limits _{k=1}^{\infty} x_kp_k (要求此级数绝对收敛) \quad (1) \\ 为X的数学期望(或均值).$

$例2.设X服从参数为p的(0-1)分布, 求X的数学期望.$
$解：X的分布律为$
$\begin{array}{l | c c } X & 0 & 1 & \\ \hline Y & 1 - p & p & \end{array}$
$E(X) = 0 \times (1-p) + 1 \times p = p.$

$例3.设X \sim B(n, p),求E(X).$
$解: X的分布律为:\\ p_k = P\lbrace X = k \rbrace = C_n^k p^k(1-p)^{n-k}, k = 0, 1, 2, \cdots, n \\ E(X) = \sum \limits _ {k = 0} ^ n kp_k \\ = \sum \limits _ {k=1}^n k \dfrac{n!}{k!(n-k)!} p^k(1-p)^{n-k} \\ = np \sum \limits_ {k=1}^n C_{n-1}^{k-1} p^{k-1}(1-p)^{(n-1)-(k-1)} \\ = np(p + 1 - p)^{n-1} \\ = np$

$例4.设X \sim \pi(\lambda),求E(X).$
$解: p_k = P\lbrace X = k \rbrace = \dfrac{\lambda^k}{k!}e^{-\lambda}, k = 0, 1, 2, \cdots \\ E(X) = \sum \limits _ {k=0}^{\infty} kp_k \\ = \sum \limits _{k=1}^{\infty} k \dfrac{\lambda^k}{k!}e^{-\lambda} \\ = \lambda e^{-\lambda} \sum \limits _ {k=1}^{\infty} \dfrac{\lambda^{k-1}}{(k-1)!} \\ = \lambda e^{-\lambda} e^{\lambda} \\ = \lambda$

$例5.已知10件产品中有2件次品,求任取3件中次品数的数学期望.$
$解:以X表示任取3件中次品的个数,可取值为0, 1, 2,其分布律为:\\ P\lbrace X = 0 \rbrace = \dfrac{\dbinom{8}{3}}{\dbinom{10}{3}} = \dfrac{7}{15}, \\ P\lbrace X = 1 \rbrace = \dfrac{\dbinom{8}{2}\dbinom{2}{1}}{\dbinom{10}{3}} = \dfrac{7}{15}, \\ P\lbrace X = 2 \rbrace = \dfrac{\dbinom{8}{2}\dbinom{2}{2}}{\dbinom{10}{3}} = \dfrac{1}{15}, \\ 因此 \quad E(X) = 0 \times \dfrac{7}{15} + 1 \times \dfrac{7}{15} + 2 \times \dfrac{1}{15} = \dfrac{3}{5}$

$\color{blue}{二、连续型随机变量的数学期望}$

$定义2.设连续型随机变量X的概率密度为f(x),则称 \\ E(X) = \int_{-\infty}^{+\infty} xf(x) dx (要求此积分绝对收敛) \quad (2) \\ 为X的数学期望(或均值).$

$例6.设X在[a, b]上服从均匀分布,求E(X).$
$解:X的概率密度为$
$f(x) = \left \{ \begin{array}{l} \dfrac{1}{b - a}, a \leq x < b, \\ 0 , 其它 \end{array} \right.$
$E(X) = \int_{-\infty}^{+\infty} xf(x) dx = \int_a^b x \dfrac{1}{b - a} dx = \dfrac{a + b}{2}$

$例7.设X服从参数为\lambda 的指数分布,求E(X).$
$解:X的密度函数为$
$f(x) = \left \{ \begin{array}{l} \lambda e^{-\lambda x}, x \geq 0, \\ 0, 其它. \end{array} \right.$
$E(X) = \int_{-\infty}^{+\infty} xf(x) dx = \int_0^{+\infty} x \lambda e^{-\lambda x} dx =\dfrac{1}{\lambda}$

