Section02_二维随机变量

二维离散型随机变量

1. 定义 取值有限对或可列对
2. 联合分布律
   1. $\mathbb{P}\{X=x_{i},Y = y_{j}\} = p_{ij},\quad i,j= 1,2,\cdots$
   2. $$\begin{array}{c|ccccc|c} & y_{1} & y_{2} & \cdots & y_{j} & \cdots & \\ \hline x_{1} & p_{11} & p_{12} & \cdots & p_{1j} & \cdots & p_{1\cdot} \\ x_{2} & p_{21} & p_{22} & \cdots & p_{2j} & \cdots & p_{2\cdot}\\ \vdots & \vdots & \vdots & \ddots &\vdots & \vdots & \vdots \\ x_{i} & p_{i1} & p_{i2} & \cdots & p_{ij} & \cdots & p_{i\cdot}\\ \vdots & \vdots & \vdots & \vdots &\vdots & \vdots & \vdots \\ \hline & p_{\cdot 1} & p_{\cdot 2} & \cdots & p_{\cdot j} & \cdots \end{array}$$
      • 联合概率分布 $\mathbb{P}\{X=x_{i}, Y = y_{j}\} = p_{ij}$
      • 边缘概率分布
        • $\displaystyle \mathbb{P}\{X = x_{i}\} = p_{i\cdot} = \sum_{j=1}^{\infty}\mathbb{P}\{X=x_{i}, Y = y_{j}\}$
        • $\displaystyle \mathbb{P}\{Y = y_{j}\} = p_{\cdot j} = \sum_{i=1}^{\infty}\mathbb{P}\{X=x_{i}, Y = y_{j}\}$
3. 求概率 (找点求概率之和) $$\mathbb{P}\{(X,Y)\in \mathbb{D}\} = \mathop{\sum\sum}_{(x_{i},y_{j})\in \mathbb{D}} \mathbb{P}\{X = x_{i}, Y = y_{j}\}$$
4. 边缘分布律 $$\begin{split} \mathbb{P}\{X = x_{i}\} & = \mathbb{P}\{X=x_{i}, Y < +\infty\} \\ & = \sum_{j=1}^{\infty}\mathbb{P}\{X=x_{i}, Y = y_{j}\} \\ & = \sum_{j=1}^{\infty}p_{ij} \\ & \triangleq p_{i\cdot} \quad i = 1,2,\cdots \\\\ \mathbb{P}\{Y = y_{j}\} & = \mathbb{P}\{X< +\infty, Y = y_{j}\} \\ & = \sum_{i=1}^{\infty}\mathbb{P}\{X=x_{i}, Y = y_{j}\} \\ & = \sum_{i=1}^{\infty}p_{ij} \\ & \triangleq p_{\cdot j} \quad j = 1,2,\cdots \end{split}$$
   • 注 联合分布律 $\substack{\rightarrow\\ \nleftarrow}$ 边缘分布律
5. 条件分布律
   1. 若 $\mathbb{P}\{X=x_{i}\} = p_{i\cdot} > 0$，则 $\mathbb{P}\{Y = y_{j}\vert X = x_{i}\}= \frac{\mathbb{P}\{X=x_{i}, Y = y_{j}\}}{\mathbb{P}\{X= x_{i}\}} = \frac{p_{ij}}{p_{i\cdot}}$ 或 $Y\vert X = x_{i}\sim \left( \begin{matrix} y_{1} & y_{2} & \cdots & y_{j} & \cdots \\ \frac{p_{i1}}{p_{i\cdot}} & \frac{p_{i2}}{p_{i\cdot}} & \cdots & \frac{p_{ij}}{p_{i\cdot}} & \cdots \\ \end{matrix} \right)$ 称为 $X= x_{i}$ 条件下，$Y$ 的分布律
   2. 若 $\mathbb{P}\{Y=y_{j}\} = p_{\cdot j } > 0$，则 $\mathbb{P}\{X = x_{i}\vert Y = y_{j}\}= \frac{\mathbb{P}\{X=x_{i}, Y = y_{j}\}}{\mathbb{P}\{Y= y_{j}\}} = \frac{p_{ij}}{p_{\cdot j}}$ 或 $X \vert Y = y_{j}\sim \left( \begin{matrix} x_{1} & x_{2} & \cdots & x_{j} & \cdots \\ \frac{p_{1j}}{p_{\cdot j}} & \frac{p_{2j}}{p_{\cdot j}} & \cdots & \frac{p_{ij}}{p_{\cdot j}} & \cdots \\ \end{matrix} \right)$ 称为 $Y= y_{j}$ 条件下，$X$ 的分布律

