概率统计D 03.01 多维随机变量及其分布A

$\color{blue}{第三章 多维随机变量及其分布}$

$\color{blue}{\S 3.1 二维随机变量}$

$\color{blue}{一、二维随机变量及其分布函数}$

$定义:设随机试验E的基本空间为\Omega, X和Y是定义在\Omega上的两个随机变量,\\ 由它们构成的向量(X,Y)叫做二维随机变量.$
$\qquad 二维随机变量(X, Y)可以看作是xoy面上的随机点,它们的取值是\\ xoy面上的一个定点(x, y).(X, Y)可能落在xoy面上的有限个点处,也可\\ 能落在xoy面上的某个区域内的所有点上.我们把二维随机变量分成离散型\\ 和连续型两类.$

$定义:设(X,Y)为二维随机变量,对任意实数x,y,二元函数\\ \qquad F(x, y) = P\lbrace X \leq x, Y \leq y \rbrace \\ 称为二维随机变量(X,Y)的分布函数,或称为X与Y的联合分布函数.$
$注:①规定\lbrace X \leq x, Y \leq y \rbrace表示\lbrace X \leq x \rbrace 与 \lbrace Y \leq y \rbrace 的积事件.$
$②分布函数F(x, y)在点(x, y)处的值,就是(X,Y)的取值落在矩形\\ -\infty < X \leq x, -\infty < Y \leq y 上的概率.$

$二维随机变量(X,Y)的分布函数F(x, y)具有性质:$
$①0 \leq F(x, y) \leq 1,且对任意x,y有\\ F(-\infty, y) = 0,F(x, -\infty) = 0, \\ F(-\infty, -\infty) = 0, F(+\infty, +\infty) = 1$
$②F(x, y)是变量x和y的单调不减函数.$
$③F(x, y)关于x右连续,关于y也右连续.$
$④(X,Y)落在矩形区域x_1 < X \leq x_2, y_1 < Y \leq y_2上的概率为\\ P\lbrace x_1 < X \leq x_2, y_1 < Y \leq y_2 \rbrace = F(x_2, y_2) - F(x_1, y_2) - F(x_2, y_1) + F(x_1, y_1)$

$\color{blue}{二、二维离散随机变量}$

$定义:若二维随机变量(X, Y)所有可能取的值是有限对或可列无穷多对,\\ 则称(X, Y)为二维离散型随机变量.$

$设二维离散型随机变量(X, Y)所有可能取值为(x_i, y_j), i = 1, 2, \cdots, \\ j = 1, 2, \cdots, 记\\ p_{ij} = P\lbrace X = x_i, Y = y_j \rbrace \quad i, j = 1, 2, \cdots \quad (*) \\ 且有 \\ p_ {ij} \geq 0, \sum \limits _ {i,j} p_ {ij} = 1 \\ 则称(*)式为(X, Y)的概率分布或X与Y的联合分布律.$

$$\begin{array}{c c | c c } & Y & y_1 & y_2 & y_3 & \cdots \\ X & & & & & & \\ \hline x_1 & & P_{11} & P_{12} & P_{13} & \cdots \\ x_2 & & P_{21} & P_{22} & P_{23} & \cdots \\ x_3 & & P_{31} & P_{32} & P_{33} & \cdots \\ \vdots & & \vdots & \vdots & \vdots & \\ \end{array}$$

$例1.设试验E为掷一颗骰子,观察出现的点数,定义两个随机变量如下:\\ \qquad X表示骰子出现的点数. \\ \qquad Y = \Big \lbrace \begin{array}{l}1,当出现奇数点时, \\ 2,当出现偶数点时. \end{array} \\ 试求X与Y的联合分布律.$
$解: (X,Y)可能取值为(1, 1),(1,2),(2,1), (2, 2), (3,1), (3, 2), (4, 1), (4,2), (5, 1), (5, 2), (6, 1), (6, 2).$
$P_{11} = P\lbrace X = 1, Y = 1 \rbrace = P\lbrace X = 1 \rbrace P\lbrace Y = 1 | X = 1 \rbrace = \dfrac{1}{6} \times 1 = \dfrac{1}{6}, \\ P_{12} = P\lbrace X = 1, Y = 2 \rbrace = P\lbrace X = 1 \rbrace P\lbrace Y = 2 | X = 1 \rbrace = \dfrac{1}{6} \times 0 = 0$
$同理\\ P_{22} = P_{31} = P_{42} = P_{51} = P_{62} = \dfrac{1}{6}, \\ P_{21} = P_{32} = P_{41} = P_{52} = P_{61} = 0$

$$\begin{array}{c c | c c } & Y & 1 & 2 \\ X & & & & & & \\ \hline 1 & & 1/6 & 0 \\ 2 & & 0 & 1/6 \\ 3 & & 1/6 & 0 \\ 4 & & 0 & 1/6 \\ 5 & & 1/6 & 0 \\ 6 & & 0 & 1/6 \\ \end{array}$$