$例8.设X \sim N(\mu, \sigma^2),求E(X).$
$解:f(x) = \dfrac{1}{\sqrt{2 \pi} \sigma} e^ {-\frac{(x - \mu)^2}{2 \sigma^2}}, -\infty < x < +\infty$
$E(X) = \int_{-\infty}^{+\infty} xf(x) dx = \int_{-\infty}^{+\infty} \dfrac{x}{\sqrt{2 \pi} \sigma} e^{-\frac{(x - \mu)^2}{2 \sigma^2}} dx \\ \overset{令 \frac{x - \mu}{\sigma} = t}{=} \dfrac{1}{\sqrt{2 \pi}} \int_{-\infty}^{+\infty} (\sigma t + \mu) e^{-\frac{t^2}{2}} dt \\ = \dfrac{\sigma}{\sqrt{2 \pi}} \int_{-\infty}^{+\infty} t e^{-\frac{t^2}{2}} dt + \mu \int_{-\infty}^{+\infty} \dfrac{1}{\sqrt{2 \pi}} e^{-\frac{t^2}{2}} dt \\ = \mu$

$\color{blue}{三、随机变量的函数的数学期望}$

$定理1.设随机变量Y是随机变量X的函数:Y = g(X); \\ (1)若X为离散型,且分布律为p_k = P\lbrace X = x_k \rbrace, k=1,2,\cdots \\ 则 E(Y) = E[g(X)] = \sum \limits _ {k=1}^{\infty} g(x_k) p_k \quad (3) \\ (2) 若X为连续型,其密度为f(x),则\\ E(Y) = E[g(X)] = \int_{-\infty}^{+\infty} g(x)f(x) dx \quad (4)$

$例9.设X的分布律为$
$\begin{array}{l | c c } X & -2 & -1 & 0 & 1/2 & 1 & \\ \hline P & 1/6 & 1/3 & 1/4 & 1/12 & 1/6 & \end{array}$
$求E(X^2),E(aX + b).$
$解: E(X^2) = (-2)^2 \times \dfrac{1}{6} + (-1)^2 \times \dfrac{1}{3} + 0^2 \times \dfrac{1}{4} + (\dfrac{1}{2})^2 \times \dfrac{1}{12} + 1^2 \times \dfrac{1}{6} = \dfrac{19}{16} \\ E(aX + b) = (-2a + b) \times \dfrac{1}{6} + (-a+b) \times \dfrac{1}{3} + b \times \dfrac{1}{4} + (\dfrac{1}{2} a + b) \times \dfrac{1}{12} + (a+b) \times \dfrac{1}{6} = \dfrac{11}{24}a + b$

$例10.设X \sim N(0, 1),求E(X^2).$
$解: E(X^2) = \int_{-\infty}^{+\infty} x^2 \varphi(x) dx = \dfrac{1}{\sqrt{2 \pi}}\int_{-\infty}^{+\infty} x^2 e^{-\frac{x^2}{2}} dx = 1$

$例11.设X在区间(0, a)上服从均匀分布,求Y = kX^2的数学期望.$
$解:X的密度为f(x) = \left \{ \begin{array}{l} \dfrac{1}{a}, 0 < x < a, \\ 0 , 其它 \end{array} \right.$
$E(Y) = E(kX^2) = \int_{-\infty}^{+\infty} kx^2 f(x) dx = \int_0^a kx^2 \dfrac{1}{a} dx = \dfrac{1}{3}ka^2$

$例12.设X的概率密度为$
$f(x) = \left \{ \begin{array}{l} \dfrac{1}{2} \cos{x}, -\dfrac{\pi}{2} < x < \dfrac{\pi}{2}, \\ 0 , 其它 \end{array} \right.$
$Y = \sin{X}, Z = \cos{X},求E(Y),E(Z).$
$解:E(Y) = E(\sin{x}) = \int_{-\infty}^{+\infty} \sin{x} f(x) dx \\ = \int_{-\frac{\pi}{2}}^{+\frac{\pi}{2}} \sin{x} \dfrac{1}{2} \cos{x} dx = 0$
$E(Z) = E(\cos{X}) = \int_{-\infty}^{+\infty} \cos{x}f(x) dx \\ = \int_{-\frac{\pi}{2}}^{+\frac{\pi}{2}} \cos{x} \dfrac{1}{2} \cos{x} dx \\ = \int_0^{\frac{\pi}{2}} \dfrac{1 + \cos{2x}}{2} dx \\ = \dfrac{\pi}{4}$