例题

1. 已知随机变量 $X,Y$ 同分布，$X\sim \left( \begin{matrix} 0 & 1 \\ \frac{1}{4} & \frac{3}{4} \end{matrix} \right)$，且 $\mathbb{P}\{XY\ne 1\} = \frac{3}{8}$，则 $\mathbb{P}\{X+Y\le 1\} =$          $$\begin{array}{ll} \because & X,Y \text{同分布} \\ \therefore & Y\sim \left( \begin{matrix} 0 & 1 \\ \frac{1}{4} & \frac{3}{4} \end{matrix} \right)\\ \because & \mathbb{P}\{X+Y\le 1\} = 1 - \mathbb{P}\{X=1,Y=1\} = \frac{3}{8}\\ \therefore & \mathbb{P}\{X=1,Y=1\} = \frac{5}{8} \\ \therefore & \begin{array}{c|cc|c} & Y = 0 & Y = 1 \\ \hline X =0 & \color{#D0104C}\frac{1}{8} & \color{#D0104C}\frac{1}{8} & \frac{1}{4}\\ X =1 & \color{#D0104C}\frac{1}{8} & \frac{5}{8} & \frac{3}{4} \\ \hline & \frac{1}{4} & \frac{3}{4} \end{array} \\ \therefore & \mathbb{P}\{X+Y\le 1\} = 1 - \mathbb{P}\{X+Y > 1\}\\ & = 1- \mathbb{P}\{X=1,Y=1\} = \frac{3}{8}\\ \end{array}$$

二维连续型随机变量

1. 定义 若 $(X,Y)$ 的联合分布函数: $$\begin{split} F(x,y) & = \mathbb{P}\{X\le x, Y\le y\} \\ & = \int_{-\infty}^{x}\int_{-\infty}^{y}f(u,v)\cdot dudv \end{split}$$
   • 其中 $f(x,y)$ 非负可积，称 $(X,Y)$ 为二维连续型随机变量，$f(x,y)$ 为密度函数
   • 注 $F(x,y)$必处处连续，$\frac{\partial^{2}{F(x,y)}}{\partial{x}\partial{y}} = f(x,y)$
2. $f(x,y)$ 为密度 $\Leftrightarrow$ $\begin{cases} f(x,y)\le 0 \\ \displaystyle \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}f(x,y)\cdot dxdy = 1 \end{cases}$
   • 且$f(x,y)$不唯一，边界可忽略
   • $(X,Y)$ 取 $f(x,y)>0$ 的 $(x,y)$
3. 求概率 $$\mathbb{P}\{(X,Y)\in \mathbb{D}\} = \iint\limits_{\mathbb{D}}f(x,y)\cdot d\sigma$$
4. 边缘密度 $$\begin{array}{ll} & \displaystyle f_{X}(x) = \int_{-\infty}^{+\infty}f(x,y)\cdot dy \\ & \displaystyle f_{Y}(y) = \int_{-\infty}^{+\infty}f(x,y)\cdot dx \\ \end{array}$$
5. 条件密度
   • $f_{X}(x)\ne 0$ 时，称 $f_{Y\vert X}(y\vert x) = \frac{f(x,y)}{f_{X}(x)}$ 为 $X=x\ (f_{X}(x)>0)$ 的条件下，$Y$ 的密度
   • $f_{Y}(y)\ne 0$ 时，称 $f_{X\vert Y}(x\vert y) = \frac{f(x,y)}{f_{Y}(y)}$ 为 $Y=y\ (f_{Y}(y)>0)$ 的条件下，$X$ 的密度
   • 注 $\displaystyle \int_{-\infty}^{+\infty}f_{Y\vert X}(y\vert x)\cdot dy = \int_{-\infty}^{+\infty}\frac{f(x,y)}{f_{X}(x)}\cdot dy = \frac{f_{X}(x)}{f_{X}(x)} = 1$