$二维离散型随机变量(X,Y)的分布函数为F(x,y) = \sum \limits _ {x_i \leq x , y_i \leq y} p_{ij}$

$\color{blue}{三、二维连续型随机变量}$

$定义:设二维随机变量(X,Y)的分布函数为F(x, y), 如果存在非负函数f(x, y),\\ 使对任意x, y有 \\ F(x, y) = \int_{-\infty}^{x} \int_{-\infty}^{y} f(x, y) dxdy \\ \Big( = \int_{-\infty}^x \int_{-\infty}^y f(u, v) dudv \Big) . \\ 则称(X, Y)为二维连续型随机变量,称f(x, y)为(X, Y)的概率密度\\ 或X、Y的联合概率密度.$

$性质1. f(x, y) \geq 0$
$性质2. \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} f(x, y) dxdy = 1$
$性质3.在f(x, y)的连续点处有f(x, y) = \dfrac{\partial ^2 F(x, y)}{\partial x \partial y}$
$性质4.设G为xoy面上一个区域, 点(X, Y)落在G内的概率为\\ P\lbrace (X, Y) \in G \rbrace = \iint \limits _{G} f(x, y) dxdy$

$例2.设二维随机变量(X,Y)的概率密度为$
$f(x, y) = \left \{ \begin{array}{l} ke^ {-(2x + 3y)}, x \geq 0, y \geq 0, \\ 0, 其他. \end{array} \right.$
$(1).求k; (2) 求分布函数F(x, y); (3) 求P\lbrace X > Y \rbrace .$
$解:(1) 由\int_{-\infty}^{+\infty} f(x, y) dxdy = 1,而 \\ \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} f(x, y) dxdy \\ = \int_{0}^{+\infty} \int_0^{+\infty} ke^{-(2x + 3y)} dxdy \\ = k\int_0^{+\infty}e^{-2x} dx \int_0^{+\infty} e^{-3y}dy = \dfrac{k}{6} \\ 则有 k = 6.$
$(2) 当x > 0, y > 0时, \\ F(x, y) = \int_{-\infty}^x \int_{-\infty}^y f(x, y) dxdy \\ = \int_0^x \int_0^y 6e^{-(2x+3y)} dxdy \\ = (1 - e^{-2x})(1 - e^{-3y}). \\ 对于其它点(x, y), 由于f(x, y) = 0,则F(x, y) = 0. \\ 于是$
$F(x, y) = \left \{ \begin{array}{c} (1 - e^{-2x})(1 - e^{-3y}), x > 0, y > 0, \\ 0, 其它. \end{array} \right.$
$(3) P\lbrace X > Y \rbrace \\ = P\lbrace (X, Y) \in G \rbrace \\ = \iint \limits _{G} f(x, y) dxdy \\ = \int_0^{+\infty} dx \int_0^x 6e^{-(2x + 3y)}dy \\ = \dfrac{3}{5}$

$\color{blue}{四、均匀分布和正态分布}$

$\color{blue}{1.均匀分布}$

$设D为xoy面上的有界区域,其面积为S,\\ 如果二维随机变量(X, Y)具有概率密度$
$f(x, y) = \left \{ \begin{array}{l} \dfrac{1}{S}, (x, y) \in D, \\ 0 , 其它 \end{array} \right.$
$则称(X, Y)在区域D上服从均匀分布.$