$定理2.设随机变量Z是X、Y的函数Z= g(x,y); \\ (1) 若(X,Y)为二维连续型随机变量,联合密度为f(x, y),则\\ E(Z) = E[g(X, Y)] = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} g(x, y) f(x, y) dxdy \quad (5) \\ (2) 若(X, Y)为二维离散型随机变量,联合分布律为:\\ p_{ij} = P\lbrace X = x_i, Y = y_j \rbrace, i, j = 1, 2, \cdots \\ 则E(Z) = E[g(X,Y)] = \sum \limits _ {j=1}^{\infty} \sum \limits _ {i=1}^{\infty}g(x_i, y_i)p_{ij}$

$例13.设(X, Y)的联合密度为$
$f(x, y) = \left \{ \begin{array}{l} x + y, 0 \leq x \leq 1, 0 \leq y \leq 1, \\ 0, 其它 \end{array} \right.$
$求E(X), E(XY).$
$解: E(X) = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} xf(x, y) dxdy \\ = \int_0^1 dx \int_0^1 x(x + y) dy = \dfrac{7}{12} \\ E(XY) = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} xyf(x,y) dxdy \\ = \int_0^1 dx \int_0^1 xy(x+y) dy = \dfrac{1}{3}$

$\color{blue}{四、数学期望的性质(设E(X)、E(Y)存在)}$

$性质1. 设C为常数,则有E(C) = C.$
$性质2. E(C \cdot X) = C \cdot E(X).$
$性质3. E(X \pm Y) = E(X) \pm E(Y).$
$证:只对连续型随机变量的情形来证明:\\ E(X +Y) = E(X) + E(Y), 离散型的证明从略.$
$设(X,Y)的概率密度为f(x, y),则有 \\ E(X + Y) = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} (x + y) f(x, y) dxdy \\ = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} x f(x, y) dxdy + \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} y f(x, y) dxdy \\ = E(X) + E(Y)$

$性质4.若X、Y相互独立,则E(XY) = E(X)E(Y).$
$证:只对连续型加以证明.\\ 设(X,Y)的联合密度为f(x, y),关于X、Y的边缘密度分别\\ 为f_X(x)、f_Y(y).则有f(x, y) = f_X(x) f_Y(y),于是\\ E(XY) = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} xyf(x, y) dxdy \\ = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty}xyf_X(x) f_Y(y) dxdy \\ = \int_{-\infty}^{+\infty} xf_X(x) dx \int_{-\infty}^{+\infty} yf_Y(y) dy \\ = E(X)E(Y)$

$例14.设X与Y独立,$
$f_X(x) = \left \{ \begin{array}{l} 2x, 0 < x < 1, \\ 0, 其它 \end{array} \right. f_Y(y) = \left \{ \begin{array}{l} e^{-(y-2)}, y > 2 , \\ 0 , 其它, \end{array} \right.$
$求E(XY).$
$解:E(XY) = E(X)E(Y) \\ = \int_{-\infty}^{+\infty} xf_X(x) dx \int_{-\infty}^{+\infty} yf_Y(y) dy \\ = \int_0^1 x \cdot 2x dx \int_2^{+\infty} ye^{-(y-2)} dy \\ = 2$

$思考题:是否任何一个随机变量都存在数学期望?请研究随机变量X,\\ 其概率密度为f(x) = \dfrac{1}{\pi} \dfrac{1}{1 + x^2}, -\infty < x < +\infty$

$\color{blue}{\S 4.2 方差}$

$\color{blue}{一、方差的定义}$

$定义3.D(X) = E\lbrace [X - E(X)]^2\rbrace \qquad (6) \\ 称为随机变量X的方差,称\sqrt{D(X)}为X的均方差或标准差.$

$\color{blue}{二、方差的计算公式}$

$1.设X为离散型随机变量，分布律为$
$\begin{array}{l | c c } X & x_1 & x_2 & \cdots & x_k & \cdots & \\ \hline P & p_1 & p_2 & \cdots & p_k & \cdots & \end{array}$
$则D(X) = \sum \limits _{k=1}^{\infty}[x_k - E(X)]^2p_k \qquad (7)$

$2.设X为连续型随机变量,概率密度为f(x), 则\\ D(X) = \int_{-\infty}^{+\infty} [x - E(X)]^2 f(x) dx \qquad (8)$