例题

1. 设二维随机变量 $(X,Y)$ 的概率密度函数为 $f(x,y) = \begin{cases} k(6-x-y), & 0<x<2,2<y<4 \\ 0, & \text{其他} \end{cases}$，求：
   1. 常数 $k$
   2. $\mathbb{P}\{X<1.5\}$
   3. $\mathbb{P}\{X+Y\le 4\}$
   4. $(X,Y)$ 的联合分布函数值 $F(1,5)$ $$\begin{array}{ll} \because & \displaystyle \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}f(x,y)\cdot dxdy = 1 \\ \therefore & \displaystyle \int_{0}^{2}dx \int_{2}^{4}f(x,y)\cdot dy = \int_{0}^{2}k(6 - 2x)\cdot dx \\ & \displaystyle = k(6x - x^{2})\vert^{2}_{0} = 1 \\ \therefore & k = \frac{1}{8} \\ \\ \because & \mathbb{P}\{X<1.5\} = \mathbb{P}\{0<X<1.5, 2<Y< 4\} \\ \therefore & \displaystyle \mathbb{P}\{X<1.5\} = \int_{0}^{1.5} dx \int_{2}^{4}f(x,y)\cdot dx = \frac{1}{8}(6x - x^{2})\vert_{0}^{1.5} = \frac{27}{32} \\\\ \because & \displaystyle \mathbb{P}\{X+Y \le 4\} = \iint\limits_{x+y\le 4} f(x,y)\cdot d\sigma = \int_{0}^{2}dx \int_{2}^{4-x}f(x,y)\cdot dy\\ \therefore & \mathbb{P}\{X+Y\le 4\} = \frac{2}{3}\\ \\ \because & F(1,5) = \mathbb{P}\{X\le 1, Y\le 5\} = \mathbb{P}\{0 < X < 1, 2< Y <4\}\\ \therefore & \displaystyle F(1,5) = \int_{0}^{1}dx \int_{2}^{4} f(x,y)\cdot dy = \frac{5}{8} \\ \end{array}$$

      

2. 设二维随机变量 $(X,Y)$ 的概率密度为 $f(x,y) = \begin{cases} 6, & 0<x<1, x^{2} < y < x \\ 0, & \text{其他} \end{cases}$，求 $f_{X}(x)$ 和 $f_{Y}(y)$

   

$$\begin{array}{ll} & \displaystyle f_{X}(x) = \int_{-\infty}^{+\infty}f(x,y) \cdot dy = \int_{x^{2}}^{x}f(x,y)\cdot dx \\ & \displaystyle f_{Y}(y) = \int_{-\infty}^{+\infty}f(x,y) \cdot dx = \int_{y}^{\sqrt{y}}f(x,y)\cdot dx \\ \because & f(x,y) = \begin{cases} 6, & 0<x<1,x^{2}<y<x \\ 0, & \text{其他} \end{cases} \\ \therefore & f_{X}(x) = \begin{cases} 6(x - x^{2}), & 0 < x < 1 \\ 0, & \text{其他} \end{cases} \\ & f_{Y}(y) = \begin{cases} 6(\sqrt{y}-y), & 0 < y < 1 \\ 0, & \text{其他} \end{cases} \end{array}$$

1. 设二维随机变量 $(X,Y)$ 的概率密度为 $f(x,y) = \begin{cases} 6, & 0<x<1, x^{2} < y < x \\ 0, & \text{其他} \end{cases}$，求 $f_{X\vert Y}(x\vert y)$ 和 $f_{Y\vert X}(y\vert x)$ $$\begin{array}{ll} \because & f_{X\vert Y}(x\vert y) = \frac{f(x,y)}{f_{Y}(y)}; f_{Y\vert X}(y\vert x) = \frac{f(x,y)}{f_{X}(x)} \\ \therefore & f_{X\vert Y}(x\vert y) = \begin{cases} \frac{1}{\sqrt{y}-y}, & y< x< \sqrt{y} \\ 0, & \text{其他} \end{cases} \\ & f_{Y\vert X}(y\vert x) = \begin{cases} \frac{1}{x-x^{2}} & x^{2}<y<x \\ 0, & \text{其他} \end{cases} \end{array}$$

补充 离连型二维随机变量

• 例 设 $X\sim \left( \begin{matrix} 1 & 2 & 3 \\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \end{matrix} \right)$，在 $X=i,\quad(i=1,2,3)$ 的条件下，$Y\sim U(0,i)$，求 $\mathbb{P}\{Y\le 1\}$

  • 分析 由题可得图

    graph LR; id1(start); id2.1(X1); id2.2(X2); id2.3(X3); id3(Y); id1 --1/3--> id2.1 --"U(0,1)"--> id3 id1 --1/3--> id2.2 --"U(0,2)"--> id3 id1 --1/3--> id2.3 --"U(0,3)"--> id3

    $$\begin{array}{ll} \therefore & \mathbb{P}\{Y\le 1\} = \sum\limits_{i=1}^{3}\mathbb{P}\{X=i\}\mathbb{P}\{Y\le 1\vert X=i\} \\ & = \sum\limits_{i=1}^{3}\frac{1}{3} \frac{1}{i} = \frac{11}{18} \end{array}$$

  • 注 碰到离连型随机变量求概率，总是视==离散型的取值情况为完备事件组==，使用==全概率公式求解==