$例3.设二维随机变量(X, Y)在D = \lbrace (x, y) | 0 \leq y \leq x, 0 \leq x \leq 1 \rbrace \\ 上服从均匀分布,求: \\ (1) (X, Y)的概率密度; \\ (2) P\lbrace \dfrac{1}{2} < X < \dfrac{3}{4}, 0 < Y < \dfrac{3}{4} \rbrace .$
$解:(1) 区域D的面积 S = \dfrac{1}{2} \times 1 \times 1 = \dfrac{1}{2},因此(X, Y)的概率密度为$
$f(x, y) = \left \{ \begin{array}{l} 2 , (X, Y) \in D \\ 0, 其它 \end{array} \right.$
$(2) 记区域G = \lbrace (x, y) | \dfrac{1}{2} < x < \dfrac{3}{4}, 0 < y < \dfrac{3}{4} \rbrace,$
$G_1 = \lbrace (x, y) | 0 < y < x, \dfrac{1}{2} < x< \dfrac{3}{4} \rbrace, \\ G_2 = \lbrace (x, y) | x \leq y < \dfrac{3}{4}, \dfrac{1}{2} < x < \dfrac{3}{4} \rbrace.$
$则有\\ P\lbrace \dfrac{1}{2} < X < \dfrac{3}{4}, 0 < Y < \dfrac{3}{4} \rbrace = P\lbrace (X, Y) \in G\rbrace \\ = \iint \limits _{G} f(x,y) dxdy \\ = \iint \limits _ {G_ 1 } 2 dxdy + \iint \limits _ { G_ 2} 0 dxdy \\ = \dfrac{5}{16}$

$\color{blue}{2.正态分布}$

$设二维随机变量(X,Y)的概率密度为$
$f(x, y) = \dfrac{1}{2 \pi \sigma_1 \sigma_2 \sqrt{1 - \rho^2}} e^{\Big {\lbrace} \frac{-1}{2(1 - \rho^2)}\big[\frac{(x - \mu_1)^2}{\sigma_1^2} - 2\rho \frac{(x - \mu_1)(y - \mu_2)}{\sigma_1 \sigma_2} + \frac{(y - \mu_2)^2}{\sigma_2^2}\big] \Big {\rbrace} }$
$-\infty < x < +\infty, -\infty < y < +\infty$
$其中\mu_1, \mu_2, \sigma_1, \sigma_2, \rho 均为常数,且\sigma_1 > 0, \sigma_2 > 0, -1 < \rho < 1, \\ 则称(X, Y)服从参数为\mu_1, \mu_2, \sigma_1, \sigma_2, \rho的二维正态分布,\\ 记作(X,Y) \sim N(\mu_1, \mu_2, \sigma_1^2, \sigma_2^2, \rho).$

$例4.设二维随机变量(X, Y)的概率密度为$
$f(x, y) = \dfrac{1}{2 \pi \sigma^2} e^{-\frac{1}{2\sigma^2}(x^2 + y^2)}, -\infty < x, y < + \infty \\ G = \lbrace(x, y) | x^2 + y^2 \leq \sigma^2 \rbrace, \\ 求P\lbrace (X, Y) \in G \rbrace (P\lbrace X^2 + Y^2 \leq \sigma^2 \rbrace).$
$解:P\lbrace (X, Y) \in G \rbrace \\ =\iint \limits _{G} f(x, y) dxdy \\ = \iint \limits _{G} \dfrac{1}{2 \pi \sigma^2} e^{-\frac{1}{2\sigma^2}(x^2 + y^2)} dxdy \\ = \dfrac{1}{2 \pi \sigma^2} \int_0^{2 \pi} d\theta \int_0^{\sigma} e^{-\frac{r^2}{2\sigma^2}} rdr \\ = 1 - e^{-\frac{1}{2}}$

$\color{blue}{\S 3.2 边缘分布}$

$\color{blue}{一、二维随机变量(X,Y)的边缘分布函数}$

$设(X,Y)的分布函数为F(x, y),关于X的边缘分布函数为\\ F_X(x) = P\lbrace X \leq x \rbrace = P\lbrace X \leq x, Y < +\infty \rbrace \\ = F(x, +\infty) = \lim \limits _{y \rightarrow +\infty} F(x, y) .$
$关于Y的边缘分布函数为$
$F_Y(y) = F(+\infty, y) = \lim \limits _{x \rightarrow +\infty} F(x, y).$