$3.D(X) = E(X^2) - [E(X)]^2 \qquad (9)$
$证明: D(X) = E[X - E(X)]^2 \\ = E\lbrace X^2 - 2XE(X) + [E(X)]^2 \rbrace \\ = E(X^2) - E[2XE(X)] + E[E(X)]^2 \\ = E(X^2) - 2E(X)E(X) + [E(X)]^2 \\ = E(X^2) - [E(X)]^2$

$例1.设X服从参数p的(0-1)分布,求D(X).$
$解\begin{array}{l| c c } X & 0 & 1 & \\ \hline P & 1-p & p & \end{array}, E(X) = p.$
$E(X^2) = 0^2 \times (1 - p) + 1^2 \times p = p, \\ D(X) = E(X^2) - [E(X)]^2 = p - p^2 = p(1 - p)$

$例2.设X \sim \pi(\lambda),求D(X).$
$解：p_k = P\lbrace X = k \rbrace = \dfrac{\lambda^k}{k!}e^{-\lambda}, k = 1, 2, \cdots, E(X) = \lambda, \\ E(X^2) = E[X(X-1) + X] = E[X(X-1)] + E(X), \\ E[X(X-1)] = \sum \limits _ {k = 0} ^ {\infty} k(k-1) \dfrac{\lambda^k e^ {-\lambda} } {k!} \\ = \lambda^2 e^{-\lambda} \sum \limits _{k=2}^{\infty} \dfrac{\lambda ^{k-2} }{(k-2)!} \\ = \lambda^2 e^{-\lambda} e^{\lambda} = \lambda^2 \\ E(X^2) = \lambda^2 + \lambda , \\ D(X) = E(X^2) - [E(X)]^2 = \lambda^2 + \lambda - \lambda^2 = \lambda$

$例3 设X在区间(a, b)上服从均匀分布,求D(X)$
$解:E(X) = \dfrac{a +b}{2} \\ D(X) = E(X^2) - [E(X)]^2 \\ = \int_{-\infty}^{+\infty} x^2f(x) dx - (\dfrac{a + b}{2})^2 \\ = \int_{a}^{b} \dfrac{x^2}{b - a} dx - (\dfrac{a + b}{2})^2 \\ = \dfrac{a^2 + ab + b^2}{3} - (\dfrac{a + b}{2})^2 \\ = \dfrac{(b-a)^2}{12}$

$例4.设X服从参数为\lambda 的指数分布,求D(X).$
$解：f(x) = \left \{ \begin{array}{l} \lambda e^{-\lambda x}, x > 0 \\ 0 , x \leq 0 \end{array} \right.$
$E(X) = \dfrac{1}{\lambda}$
$D(X) = E(X^2) - (E(X)]^2 \\ = \int_{-\infty}^{+\infty} x^2f(x) dx - (E(X)]^2 \\ = \int_{0}^{+\infty} x^2\lambda e^{-\lambda x} dx - \dfrac{1}{\lambda^2} \\ = 2\dfrac{1}{\lambda^2} - \dfrac{1}{\lambda^2} \\ = \dfrac{1}{\lambda^2}$

$例5.设X \sim N(\mu, \sigma^2),求D(X).$
$解: D(X) = E\lbrace [X - E(X)]^2 \rbrace \\ = \int_{-\infty}^{+\infty} (x - \mu)^2 \dfrac{1}{\sqrt{2 \pi} \sigma }e^{-\frac{(x-\mu)^2}{2\sigma^2}} dx \\ \overset{t = \frac{x - \mu}{\sigma}}{=} \int _{-\infty}^{+\infty} t^2 \sigma^2 \dfrac{1}{\sqrt {2 \pi} \sigma } e^{-\frac{t^2}{2}} \sigma dt \\ = \sigma^2$

$\color{blue}{三、方差的性质}$

$性质1.设C为常数,则D(C) = 0.$
$证:D(C) = E[C - E(C)]^2 = E[C - C]^2 = E(0) = 0$

$性质2. D(CX) = C^2D(X).$
$证:D(CX) = E[CX - E(CX)]^2 = E\lbrace C^2[X - E(X)]^2 \rbrace = C^2D(X).$

$性质3. 设X与Y相互独立, 则有: D(X \pm Y) = D(X) + D(Y)$
$证: D(X \pm Y) = E\lbrace [(X \pm Y) - E(X \pm Y)]^2 \rbrace \\ = E\lbrace[X - E(X)] \pm [Y - E(Y)] \rbrace ^2 \\ = E \lbrace [X - E(X)]^2 \pm 2[X - E(X)][Y-E(Y)] + [Y - E(Y)]^2 \rbrace \\ =E[X-E(X)]^2 + E[Y-E(Y)]^2 \pm 2E\lbrace [X-E(X)][Y-E(Y)]\rbrace \\ = D(X) + D(Y) \pm 2[E(X) - E(X)][E(Y) - E(Y)] \\ = D(X) + D(Y)$

$例6.设X \sim B(n, p), 求D(X).$
$解:设X_i服从参数为p的(0-1)分布,且X_1, X_2, \cdots, X_n相互独立,\\ 则X= X_1 + X_2 + \cdots + X_n \sim B(n, p).于是 \\ D(X) = D(\sum \limits _ {i=1}^{n} X_i) \\ = \sum \limits _ {i=1}^n D(X_i) = \sum \limits _{i=1}^n p(1-p) \\ = np(1-p)$

$例7.设X与Y相互独立,D(X) = 2,Y \sim B(10, 0.2), 求D(3X - 5Y).$
$解: D(3X - 5Y) = D(3X) + D(5Y) \\ = 3^2D(X) + 5^2D(Y) \\ = 9 \times 2 + 25 \times 10 \times 0.2 \times 0.8 \\ = 58$

$例8.设E(X)、D(X)均存在,且D(X) > 0, \\ X^ {* } = \dfrac{X - E(X)}{\sqrt{D(X)}}, 求E(X^ {* }), D(X^ {*}).$
$\color{blue}{解: E(X^ {*}) = E(\dfrac{X - E(X)}{\sqrt{D(X)}}) \\ = \dfrac{1}{\sqrt{D(X)}} E[X - E(X)] \\ = \dfrac{1}{\sqrt{D(X)}} [E(X) - E(X)] \\ = 0 \\ D(X^ {*}) = D(\dfrac{X - E(X)}{\sqrt{D(X)}}) \\ = \dfrac{1}{D(X)} D[X - E(X)] \\ = \dfrac{1}{D(X)}[D(X) - D(E(X))] \\ = 1 \\ 称X^ {*} = \dfrac{X - E(X)}{\sqrt{D(X)}}为X的标准化随机变量. }$

$例9.设X_1, X_2, \cdots, X_n相互独立,并且具有相同的\\ 期望\mu 与方差\sigma^2，\bar X = \dfrac{1}{n}\sum \limits _{i=1}^n X_i, 求E(\bar X), D(\bar X), \bar X ^ {*} .$
$解: E(\bar X) = E(\dfrac{1}{n} \sum \limits _ {i=1}^n X_i) = \dfrac{1}{n} \sum \limits _ {i=1}^n E(X_i) = \dfrac{1}{n} \sum \limits _ {i=1}^n \mu = \mu \\ D(\bar X) = D(\dfrac{1}{n} \sum \limits _ {i=1} ^n X_i) = \dfrac{1}{n^2} \sum \limits _ {i=1} ^ n D(X_i) = \dfrac{1}{n^2} \sum \limits _ {i=1}^n \sigma^2 = \dfrac{\sigma^2}{n} \\ \bar X^ {*} = \dfrac{\bar X - E(\bar X)}{\sqrt{D \bar X}} = \dfrac{\bar X - \mu}{\sigma} \sqrt{n}$

$\color{blue}{\S 4.3 协方差与相关系数}$

$定义4.称E\lbrace [X - E(X)][Y-E(Y)]\rbrace为X与Y的协方差,\\ 记作 Cov(X,Y) = E\lbrace[X - E(X)][Y - E(Y)] \rbrace. \quad (10) \\ 称 \rho_{XY} = \dfrac{Cov(X,Y)}{\sqrt{D(X)}\sqrt{D(Y)}} \quad (11) 为X与Y的相关系数或标准协方差.$
$\qquad Cov(X,Y) = E(XY) - E(X)E(Y) \qquad (12)$
$\qquad D(X \pm Y) = D(X) + D(Y) \pm 2Cov(X, Y)$