$\color{blue}{一、二维离散型随机变量的边缘分布律}$

$设(X, Y)的联合分布律为\\ P\lbrace X = x_i, Y = y_i \rbrace = p_{ij} (i, j = 1, 2, \cdots), \\ 则 \\ P\lbrace X = x_i \rbrace = P\lbrace X = x_i, Y < +\infty \rbrace \\ = P\lbrace X = x_i, \sum \limits _ { j = 1 }^ { \infty }(Y = y_j) \rbrace \\ = P\Big( \sum \limits _ {j = 1}^ {\infty} \lbrace X = x_i, Y = y_i \rbrace \Big) \\ = \sum \limits _ {j = 1}^{\infty} P\lbrace X = x_i, Y = y_i \rbrace \\ = \sum \limits _ {j = 1}^{\infty} p_ {ij} = p_{\color{green}{i\cdot}}$
$即关于X的边缘分布律为$
$P\lbrace X = x_i \rbrace = p_{\color{green}{i\cdot}} = \sum \limits _ {j = 1} ^{\infty} p_{ij} (i = 1, 2, \cdots)$
$关于Y的边缘分布律为$
$P\lbrace Y = y_j \rbrace = p_{\color{green}{\cdot j}} = \sum \limits _ {i = 1} ^{\infty} p_{ij} (j = 1, 2, \cdots)$

$例1.设(X, Y)的联合分布律为$

$$\begin{array}{c c | c c } & Y & 1 & 2 \\ X & & & & & & \\ \hline 1 & & 1/6 & 0 \\ 2 & & 0 & 1/6 \\ 3 & & 1/6 & 0 \\ 4 & & 0 & 1/6 \\ 5 & & 1/6 & 0 \\ 6 & & 0 & 1/6 \\ \end{array}$$
$求关于X、Y的边缘分布律.$
$解:\\ P\lbrace X = 1 \rbrace = p_{1\cdot} = p_{11} + p_{12} = \dfrac{1}{6} + 0 = \dfrac{1}{6}, \\ P\lbrace X = 2 \rbrace = p_{2\cdot} = p_{21} + p_{22} = 0 + \dfrac{1}{6}= \dfrac{1}{6}, \\ \cdots \quad \cdots \quad \cdots \\ P\lbrace X = 6 \rbrace = p_{6\cdot} = p_{61} + p_{62} = 0 + \dfrac{1}{6} = \dfrac{1}{6},$
即关于X的边缘分布律为
$$\begin{array}{l |c c }X & 1 & 2 & 3 & 4 & 5 & 6 & \\ \hline P & 1/6 & 1/6 & 1/6 & 1/6 & 1/6 & 1/6 \end{array}$$
$P\lbrace Y = 1 \rbrace = p_{\cdot 1} \\ = p_{11} + p_{21} + p_{31} + p_{41} + p_{51} + p_{61} \\ = \dfrac{1}{6} + 0 + \dfrac{1}{6} + 0 + \dfrac{1}{6} + 0 = \dfrac{1}{2}, \\ P\lbrace Y = 2 \rbrace = p_{\cdot 2} \\ = p_{12} + p_{22} + p_{32} + p_{42} + p_{52} + p_{62} \\ = 0 + \dfrac{1}{6} + 0 + \dfrac{1}{6} + 0 + \dfrac{1}{6}= \dfrac{1}{2}, \\ $
即关于Y的边缘分布律为
$$\begin{array}{l |c c }Y & & 1 & 2 & \\ \hline P & & 1/2 & 1/2 \end{array}$$

$例2.设(X, Y)的联合分布律为$

$$\begin{array}{c c | c c c c | c c } & Y & 1 & 2 & 3 & 4 & Y的边缘分布律 & & & \\ X & & & & & & \\ \hline 1 & & 1/4 & 1/8 & 1/12 & 1/16 & \\ 2 & & 0 & 1/8 & 1/12 & 1/16 & \\ 3 & & 0 & 0 & 1/12 & 1/16 & \\ 4 & & 0 & 0 & 0 & 1/16 & \\ \hline X的边缘分布律 & & & & & & \\ \end{array}$$

解:

$$\begin{array}{c c | c c c c | c c } & Y & 1 & 2 & 3 & 4 & Y的边缘分布律 & & & \\ X & & & & & & \\ \hline 1 & & 1/4 & 1/8 & 1/12 & 1/16 & 25/48 \\ 2 & & 0 & 1/8 & 1/12 & 1/16 & 13/48 \\ 3 & & 0 & 0 & 1/12 & 1/16 & 7/48 \\ 4 & & 0 & 0 & 0 & 1/16 & 1/16 \\ \hline X的边缘分布律 & & 1/4 & 1/4 & 1/4 & 1/4 & \\ \end{array}$$
X的边缘分布律
$$\begin{array}{l |c c }X & & 1 & 2 & 3 & 4 & & \\ \hline P & & 1/4 & 1/4 & 1/4 & 1/4 \end{array}$$