$性质1:\\ Cov(X, Y) = Cov(Y, X). \\ Cov(aX, bY) = ab Cov(X, Y). (a,b为常数) \\ Cov(X_1 + X_2, Y) = Cov(X_1, Y) + Cov(X_2, Y)$
$性质2: |\rho_{XY}| \leq 1, |Cov(X,Y) \leq \sqrt{D(X)D(Y)}$
$性质3: 若X与Y相互独立,则Cov(X,Y) = 0, \rho_{XY} = 0$
$性质4:|\rho_{XY}| = 1的充分必要条件是:存在常数a, b,\\ 使得P\lbrace Y = aX + b \rbrace = 1 \\ 当\rho_{XY} = 0时, 称X与Y不相关.$

$例1.设二维随机变量(X,Y)的概率分布为:$
$\begin{array}{l c | c c } & Y & -1 & 0 & 1 & \\ X & & & & & \\ \hline -1 & & 1/8 & 1/8 & 1/8 & \\ 0 & & 1/8 & 0 & 1/8 & \\ 1 & & 1/8 & 1/8 & 1/8 & \\ \hline \end{array}$
$证明X与Y不相关,但X与Y不相互独立.$
$证:(X,Y)关于X与Y的边缘分布为$
$\begin{array}{l | c c }\hline X & -1 & 0 & 1 & \\ \hline P & 3/8 & 2/8 & 3/8 & \\ \hline \end{array}$
$\begin{array}{l | c c }\hline Y & -1 & 0 & 1 & \\ \hline P & 3/8 & 2/8 & 3/8 & \\ \hline \end{array}$

$E(X) = -1 \times \dfrac{3}{8} + 0 \times \dfrac{2}{8} + 1 \times \dfrac{3}{8} = 0 = E(Y) \\ E(XY) = \sum \limits _ {i, j} x_i y_j p _ {ij} = 0 \\ 于是有 Cov(X, Y) = E(XY) - E(X)E(Y) = 0 \\ 因此\rho_{XY} = 0, 即X与Y不相干.$
$由于 P\lbrace X = -1, Y = -1 \rbrace = \dfrac{1}{8}, \\ P\lbrace X =-1\rbrace P\lbrace Y = -1 \rbrace = \dfrac{3}{8} \times \dfrac{3}{8} = \dfrac{9}{64}, \\ 即有P\lbrace X = -1, Y = -1 \rbrace \neq P\lbrace X = -1\rbrace P\lbrace Y = -1 \rbrace, \\ 所以X与Y不相互独立.$

$例2.设(X,Y)的联合概率密度为$
$f(x, y) = \left \{ \begin{array}{l} \dfrac{1}{\pi}, x^2 + y^2 \leq 1, \\ 0, 其它 \end{array} \right.$
$验证X与Y不相关, 但不相互独立.$
$解:D:x^2 + y^2 \leq 1 \\ E(X) = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} xf(x, y) dxdy \\ = \iint \limits _{D} xf(x,y) dxdy \\ = \dfrac{1}{\pi} \iint \limits _{D} xdxdy \\ = \dfrac{1}{\pi} \int_0^{2 \pi} d\theta \int_0^1 r^2 \cos{\theta} dr \\ = 0 \\ 同理E(Y) = 0, E(XY) = 0,于是 \\ Cov(X, Y) = E(XY) - E(X)E(Y) = 0 \\ 从而有\rho_{XY} = 0, 即X与Y不相关.$
$f_X(x) = \int_{-\infty}^{+\infty} f(x, y) dy = \left \{ \begin{array}{l} \dfrac{2}{\pi}\sqrt{1-x^2}, |x| \leq 1, \\ 0, 其它 \end{array} \right.$
$f_Y(y) = \int_{-\infty}^{+\infty} f(x, y) dx = \left \{ \begin{array}{l} \dfrac{2}{\pi}\sqrt{1 - y^2}, |y| \leq 1 \\ 0, 其它 \end{array} \right.$
$当x = 0, y = 0时,f_X(0)f_Y(0) = \dfrac{4}{\pi^2}, 而f(0, 0) = \dfrac{1}{\pi}, \\ 既有f(0, 0) \neq f_X(0)f_Y(0), 所以X与Y不相互独立.$

$例3.设二维随机变量(X,Y)的概率密度为$
$f(x, y) = \left \{ \begin{array}{l} 1, |y| < x, 0 < x < 1, \\ 0, 其它 \end{array} \right.$
$证明:X与Y不相关,但不相互独立.$
$证:E(X) = \int _ {-\infty}^{+\infty} \int_{-\infty}^{+\infty} xf(x, y) dxdy \\ = \int_0^1 dx \int_{-x}^{x} x \cdot 1 dy \\ = \dfrac{2}{3}, \\ E(Y) = \int_ {-\infty}^{+\infty} \int_{-\infty}^{+\infty} y f(x, y) dxdy \\ = \int_0^1 dx \int _ {-x}^{x} ydy \\ = 0, \\ E(XY) = \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} xyf(x, y) dxdy \\ = \int_0^1 dx \int_{-x}^{x} xy dy = 0 \\ 从而有Cov(X,Y) = 0, \rho_{XY} = 0,即X与Y不相关.$

$由于$
$f_X(x) = \int_{-\infty}^{+\infty} f(x, y) dy = \left \{ \begin{array}{l} 2x, 0 < x < 1, \\ 0 , 其它 \end{array} \right.$
$f_Y(y) = \int_{-\infty}^{+\infty} f(x, y) dx = \left \{ \begin{array}{l} 1 + y, -1 < y < 0, \\ 1 - y, 0 \leq y < 1, \\ 0, 其它 \end{array} \right.$
$而f(\dfrac{1}{2}, \dfrac{1}{4}) = 1, f_X(\dfrac{1}{2}) = 1, f_Y(\dfrac{1}{4}) = \dfrac{3}{4}, \\ 可见f(\dfrac{1}{2}, \dfrac{1}{4}) \neq f_X(\dfrac{1}{2}) f_Y(\dfrac{1}{4}), 所以X和Y不相互独立.$

$例4.设(X, Y) \sim N(\mu_1, \mu_2, \sigma_1^2, \sigma_2^2, \rho), 即(X, Y)的联合密度为$
$f(x, y) = \dfrac{1}{2\pi \sigma_1 \sigma_2 \sqrt{1 - \rho^2}} exp \Big \lbrace \dfrac{-1}{2(1-\rho^2)} \big [\dfrac{(x-\mu_1)^2}{\sigma_1^2} - 2\rho \dfrac{(x-\mu_1)(y-\mu_2)}{\sigma_1 \sigma_2} + \dfrac{(y-\mu_2)^2}{\sigma_2^2} \big ] \Big \rbrace$
$求\rho_{XY}.$
$解:由于X \sim N(\mu_1, \sigma_1^2), Y \sim N(\mu_2, \sigma_2^2),而\\ Cov(X, Y) = E\lbrace [X - E(X)][Y - E(Y)]\rbrace \\ = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} (x - \mu_1)(y - \mu_2) f(x, y) dxdy \\ = \rho \sigma_1 \sigma_2 \\ 因此, \rho(X, Y) = \dfrac{Cov(X, Y)}{\sqrt{D(X)D(Y)}} = \rho$

$\color{blue}{\S 4.4 矩}$

$定义5.设X与Y是两个随机变量,称E(X^k)为X的k阶原点矩;\\ 称E\lbrace [E - E(X)]^k\rbrace 为X的k阶中心矩;称E(X^kY^l)为X与\\ Y的k+l阶混合原点矩;称E\lbrace[X - E(X)]^k[Y-E(Y)]^l\rbrace \\ 为X与Y的k+l阶混合中心矩.$

$随机变量X的标准化随机变量X^{*} = \dfrac{X - E(X)}{\sqrt{D(X)}}的3阶原点矩\\ E(X^{*3}) = E\Big[\dfrac{X-E(X)}{\sqrt{D(X)}} \Big]^3 = \dfrac{E[X - E(X)]^3}{[D(X)]^{\frac{3}{2}} }\\ 称为X的偏度;X^{*}的4阶原点矩\\ E(X^{*4}) = E \Big[ \dfrac{X - E(X)}{\sqrt{D(X)}} \Big] ^4 = \dfrac{E[X - E(X)]^4}{[D(X)]^2} \\ 称为X的峰度.$