Y的边缘分布律

$$\begin{array}{l |c c }Y & & 1 & 2 & 3 & 4 & & \\ \hline P & & 25/48 & 13/48 & 7/48 & 3/48 \end{array}$$

$\color{blue}{三、二维连续型随机变量的边缘概率密度}$

$设(X,Y)的联合概率密度为f(x, y), 则关于X的边缘分布函数为\\ F_X(x) = F(x, +\infty) = \int_{-\infty}^{x} \Big( \int_{-\infty}^{+\infty} f(x, y) dy \Big) dx \\ 关于X的边缘概率密度为\\ f_X(x) = \int_{-\infty}^{+\infty} f(x, y) dy (-\infty < x < +\infty)$
$同理,关于Y的边缘概率密度为\\ f_Y(y) = \int_{-\infty}^{+\infty} f(x, y) dx (-\infty < y < +\infty)$

$例3.设(X,Y)在由曲线y = x^2 与 y = x围成的区域D\\ 上服从均匀分布,求关于X,Y的边缘密度.$
$解: S_D = \int_0^1 (x - x^2) dx = [\dfrac{1}{2}x^2 - \dfrac{1}{3}x^3]_0^1 = \dfrac{1}{6}$
$密度函数f(x, y) = \left \{ \begin{array}{l}6, (x^2 \leq y \leq x, 0 \leq x \leq 1) \\ 0, 其它 \end{array} \right.$
$f_X(x) = \int_{-\infty}^{+\infty} f(x, y) dy \\ = \int_ {-\infty}^{x^2} 0 dy + \int_{x^2}^{x} 6 dy + \int_x^{+\infty} 0 dy \\ = 0 + 6(x - x^2) + 0 \\ = 6(x - x^2)$
$f_X(x)= \left \{ \begin{array}{l} 6(x - x^2) , 0 \leq x \leq 1, \\ 0, 其它 \end{array} \right.$
$f_Y(y) = \int_{-\infty}^{+\infty} f(x, y) dx \\ = \int_{-\infty}^{y}0 dy + \int_{y}^{\sqrt{y}} 6 dx + \int_{\sqrt{y}}^{+\infty} 0 dx \\ = 6(\sqrt{y} - y)$
$f_Y(y)= \left \{ \begin{array}{l} 6(\sqrt{y} - y) , 0 \leq y \leq 1, \\ 0, 其它 \end{array} \right.$

$练习:1.设(X, Y)在由x = 0, y = 0, x + y = 1围成的区域上\\ 服从均匀分布,求关于X、Y的边缘分布.$
$\Big( f_X(x) = \left \{ \begin{array}{l} 2(1 - x), 0 < x < 1, \\ 0, 其它 \end{array} \right. , f_Y(y) = \left \{ \begin{array}{l} 2(1 - y), 0 < y < 1, \\ 0 , 其它. \end{array} \right. \Big)$

$练习:2.设(X, Y) \sim N(\mu_1, \mu_2, \sigma_1^2, \sigma_2^2, \rho), 即\\ f(x, y) = \dfrac{1}{2\pi \sigma_1 \sigma_2 \sqrt{1 - \rho^2}} e^{\Big \lbrace \frac{-1}{2(1 - \rho^2)} \big[ \frac{(x - \mu_1)^2}{\sigma_1^2} - 2\rho \frac{(x -\mu_1)(y - \mu_2)}{\sigma_1 \sigma_2} + \frac{(y - \mu_2)^2}{\sigma_2^2} \big] \Big \rbrace } \\ (-\infty < x, y < +\infty)$
$关于X、Y的边缘密度分别为\\ f_X(x) = \dfrac{1}{\sqrt{2 \pi} \sigma_1} e^{-\frac{(x - \mu_1)^2}{2\sigma_1^2} }, -\infty < x < +\infty$
$f_Y(y) = \dfrac{1}{\sqrt{2 \pi} \sigma_2} e^{-\frac{(y - \mu_2)^2}{2\sigma_2^2}} , -\infty < y < +\infty$
$即X \sim N(\mu_1, \sigma_1^2), Y \sim N(\mu_2, \sigma_2^2).$